Let $\vartriangle ABC$ be an acute triangle with $AB > AC$. Let $P$ be the foot of the altitude from $C$ to $AB$ and let $Q$ be the foot of the altitude from $B$ to $AC$. Let $X$ be the intersection of $PQ$ and $BC$. Let the intersection of the circumcircles of triangle $\vartriangle AXC$ and triangle $\vartriangle PQC$ be distinct points: $C$ and $Y$ . Prove that $PY$ bisects $AX$.
Problem
Source: 2020 NZMO Round 1 p6
Tags: geometry, bisects segment
EulersTurban
12.01.2021 12:50
[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -8.30297365286647, xmax = 9.81224820356545, ymin = -4.314182898113443, ymax = 7.004870758585171; /* image dimensions */ /* draw figures */ draw((2.44,4.32)--(-4.96,0.74), linewidth(0.8) + blue); draw((-4.96,0.74)--(1.18,-1.76), linewidth(0.8) + blue); draw((1.18,-1.76)--(2.44,4.32), linewidth(0.8) + blue); draw((-0.9645755618825507,2.672921552494658)--(2.7590046309057836,-2.402917194994212), linewidth(0.8) + blue); draw((2.7590046309057836,-2.402917194994212)--(1.18,-1.76), linewidth(0.8) + blue); draw(circle((-1.89,-0.51), 3.314724724618924), linewidth(0.8) + red); draw(circle((3.219261176066693,0.9879491641703895), 3.421960074734604), linewidth(0.8) + red); draw((2.44,4.32)--(2.7590046309057836,-2.402917194994212), linewidth(0.8) + blue); draw((-0.9645755618825507,2.672921552494658)--(2.5995023154528916,0.9585414025028942), linewidth(0.8) + blue); draw((0.019060352623476606,2.199776479719522)--(1.18,-1.76), linewidth(0.8) + blue); draw((0.019060352623476606,2.199776479719522)--(2.7590046309057836,-2.402917194994212), linewidth(0.8) + blue); draw((0.019060352623476606,2.199776479719522)--(2.44,4.32), linewidth(0.8) + blue); draw(circle((-2.9303810980546503,-2.672880418087165), 5.695787049628996), linewidth(0.8) + linetype("4 4") + red); draw(circle((0.38643824402263777,4.222557819618011), 2.0558722878939784), linewidth(0.8) + linetype("4 4") + red); /* dots and labels */ dot((2.44,4.32),dotstyle); label("$A$", (2.483279360995934,4.435587658686009), NE * labelscalefactor); dot((-4.96,0.74),dotstyle); label("$B$", (-4.916729567285079,0.8599033445406308), NE * labelscalefactor); dot((1.18,-1.76),dotstyle); label("$C$", (1.2282378467594743,-1.6383396696470338), NE * labelscalefactor); dot((-0.9645755618825507,2.672921552494658),linewidth(4pt) + dotstyle); label("$P$", (-0.914804738870707,2.7661456444658157), NE * labelscalefactor); dot((1.423921149556466,-0.5829836592830834),linewidth(4pt) + dotstyle); label("$Q$", (1.4768781467497163,-0.4898582839778231), NE * labelscalefactor); dot((2.7590046309057836,-2.402917194994212),linewidth(4pt) + dotstyle); label("$X$", (2.802959746697674,-2.3132204839062607), NE * labelscalefactor); dot((0.019060352623476606,2.199776479719522),linewidth(4pt) + dotstyle); label("$Y$", (0.0679164468050115,2.292545073055832), NE * labelscalefactor); dot((2.5995023154528916,0.9585414025028942),linewidth(4pt) + dotstyle); label("$M$", (2.649039560989429,1.0493435731046243), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Nice quick exercise $\color{black}\rule{25cm}{1pt}$ Let $M$ be the intersection of $PY$ with $AX$. Since we have that $\angle YPQ=\angle YCQ=\angle YCA=\angle YXA$ we have that $AX$ is tangent to the circumcircle of $PXY$. But also notice that we have that $\angle YAX = \angle YCB = \angle YPA$, thus we have that $AX$ is tangent to the circumcircle of $PYA$. Thus by power of a point we have that $MX ^2=MY.MP=MA^2$. This implies that $MX=MA$.