Competition problem — Geometry — Maths Olympiad Prep

Geometry Difficulty 5.2 AIME, harder Prove it United States

Problem:

Let
- $P$ be a point inside a triangle $\triangle ABC$ ,
- $\triangle DEF$ be the pedal triangle of $P$ , i.e., let $D, E, F$ be the feet of the altitudes from $P$ to $BC, CA, AB$ , respectively,
- $I$ be the incenter of $\triangle ABC$ , and
- $\triangle XYZ$ be the Cevian triangle of $I$ , i.e., $X, Y, Z$ be the intersections of $AI, BI, CI$ with $BC, CA, AB$ , respectively.
Show that there is a triangle with side lengths $PD, PE$ , and $PF$ if and only if $P$ is inside $\triangle XYZ$ .

Solution

Solution:

While there is a synthetic solution, we present a solution using (unnormalized) barycentric coordinates.
Let $P = (x, y, z)$ in barycentric coordinates. We know that $PD = \frac{2[\triangle BCD]}{a} = 2[\triangle ABC] \frac{x}{a}$ , so by SSS similarity if $PD, PE, PF$ forms a triangle, so do $x/a, y/b, z/c$ . The equation $\frac{x}{a} = \frac{y}{b} + \frac{z}{c}$ is a line. However, we already know two points on the line, $Y = (a, 0, c)$ and $Z = (a, b, 0)$ . Thus, $\frac{x}{a} < \frac{y}{b} + \frac{z}{c}$ is some side of line $YZ$ . We know that this side does not contain $A = (1, 0, 0)$ , so the region of $\triangle ABC$ on the side of the line must be $BCYZ$ . Intersecting this with the other regions ( $CAZX$ and $ABXY$ ) gives the desired.

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