HMMT February

Geometry Difficulty 5.2 AIME, harder Prove it United States

Problem:
Let $\triangle ABC$ be an equilateral triangle with side length $4$ . Across all points $P$ inside triangle $\triangle ABC$ satisfying $[PAB] + [PAC] = [PBC]$ , compute the minimum possible length of $PA$ .

(Here, $[XYZ]$ denotes the area of triangle $\triangle XYZ$ .)

Solution

Solution:

The area condition implies $[ABC] = 2[PBC]$ . Hence, $P$ lies on the $A$ -midline of $\triangle ABC$ . Therefore, the minimum possible value of $PA$ is the distance from $A$ to this midline. This is achieved by taking $P$ to be the foot of the perpendicular from $A$ to the $A$ -midline. This distance is half the altitude of $ABC$ , which has side length $4$ , so the answer is $\frac{1}{2} (2\sqrt{3}) = \boxed{\sqrt{3}}$ .

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

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