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An easy geometry from iran

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Source: IRAN 2017 2nd round p6

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824 posts

andria

#1 h Apr 21, 2017, 11:41 AM • 6 Y

Y by Ankoganit, buratinogigle, doxuanlong15052000, nguyendangkhoa17112003, Adventure10, Mango247

Let $ABC$ be a triangle and $X$ be a point on its circumcircle. $Q,P$ lie on a line $BC$ such that $XQ\perp AC , XP\perp AB$ . Let $Y$ be the circumcenter of $\triangle XQP$ .
Prove that $ABC$ is equilateral triangle if and if only $Y$ moves on a circle when $X$ varies on the circumcircle of $ABC$ .

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#2 h Apr 22, 2017, 12:56 AM • 3 Y

Y by Ankoganit, Adventure10, Mango247

Denote by $A_O$ antipode of $A$ WRT $(ABC)$ (with center $O$ ) and by $A_H$ and $H_A$ intersection point of $A-$ altitude in $\triangle ABC$ with $(ABC)$ and $BC$ , respectively. We'll consider $4$ cases:

If $X\equiv B$ , then $Y\equiv B$ .
If $X\equiv C$ , then $Y\equiv C$ .
If $X\equiv A_O$ , then $Y\equiv O$ .
If $X\equiv A_H$ , then note that $\angle H_AA_HQ=\angle ACH_A=\angle H_AA_HB$ , thus $Q$ is reflection of $B$ WRT $H_A$ , similarly for $C$ and $P$ . Thus reflecting $(A_HPQ)$ over $AA_H$ we get $(ABC)$ , thus $Y$ is reflection of $O$ WRT $AA_H$ .

Thus if $O'$ is reflection of $O$ WRT $AA_H$ , then $B$ , $C$ , $O$ and $O'$ lie on circle $\Longleftrightarrow \triangle ABC$ is isosceles with $AB=AC$ .
Now in that isosceles triangle consider case when $X\equiv A$ . Note that $Y\in (BOC)\Longleftrightarrow YB=AB=AC=YC$ (because of symmetry) i.e. $Y$ is reflection of $A$ in $BC$ , but as reflection of $A$ in $BC$ lies on $(BHC)$ as well it follows that $(BOC)\equiv (BHC)$ i.e. since $AB=AC$ we have $H\equiv O\Longleftrightarrow \triangle ABC$ is equilateral.

This post has been edited 1 time. Last edited by mihajlon, Apr 22, 2017, 12:39 PM
Reason: Small typo...

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824 posts

andria

#3 h Apr 22, 2017, 2:29 PM • 3 Y

Y by seyyed_khandan, Adventure10, Mango247

I found 3 generalizations:

1)
if $ABC$ is arbitrary then the locus of $Y$ is a conic with focus $F_1,F_2$ with center $S$ which lies on $AO$ .
$[asy] import graph; size(13.3724559250421cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = -40.28737391388937, xmax = 97.08508201115272, ymin = -39.023809374011826, ymax = 25.452733585043447; /* image dimensions */ pen uuuuuu = rgb(0.26666666666666666,0.26666666666666666,0.26666666666666666); pen evefev = rgb(0.8980392156862745,0.9372549019607843,0.8980392156862745); pen qqwuqq = rgb(0.,0.39215686274509803,0.); filldraw((16.198633778972557,5.5851831491241155)--(15.348467135642982,6.853036878625743)--(14.080613406141355,6.002870235296166)--(14.93078004947093,4.7350165057945395)--cycle, evefev, qqwuqq); /* draw figures */ draw(circle((8.034726851046631,8.006078882190646), 12.235783169215052)); draw(circle((26.035883029975647,-12.879065120450552), 21.96334370761289)); draw(shift((13.62214242448413,-12.518172409781231))*rotate(120.1051946201011)*xscale(23.693702738075576)*yscale(10.984768326296559)*unitcircle); draw((2.9154765272611467,14.058561048114557)--(20.266659491480425,8.313021293624193)); draw((20.266659491480425,8.313021293624193)--(43.31289216421736,0.6816657667128443)); draw((20.266659491480425,8.313021293624193)--(14.93078004947093,4.7350165057945395)); draw((14.93078004947093,4.7350165057945395)--(8.620701828524002,0.5037570983246745)); draw((2.9154765272611467,14.058561048114557)--(4.820691264685021,19.81219445902825)); draw((4.820691264685021,19.81219445902825)--(-1.5902983907090278,0.45139299463630034)); draw((4.820691264685021,19.81219445902825)--(14.93078004947093,4.7350165057945395)); draw((14.93078004947093,4.7350165057945395)--(17.73672777906071,0.5505059493530675)); draw((8.620701828524002,0.5037570983246745)--(-1.5902983907090278,0.45139299463630034)); draw((-1.5902983907090278,0.45139299463630034)--(17.73672777906071,0.5505059493530675)); draw((17.73672777906071,0.5505059493530675)--(43.31289216421736,0.6816657667128443)); label("$A$",(3.536526378593938,23.14999990793433),SE*labelscalefactor); label("$B$",(-3.947358072010765,1.561871685036358),SE*labelscalefactor); label("$C$",(17.209007586429454,0.050702709433500005),SE*labelscalefactor); label("$O$",(7.854152023173574,10.916727248292146),SE*labelscalefactor); label("$F_1$",(3.0328033867263136,6.167339039254593),SE*labelscalefactor); label("$F_2$",(23.757406480708568,-26.934457569188968),SE*labelscalefactor); label("$Q$",(8.50179586986052,0.9142278383494189),SE*labelscalefactor); label("$P$",(43.90632615541354,3.0010802332295565),SE*labelscalefactor); label("$X$",(20.66310810209316,11.996133659437046),SE*labelscalefactor); label("$Y$",(26.923665286733634,-10.599440547196167),SE*labelscalefactor); label("$S$",(11.883935958114568,-12.470411659847324),SE*labelscalefactor); draw((4.820691264685021,19.81219445902825)--(13.622142424484128,-12.51817240978123), red); /* dots and labels */ dot((4.820691264685021,19.81219445902825),linewidth(3.pt)); dot((-1.5902983907090278,0.45139299463630034),linewidth(3.pt)); dot((17.73672777906071,0.5505059493530675),linewidth(3.pt)); dot((20.266659491480425,8.313021293624193),linewidth(3.pt)); dot((8.620701828524002,0.5037570983246745),linewidth(3.pt) + uuuuuu); dot((43.31289216421736,0.6816657667128443),linewidth(3.pt) + uuuuuu); dot((26.035883029975647,-12.879065120450552),linewidth(3.pt) + uuuuuu); dot((8.034726851046631,8.006078882190646),linewidth(3.pt) + uuuuuu); dot((2.9154765272611467,14.058561048114557),linewidth(3.pt) + uuuuuu); dot((14.93078004947093,4.7350165057945395),linewidth(3.pt) + uuuuuu); dot((24.15224721550223,-30.679761198983),linewidth(3.pt) + uuuuuu); dot((3.0920376334660293,5.643416379420541),linewidth(3.pt) + uuuuuu); dot((13.622142424484128,-12.51817240978123),linewidth(3.pt) + uuuuuu); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$

2)
If there exist 5 distinct points $X_1,X_2,X_3,X_4,X_5$ such that $Y_1,Y_2,Y_3,Y_4,Y_5$ lie on a circle then $ABC$ must be equaliteral.

3)
The locus of the second intersection of circumcircles $YPC,YQB$ is a circle.

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