This solution is from Etienne Chan, age 15, 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 = √(1+1/4) = √5/2.

The semi perimeter of ABP = (1 + 1/2 + √5/2)/2 = [3+√5]/4. Using the formula A = rs for the area of the triangle, the area of ABP = r1[3+√5]/4 = 1/4. This gives r1 = 1/[3 + √5] = [3 − √5]/4 so r1 = r2 = [3 − √5]/4.

The area of BPC is 1 - 1/4 - 1/4 = 1/2 and PB = PC = √5/2.

The semi-perimeter of BPC = [1+ √5/2 + √5/2]/2 = [1 + √5]/2. From the area of triangle BPC we get r3[1 + √5]/2 = 1/2 so r3 = 1/[1 + √5] = [√5 − 1]/4.

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


r2 =  1−p
2
 ×  2
[2 − p + √(p2 − 2p + 2)]
 =  1
2
 [2 − p − √(p2 − 2p + 2)].

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


r1 =  p
2
 ×  2
[1 + p + √(p2 + 1)]
 =  1
2
 [1 + p − √(p2 + 1)].

Similarly
r3 =  1
2
 ×  2
[1 + √(p2 + 1) + √(p2 −2p + 2)]
 =  1
2
 [1 + √(p2 + 1) − √(p2 −2p +2)][√(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.