Problem 1. A right-angled trapezoid with area 10 and altitude 4 is divided into two circumscribed trapezoids by a line parallel to its bases. Find their inradii.

Oleg Mushkarov
1. Let $A B C D$ be a right-angled trapezoid with area 10 and altitude $A D=4$. Let the line $M N \| A B, M \in A D, N \in B C$, divides it into two circumscribed trapezoids. Set $A B=a, C D=b$ $(a>b), M N=c, A M=h_{1}$ and $D M=h_{2}$. Then $h_{1}^{2}+(a-c)^{2}=\left(a+c-h_{1}\right)^{2}$ and we get that $h_{1}=\frac{2 a c}{a+c}$.

![](https://cdn.mathpix.com/cropped/2024_06_03_f8ccd71b4db367e7c2a9g-072.jpg?height=427&width=418&top_left_y=376&top_left_x=1204)

Analogously $h_{2}=\frac{2 b c}{b+c}$ which implies that $\frac{h_{1}}{h_{2}}=\frac{a(b+c)}{b(a+c)}$. On the other hand, $\frac{h_{1}}{h_{2}}=\frac{a-c}{c-b}$ and therefore $(a+b)\left(a b-c^{2}\right)=0$. Hence $c^{2}=a b$ and we get that

$$
h_{1}=\frac{2 a \sqrt{b}}{\sqrt{a}+\sqrt{b}}, \quad h_{2}=\frac{2 b \sqrt{a}}{\sqrt{a}+\sqrt{b}}
$$

Then $\sqrt{a b}=2$ and $a+b=5$, i.e., $a$ and $b$ are the roots of the equation $x^{2}-5 x+4=0$. Consequently $a=4, b=1$ and the radii equal $\frac{h_{1}}{2}=\frac{4}{3}$ and $\frac{h_{2}}{2}=\frac{2}{3}$.