IMO 1968 SL 9

Origin: ROM

Problem

Let $ABC$ be an arbitrary triangle and $M$ a point inside it. Let $d_a, d_b, d_c$ be the distances from $M$ to sides $BC, CA, AB$; $a, b, c$ the lengths of the sides respectively, and $S$ the area of the triangle $ABC$. Prove the inequality

$$abd_ad_b + bcd_bd_c + cad_cd_a \leq \frac{4S^2}{3}.$$

Prove that the left-hand side attains its maximum when $M$ is the centroid of the triangle.

Solution

We note that $S_a = \frac{ad_a}{2}, S_b = \frac{bd_b}{2},$ and $S_c = \frac{cd_c}{2}$ are the areas of the triangles $MBC, MCA,$ and $MAB$ respectively. The desired inequality now follows from

$$S_aS_b + S_bS_c + S_cS_a \leq \frac{1}{3}(S_a + S_b + S_c)^2 = \frac{S^2}{3}.$$

Equality holds if and only if $S_a = S_b = S_c$, which is equivalent to $M$ being the centroid of the triangle.