It's High Time - It's Pie Time!

Pyrgic Puzzle 3 (29th January 2011) reads as follows:

One of Ma Rainey's circular blueberry pies fits snugly into a Pie Thagoras™— a baking tray in the form of a right-angled triangle of base 1 unit and hypotenuse 2. The pie is cut with 3 straight-line cuts each joining the centre of the pie to a side of the tray which it meets perpendicularly. A gourmand stands at each apex and takes the piece of pie nearest him. What are their shares?

Who would divide pies in this way? Only in a puzzle! But this puzzle was simpler than some readers assumed. There is, for example, no need to find the radius of the circle, though that is easily done in the case of such a right-angled triangle. How do we know this? It is clear that the fraction you get is independent of the absolute size of the pie.

Nor is there a need to presume a priori that connecting the centre of the circle to the apices of the triangle necessarily bisects the angles of the triangle though of course it does. All that comes out in the wash.

The first step is to realise that the right-angled triangle is half of an equilateral triangle, so that its angles are 30° and 60°.

Join the centre of the circle to the vertices hosting the acute angles.

It is then easy to show that ABC and DBC are congruent triangles (the same triangle flipped over). We know this as they are right-angled, they share a hypotenuse and have the same length for one of the other sides. By Pythagoras, the other side must be the same. As these triangles are congruent, corresponding angles are the same. That means that angle CBD is 15°, and the angle DCB is 75°. So the share of the gourmand at B is twice this, namely 150°. This constitutes 150°/360° = 150/360 = 15/36 = 5/12 of the pie.

The second share taken by the gourmand at the right angle is clearly 1/4.

The third share (by subtraction) must be 1 - 5/12 - 1/4 = 1/3.

This last result can be checked in the same way by considering congruent but flipped triangles at the other acute angle.

Points to Ponder

a) What if the pie tray had angles P, Q and π/2?

b) What if the pie tray were not a right-angled triangle?