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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.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]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
Point satisfies triple property
62861   36
N 24 minutes ago by cursed_tangent1434
Source: USA Winter Team Selection Test #2 for IMO 2018, Problem 2
Let $ABCD$ be a convex cyclic quadrilateral which is not a kite, but whose diagonals are perpendicular and meet at $H$. Denote by $M$ and $N$ the midpoints of $\overline{BC}$ and $\overline{CD}$. Rays $MH$ and $NH$ meet $\overline{AD}$ and $\overline{AB}$ at $S$ and $T$, respectively. Prove that there exists a point $E$, lying outside quadrilateral $ABCD$, such that
[list]
[*] ray $EH$ bisects both angles $\angle BES$, $\angle TED$, and
[*] $\angle BEN = \angle MED$.
[/list]

Proposed by Evan Chen
36 replies
1 viewing
62861
Jan 22, 2018
cursed_tangent1434
24 minutes ago
Inspired by 2025 Xinjiang
sqing   1
N 31 minutes ago by sqing
Source: Own
Let $ a,b,c >0  . $ Prove that
$$  \left(1+\frac {a} { b}\right)\left(2+\frac {a}{ b}+\frac {b}{ c}\right) \left(1+\frac {a}{b}+\frac {b}{ c}+\frac {c}{ a}\right)  \geq 12+8\sqrt 2 $$
1 reply
sqing
Yesterday at 5:32 PM
sqing
31 minutes ago
Inspired by 2025 Beijing
sqing   4
N 38 minutes ago by pooh123
Source: Own
Let $ a,b,c,d >0  $ and $ (a^2+b^2+c^2)(b^2+c^2+d^2)=36. $ Prove that
$$ab^2c^2d \leq 8$$$$a^2bcd^2 \leq 16$$$$ ab^3c^3d \leq \frac{2187}{128}$$$$ a^3bcd^3 \leq \frac{2187}{32}$$
4 replies
sqing
Yesterday at 4:56 PM
pooh123
38 minutes ago
Find the minimum
sqing   27
N an hour ago by sqing
Source: Zhangyanzong
Let $a,b$ be positive real numbers such that $a^2b^2+\frac{4a}{a+b}=4.$ Find the minimum value of $a+2b.$
27 replies
sqing
Sep 4, 2018
sqing
an hour ago
Sharygin 2025 CR P15
Gengar_in_Galar   7
N an hour ago by Giant_PT
Source: Sharygin 2025
A point $C$ lies on the bisector of an acute angle with vertex $S$. Let $P$, $Q$ be the projections of $C$ to the sidelines of the angle. The circle centered at $C$ with radius $PQ$ meets the sidelines at points $A$ and $B$ such that $SA\ne SB$. Prove that the circle with center $A$ touching $SB$ and the circle with center $B$ touching $SA$ are tangent.
Proposed by: A.Zaslavsky
7 replies
Gengar_in_Galar
Mar 10, 2025
Giant_PT
an hour ago
Moscow Geometry Problems
nataliaonline75   1
N 2 hours ago by MathLuis
Source: MMO 2003 10.4
Let M be the intersection point of the medians of ABC. On the perpendiculars dropped from M to sides BC, AC, AB, points A1, B1, C1 are taken, respectively, with A1B1 perpendicular to MC and A1C1 perpendicular to MB. prove that M is the intersections pf the medians in A1B1C1.
Any solutions without vectors? :)
1 reply
nataliaonline75
Jul 9, 2024
MathLuis
2 hours ago
Another OI geometry problem
chengbilly   7
N 2 hours ago by MathLuis
Source: own
Let $ABC$ be a triangle with incenter $I$ and circumcenter $O$. $H$ the orthocenter of triangle $BIC$ . The incircle of $ABC$ touches side $AC,AB$ at $E,F$ respectively. Suppose that $\odot (AEF)$ and $\odot (AIO)$ intersects $\odot (ABC)$ at $S$ and $T$ respectively (differ from $A$). Prove that $T,H,I,S$ lies on a circle.

($\odot (ABC)$ denote the circumcircle of $ABC$)
7 replies
chengbilly
Apr 14, 2021
MathLuis
2 hours ago
USA GEO 2003
dreammath   21
N 4 hours ago by lpieleanu
Source: TST USA 2003
Let $ABC$ be a triangle and let $P$ be a point in its interior. Lines $PA$, $PB$, $PC$ intersect sides $BC$, $CA$, $AB$ at $D$, $E$, $F$, respectively. Prove that
\[ [PAF]+[PBD]+[PCE]=\frac{1}{2}[ABC]  \]
if and only if $P$ lies on at least one of the medians of triangle $ABC$. (Here $[XYZ]$ denotes the area of triangle $XYZ$.)
21 replies
dreammath
Feb 16, 2004
lpieleanu
4 hours ago
Integers on a cube
Rushil   6
N 5 hours ago by SomeonecoolLovesMaths
Source: Indian RMO 2004 Problem 2
Positive integers are written on all the faces of a cube, one on each. At each corner of the cube, the product of the numbers on the faces that meet at the vertex is written. The sum of the numbers written on the corners is 2004. If T denotes the sum of the numbers on all the faces, find the possible values of T.
6 replies
Rushil
Feb 28, 2006
SomeonecoolLovesMaths
5 hours ago
Tangents to circle concurrent on a line
Drytime   9
N 6 hours ago by Autistic_Turk
Source: Romania TST 3 2012, Problem 2
Let $\gamma$ be a circle and $l$ a line in its plane. Let $K$ be a point on $l$, located outside of $\gamma$. Let $KA$ and $KB$ be the tangents from $K$ to $\gamma$, where $A$ and $B$ are distinct points on $\gamma$. Let $P$ and $Q$ be two points on $\gamma$. Lines $PA$ and $PB$ intersect line $l$ in two points $R$ and respectively $S$. Lines $QR$ and $QS$ intersect the second time circle $\gamma$ in points $C$ and $D$. Prove that the tangents from $C$ and $D$ to $\gamma$ are concurrent on line $l$.
9 replies
Drytime
May 11, 2012
Autistic_Turk
6 hours ago
Parallelogram in the Plane
Taco12   8
N 6 hours ago by lpieleanu
Source: 2023 Canada EGMO TST/2
Parallelogram $ABCD$ is given in the plane. The incircle of triangle $ABC$ has center $I$ and is tangent to diagonal $AC$ at $X$. Let $Y$ be the center of parallelogram $ABCD$. Show that $DX$ and $IY$ are parallel.
8 replies
Taco12
Feb 10, 2023
lpieleanu
6 hours ago
Problem 4 of RMO 2006 (Regional Mathematical Olympiad-India)
makar   7
N Yesterday at 7:56 PM by SomeonecoolLovesMaths
Source: Combinatorics (Box Principle)
A $ 6\times 6$ square is dissected in to 9 rectangles by lines parallel to its sides such that all these rectangles have integer sides. Prove that there are always two congruent rectangles.
7 replies
makar
Sep 13, 2009
SomeonecoolLovesMaths
Yesterday at 7:56 PM
IMO Shortlist 2011, G5
WakeUp   72
N Yesterday at 7:27 PM by ItsBesi
Source: IMO Shortlist 2011, G5
Let $ABC$ be a triangle with incentre $I$ and circumcircle $\omega$. Let $D$ and $E$ be the second intersection points of $\omega$ with $AI$ and $BI$, respectively. The chord $DE$ meets $AC$ at a point $F$, and $BC$ at a point $G$. Let $P$ be the intersection point of the line through $F$ parallel to $AD$ and the line through $G$ parallel to $BE$. Suppose that the tangents to $\omega$ at $A$ and $B$ meet at a point $K$. Prove that the three lines $AE,BD$ and $KP$ are either parallel or concurrent.

Proposed by Irena Majcen and Kris Stopar, Slovenia
72 replies
WakeUp
Jul 13, 2012
ItsBesi
Yesterday at 7:27 PM
Reflections and midpoints in triangle
TUAN2k8   2
N Yesterday at 6:35 PM by Double07
Source: Own
Given an triangle $ABC$ and a line $\ell$ in the plane.Let $A_1,B_1,C_1$ be reflections of $A,B,C$ across the line $\ell$, respectively.Let $D,E,F$ be the midpoints of $B_1C_1,C_1A_1,A_1B_1$, respectively.Let $A_2,B_2,C_2$ be the reflections of $A,B,C$ across $D,E,F$, respectively.Prove that the points $A_2,B_2,C_2$ lie on a line parallel to $\ell$.
2 replies
TUAN2k8
Yesterday at 1:34 PM
Double07
Yesterday at 6:35 PM
Similar triangles formed by angular condition
Mahdi_Mashayekhi   5
N Apr 21, 2025 by sami1618
Source: Iran 2025 second round P3
Point $P$ lies inside of scalene triangle $ABC$ with incenter $I$ such that $:$
$$ 2\angle ABP = \angle BCA , 2\angle ACP = \angle CBA $$Lines $PB$ and $PC$ intersect line $AI$ respectively at $B'$ and $C'$. Line through $B'$ parallel to $AB$ intersects $BI$ at $X$ and line through $C'$ parallel to $AC$ intersects $CI$ at $Y$. Prove that triangles $PXY$ and $ABC$ are similar.
5 replies
Mahdi_Mashayekhi
Apr 19, 2025
sami1618
Apr 21, 2025
Similar triangles formed by angular condition
G H J
G H BBookmark kLocked kLocked NReply
Source: Iran 2025 second round P3
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Mahdi_Mashayekhi
696 posts
#1 • 2 Y
Y by Rounak_iitr, Parsia--
Point $P$ lies inside of scalene triangle $ABC$ with incenter $I$ such that $:$
$$ 2\angle ABP = \angle BCA , 2\angle ACP = \angle CBA $$Lines $PB$ and $PC$ intersect line $AI$ respectively at $B'$ and $C'$. Line through $B'$ parallel to $AB$ intersects $BI$ at $X$ and line through $C'$ parallel to $AC$ intersects $CI$ at $Y$. Prove that triangles $PXY$ and $ABC$ are similar.
Z K Y
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gghx
1072 posts
#2
Y by
Let $S=BP\cap CI$ and $T=BI\cap CP$. Note that $$\angle BPC=\angle ABP+\angle ACP=\angle ACI+\angle ABI=\angle BIC,$$hence $BPIC$ is cyclic.

We now prove that $S$ is the circumcenter of $\triangle B'IX$. This is true because $$\angle SB'I=\angle BAI + \angle ABP=\frac{1}{2}(\angle A + \angle C)=\angle IAC+\angle ACI = \angle SIB',$$hence $SI=SB'$. Furthermore, $$\angle ISB'=180^\circ-\angle A - \angle C=\angle B=2\angle ABI=2\angle B'XI,$$hence $S$ is the circumcenter of $\triangle B'IX$ as desired.

Now, $\angle SXI=\angle SIX=\angle SIT=\angle SPT$, hence $SXTP$ is cyclic. Similarly, $SYTP$ is cyclic, so $SXTPY$ is cyclic.

We are now ready to finish the question. We have $$\angle YPX=\angle XSI=180^\circ-2\angle SIX=\angle A$$and $$\angle YXP=180^\circ-\angle PSI=\angle ABP+\angle ACI=\angle C,$$so triangles $PXY$ and $ABC$ are similar.
Z K Y
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ItzsleepyXD
149 posts
#3
Y by
quite similar (or same IDK) to @above

since $\triangle ABB' \sim \triangle ACI$ and $\triangle ACC' \sim \triangle ABI$
implies that $\triangle BB'I \sim \triangle CIC'$

Let $E = BB' \cap IY , F= CC' \cap IX$ .
So $EB'=EI , FC'=FI$
Known that $\angle B'EI = \angle ABC = 2 \angle B'XI$ implies that $E =$ center of $(B'IX)$.
so $ \angle EXI = \angle EIX = \angle EPF$ so $E,X,F,P$ concyclic.
same as $E,Y,F,P$ concyclic .
so $E,Y,X,F,P$ concyclic.
thus $\angle PXY = \angle PEY = \angle ABC$ and $ \angle PYX = 180^{\circ} - \angle PFX = \angle ACB$
Conclude that $\triangle PXY \sim \triangle ABC$ . done $\square$
Z K Y
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mathuz
1525 posts
#4
Y by
Consider the intersection $C'Y\cap B'X = O(.)$. Then $O$ is the circumcenter of $PB'C'$, and it suffices to show that $O$ lies on the circumcircle of $PXY$.
Z K Y
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bin_sherlo
733 posts
#5
Y by
Let $BP\cap CI=U,PC\cap BI=V,AI\cap BC=D$. Note that $\measuredangle CPB=90+\frac{\measuredangle A}{2}=\measuredangle CIB$ hence $B,I,C,P$ are concyclic.
By Menelaus at $B'UBDCI$ we get $\frac{B'B}{IC}=\frac{\sin \measuredangle C}{\sin \measuredangle B}$. Hence $BX=BB'.\frac{\sin \frac{\measuredangle C}{2}}{\sin \frac{\measuredangle B}{2}}=\frac{BB'.IB}{IC}=\frac{\sin \measuredangle C}{\sin \measuredangle B}.IB$ Also $\frac{BV}{BP}=\frac{\cos \frac{\measuredangle A}{2}}{\sin \measuredangle C}$ thus,
\[\frac{BX.BV}{BP}=\frac{BI}{\sin \measuredangle B}.\cos \frac{\measuredangle A}{2}=BU\]which implies $X\in (PUV)$. Similarily $Y\in (PUV)$. Hence $\measuredangle XYP=\measuredangle XVC=\measuredangle C$ and $\measuredangle PXY=\measuredangle PUY=\measuredangle B$ as desired.$\blacksquare$
Z K Y
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sami1618
913 posts
#6
Y by
Here is a different approach rephrasing the problem in terms of reference triangle $PB'C'$. When I first drew the diagram the points $P$, $X$, and $Y$ were all very close together so this solution was motivated by drawing a diagram consisting of the points other than $A$, $B$, and $C$. :)
It is not hard to show that $\angle B'PC'=90^{\circ}-\tfrac{1}{2}\angle A$, $\angle PB'C'=90^{\circ}-\tfrac{1}{2}\angle B$, and $\angle PC'B'=90^{\circ}-\tfrac{1}{2}\angle C$. Since $\angle BPC=\angle BIC$, it must be that $I$ lies on the interior of segment $B'C'$. Because of this, it is also not hard to see that $X$, $Y$, and $P$ all lie on the same side of segment $AI$.
[asy]

import geometry;

size(10cm);
pair A = dir(110);
pair B = dir(200);
pair C = dir(340);
pair I = incenter(A, B, C);
pair D = foot(I, B, C);
pair Ep=B+C-D;
pair J=I+Ep-D;
pair P=isogonalconjugate(triangle(A,B,C),J);
pair Bp=intersectionpoint(line(B,P),line(A,I));
pair Cp=intersectionpoint(line(C,P),line(A,I));
pair X=intersectionpoint(line(Bp,B+Bp-A), line(B,I));
pair Y=intersectionpoint(line(Cp,C+Cp-A), line(C,I));
pair O=intersectionpoint(line(Cp,Y), line(Bp,X));
point[] Ap=intersectionpoints(circle(Bp,O,Cp),line(O,I));
pair Ap=Ap[1];
pair Op=circumcenter(Ap,Bp,Cp);
pair M_a=2*Op-O;
point[] N_c=intersectionpoints(circle(Bp,O,Cp),line(P,Cp));
pair N_c=N_c[0];
point[] N_b=intersectionpoints(circle(Bp,O,Cp),line(P,Bp));
pair N_b=N_b[0];


draw(A--B--C--cycle, black);
draw(B--Bp,black); draw(C--P,black); draw(A--Cp,black); draw(B--I,black); 
draw(C--Y,black); draw(Bp--X); draw(Cp--Y);  draw(P--Bp--Cp--cycle, black);


dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$I$", I, dir(50));

dot("$P$", P, dir(140));
dot("$B'$", Bp, dir(40));
dot("$C'$", Cp, dir(290));
dot("$X$", X, dir(310));
dot("$Y$", Y, dir(110));
[/asy]
Now notice that $\triangle B'IX\sim\triangle AIB$ and $\triangle C'IY\sim \triangle AIC$. We will now focus on the acute reference triangle $PB'C'$. Let $O$ be the circumcenter of $PB'C'$. Let $X'$ be the second intersection of line $B'O$ with the circumcircle of triangle $POC'$ and let $Y'$ be the second intersection of line $C'O$ with the circumcircle of triangle $POB'$. It is easy to show that $\triangle PX'C'\sim$ $\triangle PB'Y'\sim$ $\triangle ABC$. Now notice that $\triangle B'C'X'\sim$ $\triangle AIB\sim$ $\triangle B'IX$. Thus $X$ is the point on segment $B'X'$ with $IX\parallel C'X'$. Similarly, $Y$ is the point along segment $C'Y'$ with $IY\parallel B'Y'$.
[asy]
import geometry;

size(10cm);

pair P=dir(100);
pair Bp=dir(270+55);
pair Cp=dir(270-55);
pair O=(0,0);
pair I=intersectionpoint(line(O,P),line(Bp,Cp));
pair Xp[]=intersectionpoints(line(Bp,O),circle(P,O,Cp));
pair Xp=Xp[1];
pair Yp[]=intersectionpoints(line(Cp,O),circle(P,O,Bp));
pair Yp=Yp[1];
pair X=intersectionpoint(line(I,I+Xp-Cp), line(Bp,Xp));
pair Y=intersectionpoint(line(I,I+Yp-Bp), line(Cp,Yp));

fill(P--Xp--Cp--cycle,palered+white);
fill(P--Yp--Bp--cycle,palered+white);
fill(P--X--Y--cycle, palered);

draw(P--X--Y--cycle);
draw(P--Bp--Cp--cycle);
draw(circle(P,O,Cp));
draw(circle(P,O,Bp));
draw(Bp--Xp); draw(Cp--Yp);
draw(X--I--Y);
draw(Cp--Xp--P--Yp--Bp);

dot("P",P,2*dir(P));
dot("B'",Bp,dir(270));
dot("C'",Cp,dir(270));

dot("O",O,dir(270));
dot("I",I,dir(270));
dot("X'",Xp,dir(150));
dot("Y'",Yp,dir(30));
dot("X",X,dir(220));
dot("Y",Y,dir(-40));
[/asy]
Then we have that $$\frac{X'X}{XB'}=\frac{C'I}{IB'}=\frac{C'Y}{YY'}.$$Thus triangle $PXY$ is a linear combination of triangle $PX'C'$ and $PB'Y'$. Since these triangles are both similar to $ABC$, it is a well-known result that $PXY$ must be similar to $ABC$ as well.
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