2021 ELMO Problem 1

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5289 posts

#1 h Jun 24, 2021, 7:01 AM • 9 Y

Y by PRMOisTheHardestExam, A-Thought-Of-God, starchan, centslordm, megarnie, tiendung2006, itslumi, Rounak_iitr, ItsBesi

In $\triangle ABC$ , points $P$ and $Q$ lie on sides $AB$ and $AC$ , respectively, such that the circumcircle of $\triangle APQ$ is tangent to $BC$ at $D$ . Let $E$ lie on side $BC$ such that $BD = EC$ . Line $DP$ intersects the circumcircle of $\triangle CDQ$ again at $X$ , and line $DQ$ intersects the circumcircle of $\triangle BDP$ again at $Y$ . Prove that $D$ , $E$ , $X$ , and $Y$ are concyclic.

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MrOreoJuice

594 posts

MrOreoJuice

#2 h Jun 24, 2021, 7:01 AM • 15 Y

Y by CANBANKAN, PRMOisTheHardestExam, mathisawesome2169, centslordm, MathThm, BVKRB-, megarnie, 636510, Kagebaka, TETris5, Nuterrow, Mathlover_1, paccongnong19hy, Aryan-23, NonoPL

Solution
LOL

This post has been edited 2 times. Last edited by MrOreoJuice, Oct 15, 2021, 7:01 AM
Reason: cleaned up the solution by a bit and added a small remark :P

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pad

1671 posts

pad

#3 h Jun 24, 2021, 7:01 AM • 6 Y

Y by mormonmath, CANBANKAN, PRMOisTheHardestExam, centslordm, megarnie, Rounak_iitr

Diagram

Claim: We have $\overline{CX}\parallel \overline{AB}$ and $\overline{BY}\parallel \overline{AC}$ .

Proof: We have
$\begin{align*} \angle DXC &= 180^\circ - \angle DQC = \angle QDC+\angle ACB \\ &= \angle DAC+\angle ACB = 180^\circ-\angle ADC=\angle ADB. \end{align*}$ Hence
$\begin{align*} \angle DCX &= 180^\circ-\angle CDX-\angle DXC \\ &= 180^\circ - \angle PDB - \angle ADB \\ &= 180^\circ-\angle PAD-\angle ADB \\ &=\angle ABC. \end{align*}$ This proves the claim. $\blacksquare$

The key is to now define $A'=\overline{CX}\cap \overline{BY}$ . By the claim, $ACA'B$ is a parallelogram. Hence $M\in \overline{AA'}$ . Now think of the problem in terms of $\triangle A'BC$ . Notice $X$ is the point on $\overline{A'C}$ such that $\angle DXC=\angle ADB=\angle A'EC$ , where the last equality follows by reflection over $M$ . Hence $\triangle DXC\sim \triangle A'EC$ , so $DXA'E$ cyclic. Similarly, $DEYA'$ is cyclic. Combining, in particular $DEXY$ is cyclic.

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lilavati_2005

357 posts

lilavati_2005

#5 h Jun 24, 2021, 7:06 AM • 5 Y

Y by CANBANKAN, PRMOisTheHardestExam, centslordm, A-Thought-Of-God, Rounak_iitr

Let $A'$ be the reflection of $A$ over $M$ . We know that $ABA'C$ and $ADA'E$ are parallelograms.
$\angle AQD = \angle BPD = 180 - \angle BYD \Longrightarrow BY\parallel AC$ $\angle APD = \angle CQD = 180 - \angle DXC \Longrightarrow CX \parallel AB$ Hence $BY\cap CX = A'$ .
$\angle BAC = 180-\angle PDQ = 180-\angle YDX=\angle BA'C \Longrightarrow A' \text{ lies on } (DYX).$ $\angle DEA' = \angle ADE = \angle APD = \angle BYD = 180-\angle DYA' \Longrightarrow E \text{ lies on } (DYA').$ Hence $E$ lies on $(DYX).$

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bora_olmez

277 posts

bora_olmez

#7 h Jun 24, 2021, 7:12 AM • 4 Y

Y by CANBANKAN, PRMOisTheHardestExam, centslordm, ehuseyinyigit

Essentially the same compared to the solutions above (and probably many below as well - in the future) - posting for storage.

By Reim's Theorem on the circumcircles of $PDQA$ and $DQCX$ and using that $A,Q,C$ and $P,D,X$ are collinear, we get that $PA \parallel$ $XC$ and therefore $AB \parallel CX$ and similarly, $BY \parallel AC$ .
Let $M$ be the midpoint of $BC$ and let $S$ be the point such that $ABSC$ is a parallelogram. Notice that there for $S \in BY$ and $S \in CX$ as $BY \parallel AC$ and $AB \parallel CX$ .
Notice that $M$ lies on $AS$ and $ME = MD$ as $BD = EC$ meaning that $SEAD$ is also a paralleogram.
Then $\angle SED = \angle EDA = \angle DAC + \angle DCA = \angle DQC + \angle DCQ = \angle DXC$ meaning that $S,E,D,X$ are conyclic. Analogously , $S,Y,E,D$ are also conyclic meaning that in conclusion $D,E,X,Y$ are conyclic. $\square$

This post has been edited 1 time. Last edited by bora_olmez, Jun 25, 2021, 7:19 PM

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Imayormaynotknowcalculus

974 posts

Imayormaynotknowcalculus

#8 h Jun 24, 2021, 7:15 AM • 6 Y

Y by a_n, starchan, CANBANKAN, centslordm, Mango247, Mango247

Note that after completing the parallelogram, you can also finish by showing (with some length chasing) that $\operatorname{Pow}_{(XDY)}(B)=\operatorname{Pow}_{(XDY)}(C)$ .

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starchan

1603 posts

starchan

#9 h Jun 24, 2021, 7:28 AM • 2 Y

Y by CANBANKAN, centslordm

Imayormaynotknowcalculus wrote:

Note that after completing the parallelogram, you can also finish by showing (with some length chasing) that $\operatorname{Pow}_{(XDY)}(B)=\operatorname{Pow}_{(XDY)}(C)$ .

Exactly my approach:(probably?)
Sol

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MatBoy-123

396 posts

MatBoy-123

#10 h Jun 24, 2021, 7:40 AM • 5 Y

Y by CANBANKAN, centslordm, samrocksnature, TechnoLenzer, StarLex1

Mine is an Inversive Solution -

We start with a claim -
Claim - $(BPD)$ & $(CDQ)$ are tangent at $D$ .
Proof -Consider a tangent $l_a$ at $D$ to $(BPD)$ and let $\measuredangle(l_a , MN)$ deontes the angle between the tangent $l_a$ and the chord $MN$ .

Now by Reims Theorem , $PY \parallel QX$ , so $\measuredangle(l_a , DX) =\measuredangle(l_a ,PD) = \measuredangle PBD = \measuredangle PYD = \measuredangle XQD$ $\blacksquare$

Redefine $Y$ as $DQ \cap (DEX)$ , so we have to prove that $Y$ lies on $(DPB)$
Now we invert around $D$ with arbitrary radius , call this mapping as $\Psi$ , and denote the image of object $K$ by $K'$ .
Note that $\begin{align*} \Psi: (APQD) \to \overline {A'P'Q'} \\ \Psi: (PBD) \to \overline {P'B'} \\ \Psi: (CQXD) \to \overline {C'X'Q'} \\ \Psi: (EXYD) \to \overline {E'X'Y'} \\ \Psi: Y \to \overline{P'B'} \cap \overline{DQ'} \\ \end{align*}$
In the inverted sketch note that $B'P' \parallel C'Q'$ as $(BPD)$ & $(CDQ)$ are tangent at $D$ , also $P'Q' \parallel C'B'$ , due to tangency given in question . So we got a rectangle $P'Q'C'B'$ .
So in the inverted picture we have to prove that $Y'$ lies on $P'D'$ , which is equivalently to prove that $\triangle B'E'Y' \sim C'E'X'$ , but note that $\measuredangle E'B'Y' = \measuredangle E'C'X' = 90$ , hence we have to prove that $\frac{B'E'}{C'E'} = \frac{B'Y'}{C'X'}$ , but we have $BD = CE$ which is equivalently having $C'D. E'D = B'D.C'E'$ , and we also have $D'E' = D'B' + B'E'$ , by putting this value and doing some computations we get $\frac{B'E'}{C'E'} = \frac{B'D'.D'E'}{C'E'.C'D'}$ , but from $C'D. E'D = B'D.C'E'$ we got this equal to ${\frac{D'E'}{C'E'}}^2$ , so now we have to prove that $({\frac{D'E'}{C'E'}})^2 = \frac{B'Y'}{C'X'}$
but from $\triangle D'B'Y' \sim D'C'Q'$ and $\triangle D'B'P' \sim D'C'X'$ , we get $\frac{B'Y'}{C'X'} = \frac{C'Q'. B'P'}{{C'X'}^2}$ , but as $C'Q' = B'P'$ from rectangle $P'Q'C'B'$ , we got this equal to $({\frac{B'P'}{C'X'}})^2$ , so we have to prove -
$\frac{D'E'}{C'E'} = \frac{B'P'}{C'X'}$ , now using Inversion distance formula this is equivalently to prove $\frac{CX}{CE} = \frac{DX.BP}{DP.BD}$ , in the reinverted picture.

Now note that as $BD = CE$ , this is equivalently to prove that $CX . DP = DX. BP$ , but this follows from the fact $\triangle BDP \sim CXD$ . $\blacksquare$

Original Diagram -

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -7.05, xmax = 12.51, ymin = -11.07, ymax = 1.97; /* image dimensions */ pen ffvvqq = rgb(1,0.3333333333333333,0); pen qqwuqq = rgb(0,0.39215686274509803,0); pen bfffqq = rgb(0.7490196078431373,1,0); pen qqffff = rgb(0,1,1); draw((-0.2,1.88)--(-5.42,-4.26)--(7.98,-4.76)--cycle, linewidth(0.5) + ffvvqq); /* draw figures */ draw((-0.2,1.88)--(-5.42,-4.26), linewidth(0.5) + ffvvqq); draw((-5.42,-4.26)--(7.98,-4.76), linewidth(0.5) + ffvvqq); draw((7.98,-4.76)--(-0.2,1.88), linewidth(0.5) + ffvvqq); draw(circle((0.19930503825233759,-1.28120070027316), 3.1863198805175252), linewidth(0.5) + blue); draw(circle((4.007759537127925,-4.948044404971607), 3.9766889484996537), linewidth(0.5) + qqwuqq); draw(circle((-2.6710397474259597,-4.129865231015716), 2.7520387948447462), linewidth(0.5) + qqwuqq); draw((-2.9849537476134755,-1.3957885077292604)--(4.461364456383953,-8.898778206475576), linewidth(0.5) + bfffqq); draw((-2.233188374984234,-6.846849534242625)--(3.3750654351639153,-1.022009106294425), linewidth(0.5) + bfffqq); draw(circle((1.1608410608336417,-7.7734595696584154), 3.4870900072901767), linewidth(0.5) + linetype("4 4") + qqffff); /* dots and labels */ dot((-0.2,1.88),dotstyle); label("$A$", (-0.13,1.65), NE * labelscalefactor); dot((-5.42,-4.26),dotstyle); label("$B$", (-5.35,-4.05), NE * labelscalefactor); dot((7.98,-4.76),dotstyle); label("$C$", (8.05,-4.55), NE * labelscalefactor); dot((-2.9849537476134755,-1.3957885077292604),dotstyle); label("$P$", (-2.91,-1.19), NE * labelscalefactor); dot((0.06057838829875983,-4.464499193593237),dotstyle); label("$D$", (0.15,-4.27), NE * labelscalefactor); dot((3.3750654351639153,-1.022009106294425),linewidth(1pt) + dotstyle); label("$Q$", (3.45,-0.87), NE * labelscalefactor); dot((2.504638229241978,-4.555695456314999),dotstyle); label("$E$", (2.59,-4.35), NE * labelscalefactor); dot((4.461364456383953,-8.898778206475576),linewidth(1pt) + dotstyle); label("$X$", (4.55,-8.73), NE * labelscalefactor); dot((-2.233188374984234,-6.846849534242625),linewidth(1pt) + dotstyle); label("$Y$", (-2.15,-6.69), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Inverted Picture-
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -14.67, xmax = 14.67, ymin = -9.06, ymax = 9.06; /* image dimensions */ pen ttzzqq = rgb(0.2,0.6,0); draw((-3.06,2.94)--(3.84,2.96)--(3.8518842941878244,-1.1400814947994555)--(-3.02,-1.16)--cycle, linewidth(0.5) + red); /* draw figures */ draw((-3.06,2.94)--(3.84,2.96), linewidth(0.5)); draw((-3.06,2.94)--(3.84,2.96), linewidth(0.5) + red); draw((3.84,2.96)--(3.8518842941878244,-1.1400814947994555), linewidth(0.5) + red); draw((3.8518842941878244,-1.1400814947994555)--(-3.02,-1.16), linewidth(0.5) + red); draw((-3.02,-1.16)--(-3.06,2.94), linewidth(0.5) + red); draw((-9.480003696671316,-1.1787246483961487)--(3.8704057936921346,-7.529998823786401), linewidth(0.5) + ttzzqq); draw((-9.480003696671316,-1.1787246483961487)--(3.8518842941878244,-1.1400814947994555), linewidth(0.5) + red); draw((3.8518842941878244,-1.1400814947994555)--(3.8704057936921346,-7.529998823786401), linewidth(0.5) + blue); draw((-3.02,-1.16)--(-3.039180121421116,-4.242858109714994), linewidth(0.5) + blue); /* dots and labels */ dot((-3.06,2.94),dotstyle); label("$P'$", (-2.94,3.24), NE * labelscalefactor); dot((3.84,2.96),dotstyle); label("$Q'$", (3.96,3.27), NE * labelscalefactor); dot((-3.02,-1.16),dotstyle); label("$B'$", (-2.91,-0.87), NE * labelscalefactor); dot((3.8518842941878244,-1.1400814947994555),linewidth(4pt) + dotstyle); label("$C'$", (3.96,-0.9), NE * labelscalefactor); dot((-9.480003696671316,-1.1787246483961487),dotstyle); label("$E'$", (-9.36,-0.87), NE * labelscalefactor); dot((3.8704057936921346,-7.529998823786401),dotstyle); label("$X'$", (3.99,-7.23), NE * labelscalefactor); dot((-3.039180121421116,-4.242858109714994),linewidth(4pt) + dotstyle); label("$Y'$", (-2.91,-3.99), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

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Ainulindale

32 posts

Ainulindale

#11 h Jun 24, 2021, 7:44 AM • 3 Y

Y by CANBANKAN, centslordm, starchan

A cleaner length finish, with the diagram as in #2:

Since $BD$ is tangent to $(APD)$ , we have that $BP\cdot BA=BD^2$ , so $BP = \frac{BD^2}{BA} = \frac{BD^2}{CF}$ since $CF=AB$ .

Then as $AB\parallel CX$ , we note that $\triangle CXD\sim \triangle BPD$ and hence $CX=\frac{CD}{BD} \cdot BP$ . Combining,
$CX\cdot CF = \frac{CD}{BD} \cdot \frac{BD^2}{CF} \cdot CF = CD\cdot BD$ which implies $E$ lies on $(FXD)$ .

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lneis1

243 posts

lneis1

#12 h Jun 24, 2021, 7:52 AM • 3 Y

Y by starchan, centslordm, Mango247

Storage

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A-Thought-Of-God

454 posts

A-Thought-Of-God

#13 h Jun 24, 2021, 7:55 AM • 5 Y

Y by Aimingformygoal, centslordm, Mango247, Mango247, Mango247

This is what I submitted. :maybe:

Define $K \equiv \overline{BY} \cap \overline{CX}$ . We begin with the following claim.

Claim. We have, $ABKC$ a paralleogram.
Proof. We show that opposite sides are parallel. Indeed,
$\angle DBY = \angle PBY - \angle PBD = PDQ - \angle ABC = 180^\circ - \angle BAC - \angle ABC = \angle C$ so

$\angle ABK + \angle BAC = \angle PBD + \angle DBY + \angle BAC = \angle B + \angle C + \angle A = 180^\circ$ implying that $\overline{AC} \parallel \overline{BK}$ . Similarly, $\overline{AB} \parallel \overline{CK}$ , which establishes our claim. $\square$

Now since $AB=KC$ and $BD=CE$ , it is easy to see that $\triangle ABD \sim \triangle KCE$ . Also, $\angle YKX + \angle YDX = \angle BKC + \angle PDQ = \angle A + 180^\circ - \angle A = 180^\circ$ implying that $YDXK$ is cyclic. It suffice to show that $E$ lie on this circle. Indeed, $\angle CEK = \angle BDA = \angle BPD = \angle DYK$ where first equality follows from similarity, while second follows from the similarity of $\triangle BDA$ and $\triangle BPD$ . This implies that $E \in \odot(YDXK)$ , which finishes the problem. $\blacksquare$

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Aimingformygoal

126 posts

Aimingformygoal

#14 h Jun 24, 2021, 8:03 AM • 2 Y

Y by A-Thought-Of-God, centslordm

Solution

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rafaello

1079 posts

rafaello

#15 h Jun 24, 2021, 8:06 AM • 2 Y

Y by PRMOisTheHardestExam, centslordm

By Reim's theorem, we obtain that $CX\parallel AB$ and $BY\parallel AC$ . Let $A'$ be the reflection of $A$ over $M$ , the midpoint of $BC$ . Thus, $X,Y$ lie on $A'C$ , $A'B$ respectively. By Reim again, we obtain that $A'DXY$ is cyclic. $E$ also lies on that circle, since $\measuredangle DEA'=\measuredangle EDA=\measuredangle DPA=\measuredangle DXA'$ , where we used tangent-chord theorem.

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SerdarBozdag

892 posts

SerdarBozdag

#16 h Jun 24, 2021, 8:09 AM • 1 Y

Y by centslordm

The same solution:

Let $BY\cap CX=T$ . We will show that $X$ and $Y$ are on the circumcircle of $DET$ . Because of the circles we have $\angle A=\angle QDX=180-\angle QCX$ and $\angle A=\angle PDY=180-\angle PBY$ . These equalities show that $ABTC$ is a parallelogram.

$\textbf{Claim:}$ $\triangle DXC \sim \triangle ADB$ .

$\textbf{Proof:}$ $\angle ABD=\angle ECT=\angle DCX$ because $ABTC$ is a parallelogram.

$\angle ADB=180-\angle ADC=180-\angle DQC=\angle DXC$ . Second equality comes from tangecy condition and the third one comes form the fact that $DQCX$ is cyclic.

$\angle DAB=\angle CDX$ and $\angle ADB=\angle DXC$ finishes the proof. $\blacksquare$

Note that $\angle BAD=\angle CDX$ from the claim. Because $ABTC$ is a parallelogram we have , $AB=CT$ . With $\angle ABD=\angle ECT$ and $BD=CE$ , we can say that $\triangle BDA \cong \triangle CET \implies \angle ETC=\angle BAD$ .
Thus we have $\angle ETC=\angle DAB=\angle CDX$ which implies $DEXT$ is a cyclic quadrilateral. By symmetry $DEYT$ is also a cyclic quadrilateral. In conclusion; $D$ , $E$ , $X$ , $Y$ are concyclic. $\square$

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PRMOisTheHardestExam

409 posts

PRMOisTheHardestExam

#17 h Jun 24, 2021, 8:09 AM • 1 Y

Y by centslordm

Imayormaynotknowcalculus wrote:

Note that after completing the parallelogram, you can also finish by showing (with some length chasing) that

$\operatorname{Pow}_{(XDY)}(B)=\operatorname{Pow}_{(XDY)}(C)$

.

But why length bash such an elegant problem :woozy_face:, my solution is essentially the same parallelogram + angle chase thingy : D

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nmoon_nya

26 posts

nmoon_nya

#62 h Nov 5, 2023, 11:46 AM • 1 Y

Y by PRMOisTheHardestExam

$\angle{BPY} = \angle{BDY} = \angle{CEQ} = \angle{CXQ}$ , $\angle{PBD} = \angle{PYQ}$ . So $\triangle{BDP}$ and $\triangle{YQP}$ are similar by AAA so $QY : DY = BD : BP$ . Also $\angle{BYP} = \angle{BDP} = \angle{CEX} = \angle{CQX}$ so $\triangle{BPY}$ and $\triangle{CXQ}$ are similar by AAA and $BP : CX = YP : QX$ .
Moreover $BD = CE$ so $CX : CE = QX : QY$ .And $\angle{XCE} = \angle{XQY}$ so $\triangle{CXE}$ so $\triangle{QXY}$ are similar by SAS and $\angle{CEX} = \angle{DYX}$ . Then we are done.

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cj13609517288

1888 posts

cj13609517288

#63 h Dec 30, 2023, 6:31 PM

Y by

Used the 10% hint on ARCH.

Assume WLOG that $B,D,E,C$ are collinear in that order. Note that by Reim's theorem, $BY\parallel AC$ and $CX\parallel AB$ . Thus if $BY$ and $CX$ meet at $Z$ , quadrilateral $ABZC$ is a parallelogram. We have
$\angle CDX=\angle PDB=\angle PAD.$ Similarly, $\angle BDY=\angle QAD$ , so
$\angle YDX=180^{\circ}-\angle BDY-\angle CDX =180^{\circ}-\angle QAD-\angle PAD =180^{\circ}-\angle BAC =180^{\circ}-\angle BZC.$ Therefore, quadrilateral $YDXZ$ is cyclic. But
$\angle EZX=\angle EZC=\angle DAB=\angle PAD=\angle CDX=\angle EDX,$ so quadrilateral $EDZX$ is cyclic. Therefore, $E$ and $Y$ are both on $(DXZ)$ , so quadrilateral $DEXY$ is cyclic. $\blacksquare$

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Leo.Euler

577 posts

Leo.Euler

#64 h Mar 1, 2024, 4:23 AM • 1 Y

Y by GeoKing

Bit hard for a p1?

First, note that since $(APDQ)$ is cyclic, $\angle BYD = 180^{\circ} - \angle BPD = \angle APD = 180^{\circ} - \angle AQD = \angle DQC,$ which implies $BY \parallel AC$ . Analogously, $CX \parallel AB$ . Let $A' = BY \cap CX$ , and realize by our prior observation that $A'=B+C-A$ . By easy angle chasing using the fact that $ABA'C$ is a parallelogram, $(A'XDY)$ is cyclic.

Let $N$ denote the intersection of $(DXY)$ and $BC$ . Now using the tangency condition, we have $\angle BAD = \angle BDP = 180^{\circ} - \angle BDX = \angle NA'C$ . Similarly, $\angle NA'B = \angle CAD$ . Thus, upon reflection about $M$ , we have $N=E$ , as desired.

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P2nisic

406 posts

P2nisic

#65 h Jul 7, 2024, 10:55 PM • 1 Y

Y by GeoKing

reaganchoi wrote:

In $\triangle ABC$ , points $P$ and $Q$ lie on sides $AB$ and $AC$ , respectively, such that the circumcircle of $\triangle APQ$ is tangent to $BC$ at $D$ . Let $E$ lie on side $BC$ such that $BD = EC$ . Line $DP$ intersects the circumcircle of $\triangle CDQ$ again at $X$ , and line $DQ$ intersects the circumcircle of $\triangle BDP$ again at $Y$ . Prove that $D$ , $E$ , $X$ , and $Y$ are concyclic.

$<YBP=<PDQ=180-<A$ so $YB//AC$ similar $XC//AB$ let $A'$ be the symmetric poin of $A$ we respect the midpoint of $BC$ then we have:
$<XMY=<A=180-<XDY$ so $X,D,Y,A'$ are concyclic.
$<DEA'=<ADC=<APD=<BYD$ so $E$ belongs to $(XDYA')$
We use that $ADA'E$ is parallhlogram

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Z4ADies

63 posts

Z4ADies

#66 h Aug 28, 2024, 7:35 PM

Y by

Solution using Ptolemy's sine lemma (no need to any syntehtic constructions).

Let $\angle BAD=\angle PQD=\angle PDB=\angle PYB=\angle CDX=\alpha$ and $\angle CAD=\angle QPD=\angle QDC=\angle PDY=\beta$ .
Ptolemy sine to quadrilaterals $PBYD$ and $QDXC$ from pencil $D$ .
$(....1)$ $DP\sin(\beta)+DY\sin(\alpha)=DB\sin(\alpha+\beta)$ .
$(....2)$ $DQ\sin(\alpha)+DX\sin(\beta)=DC\sin(\alpha+\beta)$ .
If $DEXY$ is cyclic, then $DE\sin(\alpha+\beta)+DY\sin(\alpha)=DX\sin(\beta)(?)$ .
Let's write $(....2)$ as $DQ\sin(\alpha)+DX\sin(\beta)=DE\sin(\alpha+\beta)+DB\sin(\alpha+\beta)$ more precisely $DQ\sin(\alpha)+DX\sin(\beta)=DE\sin(\alpha+\beta)+DP\sin(\beta)+DY\sin(\alpha)$ if we prove $DQ\sin(\alpha)=DP\sin(\beta)$ we will finish the problem.
So,sine thrm to $\triangle DPQ$ gives us $DQ\sin(\alpha)=DP\sin(\beta)$ .

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Saucepan_man02

1319 posts

Saucepan_man02

#67 h Oct 30, 2024, 11:12 AM

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Angle-Chase:

By Reims, $BY \parallel AC$ and $CX \parallel AB$ .
Extend it to point $F$ such that $ABFC$ is a parallelogram.
Notice that: $\angle YDX = \angle PDQ = 180^\circ - \angle PAQ=180^\circ \angle YFX$ Thus, $YDXF$ is cyclic. Note that: $\triangle ABD \cong \triangle FCE$ . Thus, we have: $AD \parallel BF$ . Therefore: $\angle FBD = \angle ADE = \angle AQD = \angle DXC$ which implies $FEDX$ is also cyclic and we are done.

This post has been edited 1 time. Last edited by Saucepan_man02, Oct 30, 2024, 11:13 AM
Reason: Reims

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quantam13

109 posts

quantam13

#68 h Dec 21, 2024, 12:56 PM • 1 Y

Y by adrian_042

Solved with adrian_042 :trampoline:

Solution

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quantam13

109 posts

quantam13

#69 h Jan 24, 2025, 3:13 AM

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redacted accidental post

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cherry265

16 posts

cherry265

#70 h Jan 26, 2025, 2:07 AM

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Let $A’$ be the point such that $ABA’C$ is a parallelogram. From here show $DEXA’Y$ cyclic pentagon by angle and length chase.

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ItsBesi

141 posts

ItsBesi

#72 h Feb 27, 2025, 2:00 PM

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Here is my solution I found on 16.04.2024 (yes almost a year), I solved this one with Ismayil

Solution:

Let $\odot(APQ)=\Omega, \odot(BPYD)=\omega_B , \odot(CQXD)=\omega_C$ and let $BY \cap CX=\{Z\}$

Claim: Quadrilateral $\square ABCZ$ is a parallelogram

Proof:

Now by Reim's Theorem we have that :

$AQ \parallel BY \implies AC \parallel BZ$

Similarly $AP \parallel CX \implies AB \parallel CZ$

Now since $AC \parallel BZ$ and $AB \parallel CZ$ we get that the quadrilateral $ABCZ$ is a parallelogram $\square$

Claim: $\triangle BAD \cong \triangle CZE$

Proof:

From the parallelogram we have: $\boxed{AB=CZ} , \angle BAC=\angle BZC$

So $\angle ABD \equiv \angle ABC=\angle BCZ \equiv \angle ECX \implies \boxed{\angle ABD=\angle ECZ}$

So by combining $AB=CZ , BD=CE$ and $\angle ABD=\angle ECZ$ we get that triangle $\triangle BAD, \triangle CZE$ are congruent

$\iff \triangle BAD \cong \triangle CZE$ $\square$

Claim: Points $Y$ , $D$ , $E$ and $Z$ are concyclic

Proof:

From previous claim we have: $\angle ZEC=\angle ADB$

$\angle DEZ=180-\angle ZEC=180-\angle ADB=\angle ADC \stackrel{\Omega}{=} \angle APD=180-\angle BPD \stackrel{\omega_B}{=} \angle BYD=180-\angle DYZ \implies$

$\angle DEZ=180-\angle DYZ \implies \angle DEZ+\angle DYZ=180 \implies$ Points $Y$ , $D$ , $E$ and $Z$ are concyclic $\square$

Claim: Points $Y$ , $D$ , $X$ and $Z$ are concyclic

Proof:

$\angle YDX=\angle PDQ \stackrel{\Omega}{=}180-\angle PAQ \equiv 180-\angle BAC=180-\angle BZC \equiv 180-\angle YZX \implies \angle YDX=180-\angle YZX \implies$

$\angle YDX+\angle YZX=180 \implies$ Points $Y$ , $D$ , $X$ and $Z$ are concyclic $\square$

Claim: Points $D$ , $E$ , $X$ and $Y$ are concyclic

Proof:

Since $\square YDEZ$ is cyclic and $YDXZ$ is also cyclic we get that points $Y$ , $D$ , $E$ , $X$ , $Z$ all lie on a circle hence

Points $D$ , $E$ , $X$ and $Y$ are concyclic $\blacksquare$

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Retemoeg

55 posts

Retemoeg

#73 h Feb 27, 2025, 4:30 PM

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$CX$ intersects $BY$ at $F$ . Since $APDQ$ and $BYDP$ are cyclic, $\angle BYD = \angle APD = \angle CQD$ implying $BY \parallel CA$ . Similarly, $CX \parallel BA$ . Thus, $ACFB$ is a paralellogram, thus $\angle BAD = \angle CFE$ cuz of symmetry. Now, because $DB$ is tangent to $(APDQ)$ :
$\angle CDX = \angle BDP = \angle BAD = \angle CFE$ Hence $DEXF$ is cyclic. Similarly, $DYXF$ is cyclic so $DEXY$ is cyclic.

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dolphinday

1319 posts

dolphinday

#74 h Mar 11, 2025, 8:08 PM • 1 Y

Y by peace09

By Reim, we have $CX \parallel AB$ and $BY \parallel AC$ . Let $CX \cap BY = Z$ . Then $ABZC$ is a parallelogram so $\measuredangle{BAC} = \measuredangle{CZB} = \measuredangle{YDX} = \measuredangle{QDP} = \measuredangle{PAQ} \implies Z \in (DYX)$ .

Then by symmetry we have $\measuredangle{CEZ} = \measuredangle{BDA} = \measuredangle{DQA} = \measuredangle{DPA} = \measuredangle{DPB} = \measuredangle{DYB} = \measuredangle{DYZ}$ . Then $\measuredangle{DYZ} = \measuredangle{CEZ} = \measuredangle{DEZ}$ implies that $E \in (DYZ) = (DYX)$ as desired.

wrote D \in DYX initially lmao

This post has been edited 3 times. Last edited by dolphinday, Mar 15, 2025, 6:30 PM

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EpicBird08

1745 posts

EpicBird08

#75 h Mar 12, 2025, 1:32 PM

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This problem is so cool. Similar in spirit to this problem (has spoilers).

First, we will show that $CX \parallel AB.$ Indeed, we have $\measuredangle ACX = \measuredangle QCX = \measuredangle QDX = \measuredangle QDP = \measuredangle QAP = \measuredangle CAB.$ Similarly, $BY \parallel AC.$ Thus $BY$ and $CX$ intersect at the point $A'$ such that $ABA'C$ is a parallelogram.

Now, we know that $ADA'E$ is a parallelogram by the given condition, so $A'E \parallel AD,$ giving $\measuredangle DE'A = \measuredangle EDA = \measuredangle DPA = \measuredangle DPB = \measuredangle DYB = \measuredangle DYA',$ so $D,E,A',Y$ are concyclic. Similarly, $D,E,A',X$ are concyclic, finishing the problem.

Motivation

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blueprimes

326 posts

blueprimes

#76 h Apr 1, 2025, 1:11 AM

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Let $BY \cap CX = T$ . Define $\angle A, \angle B, \angle C$ wrt $\triangle ABC$ .

Claim 1: $D, X, T, Y,$ are concyclic.
We have $\angle XDY = \angle PDQ = 180^\circ - \angle A$ . On the other hand, note that $\angle PDY = \angle QDX = \angle A$ . Thus, we have
$\angle DBY = \angle DPY = 180^\circ - \angle PDY - \angle PYD = 180^\circ - \angle A - \angle B = \angle C.$ Similarly, $\angle DCX = \angle B$ . Then $\angle XTY = 180^\circ - \angle DBY - \angle DCX = \angle A$ which proves the claim.

Claim 2: $D, E, X, T,$ are concyclic.
Our angle equalities in the last claim imply $T$ is the reflection of $A$ over the midpoint of $BC$ , so in fact $AD \parallel TE$ . Thus,
$\angle DXT = 180^\circ - \angle BYD = 180^\circ - \angle BPD = \angle BAD + \angle B = \angle ADE = \angle DET$ as wanted.

Therefore, $D, E, X, Y$ are concyclic as needed.

This post has been edited 2 times. Last edited by blueprimes, Apr 1, 2025, 2:05 AM

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Giant_PT

24 posts

Giant_PT

#77 h Apr 1, 2025, 2:47 AM

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Let $E'$ be the intersection with $(DXY)$ and $BC$ that is not $D$ . Also, Let $F$ be the second intersection of $(BPD)$ and $PQ$ and $G$ be the second intersection of $CQD$ and $PQ$ . We prove that $E=E'$

Claim 1: $BFGC$ is cyclic
By angle chasing we have,
$\measuredangle PFB=\measuredangle PDB=\measuredangle PQD=\measuredangle GQD=\measuredangle GCD,$ which proves the claim.

Claim 2: $BF\parallel DX$ and $CG\parallel DY$
Again by straightforward angle chasing, we have
$\measuredangle BDX =\measuredangle CDQ=\measuredangle DPQ=\measuredangle DPF=\measuredangle DBF,$ which proves that $BF\parallel DX.$ Similarly, we can prove that $CG\parallel DY.$ Therefore, the claim is proven.

Claim 3: $(BFGC)$ and $(DXY)$ are concentric.
Since quadrilaterals $BFDX$ and $CGDY$ are concyclic, and $BF\parallel DX$ and $CG\parallel DY$ by claim 2, they are clearly isosceles trapezoids. Then we can see that perpendicular bisectors of $\overline{BF}$ and $\overline{DX}$ are the same, and perpendicular bisectors of $\overline{CF}$ and $\overline{DY}$ are the same. Therefore $(BFGC)$ and $(DXY)$ are concentric as wanted.

$(BFGC)$ and $(DXY)$ being concentric is enough to prove that $BD=E'C$ since it implies that the two circles' center lie on the perpendicular bisector of $BC.$ Therefore, $E=E'$ as wanted.

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Reason: Typo

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