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 $AD, r_{1} = r_{2}$ because they sit on congruent triangles. The area of triangle $ABP = AP \times AP / 2 = 1 / 4 . PB = \sqrt{(} 1 + 1 / 4) = \sqrt{5} / 2 .$

The semi perimeter of $ABP = (1 + 1 / 2 + \sqrt{5} / 2) / 2 = [3 + \sqrt{5}] / 4 .$ Using the formula $A = rs$ for the area of the triangle, the area of $ABP = r_{1} [3 + \sqrt{5}] / 4 = 1 / 4 .$ This gives $r_{1} = 1 / [3 + \sqrt{5}] = [3 - \sqrt{5}] / 4$ so $r_{1} = r_{2} = [3 - \sqrt{5}] / 4 .$

The area of BPC is 1 - 1/4 - 1/4 = 1/2 and $PB = PC = \sqrt{5} / 2 .$

The semi-perimeter of $BPC = [1 + \sqrt{5} / 2 + \sqrt{5} / 2] / 2 = [1 + \sqrt{5}] / 2 .$ From the area of triangle $BPC$ we get $r_{3} [1 + \sqrt{5}] / 2 = 1 / 2$ so $r_{3} = 1 / [1 + \sqrt{5}] = [\sqrt{5} - 1] / 4 .$

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

$r_{2} = \frac{1 - p}{2} \times \frac{2}{[2 - p + \sqrt{(} p^{2} - 2 p + 2)]} = \frac{1}{2} [2 - p - \sqrt{(} p^{2} - 2 p + 2)] .$

Similary with $DPC$ . The area of $DPC = p / 2$ and the length $PC = \sqrt{(} p^{2} + 1) .$ The semi perimeter of $PC = [1 + p + \sqrt{(} p^{2} + 1)] / 2 .$ So

$r_{1} = \frac{p}{2} \times \frac{2}{[1 + p + \sqrt{(} p^{2} + 1)]} = \frac{1}{2} [1 + p - \sqrt{(} p^{2} + 1)] .$

Similarly

$r_{3} = \frac{1}{2} \times \frac{2}{[1 + \sqrt{(} p^{2} + 1) + \sqrt{(} p^{2} - 2 p + 2)]} = \frac{1}{2} [1 + \sqrt{(} p^{2} + 1) - \sqrt{(} p^{2} - 2 p + 2)] [\sqrt{(} p^{2} + 1) - p] .$

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