$\left[\begin{array}{l}\text { Tangent Circles } \\ \text { [Pythagorean Theorem (direct and inverse).] }\end{array}\right]$

Two circles with radii $\sqrt{19}$ and $\sqrt{76}$, touching each other externally, are inscribed in a semicircle (i.e., each circle touches this semicircle and its diameter). Find the radius of the semicircle.
The line of centers of touching circles passes through the point of their tangency. Formulate an equation in terms of the desired radius.

## Solution

Let $O, O_{1}$ and $O_{2}$ be the centers of the semicircle and the circles, with radii denoted as $R, r_{1}$, and $r_{2}$, respectively. Let $A$ and $B$ be the points of tangency of the corresponding circles with the diameter of the semicircle, and $C$ be the projection of point $O_{1}$ onto $O_{2} B$. Since $r_{1}=\sqrt{19}$ and $r_{2}=2 \sqrt{19}$, we have $r_{2}=2 r_{1}$. Let $r_{1}=r$. Then $r_{2}=2 r$, $O_{1} O_{2}=r_{1}+r_{2}=3 r$, and $O_{2} C=2 r-r=r$. From the right triangle $O_{1} C O_{2}$, by the Pythagorean theorem, we find that

$$
O_{1} C=\sqrt{O_{1} O_{2}^{2}-O_{2} C^{2}}=\sqrt{9 r^{2}-r^{2}}=2 r \sqrt{2}
$$

Let $A O=x$ and $B O=y$. Since $O O_{1}=R-r$ and $O O_{2}=R-2 r$, from the right triangles $O A O_{1}$ and $O B O_{2}$, we get

$$
(R-r)^{2}=r^{2}+x^{2} \text { and }(R-2 r)^{2}=4 r^{2}+y^{2}
$$

from which $x^{2}=(R-r)^{2}-r^{2}$ and $y^{2}=(R-2 r)^{2}-4 r^{2}$, and since $x+y=A B=2 r \sqrt{2}$, we obtain the equation

$$
\sqrt{R^{2}-2 R r}+\sqrt{R^{2}-4 R r}=2 r \sqrt{2}
$$

By making the substitution $t=\frac{R}{r}$, we get the irrational equation

$$
\sqrt{t^{2}-2 t}=2 \sqrt{2}-\sqrt{t^{2}-4 t}
$$

After squaring both sides, we get the equation $2 \sqrt{2} \sqrt{t^{2}-4 t}=4-t$, from which it follows that $t=4$. Thus, $R=4 r=4 \sqrt{19}$.

## Answer

$4 \sqrt{19}$.