Let $P$ be a point inside a triangle $ABC$ such that $\angle PAC= \angle PCB$. Let the projections of $P$ onto $BC$, $CA$, and $AB$ be $X,Y,Z$ respectively. Let $O$ be the circumcenter of $\triangle XYZ$, $H$ be the foot of the altitude from $B$ to $AC$, $N$ be the midpoint of $AC$, and $T$ be the point such that $TYPO$ is a parallelogram. Show that $\triangle THN$ is similar to $\triangle PBC$. Proposed by Sammy Luo
Problem
Source: ELMO 2014 Shortlist G9, by Sammy Luo
Tags: geometry, parallelogram, circumcircle, geometric transformation, homothety, trigonometry, geometry proposed
XmL
25.07.2014 04:56
Let the line through $P,X$ parallel to $AC$ meet $BH,AB$ at $H',A'$ respectively. Using the angle equality condition and some angle chasing, we and get $\angle ZYX=\angle A\Rightarrow \angle ZOX=2\angle BA'X$, hence $A'$ lies on $(O)$. Now since $H'$ lies on $(BXPZ)$ because $\angle BH'P=90$, therefore $\angle A'XH'=\angle XH'P= 90-\angle BZX=\angle A'XO\Rightarrow X,O,H'$ are collinear and $\angle YHN=\angle OH'P=\angle CBP$. Let the pedal feet from $O$ to $AC$ be $L$, then $\triangle THL\sim \triangle PBX$. Hence it now suffices to prove that $\frac {HL}{LN}=\frac {BX}{XC}$. This is obvious, however, as we apply a homothety at $B$ that maps $X,A',O\Rightarrow C,A,O'$, since $CO'=AO'$, hence $NO'\parallel OL\parallel BH$ and $\frac {HL}{LN}=\frac {BO}{OO'}=\frac {BX}{XC}$ and we done.
Arab
26.07.2014 13:26
Let $Q$ be the pedal of $P$ on $BH$,then $B,X,Q,P,Z$ are concyclic,so $\angle QXZ=\angle QBZ=\angle ABH=90^\circ-\angle BAC$. Since $\angle XYZ=\angle XYP+\angle ZYP=\angle XCP+\angle BAP=\angle CAP+\angle BAP=\angle BAC$, we obtain that $\angle OXZ=90^\circ-\angle XYZ=90^\circ-\angle BAC$,and hence $O,Q,X$ are collinear. $\triangle TYH\cong \triangle OPQ$ implies that $\angle THN=\angle THY=\angle OQP=\angle PBC$. Denote by $R'$ the circumradii of $\triangle XYZ$,then $\frac{AC}{BC}=\frac{\sin \angle ABC}{\sin \angle BAC}=\frac{\sin \angle XBZ}{\sin \angle XYZ}=\frac{2R'}{PB}$,so $\frac{PB}{BC}=\frac{2R'}{AC}$. Let $OZ$ intersects the circle with diameter $PB$ again at $S$,then $QS\parallel XZ$ due to $OX=OZ$,and $O'$ the circumcenter of $\triangle ABC$ lies on $BS$ since $BS,BH$ are isogonal with respect to $\angle ABC$.Let the pedal of $O'$ on $BH$ be $M$,then $O'M=HN$. Consequently,assume the circumradii of $\triangle ABC$ is $R$,and we obtain that \begin{align*}\frac{TH}{R'}& =\frac{OQ}{R'}\\& =\frac{ZQ\cdot \sin \angle OZQ}{ZX\cdot \sin \angle OZX}\\& =\frac{\sin \angle ABH\cdot \sin \angle QBS}{\sin \angle ABC\cdot \cos \angle BAC}\\& =\frac{\cos \angle BAC\cdot \frac{HN}{R}}{\frac{AC}{2R}\cdot \cos \angle BAC}\\& =\frac{2HN}{AC}\end{align*} which follows that $\frac{TH}{HN}=\frac{2R'}{AC}=\frac{PB}{BC}$,and hence $\triangle THN\sim \triangle PBC$,as desired. $Q.E.D.$