10.5. Petya and Vasya have equal paper right-angled triangles with legs $a$ and $b$. The boys want to cut out the largest possible square from each triangle, such that one vertex of Petya's square coincides with the right angle of the triangle, and one side of Vasya's square lies on the hypotenuse. a) Find the sizes of both squares; b) Is Petya's square always larger than Vasya's (i.e., for any legs)?
Answer: a) $\frac{a b}{a+b}$ and $\frac{a b \sqrt{a^{2}+b^{2}}}{a^{2}+a b+b^{2}}$; b) always. Solution. a) The side of Petya's square is calculated in problem 9.5. Let $x$ be the side of Vasya's square. Then the hypotenuse is divided into three segments of lengths, respectively, $\quad x \operatorname{tg} A, \quad x$ and $x \operatorname{tg} B$. From this, we get the equation $x \frac{a}{b}+x+x \frac{b}{a}=\sqrt{a^{2}+b^{2}} \Leftrightarrow x=\frac{a b \sqrt{a^{2}+b^{2}}}{a^{2}+a b+b^{2}}$.

b) We will prove the inequality $\quad \frac{a b \sqrt{a^{2}+b^{2}}}{a^{2}+a b+b^{2}}<\frac{a b}{a+b} \Leftrightarrow(a+b) \sqrt{a^{2}+b^{2}}<a^{2}+a b+b^{2} . \quad$ Let $u=a+b, v=a b, \quad$ then the inequality can be written as $\quad u \sqrt{u^{2}-2 v}<u^{2}-v \Leftrightarrow$ $u^{2}\left(u^{2}-2 v\right)<u^{4}-2 u^{2} v+v^{2} \Leftrightarrow 0<v^{2}$.

## 11th grade