2. On the legs $a, b$ of a right triangle, the centers of two circles $k_{a}, k_{b}$ lie in sequence. Both circles touch the hypotenuse and pass through the vertex opposite the hypotenuse. Let the radii of the given circles be $\varrho_{a}, \varrho_{b}$. Determine the greatest positive real number $p$ such that the inequality

$$
\frac{1}{\varrho_{a}}+\frac{1}{\varrho_{b}} \geqq p\left(\frac{1}{a}+\frac{1}{b}\right)
$$

holds for all right triangles.
2. Let the vertices of the given triangle be $A, B, C$ such that vertices $A, B$ lie opposite the legs of lengths $a, b$.

First, we calculate the radii of the two circles $k_{a}$ and $k_{b}$. Let $A^{\prime}$ be the image of point $A$ under reflection across the line $BC$. The circle $k_{a}$ is inscribed in triangle $A^{\prime} A B$ (Fig.1). The isosceles triangle $A B A^{\prime}$ has a perimeter $o=2(b+c)$ and an area $S=a b$. For the radius $\varrho_{a}$ of circle $k_{a}$, we have, according to the known formula,

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

Similarly, we calculate the radius of circle $k_{b}$: it turns out to be $\varrho_{b}=a b /(a+c)$.

![](https://cdn.mathpix.com/cropped/2024_04_17_c28e9ab487838f076aabg-3.jpg?height=365&width=1056&top_left_y=1234&top_left_x=500)

Fig. 1

For a number $p$ and for any right triangle $\mathrm{s}$ with legs $a, b$ and hypotenuse $c$, the following must hold:

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

Since in the case $a=b$ the last expression has the value $1+\sqrt{2}$, any number $p$ that satisfies the inequality must satisfy $p \leqq 1+\sqrt{2}$. If we now show that for any two positive values $a, b$ the inequality

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

holds, the above derived inequality will mean that $p=1+\sqrt{2}$ is the desired real number (and the problem will be solved).

The inequality (1) for any positive $a, b$ can be easily transformed by equivalent manipulations into an inequality that is clearly true:

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

Instead of such a verification, it was possible to use the Cauchy-Schwarz inequality $2\left(a^{2}+b^{2}\right) \geqq(a+b)^{2}$ or the inequality between the quadratic and arithmetic means

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

Both of these classical inequalities are clearly just different formulations of inequality (1).

Answer. The desired number $p$ has the value $1+\sqrt{2}$.

Note. The sizes of the radii $\varrho_{a}$ and $\varrho_{b}$ can also be calculated differently: By expressing the sine of angle $A B C$ twice from the right triangles $S_{a} B T$ and $A B C$ (Fig.2) we get

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

from which it follows that $\varrho_{a}=a b /(b+c)$. Similarly, we calculate $\varrho_{b}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_c28e9ab487838f076aabg-4.jpg?height=363&width=694&top_left_y=1343&top_left_x=681)

Fig. 2

For a complete solution, award 6 points, of which 2 points for calculating the sizes of the radii $\varrho_{a}$ and $\varrho_{b}$, 1 point for finding the value $p=1+\sqrt{2}$, and 3 points for proving the inequality from the problem for $p=1+\sqrt{2}$, with properly named classical inequalities not needing to be proven.