Problem 1727

1. Restate the problem and setup:
We are given a triangle $\triangle ABC$ with a point $M$ on side $AB$. In triangle $\triangle ACM$, $I_1$ is the incenter and $J_1$ is the excenter opposite $CM$. In triangle $\triangle BCM$, $I_2$ is the incenter and $J_2$ is the excenter opposite $CM$. We need to prove that the line passing through the midpoints of segments $I_1I_2$ and $J_1J_2$ is perpendicular to $AB$.

2. Introduce the incentric bazooka:
We start with a claim that will be crucial for solving the problem. We name it the incentric bazooka.
$$\text{Claim: } \angle I_1DB = \angle I_2DI$$
where $I$ is the incenter of $\triangle ABC$ and $D$ is the point where the incircle of $\triangle ABC$ touches $BC$.

3. Prove the claim:
Let $I$ be the incenter of $\triangle ABC$. The incircle of $\triangle ABC$ touches $BC$ at $D$. Let the incircles of $\triangle ABM$ and $\triangle ACM$ touch $BC$ at $X$ and $Y$ respectively.

We use the following notation for lengths:
$$BC = a, \quad AC = b, \quad AB = c, \quad AM = d, \quad BM = m, \quad CM = n, \quad s = \frac{a+b+c}{2}$$
$$I_1X = r_1, \quad I_2Y = r_2$$

4. Show similarity of triangles:
We claim that $\triangle I_1XD \sim \triangle DYI_2$. Note that $\angle I_1XD = 90^\circ = \angle I_2YD$.

5. Prove the product of segments:
We need to show that $r_1 \cdot r_2 = DX \cdot DY$:
$$r_1 \cdot r_2 = \left( \frac{m+d-c}{2} \tan \frac{\angle AMB}{2} \right) \cdot \left( \frac{n+d-b}{2} \tan \frac{\angle AMC}{2} \right) = \frac{m+d-c}{2} \cdot \frac{n+d-b}{2} = MX \cdot MY$$
The last step follows from the identity $\tan \theta = \tan (90^\circ - \theta)$.

6. Calculate segments:
$$DX = BD - BX = \frac{a+c-b}{2} - \frac{m+c-d}{2} = \frac{n+d-b}{2} = MY$$
$$DY = DM + MY = DM + DX = MX$$
This implies $r_1 \cdot r_2 = DX \cdot DY$, which gives us $\triangle I_1DX \sim \triangle DI_2Y$:
$$\implies \angle I_1DB = \angle DI_2Y = \angle IDI_2$$
$\blacksquare$

7. Return to the main problem:
By the incentric bazooka, we have that $I_1I_2DM$ is cyclic. By a similar argument, we can also show that $J_1J_2DM$ is cyclic. This implies that $D$ is the Miquel point of $I_1I_2J_1J_2$.

8. Midpoints and perpendicularity:
Let $M_1$ and $M_2$ be the midpoints of $I_1I_2$ and $J_1J_2$ respectively. Notice that $M_1$ and $M_2$ are the centers of $\odot(I_1I_2MD)$ and $\odot(J_2J_1MD)$ (since $I_1I_2$ and $J_1J_2$ are diameters). This implies $M_1M_2 \perp DM$ since $DM$ is the radical axis.

9. Conclusion:
Therefore, the line passing through the midpoints of segments $I_1I_2$ and $J_1J_2$ is perpendicular to $AB$.

$\blacksquare$