Problem 1502

Let $\alpha = B C$, $\beta = C A$, $\gamma = A B$. Without loss of generality we may assume that $\beta \geq \gamma$. Let $M$ be the midpoint of $B C$, and let $A'$, $I'$, and $G'$ be the orthogonal projections of $A$, $I$ and $G$ on $M B$.

We have
$$M I' = M B - B I' = \frac{\alpha}{2} - (s - \beta) = \frac{\beta - \gamma}{2},$$
and because
$$\gamma^2 - A' B^2 = \beta^2 - (\alpha - A' B)^2,$$
we have
$$\begin{aligned} & M G' = \frac{1}{3} M A' = \frac{1}{3}(M B - A' B) \\ = & \frac{1}{3}\left(\frac{\alpha}{2} - \frac{\alpha^2 - \beta^2 + \gamma^2}{2 \alpha}\right) = \frac{\beta^2 - \gamma^2}{6 \alpha} . \end{aligned}$$
We conclude that $I G \perp B C$ is equivalent to $I'$ coincides to $G'$, that is $M I' = M G'$. The last relation is equivalent to
$$\frac{\beta - \gamma}{2} = \frac{\beta^2 - \gamma^2}{6 \alpha}$$
or
$$(\beta - \gamma)(\beta + \gamma - 3 \alpha) = 0$$
and the conclusion follows.