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Equal lengths and concurrency on circle

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Source: Japan Mathematical Olympiad Finals 2018 Q2

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179 posts

#1 h Feb 13, 2018, 2:05 PM • 6 Y

Y by Davi-8191, nguyendangkhoa17112003, Bachsonata3, tiendung2006, Adventure10, Rounak_iitr

Given a scalene triangle $\triangle ABC$ , $D,E$ lie on segments $AB,AC$ respectively such that $CA=CD, BA=BE$ . Let $\omega$ be the circumcircle of $\triangle ADE$ . $P$ is the reflection of $A$ across $BC$ , and $PD,PE$ meets $\omega$ again at $X,Y$ respectively. Prove that $BX$ and $CY$ intersect on $\omega$ .

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544 posts

lminsl

#2 h Feb 13, 2018, 3:27 PM • 7 Y

Y by rkm0959, Mobashereh, yayitsme, Bachsonata3, AllanTian, Adventure10, Mango247

Nice problem

Lemma(Well-known) : Given $\Delta XYZ$ with $X$ -excenter $I$ . $\odot (IYZ)$ meets $XY, XZ$ at $P,Q$ respectively. Then $XP=XZ$ and $XY=XQ$ .

Back to the main problem : Let $R=BX \cap CY$ .
Since $A$ is the $P$ -excenter of $\Delta PDE$ , $D,Y$ and $E,X$ are symmetric with respect to $AP$ .
Let $B' \in BC$ be a point such that $AP$ is the perpendicular bisector of $BB'$ . Then $\angle YB'D=\angle B'BD= \angle ADY= \angle YEC \rightarrow (B'YEC)$ is cyclic. Hence $\angle RYD=\angle RCB'=\angle YEB'=\angle DXB=\angle RXD$ , so $R$ lies on $\odot (XYD)$ , as desired.

This post has been edited 1 time. Last edited by lminsl, Feb 13, 2018, 3:28 PM

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#3 h Feb 13, 2018, 4:11 PM • 5 Y

Y by lminsl, Mobashereh, Bachsonata3, Adventure10, Mango247

You can also just notice that $DE$ and $BC$ are antiparallel - this gives us the desired symmetry wrt $AP$ . (It also erases $P$ )

Then, set $Z=(ABC) \cap \omega$ . We can show that $Z$ lies on both $BX$ and $CY$ with easy angle chase.

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Bx01

#4 h Feb 15, 2018, 8:44 PM • 3 Y

Y by R8450932, Bachsonata3, Adventure10

Let $H$ orthocenter of $\triangle ABC$ is the center of $\omega$ and $\Delta$ be the line passing through $A,H,P$
Lemma $1$ : $BHECPD$ is a cyclic hexagon
Proof $1$ : $\angle AEB=\angle A=\angle ADC=\angle BPC=180-\angle BHC$
Lemma $2$ : $D,Y$ are symmetric wrt $\Delta$
Proof $2$ : $\angle AYD=\angle AED=\angle B=\angle PBC=\angle PEC=\angle YEC=\angle ADY$
Lemma $3$ : $X,E$ are symmetric wrt $\Delta$
Proof $3$ : $\angle EXP=\angle DYP=\angle YDP=\angle XEY$
We deduce, $DY\parallel BC \parallel XE$
Now, Let $K=(\omega ) \cap (XB)$ and $C'=(EA)\cap(KY)$ ,
By Pascal on $XEADYK$ , we obtain $(XE)\parallel (DY) \parallel (BC')$
we deduce that $C'$ is the intersection of $EA$ and the parallel to $XE$ in $B$ ,which is $C$
Conclusion: $K=(\omega) \cap XB \cap YC$

This post has been edited 1 time. Last edited by Bx01, Feb 16, 2018, 2:40 PM

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#6 h May 1, 2018, 4:34 AM • 7 Y

Y by RAMUGAUSS, Kagebaka, Bachsonata3, myh2910, Elnuramrv, Adventure10, Mango247

Nobody seems to have posted this solution yet, so....

Well, one may notice that triangles $ABC$ and $AXY$ are similar. Yes, that is because $\angle AXY = \angle AEY = \angle PEC = \angle PBC = \angle ABC$

Where $\angle PEC = \angle PBC$ follows from the fact that $PBCDE$ is cyclic because $\angle BEC = 180 -\angle BEA = 180 - \angle BAC = 180 - \angle BPC$ and the conclusion follows. But then this would imply that $A$ is the center of the spiral similarity which sends $XY$ to $BC$ , and as such we can show that the circumcircle of $ABC$ would intersect $\omega$ at a point which lies on both $BX$ and $CY$ . This is what we want to prove.

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756 posts

#7 h May 1, 2018, 4:39 PM • 3 Y

Y by Bachsonata3, Adventure10, Mango247

I guess this problem can also be approached using Barycentrics.

Here is what I have been able to obtain:
$\longrightarrow$ Set $A=(1,0,0)$ etc...
And since $D \in AB \implies D= (\alpha,1-\alpha,0)$ for some $\alpha \in \mathbb{R}$ .
We also know that $CA = CD$ so using Distance formula, we obtain $|CA|^2 = b^2= |CD|^2 = a^2(1-\alpha) + b^2\alpha + c^2(\alpha)(1-\alpha) \iff 0 = (\alpha -1) [ -a^2 +b^2 + c^2\alpha] \implies \alpha = \dfrac{a^2-b^2}{c^2}$ .
Similarly, we obtain $E= (a^2-c^2: 0 : c^2 +b^2 -a^2)$ .
And it is easy to obtain that $P= ( -a^2:S_c : S^2_b)$ .
Now one can evaluate that $(ADE)$ has the equation $-\sum_{cyc} a^2yz + (c^2xy + b^2yz)(x+y+z) = 0$ .

From here we can proceed to find $X,Y$ and then the desired result.
I would like someone to post a complete solution using Bary.( I cannot do so since I am new to application of Bary.

)

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#8 h Jan 31, 2019, 10:58 AM • 3 Y

Y by Bachsonata3, Adventure10, Mango247

Notice, $\angle BPC= \angle A=\angle BEA=180^{\circ}- \angle BEC =\angle CDA=180^{\circ} -\angle BDC \implies \text{ Points } P,C,E,D,B \text{ are concyclic}$ , now again angle chase, $\angle ADY=180^{\circ} -\angle AEY=\angle PEC=\angle PBC=\angle B \implies$ $BC||DY$ and Trivial to see $\Delta AED \sim \Delta ABC$ $\implies$ $\angle ADY$ $=$ $\angle AYD$ $\implies$ $AD$ $=$ $AY$ $\Longrightarrow$ $AD$ $\text{ bisects }$ $\angle XDY$ $\implies$ $XE || DY || BC$ ,

Define: $BX \cap \odot (ADE) =T$ , and let $TY \cap AE =Z$ , then, Applying Pascal's Theorem on $AEXTYD$ , $\implies Z \in BC$ $\implies \boxed{ Z=C}$ , hence, $CY, BX$ concur at $T$ on $\omega$ (or) $\odot (ADE)$

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1618 posts

#9 h Feb 21, 2019, 3:28 PM • 3 Y

Y by Bachsonata3, Adventure10, Mango247

Here is my solution for this problem
Solution
Let $H$ be orthocenter of $\triangle$ $ABC$
Since: $BH$ $\perp$ $AC$ and $AB$ = $BE$ so: $BH$ is perpendicular bisector of $AE$
Similarly: $CH$ is perpendicular bisector of $AD$
Then: $H$ is circumcenter of $\triangle$ $ADE$
Let $K$ $\equiv$ $DE$ $\cap$ $BC$ , $Z$ be second intersection of $AK$ and ( $H$ ; $HA$ )
We have: $\widehat{BDE}$ = $\dfrac{1}{2}$ $\widehat{AHE}$ = $90^0$ $-$ $\dfrac{1}{2}$ $\widehat{HAE}$ = $\widehat{ECB}$
So: $B$ , $E$ , $C$ , $D$ lie on a circle
Then: $\overline{KE} . \overline{KD} = \overline{KB} . \overline{KC} = \overline{KA} . \overline{KZ}$ or $Z$ $\in$ ( $ABC$ )
Since: $P$ is reflection of $A$ through $BC$ then $P$ $\in$ ( $BHC$ )
But: $\widehat{BAC}$ = $\widehat{AEB}$ = $\widehat{BPC}$ = $\widehat{BDC}$
Hence: $B$ , $D$ , $P$ , $C$ lie on a circle; $B$ , $E$ , $C$ , $P$ lie on a circle or $H$ , $B$ , $F$ , $P$ , $C$ , $E$ lie on a circle
Let $F$ be second intersection of $AP$ and ( $H$ , $HA$ )
We have: $\stackrel\frown{FD}$ = $\widehat{DHP}$ = $\widehat{DEP}$ = $\dfrac{1}{2}$ $\stackrel\frown{DY}$
So: $F$ is midpoint of $\stackrel\frown{DY}$ or $AP$ is perpendicular bisector $DY$
Then: $PD$ = $PY$
It leads to: $PX$ = $PE$ or $XD$ = $YE$
Hence: $\stackrel\frown{XD}$ = sđ $\stackrel\frown{YE}$
So: $\stackrel\frown{AX}$ = $\stackrel\frown{AE}$
Therefore: $\widehat{BZK}$ = $\widehat{ACB}$ = $\widehat{BDE}$ = $\dfrac{1}{2}$ $\stackrel\frown{AE}$ = $\dfrac{1}{2}$ $\stackrel\frown{AX}$ = $\widehat{XZK}$
But: $B$ , $X$ lie on a same side with $ZK$ then $X$ , $B$ , $Z$ are collinear
Similarly: $C$ , $Y$ , $Z$ are collinear
Then: $CY$ and $BX$ intersect at a point on ( $ADE$ )

Attachments:

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#10 h Feb 15, 2021, 11:37 PM • 1 Y

Y by Bachsonata3

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We can easily remove $F$ by a simple trick
by inverse of pascal on $AEXFYD$ we need to prove that lines $XE,DY,BC$ are either collinear or parallel.
by angle chaseing we get that $BDECP$ is cyclic.
from now on, $\angle XEY=180^{\circ}-\angle PEC-\angle AEX=180^{\circ}-\angle B-\angle C=\angle A=\angle PXE$ so simply $XDYE$ is trapezoid.
So $XE||DY||BC$ so we are done. $\blacksquare$

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#11 h Mar 3, 2021, 5:48 PM • 2 Y

Y by Bachsonata3, Mango247

mofumofu wrote:

Given a scalene triangle $\triangle ABC$ , $D,E$ lie on segments $AB,AC$ respectively such that $CA=CD, BA=BE$ . Let $\omega$ be the circumcircle of $\triangle ADE$ . $P$ is the reflection of $A$ across $BC$ , and $PD,PE$ meets $\omega$ again at $X,Y$ respectively. Prove that $BX$ and $CY$ intersect on $\omega$ .

a solution using spiral similarity

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L567

#12 h Apr 18, 2021, 6:43 AM

Y by

Let $H$ be the orthocenter of $\triangle ABC$ . Then since $\angle BHC = 180 - \angle BAC = 180 - \angle BEA = \angle BEC$ , $E$ lies on $(BHC)$ and similarly, we get that $D$ also lies on $(BHC)$ . Since its well known that $(BHC)$ is just the reflection of $(ABC)$ over $BC$ , it follows that $P$ also lies on $(BHC)$ and so we have that $BHECPD$ is cyclic.

Now, $\angle AXY = 180 - \angle AEY = 180 - (\angle AEB + \angle BEY)$

$= 180 - (\angle ABC + \angle BEP) = (180 - \angle ABC) - \angle BCP = \angle ABC + \angle ACB - \angle ACB = \angle ABC$

Similarly, we get that $\angle AYX = \angle ACB$ and so $\triangle ABC \sim \triangle AXY$ and so $A$ is the center of spiral similarity taking $BC$ to $XY$ and so $BX, CY$ intersect on the intersection of $(AXY), (ABC)$ , which obviously is on $\omega$

This post has been edited 2 times. Last edited by L567, Apr 18, 2021, 6:44 AM

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JustKeepRunning

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JustKeepRunning

#13 h Aug 10, 2021, 6:16 PM

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A very instructive problem for spiral sim!

Notice that by sprial sim, it suffices to show that there is a spiral similarity at $A$ that maps $XY$ to $BC$ (it is obvious that $XC$ and $YB$ do intersect, so we know that if $A$ is the sprial sim, then the intersection of $BX$ and $YC$ will be on $(AXY)$ , or $\omega$ ).

Notice that because $P$ is the reflection of $A$ in $BC,$ we get that $\angle BPC=\angle BAC,$ so we have that $DBPC$ and $BPCE$ are cyclic. We then have that $\angle AXY=\angle PEC=\angle PBC=\angle ABC$ and $\angle AYX=\angle ADX=\angle BDP=\angle BCP=\angle ACB$ . We are done!

Remarks: At first, I spent a lot of time trying to do something with parallel lines with the $A$ symmedian and its intersection with $\omega,$ and somehow use the midpoint of the symmedian as the sprial center mapping $DA$ to $AE,$ and using the midpoint as a parallel to the reflection (similar to 2018 J3). But in the end, I couldn't make it work. Did anyone use this approach?

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third_one_is_jerk

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third_one_is_jerk

#14 h Sep 7, 2021, 7:59 PM • 3 Y

Y by Mango247, Mango247, Mango247

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Let $S = \omega \cap \odot(ABC)$ and H be the orthocenter of triangle $ABC$ . Then $S$ is the center of the spiral similarity that maps $\overline{DE}$ to $\overline{BC}$ . We are going to prove that $S = BX \cap CY$ .
Consider a homothety of factor $1/2$ with center $A$ . We can easily verify that this maps $D, P, E$ to the feet of perpendiculars from $A, B, C$ on $BC, CA, AB$ , respectively. Call these points $A_1, B_1, C_1$ , respectively. Furthermore, $\omega$ maps to $\odot(AC_1HB_1)$ . Notice that $A$ is the $A_1$ -excenter and $H$ the incenter of triangle $A_1B_1C_1$ . Let $X'$ be the image of $X$ under the homothety. Then from the incenter-excenter lemma on triangle $A_1B_1C_1$ we have $AH$ perpendicular to $B_1X'$ . Thus $AH$ perpendicular to $EX$ . Since, $\measuredangle SXE = \measuredangle SDE = \measuredangle SBC$ , we have $S, X, B$ collinear. We can similarly prove that $S, Y, C$ are collinear. $\blacksquare$ .

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#15 h Apr 16, 2022, 5:48 AM

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Solution with jerr bear:

Let $N = BC\cap XY;$ it suffices to show that $A$ is the Miquel Point of $BCXY.$ Observe that by the length conditions given, $\measuredangle AEB, \measuredangle CDA,$ and $\measuredangle CPB$ are all equal to $\measuredangle BAC$ so $BECPD$ is cyclic. This implies that
$\measuredangle YPD = \measuredangle CPD - \measuredangle CPE = \measuredangle CBD - \measuredangle CBE = \measuredangle CBA - (\measuredangle CBA + 2\measuredangle BAC) = -2\measuredangle BAC,$ but since $\measuredangle DYP = \measuredangle DAE = \measuredangle BAC,$ it follows that $\measuredangle PDY = \measuredangle BAC$ so $YD, CD$ are isogonal in $\angle PDA.$ Finally, this means that $\measuredangle NXA = \measuredangle YDA = \measuredangle PDC = \measuredangle CBA = \measuredangle NBA,$ whence $ABNX$ is cyclic. By symmetry, $ACNY$ is also cyclic, so we're done. $\blacksquare$

This post has been edited 2 times. Last edited by Kagebaka, Apr 16, 2022, 9:25 PM

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#16 h Apr 16, 2022, 8:18 PM

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Clearly $B,C,D,E,P$ are concyclic and by spiral similarity $\odot (AXDYE)\mapsto \odot (BPCDE)$ $AYD\stackrel{+}{\sim} CPD, AXE\stackrel{+}{\sim} BPE,ABE\stackrel{+}{\sim} DCA\implies \measuredangle BAX=\measuredangle CAY,$ $\frac{|AB|}{|AX|}=\frac{|AB|}{|AE|}=\frac{|AC|}{|AD|}=\frac{|AC|}{|AY|}\implies AXB\stackrel{+}{\sim} AYC.$ Therefore $A$ is the Miquel point of $BXYC\implies BX\cap CY\in \odot (AXY)$ as desired.

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#17 h Sep 3, 2022, 6:37 PM

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Let $Z$ be the intersection point. Observe that $\angle ADC = \angle BEC = 180^\circ - A \text{ and } \angle BPC = A,$ so $B, D, E, C, P$ are all concyclic.

Claim. [main observation] $PD=PY$ (and similarly, $PE=PX$ ).

It suffices to show that $\triangle DPY \sim \triangle DCA$ . So $\angle DYP = 180^\circ- \angle DYE= \angle A,$ and $\angle DPE = \angle DCE$ , so this conclusion follows. $\blacksquare$

Claim. [finishing touches] $Z$ lies on $(ABC)$ .

We will show that $\triangle AXB \sim \triangle AYC$ or $\triangle AXE \sim \triangle ABC$ by spiral similarity; this is evident as both are similar to $\triangle AED$ . $\blacksquare$

So $\angle BZC = \angle DAE = \angle XAY$ and $Z \in \omega$ , as required.

This post has been edited 3 times. Last edited by HamstPan38825, Sep 3, 2022, 6:39 PM

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eibc

#18 h Aug 26, 2023, 7:57 PM

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To start off, note that
$\measuredangle BPC = \measuredangle CAB = \measuredangle ADC = \measuredangle BDC,$ so $PBDC$ is cyclic. Similarly, $PCEB$ is cyclic, so their two circumcircles coincide and all five points lie on one circle. But since
$\measuredangle AXY = \measuredangle AEY = \measuredangle CEP = \measuredangle CBP = \measuredangle ABC$ and similarly $\measuredangle XYA = \measuredangle BCA$ , we see that $A$ is the center of the spiral similarity which takes $\overline{XY}$ to $\overline{BC}$ . Thus $\overline{BX} \cap \overline{CY}$ lies on $(AXY)$ , as desired.

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#19 h Aug 27, 2023, 3:08 AM

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https://i.ibb.co/qYNY3Ks/2018-JMO-P3.png

Let $(ADE)\cap (ABC)=Z,$ we will prove that $BX\cap CY=Z$

We have $\angle AEB= \angle ADC=\angle DAC=\angle BPC \Rightarrow BDECP$ is cyclic

Hence $\angle AZY=\angle YEC=\angle CBP=\angle ABC=\angle AZC\Rightarrow \overline{Z,Y,C}$

Similarly we get $\overline{Z,X,B},$ thus we are done.

This post has been edited 2 times. Last edited by trinhquockhanh, Aug 27, 2023, 3:22 AM

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#20 h Dec 26, 2023, 10:19 PM

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Note that $PBCDE$ being cyclic implies
$\measuredangle ABC = \measuredangle CBP = \measuredangle CEP = \measuredangle AXY,$ $\measuredangle BCA = \measuredangle PCB = \measuredangle PDB = \measuredangle XYA.$
Hence there exists a spiral similarity mapping $XY \mapsto BC$ , or $BX \cap CY \in (AXY) = \omega$ . $\blacksquare$

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#21 h Mar 14, 2024, 12:20 PM

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$DE\cap BC=T, AX\cap BC=K,AY\cap BC=L$
$B$ is the circumcenter of $AEP$ and $C$ is the circumcenter of $ADP$ hence $\angle APE=90-\angle A=\angle DPA$
$\angle DCE=180-2\angle A=\angle DBE\implies B,C,D,E$ are cyclic.
$\angle YDP=180-\angle DYE=\angle A=90-\angle APY\implies AP\perp DY$
Similarily $AP\perp EX$
And $AP\perp BC\perp DY\parallel EX\implies BC\parallel DY\parallel EX$
We have $AD=AY, AB=AL, AX=AE, AK=AC$
$\angle TKA=180-\angle C=\angle TDA\implies T,K,D,A$ are cyclic.
$\angle DTK=\angle DAK=\angle EAL\implies A,E,L,T$ are cyclic.
By inversion centered at $A$ with radius $\sqrt{AD.AB}$ , we get
$D \leftrightarrow B, Y \leftrightarrow L, X\leftrightarrow K, E \leftrightarrow C$
$BX\rightarrow (ADKT)$
$CY\rightarrow (AELT)$
$(ADE)\rightarrow TBC$
So $BX, CY, (ADE)$ pass through $T^*$ as desired. $\blacksquare$

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cursed_tangent1434

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cursed_tangent1434

#22 h Aug 7, 2024, 12:21 AM

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Very straightforward. Note that
$\measuredangle XYA = \measuredangle XDA = \measuredangle PDA = 2\measuredangle CDA = \measuredangle BCA$ and
$\measuredangle AXY = \measuredangle AEY = \measuredangle AEP = 2\measuredangle AEB = \measuredangle ABC$ Thus, $\triangle AXY \sim \triangle ABC$ . This means $A$ is the center of the spiral similarity mapping $XY$ to $BC$ . But then it is well known that $Z=\overline{BX} \cap \overline{CY}$ is the second intersection of $(AXY)$ and $(ABC)$ which means $Z$ indeed lies on $\omega$ as claimed.

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#23 h Oct 22, 2024, 4:26 PM

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Note that $BDECP$ is cyclic because $180^{\circ}-\angle BDC=180^{\circ}-\angle BEC=\angle BPC=\angle A$ . Then note that since $CD=CA=CP$ ,
$\angle ABC=\angle DBC=\angle DPC=\angle PDC=\angle PEC=\angle ADY,$ so $BC$ and $DY$ are parallel. Similarly, these are both parallel to $EX$ as well.

Now I claim that $BX$ and $CY$ meet at the Miquel point of $BDEC$ , say $M$ . We will show that $BXM$ are collinear, and similarly $CYM$ will be collinear and we will be done.

But indeed, $180^{\circ}-\angle EXM=\angle EDM=\angle CBM$ . Since $XE$ and $BC$ are parallel, we get that $\angle BMX=180^{\circ}$ , as desired. $\blacksquare$

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kotmhn

#24 h Oct 27, 2024, 1:50 PM

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Angle chase solution:
Let $Z= (ABC)\cap (ADE)$

Claim: $BDECP$ is cyclic
Observe that
$\angle BDC = 180 - A = \angle BEC$ Hence $BDEC$ cyclic.
Also $P$ is the reflection of $A$ about $BC$ we get
$\angle BPC = \angle BAC = \angle BEA = 180 - \angle BEC$ Hence $BECP$ is cyclic. As $(BECP)$ and $(BDEC)$ share three point , namely $B,E,P$ , the circle are same.
Hence the claim is proved.
Call the circle $(BDECP)$ , $\Omega$ .
Next
$\measuredangle AEX = \measuredangle ADX = \measuredangle BDP \stackrel{\Omega}{=} \measuredangle BCP = \measuredangle ACB$ And
$\measuredangle PEC = \measuredangle PBC = CBA$ Finally
$\begin{align*}\measuredangle XZY & \equiv \measuredangle XZC \\& = \measuredangle XYE \\& = -\measuredangle AEX - \measuredangle YEC \\& \equiv -\measuredangle AEX - \measuredangle PEC \\& = -\measuredangle ACB - \measuredangle CBA \\& = \measuredangle BAC \\ & \stackrel{\Omega}{=} \measuredangle BZC \end{align*}$ Hence $B-Z-X$ .
Final Finally
$\begin{align*} \measuredangle AZY & = \measuredangle AEY \\ & \equiv \measuredangle PEC \\ & = \measuredangle CBA \\ & \stackrel{\Omega}{=} - \measuredangle AZC \end{align*}$
Hence we $C-Y-Z$ .
And we are done, as $Z$ is the required opint

This post has been edited 1 time. Last edited by kotmhn, Oct 27, 2024, 1:54 PM

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#25 h Feb 14, 2025, 2:00 PM

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Rename $D$ to $F$
It is fairly obvious that $H$ is the circumcenter $AEF$
Note that $BFHECP$ is cyclic since $\angle BFE = 180 -\angle AFE = 180 - \angle C$ and $\angle BFH = 180 - \angle HFA = 180 - \angle BAD = 180 - \angle HCB$ and $P$ lies on $(HBC).$
Now, observe that $C$ lies on the perp bisector of $AF$ so $ACF$ is isosceles
Next, notice that angle $FYP = 180 - FYE = FAE$ and $FPY = FPE = FCE = FCA$
Thus, $FP = FY$
Thus, there is spiral centered at $A$ mapping $XB$ to $YC$
This implies $A$ is the Miquel point of quad $XBCY$
However, $G = XB cap YC$ , so $GAXY$ is cyclic.

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#26 h Mar 13, 2025, 6:30 AM

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Let $Z = BX \cap CY$ and $H$ be the orthocenter of $\triangle A BC$ .

Claim: $BPCDE$ cyclic.
Proof. Angle chase with isosceles triangles.

Claim: $\triangle ABX \sim \triangle AYC$ .
Proof. Note $(ADE)$ symmetric about $AP$ as $(ADE)$ is the image of $(AH)$ under a homothety with scale factor $2$ . Therefore $\angle XAB = \angle YAC$ by symmetry. Also $AX/AY = AX/AD = \sin \angle AEX/\sin \angle AXP = \sin \angle ADX/\sin \angle AED = \sin \angle C/\sin \angle B = AB/AC$ proving the similarity.

Thus $A$ is the Miquel point of $BCXY$ . However, then $Z \in \omega$ follows immediately.

This post has been edited 1 time. Last edited by Shreyasharma, Mar 13, 2025, 6:30 AM

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