Competition problem — Geometry — Maths Olympiad Prep

Library / /37 of 61

Geometry Difficulty 5.9 AIME, harder Prove it Ibero-American Mathematical Olympiad

Problem:

$ABC$ is an equilateral triangle. $P$ is a variable interior point such that $\angle APC = 120^\circ$ . The ray $CP$ meets $AB$ at $M$ , and the ray $AP$ meets $BC$ at $N$ . What is the locus of the circumcenter of the triangle $MBN$ as $P$ varies?

Solution

Solution:

$\angle MPN = \angle APC = 120^\circ$ and $\angle MBN = 60^\circ$ , so $MBNP$ is cyclic, in other words, $P$ lies on the circumcircle of $BMN$ .

$P$ also lies on the circle $AGC$ , so $\angle CPG = \angle CAG$ (if $P$ is on the same side of $AG$ as $A$ ) $= 30^\circ = \angle MBG$ . So $PMBG$ is cyclic. In other words, $G$ also lies on the circumcircle of $BMN$ . If $P$ lies on the other side, the same conclusion follows from considering $\angle APG$ .

Since $B$ and $G$ lie on the circumcircle, the center $O$ must lie on the perpendicular bisector of $BG$ . But it is clear that the extreme positions of $O$ occur when $P$ is at $A$ and $B$ and that these are the feet of the perpendiculars from $A$ and $B$ to the perpendicular bisector.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

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