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HMMT February Guts Round, 2024

Geometry Difficulty 5.2 AIME, harder Prove it United States

Problem:
Let $ABCD$ be a rectangle such that $AB = 20$ and $AD = 24$ . Point $P$ lies inside $ABCD$ such that triangles $PAC$ and $PBD$ have areas $20$ and $24$ , respectively. Compute all possible areas of triangle $PAB$ .

Solution

Solution:

There are four possible locations of $P$ as shown in the diagram. Let $O$ be the center. Then, $[PAO] = 10$ and $[PBO] = 12$ . Thus, $[PAB] = [AOB] \pm [PAO] \pm [PBO] = 120 \pm 10 \pm 12$ , giving the four values $98, 118, 122$ , and $142$ .

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