In triangle $ABC$, arbitrary points $P,Q$ lie on side $BC$ such that $BP=CQ$ and $P$ lies between $B,Q$.The circumcircle of triangle $APQ$ intersects sides $AB,AC$ at $E,F$ respectively.The point $T$ is the intersection of $EP,FQ$.Two lines passing through the midpoint of $BC$ and parallel to $AB$ and $AC$, intersect $EP$ and $FQ$ at points $X,Y$ respectively. Prove that the circumcircle of triangle $TXY$ and triangle $APQ$ are tangent to each other. Proposed by Iman Maghsoudi
Problem
Source: Iranian TST 2017, first exam day 2, problem 5
Tags: geometry, Iran, Iranian TST, circumcircle
06.04.2017 19:26
could you post the problems in contest collections?Thanks!
06.04.2017 19:28
guangzhou-2015 wrote: could you post the problems in contest collections?Thanks! I don't think I can. How to?
06.04.2017 19:28
any solution
06.04.2017 19:30
you can see the top topic
06.04.2017 19:30
06.04.2017 19:30
and where is P4
06.04.2017 19:37
guangzhou-2015 wrote: and where is P4 Just a few minutes...
06.04.2017 20:03
My solution: Let $M$ is midpoit of $BC$, $S=(APQ)\cap AM (S\neq A)$. We have $\angle PSM=\angle PSA=\angle PQA=\angle PEB=\angle PXM$ $\Rightarrow S\in (BMX)$. Similarly, $S\in (CMY)$ $\Rightarrow S=(BMX)\cap (CMY)\cap (TXY)$(Miquel's theorem). On ther other hand, $\frac{MX}{BE}=\frac{MP}{BP}=\frac{MQ}{CQ}=\frac{MY}{CF}$ (1). And, $BE.BA=BP.BQ=CQ.CP=CF.CA$ (2). From (1), (2)$\Rightarrow \frac{MX}{AC}=\frac{MY}{AB}$ $\Rightarrow \Delta MXY\sim \Delta ACB$ (3). So, $\angle QSY=\angle QMY=\angle ACB=\angle MXY(\because (3))=\angle MXS+\angle YXS=\angle MPS+\angle YXS$ $\Rightarrow$ $ (SPQ)$ and $(SXY)$ are tangent at $S$. DONE
06.04.2017 20:12
guangzhou-2015 wrote: could you post the problems in contest collections?Thanks! https://artofproblemsolving.com/community/c435323_iran_tst_2017
06.04.2017 20:14
A generalization: Let $ABC$ be triangle and the points $P, Q$ lies on the side $BC$ s.t $B, C, P, Q$ are all different. The circumcircles of triangles $ABP$ and $ACQ$ intersect again at $G$. $AG$ intersects $BC$ at $M$. The circumcircle of triangle $APQ$ intersects $AB, AC$ again at $E, F$ , respectively. $EP$ and $FQ$ intersect at $T$. The lines through $M$ and parallel to $AB, AC$, intersect $EP, FQ$ at $X, Y$, respectively. Prove that the circumcircles of triangle $TXY$ and $APQ$ are tangent to each other.
06.04.2017 20:31
Consider the Miquel point of $\triangle MXY$. Denote it $J$. Then $\angle PJQ=\angle PJM+\angle QJM=\angle PXM+\angle QYM=180-\angle PEA+180-\angle AFQ=\angle APQ+\angle AQP=180-\angle PAQ$, so $J$ is on the circumcircle of $\triangle APQ$. Since $\angle PJM=\angle PQA=180-\angle PEA$ we get that $A-M-J$ are collinear.Using $BP=CQ$, we get easily by P.O.P that $\triangle MXY\sim \triangle ACB$. So $\angle ABP=\angle MYX$ Now let's take K a point on the tangent from J to the circumcircle of $APQ$(to make the angle chasing easier). It is enough to prove that $\angle KJX=\angle JYX$. A "SHORT" angle chasing gives: $\angle KJX=\angle XJM-\angle KJM=180-\angle BPE-\angle KJP-\angle PJM=(180-\angle BPE-\angle PXM)-\angle PAJ=(180-\angle BPE-\angle BEP)-\angle MQJ=\angle ABP-\angle MYJ$ and since $\angle ABP=\angle MYX$ we get that $\angle KJX=\angle MYX-\angle MYJ=\angle JYX$, which we wanted to prove.
09.04.2017 08:27
My solution of the generalization of LeVietAn:Denote $\odot (APQ)$ as $\Gamma$.Let $\{D,A \}= AM \cap \Gamma,$ $K=QD \cap MX,L=PD \cap MY,\{ S,D \}=TD \cap \Gamma.$ $\measuredangle DPT= \measuredangle MAB= \measuredangle GPQ,\measuredangle DQT= \measuredangle MAC= \measuredangle GQP \Rightarrow $ $G,T$ are isogonal conjugates WRT $\triangle ABC$ $\Rightarrow \measuredangle KDT= \measuredangle ADP= \measuredangle BET= \measuredangle MXP \Rightarrow X,K,D,T$ are concyclic,analogously $Y,L,D,T$ are concyclic ......(1) $\measuredangle DMX= \measuredangle MAB= \measuredangle DPT \Rightarrow $ $P,X,D,M$ are concyclic $\Rightarrow MXD= \measuredangle MPD \Rightarrow \measuredangle KTD = \measuredangle QSD \Rightarrow KT \parallel QS$,analogously $LT \parallel PS$ $\Longrightarrow T,K,L,D$ are concyclic......(2),and $\odot (TKLD)$ is tangent to $\Gamma$. From (1)(2) we can know $T,X,Y,K,L,D$ are concyclic.
24.04.2017 13:51
Solution without Miquel: Let $K,L$ be the intersections of $MX, MY$ with $(TXY)$. As noted above, $\triangle ACB ~\triangle MXY~\triangle MLK$, and therefore $KL||BC$. If $D'$ is point on $(APQ)$ s.t. $AD'||BC$, angle chasing gives us two useful facts. $\textbf{Claim 1:} $$\triangle D'EF \sim \triangle ACB$ $\angle D'EF=\angle D'AC=\angle ACB$, and similarly $\angle ABC=\angle D'FE$. $\textbf{Claim 2:} $$\triangle TKL \sim \triangle D'QP$ $\angle D'QP=\angle APQ=\angle QFC=\angle QYM=\angle TKL$ and similarly $\angle TLK=\angle D'PQ$ As $KL||BC$, these two triangles are in fact homotetic, so let $U$ be their center of homothety. Clearly, it's enough to have $U \in (APQ)$, in order to conclude the wanted result. If $N$ is the midpoint of $KL$, and $D=EF \cap BC$, we see that $U\in AM$($A, M, N$ are collinear), so we only need the fact that $AM$ and $DT$ intersect on $(APQ)$. $\textbf{Claim 3:} $If $R=QE\cap PF$, $D', R, T$ are collinear. It is well-known(Brokard) that $TQ$ is the polar of $Q$ w.r.t. $(APQ)$, so if we prove that $DD'$ touches $(APQ)$, we prove the claim. Employ Menelaus on $AEF$, with $B,C,D$ collinear, and statement $\frac {BE}{CF}=\frac {AC}{AB}$, to get $\frac{DE}{DF}=\frac{AC^2}{AB^2}$. But, it is well-known that tangent from $D'$ to $(D'EF)$ divides $EF$ in the very same ratio(isogonal conjugate theorem, or few sine law applications). Now we are ready to show that $U$ lies on $(APQ)$. Let $V$ be the touchpoint of other tangent from $D$ to $(APQ)$. Then $(D', V; E, F)$ are harmonic, and by projecting on $BC$ through $A$, we get $A, M, V$ collinear, so $U\equiv V\in (APQ)$.
15.08.2017 17:21
Nice problem! bgn wrote: In triangle $ABC$, arbitrary points $P,Q$ lie on side $BC$ such that $BP=CQ$ and $P$ lies between $B,Q$.The circumcircle of triangle $APQ$ intersects sides $AB,AC$ at $E,F$ respectively.The point $T$ is the intersection of $EP,FQ$.Two lines passing through the midpoint of $BC$ and parallel to $AB$ and $AC$, intersect $EP$ and $FQ$ at points $X,Y$ respectively. Prove that the circumcircle of triangle $TXY$ and triangle $APQ$ are tangent to each other. Proposed by Iman Maghsoudi Let $M$ be the midpoint of $BC$. Let line $AM$ meet $(APQ)$ again at $Z$. We'll show that $(TXY)$ and $(APQ)$ are tangent at $Z$. Indeed, construct isosceles trapezoid $ABCD$ with $AD \parallel BC$. Note that $$(EF, DZ) \overset{A}{=} (BC, M\infty)=-1$$and $(PQ, DZ)=-1$ as well. Consequently, line $ZD$ passes through $U=EE \cap FF$ and $V=PP \cap QQ$, so $T, Z, D$ are collinear since $T, U, V$ are collinear. Note that $\angle PZM=\angle PQA=\angle PEB=\angle PXM$ so $Z$ lies on $(MPX)$. Likewise, $Z$ lies on $(MQY)$, hence it also lies on $(TXY)$. Meanwhile, $T$ lies on $ZD$ so $$\angle TZQ=\angle AEP \implies \angle PZX=\angle PQZ+\angle ZTX$$Evidently, $(TXY)$ and $(APQ)$ are tangent at $Z$ as desired. $\blacksquare$
15.08.2017 18:56
Sorry for asking a silly question but I'm interested in the backward approach of this problem for the solutions@ above When seeing this problem what do you think of first Does the idea of two circles (circumcircles) being tangent to each other direct your proof a certain route? Thanks
15.08.2017 19:42
Math1331Math wrote: Sorry for asking a silly question but I'm interested in the backward approach of this problem for the solutions@ above When seeing this problem what do you think of first Does the idea of two circles (circumcircles) being tangent to each other direct your proof a certain route? Thanks I assume this question is asked to me. Let's see. One usual approach I have while trying to show two circles are tangent is to find a point on both of them and draw tangent at that point to one and show it is tangent to the other. For (1), after a few guesses, it turns out that the Miquel point of $MXY$ in $\triangle TPQ$ lies on both $(TXY)$ and $(APQ)$. According to the (2), we need to show $$\angle PZX=\angle PQZ+\angle ZTX.$$Working backwards from here, it is sufficient to show $T, Z, D$ are collinear. This succumbs to standard projective tools. I think the key step is (1), which can be a hit or miss, so yes, having a good diagram would be helpful here
15.08.2017 19:59
anantmudgal09 wrote: Math1331Math wrote: Sorry for asking a silly question but I'm interested in the backward approach of this problem for the solutions@ above When seeing this problem what do you think of first Does the idea of two circles (circumcircles) being tangent to each other direct your proof a certain route? Thanks I assume this question is asked to me. Let's see. One usual approach I have while trying to show two circles are tangent is to find a point on both of them and draw tangent at that point to one and show it is tangent to the other. For (1), after a few guesses, it turns out that the Miquel point of $MXY$ in $\triangle TPQ$ lies on both $(TXY)$ and $(APQ)$. According to the (2), we need to show $$\angle PZX=\angle PQZ+\angle ZTX.$$Working backwards from here, it is sufficient to show $T, Z, D$ are collinear. This succumbs to standard projective tools. I think the key step is (1), which can be a hit or miss, so yes, having a good diagram would be helpful here Thank you for your response! It is not just you but others as well, even those who didn't write a solution can contribute
02.08.2018 04:49
Let H is the Miquel point of PQT wrt X,Y,M->PHQ=PHM+MHQ=PYM+MXQ=PEB+QFC=180-APQ then APHQ is cyclic. We also have MX/MY =MX/MQ.MP/MY=CF/CQ.BP/BE=AC/AB and YMX=GMD=BAC then MYX~BAC so ACB=XYM=XYH+HYM=XYH+HPQ then (TXY) tangent to (APQ)
Attachments:

26.05.2019 11:04
Here is my solution for this problem Solution Let $M$ be midpoint of $BC$; $N$ $\equiv$ $AM$ $\cap$ $(APQ)$ $(N \not \equiv A)$ We have: $(XM; XP) \equiv (EB; EP) \equiv (QA; QP) \equiv (NA; NP)$ (mod $\pi$) So: $M$, $N$, $X$, $P$ lie on a circle Similarly: $M$, $N$, $Y$, $Q$ lie on a circle Then: $N$ is Miquel point of 3 point $M$, $X$, $Y$ with respect to $\triangle TPQ$ or $N$, $X$, $T$, $Y$ lie on a circle We also have: $BE . AB = BP . BQ = CP . CQ = CF . AC$ So: $\dfrac{AB}{AC} = \dfrac{CF}{BE}$ But: $\dfrac{BE}{MX} = \dfrac{BP}{MP} = \dfrac{CQ}{MQ} = \dfrac{CF}{MY}$ then: $\dfrac{MY}{MX} = \dfrac{CF}{BE} = \dfrac{AB}{AC}$ Combine with: $(MY; MX) \equiv (AC; AB) \equiv - (AB; AC)$ (mod $\pi$), we have: $\triangle ABC$ $\stackrel{-}{\sim}$ $\triangle MYX$ Through $N$ draw tangent $Nx$ to $(APQ)$ We have: $(Nx; NX) \equiv (Nx; NP) + (NP; NX) \equiv (QN; QP) + (MP; MX) \equiv (YN; YM) + (BC; BA)$ $\equiv (YN; YM) + (YM; YX) \equiv (YN; YX)$ (mod $\pi$) So: $Nx$ tangents $(TXY)$ at $N$ or $(TXY)$ tangents $(APQ)$ at $N$
22.11.2019 01:30
Solution with Aditya Khurmi, Anshul Guha, Anushka Aggarwal, Brandon Chen, Derek Liu, Dhrubajyoti Ghosh, Ethan Zhou, Max Lu, Paul Hamrick, Robin Son, Robu Vlad, Samuel Wang: We let line $AM$ meet $(APQ)$ again at $D$. We claim $D$ is the desired tangency point. Claim: Point $D$ lies on $(PMX)$, $(QMY)$ and $(XMT)$. Proof. The first two are by Reim's theorem, (e.g.\ $\measuredangle DMX = \measuredangle DAE = \measuredangle DPE = \measuredangle DPX$) and the third is by Miquel theorem on $\triangle TPQ$. $\blacksquare$ [asy][asy] import graph; size(8cm); pair A = (3.,6.84), B = (-1.,-4.), C = (15.,-4.), P = (3.74,-4.), Q = (10.26,-4.), F = (12.550840976254428,-1.7875930152165012), T = (6.424628766327579,-7.704055495151994), M = (7.,-4.), X = (5.900192950707685,-6.980477103582171), Y = (8.684442704095057,-5.521613242699201), D = (7.318416605745377,-4.862909001569971); pair E = extension(B,A,P,T); draw(A--B--C--cycle, lightred); draw(circumcircle(A,B,C), lightred); draw(circumcircle(A,P,Q), orange); draw(circumcircle(P,M,X), lightblue); draw(circumcircle(Q,M,Y), lightblue); draw(circumcircle(T,X,Y), lightgreen); draw(E--T--F, blue); draw(X--M--Y, red); dot("$A$", A, dir(110)); dot("$B$", B, dir(210)); dot("$C$", C, dir(330)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$P$", P, dir(P)); dot("$Q$", Q, dir(315)); dot("$X$", X, dir(260)); dot("$Y$", Y, dir(Y)); dot("$M$", M, dir(45)); dot("$T$", T, dir(-90)); dot("$D$", D, dir(D)); [/asy][/asy] Claim: $\triangle MXY \sim \triangle ACB$. Proof. We have $\angle XMY = \angle BAC$ and $MX / BE = MP / PB = MQ / QC = MY / CF$ but also by power of a point $BE / CF = AB / AC$. $\blacksquare$ Now $\measuredangle MXY = \measuredangle ACB$, and one last angle chase: \[ \measuredangle QDY= \measuredangle QMY = \measuredangle ACB = \measuredangle MXY = \measuredangle MXD + \measuredangle DXY = \measuredangle MPD + \measuredangle DXY \]which gives the needed tangency.
05.01.2020 10:55
Solved with Th3Numb3rThr33. [asy][asy] size(7cm); defaultpen(fontsize(10pt)); pen pri=red; pen sec=orange; pen tri=fuchsia; pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pair A,B,C,Ap,M,P,Q,O,D,EE,F,Z,T,X,Y,Tp; A=dir(115); B=dir(190); C=dir(350); Ap=B*C/A; M=(B+C)/2; P=(3B+2M)/5; Q=B+C-P; O=circumcenter(A,P,Q); D=2*foot(O,A,M)-A; EE=2*foot(O,A,B)-A; F=2*foot(O,A,C)-A; Z=extension(B,C,EE,F); T=extension(EE,P,F,Q); X=extension(EE,P,M,M+B-A); Y=extension(F,Q,M,M+C-A); Tp=2*foot(circumcenter(T,X,Y),A,D)-D; filldraw(circumcircle(P,M,D),tfil,tri); filldraw(circumcircle(Q,M,D),tfil,tri); draw(A--Tp--T,tri); filldraw(circumcircle(T,X,Y),sfil,sec); draw(Ap--T,sec); draw(EE--T--F,sec); filldraw(circumcircle(A,P,Q),fil,pri); draw(Ap--Z--D,pri+Dotted); draw(EE--Z--C,pri); filldraw(A--B--C--cycle,fil,pri); dot("$A$",A,NW); dot("$B$",B,SW); dot("$C$",C,S); dot("$A'$",Ap,NE); dot("$M$",M,SW); dot("$P$",P,dir(260)); dot("$Q$",Q,SE); dot("$D$",D,dir(230)); dot("$E$",EE,W); dot("$F$",F,dir(35)); dot("$Z$",Z,E); dot("$T$",T,SW); dot("$X$",X,dir(220)); dot("$Y$",Y,dir(280)); dot("$T'$",Tp,SE); [/asy][/asy] Let $\overline{AM}$ intersect $(APQ)$ again at $D$, let $A'$ be the point such that $ABCA'$ is an isosceles trapezoid, and let $Z=\overline{BC}\cap\overline{EF}$. Denote by $\infty$ the point at infinity along $\overline{BC}$. Note that \begin{align*} -1&=(BC;M\infty)\stackrel A=(EF;DA'),\\ -1&=(PQ;M\infty)\stackrel A=(PQ;DA'). \end{align*}Consequently $\overline{A'D}$ is the polar of $Z$, which passes through $T$ by Brokard's theorem. By $\measuredangle PXM=\measuredangle PEA=\measuredangle PDA=\measuredangle PDM$ and $\measuredangle QYM=\measuredangle QDM$, we have $PMDX$ and $QMDY$ cyclic. Let $T'$ be the point on $\overline{AD}$ such that $\overline{TT'}\parallel\overline{BC}$. By homothety at $D$, we know $(DTT')$ and $(APQ)$ are tangent, so it suffices by symmetry to show that $DXTT'$ is cyclic. This follows readily from $\measuredangle PXT=\measuredangle DXP=\measuredangle DMP=\measuredangle DAA'=\measuredangle DT'T$, and we are done.
18.07.2020 11:01
Same idea as the one used in Iran TST 2016/3/2 and China TST 2015/2/3, hence solved fairly easily . Let $M$ be the midpoint of $BC$. Let $Z$ be the intersection of circles $(PXM)$ and $(PYM)$. By Miquel's theorem, $Z$ lies on $(TXY)$ CLAIM 1. $Z$ lies on $(APQ)$ Proof. This is just angle chasing. Notice that \begin{align} \angle PZX+\angle QZY=\angle PMX+\angle QMY=B+C\\ \angle XZY=180^{\circ}-\angle PTY=\angle BPE+\angle FQC=\angle EAQ+\angle CAP=\angle A+\angle APQ \end{align}Adding them we have $$\angle PZQ=180^{\circ}-\angle APQ$$ CLAIM 2. $\triangle MXY\sim\triangle ACB$ Proof. Notice that $$BE\times BA=BP\times BQ$$$$CF\times CA=CP\times CQ$$since the right hand side are the same, $$\frac{BE}{CF}=\frac{CA}{BA}$$Moreover $$\frac{MX}{BE}=\frac{MP}{BP}=\frac{MQ}{QC}=\frac{MY}{FC}$$Hence $$\frac{MX}{MY}=\frac{BE}{CF}=\frac{CA}{BA}$$Together with $\angle XMY=180^{\circ}-\angle B-\angle C=\angle A$, we justify our claim. Now $\angle MQZ+\angle ZYX=\angle MYX=\angle ABC=\angle PMX=\angle PZX$ hence the two circles are tangent at $Z$.
22.10.2021 09:55
Here's my solution: Let $Z = \overline{AM} \cup \odot{(APQ)}$. I'll show, that $Z$ is the desired point of tangency. Claim 1. $MPXZ$ and $QMZY$ are cyclic.
Claim 2. $\triangle{MXY} \sim \triangle{ACB}$.
From previous claim we have $$\measuredangle{XYM} = \measuredangle{ABC}$$therefore: $$\measuredangle{XTZ} + \measuredangle{ZAP} = \measuredangle{XYZ} + \measuredangle{ZQP} = \measuredangle{XYZ} + \measuredangle{ZQM} = \measuredangle{XYZ} + \measuredangle{ZYM} = \measuredangle{XYM} = \measuredangle{ABC} = \measuredangle{EBP} = \measuredangle{XMP} = \measuredangle{XZP}$$Hence, $\odot{TXY}$ and $\odot{APQ}$ are tangent to each other. $\blacksquare$
Attachments:

17.04.2023 20:05
solved with mxlcv
04.12.2023 07:00
Let $D = \overline{AM} \cap (PMX)$. I claim $D$ is the desired tangency point. The majority of the problem is characterizing $D$. First, note that $ \measuredangle PDA = \measuredangle PXM = \measuredangle PEA$, so $D$ lies on $(APQ)$. Also, $\measuredangle MDQ = \measuredangle MYQ$ by the same logic, so $D$ lies on $(MQY)$. Finally, by Miquel theorem, $D$ lies on $(TXY)$ too. Now, notice $$\frac{MX}{MY} = \frac{\sin \angle MPX}{\sin \angle MQY} = \frac{\sin \angle BAM}{\sin \angle CAM} = \frac{AC}{AB}$$and $\angle XMY = \angle A$ by the two parallel lines, hence $\triangle MYX \sim \triangle ABC$. To finish, this implies $$\angle MQD + \angle DYX = \angle MYX = \angle B = \angle PMX = \angle PDX$$and thus the tangent to $(APQ)$ at $D$ is also tangent to $(TXY)$. This implies the result.
27.12.2023 07:44
Let $D=AM\cap (APQ)$. We claim that $D$ is the tangency point. Claim: $D$ lies on $(PMX)$ and $(QYM)$. Proof: We can angle-chase: \[\angle MDP=180-\angle PEA=\angle PXM,\]\[\angle MDQ=180-\angle AFQ=\angle MYQ,\]as desired $\square$ Claim: $D$ lies on $(TXY)$. Proof: We can angle-chase: \begin{align*} \angle XTY&=180-\angle XPM-\angle YQM\\ &=180-(360-\angle MDX-\angle MDY)\\ &=180-\angle XDY,\\ \end{align*}as desired $\square$ Claim: $\triangle MYX\sim \triangle ABC$. Proof: We can say through similar triangles: \[\frac{BE}{XM}=\frac{BP}{PM}=\frac{CQ}{QM}=\frac{FC}{MY},\]\[\frac{XM}{YM}=\frac{BE}{FC}.\] Also, by the given condition, $B$ and $C$ have the same power with respect to $(APQ)$. Therefore: \[\frac{AC}{BC}=\frac{BE}{CE}=\frac{XM}{YM}.\]Also, using the parallel lines, it is clear that $\angle XMY=\angle BAC$. Thus by $SAS$, we are done $\square$ We now finish by angle-chase: \[\angle QDY=\angle QMY=\angle ACB=\angle MXY=\angle DTY+\angle DAQ,\]as desired $\blacksquare$
10.01.2024 06:17
Let $D= \overline{AM} \cap (APQ)$. Notice that \[\angle PDM = \angle BEP = \angle PXM,\] and \[\angle QDM = \angle CFQ = \angle QYM.\] Hence, $D$ lies on $(MXP)$ and $(MYQ)$. On top of this, $D$ is the Miquel Point of $\triangle TPQ$, meaning $D$ also lies on $(TXY)$. We also have \[\frac{BE}{MX} = \frac{BP}{PM} = \frac{CQ}{QM} = \frac{CF}{MY}\]\[\implies \frac{MY}{MX} = \frac{CF}{BE}.\] Power of a point gives \[BE \cdot BA = BE \cdot BQ = CQ \cdot BP = CF \cdot CA\]\[\implies \frac{AB}{BC} = \frac{CF}{BE} = \frac{MY}{MX}.\] Since the parallel lines give $\angle YMX = \angle BAC$, we have $\triangle ABC \sim \triangle MYX$ from SAS similarity. Let $S$ be the point at infinity on the tangent to $(APQ)$ at $D$ on the side containing $B$. Note that \begin{align*} \angle PQD + \angle DYX &= \angle MYX = \angle EBP \\ &= \angle PMX = \angle PDX = \angle PDS + \angle XDS. \end{align*} Because $\angle PQD = \angle PDS$, we must have $\angle DYX = \angle XDS$, meaning that the tangent is also tangent to $(TXY)$ at point $D$. $\square$
20.02.2024 00:16
Define $M$, $N$, and $L$ as the midpoints of $BC$, $CA$, and $AB$, respectively, and $K = AM \cap (APQ)$. Also note that $A' = (APQ) \cap (ABC)$ is the reflection of $A$ over the perpendicular bisector of $BC$ by symmetry. Then we have: We notice that $KMPX$ is cyclic, and similarily $KMQY$, as \[\measuredangle KMX = \measuredangle KAE = \measuredangle KPX.\] Miquel on $\triangle TPQ$ then tells us $TXKY$ is also cyclic. We have $\triangle ABC \sim \triangle MYX$, as $\measuredangle YMX = \measuredangle CAB$ and \[\frac{MX}{MY} = \frac{BE \cdot \frac{BP}{MP}}{CF \cdot \frac{CQ}{MQ}} = \frac{\frac{BP \cdot BQ}{AB} \cdot \frac{BP}{MP}}{\frac{CQ \cdot CP}{AC} \cdot \frac{CQ}{MQ}} = \frac{AC}{AB}.\] To prove the desired circles are tangent, we claim there exists a homothety at $K$ mapping $A' \mapsto T$. We first see that they are collinear, as \begin{align*} \measuredangle A'KY &= \measuredangle AKQ + \measuredangle QKY = \measuredangle PQA + \measuredangle QMY = \measuredangle PEB + \measuredangle BCA \\ &= \measuredangle PXM + \measuredangle MXY = \measuredangle PXY = \measuredangle TKY. \end{align*} Finally, the corresponding arcs are equal in measure since \[\frac{\overarc{TK}}{2} = \measuredangle TYK = \measuredangle QYK = \measuredangle PMA = \measuredangle A'AK = \frac{\overarc{A'K}}{2}. \quad \blacksquare\]
12.04.2024 11:20
First we will need the following obesrvation: Claim: $\triangle MYX \sim \triangle ABC$ Proof: Obviously $\angle XMY=\angle BAC$ and so we want to prove $\frac{MY}{MX}=\frac{AB}{AC}$ but for this just use $\frac{MY}{CF}=\frac{MQ}{QC}=\frac{MP}{PB}=\frac{MX}{BE}$ and also $CF\cdot CA=CQ\cdot CP=BP\cdot BQ=BE\cdot BA$ $\blacksquare$ The following is the crucial claim: Lemma:Let ${H}=(PMX)\cap (QMY)$ then $H\in (PAQ)\cap (XTY)\cap AM$ Proof: First $\angle PHQ=\angle PHM+\angle MHQ=\angle PXM+\angle MYQ=\angle BEP+\angle QFC=\angle AQP+\angle APQ=180^{\circ}-\angle PAQ$ so $H\in (PAQ)$. Next note $H$ is Miquel for $\triangle PTQ$ with $M,X,Y$ so $H\in (TXY)$ For the last part let's see that $\angle HMX=\angle HPX=180^{\circ}-\angle EPH=\angle BAH$ but also $\angle HMX=\angle (HM,MX)=\angle (HM,AB)=\angle (AH,AB)$ so $AH\parallel HM$ meaning $A,H,M$ are collinear. The lemma is proved. $\blacksquare$ Coming back to our problem it is enough to prove $\angle PHX=\angle PQH+\angle HYX$ (because after that we will get the circles tangent to $H$). But for this just see that : $\angle PQH+\angle HYX=\angle MYH+\angle HYX=\angle MYX=\angle ABC=\angle PMX=\angle PHX$ which is what we wanted. The problem is solved.
12.04.2024 13:39
Let $M$ be the midpoint of $BC$. Then, let $S$ be the second intersection of $(APQ)$ and $(ABC)$ and $R = \overline{AM} \cap \overline{RS}$. First, note that, $AS$ is the radical axis of circles $(APQ)$ and $(ABC)$. Thus, it is perpendicular to the line joining the centers of these two circles. But the line joining these two circles is the perpendicular bisector of $BC$ (since the perpendicular bisector of $PQ$ is also the perpendicular bisector of $BC$ as $MP = MB-BP=MC-CQ=MQ$). So, this means that $AS$ is perpendicular to the perpendicular bisector of $BC$ from which it follows that $AS \parallel BC$. Now, we can prove our first claim. Claim : $R$ lies on $(APQ)$. Proof : Let $R' = \overline{AM} \cap (APQ)$. Then, note that, \[-1=(PQ;MP_\infty)\overset{A}{=}(PQ;R'S)\]Further, let $T' = \overline{R'S} \cap \overline{PE}$ and $Q' = \overline{T'F}\cap (APQ)$. Then, \[-1=(BC;MP_\infty)\overset{A}{=}(EF;R'S)\overset{T'}{=}(PQ';SR')\]Thus, $(PQ';SR')=(PQ;SR')$ from which it follows that $Q'=Q$. Thus, $T'=T$ from which it follows that $R'=R$ which proves the claim. Once we have this redefinition of $R$ we can obtain the following claim. Claim : Quadrilaterals $PMRX$ and $QMRY$ are cyclic. Proof : Note that, \[\measuredangle PXM = \measuredangle EXM = \measuredangle PEB = \measuredangle PRA = \measuredangle PRM \]from which it is clear that $PMRX$ is cyclic. Similarly, we can show that $QMRY$ is cyclic. Claim : $R$ also lies $(TXY)$. Proof : Due to the cyclic quadrilaterals of the previous claim we note that, \[\measuredangle YRX = \measuredangle MRX + \measuredangle YRM = \measuredangle MPX + \measuredangle YCM = \measuredangle CTP = \measuredangle YTX \]from which it is clear that $TYRX$ is cyclic as claimed. Now, we are left with establishing tangency. This is a simple angle chase. Let $T'=\overline{AR} \cap (TXY)$. Then, \[\measuredangle TT'R = \measuredangle PXR = \measuredangle PMA = \measuredangle SAM = \measuredangle SAT' \]Thus, $TT' \parallel AS$. This means $\triangle TRT'$ and $\triangle ARS$ are homothetic which implies that circles $(SAR)$ and $(TT'R)$ are tangent at $R$. Thus, we have that the circumcircles of triangle $TXY$ and triangle $APQ$ are tangent to each other as required.
08.06.2024 04:54
Let $G$ be the intersection of $(MPX)$ and $(MQY)$. From Miquel's theorem, $G$ also lies on $(TXY)$. We claim that $G$ is the required tangency point. First, we prove that $G$ lies on $(APQ)$. This follows from angle chasing: \begin{align*} \measuredangle PGQ &= \measuredangle PGM + \measuredangle MGQ \\ &= \measuredangle PXM + \measuredangle MYQ \\ &= \measuredangle PEB + \measuredangle CFQ \\ &= \measuredangle PQA + \measuredangle APQ \\ &= \measuredangle PAQ. \end{align*} Now to prove that $(GXTY)$ and $(APGQ)$ are tangent, it suffices to prove that $\measuredangle QGY = \measuredangle MXG + \measuredangle GXY$, or equivalently that $\measuredangle QMY = \measuredangle MXY$ or that $\triangle MXY \sim \triangle ACB$. We will prove that $\frac{MX}{MY} = \frac{AC}{AB}$. From POP, $$AB \cdot BE = BP \cdot BQ = CQ \cdot CP = AC \cdot CF,$$so $\frac{AC}{AB} = \frac{BE}{CF}$. From similar triangles, $$\frac{MX}{BE} = \frac{MP}{PB} = \frac{MQ}{QC} = \frac{MY}{FC},$$so $\frac{MX}{MY} = \frac{BE}{FC}$. Therefore, $\frac{MX}{MY} = \frac{AC}{AB}$, as desired.
21.09.2024 05:03
The common technique of expressing the tangency point as the intersection of two circles Let $(MPX)$ and $(MQY)$ meet again at $Z$. Then $Z$ lies on $(APQ)$ as $$\angle PZQ=\angle PZM\;+\;\angle QZM=\angle PXM\;+\;\angle QYM=\angle PEB\;+\;\angle QFC=\angle PQA\;+\;\angle QPA=180^{\circ}\:-\:\angle PAQ$$Also $Z$ lies on $(XTY)$ by Miquel's theorem. Notice that as $\frac{MX}{MY}=\frac{BE}{CF}=\frac{AC}{AB}$, the last one following from PoP. As $\angle XMY=\angle BAC$ we have that $\Delta ABC\sim \Delta MYX$. We are done as, $$\angle ZPQ\;+\;\angle ZXY=\angle ZXM\;+\;\angle ZXY=\angle MXY=\angle ACB=\angle QMY=\angle QZY$$
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