This solution is from Etienne of Parramatta Highschool, NSW Australia

I need to prove the area of a triangle A is given by A = rs where r is the radius of the incircle (inscribed circle) and s is the semi-perimeter (half the perimeter (a+b+c)/2). Let a , b , c be the lengths of the sides of triangle ABC . Join the incentre I of the triangle to the 3 corners. From I drop 3 perpendiculars to each of the sides, each of these has a length of r . The areas of ABI , BCI , CAI are cr/2 , ar/2 and br/2 respectively. They sum up to give the area of triangle ABC

Area of ABC = cr/2 + ar/2 + br/2 = r(a+b+c)/2 = rs .

Back to the question!

When P is the midpoint of AD, r1 = r2 because they sit on congruent triangles. The area of triangle ABP = AP ×AP/2 = 1/4. PB = Ö(1+1/4) = Ö5/2.

The semi perimeter of ABP = (1 + 1/2 + Ö5/2)/2 = [3+Ö5]/4. Using the formula A = rs for the area of the triangle, the area of ABP = r1[3+Ö5]/4 = 1/4. This gives r1 = 1/[3 + Ö5] = [3 - Ö5]/4 so r1 = r2 = [3 - Ö5]/4.

The area of BPC is 1 - 1/4 - 1/4 = 1/2 and PB = PC = Ö5/2.

The semi-perimeter of BPC = [1+ Ö5/2 + Ö5/2]/2 = [1 + Ö5]/2. From the area of triangle BPC we get r3[1 + Ö5]/2 = 1/2 so r3 = 1/[1 + Ö5] = [Ö5 - 1]/4.

Now suppose the lengths AP and PD are 1-p and p . The area of APB = (1-p)/2. The length PB = Ö(1 + p2 - 2p + 1) = Ö(p2 - 2p + 2) and the semi perimeter of APB = [1 + (1-p) + Ö(p2 - 2p + 2)]/2. So, using the area formula again,


r2 = 1-p
2
 × 2
[2 - p + Ö(p2 - 2p + 2)]
 = 1
2
 [2 - p - Ö(p2 - 2p + 2)].

Similary with DPC . The area of DPC = p/2 and the length PC = Ö(p2 + 1). The semi perimeter of PC = [1 + p + Ö(p2 + 1)]/2. So


r1 = p
2
 × 2
[1 + p + Ö(p2 + 1)]
 = 1
2
 [1 + p - Ö(p2 + 1)].

Similarly
r3 = 1
2
 × 2
[1 + Ö(p2 + 1) + Ö(p2 -2p + 2)]
 = 1
2
 [1 + Ö(p2 + 1) - Ö(p2 -2p +2)][Ö(p2 +1)- p].

To finish this question you need to use the formulae found above which give the radii r1 , r2 and r3 in terms of p and then consider how these values change as p varies from 0 to 1. For example r1 varies from 0 to approximately 0.293 as p increases from 0 to 1. In order to discover whether the ratio of the radii r1 : r2 : r3 can ever take the value 1 : 2 : 3 you could plot on the same axes the graphs of r1 , r2 and r3 as p varies from 0 to 1.