\section*{Problem 5 - 131035}

Given a circle \(k\) with diameter \(A B\) of length \(d\). In this circle, there are two circles \(k_{1}\) and \(k_{2}\) such that they touch \(k\) from the inside at points \(A\) and \(B\) respectively, and touch each other from the outside at a point \(M\), so that \(A M + M B = A B\) holds. Here, \(A M \geq M B\).

The area of the colored region is equal to the difference between the area of \(k\) and the sum of the areas of \(k_{1}\) and \(k_{2}\).

Determine the length of \(A M\) for which the area of this colored region is maximized!

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1334.jpg?height=439&width=488&top_left_y=1985&top_left_x=1275)
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Let the length of the segment \(A M\) be \(x\). Then the area of the colored region is

\[
A=A_{k}-A_{k_{1}}-A_{k_{2}}=\frac{\pi}{4} d^{2}-\frac{\pi}{4} x^{2}-\frac{\pi}{4}(d-x)^{2}
\]

The area \(A\) is thus a function of \(x\). Simplifying yields:

\[
\begin{aligned}
A & =-\frac{\pi}{2} x^{2}+\frac{\pi d}{2} x=-\frac{\pi}{2}\left(x^{2}-d x\right)=-\frac{\pi}{2}\left(\left(x-\frac{d}{2}\right)^{2}-\frac{d^{2}}{4}\right) \\
& =-\frac{\pi}{2}\left(x-\frac{d}{2}\right)^{2}+\frac{\pi d^{2}}{8}
\end{aligned}
\]

The area \(A\) is maximized when \(\left(x-\frac{d}{2}\right)^{2}\) is zero, i.e., for \(x=\frac{d}{2}\). The area of the colored region is thus maximized when the point of tangency \(M\) is at the center of the circle \(k\), i.e., \(A M=\frac{d}{2}\).