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Annual Harvard-MIT Mathematics Tournament

Geometry Difficulty 5.7 AIME, harder Prove it United States

Problem:

Let $\triangle ABC$ be a scalene triangle. Let $h_{a}$ be the locus of points $P$ such that $|PB - PC| = |AB - AC|$ . Let $h_{b}$ be the locus of points $P$ such that $|PC - PA| = |BC - BA|$ . Let $h_{c}$ be the locus of points $P$ such that $|PA - PB| = |CA - CB|$ . In how many points do all of $h_{a}$ , $h_{b}$ , and $h_{c}$ concur?

Solution

Solution:

Answer: 2 The idea is similar to the proof that the angle bisectors concur or that the perpendicular bisectors concur. Assume WLOG that $BC > AB > CA$ . Note that $h_{a}$ and $h_{b}$ are both hyperbolas. Therefore, $h_{a}$ and $h_{b}$ intersect in four points (each branch of $h_{a}$ intersects exactly once with each branch of $h_{b}$ ). Note that the branches of $h_{a}$ correspond to the cases when $PB > PC$ and when $PB < PC$ . Similarly, the branches of $h_{b}$ correspond to the cases when $PC > PA$ and $PC < PA$ .
If either $PA < PB < PC$ or $PC < PB < PA$ (which each happens for exactly one point of intersection of $h_{a}$ and $h_{b}$ ), then $|PC - PA| = |PC - PB| + |PB - PA| = |AB - AC| + |BC - BA| = |BC - AC|$ , and so $P$ also lies on $h_{c}$ . So, exactly two of the four points of intersection of $h_{a}$ and $h_{b}$ lie on $h_{c}$ , meaning that $h_{a}$ , $h_{b}$ , and $h_{c}$ concur in two points.

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