[ Tangent Circles ]

The legs of a right triangle are 6 and 8. On all its sides, semicircles are constructed with the sides as diameters, lying outside the triangle. Find the radius of the circle that is tangent to the constructed semicircles.

#
Let $O$ be the center of the desired circle, and $R$ its radius. The distances from point $O$ to the midpoints of the sides of the given triangle are $R-3, R-4$, and $R-5$.

## Solution

Let $O$ be the center of the desired circle, $R$ its radius, and $M, N$, and $K$ the midpoints of sides $BC, AC$, and $AB$ of triangle $ABC (BC=6, AC=8, AB=10)$.

Since $M, N$, and $K$ are the centers of semicircles with radii 3, 4, and 5, respectively, we have

$$
O M=R-3, O N=R-4, O K=R-5
$$

Let $P$ and $Q$ be the projections of point $O$ onto $MK$ and $NK$. Then

$$
O K^{2}-K P^{2}=O M^{2}-M P^{2}, O N^{2}-N Q^{2}=O K^{2}-K Q^{2},
$$

or

$$
(R-5)^{2}-K P^{2}=(R-3)^{2}-(4-K P)^{2},(R-4)^{2}-(3-K Q)^{2}=(R-5)^{2}-K Q^{2}.
$$

From these equations, we find that $K P=4-\frac{1}{2} R$ and $K Q=3-\frac{1}{3} R$. By the Pythagorean theorem in triangle $OPK$, we have

$$
O K^{2}=O P^{2}+P K^{2}=K Q^{2}+K P^{2} \text {, or }(R-5)^{2}=\left(3-\frac{1}{3} R\right)^{2}+\left(4-\frac{1}{2} R\right)^{2}
$$

Simplifying, we get the equation

$$
\frac{23}{36} R^{2}-4 R=0
$$

From this, we find that $R=\frac{144}{23}$.

## Answer

$\frac{144}{23}$.