Midpoint, foot, pedal triangle, parallelogram

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Source: ELMO 2014 Shortlist G9, by Sammy Luo

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#1 h Jul 24, 2014, 1:35 PM • 2 Y

Y by narutomath96, Adventure10

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 $\triangle PBC$ .

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#2 h Jul 25, 2014, 1:56 AM • 3 Y

Y by narutomath96, BobaFett101, Adventure10

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.

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#3 h Jul 26, 2014, 10:26 AM • 2 Y

Y by narutomath96, Adventure10

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.

$Q.E.D.$

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