Let $P$ be a point in the interior of the equilateral triangle $\triangle ABC$ such that $\sphericalangle{APC}=120^\circ$. Let $M$ be the intersection of $CP$ with $AB$, and $N$ the intersection of $AP$ and $BC$. Find the locus of the circumcentre of the triangle $MBN$ as $P$ varies.
Problem
Source: Spanish Communities
Tags: geometry, circumcircle, symmetry, perpendicular bisector, geometry unsolved
grobber
15.04.2006 00:31
Let $O$ be the center of $ABC$. The locus is a segment situated on the perpendicular bisector of $BO$, but I'll just show that the circumcenter of $BMN$ lies on this perpendicular bisector. The bounds of the locus can easily be determined by the restriction of $P$ to the interior of $ABC$. $BNPM$ is cyclic, because $\angle MBN=\frac\pi 3,\angle\MPN=\frac{2\pi}3$. From here we get $\angle BMC=\angle CNA$. We also have $\angle MBC=\angle NCA$, and $BC=CA$, so $MBC,NCA$ are congruent. In particular, $NC=MB$. This means that $\angle MON=\frac{2\pi}3$, so $O$ lies o the circumcircle of $BMN$, and thus the center of this circle lies on the perpendicular bisector of $BO$.
Joao Pedro Santos
10.03.2010 01:16
Let $ O$ be the center of $ [ABC]$. From $ \angle APC = 120^{\circ}$, we get $ P$ lies on the circumcircle of $ [AOC]$, and $ \angle MPN = 120^{\circ}$, so $ P$ also lies on the circumcircle of $ [BMN]$. Lemma: O lies on the circumcircle of $ [BMN]$. Proof: If $ P = O$, it's trivial, so suppose $ P\ne O$ and, WLOG, $ P$ belongs to the small arc $ AO$. So we have $ \angle POB = \angle AOB - \angle AOP = 120^{\circ} - \angle ACP = \angle AMC = \angle PNB$, so $ O$ lies on the circumcircle of $ [BMN]$, QED. Let $ D$ and $ E$ be the circumcenters of $ [AOB]$ and $ [BOC]$. By symmetry, $ DE$ is the perpendicular bisector of $ [OB]$, therefore the circumcircle lies on line $ DE$. Now we shall prove the desired locus is the line segment $ [DE]$, without points $ D$ and $ E$. Let $ Q$ be a point on line $ DE$ and let $ P$ be one of the intersections of the circumference with center $ Q$ and radius $ \overline{QO} = \overline{QB}$ with the circumcircle of $ AOC$. Note $ P$ must lie inside $ [ABC]$. If $ Q = D$ we have $ P = A$. If $ Q = E$, we have $ P = B$. If $ Q$ doesn't lie on line segment $ [DE]$, $ P$ lies outside $ [ABC]$. Let $ M = CP\cap AB$ and $ N = AP\cap BC$. Now it suffices to prove $ M$ and $ N$ lie on the circumference with center $ Q$ and radius $ \overline{QO} = \overline{QB}$. We have $ \angle MPN = \angle APC = 120^{\circ}$, so $ B,M,N,P$ lie on the same circumference. If $ Q$ is the midpoint of $ [DE]$, then $ O = P$ and $ Q$ is also the midpoint of $ OB$, so we have what we want. If $ Q$ is not the midpoint of $ [DE]$, then $ O\ne P$. Using the Lemma, $ O$ lies on the circumference which contains $ B,M,N,P$, so that circumference is the circumcircle of $ [BPO]$, and it's center is $ Q$, so we have what we want.