
2. Two circles $k_{a}$ and $k_{b}$ are given, whose centers lie on the legs of lengths a and b, respectively, of a right triangle. Both circles are tangent to the hypotenuse of the triangle pass through the vertex opposite to the hypotenuse. Denote the radii of the circles by $\rho_{a}$ and $\rho_{b}$. Find the greatest real number $p$ such that the inequality

$$
\frac{1}{\rho_{a}}+\frac{1}{\rho_{b}} \geqslant p\left(\frac{1}{a}+\frac{1}{b}\right)
$$

holds for all right triangles.


Solution. Denote the vertices of the triangle by $A, B, C$, with $A, B$ lying opposite the legs with lengths $a$ and $b$, respectively.

Let us first determine the magnitudes of the two radii $\rho_{a}$ and $\rho_{b}$. Denote by $A^{\prime}$ the image of $A$ in the symmetry with respect to the line $B C$. The circle $k_{a}$ is the incircle of the triangle $A^{\prime} A B$ (Fig. 1). The isosceles triangle $A B A^{\prime}$ has perimeter $o=2(b+c)$ and area $S=a b$, so by the familiar formula the radius $\varrho_{a}$ of its incircle $k_{a}$ is

$$
\varrho_{a}=\frac{2 S}{o}=\frac{a b}{b+c} .
$$

Similarly $\varrho_{b}=a b /(a+c)$.

![](https://cdn.mathpix.com/cropped/2024_04_17_5885ae9551045f8982ddg-13.jpg?height=261&width=761&top_left_y=1032&top_left_x=318)

Fig. 1

The number $p$ is to satisfy

$$
p \leqslant \frac{\frac{1}{\varrho_{a}}+\frac{1}{\varrho_{b}}}{\frac{1}{a}+\frac{1}{b}}=\frac{\frac{b+c}{a b}+\frac{a+c}{a b}}{\frac{a+b}{a b}}=\frac{a+b+2 c}{a+b}=1+\frac{2 c}{a+b}=1+\frac{2 \sqrt{a^{2}+b^{2}}}{a+b}
$$

for any right triangle with legs $a, b$.

Since for $a=b$ the last expression equals $1+\sqrt{2}$, any such number $p$ must satisfy $p \leqslant 1+\sqrt{2}$. If we can show that

$$
\frac{2 \sqrt{a^{2}+b^{2}}}{a+b} \geqslant \sqrt{2}
$$

for any positive $a$ and $b$, it will follow that $p=1+\sqrt{2}$ is the solution of the problem.

The inequality (1) for arbitrary positive $a, b$ is easily shown to be equivalent to an inequality whose validity is evident:

$$
\begin{aligned}
2 \sqrt{a^{2}+b^{2}} & \geqslant \sqrt{2}(a+b), \\
4\left(a^{2}+b^{2}\right) & \geqslant 2(a+b)^{2}, \\
4 a^{2}+4 b^{2} & \geqslant 2 a^{2}+4 a b+2 b^{2}, \\
2(a-b)^{2} & \geqslant 0 .
\end{aligned}
$$

Alternatively, one can appeal to Cauchy's inequality $2\left(a^{2}+b^{2}\right) \geqslant(a+b)^{2}$, or to the inequality between the quadratic and arithmetic mean

$$
\sqrt{\frac{a^{2}+b^{2}}{2}} \geqslant \frac{a+b}{2}
$$

Answer. The sought real number $p$ is $1+\sqrt{2}$.

Remark. The magnitude of the radii $\varrho_{a}, \varrho_{b}$ can be derived also in another way: expressing the sine of the angle $A B C$ in two ways from the right triangles $S_{a} B T$ a $A B C$ (Fig. 2), we get

$$
\frac{\varrho_{a}}{a-\varrho_{a}}=\frac{b}{c}
$$

implying that $\varrho_{a}=a b /(b+c)$. Similarly for $\varrho_{b}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_5885ae9551045f8982ddg-14.jpg?height=264&width=502&top_left_y=1083&top_left_x=512)

Fig. 2
