Romania — NMO Selection Tests for the Balkan and International Mathematical Olympiads

Let $a$, $h_A$ and $\Delta$ denote the length of the side $BC$, the length of the altitude from $A$ in the triangle $ABC$, and the area of the triangle $ABC$, respectively. Consider the circle internally tangent at $A$ to the circle $ABC$ and tangent at $T$ to the line $BC$. Then $2R_A \ge AT \ge h_A = 2\Delta/a$, and equality holds throughout if and only if the triangle $ABC$ is isosceles with apex at $A$. Similar inequalities hold for $R_B$ and $R_C$, and the conclusion follows at once; equality holds if and only if the triangle $ABC$ is equilateral.

In the notation in Solution 1, we shall prove that $1/R_A = (a/\Delta) \cos^2(B/2 - C/2)$. Similar formulae hold for $R_B$ and $R_C$, and the conclusion follows at once; equality holds if and only if the triangle $ABC$ is equilateral.
To prove the above formula for $R_A$, let $A'$ be the orthogonal projection of $A$ on the line $BC$, let $O$ be the circumcentre of the triangle $ABC$, and let $O_A$ be the centre of the circle tangent at $A$ to the circle $ABC$ and tangent at $T$ to the line $BC$. Since the points $A$, $O_A$ and $O$ are collinear, the angle $A'AO_A$ is congruent to the absolute value of the difference of the internal angles of the triangle $ABC$ at $B$ and $C$, so $\cos(B - C) = \cos(\angle A'AO_A) = (AA' - O_AT)/O_A A = (h_A - R_A)/R_A = h_A/R_A - 1 = 2\Delta/(aR_A) - 1$, whence the desired formula.