Concyclicity on some quadrilateral from isosceles triangle

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18 posts

#1 h Jul 17, 2019, 12:11 PM • 12 Y

Y by Me.021, mathematicsy, zunaid_cnd, megarnie, ImSh95, Adventure10, Mango247, Rounak_iitr, deplasmanyollari, ohhh, Tastymooncake2, ehuseyinyigit

Let $ABC$ be a triangle with $AB=AC$ , and let $M$ be the midpoint of $BC$ . Let $P$ be a point such that $PB<PC$ and $PA$ is parallel to $BC$ . Let $X$ and $Y$ be points on the lines $PB$ and $PC$ , respectively, so that $B$ lies on the segment $PX$ , $C$ lies on the segment $PY$ , and $\angle PXM=\angle PYM$ . Prove that the quadrilateral $APXY$ is cyclic.

This post has been edited 1 time. Last edited by bluecarneal, Aug 14, 2019, 2:05 PM
Reason: fix typo

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a1267ab

223 posts

a1267ab

#2 h Jul 17, 2019, 12:17 PM • 7 Y

Y by mathsworm, shalomrav, ImSh95, Adventure10, Tastymooncake2, ehuseyinyigit, Sandro175

Inverting a G2 is a bit overkill, but I hope that someone finds this solution informative.

We invert about $P$ . The new problem is as follows:

Quote:

Let $PBC$ be a triangle and $\Gamma=(PBC)$ . The $P$ -Apollonius circle intersects $\Gamma$ again at $M$ and the tangent to $\Gamma$ at $P$ again at $A$ . Points $X$ and $Y$ lie on segments $PB$ and $PC$ respectively so that $\angle PMX=\angle PMY$ . Show that $A, X, Y$ are collinear.

$[asy] size(12cm); pair P=dir(125); pair B=dir(200); pair C=dir(-20); pair A= 2*extension(B, C, P, P+dir(90)*P)-P; pair M=2*foot(origin, A, -P)+P; pair X=P+0.6*(B-P); pair Y=extension(A, X, P, C); pair Z=extension(A, X, P, M); draw(anglemark(P, M, X, t=10),green); draw(anglemark(Y, M, P, t=10),green); draw(unitcircle); draw(P--B--C--cycle); draw(M--A); draw(circumcircle(P, A, M), dotted); draw(A--P--M); draw(X--M--Y, red); draw(A--Y, dashed+blue); string[] names = {"$P$", "$B$", "$C$", "$A$", "$M$", "$X$", "$Y$", "$Z$"}; pair[] pts = {P, B, C, A, M, X, Y, Z}; pair[] labels = {dir(90), B, C, A, M, dir(120), dir(45), Z}; for(int i=0; i<names.length; ++i){ dot(names[i], pts[i], dir(labels[i])); } [/asy]$
Let $Z=PM\cap AX$ and $Y'=AX\cap PC$ . Note that $\angle PMA=90^{\circ}$ by the inversion. Since $MP$ bisects $\angle XMY$ , we have
$(MA, MZ; MX, MY)=-1=(P, M; B, C)\stackrel{P}{=}(A, Z; X, Y').$ Therefore $Y=Y'$ , which solves the problem.

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v_Enhance

6874 posts

v_Enhance

#3 h Jul 17, 2019, 1:07 PM • 26 Y

Y by SHREYAS333, MathBoy23, Purple_Planet, Smkh, v4913, lambda5, mijail, HamstPan38825, Executioner230607, Elyson, phongzel34-vietnam, MatBoy-123, math31415926535, CyclicISLscelesTrapezoid, rayfish, Infinityfun, Ibrahim_K, Chokechoke, ImSh95, Adventure10, Mango247, rafid149, Rounak_iitr, Tastymooncake2, ehuseyinyigit, Funcshun840

Let $S = \overline{XM} \cap \overline{PY}$ , $T = \overline{YM} \cap \overline{PX}$ , so that quadrilateral $XYST$ is cyclic by the given angles. We denote by $O$ the center of this circle.
$[asy]size(250); defaultpen(fontsize(11pt)); pair D = dir(130); pair E = -conj(D); pair M = midpoint(D--E); pair X = dir(230); pair Y = dir(25); pair S = -X+2*foot(origin, X, M); pair T = -Y+2*foot(origin, Y, M); pair B = extension(X, T, D, E); pair C = extension(Y, S, D, E); pair P = extension(X, T, S, Y); pair A = 1/conj(M); pair O = origin; markscalefactor *= 0.6; draw(anglemark(M, X, P), lightblue); draw(anglemark(P, Y, M), lightblue); filldraw(A--B--C--cycle, invisible, red); draw(P--X--S, lightblue); draw(P--Y--T, lightblue); draw(D--E, red); filldraw(unitcircle, invisible, heavycyan); draw(O--A, red); draw(P--A, dotted+red); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$M$", M, dir(295)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(Y)); dot("$S$", S, dir(S)); dot("$T$", T, dir(T)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$P$", P, dir(P)); dot("$A$", A, dir(A)); dot("$O$", O, dir(-90)); /* TSQ Source: D = dir 130 E = -conj(D) M = midpoint D--E R295 X = dir 230 Y = dir 25 S = -X+2*foot origin X M T = -Y+2*foot origin Y M B = extension X T D E C = extension Y S D E P = extension X T S Y A = 1/conj(M) O = origin R-90 !markscalefactor *= 0.6; anglemark M X P lightblue anglemark P Y M lightblue A--B--C--cycle 0.1 lightred / red P--X--S lightblue P--Y--T lightblue D--E red unitcircle 0.1 lightcyan / heavycyan O--A red P--A dotted red */ [/asy]$
Claim: We have $\overline{OM} \perp \overline{BC}$ . Hence $\overline{PA} \perp \overline{AOM}$ .

Proof. Follows by butterfly theorem if one extends $\overline{BC}$ to a chord of the circle, since then $M$ is the midpoint of that chord. $\blacksquare$

Thus, since $M \in \overline{OA}$ and $\overline{PA} \perp \overline{AOM}$ , we conclude $A$ coincides with the Miquel point of quadrilateral $STXY$ . Therefore $PAXY$ is cyclic.

This post has been edited 1 time. Last edited by djmathman, Jul 18, 2019, 3:42 PM
Reason: adjusted diagram a bit (@Evan: feel free to change again if you want)

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fastlikearabbit

28322 posts

fastlikearabbit

#4 h Jul 17, 2019, 1:11 PM • 4 Y

Y by AbodeMokayed, ImSh95, Adventure10, Mango247

Take $X$ and $Y$ as the intersection points of a circle $\Omega$ passing through $A$ and $P$ . I shall prove that $\angle{PXM}=\angle{PYM}$ .
Let $Q$ be the antipodal of $P$ in $\Omega$ . Then $Q \in AM$ and $\angle{CYQ}=\angle{CMQ}=\angle{BMQ}=\angle{BXQ}=90^{\circ}$ , implying that both $CMQY$ and $BXQM$ are cyclic. Therefore, $=\angle{PYM}=\angle{CYM}=\angle{CQM}=\angle{BQM}=\angle{BXM}=\angle{PXM}.$ The conclusion follows.

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psi241

49 posts

psi241

#5 h Jul 17, 2019, 1:13 PM • 4 Y

Y by Infinityfun, ImSh95, Adventure10, Mango247

Let $X_1, Y_1$ be reflections of $B,C$ across $X,Y$ respectively. The angle condition gives
$\angle BX_1C = \angle BXM = \angle CYM = \angle BY_1C\implies B,C,X_1,Y_1\text{ concyclic.}$ Let $P_1$ be the reflection of $P$ across $A$ . Observe that $\triangle PX_1Y_1\stackrel{-}{\sim}\triangle PCB\stackrel{-}{\sim}\triangle P_1BC$ . Thus Gliding Principle gives $\triangle AXY\stackrel{-}{\sim}\triangle PCB$ , hence we can conclude that
$\angle AXY = \angle PCB = \angle YPA\implies A,P,X,Y\text{ concyclic}$ as desired.

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Leartia

93 posts

Leartia

#6 h Jul 17, 2019, 1:22 PM • 2 Y

Y by ImSh95, Adventure10

Perform inversion around the circle with center $M$ and radius $MC.$ We obtain the following equivalent problem: " $M$ - midpoint of $BC$ $P$ a random point (not on $BC$ ). Let $\omega_1=(PBM)$ and $\omega_2=(PCM)$ . Let $D$ be the antipode of $M$ wrt $\omega_2$ . $A$ is a point on $PD$ such that $AB=AC$ .Let $X \in \omega_1$ and $Y \in \omega_2$ such that $\angle XPM=\angle YPM$ . Prove that $APXY$ is cyclic"

$\angle XBC=\angle YCB$ $\Rightarrow$ $BX,CY$ , and $AM$ are concurrent at $L.$

$\angle PMB=\angle PAM=180-\angle PMC=180-\angle PYC=\angle PYL$ $\Rightarrow$ $APLY$ is cyclic. Symmetrically, $APXL$ is also cyclic. This implies that $APXY$ is cyclic.

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MarkBcc168

1595 posts

MarkBcc168

#7 h Jul 17, 2019, 1:58 PM • 38 Y

Y by Pluto1708, lminsl, Not_real_name, amar_04, Richangles, samuel, Arhaan, DashTheSup, rashah76, Purple_Planet, DUSHYANT, ApraTrip, hsiangshen, khina, Executioner230607, Elyson, math31415926535, megarnie, CyclicISLscelesTrapezoid, rayfish, Infinityfun, Chokechoke, BorivojeGuzic123, ImSh95, IAmTheHazard, Aryan-23, Mathlover_1, everythingpi3141592, Philomath_314, Adventure10, Mango247, Kingsbane2139, Tastymooncake2, EpicBird08, Om245, eduD_looC, Rounak_iitr, GreenTea2593

Easy for a G2
$[asy] unitsize(2.7cm); defaultpen(fontsize(10pt)); pair P = dir(110); pair B = dir(230); pair C = dir(310); pair P1 = dir(70); pair A = (P+P1)/2; pair M = (B+C)/2; pair Q = 1.4*M-0.4*A; pair X = foot(Q,P,B); pair Y = foot(Q,P,C); dot(P);dot(B);dot(C);dot(A);dot(Q);dot(X);dot(Y);dot(M); draw(X--P--Y); draw(P--B--C--cycle,linewidth(1.5)); draw(X--Q--A--P); draw(Y--Q,dashed); draw(X--M--Y); draw(B--Q--C); draw(circumcircle(A,X,Y),dotted); markscalefactor = 0.02; draw(anglemark(M,X,P)); draw(anglemark(P,Y,M)); draw(anglemark(M,Q,B)); draw(anglemark(C,Q,M)); label("$A$",A,N); label("$B$",B,W); label("$C$",C,NE); label("$P$",P,dir(P)); label("$Q$",Q,S); label("$X$",X,SW); label("$Y$",Y,SE); label("$M$",M,2*NE); [/asy]$
Let $Q$ be the point on $AM$ such that $\angle QXB=90^{\circ}$ . Notice the cylic quadrilateral $BMQX$ , thus
$\angle CYM = \angle BXM = \angle BQM = \angle CQM\implies CYQM\text{ concyclic.}$ Thus $\angle CYQ=90^{\circ}$ or $A,X,Y,P,Q$ lies on circle with diameter $PQ$ .

This post has been edited 1 time. Last edited by MarkBcc168, Aug 2, 2019, 3:32 AM

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a1267ab

223 posts

a1267ab

#8 h Jul 17, 2019, 2:10 PM • 9 Y

Y by Pluto1708, Wizard_32, anantmudgal09, ImSh95, Adventure10, Mango247, Tastymooncake2, Assassino9931, ehuseyinyigit

Daniel Zhu notes that this problem and USA TSTST 2019/5 can be simultaneously generalized as follows:

Quote:

Let $ABC$ be a triangle with point $M$ on segment $BC$ . Points $K$ , $X$ , and $Y$ are chosen so that $X$ and $Y$ lie on the extensions of $AB$ and $AC$ respectively, $K$ lies inside $ABC$ , and
$\angle BXM=\angle CYM=\angle KBC=\angle KCB.$ Line $MK$ meets the parallel from $A$ to $BC$ at $P$ . Show that $APXY$ is cyclic.

If you found the angle chasing solution to either problem, then this should be a breeze.

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liberator

95 posts

liberator

#10 h Jul 17, 2019, 4:59 PM • 43 Y

Y by babu2001, Ankoganit, Kayak, TheUltimate123, RudraRockstar, anantmudgal09, e_plus_pi, Wizard_32, AlastorMoody, Fermat_Theorem, aops29, DPS, blacksheep2003, math_pi_rate, AopsUser101, Aryan-23, Mike_The_Pro_420xx, Limerent, hsiangshen, Flash_Sloth, Kanep, tree_3, amar_04, quirtt, OlympusHero, HamstPan38825, Elyson, BrainiacVR, rama1728, CyclicISLscelesTrapezoid, ike.chen, Project_Donkey_into_M4, rayfish, ImSh95, Creeper1612, sabkx, Adventure10, Mango247, Mango247, Mango247, Tastymooncake2, shanelin-sigma, Goodguy8809

Code golf solution

The next solution drew the following reaction from graders when I submitted a version of it.

So consider yourself warned

For the memes

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hukilau17

283 posts

hukilau17

#11 h Jul 17, 2019, 8:05 PM • 9 Y

Y by RudraRockstar, blacksheep2003, recoco, amogususususus, EpicBird08, ImSh95, Adventure10, Mango247, Tastymooncake2

Strange but effective complex bash

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JustKeepRunning

2958 posts

JustKeepRunning

#12 h Jul 17, 2019, 8:27 PM • 2 Y

Y by ImSh95, Adventure10

liberator wrote:

Code golf solution

The next solution drew the following reaction from graders when I submitted a version of it.

So consider yourself warned

For the memes

What does $\text{[D]DIT}$ mean?

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AIME12345

1129 posts

AIME12345

#13 h Jul 20, 2019, 10:54 AM • 4 Y

Y by yayups, dchen, ImSh95, Adventure10

Solution

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BOBTHEGR8

272 posts

BOBTHEGR8

#14 h Jul 29, 2019, 5:33 AM • 3 Y

Y by ImSh95, Adventure10, Mango247

I only used MPT!!!!
Proof

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lukamacho2001

12 posts

lukamacho2001

#15 h Sep 10, 2019, 2:48 PM • 5 Y

Y by Pluto1708, itslumi, ImSh95, Adventure10, ehuseyinyigit

Nineth problem in this pdf

Attachments:

30 Geometry problems solved by using complex numbers.pdf (364kb)

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Reason: There was a mistake in the pdf

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PUjnk

71 posts

PUjnk

#16 h Jan 1, 2020, 3:16 PM • 4 Y

Y by AlastorMoody, howcanicreateacccount, ImSh95, Adventure10

A slightly different solution :
Let XM intersect PC at Z and YM intersect PB at W. Clearly XYZW is cyclic.

CLAIM : A is the Miguel point wrt. XYZW.

Proof: We prove XY and ZW intersect on PA. Let PM intersect ZW at Q and let XY intersect ZW at R.
Then -1=(Z,Y;Q,R)=(PC,PB;PM,PR). So PR parallel to BC.
So R lies on PA. So A is the foot of altitude from M to PR. Thus A is the Miquel point of (XYZW) [Since (XYZW) is cyclic].

XW intersects YZ at P so A lies on (PXY) proving the result.

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shendrew7

794 posts

shendrew7

#70 h Feb 18, 2024, 10:47 PM

Y by

The angle condition tells us that $(BMX)$ and $(CMY)$ congruent by LOS, so if we suppose they intersect for a second time at $N$ , the two circles are symmetric about $MN$ . From here, we can see
$\angle PAN = \angle PYN = \angle PXN = 90 \implies APXYN \text{ cyclic.} \quad \blacksquare$

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dolphinday

1323 posts

dolphinday

#72 h May 24, 2024, 4:19 PM • 1 Y

Y by cursed_tangent1434

Solved with cursed_tangent1434(who also did the asy).
Let $Z = (BXM) \cap AM$ . Note that $Z \in (PAX)$ since $\measuredangle XZA = \measuredangle XZM = \measuredangle XBM = \measuredangle XPA$ . We also have $Z \in (CYM)$ since $\angle CYM = \angle BXM = \angle BZM = \angle MZC$ . Since $\angle ZMC = \angle ZYC = \angle ZYP = ZAP = 90^\circ$ we have $Z \in (PAY)$ and it follows that $X$ , $Y \in (PAZ)$ as desired.

$[asy] import geometry; size(10cm); defaultpen(fontsize(9pt)); pen pri; pri=RGB(24, 105, 174); pen sec; sec=RGB(217, 165, 179); pen tri; tri=RGB(126, 123, 235); pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pair A,B,C,X,Y,P,Z,M; pair A = dir(90); pair B = dir(210); pair C = dir(330); pair M = midpoint(B--C); P=A-1; Z=(0, -1); pair X = intersectionpoints(circle(B,M,Z),line(P,B))[0]; pair Y = intersectionpoints(circle(C,M,Z),line(P,C))[1]; filldraw((path)(A--B--C--cycle), white+0.1*pri, pri); draw(A--P); draw(circumcircle(B,M,Z), sec+dashed); draw(circumcircle(Z,M,C), sec+dashed); draw(P--X); draw(P--C); draw(A--Z); filldraw(circumcircle(A,X,Y), tfil, tri); dot("$A$",A,dir(90)); dot("$B$",B,dir(210)); dot("$C$",C,dir(260)); dot("$M$",M,dir(210)); dot("$P$",P,dir(180)); dot("$X$",X,dir(260)); dot("$Y$",Y,dir(260)); dot("$Z$",Z,dir(260)); [/asy]$

This post has been edited 1 time. Last edited by dolphinday, May 24, 2024, 4:20 PM

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EpicBird08

1749 posts

EpicBird08

#73 h May 25, 2024, 3:29 PM

Y by

Hmmmm

By the Law of Sines, note that $(BMX)$ and $(CMY)$ have the same radius, so their radical axis by symmetry is just the perpendicular bisector of $BC,$ which $A$ lies on. Let the two circles intersect again at point $N.$ Note that $\measuredangle PXN = \measuredangle BXN = \measuredangle BMN = 90^\circ$ and similarly $\measuredangle PYN = 90^\circ.$ We also know that $\measuredangle PAN = 90^\circ$ since $AP \parallel BC.$ Thus $\measuredangle PXN = \measuredangle PAN = \measuredangle PYN = 90^\circ,$ which implies the result.

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lnzhonglp

120 posts

lnzhonglp

#74 h Jun 6, 2024, 7:03 AM

Y by

Let $(BMX)$ meet line $AM$ at $G$ . Then $\angle BXM = \angle BGM = \angle MGC = \angle MYC$ , so $MGYC$ is cyclic. We have that $\angle PXG = \angle PAG = 90^\circ$ , so $PAGX$ is cyclic. Similarly, $\angle PXG = \angle PAG = 90^\circ$ , so $PAYG$ is cyclic. Therefore, $X$ and $Y$ both lie on the circumcircle of $PAG$ , so $PAXY$ is cyclic.

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fearsum_fyz

48 posts

fearsum_fyz

#75 h Jun 26, 2024, 5:40 PM

Y by

Claim: The circumcircles of $\Delta{BMX}$ and $\Delta{CMY}$ intersect on $\overrightarrow{\rm AM}$ .
Proof: Let $D'$ be the second intersection of these circles. We have to show that $A$ is on $MD'$ , the radical axis, so it would suffice to show that $A$ has equal powers with respect to $(BMX)$ and $(CMY)$ . However, by the extended law of sines in $\Delta{BMX}$ and $\Delta{CMY}$ , they have equal circumradii. Hence it is only required to show that $AE^2 = AF^2$ where $E$ and $F$ are the circumcenters of $\Delta{BMX}$ and $\Delta{CMY}$ . Notice that $ME = MF$ (equal circumradii) and $\angle{AME} = 90^{\circ} + \angle{BME} = 90^{\circ} + \angle{CMF} = \angle{AMF}$ , so $\Delta{AME} \cong \Delta{AMF}$ by SAS, hence $AE = AF$ as desired.

Now let us call this intersection $D$ . Let $\measuredangle$ denote directed angles modulo $180^{\circ}$ .
Notice that $\measuredangle{ADX} = \measuredangle{MDX} = \measuredangle{MBX} = \measuredangle{MBP} = \measuredangle{APX}$ , hence $A, P, X, D$ are concyclic.
Similarly, $A, P, Y, D$ are concyclic.
Hence, $A, P, X, Y$ are concyclic as desired.

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P2nisic

406 posts

P2nisic

#76 h Jul 2, 2024, 1:27 PM

Y by

Neothehero wrote:

Let $ABC$ be a triangle with $AB=AC$ , and let $M$ be the midpoint of $BC$ . Let $P$ be a point such that $PB<PC$ and $PA$ is parallel to $BC$ . Let $X$ and $Y$ be points on the lines $PB$ and $PC$ , respectively, so that $B$ lies on the segment $PX$ , $C$ lies on the segment $PY$ , and $\angle PXM=\angle PYM$ . Prove that the quadrilateral $APXY$ is cyclic.

Make the circles $(BXM),(CYM)$ then since $\angle PXM=\angle PYM$ and $MB=MC$ we get that the two circle are equal.
Let $F=(MBX)\cap (MCY)$ then we have that $\angle MBF=\angle MCF$ so $A,M,F$ are collinear.

Now $\angle PXF=180-\angle BMF=180-90=90=\angle PYF=\angle PAF$ so $A,P,X,Y,F$ lie on the same circle.

Attachments:

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AngeloChu

470 posts

AngeloChu

#77 h Jul 6, 2024, 12:02 AM

Y by

construct circles $BXM$ and $CYM$
they intersect on $AM$ since $BXM=CYM$ and $BM=CM$
let that point be $O$
we also have that $PAM=AMB=AMC=CYO=BXO$ since cyclic quadrilaterals exist and $BMOX$ and $CMOY$ are cyclic
then, $APXO$ and $APYO$ are cyclic, and $X$ and $Y$ both lie on circle $APO$
thus, $APXY$ are concyclic

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Martin2001

137 posts

Martin2001

#78 h Aug 16, 2024, 11:52 PM

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Let $K= \overline{MY} \cap \overline{PB}, L=\overline{MX} \cap \overline {PC}.$ Then $A$ is the Miquel point of cyclic quadrilateral $YXKL$ by butterfly. Therefore $APXY$ is also cyclic $.\blacksquare$

This post has been edited 1 time. Last edited by Martin2001, Aug 16, 2024, 11:52 PM

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kotmhn

58 posts

kotmhn

#79 h Sep 18, 2024, 8:08 PM

Y by

Let $P = XM \cap PC \text{ } Q = YM \cap PB \text{ and } Z = (BXM) \cap AM$

By the given angle We get that $XPQY$ is cyclic. Observe that by reim's we have $PAZX$ cyclic.

Also, as $ABC$ is isosceles so is $BZC$ then we have
$\measuredangle MYC = \measuredangle BXM = \measuredangle BZM = \measuredangle MZC$ Where the last equality is due to the isosceles condition. So $YCMZ$ is cyclic hence
$\measuredangle ZYP = \measuredangle ZYC = \measuredangle ZMC = 90^{\circ} = \measuredangle ZAP$ Hence we get that $ZYAP$ is cyclic.
Combining this with the above para we get that $PAYZX$ is cyclic proving the required result.

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Kappa_Beta_725

24 posts

Kappa_Beta_725

#80 h Oct 10, 2024, 6:56 AM

Y by

Inversion + Harmonics! What else can one ask for?

Invert about $P$ with radius $\sqrt{PB\cdot PC}$ then reflect about the angle bisector of $BPC$ . Observe that $M*$ is the intersection of the symmedian and the circumcircle of $(PB^*C^*)$ , lines $X^*M^*$ and $Y^*M^*$ are reflections over the symmedian and $X^*\in PB^*$ and $Y^* \in PC^*$ . Let $\ell$ be the line through $M^*$ perpendicular to $PM^*$ . Observe that we need to prove that we need to prove that $X^*Y^*\cap PP$ is the same as $X^*Y^* \cap \ell$ . Observe that $(X^*Y^*\cap PP, X^*Y^* \cap PM^*; X^*, Y^*)\overset{P}{=} (B^*C^* \cap PP, BC \cap PM^*; B^*, C^* )=-1$ (harmonic quadrilateral) and $(X^*Y^*\cap \ell, X^*Y^* \cap PM^*; X^*, Y^*)=-1$ (angle bisector). $\blacksquare$

Edit: As it turns out this question could have been angle chased +Pop. But I think I have forgotten to do both of those things.

This post has been edited 1 time. Last edited by Kappa_Beta_725, Oct 10, 2024, 6:58 AM

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Saucepan_man02

1323 posts

Saucepan_man02

#81 h Oct 31, 2024, 3:37 AM

Y by

Here are few proofs:

Proof-1: (Angle-Chase)
Let $Z = (BMX) \cap (CMY)$ with $Z \neq M$ . Note that: $(BMZX)$ is cyclic and $AP \parallel BM$ which implies $APXZ$ to be cyclic. Similarly, $APYZ$ is also cyclic. Therefore, $APXYZ$ is cyclic and we are done.

Proof-2: (Config-Geo)

Let $S=MX \cap CY$ and $T=MY \cap BX$ . Note that: $STXY$ is cyclic. Extend $BC$ into chord $B'C'$ of circle $(STXY)$ and $O$ denote the center of $(STXY)$ . As $M$ is midpoint of $BC$ , by Butter-fly theorem $M$ is the midpoint of $B'C'$ . Therefore, $OM \perp BC$ and $OM \perp AP$ .
Notice that: $A$ is the Miquel point of quadrilateral $STXY$ which implies $APXY$ to be cyclic and we are done.

This post has been edited 1 time. Last edited by Saucepan_man02, Oct 31, 2024, 3:37 AM
Reason: Latex

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ezpotd

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ezpotd

#82 h Dec 5, 2024, 4:50 PM

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Let $(BXM), (CYM)$ meet at $K$ . I claim $(PK)$ contains $A,X,Y$ , it is easy to see that $\angle PXM = \angle PYM$ implies $\angle BXM = \angle CYM$ implies $\angle BKM = \angle CKM$ , so $K$ lies on the perpendicular bisector of $BC$ , and so does $A$ , so $AK \perp BC$ and thus $AK \perp PA$ , as desired. Likewise, this implies $KM \perp BC$ , so $\angle PXK = \angle BXK = 180 - \angle BMK = 90$ , likewise $\angle PYK = 90$ , so we are done.

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SimplisticFormulas

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SimplisticFormulas

#83 h Dec 7, 2024, 1:49 PM

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Solution:
Let $PB \cap \odot (APY)=X’ \neq P$ . Let $AM \cap \odot (APY)= N \neq A$ .
Note that $\angle PAN=90=\angle PX’N=\angle BMN$ , so $BMX’N$ must be cyclic. Also, $\angle PYN=\angle CYN=\angle CMN$ , which gives us that $YCMN$ is also cyclic. Now finally note that $\angle PX’M=\angle BX’M=\angle BNM=\angle CNM=\angle CYM=\angle PYM=\angle PXM$ , so $X’ \equiv X$ , and we are done. $\blacksquare$

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lelouchvigeo

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lelouchvigeo

#84 h Dec 11, 2024, 6:28 AM

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Nice problem, took me some time

Sol

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Ilikeminecraft

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Ilikeminecraft

#86 h Feb 9, 2025, 12:01 AM

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Draw $(BDM), (CME).$ Let the intersection be $K.$
By the given condition, we have that $AM$ bisects $\angle BKC.$ Thus, $\angle PDK = 180 - \angle BMK = 90,$ and similarly, $\angle PEK = 90.$ However, since $\angle PAK = 90,$ we get that $PAEKD$ is cyclic, which finishes.

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