All-Soviet-Union Mathematical Olympiad

Solution:

Denote the side lengths by $a$, $b$, $c$ and the corresponding median lengths by $m_a$, $m_b$, $m_c$. The six small triangles all have equal area. [Let the areas be $t_1, \ldots, t_6$. It is obvious that the adjacent pairs have equal height and equal base, so we have $t_1 = t_2$, $t_3 = t_4$, $t_5 = t_6$. The three on each side of a median sum to the same area, so $t_1 + t_2 + t_3 = t_4 + t_5 + t_6$, $t_1 + t_5 + t_6 = t_2 + t_3 + t_4$. Subtracting gives $t_1 = t_4$. Similarly, $t_2 = t_5$ and we are home.] So by the usual result that twice the area of a triangle equals its perimeter times its inradius, we conclude that the perimeters of four of the small triangles are equal.

Two of them must share a side of the original triangle. Suppose it is $a$. Then we have: $a/2 + m_a/3 + 2m_b/3 = a/2 + m_a/3 + 2m_c/3$. So $m_b = m_c$. That implies that $b = c$. [Because the triangle formed by the centroid and side $a$ is isosceles, so the median is perpendicular to the side, so the main triangle is isosceles.]

Using the facts that $b = c$ and $m_b = m_c$, we see that two of the remaining small triangles have perimeter $b/2 + m_b$ and two have perimeter $b/2 + m_b/3 + 2m_a/3$. So there are two cases to consider. In the first case $a/2 - m_a/3 = b/2 - m_b/3$. That implies $a = b$, since if $a < b$, $m_a > m_b$ (consider the triangle formed by the centroid and the side $c$). So the triangle is equilateral.

The second case is harder. We have: $a/2 + m_a/3 + 2m_b/3 = b/2 + m_b$, and hence $a/2 + m_a/3 = b/2 + m_b/3$ (*). Take the angle between $a$ and $b$ to be $\theta$. Then $m_a = b \sin \theta$, $a = 2b \cos \theta$, and $m_b^2 = b^2/4 + a^2 - ab \cos \theta = b^2/4 + 2b^2 \cos^2 \theta$. We can now use (*) to get an equation for $\theta$. First we square (*) to get: $m_b^2 = (3a/2 - 3b/2 + m_a)^2$. We divide out the factor $b^2$ to get: $1/4 + 2\cos^2 \theta = 3 1/4 + 8\cos^2 \theta - 9\cos \theta + 3\sin \theta (2\cos \theta - 1)$. Squaring, so that we can use $\sin^2 \theta = 1 - \cos^2 \theta$, and writing $c = \cos \theta$, we get: $(1 - c^2)(4c^2 - 4c + 1) = 4c^4 - 12c^3 + 13c^2 - 6c + 1$. Hence $8c^4 - 16c^3 + 10c^2 - 2c = 0$. Factorizing: $c(c - 1)(2c - 1)^2 = 0$. $c = 0$ and $c = 1$ give degenerate triangles, so we must have $c = 1/2$ and hence the triangle is equilateral.