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Source: 2019 Taiwan TST Round2 Day1 P2

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191 posts

#1 h Apr 12, 2019, 11:35 AM • 1 Y

Y by Adventure10

Let $ABC$ be a scalene triangle and $I$ be its incenter, $\Omega$ be its circumcircle. Let $M$ be the midpoint of $BC$ . The incircle $\omega$ touches $CA,AB$ at $E,F$ , respectively. Suppose the line $EF$ intersects $\Omega$ at two points $P,Q$ , and $R$ be the point on the circumcircle $\Gamma$ of triangle $MPQ$ such that $MR$ is perpendicular to $PQ$ . Prove that $AR,\Gamma,\omega$ intersects at a point.

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ABCCBA

#2 h Apr 12, 2019, 12:40 PM • 2 Y

Y by M87, Adventure10

See https://artofproblemsolving.com/community/c6h1740472p11312598 (in post #5)

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2125 posts

#3 h Apr 21, 2019, 5:34 PM • 7 Y

Y by hansu, math_pi_rate, AmirKhusrau, amar_04, SenatorPauline, Adventure10, Mango247

Taiwan TST 2019 Round 2 Day 1 P2 wrote:

Let $ABC$ be a scalene triangle and $I$ be its incenter, $\Omega$ be its circumcircle. Let $M$ be the midpoint of $BC$ . The incircle $\omega$ touches $CA,AB$ at $E,F$ , respectively. Suppose the line $EF$ intersects $\Omega$ at two points $P,Q$ , and $R$ be the point on the circumcircle $\Gamma$ of triangle $MPQ$ such that $MR$ is perpendicular to $PQ$ . Prove that $AR,\Gamma,\omega$ intersects at a point.

Solution: Let $\Delta M_AM_BM_C$ be the circumcevian triangle of $I$ WRT $\Delta ABC$ . $G$ be the midpoint of $M_BM_C$ . Let $\Delta DEF$ be the contact triangle WRT $\Delta ABC$ . Let $M$ be midpoint of $BC$ and $M_{BC}$ be the intersection of $MM_A$ and $\odot (ABC)$ . Let $AD$ $\cap$ $\odot (ABC)$ $=$ $D'$ and $AK$ be the $A-$ symmedian chord in $\odot (ABC)$ . Let the parallel to $BC$ through $A$ intersect $\odot (ABC)$ $=$ $A'$
$-1=(A, K ; B,C) \overset{A_1}{=} (\infty_{BC}, A'K \cap BC ; B, C) \implies K - M -A'$ Hence, $AK,$ $A'K$ are isogonal WRT $\angle BKC$ $\implies$ $\angle D'KA'$ $=$ $\angle D'AA'$ $=$ $\angle D'DM$ $\implies$ $D,K$ $,D',M$ concyclic. Let $KD'$ $\cap$ $BC$ $=$ $N_1$
$-1=(A,K;B,C) \overset{D'}{=} (D,N_1;B,C)$ Let $EF \cap BC=N_2$ . By Menelaus'
$\frac{CN_2}{BN_2} \cdot \frac{BF}{AF} \cdot \frac{AE}{CE}=1 \implies \frac{BD}{BN_2} =\frac{CD}{CN_2} \implies N_1 \equiv N_2$ Hence, $EF, KD', BC$ are concurrent, say at $N$
$NP \cdot NQ = NB \cdot NC = NK \cdot ND' = ND \cdot NM \implies PDMQ \text{ is cyclic}$ Let $A'D \cap \odot (ABC)=T$ , then $T$ is the $A-$ mixtilinear incircle touch point WRT $\Delta ABC$ . Then, obviously $T$ $-$ $I$ $-$ $G$ $-$ $M_{BC}$ . Since, $M_BM_C$ is the perpendicular bisector of $AI$ $\implies$ $IG$ $=$ $GM_ {BC}$ . Let $G'$ be the projection of $G$ on $BC$ $\implies$ $DG'$ $=$ $G'M$ $\implies$ $G$ lies on the perpendicular bisector of $DM$ . Since, $M_BM_CQP$ is an isosceles trapezium $\implies$ $G$ lies on the perpendicular bisector of $PQ$ $\implies$ $G$ is the center of $\odot (PDMQ)$ . Recall, $TI$ bisects $\angle BTC$ and, $AT$ and $TD$ are isogonal WRT $\angle BTC$ $\implies$ $AT$ is the reflection of $TD$ over $TI$ . Let $AT$ $\cap$ $(I)$ $=D_1$ $\implies$ $D_1$ is the reflection of $D$ over $IT$ , the perpendicular bisector of $DD_1$ and Since, $G$ $\in$ $IT$ $\implies$ $D_1$ $\in$ $\odot (PDMQ)$ $\implies$ $(I)$ $\cap$ $\odot (PDMQ)$ $=$ $D_1$ $\cancel{=}$ $D$ $\implies$ $A$ $-$ $D_1$ $-$ $T$ $\implies$ $\Delta TDD_1$ is an isosceles triangle. Let $ID$ $\cap$ $EF$ $=$ $J$ $\implies$ $J$ $\in$ $AM$ and also $TM_A$ $||$ $DD_1$ . Let $MR$ $\cap$ $PQ$ $=H$ and $AT$ $\cap EF$ $=$ $L$ , Then $JDMH$ is cyclic and from Nice concyclic Points (Thanks to ABCCBA & math_pi_rate), we have $LD_1DJ$ is cyclic $\implies$ $\angle RD_1D$ $=$ $\angle DJH$ $=$ $\angle AD_1D$ $\implies$ $R$ $-$ $A$ $-$ $D_1$

This post has been edited 1 time. Last edited by AlastorMoody, Apr 21, 2019, 5:34 PM

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466 posts

#4 h Apr 21, 2019, 8:19 PM • 1 Y

Y by Adventure10

Instructive mixtilinear review.

Let $D$ be the incircle touch point on $BC$ so that $D \in \odot(MPQ)$ (proof). Let $T$ be the A-mixtilinear touchpoint and $L$ be the midpoint of arc $BAC$ so that $T, I, L$ collinear. Let $J$ be the foot from $M$ to $EF$ so that $J \in TIL$ (proof). Let $\odot(MPQ)$ meet incircle again at a point $K$ so that $K \in AT$ (proof). Now define $R' = TA \cap MJ$ . If we show that $R' \in \odot(MPQ)$ then $R=R'$ and we'd be done. But note that
$\angle TLM = TAI = \angle TR'N,$ so $TMLR'$ is cyclic. Then $JR' \cdot JM = JL \cdot JT = JP \cdot JQ$ . Thus $MPQR'$ is cyclic and we conclude.

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1916 posts

#5 h Jun 9, 2020, 2:32 PM • 3 Y

Y by Aimingformygoal, GeoMetrix, Bumblebee60

Taiwan TST 2019 Round 2 wrote:

Given a triangle $\triangle{ABC}$ . Denote its incircle and circumcircle by $\omega, \Omega$ , respectively. Assume that $\omega$ tangents the sides $AB, AC$ at $F, E$ , respectively. Then, let the intersections of line $EF$ and $\Omega$ to be $P,Q$ . Let $M$ to be the mid-point of $BC$ . Take a point $R$ on the circumcircle of $\triangle{MPQ}$ , say $\Gamma$ , such that $MR \perp EF$ . Prove that the line $AR$ , $\omega$ and $\Gamma$ intersect at one point.

Let $X$ be the midpoint of $\widehat{BAC}$ and $X^*$ denote it's antipode WRT $\Delta ABC$ . Let $\odot(DPQ)\cap\odot(I)=A'$ . Let $\overline{PQ}\cap\overline{BC}=\{V\}$ . Then $VP\cdot VQ=VB\cdot VC=VD\cdot VM$ from Ma'laurin. So, $\{D\}\in\odot(MPQ)$ . Then from OGM- (#109) we get that $\overline{AA'}$ is the $A-$ Mixtillinear Cevian of $\Delta ABC$ $(\bigstar)$ . Let $\{I\}$ denote the Incenter of $\Delta ABC$ . So from here (#1,#2) we get that $\overline{X-R-I-T}$ where $\{T\}$ is the $A-$ Mixtillinear Incircle touchpoint with $\odot(ABC)$ and from $(\bigstar)$ we get $\overline{A-A'-T}$ . Let $\overline{AT}\cap\overline{MR}=K$ . Then by Reims $KTXX^*$ is cyclic. If $\overline{MR}\cap\odot(MPQ)=K^*$ . Then $MR\cdot MK^*=RP\cdot RQ=RX\cdot RT\implies K^*\in\odot(XX^*T)\implies K^*\equiv K$ . $\blacksquare$

This post has been edited 1 time. Last edited by amar_04, Jun 9, 2020, 2:34 PM

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#6 h Sep 13, 2020, 6:24 AM • 1 Y

Y by tiendung2006

Here is a solution without involving mixtillinear circle, an official solution provided by me

By the proof above we have $D \in \odot(MPQ)$ . Let $T$ be the second intersection of $\Omega$ and $\odot(AEF)$ , then $TD, AI, \Omega$ concur at the minor arc of $BC$ .

Suppose that the second intersection of $\Gamma$ and $\omega$ is $K$ , and the radical axis of $\odot(AEF)$ and $\Gamma$ is $\ell$ . Consider three circles $\odot(AEF), \Gamma, \Omega$ with using radical axis theorem, we have $\ell, DK, EF$ are collinear. Moreover, consider $\odot(AEF), \Gamma, \Omega$ with using radical axis theorem, we have $\ell, PQ, AT$ are collinear. Hence $\ell, DK, EF, AT$ are collinear and $A, T, K, D$ are collinear.

Finally,
$\measuredangle AKD = \measuredangle ATD = \measuredangle ATN = \measuredangle (AN, BC) = \measuredangle RMD = \measuredangle RKD,$ i.e. $R, A, K$ are collinear, as desired.

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