HMMT February

Geometry Difficulty 5.1 AIME, harder Find the answer United States

Problem:

Let $ABC$ be a triangle whose incircle has center $I$ and is tangent to $\overline{BC}$ , $\overline{CA}$ , $\overline{AB}$ , at $D, E, F$ . Denote by $X$ the midpoint of major arc $\widehat{BAC}$ of the circumcircle of $ABC$ . Suppose $P$ is a point on line $XI$ such that $\overline{DP} \perp \overline{EF}$ .
Given that $AB=14$ , $AC=15$ , and $BC=13$ , compute $DP$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Solution:

$\boxed{\dfrac{4 \sqrt{5}}{5}}$

Let $H$ be the orthocenter of triangle $DEF$ . We claim that $P$ is the midpoint of $\overline{DH}$ . Indeed, consider an inversion at the incircle of $ABC$ , denoting the inverse of a point with an asterisk. It maps $ABC$ to the nine-point circle of $\triangle DEF$ . According to $\angle IAX = 90^\circ$ , we have $\angle A^* X^* I = 90^\circ$ . Hence line $XI$ passes through the point diametrically opposite to $A^*$ , which is the midpoint of $\overline{DH}$ , as claimed.

The rest is a straightforward computation. The inradius of $\triangle ABC$ is $r = 4$ . The length of $EF$ is given by $EF = 2 \dfrac{AF \cdot r}{AI} = \dfrac{16}{\sqrt{5}}$ . Then,
$DP^2 = \left(\frac{1}{2} DH\right)^2 = \frac{1}{4}\left(4r^2 - EF^2\right) = 4^2 - \frac{64}{5} = \frac{16}{5}$
Hence $DP = \dfrac{4 \sqrt{5}}{5}$ .

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