Nice geometry: EMMO 2016, Jr. 5

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Nice geometry: EMMO 2016, Jr. 5

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Source: Based largely on a problem of Tran Quang Hung (buratinogigle) and its solution by Telv.

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#1 h May 7, 2016, 3:40 PM • 4 Y

Y by buratinogigle, adityaguharoy, Adventure10, Mango247

Let $\triangle ABC$ be a triangle with circumcenter $O$ and circumcircle $\Gamma.$ The point $X$ lies on $\Gamma$ such that $AX$ is the $A$ - symmedian of triangle $\triangle ABC.$ The line through $X$ perpendicular to $AX$ intersects $AB,AC$ in $F,E,$ respectively. Denote by $\gamma$ the nine-point circle of triangle $\triangle AEF,$ and let $\Gamma$ and $\gamma$ intersect again in $P \neq X.$ Further, let the tangent to $\Gamma$ at $A$ meet the line $BC$ in $Y,$ and let $Z$ be the antipode of $A$ with respect to circle $\Gamma.$ Prove that the points $Y,P,Z$ are collinear.
Notes: 1. The $A$ -symmedian of triangle $\triangle ABC$ is the reflection of the $A$ -median in the $A$ -angle bisector.
2. The antipode of a point with respect to a circle is the point on the circle diametrically opposite to it.

Proposed by Adithya Bhaskar

This post has been edited 2 times. Last edited by AdithyaBhaskar, May 7, 2016, 4:55 PM

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#2 h May 7, 2016, 4:52 PM • 3 Y

Y by AdithyaBhaskar, Adventure10, Mango247

AdithyaBhaskar wrote:

The point $X$ lies on $\Gamma$ such that $AP$ is the $A$ - symmedian of triangle $\triangle ABC.$

?

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652 posts

#3 h May 7, 2016, 4:54 PM • 3 Y

Y by aopser123, Adventure10, Mango247

Sorry, edited.

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#4 h May 7, 2016, 8:25 PM • 9 Y

Y by aopser123, Ankoganit, AdithyaBhaskar, biomathematics, buratinogigle, brokendiamond, mijail, Adventure10, Mango247

Obviously $Z \in EF$ and since $AX$ is the polar of $Y$ WRT $\Gamma,$ then $AX$ is perpendicular to $OY$ at $S.$ Therefore, $OY$ is the A-midline of $\triangle AEF,$ cutting $AE,AF$ at their midpoints $U,V$ $\Longrightarrow$ $\odot(XUV)$ is 9-point circle of $\triangle AEF.$

Let $P^*$ be the 2nd intersection of $YZ$ with $\Gamma.$ Then $\angle XP^*Z=\angle XAZ=\angle AYO=\angle AP^*S=\angle XYO$ $\Longrightarrow$ $\odot(P^*XY)$ is tangent to $OY$ and $\angle SP^*X=\angle AP^*Z=90^{\circ}.$ Thus if $M \equiv P^*X \cap OY,$ we have $MS^2=MP^* \cdot MX=MY^2$ $\Longrightarrow$ $M$ is midpoint of $YS.$ Together with $(S,U,V,Y)=-1,$ we have $MS^2=MY^2=MU \cdot MV$ $\Longrightarrow$ $MP^*\cdot MX=MU \cdot MV$ $\Longrightarrow$ $P^* \in \odot(XUV)$ $\Longrightarrow$ $P \equiv P^*$ $\Longrightarrow$ $P,Y,Z$ are collinear.

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#5 h May 23, 2016, 7:42 PM • 1 Y

Y by Adventure10

Here is a quick sketch: a standard $\sqrt{bc}$ inversion reduces this to an intermixing of Russian MO 2015 10th grade P7 and IMO Shortlist 2011 G4.

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#6 h May 24, 2016, 12:18 AM • 2 Y

Y by mhq, Adventure10

Let $M,N$ be the midpoints of $AE,AF$ . Note that since $(A,X;B,C)$ is harmonic, $YX$ is tangent to $\Gamma$ , so $MN$ , the perpendicular bisector of $AX$ , is the line $OY$ . Now let $MN$ intersect $AX$ at $D$ , and note that $(D,Y;M,N)=(AD,AY;AC,AB)=-1$ . Consider an inversion about $Y$ preserving $\Gamma$ . It fixes $A,X$ and swaps $B,C$ , and we wish to show it swaps $P,Z$ . It suffices to show that the image of $\gamma$ passes through $Z$ . Let $M',N',D'$ be the inverses of $M,N,D$ . Since $(Y,D;M,N)$ is harmonic, $D'$ is the midpoint of $M'N'$ . But since $D'$ is the midpoint of $AX$ , $D'=O$ . Thus the center of $\gamma'$ lies on the perpendicular bisector of $M'N'$ , which, since $XZ\parallel MN$ , is the perpendicular bisector of $XZ$ . This implies $\gamma'$ passes through $Z$ , as desired.

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