\section*{Problem 6 - 290936}

Let \(k_{1}\) and \(k_{2}\) be two circles that touch each other externally. For their radii \(r_{1}\) and \(r_{2}\), it holds that \(r_{1}>r_{2}\).

A line that touches \(k_{1}\) and \(k_{2}\) at two different points is denoted by \(t\). The tangent to \(k_{1}\) that is parallel to \(t\) and different from \(t\) is denoted by \(u\).

Determine the radius \(r_{3}\) of the circle \(k_{3}\) that touches \(u\) and also touches \(k_{1}\) and \(k_{2}\) externally, in terms of \(r_{1}\) and \(r_{2}\)!
Let \(M_{1}, M_{2}, M_{3}\) be the centers of the circles and \(a, b, c\) the projections of the segments \(M_{1} M_{2}, M_{1} M_{3}\) and \(M_{2} M_{3}\) onto the tangents. Then, using Pythagoras, we get

\[
\begin{aligned}
a^{2}+\left(r_{1}-r_{2}\right)^{2} & =\left(r_{1}+r_{2}\right)^{2} \\
b^{2}+\left(r_{1}-r_{3}\right)^{2} & =\left(r_{1}+r_{3}\right)^{2} \\
c^{2}+\left(2 r_{1}-r_{2}-r_{3}\right)^{2} & =\left(r_{2}+r_{3}\right)^{2}
\end{aligned}
\]

From the first two equations, we obtain \(a=2 \sqrt{r_{1} r_{2}}\) and \(b=2 \sqrt{r_{1} r_{3}}\). With \(c=|a-b|\), the third equation then becomes

\[
4\left(\sqrt{r_{1} r_{2}}-\sqrt{r_{1} r_{3}}\right)^{2}+\left(2 r_{1}-r_{2}-r_{3}\right)^{2}=\left(r_{2}+r_{3}\right)^{2}
\]

After expanding and combining the terms, this is equivalent to \(4 r_{1}^{2}=8 r_{1} \sqrt{r_{2} r_{3}}\) and thus

\[
r_{1}^{2}=4 r_{2} r_{3} \Rightarrow r_{3}=\frac{r_{1}^{2}}{4 r_{2}}
\]

\subsection*{6.32 XXX. Olympiad 1990}

\subsection*{6.32.1 First Round 1990, Grade 9}