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Berkeley Math Circle: Monthly Contest 1

Geometry Difficulty 4.8 AIME Prove it United States

Problem:

In $\triangle ABC$ , suppose the incircle has center $I$ and is tangent to $BC$ at $D$ , and the $A$ excircle has center $I_{a}$ and is tangent to $BC$ at $D'$ . Show that $ID'$ and $I_{a}D$ intersect on the altitude from $A$ to $BC$ .

Solution

Solution:

The intersection point is the midpoint of the altitude.

Let $E$ be the point on the incircle diametrically opposite from $D$ . Then the homothety centered at point $A$ which takes the incircle to the $A$ -excircle takes $E$ to $D'$ , so $A$ , $D'$ , and $E$ are collinear. Since $DE$ is parallel to the altitude of $\triangle ABC$ and $AI$ bisects $DE$ , it also bisects the altitude.

Similarly, let $E'$ be the point on the $A$ -excircle diametrically opposite from $D'$ . Then the same homothety as before takes $D$ to $E'$ , so $A$ , $D$ , and $E'$ are collinear. Since $D'E'$ is parallel to the altitude of $\triangle ABC$ and $DI_{a}$ bisects $D'E'$ , it bisects the altitude as well. Thus, both lines pass through the midpoint of the altitude.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.