A circle containing Miquel point

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Source: 2018 Taiwan TST Round 2, Test 1, Problem 1

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2312 posts

#1 h Apr 13, 2018, 6:38 PM • 18 Y

Y by buratinogigle, Smita, Assmit, SAUDITYA, anantmudgal09, Mathuzb, fastlikearabbit, Ankoganit, doxuanlong15052000, AmirKhusrau, GeoMetrix, thczarif, amar_04, enhanced, Gaussian_cyber, AFSA, Adventure10, Mango247

Given a $\triangle ABC$ with circumcircle $\Omega$ and a point $P.$ Let $D$ be the second intersection of $AP$ with $\Omega,$ $E, F$ be the intersection of $BP, CP$ with $CA, AB,$ respectively $,$ $M$ be the intersection of $\odot (AEF)$ with $\Omega,$ $T$ be the intersection of the tangent of $\Omega$ at $B,C$ and $U$ be the second intersection of $TD$ with $\Omega.$ Prove that the reflection of $U$ in $BC$ lies on $\odot (DMP).$

Proposed by Telv Cohl

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#2 h Apr 14, 2018, 7:59 AM • 4 Y

Y by BogdanB, mijail, Adventure10, Mango247

Let $EF \cap BC \equiv S$ , $CF \cap \odot SBMF \equiv K$ , $BE \cap \odot SCME \equiv L$ and $R$ be the reflection of $U$ across $BC$
1. $U \equiv AS \cap \odot ABC$
See that if $U' \equiv AS \cap \odot ABC$ then $U'BDC$ is harmonic but $UBDC$ is harmonic as well $=> U' \equiv U$
2. $K,L \in \odot DMP$
See that $\angle MKP = \angle MKF = \angle MBF = \angle MBA = \angle MDP$ .
3. $R \equiv \odot SBL \cap \odot SCK$
See that $\angle SRB = \angle SUB = \angle BCA = \angle SCE = \angle SLE = \angle SLB$
4. $R \in \odot KPL$
See that $\angle KRL = \angle KRS + \angle SRL = \angle KCS + \angle SBL = \angle KPL$

This post has been edited 2 times. Last edited by SAUDITYA, Apr 15, 2018, 4:13 PM

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#3 h Apr 14, 2018, 3:04 PM • 13 Y

Y by anantmudgal09, Wizard_32, Math-Ninja, 62861, AlastorMoody, amar_04, GeoMetrix, Ya_pank, Kagebaka, Aryan-23, AFSA, Adventure10, Mango247

Fun fact:
Instead of naming the random point $P$ as it always was, a genius changed it to $O$ in order to form a strange word.
As a result, my friend ltf0501 and I thought $O$ is the circumcenter of $ABC$ throughout the contest, we solved the case that $O$ is the circumcenter and only got a small partial score.

This post has been edited 1 time. Last edited by houkai, Apr 14, 2018, 3:18 PM

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#6 h Apr 15, 2018, 10:35 AM • 5 Y

Y by CZRorz, tworigami, AmirKhusrau, amar_04, Adventure10

Let $X \equiv EF \cap BC$ and let $CF$ and $BE$ cut $\odot(XBF)$ and $\odot(XCE)$ again at $Y$ and $Z,$ respectively. We have $\angle MYP=\angle MBF=\angle MDP$ $\Longrightarrow$ $Y \in \odot (DMP)$ and similarly $Z \in \odot(DMP).$ By Menelaus' theorem for $\triangle PBC$ cut by $\overline{EFX},$ keeping in mind $\triangle BCY \sim \triangle FCX$ and $\triangle CBZ \sim \triangle EBX,$ we obtain:

$CX \cdot \frac{PF}{CF}=BX \cdot \frac{PE}{BE} \Longrightarrow CX \left (1-\frac{CP}{CF} \right)=BX \left (1-\frac{BP}{BE} \right) \Longrightarrow CX-BX=BC=\frac{CP}{CF} \cdot CX-\frac{BP}{BE} \cdot BX \Longrightarrow$

$BC=CP \cdot \frac{CY}{BC}-BP \cdot \frac{BZ}{BC} \Longrightarrow BC^2=CP \cdot CY-BZ \cdot BP.$

The latter implication means, in other words, that the powers of $B$ and $C$ WRT $\odot(DMP)$ add up to $BC^2$ $\Longrightarrow$ $B,C$ are conjugate points WRT $\odot(DMP)$ $\Longrightarrow$ $\odot(DMP)$ is orthogonal to the circle $(L)$ with diameter $\overline{BC}$ (very nice fact !!).

Let $LD$ cut $\odot(DMP)$ again at $V$ and let $UL$ cut $\Omega$ again at $D'.$ By obvious symmetry we have $DD' \parallel BC$ and $LU,LD$ are symmetric WRT $BC.$ Since $(L) \perp \odot(DMP),$ then $LB^2=LD \cdot LV$ $\Longrightarrow$ $\angle BVL=\angle CBD=\angle BCD'=\angle BUL.$ Consequently $\triangle UBL \cong \triangle VBL$ are symmetric WRT $BC,$ i.e. $V$ is is reflection of $U$ on $BC$ and the conclusion follows.

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#7 h Apr 15, 2018, 4:26 PM • 11 Y

Y by Mathuzb, brokendiamond, nhusanboev, nguyenhaan2209, AmirKhusrau, amar_04, enhanced, ILOVEMYFAMILY, Adventure10, Mango247, nguyenloc1712

Official solution :

Let $V$ be the reflection of $U$ in $BC,$ $X$ be the intersection of $BC$ with $EF,$ and let $Y,Z$ be the second intersection of $BE, CF$ with $\odot (CEM), \odot (BFM),$ respectively. Since $M$ is the Miquel point of $BCEF,$ so $X$ lies on $\odot (BFM), \odot (CEM).$ Note that $BDCU$ is harmonic, so $A(B,C;D,U) = -1 = A(B,C;P,X)$ $\Longrightarrow$ $X \in AU.$

Since $\measuredangle BYX = \measuredangle ACB = \measuredangle XUB = \measuredangle BVX,$ so $X \in \odot (BVY).$ Similarly, $X$ lies on $\odot (CVZ),$ so $\measuredangle YVZ = \measuredangle YVX + \measuredangle XVZ = \measuredangle YBX + \measuredangle XCZ = \measuredangle YPZ,$ $\Longrightarrow$ $P, V, Y, Z$ are concyclic. On the other hand, from $\measuredangle MYP = \measuredangle MCA = \measuredangle MDP$ we get $Y \in \odot (DMP).$ Similarly, $Z$ lies on $\odot (DMP),$ so we conclude that $D, M, P, V, Y, Z$ are concyclic. $\qquad \blacksquare$

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#8 h Apr 16, 2018, 4:38 PM • 4 Y

Y by Superguy, amar_04, Adventure10, Mango247

TelvCohl wrote:

Given a $\triangle ABC$ with circumcircle $\Omega$ and a point $P.$ Let $D$ be the second intersection of $AP$ with $\Omega,$ $E, F$ be the intersection of $BP, CP$ with $CA, AB,$ respectively $,$ $M$ be the intersection of $\odot (AEF)$ with $\Omega,$ $T$ be the intersection of the tangent of $\Omega$ at $B,C$ and $U$ be the second intersection of $TD$ with $\Omega.$ Prove that the reflection of $U$ in $BC$ lies on $\odot (DMP).$

Proposed by Telv Cohl

Let $S$ be the midpoint of $\overline{BC}$ and $R$ be the reflection of $A$ in $S$ . Now we consider:

Lemma. Fix a line $\ell$ passing through $R$ and let $Q \in \ell$ . Let $E'=\overline{BQ} \cap \overline{CA}$ and $F'=\overline{CQ} \cap \overline{BA}$ . Then $\odot(AE'F')$ passes through a fixed point on $\Omega$ ; which is the inverse of $L=\ell \cap \odot(BRC)$ in $\odot(BC)$ .

(Proof) Let $W=\ell \cap \overline{BC}$ and $N, K$ be points on $\overline{BC}$ with $\overline{QN} \parallel \overline{CA}$ and $\overline{QK} \parallel \overline{BA}$ . Then, $CE'=\tfrac{LN \cdot BC}{BN}$ and $BF'=\tfrac{LK\cdot BC}{CK}$ . Note that $\tfrac{BW}{WN}=\tfrac{WR}{WQ}$ and $\tfrac{CW}{WK}=\tfrac{WR}{WQ}$ proving that $\tfrac{BF'}{CE'}$ is independent of point $Q$ . Thus, $\odot(AE'F')$ passes through a fixed point. For $Q=L$ we can angle-chase to show that this point is the desired inverse. $\blacksquare$

Let $V$ be the reflection of $U$ in $\overline{BC}$ . Then $D$ is the $V$ -HM point in $\triangle VBC$ proving $S \in \overline{VD}$ and $SD \cdot SV=SB^2$ . Now construct $M^{*}$ , the inverse of $M$ in $\odot(BC)$ and let $M'$ be its reflection in $\overline{BC}$ . Reflect $P$ in $\overline{BC}$ to get point $P'$ . By the lemma, $\overline{PM^{*}}$ passes through point $R$ hence $\overline{M'P'}$ passes through the point $A'$ such that $ABCA'$ is an isosceles trapezoid. Then $\angle SM^{*}P=\angle SM'P'=\angle ADM=\angle PDM$ thus proving $MM^{*}PDV$ is cyclic. $\blacksquare$

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#9 h Mar 27, 2020, 3:34 PM • 5 Y

Y by AmirKhusrau, sameer_chahar12, SpecialBeing2017, mueller.25, amar_04

Beautiful configuration. Here goes my solution.

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label("$R$", (144.24864860273215,58.093500358721606), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Proof: Let $U'$ denote the reflection of $U$ in $\overline{BC}$ and let $\overline{AP}\cap \overline{BC}=J$ . Let $\overline{AT} \cap \Omega=I$ .Let $\overline{FE} \cap \overline{BC}=K$ . Define $R=\overline{BE} \cap \odot(DMP)$ and $L=\overline{CF} \cap \odot(DMP)$ . We break the problem into several claims.

Claim 1: $\overline{AU},\overline{BC},\overline{ID},\overline{EF}$ concur at $K$ .

Proof: Clearly $(A,I;B,C)=-1$ and also $(U,D;B,C)=-1$ . Now notice that if $\overline{AU} \cap \overline{BC}=K'$ then $-1=(U,D;B,C)\overset{A}{=}(B,C,J,K')$ which implies that $K'=K$ or that $\overline{AU} \cap \overline{BC}=K$ . Similiarly obtain that $\overline{ID} \cap \overline{BC}=K$ with which we are done $\blacksquare$

Claim 2: $(BFMKL)$ and $(MECKR)$ are cyclic.

Proof: Notice that $M$ is the miquel point of the complete quad $BCEF$ $\implies$ $(KBFM)$ is cyclic. Now observe that $\angle MBF=\angle MBA=\angle MDA=\angle MDP=\angle MLP=\angle MLF$ with which we have that $(MFBL)$ is cyclic $\implies$ $(BFMKL)$ is cyclic. In a similiar fashion we can prove that $(MECKR)$ is cyclic with which we are done $\blacksquare$

Claim 3: $U' \in \odot(CKL)$ and also $U' \in \odot(BKR)$ .

Proof: We'll only prove the first part because the other part can be proved in a similiar way. Notice that $\angle CU'K=\angle CUK=\angle ABC=\angle FBK=\angle FLK=\angle CLK$ with which we are done $\blacksquare$

Now back to the main problem. Notice that $\angle PLU'=\angle CLU'=\angle CKU'=\angle BKU'=\angle BRU'=\angle PRU'$ with which we have that $U'\in \odot(PLR)$ but notice that $\odot(PLR)$ is nothing but $\odot(PMD)$ by definition with which we are done $\square$

This post has been edited 7 times. Last edited by GeoMetrix, Mar 27, 2020, 3:40 PM

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#10 h Mar 28, 2020, 2:08 AM • 1 Y

Y by Mango247

@above what is motivation for adding $R$ and $L$ ?

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#11 h Mar 28, 2020, 9:46 AM • 4 Y

Y by mueller.25, sameer_chahar12, FISHMJ25, amar_04

Ok so i'll add the motivation here.

As anyone might be able to notice angle chasing with only the points $P,M,D$ in the diagram was really difficult since computing $\angle PU'D$ seemed really difficult (atleast to me). The only way i could think of utilizing the point $U'$ was the condition that $\angle CKU'=\angle CKU$ since $U'$ is the reflection of $U$ in $\overline{BC}$ . The other way possible would have been to reflect some point in $\overline{BC}$ but that seemed as a really remote idea since it added a lot of points in the diagram which didnt have too many properties. So i started exploring what this angle $\angle CKU'$ was equal to. I began by adding the circumcircle of $\triangle{CKU'}$ (here's an advice- if u want to explore suppose $\angle XYZ$ then the first thing u should do is add some circle that has this angle inscribed in them). I defined it's intersection with $\odot(PMD)$ as $L$ . One can even notice that due to the symmetry of the problem the claims that hold for $E$ must also hold for $F$ . So i even added the the point $R=\odot(BU'K) \cap \odot(DMP)$ . Now i started working backwards. From here a natural guess would be to show that $PU'LR$ is cyclic since we have defined $L,R$ such that $L,R$ lie on $\odot(DMP)$ ! Notice that $\angle CLU'=\angle CKU'=\angle BKU'=\angle BRU'$ . So what we are just left to show is that $\overline{CL} \cap \overline{BR}=P$ . I always prefer proving cyclic quads insted of collinearity(yeah that's just me). So i started following a phantom point approach and defined $L,R$ in the opposite way i.e. $L =\overline{CF} \cap \odot(DMP)$ and $R=\overline{BE} \cap \odot(DMP)$ . And now you notice that a bunch of angle chasing arguements finish the problem!

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#12 h Mar 30, 2020, 12:27 PM • 4 Y

Y by AlastorMoody, amar_04, ywq233, Mango247

So I was tricked into proving some extra things in the name of this problem

. Anyway, I'll stick my solution for the actual part here:

Taiwan TST, restated with notation from other thread, wrote:

Given a $\triangle ABC$ with circumcircle $\Gamma$ and a point $O.$ Let $F$ be the second intersection of $AO$ with $\Gamma,$ $D, E$ be the intersection of $CO, BO$ with $AB, CA,$ respectively $,$ $I$ be the intersection of $\odot (ADE)$ with $\Gamma,$ $T$ be the intersection of the tangent of $\Omega$ at $B,C$ and $U$ be the second intersection of $TF$ with $\Omega.$ Prove that the reflection $G$ of $U$ in $BC$ lies on $\odot (FIO).$

Let $K=DE \cap BC$ and $P=AO \cap BC$ . Since $I$ is the center of spiral similarity which takes $BD$ to $CE$ , so $K \in \odot (CIE)$ . Fix $F$ (which automatically fixes point $G$ ), and animate $O$ linearly on line $AF$ . Since $(B,C;P,K)=-1$ , and $P$ is fixed, so that means that point $K$ is also fixed. Also, $O \mapsto E$ is a projective map. And, as $CEIK$ is cyclic, so inversion at $C$ gives that $E \mapsto I$ is also a projective map (Since after inversion at $C$ , point $E$ still moves projectively on a line, while $I$ is now simply related to $E$ by a perspectivity from a fixed point $K$ to a fixed line). Thus, it suffices to show that $I,O,F,G$ are concyclic for $3$ positions of $O$ (the fact that $3$ is sufficient follows from inversion at $F$ , since then $O$ and $I$ both move with degree $1$ on fixed lines, while now we are simply supposed to show a collinearity). For $O=A$ , we have $I=A$ making the result obvious. Now consider $O=BG \cap AF$ . Then $E=BG \cap AC$ and $I=\odot (ABC) \cap \odot (CEK)$ . We first show that that $KCGE$ is cyclic. Note that we have $-1=(B,C;P,K) \overset{A}{=} (B,C;F,\Gamma) \cap AK) \Rightarrow U \in AK$ This gives that $\measuredangle GKC=\measuredangle CKU=\measuredangle CBU+\measuredangle BUK=\measuredangle GBC+\measuredangle BUA=\measuredangle EBC+\measuredangle BCE=\measuredangle GEC$ This means that $I,C,G,E,K$ are concyclic. Thus, we have $\measuredangle OGI=\measuredangle EGI=\measuredangle ECI=\measuredangle ACI=\measuredangle AFI=\measuredangle OFI$ giving the desired result. Similarly, the result holds for $O=CG \cap AF$ . $\blacksquare$

EDIT: As stated by @below, there were some flaws in one of the cases taken. I have corrected the solution.

This post has been edited 3 times. Last edited by math_pi_rate, Mar 30, 2020, 8:21 PM

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#14 h Mar 30, 2020, 7:18 PM

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@above but the case $O=F$ is not simple because claim that $O$ belongs to $(FGI)$ is equivalent to $AF$ being tangent to $(FGI)$ .

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RC.

#16 h Mar 31, 2020, 9:11 AM • 1 Y

Y by AmirKhusrau

I do not see a similar solution above.
Let N and N' resp. be the midpoint of BC and EF and $MN \cap \Omega = R$ . Clearly, it suffices to prove that $\odot DMP$ and $\odot (BC)$ are orthogonal or that $M-$ HM point (say $H_M$ ) in $\Delta MBC$ lies on $\odot DMP$ .
Claim: $NN' || AR$ Proof:- Similar to the existence of simson wallace line. Then $NN'$ is the NEwton- gauss line, so bisects $AP$ and also N lies midway b/w $\overline{H_MR}$ so $H_MP || NN' || AR \Rightarrow MPH_MD$ is cyclic by reims, done.

This post has been edited 1 time. Last edited by RC., Mar 31, 2020, 9:14 AM

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