In the right isosceles triangle $A O B$, the points $P, Q$, and $S$ are chosen on the sides $O B, O A$, and $A B$, respectively, such that $P Q R S$ is a square. If the lengths of $O P$ and $O Q$ are $a$ and $b$, respectively, and the area of the square $P Q R S$ is $2 / 5$ of the area of the triangle $A O B$, determine the value of $a / b$.

![](https://cdn.mathpix.com/cropped/2024_05_01_82072e76ed187b592b1cg-09.jpg?height=439&width=429&top_left_y=592&top_left_x=901)
Solution

Let $C$ be the foot of the perpendicular from point $S$ to segment $O B$. The triangles $S P C$ and $P Q O$ have the same angles, since

$$
\begin{aligned}
\angle C P S & =\angle 180^{\circ}-\angle S P Q-\angle O P Q \\
& =90^{\circ}-\angle O P Q \\
& =\angle P Q O
\end{aligned}
$$

Since $P S=P Q$, these triangles are congruent by the A.L.A. case. Thus $P C=b, C S=a$. Since $B S C$ is an isosceles right triangle, we get $O B=2 a+b$. Consequently, the area of triangle $A O B$ is $(2 a+b)^{2} / 2$. By the Pythagorean Theorem, the area of square $P Q R S$ is $P Q^{2}=a^{2}+b^{2}$. Therefore,

$$
\begin{aligned}
\frac{A_{P Q R S}}{A_{O A B}} & =\frac{2}{5} \\
5\left(a^{2}+b^{2}\right) & =(2 a+b)^{2} \\
a^{2}+4 b^{2} & =4 a b \\
(a-2 b)^{2} & =0
\end{aligned}
$$

Thus, $a=2 b$ and the desired ratio is $a / b=2$.