HMMT February, 2019

Geometry Difficulty 5.2 AIME, harder Prove it United States

Problem:

Given two distinct points $A$ , $B$ and line $\ell$ that is not perpendicular to $A B$ , what is the maximum possible number of points $P$ on $\ell$ such that $A B P$ is an isosceles triangle?

Solution

Solution:

In an isosceles triangle, one vertex lies on the perpendicular bisector of the opposite side. Thus, either $P$ is the intersection of $A B$ and $\ell$ , or $P$ lies on the circle centered at $A$ with radius $A B$ , or $P$ lies on the circle centered at $B$ with radius $A B$ . Each circle-line intersection has at most two solutions, and the line-line intersection has at most one, giving $5$ . This can be easily constructed by taking any $\overline{A B}$ , and taking $\ell$ that isn't a diameter but intersects both relevant circles twice.

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