\section*{Exercise 4 - 221014}

Let \(r\) be the radius of the circumcircle of a regular decagon \(P_{1} P_{2} \ldots P_{10}\) and \(s\) the length of one side of this decagon.

Calculate \(s\) in terms of \(r\)!
Two adjacent vertices \(P_{1}, P_{2}\) of the decagon form, together with the center \(M\) of the circumcircle, the vertices of an isosceles triangle with side lengths

\[
P_{1} M = P_{2} M = r \quad \text{and} \quad P_{1} P_{2} = s
\]

For the angle \(\angle P_{1} M P_{2}\), since it is one-tenth of a full angle, \(\angle P_{1} M P_{2} = 36^{\circ}\). Consequently, by the interior angle theorem and the base angle theorem applied to triangle \(P_{1} M P_{2}\),

\[
\angle M P_{1} P_{2} = \angle M P_{2} P_{1} = \frac{1}{2} \left(180^{\circ} - 36^{\circ}\right) = 72^{\circ}
\]

The angle bisector of \(\angle M P_{1} P_{2}\) intersects the opposite side \(M P_{2}\) at a point \(A\). For this point, \(\angle M P_{1} A = \angle A P_{1} P_{2} = 36^{\circ}\) and therefore, by the interior angle theorem applied to triangle \(P_{1} P_{2} A\),

\[
\angle P_{1} A P_{2} = 180^{\circ} - 36^{\circ} - 72^{\circ} = 72^{\circ}
\]

Thus, triangles \(P_{1} M A\) and \(P_{1} P_{2} A\) are isosceles with \(A M = A P_{1} = P_{1} P_{2} = s\). It follows that \(A P_{2} = M P_{2} - A M = r - s\).

Furthermore, by the main similarity theorem, triangles \(M P_{1} P_{2}\) and \(P_{1} P_{2} A\) are similar. Therefore, \(P_{1} M : P_{1} P_{2} = P_{1} P_{2} : A P_{2}\), i.e., \(\frac{r}{s} = \frac{s}{r - s}\) and hence \(s^{2} + r s - r^{2} = 0\) and since \(s > 0\),

\[
\begin{gathered}
s_{1,2} = -\frac{r}{2} \pm \sqrt{\frac{1}{4} r^{2} + r^{2}} = -\frac{r}{2} \pm \frac{r}{2} \sqrt{5} \\
s = \frac{r}{2} (\sqrt{5} - 1) \approx 0.618 r
\end{gathered}
\]

Solutions of the First Round 1984 adopted from [5]

\subsection*{7.24.2 Second Round 1982, Class 10}