\section*{Problem 1 - 161031}

In a trapezoid \(A B C D\) with \(A B \perp C D\) and \(A B > C D\), let \(a\) be the length of the sides \(B C\), \(C D\), and \(D A\).

Circles with the same radius are drawn around the vertices such that the circle around \(A\) intersects side \(A B\) at \(H\) and side \(A D\) at \(E\), the circle around \(B\) intersects side \(A B\) at \(I\) and side \(B C\) at \(F\), the circle around \(C\) intersects side \(B C\) at \(F\) and side \(C D\) at \(G\), and the circle around \(D\) intersects side \(C D\) at \(G\) and side \(A D\) at \(E\).

The semicircle constructed over \(H I\) touches the circles drawn around \(C\) and \(D\) externally at points \(N\) and \(P\). Determine the length of side \(A B\)!
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Since \(A E=E D\), the radius of the four circles drawn around the corner points is \(\frac{a}{2}\).

Let \(x\) be the length of the side \(A B\) to be determined, then \(H I=x-a\) because due to \(A B>C D\), the point \(H\) lies between \(A\) and \(I\).

Let \(M\) be the midpoint of \(A B\) and thus also of \(H I\), so \(H M=\frac{x-a}{2}\) is the radius of the semicircle over \(H I\).

Since \(M, N\), and \(C\) lie on the same line, it follows

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1378.jpg?height=363&width=640&top_left_y=1029&top_left_x=274)

\[
M C=\frac{a}{2}+\frac{x-a}{2}=\frac{x}{2}
\]

Similarly, \(M D=\frac{x}{2}\), so \(M A=M B=M C=M D\).

Thus, the triangles \(A M D, D M C\), and \(C M B\) are isosceles and, due to the SSS congruence criterion, congruent. The angles \(\angle A M D, \angle D M C\), and \(\angle C M B\) are therefore also congruent and since \(\angle A M D+\angle D M C+\angle C M B=180^{\circ}\), each of them is \(60^{\circ}\).

Therefore, the mentioned triangles are equilateral and thus \(M A=M B=a\). The side \(A B\) therefore has a length of \(2 a\).