  GO TO... Contents Copyright BUY THE BOOK

Problem 6:  The regular Reuleaux heptagon ($✓$) 1987 Specimen Paper I The diagram shows a British 50 pence coin. The seven arcs $AB$, $BC$, $\dots$ , $FG$, $GA$ are of equal length and each arc is formed from the circle of radius $a$ having its centre at the vertex diametrically opposite the mid-point of the arc. Show that the area of the face of the coin is

$\frac{{a}^{2}}{2}\left(\pi -7tan\frac{\pi }{14}\right).$

The ﬁrst difficulty with this elegant problem is drawing the diagram. However, you can simplify both the drawing and the solution by restricting your attention to just one sector of a circle of radius $a$.

The ﬁgure sketched above has constant diameter; it can roll between two parallel lines without losing contact with either. (This looks plausible and you can verify it by sellotaping some 50 pence coins together, but a solid proof is not very easy.) The distance between these lines is the diameter of the ﬁgure. Like a circle, the 50 pence piece has circumference equal to $\pi$ times the diameter, which is in fact always true for a ﬁgure with constant diameter.

Reuleaux polygons are general polygons of constant diameter, and a heptagon has seven sides like, for example, British 50 and 20 pence pieces, Botswanan 50 thebe coins and Jordanian half dinars. The shield on Lancia cars is a Reuleaux triangle.

Solution to problem 6 In the ﬁgure, the point $O$ is equidistant from each of three vertices $A$, $B$ and $E$. The plan is to ﬁnd the area of the sector $AOB$ by calculating the area of $AEB$ and subtracting the areas of the two congruent isosceles triangles $OBE$ and $OAE$. The required area is 7 times this.

First we need angle $\angle AEB$. We know that $\angle AOB=\frac{2}{7}\pi$ and hence $\angle BOE=\frac{1}{2}\left(2\pi -\frac{2}{7}\pi \right)=\frac{6}{7}\pi$ (using the sum of angles round the point $O$). Finally,

$\angle AEB=2\angle OEB=2×\frac{1}{2}\left(\pi -\frac{6}{7}\pi \right)=\frac{1}{7}\pi \phantom{\rule{0.3em}{0ex}},$

using the sum of angles of an isosceles triangle.16

Now $ABE$ is a sector of a circle of radius $a$, so its area is

$\pi {a}^{2}×\frac{\frac{1}{7}\pi }{2\pi }=\frac{\pi {a}^{2}}{14}.$

The area of triangle $OBE$ is $\frac{1}{2}BE×$ height, i.e.

$\frac{1}{2}a×\frac{a}{2}tan\angle OBE=\frac{{a}^{2}}{4}tan\frac{\pi }{14}.$

The area of the coin is therefore

$7×\left(\frac{\pi {a}^{2}}{14}-2×\frac{{a}^{2}}{4}tan\frac{\pi }{14}\right),$

which reduces to the given answer.

Post-mortem

It is simple now to calculate the area of a regular $n$-sided Reuleaux polygon. You should of course ﬁnd that the area tends to that of a circle of the same diameter as $n\to \infty \phantom{\rule{0.3em}{0ex}}$.

16 After the ﬁrst edition, a correspondent pointed out that the following argument gives angle $EAB$ more quickly: clearly $A$, $B$ and $E$ all lie on a circle with centre $O$; the angle at $O$ subtended by the chord $AB$ is $\frac{2}{7}\pi$, so the angle at the circumference is $\frac{1}{7}\pi$. Obvious, really — can’t think why I didn’t see it.