Strange angle condition and concyclic points

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544 posts

#1 h Jul 16, 2019, 12:11 PM • 31 Y

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In triangle $ABC$ , point $A_1$ lies on side $BC$ and point $B_1$ lies on side $AC$ . Let $P$ and $Q$ be points on segments $AA_1$ and $BB_1$ , respectively, such that $PQ$ is parallel to $AB$ . Let $P_1$ be a point on line $PB_1$ , such that $B_1$ lies strictly between $P$ and $P_1$ , and $\angle PP_1C=\angle BAC$ . Similarly, let $Q_1$ be the point on line $QA_1$ , such that $A_1$ lies strictly between $Q$ and $Q_1$ , and $\angle CQ_1Q=\angle CBA$ .

Prove that points $P,Q,P_1$ , and $Q_1$ are concyclic.

Proposed by Anton Trygub, Ukraine

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lminsl

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lminsl

#4 h Jul 16, 2019, 12:37 PM • 20 Y

Y by TUGUMEANDREW, Assassino9931, jeteagle, Wizard_32, pavel kozlov, JustKeepRunning, HolyMath, Gumnaami_1945, jnzm02, myh2910, Kagebaka, k.ata.tkm, AllanTian, Illuzion, ImSh95, rayfish, Adventure10, Mango247, ihatemath123, Rounak_iitr

Let $R$ be the intersection of $AA_1$ , $BB_1$ . Let $\Omega$ denote the circumcircle of triangle $ABC$ , and let $A_0=AA_1 \cap \Omega, B_0=BB_1 \cap \Omega$ .

Step 1. The points $P,Q, A_0, B_0$ are concyclic.
Since $PQ \parallel BC$ , $\angle PQR=\angle ABB_0=\angle PA_0B_0$ . The conclusion follows.

Step 2. The points $P,Q, B_0, P_1$ are concyclic, hence $P_1$ and $Q_1$ lies in $\odot(PQA_0B_0)$ .

Note that $\angle B_1B_0C=\angle BAC=\angle B_1P_1C$ , or $(B_0B_1P_1C)$ are concyclic. Hence $\angle B_0P_1P=\angle B_1CB_0=\angle ABB_0=\angle PQB_0,$ or $P, Q, B_0, P_1$ are cyclic. This concludes the proof. $\square$

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mofumofu

179 posts

mofumofu

#5 h Jul 16, 2019, 12:42 PM • 14 Y

Y by Flash_Sloth, zhangruichong, jeteagle, Wizard_32, JustKeepRunning, HolyMath, Not_real_name, ImSh95, ike.chen, Quidditch, CyclicISLscelesTrapezoid, SADAT, Adventure10, Rounak_iitr

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(13cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -7.63900015685, xmax = 14.779588377, ymin = -2.38665690592, ymax = 8.57946862605; /* image dimensions */ /* draw figures */ draw((1.,4.96)--(-0.798898966671,-0.708008219337)); draw((1.,4.96)--(7.68847429913,-0.833514663193)); draw((7.68847429913,-0.833514663193)--(-0.798898966671,-0.708008219337)); draw(circle((5.34391751537,3.21734977311), 4.68043258616), green); draw(circle((0.279066234699,2.06933899636), 2.97920568808), green); draw(circle((1.75406921626,4.18630229154), 4.07948863865), blue); draw(circle((3.47180131167,1.0560363319), 4.62068545199), red); draw((3.88543194857,7.66473878747)--(2.88060830443,-0.762418678059)); draw((-2.28184506812,3.59163983096)--(1.27443840444,-0.738667551996)); draw((-1.14173984034,1.31288664799)--(7.68847429913,-0.833514663193)); draw((-0.798898966671,-0.708008219337)--(5.73321976419,5.08551776613)); draw((0.386107411678,0.343008384736)--(3.00779285736,0.304240392932)); /* dots and labels */ dot((1.,4.96),dotstyle); label("$C$", (1.0680093857,5.04959989258), NE * labelscalefactor); dot((7.68847429913,-0.833514663193),dotstyle); label("$B$", (7.75122752106,-0.739384830301), NE * labelscalefactor); dot((-0.798898966671,-0.708008219337),dotstyle); label("$A$", (-0.736145744742,-0.613878386444), NE * labelscalefactor); dot((3.3230046287,2.94782789959),dotstyle); label("$A_1$", (3.38987859704,3.04149679088), NE * labelscalefactor); dot((-0.22800366956,1.09078053776),dotstyle); label("$B_1$", (-0.171366747388,1.19027674399), NE * labelscalefactor); dot((0.386107411678,0.343008384736),dotstyle); label("$P$", (0.456165471895,0.437238080855), NE * labelscalefactor); dot((3.00779285736,0.304240392932),dotstyle); label("$Q$", (3.0761124874,0.405861469891), NE * labelscalefactor); dot((1.27443840444,-0.738667551996),dotstyle); label("$R$", (1.33471057889,-0.645254997408), NE * labelscalefactor); dot((2.88060830443,-0.762418678059),dotstyle); label("$S$", (2.95060604354,-0.66094330289), NE * labelscalefactor); dot((-2.28184506812,3.59163983096),dotstyle); label("$P_1$", (-2.22653476554,3.68471731564), NE * labelscalefactor); dot((3.88543194857,7.66473878747),dotstyle); label("$Q_1$", (3.9546575944,7.76367674098), NE * labelscalefactor); dot((-1.14173984034,1.31288664799),dotstyle); label("$E$", (-1.08128846535,1.40991302074), NE * labelscalefactor); dot((5.73321976419,5.08551776613),dotstyle); label("$D$", (5.7901893358,5.17510633644), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Let $AA_1,BB_1$ intersect the circumcircle of $\triangle ABC$ again at $D,E$ respectively, and $PB_1,QA_1$ meet $BC$ again at $R,S$ respectively. It is clear that $A,R,C,P_1$ and $B,S,C,Q_1$ are cyclic. By Reim's theorem, we know that $P,Q,D,E$ are cyclic. Taking power of a point at $B_1$ , we get $B_1B\cdot B_1E=B_1A\cdot B_1C=B_1R\cdot B_1P_1$ . Also, since $\triangle B_1PQ\sim \triangle B_1RB$ , we get $B_1B\cdot B_1P=B_1R\cdot B_1Q$ . Combining the two gives us $B_1P_1\cdot B_1P= B_1Q\cdot B_1E$ , whence $P_1\in (PQDE)$ . Similarly $Q_1\in (PQDE)$ are we are done.

This post has been edited 4 times. Last edited by mofumofu, Jul 17, 2019, 12:39 PM

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nguyenhaan2209

111 posts

nguyenhaan2209

#6 h Jul 16, 2019, 12:47 PM • 5 Y

Y by top1csp2020, jeteagle, ImSh95, Pranav1056, Adventure10

$PB_1$ cuts $CB$ , $AB$ at $E$ , $D$ , $QA_1$ cuts $CA$ , $AB$ at $G$ , $F$ . Apply Pappus for ${B_1GA}$ and ${A_1BE}$ we have $B_1B$ - $A_1G$ = $Q$ , $B_1E$ - $A_1A$ = $P$ , $GE$ - $AB$ collinear or parallel but $PQ$ // $AB$ thus $GE//AB$ . It's obvious that $P_1$ , $Q_1$ also lies on $(AGE)$ thus by Reim we have the conclusion!

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Seicchi28

252 posts

Seicchi28

#7 h Jul 16, 2019, 12:48 PM • 6 Y

Y by a_simple_guy, jeteagle, Wizard_32, CrazyMathMan, ImSh95, Adventure10

Nice problem

Let $A_1Q \cap AC = X$ , $B_1P \cap BC = Y$ , and extension of $PQ$ meets $AC,BC$ at $M,N$ respectively.
From the condition of the problem, $MN||AB$ .

Lemma $XY||AB||MN$ .
Proof

Construct the circumcircle $\omega$ of $\bigtriangleup CXY$ . Denote $P'$ as the second intersection of $YP$ and $\omega$ . Since $\angle PP'C = \angle YP'C = \angle YXC = \angle BAC,$ therefore $P'=P_1$ . Similarly, $Q'=Q_1$ . So, $C,X,Y,P_1,Q_1$ are concyclic and
$\angle Q_1P_1P = \angle Q_1P_1Y = \angle Q_1XY = \angle Q_1QP.$
Done. $\blacksquare$

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JustKeepRunning

2958 posts

JustKeepRunning

#9 h Jul 16, 2019, 12:56 PM • 2 Y

Y by ImSh95, Adventure10

mofumofu wrote:

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -7.63900015685, xmax = 14.779588377, ymin = -2.38665690592, ymax = 8.57946862605; /* image dimensions */ /* draw figures */ draw((1.,4.96)--(-0.798898966671,-0.708008219337)); draw((1.,4.96)--(7.68847429913,-0.833514663193)); draw((7.68847429913,-0.833514663193)--(-0.798898966671,-0.708008219337)); draw(circle((5.34391751537,3.21734977311), 4.68043258616), green); draw(circle((0.279066234699,2.06933899636), 2.97920568808), green); draw(circle((1.75406921626,4.18630229154), 4.07948863865), blue); draw(circle((3.47180131167,1.0560363319), 4.62068545199), red); draw((3.88543194857,7.66473878747)--(2.88060830443,-0.762418678059)); draw((-2.28184506812,3.59163983096)--(1.27443840444,-0.738667551996)); draw((-1.14173984034,1.31288664799)--(7.68847429913,-0.833514663193)); draw((-0.798898966671,-0.708008219337)--(5.73321976419,5.08551776613)); draw((0.386107411678,0.343008384736)--(3.00779285736,0.304240392932)); /* dots and labels */ dot((1.,4.96),dotstyle); label("$C$", (1.0680093857,5.04959989258), NE * labelscalefactor); dot((7.68847429913,-0.833514663193),dotstyle); label("$B$", (7.75122752106,-0.739384830301), NE * labelscalefactor); dot((-0.798898966671,-0.708008219337),dotstyle); label("$A$", (-0.736145744742,-0.613878386444), NE * labelscalefactor); dot((3.3230046287,2.94782789959),dotstyle); label("$A_1$", (3.38987859704,3.04149679088), NE * labelscalefactor); dot((-0.22800366956,1.09078053776),dotstyle); label("$B_1$", (-0.171366747388,1.19027674399), NE * labelscalefactor); dot((0.386107411678,0.343008384736),dotstyle); label("$P$", (0.456165471895,0.437238080855), NE * labelscalefactor); dot((3.00779285736,0.304240392932),dotstyle); label("$Q$", (3.0761124874,0.405861469891), NE * labelscalefactor); dot((1.27443840444,-0.738667551996),dotstyle); label("$R$", (1.33471057889,-0.645254997408), NE * labelscalefactor); dot((2.88060830443,-0.762418678059),dotstyle); label("$S$", (2.95060604354,-0.66094330289), NE * labelscalefactor); dot((-2.28184506812,3.59163983096),dotstyle); label("$P_1$", (-2.22653476554,3.68471731564), NE * labelscalefactor); dot((3.88543194857,7.66473878747),dotstyle); label("$Q_1$", (3.9546575944,7.76367674098), NE * labelscalefactor); dot((-1.14173984034,1.31288664799),dotstyle); label("$E$", (-1.08128846535,1.40991302074), NE * labelscalefactor); dot((5.73321976419,5.08551776613),dotstyle); label("$D$", (5.7901893358,5.17510633644), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Let $AA_1,BB_1$ intersect the circumcircle of $\triangle ABC$ again at $D,E$ respectively, and $PB_1,QA_1$ meet $BC$ again at $R,S$ respectively. It is clear that $A,R,C,P_1$ and $B,S,C,Q_1$ are cyclic. By Reim's theorem, we know that $P,Q,D,E$ are cyclic. Taking power of a point at $B_1$ , we get $B_1B\cdot B_1E=B_1A\cdot B_1C=B_1R\cdot B_1P_1$ . Also, since $\triangle B_1PQ\sim \triangle B_1RB$ , we get $B_1B\cdot B_1P=B_1R\cdot B_1Q$ . Combining the two gives us $B_1P_1\cdot B_1P= B_1Q\cdot B_1E$ , whence $P_1\in (PQDE)$ . Similarly $Q_1\in (BQDE)$ are we are done.

How do you import a graph form geogebra?

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Wictro

119 posts

Wictro

#10 h Jul 16, 2019, 12:57 PM • 3 Y

Y by ImSh95, Adventure10, Mango247

Seicchi28 wrote:

Nice problem

Let $A_1Q$ meet $AC$ at $X$ , $B_1P$ meet $BC$ at $Y$ , extension of $PQ$ meets $AC,BC$ at $M,N$ respectively.
From the question, $MN||AB$ .

Lemma

Now by the lemma, create the circumcircle $\omega$ of $\bigtriangleup CXY$ . Denote $P'$ as the second intersection of $YP$ and $\omega$ . Since $\angle PP'C = \angle YP'C = \angle YXC = \angle BAC,$ therefore $P'=P_1$ . Similarly, $Q'=Q_1$ . So, $C,X,Y,P_1,Q_1$ are concyclic and
$\angle Q_1P_1P = \angle Q_1P_1Y = \angle Q_1XY = \angle Q_1QP.$ Done.

Exactly how I did it. Though I proved the lemma with Pappus and then Desargues on perspective triangles.

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TUGUMEANDREW

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TUGUMEANDREW

#13 h Jul 16, 2019, 1:44 PM • 2 Y

Y by ImSh95, Adventure10

Since Problem 2 was geometry and problem 2 is medium level, is there a chance that Problem 4 won't be geometry?

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richrow12

411 posts

richrow12

#14 h Jul 16, 2019, 1:54 PM • 2 Y

Y by ImSh95, Adventure10

Seicchi28 wrote:

Nice problem

Let $A_1Q \cap AC = X$ , $B_1P \cap BC = Y$ , and extension of $PQ$ meets $AC,BC$ at $M,N$ respectively.
From the condition of the problem, $MN||AB$ .

Lemma $XY||AB||MN$ .
Proof

Construct the circumcircle $\omega$ of $\bigtriangleup CXY$ . Denote $P'$ as the second intersection of $YP$ and $\omega$ . Since $\angle PP'C = \angle YP'C = \angle YXC = \angle BAC,$ therefore $P'=P_1$ . Similarly, $Q'=Q_1$ . So, $C,X,Y,P_1,Q_1$ are concyclic and
$\angle Q_1P_1P = \angle Q_1P_1Y = \angle Q_1XY = \angle Q_1QP.$
Done. $\blacksquare$

Just the same but my proof of the lemma as follows:

Consider two lines $(P, B_1, Y)$ and $(Q, A_1, X)$ . By Pappus's theorem points $S=PA_1\cap QB_1$ , $T=PX\cap QY$ and $C=B_1Y\cap A_1X$ are collinear. Now, consider trinagles $APX$ and $BQY$ . We claim that they are perspective. Indeed, $AP\cap BQ = S$ , $AX\cap BY=C$ and $PX\cap QY=T$ are colinear. Hence, by Desargues's theorem lines $AB$ , $PQ$ and $XY$ are concurrent. Since $AB\parallel PQ$ we have $AB\parallel PQ\parallel XY$ .

TUGUMEANDREW wrote:

Since Problem 2 was geometry and problem 2 is medium level, is there a chance that Problem 4 won't be geometry?

Usually (at least in recent years) problems 1,2,4,5 have different topics (ACGN).

This post has been edited 1 time. Last edited by richrow12, Jul 16, 2019, 1:56 PM

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-[]-

97 posts

-[]-

#15 h Jul 16, 2019, 2:01 PM • 5 Y

Y by richrow12, ImSh95, carefully, Adventure10, Mango247

Finally, the first medium geometry problem since the introduction of the new protocol! So surprised that they didn't vote for G1 this time.

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Math-wiz

6107 posts

Math-wiz

#16 h Jul 16, 2019, 2:01 PM • 3 Y

Y by ImSh95, Adventure10, Mango247

TUGUMEANDREW wrote:

Since Problem 2 was geometry and problem 2 is medium level, is there a chance that Problem 4 won't be geometry?

P6 can be a Geo. P4,5 would be NT and Combi

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MS_Kekas

275 posts

MS_Kekas

#17 h Jul 16, 2019, 2:02 PM • 65 Y

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Lol, this is my problem! Didn't expect to see this at P2

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AngleChasingXD

109 posts

AngleChasingXD

#18 h Jul 16, 2019, 2:03 PM • 5 Y

Y by jeteagle, ImSh95, DashTheSup, Adventure10, Mango247

Beautiful problem!!!

My solution is kinda long, especially because I used Moving Points during it.

Let $B_1P_1$ cut $AB$ at $U$ and $Q_1A_1$ cut $AB$ at $V$ . Moreover, let $L$ be the intersection point of lines $B_1P_1$ and $BC$ and $K$ the intersection point of lines $Q_1A_1$ and $AC$ .

Since $\angle PP_1C\equiv \angle BAC$ , we conclude that $AP_1CU$ is cyclic. Similarly, since $\angle QQ_1C\equiv \angle ABC$ , $BQ_1CV$ is cyclic, too. Keep these in mind as (1).

The conclusion is equivalent to $\angle QPP_1\equiv \angle QQ_1P_1$ . However, since $PQ\parallel AB$ , we have $\angle QPP_1\equiv \angle AUP_1$ , so it suffices to prove $\angle AUP_1\equiv \angle QQ_1P_1$ , which is equivalent to $UVP_1Q_1$ being cyclic.

$\textbf{Claim.}$ We have that $KL\parallel AB$ .

$\textbf{Proof:}$ We show this by the Moving Points method. Fix $\Delta ABC$ , points $A_1$ , $B_1$ and animate $P$ on $AA_1$ . Consider $L'$ on $AB$ such that $KL'\parallel AB$ . We prove that the map $L\to P\to Q\to K\to L'~~by~~\overline{BC}\to \overline{AA_1}\to \overline{BB_1}\to \overline{AC}\to \overline{BC}$ is projective, thus it would ve enough to check $L=L'$ for $3$ distinct positions of $P$ .
$\textbf{(1)}$ $L\to P$ by $\overline{BC}\to \overline{AA_1}$ is projective since it is a perspectivity through $B_1$ from $BC$ to $AA_1$ .
$\textbf{(2)}$ $P\to Q$ by $\overline{AA_1}\to \overline{BB_1}$ is projective since it preserves linear motion.
$\textbf{(3)}$ $Q\to K$ by $\overline{BB_1}\to \overline{AC}$ is projective since it is a perspectivity through $A_1$ from $BB_1$ to $AC$ .
$\textbf{(4)}$ $K\to L'$ by $\overline{AC}\to \overline{BC}$ is projective since it preserves linear motion.

Now it suffices to check for $3$ distinct positions of $P$ the desired parallelism.
$\textbf{(1)}$ When $P\in AA_1\cap BB_1$ , we actually have $Q=P$ , $L=B$ and $K=A$ so there is nothing to prove.
$\textbf{(2)}$ When $P=A$ , $Q=B$ so $C=K=L$ , again nothing to prove.
$\textbf{(3)}$ When $P$ is the point at infinity on $\overline{AA_1}$ , $Q$ is the point at infinity on $\overline{BB_1}$ so $B_1L\parallel AA_1$ and $A_1K\parallel BB_1$ . Then apply Thales' Lemma to obtain $CB_1\cdot CA_1=CK\cdot CB$ and $CA_1\cdot CB_1=CL\cdot CA$ , hence $CK\cdot CB=CL\cdot CA$ so indeed $KL\parallel BC$ . $\blacksquare$

Now we are ready to finish. From (1) and Power of a Point Theorem we get $PB_1\cdot B_1U=AB_1\cdot CB_1$ . By our lemma and Thales' Lemma, $B_1K\cdot B_1U=B_1L\cdot B_1A$ . Divide these two last relations in order to get $PB_1\cdot B_1L=CB_1\cdot B_1K$ , which again by Power of a Point leads to $CP_1KL$ being cyclic. In a similar manner one proves the cyclicity of $CQ_1KL$ , so the pentagon $CP_1KLQ_1$ is cyclic. But then $\angle KQ_1P_1\equiv \angle KP_1C$ , and since by (1) we have $\angle KCP_1\equiv \angle AUP_1$ , we arrive at $\angle AUP_1\equiv \angle VQ_1P_1$ , hence $UVQ_1P_1$ is cyclic and done.

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prague123

230 posts

prague123

#19 h Jul 16, 2019, 2:04 PM • 2 Y

Y by ImSh95, Adventure10

-[]- wrote:

Finally, the first medium geometry problem since the introduction of the new protocol! So surprised that they didn't vote for G1 this time.

This probably just means that G1 and G2 got eliminated as known problems, and that G3 contained traces of combinatorial geometry.

This post has been edited 1 time. Last edited by prague123, Jul 16, 2019, 2:13 PM

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Math-wiz

6107 posts

Math-wiz

#20 h Jul 16, 2019, 2:04 PM • 2 Y

Y by ImSh95, Adventure10

MS_Kekas wrote:

Lol, this is my problem! Didn't expect to see this at P2

Wowww. Where are you from?

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GrantStar

821 posts

GrantStar

#153 h Sep 27, 2023, 1:20 AM • 1 Y

Y by OronSH

Define $X=QA_1\cap AB$ , $Y=PB_1\cap AB$ , $S= AQ\cap PB$ , and $T=QX\cap PY.$

Claim: $P_1Q_1XY$ is cyclic
Proof. Notice that by radical axis this is equivalent to showing $T$ is on the radical axis of $(BXCP_1),(AYCQ_1)$ . First, pappus gives $C$ , $S$ , $T$ are collinear. Then, DDIT on the complete quad formed by lines $AB$ , $PQ$ , $AQ$ , $BP$ gives an involution swapping pairs $(TA,TY)$ $(TX,TB)$ and $(TS,T\infty)$ . Projecting onto $AB$ gives an involution swpaping $(A,Y)$ $(B,X)$ $(\infty, F'=AB\cap TS)$ But if the radical axis from above hits $AB$ at $F$ , then $FY\cdot FA=FX\cdot FB$ by powers so the involution swapping $(A,Y)$ and $(B,X)$ is an inversion at $F$ so $(F,\infty)$ is an involute pairing and $F=F'$ . Thus line $CST$ is the desired radical axis. $\blacksquare$

Reim's theorem now finishes.

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Zhaom

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Zhaom

#154 h Sep 27, 2023, 11:24 PM • 1 Y

Y by TestX01

lengths

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IAmTheHazard

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IAmTheHazard

#155 h Dec 6, 2023, 11:02 PM • 1 Y

Y by pokpokben

Let $\overline{AA_1}$ and $\overline{BB_1}$ intersect $(ABC)$ again at $D$ and $E$ respectively, so the angle condition implies $CA_1DQ_1$ and $CB_1EP_1$ are cyclic. By Reim's, $PQDE$ is cyclic as well, since $ABDE$ is. I claim that $PQDQ_1$ is cyclic. Indeed, note that
$\measuredangle DQ_1Q=\measuredangle DAB=\measuredangle DBC=\measuredangle DPQ;$ $PQEP_1$ cyclic follows similarly. Combining these, it follows that $PQDEP_1Q_1$ is cyclic as desired. $\blacksquare$

Remark: wow.

The author of this problem seems to always come up with these really clean, fresh geo problems (JMO 2023/6 is another example)

Remark (motivation): The construction of $D$ and $E$ is motivated by trying to make sense of the strange angle condition: I personally noticed that as $\overline{PQ}$ varies, $P_1$ and $Q_1$ lie on fixed circles, which naturally led me to figure out what these circles were by introducing these points. Then drawing an accurate diagram makes it clear how to proceed.

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asdf334

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asdf334

#156 h Feb 6, 2024, 2:48 PM • 1 Y

Y by CT17

pretty straightforward

$U=PQ\cap BC$
$V=PQ\cap AC$
$R=PB_1\cap QA_1$
$X$ is the center of homothety sending $PQ$ to $UV$
$S=AQ\cap BP$

Notice that $PVP_1C$ and $QUQ_1C$ are cyclic.

By Monge, we have $C,S,X$ collinear.
By Pappus on $A_1PA$ and $B_1QB$ we have that $R,C,S$ are collinear.

Notice that $CX$ is the radical axis of $(PVC)$ and $(QUC)$ .

Thus $R$ lies on the radical axis. Put another way, $RP\cdot RP_1=RQ\cdot RQ_1$ . Thus $P,Q,P_1,Q_1$ are concyclic. Done.

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Ywgh1

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Ywgh1

#157 h Mar 29, 2024, 6:30 AM

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A Plain figure with no additions
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#158 h Mar 29, 2024, 10:40 AM • 1 Y

Y by Rounak_iitr

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(14.473791467295925cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(12); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -3.722705371673187, xmax = 36.75108609562274, ymin = -7.619324939355466, ymax = 10.565718542968266; /* image dimensions */ pen qqttzz = rgb(0.,0.2,0.6); pen qqzzcc = rgb(0.,0.6,0.8); pen qqttqq = rgb(0.,0.2,0.); pen qqqqcc = rgb(0.,0.,0.8); pen qqwwtt = rgb(0.,0.4,0.2); pen wwzzff = rgb(0.4,0.6,1.); pen cqcqcq = rgb(0.7529411764705882,0.7529411764705882,0.7529411764705882); pen qqttcc = rgb(0.,0.2,0.8); pen eqeqeq = rgb(0.8784313725490196,0.8784313725490196,0.8784313725490196); pen zzccff = rgb(0.6,0.8,1.); draw(circle((8.506812245675057,2.9604181885520444), 4.140885015934622), linewidth(1.) + linetype("2 2") + wwzzff); /* draw figures */ draw(circle((10.,0.), 5.604319762468949), linewidth(1.) + qqttzz); draw((4.583966606974815,-1.4404798803302028)--(15.506441446309776,-1.0428339265491575), linewidth(1.) + qqzzcc); draw((15.506441446309776,-1.0428339265491575)--(8.202344722423005,5.30818570728263), linewidth(1.) + qqttqq); draw((4.583966606974815,-1.4404798803302028)--(8.202344722423005,5.30818570728263), linewidth(1.) + qqttqq); draw((6.008784688561027,1.2169595715492947)--(15.506441446309776,-1.0428339265491575), linewidth(1.) + qqqqcc); draw((4.583966606974815,-1.4404798803302028)--(10.151186194117859,3.6136393563995135), linewidth(1.) + qqqqcc); draw(circle((13.426534840451424,3.940743903404536), 5.4001906888830495), linewidth(1.) + qqwwtt); draw((9.024571191585899,7.068806484358715)--(8.202344722423005,5.30818570728263), linewidth(1.)); draw(circle((5.045453049041576,2.6564398586017166), 4.122829257131737), linewidth(1.) + qqwwtt); draw((6.426784885318526,6.540978884219707)--(8.202344722423005,5.30818570728263), linewidth(1.)); draw((11.714520493171745,-1.1808834113273448)--(9.024571191585899,7.068806484358715), linewidth(1.) + qqttzz); draw((5.803629606185249,-1.3960765684666794)--(6.426784885318526,6.540978884219707), linewidth(1.) + qqttzz); draw((5.893480461757436,-0.2516567953731126)--(11.346795115678864,-0.053122256385905266), linewidth(1.) + qqzzcc); draw((5.803629606185249,-1.3960765684666794)--(8.851538800117314,1.071689929142913), linewidth(1.) + cqcqcq); draw((6.008784688561027,1.2169595715492947)--(4.613952109782934,1.5488344405675873), linewidth(1.) + qqttcc); draw((10.151186194117859,3.6136393563995135)--(11.9558341730438,5.251962746207756), linewidth(1.) + qqttcc); draw((11.9558341730438,5.251962746207756)--(9.024571191585899,7.068806484358715), linewidth(1.) + cqcqcq); draw((4.613952109782934,1.5488344405675873)--(6.426784885318526,6.540978884219707), linewidth(1.) + cqcqcq); draw((4.613952109782934,1.5488344405675873)--(11.9558341730438,5.251962746207756), linewidth(1.) + eqeqeq); draw((11.346795115678864,-0.053122256385905266)--(11.9558341730438,5.251962746207756), linewidth(1.) + eqeqeq); draw((5.893480461757436,-0.2516567953731126)--(4.613952109782934,1.5488344405675873), linewidth(1.) + eqeqeq); draw((11.346795115678864,-0.053122256385905266)--(6.426784885318526,6.540978884219707), linewidth(1.) + cqcqcq); draw((4.613952109782934,1.5488344405675873)--(8.202344722423005,5.30818570728263), linewidth(1.) + qqttzz); draw(circle((5.910994086898991,3.903056024991948), 2.6878759567678383), linewidth(1.) + zzccff); draw((4.583966606974815,-1.4404798803302028)--(4.613952109782934,1.5488344405675873), linewidth(1.) + qqttzz); draw((4.613952109782934,1.5488344405675873)--(5.803629606185249,-1.3960765684666794), linewidth(1.) + qqttzz); draw(circle((10.082420780384243,5.502476635340945), 1.890088608622575), linewidth(1.) + zzccff); /* dots and labels */ dot((15.506441446309776,-1.0428339265491575),dotstyle); label("$B$", (15.615666827081196,-0.779817434941671), NE * labelscalefactor); dot((4.583966606974815,-1.4404798803302028),dotstyle); label("$A$", (4.699276418075602,-1.1821414057895412), NE * labelscalefactor); dot((8.202344722423005,5.30818570728263),dotstyle); label("$C$", (8.320192155706449,5.576901304454677), NE * labelscalefactor); 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dot((6.426784885318526,6.540978884219707),linewidth(4.pt) + dotstyle); label("$P'$", (6.523145085919288,6.757051618941763), NE * labelscalefactor); dot((8.851538800117314,1.071689929142913),linewidth(4.pt) + dotstyle); label("$N$", (8.963910509063044,1.285445615410729), NE * labelscalefactor); dot((4.613952109782934,1.5488344405675873),linewidth(4.pt) + dotstyle); label("$E$", (4.726098016132126,1.7682343804281733), NE * labelscalefactor); dot((11.9558341730438,5.251962746207756),linewidth(4.pt) + dotstyle); label("$F$", (12.075215883619922,5.469614912228579), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Sketch of solution:

1) Let $AA_1$ meets circle $(ABC)$ at $F$ , and let $BB_1$ meets circle $(ABC)$ at $E$ .

2) By Reim theorem you have that $EFPQ$ cyclic.

3) By some angle chase you show $EB_1P'C$ cyclic, and similarly $FA_1Q'C$ .

By simple angle chase you will se that $EPQP'$ cyclic, and similarly $FPQQ'$ as desired.

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Mr.Sharkman

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Mr.Sharkman

#160 h Apr 20, 2024, 7:04 PM

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Solution

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ihatemath123

3449 posts

ihatemath123

#165 h Aug 27, 2024, 4:17 AM • 2 Y

Y by Rounak_iitr, OronSH

WHAT IN THE WORLD IS THIS :huuh:

Define $Q_2$ and $P_2$ as the intersection of side $AB$ with lines $QQ_1$ and $PP_1$ respectively, and define $X$ as the intersection of lines $QQ_1$ and $PP_1$ .

Claim: The problem is affine.
Proof: The angle conditions imply that $CQ_1 BQ_2$ and $CP_1 A P_2$ are cyclic – it suffices to show that $X$ lies on the radical axis of the two circles. Let $Y$ be the intersection of lines $CX$ and $AB$ – it is equivalent to the original problem statement to show that $YQ_2 \cdot YB = YP_2 \cdot YA$ . Now, points $P_1$ and $Q_1$ don't matter anymore; the definition of all points is affine, and what we're asked to prove is affine.

Now we will solve the original problem (adding points $P_1$ and $Q_1$ back) with the additional assumption that $\overline{AA_1}$ and $\overline{BB_1}$ are altitudes. Since $BA_1 B_1A$ is cyclic, by Reim's theorem, it follows that $A_1 QPB_1$ too is cyclic.

Claim: We have $\overline{P_1Q_1} \parallel \overline{A_1 B_1}$ .
Proof: Let $H$ be the orthocenter of $\triangle A_1 B_1 C$ . Since $\angle CP_1 B_1 = \angle A = 180^{\circ} - \angle CHB_1$ , it follows that $P_1$ lies on $(B_1 H C)$ . Similarly, $Q_1$ lies on the congruent circle $(A_1 H C)$ . Angle chasing shows that $\angle Q_1 A_1 C = \angle P_1 B_1 C$ – the arcs are in congruent circles, so $CP_1 = CQ_1$ and by symmetry, the claim follows.

Since $PQA_1 B_1$ is cyclic, by Reim once more, $PQP_1 Q_1$ is cyclic.

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Rounak_iitr

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Rounak_iitr

#168 h Oct 9, 2024, 6:52 AM • 1 Y

Y by radian_51

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(14.473791467295925cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.4) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -33.90294294377877, xmax = 61.659999263181504, ymin = -41.602313770778665, ymax = 14.038772683926569; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); pen ffdxqq = rgb(1,0.8431372549019608,0); pen ffqqtt = rgb(1,0,0.2); pen ttttff = rgb(0.2,0.2,1); pen ttffqq = rgb(0.2,1,0); pen ffwwqq = rgb(1,0.4,0); pen qqzzff = rgb(0,0.6,1); pen ffqqff = rgb(1,0,1); pen xfqqff = rgb(0.4980392156862745,0,1); draw((-5,10)--(-16.24999735332854,-20.518583701619058)--(20.270971501626907,-20.662633307357392)--cycle, linewidth(1.) + zzttqq); /* draw figures */ draw((-5,10)--(-16.24999735332854,-20.518583701619058), linewidth(1.) + ffdxqq); draw((-16.24999735332854,-20.518583701619058)--(20.270971501626907,-20.662633307357392), linewidth(1.) + ffdxqq); draw((20.270971501626907,-20.662633307357392)--(-5,10), linewidth(1.) + ffdxqq); draw(circle((2.0525255237098183,-9.932579284308634), 21.143458401780634), linewidth(1.) + ffqqtt); draw((-5,10)--(-4.913765920945284,-29.895460256151182), linewidth(1.) + ttttff); draw((-16.24999735332854,-20.518583701619058)--(16.300614749971015,5.689129539899033), linewidth(1.) + ttffqq); draw((xmin, 2.712763633903391*xmin + 12.457690285813076)--(xmax, 2.712763633903391*xmax + 12.457690285813076), linewidth(0.4) + linetype("0 3 4 3") + white); /* line */ draw((-4.976134034432313,-1.041385160223868)--(-10.428035306194664,-15.831104665892425), linewidth(1.) + ffwwqq); draw((-4.976134034432313,-1.041385160223868)--(31.423730105365635,-7.674407099508636), linewidth(1.) + ttffqq); draw((31.423730105365635,-7.674407099508636)--(16.300614749971015,5.689129539899033), linewidth(1.) + qqzzff); draw((31.423730105365635,-7.674407099508636)--(20.270971501626907,-20.662633307357392), linewidth(1.) + qqzzff); draw((-10.428035306194664,-15.831104665892425)--(14.190417657765865,-36.92397129981129), linewidth(1.) + ffqqtt); draw((14.190417657765865,-36.92397129981129)--(20.270971501626907,-20.662633307357392), linewidth(1.) + ffqqtt); draw((16.300614749971015,5.689129539899033)--(20.270971501626907,-20.662633307357392), linewidth(1.) + ttffqq); draw((-4.913765920945284,-29.895460256151182)--(20.270971501626907,-20.662633307357392), linewidth(1.) + red); draw(circle((18.120084891832406,-7.490787774969996), 13.30491232680282), linewidth(1.) + ffwwqq); draw(circle((7.6504156543471025,-25.202159646033778), 13.412148570376223), linewidth(1.) + ffqqff); draw(circle((11.275454012939369,-15.42082374137463), 21.699824146074572), linewidth(1.) + linetype("4 4") + blue); draw((-5,10)--(16.300614749971015,5.689129539899033), linewidth(1.) + xfqqff); /* dots and labels */ dot((-5,10),dotstyle); label("$A$", (-5.708132318748717,10.857499803668759), NE * labelscalefactor); dot((-16.24999735332854,-20.518583701619058),dotstyle); label("$B$", (-17.7470669440381,-21.329496396586734), NE * labelscalefactor); dot((20.270971501626907,-20.662633307357392),dotstyle); label("$C$", (21.176742414410473,-21.89089749310282), NE * labelscalefactor); dot((-4.913765920945284,-29.895460256151182),dotstyle); label("$S$", (-4.647708025329445,-29.251489647424812), NE * labelscalefactor); dot((16.300614749971015,5.689129539899033),dotstyle); label("$T$", (16.560777843055995,6.303913131927187), NE * labelscalefactor); dot((-4.933937573159862,-20.56321762018564),dotstyle); label("$A_1$", (-6.706178712555091,-21.641385894651226), NE * labelscalefactor); dot((5.632094760585414,-2.9004942652053733),dotstyle); label("$B_1$", (5.0208664146698,-1.3061906208463983), NE * labelscalefactor); dot((-4.976134034432313,-1.041385160223868),dotstyle); label("$P$", (-6.331911314877701,-0.9319232231690088), NE * labelscalefactor); label("$h$", (0.8415471406056099,12.541703093217011), NE * labelscalefactor,white); dot((-10.428035306194664,-15.831104665892425),dotstyle); label("$Q$", (-11.821166480812755,-15.902619130264586), NE * labelscalefactor); dot((31.423730105365635,-7.674407099508636),dotstyle); label("$P_1$", (31.656229549377397,-7.044957385233036), NE * labelscalefactor); dot((14.190417657765865,-36.92397129981129),dotstyle); label("$Q_1$", (14.190417657765856,-38.35866299090795), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Let, $T=BB_1\cap (ABC)$ and $S=AA_1\cap (ABC).$

$\color{red}\textbf{Claim:-}$ $P_1,T,B_1,C$ and $Q_1,C,A,S$ are concyclic points.

$\color{blue}\textbf{Proof:-}$ Since, the points $A,B,C,T$ are cyclic and $PQ||AB$ we get some angle conditions $\angle TCB_1=\angle ABT=\angle PQB\implies\angle BAC=\angle BTC=\angle PP_1C$ Therefore the points $P_1,T,B_1,C$ are concyclic. Now From second Intersection of points we get, $\angle CBA=\angle ASC=\angle A_1SC=\angle CQ_1A_1$ Therefore the points $Q_1,C,A,S$ are concyclic also. Now By Simple Angle chase we get, $\angle PST=\angle ABT=\angle PQT$ Therefore the points $P,Q,S,Q_1,P_1,T$ are Concyclic points. :love:

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SimplisticFormulas

125 posts

SimplisticFormulas

#169 h Dec 10, 2024, 11:06 AM • 1 Y

Y by radian_51

One small construction and it all comes crashing down.

Let $AA_{1}$ meet $\odot (ABC)$ again in $X$ and $BB_{1}$ meet $\odot (ABC)$ again in $Y$ .

CLAIM 1: $Q,X,Q_{1},C$ and $P,Y,P_{1},C$ are concyclic
PROOF: Indeed, note that $\angle QQ_{1}C=\angle ABC=\angle QXC$ . Similarly, $\angle PP_{1}C=\angle BAC=\angle PYC$

CLAIM 2: $P,Q,X,Q_{1},P_{1},Y$ are concyclic
PROOF: Note that $\angle QQ_{1}X=\angle QCX=\angle BCX=\angle BAX$ $=\angle QPX=\angle BYX=\angle QYX$
and $\angle PP_{1}Y=\angle PCY=\angle ACY=\angle ABY$ $=\angle PQY=\angle AXY=\angle PXY$ , and we are done. $\blacksquare$

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cursed_tangent1434

650 posts

cursed_tangent1434

#170 h Mar 7, 2025, 5:44 PM • 2 Y

Y by radian_51, reni_wee

Let $X$ and $Y$ denote the intersections of lines $A_1Q$ and $B_1P$ with line $BC$ respectively. We can start off by observing the following.

Claim : Quadrilaterals $CPNP_1$ and $CQMQ_1$ are cyclic.

Proof : First note that the given angle condition implies,
$\measuredangle XBC = \measuredangle ABC = \measuredangle QQ_1C = \measuredangle XQ_1C$ which implies that $CQ_1BX$ is cyclic. A similar argument shows that $CP_1AX$ is also cyclic. But then, by Reim's Theorem since $MN \parallel BC$ it follows that quadrilaterals $CQMQ_1$ and $CPNP_1$ are indeed cyclic, as desired.

Thus, letting $K$ be the intersection of segments $PB_1$ and $QC_1$ it suffices to show that $K$ lies on the radical axis of circles $(CPN)$ and $(CQM)$ .

We define the points $R=BQ \cap AP$ and $T=BP \cap AQ$ . Applying Dual Desargue's involution theorem on self-intersecting quadrilaterals $PBQA$ and $PA_1QB_1$ at $C$ , it follows that the pairs $(CP,CQ);(CA,CB)$ and $(CR,CT)$ and also the pairs $(CP,CQ);(CB_1,CA_1)$ and $(CR,CK)$ . Since involutions are uniquely determined by two pairings this implies that lines $CK$ and $CT$ must coincide. Thus, points $C$ , $K$ and $T$ are collinear.

With this observation in hand, we now simply desire the following claim which we shall handle using the method of moving points.

Claim : Point $T$ lies on the radical axis of circles $(CPN)$ and $(CQM)$ .

Proof : Fix the line $MN$ and denote it by $\ell$ . Let $Q$ be an arbitrary fixed point on $\ell \parallel BC$ . Denote the fixed circle $(CQM)$ by $\omega$ . We animate $P$ on line $\ell$ . First note that the map,
$\ell \to \omega \to \mathcal{P}_C \to \overline{AQ} \text{ by } P \to S \to \overline{CS} \to T'$ is projective as it is a composition of perspectivities and second intersection of two circles with one intersection a fixed point. Also the map,
$\ell \to \mathcal{P}_B \to \overline{AQ} \text{ by } P \to \overline{PB} \to T$ is projective since this is simply a composition of perspectivities. Now, it suffices to check that these maps coincide for three specific cases.

If $P=M$ (similarly $P=N$ ) then $S=M$ and thus $T' = AQ \cap CS=AQ \cap MB = T$ .
If $P=Q$ then clearly points $T$ and $T'$ are also both just $P$ .

Having checked it for three cases, the claim is proved.

Thus, point $T$ lies on the radical axis of circles $(CPN)$ and $(CQM)$ and so does $C$ . We showed that $K$ lies on $CT$ so we are done.

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cj13609517288

1926 posts

cj13609517288

#171 h Apr 4, 2025, 8:20 PM • 1 Y

Y by radian_51

what being stubborn feels like:

Define $P_2$ and $Q_2$ on line $AB$ such that $AP_1CP_2$ and $BQ_1CQ_2$ are cyclic. Then $P$ lies in $P_1P_2$ and $Q$ lies on $Q_1Q_2$ .

Now define $X=AA_1\cap BB_1$ , $R=PQ_2\cap P_2Q$ , and $Y=PP_2\cap QQ_2$ .

By Reim's or whatever, we want to prove that $P_1P_2Q_2Q_1$ is cyclic. By converse radical center, this is equivalent to proving that $Y$ lies on the radical axis of $(AP_1CP_2)$ and $(BQ_1CQ_2)$ , say $\ell$ .

Let $Z=\ell\cap AB$ . Then $(ZP_2)(ZA)=(ZQ_2)(ZB)$ . In particular, there exists an involution on line $AB$ swapping $(A,P_2),(B,Q_2),(Z,\infty_{AB})$ .

Use DDIT on self-intersecting quadrilateral $PQP_2Q_2$ through $X$ . We end up getting that $Z$ lies on $XR$ .

Recall that we want to prove $C,Y,Z$ collinear. Let's use barycentric coordinates of triangle $XPQ$ .

I did this barybash successfully on paper, so I will provide a summary here. We have
$\begin{align*} A &= (-p,1+p,0) \\ B &= (-p,0,1+p) \\ P_2 &= (-p,s,1+p-s) \\ Q_2 &= (-p,t,1+p-t) \\ R &= (-p:s:1+p-t) \\ Z &= (-p(1+p+s-t):s(1+p):(1+p-t)(1+p)) \\ Y &= (-p:t:1+p-s) \\ A_1 &= (-p:t:0) \\ B_1 &= (-p:0:1+p-s). \end{align*}$ Now $C=AB_1\cap A_1B$ . We have
$\begin{align*} \frac{x}{p}+\frac{y}{1+p}+\frac{z}{1+p-s} &= 0 \\ \frac{x}{p}+\frac{y}{t}+\frac{z}{1+p} &= 0. \end{align*}$ Let $x=-p$ , then after some algebra we get
$C=(-p(t+tp-ts-(1+p)^2:-st(1+p):-(1+p-s)(1+p-t)(1+p)).$ Thus it suffices to verify that the following determinant is zero:
$\begin{vmatrix} -p & t & 1+p-s \\ -p(1+p+s-t) & s(1+p) & (1+p-t)(1+p) \\ -p(t+tp-ts-(1+p)^2) & -st(1+p) & -(1+p-s)(1+p-t)(1+p). \end{vmatrix}$ We will do a few row/column operations. Again, I did all of this out on paper but I will only provide a summary here.

Divide column 1 by $-p$ .
Add $1+p-s$ times row 2 to row 3. (The bottom-left element will work out to be $-s^2$ .)
Divide row 3 by $s$ .
Subtract column 3 from column 2, then negate column 2.
Divide column 2 by $1+p-s-t$ .
Add row 3 to row 2.
Subtract column 2 from column 1.

Now the main diagonal should have all zeroes. We need to verify
$(1)(1+p-t)(1+p)(-s+1+p)+(1+p-s)(1+p-t)(-1-p)=0$ which is true! $\blacksquare$

Remark. I just realized that my initial attempt at a homography does in fact work and I just messed up. Here is how the homography would have worked in place of the barybash:

We drop the $PQ\parallel AB$ condition as it is no longer necessary. Take a homography sending $CB_1XA_1$ to a square. Then line $AB$ is the line at infinity, so $PY\parallel RQ$ and $QY\parallel RP$ . Therefore, $RPYQ$ is now a parallelogram, and we want to prove that $CY$ and $XR$ are parallel. So here's the rephrased problem:

Let $CB_1XA_1$ be a square and let $P$ and $Q$ lie on $A_1X$ and $B_1X$ , respectively. Let $Y=A_1Q\cap B_1P$ , and let $R$ be the point such that $YPRQ$ is a parallelogram. Prove that $CY$ and $XR$ are parallel.

Let $X'$ be the point such that $XPX'Q$ is a rectangle. Then it suffices to show that $X'$ lies on $CY$ . So the problem actually reduces to the following:

Let $ABCDEFGHI$ be a $3\times 3$ grid of points named in reading order, with perpendicular but not necessarily equally-spaced gridlines. Prove that $BD,CG,FH$ concur.

At this point, MMP on where $B$ is on $AC$ (therefore $E$ and $H$ also move linearly) just works (we need to check two cases since $BD\cap FH$ has degree one by Zack's lemma, and $B=A,B=C$ just work).

This post has been edited 3 times. Last edited by cj13609517288, Apr 4, 2025, 10:00 PM

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reni_wee

63 posts

reni_wee

#172 h May 3, 2025, 6:21 PM

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Let $\omega$ be the circumcircle of $\triangle ABC$ , $A_{2}$ be the intersection of $\omega$ and $AA_{1}$ , $B_{2}$ be the intersection of $\omega$ and $BB_{1}$ .
Claim: $P_{1}CB_{1}B_{2}$ and $Q_{1}CA_{1}A_{2}$ are cyclic quadrilaterals.
Proof
$\angle CP_{1}B_{1} = \angle CAB =\angle CB_{2}B_{1}$ Hence $P_{1}CB_{1}B_{2}$ is cyclic. Analogously $Q_{1}CA_{1}A_{2}$ is cyclic as well.

Claim: $Q_{1}B_{2}PQA_{2}$ and $P_{1}A_{2}QPB_{2}$ are cyclic quadrilaterals.
Proof:-
$\angle A_{1}PQ = \angle A_{1}AB = \angle A_{2}B_{2}B = \angle A_2CA_1 = \angle A_2Q_1A_1$ Hence $Q_{1}B_{2}PQA_{2}$ is cyclic. Analogously $P_{1}A_{2}QPB_{2}$ is cyclic as well.

Therefore $P_1$ and $Q_1$ lie on the circumcircle of $B_2A_2QP$ which implies that $P_1Q_1QP$ is cyclic.

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Giant_PT

50 posts

Giant_PT

#173 h May 15, 2025, 1:29 AM

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My solution is really bad, but can anyone check if this actually works? I am new to bary

.
Let $R=AA_1\cap BB_1$ , $D=AB\cap PB_1$ , $E=AB\cap QA_1$ , $S=BC\cap PB_1$ , and $T=AC\cap QA_1$ .
Claim 1: $AP_1CD$ and $BQ_1CE$ are concyclic
Through angle chasing,
$\measuredangle DAC = \measuredangle BAC = \measuredangle PP_1C = \measuredangle DP_1C,$ which implies that $AP_1CD$ is concyclic. We can similarly prove that $BQ_1CE$ is concyclic, thus finishing the claim $\square$ .

Claim 2: $AB\parallel TS$
Set $\triangle ABC$ as the reference triangle and let $R=(p,q,r)$ where $p+q+r=1$ . Also, let line $PQ$ be $z=k$ where $z$ is homogenized. Then clearly, $A_1=(0:q:r)$ , and $B_1=(p:0:r).$ Since line $AA_1$ and $BB_1$ are $\frac{q}{r}=\frac{y}{z}$ and $\frac{p}{r}=\frac{x}{z}$ respectively, we can now calculate $P$ and $Q$ to be $(1-\frac{qk}{r}-k,\frac{qk}{r},k)$ and $(\frac{pk}{r},1-\frac{pk}{r}-k,k)$ respectively. Now calculating for line $QA_1$ , we have,
$\begin{vmatrix} x & y & z \\ \frac{pk}{r} & 1-\frac{pk}{r}-k & k \\ 0 & q & r \end{vmatrix}=0 \implies x(r-pk-rk-qk)-y(pk)+z(\frac{pqk}{r})=0 \implies x(r-k)-y(pk)+z(\frac{pqk}{r})=0.$ Now we can find that $T = (\frac{pqk}{r}: 0: k-r)$ . Exactly the same procedure on $PB_1$ yields $S = (0:\frac{pqk}{r}:k-r).$ This clearly shows that $AB\parallel TS,$ thus finishing the claim $\square$ .

Claim 3: $CP_1TSQ_1$ is concyclic.
By angle chasing,
$\measuredangle TSP_1 = \measuredangle ADP_1 = \measuredangle ACP_1 = \measuredangle TCP_1,$ which proves that $CP_1TS$ is concyclic. We can similarly show that $TSQ_1C$ is concyclic, and since they share 3 points, we see that these circles are in fact the same, which finishes the claim $\square$ .

Claim 3 shows that lines $TS$ and $P_1Q_1$ are antiparallel about the angle formed between lines $PB_1$ and $QA_1$ . Since we know that $PQ\parallel AB\parallel TS$ , we see that $PQ$ is antiparallel to $P_1Q_1$ as well, which finishes the proof $\square$ .

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Aiden-1089

302 posts

Aiden-1089

#174 h May 27, 2025, 8:27 PM

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Let $D=PP_1 \cap AB$ and $E=QQ_1 \cap AB$ . Clearly $(ADCP_1)$ and $(BECQ_1)$ are concyclic.
Let $X=AC \cap QQ_1, Y=BC \cap PP_1, Z=PP_1 \cap QQ_1$ .
Now $(Y,D;P,Z) \stackrel{A_1}{=} (B,D;A,E) = (A,E;B,D) \stackrel{B_1}{=} (X,E;Q,Z)$ . But $PQ // AB$ , so we get $XY // AB$ .

Note that by a negative homothety at $C$ we get that $(CXY)$ is tangent to $(ABC)$ .
Also note that $\angle ACP_1 = \angle ADY = \angle ABC + \angle P_1YC$ , so $(CP_1Y)$ is tangent to $(ABC)$ . Similarly $(CQ_1X)$ is tangent to $(ABC)$ , so $(CP_1YXQ_1)$ concyclic.

Since $(P_1Q_1XY)$ concyclic and $PQ // XY$ , we get that $(P_1PQQ_1)$ concyclic by Reim's theorem. $\square$

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