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Source: USA Team Selection Test 2007

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3595 posts

#1 h Dec 8, 2007, 1:35 AM • 5 Y

Y by Davi-8191, mathematicsy, Adventure10, Mango247, and 1 other user

Circles $\omega_1$ and $\omega_2$ meet at $P$ and $Q$ . Segments $AC$ and $BD$ are chords of $\omega_1$ and $\omega_2$ respectively, such that segment $AB$ and ray $CD$ meet at $P$ . Ray $BD$ and segment $AC$ meet at $X$ . Point $Y$ lies on $\omega_1$ such that $PY \parallel BD$ . Point $Z$ lies on $\omega_2$ such that $PZ \parallel AC$ . Prove that points $Q,X,Y,Z$ are collinear.

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2417 posts

K81o7

#2 h Dec 8, 2007, 2:45 AM • 4 Y

Y by mathematicsy, Adventure10, Mango247, and 1 other user

I recall failing to solve this problem in DC. The key lies in finding
Click to reveal hidden text

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153 posts

#3 h Dec 8, 2007, 8:52 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

$\angle CXD = \angle CDB - \angle XCD$ and $\angle AQB = \angle BQP + \angle AQP = (180 - \angle CDB)+ \angle XCD$ so $\angle CXD + \angle AQB = 180$ so $AQBX$ is cyclic quadrilateral. $\angle QYP = \angle QAP = \angle QXB$ and $PY\parallel BX$ so $Q,Y,X$ are collinear. Analogously $Q,Z,X$ are collinear

This post has been edited 1 time. Last edited by Litlle 1000t, Dec 8, 2007, 11:04 AM

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#4 h Dec 8, 2007, 10:36 AM • 2 Y

Y by Adventure10, Mango247

K81o7 wrote:

the cyclic quadrilateral, $ABQX$

This follows directly from the fact that $Q$ is the Miquel point

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240 posts

#5 h Dec 16, 2007, 4:54 AM • 4 Y

Y by mathematicsy, Adventure10, Mango247, and 1 other user

I think I've got a slightly different solution.
$\angle XCQ = \angle ACQ = \angle BPQ = \angle BDQ$ , so $DXCQ$ is a cyclic quad.
Then $\angle QXD = \angle QCD = \angle QCP = \angle QYP$ . Thus $\angle QXD = \angle QYP$ , and since $XB$ and $YP$ are parallel, it follows that $Q,X,Y$ are collinear. Similarly, $Q,X,Z$ are collinear.

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#6 h Apr 23, 2013, 4:51 PM • 7 Y

Y by va2010, myh2910, AllanTian, HamstPan38825, Adventure10, Mango247, and 1 other user

$Q$ is the unique center of the spiral similarity sending $CD$ to $AB$ , so it is also the unique center of the spiral similarity sending $CA$ to $DB$ ; hence $DXQC$ and $AXQB$ are concyclic. Let $Z' = XQ \cap \omega_2$ ; then $\measuredangle CAB = \measuredangle XAB = \measuredangle XQB = \measuredangle Z'QB = \measuredangle Z'PB \implies PZ' \parallel AC \implies Z=Z'$ and $Y=Y'$ is similar.

$[asy]/* DRAGON 0.0.9.6 Homemade Script by v_Enhance. */ import olympiad; import cse5; size(11cm); real lsf=0.8000; real lisf=2011.0; defaultpen(fontsize(10pt)); /* Initialize Objects */ pair A = (-2.0, 2.0); pair B = (4.031165686870916, 3.133944005249443); pair C = (1.0, -2.0); pair D = (1.0, 0.0); pair P = IntersectionPoint(A--B,Line(C,D,0,lisf)); pair X = IntersectionPoint(A--C,Line(B,D,0,lisf)); pair O_1 = circumcenter(A,P,C); pair O_2 = circumcenter(P,D,B); pair Q = 2*foot(P,relpoint(O_1--O_2,0.5-10/lisf),relpoint(O_1--O_2,0.5+10/lisf))-P; pair Y_prime = (2)*(foot(O_1,X,Q))-Q; pair Z_prime = (2)*(foot(O_2,Q,X))-Q; /* Draw objects */ draw(circumcircle(A,P,C), rgb(0.7,0.8,0.9) + linewidth(1.0)); draw(circumcircle(B,P,D), rgb(0.7,0.8,0.9) + linewidth(1.0)); draw(A--B, rgb(0.4,0.4,0.4)); draw(C--P, rgb(0.4,0.4,0.4)); draw(A--C, rgb(0.5,0.5,0.5) + linewidth(1.0) + linetype("4 4")); draw(B--X, rgb(0.5,0.5,0.5) + linetype("4 4")); draw(P--Y_prime, rgb(0.5,0.5,0.5) + linetype("4 4")); draw(P--Z_prime, rgb(0.5,0.5,0.5) + linewidth(1.0) + linetype("4 4")); draw(Y_prime--Z_prime, rgb(0.4,0.8,1.0)); draw(circumcircle(C,Q,D), rgb(0.7,0.8,0.9) + linewidth(1.0)); draw(circumcircle(Q,A,B), rgb(0.7,0.8,0.9) + linewidth(1.0)); /* Place dots on each point */ dot(A); dot(B); dot(C); dot(D); dot(P); dot(X); dot(Q); dot(Y_prime); dot(Z_prime); /* Label points */ label("$A$", A, lsf * dir(150)); label("$B$", B, lsf * dir(45)); label("$C$", C, lsf * dir(-90)); label("$D$", D, lsf * dir(60) * 1.414); label("$P$", P, lsf * dir(100)); label("$X$", X, lsf * dir(225)); label("$Q$", Q, lsf * dir(-45)); label("$Y'$", Y_prime, lsf * dir(225)); label("$Z'$", Z_prime, lsf * dir(-45)); [/asy]$

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355 posts

#7 h May 11, 2013, 4:58 PM • 1 Y

Y by Adventure10

v_Enhance wrote:

... Let $Z' = XQ \cap \omega_2$ ; then $\langle CAB = \langle XAB = \langle XQB = \langle Z'QB = \langle Z'PB \implies PZ' \parallel AC \implies Z=Z'$ ...

I've noticed that some people on the forum use Reim's theorem
in particular if we look at circles $AXBQ$ and $PBQZ'$ we have $PZ'||AX$ .

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1099 posts

jlammy

#8 h Aug 6, 2014, 2:00 PM • 2 Y

Y by Adventure10, Mango247

$[asy] /* Borrowed then adapted from v_enhance, thanks */ import olympiad; import cse5; size(8cm); real lsf=0.8000; real lisf=2011.0; defaultpen(fontsize(11pt)); /* Initialize Objects */ pair A = (-2.0, 2.0); pair B = (4.031165686870916, 3.133944005249443); pair C = (1.0, -2.0); pair D = (1.0, 0.0); pair P = IntersectionPoint(A--B,Line(C,D,0,lisf)); pair X = IntersectionPoint(A--C,Line(B,D,0,lisf)); pair O_1 = circumcenter(A,P,C); pair O_2 = circumcenter(P,D,B); pair Q = 2*foot(P,relpoint(O_1--O_2,0.5-10/lisf),relpoint(O_1--O_2,0.5+10/lisf))-P; pair Y= (2)*(foot(O_1,X,Q))-Q; pair Z = (2)*(foot(O_2,Q,X))-Q; /* Draw objects */ draw(circumcircle(A,P,C), green); draw(circumcircle(B,P,D), blue); draw(A--B); draw(X--D--Q--C--cycle); draw(Y--P--Z); draw(X--A); draw(C--P); draw(D--B); draw(Y--Z, dashed); draw(circumcircle(X,D,C), red); /* Place dots on and label each point */ dot(A); label("$A$", A, lsf * dir(150)); dot(B); label("$B$", B, lsf * dir(45)); dot(C); label("$C$", C, lsf * dir(-90)); dot(D); label("$D$", D, lsf * dir(160) * 1.414); dot(P); label("$P$", P, lsf * dir(100)); dot(X); label("$X$", X, lsf * dir(225)); dot(Q); label("$Q$", Q, lsf * dir(-45)); dot(Y); label("$Y$", Y, lsf * dir(225)); dot(Z); label("$Z$", Z, lsf * dir(-45));[/asy]$
$Q$ is the Miquel point of $CXA, CDP, XDB, APB$ , so $CXDQ$ is cyclic. By Reim's theorem on $\omega_1$ and $(CXDQ)$ , $Q,X,Y$ are collinear; by Reim's theorem on $\omega_2$ and $(CXDQ)$ , $Z,Q,X$ are collinear, and the result follows.

Note

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239 posts

#9 h Oct 5, 2015, 3:22 PM • 1 Y

Y by Adventure10

Let $J= YP \cap \omega_2$ .
Let $L= YP \cap \omega_1$ .
Now, $\measuredangle ZQB = \measuredangle ZPB = - \measuredangle BPZ = \measuredangle ZPA = \measuredangle LPA =\measuredangle PAC =\measuredangle XAP$ Now, $\measuredangle YQB = \measuredangle YQP + \measuredangle PQB = \measuredangle YCP + \measuredangle PJB = \measuredangle YCP + \measuredangle DPJ = \measuredangle YCP + \measuredangle CPY = - \measuredangle PYC = - \measuredangle PAC = \measuredangle XAP$ Finally observe that $Q$ is the $Miquel$ $Point$ of the complete quadrilateral formed by the vertices $X,D,P,A$ .
$\implies XAQB concyclic.$ .
So, $\measuredangle XQB = \measuredangle XAB = \measuredangle XAP$ .
So , $\measuredangle ZQB = \measuredangle YQB = \measuredangle XQB$ This leads us to the desired conclusion.

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2079 posts

pi37

#10 h Oct 6, 2015, 12:24 AM • 2 Y

Y by Adventure10, Mango247

By spiral similarity $Q$ maps $DB$ to $CA$ so $QCXD$ is cyclic. If $QX$ intersects $\omega_1$ at $Y'$ , $PY'$ and $DX$ are both antiparallel to $QC$ so $PY'\parallel DX$ , $Y'=Y$ , and $Y,Z$ lie on $QX$ .

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563 posts

AMN300

#11 h Apr 2, 2016, 7:37 PM • 1 Y

Y by Adventure10

$Q$ is the Miquel point of complete quadrilateral $PBXC$ , so $QDXC$ and $ADBX$ are cyclic.

Let $QX \cap w_1 \equiv Y'$ . Angle chasing, we have $\angle Y'QC = \angle Y'AC = \angle Y'PD$ . By the above and $QDXC$ cyclic, we have that $\angle Y'QC = \angle CDX = \angle Y'PD$ so $Y'P \parallel DX$ , and similarly for $Z'$ . The conclusion follows.

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471 posts

eshan

#12 h Jul 3, 2016, 12:06 PM • 4 Y

Y by Dilshodbek, m2017m, Adventure10, Mango247

Maybe this solution is already given above, but still I'm writing it down.

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(15.cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = -4.716666666666658, xmax = 17.6, ymin = -1.7266666666666675, ymax = 8.073333333333334; /* image dimensions */ pen xdxdff = rgb(0.49019607843137253,0.49019607843137253,1.); pen ffxfqq = rgb(1.,0.4980392156862745,0.); pen ffqqff = rgb(1.,0.,1.); /* draw figures */ draw(shift((3.185458267893191,3.1111558596963573)) * scale(3.1559227963577987, 3.1559227963577987)*unitcircle, linewidth(1.6) + red); draw(shift((5.7380030468626,4.431102969733766)) * scale(2.628399180397597, 2.628399180397597)*unitcircle, linewidth(1.6) + blue); draw((3.8,6.206666666666669)--(5.453214154262895,0.9163593004666906), linewidth(1.6) + red); draw((3.8,6.206666666666669)--(2.4181126438879694,0.04994214780186379), linewidth(1.6) + red); draw((5.155831433496555,1.8679879773650003)--(3.2270742084802806,3.6541096492381664), linewidth(1.6) + blue); draw((2.4181126438879694,0.04994214780186379)--(5.453214154262895,0.9163593004666906), linewidth(1.6) + red); draw((3.8,6.206666666666669)--(6.446607586458077,6.962182158139266), linewidth(1.6) + xdxdff); draw((6.224733389568109,3.9612472849787848)--(3.8,6.206666666666669), linewidth(1.6) + ffxfqq); draw((5.453214154262895,0.9163593004666906)--(6.011390333566628,1.075699412915823), linewidth(1.6) + linetype("4 4") + red); draw((6.011390333566628,1.075699412915823)--(5.155831433496555,1.8679879773650003), linewidth(1.6) + linetype("4 4") + blue); draw((6.011390333566628,1.075699412915823)--(6.446607586458077,6.962182158139266), linewidth(1.6) + dotted + green); draw((5.155831433496555,1.8679879773650003)--(6.06666666666666,1.823333333333333), linewidth(1.6) + linetype("2 2") + ffqqff); draw((3.2270742084802806,3.6541096492381664)--(6.06666666666666,1.823333333333333), linewidth(1.6) + linetype("2 2") + ffqqff); draw(shift((3.916839652865845,1.6064204600289003)) * scale(2.1607423224144324, 2.1607423224144324)*unitcircle, linewidth(1.6) + linetype("2 2") + ffqqff); draw(shift((5.593442797870414,1.4824607634596496)) * scale(0.5832108871504142, 0.5832108871504142)*unitcircle, linewidth(1.6) + linetype("2 2") + ffqqff); draw((5.453214154262895,0.9163593004666906)--(6.06666666666666,1.823333333333333), linewidth(1.6) + linetype("2 2") + red); label("X = X'",(6.133333333333323,1.0566666666666664),SE*labelscalefactor,green); draw((2.4181126438879694,0.04994214780186379)--(6.06666666666666,1.823333333333333), linewidth(1.6) + linetype("2 2") + ffqqff); /* dots and labels */ dot((3.8,6.206666666666669),red); label("$P$", (3.783333333333327,6.506666666666668), NE * labelscalefactor,red); dot((6.06666666666666,1.823333333333333),red); label("$Q$", (6.2,1.6233333333333333), NE * labelscalefactor,red); label("$\omega_1$", (1.1333333333333315,5.906666666666667), NE * labelscalefactor,red); label("$\omega_2$", (5.4,7.24), NE * labelscalefactor,blue); dot((5.453214154262895,0.9163593004666906),red); label("$A$", (5.4,0.573333333333333), NE * labelscalefactor,red); dot((5.155831433496555,1.8679879773650003),blue); label("$B$", (4.833333333333325,1.69), NE * labelscalefactor,blue); dot((2.4181126438879694,0.04994214780186379),red); label("$C$", (2.2833333333333297,-0.2933333333333338), NE * labelscalefactor,red); dot((3.2270742084802806,3.6541096492381664),blue); label("$D$", (2.966666666666662,3.6233333333333335), NE * labelscalefactor,blue); dot((6.446607586458077,6.962182158139266),xdxdff); label("$Z$", (6.516666666666656,7.123333333333334), NE * labelscalefactor,xdxdff); dot((6.224733389568109,3.9612472849787848),ffxfqq); label("$Y$", (6.4,3.923333333333334), NE * labelscalefactor,ffxfqq); dot((6.011390333566628,1.075699412915823),green); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* re-scale y/x */ currentpicture = yscale(1.1386054421768685) * currentpicture; /* end of picture */[/asy]$

First, we prove that $X,Q,Z$ are collinear $\iff XAQB$ is cyclic.

Let $ZQ\cap AC=X'.$

Now, $\measuredangle X'AB=\measuredangle BAC=\measuredangle APZ=\measuredangle BQX'.$

So $X'ABQ$ is cyclic.

Thus we need to prove $X=X'$

Note that $\measuredangle ABX=\measuredangle PBD=\measuredangle BPY=\measuredangle APY=\measuredangle AQX'=\measuredangle ABX'$

This means $X=X'\implies X,Q,Z$ are collinear.

So it remains to prove $X,Q,Y$ are collinear.

Note that $\measuredangle AQX=\measuredangle APY=\measuredangle AQY\implies X,Q,Y$ are collinear.

Thus, $Q,X,Y,Z$ are collinear. $\blacksquare$

This post has been edited 3 times. Last edited by eshan, Jul 3, 2016, 12:19 PM

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348 posts

Delray

#13 h May 23, 2017, 8:42 PM • 2 Y

Y by Adventure10, Mango247

My proof is literally the exact same as everyone else's, but here goes.

Let $QX\cap w_1=Y'$ . It suffices to show that $PY' \parallel BD$ . We have $\measuredangle{APY'}=\measuredangle{APC}-\measuredangle{Y'PC}=\measuredangle{BPC}-\measuredangle{Y'QC}=\measuredangle{BPD}-\measuredangle{XQC}$ Now observe that $Q$ is the Miquel Point of complete quadrilateral $AXDP$ , since $Q=w_1 \cap w_2$ , so $DXCQ$ is cyclic. We now have
$\measuredangle{BPD}-\measuredangle{XQC}=\measuredangle{BPD}-\measuredangle{XDC}=\measuredangle{BPD}-\measuredangle{BDP}=-\measuredangle{DPB}-\measuredangle{BDP}=\measuredangle{PBD}$ Hence $PY' \parallel BD$ . An identical argument shows that $PZ' \parallel AC$ , where $Z'$ is defined analogously to $Y'$ . It follows that $Q,X,Y$ and $Z$ are collinear, as desired. $\square$

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#14 h Oct 29, 2017, 11:18 PM • 1 Y

Y by Adventure10

Note that $\ Q$ is the Miquel Point of the quadrilateral $\ PDXA$ , therefore $\ Q$ is the center of the unique spiral similarity sending $\ PD$ to $\ AX$ , hence $\angle{AQX}=\angle{PQD}=\angle{PBD}=\angle{ABX}=\angle{APY}=\angle{AQY}$ , implying that $\ Q,X$ and $\ Y$ are collinear.
In addition, $\angle{BQZ}=\angle{BPZ}=\angle{PAZ}=\angle{PAQ}+\angle{QAC}=\angle{QCP}+\angle{QPC}=\angle{QXD}+\angle{QBD}=180\ -\angle{BQX}$ , thus $\ X,Z,Q$ are collinear.
Finally, the points $\ X,Y,Z$ and $\ Q$ are collinear.

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328 posts

#15 h Mar 15, 2018, 1:12 PM • 2 Y

Y by adihaya, Adventure10

Since $Q$ is the miquel point of complete quadrilateral $PDXA,$ quadrilateral $ABQX$ is cyclic.
But since $\measuredangle AQY=\measuredangle APY=\measuredangle ABX=\measuredangle AQX;$ we get that $Q,X$ and $Y$ are collinear. Similarly $Q,X$ and $Z$ are collinear, and the result follows. $\square$

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2125 posts

#16 h Mar 19, 2020, 1:52 PM • 1 Y

Y by gamerrk1004

Solution

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#17 h Jun 28, 2020, 3:52 AM

Y by

Note that $Q$ is the miquel point of $APDX$ , so $AXQB$ and $CXDQ$ are cyclic.

$\angle DQZ = \angle DPZ = \angle CPZ = \angle ACP = \angle XCD = \angle DQX,$ so $X,Z,Q$ are collinear.

$\angle AQX = \angle ABX = \angle BPX = \angle APY = \angle AQY,$ so $Y,X,Q$ are collinear.

This post has been edited 1 time. Last edited by mathlogician, Jun 28, 2020, 1:37 PM

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GeronimoStilton

1521 posts

GeronimoStilton

#18 h Nov 9, 2020, 3:27 PM

Y by

Identify $Q$ as the Miquel point of $ACDB$ . Let $Y'$ denote the second intersection of $\omega_1$ and $XQ$ , note that by the spiral similarity circles,
$\measuredangle Y'PC=\measuredangle Y'QC=\measuredangle XQC=\measuredangle XDC,$ hence $Y'=Y$ . Then, $Q,X,Y$ are collinear, so we are done by symmetry.

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#19 h May 1, 2021, 8:20 AM

Y by

N.T.TUAN wrote:

Circles $\omega_1$ and $\omega_2$ meet at $P$ and $Q$ . Segments $AC$ and $BD$ are chords of $\omega_1$ and $\omega_2$ respectively, such that segment $AB$ and ray $CD$ meet at $P$ . Ray $BD$ and segment $AC$ meet at $X$ . Point $Y$ lies on $\omega_1$ such that $PY \parallel BD$ . Point $Z$ lies on $\omega_2$ such that $PZ \parallel AC$ . Prove that points $Q,X,Y,Z$ are collinear.

Invert the problem about $P$ . Let $T'$ denote the image of point $T$ under the given inversion. Then our problem is

Inverted form of given problem wrote:

$A',B',C',D',P$ are points in a plane such that $P=C'D' \cap A'B'$ . Let the lines $A'C'$ and $B'D'$ intersect at $Q'$ . Let $X'=(A'C'P) \cap (B'D'P)$ . Let the tangents from $P$ to $(A'C'P), (B'D'P)$ intersect $B'D',A'C'$ at $Z',Y'$ respectively. Prove that $Y'X'PZ'Q'$ is cyclic

Observe that $\angle X'C'Y' =\angle X'P'B' =\angle X'D'B'$ Hence $X'C'Q'D'$ is cyclic.
Note that $\begin{align*} \angle PZ'D' =\angle Z'PC' -\angle Z'D'P &=\angle PX'C' -\angle Z'D'P \\ &= \angle PX'C' -\angle Q'X'C' \\&= \angle PX'Q' \end{align*}$ Hence $PX'Q'Z'$ is cyclic. Similarly we can conclude that $PX'Y'Q'$ is cyclic. Hence we are done.

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548 posts

#20 h May 6, 2021, 4:47 AM

Y by

Obviously $Q$ is the Miquel Point, which means that $BAXQ$ and $CXDQ$ are cyclic. I will first show that $Q, Z, X$ are collinear. Note that
$\measuredangle QZP=\measuredangle QDP=\measuredangle QDC=\measuredangle QXC=\measuredangle QXA.$ Since $PZ$ and $AC$ are parallel, $Q, Z, X$ must be collinear.

Now I will show that $Q, X, Y$ are collinear. We have
$\measuredangle QXB=\measuredangle QAB=\measuredangle QAP=\measuredangle QYP.$ Since $PY$ and $BD$ are parallel, $Q, X, Y$ must be collinear.

Combining the facts that $Q, Z, X$ are collinear and $Q, X, Y$ are collinear, we have that $Q,X,Y,Z$ are collinear.

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1002 posts

Afo

#21 h May 12, 2021, 4:07 PM • 1 Y

Y by Mango247

https://i.ibb.co/gtS2NjD/aops5.png

Solution.

Claim 1. $XCQD$ is cyclic.
Proof.
We show that $\angle XCQ + \angle XDQ = \pi \iff \angle XCQ = \pi - \angle XDQ$ .
$\angle XCQ = \angle ACQ = \pi-\angle APQ = \angle BPQ = \angle BDQ = \pi -\angle XDQ$
Claim 2. $Y,X,Q$ collinear.
Proof.
We show that $\angle XQP = \angle YQP$ .
$\angle XQP = \angle XQD + \angle DQP = \angle XCD + \angle DBP$ $=\angle ACP + \angle YPA = \angle YQP$
Claim 3. $X,Q,Z$ collinear.
Proof.
We show that $\angle XQP+\angle PQZ = \pi\iff \angle PQZ = \pi - \angle XQP$ .
$\angle PQZ = \pi - (\angle PBD + \angle DBZ)=\pi -( \angle APY + \angle DPZ)$ $=\pi -( \angle APY + \angle DCA)=\pi -( \angle APY + \angle PCA)=\pi -\angle XQP$

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#22 h Jul 31, 2021, 9:26 PM

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Consider complete quadrilateral $APDX$ with $AP \cap DX = B$ and $AX \cap DP = C$ . Since $Q = (CAP) \cap (BDP)$ , we know it's the Miquel Point of $APDX$ . Thus, $CXDQ$ and $BAXQ$ are cyclic.

Claim: $X, Q, Z$ are collinear.

Proof. Notice $\measuredangle XQD = \measuredangle XCD = \measuredangle ACP = \measuredangle ZPC = \measuredangle ZPD = \measuredangle ZQD$ which obviously suffices. $\square$

Claim: $AY \parallel BZ$ .

Proof. Let $BX \cap AY = E$ . Then, $\measuredangle AEB = \measuredangle AYP = \measuredangle ACP = \measuredangle ZPC = \measuredangle ZPD = \measuredangle ZBD = \measuredangle ZBE$ as desired. $\square$

Thus, $\measuredangle PQY = \measuredangle PAY = \measuredangle BAY = \measuredangle ABZ = \measuredangle PBZ = \measuredangle PQZ$ implying $Q, Y, Z$ are collinear. Combining the two collinearities yields the desired result. $\blacksquare$

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#23 h Nov 22, 2021, 7:21 PM • 1 Y

Y by centslordm

Notice that $Q$ is the Miquel point of complete quadrilateral $BPCAXD.$ Let $Y'=\overline{PQ}\cap\overline{ACP}.$ Since $CAPQ$ and $BXAQ$ are cyclic, $\measuredangle BPY'=\measuredangle APY'=\measuredangle AQY'=\measuredangle ABX,$ so $\overline{PY'}\parallel\overline{BD}.$ Thus, $Y'=Y$ and $Y$ is on $\overline{PQ}.$ Similarly, $X$ is also on $\overline{PQ}.$ $\square$

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#24 h Jul 10, 2022, 11:21 PM

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solution

This post has been edited 2 times. Last edited by channing421, Sep 20, 2022, 9:18 PM

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awesomeming327.

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#25 h Jul 11, 2022, 10:50 PM • 3 Y

Y by Mango247, Mango247, Mango247

We have $XCQD$ cyclic because $\angle XCQ=\angle ACQ=\angle QPB=\angle QDB.$

We have $\angle DXQ=\angle DCQ=\angle PCQ=\angle PYQ$ and $PY \parallel BD$ implies $Y,X,Q$ collinear.

Similarly, $\angle QXC=\angle QDC=\angle PZQ$ and $XC\parallel PZ$ implies $X,Q,Z.$ The result follows.

Attachments:

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#26 h Jun 13, 2023, 5:00 PM • 1 Y

Y by HoripodoKrishno

$[asy]/* DRAGON 0.0.9.6 Homemade Script by v_Enhance. */ import olympiad; import cse5; size(11cm); real lsf=0.8000; real lisf=2011.0; defaultpen(fontsize(10pt)); /* Initialize Objects */ pair A = (-2.0, 2.0); pair B = (4.031165686870916, 3.133944005249443); pair C = (1.0, -2.0); pair D = (1.0, 0.0); pair P = IntersectionPoint(A--B,Line(C,D,0,lisf)); pair X = IntersectionPoint(A--C,Line(B,D,0,lisf)); pair O_1 = circumcenter(A,P,C); pair O_2 = circumcenter(P,D,B); pair Q = 2*foot(P,relpoint(O_1--O_2,0.5-10/lisf),relpoint(O_1--O_2,0.5+10/lisf))-P; pair Y_prime = (2)*(foot(O_1,X,Q))-Q; pair Z_prime = (2)*(foot(O_2,Q,X))-Q; /* Draw objects */ draw(circumcircle(A,P,C), rgb(0.7,0.8,0.9) + linewidth(1.0)); draw(circumcircle(B,P,D), rgb(0.7,0.8,0.9) + linewidth(1.0)); draw(A--B, rgb(0.4,0.4,0.4)); draw(C--P, rgb(0.4,0.4,0.4)); draw(A--C, rgb(0.5,0.5,0.5) + linewidth(1.0) + linetype("4 4")); draw(B--X, rgb(0.5,0.5,0.5) + linetype("4 4")); draw(P--Y_prime, rgb(0.5,0.5,0.5) + linetype("4 4")); draw(P--Z_prime, rgb(0.5,0.5,0.5) + linewidth(1.0) + linetype("4 4")); draw(Y_prime--Z_prime, rgb(0.4,0.8,1.0)); draw(circumcircle(C,Q,D), rgb(0.7,0.8,0.9) + linewidth(1.0)); draw(circumcircle(Q,A,B), rgb(0.7,0.8,0.9) + linewidth(1.0)); /* Place dots on each point */ dot(A); dot(B); dot(C); dot(D); dot(P); dot(X); dot(Q); dot(Y_prime); dot(Z_prime); /* Label points */ label("$A$", A, lsf * dir(150)); label("$B$", B, lsf * dir(45)); label("$C$", C, lsf * dir(-90)); label("$D$", D, lsf * dir(60) * 1.414); label("$P$", P, lsf * dir(100)); label("$X$", X, lsf * dir(225)); label("$Q$", Q, lsf * dir(-45)); label("$Y'$", Y_prime, lsf * dir(225)); label("$Z'$", Z_prime, lsf * dir(-45)); [/asy]$
(diagram stolen from @vEnhance)

I tried removing the point $X$ first to show that $\overline{Y-Q-Z}$ were collinear and my goodness, how badly stuck was I back then... :stretcher:

:stretcher:

But fortunately, spiral sim gotchme!

Firstly note that $\measuredangle QCA=\measuredangle QPA=\measuredangle QPB=\measuredangle QDB$ and $\measuredangle CAQ=\measuredangle CPQ=\measuredangle DPQ=\measuredangle DBQ$ , and so, $Q$ is the center of the spiral similarity taking $CA\mapsto DB$ .

So we thus get that $QCXD$ and $QXAB$ are cyclic. Now we have $\measuredangle AQX=\measuredangle ABX=\measuredangle PBD=\measuredangle ABD=\measuredangle APY=\measuredangle AQY$ which gives that $\overline{Q-X-Y}$ are collinear and similarly $\overline{Q-X-Z}$ are also collinear and we are done.

This post has been edited 2 times. Last edited by kamatadu, Jun 15, 2023, 9:19 AM

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#27 h Jan 12, 2024, 3:29 PM

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first, by spiral similarity $CQD$ and $AQB$ are similar
from that, we also get that $AQC$ and $BQD$ are similar, and angles $QAX$ and $QBX$ , so $ABQX$ is a cyclic quadrilateral.
then, if $BQZ=BAX$ then $XQZ$ are collinear, but $BQZ=BPZ=BAX$ because $PZ||AC$ , so $XQZ$ are collinear
lastly, we need $PYQ=BXQ$ to prove $Y$ is collinear with $XQZ$ , but because we have $PYQ=PAQ=BXQ$ , $Y$ is collinear with $XQZ$
therefore, $Q,X,Y$ , and $Z$ are collinear

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#28 h Jun 23, 2024, 3:58 PM

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Take the quadrilateral $APDX$ . We know that $Q$ is the Miquel point of this quadrilateral, since $Q$ is the
second intersection of $(BPD)$ and $(CAP)$ . Thus, $AXQB$ is cyclic.

Then, we have (using directed angles)
$\angle QYP = \angle QAP = \angle QAB = \angle QXB.$ Since $\overline{XB} || \overline{YP}$ , we must have that $Q$ , $X$ , and $Y$ are collinear.

Similarly,
$\angle QZP = \angle QBP = \angle QBA = \angle QXA,$ so points $Q$ , $X$ , and $Z$ are collinear.

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