Incenter geo, prove that AR is parallel to BC

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Incenter geo, prove that AR is parallel to BC

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Source: IMOC 2021 G3

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#1 h Aug 11, 2021, 8:39 AM • 2 Y

Y by centslordm, tiendung2006

Let $I$ be the incenter of the acute triangle $\triangle ABC$ , and $BI$ , $CI$ intersect the altitude of $\triangle ABC$ through $A$ at $U$ , $V$ , respectively. The circle with $AI$ as a diameter intersects $\odot(ABC)$ again at $T$ , and $\odot(TUV)$ intersects the segment $BC$ and $\odot(ABC)$ at $P$ , $Q$ , respectively. Let $R$ be another intersection of $PQ$ and $\odot(ABC)$ . Show that $AR\parallel BC$ .

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MP8148

#2 h Aug 11, 2021, 10:06 AM • 1 Y

Y by centslordm

Let $S = \overline{AT} \cap \overline{BC}$ . By ratio lemma, if $f(X) = \frac{XB}{XC}$ for any $X$ , the desired result is equivalent to $f(P) = f(Q)f(R) = \frac{f(Q)}{f(A)} = \frac{f(Q)f(T)}{f(S)} \iff f(P)f(S) = f(Q)f(T),$ or that $TPQS$ is cyclic. Since $\angle AIV = \angle IUV = 90^\circ - \tfrac 12 \angle B$ , $\overline{AI}$ is tangent to $(IVU)$ . It is well known that $\angle AIS = 90^\circ$ , so $\overline{AI}$ is also tangent to $(ITS)$ . Thus $AT \cdot AS = AI^2 = AV \cdot AU,$ meaning $STVU$ is cyclic, as desired.

This post has been edited 1 time. Last edited by MP8148, Aug 11, 2021, 10:06 AM

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#3 h Aug 15, 2021, 9:18 AM • 1 Y

Y by centslordm

First of all observe that $AR||BC \iff BC\cap AT,T,U,V$ are concyclic. This means that it's enough to prove that $AT\cap BC=E$ lies on $(TUV)$ .
Radical center of $(AI),(BIC),(ABC)$ is $E$ , becuse $ET\cdot EA=EB\cdot EC$ .
It's well-known that $(AI)$ touches $(BIC)$ $\implies$ $EI$ touches $(AI)$
Easy angle chasing shows that $\angle VUI=\angle VIA \implies AI^2=AV\cdot AU$
$T$ is foot of perrpendicular in right triangle $AIE \implies AI^2=AT\cdot AE$
Finally, $AT\cdot AE=AI^2=AV\cdot AU \implies TVUE$ is cyclic $\blacksquare$

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#4 h Aug 17, 2021, 5:14 PM • 2 Y

Y by centslordm, i3435

$[asy] import olympiad; size(12cm);defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta; pair O,A,B,C,I,K,U,V,M,T,D,L,P,Q,R; O=(0,0);A=dir(120);B=dir(200);C=dir(340);path w=circumcircle(A,B,C); I=incenter(A,B,C);K=foot(A,B,C);U=extension(B,I,A,K);V=extension(C,I,A,K);M=extension(O,midpoint(B--C),A,I);D=foot(I,B,C); T=intersectionpoints(w,M--100D-99M)[0];L=extension(A,T,B,C);path q=circumcircle(T,U,V); Q=intersectionpoints(q,w)[1];P=intersectionpoints(q,C--100B-99C)[1]; R=intersectionpoints(w,Q--100P-99Q)[0]; draw(A--B--C--cycle,deep);draw(w,deep); draw(K--A,deep);draw(A--L,deep); draw(M--T,deep);draw(M--A,deep);draw(q,heavyblue);draw(L--B,deep); draw(R--Q,heavymagenta);draw(A--R,heavymagenta);draw(C--V,deep);draw(B--I,deep); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$I$",I,dir(I)); dot("$U$",U,dir(U)); dot("$V$",V,dir(V)); dot("$M$",M,dir(M)); dot("$D$",D,dir(D)); dot("$T$",T,dir(T)); dot("$L$",L,dir(L)); dot("$Q$",Q,dir(Q)); dot("$P$",P,dir(P)); dot("$R$",R,dir(R)); [/asy]$

Let $M$ be the midpoint of arc $BC$ not containing $A$ . Let $D=TM\cap BC$ , let $L=AT\cap BC$ .

Claim. $AI$ is tangent $(UVI)$ .
Proof. Obvious due to angle chase.
$\measuredangle VIA=\measuredangle CIA=90^\circ+\measuredangle CBI= \measuredangle VUB=\measuredangle VUI.\quad \square$
Claim. $AI$ is tangent $(TID)$ .
Proof. Obvious due to inversion at $M$ with radius $MB=MI$ . $\square$

Note that $L$ lies on $(TID)$ , hence by PoP, $LTUV$ is cyclic.

Now, $\measuredangle PLT=\measuredangle PQT=\measuredangle RQT=\measuredangle RAT,$ thus $AR\parallel BC$ .

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i3435

#5 h Aug 17, 2021, 8:15 PM

Y by

$\overline{AR}||\overline{BC}$ is equivalent to $T,U,V,\overline{AT}\cap \overline{BC}$ concyclic by angle chase (it's a general version of shooting lemma). Let $T_A$ be the point on $\overline{BC}$ such that $\angle AIT=90$ . Then radical axis on $(AI),(BIC),(ABC)$ means that $A-T-T_A$ .

Let $D$ be the foot from $I$ to $\overline{BC}$ . $\measuredangle T_ATI=\measuredangle ITA=90=\measuredangle IDT_A$ , so $T_A,T,I,D$ cyclic. Let $(T_ATID)\cap \overline{AV}=U',V'$ . Let $T_A'$ be a point such that $\overline{IT_A'}||\overline{BC}$ . Then $\overline{IU'},\overline{IV'}$ are isogonal in $\angle T_AIT_A'$ . $\overline{IT_A},\overline{IT_A'}$ , as well as $\overline{IA},\overline{ID}$ , are isogonal. Projecting this involution onto $\overline{AV}$ , we have $(A,P_{\infty});(U,V);(U',V')$ are pairs under an involution. Thus $AU\cdot AV=AU'\cdot AV'=AT\cdot AT_A$ as desired.

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#6 h Jan 3, 2022, 8:53 AM

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$X = \overline{AT} \cap \overline{(TVU)}$
$X,B,C$ Collinear.

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