Concurrency in Parallelogram

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amuthup

779 posts

amuthup

#1 h Jul 12, 2022, 12:25 PM • 9 Y

Y by itslumi, YesToDay, Ahmed102, GeNuAlCo_pi, Mango247, deplasmanyollari, ItsBesi, ehuseyinyigit, cubres

Let $ABCD$ be a parallelogram with $AC=BC.$ A point $P$ is chosen on the extension of ray $AB$ past $B.$ The circumcircle of $ACD$ meets the segment $PD$ again at $Q.$ The circumcircle of triangle $APQ$ meets the segment $PC$ at $R.$ Prove that lines $CD,AQ,BR$ are concurrent.

This post has been edited 1 time. Last edited by amuthup, Jul 15, 2022, 4:57 PM

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Assassino9931

1408 posts

Assassino9931

#3 h Jul 12, 2022, 12:39 PM • 6 Y

Y by PRMOisTheHardestExam, GeNuAlCo_pi, Philomath_314, Stuffybear, lomta, cubres

It turns out that only angle chasing is enough (no similarities, no radical axis etc., but only cyclic quads), but I'll let somebody else post that.

Let $AQ \cap CD = T$ , with the aim to show that $B$ , $R$ and $T$ are collinear. Observe that $\angle QRC = \angle QAP = \angle ATC = \angle QTC$ and so $CTRQ$ is cyclic, implying $\angle CRT = \angle CQT = \angle ADC = \angle ABC$ . So if we show that $\angle BRC = \angle PBC$ we shall be done. The latter is equivalent to $\triangle PBC \sim \triangle BRC$ and hence to $BC^2 = CR \cdot CP$ . But since $AC = BC$ , we reduce to $AC^2 = CR \cdot CP$ and hence to $\triangle ACR \sim \triangle PCA$ . Now it remains to observe that $\angle ACR = \angle ACP$ and $\angle CAR = \angle CAQ + \angle QAR = \angle CDQ + \angle QPR = \angle APD + \angle DPC = \angle APC$ and we are done.

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v_Enhance

6882 posts

v_Enhance

#5 h Jul 12, 2022, 12:59 PM • 7 Y

Y by HamstPan38825, PHSH, I.owais, shujingsheng, GeNuAlCo_pi, Rounak_iitr, cubres

Define $X = \overline{AQ} \cap \overline{CD}$ and let $R' = \overline{BX} \cap \overline{PC}$ . We wish to show $APR'Q$ is cyclic.
Let $M$ be the center of the parallelogram.
Claim: $M$ , $Q$ , $R'$ are collinear.
Proof. By Pappus theorem on $\overline{ABP}$ and $\overline{DCX}$ . $\blacksquare$

$[asy] size(12cm); pair A = dir(90); pair D = dir(240); pair C = A*A/D; pair Q = dir(20); pair B = A+C-D; pair P = extension(D, Q, A, B); pair M = midpoint(A--C); pair X = extension(A, Q, C, D); pair Rp = extension(P, C, B, X); filldraw(unitcircle, invisible, blue); filldraw(A--B--C--D--cycle, invisible, red); draw(A--C, red); draw(B--D, red); draw(C--X, red); draw(B--P, red); draw(A--X, lightblue); draw(B--X, lightblue); draw(C--P, lightblue); draw(M--Rp, deepgreen); draw(D--P, lightblue); draw(Q--C, lightblue); pair Z = 2*A-P; pair W = -D+2*foot(origin, D, Z); draw(Q--W, deepgreen); draw(D--Z--A, orange); dot("$A$", A, dir(A)); dot("$D$", D, dir(D)); dot("$C$", C, dir(C)); dot("$Q$", Q, dir(72)); dot("$B$", B, dir(B)); dot("$P$", P, dir(P)); dot("$M$", M, dir(190)); dot("$X$", X, dir(X)); dot("$R'$", Rp, dir(Rp)); dot("$Z$", Z, dir(Z)); dot("$W$", W, dir(W)); /* TSQ Source: !size(12cm); A = dir 90 D = dir 240 C = A*A/D Q = dir 20 R72 B = A+C-D P = extension D Q A B M = midpoint A--C R190 X = extension A Q C D R' = extension P C B X unitcircle 0.1 lightcyan / blue A--B--C--D--cycle 0.1 lightred / red A--C red B--D red C--X red B--P red A--X lightblue B--X lightblue C--P lightblue M--Rp deepgreen D--P lightblue Q--C lightblue Z = 2*A-P W = -D+2*foot origin D Z Q--W deepgreen D--Z--A orange */ [/asy]$
Let $W$ be the point for which $AQCW$ is harmonic. Let $Z = \overline{DW} \cap \overline{AB}$ .
Claim: $\triangle ZAD \cong \triangle PAC$ .
Proof. Projecting from $D$ we get $-1 = (AC;QW) = (A\infty;PZ)$ so $A$ is the midpoint of $\overline{PZ}$ . Since $AD=AC$ and $\measuredangle ZAD = \measuredangle CAP$ , we're done. $\blacksquare$
To finish, $\begin{align*} \measuredangle AQR' &= \measuredangle AQM = \measuredangle WQC = \measuredangle WDC = \measuredangle WDA + \measuredangle ADC \\ &= \measuredangle ACP + \measuredangle DCA = \measuredangle DCP = \measuredangle APR'. \end{align*}$
Remark: [Motivation] The Pappus step allows one to erase many points in the picture. After this step, rewrite the conclusion as $\measuredangle AQM = \measuredangle APC$ ; then the points $B$ , $R'$ , and $X$ may now be deleted.

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VicKmath7

1391 posts

VicKmath7

#6 h Jul 12, 2022, 1:04 PM • 3 Y

Y by GeNuAlCo_pi, ehuseyinyigit, cubres

Angle chase sol sketch: Let $AQ$ and $BR$ meet at $X$ , I will prove that $CX||AB$ . This is equivalent to $CQRX$ being cyclic. Note that $\angle ARC= \angle AQD= \angle ACD= \angle ABC$ , so $ABRC$ is cyclic. Hence $\angle CRX= \angle BAC = \angle ADC = \angle CQX$ so we are done.

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Kamran011

678 posts

Kamran011

#7 h Jul 12, 2022, 1:07 PM • 3 Y

Y by PRMOisTheHardestExam, FriIzi, cubres

Let $T = AC\cap BD$ . Note that $\angle QPA = \angle QDC = \angle QAC\to AC$ touches $(AQRP)$
As well as $\angle ACQ = \angle ADQ = \angle ADC - \angle CDQ = \angle CBA - \angle CAQ = \angle CAB - \angle CAQ = \angle QAP = \angle CKQ$
$\to AC$ touches $(CQRK).\hspace{10 cm}$
Now, we have $T\in QR$ since $QR$ bisects $AC$ and $R\in BK$ by Pappus' Theorem.

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DottedCaculator

7358 posts

DottedCaculator

#8 h Jul 12, 2022, 1:12 PM • 5 Y

Y by PROA200, s18_kanimet_bursumankulov, rafid149, Frank25, cubres

$[asy] unitsize(1cm); pair D, C, A, B, Q, P, R, X; D=(0,0); A=(2,5); C=(4,0); B=A+C-B; P=(10,5); Q=intersectionpoints(D--P,circumcircle(A,D,C))[1]; R=intersectionpoints(C--P,circumcircle(A,Q,P))[0]; X=extension(A,Q,B,R); draw(circumcircle(A,C,D)); draw(circumcircle(A,Q,P)); draw(A--D--X--C--A--P--C--Q--R--B--X--D--P--A--X); label("$D$", D, SW); label("$C$", C, SE); label("$A$", A, NE); label("$B$", B, N); label("$Q$", Q, NNE); label("$P$", P, SE); label("$R$", R, ESE); label("$X$", X, SE); [/asy]$

We claim that $ABRC$ is cyclic. We will first show that $AC$ is tangent to the circumcircle of $APQ$ . We have
$\begin{align*} \angle QAC&=\angle QDC\\ &=\angle QPA, \end{align*}$ so $AC$ is tangent to the circumcircle of $APQ$ . Now, we have
$\begin{align*} \angle ABC&=\angle CAB\\ &=180^{\circ}-\angle PRA\\ &=\angle ARC. \end{align*}$ This means that $ABRC$ is cyclic. If $AQ$ and $CD$ intersect at $X$ , then we have
$\begin{align*} \angle RCX&=\angle RPA\\ &=180^{\circ}-\angle AQR\\ &=\angle RQX. \end{align*}$ Therefore, $RQCX$ is also cyclic. Since $AD=BC=AC$ , we get $\angle CDA=\angle ACD$ , so we get
$\begin{align*} \angle CRX&=\angle CQX\\ &=180^{\circ}-\angle AQC\\ &=\angle CDA\\ &=\angle ACD\\ &=\angle CAB\\ &=180^{\circ}-\angle BRC. \end{align*}$ Therefore, $B$ , $R$ , and $X$ are collinear, so $AQ$ , $BR$ , and $CD$ concur at $X$ .

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isaacmeng

113 posts

isaacmeng

#9 h Jul 12, 2022, 1:24 PM • 2 Y

Y by Ahmed102, cubres

ISL 2021 G1. Let $ABCD$ be a parallelogram such that $AC=BC$ . A point $P$ is chosen on the extension of segment $AB$ beyond $B$ . Define $Q=(ACD)\cap PD$ and $R=(APQ)\cap PC$ . Prove that $CD$ , $AQ$ , and $BR$ are concurrent.

Solution. Define $E=AQ\cap CD$ . Use directed angles modulo $360^\circ$ .

Claim 1. $C$ , $Q$ , $R$ , $E$ are concyclic.

Proof. We have $\begin{align*}\measuredangle CEQ&=180^\circ-\measuredangle EAD-\measuredangle ADE\\&=180^\circ-(\measuredangle QAC+\measuredangle CAD)-(\measuredangle ADQ+\measuredangle QDC)\\&=180^\circ-\measuredangle QAC-\measuredangle QDC-\measuredangle CAD-\measuredangle BAQ\\&=180^\circ-2\measuredangle QAC-(180^\circ-\measuredangle ADC-\measuredangle DCA)-\measuredangle BAQ\\&=\measuredangle ADC+\measuredangle DCA-2\measuredangle QAC-\measuredangle BAQ\\&=2\measuredangle ADC-2\measuredangle QAC-\measuredangle BAQ\\&=2\measuredangle ADQ+2\measuredangle QDC-2\measuredangle QAC-\measuredangle BAQ\\&=2\measuredangle ADQ-\measuredangle BAQ\\&=\measuredangle BAQ\\&=\measuredangle CRQ.\end{align*}$
Claim 2. $B$ , $R$ , $E$ are collinear.

Proof. We have $\begin{align*}\measuredangle QRB&=180^\circ-\measuredangle CRQ-\measuredangle ERC\\&=180^\circ-\measuredangle PAQ-\measuredangle EQC\\&=180^\circ-\measuredangle PAQ-\measuredangle ADC\\&=180^\circ-\measuredangle ACQ-\measuredangle DCA\\&=\measuredangle QCE\\&=180^\circ-\measuredangle ERQ.\end{align*}$
Now, Claim 2 implies the result.

$[asy] size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.195830202854993, xmax = 18.824455296769358, ymin = -6.88018031555221, ymax = 5.561592787377907; /* image dimensions */ /* draw figures */ draw((4.37,-1.48)--(9.429564356435643,-1.4836435643564359), linewidth(0.8)); draw((5.65,1.88)--(10.709564356435642,1.8763564356435638), linewidth(0.8)); draw((9.429564356435643,-1.4836435643564359)--(5.65,1.88), linewidth(0.8)); draw((5.65,1.88)--(3.733168316831683,-3.151683168316832), linewidth(0.8)); draw((3.733168316831683,-3.151683168316832)--(10.709564356435642,1.8763564356435638), linewidth(0.8)); draw(circle((8.179089667867803,0.9165372693836932), 2.7063915055664625), linewidth(0.8)); draw(circle((3.7352424196126757,-0.2715209217572098), 2.8801629933754995), linewidth(0.8)); draw((3.733168316831683,-3.151683168316832)--(9.429564356435643,-1.4836435643564359), linewidth(0.8)); draw((5.65,1.88)--(7.737054951313305,-5.926480752802572), linewidth(0.8)); draw((7.737054951313305,-5.926480752802572)--(10.709564356435642,1.8763564356435638), linewidth(0.8)); draw((7.737054951313305,-5.926480752802572)--(4.37,-1.48), linewidth(0.8) + linetype("2 2")); /* dots and labels */ dot((5.65,1.88),dotstyle); label("$A$", (5.706498873027804,2.0304132231404948), NE * labelscalefactor); dot((4.37,-1.48),dotstyle); label("$B$", (4.429263711495121,-1.335477084898571), NE * labelscalefactor); dot((9.429564356435643,-1.4836435643564359),dotstyle); label("$C$", (9.493125469571757,-1.335477084898571), NE * labelscalefactor); dot((10.709564356435642,1.8763564356435638),linewidth(4pt) + dotstyle); label("$D$", (10.77036063110444,2.0003606311044315), NE * labelscalefactor); dot((3.733168316831683,-3.151683168316832),dotstyle); label("$P$", (3.798159278737796,-3.0033959429000725), NE * labelscalefactor); dot((6.468215171114042,-1.1804757104902723),linewidth(4pt) + dotstyle); label("$Q$", (6.53294515401954,-1.0650037565740034), NE * labelscalefactor); dot((5.290538122451863,-2.6956484235895024),linewidth(4pt) + dotstyle); label("$R$", (5.345867768595046,-2.582659654395189), NE * labelscalefactor); dot((7.737054951313305,-5.926480752802572),linewidth(4pt) + dotstyle); label("$E$", (7.79515401953419,-5.813313298271971), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$

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Hopeooooo

819 posts

Hopeooooo

#10 h Jul 12, 2022, 1:35 PM • 5 Y

Y by Mango247, Upwgs_2008, H_Taken, Math_.only., cubres

The circumcircle $DCR$ is $w$ and $BR$ Intersects $w$ at $F$ .
Claim: $D,A,F$ are collinear.
$\angle{PAF}=\angle{PRF}=\angle{CDF}$ and $CD||AB$ implies the result.
Radical axis theorem on $(CDFR),(AQRP),(AQCD)$

This post has been edited 4 times. Last edited by Hopeooooo, Jul 12, 2022, 1:38 PM
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MrOreoJuice

595 posts

MrOreoJuice

#11 h Jul 12, 2022, 1:42 PM • 2 Y

Y by PHSH, cubres

Cute
$\measuredangle ARC = \measuredangle ARP = \measuredangle AQP = \measuredangle AQD = \measuredangle ACD = \measuredangle ABC$ meaning $ABRC$ is cyclic. Let $S= \overline{AD} \cap \overline{BR}$ .
$\measuredangle SAP = \measuredangle ABC = \measuredangle CAB = \measuredangle CRB = \measuredangle SRP$ meaning $SAQRP$ is cyclic.
$\measuredangle SRC = \measuredangle BRC = \measuredangle BAC = \measuredangle ADC = \measuredangle SDC$ meaning $SRCD$ is cyclic. Finish by radical axis theorem on $(SAQR), (AQCD) , (SRCD)$ .

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JAnatolGT_00

559 posts

JAnatolGT_00

#12 h Jul 12, 2022, 2:01 PM • 1 Y

Y by cubres

Let $S=AD\cap BR.$ Note that $\measuredangle ABC=\measuredangle ACD=\measuredangle AQD=\measuredangle ARC\implies R\in \odot (ABC).$ Next $\measuredangle SDP=$ $=\measuredangle ADC=$ $\measuredangle SRP\implies S\in \odot (APQ)\cap \odot (CDR).$ Radical axis on $\odot (APQRS),\odot (ACDQ),\odot (CDRS)$ finish.

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ppanther

160 posts

ppanther

#13 h Jul 12, 2022, 2:17 PM • 1 Y

Y by cubres

Angles are directed. We have
$\angle ARC = \angle ARP = \angle AQP = \angle AQD = \angle ACD = \angle ABC,$ so $BRCA$ is cyclic. Let ray $\overline{RB}$ intersect $(APQ)$ at $E$ . We claim that $E$ lies on $\overline{AB}$ . Indeed,
$\angle EAP = \angle ERP = \angle BRC = \angle BAC = \angle DAB.$ Finally, as
$\angle ERC = \angle BRC = \angle BAC = \angle EDC,$ quadrilateral $ERCD$ is cyclic. Considering the radical center of $(ERQA)$ , $(AQCD)$ , and $(ERCD)$ establishes the concurrency.

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lneis1

243 posts

lneis1

#14 h Jul 12, 2022, 2:21 PM • 2 Y

Y by Mango247, cubres

Let $\angle ABC=x$
Note that, $AB$ is tangent to $(ACD)$ since $\angle BAC=\angle ADC$ .
Similarly $AC$ is tangent to $(AQRP)$ since $\angle PQA=180-\angle AQD=180-\angle ACD=180-\angle CAP=180-x$

Note that $\angle ARP=180- \angle ARP=180- \angle AQP=x$ , hence $R \in (ABC)$ .

Let $AQ \cap CD=E$
Then since $AB || CD \implies \angle RCE=\angle RPA=\angle RQE$ where the last equality holds because $Q \in (ARP)$
Hence $E \in (RCQ)$

Note that $\angle ERC=x$ , since $\angle ERC=\angle CQE=180-\angle AQC=180-(180-x)$
But since $R \in (ABC)$ we have $\angle BRC=180-x$ Hence we get $B-R-E$

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BVKRB-

322 posts

BVKRB-

#15 h Jul 12, 2022, 2:37 PM • 2 Y

Y by adorefunctionalequation, cubres

Yay I can solve something now

Let $RB \cap \odot(APQ) = X$
$\measuredangle ARC = \measuredangle ARP = \measuredangle AQP = \measuredangle AQD = \measuredangle ACD = \measuredangle ABC \implies \odot(ABRC)$ This implies that $D-A-X$ , Now $\measuredangle DXR = \measuredangle AXR = \measuredangle APR = \measuredangle DCR \implies \odot(CDXR)$ Now radical centre on $\odot(AQDC),\odot(AQRP),\odot(CDXR)$ finishes the problem

$\blacksquare$

v_Enhance wrote:

Define $X = \overline{AQ} \cap \overline{CD}$ and let $R' = \overline{BX} \cap \overline{PC}$ . We wish to show $APR'Q$ is cyclic.
Let $M$ be the center of the parallelogram.
Claim: $M$ , $Q$ , $R'$ are collinear.
Proof. By Pappus theorem on $\overline{ABP}$ and $\overline{DCX}$ . $\blacksquare$

$[asy] size(12cm); pair A = dir(90); pair D = dir(240); pair C = A*A/D; pair Q = dir(20); pair B = A+C-D; pair P = extension(D, Q, A, B); pair M = midpoint(A--C); pair X = extension(A, Q, C, D); pair Rp = extension(P, C, B, X); filldraw(unitcircle, invisible, blue); filldraw(A--B--C--D--cycle, invisible, red); draw(A--C, red); draw(B--D, red); draw(C--X, red); draw(B--P, red); draw(A--X, lightblue); draw(B--X, lightblue); draw(C--P, lightblue); draw(M--Rp, deepgreen); draw(D--P, lightblue); draw(Q--C, lightblue); pair Z = 2*A-P; pair W = -D+2*foot(origin, D, Z); draw(Q--W, deepgreen); draw(D--Z--A, orange); dot("$A$", A, dir(A)); dot("$D$", D, dir(D)); dot("$C$", C, dir(C)); dot("$Q$", Q, dir(72)); dot("$B$", B, dir(B)); dot("$P$", P, dir(P)); dot("$M$", M, dir(190)); dot("$X$", X, dir(X)); dot("$R'$", Rp, dir(Rp)); dot("$Z$", Z, dir(Z)); dot("$W$", W, dir(W)); /* TSQ Source: !size(12cm); A = dir 90 D = dir 240 C = A*A/D Q = dir 20 R72 B = A+C-D P = extension D Q A B M = midpoint A--C R190 X = extension A Q C D R' = extension P C B X unitcircle 0.1 lightcyan / blue A--B--C--D--cycle 0.1 lightred / red A--C red B--D red C--X red B--P red A--X lightblue B--X lightblue C--P lightblue M--Rp deepgreen D--P lightblue Q--C lightblue Z = 2*A-P W = -D+2*foot origin D Z Q--W deepgreen D--Z--A orange */ [/asy]$
Let $W$ be the point for which $AQCW$ is harmonic. Let $Z = \overline{DW} \cap \overline{AB}$ .
Claim: $\triangle ZAD \cong \triangle PAC$ .
Proof. Projecting from $D$ we get $-1 = (AC;QW) = (A\infty;PZ)$ so $A$ is the midpoint of $\overline{PZ}$ . Since $AD=AC$ and $\measuredangle ZAD = \measuredangle CAP$ , we're done. $\blacksquare$
To finish, $\begin{align*} \measuredangle AQR' &= \measuredangle AQM = \measuredangle WQC = \measuredangle WDC = \measuredangle WDA + \measuredangle ADC \\ &= \measuredangle ACP + \measuredangle DCA = \measuredangle DCP = \measuredangle APR'. \end{align*}$
Remark: [Motivation] The Pappus step allows one to erase many points in the picture. After this step, rewrite the conclusion as $\measuredangle AQM = \measuredangle APC$ ; then the points $B$ , $R'$ , and $X$ may now be deleted.

It seems that vEnhance has forgotten simple thinking :rotfl:

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Leartia

93 posts

Leartia

#16 h Jul 12, 2022, 2:42 PM • 1 Y

Y by cubres

Let $E\neq R$ be such that $E=BR\cap (APQ)$ and let $F=BC\cap PE.$
By parallelism we have $AD=AC=BC.$
Since $\angle BAC=\angle ADC$ and $\angle PAC=180^{\circ}-\angle AQP=\angle AQD=\angle ACD.$ we have that $AP$ is tangent with $(ADC)$ and $AC$ is tangent with $(APQ).$
Let $\angle BAC=\alpha.$
We have $\angle CDA=\angle CBA=\angle CAB=\angle ACD$ , $\angle CAB=\angle CAP=\angle AEP,$ and $\angle CBA=\angle FBP.$
Since $\angle ABF=180^{\circ}- \alpha$ we have that $A,B,F,E$ are concyclic.
Therefore, $\angle FAP=\angle FAB=\angle FEB=\angle PER=\angle PAR.$
We also have $\angle CPA=\angle CAR=\angle AER=\angle AFC$ hence $A,C,P,F$ are concyclic.
So we get $\angle BAE=\angle CFP=\angle CAP=\alpha$ so $P, A,$ and $E$ are collinear.
Since $\angle PRE=\angle PAE=\alpha$ we get that $\angle EDC+\angle CRE=180^{\circ}$ therefore $D,C,R,$ and $E$ are concyclic.
Applying the radical axis theorem on $(AQCD), (AQRE)$ and $(DCRE)$ we get that $DC, AQ,$ and $RE$ concur. \qed

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Mahmood.sy

26 posts

Mahmood.sy

#17 h Jul 12, 2022, 3:13 PM • 2 Y

Y by Muaaz.SY, cubres

let $L$ is the intersection of $(CD)$ & $(BR)$
claim_1: $ACRB$ is cyclic
proof : $\angle AQD=\angle ACD=\angle ADC=\angle ABC$
$\rightarrow \angle AQP=180^o -\angle ABC$
$angle ARP=\angle AQP$ $\rightarrow \angle ARC=\angle ABC$ so $ACRB$ is cyclic
also $\angle LCB=\angle ABC=\angle CAB$ so $LC$ is tangent to the circle of $ACRB$
$\rightarrow LC^2=LR.LB$
let $T$ is the intersection of $BR$ and the circum circle of traingle $APQ$
claim_2: traingles $ABT$ & $LRC$ are similiarity
proof: $\angle CLR=\angle BTA$ by paralle
$\angle LRC=\angle PRT=\angle BAT$ so we proof our claim , this mean that $LR.BT=LC.AB=LC.CD$
$\rightarrow LC^2+LC.CD=LR.LB+LR.BT \rightarrow LC.LD=LR.LT$
this mean that $L\in(AQ)$ , so we done

This post has been edited 2 times. Last edited by Mahmood.sy, Jul 12, 2022, 3:16 PM

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MrPopo9959

2 posts

MrPopo9959

#90 h Sep 19, 2024, 9:13 PM • 1 Y

Y by cubres

First AoPS Post

Let $X=\overline{RB} \cap \overline{DA}$

I claim that $X$ lies on $(AQRP)$ . Simple angle chasing gives that $\overline{CA}$ is tangent to $(APQR)$ . Hence $CB^{2}=CA^{2}=CR \cdot CP$ by power of a point. Therefore $\measuredangle RPA= \measuredangle CPB= \measuredangle CBR= \measuredangle AXR$ as desired.

Now note that $\measuredangle XRC= \measuredangle XRP =\measuredangle XAB =\measuredangle XDC$ Hence $RXDC$ is cyclic. Hence by Radical Axis Theorem, $\overline{XBR}$ , $\overline{AQ}$ , and $\overline{BC}$ are concurrent.

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ezpotd

1342 posts

ezpotd

#91 h Sep 28, 2024, 5:43 AM • 1 Y

Y by cubres

Claim 1: $ABCR$ cyclic.
Proof: Consider $\angle ABC = \angle ACD = \angle AQD = 180 - \angle AQP = 180 - \angle ARP = \angle ARC$ .

Claim 2: $AD \cap BR = K$ lies on $(AQR)$
Consider $\angle APR = \angle APC = 180 - \angle PAC - \angle PCA = 180 - \angle BCA - \angle PBR = 180 - \angle KAB - \angle KBA = \angle AKR$ , as desired.

Claim 3: $(KCDR)$ cyclic.
Proof: Consider $\angle KDC + \angle CRK = \angle ADC + \angle BRC = \angle CAB + (180 - \angle CAB) = 180$ .

To finish, apply the radical axis theorem on $(AQCD), (AQKR), (KRCD)$ , to get $CD, AQ, KR$ concur, and since $K,B,R$ are collinear, we are done.

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ehuseyinyigit

849 posts

ehuseyinyigit

#92 h Sep 28, 2024, 12:37 PM • 1 Y

Y by cubres

So many concyclic quadrilaterals and paralellism implies that Reim's Theorem would be useful here.

Let $BD\cap PC=H$ and $AQ\cap CD=J$ .
Claim: $C,H,Q,R$ and $J$ are concyclic.
Proof:
Since $\angle QRC=\angle PAQ=\angle QJC$ implies points $C,Q,R$ and $J$ are collinear. On the other hand
$\angle QHC=\pi-\angle BHQ=\pi-\angle PAQ=\angle PRQ$ Thus, $H\in (CQRJ)$ .

Claim: $ABQH$ and $ABRC$ are concyclic quads.
Proof:
$AC=BC$ is given. Notice that $\angle AQH=\angle ACD=\angle ABH$ implies points $A,B,Q$ and $H$ lie on a circle. On the other hand, PoP gives
$PB\cdot PA=PQ\cdot PH=PR\cdot PC\Longleftrightarrow R\in (ABC)$ Claim: Points $B$ , $R$ and $J$ are collinear.
Proof:
Observe that $AB\parallel CJ$ and common chord of $(ABRC)$ and $(CHQRJ)$ is $CR$ . By Converse of Reim's Theorem, $B$ , $R$ and $J$ are collinear, as desired.

This post has been edited 1 time. Last edited by ehuseyinyigit, Sep 28, 2024, 12:37 PM

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shendrew7

815 posts

shendrew7

#93 h Oct 26, 2024, 5:25 AM • 1 Y

Y by cubres

Let $AQ \cap BR = K$ . It suffices to show $C$ , $D$ , $K$ collinear, or $KC \parallel AB$ :

Notice $ABCR$ is cyclic, as
$\measuredangle CBA = \measuredangle DCA = \measuredangle DQA = \measuredangle PQA = \measuredangle PRA = \measuredangle CRA.$
Notice $CQRK$ is cyclic, as
$\measuredangle KRC = \measuredangle BRP = \measuredangle BAC = \measuredangle ADC = \measuredangle KQC.$
Hence we get the desired from
$\measuredangle RCK = \measuredangle RQK = \measuredangle RPA. \quad \blacksquare$

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Ianis

429 posts

Ianis

#94 h Oct 26, 2024, 2:33 PM • 1 Y

Y by cubres

I am not going to read $81$ posts, so this is probably nothing new, but here it goes.

Let $E=BR\cap AD$ . Observe that $\measuredangle QAC=\measuredangle QDC=\measuredangle QPA,$ so $CA$ is tangent to $(AQRP)$ , which gives $CB^2=CA^2=CR\cdot CP.$ Hence $CB$ is tangent to $(BPR)$ . Observe now that $\measuredangle EAP=\measuredangle EAB=\measuredangle CBA\underset{(1)}{=}\measuredangle BRP=\measuredangle ERP,$ where $(1)$ holds because $CB$ is tangent to $(BRP)$ . Hence $AQRPE$ is cyclic. Finally, we have $\measuredangle EDC=\measuredangle EAP=\measuredangle ERP=\measuredangle ERC,$ so $CDER$ is cyclic. Hence the three lines in question are the radical axes of $(AQCD)$ , $(AQRPE)$ and $(CDER)$ , so they are concurrent (*).

(*) If $AQ\parallel CD$ then $Q=P$ , so $PA$ is the exterior angle bisector of $\angle CPD$ , so $PC=PD$ , which gives $P=A$ ; similarly, if $BR\parallel AD$ then $P=B$ ; both of these contradict the fact that $P$ lies on the extension of ray $AB$ past $B$ .

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Saucepan_man02

1381 posts

Saucepan_man02

#95 h Oct 31, 2024, 7:48 AM • 1 Y

Y by cubres

Heres the angle-chase:

Let $X= AQ \cap BR$ .

Claim: $ABCR$ is cyclic.
Proof: Notice that: $\angle ARC = 180^\circ - \angle ARP = 180^\circ - \angle AQP = \angle AQD = \angle ACD = \angle ABC.$
Thus, we have: $\angle CRX = \angle CAB =\angle ADC = \angle CQX$ which implies $CQRX$ to be cyclic. Therefore, by Reims, $AP \parallel CX$ and we conclude.

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Bonime

38 posts

Bonime

#96 h Dec 29, 2024, 10:10 PM • 1 Y

Y by cubres

There´s nothing new, just to record my solution somewhere.

Let $K=AQ\cap CD, \angle QDC=\alpha, \angle QDA=\beta$ and $\angle CPA=\theta$ . Since $\angle QPA=\alpha=\angle QAC$ , $CA$ is tangent to $(APQ)$ . Therefore $CA^2=CB^2=CR\cdot CP$ , so $CB$ is tangent $(BPR)$ and $\angle RBC=\theta=\angle RCK$ . Now, note that $\angle CAB=\angle ACD$ , so $BA$ is tangent to $(ACD)$ . Hence $\beta=\angle PAQ=\angle QKC=\angle QRC \Rightarrow \text{$QCKR$ is cyclic} \Rightarrow \angle RKQ=\angle RCQ=180-(\alpha+2\beta+\theta)$ But $\angle RBC+\angle BCK+\angle BKR=(\theta +(\alpha +\beta)+(180-(\alpha+\beta))=180 \Rightarrow \text{B,R,K are collinear} \blacksquare$

This post has been edited 2 times. Last edited by Bonime, Jan 10, 2025, 12:42 AM
Reason: typo

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redred123

32 posts

redred123

#97 h Jan 10, 2025, 10:58 PM • 1 Y

Y by cubres

sketch
Click to reveal hidden text

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Ilikeminecraft

706 posts

Ilikeminecraft

#98 h Mar 13, 2025, 6:27 AM • 1 Y

Y by cubres

Let $E = AD\cap(APQ).$
Note that $AC$ is tangent to $(PQR)$ as $\angle CAP = \angle DCA = \angle DQA = 180-\angle PQA.$ Also note that since it is a parallelogram, we have $AC = BC = AD.$ Especially, we get $\angle CDA = \angle DCA = \angle BAC,$ which tells us $AB$ is tangent to $(ACD).$
Claim: $RBE$ are collinear
Proof: Redefine $B$ as the intersection of $RE$ and the line through $C$ parallel to $DE.$ We show that $P,B,A$ are collinear. Note that $\angle BCA = \angle CAD = 180 - \angle CAE = \angle BRA,$ so $CRBA$ is cyclic.
Note that $\angle CBA = \angle CRA =180-\angle PRA = \angle DQA = \angle CDA = 180 - \angle DAB,$ which implies $AB\parallel CD,$ but $PA\parallel CD$ so we are done.

Claim: $CDER$ is cyclic
Proof: Simply observe $\angle CDA = \angle BAE = \angle PRE = 180 - \angle CRE.$

By radax, we are done.

This post has been edited 4 times. Last edited by Ilikeminecraft, Mar 13, 2025, 6:37 AM

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joshualiu315

2535 posts

joshualiu315

#99 h Mar 15, 2025, 12:40 AM • 1 Y

Y by cubres

Let $T = \overline{AQ} \cap \overline{CD}$ . We wish to show that $T$ lies on $\overline{BR}$ .

Claim: Points $A,B,R,C$ are concyclic.

Proof: Note that

$\begin{align*} \angle CRA = 180^\circ - \angle PRA &= 180^\circ- \angle PQA \\ &= \angle AQD = \angle ACD = \angle CBA. \ \square \end{align*}$

On top of this, we can find another cyclic quadrilateral.

Claim: Points $C,Q,R,T$ are concyclic.

Proof: Some angle chasing yields

$\angle CRQ = 180^\circ - \angle QRP = \angle QAP = \angle CTQ. \ \square$

Combining all of the information we have, we get

$\begin{align*} \angle CRT = \angle CQT &= \angle CDA = \angle CBA \\ &= \angle CAB = 180^\circ - \angle CRB, \end{align*}$
so we have $\angle CRT + \angle CRB = 180^\circ$ , as desired. $\blacksquare$

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scannose

1030 posts

scannose

#100 h Mar 18, 2025, 2:00 PM • 1 Y

Y by cubres

tfw i read the problem statement as ab = ac first and was wondering why the claim was false

no latex warning

This post has been edited 1 time. Last edited by scannose, Mar 18, 2025, 2:47 PM

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happypi31415

769 posts

happypi31415

#101 h May 3, 2025, 9:02 PM • 1 Y

Y by cubres

Very nice problem! There are a ton of cyclic quads and tangencies to work with, and it was satisfying to put them together

Solution

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Maximilian113

583 posts

Maximilian113

#102 h May 3, 2025, 10:58 PM • 1 Y

Y by cubres

Beautiful!

Let $K=(APRQ) \cap RB.$ Note that $\angle ARC = 180^\circ - \angle ARP = \angle AKP = 180^\circ - \angle AQP = \angle AQD = \angle ACD = \angle BAC = \angle ABC,$ so $ABRC$ is cyclic. Thus $\angle KAB = \angle KRP = \angle BAC = \angle ABC \implies AK \parallel BC,$ so $K, A, D$ is collinear. Hence $\angle KDC = \angle KAP = \angle KRP = 180^\circ - \angle KRC,$ so $KDCR$ is cyclic. The result follows from Radical Axis on $(KDCR), (ADCQ), (KAQR).$ QED

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Rayvhs

44 posts

Rayvhs

#103 h May 16, 2025, 1:53 PM • 1 Y

Y by cubres

Here's a very simple solution.
Let $DC \cap AQ = T$ and $AQ \cap PR = N$ .

Note that
$\angle CBA = \angle CAB = \angle ADC = \angle CQT.$
Also, $ABRC$ is cyclic because $\angle ABC = \angle ACD = \angle AQD$ , so
$\angle AQP = 180^\circ - \angle AQD = 180^\circ - \angle ARP,$ which implies $\angle ARC = \angle AQD = \angle ABC$ .

Therefore,
$\angle BRP = \angle CAB = \angle CQT.$
Since we want to prove that $T$ , $R$ , $B$ are collinear (i.e., $\angle BRP = \angle CRT$ ), we will prove that $CQRT$ is cyclic.

Since $CT \parallel AP$ , we have
$\frac{NP}{NA} = \frac{NC}{NT}.$ But $AQRP$ is cyclic, so
$\frac{NP}{NA} = \frac{NQ}{NR}.$
Thus,
$\frac{NQ}{NR} = \frac{NC}{NT} \Rightarrow NC \cdot NR = NT \cdot NQ,$ which shows that $QCTR$ is cyclic.

Therefore,
$\angle CRT = \angle CQT = \angle CAB = \angle BRP,$ and we're done.

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ItsBesi

152 posts

ItsBesi

#104 h Jun 13, 2025, 10:03 PM • 1 Y

Y by cubres

Solved this exactly one year ago but since I didn't remember the solution I thought would be nice to solve it again. :blush:

Let $DC \cap AQ=\{X \}$

Claim: Points $Q$ , $C$ , $X$ and $R$ are concyclic
Proof:
$\angle RQX = 180- \angle AQR \stackrel{(APQR)}{=} \angle APR \equiv APC \stackrel{AP \parallel DX}{=} \angle PCX \equiv RCX \implies \angle RQX=\angle RPX \implies$
Points $Q$ , $C$ , $X$ and $R$ are concyclic $\square$

Let $BC \cap BD = \{G\}$
Claim: Points $A$ , $Q$ , $G$ and $B$ are concyclic
Proof:
$\angle ABG \equiv \angle ABC \stackrel{ABCD}{=} \angle ADC \stackrel{AD=BC=AC}{=} \angle ACD \stackrel{(AQCD)}{=} \angle AQD = 180- \angle AQG \implies \angle ABG=180-\angle AQG \implies$
$\angle ABG+\angle AQG=180 \implies$ Points $A$ , $Q$ , $G$ and $B$ are concyclic $\square$

Claim: $G \in (QCXR)$
Proof:
$\angle CGQ= 180 -\angle QGB \stackrel{(AQGB)}{=} \angle QAB \equiv XAB \stackrel{DX \parallel AB}{=} \angle AXC \equiv \angle QXC \equiv \angle CXQ \implies \angle CGQ=\angle CXQ \implies$
Points $C,G,X$ and $Q$ are concyclic $\iff$ $G \in (QCXR)$ $\square$

Claim: Points $A$ , $C$ , $R$ and $B$ are concyclic
Proof:
From Power of the Point (POP) we get:
$PR \cdot PC=Pow(P , \odot(QCXRG))=PG \cdot PQ = Pow(P , \odot(AQGB))=PB \cdot PA \implies PR \cdot PC=PB \cdot PA$
Which by the converse of POP we get that: Points $A$ , $C$ , $R$ and $B$ are concyclic $\square$

Claim: Points $\overline{X-R-B}$ are collinear
Proof:
$\angle XRC \stackrel{(QCXR)}{=} \angle XQC = 180-\angle AQC \stackrel{(AQCD)}{=} \angle ADC \stackrel{ABCD}{=} \angle ABC \stackrel{CA=CB}{=} \angle CAB \stackrel{(ACRB)}{=} 180-\angle CRB \implies$
$\angle XRC=180-\angle CRB \implies \angle XRC+\angle CRB=180 \implies$ Points $\overline{X-R-B}$ are collinear $\blacksquare$

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