# V. Geometry

*Those who fear to experiment with their hands will never know anything.*

George Sarton (1884-1956)

*Mathematical truth is not determined arbitrarily by the rules of some ‘man-made’ formal system, but has an absolute nature and lies beyond any such system of specifiable rules.*

Roger Penrose (1930– )

Geometry is in many ways the most natural branch of elementary mathematics through which to convey “the essence” of the discipline.

- The underlying subject matter is rooted in seeing, moving, doing, drawing, making, etc., and so is accessible to everyone.
- At secondary level this practical experience leads fairly naturally to a semi-formal treatment of “geometry as a mental universe”
- — a universe that is bursting with surprising facts, whose statements can be easily understood; and
- — which has a clear logical structure, in terms of which the proofs of these facts are accessible, if sometimes tantalisingly elusive.

This combination of elusive problems to be solved and the steady accumulation of proven results has provided generations of students with their first glimpse of serious mathematics. All readers can imagine the kind of experiences which lie behind the first bullet point above: many of the problems we have already met (such as Problems **4**, **19**, **20**, **26**, **27**, **28**, **29**, **30**, **31**, **37**, **38**, **39**) do not depend on the “semi-formal treatment” referred to in the second bullet point, so can be tackled by anyone who is interested - *provided they accept the importance of learning to construct their own diagrams* (in the spirit of the George Sarton quotation).

*The hand is the cutting edge of the mind.*

Jacob Bronowski (1908-1974)

But there is a catch - which explains why the present chapter appears so late in the collection. For many problems to successfully convey “the essence of mathematics” there has to be some shared understanding of what constitutes a solution. And in geometry, many solutions require the construction of a **proof**. Yet many readers will never have experienced a coherent “semi-formal treatment” of elementary geometry in the spirit of the second bullet point. Hence in Problems **3**(c), **18**, **21**, **32**, **34**, **36** we committed the cardinal sin of leading the reader by the nose - breaking each problem into steps in order to impose a logical structure. This may have been excusable in Chapter 1; but in a chapter explicitly devoted to geometry, the underlying challenge has to be faced head on: that is, the raw experience of the *hand* has to be refined to provide a deductive structure for the *mind*.

As in Chapter 1, some of the problems listed from Section 5.3 onwards can be tackled without worrying too much about the logical structure of elementary geometry. But in many instances, the “essence” that is captured by a problem requires that the problem be seen within an agreed logical hierarchy - a sequencing of properties, results, and methods, which establishes *what* is a consequence of *what* - and hence, what can be used as part of a solution. In particular, we need to construct proofs that avoid circular reasoning.

If

Bis a consequence ofA, or ifBis equivalent toA,then a ‘proof’ ofAwhich makes use ofBis at best dubious, and may well be a delusion.

The need to avoid such circular reasoning arose already in Problem **21** (the converse of Pythagoras’ Theorem), where we felt the need to state explicitly that it would be inappropriate to use the Cosine Rule: (see Problem **192** below).

Such concerns may explain why this chapter on geometry is the last of the chapters relating to elementary ‘school mathematics’, and why we begin the chapter with

- an apparent digression (Section 5.1), and
- an outline of elementary Euclidean geometry (Section 5.2).

Those with a strong background in geometry may choose to skip these sections on a first reading, and move straight on to the problems which start in Section 5.3. But they may then fail to see how the cumulative architecture of Section 5.2 conveys a rather different aspect of the “essence of mathematics”, deriving not just from the individual problems, but from the way a carefully crafted, systematic arrangement of simple “bricks” can create a much more significant mathematical structure.

## 5.1. Comparing geometry and arithmetic

The opening quotations remind us that the mental universe of formal mathematics draws much of its initial inspiration from human perception and activity - activity which starts with infants observing, moving around, and operating with objects in time and in space. Many of our earliest pre-mathematical experiences are quintessentially proto-geometrical. We make sense of visual inputs; we learn to recognise faces and objects; we crawl around; we learn to look ‘behind’ and ‘underneath’ obstructions in search of hidden toys; we sort and we build; we draw and we make; etc.. However, for this experience to develop into *mathematics,* we then need to

- identify certain semi-formal “objects” (points, lines, angles, triangles),
- pinpoint the key relations between them (bisectors, congruence, parallels, similarity), and then
- develop the associated language that allows us to encapsulate insights from prior experience into a coherent framework for calculation and deduction.

Too little attention has been given to achieving a consensus as to how this transition (from *informal experience,* to *formal reasoning*) can best be established for beginners in elementary geometry. In contrast, number and arithmetic move much more naturally

- from our early experience of time and quantity
- to the notation, the operations, the calculational procedures, and the rules of formal arithmetic and algebra.

Counting is rooted in the idea of a *repeated unit* - a notion that may stem from the ever-present, regular heartbeat that envelops every embryo (where the beat is presumably *felt* long before it is *heard*). Later we encounter repeated units with longer time scales (such as the cycles of day and night, and the routines of feeding and sleeping). The first months and years of life are peppered with instances of numerosity, of continuous quantity, of systematic ordering, of sequences, of combinations and partitions, of grouping and replicating, and of relations between quantities and operations - experiences which provide the raw material for the mathematics of number, of place value, of arithmetic, and later of ‘internal structure’ (or algebra).

The need for political communities to construct a formal school curriculum linking early infant experience and elementary formal mathematics is a recent development. Nevertheless, in the domain of number, quantity, and arithmetic (and later algebra), there is a surprising level of agreement about the steps that need to be incorporated - even though the details may differ in different educational systems and in different classrooms. For example:

- One must somehow establish the idea of a
*unit*, which can be replicated to produce larger numbers, or*multiples.* - One must then group units relative to a chosen base (e.g. 10), iterate this grouping procedure (by taking “ten tens”, and then “ten hundreds”), and use
*position*to create*place value*notation. - One must introduce “0” - both as a number in its own right, and as a placeholder for expressing numbers using place value.
- One can then use combinations and differences, multiples and sharing (and partitions), to develop
*arithmetic.* - At some stage one introduces subunits (i.e.
*unit fractions*) and submultiples (i.e. multiples of these subunits) to produce*general fractions*; one can then use*equivalence*and common submultiples to extend arithmetic to fractions. - If we restrict to
*decimal fractions,*then our ideas of place value for integers can be extended to the right of the decimal point to produce*decimals.* - At every stage we need to
— relate these ideas to

*quantities,*— require pupils to interpret and solve

*word problems,*and— cultivate both mental arithmetic and standard written algorithms

. - Towards the end of primary school, attention begins to move beyond bare hands computation, to consciously exploit internal
*structure*in preparation for algebra.

Our early *geometrical* experience is just as natural as that relating to number; but it is more subtle. And there is as yet no comparable consensus about the path that needs to be followed if our primitive geometrical experience is to be formalised in a useable way.

The 1960s saw a drive to modernise school mathematics, and at the same time to make it accessible to all. Elementary geometry certainly needed a re-think. But the reformers in most countries simply dismissed the traditional mix (e.g. in England, where one found a blend of technical drawing, Euclidean, and coordinate geometry in different proportions for different groups of students) in favour of more modern-sounding alternatives. Some countries favoured a more abstract, deductive framework; some tried to exploit motion and transformations; some used matrices and groups; some used vectors and linear algebra; some even toyed with topology. More recently we have heard similarly ambitious claims on behalf of dynamic geometry software. And although each approach has its attractions,

noneof the alternatives has succeeded in helpingmorestudents to visualise, to reason, and to calculate effectively in geometrical settings.

At a much more advanced level, geometry combines

- with abstract algebra (where the approach proposed by Felix Klein (1849-1925) shows how to identify each geometry with a group of transformations), and
- with analysis and linear algebra (where, following Gauss (1777-1855), Riemann (1826-1866) and Grassmann (1809-1877), calculus, vector spaces, and later topology can be used to analyse the geometry of surfaces and other spaces).

However, these subtle formalisms are totally irrelevant for beginners, who need an approach

- based on concepts which are relatively familiar
*(points, lines, triangles*etc.), and - whose basic properties can be formulated relatively simply.

The subtlety and flexibility of dynamic geometry software may be hugely impressive; but if students are to harness this power, they need *prior *mastery of some simple, semi-formal framework, together with the associated language and modes of reasoning. Despite the lack of an accepted consensus, the experience of the last 50 years would seem to suggest that the most relevant framework for beginners at secondary level involves some combination of:

*static,*relatively traditional Euclidean geometry, and- Cartesian, or coordinate (analytic) geometry.

## 5.2. Euclidean geometry: a brief summary

*Philosophy is written in this grand book - I mean the universe - which stands continually open to our gaze, but it cannot be understood unless one first learns to comprehend the language and to interpret the characters in which it is written. It is written in the language of mathematics, and its characters are triangles, circles, and other geometrical figures, without which it is humanly impossible to understand a single word of it; without these, one is wandering about in a dark labyrinth.*

Galileo Galilei (1564-1642)

This section provides a detailed, but compressed, outline of an initial formalisation of school geometry - of a kind that one would like good students and all teachers to appreciate. It is unashamedly a *semi*-formal approach for beginners, **not** a strictly formal treatment (such as that provided by David Hilbert (1862-1943) in his 1899 book *Foundations of Geometry,* or in the more detailed exposition by Edwin Moise (1918-1998) *Elementary Geometry from an Advanced Standpoint,* published in 1963). In particular:

- we work with relatively informal notions of
*points, lines,*and*angles*in the plane; - we focus attention on certain simple issues which really matter at school level (such as how points, lines, line segments, and angles are referred to; the notion of a triangle as an
*ordered*triple of vertices; the fact that the vertices of a quadrilateral must be labelled cyclically; etc.); - we limit the formal deductive structure to just three central criteria, namely the criteria for
*congruence*, for*parallels*, and for*similarity*, and show how they allow one to develop results and methods in a logical sequence.

We begin with the intuitive idea of *points* and *lines* in the plane. Two points *A, B* determine

- the
*line segment*(with endpoints*AB**A*and*B*), and - the
*line AB*(which extends the line segmentin both directions - beyond*AB**A*, and beyond*B*).

Three points *A*, *B*, *C* determine an angle ∠*ABC* (between the two line segments __ BA__ and

__).__

*BC*We can then begin to build more complicated figures, such as

- a triangle
*ABC*(with three vertices*A, B, C*; three sides,*AB*,*BC*; and three angles ∠*CA**ABC*at the vertex*B*, ∠*BCA*at*C*, and ∠*CAB*at*A*), - a quadrilateral
*ABCD*(with four vertices*A, B, C, D;*and four sides,__AB__,__BC__,__CD__which meet only at their endpoints).__DA__

And so on. Two given points *A, B* also allow us to construct the *circle* with centre *A*, and passing through B (that is, with *radius AB).*

This very limited beginning already opens up the world of *ruler and compasses constructions.* In particular, given a line segment * AB* one can draw:

- the circle with centre
*A*, and passing through*B*, and - the circle with centre
*B*, and passing through*A*.

If the two circles meet at *C*,

- then
=__AB__(radii of the first circle), and__AC__=__BA__(radii of the second circle).__BC__

Hence we have constructed the *equilateral triangle* Δ*ABC* on the given segment * AB*. This construction is the very first proposition in Book 1 of the

*Elements*of Euclid (flourished c. 300 BC). Euclid’s second proposition is presented next as a problem.

**Problem 137** Given three points *A, B, C,* show how to construct - without measuring - a point *D* such that the segments * AB* and

*are equal (in length).*

__CD__Problem **137** looks like a simple starter (where the only available construction is to produce the third vertex of an equilateral triangle on a given line segment). However, to produce a valid solution requires a clear head and a degree of ingenuity.

Given two points *A, B,* the process of constructing an equilateral triangle Δ*ABC* illustrates how we are allowed to construct new points from old.

- Whenever we construct two lines or circles that cross, the points where they cross (such as the point C in the above construction of the equilateral triangle Δ
*ABC*) become available for further constructions. So, if points*A*and*B*are given, then once*C*has been constructed, we may proceed to draw the lines*AC*and*BC*.

However, the fact that we can construct a line segment * AB* does not allow us to ‘measure’ the segment with a ruler, and then to use the resulting measurement to ‘copy’ the segment

*to the point C in order to construct the required point D such that*

__AB__*=*

__AB__*. The “ruler” in*

__CD__*ruler and compasses constructions*is used only to draw the line through two known points - not to measure. (Measuring is an

*approximate*physical action, rather than an

*exact*“mental construction”, and so is not really part of mathematics.) Hence in Problem

**137**we have to find another way to produce a copy

*of the segment*

__CD__*starting at the point*

__AB__*C*. Similarly, we can construct the circle with centre

*A*and passing through

*B*, but this does not allow us to use the pair of compasses to transfer distances physically (e.g. by picking up the compasses from

*and placing the compass point at*

__AB__*C*, like using the old geometrical drawing instrument that was called

*a pair of dividers*). In seeking the construction required in Problem

**137**, we are restricted to “exact mental constructions” which may be described in terms of:

- drawing (or constructing) the line joining any two known points,
- constructing the circle with centre at a known point and passing through a known point, and
- obtaining a new point
*D*as the intersection of two constructed lines or circles (or of a line and a circle).

If on the line *AB*, the point *X* lies between *A* and *B,* then we obtain a *straight angle* *∠AXB* at *X* (or rather *two* straight angles at *X* - one on each side of the line *AB*). If we assume that all straight angles are equal, then it follows easily that “vertically opposite angles are always equal”.

**Problem 138** Two lines AB and *CD* cross at *X*, where *X* lies between *A* and *B* and between *C* and *D*. Prove that ∠*AXC*=∠*BXD*.

Define a *right angle* to be ‘half a straight angle’. Then we say that two lines which cross at a point *X* are *perpendicular* if an angle at *X* is a right angle (or equivalently, if all four angles at *X* are equal). The next step requires us to notice two things - partly motivated by experience when coordinating hand, eye and brain to construct, and to think about, physical structures.

- First we need to recognise that
*triangles*hold the key to the analysis of more complicated shapes. - Then we need to realise that triangles in different positions can still be “equal”, or
*congruent*- which then focuses attention on the**minimal**conditions under which two triangles can be guaranteed to be congruent.

The first of these two bullet points has an important consequence - namely that solving any problem in 2- or in 3-dimensions generally reduces to working with **triangles**. In particular, solving problems in 3-dimensions reduces to working in some 2-dimensional *cross-section* of the given figure (since three points not only determine a triangle, but also determine the plane in which that triangle lies). It follows that 2-dimensional geometry holds the key to solving problems in 3-dimensions, and that **working with triangles is central in all geometry**.

The second bullet point forces us to think carefully about:

- what we mean by a
*triangle*(and in particular, to understand why Δ*ABC*and Δ*BCA*are in some sense*different*triangles, even though they use the same three vertices and sides), and - what it means for two triangles to be “the same”.

A triangle Δ*ABC* incorporates six pieces of data, or information: the three sides * AB*,

*,*

__BC__*and the three angles ∠*

__CA__*ABC*, ∠

*BCA*, ∠

*CAB*. We say that two (ordered) triangles Δ

*ABC*and Δ

*AʹBʹCʹ*are

*congruent*(which we write as

$$\Delta ABC\equiv \Delta {A}^{\prime}{B}^{\prime}{C}^{\prime},$$

where the order in which the vertices are listed matters) if their sides and angles “match up” in pairs, so that

$$\begin{array}{l}\underset{\_}{AB}=\underset{\_}{{A}^{\prime}{B}^{\prime}},\underset{\_}{BC}={\underset{\_}{{B}^{\prime}{C}^{\prime}}}^{\prime},\underset{\_}{CA}=\underset{\_}{{C}^{\prime}{A}^{\prime}},\\ \angle ABC=\angle {A}^{\prime}{B}^{\prime}{C}^{\prime},\angle BCA=\angle {B}^{\prime}{C}^{\prime}{A}^{\prime},\angle CAB=\angle {C}^{\prime}{A}^{\prime}{B}^{\prime}.\end{array}$$

As a result of drawing and experimenting with our hands, our minds may realise that certain subsets of these six conditions suffice to imply the others. In particular:

**SAS-congruence criterion:** if

$$\underset{\_}{AB}=\underset{\_}{{A}^{\prime}{B}^{\prime}},\text{\hspace{1em}}\angle ABC=\angle {A}^{\prime}{B}^{\prime}{C}^{\prime},\text{\hspace{1em}}\underset{\_}{BC}=\underset{\_}{{B}^{\prime}{C}^{\prime}},$$

then

$$\Delta ABC\equiv \Delta {A}^{\prime}{B}^{\prime}{C}^{\prime}$$

(where the name “SAS” indicates that the three listed match-ups occur in the specified order *S* (side), A (angle), *S* (side) as one goes round each triangle).

**SSS-congruence criterion:** if

$$\underset{\_}{AB}=\underset{\_}{{A}^{\prime}{B}^{\prime}},\text{\hspace{1em}}\underset{\_}{BC}=\underset{\_}{{B}^{\prime}{C}^{\prime}},\text{\hspace{1em}}\underset{\_}{CA}=\underset{\_}{{C}^{\prime}{A}^{\prime}},$$

then

$$\Delta ABC\equiv \Delta {A}^{\prime}{B}^{\prime}{C}^{\prime}.$$

**ASA-congruence criterion:** if

$$\angle ABC=\angle {A}^{\prime}{B}^{\prime}{C}^{\prime},\text{\hspace{1em}}\underset{\_}{BC}=\underset{\_}{{B}^{\prime}{C}^{\prime}},\text{\hspace{1em}}\angle BCA=\angle {B}^{\prime}{C}^{\prime}{A}^{\prime},$$

then

$$\Delta ABC\equiv \Delta {A}^{\prime}{B}^{\prime}{C}^{\prime}.$$

If in a given triangle Δ*ABC* we have * AB* =

*, then we say that Δ*

__AC__*ABC*is

*isosceles*with

**apex A**, and

**base**

__BC__*(iso*= same, or equal;

*sceles*= legs).

**Problem 139** Let Δ*ABC* be an isosceles triangle with apex *A*. Let *M* be the midpoint of the base * BC*. Prove that Δ

*AMB*= Δ

*AMC*and conclude that

*AM*is perpendicular to the base

*.*

__BC__**Problem 140** Construct two non-congruent triangles, Δ*ABC* and Δ*AʹBʹCʹ*, where $\angle BCA=\angle {B}^{\prime}{C}^{\prime}{A}^{\prime}={30}^{\xb0},\left|\underset{\_}{CA}\right|=\left|\underset{\_}{{C}^{\prime}}\right|=\sqrt{3},\text{\hspace{0.17em}}\left|\underset{\_}{AB}\right|=\left|\underset{\_}{{A}^{\prime}{B}^{\prime}}\right|=1.$

Conclude that there is in general no “ASS-congruence criterion”.

The congruence criteria allow one to prove basic results such as:

**Claim** In any isosceles triangle Δ*ABC* with apex *A* (i.e. with * AB* =

*), the two base angles ∠*

__AC__*B*and ∠

*C*are equal.

**Proof 1** Let *M* be the midpoint of * BC*.

Then Δ*AMB* = Δ*AMC* (by the SSS-congruence criterion, since

* AM* =

*,*

__AM__* MB* =

*(by construction of*

__MC__*M*as the midpoint)

* BA* =

*(given)).*

__CA__$\therefore \angle B=\angle ABM=\angle ACM=\angle C.$ QED

**Proof 2** Δ*BAC* = Δ*CAB* (by the SAS-congruence criterion, since *BA*

* BA* =

*(given),*

__CA__∠*BAC* = ∠*CAB* (same angle),

* AC* =

*(given),*

__AB__$\therefore \angle B=\angle ABC=\angle ACB=\angle C.$ QED

We also have the converse result:

**Claim** In any triangle Δ*ABC*, if the base angles ∠*B* and ∠*C* are equal, then the triangle is isosceles with apex *A* (i.e. * AB* =

*).*

__AC__**Proof** Δ*ABC* = Δ*ACB* (by the ASA-congruence criterion, since

∠ABC = ∠ACB (given),

* BC* =

*, and*

__CB__∠*B*AC = ∠*C*AB (given)).

$\therefore \underset{\_}{AB}=\underset{\_}{AC}.$ QED

**Problem 141**

(i) A circle with centre *O* passes through the point *A*. The line *AO* meets the circle again at *B*. If *C* is a third point on the circle, prove that ∠*ACB* is equal to ∠*CAB* + ∠*CAB*.

(ii) Conclude that, if the angles in Δ*ABC* add to a straight angle, then ∠*ACB* is a right angle.

Once we introduce the parallel criterion, and hence can prove that the three angles in any triangle add to a straight angle, Problem **141** will guarantee that “the angle subtended on the circumference by a diameter is always a right angle”.

**Problem 142** Show how to implement the basic ruler and compasses constructions:

(i) to construct the midpoint *M* of a given line segment * AB*;

(ii) to bisect a given angle ∠*ABC*;

(iii) to drop a perpendicular from *P* to a line *AB* (that is, to locate *X* on the line *AB,* so that the two angles that *PX* makes with the line *AB* on either side of *PX* are equal).

Prove that your constructions do what you claim.

**Problem 143** Given two points *A* and *B*.

(a) Prove that each point *X* on the perpendicular bisector of * AB* is equidistant from

*A*and from

*B*(that is, that

*=*

__XA__*).*

__XB__(b) Prove that, if *X* is equidistant from *A* and from *B*, then *X* lies on the perpendicular bisector of * AB*.

Problem **143** shows that, given a line segment * AB*, the perpendicular bisector of

*is the locus of all points*

__AB__*X*which are equidistant from

*A*and from

*B*. This observation is what lay behind the construction of the circumcentre of a triangle (back in Chapter 1, Problem

**32**(a)):

Given any Δ*ABC*.

Let *O* be the point where the perpendicular bisectors of * AB* and

*meet.*

__BC__Then * OA* =

__OB__and * OB* =

*.*

__OC__$\therefore \underset{\_}{OA}=\underset{\_}{OB}=\underset{\_}{OC}.$

Hence *O* is the centre of a circle passing through all three vertices *A, B, C*.

Moreover *O* also lies on the perpendicular bisector of * CA*.

This circle is called the *circumcircle* of Δ*ABC*, and *O* is called the *circumcentre* of Δ*ABC*. As indicated back in Problem **32**, the radius of the circumcircle of Δ*ABC* (called the *circumradius* of the triangle) is generally denoted by *R.* Later we will meet other circles and “centres” associated with a given triangle Δ*ABC*.

Before moving on it is worth extending Problem **143** to three dimensions

**Problem 144** Given any two points *N*, *S* in 3D space, prove that the locus of all points *X* which are equidistant from *N* and from *S* form the plane perpendicular to the line *NS* and passing through the midpoint *M* of * NS*.

The next two fundamental results are often neglected.

**Problem 145** Given any Δ*ABC*, if we extend the side * BC* beyond

*C*to a point

*X*, then the “exterior angle” ∠

*ACX*at

*C*is greater than each of the “two interior opposite angles” ∠

*A*and ∠

*B*.

**Problem 146**

(a) If in Δ*ABC* we have * AB* >

*, then ∠*

__AC__*> ∠*

__ACB__*. (“In any triangle, the larger angle lies opposite the longer side.”)*

__ABC__(b) If in Δ*ABC* we have ∠* ACB* > ∠

*, then*

__ABC__*>*

__AB__*. (“In any triangle, the longer side lies opposite the larger angle.”)*

__AC__(c) **(The triangle inequality)** Prove that in any triangle Δ*ABC*,

$$\underset{\_}{AB}+\underset{\_}{BC}>\underset{\_}{AC}.$$

The results in Problems **145** and **146** have surprisingly many consequences. For example, they allow one to prove the converse of the result in Problem **141**

**Problem 147** Suppose that in Δ*ABC*, ∠*C* = ∠*A* + ∠*B*. Prove that *C* lies on the circle with diameter * AB*.

(In particular, if the angles of Δ*ABC* add to a straight angle, and ∠*ACB* is a right angle, then *C* lies on the circle with diameter * AB*.

We come next to a result whose justification is often fudged. At first sight it is unclear how to begin: there seems to be so little information to work with - just two points and a line through one of the points.

**Problem 148** A circle with centre *O* passes through the point *P*. Prove that the tangent to the circle at *P* is perpendicular to the radius * OP*.

Problem **148** is an example of a result which implies its own converse - though in a backhanded way. Suppose a circle with centre *O* passes through the point *P*. If *OP* is perpendicular to a line *m* passing through *P*, then *m* must be tangent to the circle (because we know that the tangent at *P* is perpendicular to *OP*, so the angle between *m* and the tangent is “zero”, which forces *m* to be equal to the tangent). This converse will be needed later, when we meet the *incircle*.

**Problem 149** Let *P* be a point and *m* a line not passing through *P*. Prove that, among all possible line segments * PX* with

*X*on the line

*m*, a perpendicular from

*P*to the line

*m*is the shortest.

The result in Problem **149** allows us to define the “distance” from *P* to the line *m *to be the length of any perpendicular from *P* to *m*. (As far as we know at this stage of the development, there could be more than one perpendicular from *P* to *m*.)

Note that all the results mentioned so far have avoided using the Euclidean “parallel criterion” (or - equivalently - the fact that the three angles in any triangle add to a straight angle). So results proved up to this point should still be “true” in any geometry where we have points, lines, triangles, and circles satisfying the congruence criteria - whether or not the geometry satisfies the Euclidean “parallel criterion”.

The idea that there is only one “shortest” distance from a point to a line may seem “obvious”; but it is patently *false* on the sphere, where *every* line (i.e. ‘great circle’) from the North pole *P* to the equator is perpendicular to the equator (and all these lines have the same “length”). The proof that there is just one such perpendicular from *P* to *m* depends on the *parallel criterion* (see below) - a criterion which fails to hold for geometry on the sphere.

Euclid’s *Elements* started with a few basic axioms that formalised the idea of ruler and compasses constructions. He then added a simple axiom that allowed one to compare angles in different locations. He made the forgivable mistake of omitting an axiom for congruence of triangles - imagining that it can be *proved.* (It can’t.) However he then stated, and carefully developed the consequences of, a much more subtle axiom about parallel lines (two lines *m, n* in the plane are said to be *parallel* if they never meet, no matter how far they are extended). For reasons that remain unclear, instead of appreciating that Euclid's “parallel postulate” constituted a profound insight into the foundations of geometry, mathematicians in later ages saw the complexity of Euclid’s postulate as some kind of flaw, and so tried to show that it could be derived from the other, simpler postulates. The attempt to “correct” this perceived flaw became a kind of Holy Grail.

The story is instructive, but too complicated to summarise accurately here. The situation was eventually clarified by two nineteenth century mathematicians (more-or-less at the same time, but working independently). In the revolutionary, romantic spirit of the nineteenth century, János Bolyai (Hungarian: 1802-1860) and Nikolai Lobachevski (Russian: 1792-1856) each allowed himself to consider what would happen if one adopted a *different *assumption about how “parallel lines” behave. Both discovered that one can then derive an apparently coherent theory of a completely novel kind, with its own beautiful results: that is, a geometry which seemed to be internally “consistent” - but different from Euclidean geometry. Lobachevski published brief notes of his work in 1829-30 (in Kazan); Bolyai knew nothing of this and published incomplete notes of his researches in 1832. Lobachevski published a more detailed booklet in 1840.

Neither mathematician got the recognition he might have anticipated, and it was only much later (largely after their deaths) that others realised how to show that the fantasy world they had each dreamt up was just as “internally consistent” as traditional Euclidean geometry. The story is further complicated by the fact that the dominant mathematician of the time - namely Gauss (1777-1855) - claimed to have proved something similar (and he may well have done so, but exactly what he knew has to be inferred from cryptic remarks in occasional letters, since he published nothing on the subject). If there is a moral to the story, it could be that success in mathematics may not be recognised, or may only be recognised after one’s death: so those who spend their lives exploring the mathematical universe had better appreciate the delights of the mathematical journey, rather than being primarily motivated by a desire for immediate recognition and acclaim!

Two lines *m, n* in the plane are said to be *parallel* if they never meet - no matter how far they are extended. We sometimes write this as “*m*||*n*”.

Given two lines *m, n* in the plane, a third line *p* which crosses both *m* and *n* is called a *transversal* of *m* and *n*.

Parallel criterion:Given two linesmandn, if some transversalpis such that the two “internal” angles on one side of the linep(that is the two angles thatpmakes withmand withn, and which lie between the two linesmandn) add toless thana straight angle, then the linesmandnmust meet on that side of the linep.

If the internal angles on one side of *p* add to *more than* a straight angle, then internal angles on the *other side* of *p* add to less than a straight angle, so the lines *m* and *n* must meet on the other side of *p*. It follows.

- that two lines
*m*and*n*are parallel precisely when the two internal angles on one side of a transversal add to**exactly**a straight angle.

Parallel lines can be thought of as “all having the same direction”; so it is convenient to insist that “every line is parallel to itself” (even though it has lots of points in common with itself). It then follows

- that, given three lines
*k, m, n*, if*k*is parallel to*m*and*m*is parallel to*n*, then*k*is parallel to*n*; and - that given a line
*m*and a point*P*, there is a unique line*n*through*P*which is parallel to*m*.

All this then allows one

- to conclude that, if
*m*and*n*are any two lines, and*p*is a transversal, then*m*and*n*are parallel if and only if*alternate angles*are equal (or equivalently, if and only if*corresponding angles*are equal); and - to extend the basic ruler and compasses constructions to include the construction:
“given a line

*AB*and a point*P*, construct the line through*P*which is parallel to*AB*”(namely, by first constructing the line

*PX*through*P*, perpendicular to*AB*, and then the line through*P*, perpendicular to*PX*).

One can then prove the standard result about the angles in any triangle.

**Claim** The angles in any triangle Δ*ABC* add to a straight angle.

**Proof** Construct the line *m* through *A* that is parallel to *BC*. Then *AB* and *AC* are transversals, which cross both the line *m* and the line *BC*, and which make three angles at the point *A* on *m*:

- one being just the angle ∠A in the triangle Δ
*ABC*, - one being equal to ∠
*B*(alternate angles relative to the transversal*AB*) and - one being equal to ∠
*C*(alternate angles relative to the transversal*AC*).

The three angles at *A* clearly add to a straight angle, so the three angles ∠*A*, ∠*B*, ∠*C* also add to a straight angle. QED

Once we know that the angles in any triangle add to a straight angle, we can prove all sorts of other useful facts. One is a simple reformulation of the above **Claim**.

**Problem 150** Given any triangle Δ*ABC*, extend * BC* beyond

*C*to a point

*X*. Then the exterior angle

$$\angle XCA=\angle A+\angle B.$$

(“In any triangle, each exterior angle is equal to the sum of the two interior opposite angles.”)

Another important consequence is the result which underpins the sequence of “circle theorems”.

**Problem 151** Let *O* be the circumcentre of Δ*ABC*. Prove that

$$\angle AOB=2\xb7\angle ACB.$$

Problem **151** implies that

“the angles subtended by any chord

on a given arc of the circle are all equal”,AB

and are equal to exactly one half of the angle subtended by * AB* at the centre

*O*of the circle. This leads naturally to the familiar property of

*cyclic quadrilaterals.*

**Problem 152** Let *ABCD* be a quadrilateral inscribed in a circle (such a quadrilateral is said to be *cyclic,* and the four vertices are said to be *concyclic *- that is, they lie together on the same circle). Prove that opposite angles (e.g. ∠*B* and ∠D) must add to a straight angle. (Two angles which add to a straight angle are said to be *supplementary.*)

These results have lots of lovely consequences: we shall see one especially striking example in Problem **164**. Meantime we round up our summary of the "circle theorems".

**Problem 153** Suppose that the line *XAY* is tangent to the circumcircle of Δ*ABC* at the point *A*, and that *X* and *C* lie on opposite sides of the line *AB.* Prove that ∠*XAB* = ∠*ACB*.

**Problem 154**

(a) Suppose *C, D* lie on the same side of the line *AB.*

(i) If *D* lies inside the circumcircle of Δ*ABC*, then ∠*ADB* > ∠*ACB*.

(ii) If *D* lies outside the circumcircle of Δ*ABC*, then ∠*ADB* < ∠*ACB*.

(b) Suppose *C*, *D* lie on the same side of the line AB, and that ∠*ACB* = ∠*ADB*. Then *D* lies on the circumcircle of Δ*ABC*.

(c) Suppose that *ABCD* is a quadrilateral, in which angles ∠*B* and ∠D are supplementary. Then *ABCD* is a cyclic quadrilateral.

Another result which follows now that we know that the angles of a triangle add to a straight angle is a useful additional congruence criterion - namely the **RHS-congruence criterion**. This is a ‘limiting case’ of the failed ASS-congruence criterion (see the example in Problem **140**). In the failed ASS criterion the given data correspond to two *different* triangles - one in which the angle opposite the first specified side (the first “S” in “ASS”) is acute, and one in which the angle opposite the first specified side is obtuse. In the RHS-congruence criterion, the angle opposite the first specified side is a *right angle,* and the two possible triangles are in fact congruent.

**RHS-congruence criterion:** If ∠*ABC* and ∠*AʹBʹCʹ* are both right angles, and * BC* =

*,*

__BʹCʹ__*=*

__CA__*, then*

__CʹAʹ__$$\Delta ABC\equiv \Delta {A}^{\prime}{B}^{\prime}{C}^{\prime}.$$

**Proof** Suppose that * AB* =

*. Then*

__AʹBʹ__* AB* =

__AʹBʹ__∠*ABC* = ∠*AʹBʹCʹ*,

* BC* =

__BʹCʹ__Hence we may apply the SAS-congruence criterion to conclude that Δ*ABC* = Δ*AʹBʹCʹ*.

If on the other hand $AB\ne \underset{\_}{{A}^{\prime}{B}^{\prime}}$
, we may suppose that * BA* >

*. Now construct*

__BʹAʹ__*Aʹʹ*on

*BA*such that

*=*

__BAʹʹ__*. Then*

__BʹAʹ__* AʹʹB* =

*,*

__AʹBʹ__∠AʹʹBC = ∠AʹBʹCʹ,

* BC* =

*,*

__BʹCʹ__$\therefore \Delta {A}^{\prime \prime}BC\equiv \Delta {A}^{\prime}{B}^{\prime}{C}^{\prime}$ (by SAS-congruence).

Hence * AʹʹC* =

*=*

__AʹCʹ__*AC*, so Δ

*CAAʹʹ*is isosceles.

$$\therefore \angle C{A}^{\prime \prime}A=\angle CA{A}^{\prime \prime}.$$

However, ∠*CAʹʹA* > ∠*CBA* (since the exterior angle ∠*CAʹʹA* in Δ*CBAʹʹ* must be greater than the interior opposite angle ∠*CBA*, by Problem **145**).

But then the two base angles in the isosceles triangle Δ*CAAʹʹ* are each greater than a right angle - so the angle sum of Δ*CAAʹʹ* is greater than a straight angle, which is impossible. Hence this case cannot occur. QED

RHS-congruence seems to be needed to prove the basic result (Problem **161** below) about the area of parallelograms, and this is then needed in the proof of Pythagoras’ Theorem (Problem **18**). In one sense RHS-congruence looks like a special case of SSS-congruence (as soon as two pairs of sides in two right angled triangles are equal, Pythagoras’ Theorem guarantees that the third pair of sides are also equal). However this observation cannot be used to justify RHS-congruence if RHS-congruence is needed to justify Pythagoras’ Theorem.

**Problem 155** Given a circle with centre *O,* let *Q* be a point outside the circle, and let *QP*, *QP'* be the two tangents from Q, touching the circle at *P* and at *P'*. Prove that * QP* =

__QPʹ__, and that the line

*OQ*bisects the angle ∠

*PQP*ʹ.

**Problem 156** You are given two lines *m* and *n* crossing at the point *B.*

(a) If *A* lies on *m* and *C* lies on *n*, prove that each point *X* on the bisector of angle ∠*ABC* is equidistant from *m* and from *n*.

(b) If *X* is equidistant from *m* and from *n*, prove that *X* must lie on one of the bisectors of the two angles at *B*.

Problem **156** shows that, given two lines *m* and *n* that cross at *B*, the bisectors of the two pairs of vertically opposite angles formed at B form the *locus* of **all** points *X* which are equidistant from the two lines *m* and *n*. This allows us to mimic the comments following Problem **143** and so to construct the *incentre* of a triangle.

Given any Δ

ABC, letIbe the point where the angle bisectors of ∠ABCand ∠BCAmeet.Let the perpendiculars from

Ito the three sidesAB, BC,CAmeet the sides atP, Q,Rrespectively. Then

=IP(sinceIQIlies on the bisector of ∠ABC) and

=IQ(sinceIRIlies on the bisector or ∠BCA).Hence the circle which has centre

Iand which passes throughPalso passes throughQandR.Moreover,

Ialso lies on the bisector of ∠CAB; and since the radii,IP,IQare perpendicular to the sides of the triangle, the circle is tangent to the three sides of the triangle (by the comments following ProblemIR148)

This circle is called the *incircle* of Δ*ABC*, and *I* is called the *incentre* of Δ*ABC*. The radius of the incircle of Δ*ABC* is called the *inradius,* and is generally denoted by *r*.

A quadrilateral *ABCD* in which *AB* ∥ *DC* and *BC* ∥ *AD* is called a *parallelogram.* A parallelogram *ABCD* with a right angle is a *rectangle. *A parallelogram *ABCD* with * AB* =

*is called a*

__AD__*rhombus.*A rectangle which is also a rhombus is called a

*square.*

**Problem 157** Let *ABCD* be a parallelogram.

(i) Prove that Δ*ABC* = ΔCDA, so that each triangle has area exactly half of area(*ABCD*).

(ii) Conclude that opposite sides of *ABCD* are equal in pairs and that opposite angles are equal in pairs.

(iii) Let *AC* and *BD* meet at *X*. Prove that *X* is the midpoint of both * AC* and

*.*

__BD__**Problem 158** Let *ABCD* be a parallelogram with centre *X* (where the two main diagonals * AC* and

*meet), and let*

__BD__*m*be any straight line passing through the centre. Prove that

*m*divides the parallelogram into two parts of equal area.

We defined a parallelogram to be “a quadrilateral *ABCD* in which *AB* ∥ *DC* and *BC* ∥ *AD* however, in practice, we need to be able to recognise a parallelogram even if it is not presented in this form. The next result hints at the variety of other conditions which allow us to recognise a given quadrilateral as being a parallelogram “in mild disguise”.

**Problem 159**

(a) Let *ABCD* be a quadrilateral in which *AB* ∥ *DC*, and * AB* =

*. Prove that*

__DC__*BC*∥

*AD*, and hence that

*ABCD*is a parallelogram.

(b) Let *ABCD* be a quadrilateral in which * AB* =

*and*

__DC__*=*

__BC__*. Prove that*

__AD__*AB*∥

*DC*, and hence that

*ABCD*is a parallelogram.

(c) Let *ABCD* be a quadrilateral in which ∠A = ∠*C* and ∠*B* = ∠D. Prove that *AB* ∥ *DC* and that *BC* ∥ *AD*, and hence that *ABCD* is a parallelogram.

The next problem presents a single illustrative example of the kinds of things which we know in our bones must be true, but where the reason, or proof, may need a little thought.

**Problem 160** Let *ABCD* be a parallelogram. Let *M* be the midpoint of * AD* and

*N*be the midpoint of

__BC__. Prove that MN || AB, and that

*MN*passes through the centre of the parallelogram (where the two diagonals meet).

**Problem 161** Prove that any parallelogram *ABCD* has the same area as the rectangle on the same base * DC* and “with the same height” (i.e. lying between the same two parallel lines

*AB*and

*DC*).

The ideas and results we have summarised up to this point provide exactly what is needed in the proof of Pythagoras’ Theorem outlined back in Chapter 1, Problem **18**. They also allow us to identify two more “centres” of a given triangle Δ*ABC*.

**Problem 162** Given any triangle Δ*ABC*, draw the line through A which is parallel to *BC*, the line through *B* which is parallel to *AC*, and the line through *Cʹ* which is parallel to *AB.* Let the first two constructed lines meet at *C,* the second and third lines meet at *Aʹ*, and the first and third lines meet at *Bʹ*.

(a) Prove that *A* is the midpoint of * BʹCʹ*, that

*B*is the midpoint of

*and that*

__CʹAʹ__*C*is the midpoint of

*.*

__AʹBʹ__(b) Conclude that the perpendicular from *A* to *BC*, the perpendicular from *B* to *CA*, and the perpendicular from *C* to *AB* all meet in a single point *H*. (*H* is called the *orthocentre* of Δ*ABC*.)

Let the foot of the perpendicular from *A* to *BC* be *P*, the foot of the perpendicular from *B* to *CA* be *Q*, and the foot of the perpendicular from *C* to *AB* be *R.* Then ΔPQR is called the *orthic triangle* of Δ*ABC*. The “circle theorems” (especially Problems **151** and **154**(c)) lead us to discover that this triangle has two quite unexpected properties. As a partial preparation for one of the properties we digress slightly to introduce a classic problem.

**Problem 163** My horse is tethered at *H* some distance away from my village *V*. Both *H* and *V* are on the same side of a straight river. How should I choose the shortest route to lead the horse from *H* to *V*, if I want to water the horse at the river en route?

**Problem 164** Let Δ*ABC* be an acute angled triangle.

(a) Prove that, among all possible triangles Δ*PQR* inscribed in Δ*ABC*, with *P* on *BC*, *Q* on *CA*, *R* on *AB,* the orthic triangle is the one with the shortest perimeter.

(b) Suppose that the sides of Δ*ABC* act like mirrors. A ray of light is shone along one side of the orthic triangle *PQ,* reflects off *CA*, and the reflected beam then reflects in turn off *AB.* Where does the ray of light next hit the side *BC*? (Alternatively, imagine the sides of the triangle as billiard table cushions, and explain the path followed by a ball which is projected, without spin, along *PQ*.)

We come next to the fourth among the standard “centres of a triangle”.

**Problem 165** Given Δ*ABC*, let *L* be the midpoint of the side * BC*. The line

*AL*is called a

*median*of Δ

*ABC*. (It is not at all obvious, but if we imagine the triangle as a lamina, having a uniform thickness, then Δ

*ABC*would exactly balance if placed on a knife-edge running along the line AL.) Let

*M*be the midpoint of the side

*, so that*

__CA__*BM*is another median of Δ

*ABC*. Let

*G*be the point where

*AL*and

*BM*meet.

(a) (i) Prove that Δ*ABL* and Δ*ACL* have equal area. Conclude that Δ*ABG* and Δ*ACG* have equal area.

(ii) Prove that Δ*BCM* and Δ*BAM* have equal area. Conclude that Δ*BCG* and Δ*BAG* have equal area.

(b) Let *N* be the midpoint of * AB*. Prove that

*CG*and

*GN*are the same straight line (i.e. that ∠

*CGN*is a straight angle). Hence conclude that the three medians of any triangle always meet in a point

*G*.

The point where all three medians meet is called the *centroid* of the triangle. For the geometry of the triangle, this is all you need to know. However, it is worth noting that the centroid is the point that would be the ‘centre of gravity’ of the triangle if the triangle is thought of as a thin lamina with a uniform distribution of mass.

Next we revisit, and reprove in the Euclidean spirit, a result that you proved in Problem **95** using coordinates - namely the *Midpoint Theorem*.

**Problem 166 (The Midpoint Theorem)** Given any triangle Δ*ABC*, let *M* be the midpoint of the side * AC*, and let

*N*be the midpoint of the side

*. Draw in*

__AB__*and extend it beyond*

__MN__*N*to a point

*M*ʹ such that

*=*

__MN__*.*

__NMʹ__(a) Prove that Δ*ANM* = Δ*BNM*ʹ.

(b) Conclude that * BM*ʹ =

*and that*

__CM__*ʹ ||*

__BM__*CM*.

(c) Conclude that *MMʹBC* is a parallelogram, so that __ CB__ =

__. Hence__

*MM*ʹ*is parallel to*

__MN__*and half its length.*

__CB__The *Midpoint Theorem* can be reworded as follows:

Given Δ

AMN.Extend

toAMCsuch that=AMand extendMCtoANBsuchthat

=AN.NBThen CB || MN and

= 2.CB.MN

This rewording generalizes SAS-congruence in a highly suggestive way, and points us in the direction of “SAS-similarity”.

SAS-similarity(x2): if= 2 .AʹBʹ, ∠ABBAC= ∠BʹAʹCʹ, and

= 2 .AʹCʹ, thenAC

= 2 .BʹCʹ, ∠BCABC= ∠AʹBʹCʹ, and ∠BAC= ∠BʹAʹCʹ.

ProofExtendABto the point Bʹʹ such thatABʹʹ=AʹBʹ, and extendACto the pointCʹʹsuch that=ACʹʹ. Then ΔAʹCʹBʹʹACʹʹ= ΔBʹAʹCʹ(by SAS-congruence), so=BʹʹCʹʹ, ∠BʹCʹBʹʹCʹʹA= ∠BʹCʹAʹ, ∠CʹʹBʹʹA= ∠CʹBʹAʹ. By construction we have= 2 .ABʹʹandAB= 2 .ACʹʹ. Hence (by the Midpoint Theorem):AC= 2 .BʹʹCʹʹ(soBC= 2 .BʹCʹ), andBCBC||BʹʹCʹʹ(so ∠BCA= ∠BʹʹCʹʹAand ∠CBA= ∠CʹʹBʹʹA).∴ ∠

BʹʹCʹʹA= ∠BʹCʹAʹ= ∠BCA,and ∠

CʹʹBʹʹA= ∠CʹBʹAʹ= ∠CBA. QED

The SAS-similarity ($\times 2$) interpretation of the Midpoint Theorem is like the SAS-congruence criterion in that one pair of corresponding angles in Δ*BAC* and Δ*BʹAʹCʹ* are equal, while the sides on either side of this angle in the two triangles are related; but instead of the two pairs of corresponding sides being equal, the sides of Δ*BʹAʹCʹ* are double the corresponding sides of Δ*BAC*.

In general we say that

Δ

ABCissimilarto ΔAʹBʹCʹ(written as ΔABC~ ΔAʹBʹCʹ) with scale-factormif each angle of ΔAʹBʹCʹis equal to the corresponding angle of ΔABC, and if corresponding sides are all in the same ratio:

$$\underset{\_}{{A}^{\prime}{B}^{\prime}}:\underset{\_}{AB}=\underset{\_}{{B}^{\prime}{C}^{\prime}}:\underset{\_}{BC}=\underset{\_}{{C}^{\prime}{A}^{\prime}}:\underset{\_}{CA}=m:1.$$

If two triangles Δ*AʹBʹC*ʹ and Δ*ABC* are similar, with (linear) scale factor Δ*AʹBʹC*ʹ, then the ratio between their areas is

$$area\left(\Delta {A}^{\prime}{B}^{\prime}{C}^{\prime}\right):area(\Delta ABC)={m}^{2}:1.$$

Two similar triangles Δ *ABC* and Δ *A'B'C'* give rise to six matching pairs:

- the three pairs of corresponding angles (which are equal in pairs), and
- the three pairs of corresponding sides (which are in the same ratio).

In the case of congruence, the **congruence criteria** tell us that we do not need to check all six pairs to guarantee that two triangles are congruent: these criteria guarantee that certain *triples* suffice. The **similarity criteria** guarantee much the same for similarity.

Suppose we are given triangles Δ*ABC*, Δ*AʹBʹC*ʹ.

**AAA-similarity: If**

$$\angle ABC=\angle {A}^{\prime}{B}^{\prime}{C}^{\prime},\angle BCA=\angle {B}^{\prime}{C}^{\prime}{A}^{\prime},\angle CAB=\angle {C}^{\prime}{A}^{\prime}{B}^{\prime},$$

then

$$\underset{\_}{{A}^{\prime}{B}^{\prime}}:\underset{\_}{AB}=\underset{\_}{{B}^{\prime}{C}^{\prime}}:\underset{\_}{BC}=\underset{\_}{{C}^{\prime}{A}^{\prime}}:\underset{\_}{CA},$$

so the two triangles are similar.

**SSS-similarity**: If

$$\underset{\_}{{A}^{\prime}{B}^{\prime}}:\underset{\_}{AB}=\underset{\_}{{B}^{\prime}{C}^{\prime}}:\underset{\_}{BC}=\underset{\_}{{C}^{\prime}{A}^{\prime}}:\underset{\_}{CA},$$

then

$$\angle ABC=\angle {A}^{\prime}{B}^{\prime}{C}^{\prime},\angle BCA=\angle {B}^{\prime}{C}^{\prime}{A}^{\prime},\angle CAB=\angle {C}^{\prime}{A}^{\prime}{B}^{\prime},$$

so the two triangles are similar.

**SAS-similarity**: If

$$\underset{\_}{{A}^{\prime}{B}^{\prime}}:\underset{\_}{AB}=\underset{\_}{{A}^{\prime}{C}^{\prime}}:\underset{\_}{AC}=m:1$$

and

$$\angle {B}^{\prime}{A}^{\prime}{C}^{\prime}=\angle BAC,$$

then

$$\underset{\_}{{B}^{\prime}{C}^{\prime}}:\underset{\_}{BC}=\underset{\_}{{A}^{\prime}{B}^{\prime}}:\underset{\_}{AB}={\underset{\_}{A}}^{\prime}{C}^{\prime}:\underset{\_}{AC}$$

and

$$\angle {A}^{\prime}{B}^{\prime}{C}^{\prime}=\angle ABC,\text{\hspace{1em}}\angle {B}^{\prime}{C}^{\prime}{A}^{\prime}=\angle BCA,$$

so the two triangles are similar.

Our rewording of the Midpoint Theorem gave rise to a version of the third of these criteria, with *m* = 2.

AAA-similarity in right angled triangles is what makes trigonometry possible. Suppose that two triangles Δ*ABC*,ΔAʹBʹCʹ have right angles at A and at Aʹ. If ∠ABC,∠AʹBʹCʹ, then (since the angles in each triangle add to two right angles) we also have ∠*B*CA,∠*B*ʹCʹAʹ. It then follows (from AAA-similarity) that

$$\underset{\_}{{A}^{\prime}{B}^{\prime}}:\underset{\_}{AB}=\underset{\_}{{B}^{\prime}{C}^{\prime}}:\underset{\_}{BC}=\underset{\_}{{C}^{\prime}{A}^{\prime}}:\underset{\_}{CA},$$

so the trig ratio in Δ*ABC*

$$\mathrm{sin}B=\frac{\underset{\_}{AC}}{\underset{\_}{BC}}$$

has the same value as the corresponding ratio in Δ*AʹBʹCʹ*

$$\mathrm{sin}{B}^{\prime}=\frac{\underset{\_}{{A}^{\prime}{C}^{\prime}}}{\underset{\_}{{B}^{\prime}{C}^{\prime}}}.$$

Hence this ratio *depends only on the angle B,* and not on the triangle in which it occurs. The same holds for cos ∠*B* and for tan ∠*B*.

The art of solving geometry problems often depends on looking for, and identifying, similar triangles hidden in a complicated configuration. As an introduction to this, we focus on three classic properties involving circles, where the figures are sufficiently simple that similar triangles should be fairly easy to find.

**Problem 167** The point *P* lies outside a circle. The tangent from *P* touches the circle at *T*, and a secant from P cuts the circle at *A* and at *B*. Prove that $\underset{\_}{PA}\times \underset{\_}{PB}={\underset{\_}{PT}}^{2}$
.

**Problem 168** The point *P* lies outside a circle. Two secants from *P* meet the circle at A, B and at C, *D* respectively. Prove in two different ways that

$$\underset{\_}{PA}\times \underset{\_}{PB}=\underset{\_}{PC}\times \underset{\_}{PD}.$$

**Problem 169** The point *P* lies inside a circle. Two secants from *P* meet the circle at *A*, *B* and at *C*, *D* respectively. Prove in two different ways that

$$\underset{\_}{PA}\times \underset{\_}{PB}=\underset{\_}{PC}\times \underset{\_}{PD}.$$

We end our summary of the foundations of Euclidean geometry by deriving the familiar formula for the area of a trapezium and its 3-dimensional analogue, and a formulation of the similarity criteria which is often attributed to Thales (Greek 6^{th} century BC).

**Problem 170** Let *ABCD* be a trapezium with *AB* || *DC*, in which *AB* has length *a* and *DC* has length *b.*

(a) Let *M* be the midpoint of * AD* and let

*N*be the midpoint of

*. Prove that*

__BC__*MN*|| AB and find the length of

*.*

__MN__(b) If the perpendicular distance between *AB* and *DC* is *d,* find the area of the trapezium *ABCD*.

**Problem 171** A pyramid *ABCDE*, with apex *A* and square base *BCDE* of side length *b*, is cut parallel to the base at height *d* above the base, leaving a frustum of a pyramid, with square upper face of side length *a.* Find a formula for the volume of the resulting solid (in terms of *a*, *b*, and *d*).

The following general result allows us to use “equality of ratios of line segments” whenever we have three parallel lines (without first having to conjure up similar triangles).

**Problem 172 (Thales’ Theorem)** The lines *AA'* and *BB'* are parallel. The point *C* lies on the line *AB*, and *C'* lies on the line *A'B'* such that *CCʹ* || BBʹ. Prove that * AB* :

*=*

__BC____AʹBʹ__:

__BʹCʹ__.

Under certain conditions, the similarity criteria guarantee the equality of ratios of sides of two triangles. Thales’ Theorem extends this “equality of ratios” to line segments which arise whenever two lines cross three parallel lines. One of the simplest, but most far-reaching, applications of this result is the tie-up between geometry and algebra which lies behind ruler and compasses constructions, and which underpins Descartes’ (1596-1650) re-formulation of geometry in terms of coordinates (see Problem **173**).

Thales (c. 620-c. 546 BC) was part of the flowering of Greek thought having its roots in Milesia (in the south west of Asia Minor, or modern Turkey). Thales seems to have been interested in almost everything - philosophy, astronomy, politics, and also geometry. In Britain his name is usually attached to the fact that the angle subtended by a diameter is a right angle. On the continent, his name is more strongly attached to the result in Problem **172**. His precise contribution to geometry is unclear - but he seems to have played a significant role in kick-starting what became (300 years later) the polished version of Greek mathematics that we know today.

Thales’ contributions in other spheres were perhaps even more significant than in geometry. He seems to have been among the first to try to “explain” phenomena in reductionist terms - identifying “water” as the single “element”, or first principle, from which all substances are derived. Anaximenes (c. 586-c. 526 BC) later argued in favour of “air” as the first principle. These two elements, together with “fire” and “earth”, were generally accepted as the four Greek “elements” - each of which was supposed to contribute to the construction of observed matter and change in different ways.

**Problem 173** To define “length”, we must first decide which line segment is deemed to have unit length. So suppose we are given line segments * XY* of length 1,

*of length*

__AB__*a,*(i.e.

*:*

__AB__*= a : 1), and*

__XY__*of length*

__CD__*b*.

(a) Use Problem **137** to construct a segment of length *a + b,* and if $a\u2a7eb$, a segment of length *a — b.*

(b) Show how to construct a line segment of length *ab* and a segment of length $\frac{a}{b}$
.

(c) Show how to construct a line segment of length $\sqrt{a}$ .

## 5.3. Areas, lengths and angles

**Problem 174** A rectangular piece of fruitcake has a layer of icing on top and down one side to form a larger rectangular slab of cake (as shown in Figure 3).

Figure 3: Icing on the cake

Describe how to make a single straight cut so as to divide both the fruitcake and the icing exactly in half. (The thickness of the icing on top is not necessarily the same as the thickness down the side.)

**Problem 175**

(a) What is the angle between the two hands of a clock at 1:35? Can you find another time when the angle between the two hands is the same as this?

(b) How many times each day do the two hands of a clock ‘coincide’? And at what times do they coincide?

(c) If we add a second hand, how many times each day do the three hands coincide?

**Problem 176** The twelve hour marks for a clock are marked on the circumference of a unit circle to form the vertices of a regular dodecagon *ABCDEFGHIJKL.* Calculate exactly (i.e. using Pythagoras’ Theorem rather than trigonometry) the lengths of all the possible line segments joining two vertices of the dodecagon.

**Problem 177** Consider the lattice of all points *(k,m,n)* in 3-dimensions with integer coordinates *k*, *m*, *n*. Which of the following distances can be realised between lattice points?

$$\sqrt{1},\sqrt{2},\sqrt{3},\sqrt{4},\sqrt{5},\sqrt{6},\sqrt{7},\sqrt{8},\sqrt{9},\sqrt{10},\sqrt{11},\sqrt{13},\sqrt{13},\sqrt{15},\sqrt{16},\sqrt{16},\sqrt{17}$$

**Problem 178**

(a) Five vertices *A, B, C, D*, *E* are arranged in cyclic order. However instead of joining each vertex to its two immediate neighbours to form a convex pentagon, we join each vertex to the *next but one* vertex to form a pentagonal star, or pentagram *ACEBD.* Calculate the sum of the five “angles” in any such pentagonal star.

(b) There are two types of 7-gonal stars. Calculate the sum of the angles at the seven vertices for each type.

(c) Try to extend the previous two results (and the proofs) to arbitrary *n*-gonal stars.

**Problem 179**

(a) A regular pentagon *ABCDE* with edges of length 1 is surrounded in the plane by five new regular pentagons - *ABLMN* joined to * AB*,

*BCOPQ*joined to

*, and so on.*

__BC__(i) Prove that *M, N*, *X*, *Y* lie on a line.

(ii) Prove that *MPSVY* is a regular pentagon.

(iii) Find the edge length of this larger surrounding regular pentagon.

(b) Given a regular pentagon *MPSVY*, with edge length 1, draw the five diagonals to form the pentagram *MSYPV*. Let *PY* meet *MV* at *A,* and *MS* at *B*; let *PV* meet *MS* at *C* and *SY* at *D;* and let *SY* meet *VM* at *E*.

(i) Prove that *ABCDE* is a regular pentagon.

(ii) Prove that *A*, *B*, and *M* are three vertices of a regular pentagon *ABLMN*, where *L* lies on * MP* and

*N*lies on

*.*

__MY__(iii) Find the edge length of the regular pentagon *ABCDE*.

## 5.4. Regular and semi-regular tilings in the plane

In Problem **36** we saw that a regular n-gon has a *circumcentre O.* If we join each vertex to the point *O*, we get *n* triangles, each with angle sum π. Hence the total angle sum in all *n* triangles is πn. Since the *n* angles around the point *O* add to 2π, the angles of the regular *n*-gon itself have sum (n-2)π. Hence each angle of the regular *n*-gon has size $\left(1-\frac{2}{n}\right)\pi $
.(In the next chapter you will prove the general result that the sum of the angles in **any** *n*-gon is equal to (n-2)π radians.)

**Problem 180** A *regular tiling* of the plane is an arrangement of identical regular polygons, which fit together edge-to-edge so as to cover the plane with no overlaps.

(a) Prove that if a regular tiling of the plane with p-gons is possible, then *p =* 3,4, or 6.

(b) Prove that a regular tiling of the plane exists for each of the values in (a).

We refer to the arrangement of tiles around a vertex as the *vertex figure.* In a regular tiling all vertex figures are automatically identical, so it is natural to refer to the tiling in terms of its vertex figure. When *p =* 3, exactly *q =* 6 tiles fit together at each vertex, and we abbreviate “six equilateral triangles” as 3^{6}. In the same way we denote the tiling whose vertex figure consists of “four squares” as 4^{4}, and the tiling whose vertex figure consists of “three regular hexagons” as 6^{3}.

The natural approach in part (a) of Problem **180** is first to identify which *vertex figures* have no gaps or overlaps - giving a *necessary* condition for a regular tiling to exist. It is tempting to stop there, and to **assume** that this obvious *necessary* condition is also *sufficient.* The temptation arises in part because 2-dimensional regular tilings are so familiar. But it is important to recognize the distinction between a necessary and a sufficient condition; so the temptation should be resisted, and a construction given.

The procedure hidden in the solution to Problem **180** illustrates a key strategy, which dates back to the ancient Greeks, and which is called the *method of analysis*.

- First, we
*imagine*that we have a typical solution to the problem - perhaps by giving it a name (even though we do not yet know anything about such a solution). - We then use the given conditions to deduce features which any such solution must
*necessarily*have. - And we continue deriving more and more necessary conditions until we believe our list of derived conditions may also be
*sufficient*. - Finally we show that any configuration which satisfies our final derived list of necessary conditions is in fact a solution to the original problem, so that the list of necessary conditions is in fact
*sufficient*, and we have effectively pinned down all possible solutions.

This is what we did in a very simple way in the solution to Problem **180**: the condition on vertex figures gave an evident necessary condition, which turned out to be sufficient to guarantee that such a tiling exists. The same general strategy guided our classification of *primitive Pythagorean triples *back in Problem **23**.

In the seventeenth century, this ancient Greek strategy was further developed by Fermat (1601-1665), and by Descartes (1596-1650). For example, Fermat left very few proofs; but his proof that the equation

$${x}^{4}+{y}^{4}={z}^{4}$$

has no solutions in positive integers *x*, *y, z* illustrated the method:

- Fermat started by supposing that a solution exists, and concluded that (
*x*^{2},*y*^{2},*z*^{2}) would then be a Pythagorean triple. - The known formula for such Pythagorean triples then allowed him to derive even stronger necessary conditions on
*x*,*y*,*z*. - These conditions were so strong they could never be satisfied!

Descartes developed a “method”, whereby hard geometry problems could be solved by translating them into algebra - essentially using the *method of analysis*.

- Faced with a hard problem, Descartes first imagined that he had a point, or a locus, or a curve of the kind required for a solution.
- Then he introduced coordinates “x” and “y” to denote unknowns that were linked in the problem to be solved, and interpreted the given conditions as
*equations*which the unknowns x and*y*would have to satisfy (i.e. as*necessary*constraints). - The solutions to these equations then corresponded to possible solutions of the original problem.
- Sometimes the algebra did not quite generate a
*sufficient*condition, giving rise to “pseudo-solutions” (values of x that satisfy the necessary conditions, but which did not correspond to actual solutions). So it was important to check each apparent solution - exactly as we did in Problem**180**(b), where we checked that we can construct tilings for each of the vertex figures that arise in part (a).

The importance of the final step in this process (checking that the list of necessary constraints is also sufficient) is underlined in the next problem where we try to classify certain “almost regular” tilings.

**Problem 181** A *semi-regular tiling* of the plane is an arrangement of regular polygons (not necessarily all identical), which fit together edge-to-edge so as to cover the plane without overlaps, and such that the arrangements of tiles around any two vertices are congruent.

(a) (i) Refine your argument in Problem **180**(a) to list all possible vertex figures in a semi-regular tiling.

(ii) Try to find additional necessary conditions to eliminate vertex figures which cannot be realized, until your list of necessary conditions seems likely to be sufficient.

(b) The necessary conditions in part (a) give rise to a finite list of possible vertex figures. Construct all possible tilings corresponding to this list of possible vertex figures.

Semi-regular tilings are often called *Archimedean* tilings. The reason for this name remains unclear. Pappus (c. 290-c. 350 AD), writing more than 500 years after the death of Archimedes (d. 212 BC), stated that Archimedes classified the semi-regular *polyhedra.* Now the classification of semi-regular polyhedra (Problem **190**) uses a similar approach to the classification of planar tilings, except that the sum of the angles at each vertex has sum *less than* (rather than exactly equal to) 360°. So it may be that the semi-regular tilings are named after Archimedes simply because he did something similar for polyhedra; or it may be that, since inequalities are harder to control than equalities, someone inferred (perhaps dodgily) that Archimedes must have known about semi-regular tilings as well as about semi-regular polyhedra. Whatever the reason, tilings and polyhedra have fascinated mathematicians, artists and craftsmen for all sorts of unexpected reasons - as indicated by:

- the fact that the classification and construction of the five regular polyhedra appear as the culmination of the thirteen books of
*Elements*by Euclid (flourished c. 300 BC); - the ancient Greek attempt to link the five regular polyhedra with the four elements (earth, air, fire, and water) and the cosmos;
- the ceramic tilings to be found in Islamic art - for example, on the walls of the
*Alhambra*in Grenada; - the book
*De Divina Proportions*by Luca Pacioli (c. 1445-1509), and the continuing fascination with the Golden Ratio; - the geometric sketches of Leonardo da Vinci (1452-1519);
- the work of Kepler (1571-1630) who used the regular polyhedra to explain his bold theoretical cosmology in the
*Astronomia Nova*(1609).

## 5.5. Ruler and compasses constructions for regular polygons

Euclid’s *Elements* include methods for constructing the regular polygons that are required for the construction of the regular polyhedra (see Section 5.6). In one sense, Euclid is thoroughly modern: he is reluctant to work with entities that cannot be *constructed.* And for him, geometrical construction means construction “using ruler and compasses” only.

For each regular polygon, there are two related (and sometimes very different) construction problems:

- given two points
*A*and*B,*construct the regular n-gon with__AB__as an*edge*of the regular polygon; - given two points O and A, construct the regular n-gon
*ABCD*. . . inscribed in the circle with centre O and passing through A, that is with*circumradius*__OA__.

Before Problem **137** we saw how to construct an equilateral triangle *ABC *given the points A, B. And in Problem **36** we saw that every regular polygon has a circumcentre *O*.

**Problem 182** Given points *O, A,* show how to construct the regular 3-gon *ABC* with circumcentre O.

**Problem 183**

(a) Given two points *O*, *A*, show how to construct a regular 4-gon *ABCD *with circumcentre *O*.

(b) Given points *A*, *B*, show how to construct a regular 4-gon *ABCD.*

**Problem 184**

(a)(i) Given two points *O*, *A*, show how to construct a regular 6-gon *ABCDEF* with circumcentre *O*.

(ii) Given two points *O*, *A*, show how to construct a regular 8-gon *ABCDEFGH* with circumcentre *O*.

(b)(i) Given points *A*, *B*, show how to construct a regular 6-gon *ABCDEF*.

(ii) Given points *A*, *B*, show how to construct a regular 8-gon *ABCDEFGH*.

**Problem 185**

(a) (i) Given two points *O*, *A*, show how to construct a regular 5-gon *ABCDE* with circumcentre *O*.

(ii) Given points *O*, *A*, show how to construct a regular 10-gon *ABCDEFGHIJ* with circumcentre *O*.

(b) (i) Given points *A*, *B*, show how to construct a regular 5-gon *ABCDE*.

(ii) Given points *A*, *B*, show how to construct a regular 10-gon *ABCDEFGHIJ*.

We shall not prove it here, but it is impossible to construct a regular 7-gon, or a regular 9-gon, or a regular 11-gon using ruler and compasses. All constructions with ruler and compasses come down to two moves:

- if
*a*is a known length, then $\sqrt{a}$ can be constructed (see Problem**173**(c)); - if an
*n*-gon can be constructed, then the sides can be bisected to produce a 2n-gon.

Put slightly differently, all ruler and compasses constructions involve solving linear or quadratic equations, so the only new points, or lengths we can construct are those which involve iterated square roots of expressions or lengths which were previously known.

This iterated extraction of square roots is linked to a fact first proved by Gauss (1777-1855), namely that the only regular p-gons (where *p* is a prime) that can be constructed are those where p is a Fermat prime - that is, a prime of the form p = *2 ^{k} +* 1 (in which case k has to be a power of 2: see Problem

**118**). Gauss proved (as a teenager, though it was first published in his book

*Disquisitiones arithmeticae*in 1801):

a regular n-gon can be constructed with ruler and compasses if and only if n has the form$${2}^{m}\xb7{p}_{1}\xb7{p}_{2}\xb7{p}_{3}\cdots {p}_{k},$$

where p

_{1},p_{2},p_{3},... ,p_{k}are distinct Fermat primes.

As we noted in Chapter 2, the only known Fermat primes are the five discovered by Fermat himself, namely 3, 5, 17, 257, and 65 537.

## 5.6. Regular and semi-regular polyhedra

We have seen how regular polygons sometimes fit together edge-to-edge in the plane to create tilings of the whole plane. When tiling the plane, the angles of polygons meeting edge-to-edge around each vertex must add to 360°, or two straight angles. If the angles at a vertex add to *less than* 360°, then we are left with an empty gap and two free edges; and when these two free edges are joined, or glued together, the vertex figure rises out of the plane and becomes a 3-dimensional corner, or *solid angle.*

To form such a corner we need at least three polygons, or faces - and hence at least three edges and three faces meet around each vertex. For example, three squares fit nicely together in the plane, but leave a 90° gap. When the two spare edges are glued together, the result is to form a corner of a cube, where we have a vertex figure consisting of three regular 4-gons: so we refer to this vertex figure as 4^{3}.

Given a 3-dimensional corner, it may be possible to extend the construction, repeating the same vertex figure at every vertex. The resulting shape may then ‘close up’ to form a *convex* polyhedron. The assumption that in each vertex figure, the angles meeting at that vertex add to less than 360°, means that all the corners then project outwards - which is roughly what we mean when we say that the polyhedron is “convex”.

A *regular polygon* is an arrangement of finitely many congruent line segments, with two line segments meeting at each vertex (and never crossing, or meeting internally), and with all vertices alike; a regular polygon can be inscribed in a circle (Problem **36**), and so encloses a convex subset of the plane. In the same spirit, a *regular polyhedron* is an arrangement of finitely many congruent regular polygons, with two polygons meeting at each edge, and with the same number of polygons in a single cycle around every vertex, enclosing a convex subset of 3-dimensional space (i.e. the polyhedron separates the remaining points of 3D into those that lie ‘inside’ and those that lie ‘outside’, and the line segment joining any two points of the polyhedral surface contains no points lying outside the polyhedron).

The important constraints here are the assumptions: that the polygons meet edge-to-edge with exactly two polygons meeting at each edge; that the same number of polygons meet around every vertex; and that the overall number of polygons, or faces, is *finite.* The assumption that the figure is convex should be seen as a temporary additional constraint, which means that the angles in polygons meeting at each vertex have sum less than 360°.

**Problem 186** A vertex figure is to be formed by fitting regular p-gons together, edge-to-edge, for a fixed *p.* If there are *q* of these p-gons at a vertex, we denote the vertex figure by *p*^{q}. If the angles at each vertex add to less than 360°, prove that the only possible vertex figures are 3^{3}, 3^{4}, 3^{5}, 4^{3}, 5^{3}.

The vertex figure 4^{3} is realized by the way the positive axes meet at the vertex (0,0,0), where

- the unit square (0, 0,0), (1, 0,0), (1,1,0), (0,1,0) in the xy-plane (with equation z = 0) meets
- the unit square (0,0,0), (1,0,0), (1,0,1), (0,0,1) in the xz-plane (with equation y = 0), and
- the unit square (0, 0, 0), (0,1,0), (0,1,1), (0, 0,1) in the yz-plane (with equation x = 0).

If we include an eighth vertex (1, 1, 1), and

- the unit square (0, 0, 1), (1, 0, 1), (1, 1, 1), (0, 1, 1) in the plane with equation z = 1,
- the unit square (0,1,0), (1,1,0), (1,1,1), (0,1,1) in the plane with equation y = 1,
- the unit square (1,0,0), (1,1,0), (1,1,1), (1,0,1) in the plane with equation x = 1,

we see that all eight vertices have the same vertex figure 4^{3}. Hence the possible vertex figure 4^{3} in Problem **186** arises as the vertex figure of a regular polyhedron - namely the *cube.*

If we select the four vertices whose coordinates have odd sum *A* = (1, 0, 0), *B =* (0,1, 0), *C =* (0, 0,1), *D =* (1,1,1), then the distance between any two of these vertices is equal to $\sqrt{2}$, so each triple of vertices (such as (1, 0, 0), (0,1,0), (0, 0,1)) defines a regular 3-gon *ABC*, with three such 3-gons meeting at each vertex of *ABCD.* Hence the possible vertex figure 3^{3} in Problem **186** arises as the vertex figure of a regular polyhedron - namely the regular *tetrahedron* (*tetra* = four; *hedra* = faces).

**Problem 187** With *A =* (1,0,0) etc. as above, write down the coordinates of the six midpoints of the edges of the regular tetrahedron *ABCD* (or equivalently, the six centres of the faces of the original cube). Each edge of the regular tetrahedron meets four other edges of the regular tetrahedron (e.g. * AB* meets

*and*

__AC__*at one end, and*

__AD__*and*

__BC__*at the other end). Choose an edge*

__BD__*and its midpoint*

__AB__*P*. Calculate the distance from

*P*to the midpoints

*Q*,

*R, S, T*of the four edges which

*meets (namely the midpoints of*

__AB__*,*

__AB__*,*

__AD__*,*

__BD__*respectively). Confirm that the triangles ΔPQR, ΔPRS, ΔPST, ΔPTQ are all regular 3-gons, and that the vertex figure at*

__BC__*P*is of type 3

^{4}. Conclude that the possible vertex figure 3

^{4}in Problem

**186**arises as the vertex figure of a regular polyhedron

*PQRSTU*- namely the

*regular octahedron*(

*octa*= eight;

*hedra*= faces).

**Problem 188**

(a) A *regular tetrahedron ABCD* has edges of length 2, and sits with its base *BCD* on the table. Find the height of *A* above the base.

(b) A *regular octahedron ABCDEF* has four triangles meeting at each vertex.

(i) Let the four triangles which meet at *A* be *ABC*, *ACD, ADE, AEB. *Prove that *BCDE* must be a square.

(ii) Suppose that all the triangles have edges of length 2, and that the octahedron sits with one face *BCF* on the table - next to the regular tetrahedron from part (a). Which of these two solids is the taller?

**Problem 189** Let *O* = (0, 0, 0), *A* = (1,0,0), *B* = (0,1,0), *C* = (0,0,1) be four vertices of the cube as described after Problem **186** above. Draw equal and parallel line segments (initially of unknown length 1 — 2*a)* through the centres of each pair of opposite faces - running in the three directions parallel to *OA*, or to *OB*, or to *OC*

- from $N=\left(a,\frac{1}{2},0\right)$ to $P=\left(1-a,\frac{1}{2},0\right)$ and from $Q=\left(a,\frac{1}{2},1\right)$ to $R=\left(1-a,\frac{1}{2},1\right)$
- from $S=\left(\frac{1}{2},0,a\right)$ to $T=\left(\frac{1}{2},0,1-a\right)$ and from $U=\left(\frac{1}{2},1,a\right)$ to $V=\left(\frac{1}{2},1,1-a\right)$
- from $W=\left(0,a,\frac{1}{2}\right)$ to $X=\left(0,1-a,\frac{1}{2}\right)$ and from $Y=\left(1,a,\frac{1}{2}\right)$ to $Z=\left(1,1-a,\frac{1}{2}\right)$ .

Figure 4: Construction of the regular icosahedron.

These are to form all 12 vertices and six of the 30 edges (of length 1 — 2a) of a polyhedron, see Figure 4. The other 24 edges join each of these 12 vertices to its four natural neighbours on adjacent faces of the cube - to form the 20 triangular faces of the polyhedron: for example,

Njoins: toS; toW; toX; and toU.

(i) Prove that * NS* =

*=*

__NW__*=*

__NX__*and calculate the length of*

__NU__

*.*__NS__(ii) Choose the value of the parameter a to guarantee that * NP* =

*, so that the five triangular faces meeting at the vertex N are all equilateral triangles, and each vertex figure of the resulting polyhedron then has vertex figure 3*

__NS__^{5}.

The polyhedron is called the *regular icosahedron (icosa* = twenty, *hedra* = faces).

In the paragraph before Problem **187** we constructed the *dual* of the cube by marking the circumcentre of each of the six square faces of the cube, and then joining each circumcentre to its four natural neighbours. We now construct the *dual* of the regular icosahedron in exactly the same way. Each of the 20 circumcentres of the 20 triangular faces of a regular icosahedron has three natural neighbours (namely the circumcentres of the three neighbouring triangular faces). If we construct the 30 edges joining these 20 circumcentres, the five circumcentres of the five triangles in each vertex figure of the regular icosahedron form a regular pentagon, which becomes a face of the dual polyhedron - so we get 12 regular pentagons (one for each vertex of the regular icosahedron), with three pentagons meeting at each vertex of the dual polyhedron to give a vertex figure 5^{3} at each of the 20 vertices, which form a *regular dodecahedron*.

Hence each of the five possible vertex figures in Problem **186** can be realised by a regular polyhedron. These are sometimes called the *Platonic solids *because Plato (c. 428-347 BC) often used them as illustrative examples in his writings on philosophy.

Constructing the five regular polyhedra is part of the essence of mathematics for everyone. In contrast, what comes next (in Problem 190) may be viewed as “optional” at this stage. The ideas are worth noting, but the details may be best postponed for a rainy day.

Just as you classified semi-regular *tilings* in Section 5.4, so one can look for semi-regular *polyhedra.* A polyhedron is semi-regular if all of its faces are regular polygons (possibly with differing numbers of edges), fitting together edge-to-edge, with exactly the same ring of polygons around each vertex - the *vertex figure* of the polyhedron. Problem **190** uses “the method of analysis” - combining simple arithmetic, inequalities, and a little geometric insight - to achieve a remarkable complete classification of semi-regular polyhedra. There are usually said to be thirteen individual semi-regular polyhedra (excluding the five regular polyhedra); but one of these has a vertex figure that extends to a polyhedron in two different ways - each being the reflection of the other. There are in addition two infinite families - namely

- the n-gonal
*prisms,*which consist of two parallel regular n-gons, with the top one positioned exactly above the bottom one, the two being joined by a belt of n squares (so with vertex figure n . 4^{2}); and - the n-gonal
*antiprisms,*which consist of two parallel regular n-gons, but with the top n-gon turned through an angle of n radians relative to the bottom one, the two being joined by a belt of 2n equilateral triangles (so with a vertex figure n . 3^{3}).

Notice that the cube can also be interpreted as being a “4-gonal prism”, and the regular octahedron can be interpreted as being a “3-gonal antiprism”. Those interested in regular and semi-regular polyhedra are referred to the classic book *Mathematical models* by H.M. Cundy and A.P. Rollett.

**Problem 190** Find possible combinations of three or more regular polygons whose angles add to less than 360°, and hence derive a complete list of possible vertex figures for a (convex) semi-regular polyhedron. Try to eliminate those putative vertex figures that cannot be extended to a semi-regular polyhedron.

## 5.7. The Sine Rule and the Cosine Rule

Where given information, or a specified geometrical construction, determines an angle or length uniquely, it is sometimes - but not always - possible to find this angle or length using simple-minded angle-chasing and congruence.

**Problem 191**

(a) In the quadrilateral *ABCD* the two diagonals * AC* and

*cross at*

__BD__*X*. Suppose

*=*

__AB__*, ∠*

__BC__*B*AC = 60°, ∠DAC = 40°, ∠

*B*XC = 100°.

(i) Calculate (exactly) ∠ADB and ∠*C*BD.

(ii) Calculate ∠*B*DC and ∠ACD.

(b) In the quadrilateral *ABCD* the two diagonals * AC* and

*cross at*

__BD__*X*. Suppose

*=*

__AB__*, ∠*

__BC__*B*AC = 70°, ∠DAC = 40°, ∠

*B*XC = 100°.

(i) Calculate (exactly) the size of ∠*B*DC + ∠ACD.

(ii) Explain how we can be sure that ∠*B*DC and ∠ACD are uniquely determined, even though we cannot calculate them immediately.

If it turns out that the simplest tools do not allow us to determine angles and lengths, this is usually because we are only using the most basic properties: the congruence criteria, and the parallel criterion. The general art of ‘solving triangles’ depends on the similarity criterion (usually via trigonometry). And the two standard techniques for ‘solving triangles’ that go beyond “angle-chasing” and congruence are the *Sine Rule,* which was established back in Problem **32** (and its consequences, such as the area formula $\frac{1}{2}ab\mathrm{sin}C$
- see Problem **33**), and the *Cosine Rule.*

The next problem invites you to use Pythagoras’ Theorem to prove the Cosine Rule - an extension of Pythagoras’ Theorem which applies to *all *triangles *ABC* (including those where the angle at *C* may not be a right angle).

**Problem 192 (The Cosine Rule)** Given Δ*ABC*, let the perpendicular from *A* to *BC* meet *BC* at *P*. If *P = C,* then we know (by Pythagoras’ Theorem) that *c ^{2} = a^{2} + b^{2}.* Suppose $P\ne C$
.

(i) Suppose first that *P* lies on the line segment * CB*, or on

*extended beyond B. Express the lengths of*

__CB__*and*

__PC__*in terms of b and ∠*

__AP__*C*. Then apply Pythagoras’ Theorem to A ΔAPB to conclude that

$${c}^{2}={a}^{2}+{b}^{2}-2ab\mathrm{cos}C.$$

(ii) Suppose next that *P* lies on the line segment * BC* extended beyond

*C*. Prove once again that

$${c}^{2}={a}^{2}+{b}^{2}-2ab\mathrm{cos}C.$$

**Problem 193** Go back to the configuration in Problem **191**(b). The required angles are unaffected by scaling, so we may choose * AB* =

*= 1. Devise a strategy using the Sine Rule and the Cosine Rule to calculate ∠*

__BC__*B*DC and ∠ACD exactly.

It is worth reflecting on what the Cosine Rule really tells us:

(i) if in a triangle, we know any two sides *(a* and *b)* and the included angle (C), then we can calculate the third side (c); and

(ii) if we know all three sides (a, b, c), then we can calculate any angle (say C).

Hence if we know three sides, or two sides and the angle between them, we can work out all of the angles. The Sine Rule then complements this by ensuring that:

(iii) if we know any side and two angles (in which case we also know the third angle), then we can calculate the other two sides; and

(iv) if we know any angle *A*, and two sides - one of which is the side *a* opposite A, then we can calculate (one and hence) both the other angles (and hence the third side).

The upshot is that once a triangle is uniquely determined by the given data, we can “solve” to find all three sides and all three angles.

Trigonometry has a long and very interesting history (which is not at all easy to unravel). Euclid (flourished c. 300 BC) understood that corresponding sides in similar figures were “proportional”. And he stated and proved the generalization of Pythagoras’ Theorem, which we now call the Cosine Rule; but he did this in a theoretical form, without introducing cosines. Euclid’s versions for acute-angled and obtuse-angled triangles involved correction terms with opposite signs, so he proved them separately *(Elements*, Book II, Propositions 12 and 13).

However, the development of trigonometry as an effective theoretical and practical tool seems to have been due to Hipparchus (died c. 125 BC), to Menelaus (c. 70-130 AD), and to Ptolemy (died 168 AD). Once trigonometry moved beyond the purely theoretical, the combination of

- the (exact) language of trigonometry, together with the Sine Rule and the Cosine Rule, and
- (approximate) “tables of trigonometric ratios” (nowadays replaced by calculators)

liberated astronomers, and later engineers, to calculate lengths and angles efficiently, and as accurately as they required.

In mathematics we either work with *exact* values, or we have to control errors precisely. But trigonometry can still be a valuable *exact* tool, provided we remember the lessons of working with fractions such as $\frac{2}{3}$ or with surds such as , or with constants such as $\sqrt{2}$, and resist the temptation to replace them by some unenlightening approximate decimal. We can replace ${\mathrm{cos}}^{-1}\left(\frac{1}{2}\right)\left(=\frac{\pi}{3}\right)$ and ${\mathrm{cos}}^{-1}\left(-\frac{1}{2}\right)\left(=\frac{2\pi}{3}\right)$ by their exact values; but in general we need to be willing to work with, and to think about, *exact forms* such as “${\mathrm{cos}}^{-1}\left(\frac{1}{3}\right)$” and “${\mathrm{cos}}^{-1}\left(-\frac{1}{3}\right)$”
, without switching to some approximate evaluation.

**Problem 194**

(a) Let *ABCD* be a regular tetrahedron with edges of length 2. Calculate the (exact) angle between the two faces *ABC* and *DBC*.

(b) We know that in 2D five equilateral triangles fit together *at a point* leaving just enough of an angle to allow a sixth triangle to fit. How many identical regular tetrahedra can one fit together, without overlaps *around an edge, *so that they all share the edge * BC* (say)?

**Problem 195**

(a) Let *ABCDEF* be a regular octahedron with vertices *B, C, D, E* adjacent to A forming a square *BCDE,* and with edges of length 2. Calculate the (exact) angle between the two faces *ABC* and *FBC*.

(b) How many identical regular octahedra can one fit together *around an edge*, without overlaps, so that they all share the edge * BC* (say)?

**Problem 196** Go back to the scenario of Problem **188**, with a regular tetrahedron and a regular octahedron both having edges of length 2, and both having one face flat on the table. Suppose we slide the tetrahedron across the table towards the octahedron. What unexpected phenomenon is guaranteed by Problems **194**(a) and **195**(a)?

**Problem 197** Consider the cube with edges of length 2 running parallel to the coordinate axes, with its centre at the origin (0,0,0), and with opposite corners at (1,1,1) and (—1, —1, —1). The *x-*, *y-*, and *z*-axes, and the *xy-*, *yz-,* and *zx*-planes cut this cube into eight unit cubes - one sitting in each octant.

(i) Let *A* = (0,0,1), *B* = (1, 0, 0), *C* = (0,1,0), *W* = (1,1,1). Describe the solid *ABCW*.

(ii) Let *D* = (—1, 0, 0), *X* = (—1,1,1). Describe the solid *ACDX*.

(iii) Let *E* = (0, —1,0), *Y* = (—1, —1,1). Describe the solid *ADEY*.

(iv) Let *Z* = (1, —1,1). Describe the solid *AEBZ*.

(v) Let *F* = (0,0, —1) and repeat steps (i)-(iv) to obtain the four mirror image solids which lie beneath the *xy*-plane.

(vi) Describe the solid *ABCDEF* which is surrounded by the eight identical solids in (i)-(v).

**Problem 198** Consider a single face *ABCDE* of the regular dodecahedron, with edges of length 1, together with the five pentagons adjacent to it - so that each of the vertices *A*, *B*, *C*, *D*, *E* has vertex figure 5^{3}. Each vertex figure is rigid, so the whole arrangement of six regular pentagons is also rigid. Let *V*, *W*, *X*, *Y*, *Z* be the five vertices adjacent to *A*, *B*, *C*, *D*, *E* respectively. Calculate the dihedral angle between the two pentagonal faces that meet at the edge * AB*.

**Problem 199** Suppose a regular icosahedron (Problem **189**) has edges of length 2. Position vertex A at the ‘North pole’, and let *BCDEF* be the regular pentagon formed by its five neighbours.

(a)(i) Calculate the exact angle between the two faces *ABC* and *ACD.*

(ii) How many identical regular icosahedra can one fit together, without overlaps, around a single edge?

(b) Let C be the circumcircle of *BCDEF*, and let *O* be the circumcentre of this regular pentagon.

(i) Prove that the three edge lengths of the right-angled triangle ΔBOA are the edge lengths of the regular hexagon inscribed in the circle *C*, of the regular 10-gon inscribed in the circle *C*, and of the regular 5-gon inscribed in the circle *C*.

(ii) Calculate the distance separating the plane of the regular pentagon *BCDEF*, and the plane of the corresponding regular pentagon joined to the ‘South pole’.

Notice that Problem **199**(b) shows that the regular icosahedron can be ‘constructed’ in the Euclidean spirit: part (b)(i) is essentially Proposition 10 of Book XIII of Euclid’s *Elements*, and part (b)(ii) is implicit in Proposition 16 of the Book XIII. Once we are given the radius * OB*, we can:

- construct the regular pentagon
*BCDEF*in the circle C; - bisect the sides of the regular pentagon and hence construct the regular 10-gon
*BVC . . .*in the same circle; - construct the vertical perpendicular at O, and transfer the length
to the point__BV__*O*to determine the vertex*A*directly above*O*; - transfer the radius
to the vertical perpendicular at__OB__*O*to determine the plane directly below*O*, and hence construct the lower regular pentagon; etc..

It may be worth commenting on a common confusion concerning the regular icosahedron. Each regular polyhedron has a circumcentre, with all vertices lying on a corresponding sphere. If we join any triangular face of the regular icosahedron to the circumcentre *O*, we get a tetrahedron. These 20 tetrahedra are all congruent and fit together exactly at the point *O* “without gaps or overlaps”. But they are **not** *regular* tetrahedra: the circumradius is *less than* the edge length of the regular icosahedron.

**Problem 200** Prove that the only regular polyhedron that tiles 3D (without gaps or overlaps) is the cube.

In one sense the result in Problem **200** is disappointing. However, since we know that there are all sorts of interesting 3-dimensional arrangements related to crystals and the way atoms fit together, the message is really that we need to look *beyond* regular tilings. For example, the construction in Problem **197** shows how the familiar *regular* tiling of space with cubes incorporates a *semi-regular* tiling of space with eight regular tetrahedra and two regular octahedra at each vertex.

## 5.8. Circular arcs and circular sectors

Length is defined for *straight* line segments, and area is defined in terms of *rectangles*; neither measure is defined for shapes with curved boundaries - unless, that is, they can be cunningly dissected and the pieces rearranged to make a straight line, or a rectangle.

Figure 5: Dumbbell.

**Problem 201** Four identical semicircles of radius 1 fit together to make the dumbbell shape shown in Figure 5. Find the exact area enclosed without using the formula for the area of a circle.

In general, making sense of length and area for shapes with curved boundaries requires us to combine a little imagination with what we know about straight line segments and polygons. Our goal here is to lead up to the familiar results for the length of circular arcs and the area of circular sectors. But first we need to explore the perimeter and area of regular polygons, and the surface area of prisms and pyramids.

As so often in mathematics, to make sense of the perimeter and area of regular polygons we need to look beyond their actual values (which will vary according to the size of the polygon), and instead interpret these values as a function of some normalizing parameter - such as the *radius.* The calculations will be simpler if you first prove a general result.

**Problem 202**

(a) A regular n-gon and a regular 2n-gon are inscribed in a circle of radius 1. The regular n-gon has edges of length *s _{n} =* s, while the regular 2 n-gon has edges of length s

_{2n}=

*t.*Prove that

$${t}^{2}=2-\sqrt{4-{s}^{2}}.$$

(b) A regular “2-gon” inscribed in the unit circle is just a diameter (repeated twice), so has two identical edges of length s_{2} = 2. Use the result in part (a) to calculate the edge length s_{4} of a regular 4-gon, and the edge length s_{8} of a regular 8-gon inscribed in the same circle.

(c) A regular 6-gon inscribed in the unit circle has edge length s_{6} equal to the radius 1. Use the result in part (a) to calculate the edge length s_{3} of a regular 3-gon inscribed in the unit circle, and the edge length s_{12} of a regular 12-gon inscribed in the unit circle.

(d) In Problem **185** we saw that a regular 5-gon inscribed in the unit circle has edge length

$${s}_{5}=\frac{\sqrt{10-2\sqrt{5}}}{2}.$$

Use the result in part (a) to calculate the edge length s_{10} of a regular 10-gon inscribed in the same circle.

**Problem 203**

(a) A regular n-gon is inscribed in a circle of radius r.

(i) Find the exact perimeter p_{n} (in surd form): when n = 3; when n = 4; when n = 5; when n = 6; when n = 8; when n = 10; when n = 12.

(ii) Check that, for each n:

$${p}_{n}={c}_{n}\times r$$

for some constant c_{n}, where

$${c}_{3}<{c}_{4}<{c}_{5}<{c}_{6}<{c}_{8}<{c}_{10}<{c}_{12}\cdots $$

(b) A regular n-gon is circumscribed about a circle of radius r.

(i) Find the exact perimeter *P _{n}* (in surd form): when n = 3; when n = 4; when n = 5; when n = 6; when n = 8; when n = 10; when n = 12.

(ii) Check that, for each n:

$${p}_{n}={c}_{n}\times r$$

for some constant *C*_{n}, where

$${c}_{3}<{c}_{4}<{c}_{5}<{c}_{6}<{c}_{8}<{c}_{10}<{c}_{12}\cdots $$

(c) Explain why c_{12} < C_{12}.

It follows from Problem **203** that

- the perimeters p
_{n}and*P*of regular_{n}*n*-gons inscribed in, or circumscribed about, a circle of radius r all have the same form:$$(inscribed)\text{\hspace{0.17em}}{p}_{n}={c}_{n}\times r;(circumscribed){P}_{n}={C}_{n}\times r.$$

- The perimeters of inscribed regular
*n*-gons all increase with n, but remain less than the perimeter of the circle, while - the perimeters of the circumscribed regular
*n*-gons all decrease with*n*, but remain greater than the perimeter of the circle. - the perimeter
*P*of the circle appears to have the form P =*K × r,*where the ratio$$K=\frac{\text{perimeter}}{\text{radius}}$$

satisfies

$${c}_{3}<{c}_{4}<{c}_{5}<{c}_{6}<{c}_{8}<{c}_{10}<\cdots <K<\cdots <{C}_{10}<{C}_{8}<{C}_{6}<{C}_{5}<{C}_{4}<{C}_{3}.$$

In particular, the value of the constant K lies somewhere between c

_{42}*=*6.21 . . . and C12 = 6.43 . . . . If we now define the quotient K to be equal to “2π”, we see that$$(perimeter\text{\hspace{0.17em}}of\text{\hspace{0.17em}}circle\text{\hspace{0.17em}}of\text{\hspace{0.17em}}radius\text{\hspace{0.17em}}r)=2\pi r,$$

where π denotes some constant lying between 3.1 and 3.22

In this spirit one might reinterpret the first two bullet points as defining two sequences of constants "π

_{n}" and "Π_{n}" for $n\u2a7e3$, such that - (perimeter of a regular n-gon with circumradius r) = 2π
_{n}r, where$${\pi}_{3}=\frac{3\sqrt{3}}{2}=2.59\cdots ,{\pi}_{4}=2\sqrt{2}=2.82\cdots ,{\pi}_{5}=\frac{5\sqrt{10-2\sqrt{5}}}{4}=2.93\cdots ,{\pi}_{6}=3,$$

etc.,

and

- (perimeter of a regular n-gon with inradius r)= 2Π
_{n}r, where$${\Pi}_{3}=3\sqrt{3}=5.19\cdots ,{\Pi}_{4}=4,{\Pi}_{5}=5\sqrt{5-2\sqrt{5}}=3.63\cdots ,{\Pi}_{6}=2\sqrt{3}=3.46\cdots ,$$

etc.,

Moreover

$${\pi}_{3}<{\pi}_{4}<{\pi}_{5}<{\pi}_{6}<{\pi}_{8}<\cdots <\pi <\cdots <{\Pi}_{8}<{\Pi}_{6}<{\Pi}_{5}<{\Pi}_{4}<{\Pi}_{3}.$$

Hence

**Problem 204** Find the exact length (in terms of π)

(i) of a semicircle of radius r;

(ii) of a quarter circle of radius r;

(iii) of the length of an arc of a circle of radius r that subtends an angle *θ* radians at the centre.

In the next problem we follow a similar sequence of steps to conclude that the quotient

$$L=\frac{\text{areaofcircleofradius}r}{{r}^{2}}$$

is also constant. The surprise lies in the fact that this different constant is so closely related to the previous constant *K*.

**Problem 205**

(a) A regular n-gon is inscribed in a circle of radius *r*.

(i) Find the exact area *a _{n}* (in surd form): when n = 3; when n = 4; when n = 5; when n = 6; when n = 8; when n = 10; when n = 12.

(ii) Check that, for each n:

$${a}_{n}={d}_{n}\times {r}^{2}$$

for some constant d_{n}, where

$${d}_{3}<{d}_{4}<{d}_{5}<{d}_{6}<{d}_{8}<{d}_{10}<{d}_{12}\cdots $$

(b) A regular n-gon is circumscribed about a circle of radius *r*.

(i) Find the exact area A_{n} (in surd form): when n = 3; when n = 4; when n = 5; when n = 6; when n = 8; when n = 10; when n = 12.

(ii) Check that, for each n:

$${A}_{n}={D}_{n}\times {r}^{2}$$

for some constant D_{n}, where

$${D}_{3}>{D}_{4}>{D}_{5}>{D}_{6}>{D}_{8}>{D}_{10}>{D}_{12}\cdots $$

(c) Explain why d_{12} < D_{12}.

It follows from Problem **205** that

*a*and A

_{n}_{n}of regular n-gons inscribed in, or circumscribed about, a circle of radius r all have the same form:

$$(inscribed)\text{\hspace{0.17em}}{a}_{n}={d}_{n}\times {r}^{2};\text{\hspace{0.17em}}(circumscribed)\text{\hspace{0.17em}}{A}_{n}={D}_{n}\times {r}^{2}.$$

- The areas of inscribed regular n-gons all increase with n, but remain less than the area of the circle, while
- the areas of the circumscribed regular
*n*-gons all decrease with*n*, but remain greater than the area of the circle, whence - the area A of the circle appears to have the form A =
*L × r*where the ratio^{2},$$L=\frac{\text{areaofcircleofradius}r}{\text{radiussquared}}$$

satisfies

$${d}_{3}<{d}_{4}<{d}_{5}<{d}_{6}<{d}_{8}<{d}_{10}\cdots <L<\cdots <{D}_{10}<{D}_{8}<{D}_{6}<{D}_{4}<{D}_{3}.$$

In particular, the value of

*L*lies somewhere between d_{12}*=*3 and ${D}_{12}=12(2-\sqrt{3})=3.21\cdots $ . The surprise lies in the fact that the constant L is exactly half of the constant*K*- that is, L = π, so$$(area\text{\hspace{0.17em}}of\text{\hspace{0.17em}}circle\text{\hspace{0.17em}}of\text{\hspace{0.17em}}radius\text{\hspace{0.17em}}r)=\pi {r}^{2}.$$

The next problem offers a heuristic explanation for this surprise.

Figure 6: Circle cut into 8 slices.

**Problem 206** A regular 2n-gon *ABCDE . . .* is inscribed in a circle of radius r. The 2n radii *OA, OB,...* joining the centre *O* to the 2n vertices cut the circle into 2n sectors, each with angle $\frac{\pi}{n}$
(Figure 6).

These 2n sectors can be re-arranged to form an “almost rectangle”, by orienting them alternately to point “up” and “down”. In what sense does this “almost rectangle” have “height = r” and πr?

**Problem 207**

(a) Find a formula for the surface area of a right cylinder with height *h* and with circular base of radius *r*.

(b) Find a similar formula for the surface area of a right prism with height *h*, whose base is a regular n-gon with inradius *r*.

**Problem 208**

(a) Find the exact area (in terms of π)

(i) of a semicircle of radius *r*;

(ii) of a quarter circle of radius *r*;

(iii) of a sector of a circle of radius r that subtends an angle *θ* radians at the centre.

(b) Find the area of a sector of a circle of radius 1, whose total perimeter (including the two radii) is exactly half that of the circle itself.

**Problem 209**

(a) Find a formula for the surface area of a right circular cone with base of radius r and slant height *l*.

(b) Find a similar formula for the surface area of a right pyramid with apex A whose base *BCDE*. . . is a regular n-gon with inradius *r*.

**Problem 210**

(a) Find an expression involving “$\mathrm{sin}\frac{\pi}{n}$ ” for the ratio

$$\frac{\text{perimeterofinscribedregular}n\text{-gon}}{\text{perimeterofcircumscribedcircle}}.$$

(b) Find an expression involving “$\mathrm{tan}\frac{\pi}{n}$ ” for the ratio

$$\frac{\text{perimeterofcircumscribedregular}n\text{-gon}}{\text{perimeterofinscribedcircle}}.$$

**Problem 211**

(a) Find an expression involving “$\mathrm{sin}\frac{2\pi}{n}$ ” for the ratio

$$\frac{\text{areaofinscribedregular}n-\text{gon}}{\text{areaofcircumscribedcircle}}.$$

(b) Find an expression involving “$\mathrm{tan}\frac{\pi}{n}$ ” for the ratio

$$\frac{area\text{\hspace{0.17em}}of\text{\hspace{0.17em}}circumscribed\text{\hspace{0.17em}}regular\text{\hspace{0.17em}}n-\text{gon}}{area\text{\hspace{0.17em}}of\text{\hspace{0.17em}}inscribed\text{\hspace{0.17em}}circle}.$$

## 5.9. Convexity

This short section presents a simple result which to some extent justifies the assumptions made in the previous section - namely that the perimeter (or area) of a regular n-gon inscribed in a circle is less than the perimeter (or area) of the circle, and of the circumscribed regular *n*-gon.

**Problem 212** A convex polygon P_{1} is drawn in the interior of another convex polygon P_{2}.

(a) Explain why the area of *P _{1}* must be less than the area of P

_{2}.

(b) Prove that the perimeter of *P _{1}* must be less than the perimeter of P

_{2}.

## 5.10. Pythagoras’ Theorem in three dimensions

Pythagoras’ Theorem belongs in 2-dimensions. But does it generalise to 3-dimensions? The usual answer is to interpret the result in terms of coordinates.

**Problem 213**

(a) Construct a right angled triangle that explains the standard formula for the distance from P = *(a, b)* to Q *=* (d, e).

(b) Use part (a) to derive the standard formula for the distance from P =*(a, b, c)* to Q = *(d, e, f*).

This extension of Pythagoras’ Theorem to 3-dimensions is extremely useful, but not very profound. In contrast, the next result is more intriguing, but seems to be a complete uke of limited relevance. In 2D, a right angled triangle is obtained by

- taking one corner A of a rectangle
*ABCD,*together with its two neighbours B and D; - then “cutting off the corner” to get the triangle
*ABD.*

This suggests that a corresponding figure in 3D might be obtained by

*A*of a cuboid, together with its three neighbours

*B, C*,

*D*;

*ABCD,*with the right angled triangle Δ

*ABC*as base, and with apex

*D*.

The obvious candidate for the “3D-hypotenuse” is then the sloping face *BCD,* and the three right angled triangles Δ*ABC*, ΔACD, ΔADB presumably correspond to the ‘legs’ (the shorter sides) of the right triangle in 2D.

**Problem 214** You are given a pyramid *ABCD* with all three faces meeting at *A* being right angled triangles with right angles at A. Suppose * AB* = b,

*= c,*

__AC__*= d.*

__AD__(a) Calculate the areas of Δ*ABC*, ΔACD, ΔADB in terms of b, c, d.

(b) Calculate the area of ΔBCD in terms of b, c, d.

(c) Compare your answer in part (b) with the sum of the squares of the three areas you found in part (a).

More significant (e.g. for navigation on the surface of the Earth) and more interesting than Problem **214** is to ask what form Pythagoras’ Theorem takes for “lines on a sphere”.

For simplicity we work on a unit sphere. We discovered in the run-up to Problem **34** that lines, or shortest paths, on a sphere are arcs of great circles. So, if the triangle Δ*ABC* on the unit sphere is right angled at *A*, we may rotate the sphere so that the arc *AB* lies along the equator and the arc

*runs up a circle of longitude. It is then clear that, once the lengths c,*

__AC__*b*of

*and*

__AB__*are known, the locations of*

__AC__*B*and

*C*are essentially determined, and hence the length of the arc

*on the sphere is determined. So we would like to have a simple formula that would allow us to calculate the length of the arc*

__BC__*directly in terms of*

__BC__*c*and

*b*.

**Problem 215** Given a spherical triangle Δ*ABC* on the unit sphere with centre *O,* such that ∠*B*AC is a right angle, and such that *AB* has length c, and

*has length*

__AC__*b*.

(a) We have (rightly) referred to *b* and *c* as ‘lengths’. But what are they really?

(b) We want to know how the inputs *b* and *c* determine the value of the length *a* of the arc *BC;* that is, we are looking for a function with inputs b and c, which will allow us to determine the value of the "output"

*a*. Think about the answer to part (a). What kind of standard functions do we already know that could have inputs

*b*and

*c*?

(c) Suppose $c=0\ne b$
. What should the output *a* be equal to? (Similarly if $b=0\ne c$.) Which standard function of *b* and of *c* does this suggest is involved?

(d) (i) Suppose $\angle B=\angle C=\frac{\pi}{2}$
, what should the output *a* be equal to?

(ii) Suppose $\angle B=\frac{\pi}{2}$
, but ∠*C* (and hence *c*) is unconstrained. The output *a* is then determined - but the formula must give this fixed output for different values of *c*. What does this suggest as the “simplest possible” formula for *a*?

The answers to Problem **215** give a pretty good idea what form Pythagoras’ Theorem must take on the unit sphere. The next problem proves this result as a simple application of the familiar 2D Cosine Rule.

**Problem 216** Given any triangle Δ*ABC* on the unit sphere with a right angle at the point *A*, we may position the sphere so that A lies on the equator, with * AB* along the equator and

*up a circle of longitude. Let*

__AC__*O*be the centre of the sphere and let

**T**be the tangent plane to the sphere at the point

*A*. Extend the radii

*and*

__OB__*to meet the plane*

__OC__**T**at Bʹ and Cʹ respectively.

(a) Calculate the lengths of the line segments __ABʹ__ and __ACʹ__, and hence of __BʹCʹ__.

(b) Calculate the lengths of __OBʹ__ and __OCʹ__, and then apply the Cosine Rule to ΔBʹOCʹ to find an equation linking b and c with ∠*B*ʹOCʹ(=a).

When “solving triangles” on the sphere the same principles apply as in the plane: right angled triangles hold the key - but Pythagoras’ Theorem and trig in right angled triangles must be extended to obtain variations of the Sine Rule and the Cosine Rule for *spherical* triangles. The corresponding results on the sphere are both similar to, and intriguingly different from, those we are used to in the plane. For example, there are two forms of the Cosine Rule extending the result in Problem **216**.

**Problem 217** Given a (not necessarily right angled) triangle *Δ ABC* on the unit sphere, apply the same proof as in Problem

**216**to show (with the usual labelling) that:

$$\mathrm{cos}a=\mathrm{cos}b\xb7\mathrm{cos}c+\mathrm{sin}b\xb7\mathrm{sin}c\xb7\mathrm{cos}A$$

The other form of the Cosine Rule is “dual” to that in Problem **217** (with arcs and angles interchanged, and with an unexpected change of sign) - namely:

$$\mathrm{cos}A=-\mathrm{cos}B\xb7\mathrm{cos}C+\mathrm{sin}B\xb7\mathrm{sin}C\xb7\mathrm{cos}a$$

The next two problems derive a version of the Sine Rule for spherical triangles.

**Problem 218** Let Δ*ABC* be a triangle on the unit sphere with a right angle at *A*. Let *Aʹ* lie on the arc BA produced, and *Cʹ* lie on the arc *BC* produced so that *ΔAʹBCʹ* is right angled at Aʹ. With the usual labelling (so that x denotes the length of the side of a triangle opposite vertex X, with arc *AC* = b, arc *BC* = *a,* arc BCʹ = *aʹ,* and arc AʹCʹ = *bʹ,* prove that:

$$\frac{\mathrm{sin}b}{\mathrm{sin}a}=\frac{\mathrm{sin}{b}^{\prime}}{\mathrm{sin}{a}^{\prime}}$$

**Problem 219** Let Δ*ABC* be a general triangle on the unit sphere with the usual labelling (so that *x* denotes the length of the side of a triangle opposite vertex *X*, and *X* is used both to label the vertex and to denote the size of the angle at *X*). Prove that:

$$\frac{\mathrm{sin}a}{\mathrm{sin}A}=\frac{\mathrm{sin}b}{\mathrm{sin}B}=\frac{\mathrm{sin}c}{\mathrm{sin}C}.$$

It is natural to ask (cf Problem **32**):

“If the three ratios in Problem

219are all equal, what is it that they are all equal to?”

The answer may not at first seem quite as nice as in the Euclidean 2-dimensional case: one answer is that they are all equal to

$$\frac{\mathrm{sin}a\xb7\mathrm{sin}b\xb7\mathrm{sin}c}{\text{volumeofthetetrahedron}OABC}.$$

Notice that this echoes the result in the Euclidean plane, where the three ratios in the Sine Rule are all equal to *2R,* and

$$2R=\frac{abc}{2(\text{areaof}\Delta ABC)}.$$

## 5.11. Loci and conic sections

This section offers a brief introduction to certain classically important *loci *in the plane. The word *locus* here refers to the set of all points satisfying some simple geometrical condition; and all the examples in this section are based on the notion of distance from a point and from a line.

Given a point *O* and a positive real *r*, the locus of points at distance *r* from *O* is precisely the circle of radius *r* with centre O. If *r* < 0, then the locus is empty; while if *r* = 0, the locus consists of the point *O* alone.

Given a line *m* and a positive real *r*, the locus of all points at distance *r* from the line m consists of a pair of parallel lines - one either side of the line *m*. Given a circle of radius *r*, and a positive real number *d < r*; the locus of points at distance d from the circle consists of two circles, each concentric with the given circle (one inside the given circle and one outside). If *d > r*, the locus consists of a single circle outside the given circle.

Given two points *A* and *B*, the locus of points which are equidistant from *A* and from *B* is precisely the perpendicular bisector of the line segment * AB*. And given two lines

*m*,

*n*the locus of points which are equidistant from

*m*and from

*n*takes different forms according as

*m*and

*n*are, or are not, parallel.

- If
*m*and*n*are parallel, then the locus consists of a single line parallel to*m*and*n*and half way between them. - If
*m*and*n*meet at*X*(say), then the locus consists of the pair of perpendicular lines through*X*, that bisect the four angles at*X*.

**Problem 220** Given a point *F* and a line *m*, choose *m* as the *x*-axis and the line through *F* perpendicular to *m* as the *y*-axis. Let *F* have coordinates (0, 2a).

(i) Find the equation that defines the locus of points which are equidistant from *F* and from *m*.

(ii) Does the equation suggest a more natural choice of axes - and hence a simpler equation for the locus?

The locus, or curve, in Problem **220** is called a *parabola;* the point *F* is called the *focus* of the parabola, and the line *m* is called the *directrix.* In general, the ratio

"the distance from

XtoF” : “the distance fromXtom"

is called the *eccentricity* of the curve. Hence the parabola has eccentricity *e* = 1.

The parabola has many wonderful properties: for example, it is the path followed by a projectile under the force of gravity; if viewed as the surface of a mirror, a parabola reflects the sun's rays (or any parallel beam) to a single point - the focus *F*. Since the only variable in the construction of the parabola is the distance “2a” between the focus and the directrix, we can scale distances to see that any two different-looking parabolas must in fact be *similar to one another* - just as with any two circles. (It is hard not to infer from the graphs that *y* = 10*x*^{2} is a “thin” parabola, and that $y=\left(\frac{1}{10}\right){x}^{2}$ is a “fat” parabola. But the first can be rewritten in the form 10*y*=(10*x*)^{2}, and the second can be rewritten in the form $\left(\frac{y}{10}\right)={\left(\frac{x}{10}\right)}^{2}$
, so each is a re-scaled version of *Y = X*^{2}.)

So far we have considered loci defined by some pair of distances being equal, or in the ratio 1 : 1. More interesting things begin to happen when we consider conditions in which two distances are in a fixed ratio other than 1 : 1.

**Problem 221**

(a) Given two points *A*, *B*, with * AB* = 6. Find the locus of all points

*X*such that

*:*

__AX__*= 2:1.*

__BX__(b) Given points *A*, *B*, with * AB* = 2

*b*and a positive real number

*f*. Find the locus of all points

*X*such that

*:*

__AX__*=*

__BX__*f*: 1.

**Problem 222**

(a) Given points *A*, *B*, with * AB* = 2

*c*and a real number

*a*>

*c*. Find the locus of all points

*X*such that

*+*

__AX__*= 2*

__BX__*a*.

(b) Given a point *F* and a line *m*, find the locus of all points *X* such that the ratio

distance from

Xto the pointF: distance fromXto the linem

is a positive constant *e* < 1.

(c) Prove that parts (a) and (b) give different ways of specifying the same curve, or locus.

**Problem 223**

(a) Given points *A*, *B*, with * AB* = 2

*c*, and a positive real number

*a*. Find the locus of all points

*X*such that |

*-*

__AX__*| = 2*

__BX__*a*.

(b) Given a point *F* and a line *m*, find the locus of all points *X* such that the ratio

distance from

Xto the pointF: distance fromXto the linem

is a constant e > 1.

(c) Prove that parts (a) and (b) give different ways of specifying the same curve, or locus.

Problem **221** is sometimes presented in the form of a mild joke.

Two dragons are sleeping, one at

Aand one atB. DragonAcan run twice as fast as dragonB. A specimen ofhomo sapiensis positioned on the line segment, twice as far fromABAas fromB, and cunningly decides to crawl quietly away, while maintaining the ratio of his distances fromAand fromB(so as to make it equally difficult for either dragon to catch him should they wake).

The locus that emerges generally comes as a surprise: if the man sticks to his imposed restriction, by moving so that his position *X* satisfies * XA* = 2 .

*, then he follows a circle and lands back where he started! The circle is called the*

__XB__*circle of Apollonius,*and the points

*A*and

*B*are sometimes referred to as its foci.

The locus in Problem **222** is an *ellipse* - with *foci* A (or *F* = (-ae, 0)) and *B* (= (ae, 0)), and with directrix *m* (the line $y=-\frac{a}{e}$
; the line *y* = a is the second directrix of the ellipse). The “focus-focus” description in part (a) is symmetrical under reflection in both the line *AB* and the perpendicular bisector of * AB*. The “focus-directrix” description in (b) is clearly symmetrical in the line through

*F*perpendicular to

*m*; but it is a surprise to find that the equation

$$\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{a}^{2}\left(1-{e}^{2}\right)}=1$$

is also symmetrical under reflection in the *y*-axis. If we set b^{2} = a^{2}(1 - e^{2}), the equation takes the form

$$\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1,$$

which crosses the x-axis when *x* = ±*a*, and crosses the y-axis when *y* = +*b*. In its standard form, we usually choose coordinates so that *b* < *a*: the line segment from (-*a*, 0) to (*a*, 0) is then called the *major axis,* and half of it (say from (0,0) to (*a*, 0)) - of length *a* - is called the *semi-major axis*; the line segment from (0, -*b*) to (0, *b*) is called the *minor axis,* and half of it (say from (0, 0) to (0, *b*)) - of length *b* - is called the *semi-minor axis.*

The form of the equation shows that an ellipse is obtained from a unit circle by stretching by a factor “*a*” in the *x*-direction, and by a factor “*b*” in the y-direction. This implies that the area of an ellipse is equal to π*ab* (since each small s by s square that arises in the definition of the “area” of the unit circle gets stretched into an “as by bs rectangle”). However, the equation tells us nothing about the *perimeter* of an ellipse. Attempts to pin down the *perimeter* of an ellipse gave rise in the 18^{th} century to the subject of “elliptic integrals”.

Like parabolas, ellipses arise naturally in many important settings. For example, Kepler (1571-1630) discovered that the planetary orbits are not circular (as had previously been believed), but are ellipses - with the Sun at one focus (a conjecture which was later explained by Isaac Newton (1642-1727)). Moreover, the tangent to an ellipse at any point *X* is equally inclined to the two lines *XA* and *XB,* so that a beam emerging from one focus is reflected at every point of the ellipse so that all the reflected rays pass through the other focus.

The curve in Problem **223** is a *hyperbola* - with *foci* A (or *F = (-ae,* 0)) and *B* (= *(ae,* 0)), and with *directrix m* (the line $y=-\genfrac{}{}{0.1ex}{}{a}{e}$; the line $y=\genfrac{}{}{0.1ex}{}{a}{e}$ is the second directrix of the hyperbola). The “focus-focus” description in part (a) is symmetrical under reflection in both the line *AB* and the perpendicular bisector of * AB*. The “focus-directrix” description in (b) is clearly symmetrical in the line through

*F*perpendicular to

*m*; but it is a surprise to find that the equation

$$\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{a}^{2}\left({e}^{2}-1\right)}=1$$

is also symmetrical under reflection in the *y*-axis. Like parabolas and ellipses, hyperbolas arise naturally in many important settings - in mathematics and in the natural sciences.

All these loci were introduced and studied by the ancient Greeks without the benefit of coordinate geometry and equations. They were introduced as planar *cross-sections of a cone* - that is, as natural extensions of straight lines and circles (since the doubly infinite cone is the surface traced out when one rotates a line about an axis through a point on that line). The equivalence of the focus-directrix definition in Problems **220**, **222**, and **223** and cross sections of a cone follows from the next problem. All five constructions in Problem **224** work with the doubly-infinite cone, which we may represent as *x*^{2} +*y*^{2} = (*rz*)^{2} - although this representation is not strictly needed for the derivations. The surface of the double cone extends to infinity in both directions, and is obtained by taking the line *y* = *rz* in the *yz*-plane (where *r* > 0 is constant), and rotating it about the *z*-axis. Images of this rotated line are called *generators* of the cone; and the point they all pass through (i.e. the origin) is called the *apex* of the cone.

**Problem 224** (**Dandelinʹs spheres**: Dandelin (1794-1847))

(a) Describe the cross-sections obtained by cutting such a double cone by a horizontal plane (i.e. a plane perpendicular to the *z*-axis). What if the cutting plane is the *xy*-plane?

Figure 7: Conic sections.

(b) (i) Describe the cross-section obtained by cutting such a cone by a vertical plane through the origin, or apex.

(ii) What cross-section is obtained if the cutting plane passes through the apex, but is not vertical?

(c) Give a *qualitative* description of the curve obtained as a cross-section of the cone if we cut the cone by a plane which is parallel to a generator: e.g. the plane *y - rz = c.*

(i) What happens if *c* = 0?

(ii) Now assume the cutting plane is parallel to a generator, but does not pass through the apex of the cone - so we may assume that the plane cuts only the bottom half of the cone. Insert a small sphere inside the bottom half of the cone and above the cutting plane, and inflate the sphere as much as possible - until it touches the cone around a horizontal circle (the “contact circle with the cone”), and touches the plane at a single point *F*. Let the horizontal plane of the “contact circle with the cone” meet the cutting plane in the line *m*. Prove that each point of the cross-sectional curve is equidistant from the point *F* and from the line *m* - and so is a parabola.

(d) (i) Give a *qualitative* description of the curve obtained as a cross-section of the cone if we cut the cone by a plane which is *less steep* than a generator, but does not pass through the apex - and so cuts right across the cone.

(ii) We may assume that the plane cuts only the bottom half of the cone. Insert a small sphere inside the bottom half of the cone and above the cutting plane (i.e. on the same side of the cutting plane as the apex of the cone), and inflate the sphere as much as possible - until it touches the cone around a horizontal circle, and touches the plane at a single point *F*. Let the horizontal plane of the contact circle meet the cutting plane in the line *m*. Prove that, for each point *X* on the cross-sectional curve, the ratio

"distance from

XtoF” : “distance fromXtom” =e: 1

is constant, with *e* < 1, and so is an ellipse.

Figure 8: The conic section arising in Problem **224**(d).

(e) (i) Give a *qualitative* description of the curve obtained as a cross-section if we cut the cone by a plane which is *steeper* than a generator, but does not pass through the apex (and hence cuts both halves of the cone)?

(ii) We can be sure that the plane cuts the bottom half of the cone (as well as the top half). Insert a small sphere inside the bottom half of the cone and on the same side of the cutting plane as the apex, and inflate the sphere as much as possible - until it touches the cone around a horizontal circle, and touches the plane at a single point *F*. Let the horizontal plane of the contact circle meet the cutting plane in the line *m*. Prove that, for each point *X* on the cross-sectional curve, the ratio

“distance from

XtoF” : “distance fromXtom” =e: 1

is constant, with *e* > 1, and so is a hyperbola.

Problem **224** reveals a remarkable correspondence. It is not hard to show algebraically that any quadratic equation in two variables *x*, *y* represents either a point, or a pair of crossing (possibly identical) straight lines, or a parabola, or an ellipse, or a hyperbola: that is, by changing coordinates, the quadratic equation can be transformed to one of the standard forms obtained in this section. Hence, *the possible quadratic curves* are precisely the same as the possible cross-sections of a cone. This remarkable equivalence is further reinforced by the many natural contexts in which these conic sections arise.

## 5.12. Cubes in higher dimensions

This final section on elementary geometry seeks to explore fresh territory by going beyond three dimensions. Whenever we try to jump up to a new level, it can help to first take a step *back* and 'take a longer run up'. So please be patient if we initially take a step or two backwards.

We all know what a unit 3D-cube is. And - going backwards - it is not hard to guess what is meant by a unit “2D-cube": a unit 2D-cube is just another name for a unit *square.* It is then not hard to notice that a unit 3D-cube can be constructed from two unit 2D-cubes as follows:

- first position two unit 2D-cubes 1 unit apart in 3D space, with one directly above the other;
- make sure that each vertex of the lower 2D-cube is directly beneath a vertex of the upper 2D-cube;
- then join each vertex of the upper 2D-cube to the corresponding vertex below it.

Perhaps a unit 2D-cube can be constructed in a similar way from “unit lD-cubes"! This idea suggests that a unit “1D-cube” is just another name for a unit *line segment*.

Take the unit lD-cube to be the line segment from 0 to 1:

- position two such 1D-cubes in 2D (e.g. one joining (0,0) to (1, 0), and the other joining (0,1) to (1,1));
- check that each vertex of the lower 1D-cube is directly beneath a vertex of the upper 1D-cube;
- then join corresponding pairs of vertices - one from the upper 1D-cube and one from the lower 1D-cube ((0, 0) to (0,1), and (1,0) to (1,1)) - to obtain a unit 2D-cube.

Having taken a step back, we repeat (and reformulate) the previous construction of a 3D-cube: ^{o}

- position two such unit 2D-cubes in 3D: with one 2D-cube joining (0, 0, 0) to (1,0,0), then to (1,1, 0), then to (0,1, 0) and back to (0, 0, 0), and the other 2D-cube joining (0,0,1) to (1, 0,1), then to (1,1,1), then to (0,1,1), and back to (1,0, 0);
- with one 2D-cube directly above the other,
- then join corresponding pairs of vertices - one from the upper 2D-cube and one from the lower 2D-cube ((0,0,0) to (0,0,1), and (1, 0,0) to (1, 0,1), and (1,1, 0) to (1,1,1), and (0,1, 0) to (0,1,1)) - to obtain a unit 3D-cube.

To sum up: a unit cube in 1D, or in 2D, or in 3D:

- has as “vertices” all points whose coordinates are all “0s or 1s” (in 1D, or 2D, or 3D)
- has as “edges” all the unit segments (or unit 1D-cubes) joining vertices whose coordinates differ in exactly one place
- and a unit 3D-cube has as “faces” all the unit 2D-cubes spanned by vertices with a constant value (0 or 1) in one of the three coordinate places (that is, for the unit 3D-cube: the four vertices with
*x*= 0, or the four vertices with*x*= 1; or the four vertices with*y*= 0, or the four vertices with*y*= 1; or the four vertices with*z*= 0, or the four vertices with*z*= 1).

A 3D-cube is surrounded by six 2D-cubes (or faces), and a 2D-cube is surrounded by four 1D-cubes (or faces). So it is natural to interpret the two end vertices of a 1D-cube as being '0D-cubes'. We can then see that a cube in any dimension is made up from cubes of smaller dimensions. We can also begin to make a reasonable guess as to *what we might expect to find in a '4D-cuhe'.*

**Problem 225**

(a) (i) How many vertices (i.e. 0D-cubes) are there in a 1D-cube?

(ii) How many edges (i.e. 1D-cubes) are there in a 1D-cube?

(b) (i) How many vertices (or 0D-cubes) are there in a 2D-cube?

(ii) How many “faces” (i.e. 2D-cubes) are there in a 2D-cube?

(iii) How many edges (i.e. 1D-cubes) are there in a 2D-cube?

(c) (i) How many vertices (or 0D-cubes) are there in a 3D-cube?

(ii) How many 3D-cubes are there in a 3D-cube?

(iii) How many edges (i.e. 1D-cubes) are there in a 3D-cube?

(iv) How many “faces” (i.e. 2D-cubes) are there in a 3D-cube?

(d) (i) How many vertices (or 0D-cubes) do you expect to find in a 4D-cube?

(ii) How many 4D-cubes do you expect to find in a 4D-cube?

(iii) How many edges (i.e. lD-cubes) do you expect to find in a 4D-cube?

(iv) How many “faces” (i.e. 2D-cubes) do you expect to find in a 4D-cube?

(v) How many 3D-cubes do you expect to find in a 4D-cube?

**Problem 226**

(a) (i) Sketch a unit 2D-cube as follows. Starting with two unit lD-cubes - one directly above the other. Then join up each vertex in the upper lD-cube to the vertex it corresponds to in the lower lD-cube (directly beneath it).

(ii) Label each vertex of your sketch with coordinates (*x, y*) (*x,y =* 0 or 1) so that the lower 2D-cube has the equation “*y* = 0” and the upper 2D-cube has the equation “*y* = 1”.

(b) (i) Sketch a unit 3D-cube, starting with two unit 2D-cubes - one directly above the other. Then join up each vertex in the upper 2D-cube to the vertex it corresponds to in the lower 2D-cube (directly beneath it).

(ii) Label each vertex of your sketch with coordinates (*x, y, z*) (where each *x,y,z =* 0 or 1) so that the lower 2D-cube has the equation “*z* = 0” and the upper 2D-cube has the equation “*z* = 1”.

(c) (i) Now sketch a unit 4D-cube in the same way - starting with two unit 3D-cubes, one “directly above” the other.

[**Hint**: In part (b) your sketch was a projection of a 3D-cube onto 2D paper, and this forced you to represent the lower and upper 2D-cubes as rhombuses rather than genuine 2D-cubes (unit squares). In part (c) you face the even more difficult task of representing a 4D-cube on 2D paper; so you must be prepared for other “distortions”. In particular, it is almost impossible to see what is going on if you try to physically position one 3D-cube “directly above” the other on 2D paper. So start with the “upper” unit 3D-cube towards the top right of your paper, and then position the “lower” unit 3D-cube not directly below it on the paper, but below and slightly to the left, before pairing off and joining up each vertex of the upper 3D-cube with the corresponding vertex in the lower 3D-cube.]

(ii) Label each vertex of your sketch with coordinates (*w, x, y, z*) (where each *w,x,y, z* = 0 or 1) so that the lower 3D-cube has the equation “*z* = 0” and the upper 3D-cube has the equation “*z* = 1”.

**Problem 227** The only possible path along the edges of a 2D-cube uses each vertex once and returns to the start after visiting all four vertices.

(a) (i) Draw a path along the edges of a 3D-cube that visits each vertex exactly once and returns to the start.

(ii) Look at the sequence of coordinate triples as you follow your path. What do you notice?

(b) (i) Draw a path along the edges of a 4D-cube that visits each vertex exactly once and returns to the start.

(ii) Look at the sequence of coordinate 4-tuples as you follow your path. What do you notice?

## 5.13. Chapter 5: Comments and solutions

**137. Note:** The spirit of constructions restricts us to:

- drawing the line joining two known points
- drawing the circle with centre a known point and passing through a known point.

The whole thrust of this first problem is to find some way to “jump” from *A* (or *B*) to *C*. So the problem leaves us with very little choice; * AB* is given, and A and

*B*are more-or-less indistinguishable, so there are only two possible 'first moves’ - both of which work with the line segment

*(or*

__AC__*).*

__BC__Join * AC*.

Then construct the point *X* such that Δ*ACX* is equilateral (i.e. use Euclid's *Elements*, Book I, Proposition 1). Construct the circle with centre *A* which passes through *B*; let this circle meet the line *AX* at the point *Y*, where either

(i) *Y* lies on the segment * AX* (if $\underset{\_}{AB}\u2a7d\underset{\_}{AC}$
), or

(ii) *Y* lies on * AX* produced (i.e. beyond X, if

*>*

__AB__*).*

__AC__In each case, * AY* =

*. Finally construct the circle with centre*

__AB__*X*which passes through

*Y*. In case (i), let the circle meet the line segment

*XC*at

*D*; in case (ii), let the circle meet

*produced (beyond*

__CX__*X*) at

*D*.

In case (i), * CX* =

*+*

__CD__*; therefore*

__DX__$$\begin{array}{l}\underset{\_}{CD}\text{\hspace{0.17em}}=\underset{\_}{CX}-\underset{\_}{DX}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\underset{\_}{AX}-\underset{\_}{YX}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\underset{\_}{AY}=\underset{\_}{AB}.\end{array}$$

In case (ii),

$$\begin{array}{l}\underset{\_}{CD}\text{\hspace{0.17em}}=\underset{\_}{CX}+\underset{\_}{XD}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\underset{\_}{AX}+\underset{\_}{XY}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\underset{\_}{AY}=\underset{\_}{AB}.\end{array}$$ QED

**138.**

**139.**

* AM* =

__AM__* MB* =

*(by construction of the midpoint*

__MC__*M*)

* BA* =

*(given).*

__CA__$\therefore \Delta AMB\equiv \mathrm{\Delta}AMC$ (by SSS-congruence)

$\therefore \angle AMB=\angle AMC$, so each angle is exactly half the straight angle *=BMC*. Hence *AM* is perpendicular to *BC*. QED

**140.** Let *ABCDEF* be a regular hexagon with sides of length 1. Then Δ*ABC *(formed by the first three vertices) satisfies the given constraints, with *∠ABC = *120°.

Let *∠B'C'D* be an equilateral triangle with sides of length 2, and with *A'* the midpoint of *B'D*. Then *∠A'B'C'* satisfies the given constraints with angle *∠A'B'C' =* 60°.

**141**.

(i) Join * CO.* Then Δ

*ACO*is isosceles (since

*) and Δ*

__OA__=__OC__*BCO*is isosceles (since

*OB = OC*).

Hence *∠OAC = ∠OCA = x* (say), and ∠*OBC = ∠OCB = y* (say).

So *∠C = x* + *y* (and *∠A* +*∠B*+*∠C= x* +(x+*y)+y = 2(x* + y)). QED

(ii) If *∠A*+*∠B*+*∠C =2(x*+*y)* is equal to a straight angle, then *∠C= x* + *y* is half a straight angle, and hence a right angle. QED

**142.**

(i) Draw the circle with centre *A* and passing through *B*, and the circle with centre *B* passing through *A*. Let these two circles meet at *C* and *D*.

Wherever the midpoint *M* of * AB* may be, we know from Euclid Book I, Proposition 1 and Problem

**139**:

that Δ

ABCis equilateral, and thatCMis perpendicular toAB, and that ΔABDis equilateral, and thatDMis perpendicular toAB.

Hence *CMD* is a straight line.

So if we join *CD*, then this line cuts * AB* at its midpoint

*M*. QED

(ii) We may suppose that * BA* ≤

*.*

__BC__Then the circle with centre *B*, passing through *A* meets * BC* internally at

*Aʹ*(say).

Let the circle with centre *A* and passing through *B* meet the circle with centre *Aʹ* and passing through *B* at the point *D*.

**Claim** *BD* bisects *∠ABC*.

**Proof**

* BA* =

*(radii of the same circle with centre*

__BAʹ__*B*)

* AD* =

*(radii of the same circle with centre*

__AB__*A*)

= * AʹB*=

*(radii of same circle with centre*

__AʹD__*Aʹ*)

* BD* =

__BD__Hence Δ* BAD* ≡ Δ *BAʹD* (by SSS-congruence).

∴ *∠ABD =∠AʹBD*. QED

(iii) Suppose first that * PA* =

*. Then the circle with centre*

__PB__*P*and passing through

*A*meets the line

*AB*again at

*B*. If we construct the midpoint

*M*of

*as in part (i), then*

__AB__*PM*will be perpendicular to

*AB*.

Now suppose that one of * PA* and

*is longer than the other. We may suppose that*

__PB__*>*

__PA__*, so*

__PB__*B*lies inside the circle with centre

*P*and passing through

*A*. Hence this circle meets the line

*AB*again at

*Aʹ*where

*B*lies between

*A*and

*Aʹ*. If we now construct the midpoint

*M*of

*as in part (i), then*

__AAʹ__*PM*will be perpendicular to

*AAʹ*, and hence to

*AB*. QED

**143.** Let *M* be the midpoint of * AB*.

(a) Let *X* lie on the perpendicular bisector of * AB*.

∴ Δ *XMA* ≡ Δ *XMB* (by SAS congruence, since * XM* =

*, ∠XMA =∠XMB,*

__XM__*=*

__MA__*)*

__MB__∴ * XA* =

*.*

__XB__(b) If *X* is equidistant from *A* and from *B*, then Δ *XMA* ≡ Δ *XMB* (by SSS-congruence, since * XM* =

*,*

__XM__*=*

__MA__*,*

__MB__*=*

__AX__*).*

__BX__∴ ∠*XMA* =∠*XMB*, so each must be exactly half a straight angle.

∴ *X* lies on the perpendicular bisector of * AB*. QED

**144.** Let *X* lie on the plane perpendicular to * NS*, through the midpoint

*M*.

∴ Δ *XMN* ≡ Δ *XMS* (by SAS congruence, since * XM* =

*, ∠*

__XM__*XMN*=∠

*XMS*,

*=*

__MN__*)*

__MS__∴ * XN* =

*.*

__XS__Let *X* be equidistant from *N* and from *S*, then Δ *XMN* ≡ Δ *XMS* (by SSS-congruence, since * XM* =

*,*

__XM__*=*

__MN__*,*

__MS__*=*

__NX__*).*

__SX__∴ ∠*XMN* =∠*XMS*, so each must be exactly half a straight angle.

∴ *X* lies on the perpendicular bisector of * NS*. QED

**145.** Let *M* be the midpoint of * AC*. Join

*BM*and extend the line beyond

*M*to the point

*D*such that

*=*

__MB__*. Join*

__MD__*CD*. Then Δ

*AMB*≡ Δ

*CMD*(by SAS-congruence, since

* AM* =

*(by construction of the midpoint*

__CM__*M*)

∠*AMB* =∠*CMD* (vertically opposite angles)

* MB* =

*(by construction)).*

__MD__∴ ∠*DCM* =∠*BAM*.

Now

$$\begin{array}{l}\angle ACX=\angle DCM+\angle DCX\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}>DCM\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\angle BAM\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\angle A.\end{array}$$Hence∠*ACX* >∠*A*.

Similarly, we can extend * AC* beyond

*C*to a point

*Y*. Let

*N*be the midpoint of

*. Join*

__BC__*AN*and extend the line beyond

*N*to the point

*E*such that

*=*

__NA__*.*

__NE__Join *CE*.

Then Δ *BNA* ≡ Δ *CNE* (again by SAS-congruence).

∴ ∠*BCY* >∠*BCE* =∠*CBA* =∠*B*. QED

**146.**

(a) Suppose that * AB* >

*.*

__AC__Let the circle with centre *A*, passing through *C*, meet * AB* (internally) at

*X*.

Then Δ *ACX* is isosceles, so∠*ACX* =∠*AXC*.

By Problem **145**, ∠*AXC* >∠*ABC*.

∴ ∠*ACB*; (>∠*ACX* =∠*AXC*) >∠*ABC*. QED

(b) Suppose the conclusion does not hold. Then either

(i) * AB* =

*, or*

__AC__(ii) * AC* >

*.*

__AB__(i) If * AB* =

*, then Δ*

__AC__*ABC*is isosceles, so∠

*ACB*=∠

*ABC*– contrary to assumption.

(ii) If * AC* >

*, then∠*

__AB__*ABC*>∠

*ACB*(by part (a)) – again contrary to assumption.

Hence, if∠*ACB* >∠*ABC*, it follows that * AB* >

*. QED*

__AC__(c) Extend * AB* beyond

*B*to the point

*D*, such that

*=*

__BD__*.*

__BC__Then Δ *BDC* is isosceles with apex *B*, so∠*BDC* =∠*BCD*.

Now

∠*ACD* =∠*ACB* +∠*BCD* >∠*BCD* =∠*BDC*.

Hence, by part (b), * AD* >

*.*

__AC__By construction,

=AD+AB=BD+AB,BC

so * AB* +

*>*

__BC__*. QED*

__AC__**147.** Suppose∠*C* =∠*A* +∠*B*, but that *C* does not lie on the circle with diameter * AB*. Then

*C*lies either inside, or outside the circle. Let

*O*be the midpoint of

*.*

__AB__(i) If *C* lies outside the circle, then * OC* >

*=*

__OA__*.*

__OB__∴ ∠*OAC* >∠*OCA* and∠*OBC* >∠*OCB* (by Problem **146**(a)).

∴ ∠*C* =∠*OCA* +∠*OCB* <∠*A* +∠*B* – contrary to assumption.

Hence *C* does not lie outside the circle.

(ii) If *C* lies inside the circle, then * OC* <

*=*

__OA__*.*

__OB__∴ ∠*OAC* < ∠*OCA* and∠*OBC* < ∠*OCB* (by Problem **146**(a)).

∴ ∠*C* =∠*OCA* +∠*OCB* >∠*A* +∠*B* – contrary to assumption.

Hence *C* does not lie inside the circle.

Hence *C* lies on the circle with diameter * AB*. QED

**148.** Suppose, to the contrary, that *OP* is **not** perpendicular to the tangent at *P*.

Drop a perpendicular from *O* to the tangent at *P* to meet the tangent at *Q*. Extend * PQ* beyond

*Q*to some point

*X*. Then

*∠OQP*and

*∠OQX*are both right angles. Since

*Q*(≠

*P*) lies on the tangent,

*Q*lies outside the circle, so

*>*

__OQ__*. Hence (by Problem*

__OP__**146**(a))∠

*OPQ*>∠

*OQP*= ∠

*OQX*– contrary to the fact that∠

*OQX*>∠

*OPQ*(by Problem

**145**). QED

**149.** Let *Q* lie on the line *m* such that * PQ* is perpendicular to

*m*.

Let *X* be any other point on the line *m*, and let *Y* be a point on *m* such that *Q* lies between *X* and *Y*.

Then *∠PQX* and *∠PQY* are both right angles.

Suppose that * PX* <

*.*

__PQ__Then∠*PQY* =∠*PQX* < ∠*PXQ* (by Problem **146**(a)), which contradicts Problem **145** (since *∠PQY* is an exterior angle of Δ *PQX*).

Hence * PX* ≥

*as required. QED*

__PQ__**150**.∠*A* +∠*B* +∠*C*, and∠*XCA* +∠*C* are both equal to a straight angle. So∠A +∠B =∠XCA.

**151.** Join * OA*,

*,*

__OB__*. Since these radii are all equal, this produces three isosceles triangles. There are*

__OC__**five**cases to consider.

(i) Suppose first that *O* lies on *AB*. Then *AB* is a diameter, so *∠ACB* is a right angle (by Problem **141**), *∠AOB* is a straight angle, and the result holds.

(ii) Suppose *O* lies on *AC*, or on *BC*. These are similar, so we may assume that *O* lies on *AC*.

Then Δ *OBC* is isosceles, so∠*OBC* =∠*OCB*.

*∠AOB* is the exterior angle of Δ *OBC*, so

∠

AOB=∠OBC+∠OCB= 2· ∠ACB

(by Problem **150**).

(iii) Suppose *O* lies inside Δ *ABC*.

Δ *OAB*, Δ *OBC*, Δ *OBC* are isosceles, so let ∠*OAB* =∠*OBA* = *x*, ∠*OBC* =∠*OCB* = *y*, ∠*OBC* =∠*OAC* = *z*.

Then∠*ACB = y + z*, ∠*ABC = x + y*, ∠*BAC = x + z*.

The three angles of Δ *ABC* add to a straight angle, so 2(*x + y + z*) equals a straight angle.

Hence, in Δ *OBA*,

∠AOB = 2(x + y + z) - (∠

OAB+∠OBA) = 2(y + z)= 2· ∠ACB.

(iv) Suppose *O* lies outside Δ ABC with *O* and *B* on opposite sides of *AC*.

Δ *OAB*, Δ *OBC*, Δ *OBC* are isosceles, so let ∠*OAB* =∠*OBA* = *x*, ∠*OBC* =∠*OCB* = *y*, ∠*OBC* =∠*OAC* = *z*.

Then*∠ACB = y - z, ∠ABC = x + y, ∠BAC = x - z*.

The three angles of Δ *ABC* add to a straight angle, so *2x + 2y - 2z* equals a straight angle.

Hence

2x + 2y - 2z =∠AOB +∠

OAB+∠OBA=∠AOB + 2x,

so∠AOB = 2y - 2z = 2· ∠ACB.

(v) Suppose *O* lies outside Δ *ABC* with *O* and *B* on the same side of *AC*.

Δ *OAB*, Δ *OBC*, Δ *OBC* are isosceles, so let ∠*OAB* =∠*OBA* = x, ∠*OBC* =∠*OCB* = y, ∠*OBC* =∠*OAC* = z.

Then*∠ACB = y + z, ∠ABC = y - x, ∠BAC = z - x*.

The three angles of *ABC* add to a straight angle, so *2y + 2z - 2x* equals a straight angle.

Since *C* lies on the minor arc relative to the chord * AB*, we need to interpret “the angle subtended by the chord

*at the centre*

__AB__*O*” as the

*reflex*angle

**outside**the triangle Δ

*AOB*– which is equal to “

*2x*more than a straight angle” , so ∠

*AOB = 2y + 2z = 2· ∠ACB*. QED

**152.** The chord * AB* subtends angles at

*C*and at

*D*on the same arc. Similarly

*subtends angles at*

__BC__*A*and at

*D*on the same arc. Hence (by Problem

**151**)

∠*ACB* =∠*ADB*, and∠*BAC* =∠*BDC*.

Hence

∠ADC =∠ADB +∠BDC =∠ACB +∠BAC, so ∠*ADC* +∠*ABC* equals the sum of the three angles in Δ *ABC*. QED

**153.** Let *O* be the circumcentre of Δ ABC, and let∠XAB = x.

Then *∠XAO* is a right angle, and Δ *OAB* is isosceles. There are two cases.

(i) If *X* and *O* lie on opposite sides of *AB*, then∠*OBA* =∠*OAB* = 90° - x.

∴ ∠AOB = 2x.

∴ ∠ACB = *X* =∠XAB.

(ii) If *X* and *O* lie on the same side of *AB*, then *Y* and *O* lie on opposite sides of *AB* and of *AC*. Hence we can apply case (i) (with *Y* in place of *X*, *AC* in place of *AB*, *∠YAC* in place of *∠XAB*, and *∠ABC* in place of *∠ACB*) to conclude that∠YAC =∠ABC. Hence

∠XAB +∠BAC +∠YAC =∠XAB + ∠BAC +∠ABC

are both straight angles, so ∠XAB =∠ACB (since the three angles of Δ ABC also add to a straight angle). QED

**154.**

(a) (i) Extend *AD* to meet the circumcircle at *X*. Then (applying Problem **145** to Δ *DXB*), the exterior angle∠ADB > ∠*DXB* = ∠*AXB* = ∠*ACB*.

(ii) We are told that the points *C*, *D* lie “on the same side of the line *AB*”. This “side” of the line *AB* (or “half-plane” ) is split into three parts by the half-lines “* AC* produced beyond

*C*” and “

*produced beyond*

__BC__*C*” . There are two very different possibilities.

Suppose first that *D* lies in one of the two overlapping wedge-shaped regions “between * AB* produced and

*produced” or “between*

__AC__*produced and*

__BA__*produced”. Then either*

__BC__*DA*or

*DB*cuts the arc

*AB*at a point

*X*(say), and ∠ACB = ∠AXB > ∠ADB (by Problem

**145**applied to Δ AXD or to Δ BXD).

The only alternative is that *D* lies in the wedge shaped region outside the circle at the point *C*, lying “between * AC* produced and

*produced”. Then*

__BC__*C*lies inside Δ ADB, so

*C*lies inside the circumcircle of Δ ADB. Hence part (i) implies that ∠ACB < ∠ADB as required. QED

(b) If *D* does not lie on the circumcircle of Δ ABC, then it lies either inside, or outside the circle – in which case ∠ADB ≠ ∠ACB (by part (a)).

(c) *∠D* is less than a straight angle, so *D* must lie outside Δ ABC. Moreover, *B*, *D* lie on opposite sides of the line *AC* (since the edges * BC*,

*do not cross internally, and neither do the edges*

__DA__*,*

__CD__*). Consider the circumcircle of Δ ABC, and let*

__AB__*X*be any point on the circle such that

*X*and

*D*lie on the same side of the line

*AC*. Then

*ABCX*is a cyclic quadrilateral, so

*∠ABC*and

*∠CXA*are supplementary. Hence∠CXA =∠CDA =∠D. But then

*D*lies on the circle (by part (b)).

**155.** Δ OPQ ≡ Δ OPʹQ (by RHS-congruence: ∠OPQ =∠OPʹQ are both right angles, * OP* =

*,*

__OPʹ__*=*

__OQ__*)*

__OQ__∴ * QP* =

*, and∠QOP =∠QOPʹ. QED*

__QPʹ__**156.**

(a) Let *X* be any point on the bisector of *∠ABC*. Drop the perpendiculars *XY* from *X* to *AB* and *XZ* from *X* to *CB*. Then:∠XYB =∠XZB are both right angles by construction; and∠XBY =∠XBZ since *X* lies on the bisector of *∠YBZ*; hence∠BXY =∠BXZ (since the three angles in each triangle have the same sum; so as soon as two of the angles are equal in pairs, the third pair must also be equal). Hence Δ BXY ≡ Δ BXZ (by ASA-congruence:∠YBX =∠ZBX, * BX* =

*, ∠BXY =∠BXZ.)*

__BX__∴ * XY*=

*. QED*

__XZ__(b) Suppose *X* is equidistant from *m* and from *n*. Drop the perpendiculars *XY* from *X* to *m*, and *XZ* from *X* to *n*.

Then Δ BXY ≡ Δ BXZ (by RHS-congruence:

∠BYX =∠BZX are both right angles * XY* =

*, since we are assuming*

__XZ__*X*is equidistant from

*m*and from

*n*

*=*

__BX__*).*

__BX__Hence∠XBY =∠XBZ, so *X* lies on the bisector of *∠YBZ*. QED

**157.**

(i) Join * AC*. Then Δ ABC ≡ Δ CDA by ASA-congruence:

∠BAC =∠DCA (alternate angles, since *AB* || DC)

* AC* =

__CA__∠ACB =∠CAD (alternate angles, since CB || DA).

In particular, Δ ABC and Δ CDA must have equal area, and so each is exactly half of *ABCD*. QED

**Note:** Once we prove (Problem **161** below) that a parallelogram has the same area as the rectangle on the same base and lying between the same pair of parallels (whose area is equal to “base × height” ), the result in part (i) will immediately translate into the familiar formula for the area of the triangle

(ii) Δ ABC ≡ Δ CDA by part (i). Hence * AB* =

*,*

__CD__*=*

__BC__*; and∠B =∠ABC =∠CDA =∠D.*

__DA__To show that∠A =∠C, we can either copy part (i) after joining *BD* to prove that Δ BCD ≡ Δ DAB, or we can use part (i) directly to note that∠A =∠BAC +∠DAC =∠DCA +∠BCA =∠C. QED

(iii) Δ AXB ≡ Δ CXD by ASA-congruence:

∠XAB =∠XCD (alternate angles, since *AB* || DC)

* AB* =

*(by part (ii))*

__CD__∠XBA =∠XDC (alternate angles, since *AB* || DC).

Hence * XA* (in Δ AXB) =

*(in Δ CXD), and*

__XC__*(in Δ AXB) =*

__XB__*(in Δ CXD). QED*

__XD__**158.** We may assume that *m* cuts the opposite sides *AB* at *Y* and *DC* at *Z*.

Δ XYB ≡ Δ XZD by ASA-congruence:

∠YXB =∠ZXD (vertically opposite angles)

* XB* =

*(by Problem*

__XD__**157**(iii))

∠XBY =∠XDZ (alternate angles).

Therefore

area(YZCB) = area(Δ BCD) - area(Δ XZD) + area(Δ XYB)

= area(Δ BCD)

$=\frac{1}{2}area(ABCD)$.

QED

**159.**

(a) Join *AC*. Then Δ ABC ≡ Δ CDA by SAS-congruence:

* BA* =

*(given)*

__DC__∠BAC =∠DCA (alternate angles, since *AB* || DC)

* AC* =

*.*

__CA__Hence∠BCA =∠DAC, so AD || BC as required. QED

(b) Join *AC*. Then Δ ABC≡ Δ CDA by SSS-congruence:

* AB* =

*(given)*

__CD__* BC* =

*(given)*

__DA__* CA* =

*.*

__AC__Hence∠BAC =∠DCA, so *AB* || DC; and∠BCA =∠DAC, so BC || AD.

QED

(c)∠A +∠B +∠C +∠D = 2 ∠A + 2 ∠B = 2 ∠A + 2 ∠D are each equal to two straight angles.

∴ ∠A +∠B is equal to a straight angle, so AD || BC; and∠A +∠D is equal to a straight angle, so *AB* || DC. QED

**Note:** The fact that the angles in a quadrilateral add to two straight angles is proved in the next chapter. However, if preferred, it can be proved here directly. If we imagine pins l*OBC*ted at *A*, *B*, *C*, *D*, then a string tied around the four points defines their “convex hull” – which is either a 4-gon (if the string touches all four pins), or a 3-gon (if one vertex is inside the triangle formed by the other three). In the first case, either diagonal (*AC* or *BD*) will split the quadrilateral internally into two triangles; in the second case, one of the three ‘edges’ joining vertices of the convex hull to the internal vertex cannot be an edge of the quadrilateral, and so must be a diagonal, which splits the quadrilateral internally into two triangles.

**160.** * AM* =

*(by construction of*

__MD__*M*as the midpoint), and

*=*

__BN__*.*

__NC__∴* AM* =

*(since*

__BN__*=*

__AD__*by Problem*

__BC__**157**(ii)).

∴ *ABNM* is a parallelogram (by Problem **159**(a)), so MN || AB.

Let *AC* cross *MN* at *Y*.

Then Δ AYM ≡ Δ CYN (by ASA-congruence, since

∠YAM =∠YCN (alternate angles, since AD || BC)

=AMCN∠AMY =∠CNY (alternate angles, since AD || BC).

Hence * AY* =

*, so*

__CY__*Y*is the midpoint of

*–the centre of the parallelogram (where the two diagonals meet (by Problem*

__AC__**157**(iii))). QED

**161.**

**Note:** In the easy case, where the perpendicular from *A* to the line *DC* meets the side * DC* internally at

*X*, it is natural to see the parallelogram

*ABCD*as the “sum” of a trapezium

*ABCX*and a right angled triangle Δ AXD. If the perpendicular from

*B*to

*DC*meets

*DC*at

*Y*, then Δ AXD ≡ Δ BYC. Hence we can rearrange the two parts of the parallelogram

*DCBA*to form a rectangle

*XYBA*.

However, a general proof cannot assume that the perpendicular from *A* (or from *B*) to *DC* meets * DC* internally. Hence we are obliged to think of the parallelogram in terms of

**differences**. This is a strategy that is often useful, but which can be surprisingly elusive.

Draw the perpendiculars from *A* and *B* to the line *CD*, and from *C* and *D* to the line *AB*. Choose the two perpendiculars which, together with the lines *AB* and *CD* define a rectangle that completely contains the parallelogram *ABCD* (that is, if *AB* runs from left to right, take the left-most, and the right-most perpendiculars). These will be either the perpendiculars from *B* and from *D*, or the perpendiculars from *A* and from *C* (depending on which way the sides *AC* and *BD* slope).

Suppose the chosen perpendiculars are the one from *B* – meeting the line *DC* at *P*, and the one from *D* – meeting the line *AB* at *Q*.

Then BP || QD (by Problem **159**(c)), so *BQDP* is a parallelogram with a right angle – and hence a rectangle. Hence * BQ* =

*, and*

__PD__*=*

__BP__*(by Problem*

__QD__**157**(ii)).

Δ QAD ≡ Δ PCB (by RHS-congruence), so each is equal to half the rectangle on base * PC* and height

*. Hence*

__PB__area(ABCD) = area(rectangle; BQDP)

- area(rectangle on base __PC__ with height * PB*)

= area(rectangle on base * CD* with height

*).*

__DQ__QED

**162.**

(a) *ABCBʹ* is a parallelogram, so ABʹ|| CB and * ABʹ* =

*.*

__BC__Similarly, *BCACʹ* is a parallelogram, so ACʹ|| CB and * ACʹ*=

*.*

__CB__Hence * BʹA* =

*, so*

__ACʹ__*A*is the midpoint of

*.*

__BʹCʹ__Similarly *B* is the midpoint of * CʹAʹ*, and

*C*is the midpoint of

*.*

__AʹBʹ__(b) Let *H* be the circumcentre of Δ AʹBʹCʹ – that is, the common point of the perpendicular bisectors of * AʹBʹ*,

*, and*

__BʹCʹ__*. Then*

__CʹAʹ__*H*is a common point of the three perpendiculars from

*A*to

*BC*, from

*B*to

*CA*, and from

*C*to

*AB*.

**163.** Consider any path from *H* to *V*. Suppose this reaches the river at *X*. The shortest route from *H* to *X* is a straight line segment; and the shortest route from *X* to *V* is a straight line segment.

If we reflect the point *H* in the line of the river, we get a point *Hʹ*, where *HHʹ* is perpendicular to the river and meets the river at *Y* (say).

Then Δ HXY ≡ Δ HʹXY (by SAS-congruence, since * HY* =

*, ∠HYX =∠HʹYX, and*

__HʹY__*=*

__YX__*).*

__YX__Hence * HX* =

*, so the distance from*

__HʹX__*H*to

*V*via

*X*is equal to

*+*

__HX__*=*

__XV__*+*

__HʹX__*, and this is shortest when*

__XV__*Hʹ*,

*X*, and

*V*are collinear.(So to find the shortest route, reflect

*H*in the line of the river to

*Hʹ*, then draw HʹV to cross the line of the river at

*X*, and travel from

*H*to

*V*via

*X*.)

**164.**

(a) Let Δ PQR be any triangle inscribed in Δ ABC, with *P* on *BC*, *Q* on *CA*, *R* on *AB* (not necessarily the orthic triangle). Let *Pʹ* be the reflection of *P* in the side *AC*, and let Pʹʹ be the reflection of *P* in the side *AB*. Then * PQ* =

*, and*

__PʹQ__*=*

__PR__*(as in Problem*

__PʹʹR__**163**).

∴ * PQ* +

*+*

__QR__*=*

__RP__*+*

__PʹQ__*+*

__QR__*.*

__RPʹʹ__Each choice of the point *P* on * AB* determines the positions of

*Pʹ*and Pʹʹ. Hence the shortest possible perimeter of Δ PQR arises when PʹQRPʹʹ is a straight line. That is, given a choice of the point

*P*, choose

*Q*and

*R*by:

– constructing the reflections *Pʹ* of *P* in *AC*, and Pʹʹ of P in *AB*;

– join * PʹPʹʹ*and let

*Q*,

*R*be the points where this line segment crosses

*AC*,

*AB*respectively.

It remains to decide how to choose *P* on * BC* so that

*is as short as possible.*

__PʹPʹʹ__The key here is to notice that *A* lies on both *AC* and on *AB*.

∴ * AP* =

*, and*

__APʹ__*=*

__AP__*, so Δ APʹPʹʹis isosceles.*

__APʹʹ__Also∠PAC =∠PʹAC, and∠PAB =∠PʹʹAB.

∴ ∠PʹAPʹʹ= 2· ∠A.

Hence, for each position of the point *P*, Δ APʹPʹʹ is isosceles with apex angle equal to 2· ∠A. Any two such triangles are similar (by SAS-similarity).

Hence the triangle Δ APʹPʹʹ with the shortest “base” * PʹPʹʹ*occurs when the legs

*and*

__APʹ__*are as short as possible. But*

__APʹʹ__*=*

__AP__*=*

__APʹ__*, so this occurs when*

__APʹʹ__*is as short as possible – namely when*

__AP__*is perpendicular to*

__AP__*BC*.

Since the same reasoning applies to *Q* and to *R*, it follows that the required triangle Δ PQR must be the orthic triangle of Δ ABC. QED

(b) Let Δ PQR be the orthic triangle of Δ ABC, with *P* on *BC*, *Q* on *CA*, *R* on *AB*. Let *H* be the orthocenter of Δ ABC.

*∠BPH* and *∠BRH* are both right angles, so add to a straight angle. Hence (by Problem **154**(c)), *BPHR* is a cyclic quadrilateral. Similarly *CPHQ* and *AQHR* are cyclic quadrilaterals.

In the circumcircle of *CPHQ*, we see that the initial “angle of incidence” ∠CQP =∠CHP. Also∠CHP =∠AHR (vertically opposite angles); and in the circumcircle of AQHR, ∠AHR =∠AQR.

Hence∠CQP =∠AQR, so a ray of light which traverses * PQ* will reflect at

*Q*along the line

*. Similarly one can show that∠ARQ =∠BRP, so that the ray will then reflect at*

__QR__*R*along

*; and∠BPR =∠CPQ, so the ray will then reflect at*

__RP__*P*along

*. QED*

__PQ__**165.**

(a) (i) Triangles Δ ABL and Δ ACL have equal bases * BL* =

*, and the same apex*

__CL__*A*– so lie between the same parallels. Hence they have equal area (by Problems

**157**and

**161**).

Similarly, Δ GBL and Δ GCL have equal bases * BL* =

*, and the same apex*

__CL__*G*– so have equal area.

Hence the differences Δ ABG = Δ ABL - Δ GBL and Δ ACG = Δ ACL - Δ GCL have equal area.

(ii) [Repeat the solution for part (i) replacing *A*, *B*, *C*, *L*, *G* by *B*, *C*, *A*, *M*, *G*.]

(b) Δ ABL and Δ ACL have equal area (as in (a)(i)). Similarly Δ GBL and Δ GCL have the same area – say *x* (since * BL* =

*). Hence Δ ABG and Δ ACG have equal area.*

__CL__In the same way Δ BCM and Δ BAM have equal area; and Δ GCM and Δ GAM have the same area – say *y* (since CM = AM). Hence Δ BCG and

Δ BAG have equal area.

But then Δ ABG = Δ ACG = Δ BCG and Δ ACG = Δ AMG + Δ CMG = 2y, Δ BCG = Δ BLG + Δ CLG = 2x. Hence *X* = y, so Δ AMG, Δ CMG, Δ CLG, Δ BLG all have the same area *x*, and Δ ABG has area *2x*.

The segment * GN* divides Δ ABG into two equal parts (Δ ANG andΔ BNG), so each part has area

*x*.

Hence Δ CAG + Δ ANG has the same area (*3x*) as Δ CAN. Hence *∠CGN* is a straight angle, and the three medians *AL*, *BM*, *CN* all pass through the point *G*.

**Note:** At first sight, the ‘proof’ of the result in (b) using vectors seems considerably easier.(If *A*, *B*, *C* have position vectors **a**, **b**, **c** respectively, then *L* has the position vector
$\frac{1}{2}$(**b** + **c**), and *M* has position vector $\frac{1}{2}$(**c** + **a**), and it is easy to see that *AL* and *BM* meet at *G* with position vector $\frac{1}{3}$(**a** + **b** + **c**). One can then check directly that *G* lies on *CN*, or notice that the symmetry of the expression $\frac{1}{3}$(**a** + **b** + **c**) guarantees that *G* is also the point where *BM* and *CN* meet.

The inscrutable aspect of this ‘proof’ lies in the fact that all the geometry has been silently hidden in the algebraic assumptions which underpin the unstated axioms of the *2*-dimensional vector space, and the underlying field of real numbers. Hence, although the vector ‘proof’ may seem simpler, the two different appr*OAC*hes cannot really be compared.

**166.**

(a) * AN* =

*(by construction of*

__BN__*N*as the midpoint of

*)*

__AB__∠ANM =∠BNMʹ (vertically opposite angles)

* NM* =

*(by construction).*

__NMʹ__∴ Δ ANM ≡ Δ BNMʹ (by SAS-congruence). QED

(b) * BMʹ* =

*=*

__AM__*.*

__CM__∠NAM =∠NBMʹ (since Δ ANM ≡ Δ BNMʹ)

∴ BMʹ || MA (i.e. BMʹ || CM). QED

(c) BMʹMC is a quadrilateral with opposite sides * CM*,

*equal and parallel.*

__BMʹ__∴ BMʹMC is a (by Problem **158**(a)). QED

**167.** Since *A* and *B* are interchangeable in the result to be proved, we may assume that *A* is the point on the secant that lies between *P* and *B*.

In order to make deductions, we have to create triangles – so join *AT* and *BT*. This creates two triangles: Δ PAT and Δ PTB, in which we see that:

∠TPA =∠BPT,

∠PTA =∠PBT (by Problem **153**),

∴ ∠PAT = ∠PTB (since the three angles in each triangle add to a straight angle).

Hence Δ PAT ~ Δ PTB (by AAA-similarity).

∴ * PT*:

*=*

__PB__*:*

__PA__*, or*

__PT__*×*

__PA__*=*

__PB__

__PT__^{2}. QED

**168.** Since *A*, *B* are interchangeable in the result to be proved, and *C*, *D* are interchangeable, we may assume that *A* lies between *P* and *B*, and that *C* lies between *P* and *D*.

(i) Let the tangent from *P* to the circle touch the circle at *T*.

Then Problem **167** guarantees that * PA* ×

*=*

__PB__

__PT__^{2}.

Replacing the secant *PAB* by *PCD* shows similarly that * PC* ×

*=*

__PD__

__PT__^{2}.

Hence * PA* ×

*=*

__PB__*×*

__PC__*. QED*

__PD__(ii) **Note:** The first proof is so easy, one may wonder why anyone would ask for a second proof. The reason lies in Problem **169** – which looks very much like Problem **168**, but with *P* *inside* the circle. Hence, when we come to the next problem, the easy appr*OAC*h in (i) will not be available, so it is worth looking for another proof of **168** which has a chance of generalizing.

Notice that * AC* is a chord which links the two secants

*PAB*and

*PCD*. So join

*and*

__AD__*.*

__CB__Then Δ PAD ~ Δ PCB (by AAA-similarity: since∠APD =∠CPB, and∠PDA =∠PBC).

∴ * PA*:

*=*

__PC__*:*

__PD__*, or*

__PB__*×*

__PA__*=*

__PB__*×*

__PC__*. QED*

__PD__**169.** Join * AD* and

*.*

__CB__Then Δ PAD ~ Δ PCB (by AAA-similarity: since

∠APD =∠CPB (vertically opposite angles)

∠PDA =∠PBC (angles subtended by chord * AC* on the same arc)

∠PAD =∠PCB (angles subtended by a chord * BD* on the same arc)).

∴ * PA*:

*=*

__PC__*:*

__PD__*, or*

__PB__*×*

__PA__*=*

__PB__*×*

__PC__*. QED*

__PD__**170.**

(a) If a = *b*, then *ABCD* is a parallelogram (by Problem **159**(a)).

∴ * AD* =

*(by Problem*

__BC__**157**(ii)).

∴ * AM* =

*, so*

__BN__*ABNM*is a parallelogram (by Problem

**159**(a)).

∴ * MN* =

*has length*

__AB__*a*, and MN ||

*AB*(by Problem

**160**).

If a ≠ *B*, then a < *B*, or *B* < a. We may assume that a < b.

Construct the line through *b* parallel to *AD*, and let this line meet * DC* at

*Q*.

Then *ABQD* is a parallelogram, so * DQ* =

*, and*

__AB__*=*

__AD__*. Hence*

__BQ__*has length*

__QC__*B*- a.

Construct the line through *M* parallel to *QC* (and hence parallel to *BA*), and let this meet * BQ* at

*P*, and

*at*

__BC__*Nʹ*.

Then *ABPM* and *MPQD* are both parallelograms.

∴ * MP* =

*has length*

__AB__*a*, and

* BP* =

*=*

__AM__*=*

__MD__*.*

__PQ__Now Δ BPNʹ ~ Δ BQC (by AAA-similarity, since PNʹ || QC); so

* BP*:

*=*

__BQ__*:*

__BNʹ__*= 1: 2.*

__BC__Hence Nʹ = *n* is the midpoint of * BC*, MN || BC, and

*has length*

__MN__$\begin{array}{cc}\phantom{\rule{6.0em}{0ex}}& a+\genfrac{}{}{0.1ex}{}{b-a}{2}=\genfrac{}{}{0.1ex}{}{a+b}{2}.\hfill \end{array}$

(b) Suppose first that a = b. Then *ABCD* is a parallelogram (by Problem **159**(a)), so the area of *ABCD* is given by a × d ( “(length of base) × height” ). (by Problem **161**). Hence we may suppose that a < b.

**Solution 1:** Extend * AB* beyond

*B*to a point

*X*such that

*=*

__BX__*(so*

__DC__*has length a + b).*

__AX__Extend * DC* beyond

*C*to a point

*Y*such that

*=*

__CY__*(so*

__AB__*has length a + b).*

__DY__Clearly *ABCD* and *YCBX* are congruent, so each has area one half of area(*AXYD*).

Now AX || DY, and * AX* =

*, so*

__DY__*AXYD*is a parallelogram (by Problem

**159**(a)).

Hence *AXYD* has area “(length of base)× height” (by Problem **161**), so *ABCD* has area $\frac{a+b}{2}$ × d.

**Solution 2:** [We give a second solution as preparation for Problem **171**.]

Now a < *B* implies that∠BAD +∠ABC is greater than a straight angle.

[**Proof.** The line through *B* parallel to *AD* meets *DC* at *Q*, and *ABQD* is a parallelogram.

Hence * DQ* =

*, so*

__AB__*Q*lies between

*D*and

*C*, and

∠ABC =∠ABQ +∠QBC >∠ABQ.

∴ ∠BAD +∠ABC >∠BAD +∠ABQ, which is equal to a straight angle.]

So if we extend * DA* beyond

*A*, and

*beyond*

__CB__*B*, the lines meet at

*X*, where

*X*,

*D*are on opposite sides of

*AB*. Then Δ XAB ~ Δ XDC (by AAA-similarity), whence corresponding lengths in the two triangles are in the ratio

*:*

__AB__*= a: b. In particular, if the perpendicular from*

__DC__*X*to

*AB*has length

*h*, then $\frac{h}{h+d}=\frac{a}{b}$, so h(b - a) = ad.

Now

area(ABCD) = area(Δ XDC) - area(Δ XAB)

= $\frac{1}{2}b(h+d)-\frac{1}{2}ah$

= $\frac{1}{2}bd+\frac{1}{2}h(b-a)$

= $\frac{1}{2}bd+\frac{1}{2}ad$

= $\left(\frac{a+b}{2}\right)d$.

**171.** Note: The volume of a pyramid or cone is equal to

$\frac{1}{3}$ × (area of base) ×height.

There is no elementary general proof of this fact. The initial coefficient of $\frac{1}{3}$ arises because we are “adding up” , or integrating, cross-sections parallel to the base, whose area involves a square x^{2}, where *x* is the distance from the apex (just as the coefficient $\frac{1}{2}$ in the formula for the area of a triangle arises because we are integrating linear cross-sections whose size is a multiple of *x* – the distance from the apex). Special cases of this formula can be checked – for example, by noticing that a cube *ABCDEFGH* of side *s*, with base *ABCD*, and upper surface *EFGH*, with *E* above *D*, *F* above *A*, and so on, can be dissected into three identical pyramids – all with apex *E*: one with base *ABCD*, one with base *BCHG*, and one with base *ABGF*. Hence each pyramid has volume $\frac{1}{3}$s^{3}, which may be interpreted as

$\frac{1}{3}\times (areaofbase={s}^{2})\times (height=s)$.

To obtain the frustum of height *d*, a pyramid with height *h* (say) is cut off a pyramid with height h + d.

∴ volume(frustum) = $\left[\frac{1}{3}\times {b}^{2}\times (h+d)\right]-\left[\frac{1}{3}\times {a}^{2}\times h\right]$

$=\left[\frac{1}{3}\times {b}^{2}\times d\right]+\left[\frac{1}{3}\times ({b}^{2}-{a}^{2})\times h\right]$.

Let *N* be the midpoint of * BC*, and let the line

*AN*meet the upper square face of the frustum at

*M*.

Let the perpendicular from the apex *A* to the base *BCDE* meet the upper face of the frustum at *Y* and the base *BCDE* at *Z*.

ThenΔ AYM ~ Δ AZN (by AAA-similarity), so * AY*:

*=*

__AZ__*:*

__YM__*.*

__ZN__∴$\frac{h}{h+d}=\frac{a}{b},soh(b-a)=ad$.

∴ $volume(frustum)=\frac{1}{3}{b}^{2}d+\frac{1}{3}\left({b}^{2}-{a}^{2}\right)h$

$=\frac{1}{3}{b}^{2}d+\frac{1}{3}(b+a)ad$

$=\frac{1}{3}\left({b}^{2}+ab+{a}^{2}\right)d$.

**172.** Construct the line through *A* which is parallel to AʹBʹ, and let it meet the line *BBʹ* at *P*. Similarly, construct the line through *B* which is parallel to BʹCʹ and let it meet the line *CCʹ* at *Q*.

Then Δ ABP ~ Δ BCQ (by AAA-similarity), so * AB*:

*=*

__BC__*:*

__AP__*.*

__BQ__Now AAʹBʹP is a parallelogram, so * AP* =

*, and BBʹCʹQ is a parallelogram, so*

__AʹBʹ__*=*

__BQ__*.*

__BʹCʹ__∴ * AB*:

*=*

__BC__*:*

__AʹBʹ__*. QED*

__BʹCʹ__**173.**

(a) Suppose * AB* has length

*a*and

*has length*

__CD__*b*. Problem

**137**allows us to construct a point

*X*such that

*=*

__AX__*, so*

__CD__*has length*

__AX__*b*. Now construct the point

*Y*where the circle with centre

*A*and radius

*meets*

__AX__*produced (beyond*

__BA__*A*). Then

*has length a + b.*

__YB__If a > *B*, let *Z* be the point where the circle with centre *A* and passing through *X* meets the segment * AB* internallyThen

*has length a - b.*

__ZB__(b) Use Problem **137** to construct a point *U* (not on the line *CD*) such that * DU* =

*, and a point*

__XY__*V*on

*DU*(possibly extended beyond

*U*) such that

*=*

__DV__*. Hence*

__AB__*has length*

__DU__*1*and

*has length*

__DV__*a*.

Construct the line through *V* parallel to *UC* and let it meet the line *DC* at *W*. Then Δ DVW ~ Δ DUC, with scale factor * DW*:

*=*

__DC__*:*

__DV__*= a: 1. Hence*

__DU__*has length*

__DW__*ab*.

To construct a line segment of length $\frac{a}{b}$ construct *U*, *V* as above. Then let the circle with centre *D* and radius * DU* meet

*CD*at

*Uʹ*. Let the line through

*Uʹ*parallel to

*CV*meet

*DV*at

*X*. The Δ DUʹX ~ Δ DCV, with scale factor

*:*

__DX__*=*

__DV__*:*

__DUʹ__*= 1:*

__DC__*B*, so

*has length $\frac{a}{b}$.*

__DX__(c) Use Problem **137** to construct a point *G* so that * AG* =

*= 1. Then draw the circle with centre*

__XY__*A*passing through

*G*to meet

*BA*extended at

*H*. Hence

*= 1,*

__HA__*= a.*

__AB__Construct the midpoint *M* of * HB*; draw the circle with centre

*M*and passing through

*H*and

*B*.

Construct the perpendicular to * HB* at the point

*A*to meet the circle at

*K*.

Then Δ HAK ~ Δ KAB, so * AB*:

*=*

__AK__*:*

__AK__*. Hence*

__AH__*= $\sqrt{a}$.*

__AK__**174.** The key to this problem is to use Problem **158**: a parallelogram(and hence any rectangle) is divided into two congruent pieces by any straight cut through the centre. If *A* is the centre of the rectangular piece of fruitcake, and *B* is the centre of the combined rectangle consisting of “fruitcake plus icing” , then the line *AB* gives a straight cut that divides both the fruitcake and the icing exactly in two.

**175.**

(a) The minute hand is pointing exactly at “*7*” , but the hour hand has moved $\frac{7}{12}$ of the way from “*1*” to “*2*” : that is, $\frac{7}{12}$ of 30°, or 17$\frac{1}{2}$°. Hence the angle between the hands is 162$\frac{1}{2}$°.

The same angle arises whenever the hands are trying to point in opposite directions, but are off-set by $\frac{7}{12}$ of 30°, or 17$\frac{1}{2}$°. This suggests that instead of “*35* minutes after *1*” we should consider “*35* minutes before *11*” , or 10:25.

(b) The two hands coincide at midnight. The minute hand then races ahead, and the hands do not coincide again until shortly after 1:00 – and indeed after 1:05.

More precisely, in *60* minutes, the minute hand turns through 360°, so in *x* minutes, it turns through 6x°. In *60* minutes the hour hand turns through 30°, so in *x* minutes, the hour hand turns through $\frac{1}{2}$x°.

The hands overlap when

$$\frac{1}{2}x\text{}\equiv 6x\text{\hspace{1em}}\left(\mathrm{mod}360\right),$$or when $\frac{11}{2}$x is a multiple of *360*.

This occurs when *X* = 0 (i.e. at midnight); then not until

X= $\frac{720}{11}=65\frac{5}{11}$,

and the time is 5$\frac{5}{11}$ minutes past *1*: that is, after 1$\frac{1}{11}$ hours.

It occurs again 1$\frac{1}{11}$ hours, or 65$\frac{5}{11}$ minutes, later – namely at 10$\frac{10}{11}$ minutes past 2, and so on.

Hence the phenomenon occurs at midnight, and then *11* more times until noon (with noon as the *12*th time; and then *11* more times until midnight – and hence *23* times in all (including both midnight occurrences).

(c) If we add a third hand (the ‘second hand'), all three hands coincide at midnight. In *x* minutes, the second hand turns through 360x°.

We now know exactly when the hour hand and minute hand coincide, so we can check where the second hand is at these times. For example, at 5$\frac{5}{11}$ minutes past *1*, the second hand has turned through (360 × 65$\frac{5}{11}$)°, and

360 × 65$\frac{5}{11}$ ≡ 360 × $\frac{5}{11}\text{\hspace{1em}}\left(\mathrm{mod}360\right)$,

so the second hand is nowhere near the other two hands.

The k^{th} occasion when the hour hand and minute hands coincide occurs at k × 1$\frac{1}{11}$ hours after midnight, when the two hands point in a direction $\left(\frac{360k}{11}\right)$° clockwise beyond “12” . At the same time, the second hand has turned through (360 × 65$\frac{5}{11}$ × k)°, and

$360\times 65\frac{5}{11}\times k\text{}\equiv 360\times \frac{5}{11}\times k\text{\hspace{1em}}\left(\mathrm{mod}360\right)$,

or five times as far round, and these two are never equal $\text{\hspace{1em}}\left(\mathrm{mod}360\right)$.

**176.**

**Note:** One of the things that makes it possible to calculate distances exactly here is that the angles are all known exactly, and give rise to lots ofright angled triangles.

The rotational symmetry of the clockface means we only have to consider segments with one endpoint at 12 o'clock (say *A*). The reflectional symmetry in the line *AG* means that we only have to find * AB*,

*,*

__AC__*,*

__AD__*,*

__AE__*, and*

__AF__*.*

__AG__Clearly * AG* = 2. If

*O*denotes the centre of the clockface, then

*is the hypotenuse of an isosceles triangle Δ OAD with legs of length 1, so*

__AD__*= $\sqrt{2}$.*

__AD__Δ ACO is isosceles (* OA* =

*= 1) with apex angle∠AOC = 60°, so the triangle is equilateral. Hence*

__OC__*=*

__AC__*= 1.*

__OA__It follows that *ACEGIK* is a regular hexagon, so * AE* = $\sqrt{3}$ (if

*OC*meets

*AE*at

*X*, then Δ ACX is a 30-60-90 triangle, and so is half of an equilateral triangle, whence

*= $\frac{\sqrt{3}}{2})$.*

__AX__It remains to find * AB* and

*.*

__AF__Let *OB* meet *AC* at *Y*. Then * OY* = $\frac{\sqrt{3}}{2}$ (since Δ OYA is a 30-60-90 triangle).

∴ * BY*= 1 - $\frac{\sqrt{3}}{2}$,

*= $\frac{1}{2}$, so*

__AY__*= $\sqrt{2-\sqrt{3}}$.*

__AB__Finally, * AB* subtends∠AOB = 30° at the centre, whence∠

*OAB*=∠

*OBA*= 75° and∠AGB = 15°.

∴ ∠ABG = 90° so we may apply Pythagoras’ Theorem to Δ ABG to find * BG* =

*= $\sqrt{2+\sqrt{3}}$.*

__AF__**177.**

$\sqrt{1}$ is the distance from (0, 0, 0) to (1, 0, 0).

$\sqrt{2}$ is the distance from (0, 0, 0) to (1, 1, 0).

$\sqrt{3}$ is the distance from (0, 0, 0) to (1, 1, 1).

$\sqrt{4}$ is the distance from (0, 0, 0) to (2, 0, 0).

$\sqrt{5}$ is the distance from (0, 0, 0) to (2, 1, 0).

$\sqrt{6}$ is the distance from (0, 0, 0) to (2, 1, 1).

$\sqrt{7}$ cannot be realized as a distance between integer lattice points in 3D.

$\sqrt{8}$ is the distance from (0, 0, 0) to (2, 2, 0).

$\sqrt{9}$ is the distance from (0, 0, 0) to (3, 0, 0).

$\sqrt{10}$ is the distance from (0, 0, 0) to (3, 1, 0).

$\sqrt{11}$ is the distance from (0, 0, 0) to (3, 1, 1).

$\sqrt{12}$ is the distance from (0, 0, 0) to (2, 2, 2).

$\sqrt{13}$ is the distance from (0, 0, 0) to (3, 2, 0).

$\sqrt{14}$ is the distance from (0, 0, 0) to (3, 2, 1).

$\sqrt{15}$ cannot be realized as a distance between integer lattice points in 3D.

$\sqrt{16}$ is the distance from (0, 0, 0) to (4, 0, 0).

$\sqrt{17}$ is the distance from (0, 0, 0) to (4, 1, 0).

**Note:** The underlying question extends Problem **32**:

Which integers can be represented as a sum of three squares?

This question was answered by Legendre (1752–1833):

Theorem. A positive integer can be represented as a sum of three squares if and only if it is not of the form 4^{a}(8b + 7).

**178.**

(a) Let *AD* and *CE* cross at *X*. Join * DE*. Then∠AXC =∠EXD (vertically opposite angles). Hence∠CAD +∠ACE =∠ADE +∠CED, so the five angles of the pentagonal star have the same sum as the angles of Δ BED. Hence the five angles have sum $\pi $ radians.

(b) Start with seven vertices *A*, *B*, *C*, *D*, *E*, *F*, *G* arranged in cyclic order.

(i) Consider first the 7-gonal star *ADGCFBE*. Let *GD* and *BE* cross at *X* and let *GC* and *BF* cross at *Y*. Join * DE*,

*. As in part (a),*

__BG__∠BGC +∠GBF =∠BFC +∠GCF,

and

∠BGD +∠GBE =∠BED +∠GDE,

so the angles in the 7-gonal star have the same sum as the angles in Δ ADE. Hence the seven angles have sum $\pi $ radians.

(ii) Similar considerations with the 7-gonal star *ACEGBDF* show that its seven angles have sum 3$\pi $ radians.

**Note:** Notice that the three possible “stars” (including the polygon *ABCDEFG*) have angle sums $\pi $ radians, 3$\pi $ radians, and 5$\pi $ radians.

(c) if *n* = 2k + 1 is prime, we may join *A* to its immediate neighbour *B* (1-step), or to its second neighbor *C*
(2-step), … , or to its k^{th} neighbour (*k*-step), so there are *k* different stars, with angle sums

respectively.

If *n* is not prime, the situation is slightly more complicated, since, for each divisor *m* of *n*, the *km*-step stars break up into separate components.

**179.**

(a) (i) Let the other three pentagons be *CDRST*, *DEUVW*, *EAXYZ*.

At *A* we have three angles of 108°, so∠NAX = 36°.

Δ ANX is isosceles (* AN* =

*=*

__AB__*=*

__AE__*), so ∠AXN = 72°.*

__AX__Hence∠AXY +∠AXN = 180°, so *Y*, *X*, *N* lie in a straight line.

Similarly *M*, *N*, *X* lie in a straight line.

Hence *M*, *N*, *X*, *Y* lie on a straight line segment * MY* of length 1 +

*= 2 +*

__YN__*.*

__NX__(ii) In the same way it follows that *MP* passes through *L* and *Q*, that *PS* passes through *O* and *T*, etc. so that the figure fits snugly inside the pentagon *MPSVY*, whose angles are all equal to 108°. Moreover, Δ ANX ≡ Δ BQL (by SAS-congruence), so * XN* =

*, whence*

__LQ__*=*

__MP__*. Similarly*

__YM__*=*

__MP__*=*

__PS__*=*

__SV__*, so*

__VY__*MPSVY*is a regular pentagon.

(iii) In the regular pentagon *EAXYZ* the diagonal EY || AX.

Moreover *XAC* is a straight line, and ∠ACB +∠NBC = 180°, so AC || NB.

Hence YE || NB, and * YE* =

*= $(n-2)\pi ,(n-4)\pi ,(n-6)\pi ,\dots ,3\pi ,\pi $ (the Golden Ratio $\tau $), so*

__NB__*YEBN*is a parallelogram.

Hence * YN* =

*= $\frac{1+\sqrt{5}}{2}$, so*

__EB__*= 1 + ${\tau}^{2}$ .*

__YM__(b)(i)

Δ MPY ≡ Δ PSM ≡ Δ SVP ≡ Δ VYS ≡ Δ YMV

(by SAS-congruence), so

=YP=MS=PV=SYVM

Also∠PMS =∠MPY = 36°;

∴ Δ BMP has equal base angles, and so is isosceles.

Hence * BM* =

*, and∠MBP = 108° =∠ABC.*

__BP__Similarly∠AMY =∠AYM = 36°.

∴ Δ AMY ≡ Δ BPM (by ASA-congruence), so

=AY=AM=BM.BP

Δ MAB ≡ Δ PBC – by ASA-congruence:

∠BPC = 108° -∠BPM -∠CPS = 36° =∠AMB,

=MA, and ∠PBC =∠MBA =∠MAB.PB

Hence * AB* =

*.*

__BC__Continuing round the figure we see that

=AB=BC=CD=DEEA

and that

∠A =∠B =∠C =∠D =∠E.

(ii) Extend *DB* to meet *MP* at *L*, and extend *DA* to meet *MY* at *N*. Then 36° =∠DBC =∠LBM (vertically opposite angles). Hence Δ LBM has equal base angles and so is isosceles: * LM* =

*. Similarly*

__LB__*=*

__NM__*. Now Δ LBM ≡Δ NAM (by ASA-congruence, since*

__NA__*=*

__MA__*), so*

__MB__*=*

__LM__*=*

__LB__*=*

__NA__*.*

__NM__In the regular pentagon *ABCDE* we know that∠DBC = 36°; and in Δ LBM, ∠LBM =∠DBC (vertically opposite angles), so ∠MLB = 108°. Hence∠ BLP = 72° =∠LBP, so Δ PLB is isosceles: * PL* =

*. In the regular pentagon*

__PB__*MPSVY*, Δ PMA is isosceles, so

*=*

__PM__*. ∴*

__PA__*=*

__LM__*-*

__PM__*=*

__PL__*-*

__PA__*=*

__PB__*.*

__BA__Hence *ABLMN* is a pentagon with five equal sides. It is easy to check that the five angles are all equal.

(iii) We saw in (i) that the five diagonals of *MPSVY* are equal. We showed in Problem **3** that each has length $\tau ={\tau}^{2}$, and that *MPDY* is a rhombus, so * DY* =

*= 1. Similarly*

__PM__*= 1.*

__SE__Hence * SD* = $\tau $- 1, and

*=*

__DE__*-*

__SE__*= 2 - $\tau $ = ${\left({\tau}^{2}\right)}^{-1}$ .*

__SD__**180.**

(a) Since the tiles fit together “edge-to-edge” , all tiles have the same edge length. The number *k* of tiles meeting at each vertex must be at least *3* (since the angle at each vertex of a regular *n*-gon =$\left(1-\frac{2}{n}\right)\pi \text{}\pi )$, and can be at most 6 (since the smallest possible angle in a regular *n*-gon occurs when *n* = 3, and is then $\frac{\pi}{3}$
).

We consider each possible value of *k* in turn.

– If k = 6, then *n* = 3 and we have six equilateral triangles at each vertex.

– If k = 5, then we would have $\left(1-\frac{2}{n}\right)\pi =\frac{2\pi}{5},\mathrm{so}n=\frac{10}{3}$ is not an integer.

– If k = 4, then *n* = 4 and we have four squares at each vertex.

– If k = 3, then *n* = 6 and we have three regular hexagons at each vertex.

Hence *n* = 3, or *4*, or *6*.

(b)(i) If k = *n* = 4, it is easy to form the vertex figure. It may seem ‘obvious’ that it continues “to infinity” ; but if we think it is obvious, then we should explain why: (choose the scale so that the common edge length is “1” ; then let the integer lattice points be the vertices of the tiling, with the tiles as translations of the unit square formed by (0, 0), (1, 0), (1, 1), (0, 1)).

(ii) If k = 6, *n* = 3, let the vertices correspond to the complex numbers p + q$\omega $
, where *p*, *q* are integers, and where $\omega $ is a complex cube root of *1*

(that is, a solution of the equation $\omega $^{3} = 1 ≠ $\omega $, or $\omega $^{2} + $\omega $ + 1 = 0), with the edges being the line segments of length 1 joining nearest neighbours (at distance 1).

(iii) If k = 3, *n* = 6, take the same vertices as in (ii), but eliminate all those for which p + q ≡ 0 (mod 3), then let the edges be the line segments of length 1 joining nearest neighbours.

**181.**

(a)(i) As before, the number *k* of tiles at each vertex lies between 3 and 6. However, this time *k* does not determine the shape of the tiles. Hence we introduce a new parameter: the number of *t* of triangles at each vertex, which can range from 0 up to 6. The derivations are based on elementary arithmetic, for which it is easier to work with angles in degrees.

* If t = 6, then the vertex figure must be **3 ^{6}**.

* If t = 5, the remaining gap of 60° could only take a sixth triangle, so this case cannot occur.

* If t = 4, we are left with angle of 120°, so the only possible vertex figure is **3 ^{4}.6**.

* If t = 3, then we are left with an angle of 180°, so the only possible configurations are **3 ^{3}.4^{2}** (with the two squares together), or

**3**(with the two squares separated by a triangle).

^{2}.4.3.4* If t = 2, we are left with an angle of 240°, which cannot be filled with 3 or more tiles (since the average angle size would then be at most 80°, and no more triangles are allowed), so the only possible vertex figures are **3 ^{2}.4.12**,

**3.4.3.12**,

**3**,

^{2}.6^{2}**3.6.3.6**(since

**3**or

^{2}.5.n**3.5.3.n**would require a regular

*n*-gon with an angle of 132°, which is impossible).

**Note:** The compactness of the argument based on the parameter *t* is about to end. We continue the same appr*OAC*h, with the focus shifting from the parameter *t*to a new parameter *s*– namely the number of squares in each vertex figure.

* Suppose t= 1. We are left with an angle of 300°.

If s = 0, the 300° cannot be filled with 3 or more tiles (since then the average angle size would be at most 100°, and no squares can be used), so there are exactly two additional tiles. Since each tile has angle $\le $
180°, we cannot use a hexagon, so the smallest tile has at least 7 sides; and since the average of the two remaining angle sizes is 150°, the smallest tile has at most 12 sides. It is now easy to check that the only possible vertex figures are **3.7.42**, **3.8.24**, **3.9.20**, **3.10.15**, **3.12 ^{2}**.

If s = 1, we would be left with an angle of 210°, which would require two larger tiles with average angle size 105°, which is impossible.

If s = 2, the only possible vertex figures are **3.4 ^{2}.6**, or

**3.4.6.4**.

Clearly we cannot have s = 3 (or we would be left with a gap of 30°); and t = 1, s > 3 is also impossible.

* Hence we may assume that t = 0, in which case *s* is at most 4.

If s = 4, then the vertex figure is **4 ^{4}**.

If s = 3, then the remaining gap could only take a fourth square, so this case does not occur.

If s = 2, we are left with an angle of 180°, which cannot be filled.

If s = 1, we are left with an angle of 270°, so there must be exactly two additional tiles and the only possible vertex figures are **4.5.20**, **4.6.12**, or **4.8 ^{2}** (since a regular

*7*-gon would leave an angle of 141${\frac{3}{7}}^{\circ}$ ).

* Hence we may assume that t = s = 0, and proceed using the parameter *f*– namely the number of regular pentagons. Clearly *f* is at most 3, and cannot equal 3 (or we would leave an angle of 36°).

If *F* = 2, we are left with an angle of 144°, so the only vertex figure is **5 ^{2}.10**.

If *F* = 1, we are left with an angle of 252°, which requires exactly two further tiles, whose average angle is 126°; but this forces us to use a hexagon – leaving an angle of 132°, which is impossible.

* Hence we may assume that t = s = *F* = 0. So the smallest possible tile is a hexagon, and since we need at least 3 tiles at each vertex, the only possible vertex figure is **6 ^{3}**.

Hence, the simple minded necessary condition (namely that the vertex figure should have no gaps) gives rise to a list of twenty-one possible vertex figures:

**3 ^{6}**,

**3**,

^{4}.6**3**,

^{3}.4^{2}**3**,

^{2}.4.3.4**3**,

^{2}.4.12**3.4.3.12**,

**3**,

^{2}.6^{2}**3.6.3.6**,

**3.7.42**,

**3.8.24**,

**3.9.20**,

**3.10.15**,

**3.12**,

^{2}**3.4**,

^{2}.6**3.4.6.4**,

**4**,

^{4}**4.5.20**,

**4.6.12**,

**4.8**,

^{2}**5**,

^{2}.10**6**.

^{3}(ii) **Lemma**. The vertex figures **3 ^{2}.4.12**,

**3.4.3.12**,

**3**,

^{2}.6^{2}**3.7.42**,

**3.8.24**,

**3.9.20**,

**3.10.15**,

**3.4**,

^{2}.6**4.5.20**,

**5**do not extend to semi-regular tilings of the plane.

^{2}.10**Proof.** Suppose to the contrary that any of these vertex figures could be realized by a semi-regular tiling of the plane. Choose a vertex *B* and consider the tiles around vertex *B*.

In the first eight of the listed vertex figures we may choose a triangle **T** = ABC, which is adjacent to polygons of different sizes on the edges * BA* (say) and

*.*

__BC__In the two remaining vertex figures, there is a face **T** = ABC·swith an **odd** number of edges, which has the same property – namely that of being adjacent to an *a*-gon on the edge * BA* (say) and a

*b*-gon on the edge

*with a ≠ b.(For example, in the vertex figure*

__BC__**3**,

^{2}.4.12**T**= ABC is a triangle, and the faces on

*and on*

__BA__*are – in some order – a 3-gon and a 4-gon, or a 3-gon and a 12-gon.)*

__BC__In each case let the face **T** be a *p*-gon.

If the face adjacent to **T** on the edge * BA* is an

*a*-gon, and that on edge

*is a*

__BC__*b*-gon, then the vertex figure symbol must include the sequence “

**$\dots $ a.p.b$\dots $**” .

If we now switch attention from vertex *B* to the vertex *A*, then we know that *A* has the same vertex figure, so must include the sequence “**$\dots $ a.p.b$\dots $**” , so the face adjacent to the other edge of **T** at *A* must be a *b*-gon. As one traces round the edges of the face **T**, the faces adjacent to

*T* are alternately *a*-gons and *b*-gons – contradicting the fact that **T** has an **odd** number of edges.

Hence none of the listed vertex figures extends to a semi-regular tiling of the plane. QED

(b) It transpires that the remaining eleven vertex figures

3,^{6}3,^{4}.63,^{3}.4^{2}3,^{2}.4.3.43.6.3.6,3.12,^{2}3.4.6.4,4,^{4}4.6.12,4.8,^{2}6^{3}

can all be realized as semi-regular tilings(and one– namely **3 ^{4}.6** – can be realized in two different ways, one being a reflection of the other).

In the spirit of Problem **180**(b), one should want to do better than to produce plausible pictures of such tilings, by specifying each one in some canonical way. We leave this challenge to the reader.

**182.** [We construct a regular hexagon, and take alternate vertices.]

Draw the circle with centre *O* passing through *A*. The circle with centre *A* passing through *O* meets the circle again at *X* and *Y*.

The circle with centre *X* and passing through *A* and *O* meets the circle again at *B*; and the circle with centre *Y* and passing through *A* and *O*, meets the circle again at *C*.

Then Δ AOX, Δ AOY, Δ XOB, Δ YOC are equilateral, so∠AXB = 120° =∠AYC, and ∠XAB = 30° =∠YAC.

Hence Δ AXB ≡ Δ AYC, so * AB* =

*.Δ ABC is isosceles so∠B =∠C, with apex angle∠BAC =∠XAY -∠XAB -∠YAC = 60°; hence Δ ABC is equiangular and so equilateral.*

__AC__**183.**

(a) Draw the circle with centre *O* and passing through *A*.

Extend *AO* beyond *O* to meet the circle again at *C*.

Construct the perpendicular bisector of * AC*, and let this meet the circle at

*B*and at

*D*.

Then * BA* =

*(since the perpendicular bisector of*

__BC__*is the locus of points equidistant from*

__AC__*A*and from

*C*); similarly

*=*

__DA__*.*

__DC__Δ *OAB* and Δ *OCB* are both isosceles right angled triangles, so *∠ABC* is a right angle (or appeal to “the anglesubtended on the circle by the diameter * AC*” ). Similarly

*∠A*,

*∠C*,

*∠D*are right angles, so

*ABCD*is a rectangle with

*=*

__BA__*, and hence a square.*

__BC__**Note:** This construction starts with the regular 2-gon * AC* inscribed in its circumcircle, and doubles it to get a regular 4-gon, by constructing the perpendicular bisectors of the “two sides” to meet the circumcircle at

*B*and at

*D*.

(b) Erect the perpendiculars to *AB* at *A* and at *B*.

Then draw the circles with centre *A* and passing through *B*, and with centre *B* and passing through *A*.

Let these circles meet the perpendiculars to *AB* (on the same side of *AB*) at *D* and at *C*.

Then * AD* =

*, and AD || BC, so*

__BC__*ABCD*is a parallelogram (by Problem

**159**(a)), and hence a (being a parallelogram with a right angle), and so a square (since

*=*

__AB__*).*

__AD__**184.**

(a)(i) First construct the regular 3-gon *ACE* with circumcentre *O*. Then construct the perpendicular bisectors of the three sides * AC*,

*,*

__CE__*, and let these meet the circumcircle at*

__EA__*B*,

*D*,

*F*.

**Note:** Here we emphasise the general step from inscribed regular *n*-gon to inscribed regular *2n*-gon – even though this may seem perverse in the case of a regular *3*-gon (since we constructed the inscribed regular *3*-gon in Problem **182** by first constructing the regular *6*-gon and then taking alternate vertices).

(ii) First construct the regular 4-gon *ACEG* with circumcentre *O*. Then construct the perpendicular bisectors of the four sides * AC*,

*,*

__CE__*,*

__EG__*, and let these meet the circumcircle at*

__GA__*B*,

*D*,

*F*,

*H*.

(b)(i) Construct an equilateral triangle *ABO*.

Then draw the circle with centre *O* and passing through *A* and *B*. Then proceed as in **182**.

(ii) Extend *AB* beyond *B*, and let this line meet the circle with centre *B* and passing through *A* at *X*.

Now construct a square *BXYZ* on the side * BX* as in

**183**(a), and let the diagonal

*BY*meet the circle at

*C*.

Construct the circumcentre *O* of Δ ABC (the point where the perpendicular bisectors of * AB*,

*meet).*

__BC__Construct the next vertex *D* as the point where the circle with centre *O* and passing through *A* meets the circle with centre *C* and passing through *B*. The remaining points *E*, *F*, *G*, *H* can be found in a similar way.

**185.**

(a)(i) There are various ways of doing this – none of a kind that most of us might stumble upon. Draw the circumcircle with centre *O* and passing through *A*. Extend the line *AO* beyond *O* to meet thecircle again at *X*.

Construct the perpendicular bisector of * AX*, and let this meet the circle at

*Y*and at

*Z*. Construct the midpoint

*M*of

*, and join*

__OZ__*.*

__MA__Let the circle with centre *M* and passing through *A* meet the line segment * OY* at the point

*F*.

Finally let the circle with centre *A* and passing through *F* meet the circumcircle at *B*. Then * AB* is a side of the required regular 5-gon.(The vertex

*C*on the circumcircle is then obtained as the second meeting point of the circumcircle with the circle having centre

*B*and passing through

*A*. The points

*D*,

*E*can be found in a similar way.)

The proof that this construction does what is claimed is most easily accomplished by calculating lengths.

Let * OA* = 2. Then

*= $\sqrt{5}$ - 1, so*

__OF__*= $\sqrt{10-2\sqrt{5}}$ =*

__AF__*. It remains to prove that this is the correct length for the side of a regular pentagon inscribed in a circle of radius*

__AB__*2*. Fortunately the work has already been done, since Δ

*OAB*is isosceles with apex angle equal to 72°. If we drop a perpendicular from

*O*to

*AB*, then we need to check whether it is true that

But this was already shown in Problem **3**(c).

(ii) To construct a regular 10-gon *ABCDEFGHIJ*, first construct a regular 5-gon *ACEGI* with circumcentre *O*; then construct the perpendicular bisectors of the five sides, and so find *B*, *D*, *F*, *H*, *J* as the points where these bisectors meet the circumcircle.

(b)(i) The first move is to construct a line *BX* through *B* such that∠ABX = 108°. Fortunately this can be done using part (a), by temporarily treating *B* as the circumcentre, drawing the circle with centre *B* and passing through *A*, and beginning the construction of a regular 5-gon AP·s inscribed in this circle.

Then∠ABP = 72°; so if we extend the line *PB* beyond *B* to *X*, then ∠ABX = 108°. Let *BX* meet the circle with centre *B* through *A* at the point *C*. Then * BA* =

*and ∠ABC = 108°, so we are up and running.*

__BC__If we let the perpendicular bisectorsof * AB* and

*meet at*

__BC__*O*, then the circle with centre

*O*and passing through

*A*also passes through

*B*and

*C*(and the yet to be l

*OBC*ted points

*D*and

*E*). The circle with centre

*C*and passing through

*B*meets this circle again at

*D*; and the circle with centre

*A*and passing through

*B*meets the circle again at

*E*.

(ii) To construct the regular *10*-gon *ABCDEFGHIJ*, treat *B* as the point *O* in (a)(i) and construct a regular *5*-gon *AXCYZ* inscribed in the circle with centre *B* and passing through *A*.

Then∠ABC = 144°, and * BA* =

*, so*

__BC__*C*is the next vertex of the required regular

*10*-gon. We may now proceed as in (a)(ii) to first construct the circumcentre

*O*of the required regular

*10*-gon as the point where the perpendicular bisectors of

*and*

__AB__*meet, then draw the circumcircle, and finally step off successive vertices*

__BC__*D*,

*E*, $$\text{.}\text{.}\text{.}$$ of the

*10*-gon around the circumcircle.

**186.** The number *k* of faces meeting at each vertex can be at most five (since more would produce an angle sum that is too large). And k ≥ 3 (in order to create a genuine corner.

- If
*k*= 5, then the vertex figure must be**3**(or the angle sum would be>360°).^{5} - If
*k*= 4, then the vertex figure must be**3**(or the angle sum would be too large).^{4} - If
*k*= 3, the angle in each of the regular polygons must be<120°, so the only possible vertex figures are**5**,^{3}**4**, and^{3}**3**.^{3}

**187.** The respective midpoints have coordinates:

of * AB*: $\left(\frac{1}{2},\frac{1}{2},0\right)$; of

*: $\left(\frac{1}{2},0,\frac{1}{2}\right)$; of*

__AC__*: $\left(1,\frac{1}{2},\frac{1}{2}\right)$; of*

__AD__*: $\left(0,\frac{1}{2},\frac{1}{2}\right)$; of*

__BC__*: $\left(\frac{1}{2},1,\frac{1}{2}\right)$ ; of*

__BD__*: $\left(\frac{1}{2},\frac{1}{2},1\right)$ .*

__CD__∴ * PQ* = $\frac{1}{\sqrt{2}}$
=

*=*

__PR__*=*

__PS__*=*

__PT__*=*

__QR__*=*

__RS__*=*

__ST__*.*

__TQ__**188.**

(a) There are infinitely many planes through the apex *A* and the base vertex *B*. Among these planes, the one perpendicular to the base *BCD* is the one that passes through the midpoint *M* of * CD*. Let the perpendicular from

*A*to the base, meet the base

*BCD*at the point

*X*, which must lie on

*. Let*

__BM__*have length*

__AX__*h*. To find

*h*we calculate the area of Δ ABM in two ways.

First, * BM* has length $\sqrt{3}$, so area(Δ ABM) = $\frac{1}{2}\left(\sqrt{3}\times h\right)$

Second, Δ ABM is isosceles with base * AB* and apex

*M*, so has height $\sqrt{2}$.

∴ area(Δ ABM)= $\frac{1}{2}(2\times \sqrt{2})$ = $\sqrt{2}$.

If we now equate the two expressions for area(Δ ABM), we see that h = $\sqrt{\frac{8}{3}}$.

(b)(i) When constructing a regular octahedron (whether using card, or tiles such as Polydron®) one begins by arranging four equilateral triangles around a vertex such as *A*. This ‘vertex figure’ is not rigid: though we know that * BC* =

*=*

__CD__*=*

__DE__*, there is no*

__EB__*a priori*reason why the four neighbours

*B*,

*C*,

*D*,

*E*of

*A*should form a square, or a rhombus, or a planar quadrilateral. We show that these four neighbours lie in a single plane:

*B*,

*C*,

*D*,

*E*are all distance 2 from

*A*and distance 2 from

*F*, so (by Problem

**144**) they must all lie in the plane perpendicular to the line

*AF*and passing through the midpoint

*X*of

*. Moreover, Δ ABX ≡ Δ ACX (by RHS-congruence, since*

__AF__*=*

__AB__*= 2,*

__AC__*=*

__AX__*, and∠AXB =∠AXC are both right angles). Hence*

__AX__*=*

__XB__*, so*

__XC__*B*,

*C*lie on a circle in this plane with centre

*X*. Similarly Δ ABX ≡ Δ ADX ≡ Δ AEX, so

*BCDE*is a cyclic quadrilateral (and a rhombus), and hence a square – with

*X*as the midpoint of both

*and*

__BD__*.*

__CE__(ii) Let *M* be the midpoint of * BC* and

*N*the midpoint of

*.*

__DE__Then *NM* and *AF* cross at *X* and so define a single plane. In this plane, Δ ANM ≡ Δ FMN (by SSS-congruence, since * AN*=

*= $\sqrt{3}$,*

__FM__*=*

__NM__*,*

__MN__*=*

__MA__*= $\sqrt{3}$); hence ∠ANM =∠FMN, so AN || MF.*

__NF__Similarly, if *P* is the midpoint of * AE* and

*Q*is the midpoint of

*, then Δ DPQ ≡ Δ BQP, so DP || QB. Hence the top face*

__FC__*DEA*is parallel to the bottom face

*BCF*, so the height of the octahedron sitting on the table is equal to the height of Δ FMN. But this triangle has sides of lengths

*2*, $\sqrt{3}$, $\sqrt{3}$, so this height is exactly the same as the height

*h*in part (a).

**189.**

(i) Let L = $\left(\frac{1}{2},0,0\right)andM=$
$\left(\frac{1}{2},\frac{1}{2},0\right)$. Then *L* lies on the line *ST* and *M* is the midpoint of * NP*.

Δ LSM is a right angled trianglewith legs of length
* LS* =

*A*,

*= $\frac{1}{2}$, so*

__LM__*= $\sqrt{{a}^{2}+\frac{1}{4}}$.*

__MS__Δ MNS is a right angled trianglewith legs of length
* MN* = $\frac{1}{2}$ -

*A*,

*= $\sqrt{{a}^{2}+\frac{1}{4}}$.*

__MS__Hence* NS* = $\sqrt{2{a}^{2}-a+\frac{1}{2}}$.

Similarly * NU* = $\sqrt{2a2-a+\frac{1}{2}}$.

Let Lʹ = $0,\frac{1}{2}$, 0\right) and Mʹ =
$\left(0,\frac{1}{2},\frac{1}{2}\right)$. Then *Mʹ* is the midpoint of * WX* and

*Lʹ*lies immediately below

*Mʹ*on the line joining (0, 0, 0) to (0, 1, 0). In the right angled triangleΔ LʹNMʹ we find

*= $\sqrt{{a}^{2}+\frac{1}{4}}$, and in the right angled triangle Δ MʹWN we then find*

__NMʹ__*= $\sqrt{2{a}^{2}-a+\frac{1}{2}}$. Similarly*

__NW__*= $\sqrt{2{a}^{2}-a+\frac{1}{2}}$.*

__NX__(ii) * NP* =

*precisely when 1 - 2a = $\sqrt{2{a}^{2}-a+\frac{1}{2}}>0$; that is, whena = $\frac{3-\sqrt{5}}{4}$, so all edges of the polyhedron have length$\frac{\sqrt{5}-1}{2}=\tau -1$.*

__NS__**Note:** The rectangle *NPRQ* is a “1 by $\tau -1$” rectangle, and 1: $\tau $ - 1 = $\tau $: 1. Hence the regular icosahedron can be constructed from three congruent, and pairwise perpendicular, copies of a “Golden rectangle” .

**190.** We mimic the classification of possible vertex figures for semi-regular tilings.

We are assuming that the angles meeting at each vertex add to<360°, so the number *k* of faces at each vertex lies between 3 and 5. Because faces are regular, but not necessarily congruent, *k* does not determine the shape of the faces. Hence we let *t* denote the number of triangles at each vertex, which can range from 0 up to 5.

- If t = 5, then the vertex figure must be
**3**.^{5} - If t = 4, the remaining polygons have angle sum<120°, so the possible vertex figures are
**3**,^{4}**3**, and^{4}.4**3**.^{4}.5 - If t = 3, then the remaining polygons have angle sum<180°, so the only possible vertex figures are
**3**, and^{3}**3**(for any^{3}.n*n*>3). - If t = 2, then the remaining polygons have angle sum< 240°, so there are at most 2 additional faces in the vertex figure (since if there were 3 or more extra faces, the average angle size would then be at most 80°, with no more triangles allowed). If there is just 1 additional face, we get the vertex figure
**3**for any n>3. So we may assume that there are 2 additional faces – the smallest of which must then be a 4-gon or a 5-gon.^{2}.n

If the next smallest face is a 4-gon, then we get the possible vertex figures

3and^{2}.4^{2}3.4.3.4,3and^{2}.4.53.4.3.5,3and^{2}.4.63.4.3.6,3and^{2}.4.73.4.3.7,3and^{2}.4.83.4.3.8,3and^{2}.4.93.4.3.9,3and^{2}.4.103.4.3.10,3and^{2}.4.113.4.3.11.

If the next smallest face is a 5-gon, then we get the possible vertex figures

3and^{2}.5^{2}3.5.3.5,3and^{2}.5.63.5.3.6,3and^{2}.5.73.5.3.7.

**Note:** Before proceeding further it is worth deciding which among the putative vertex figures identified so far seem to correspond to semi-regular polyhedra – and then to prove that these observations are correct.

- The vertex figure
**3**corresponds to the regular icosahedron.^{5} - The vertex figure
**3**corresponds to the regular octahedron;^{4}**3**corresponds to the^{4}.4*snubcube*;**3**corresponds to the^{4}.5*snub dodecahedron*– which comes in left-handed and right-handed forms. - The vertex figure
**3**corresponds to the regular tetrahedron, and^{3}**3**(for any n>3) corresponds to the^{3}.n*n*-gonal*antiprism*. - The vertex figure
**3**for any^{2}.n*n*> 3 does not seem to arise. - The vertex figure
**3**does not seem to arise;^{2}.4^{2}**3.4.3.4**corresponds to the*cuboctahedron*;**3**and^{2}.4.5**3.4.3.5**,**3**and^{2}.4.6**3.4.3.6**,**3**and^{2}.4.7**3.4.3.7**,**3**and^{2}.4.8**3.4.3.8**,**3**and^{2}.4.9**3.4.3.9**,**3**and^{2}.4.10**3.4.3.10**,**3**and^{2}.4.11**3.4.3.11**do not seem to arise. - The vertex figure
**3**does not seem to arise, whereas^{2}.5^{2}**3.5.3.5**corresponds to the*icosidodecahedron*;**3**and^{2}.5.6**3.5.3.6**,**3**and^{2}.5.7**3.5.3.7**do not seem to arise.

To avoid further proliferation of spurious ‘putative vertex figures’ we inject a version of the **Lemma** used for tilings somewhat earlier than we did fortilings, and then apply the underlying idea to eliminate other spurious possibilities as they arise.

**Lemma.** The vertex figures

3,^{2}.4^{2}3(n>3),^{2}.n3,^{2}.4.53.4.3.5,3,^{2}.4.63.4.3.6,3,^{2}.4.73.4.3.7,3,^{2}.4.83.4.3.8,3,^{2}.4.93.4.3.9,3,^{2}.4.103.4.3.10,3,^{2}.4.113.4.3.11,3,^{2}.5^{2}3,^{2}.5.63.5.3.6,3,^{2}.5.73.5.3.7

do not arise as vertex figures of any semi-regular polyhedron.

**Proof outline.** Each of these requires that the vertex figure of any vertex *B* includes a tile **T** = ABC·swith an **odd** number of edges, for which the edge * BA* is adjacent to an

*a*-gon, the edge

*is adjacent to a*

__BC__*b*-gon (where a ≠ b), and where the subsequent faces adjacent to

**T**are forced to alternate –

*a*-gon,

*b*-gon,

*a*-gon, $\dots $ – which is impossible. QED

For the rest we introduce the additional parameter *s* to denote the number of 4-gons in the vertex figure.

Suppose t= 1. Then the remaining polygons have angle sum< 300°, so there are at most 3 additional faces in the vertex figure (since if there were 4 or more extra faces, the average angle size would be<75°, with no more triangles allowed).

If there are 3 additional faces, the average angle size is<100°, so s > 0.

- If s > 1, then the possible vertex figures are
**3.4**(which corresponds to the^{3}*rhombicuboctahedron*),**3.4**(which corresponds to the 3-gonal prism),^{2}**3.4**(which is impossible as in the Lemma) and^{2}.5**3.4.5.4**(which corresponds to the*small rhombicosidodecahedron*). - If s = 1, then the remaining faces have angle sum<210°, so there can only be one additional face, and every
**3.4.n**(n > 4) is impossible as in the Lemma. - If s = 0, then the remaining faces have angle sum < 300°, so there are exactly two other faces and the smallest face has < 12 edges, so the only possible vertex figures are
–

**3.5**.*n*(4 < n), which is impossible as in the Lemma;–

**3.6**, which corresponds to the^{2}*truncated tetrahedron*;–

**3.6**.*n*(6 < n), which is impossible as in the Lemma;–

**3.7**.*n*(6 < n), which is impossible as in the Lemma;–

**3.8**, which corresponds to the^{2}*truncated cube*;–

**3.8**.*n*(8 < n), which is impossible as in the Lemma;–

**3.9**.*n*(8 < n), which is impossible as in the Lemma;–

**3.10**, which corresponds to the^{2}*truncated dodecahedron*;–

**3.10**.*n*(10 < n), which is impossible as in the Lemma;–

**3.11**.*n*(10 < n), which is impossible as in the Lemma.

Thus we may assume that t = 0 – in which case, s<4.

- If s = 3, then the only possible vertex figure is
**4**, which corresponds to the^{3}*cube*. - If s = 2, then the remaining faces have angle sum<180°, so there is exactly one additional face, and every
**4**(n>4) corresponds to the^{2}.n*n*-gonal*prism.* - If s = 1, then the remaining faces have angle sum<270°, so there are at most 2 other faces with the smallest face having<8 edges, so the only possible vertex figures are
**4.5.n**(for 4<n<20),**4.6**,^{2}**4.6.n**(for 6<n<12),**4.7**,^{2}**4.7.n**(for 7<n<10). Among these**4.5.n**is impossible as in the Lemma;**4.6**corresponds to the^{2}*truncated octahedron*;**4.6.n**with*n*odd is impossible as in the Lemma;**4.6.8**corresponds to the*great rhombicuboctahedron*;**4.6.10**corresponds to the*great rhombicosidodecahedron*;**4.7.n**is impossible as in the Lemma. - If s = 0, there must be exactly three faces at each vertex, and the smallest must be a 5-gon, so the only possible vertex figures are
**5**, which corresponds to the^{3}*regular dodecahedron*;**5**(for n>5), which is impossible as in the Lemma;^{2}.n**5.6**, which corresponds to the^{2}*truncated icosahedron*; or**5.6.7**, which is impossible as in the Lemma.

**191.**

(a)(i)∠AXD = 100°, so∠ADB = 40°. In Δ ABD we then see that∠ABD = 40°, so Δ ABD is isosceles with * AB* =

*. Δ ABC is isosceles with a base angle∠BAC = 60°, so Δ ABC is equilateral. Hence∠CBD = 20°.*

__AD__(ii) Δ ABC is equilateral, so * AC* =

*=*

__AB__*. Hence Δ ADC is isosceles, so∠ACD = 70° =∠ADC, whence∠BDC = 70° -∠ADB = 30°.*

__AD__(b)(i) As before ∠AXD = 100°, so∠ADB = 40°.

In Δ ABD we then see that∠ABD = 30°, so Δ ABD is not isosceles (as it was in (a)).

Δ ABC is isosceles, so∠BCA = 70°, whence∠CBD = 10°.

In Δ XCD, ∠CXD = 80°, so∠BDC +∠ACD = 100°, but there is no obvious way of determining the individual summands: *∠BDC* and *∠ACD*.

(ii) No lengths are specified, so we may choose the length of * AC*. The point

*B*then lies on the perpendicular bisector of

*, and∠CAB = 70° determines the l*

__AC__*OBC*tion of

*B*exactly. The line

*AD*makes an angle of 40° with

*AC*, and

*BD*makes an angle of 80° with

*AC*, so the l

*OBC*tion of

*D*is determined. Hence, despite our failure in part (i), the angles are determined.

**192.**

(i) If *P* lies on * CB*, then

*= b cos C,*

__PC__*=*

__AP__*B*sin C, and in the right angled triangle Δ APB we have:

c

^{2}= (b sin C)^{2}+ (a -Bcos C)^{2}= a

^{2}+ b^{2}(sin^{2}C + cos^{2}C) - 2ab cos C= a

^{2}+ b^{2}- 2ab cos C.

If *P* lies on * CB* extended beyond

*B*, then

*=*

__PC__*B*cos C,

*=*

__AP__*B*sin C as before, and in the right angled triangle Δ APB we have:

c

^{2}= (b sin C)^{2}+ (b cos C -a)^{2}= (bsin C)

^{2}+ (a - bcos C)^{2}= a

^{2}+ b^{2}(sin^{2}C + cos^{2}C) - 2abcos C= a

^{2}+ b^{2}- 2abcos C.

(ii) If *P* lies on * BC* extended beyond

*C*, then

*= bcos ∠ACP= -bcos C,*

__PC__*= bsin ∠ACP = bsin C, and in the right angled triangle Δ APB we have:*

__AP__c

^{2}= (bsin C)^{2}+ (a +)PC^{2}= (bsin C)

^{2}+ (a - bcos C)^{2}= a

^{2}+ b^{2}(sin^{2}C + cos^{2}C) - 2abcos C= a

^{2}+ b^{2}- 2abcos C.

**193.** There are many ways of doing this – once one knows the Sine Rule and Cosine Rule. If we let ∠BDC = y and ∠ACD = z, then one route leads to the identity cos(z - 10°) = 2 sin 10° · sin z, from which it follows that z = 80°, y = 20°.

**194.**

(a) The angle between two faces, or two planes, is the angle one sees “end-on” – as one looks along the line of intersection of the two planes. This is equal to the angle between two perpendiculars to the line of intersection – one in each plane. If *M* is the midpoint of * BC*, then Δ ABC is isosceles with apex

*A*, so the median

*is perpendicular to*

__AM__*; similarly Δ DBC is isosceles with apex*

__BC__*D*, so the median

*is perpendicular to*

__DM__*.*

__BC__Δ MAD is isosceles with apex *M*, * MA* =

*= $\sqrt{3}$,*

__MD__*= 2, so we can use the Cosine Rule to conclude that 2*

__AD__^{2}= 3 + 3 - 2· 3· cos(∠AMD), whence cos(∠AMD) = $\frac{1}{3}$.

(b) cos(∠AMD) = $\frac{1}{3}$, and $\frac{1}{3}$ < $\frac{1}{2}$, so∠AMD > 60°; hence we cannot fit six regular tetrahedra together so as to share an edge.

Since *∠AMD* is acute, ∠AMD < 90°, so we can certainly fit four regular tetrahedra together with lots of room to spare.

We can now appeal to “trigonometric tables” , or a calculator, to see that

$\mathrm{arccos}\left(\frac{1}{3}\right)\text{<}1.24$,

that is1.24 *radians*, so five tetrahedra use up less than 6.2 radians – which is less than 2$\pi $. Hence we can fit five regular tetrahedra together around a common edge with room to spare (but not enough to fit a sixth).

**195.**

(a) The angle between the faces *ABC* and *FBC* is equal to the angle between two perpendiculars to the common edge * BC*. Since the two triangles are isosceles with the common base

*, it suffices to find the angle between the two medians*

__BC__*and*

__AM__*.*

__FM__In Problem **188** we saw that *BCDE* is a square, with sides of length 2. If we switch attention from the opposite pair of vertices *A*, *F* to the pair *C*, *E*, then the same proof shows that *ABFD* is a square with sides of length 2. Hence the diagonal * AF* = 2$\sqrt{2}$.

Now apply the Cosine Rule to Δ AMF to conclude that:

$\left(2\sqrt{2}\right)$^{2} = 3 + 3 - 2· 3 · cos(∠AMF),

so it follows that cos(∠AMF) = -$\frac{1}{3}$.

(b) cos(∠AMF) = -$\frac{1}{3}$ < 0, so∠AMF > 90°; hence we cannot fit four regular octahedra together so as to share a common edge. Moreover, -$\frac{1}{2}$ < -$\frac{1}{3}$, so∠ AMD < 120°; hence we can fit three octahedra together to share an edge with room to spare (but not enough room to fit a fourth).

**196**. The angle ∠AMD = $\mathrm{arccos}\left(\frac{1}{3}\right)$ in Problem **194** is acute, and the angle∠AMF = $\mathrm{arccos}\left(-\frac{1}{3}\right)$ in Problem **195** is obtuse. Hence these angles are supplementary; so the regular tetrahedron fits *exactly* into the wedge-shaped hole between the face *ABC* of the regular octahedron and the table.

**197.**

(i) * AB* = $\sqrt{2}$ =

*=*

__BC__*=*

__CA__*=*

__AW__*=*

__BW__*. Hence the four faces*

__CW__*ABC*,

*BCW*, CWA, WAB are all equilateral triangles, so the solid is a regular tetrahedron (or, more correctly, the surface of the solid is a regular tetrahedron).

(ii) * AC* = $\sqrt{2}$ =

*=*

__CD__*=*

__DA__*=*

__AX__*=*

__CX__*. Hence the four faces*

__DX__*ACD*,

*CDX*, DXA,

*XAC*are equilateral triangles, so the solid is a regular tetrahedron.

(iii) * AD* = $\sqrt{2}$ =

*=*

__DE__*=*

__EA__*=*

__AY__*=*

__DY__*. Hence the four faces*

__EY__*ADE*,

*DEY*, EYA, YAD are equilateral triangles, so the solid is a regular tetrahedron.

(iv) * AE* = $\sqrt{2}$ =

*=*

__EB__*=*

__BA__*=*

__AZ__*=*

__EZ__*. Hence the four faces*

__BZ__*AEB*,

*EBZ*, BZA, ZAE are equilateral triangles, so the solid is a regular tetrahedron.

(v) We get another four regular tetrahedra– such as *FBCP*, where P = (1, 1, -1).

(vi) *ABCDEF* is a regular octahedron (or, more correctly, the surface of the solid is a regular octahedron).

**Note:** The six vertices (± 1, 0, 0), (0, ± 1, 0), (0, 0, ± 1) span a regular octahedron, with a regular tetrahedron fitting exactly on each face. The resulting compound star-shaped figure is called the *stellated octahedron*. Johannes Kepler (1571–1630) made an extensive study of polyhedra and this figure is sometimes referred to as Kepler's ‘stella octangula'. It is worth making in order to appreciate the way it appears to consist of two interlocking tetrahedra.

**198.** Let the unlabeled vertex of the pentagonalface ABW-V be *P*. In the pentagon *ABWPV* the edge * AB* is parallel to the diagonal

*. Hence*

__VW__*ABWV*is an isosceles trapezium. The sides

*VA*and

*WB*(produced)meet at

*S*in the plane of the pentagon

*ABWPV*.

Δ SAB has equal base angles, so * SA* =

*.*

__SB__Hence * SV* =

*.*

__SW__Similarly BC || WX, and *WB* and *XC* meet at some point *Sʹ* on the line *WB*.

Now Δ SʹBC ≡ Δ SAB, so * SʹB* =

*=*

__SA__*. Hence Sʹ = S, and the lines*

__SB__*VA*,

*WB*,

*XC*,

*YD*,

*ZE*all meet at

*S*.

Since VW || AB, we know that Δ SAB ~ Δ SVW, with scale factor

$\tau :1$ = * VW*:

*=*

__AB__*:*

__SV__*.*

__SA__If * SA* = x, then

*X*+ 1:

*X*= $\tau :1$, so

*X*= $\tau $.

Let *M* be the midpoint of * AB* and let

*O*denote the circumcentre of the regular pentagon

*ABCDE*.

Δ *OAB* is isosceles, so *OM* is perpendicular to *AB*, and the required dihedral angle between the two pentagonal faces is *∠OMP*.

It turns out to be better to find not the dihedral angle *∠OMP*, but its supplement: namely the angle *∠SMO*.

From Δ OAM we see that

* OA*= $\frac{1}{2\mathrm{sin}{36}^{\xb0}}$,

and we know that * SA* = $\tau =2\mathrm{cos}$ 36°. One can then check that

*= $\mathrm{cot}$ 36°.*

__OS__Similarly, in Δ OAM we have

* OM* = $\frac{\mathrm{cot}{36}^{\xb0}}{2}$.

Hence in Δ OMS we have

tan(∠SMO) = 2.

Hence the required dihedral angle is equal to $\pi -\mathrm{arctan}2\approx $ 116.56°.

**199.**

(a)(i) Let *M* be the midpoint of * AC*. The angle between the two faces is equal to

*∠BMD*.

In Δ BMD, we have * BM* =

*= $\sqrt{3}$,*

__DM__*= 2$\tau $ = 1 + $\sqrt{5}$. So the Cosine Rule in Δ BMD gives:*

__BD__$6+2\sqrt{5}=3+3$ - 2· 3· cos(∠BMD),

so cos(∠BMD) = -$\frac{\sqrt{5}}{3}$,

∠$BMD=\mathrm{arccos}\left(-\frac{\sqrt{5}}{3}\right)=\pi -\mathrm{arccos}\left(\frac{\sqrt{5}}{3}\right)\approx {138.19}^{\xb0}$.

(ii) $\frac{\sqrt{5}}{3}>\frac{1}{2}$; hence∠BMD>$\frac{2\pi}{3}$, so we can fit two copies along a common edge, but not three.

(b)(i) Now let *M* denote the midpoint of * BC*, and suppose that

*OM*extended beyond

*M*meets the circumcircle at

*V*. Then

*is an edge of the regular 10-goninscribed in the circumcircle. The circumradius*

__BV__*of*

__OB__*BCDEF*(which is equal to the edge length of the inscribed regular hexagon) is

* OB*= $\frac{1}{\mathrm{sin}{36}^{\xb0}}$;

and ∠MBV = 18°,

so * BV*= $\frac{1}{\mathrm{cos}{18}^{\xb0}}$.

It is easiest to use the converse of Pythagoras’ Theorem, and to write everything first in terms of cos 36°, then (since $\tau $ $=2\mathrm{cos}36$°, so $\mathrm{cos}{36}^{\xb0}=\frac{1+\sqrt{5}}{4})$ write everything in terms of $\sqrt{5}$.

If we use sin^{2} 36° = 1 - cos^{2} 36°, and 2cos^{2} 18° - 1 = cos 36° = $\frac{1+\sqrt{5}}{4}$, then

__BV__^{2} + __OB__^{2} =
${\left(\frac{1}{\mathrm{cos}{18}^{\xb0}}\right)}^{2}+\left(\frac{1}{\mathrm{sin}{36}^{\xb0}}\right)$^{2}

= $\frac{8}{5+\sqrt{5}}+\frac{8}{5-\sqrt{5}}$

= $\frac{80}{20}$ = 2^{2}

= __AB__^{2}.

Hence, in the right angled triangle Δ AOB, we must have * OA* =

*as claimed.*

__BV__(ii) Miraculously no more work is needed. Let the vertex at the ‘south pole’ be *L*, and let the pentagon formed by its five neighbours be *GHIJK*. It helps if we can refer to the circumcircle of *BCDEF* as the ‘tropic of Cancer', and to the circumcircle of *GHIJK* as the ‘tropic of Capricorn'.

The pentagon *GHIJK* is parallel to *BCDEF*, but the vertices of the southern pentagon have been rotated through $\frac{\pi}{5}$ relative to *BCDEF*, so that *G* (say) lies on the circumcircle of the pentagon *GHIJK*, but sits directly below the midpoint of the minor arc * BC*. Let

*X*denote the point on the ‘tropic of Capricorn’ which is directly beneath

*B*. Then Δ BXG is a right angled triangle with

*=*

__BG__*= 2, and*

__AB__*=*

__XG__*is equal to the edge length of a regular 10-gon inscribed in the circumcircle of*

__BV__*GHIJK*. Hence, by the calculation in (i),

*is equal to the edge length of the regular hexagon inscribed in the same circle – which is also equal to the circumradius.*

__BX__**200.** A necessary condition for copies of a regular polyhedron to “tile 3D (without gaps or overlaps)” is that an integral number of copies should fit together around an edge. That is, the dihedral angle of the polyhedron should be an exact submultiple of 2$\pi $. Only the cube satisfies this necessary condition.

Moreover, if we take as vertices the points (p, q, r) with integer coordinates *p*, *q*, *r*, and as our regular polyhedra all possible translations of the standard unit cube having opposite corners at (0, 0, 0) and (1, 1, 1), then we see that it is possible to tile 3D using just cubes.

**201.** The four diameters form the four edges of a square *ABCD* of edge length *2*.

The protruding semicircular segments on the left and right can be cut off and inserted to exactly fill the semicircular indentations above and below.

Hence the composite shape has area exactly equal to 2^{2} = 4 square units.

**202.**

(a) If the regular *n*-gon is ACEG·sand the regular *2n*-gon is ABCDEFG·s, then * AC* = s_n = s and

*=*

__AB__*s*

_{2n}=

*t*. If

*M*is the midpoint of

*,*

__AC__*= $\frac{s}{2}$ and*

__AM__* MB* = $1-\sqrt{(1-\left(\frac{s}{2}\right)=\left(\frac{s}{2}\right)}$

^{2}.

(b) Put s = s_{2}= 2 in (1) to get

t = s_{4}= $\sqrt{2}$. Then put s = s_{4}= $\sqrt{2}$ in (1) to get t = s_{8}= $\sqrt{2-\sqrt{2}}$.

(c) Put t = s_{6}= 1 in (1) to get s = s_{3}= $\sqrt{3}$. Then put s = s_{6}= 1 to get

t = s_{12}= $\sqrt{2-\sqrt{3}}$.

(d) Put s = s_{5}= $\frac{\sqrt{10-2\sqrt{5}}}{2}$ to get

$t={s}_{10}=\sqrt{\frac{3-\sqrt{5}}{2}}$.

**203.**

(a) (i)

*n* = 3: p_{3} = 3$\sqrt{3}\times r$

*n* = 4: p_{4} = $4\sqrt{2}\times r$

*n* = 5: p_{5} = $\frac{5\sqrt{10-2\sqrt{5}}}{2}\times r$

*n* = 6: p_{6} = 6 $\times r$

*n* = 8: p_{8} = 8$\sqrt{2-\sqrt{2}}\times r$

*n* = 10: p_{10} = 5$\sqrt{6-2\sqrt{5}}\times r$

*n* = 12: p_{12} = 12$\sqrt{2-\sqrt{3}}\times r$.

(ii)

$$\begin{array}{ccc}{c}_{3}=5.19\cdots \hfill & <\hfill & {c}_{4}=5.65\cdots \hfill \\ \hfill & <\hfill & {c}_{5}=5.87\cdots \hfill \\ \hfill & <\hfill & {c}_{6}=6\hfill \\ \hfill & <\hfill & {c}_{8}=6.12\cdots \hfill \\ \hfill & <\hfill & {c}_{10}=6.18\cdots \hfill \\ \hfill & <\hfill & {c}_{12}=6.21\cdots .\hfill \end{array}$$(b)(i)

*n* = 3: P_{3} = 6$\sqrt{3}\times r$

*n* = 4: P_{4} = 8 $\times r$

*n* = 5: P_{5} = 10$\sqrt{5-2\sqrt{5}}\times r$

*n* = 6: P_{6} = 4$\sqrt{3}\times r$

*n* = 8: P_{8} = 8$\left(2\sqrt{2}-2\right)\times r$

*n* = 10: P_{10} = $4\sqrt{25-10\sqrt{5}}\times r$

*n* = 12: P_{12} = $12\left(4-2\sqrt{3}\right)\times r$.

(ii)

$$\begin{array}{ccc}{C}_{3}=10.39\cdots \hfill & >\hfill & {C}_{4}=8\hfill \\ \hfill & >\hfill & {C}_{5}=7.26\cdots \hfill \\ \hfill & >\hfill & {C}_{6}=6.92\cdots \hfill \\ \hfill & >\hfill & {C}_{8}=6.62\cdots \hfill \\ \hfill & >\hfill & {C}_{10}=6.49\cdots \hfill \\ \hfill & >\hfill & {C}_{12}=6.43\cdots .\hfill \end{array}$$(c) Let *O* be the centre of the circle of radius *r*. Let *A*, *B* lie on the circle with∠AOB = 30°.

Let *M* be the midpoint of * AB* – so that Δ

*OAB*is isosceles, with apex

*O*and height h =

*.*

__OM__Let *Aʹ* lie on * OA* produced, and let

*Bʹ*lie on

*produced, such that*

__OB__*=*

__OAʹ__*and AʹBʹ is tangent to the circle.*

__OBʹ__Then Δ *OAB* ~ Δ OAʹBʹ with Δ OAʹBʹ larger than Δ *OAB*, so the scale factor $\frac{1}{h}=2\sqrt{2-\sqrt{3}}>1$.

Hence ${P}_{12}=2\sqrt{2-\sqrt{3}}\times {p}_{12}>{p}_{12},so{C}_{12}>{c}_{12}$.

**204.** (i) $\pi r$ (ii) $\frac{\pi}{2}r$ (iii) $\theta r$

**205.**

(a) (i) Note: This could be a long slog. However we have done much of the work before: when the radius is *1*, the most of the required areas were calculated exactly back in Problem **3** and Problem **19**.

Alternatively, the area of each of the *n* sectors is equal to $\frac{1}{2}\mathrm{sin}\theta $, where $\theta $ is the angle subtended at the centre of the circle, and the exact values of the required trig functions were also calculated back in Chapter 1.

${a}_{3}=\frac{3\sqrt{3}}{4}\times $ r^{2}

${a}_{4}=2\times $ r^{2}

${a}_{5}=\frac{5}{8}\sqrt{10+2\sqrt{5}}\times $ r^{2}

${a}_{6}=\frac{3\sqrt{3}}{2}\times $ r^{2}

${a}_{8}=2\sqrt{2}\times $ r^{2}

${a}_{10}=\frac{5}{4}\sqrt{10-2\sqrt{5}}\times $ r^{2}

${a}_{12}=3\times $ r^{2}

(ii)

$$\begin{array}{ccc}{d}_{3}=1.29\cdots \hfill & <\hfill & {d}_{4}=2\hfill \\ \hfill & <\hfill & {d}_{5}=2.37\cdots \hfill \\ \hfill & <\hfill & {d}_{6}=2.59\cdots \hfill \\ \hfill & <\hfill & {d}_{8}=2.82\cdots \hfill \\ \hfill & <\hfill & {d}_{10}=2.93\cdots \hfill \\ \hfill & <\hfill & {d}_{12}=3.\hfill \end{array}$$(b)(i) Note: This could also be a long slog. However we have done much of the work before. But notice that, when the radius is *1*, the area of each of the *n* sectors is equal to half the edge length times the height (= 1); so if r=1, then the total area is numerically equal to “half the perimeter P_n” .

${A}_{3}=3\sqrt{3}\times $ r^{2}

${A}_{4}=4\times $ r^{2}

${A}_{5}=5\sqrt{5-2\sqrt{5}}\times $ r^{2}

${A}_{6}=2\sqrt{3}\times $ r^{2}

${A}_{8}=8\left(\sqrt{2}-1\right)\times $ r^{2}

${A}_{10}=2\sqrt{25-10\sqrt{5}}\times $ r^{2}

${A}_{12}=12\left(2-\sqrt{3}\right)\times $ r^{2}

(ii)

$$\begin{array}{ccc}{D}_{3}=5.19\cdots \hfill & >\hfill & {D}_{4}=4\hfill \\ \hfill & >\hfill & {D}_{5}=3.63\cdots \hfill \\ \hfill & >\hfill & {D}_{6}=3.46\cdots \hfill \\ \hfill & >\hfill & {D}_{8}=3.31\cdots \hfill \\ \hfill & >\hfill & {D}_{10}=3.24\cdots \hfill \\ \hfill & >\hfill & {D}_{12}=3.21\cdots .\hfill \end{array}$$(c) Let *O* be the centre of the circle of radius *r*. Let *A*, *B* lie on the circle with∠AOB = 30°.

Let *M* be the midpoint of * AB* – so that Δ

*OAB*is isosceles, with apex

*O*and height h =

*.*

__OM__Let *Aʹ* lie on * OA* produced, and let

*Bʹ*lie on

*produced, such that*

__OB__*=*

__OAʹ__*and AʹBʹ is tangent to the circle.*

__OBʹ__Then Δ *OAB* ~ Δ OAʹBʹ with Δ *OAB* contained in Δ OAʹBʹ, so the scale factor $\frac{1}{h}=2\sqrt{2-\sqrt{3}}>1$.

Hence 4(2 - $\sqrt{3}$)· ${a}_{12}={A}_{12}{a}_{12}$,

so ${D}_{12}{d}_{12}$.

**206.** The rearranged shape (shown in Figure 9 is an “almost rectangle” , where * OA*forms an “almost height” and

*= r. Half of the*

__OA__*2n*circular arcs such as

*AB*form the upper “width” , and the other half form the “lower width” , so each of these “almost widths” is equal to half the perimeter of the circle –namely $\pi $ r.

Figure 9: Rectification of a circle.

Hence the area of the rearranged shape is very close to

$r\times \pi r=\pi $ r^{2}.

**207.**

(a) Cut along a generator, open up and lay the surface flat to obtain: a 2$\pi $ *r* by *h* rectangle and two circular discs of radius *r*.

Hence the total surface area is

2$\pi $ r

^{2}+ 2$\pi $ rh = $2\pi r(r+h)$.

(b) The lateral surface consists of *n* rectangles, each with dimensions s_n by *h* (where s_n is the edge length of the regular *n*-gon), and hence has area P_nh = 2${\Pi}_{n}rh$.

Adding in the twoend discs (each with area $\frac{1}{2}{P}_{n}r)$ then gives total surface area $2{\Pi}_{n}r(r+h)$.

**208.**

(a) (i) $\frac{1}{2}\pi $ r^{2}; (ii) $\frac{1}{4}\pi $ r^{2}; (iii) $\frac{\theta}{2}$r^{2}.

(b) The sector has two radii of total length *2r*. Hence the circular must have length $(\pi -2)r$, and so subtends an angle $\pi $ - 2 at the centre, so has area $\frac{\pi -2}{2}$r^{2}.

**209.**

(a) Focus first on the sloping surface. If we cut along a “generator” (a straight line segment joining the apex to a point on the circumference of the base), the surface opens up and lays flat to form a sector of a circle of radius *l*. The outside arc of this sector has length 2$\pi $ *r*, so the sector angle at the centre is equal to $\frac{r}{l}$· 2$\pi $, and hence its area is $\frac{r}{l}$· $\pi $ l^{2} = $\pi $ rl.

Adding the area of the base gives the total surface area of the cone as

$\pi r(r+l)=\frac{1}{2}$· $2\pi r(r+l)$.

(b) Let *M* be the midpoint of the edge * BC* and let

*l*denote the ‘slant height’

*.*

__AM__Then the area of the *n* sloping faces is equal to $\frac{1}{2}{P}_{n}$· l, while the area of the base is equal to $\frac{1}{2}{P}_{n}$· r.

Hence the surface area is precisely$\frac{1}{2}{P}_{n}(r+l)$.

**210.**

(a) If * AB* is an edge of the inscribed regular

*n*-gon ABCD$\mathrm{...}$, and

*O*is the circumcentre, then∠AOB = $\frac{2\pi}{n}$, so

*= 2r$\mathrm{sin}\frac{\pi}{n}$. Hence the required ratio is equal to $\frac{\mathrm{sin}\frac{\pi}{n}}{\frac{\pi}{n}}$, which tends to*

__AB__*1*as

*n*tends to $\infty $.

(b) If * AB* is an edge of the circumscribed regular

*n*-gon ABCD$\mathrm{...}$, and

*O*is the circumcentre, then∠AOB = $\frac{2\pi}{n}$, so

*= 2r$\mathrm{tan}\frac{\pi}{n}$. Hence the required ratio is equal to $\frac{\mathrm{tan}\frac{\pi}{n}}{\frac{\pi}{n}}$, which tends to*

__AB__*1*as

*n*tends to $\infty $.

**211.**

(a) If * AB* is an edge of the inscribed regular

*n*-gon ABCD$\mathrm{...}$, and

*O*is the circumcentre, then∠AOB = $\frac{2\pi}{n}$, so area(Δ

*OAB*) = $\frac{1}{2}{r}^{2}\mathrm{sin}\frac{2\pi}{n}$. Hence the required ratio is equal to $\frac{\mathrm{sin}\frac{2\pi}{n}}{\frac{2\pi}{n}}$, which tends to

*1*as

*n*tends to $\infty $.

(b) If * AB* is an edge of the circumscribed regular

*n*-gon ABCD$\mathrm{...}$, and

*O*is the circumcentre, then∠AOB = $\frac{2\pi}{n}$, so area(Δ

*OAB*) = r

^{2}$\mathrm{tan}\frac{\pi}{n}$. Hence the required ratio is equal to $\frac{\mathrm{tan}\frac{\pi}{n}}{\frac{\pi}{n}}$, which tends to

*1*as

*n*tends to $\infty $.

**212.**

(a) area(P_{2}) = area(P_{1}) + area(P_{2} - P_{1}) > area(P_{1}).

(b) This general result is clearly related to the considerations of the previous section. But it is not clear whether we can really expect to prove it with the tools available. So it has been included partly in the hope that readers might come to appreciate the difficulties inherent in proving such an “obvious” result.

In the end, any attempt to prove it seems to underline the need to use “proof by induction” – for example, on the number of edges of the inner polygon. This method is not formally treated until Chapter 6, but is needed here.

• Suppose the inner polygon P_{1} = ABC has just n= 3 edges, and has perimeter p_{1}.

Draw the line through *A* parallel to *BC*, and let it meet the (boundary of the) polygon P_{2} at the points *U* and *V*. The triangle inequality (Problem **146**(c)) guarantees that the length * UV* is less than or equal to the length of the compound path from

*U*to

*V*along the perimeter of the polygon P

_{2}(keeping on the opposite side of the line

*UV*from

*B*and

*C*). So, if we cut off the part of P

_{2}on the side of the line

*UV*opposite to

*B*and

*C*, we obtain a new outer convex polygon P

_{3}, which contains P

_{1}, and whose perimeter is no larger than that of P

_{2}.

Now draw the line through *B* parallel to *AC*, and let it meet the boundary of P_{3} at points *W*, *X*. If we cut off the part of P_{3} on the side of the line *WX* opposite to *A* and *C*, we obtain a new outer convex polygon P_{4}, which contains P_{1}, and whose perimeter is no larger than that of P_{3}.

If we now do the same by drawing a line through *C* parallel to *AB*, and cut off the appropriate part of P_{4}, we obtain a final outer convex polygon P_{5}, which contains the polygon P_{1}, and whose perimeter is no larger than that of P_{4}– and hence no larger than that of the original outer polygon P_{2}.

All three vertices *A*, *B*, *C* of P_{1} now lie on the boundary of the outer polygon P_{5}, so the triangle inequality guarantees that *AB* is no larger than the length of the compound path along the boundary of P_{5} from *A* to *B* (staying on the opposite side of the line *AB* from *C*). Similarly *BC* is no larger than the length of the compound path along the boundary of P_{5} from *B* to *C*; and *CA* is no larger than the length of the compound path along the boundary of P_{5} from *C* to *A*.

Hence the perimeter p_{1} of the triangle P_{1} is no larger than the perimeter of the outer polygon P_{5}, whose perimeter was no larger than the perimeter of the original outer polygon P_{2}. Hence the result holds when the inner polygon is a triangle.

• Now suppose that the result has been proved when the inner polygon is a *k*-gon, for some k ≥ 3, and suppose we are presented with a pair of polygons P_{1}, P_{2} where the inner polygon P_{1} = ABCD·sis a convex (k+1)-gon.

Draw the line *m* through *C* parallel to *BD*. Let this line meet the outer polygon P_{2} at *U* and *V*. Cut off the part of P_{2} on the opposite side of the the line *UCV* to *B* and *D*, leaving a new outer convex polygon *P* with perimeter no greater than that of P_{2}. We prove that the perimeter of polygon P_{1} is less than that of polygon *P* – and hence less than that of P_{2}. Equivalently, we may assume that *UCV* is an edge of P_{2}.

Translate the line *m* parallel to itself, from *m* to *BD* and beyond, until it reaches a position of final contact with the polygon P_{1}, passing through the vertex *X* (and possibly a whole edge * XY*)of the inner polygon P

_{1}. Let this final contact line parallel to

*m*be

*mʹ*.

Since P_{1} is convex and k ≥qslant 3, we know that *X* is different from *B* and from *D*. As before, we may assume that *mʹ* is an edge of the outer polygon P_{2}. Cut both P_{2} and P_{1} along the line *CX* to obtain two smaller configurations, each of which consists of an inner convex polygon inside an outer convex polygon, but in which

• each of the inner polygons has at most *k* edges, and

• in each of the smaller configurations, the inner and outer polygons both share the edge * CX*.

Then (by induction on the number of edges of the inner polygon) the perimeter of each inner polygon is no larger than the perimeter of the corresponding outer polygon; so for each inner polygon, the partial perimeter running from *C* to *X* (omitting the edge * CX*) is no larger than the partial perimeter of the corresponding outer polygon running from

*C*to

*X*(omitting the edge

*). So when we put the two parts back together again, we see that the perimeter of P*

__CX___{1}is no larger than the perimeter of P

_{2}.

Hence the result holds when P_{1}is a triangle; and if the result holds whenever the inner polygon has k ≥qslant 3 edges, it also holds whenever the inner polygon has (k+ 1) edges.

It follows that the result holds whatever the number of edges of the inner polygon may be. QED

**213.**

(a) Join * PQ*. Then the lines y =

*B*and

*X*= d meet at

*R*to form the right angled triangle

*PQR*.

Pythagoras’ Theorem then implies that (d - a)^{2} + (e - b)^{2} = __PQ__^{2}.

(b) Join * PQ*. The points P = (a,

*B*, c) and

*R*= (d, e, c) lie in the plane z = c. If we work exclusively in this plane, then part (a) shows that

__PR__^{2} = (d - a)^{2} + (e - b)^{2}.

* QR* = |

*F*- c|, and Δ PRQ has a right angle at

*R*.

Hence

__PQ__^{2} = __PR__^{2} + __RQ__^{2} = (d - a)^{2} + (e - b)^{2} + (f - c)^{2}.

**214.**

(a) area(Δ ABC) = $\frac{bc}{2}$, area(Δ ACD) = $\frac{cd}{2}$, area(Δ ABD) = $\frac{bd}{2}$.

(b) [This can be a long algebraic slog. And the answer can take very different looking forms depending on how one proceeds. Moreover, most of the resulting expressions are not very pretty, and are likely to incorporate errors.

One way to avoid this slog is to appeal to the fact that the modulus of the vector product **DB × DC** is equal to the area of the parallelogram spanned by **DB**

and **DC** – and so is twice the area of Δ BCD.]

(c) However part (b) is appr*OAC*hed, it is in fact true that area(Δ $BCD{)}^{2}={\left(\frac{bc}{2}\right)}^{2}+$
${\left(\frac{cd}{2}\right)}^{2}+{\left(\frac{bd}{2}\right)}^{2}$,

so that

area(Δ ABC)^{2} + area(Δ ACD)^{2} + area(Δ ABD)^{2} = area(Δ BCD)^{2}.

**215.**

(a) *b* and *c* are indeed lengths of arcs of great circles on the unit sphere: that is, arcs of circles of radius 1 (centred at the centre of the sphere). However, back in Chapter 1 we used the ‘length’ of such circular arcs to define the **angle** (in radians) subtended by the arc at the centre. So *b* and *c* are also *angles* (in radians).

(b) The only standard functions of angles are the familiar trig functions (sin, cos).

(c) If c = 0, then the output should specify that a = *B*, so *c* should have no effect on the output. This suggests thatwe might expect a formula that involves “adding sin c” or “multiplying by cos c” .

Similarly when *B* = 0, the output should give a= c, so we might expect a formula that involved “adding sin b” or
“multiplying by cos b” .

In general, we should expect a formula in which *b* and *c* appear interchangeably (since the input pair(b, c) could equally well be replaced by the input pair (c, b) and should give the same output value of *a*).

(d)(i) If * BC* runs along the equator and∠B =∠C = $\frac{\pi}{2}$, then

*and*

__BA__*run along circles of longitude, so*

__CA__*A*must be at the North pole. Since

*A*is a right angle, it follows that a =

*B*= c = $\frac{\pi}{2}$.(This tends to rule out the idea that the formula might involve “adding sin

*B*and adding sin c” .)

(ii) Suppose that∠B = $\frac{\pi}{2}$. Since we can imagine * AB* along the equator, and since there is a right angle at

*A*, it follows that

*and*

__AC__*both lie along circles of longitude, and so meet at the North pole. Hence*

__BC__*C*will be at the North pole, so a =

*B*= $\frac{\pi}{2}$.

The inputs to any spherical version of Pythagoras’ Theorem are then *B* = $\frac{\pi}{2}$, and *c*. And *c* is not constrained, so every possible input value of *c* mustlead to the same output a = $\frac{\pi}{2}$. This tends to suggest that the formula involves some multiple of a product combining “cos b” with some function of*c*. And since the inputs “*b*” and “*c*” must appear symmetrically, we might reasonably expect some multiple of “cos *B* · cos c” .

**216.**

(a) Δ *OAB*ʹ has a right angle at *A* with∠AOBʹ =∠AOB = c. Hence * ABʹ* = tan c. Similarly

*= tan b.*

__ACʹ__Hence __BʹCʹ__^{2}= tan^{2} *B* + tan^{2} c.

(b) In Δ *OAB*ʹ we see that * OBʹ* = sec c. Similarly

*= sec b. We can now apply the Cosine Rule to Δ OBʹCʹ to obtain:*

__OCʹ__Hence cos a = cos *B* · cos c. QED

**217.** Construct the plane tangent to the sphere at *A*. Extend * OB* to meet this plane at

*Bʹ*, and extend

*to meet the plane at*

__OC__*Cʹ*.

Δ *OAB*ʹ has a right angle at *A* with∠AOBʹ =∠AOB = c. Hence * ABʹ* = tan c. Similarly

*= tan b.*

__ACʹ__Hence __BʹCʹ__^{2}= tan^{2} *B* + tan^{2} c - 2· tan *B* · tan c· cos A.

In Δ *OAB*ʹ we see that * OBʹ* = sec c. Similarly

*= sec b.*

__OCʹ__We can now apply the Cosine Rule to Δ OBʹCʹ to obtain:

$$\begin{array}{l}{\mathrm{tan}}^{2}b+{\mathrm{tan}}^{2}c-2\xb7\mathrm{tan}b\xb7\mathrm{tan}c\xb7\mathrm{cos}A\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\mathrm{sec}}^{2}b+{\mathrm{sec}}^{2}c-2\mathrm{sec}b\xb7\mathrm{sec}\left(\angle {B}^{\prime}O{C}^{\prime}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\mathrm{sec}}^{2}b+{\mathrm{sec}}^{2}c-2\mathrm{sec}b\xb7\mathrm{sec}c\xb7\mathrm{cos}a.\end{array}$$Hence cos a = cos b· cos c + sin b· sin c · cos A. QED

**218.** We show that $\frac{\mathrm{sin}b}{\mathrm{sin}a}=\mathrm{sin}B$ (where *B* denotes the angle *∠ABC* at the vertex *B*).

Construct the plane **T** which is tangent to the sphere at *B*. Let *O* be the centre of the sphere; let __OA__

produced meet the plane **T** at Aʹʹ, and let * OC* produced meet the plane

**T**at Cʹʹ.

Imagine * BA* positioned along the equator; then

*is horizontal;*

__BAʹʹ__*lies on a circle of longitude, so AʹʹCʹʹis vertical. Hence∠Cʹʹ BAʹʹ=∠B, and∠BAʹʹ Cʹʹis a right angle; so sin $B=\frac{A\prime \prime C\prime \prime}{BC\prime \prime}$.*

__AC__Δ OAʹʹCʹʹ is right angled at Aʹʹ; and∠AʹʹOCʹʹ = b; so sin *B* = $\frac{A\prime \prime C\prime \prime}{OC\prime \prime}$.

Δ *OBC*ʹʹ is right angled at *B*; and∠BOCʹʹ = a; so sin a = $\frac{BC\prime \prime}{OC\prime \prime}$.

Hence$\frac{\mathrm{sin}b}{\mathrm{sin}a}$ = sin *B* depends only on the angle at *B*, so $\frac{\mathrm{sin}b}{\mathrm{sin}a}=\frac{\mathrm{sin}{b}^{\prime}}{\mathrm{sin}{a}^{\prime}}=\mathrm{sin}B$.

**219.** We show that

$\frac{\mathrm{sin}a}{\mathrm{sin}A}=\frac{\mathrm{sin}b}{\mathrm{sin}B}$.

Position the triangle (or rather “rotate the sphere” ) so that * AB* runs along the equator, with

*leading into the northern hemisphere.*

__AC__(i) If *∠A* is a right angle, then

$\frac{\mathrm{sin}b}{\mathrm{sin}a}=\mathrm{sin}B=\frac{\mathrm{sin}B}{\mathrm{sin}A}$

by Problem **218**. Hence

$\frac{\mathrm{sin}a}{\mathrm{sin}A}=\frac{\mathrm{sin}b}{\mathrm{sin}B}$.

The same is true if *∠B* is a right angle.

(ii) If *∠A* and *∠B* are both less than a right angle, one can draw the circle of longitude from *C* to some point *X* on * AB*. One can then apply Problem

**218**to the two triangles Δ CXA and Δ CXB (each with a right angle at

*X*). Let

*x*denote the length of the

*. Then $\frac{\mathrm{sin}x}{\mathrm{sin}b}=\mathrm{sin}A,and\frac{\mathrm{sin}x}{\mathrm{sin}a}=\mathrm{sin}B$.*

__CX__Hence sin *B* · sin A = sin *X* = sin a · sin *B*, so

$\frac{\mathrm{sin}a}{\mathrm{sin}A}=\frac{\mathrm{sin}b}{\mathrm{sin}B}$.

(iii) If *∠A* (say) is greater than a right angle, the circle of longitude from *C* meets the line *BA* extended beyond *A* at a point *X* (say). If we let * CX*= x, then one can argue similarly using the triangles Δ CXA and Δ CXB to get $\frac{\mathrm{sin}x}{\mathrm{sin}a}=\mathrm{sin}B,and\frac{\mathrm{sin}x}{\mathrm{sin}b}=\mathrm{sin}A$, whence

$\frac{\mathrm{sin}a}{\mathrm{sin}A}=\frac{\mathrm{sin}b}{\mathrm{sin}B}$.

**220.** Let *X* = (x, y) be an arbitrary point of the locus.

(i) Then the distance from *X* to *m* is equal to *y*, so setting this equal to * XF* gives the equation:

y^{2} = x^{2} + (y - 2a)^{2},

or

${x}^{2}=4a(y-a)$.

(ii) If we change coordinates and choose the line y = a as a new *x*-axis, then the equation becomes x^{2} = 4aY. The curve is then tangent to the new *x*-axis at the (new) origin, and is symmetrical about the *y*-axis.

**221.**

(a) Choose the line *AB* as *x*-axis, and the perpendicular bisector of * AB* as the

*y*-axis. Then A = (-3, 0) and

*B*= (3, 0). The point

*X*= (x, y) is a point of the unknown locus precisely when

(x + 3)^{2} + y^{2} = __XA__^{2} = (2· * XB*)

^{2}= 2

^{2}((x - 3)

^{2}+ y

^{2})

that is, when

(x - 5)^{2} + y^{2} = 4^{2}.

This is the equation of a circle with centre (5, 0) and radius *r* = 4.

(b) Choose the line *AB* as *x*-axis, and the perpendicular bisector of * AB* as the

*y*-axis.

If *F* = 1, the locus is the perpendicular bisector of * AB*.

We may assume that *F* > 1 (since if *F* < 1, then * BX*:

*= f*

__AX__^{-1}: 1 and f

^{-1}> 1, so we may simplyswap the labelling of

*A*and

*B*).

Now A = (-b, 0) and *B* = (b, 0), and the point *X* = (x, y) is a point of the unknown locus precisely when

(x + b)^{2} + y^{2} = __XA__^{2} = (f· * XB*)

^{2}= f

^{2}[(x - b)

^{2}+ y

^{2}]

that is, when

x^{2}(f^{2} - 1) - 2bx(f^{2} + 1) + y^{2}(f^{2} - 1) + b^{2}(f^{2} - 1) = 0

${\left(x-\frac{b({f}^{2}+1)}{{f}^{2}-1}\right)}^{2}+{y}^{2}={\left(\frac{2fb}{{f}^{2}-1}\right)}^{2}.$

This is the equation of a circle with centre $\left(\frac{b({f}^{2}+1)}{{f}^{2}-1},0\right)$ and radius *r* = $\frac{2fb}{{f}^{2}-1}$.

**222.**

(a) Choose the line *AB* as *x*-axis, and the perpendicular bisector of * AB* as the

*y*-axis.

Then A = (-c, 0) and *B* = (c, 0). The point *X* = (x, y) is a point of the unknown locus precisely when

2a - $2a=AX+BX=\sqrt{{(x+c)}^{2}+{y}^{2}}+\sqrt{{(x-c)}^{2}+{y}^{2}}$,

that is, when

$$2a-\sqrt{{(x+c)}^{2}+{y}^{2}}=\sqrt{{(x-c)}^{2}+{y}^{2}}$$∴ 4a^{2} - 4a$\sqrt{{(x+c)}^{2}+{y}^{2}}=\sqrt{{(x-c)}^{2}+{y}^{2}}$
$\sqrt{{(x+c)}^{2}+{y}^{2}}+[{(x+c)}^{2}$

∴ a^{2} + cx = a (x + c)^{2} + y^{2}

∴ (a^{2} - c^{2})x^{2}+ a^{2}y^{2} = a^{2}(a^{2} - c^{2})

Setting $\genfrac{}{}{0.1ex}{}{c}{a}$ = e then yields the equation for the locus in the form:

$\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{a}^{2}(1-{e}^{2})}=1$.

**Note:** In the derivation of the equation we squared both sides (twice). This may introduce spurious solutions. So we should check that every solution (x, y) of the final equation satisfies the original condition.

(b) The real number e< 1 is given, so we may set the distance from *F* to *m* be $\frac{a}{e}\left(1-{e}^{2}\right)$. Choose the line through *F* and perpendicular to *m* as *x*-axis. To start with, we choose the line *m* as *y*-axis and adjust later if necessary.

Hence *F* has coordinates $\left(\frac{a}{e}\left(1-{e}^{2}\right),0\right)$, and the point *X* = (x, y) is a point of the unknown locus precisely when

${\left(x-\frac{a}{e}\left(1-{e}^{2}\right)\right)}^{2}+{y}^{2}={(ex)}^{2}$,

which can be rearranged as

$\left(1-{e}^{2}\right){x}^{2}-2\frac{a}{e}(1-{e}^{2})x+{\left(\frac{a}{e}\right)}^{2}{(1-{e}^{2})}^{2}+{y}^{2}=0$,

and further as

(1 - e^{2})$\left[{x}^{2}-2\frac{a}{e}x+{\left(\frac{a}{e}\right)}^{2}\right]+{y}^{2}={\left(\frac{a}{e}\right)}^{2}(1-{e}^{2})-{\left(\frac{a}{e}\right)}^{2}{(1-{e}^{2})}^{2}$

= $\left(\frac{a}{e}\right)$^{2}(e^{2} - e^{4})

= a^{2}(1 - e^{2})

∴${\left(x-\frac{a}{e}\right)}^{2}+\frac{{y}^{2}}{1-{e}^{2}}={a}^{2}$.

If we now move the *y*-axis to the line $x=\frac{a}{e}$ the equation takes the simpler form:

$\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{a}^{2}(1-{e}^{2})}=1$.

(c) This was done in the derivations in the solutions to parts (a) and (b).

**223.**

(a) The triangle inequality shows that, if * AX* >

*, then*

__BX__*+*

__AB__*≥qslant*

__BX__*; hence the locus is non-empty only when a $\u2a7d$ c. If a = c, then*

__AX__*X*must lie on the line

*AB*, but not between

*A*and

*B*, so the locus consists of the two half-lines on

*AB*outside

*. Hence we may assume that a < c.*

__AB__Choose the line *AB* as *x*-axis, and the perpendicular bisector of * AB* as y-axis.

Then A = (-c, 0) and *B* = (c, 0). The point *X* = (x, y) is a point of the unknown locus precisely when

$2a=|<i><u>AX</u></i>-<i><u>BX</u></i>|=\left|\sqrt{{(x+c)}^{2}+{y}^{2}}-\sqrt{{(x-c)}^{2}+{y}^{2}}\right|$.

If * AX* >

*, we can drop the modulus signs and calculate as in Problem*

__BX__**222**.

$2a+\sqrt{{(x-c)}^{2}+{y}^{2}}=\sqrt{{(x+c)}^{2}+{y}^{2}}$.

∴ $4{a}^{2}+4a\sqrt{{(x-c)}^{2}+{y}^{2}}+{(x-c)}^{2}+{y}^{2}={(x+c)}^{2}+{y}^{2}$

∴ ${a}^{2}-cx=-a\sqrt{{(x-c)}^{2}+{y}^{2}}$

∴ $({c}^{2}-{a}^{2}){x}^{2}-{a}^{2}{y}^{2}={a}^{2}({c}^{2}-{a}^{2})$

Setting $\genfrac{}{}{0.1ex}{}{c}{a}$ = e (> 1), then yields the equation for the locus in the form:

$\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{a}^{2}({e}^{2}-1)}=1$.

**Note:** In the derivation of the equation we squared both sides (twice). This may introduce spurious solutions. So we should check that every solution (x, y) of the final equation satisfies the original condition. In fact, the squaring process introduces additional solutions in the form of a second branch of the locus, corresponding precisely to points *X* where * AX* <

*.*

__BX__(b) The real number e> 1 is given, so we may set the distance from *F* to *m* be $\frac{a}{e}$(e^{2} - 1). Choose the line through *F* and perpendicular to *m* as *x*-axis. To start with, we choose the line *m* as *y*-axis and adjust later if necessary.

Hence *F* has coordinates $\left(\frac{a}{e}({e}^{2}-1),0\right)$$\left(\frac{a}{e}({e}^{2}-1),0\right)$, and the point *X* = (x, y) is a point of the unknown locus precisely when

${\left(x-\frac{a}{e}({e}^{2}-1)\right)}^{2}+{y}^{2}={(ex)}^{2}$

∴ (e^{2 - 1})x^{2} + $\frac{2a}{e}({e}^{2}-1)x-{y}^{2}={\left(\frac{a}{e}\right)}^{2}{({e}^{2}-1)}^{2}$

∴ $({e}^{2}-1)\left[{x}^{2}+\frac{2a}{e}x+{\left(\frac{a}{e}\right)}^{2}\right]-{y}^{2}={\left(\frac{a}{e}\right)}^{2}{({e}^{2}-1)}^{2}+{\left(\frac{a}{e}\right)}^{2}({e}^{2}-1)$

= ${\left(\frac{a}{e}\right)}^{2}({e}^{4}-{e}^{2})$

∴ (e^{2 - 1})$\left[{x}^{2}+\frac{2a}{e}x+{\left(\frac{a}{e}\right)}^{2}\right]-{y}^{2}={a}^{2}({e}^{2}-1)$

∴ ${\left(x+\frac{a}{e}\right)}^{2}-\frac{{y}^{2}}{{e}^{2}-1}={a}^{2}$

If we now move the *y*-axis to the line *X* = -$\frac{a}{e}$ the equation takes the simpler form:

$\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{a}^{2}({e}^{2}-1)}=1$.

(c) This was done in derivations in the solutions to parts (a) and (b).

**224.**

(a) When z = k is a constant, the equation reduces to that of a circle

x^{2} + y^{2} = (rk)^{2} in the plane z = k. When the cutting plane is the

*xy*-plane “z = 0” , the circle has radius *0*, so is a single point.

(b) (i) A vertical plane through the apex cuts the cone in a pair of generators crossing at the apex.

(ii) If the cutting plane through the apex is less steep than a generator, then it cuts the cone only at the apex.

If the cutting plane through the apex is parallel to a generator, then it cuts the cone in a generator – a single line (the next paragraph indicates that this line may be better thought of as a pair of “coincident” lines).

What happens when the cuttingplane through the apex is steeper than a generator may not be intuitively clear. One way to make sense of this is to treat the cross-section as the set of solutions of two simultaneous equations – one for the cone, and the other for the plane (say y = nz, with *n* < r). This leads to the equation

${x}^{2}=({r}^{2}-{n}^{2}){z}^{2},\text{\hspace{1em}}y=nz$,

with solution set

$x=\pm z\sqrt{{r}^{2}-{n}^{2}},\text{\hspace{1em}}xy=nz$

which specifies a pair of lines crossing at the apex.

A slightly easier way to visualize the cross-section in this case is to let the apex of the double cone be *A*, and to let *X* be any other point of the cross-section. Then the line *AX* is a generator of the cone, so lies on the cone's surface. But *A* and *X* also lie in the cutting plane – so the whole line *AX* must lie in the cutting plane. Hence the cross-section contains the whole line *AX*.

(c)(i) If the cutting plane passes through the apex and is parallel to a generator of the cone, then we saw in (b) that the cross-section is simply the generator itself.

(ii) Thus we assume that the cutting plane does not pass through the apex, and may assume that it cuts the bottom half of the cone. If *V* is the point nearest the apex where the cutting plane meets the cone, then the cross-section curve starts at *V* and becomes wider as we go down the cone. Because the plane is parallel to a generator, the plane never cuts the “other side” of the bottom half of the cone, so the cross section never “closes up” – but continues to open up wider and wider as we go further and further down the bottom half of the cone.

Let **S** be the sphere, which is inscribed in the cone above the cutting plane, and which is tangent to the cutting plane at *F*. Let **C** be the circle of contact between **S** and the cone. Let *A* be the apex of the cone, and let the apex angle of the cone be equal to 2$\theta $. Let *X* be an arbitrary point of the cross-section. To illustrate the general method, consider first the special case where *X* = V is the “highest” point of the cross-section. The line segment * VA* is tangent to the sphere

**S**, so crosses the circle

**C**at some point

*M*. Now

*lies in the cutting plane, so is also tangent to the sphere*

__VF__**S**at the point

*F*. Any two tangents to a sphere from the same exterior point are equal, so it follows that

*=*

__VM__*. Moreover,*

__VF__*is exactly equal to the distance from*

__VM__*V*to the line

*m*(since

• firstly the line *m* lies in the horizontal plane through **C** and so is on the same horizontal level as *M*, and

• secondly the shortest line VM* from *V* to *m*, runs straight up the cutting plane, which is parallel to a generator – so the angle∠MVM*= 2 $\theta $, whence * VM** =

*=*

__VM__*).*

__VF__Now let *X* be an arbitrary point of the cross-sectional curve, and use a similar argument. First the line *XA* is always a generator of the cone, so is tangent to the sphere **S**, and crosses the circle **C** at some point *Y*. Moreover, *XF* is also tangent to the sphere. Hence * XY* =

*. It remains to see that*

__XF__*is equal to the distance*

__XY__*from*

__XY*__*X*to the closest point Y* on the line

*m*– since

• firstly the two points *Y* and Y* both lie on the same horizontal level (namely the horizontal plane through the circle **C**), and

• secondly both make the same angle $\theta $ with the vertical.

Hence the cross-sectional curve is a parabola with focus *F* and directrix *m*.

(d) (i) If the cutting plane is less steep than a generator, the cross-section is a closed curve. If *V* and *W* are the highest and lowest points of intersection of the plane with the cone, then the cross-section is clearly symmetrical under reflection in the line *VW*. Intuitively it is tempting to think that the lower end near *W* should be ‘fatter’ than the upper part of the curve (giving an egg-shaped cross-section). This turns out to be false, and the correct version was known to the ancient Greeks, though the error was repeated in many careful drawings from the 14th and 15th centuries (e.g. Albrecht Dürer (1471–1528)).

(ii) The derivation is very similar to that in part (c), and we leave the reader to reconstruct it.

An alternative appr*OAC*h is to insert a second sphere **S**’ below the cutting plane, and inflate it until it makes contact with the cone along a circle **C**’ while at the same time touching the cutting plane at a point *Fʹ*. If *X* is an arbitrary point of the cross-sectional curve, then *XA* is tangent to both spheres, so meets the circle **C** at some point *Y* and meets the circle **C**’ at some point *Yʹ*. Then *Y*, *X*, *Yʹ* are collinear. Moreover, * XY* =

*(since both are tangents to the sphere*

__XF__**from the point**

*S**X*), and

*=*

__XYʹ__*(since both are tangents to the sphere*

__XFʹ__**’ from the point**

*S**X*), so

* XF* +

*=*

__XFʹ__*+*

__XY__*=*

__XYʹ__

__YYʹ__But * YY*’ is equal to the slant height of the cone between the two fixed circles

**C**and

**C**', and so is equal to a constant

*k*. Hence, the focus-focus specification in Problem

**222**shows that the cross-section is an ellipse.

(e)(i) If the cutting plane is steeper than a generator, the cross-section cuts both the bottom half and the top half of the cone to give two separate parts of the cross-section. Neither part “closes up” , so each part opens up more and more widely.

If *V* is the highest point of the cross-section on the lower half of the cone, and *W* is the lowest point of the cross-section on the upper half of the cone, then it seems clear that the cross-section is symmetrical under reflection in the line *VW*. But it is quite unclear that the two halves of the cross-section are exactly congruent (though again it was known to the ancient Greeks).

(ii) The formal derivation is very similar to that in part (c), and we leave the reader to reconstruct it.

An alternative appr*OAC*h is to copy the alternative in (d), and to insert a second sphere ** S**'in the upper part of the cone, on the same side of the cutting plane as the apex, inflate it until it makes contact with the cone along a circle

**’ while at the same time touching the cutting plane at a point**

*C**Fʹ*. If

*X* is an arbitrary point of the cross-sectional curve, then *XA* is tangent to both spheres, so meets the circle ** C** at some point

*Y*and meets the circle

**’ at some point**

*C**Yʹ*. Then

*Y*,

*X*,

*Yʹ*are collinear. Moreover,

*X*,

*F*, and

*Fʹ*all lie on the cutting plane. Now

*=*

__XY__*(since both are tangents to the sphere*

__XF__**from the point**

*S**X*), and

*’ =*

__XY__*(since both are tangents to the sphere*

__XFʹ__**’ from the point**

*S**X*). If

*X*is on the upper half of the cone, then

* XF* -

*=*

__XFʹ__*-*

__XY__*=*

__XYʹ__*.*

__YYʹ__But * YYʹ* is equal to the slant height of the cone between the two fixed circles

**and**

*C***', and so is equal to a constant**

*C**k*. Hence, the focus-focus specification in Problem

**223**shows that the cross-section is a hyperbola.

**225.**

(a)(i) 2^{1}; (ii) 1 = 2^{0}

(b) (i) 2^{2}; (ii) 1 = 2^{0}; (iii) 4 = 2 \times 2^{0} + 2^{1}

(c) (i) 2^{3}; (ii) 1 = 2^{0}; (iii) 12 = 2 \times 4 + 2^{2}; (iv) 6 = 2 \times 2^{0} + 4

(d) (i) 2^{4}; (ii) 1 = 2^{0}; (iii) 32 = 2 \times 12 + 2^{3}; (iv) 24 = 2 \times 6 + 12; (v) 8 = 2 \times 2^{0} + 6

**226.**

(c) (i)If you look carefully at the diagram shown here you should be able to see not only the upper and lower 3D-cubes, but also the four other 3D-cubes formed by joining each 2D-cube in the upper 3D-cube to the corresponding 2D-cube in the lower3D-cube.

**Note:** Once we have the 3D-cube expressed in coordinates, we can specify precisely which planes produce which cross-sections in Problem **38**. The plane *X* + y + z = 1 passes through the three neighbours of (0, 0, 0) and creates an equilateral triangular cross-section. Any plane of the form z = c (where *c* is a constant between 0 and 1) produces a square cross-section. And the plane *X* + y + z = $\frac{3}{2}$ is the perpendicular bisector of the line joining (0, 0, 0) to (1, 1, 1), and creates a regular hexagon as cross-section.

**227.**

(a) View the coordinates as (x, y, z). Start at the origin (0, 0, 0) and travel round 3 edges of the lower 2D-cube “z = 0” to the point (0, 1, 0). Copy this path of length 3 on the upper 2D-cube “z = 1” (from (0, 0, 1) to (0, 1, 1). Then join (0, 0, 0) to (0, 0, 1) and join (0, 1, 0) to (0, 1, 1). The result

(0, 0, 0), (1, 0, 0), (1, 1, 0), (0, 1, 0), (0, 1, 1), (1, 1, 1), (1, 0, 1), (0, 0, 1) (and back to (0, 0, 0))

has the property that exactly one coordinate changes when we move from each vertex to the next. This is an example of a *Gray code of length* *3*.

**Note:** How many such paths/circuits are there in the 3D-cube?We can certainly count those of the kind described here. Each such circuit has a “direction” : the 12 edges of the 3D-cube lie in one of *3* “directions” , and each such circuit contains all four edges in one of these *3* directions. Moreover this set of four edges can be completed to a circuit in exactly 2 ways. So there are 3 × 2 such circuits. In 3D this accounts for all such circuits. But in higher dimensions the numbers begin to explode (in the 4D-cube there are *1344* such circuits).

(b) View the coordinates as (w, x, y, z). Start at the origin (0, 0, 0, 0) and travel round the 8 vertices of the lower 3D-cube “z = 0” to the point (0, 0, 1, 0). Then copy this path on the upper 3D-cube “z = 1” from (0, 0, 0, 1) to (0, 0, 1, 1). Finally join (0, 0, 0, 0) to (0, 0, 0, 1) and join (0, 0, 1, 0) to (0, 0, 1, 1). The result

(0, 0, 0.0), (1, 0, 0, 0), (1, 1, 0, 0), (0, 1, 0, 0), (0, 1, 1, 0), (1, 1, 1, 0),

(1, 0, 1, 0), (0, 0, 1, 0) (0, 0, 1, 1), (1, 0, 1, 1), (1, 1, 1, 1), (0, 1, 1, 1),

(0, 1, 0, 1), (1, 1, 0, 1), (1, 0, 0, 1), (0, 0, 0, 1) (and back to (0, 0, 0, 0))

has the property that exactly one coordinate changes when we move from each vertex to the next. This is an example of a *Gray code of length* 4.

**Note:** The general construction in dimension *n* + 1 depends on the previous construction in dimension *n*, so makes use of *mathematical induction* (see Problem **262** in Chapter 6).