Strange angle condition and concyclic points

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Strange angle condition and concyclic points

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Source: IMO 2019 Problem 2

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#145 h Dec 20, 2022, 6:48 PM

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Let the intersection of $P_1P$ and $AB$ be denoted as point D, and the intersection of $Q_1Q$ and $AB$ be denoted as point E. Also let $AA_1\cap (ABC)=F$ and $BB_1\cap (ABC)=G$ .
Because $\angle PP_1C=\angle BAC$ and $\angle CB_1P_1=\angle B_1AD$ , we get $\triangle CP_1B_1 \sim \triangle B_1AD$ and $CP_1AD$ are concyclic
Because $\angle CQ_1Q=\angle CBA$ and $\angle CA_1Q_1=\angle BA_1E$ , we get $\triangle CA_1Q_1 \sim \triangle BA_1E$ and $CQ_1BE$ are concyclic.
Because $PQ \parallel AB$ , $\angle B_1QP = \angle B_1BD$ and $\angle B_1PQ = \angle B_1DB$ so $\triangle B_1PQ\sim \triangle B_1DB$
Also because $PQ \parallel AB$ , $\angle A_1PQ = \angle A_1AE$ and $\angle A_1QP = \angle A_1EA$ so $\triangle A_1PQ \sim \triangle A_1AE$
By Reim's, $FPQG$ are concyclic.
Using Power of a point of $B_1$ with respect to $CP_1AD$ , we get $CB_1 \cdot B_1A = P_1B_1 \cdot B_1D$ .
Using Power of a point of $B_1$ with respect to $(ABC)$ , we get $EB_1 \cdot B_1B = CB_1 \cdot B_1A$ .
Using Power of a point of $A_1$ with respect to $CQ_1BE$ , we get $CA_1 \cdot A_1B = Q_1A_1 \cdot A_1E$ .
Using Power of a point of $A_1$ with respect to $(ABC)$ , we get $DA_1 \cdot A_1A = CA_1 \cdot A_1B$ .
Since $\triangle B_1PQ \sim \triangle B_1DB$ , $B_1B\cdot B_1P=B_1D\cdot B_1Q$
Since $\triangle A_1PQ \sim \triangle A_1AE$ , $A_1A\cdot A_1Q=A_1E\cdot A_1P$
We can eventually get $(B_1P_1)^2= B_1Q\cdot B_1G$ and $(A_1Q_1)^2= A_1P\cdot A_1F$
These prove that $P_1$ and $Q_1$ are both on circle $FGQP$

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awesomeming327.

#146 h May 19, 2023, 11:26 PM • 1 Y

Y by Rounak_iitr

Let $AA_1$ and $BB_1$ intersect $(ABC)$ at $D$ and $E$ , respectively. The following quadrilateral are cyclic:

$PQDE$ because $\angle QPD=\angle BAD=\angle QED$ .
$B_1EP_1C$ because $\angle B_1P_1C=\angle BAC=\angle BEC$ .
$PDP_1E$ because $\angle PP_1E=\angle B_1CE=\angle ABE=\angle PDE$ .
$QDQ_1E$ similarly to $(3)$ .

Thus, $PQP_1Q_1$ is cyclic.

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#152 h Jul 15, 2023, 5:26 PM • 4 Y

Y by Amir Hossein, HoripodoKrishno, Rounak_iitr, radian_51

$[asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ /* A few re-additions (otherwise throws error due to the missing xmin) are done using bubu-asy.py. This adds back the dps, real xmin coordinates, changes the linewidth, the fontsize and adds the directions to the labellings. */ pair A = (-80,-52.9); pair B = (66.99396,-52.9); pair C = (-42.42817,70.47100); pair T = (-39.70091,-9.14504); pair P = (-57.87643,-28.87921); pair Q = (8.42014,-28.87921); pair S = (-72.68462,-28.87921); pair R = (45.68907,-28.87921); pair B_1 = (-63.67993,0.68861); pair A_1 = (-5.11751,28.40407); pair P_1 = (-78.09177,74.11437); pair Q_1 = (-21.94016,99.58748); pair X = (-90.44944,11.66663); pair Y = (32.33519,69.06855); import graph; size(13cm); pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); pen ffxfqq = rgb(1,0.49803,0); draw(A--B, linewidth(0.5) + ffxfqq); draw((-2.75727,-52.9)--(-4.63014,-55.30797), linewidth(0.5) + ffxfqq); draw((-2.75727,-52.9)--(-4.63014,-50.49202), linewidth(0.5) + ffxfqq); draw((-6.50302,-52.9)--(-8.37589,-55.30797), linewidth(0.5) + ffxfqq); draw((-6.50302,-52.9)--(-8.37589,-50.49202), linewidth(0.5) + ffxfqq); draw(B--C, linewidth(0.5)); draw(C--A, linewidth(0.5)); draw(circle((-6.50302,-7.87639), 86.19124), linewidth(0.5)); draw(circle((-65.28053,23.14823), 52.55164), linewidth(0.5) + red); draw(circle((27.05460,43.34548), 74.58988), linewidth(0.5) + red); draw(circle((-24.72814,31.10775), 68.53645), linewidth(0.5) + linetype("4 4") + blue); draw(A--Y, linewidth(0.5) + blue); draw(X--B, linewidth(0.5) + blue); draw(P--P_1, linewidth(0.5)); draw(Q--Q_1, linewidth(0.5)); draw(S--R, linewidth(0.5) + ffxfqq); draw((-9.75203,-28.87921)--(-11.62490,-31.28719), linewidth(0.5) + ffxfqq); draw((-9.75203,-28.87921)--(-11.62490,-26.47123), linewidth(0.5) + ffxfqq); draw((-13.49777,-28.87921)--(-15.37064,-31.28719), linewidth(0.5) + ffxfqq); draw((-13.49777,-28.87921)--(-15.37064,-26.47123), linewidth(0.5) + ffxfqq); draw(P_1--C, linewidth(0.5)); draw(C--Q_1, linewidth(0.5)); dot("$A$", A, W); dot("$B$", B, dir(0)); dot("$C$", C, 1.5*dir(105)); dot("$T$", T, N); dot("$Q$", Q, NE); dot("$S$", S, NW); dot("$R$", R, NE); dot("$B_1$", B_1, 2*dir(125)); dot("$A_1$", A_1, dir(0)); dot("$P_1$", P_1, NW); dot("$Q_1$", Q_1, N); dot("$X$", X, W); dot("$Y$", Y, NE); [/asy]$

Let $R=PQ\cap CB$ and $S=PQ\cap CA$ . Also let $X=BB_1\cap\odot(ABC)$ and $Y=AA_1\cap\odot(ABC)$ . Finally let $T=AA_1\cap BB_1$ .

Firstly, as we have that $PQ\parallel AB$ , we get $\measuredangle PP_1C=\measuredangle BAC=\measuredangle PSC\implies PSP_1C$ is cyclic. Similarly, $QCQ_1R$ is cyclic. Also, we have that $\measuredangle XYP=\measuredangle XYA=\measuredangle XBA=\measuredangle XQP\implies XPQY$ is cyclic.

Now also note that $\measuredangle XCS=\measuredangle XCA=\measuredangle XBA=\measuredangle XQS\implies XSQC$ is cyclic. Similarly, $YRPC$ is also cyclic.

Now then, we have that $B_1P_1\cdot B_1P=B_1C\cdot B_1S=B_1X\cdot B_1Q\implies P_1XPQ$ is cyclic. Similarly, we also get that $Q_1YQP$ is also cyclic. Combining this with the fact that $XYPQ$ is cyclic, we get that $P_1XPQYQ_1$ is cyclic which finishes.

This post has been edited 5 times. Last edited by kamatadu, Jul 15, 2023, 5:28 PM

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#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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#154 h Sep 27, 2023, 11:24 PM • 1 Y

Y by TestX01

lengths

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#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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#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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#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

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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.

This post has been edited 3 times. Last edited by Ywgh1, Mar 29, 2024, 10:44 AM

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496 posts

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

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Solution

This post has been edited 1 time. Last edited by Mr.Sharkman, Apr 20, 2024, 7:04 PM

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3443 posts

#165 h Aug 27, 2024, 4:17 AM • 2 Y

Y by Rounak_iitr, OronSH

WHAT IN THE WORLD IS THIS :huuh:

: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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455 posts

#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:

:love:

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SimplisticFormulas

95 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

595 posts

cursed_tangent1434

#170 h Mar 7, 2025, 5:44 PM • 1 Y

Y by radian_51

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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1890 posts

#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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