\section*{Problem 4 - 181024}

For a triangle \(A B C\) with \(\angle C A B=\alpha=120^{\circ}\) and \(\angle B C A=\gamma=30^{\circ}\), the length \(r\) of the circumradius is known.

Calculate the perimeter and the area of this triangle!
}

After the sentence about the sum of the interior angles in a triangle, \(\angle A B C=\beta=30^{\circ}\), and thus \(A B = A C \quad(1)\)

If \(\mathrm{M}\) is the center of the circumcircle of \(\triangle A B C\), then

\[
A M = B M = C M = r
\]

Due to (1) and (2), the perpendicular bisector of \(B C\) passes through \(A\) and \(M\), and in the isosceles triangle \(A B C\), it is the angle bisector of \(\angle C A B\). Therefore, in the isosceles triangle \(A B M\), one angle has a size of \(60^{\circ}\), so the triangle is equilateral.

The same applies to \(\triangle A C M\). Thus, \(C A B M\) is a rhombus with side length \(r\). Its diagonal \(B C\) is perpendicular to the diagonal \(A M\) and is bisected by it, so \(B C\) is twice the height of an equilateral triangle with side length \(r\). Therefore, the perimeter of \(\triangle A B C\) is

\[
C A + A B + B C = r + r + 2 \cdot \frac{r}{2} \sqrt{3} = (2 + \sqrt{3}) r
\]

The area of \(\triangle A B C\) is equal to half the area of the rhombus \(C A B M\), which is equal to the area of \(\triangle A B M\); it is thus \(\frac{r^{2}}{4} \sqrt{3}\).

Solutions of the II. Round 1978 adopted from [5]

\subsection*{7.20.3 III. Round 1978, Class 10}