\section*{Problem 3 - 061013}

In the part of the ornament shown in the figure, circles appear as basic shapes. The lengths of the radii \(r\) and \(r^{\prime}\) of the circles \(k\) and \(k^{\prime}\) are known, and it is given that \(r = r^{\prime}\).

Calculate the lengths \(r_{1}, r_{2}\) and \(r_{3}\) of the radii of the circles \(k_{1}, k_{2}\) and \(k_{3}\)!

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1231.jpg?height=486&width=848&top_left_y=2144&top_left_x=912)
If the centers of the circles \(k, k^{\prime}, k_{1}, k_{2}\) and \(k_{3}\) are denoted by \(M, M^{\prime}, M_{1}, M_{2}\) and \(M_{3}\) respectively, the point of tangency of the circles \(k\) and \(k^{\prime}\) by \(B\), and the intersection point of \(k_{1}\) with \(g_{B M_{1}}\) that is not on \(B M_{1}\) by \(A\), then it holds that

\[
\left|M M_{1}\right|=r+r_{1} \text { and } A B \perp M B \text { (Figure a) }
\]

Furthermore, \(|A B|=|M B|=r\), and thus \(\left|M_{1} B\right|=r-r_{1}\). From this, by applying the Pythagorean theorem to \(\triangle M B M_{1},\left(r+r_{1}\right)^{2}=\left(r-r_{1}\right)^{2}+r^{2}\), and thus \(4 r r_{1}=r^{2}\), i.e., because \(r>0\) : \(r_{1}=\frac{r}{4}\).

Moreover, \(\left|M M_{2}\right|=r+r_{2}\) and \(\left|M_{2} B\right|=r \cdot 2 r_{1}-r_{2}=\frac{r}{2}-r_{2}\), so by applying the Pythagorean theorem to \(\triangle M B M_{2},\left(r+r_{2}\right)^{2}=\left(\frac{r}{2}-r_{2}\right)^{2}+r^{2}\), and thus \(3 r r_{2}=\frac{r^{2}}{4}\), i.e., because \(r>0: r_{2}=\frac{r}{12}\).

Furthermore, \(\left|M_{1} M_{3}\right|=r_{1}+r_{3}\) and \(\left|M_{1} C\right|=r_{1}-r_{3}\), where \(C\) is the foot of the perpendicular from \(M_{3}\) to \(A B\) (Figure b).

By applying the Pythagorean theorem to \(\triangle M_{1} C M_{3}\), we then obtain

\[
\begin{aligned}
\left|M_{3} C\right|^{2} & =\left(r_{1}+r_{3}\right)^{2}-\left(r_{1}-r_{3}\right)^{2} \\
& =4 r_{1} r_{3}=r r_{3}
\end{aligned}
\]

thus \(\left|M_{3} C\right|=\sqrt{r r_{3}}\). Finally, \(\left|M M_{3}\right|=r+r_{3}\) and \(\left|M_{3} D\right|=r-r_{3}\), where \(D\) is the foot of the perpendicular from \(M_{3}\) to \(M B\), and \(|M D|=r-\left|M_{3} C\right|\), so \(|M D|=r-\sqrt{r r_{3}}\).
![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1232.jpg?height=448&width=1106&top_left_y=1161&top_left_x=474)

By applying the Pythagorean theorem to \(\triangle M D M_{3}\), it then holds that

\[
\begin{aligned}
\left(r+r_{3}\right)^{2} & =\left(r-r_{3}\right)^{2}+\left(r-\sqrt{r r_{3}}\right)^{2} \\
\left(r+r_{3}\right)^{2}-\left(r-r_{3}\right)^{2} & =\left(r-\sqrt{r r_{3}}\right)^{2} \\
4 r r_{3} & =r^{2}-2 r \sqrt{r r_{3}}+r r_{3}, \\
r-3 r_{3} & =2 \sqrt{r r_{3}} . \quad(\text { because } r>0) \\
r^{2}-6 r r_{3}+9 r_{3}^{2} & =4 r r_{3} \\
r_{3}^{2}-\frac{10 r r_{3}}{9}+\frac{r^{2}}{9}=0, &
\end{aligned}
\]

from which the solutions \(r_{3_{1}}\) and \(r_{3_{2}}\) are obtained as

\[
\begin{aligned}
r_{3_{1}} & =\frac{5 r}{9}+\sqrt{\left(25 r^{2}+9 r^{2}\right) \frac{1}{81}}=\frac{5 r}{9}+4 r 9=r \text { and } \\
r_{3_{2}} & =\frac{5 r}{9}-4 r 9=\frac{r}{9}
\end{aligned}
\]

Thus, either \(r_{3}=r\) or \(r_{3}=\frac{r}{9}\). Since \(r_{3}<r\), it follows that \(r_{3}=\frac{r}{9}\). In summary, it holds that:

\[
r_{1}=\frac{r}{4}, \quad r_{2}=\frac{r}{12} \quad \text { and } \quad r_{3}=\frac{r}{9}
\]