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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

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Do you have plans this summer? There are so many options to fit your schedule and goals whether attending a summer camp or taking online classes, it can be a great break from the routine of the school year. Check out our summer courses at AoPS Online, or if you want a math or language arts class that doesn’t have homework, but is an enriching summer experience, our AoPS Virtual Campus summer camps may be just the ticket! We are expanding our locations for our AoPS Academies across the country with 15 locations so far and new campuses opening in Saratoga CA, Johns Creek GA, and the Upper West Side NY. Check out this page for summer camp information.

Be sure to mark your calendars for the following events:
[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
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0 replies
jlacosta
Mar 2, 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
Sharygin 2025 CR P17
Gengar_in_Galar   5
N 4 minutes ago by HVP
Source: Sharygin 2025
Let $O$, $I$ be the circumcenter and the incenter of an acute-angled scalene triangle $ABC$; $D$, $E$, $F$ be the touching points of its excircle with the side $BC$ and the extensions of $AC$, $AB$ respectively. Prove that if the orthocenter of the triangle $DEF$ lies on the circumcircle of $ABC$, then it is symmetric to the midpoint of the arc $BC$ with respect to $OI$.
Proposed by: P.Puchkov,E.Utkin
5 replies
Gengar_in_Galar
Mar 10, 2025
HVP
4 minutes ago
Not actually combo
MathSaiyan   0
4 minutes ago
Source: PErA 2025/5
We have an $n \times n$ board, filled with $n$ rectangles aligned to the grid. The $n$ rectangles cover all the board and are never superposed. Find, in terms of $n$, the smallest value the sum of the $n$ diagonals of the rectangles can take.
0 replies
MathSaiyan
4 minutes ago
0 replies
Sharygin 2025 CR P13
Gengar_in_Galar   5
N 6 minutes ago by HVP
Source: Sharygin 2025
Each two opposite sides of a convex $2n$-gon are parallel. (Two sides are opposite if one passes $n-1$ other sides moving from one side to another along the borderline of the $2n$-gon.) The pair of opposite sides is called regular if there exists a common perpendicular to them such that its endpoints lie on the sides and not on their extensions. Which is the minimal possible number of regular pairs?
Proposed by: B.Frenkin
5 replies
Gengar_in_Galar
Mar 10, 2025
HVP
6 minutes ago
Nice, simple geo
MathSaiyan   0
7 minutes ago
Source: PErA 2025/4
Let \( ABC \) be an acute-angled scalene triangle. Let \( B_1 \) and \( B_2 \) be points on the rays \( BC \) and \( BA \), respectively, such that \( BB_1 = BB_2 = AC \). Similarly, let \( C_1 \) and \( C_2 \) be points on the rays \( CB \) and \( CA \), respectively, such that \( CC_1 = CC_2 = AB \). Prove that if \( B_1B_2 \) and \( C_1C_2 \) intersect at \( K \), then \( AK \) is parallel to \( BC \).
0 replies
MathSaiyan
7 minutes ago
0 replies
The last nonzero digit of factorials
Tintarn   1
N 8 minutes ago by AshAuktober
Source: Bundeswettbewerb Mathematik 2025, Round 1 - Problem 2
For each integer $n \ge 2$ we consider the last digit different from zero in the decimal expansion of $n!$. The infinite sequence of these digits starts with $2,6,4,2,2$. Determine all digits which occur at least once in this sequence, and show that each of those digits occurs in fact infinitely often.
1 reply
Tintarn
2 hours ago
AshAuktober
8 minutes ago
Stronger than Iran 96, not completely symmetric
MeoMayBe   2
N 9 minutes ago by CHESSR1DER
Source: Own
Let a, b, c\geq 0. Prove that
(bc+ca+ab)\left[\frac{1}{(b+c)^{2}}+\frac{1}{(c+a)^{2}}+\frac{1}{(a+b)^{2}}\right]\geq\frac{9}{4}+\frac{2abc(b-c)^{2}}{(a+b)^{2}(a+c)^{2}(b+c)}.
PS. 2 is also the best constant.
PS (2). Sorry again, I cannot type LaTeX since new members cannot share image.
2 replies
MeoMayBe
May 14, 2023
CHESSR1DER
9 minutes ago
Equilateral triangle geo
MathSaiyan   0
10 minutes ago
Source: PErA 2025/3
Let \( ABC \) be an equilateral triangle with circumcenter \( O \). Let \( X \) and \( Y \) be two points on segments \( AB \) and \( AC \), respectively, such that \( \angle XOY = 60^\circ \). If \( T \) is the reflection of \( O \) with respect to line \( XY \), prove that lines \( BT \) and \( OY \) are parallel.
0 replies
MathSaiyan
10 minutes ago
0 replies
Inspired by youthdoo
sqing   1
N 11 minutes ago by sqing
Source: Own
Let $ a,b,c $ be real numbers such that $      \frac{3}{a^2+6}+\frac{1}{b^2+2}+\frac{3}{c^2+6}=1. $ Prove that$$ab+bc+ca\leq 7$$Let $ a,b,c $ be real numbers such that $   \frac{2}{a^2+4}+\frac{3}{b^2+6}+\frac{2}{c^2+4}=1. $ Prove that$$ab+bc+ca\leq 7$$Let $ a,b,c $ be real numbers such that $    \frac{3}{a^2+6}+\frac{2}{b^2+4}+\frac{3}{c^2+6}=1. $ Prove that$$ab+bc+ca\leq 8$$Let $ a,b,c $ be real numbers such that $  \frac{3}{a^2+6}+\frac{4}{b^2+8}+\frac{3}{c^2+6}=1. $ Prove that$$ab+bc+ca\leq 10$$
1 reply
sqing
26 minutes ago
sqing
11 minutes ago
Clean number theory
MathSaiyan   0
14 minutes ago
Source: PErA 2025/2
Let $m$ be a positive integer. We say that a positive integer $x$ is $m$-good if $a^m$ divides $x$ for some integer $a > 1$. We say a positive integer $x$ is $m$-bad if it is not $m$-good.
(a) Is it true that for every positive integer $n$ there exist $n$ consecutive $m$-bad positive integers?
(b) Is it true that for every positive integer $n$ there exist $n$ consecutive $m$-good positive integers?
0 replies
MathSaiyan
14 minutes ago
0 replies
Many-solutions combogeo
MathSaiyan   1
N 14 minutes ago by Speedysolver1
Source: PErA 2025/1
Let $S$ be a set of at least three points of the plane in general position. Prove that there exists a non-intersecting polygon whose vertices are exactly the points of $S$.
1 reply
MathSaiyan
17 minutes ago
Speedysolver1
14 minutes ago
2 var inquality
sqing   6
N 18 minutes ago by ionbursuc
Source: Own
Let $ a ,  b\geq 0 $ and $ \frac{1}{a^2+1}+\frac{1}{b^2+1}\le   \frac{3}{2}. $ Show that$$ a+b+ab\geq1$$Let $ a ,  b\geq 0 $ and $ \frac{1}{a^2+1}+\frac{1}{b^2+1}\le   \frac{5}{6}. $ Show that$$ a+b+ab\geq2$$
6 replies
sqing
Today at 4:06 AM
ionbursuc
18 minutes ago
Sharygin 2025 CR P2
Gengar_in_Galar   4
N 27 minutes ago by FKcosX
Source: Sharygin 2025
Four points on the plane are not concyclic, and any three of them are not collinear. Prove that there exists a point $Z$ such that the reflection of each of these four points about $Z$ lies on the circle passing through three remaining points.
Proposed by:A Kuznetsov
4 replies
Gengar_in_Galar
Mar 10, 2025
FKcosX
27 minutes ago
100 Selected Problems Handout
Asjmaj   32
N 31 minutes ago by John_Mgr
Happy New Year to all AoPSers!
 :clap2:

Here’s my modest gift to you all. Although I haven’t been very active in the forums, the AoPS community contributed to an immense part of my preparation and left a huge impact on me as a person. Consider this my way of giving back. I also want to take this opportunity to thank Evan Chen—his work has consistently inspired me throughout my olympiad journey, and this handout is no exception.



With 2025 drawing near, my High School Olympiad career will soon be over, so I want to share a compilation of the problems that I liked the most over the years and their respective detailed write-ups. Originally, I intended it just as a personal record, but I decided to give it some “textbook value” by not repeating the topics so that the selection would span many different approaches, adding hints, and including my motivations and thought process.

While IMHO it turned out to be quite instructive, I cannot call it a textbook by any means. I recommend solving it if you are confident enough and want to test your skills on miscellaneous, unordered, challenging, high-quality problems. Hints will allow you to not be stuck for too long, and the fully motivated solutions (often with multiple approaches) should help broaden your perspective. 



This is my first experience of writing anything in this format, and I’m not a writer by any means, so please forgive any mistakes or nonsense that may be written here. If you spot any typos, inconsistencies, or flawed arguments whatsoever (no one is immune :blush: ), feel free to DM me. In fact, I welcome any feedback or suggestions.

I left some authors/sources blank simply because I don’t know them, so if you happen to recognize where and by whom a problem originated, please let me know. And quoting the legend: “The ideas of the solution are a mix of my own work, the solutions provided by the competition organizers, and solutions found by the community. However, all the writing is maintained by me.” 



I’ll likely keep a separate file to track all the typos, and when there’s enough, I will update the main file. Some problems need polishing (at least aesthetically), and I also have more remarks to add.

This content is only for educational purposes and is not meant for commercial usage.



This is it! Good luck in 45^2, and I hope you enjoy working through these problems as much as I did!

Here's a link to Google Drive because of AoPS file size constraints: Selected Problems
32 replies
Asjmaj
Dec 31, 2024
John_Mgr
31 minutes ago
Where is the equality?
AndreiVila   2
N an hour ago by MihaiT
Source: Romanian District Olympiad 2025 9.3
Determine all positive real numbers $a,b,c,d$ such that $a+b+c+d=80$ and $$a+\frac{b}{1+a}+\frac{c}{1+a+b}+\frac{d}{1+a+b+c}=8.$$
2 replies
AndreiVila
Mar 8, 2025
MihaiT
an hour ago
Variable point on the median
MarkBcc168   47
N Today at 2:36 AM by HamstPan38825
Source: APMO 2019 P3
Let $ABC$ be a scalene triangle with circumcircle $\Gamma$. Let $M$ be the midpoint of $BC$. A variable point $P$ is selected in the line segment $AM$. The circumcircles of triangles $BPM$ and $CPM$ intersect $\Gamma$ again at points $D$ and $E$, respectively. The lines $DP$ and $EP$ intersect (a second time) the circumcircles to triangles $CPM$ and $BPM$ at $X$ and $Y$, respectively. Prove that as $P$ varies, the circumcircle of $\triangle AXY$ passes through a fixed point $T$ distinct from $A$.
47 replies
MarkBcc168
Jun 11, 2019
HamstPan38825
Today at 2:36 AM
Variable point on the median
G H J
G H BBookmark kLocked kLocked NReply
Source: APMO 2019 P3
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VulcanForge
624 posts
#35
Y by
This solution contains way too many instances of "radical axis" and "Reim"
[asy]
size(11cm); pair A,B,C,D,E,P,Bp,Cp,X,Y,M,T,K,L; defaultpen(fontsize(7pt));

A=dir(124); B=dir(220); C=dir(-40); M=0.5B+0.5C; P=0.5A+0.5M; Bp=intersectionpoints(B--(B+A-M),unitcircle)[0]; Cp=intersectionpoints(C--(C+2A-2M),unitcircle)[0];

path wb, wc; wb=circumcircle(B,P,M); wc=circumcircle(C,P,M);

D=intersectionpoints(unitcircle,wb)[1]; E=intersectionpoints(unitcircle,wc)[0];

K=intersectionpoint(M--(2M-A),C--(3C-2E)); draw(A--K--E); 

X=intersectionpoints(P--(2P-D),wc)[1]; Y=intersectionpoints(P--(2P-E),wb)[1]; T=intersectionpoint((3Bp-2Cp)--Bp,B--(3B-2C));

L=intersectionpoint(Y--(2Y-B),A--M); draw(B--L--C);

draw(unitcircle); draw(A--B--C--cycle); draw(wb); draw(wc); draw(D--Cp); draw(E--Bp); draw(Cp--T--B); draw(B--K); draw(T--X); draw(D--E);

dot("$A$",A,dir(A));
dot("$B$",B,dir(225));
dot("$C$",C,dir(C));
dot("$D$",D,dir(D));
dot("$E$",E,dir(E));
dot("$B'$",Bp,dir(Bp));
dot("$C'$",Cp,dir(Cp));
dot("$X$",X,2*dir(10));
dot("$Y$",Y,1.5*dir(150));
dot("$M$",M,dir(M));
dot("$T$",T,dir(T));
dot("$P$",P,dir(140));
dot("$K$",K,1.5*dir(K));
dot("$L$",L,dir(80));
[/asy]
By radical axis theorem, we have $BD, CE$ concur at some point $K$ on $AM$. By Reim we have $BY \parallel CE$ and $CX \parallel BY$, so we have $BY,CX$ meet at a point $L$ on $AM$. By radical axis this implies $BCXY$ cyclic.

Let $B',C'$ be the points on $(ABC)$ such that $BB' \parallel CC' \parallel AM$, and let $BC$ and $B'C'$ intersect at $T$. It suffices to show that $XY$ always passes through $T$: this will imply by radical axis that $(AXY)$ always passes through the point $(AXY) \cap ABC$.

To show $XY$ passes through $T$, by radical axis it suffices to show $B'C'XY$ is cyclic, which by Reim is equivalent to $XY \parallel DE$. Consider a reflection across $M$, which sends $D,E$ to points $D',E'$ on lines $CL,BL$ respectively. Then since $BCD'E'$ is cyclic by Reim we have $XY \parallel D'E'$. But clearly $D'E' \parallel DE$, and we win.
This post has been edited 1 time. Last edited by VulcanForge, Feb 7, 2022, 10:26 PM
Z K Y
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IvoBucata
46 posts
#36
Y by
Difficult but nonetheless really beautiful problem!

Let $O$ and $H$ be point on $\Gamma $ such that $BO\parallel AM\parallel CH$. From Reim's theorem we get that $BD\parallel CY$; $BX\parallel CE$; $E-P-O-X$ and $D-P-H-Y$.

Now let $BD$ intersect $CE$ at $Q$ and $BX$ intersect $CY$ at $R$.By radical axis in circles $(BMPD);(CMPE);(BCED)$ we get that $Q$ lies on $AM$, and since $BRCQ$ is a paralellogram we get that $R$ lies on $AM$ as well. Now from angle chasing we have that the following quadrilaterals are cyclic:$XRYP; XYHO; XBCY$. From radical axis on circles $(BCOH);(XYHO); (BCXY)$ we get that $BC; XY$ and $OH$ are concurrent, say at a point $I$, which doesn't depend on $P$ since neither of $BC$ and $OH$ do. Thus we get that $IX*IY=pow_I((AXY))=IB*IC$ is fixed, so let $G$ be the point on ray $OA$ such that $IG*IA=pow_I((AXY))=IX*IY$, thus from power of a point $(AXY) $ always passes trough $G$, which doesn't depend on $P$.
Z K Y
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DottedCaculator
7303 posts
#37 • 3 Y
Y by v4913, RyanPMathematics, mathmax12
[asy]
unitsize(1cm);
pair A, B, C, M, P, D, E, X, Y, Z, B1, C1, Q;
A=(2,3sqrt(5));
B=(0,0);
C=(8,0);
M=(B+C)/2;
P=(3*M+A)/4;
dot(P);
draw(A--B--C--A--M);
draw(circumcircle(C,P,M));
draw(circumcircle(B,P,M));
draw(circumcircle(A,B,C));
D=intersectionpoints(circumcircle(B,P,M),circumcircle(A,B,C))[1];
E=intersectionpoints(circumcircle(C,P,M),circumcircle(A,B,C))[1];
B1=extension(B,B+A-M,E,P);
C1=extension(C,C+A-M,D,P);
draw(B--B1--E);
draw(C--C1--D);
X=intersectionpoints(P--C1,circumcircle(C,M,P))[1];
Y=intersectionpoints(P--B1,circumcircle(B,M,P))[1];
draw(circumcircle(A,X,Y));
Z=extension(B1,C1,B,C);
draw(C--Z--C1--Z--A);
draw(circumcircle(B,Y,B1));
draw(circumcircle(C,X,C1));
Q=intersectionpoints(A--Z,circumcircle(A,X,Y))[1];
label("$A$", A, NW);
label("$B$", B, SW);
label("$C$", C, SE);
label("$M$", M, SW);
label("$P$", P, SW);
label("$D$", D, SW);
label("$E$", E, dir(0));
label("$B_1$", B1, NW);
label("$C_1$", C1, N);
label("$X$", X, SE);
label("$Y$", Y, S);
label("$Z$", Z, SW);
label("$T$", Q, NW);
[/asy]

Let the lines through $B$ and $C$ parallel to $AM$ intersect $\Gamma$ again at $B_1$ and $C_1$, respectively. Then, $$\angle BDP=\angle BMA=\angle BCC_1=\angle BDC_1,$$so $DP$ passes through $C_1$. Similarly, $EP$ passes through $B_1$. Now, $$\angle PXC=180^{\circ}-\angle AMC=\angle BMA=\angle BCC_1,$$so the circumcircle of $CXC_1$ is tangent to $BC$ at $C$. Similarly, the circumcircle of $BXB_1$ is tangent to $BC$ at $B$, so $M$ lies on the radical axis of the two circles. Let $B_1C_1$ and $BC$ intersect at $Z$. Then, $BB_1C_1C$ is a cyclic isosceles trapezoid, so $ZC_1=ZC$ implies the circumcircle of $CXC_1$ is also tangent to $ZC_1$. Therefore, the midpoint of $B_1C_1$ also lies on the radical axis, so the radical axis is parallel to $AM$. This implies the radical axis is $AM$, so $P$ lies on the radical axis. Then, we get $PY\cdot PB_1=PX\cdot PC_1$, so $XYB_1C_1$ is cyclic. Now, consider the homothety centered at $Z$ mapping $B$ to $C$. Then, $B_1$ is mapped to $C_1$ and $Y$ is mapped to a point on the circumcircle of $CXC_1$. Therefore, $\angle ZB_1Y=\angle YXC_1$ implies $Y$ is mapped to a point on line $YX$, so $XY$ passes through $Z$. If the circumcircle of $AXY$ intersects $\Gamma$ again at $T$, then since $AXYT$, $XYB_1C_1$, and $ATB_1C_1$ are cyclic, we get $AT$, $B_1C_1$, and $XY$ are concurrent, so $T$ is the intersection of $AZ$ and $\Gamma$. Therefore, the circumcircle of $AXY$ passes through the fixed point $T$.
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IAmTheHazard
5000 posts
#38
Y by
nice problem but wtf were the test makers smoking? also I think it's fairly easy to get that the fixed point lies on $(ABC)$ by sending $P \to \infty$, in which case $X$ and $Y$ lie on $(ABC)$. I got this very quickly and then went nowhere. Guessing the actual point is still very difficult and did not happen (despite ggb) for a long time


Let $K$ and $L$ be the points on $(ABC)$ such that $BK \parallel CL \parallel AM$. Observe that $\measuredangle BDP=\measuredangle BMP=\measuredangle BCL=\measuredangle BDL$, so $D,P,L$ are collinear, and likewise so are $E,P,K$. Note that $\overline{AM}$ also bisects $\overline{KL}$.

We focus on the isosceles trapezoid $BCKL$. Draw circles $\omega_1,\omega_2$ tangent to $\overline{BC}$ and $\overline{KL}$ passing through $B,K$ and $C,L$ respectively, and note that their radical axis is $\overline{AM}$. Since $\measuredangle KYB=\measuredangle PMB=\measuredangle KBM$, $Y \in \omega_1$, and likewise $X \in \omega_2$. By radical center this implies that $KLXY$ is cyclic.

Lemma: Let $ABCD$ be an isosceles trapezoid with $\overline{AD} \parallel \overline{BC}$. with $\overline{AB} \cap \overline{CD}=T$. Let $\omega_1$ and $\omega_2$ be the circles tangent to $\overline{AB}$ and $\overline{CD}$ through $A,D$ and $B,C$ respectively. Let $\ell$ be a line through $T$ that intersects $\omega_1$ at $P_1$ and $P_2$ (with $P_1$ closer) and $\omega_2$ at $Q_1$ and $Q_2$ (with $Q_1$ closer). Then $ABP_2Q_1$ is cyclic (and symmetric results follow)
Proof: Clearly we have $TP_1P_2A \sim TQ_1Q_2B$, so $TP_1\cdot TQ_2=TP_2\cdot TQ_1$. Furthermore this product equals $TA^2\cdot TB^2$, hence $TP_2\cdot TQ_1=TA\cdot TB$ and the result follows. $\blacksquare$

This lemma (with a suitable phantom point argument) thus implies that $\overline{XY},\overline{KL},\overline{BC}$ concur, hence by radical center we have $BCXY$ cyclic, and the radical center of $(KLXY),(BCXY),(ABC)$ is $\overline{KL} \cap \overline{BC}$. Then radical center on $(BCXY),(ABC),(AXY)$ implies that the fixed point $T$ is the second intersection of the line joining $A$ with $\overline{KL} \cap \overline{BC}$, and we're done. $\blacksquare$
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CT17
1481 posts
#39
Y by
Let $DP$ and $EP$ intersect $(BCDE)$ again at $S$ and $T$. As $\measuredangle BDS = \measuredangle BDP = \measuredangle BMP$, $S$ (and similarly $T$) is fixed.

Claim: $BCXY$ and $XYST$ are cyclic.

Proof: Invert at $M$ with arbitrary radius, using $*$ to denote the image of a point. We have $X^* = (P^*D^*M)\cap P^*C^*$, $Y^* = (P^*E^*M)\cap  P^*B^*$, $S =(P^*D^*M)\cap (B^*C^*D^*E^*), T = (P^*E^*M)\cap (B^*C^*D^*E^*)$ . Now invert at $P^*$ swapping $B^*,D^*$ and $C^*,E^*$, using $'$ to denote the image of a point. Then $M'$ still lies on the $P^*-$ median of $P^*B^*C^*$, and $X' = B^*M'\cap P^*C^*$, $Y' = C^*M'\cap P^*B^*$, $S' = B^*M'\cap (B^*C^*D^*E^*)$, $T' = C^*M'\cap (B^*C^*D^*E^*)$. Since $X'Y'\parallel B^*C^*$, $B'C'X'Y'$ and $X'Y'S'T'$ are both cyclic by converse Reim as desired.

Going back to the main problem and the original diagram, $XY\cap ST\cap BC$ is fixed, so $(AXY)\cap (ABC)$ is fixed by radical axis and we're done.
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Pyramix
419 posts
#40
Y by
Let $R,S$ be the points such that $AM\parallel BS\parallel CR$ and $A,T$ be the intersections of $(AXY)$ and $(ABC)$. We show that $T$ is the required fixed point.
Note that $S,R$ are fixed points. Let $R'\in\Gamma$ be the point such that $D,P,R'$ are collinear. We have,
\[\angle RCB=\angle AMB=\angle PMB=\angle PDB=\angle R'DB=\angle R'CB\]which means that $R,R',C$ are collinear, but they also lie on $\Gamma$, so this means $R=R'$. Hence, $R,P,D$ are collinear. Similarly, $S,P,E$ are collinear.
Applying Pascal's Theorem on $(RDBSEC)$, we get $\underbrace{RD\cap SE}_{=P},DB\cap EC,\underbrace{BS\cap CR}_{=\infty_{AM}}$ are collinear. The line $P\infty_{AM}$ is simply line $AM$. So, lines $AM,BD,CE$ concur at a point, say $U$.

Claim 1. $BCXY$ is cyclic.
Proof. [Angle Chasing] Let $U'$ be the reflection of $U$ in $M$. Then, since $M$ is the mid-point of $\overline{BC}$ and $\overline{UU'}$, we have $BUCU'$ is a parallelogram. So, $CU'\parallel BD$ and $BU'\parallel CE$.
Note that
\[\angle BDR=180^\circ-\angle RDB=180^\circ-\angle RCB=\angle CBS=\angle CMA=\angle CMP=\angle CEP=\angle CXP=\angle CXR\]which means $BD\parallel XC$ and hence $C,X,U'$ are collinear. Similarly, $CE\parallel YB$ and $B,Y,U'$ are collinear. Since $BYPM$ is cyclic, we have $U'Y\cdot U'B=U'P\cdot U'M$, while since $CXPM$ is cyclic, $U'P\cdot U'M=U'X\cdot U'C$. Hence, we have $U'Y\cdot U'B=U'X\cdot U'C$, which means that $BCXY$ is cyclic, as claimed. $\blacksquare$

Claim 2. $SRXY$ is cyclic.
Proof. [Length Ratios] Let $D',E'$ be the reflections of $D,E$ in $M$. Then, $D'\in U'C$ and $E'\in U'B$. Note that
\[\frac{U'X}{U'Y}=\frac{U'B}{U'C}=\frac{UC}{UB}=\frac{UD}{UE}=\frac{U'D'}{U'E'}\]which means $XY\parallel D'E'$. However, $DE\parallel D'E'$ which means $XY\parallel DE$. So,
\[\frac{PS}{PR}=\frac{PD}{PE}=\frac{PX}{PY}\Longrightarrow PX\cdot PR=PY\cdot PS,\]which means $SRXY$ is cyclic, as claimed. $\blacksquare$

Claim 3. $AT,SR,XY,BC$ all concur at a point.
Proof. Consider the radical center $Z$ of the circles $(AXY),(BCXY),(ABC)$. The radical axis of $(AXY),(BCXY)$ is line $XY$, the radical axis of $(AXY),(ABC)$ is line $AT$, and the radical axis of $(ABC),(BCXY)$ is line $BC$. So, $Z$ is the intersection of lines $AT,XY,BC$. Let the radical center of circles $(AXY),(SRXY),(ABC)$ be $Z'$. The radical axis of $(AXY),(SRXY)$ is line $XY$, the radical center of $(AXY),(ABC)$ is line $AT$. So, $Z'=AT\cap XY$, which means $Z'=Z$. So, $Z$ also lies on the radical axis of $(SRXY),(ABC)$ which is line $SR$. Hence, $Z$ is the intersection of lines $AT,SR,XY,BC$, as desired. $\blacksquare$

Since $Z=SR\cap BC$ and $S,R,B,C$ are fixed points, it follows that $Z$ is a fixed point. Hence, $T=AZ\cap(ABC)$ is a fixed point as well. $\blacksquare$
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Ywgh1
136 posts
#42
Y by
APMO 2019 p3
HAY trio for the win.
Solved with SuperHmm7 and Ammh4 ;)
[asy]

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[/asy]

Great problem!

Here is a Sketch:

Step 1: Show that $BCXY$ cyclic.

Step 2: Show that $XY$ is parallel to $DE$.

Step 3: Construct $K$ and $L$

Step 4: Prove that $PM$, $BL$ and $CK$ are parallel.

The problem follows as this means that $K$ and $L$ are fixed which mean also implies that $Z$ is fixed by radicle axis. :)
This post has been edited 10 times. Last edited by Ywgh1, Aug 15, 2024, 7:34 PM
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GrantStar
812 posts
#44 • 1 Y
Y by dolphinday
Cute problem! We first prove a claim which reduces the problem to showing $XY\cap BC$ is fixed.

Claim: $BXCY$ is cyclic
Proof. Let $Q=BD\cap CE \cap AM$ and $Q'=2M-Q$. Then, $Q'$ is on $AM$ and $Q'B \parallel CE$. But by Reim, $CE \parallel BX$ so $Q'$ lies on $BX$. Similarly, $Q'$ lies on $CY$ so $BX, CY, AM$ concur and the claim follows by radical axis. $\blacksquare$

Now, Pascal converse gives that $DYCEXB$ are coconic. Then, Pascal on $EDBXYC$ gives $DE \cap XY$ is at infinity, so $DE\parallel XY$. Thus by Reim $XKLY$ is cyclic where $K,L$ are the second intersections of $EP, DP$ with $(ABC)$, and so by radical axis again it suffices to show that $KL\cap BC$ is fixed. But by Reim, $BK \parallel AM \parallel CL$ so they are fixed, done.

Remark: The hardest part about this is convincing yourself the fixed point lis on $(ABC)$. After that it becomes a standard ~20 mohs problem.
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L13832
250 posts
#45
Y by
Figure

MAIN CLAIM: $\odot(ABC)\cap (AXY)$ is fixed. (Kudos to the people who were able to figure this out during the exam or without geogebra.)

Let $(AXY)\cap (ABC)=F$, $BD, AM, CE$ intersect at $R$(radical-center).
Claim I: $BY \parallel CE$ and $CX \parallel BD$
Proof: Reim's Theorem

Let $BY, AM, CX$ intersect at $S$
Claim II: $BD \parallel CX$
Proof: $\measuredangle BDX=\measuredangle BDP=\measuredangle CMP=\measuredangle CXD$

Let $DX \cap (ABC)=L$ and $EY \cap (ABC)=K$. (This is the trickiest part of the solution, rest of the part is just trivial, this it is motivated because later we prove concurrence of lines using radical-axis which requires concylicity of some quadrilaterals.)We have the following claim:
Claim III: $\odot (BCXY)$
Proof: $KX\cdot KB=KP\cdot KM=KY\cdot KC$

Claim IV: $RCBS$ is a parallelogram
Proof: Direct consequence of Claim I and Claim II.

Claim V: $ XY \parallel DE$
Proof: $XY$ and $DE$ are anti-parallel to $BC$

Claim VI: $\odot (KLXY)$
Proof: This equivalent to showing $XY \parallel DE$.
\begin{align*}&\angle PDB=\angle PMB=\angle PEC\implies \odot(PDLE)\\ \implies &\angle PED=\angle PRD=\angle PSC=\angle PYX\end{align*}and we are done.

Now by applying Radical-axis on $\odot(ABC), \odot (BCXY)$and $\odot(AXY)$, we get that $AF, BC, XY$ concur.
Again applying Radical-axis on $\odot(KLXY), \odot(ABC), \odot (BCXY)$ we get that $BC, XY, KL$ concur.
Since $KL, BC$ are fixed $\implies T$ is fixed.
If $AT \cap (ABC)=F$, we have $AT \cdot TF=BT \cdot TC = XF \cdot FY\implies F$ is fixed.

Remark

p.s.
This post has been edited 7 times. Last edited by L13832, Sep 9, 2024, 4:16 PM
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Eka01
204 posts
#46
Y by
@above no i dont, will try the problem and post my soln later(If i get one).
This post has been edited 1 time. Last edited by Eka01, Aug 30, 2024, 2:18 PM
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InterLoop
247 posts
#47
Y by
yes good yes
solution

remarks
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john0512
4170 posts
#48
Y by
this problem is very :blobheart:
The main claim is the following.

Claim: $BCXY$ is cyclic, and moreover, $XY\parallel DE$.

Invert at $P$. Our claim now becomes:
Quote:
In triangle $\triangle PBC$, $M$ lies on $(PBC)$ such that $PM$ is a symmedian. Let $D$ be on $BM$ and $E$ on $CM$ such that $BCDE$ is cyclic. Let $PD$ meet $CM$ at $X$ and $PE$ meet $BM$ at $Y$. Show that $YX\parallel DE$.

Let $PM$ meet $ED$ at $S$. By Ceva, it suffices to show that $S$ is the midpoint of $ED$. However, because $MP$ is a symmedian in $\triangle MBC$, and $BCDE$ is cyclic, it flips over to a median in $\triangle EMD$. Thus, $XY\parallel ED$, so $XY\parallel ED$ in the uninverted diagram as well as $PYE$ and $PXD$ are collinear. Furthermore, in the inverted diagram, $BYXC$ is cyclic by Reim as $XY\parallel ED$, so this is true in uninverted diagram as well.

Now, let $XY\cap BC=K$. If $(AXY)\cap(ABC)=F$, then by radical axis theorem, $AFK$ are collinear.

Thus, the rest of this solution will be showing that $K$ does not depend on the choice of $P$. This will finish because then $AK\cap (ABC)$ will always lie on $(AXY)$.

Let $DP$ and $EP$ meet $(ABC)$ at $D'$ and $E'$ respectively.


Claim: $D'$ and $E'$ do not depend on the choice of $P$.


Since $PECM$ is cyclic, We have
$$\angle E'EC=\angle PMB=\angle AMB$$which does not depend on $P$. Hence, the arc measure of $E'C$ is constant, so $E'$ is also constant.

Finally, $EE'DD'$ is cyclic and $YX\parallel DE$, so by Reim, $D'E'XY$ is cyclic. By Radical Axis on $(ABC)$, $(BCXY)$, and $(D'E'XY)$, we have that $BC,XY$, and $D'E'$ concur. Thus, $K=BC\cap D'E'$, which does not depend on $P$ as $D'$ and $E'$ does not depend on $P$. We are done.
This post has been edited 2 times. Last edited by john0512, Nov 20, 2024, 4:33 AM
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bjump
962 posts
#49
Y by
Let $F= (AXY) \cap \Gamma$. Observe
$$\measuredangle BDX = \measuredangle BMP = \measuredangle CMP = \measuredangle CXP = \measuredangle CXD$$Which implies $CX \parallel BD$, by symmetry $BY \parallel EC$.

Let $BD \cap CE = G$ which lies on $AM$ by a radical axis argument. Let $BY \cap CX = H$. Since $M$ is the midpoint of $BC$ and $GBHC$ is a parallelogram we have $H \in MG = AM$. Therefore by PoP
$$HY \cdot HB = HP \cdot HM = HX \cdot HC$$Which means $BYXC$ is cyclic.

Now define $J= EP \cap \Gamma \neq E$, $K = DP \cap \Gamma \neq D$. We have
$$\measuredangle CKP = \measuredangle CKD = \measuredangle CBD = \measuredangle CBD = \measuredangle DPM = \measuredangle KPA.$$Which implies $CK \parallel AM$, by symmetry $BJ \parallel CK \parallel AM$.

Let $KC \cap EJ = L$. Now observe
$$\measuredangle LED  = \measuredangle JED = \measuredangle JBD = \measuredangle(JB, BD) = \measuredangle(LC, XC) = \measuredangle LCX = \measuredangle LDX = \measuredangle LDP.$$Which means $PD$ is tangent to $(LED)$. So $PD^2=PL \cdot PE$.

Now by PoP
$$PX \cdot PD = PY \cdot PL , PK \cdot PD = PJ \cdot PE \implies PY \cdot PJ \cdot PE \cdot PL = PD^2 PK \cdot PX \implies PY \cdot PJ = PX \cdot PK.$$Therefore $(JYXK)$ is cyclic.

To fnish note that by radical axis $AF$, $XY$, $AB$, $JK$ Concur. However $AB \cap JK$ is fixed depending on $P$. Therefore $F$ is fixed and we are finished.
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This post has been edited 1 time. Last edited by bjump, Dec 22, 2024, 3:59 PM
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awesomeming327.
1664 posts
#50
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Let $DP$ and $EP$ intersect $(ABC)$ again $C'$ and $B'$, respectively. Applying Reim's theorem, we have $BD\parallel CX$, $BY\parallel CE$, and $BB'\parallel AM\parallel CC'$. In particular, for the first two, the lines are reflections across $M$ because $M$ is the midpoint of $BC$.

Claim 1: $BYXC$ is cyclic.
By Radical Center, $BD$, $CE$, and $AM$ concur. Reflecting across $M$, we get that $CX$, $BY$, and $AM$ concur. Therefore, we are done by the converse of Radical Center.
Claim 2: $B'C'$, $XY$, and $BC$ concur.
Let $BY$, $CX$, and $AM$ intersect at $J$. Let $BD$, $CE$, and $AM$ intersect at $K$. Note that $J$ and $K$ are reflections across $M$, so $JBKC$ is a parallelogram. By Miquel Point on a triangle, $YPXJ$ and $DPEK$ are cyclic. Thus,
\[\measuredangle PYX=\measuredangle PJX=\measuredangle PJC=\measuredangle PKB=\measuredangle PKD=\measuredangle PED\]so $DE\parallel XY$. By Converse of Reim's, $B'C'XY$ is cyclic, so our claim is true by Radical Center.
In particular, $B'C'$ intersect $BC$ is independent of $P$, so $XY$ passes through a fixed point $T'$. Let $AT'$ intersect $(ABC)$ at $T$ then we have by converse of Radical Center that $ATXY$ is cyclic. We are done.
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HamstPan38825
8847 posts
#51 • 1 Y
Y by MS_asdfgzxcvb
Let $R = \overline{BD} \cap \overline{CE} \cap \overline{PM}$ be the radical center of the three drawn circles. As $\measuredangle DPM = \measuredangle DBC = \measuredangle DEC$, $DPER$ is cyclic, so by Reim's theorem, $\overline{CR} \parallel \overline{BY}$ and $\overline{BR}\parallel \overline{CY}$.

Hence we may mark a point $S$ on $\overline{AR}$ such that $SBRC$ is a parallelogram. Now let $K = \overline{EP} \cap \Gamma$ and $L = \overline{DP} \cap \Gamma$. Then $\overline{BK}$ and $\overline{CE}$ are antiparallel, so $\overline{BK} \parallel \overline{PM} \parallel \overline{CL}$ by Reim's theorem. This implies that $K$ and $L$ are fixed points.

Finally, power of a point at $S$ yields that $BCXY$ is cyclic, and $\measuredangle PDE = \measuredangle SRE = \measuredangle YSP = \measuredangle XYP$ implies $\overline{DE} \parallel \overline{YX}$. Then $\overline{XY}$ is antiparallel to $\overline{KL}$, hence by radical axis on $(KYXL)$, $(BCXY)$, and $(ABC)$, the lines $\overline{KL}$, $\overline{XY}$, $\overline{BC}$ meet at a point $T'$. Since $\overline{KL}$ and $\overline{BC}$ are fixed, $T'$ is fixed too.

Finally, taking $T = \overline{AT'} \cap (ABC)$ yields a fixed point $T$ that always lies on $(AXY)$ by power of a point. So $T$ is our desired fixed point.

Remark: Every part of this solution is not difficult except for the construction of $K$ and $L$; it's not immediately obvious that they are necessary (indeed, the problem seems to read just fine saying ``$\overline{XY}$ passes through a fixed point"), but I spent almost two hours trying to prove this claim directly to no avail. I think this problem definitely falls on the harder IMO 2/5 side or easier 3/6 side, making it much too difficult for APMO3.
This post has been edited 1 time. Last edited by HamstPan38825, Today at 2:37 AM
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