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Source: Cono Sur 2009 #3

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fprosk

#1 h Nov 17, 2015, 3:02 AM • 2 Y

Y by Adventure10, Mango247

Let $A$ , $B$ , and $C$ be three points such that $B$ is the midpoint of segment $AC$ and let $P$ be a point such that $<PBC=60$ . Equilateral triangle $PCQ$ is constructed such that $B$ and $Q$ are on different half=planes with respect to $PC$ , and the equilateral triangle $APR$ is constructed in such a way that $B$ and $R$ are in the same half-plane with respect to $AP$ . Let $X$ be the point of intersection of the lines $BQ$ and $PC$ , and let $Y$ be the point of intersection of the lines $BR$ and $AP$ . Prove that $XY$ and $AC$ are parallel.

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#2 h Nov 25, 2015, 8:40 PM • 3 Y

Y by NiltonCesar, Adventure10, Mango247

Let $O$ be the center of equilateral $\triangle APR$ and let $M$ be the midpoint of $PA.$ Since $\angle ABP=\angle AOP=120^{\circ}$ $\Longrightarrow$ $B \in \odot(POA)$ and obviously $RA,RP$ are tangents of $\odot(POA)$ $\Longrightarrow$ $BY$ is the B-symmedian of $\triangle PBA,$ isogonal of its B-median $BM$ $\Longrightarrow$ $\angle PBY=\angle ABM=\angle PCA.$ But as $\triangle PAC \cong \triangle PRQ$ are directly congruent, then $\angle PQR=\angle PCA$ $\Longrightarrow$ $\angle PQR=\angle PBY$ $\Longrightarrow$ $PQRB$ is cyclic $\Longrightarrow$ $\angle RBQ=\angle RPQ=\angle APC$ $\Longrightarrow$ $PXBY$ is cyclic $\Longrightarrow$ $\angle PYX=\angle PBX=\angle PRQ=\angle PAC$ $\Longrightarrow$ $XY \parallel AC.$

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#3 h Nov 25, 2015, 11:30 PM • 2 Y

Y by Adventure10, Mango247

Very nice, as usual, Luis! A small remark for your proof: also $BQ$ is the $B-$ symmedian of $\triangle BCP$ ; with $BC=AB$ , done!

Another idea:
A $90^\circ$ rotation about $P$ sends $\triangle PAC$ to $\triangle PRQ$ and $B$ to $T$ , midpoint of $QR$ ; since $\angle PBC=60^\circ$ and that is the rotation angle value, $T$ lies onto $AC$ and $QR\parallel PB$ . As $\triangle PTQ\equiv\triangle BTR$ (s.a.s. criterion), $PBRQ$ is an isosceles trapezoid, i.e. $\triangle BRQ\cong\triangle PQR\cong\triangle PCA$ , hence $\angle RBQ=\angle APC\ (\ 1\ )$ and $\angle CBQ=\angle APB\ (\ 2\ )$ . From $(1)$ we infer $PYBX$ cyclic, i.e. $\angle YXB=\angle APB\ (\ 3\ )$ ; with $(2)$ , done.

Best regards,
sunken rock

This post has been edited 1 time. Last edited by sunken rock, Nov 25, 2015, 11:31 PM

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