On the segment $A C$ there is a point $B$, and $A B=14, B C=28$. On the segments $A B, B C, A C$ as diameters, semicircles are constructed in the same half-plane relative to the line $A B$. Find the radius of the circle that is tangent to all three semicircles.
Apply Heron's formula or the Pythagorean theorem.

## Solution

Let points $O_{1}, O_{2}$, and $O$ be the centers of the given semicircles with diameters $A B, B C, A C$ respectively, $x$ be the radius of the desired circle, and $O_{3}$ be its center. Then

$O O_{1}=14, O O_{2}=7, O_{1} O_{3}=7+x, O O_{3}=21-x, O_{2} O_{3}=14+x$.

First method. By Heron's formula, $S_{\mathrm{OO}_{1} \mathrm{O}_{3}}=\sqrt{21 \cdot 7 x(14-x)}, S_{\mathrm{OO}_{2} \mathrm{O}_{3}}=\sqrt{21 \cdot 14 x(7-x)}$. Since $S_{O O_{1} O_{3}}: O O_{2} O_{3}=O O_{1}: O O_{2}=2: 1$, then $\frac{\sqrt{14-x}}{\sqrt{2(7-x)}}=2$. From this equation, we find that $x=6$.

![](https://cdn.mathpix.com/cropped/2024_05_06_9e7ee8b57660b9721937g-40.jpg?height=303&width=512&top_left_y=2164&top_left_x=773)

Let $K$ be the projection of point $O_{3}$ onto $A C, O K=u$. Then $(21-x)^{2}-(7+x)^{2}=u^{2}-(14-u)^{2}, 28(14-2 x)=14(2 u-14), 2(14-2 x)=2 u-14$, $(14+x)^{2}-(21-x)^{2}=(7+u)^{2}-u^{2}, 35(2 x-7)=7(2 u+7), 5(2 x-7)=2 u+7$. Subtracting, we get $14 x-63=21$, from which $x=6$.

## Answer

6.

Given a triangle with sides 10, 24, and 26. The two smaller sides are tangents to a circle, the center of which lies on the larger side.

Find the radius of the circle.

## Hint

The area of the given triangle is equal to the sum of the areas of the triangles into which it is divided by the segment connecting the vertex of the largest angle to the center of the given circle.

## Solution

Note that the given triangle is a right triangle $\left(12^{2}+5^{2}=13^{2}\right)$. Let $r$ be the desired radius.

The segment connecting the vertex of the right angle to the center of the given circle divides the triangle into two triangles. The radii of the circle, drawn to the points of tangency, are the heights of these triangles. The sum of the areas of the resulting triangles is equal to the area of the given triangle, that is, $5 r+12 r=120$. From this, $r=120 / 17$.

![](https://cdn.mathpix.com/cropped/2024_05_06_9e7ee8b57660b9721937g-41.jpg?height=314&width=486&top_left_y=898&top_left_x=795)

## Answer

$120 / 17$