Similar triangles formed by angular condition

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Similar triangles formed by angular condition

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Source: Iran 2025 second round P3

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Mahdi_Mashayekhi

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Mahdi_Mashayekhi

#1 h Apr 19, 2025, 10:52 AM • 2 Y

Y by Rounak_iitr, Parsia--

Point $P$ lies inside of scalene triangle $ABC$ with incenter $I$ such that $:$
$2\angle ABP = \angle BCA , 2\angle ACP = \angle CBA$ Lines $PB$ and $PC$ intersect line $AI$ respectively at $B'$ and $C'$ . Line through $B'$ parallel to $AB$ intersects $BI$ at $X$ and line through $C'$ parallel to $AC$ intersects $CI$ at $Y$ . Prove that triangles $PXY$ and $ABC$ are similar.

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gghx

#2 h Apr 19, 2025, 11:24 AM

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Let $S=BP\cap CI$ and $T=BI\cap CP$ . Note that $\angle BPC=\angle ABP+\angle ACP=\angle ACI+\angle ABI=\angle BIC,$ hence $BPIC$ is cyclic.

We now prove that $S$ is the circumcenter of $\triangle B'IX$ . This is true because $\angle SB'I=\angle BAI + \angle ABP=\frac{1}{2}(\angle A + \angle C)=\angle IAC+\angle ACI = \angle SIB',$ hence $SI=SB'$ . Furthermore, $\angle ISB'=180^\circ-\angle A - \angle C=\angle B=2\angle ABI=2\angle B'XI,$ hence $S$ is the circumcenter of $\triangle B'IX$ as desired.

Now, $\angle SXI=\angle SIX=\angle SIT=\angle SPT$ , hence $SXTP$ is cyclic. Similarly, $SYTP$ is cyclic, so $SXTPY$ is cyclic.

We are now ready to finish the question. We have $\angle YPX=\angle XSI=180^\circ-2\angle SIX=\angle A$ and $\angle YXP=180^\circ-\angle PSI=\angle ABP+\angle ACI=\angle C,$ so triangles $PXY$ and $ABC$ are similar.

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#3 h Apr 19, 2025, 12:27 PM

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quite similar (or same IDK) to @above

since $\triangle ABB' \sim \triangle ACI$ and $\triangle ACC' \sim \triangle ABI$
implies that $\triangle BB'I \sim \triangle CIC'$

Let $E = BB' \cap IY , F= CC' \cap IX$ .
So $EB'=EI , FC'=FI$
Known that $\angle B'EI = \angle ABC = 2 \angle B'XI$ implies that $E =$ center of $(B'IX)$ .
so $\angle EXI = \angle EIX = \angle EPF$ so $E,X,F,P$ concyclic.
same as $E,Y,F,P$ concyclic .
so $E,Y,X,F,P$ concyclic.
thus $\angle PXY = \angle PEY = \angle ABC$ and $\angle PYX = 180^{\circ} - \angle PFX = \angle ACB$
Conclude that $\triangle PXY \sim \triangle ABC$ . done $\square$

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mathuz

#4 h Apr 19, 2025, 1:14 PM

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Consider the intersection $C'Y\cap B'X = O(.)$ . Then $O$ is the circumcenter of $PB'C'$ , and it suffices to show that $O$ lies on the circumcircle of $PXY$ .

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#5 h Apr 19, 2025, 2:06 PM

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Let $BP\cap CI=U,PC\cap BI=V,AI\cap BC=D$ . Note that $\measuredangle CPB=90+\frac{\measuredangle A}{2}=\measuredangle CIB$ hence $B,I,C,P$ are concyclic.
By Menelaus at $B'UBDCI$ we get $\frac{B'B}{IC}=\frac{\sin \measuredangle C}{\sin \measuredangle B}$ . Hence $BX=BB'.\frac{\sin \frac{\measuredangle C}{2}}{\sin \frac{\measuredangle B}{2}}=\frac{BB'.IB}{IC}=\frac{\sin \measuredangle C}{\sin \measuredangle B}.IB$ Also $\frac{BV}{BP}=\frac{\cos \frac{\measuredangle A}{2}}{\sin \measuredangle C}$ thus,
$\frac{BX.BV}{BP}=\frac{BI}{\sin \measuredangle B}.\cos \frac{\measuredangle A}{2}=BU$ which implies $X\in (PUV)$ . Similarily $Y\in (PUV)$ . Hence $\measuredangle XYP=\measuredangle XVC=\measuredangle C$ and $\measuredangle PXY=\measuredangle PUY=\measuredangle B$ as desired. $\blacksquare$

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#6 h Apr 21, 2025, 2:02 PM

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Here is a different approach rephrasing the problem in terms of reference triangle $PB'C'$ . When I first drew the diagram the points $P$ , $X$ , and $Y$ were all very close together so this solution was motivated by drawing a diagram consisting of the points other than $A$ , $B$ , and $C$ .

It is not hard to show that $\angle B'PC'=90^{\circ}-\tfrac{1}{2}\angle A$ , $\angle PB'C'=90^{\circ}-\tfrac{1}{2}\angle B$ , and $\angle PC'B'=90^{\circ}-\tfrac{1}{2}\angle C$ . Since $\angle BPC=\angle BIC$ , it must be that $I$ lies on the interior of segment $B'C'$ . Because of this, it is also not hard to see that $X$ , $Y$ , and $P$ all lie on the same side of segment $AI$ .
$[asy] import geometry; size(10cm); pair A = dir(110); pair B = dir(200); pair C = dir(340); pair I = incenter(A, B, C); pair D = foot(I, B, C); pair Ep=B+C-D; pair J=I+Ep-D; pair P=isogonalconjugate(triangle(A,B,C),J); pair Bp=intersectionpoint(line(B,P),line(A,I)); pair Cp=intersectionpoint(line(C,P),line(A,I)); pair X=intersectionpoint(line(Bp,B+Bp-A), line(B,I)); pair Y=intersectionpoint(line(Cp,C+Cp-A), line(C,I)); pair O=intersectionpoint(line(Cp,Y), line(Bp,X)); point[] Ap=intersectionpoints(circle(Bp,O,Cp),line(O,I)); pair Ap=Ap[1]; pair Op=circumcenter(Ap,Bp,Cp); pair M_a=2*Op-O; point[] N_c=intersectionpoints(circle(Bp,O,Cp),line(P,Cp)); pair N_c=N_c[0]; point[] N_b=intersectionpoints(circle(Bp,O,Cp),line(P,Bp)); pair N_b=N_b[0]; draw(A--B--C--cycle, black); draw(B--Bp,black); draw(C--P,black); draw(A--Cp,black); draw(B--I,black); draw(C--Y,black); draw(Bp--X); draw(Cp--Y); draw(P--Bp--Cp--cycle, black); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$I$", I, dir(50)); dot("$P$", P, dir(140)); dot("$B'$", Bp, dir(40)); dot("$C'$", Cp, dir(290)); dot("$X$", X, dir(310)); dot("$Y$", Y, dir(110)); [/asy]$
Now notice that $\triangle B'IX\sim\triangle AIB$ and $\triangle C'IY\sim \triangle AIC$ . We will now focus on the acute reference triangle $PB'C'$ . Let $O$ be the circumcenter of $PB'C'$ . Let $X'$ be the second intersection of line $B'O$ with the circumcircle of triangle $POC'$ and let $Y'$ be the second intersection of line $C'O$ with the circumcircle of triangle $POB'$ . It is easy to show that $\triangle PX'C'\sim$ $\triangle PB'Y'\sim$ $\triangle ABC$ . Now notice that $\triangle B'C'X'\sim$ $\triangle AIB\sim$ $\triangle B'IX$ . Thus $X$ is the point on segment $B'X'$ with $IX\parallel C'X'$ . Similarly, $Y$ is the point along segment $C'Y'$ with $IY\parallel B'Y'$ .
$[asy] import geometry; size(10cm); pair P=dir(100); pair Bp=dir(270+55); pair Cp=dir(270-55); pair O=(0,0); pair I=intersectionpoint(line(O,P),line(Bp,Cp)); pair Xp[]=intersectionpoints(line(Bp,O),circle(P,O,Cp)); pair Xp=Xp[1]; pair Yp[]=intersectionpoints(line(Cp,O),circle(P,O,Bp)); pair Yp=Yp[1]; pair X=intersectionpoint(line(I,I+Xp-Cp), line(Bp,Xp)); pair Y=intersectionpoint(line(I,I+Yp-Bp), line(Cp,Yp)); fill(P--Xp--Cp--cycle,palered+white); fill(P--Yp--Bp--cycle,palered+white); fill(P--X--Y--cycle, palered); draw(P--X--Y--cycle); draw(P--Bp--Cp--cycle); draw(circle(P,O,Cp)); draw(circle(P,O,Bp)); draw(Bp--Xp); draw(Cp--Yp); draw(X--I--Y); draw(Cp--Xp--P--Yp--Bp); dot("P",P,2*dir(P)); dot("B'",Bp,dir(270)); dot("C'",Cp,dir(270)); dot("O",O,dir(270)); dot("I",I,dir(270)); dot("X'",Xp,dir(150)); dot("Y'",Yp,dir(30)); dot("X",X,dir(220)); dot("Y",Y,dir(-40)); [/asy]$
Then we have that $\frac{X'X}{XB'}=\frac{C'I}{IB'}=\frac{C'Y}{YY'}.$ Thus triangle $PXY$ is a linear combination of triangle $PX'C'$ and $PB'Y'$ . Since these triangles are both similar to $ABC$ , it is a well-known result that $PXY$ must be similar to $ABC$ as well.

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