\section*{Problem 3 - 131023}

An isosceles trapezoid \(A B C D\) with \(A B \| C D\) and \(A B=8 \mathrm{~cm}, C D=2 \mathrm{~cm}\) has an incircle with radius \(\rho\).

Calculate this incircle radius \(\rho\) !
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Since the trapezoid is isosceles, the incenter \(M\) lies on the axis of symmetry of the trapezoid. This axis of symmetry runs through the midpoints \(F\) and \(G\) of the sides \(A B\) and \(C D\).

Furthermore, \(M\) lies on the angle bisectors of the angles \(C D E\) and \(E A F\). Since \(\angle G D E + \angle E A F = 180^{\circ}\), it follows that (1) \(\angle M D G + \angle M A F = 90^{\circ}\).

Since the triangles \(M D G\) and \(M A F\) are right-angled, it follows that (2) \(\angle M D G + \angle D M G = 90^{\circ}\). From (1) and (2), it follows that \(\angle M A F = \angle D M G\). Therefore, the triangles \(M D G\) and \(M A F\) are similar.

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1330.jpg?height=289&width=349&top_left_y=1443&top_left_x=859)

Since \(D G = 1 \, \text{cm}\), \(A F = 4 \, \text{cm}\), and \(M G = M F = \rho\), it follows that \(D G : M G = M F : A F\) or \(1 \, \text{cm} : \rho = \rho : 4 \, \text{cm}\), from which \(\rho^2 = 4 \, \text{cm}^2\) and since \(\rho > 0\), we finally get \(\rho = 2 \, \text{cm}\). The inradius \(\rho\) has a length of \(2 \, \text{cm}\).

\section*{Adapted from \([5]\)}