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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Help my diagram has too many points
MarkBcc168   29
N 35 minutes ago by VideoCake
Source: IMO Shortlist 2023 G6
Let $ABC$ be an acute-angled triangle with circumcircle $\omega$. A circle $\Gamma$ is internally tangent to $\omega$ at $A$ and also tangent to $BC$ at $D$. Let $AB$ and $AC$ intersect $\Gamma$ at $P$ and $Q$ respectively. Let $M$ and $N$ be points on line $BC$ such that $B$ is the midpoint of $DM$ and $C$ is the midpoint of $DN$. Lines $MP$ and $NQ$ meet at $K$ and intersect $\Gamma$ again at $I$ and $J$ respectively. The ray $KA$ meets the circumcircle of triangle $IJK$ again at $X\neq K$.

Prove that $\angle BXP = \angle CXQ$.

Kian Moshiri, United Kingdom
29 replies
MarkBcc168
Jul 17, 2024
VideoCake
35 minutes ago
A problem with series
Pena317   1
N 37 minutes ago by venhancefan777
Source: P5, Mexico Center Regional Olympiad 2019
A serie of positive integers $a_{1}$,$a_{2}$,. . . ,$a_{n}$ is $auto-delimited$ if for every index $i$ that holds $1\leq i\leq n$, there exist at least $a_{i}$ terms of the serie such that they are all less or equal to $i$.
Find the maximum value of the sum $a_{1}+a_{2}+\cdot \cdot \cdot+a_{n}$, where $a_{1}$,$a_{2}$,. . . ,$a_{n}$ is an $auto-delimited$ serie.
1 reply
Pena317
Nov 28, 2019
venhancefan777
37 minutes ago
IMO 2009, problem 4
ZetaX   60
N 44 minutes ago by FarrukhBurzu
Let $ ABC$ be a triangle with $ AB = AC$ . The angle bisectors of $ \angle C AB$ and $ \angle AB C$ meet the sides $ B C$ and $ C A$ at $ D$ and $ E$ , respectively. Let $ K$ be the incentre of triangle $ ADC$. Suppose that $ \angle B E K = 45^\circ$ . Find all possible values of $ \angle C AB$ .

Jan Vonk, Belgium, Peter Vandendriessche, Belgium and Hojoo Lee, Korea
60 replies
ZetaX
Jul 16, 2009
FarrukhBurzu
44 minutes ago
Tennis tournament with rotating courts
v_Enhance   6
N an hour ago by Blast_S1
Source: ELMO Shortlist 2013: Problem C10, by Ray Li
Let $N\ge2$ be a fixed positive integer. There are $2N$ people, numbered $1,2,...,2N$, participating in a tennis tournament. For any two positive integers $i,j$ with $1\le i<j\le 2N$, player $i$ has a higher skill level than player $j$. Prior to the first round, the players are paired arbitrarily and each pair is assigned a unique court among $N$ courts, numbered $1,2,...,N$.

During a round, each player plays against the other person assigned to his court (so that exactly one match takes place per court), and the player with higher skill wins the match (in other words, there are no upsets). Afterwards, for $i=2,3,...,N$, the winner of court $i$ moves to court $i-1$ and the loser of court $i$ stays on court $i$; however, the winner of court 1 stays on court 1 and the loser of court 1 moves to court $N$.

Find all positive integers $M$ such that, regardless of the initial pairing, the players $2, 3, \ldots, N+1$ all change courts immediately after the $M$th round.

Proposed by Ray Li
6 replies
v_Enhance
Jul 23, 2013
Blast_S1
an hour ago
∑(a-b)(a-c)/(2a^2 + (b+c)^2) >= 0
Zhero   24
N an hour ago by RevolveWithMe101
Source: ELMO Shortlist 2010, A2
Let $a,b,c$ be positive reals. Prove that
\[ \frac{(a-b)(a-c)}{2a^2 + (b+c)^2} + \frac{(b-c)(b-a)}{2b^2 + (c+a)^2} + \frac{(c-a)(c-b)}{2c^2 + (a+b)^2} \geq 0. \]

Calvin Deng.
24 replies
Zhero
Jul 5, 2012
RevolveWithMe101
an hour ago
i am not abel to prove or disprove
frost23   8
N an hour ago by frost23
Source: made on my own
sorrrrrry
8 replies
frost23
3 hours ago
frost23
an hour ago
points on sides of a triangle, intersections, extensions, ratio of areas wanted
parmenides51   1
N an hour ago by FrancoGiosefAG
Source: Mexican Mathematical Olympiad 1997 OMM P5
Let $P,Q,R$ be points on the sides $BC,CA,AB$ respectively of a triangle $ABC$. Suppose that $BQ$ and $CR$ meet at $A', AP$ and $CR$ meet at $B'$, and $AP$ and $BQ$ meet at $C'$, such that $AB' = B'C', BC' =C'A'$, and $CA'= A'B'$. Compute the ratio of the area of $\triangle PQR$ to the area of $\triangle ABC$.
1 reply
parmenides51
Jul 28, 2018
FrancoGiosefAG
an hour ago
starting with intersecting circles, line passes through midpoint wanted
parmenides51   2
N 2 hours ago by EmersonSoriano
Source: Peru Ibero TST 2014
Circles $C_1$ and $C_2$ intersect at different points $A$ and $B$. The straight lines tangents to $C_1$ that pass through $A$ and $B$ intersect at $T$. Let $M$ be a point on $C_1$ that is out of $C_2$. The $MT$ line intersects $C_1$ at $C$ again, the $MA$ line intersects again to $C_2$ in $K$ and the line $AC$ intersects again to the circumference $C_2$ in $L$. Prove that the $MC$ line passes through the midpoint of the $KL$ segment.
2 replies
parmenides51
Jul 23, 2019
EmersonSoriano
2 hours ago
An inequality
Rushil   14
N 2 hours ago by frost23
Source: Indian RMO 1994 Problem 8
If $a,b,c$ are positive real numbers such that $a+b+c = 1$, prove that \[ (1+a)(1+b)(1+c) \geq 8 (1-a)(1-b)(1-c) . \]
14 replies
Rushil
Oct 25, 2005
frost23
2 hours ago
3 var inequality
SunnyEvan   6
N 2 hours ago by JARP091
Let $ a,b,c \in R $ ,such that $ a^2+b^2+c^2=4(ab+bc+ca)$Prove that :$$ \frac{7-2\sqrt{14}}{48} \leq \frac{a^3b+b^3c+c^3a}{(a^2+b^2+c^2)^2} \leq \frac{7+2\sqrt{14}}{48} $$
6 replies
SunnyEvan
May 17, 2025
JARP091
2 hours ago
collinearity as a result of perpendicularity and equality
parmenides51   2
N 2 hours ago by FrancoGiosefAG
Source: Mexican Mathematical Olympiad 1996 OMM P6
In a triangle $ABC$ with $AB < BC < AC$, points $A' ,B' ,C'$ are such that $AA' \perp BC$ and $AA' = BC, BB' \perp  CA$ and $BB'=CA$, and $CC' \perp AB$ and $CC'= AB$, as shown on the picture. Suppose that $\angle AC'B$ is a right angle. Prove that the points $A',B' ,C' $ are collinear.
2 replies
parmenides51
Jul 28, 2018
FrancoGiosefAG
2 hours ago
3 var inequality
JARP091   6
N 2 hours ago by JARP091
Source: Own
Let \( x, y, z \in \mathbb{R}^+ \). Prove that
\[
\sum_{\text{cyc}} \frac{x^3}{y^2 + z^2} \geq \frac{x + y + z}{2}
\]without using the Rearrangement Inequality or Chebyshev's Inequality.
6 replies
JARP091
Today at 8:54 AM
JARP091
2 hours ago
Helplooo
Bet667   1
N 2 hours ago by Lil_flip38
Let $ABC$ be an acute angled triangle.And altitudes $AD$ and $BE$ intersects at point $H$.Let $F$ be a point on ray $AD$ such that $DH=DF$.Circumcircle of $AEF$ intersects line $BC$ at $K$ and $L$ so prove that $BK=BL$
1 reply
Bet667
3 hours ago
Lil_flip38
2 hours ago
Cyclic sum of 1/(a+1/b+1)
v_Enhance   22
N 3 hours ago by Rayvhs
Source: ELMO Shortlist 2013: Problem A2, by David Stoner
Prove that for all positive reals $a,b,c$,
\[\frac{1}{a+\frac{1}{b}+1}+\frac{1}{b+\frac{1}{c}+1}+\frac{1}{c+\frac{1}{a}+1}\ge \frac{3}{\sqrt[3]{abc}+\frac{1}{\sqrt[3]{abc}}+1}. \]Proposed by David Stoner
22 replies
v_Enhance
Jul 23, 2013
Rayvhs
3 hours ago
Nice problem
Tiks   24
N Apr 10, 2025 by Nari_Tom
Source: IMO Shortlist 2000, G6
Let $ ABCD$ be a convex quadrilateral. The perpendicular bisectors of its sides $ AB$ and $ CD$ meet at $ Y$. Denote by $ X$ a point inside the quadrilateral $ ABCD$ such that $ \measuredangle ADX = \measuredangle BCX < 90^{\circ}$ and $ \measuredangle DAX = \measuredangle CBX < 90^{\circ}$. Show that $ \measuredangle AYB = 2\cdot\measuredangle ADX$.
24 replies
Tiks
Nov 2, 2005
Nari_Tom
Apr 10, 2025
Source: IMO Shortlist 2000, G6
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Tiks
1144 posts
#1 • 4 Y
Y by Adventure10, Rounak_iitr, Kingsbane2139, and 1 other user
Let $ ABCD$ be a convex quadrilateral. The perpendicular bisectors of its sides $ AB$ and $ CD$ meet at $ Y$. Denote by $ X$ a point inside the quadrilateral $ ABCD$ such that $ \measuredangle ADX = \measuredangle BCX < 90^{\circ}$ and $ \measuredangle DAX = \measuredangle CBX < 90^{\circ}$. Show that $ \measuredangle AYB = 2\cdot\measuredangle ADX$.
Z K Y
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pohoatza
1145 posts
#2 • 6 Y
Y by NewAlbionAcademy, Phie11, Adventure10, DCMaths, Om245, and 1 other user
We have that $\triangle{ADX}$ is similar with $\triangle{BCX}$, thus there exists 2 points $M,N$ on the perpendicular bisector of $(AB)$ such that $\triangle{AMN}$ is similar with $\triangle{ADX}$, and $\triangle{BMN}$ is similar with $\triangle{BCX}$.
So we have that $\frac{AD}{AM}=\frac{AX}{AN}$ and $\angle{DAM}=\angle{XAN}$, but $\triangle{ADM}$ beeing similar with $\triangle{AXN}$, we have $\frac{AD}{AX}=\frac{DM}{XN}$, and similar $\frac{BC}{BX}=\frac{CM}{XN}$, thus we have $CM=DM$, therefore $M$ lies on both perpendicular bisectors of $AB$ and $CD$, therefore $M \equiv Y$.
Now it follows easy, because $\angle{AYB}=2\angle{AYN}=2\angle{ADX}$.
Z K Y
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The QuattoMaster 6000
1184 posts
#3 • 1 Y
Y by Adventure10
Tiks wrote:
Let $ ABCD$ be a convex quadrilateral. The perpendicular bisectors of its sides $ AB$ and $ CD$ meet at $ Y$. Denote by $ X$ a point inside the quadrilateral $ ABCD$ such that $ \measuredangle ADX = \measuredangle BCX < 90^{\circ}$ and $ \measuredangle DAX = \measuredangle CBX < 90^{\circ}$. Show that $ \measuredangle AYB = 2\cdot\measuredangle ADX$.
Solution
Z K Y
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SnowEverywhere
801 posts
#4 • 1 Y
Y by Adventure10
Does this solution work? I am really unsure as to whether it works because it seems like too simple a solution especially for an ISL #6. Does anyone think that this problem is a little similar to IMO 1975 #3?

Solution

Define point $Y'$ to be on the perpendicular bisector of $AB$ and such that $\angle{AY'B}=2\angle{ADX}$. Define point $Z$ to be such that $\angle{AZY'}=\angle{BZY'}=\angle{AXD}=\angle{BXC}$. Note that since $Z$ and $Y'$ are on the perpendicular bisector of $AB$, $AY'=BY'$ and $AZ=BZ$.

By AA similarity, we have that $\triangle{AXD} \sim \triangle{AZY'}$ and $\triangle{BXC} \sim \triangle{BZY'}$. This yields by spiral similarity that $\triangle{AZX} \sim \triangle{AY'D}$ and $\triangle{BZX} \sim \triangle{BY'C}$. Therefore,

\[\frac{ZX}{DY'}=\frac{AZ}{AY'}=\frac{BZ}{BY'}=\frac{ZX}{CY'}\]
This yields that $DY'=CY'$ and therefore that $Y'=Y$ is the intersection of the perpendicular bisectors of $AB$ and $CD$. Therefore $\angle{AY'B}=\angle{AYB}=2\angle{ADX}$.
Z K Y
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mathuz
1525 posts
#5 • 2 Y
Y by Adventure10, Mango247
Let $M,N$ are midpoints of the sides $AB$ and $CD$ and $DX\cap AB=P(1)$, $AX\cap CD=Q(.)$. Then since $ \triangle AXD \sim \triangle BXC$ we get that $ \angle BXP=\angle CXQ $, $ \angle AXP=\angle DXQ $ and \[ \frac{AX}{BX}=\frac{DX}{CX}. \] So \[ \frac{AP}{BP}=\frac{DQ}{CQ} \] and the parallel lines pass through $M$ and $N$ to $DX$ and $AX$ respectively, intersect on $AD$. Analoguosly, the parallel lines pass through $M$ and $N$ to $CX$ and $BX$ respectively, intersect on $BC$. Hence $ \angle AYB=2\angle ADX$.
Z K Y
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Mathematicalx
537 posts
#6 • 2 Y
Y by Adventure10, Mango247
Dear mathuz, i dont understand how the result follows. (rest is ok)
Could you explain a bit ?
Z K Y
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Wolowizard
617 posts
#7 • 2 Y
Y by Adventure10, Mango247
My solution:
Let circumcerntres of $\triangle DAX$ and $\triangle BCX$ be $O_1,O_2$. Let $Y$ be point such that $AY=BY$ and $\angle AYB=2\angle ADX$.
Since $\angle AO_1X=2\angle ADX$ we have $\triangle O_1AX$ ~$\triangle AYB$ so $\frac{AO_1}{AY}=\frac{AX}{AB}$ and since $\angle O_1AY=\angle XAB$ we have that $\triangle AO_1Y$~$\triangle AXB$. Similary we have $\triangle BO_2Y$~$\triangle AXB$.
Now we have $O_1Y=\frac{AY}{AB}BX$ and $O_2Y=\frac{BY}{AB}AX$ and $O_1X=\frac{AX}{AB}AY$ and $O_2X=\frac{BX}{AB}BY$.

Since $\angle AO_1Y=\angle YO_2B$ and $\angle DAX=\angle CBX$ we have $\angle DO_1Y=\angle CO_2Y$. Now by Law of Cosine we have
$DY^2=O_1D^2+O_1Y^2-2O_1D\cdot O_1Y\cos(DO_1Y)=AY^2(\frac{AX^2+BX^2-2AX\cdot BXcos(DO_1Y)}{AB^2})$
and
$CY^2=O_2Y^2+O_2C^2-2O_2Y\cdot O_2Ccos(CO_2Y)=AY^2(\frac{AX^2+BX^2-2AX\cdot BXcos(DO_1Y)}{AB^2})$ which implies
$DY=CY$ so $Y$ is the intersection of perpendicular bisectors of $AB,CD$.
This post has been edited 3 times. Last edited by Wolowizard, Feb 14, 2016, 3:51 PM
Reason: Replaced X with Y
Z K Y
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anantmudgal09
1980 posts
#8 • 2 Y
Y by Adventure10, Mango247
Feels too convenient, so perhaps I missed something? :maybe:
Tiks wrote:
Let $ ABCD$ be a convex quadrilateral. The perpendicular bisectors of its sides $ AB$ and $ CD$ meet at $ Y$. Denote by $ X$ a point inside the quadrilateral $ ABCD$ such that $ \measuredangle ADX = \measuredangle BCX < 90^{\circ}$ and $ \measuredangle DAX = \measuredangle CBX < 90^{\circ}$. Show that $ \measuredangle AYB = 2\cdot\measuredangle ADX$.

WLOG let rays $\overrightarrow{BA}, \overrightarrow{CD}$ intersect. Let $M, N$ be midpoints of $\overline{AB}, \overline{CD}$ respectively. Let $P$ be the spiral center $\overline{AB} \mapsto \overline{DC}$. Construct $E, F$ outside $ABCD$ with $\triangle AEB \sim \triangle DYC$ and $\triangle AYB \sim \triangle DFC$. Observe that $\triangle AEY \sim \triangle DYF$ with spiral center $P$ again. Further, we see $k=\tfrac{EM}{MY}=\tfrac{YN}{NF}$. Let $L$ be the point on $\overline{DA}$ with $\tfrac{AL}{LD}=k$. Suppose $X$ is the point inside $ABCD$ with $\triangle LMN \sim \triangle XBC$.

Lemma. $\overline{XL} \perp \overline{DA}$ and $\triangle XAD \sim \triangle XBC$.

(Proof) By linearity, $\triangle AEY \mapsto \triangle LMN \mapsto \triangle DYF$ under spiral similarity with pivot $P$. Apply $\triangle PDA \sim \triangle PNM \sim \triangle PYE$ hence $\angle PME=\angle PLA$. Combined with $\angle PMB=\angle PLX$ we obtain $\angle XLA=\angle BME=90^{\circ}$. Now reflect $X$ in $L$ to get $X'$. Again linearity gives $\triangle X'AD \sim \triangle XBC$ and we're done. $\blacksquare$

Finally, we see $\triangle XAD \sim \triangle LMN \sim \triangle DFY$ hence $\tfrac{1}{2}\angle AYB=\angle AYE=\angle ADX$ and we're done here.
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tastymath75025
3223 posts
#9 • 2 Y
Y by Adventure10, Mango247
Assume $AB,CD$ aren't parallel; otherwise the problem doesn't make sense.

Let $X_1,X_2$ be the reflections of $X$ over $BC,AD$ and $X_3,X_4$ be the projections of $X$ onto $BC,AD$. Let $M,N$ be the midpoints of $AB,CD$. Clearly $X_2AXD, XBX_1C$ are directly similar kites, so they're also similar to their direct average $X_4MX_3N$. Since $AX_4:X_4D=BX_3:X_3C$ and $AB,CD$ aren't parallel, this implies $X_4,X_3$ are the only such points on $AD,BC$ which divide segments $AD,BC$ in the same ratio with $X_3X_4\perp MN$.

Next construct $Y_1,Y_2$ with $DYC\sim AY_1B, DY_2C\sim AYB$. Once again $AY_1BY, DYCY_2$ are directly similar kites, so for any $r$ their weighted average $r(AY_1BY)+(1-r)(DYCY_2)$ is also a kite. Since $Y_1M:MY=YN:NY_2$ we can choose $r$ so that this resulting kite contains $M,N$; then the other two vertices of the kite must lie on $AD,BC$, divide those two segments in the same ratio, and create a segment perpendicular to $MN$. This implies these other two vertices are precisely $X_4,X_3$ by our earlier work, hence $MX_4NX_3\sim Y_1AYB$ and $\angle AYB =\angle X_2DX=2\angle ADX$; the other equality follows similarly.
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lminsl
544 posts
#10 • 1 Y
Y by Adventure10
Let us deal with non-trivial cases with $AB \not \parallel CD$.

Let $\odot (ADX)$ and $\odot (BCX)$ meet at $Z$, and $\odot (ABZ)$ meets $\odot (CDZ)$ at $W$. We claim that $W$ coincides with $Y$.

Simple angle chasing gives $\angle AYB=\angle AWB$, and $\angle CYD=\angle CWD$. Also line $ZX$ is the internal angle bisector of $\angle AZB, \angle CZD$, so it suffices to prove that $ZW \perp XZ$.

Now invert the diagram with centre $Z$. Denote by $T '$ by the image of any point $T$.
Note that $X'=A'D' \cap B'C'$, and $W'=A'B' \cap C'D'$, and $ZX'$ bisects both $\angle A'ZB'$ and $\angle C'ZD'$. Assume that line $ZX'$ meets $A'B', C'D'$ at $U, V$, and a line perpendicular to $ZX'$ passing $Z$ meets $A'B', C'D'$ at $P, Q$, respectively. Then,
$$ X'(A'B', UP)=Z(A'B',UP) = -1=Z(C'D', VQ)=X(C'D', VQ),$$so $P$ and $Q$ coincides with $W'$. Thus $X'Z \perp WZ'$, which implies $W \equiv Y$. $\square$
This post has been edited 2 times. Last edited by lminsl, Oct 5, 2019, 1:48 AM
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v_Enhance
6877 posts
#11 • 2 Y
Y by v4913, Rounak_iitr
First, a long phantom point argument:
  • Fix $\triangle AXD$ first and let $O$ denote its circumcenter.
  • Let $B$ be as in the original problem.
  • Then we redefine $Y$ such that $\angle AYB = 2 \angle ADX = \angle AOX$ and $AY = BY$.
  • We then redefine $C$ such that $\angle DYC = 2 \angle DAX = \angle DOX$ and $YC = YD$.
  • We will prove $\triangle AXD \sim \triangle BXC$ (oppositely oriented). This will solve the problem.

[asy] pair A = dir(125); pair D = dir(275); pair X = dir(-5); pair O = origin; draw(unitcircle);

pair Y = 1.2*dir(-20); pair B = Y+(A-Y)*X/A;

pair C = Y+(D-Y)*X/D; filldraw(A--O--X--cycle, invisible, deepgreen); filldraw(A--B--Y--cycle, invisible, deepgreen); filldraw(D--O--X--cycle, invisible, blue); filldraw(D--C--Y--cycle, invisible, blue); filldraw(B--X--C--cycle, invisible, grey); draw(circumcircle(B, X, C), grey);

dot("$A$", A, dir(A)); dot("$D$", D, dir(D)); dot("$X$", X, dir(X)); dot("$O$", O, dir(-X)); dot("$Y$", Y, dir(Y)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C));

/* TSQ Source:

A = dir 125 D = dir 275 X = dir -5 O = origin R-X unitcircle

Y = 1.2*dir(-20) B = Y+(A-Y)*X/A

C = Y+(D-Y)*X/D A--O--X--cycle 0.1 lightgreen / deepgreen A--B--Y--cycle 0.1 lightgreen / deepgreen D--O--X--cycle 0.1 lightblue / blue D--C--Y--cycle 0.1 lightblue / blue B--X--C--cycle 0.1 yellow / grey circumcircle B X C grey

*/ [/asy]

Anyways, the main idea is that $\triangle AOX \sim \triangle AYB$ and $\triangle DOX \sim \triangle DYC$. We break symmetry now and set up as follows. Since we have a similarity $\triangle AOY \sim \triangle AXB$, there should be a single complex number $t \in {\mathbb C}$ such that \begin{align*} 	y-a &= t(0-a) \implies y=(1-t)a \\ 	b-a &= t(x-a) \implies b=t(x-a)+a. \end{align*}Next since $\frac{c-y}{d-y} = \frac{x-0}{d-0}$ we have \[ c = y + \frac{(d-y)x}{d} = \frac{d(x+y)-xy}{d}. \]Now, we simply calculate \begin{align*} 	\frac{c-b}{x-b} &= \frac{\frac{d(x+y)-xy}{d}-[t(x-a)+a]}{x-[t(x-a)+a]} \\ 	&= \frac{dx + (d-x)[(1-t)a] - dt(x-a)-da}{d(1-t)(x-a)} \\ 	&= \frac{(1-t)[(d-x)a+dx-da]}{d(1-t)(x-a)} \\ 	&= \frac{(1-t)x(d-a)}{(1-t)d(x-a)} = \frac{x(d-a)}{d(x-a)} \\ 	&= \frac{\frac1d-\frac1a}{\frac1x-\frac1a} 	= \overline{\left( \frac{d-a}{x-a} \right)} \end{align*}as desired.
This post has been edited 1 time. Last edited by v_Enhance, Apr 12, 2020, 10:09 PM
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arzhang2001
251 posts
#12
Y by
hint : let $O$ be the circumcircle of $\triangle ADX$ then use rotation and similarity properties. :jump:
also use this technique. you should began from verdict and arrive to assumptions.
This post has been edited 1 time. Last edited by arzhang2001, Apr 19, 2020, 2:02 PM
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Stormersyle
2786 posts
#13
Y by
Here's a pretty nice complex solution that nobody's posted yet.

Let $P, Q$, both inside $ABCD$, be points on the perp. bisector of $AB$ such that $\triangle{APQ}\sim \triangle{ADX}$ (note there is only one choice for $P$ and $Q$). For our setup, set $(ABP)$ be the unit circle, let $p=1$, and let $a, q, x$ be the free variables; note $b=\bar{a}=\frac{1}{a}$, and $q$ is real. We desire to prove $PC=PD$, or $|c-1|=|d-1|$.

Now, note $A$ is the spiral center of $PQ, DX$, by the spiral center formula we have $a=\frac{px-qd}{p+x-q-d}=\frac{x-qd}{1+x-q-d}$, which we can rearrange to get $d=\frac{a+ax-aq-x}{a-q}$. Thus, we have $d-1=\frac{(a-1)(x-q)}{a-q}$, so changing $a$ for $\frac{1}{a}$ in this expression, we get $c-1=\frac{(1-a)(x-q)}{1-aq}$. Thus, it suffices to prove that $|a-q|=|1-aq|$, or $(a-q)(\frac{1}{a}-q)=(1-aq)(1-\frac{q}{a})$. But multiplying both sides by $a$, we find the equation indeed holds, so we are done.
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riadok
187 posts
#14 • 2 Y
Y by zuss77, Kar-98k
Noone has posted a short and neat sollutiion using complex weighted gliding lemma, so here it is :-).

Complex Weighted Gliding Lemma: Let $A_1B_1C_1$ and $A_2B_2C_2$ be two similar triangles. Construct points $X$, $Y$, $Z$ such that triangles $A_1XA_2$, $B_1YB_2$ and $C_1ZC_2$ are similar. Then $\triangle XYZ\sim\triangle A_1B_1C_1$.
Proof sketch: As $A_1B_1C_1\sim A_2B_2C_2$ there exist linear function $f$ of complex plane mapping $A_1B_1C_1\mapsto A_2B_2C_2$. Also consider identity function $i$ as a linear function of complex plane. Then to get to points $X$, $Y$, $Z$ you just take weighted average of these functions, but average of linear functions is linear, hence $XYZ\sim A_1B_1C_1$. $\square$

Problem Proof: Denote $X_1$ reflection of $X$ by $AD$ and $X_2$ reflection of $X$ by $BC$. Note that $X_1AX\sim XBX_2$, construct $Y'$ such that $DY'C\sim X_1AX$. Note that $X_1DX\sim XCX_2$, hence if we can complex glide them to $AY'B$, hence $AY'B\sim X_1DX$. Because triangles $AY'B$ and $DY'C$ are isosceles, we get $Y'=Y$ and from similarities we got the angle condition, that problem wanted. $\square$
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pad
1671 posts
#16 • 3 Y
Y by nukelauncher, Gaussian_cyber, Rounak_iitr
Solved with nukelauncher.

Diagram (No geogebra, just Microsoft Paint!)

Let $O$ be the cirumcenter of $\triangle ADX$. Redefine $Y$ such that $\triangle AOX\sim \triangle AYB$, hence $\angle AYB=\angle AOX=2\angle ADX$. We want to show $Y$ lies on the perpendicular bisector of $\overline{CD}$. Let $(ADX)$ be the unit circle, so $O=0$. We fix $A,X,D$ and $B$ as arbitrary points, and will calculate $Y$ and $C$ based on these points.

WLOG by rotation set $x=1$, so $OX$ is the real axis. From the spiral similarity sending $\triangle AOX$ to $\triangle AYB$, we have
\[ \frac{y-a}{0-a} = \frac{b-a}{x-a} \implies y=\frac{-a(b-a)}{x-a}+a=a\cdot \frac{1-b}{1-a}\]Let $\bullet'$ denote the reflection over real axis of $\bullet$. The angle conditions imply that there exists a spiral similarity sending $A'\mapsto B$ and $D'\mapsto C$. Hence
\begin{align*}
\frac{c-x}{\bar{d}-x} = \frac{b-x}{\bar{a}-x} \implies c&=\frac{(\bar d-x)(b-x)}{\bar a-x} +x \\
&=\frac{\left(\frac1d-1\right)(b-1)}{\frac1a-1} + 1 \\
&= \frac{b/d-1/d-b+1/a}{1/a-1} \\
&= \frac{ab-a-abd+d}{d(1-a)}.
\end{align*}Since we wanted to show $Y$ is on the perpendicular bisector of $CD$, it suffices to prove that $y$ satisfies
\begin{align*}
|y-d|=|y-c|\iff (y-d)(\bar y-\bar d)=(y-c)(\bar y-\bar c). 
\end{align*}So if $Z=\tfrac{y-d}{y-c}$, we need to prove $Z\cdot \bar Z=1$. Magically, \begin{align*}     Z=\frac{y-d}{y-c}&=\frac{a\cdot\frac{1-b}{1-a}-d}{a\cdot\frac{1-b}{1-a}-\frac{ab-a-abd+d}{d(1-a)}}\\     &=\frac{a(1-b)-d(1-a)}{a(1-b)-\frac{ab-a-abd+d}d}\\     &=\frac{a-ab-d+ad}{a+\frac{a-ab}d-1}\\     &=\frac{ad-abd-d^2+ad^2}{ad+a-ab-d}\\     &=d. \end{align*}Hence $Z\cdot\bar Z=d\cdot \bar d=1$. The end.

Motivational Remarks
This post has been edited 2 times. Last edited by pad, Nov 4, 2020, 8:35 AM
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mathaddiction
308 posts
#17 • 1 Y
Y by Lcz
[asy]
size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -2.284525133697541, xmax = 1.4988894102199524, ymin = -2.1370192645104247, ymax = 0.7145926356744539;  /* image dimensions */pen qqwuqq = rgb(0,0.39215686274509803,0); pen zzttqq = rgb(0.6,0.2,0); draw(arc((-0.0013365322724531775,-0.0029500725262597838),0.0887431089269154,-67.05336014619634,-28.653452989196374)--(-0.0013365322724531775,-0.0029500725262597838)--cycle, linewidth(2) + qqwuqq); draw(arc((0.18023545231029714,-0.43181690860373295),0.0887431089269154,112.94663985380367,159.86126655150957)--(0.18023545231029714,-0.43181690860373295)--cycle, linewidth(2) + qqwuqq); draw((0.18277289490842602,0.4248335965116655)--(-0.6889109221157496,-0.7860196712682731)--(0.7567611090605788,-0.4171962836832381)--cycle, linewidth(2) + blue); draw((0.18277289490842602,0.4248335965116655)--(-0.9972925084238106,0)--(0.18023545231029714,-0.43181690860373295)--cycle, linewidth(2) + blue); draw((-0.0013365322724531775,-0.0029500725262597838)--(-0.9972925084238106,0)--(-1.4525810536799912,-1.1462058127995842)--cycle, linewidth(2) + zzttqq); draw((-0.0013365322724531775,-0.0029500725262597838)--(0.18023545231029714,-0.43181690860373295)--(0.7567611090605788,-0.4171962836832381)--cycle, linewidth(2) + zzttqq);  /* draw figures */draw((-0.9972925084238106,0)--(-0.0013365322724531775,-0.0029500725262597838), linewidth(0.8)); draw((-0.9972925084238106,0)--(0.18023545231029714,-0.43181690860373295), linewidth(0.8)); draw((0.18023545231029714,-0.43181690860373295)--(-0.0013365322724531775,-0.0029500725262597838), linewidth(0.8)); draw((-0.0013365322724531775,-0.0029500725262597838)--(0.7567611090605788,-0.4171962836832381), linewidth(0.8)); draw((-0.0013365322724531775,-0.0029500725262597838)--(-0.7836941442682793,-0.6192726960861138), linewidth(0.8)); draw((-0.9972925084238106,0)--(-1.4525810536799912,-1.1462058127995842), linewidth(0.8)); draw((-0.9972925084238106,0)--(0.18277289490842602,0.4248335965116655), linewidth(0.8)); draw((0.18277289490842602,0.4248335965116655)--(0.7567611090605788,-0.4171962836832381), linewidth(0.8)); draw((-1.4525810536799912,-1.1462058127995842)--(-0.7836941442682793,-0.6192726960861138), linewidth(0.8)); draw((-0.0013365322724531775,-0.0029500725262597838)--(0.18277289490842602,0.4248335965116655), linewidth(0.8)); draw((-1.4525810536799912,-1.1462058127995842)--(0.7567611090605788,-0.4171962836832381), linewidth(0.8)); draw((0.18277289490842602,0.4248335965116655)--(-0.6889109221157496,-0.7860196712682731), linewidth(2) + blue); draw((-0.6889109221157496,-0.7860196712682731)--(0.7567611090605788,-0.4171962836832381), linewidth(2) + blue); draw((0.7567611090605788,-0.4171962836832381)--(0.18277289490842602,0.4248335965116655), linewidth(2) + blue); draw((0.18277289490842602,0.4248335965116655)--(-0.9972925084238106,0), linewidth(2) + blue); draw((-0.9972925084238106,0)--(0.18023545231029714,-0.43181690860373295), linewidth(2) + blue); draw((0.18023545231029714,-0.43181690860373295)--(0.18277289490842602,0.4248335965116655), linewidth(2) + blue); draw((-0.0013365322724531775,-0.0029500725262597838)--(-0.9972925084238106,0), linewidth(2) + zzttqq); draw((-0.9972925084238106,0)--(-1.4525810536799912,-1.1462058127995842), linewidth(2) + zzttqq); draw((-1.4525810536799912,-1.1462058127995842)--(-0.0013365322724531775,-0.0029500725262597838), linewidth(2) + zzttqq); draw((-0.0013365322724531775,-0.0029500725262597838)--(0.18023545231029714,-0.43181690860373295), linewidth(2) + zzttqq); draw((0.18023545231029714,-0.43181690860373295)--(0.7567611090605788,-0.4171962836832381), linewidth(2) + zzttqq); draw((0.7567611090605788,-0.4171962836832381)--(-0.0013365322724531775,-0.0029500725262597838), linewidth(2) + zzttqq);  /* dots and labels */dot((-0.0013365322724531775,-0.0029500725262597838),dotstyle); label("$X$", (0.010963283878670751,0.025354489675411564), NE * labelscalefactor); dot((-0.9972925084238106,0),dotstyle); label("$D$", (-0.985917639733679,0.02831259330630874), NE * labelscalefactor); dot((0.18277289490842602,0.4248335965116655),dotstyle); label("$A$", (0.19436570899429592,0.4542795161555022), NE * labelscalefactor); dot((0.7567611090605788,-0.4171962836832381),dotstyle); label("$B$", (0.7682378133883488,-0.38878001865019324), NE * labelscalefactor); dot((0.18023545231029714,-0.43181690860373295),dotstyle); label("$A'$", (0.19140760536339874,-0.40357053680467914), NE * labelscalefactor);  dot((-1.4525810536799912,-1.1462058127995842),linewidth(4pt) + dotstyle); label("$Y$", (-0.7889109221157496,-0.7860196712682731), NE * labelscalefactor); dot((-0.6889109221157496,-0.7860196712682731),linewidth(4pt) + dotstyle); label("$C$", (-1.6782748621203722,-1.2), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
[/asy]
Let $A'$ be the reflection of $A$ in $DX$. Redefine $Y$ as the point such that $A$ is the center of spiral similarity which sends $DA'$ to $BY$, then $\angle AYB=\angle ADA'=2\angle ADX$ and $AY=YB$. Therefore, it suffices to show $YD=YC$.

Setting up the complex plane with $d=-1$ and $x=0$, then $a'=\overline{a}$.
Notice that from the spiral similarity formula we have
$$a=\frac{db-a'y}{d+b-a'-y}$$Changing the subject we have
$$y=\frac{ab+b-a-a\overline{a}}{a-\overline{a}}$$Now notice that $X$ is the center of spiral similarlity sending $DC$ to $A'B$. Therefore,
$$0=db-a'c$$Hence $$c=\frac{-b}{\overline{a}}$$Now $$y-d=\frac{ab+b-a-a\overline{a}+a-\overline{a}}{a-\overline{a}}=\frac{(b-\overline{a})(a+1)}{a-\overline{a}}$$Meanwhile
$$y-c=\frac{ab+b-a-a\overline{a}}{a-\overline{a}}+\frac{b}{\overline{a}}=\frac{a(\overline{a}+1)(b-\overline{a})}{\overline{a}(a-\overline{a})}$$It is easy to see that they have the same modulus, so we are done.
This post has been edited 1 time. Last edited by mathaddiction, Nov 19, 2020, 1:41 PM
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DottedCaculator
7357 posts
#18
Y by
complex solution without phantom points

[asy]
unitsize(2cm);
pair A, B, C, D, X, Y;
X=(0,0);
A=(-1,2);
B=(-2,-2);
C=(3,-1);
D=A*(2*foot(C/B,(0,0),(1,0))-C/B);
Y=extension((A+B)/2,circumcenter(A,B,C),(C+D)/2,circumcenter(B,C,D));
draw(A--B--C--D--A--X--D--C--X--B);
draw(A--Y--B--C--Y--D);
label("$A$", A, NW);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, NE);
label("$X$", X, W);
label("$Y$", Y, S);
[/asy]

Since $\angle ADX=\angle BCX$ and $\angle DAX=\angle CBX$, this means that $\triangle ADX\sim\triangle BCX$, so $\frac{d-x}{a-x}=\frac{\overline c-\overline x}{\overline b-\overline x}$. since $X$ is inside $ABCD$. Assume without loss of generality $x=0$. Then, we have $\overline bd=a\overline c$.
Now, since $|y-a|=|y-b|$, we must have
\begin{align*}
(y-a)(\overline y-\overline a)&=(y-b)(\overline y-\overline b)\\
-a\overline y-y\overline a+|a|^2&=-b\overline y-y\overline b+|b|^2\\
y(\overline b-\overline a)+\overline y(b-a)&=|b|^2-|a|^2.
\end{align*}Similarly, we have
$$y(\overline d-\overline c)+\overline y(d-c)=|d|^2-|c|^2.$$Therefore, we have
$$y=\frac{(|b|^2-|a|^2)(d-c)-(b-a)(|d|^2-|c|^2)}{(\overline b-\overline a)(d-c)-(b-a)(\overline d-\overline c)}.$$We need to show that $\frac{(a-y)(d-x)^2}{(y-b)(a-d)^2}$ is real. We have $$y-b=\frac{\overline a(b-a)(d-c)-(b-a)(|d|^2-|c|^2)+(b-a)(b\overline d-b\overline c)}{(\overline b-\overline a)(d-c)-(b-a)(\overline d-\overline c)}=\frac{(b-a)(\overline ad-\overline ac-d\overline d+c\overline c+b\overline d-b\overline c)}{(\overline b-\overline a)(d-c)-(b-a)(\overline d-\overline c)}.$$Similarly, we have
$$y-a=\frac{(b-a)(\overline bd-\overline bc-d\overline d+c\overline c+a\overline d-a\overline c)}{(\overline b-\overline a)(d-c)-(b-a)(\overline d-\overline c)}.$$Therefore, we have $\frac{a-y}{y-b}=-\frac{\overline bd-\overline bc-d\overline d+c\overline c+a\overline d-a\overline c}{\overline ad-\overline ac-d\overline d+c\overline c+b\overline d-b\overline c}$. Since $\overline bd=a\overline c$, we also have $b\overline d=\overline ac$. Therefore, this means that
$$\frac{a-y}{y-b}=-\frac{-\overline bc-d\overline d+c\overline c+a\overline d}{\overline ad-d\overline d+c\overline c-b\overline c},$$so $$\frac{(a-y)(d-x)^2}{(y-b)(a-d)^2}=\frac{(\overline bc+d\overline d-c\overline c-a\overline d)d^2}{(\overline ad-d\overline d+c\overline c-b\overline c)(a-d)^2}.$$Therefore, it suffices to show that $$\frac{(\overline bc+d\overline d-c\overline c-a\overline d)d^2}{(\overline ad-d\overline d+c\overline c-b\overline c)(a-d)^2}=\frac{(b\overline c+d\overline d-c\overline c-\overline ad)\overline d^2}{(a\overline d-d\overline d+c\overline c-\overline bc)(\overline a-\overline d)^2}.$$This is equivalent to
$$\frac{(\overline bc+d\overline d-c\overline c-a\overline d)^2}{(\overline ad-d\overline d+c\overline c-b\overline c)^2}=\frac{(a-d)^2\overline d^2}{d^2(\overline a-\overline d)^2}.$$We will show that $\frac{\overline bc+d\overline d-c\overline c-a\overline d}{\overline ad-d\overline d+c\overline c-b\overline c}=\frac{(a-d)\overline d}{d(\overline d-\overline a)}$. Adding one to each side gives $$\frac{\overline ad-a\overline d+\overline bc-b\overline c}{\overline ad-d\overline d+c\overline c-b\overline c}=\frac{a\overline d-\overline ad}{d(\overline d-\overline a)},$$which is equivalent to $$\frac{a\overline d-\overline ad+b\overline c-\overline bc}{d(\overline d-\overline a)+\overline c(b-c)}=\frac{a\overline d-\overline ad}{d(\overline d-\overline a)}.$$Therefore, it suffices to show $\frac{a\overline d-\overline ad}{d(\overline d-\overline a)}=\frac{b\overline c-\overline bc}{\overline c(b-c)}$. Since $d=\frac{a\overline c}{\overline b}$, this means that we have
\begin{align*}
\frac{a\overline d-\overline ad}{d(\overline d-\overline a)}&=\frac{a\frac{\overline ac}b-\overline a\frac{a\overline c}{\overline b}}{\frac{a\overline c}{\overline b}(\frac{\overline ac}b-\overline a)}\\
&=\frac{\frac cb-\frac{\overline c}{\overline b}}{\frac{\overline c}{\overline b}(\frac{c-b}b)}\\
&=\frac{\overline bc-b\overline c}{\overline c(c-b)}\\
&=\frac{b\overline c-\overline bc}{\overline c(b-c)}.
\end{align*}
Therefore, this means that $\angle AYB=2\angle ADX$.
This post has been edited 1 time. Last edited by DottedCaculator, Feb 20, 2022, 12:52 PM
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JAnatolGT_00
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#19
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Let $Y',Z$ be points on perpendicular bisector of $AB,$ such that $AY'Z\stackrel{+}{\sim} ADX,BY'Z\stackrel{+}{\sim} BCX.$
By spiral similarity $AXZ\stackrel{+}{\sim} ADY',BXZ\stackrel{+}{\sim} BCY',$ so $$\frac{|AY'|}{|Y'D|}=\frac{|AZ|}{|ZX|}=\frac{|BZ|}{|ZX|}=\frac{|BY'|}{|Y'C|}\implies |Y'C|=|Y'D|\implies Y'=Y.$$Finally $\measuredangle AYX=2\measuredangle AYZ=2\measuredangle ADX,$ as desired.
This post has been edited 1 time. Last edited by JAnatolGT_00, May 24, 2022, 6:06 PM
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awesomeming327.
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#20
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Note that $\triangle AXD\sim \triangle BXC$. There exists a unique spiral similarity that takes $DX$ onto line $EY$. Let $D\to P$ and $E\to Q$. We have that $\triangle BQP\cong \triangle AQP$ so $\triangle BQP \sim \triangle BXC$. We also have $\triangle BQX\sim \triangle BPC$ and $\triangle AQX\sim\triangle APD$. Thus,
\[\frac{DP}{CP}=\frac{AP}{BP}\cdot \frac{QX}{AQ}\cdot \frac{BQ}{QX}=1\]so $CP=DP$, implying $P=Y$ and the result follows.
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popop614
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#21
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Weird solution:
Let $O$ be the circumcenter of $\triangle AXD$, and denote by $K, Z$ the images of $D, O$ under the spiral similarity taking $AD$ to $AB$.
Now $DK/AD=BX/AX=BC/AD$ so $DK=BC$. Now in directed angles,
\[ \angle ZKD = \angle ZKA + \angle AKD = \angle ZKA + \angle ABX = \angle ZKA + \angle ABZ + \angle ZBX = \angle KAB + \angle ZBX = \angle XBC + \angle ZBX = \angle ZBC \], so \triangle ZBC \cong \triangle ZKD$, whence $Z = Y$. This finishes.
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OronSH
1748 posts
#22 • 2 Y
Y by Om245, GeoKing
Redefine $Y$ to the clawson schmidt conjugate of $X$ and let $P=(AXD)\cap(BXC)\cap(AYB)\cap(CYD).$ Angle chasing gives $PX$ bisects both $\angle APB$ and $\angle CPD.$ Now invert at $P.$ After inversion if we define $Y',Y''$ to be the intersections of the line through $P$ perpendicular to $PX$ with $AB,CD$ then prism lemma and the angle bisector harmonic bundles give $Y'=Y''=Y.$ Thus $\angle XPY=90^\circ$ so $PY$ is a bisector of $\angle APB$ and $\angle CPD$ as well. Thus $AY=BY,CY=DY$ so $Y$ is the same point as given in the problem. Now $\measuredangle AYB=\measuredangle APB=2\measuredangle APX=2\measuredangle ADX.$
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TestX01
341 posts
#23 • 3 Y
Y by GeoKing, bjump, OronSH
@above

Nice, I got the exact same solution.
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TestX01
341 posts
#25 • 2 Y
Y by OronSH, L13832
let me cook

translating unicode from word into latex on aops is annoying.

What if we take Clawson Schmidt Conjugate of $Y$!!!

Let $X'$ be the Clawson-Schmidt Conjugate of $Y$. We will prove that $X'=X. $
Let $P=(AYB\cap CPD)$. All angles are directed in this solution.
Lemma: $\angle APB$ and $\angle CPD$ share an angle bisector.
Let $E$ be the antipode of $Y$ in $(AYB)$ and let $F$ be the antipode of $Y$ in $(CPD)$. By the definition of perpendicular bisectors, $E,F$ are respective midpoints of arcs $AB, CD$ of their respective circles, not containing $Y. $
This means, by the Incentre Excentre Lemma, that $PE$ bisects $\angle APB$ and $PF$ bisects $\angle CPD.$

Yet by Thales, as $EY$ and $FY$ are diameters by symmetry, $\angle EPY=90^\circ$, and $\angle YPF=90^\circ$. Thus, $E,P,F$ are collinear, hence $\angle APB$ and $\angle CPD$ share an angle bisector.
Corollary: This angle bisector is the line through $E,P,F.$
Now consider an inversion at $P$ with radius $PY$. Let $K^*$ be the inverse of $K$. Note $Y^*=Y.$
By Clawson-Schmidt conjugation’s definition, $APYB,PYCD,APX' D,BPX' C$ are cyclic. Hence,
\[A^*,Y,B^* \quad C^*,Y,D^* \quad A^*,X'^* ,D^* \quad B^*,X'^* ,C^*\]are collinear (within each triple).
We thus have the complete quadrilateral $A^* B^* YC^* D^* X'^{*}$.
Clearly, $E^*$ is the intersection of the angle bisector of $\angle APB$ with $A^* B^*$, and similarly $F^*$ is the intersection with $C^* D^*$. Recall that by Thales, $\angle EPY=90^\circ=\angle YPF$, and since $E^* PF^*$ is the angle bisector, it is well-known that the two conditions imply :
\[(A^*,B^*,E^*,Y)=-1=(C^*,D^*,Y,F^* )\]By the Prism Lemma,$ E^*,F^*,X'^{*}$ are collinear looking at our complete quadrilateral. This means that $E,F,X'$ are collinear.
We proceed by straightforward angle relations in cyclic quadrilaterals.
\begin{align}
\angle ADX'=\angle APE=\angle EPB=\angle X' CB\\
\angle X' AD=\angle X' PD=\angle CPX'=\angle CBX'
\end{align}Hence $X'$ works.
Claim: Only one $X$ exists satisfying the properties in the problem, which is $X'$.
Let us consider the locus of points $X_1$ such $\angle DAX_1=\angle X_1 BC$, without the acute angle restraint. Construct the point $X_1'$ such that $X_1 BX_1' C$ is a parallelogram. Let $P=AD\cap BC.  $
As $P,A,D$ and $P,B,C$ are collinear,
\[\angle PAX_1'=180^\circ-\angle X_1 AD-\angle X_1' AX_1=180^\circ-\angle X_1 BX_1'-\angle CBX_1=\angle X_1' BP\]Let $\Phi$ be the function that takes the isogonal conjugate of an object with respect to $\triangle ABP.$ Then, we have
\[\angle \Phi(X_1' )AB=\angle PAX_1'=\angle X_1' BP=\angle AB\Phi(X_1' )\]This implies that $\Phi(X_1' )$ lies on $\ell_1$, which we define as the perpendicular bisector of $AB$. Since isogonal conjugation is an involution, and obviously all of the before steps hold for the converse, for any $\Phi(X_1')$ on $\ell_1, \Phi(\Phi(X_1' ))=X_1'$ which satisfies the properties in the question.
Hence, letting $\mathcal{L}_1$ be the desired locus, we have
\[\Phi(\mathcal{L}_1 )=\ell_1\]Theorem: The isogonal conjugate of a line with respect to a triangle is a circum-conic.
Let $\ell_1$ intersect $(ABP)$ at $M_1$ and $M_2$ such that $M_1$ is the midpoint of minor arc $AB$, and $M_2$ the midpoint of major arc $AB$.
Claim: $\mathcal{L}_1$ is a rectangular hyperbola.
Consider the isogonal conjugate of $M_1$. We have:
\[\angle APM_1=\angle M_1 PB=\angle M_1 AB=\angle PA\Phi(M_1 )\]By the Incenter Excenter Lemma, thus $PM_1\parallel A\Phi(M_1 )$. Similarly,
\[\angle M_1 PB=\angle APM_1=\angle ABM_1=\angle \Phi(M_1 )BP\]Hence $PM_1\parallel B\Phi(M_1 )$. Note that $P,\Phi(M_1 ),M_1$ are collinear as $PM_1$ is the angle bisector of $\angle BPA$. This means $\Phi(M_1 )$ is the point at infinity along $PM_1$, and this lies on $\mathcal{L}_1. $
Now, $P,\Phi(M_2 ),M_2$ are clearly collinear by Thales as $90^\circ+90^\circ=180^\circ. $
\[\angle M_2 AB+\angle M_1 PA=(180^\circ-\angle BPA)/2+(\angle BPA)/2=90^\circ\]Hence $PM_1\perp A\Phi(M_2 )$ which implies $P\Phi(M_2 )\parallel A\Phi(M_2 )$. Analogously, we get $B\Phi(M_2 )\parallel P\Phi(M_2 )$. This implies that $\Phi(M_2 )$ is the point at infinity along $PM_2$.

Both $\Phi(M_1 )$ and $\Phi(M_2 )$ belong to $\mathcal{L}_1$ as $M_1,M_2$ lie on $l_1$. This means that $\mathcal{L}_1$ has two distinct points at infinity (Note the parallel lines were orthogonal). As we also know that $\mathcal{L}_1$ is a conic, $\mathcal{L}_1$ must be a hyperbola.
Let $M$ be the center of the hyperbola. By definition, $M\Phi(M_1 )$ and $M\Phi(M_2 )$ are asymptotes of $\mathcal{L}_1$. But $M\Phi(M_1 )$ is parallel to $M_1 P,$ which is orthogonal to $PM_2$ by Thales as $M_1 M_2$ is a diameter of $(\triangle ABP)$. Yet $PM_2$ is parallel to $M\Phi(M_2 )$, hence
$M\Phi(M_1 )\perp M\Phi(M_2 )$
Thus $\mathcal{L}_1$ is a rectangular hyperbola.
Recall that
\[\angle PAX_1'=\angle X_1' BP\quad \Leftrightarrow\quad \angle X_1' AD=\angle CBX_1'\]Hence $X_1'$ also lies on $\mathcal{L}_1$. Note that constructing $X_1'$ is equivalent to reflection over the midpoint of $AB$ by the properties of parallelograms.
Pick a branch of the hyperbola. Reflect that branch over the midpoint of $AB$. From our previous result, this must be a component of \mathcal{L}_1. Assume for the sake of contradiction that this reflection intersects the initial branch at some point $K$, other than the midpoint of $AB$. Then $K$
is fixed upon reflection but not the center of reflection, hence lie on different half-planes: obvious contradiction.
Else, the reflection is distinct from the initial branch, yet part of $\mathcal{L}_1$, hence it is precisely the other branch of the hyperbola. This implies that the center of $\mathcal{L}_1$, $M$, is indeed the midpoint of $AB$. Note that even when we have $K$ as the midpoint of $AB$, for $K$ to lie on $\mathcal{L}_1$, it must have multiplicity $2$, one count from each branch of $\mathcal{L}_1$. Hence if we consider multiplicity, the midpoint of $AB$ is still the center of reflection hence the center of $\mathcal{L}_1$.
Hence $M$ is always the midpoint of $AB$.

Consequently, the asymptotes of $\mathcal{L}_1$ are more specifically the lines respectively parallel to, and perpendicular to the angle bisector of $\angle BPA$ through $M$.
Now consider the locus of points $X_2$ such that $\angle ADX_2=\angle X_2 CB$. From all our previous reasoning, say by relabeling:
\[A\Leftrightarrow D,B\Leftrightarrow C,X_1\Leftrightarrow X_2,X_1'\Leftrightarrow X_2'\]And so on. We have an analogous result for the locus of $X_2$, which in our original diagram is a rectangular hyperbola $\mathcal{L}_2$ with center $N$, which is the midpoint of $CD$. Its asymptotes are respectively the lines parallel to, and perpendicular to, the angle bisector of $\angle BPA$ through $N$.
The locus of points $X$ satisfying the constraints in the problem is simply the intersection of $\mathcal{L}_1$ and $\mathcal{L}_2$.
Bézout’s Theorem: If two plane algebraic curves of degrees $d_1$ and $d_2$ have no component in common, they have $d_1 d_2$ intersection points, counted with their multiplicity, and including points at infinity and points with complex coordinates.
The two rectangular hyperbolas each have degree $2$. Hence by Bézout’s Theorem, there are $4$ intersection points. First of all, the points at infinity along the angle bisector of $\angle BPA$ and the orthogonal line, both of which lie on both $\mathcal{L}_1$ and $\mathcal{L}_2$ contribute to two intersection points, and are distinct. Further, as $P,A,D$ and $P,B,C$ are collinear, $P$ also lies on $\mathcal{L}_1$ and $\mathcal{L}_2$. This gives another intersection point, which lies on a different plane to the points at infinity hence is distinct.
Note that P, and points at infinity, are outside of $ABCD$. We already know our point $X'$ from before satisfies the problem’s conditions, and since we constructed this point it must exist. If this is in the exterior of our quadrilateral, then no other point $X$ satisfying the conditions of the problem, can exist in the interior of $ABCD$ as we already have counted $4$ intersection points, hence the problem is vacuously true. Else, it is in the interior of the quadrilateral. Then, there are exactly $1+1+1+1=4$ intersection points already, which is the exact number allowed by Bézout, hence $X'$ is the unique point inside $ABCD$ that works.
In fact, this means $X'=X$ by uniqueness.
Hence, returning to our problem, by cyclic quadrilaterals,
\[\angle APE=\angle ADX\quad \quad \quad \quad \quad \angle EPB=\angle XCB\]Hence,
\[\angle AYB=\angle APB=\angle ADX+\angle XCB=2\angle ADX\]Because $\angle ADX=\angle XCB$ from the problem, and since $APYB$ is cyclic.

This concludes the problem.
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InterLoop
279 posts
#28 • 1 Y
Y by OronSH
idk what in the world above is but here's another cs sol
solution
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Nari_Tom
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#29
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Here is another nice solution using Gliding principle.

Let $P$ and $Q$ be the reflection of $X$ over $BC$ and $AD$. Let $M$ and $N$ be the midpoints of $BA$ and $CD$. Let $E$ and $F$ be the projections of $X$ to $BC$ and $AD$.

Since $BPCX$ and $ACDQ$ are similar kites By the Gliding principle we have $MENF$ is similar too.

Let $G$ and $Y'$ be the points on perpendicular bisector of $AB$ such that $BGAY'$ and $EMFN$ are similar kites. Since $\frac{BE}{EC}=\frac{AF}{AD}=\frac{GM}{MY'}$, by the gliding principle we have $\triangle BGA \sim \triangle EMF \sim \triangle CY'D$, which means that $Y'$ lies on perpendicular bisector of $CD$ $\implies$ $Y'=Y$. Conclusion follows.
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