How many triangles are there in which the measures of the angles - measured in degrees - are integers?
I. Solution: Let the angles of the triangle be $\alpha \leq \beta \leq \gamma$.

Then: $1^{\circ} \leq \alpha \leq 60^{\circ} ; \alpha \leq \beta \leq\left[\frac{180^{\circ}-\alpha}{2}\right] ;$ knowing $\alpha$ and $\beta$, $\gamma=180^{\circ}-\alpha-\beta$ is uniquely determined. The number of $\beta$ values that can be chosen for a given $\alpha$ is $\left[\frac{180^{\circ}-\alpha}{2}\right]-\alpha+1$, so the number of sought triangles is $N=$ $\sum_{\alpha=1}^{60}\left(\left[\frac{180^{\circ}-\alpha}{2}\right]-\alpha+1\right)$. Separating the terms corresponding to odd and even $\alpha$ values ($\alpha=2 k-1$, and $\alpha=2 k$ respectively):

$$
\begin{gathered}
N=\sum_{k=1}^{30}\left\{\left[\frac{180-(2 k-1)}{2}\right]-(2 k-1)+1\right\}+ \\
+\sum_{k=1}^{30}\left(\left[\frac{180-2 k}{2}\right]-2 k+1\right)= \\
=\sum_{k=1}^{30}\{([90-k]-2 k+2)+([90-k]-k+1)\}= \\
=\sum_{k=1}^{30}(183-6 k)=177+171+165+\ldots+9 \mp 3=\frac{30 \cdot 180}{2}=2700
\end{gathered}
$$

László Gehér (Zalaegerszegi Gymnasium VII. class)

II. Solution: We only need to consider the variations of two angles, as the third is determined by these. The sum of the three angles must be $180^{\circ}$. If we give $\alpha\left(\leq 178^{\circ}\right)$, $\beta$ can take any value from $1$ to $179^{\circ}-\alpha$. Therefore, if we take all possibilities, we get the sum of integers from 1 to 178. Their sum is 15931. However, in this sum, every triangle with different angles appears six times, because each of the three angles of the triangle appears in the values of $\alpha$ and each of the other two angles appears in the values of $\beta$. Isosceles triangles appear only three times, for example, the $26^{\circ}, 77^{\circ}, 77^{\circ}$ triangle, when $\alpha=26^{\circ}$, and also when $\beta=26^{\circ}$ $\left(\alpha=77^{\circ}\right)$, and finally, when $\gamma=26^{\circ}\left(\alpha=\beta=77^{\circ}\right)$. The equilateral triangle appears only once. There are 89 isosceles triangles, and the equilateral one must be counted separately. To ensure these also appear six times among the other triangles, we add $3 \cdot 88$ to the obtained number and to count the equilateral triangle six times, we add 5. Thus $15931+3 \cdot 88+5=6 \cdot 2700$, so $N=2700$.

Ottó Eisler (Bp.-i Cistercian Gymnasium V. class)