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Problem 42:  Geometry ( ) 2015 Paper II

In the triangle ABC, angle BAC = α and angle CBA = 2α, where 2α is acute, and BC = x. Show that AB = (3 4sin2α)x.

The point D is the midpoint of AB and the point E is the foot of the perpendicular from C to AB. Find an expression for DE in terms of x.

The point F lies on the perpendicular bisector of AB and is a distance x from C. The points F and B lie on the same side of the line through A and C. Show that the line FC trisects the angle ACB.


I thought it would be good to include at least one plane geometry question in this collection, so here it is. As always with geometry, the first thing to do is draw a BIG diagram. You will probably need to have quite a few tries at it.

Then you have the usual tools at your disposal: similar triangles, congruent triangles, angle-chasing, and — moving away from classical Euclidean geometry — Pythagoras, and sine and cosine rules. I leave it to you to decide what will be useful here.

Solution to problem 42


Here is a careful diagram, with a construction arc left in to show that CB = CF = x. It took me ages. But it is pretty much plain sailing now.

By the sine rule in ABC (and calculating an expression for sin3α using double angle formulae)

AB = xsin(180 3α) sinα = xsin(3α) sinα = x(3cos2αsinα sin3α) sinα = (3 4sin2α)x.

as required.

Then, since D is the mid-point of AB and BEC = 90,

DE = 1 2AB BE = 1 2x(3 4sin2α) xcos2α = 1 2x,

rather surprisingly.

For the last part, since BCA = 180 3α, we need to show that FCA = 60 α or, equivalently, FCB = 120 2α. Now BCE = 90 2α so we are done if we can show that ECF = 30. But that follows almost immediately from the DE = 1 2x and FC = x: draw the perpendicular from F to CE, which has length 1 2x, and consider the right-angled triangle thus formed.


I liked the mixture here of angle-chasing and use of the sine rule. Geometry purists, though, would be appalled; using the sine rule in a geometry problem is like using metal screws instead of wooden dowels on an antique wooden cabinet. As it happens, there is a beautiful proof of this result by pure geometry. You have to reflect the whole diagram about the perpendicular bisector of AB. For the details, see Ross Honsberger’s excellent Mathematical Chestnuts from around the World (Cambridge University Press, 2001), section 20.