Mongolian Mathematical Olympiad

Let $a$, $b$, $c$ be sides of the triangle. Then $a + b + c = 50$. Without losing generality we may assume that $a \ge b \ge c$.

Now $50 = a + b + c \le 3a$ and consequently we get $a \ge 17$. On the other hand $50 = a + b + c > 2a$ by triangle inequality and we came to conclusion $a < 25$.

It is well known fact that a triangle with sides $a \ge b \ge c$ is obtuse iff $a^2 > b^2 + c^2$. Therefore for any pair $(b, c)$ satisfying the above condition $b - c \ge 2$ implies the pair $(b - 1, c + 1)$ also satisfies. Note that from $2b \ge b + c = 50 - a$ follows $a \ge b \ge \frac{50 - a}{2}$.

Now for $a = 17, 18, \ldots, 24$ form pairs $(b, c)$ such that $a \ge b \ge \frac{50 - a}{2}$ and we shall count obtuse triangles with sides $a \ge b \ge c$.

When $a = 24$ there are $(b, c) = (24, 2), (23, 3), \ldots, (13, 13)$ pairs. We get from here 11 pairs: $(23, 3), (22, 4), \ldots, (13, 13)$.

When $a = 23$ there are $(b, c) = (23, 4), (22, 5), \ldots, (14, 13)$ pairs. We get from here 9 pairs: $(22, 5), (21, 6), \ldots, (14, 13)$.

When $a = 22$ there are $(b, c) = (22, 6), (21, 7), \ldots, (14, 14)$ pairs. We get from here 7 pairs: $(20, 8), (19, 9), \ldots, (14, 14)$.

When $a = 21$ there are $(b, c) = (21, 8), (20, 9), \ldots, (15, 14)$ pairs. We get from here 3 pairs: $(17, 12), (16, 13), (15, 14)$.

When $a = 20$ there are $(b, c) = (20, 10), (19, 11), \ldots, (15, 15)$ pairs. There are no pairs which satisfy the condition.

When $a = 19$ there are $(b, c) = (19, 12), (18, 13), \ldots, (16, 15)$ pairs. There are no pairs which satisfy the condition.

When $a = 18$ there are $(b, c) = (18, 14), (17, 15), (15, 16)$ pairs. There are no pairs which satisfy the condition.

When $a = 17$ there is only one pair: $(b, c) = (17, 16)$. But in this case we get an acute triangle.

Thus total number of triangles which satisfy the given condition is $30$.