4. In a right-angled triangle with legs $a$ and $b$ ( $a>b$ ), two identical circles are placed. The circles touch each other externally and each touches the hypotenuse and one of the legs. Find the radius of such a circle.
Solution

![](https://cdn.mathpix.com/cropped/2024_05_06_93794e1fd7f49bf16576g-2.jpg?height=1117&width=925&top_left_y=1526&top_left_x=226)

1) Let $\mathrm{O}$ and $\mathrm{Q}$ be the centers of these circles. $\mathrm{OQ}$ is parallel to the hypotenuse, since the centers of the circles are at the same distance from it and on the same side.
2) Let NM be the tangent to these circles, passing through the point of tangency of the circles with each other. NM is perpendicular to OQ (since the tangent is perpendicular to the radius drawn to the point of tangency). Therefore, NM is perpendicular to the hypotenuse.
3) Let K be the intersection point of the lines NM and BC.

Triangle MCK is similar to triangle BCA by two angles.

Triangle MNA is similar to triangle BCA by two angles.

Therefore, triangles MCK and MNA are similar.

Since the circles inscribed in these triangles are equal, the triangles are equal.

Therefore, $\mathrm{MN}=\mathrm{MC}, \mathrm{MK}=\mathrm{MA}$.

Since $\mathrm{MN}=\mathrm{MC}$, then $\mathrm{NP}=\mathrm{CR}$ ( $\mathrm{P}$ - the point of tangency of the circles, $\mathrm{R}$ - the point of tangency of the circle with the hypotenuse, the equality of the segments is explained by the fact that they are segments $\mathrm{MN}$ and MC without the radius of the inscribed circle).

Since $\mathrm{MK}=\mathrm{MA}$, then $K \mathrm{P}=\mathrm{AF}$ ( $\mathrm{F}$ - the point of tangency of the second circle with the hypotenuse, the equality of the segments is explained by the fact that they are segments MK and MA without the radius of the inscribed circle).

4) Introduce the notation $r$ - the radius of the circle, $x$ and $y$ - the lengths of the tangents drawn from the vertices of the triangles to the circles.

The length of the hypotenuse is, on the one hand, $x+y+2 r$, and on the other hand, $\sqrt{a^{2}+b^{2}}$.

Therefore, we get the equation $x+y+2 r=\sqrt{a^{2}+b^{2}}$.

Also, $B N=A B-x-y=a-x-y, K B=K C-B C=x+y-b$.

5) Since the angles at vertex $\mathrm{N}$ are vertical and equal to the angle at vertex $\mathrm{C}$ (from the similarity of triangles MNA and BCA), and angle B is right, then triangle KBN is similar to triangle $\mathrm{ABC}$.

Therefore, $\frac{K B}{A B}=\frac{B N}{B C}$

We get the equation $\frac{x+y-b}{a}=\frac{a-x-y}{b}$, from which $x+y=\frac{a^{2}+b^{2}}{a+b}$.

From the equality $x+y+2 r=\sqrt{a^{2}+b^{2}}$ we express

$$
r=\frac{1}{2} \sqrt{a^{2}+b^{2}}-\frac{1}{2}(x+y)
$$

From the obtained conditions, we find $r=\frac{1}{2} \sqrt{a^{2}+b^{2}}-\frac{1}{2} \frac{a^{2}+b^{2}}{a+b}$.

## Grading Criteria

It is proven that the line passing through the centers of the circles is parallel to the hypotenuse and that the tangent drawn to the circles at the point of their tangency is perpendicular to the hypotenuse - 1 point.

It is proven that triangles MCK and MNA (indicated in the author's notation) are equal - 2 points.

All equal tangent segments (necessary for further solution) are indicated and the equation $x+y+2 r=\sqrt{a^{2}+b^{2}}$ (indicated in the author's notation) is formulated - 1 point.

All equal tangent segments (necessary for further solution) are indicated, the similarity of triangles $K \mathrm{BN}$ and $\mathrm{ABC}$ is proven, and the equation $\frac{x+y-b}{a}=\frac{a-x-y}{b}$ (indicated in the author's notation) is formulated - 2 points.

From the equations $x+y+2 r=\sqrt{a^{2}+b^{2}}$ and $\frac{x+y-b}{a}=\frac{a-x-y}{b}$, $r$ is expressed - 1 point.

The points are summed.