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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.
[*]March 6th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar on Math Competitions from elementary through high school. Join us for an enlightening session that demystifies the world of math competitions and helps you make informed decisions about your contest journey.
[*]March 11th (Tuesday), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS Chapter Discussion MATH JAM. AoPS instructors will discuss some of their favorite problems from the MATHCOUNTS Chapter Competition. All are welcome!
[*]March 13th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar about Summer Camps at the Virtual Campus. Transform your summer into an unforgettable learning adventure! From elementary through high school, we offer dynamic summer camps featuring topics in mathematics, language arts, and competition preparation - all designed to fit your schedule and ignite your passion for learning.[/list]
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0 replies
1 viewing
jlacosta
Mar 2, 2025
0 replies
k i Suggestion Form
jwelsh   0
May 6, 2021
Hello!

Given the number of suggestions we’ve been receiving, we’re transitioning to a suggestion form. If you have a suggestion for the AoPS website, please submit the Google Form:
Suggestion Form

To keep all new suggestions together, any new suggestion threads posted will be deleted.

Please remember that if you find a bug outside of FTW! (after refreshing to make sure it’s not a glitch), make sure you’re following the How to write a bug report instructions and using the proper format to report the bug.

Please check the FTW! thread for bugs and post any new ones in the For the Win! and Other Games Support Forum.
0 replies
jwelsh
May 6, 2021
0 replies
k i Read me first / How to write a bug report
slester   3
N May 4, 2019 by LauraZed
Greetings, AoPS users!

If you're reading this post, that means you've come across some kind of bug, error, or misbehavior, which nobody likes! To help us developers solve the problem as quickly as possible, we need enough information to understand what happened. Following these guidelines will help us squash those bugs more effectively.

Before submitting a bug report, please confirm the issue exists in other browsers or other computers if you have access to them.

For a list of many common questions and issues, please see our user created FAQ, Community FAQ, or For the Win! FAQ.

What is a bug?
A bug is a misbehavior that is reproducible. If a refresh makes it go away 100% of the time, then it isn't a bug, but rather a glitch. That's when your browser has some strange file cached, or for some reason doesn't render the page like it should. Please don't report glitches, since we generally cannot fix them. A glitch that happens more than a few times, though, could be an intermittent bug.

If something is wrong in the wiki, you can change it! The AoPS Wiki is user-editable, and it may be defaced from time to time. You can revert these changes yourself, but if you notice a particular user defacing the wiki, please let an admin know.

The subject
The subject line should explain as clearly as possible what went wrong.

Bad: Forum doesn't work
Good: Switching between threads quickly shows blank page.

The report
Use this format to report bugs. Be as specific as possible. If you don't know the answer exactly, give us as much information as you know. Attaching a screenshot is helpful if you can take one.

Summary of the problem:
Page URL:
Steps to reproduce:
1.
2.
3.
...
Expected behavior:
Frequency:
Operating system(s):
Browser(s), including version:
Additional information:


If your computer or tablet is school issued, please indicate this under Additional information.

Example
Summary of the problem: When I click back and forth between two threads in the site support section, the content of the threads no longer show up. (See attached screenshot.)
Page URL: http://artofproblemsolving.com/community/c10_site_support
Steps to reproduce:
1. Go to the Site Support forum.
2. Click on any thread.
3. Click quickly on a different thread.
Expected behavior: To see the second thread.
Frequency: Every time
Operating system: Mac OS X
Browser: Chrome and Firefox
Additional information: Only happens in the Site Support forum. My tablet is school issued, but I have the problem at both school and home.

How to take a screenshot
Mac OS X: If you type ⌘+Shift+4, you'll get a "crosshairs" that lets you take a custom screenshot size. Just click and drag to select the area you want to take a picture of. If you type ⌘+Shift+4+space, you can take a screenshot of a specific window. All screenshots will show up on your desktop.

Windows: Hit the Windows logo key+PrtScn, and a screenshot of your entire screen. Alternatively, you can hit Alt+PrtScn to take a screenshot of the currently selected window. All screenshots are saved to the Pictures → Screenshots folder.

Advanced
If you're a bit more comfortable with how browsers work, you can also show us what happens in the JavaScript console.

In Chrome, type CTRL+Shift+J (Windows, Linux) or ⌘+Option+J (Mac).
In Firefox, type CTRL+Shift+K (Windows, Linux) or ⌘+Option+K (Mac).
In Internet Explorer, it's the F12 key.
In Safari, first enable the Develop menu: Preferences → Advanced, click "Show Develop menu in menu bar." Then either go to Develop → Show Error console or type Option+⌘+C.

It'll look something like this:
IMAGE
3 replies
slester
Apr 9, 2015
LauraZed
May 4, 2019
k i Community Safety
dcouchman   0
Jan 18, 2018
If you find content on the AoPS Community that makes you concerned for a user's health or safety, please alert AoPS Administrators using the report button (Z) or by emailing sheriff@aops.com . You should provide a description of the content and a link in your message. If it's an emergency, call 911 or whatever the local emergency services are in your country.

Please also use those steps to alert us if bullying behavior is being directed at you or another user. Content that is "unlawful, harmful, threatening, abusive, harassing, tortuous, defamatory, vulgar, obscene, libelous, invasive of another's privacy, hateful, or racially, ethnically or otherwise objectionable" (AoPS Terms of Service 5.d) or that otherwise bullies people is not tolerated on AoPS, and accounts that post such content may be terminated or suspended.
0 replies
dcouchman
Jan 18, 2018
0 replies
Right angles on incircle
DynamoBlaze   37
N 7 minutes ago by sangsidhya
Source: RMO 2018 P6
Let $ABC$ be an acute-angled triangle with $AB<AC$. Let $I$ be the incentre of triangle $ABC$, and let $D,E,F$ be the points where the incircle touches the sides $BC,CA,AB,$ respectively. Let $BI,CI$ meet the line $EF$ at $Y,X$ respectively. Further assume that both $X$ and $Y$ are outside the triangle $ABC$. Prove that
$\text{(i)}$ $B,C,Y,X$ are concyclic.
$\text{(ii)}$ $I$ is also the incentre of triangle $DYX$.
37 replies
1 viewing
DynamoBlaze
Oct 7, 2018
sangsidhya
7 minutes ago
Where is the equality?
AndreiVila   2
N 44 minutes 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
44 minutes ago
Truth or lie at a table
SinaQane   7
N an hour ago by Oksutok
Source: 239 2019 S4
There are $n>1000$ people at a round table. Some of them are knights who always tell the truth, and the rest are liars who always tell lies. Each of those sitting said the phrase: “among the $20$ people sitting clockwise from where I sit there are as many knights as among the $20$ people seated counterclockwise from where I sit”. For what $n$ could this happen?
7 replies
SinaQane
Jul 31, 2020
Oksutok
an hour ago
2 var inquality
sqing   4
N an hour ago by sqing
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$$ a+b+ab\geq2$$
4 replies
sqing
Today at 4:06 AM
sqing
an hour ago
k 2023 IMO
ostriches88   4
N Mar 14, 2025 by jlacosta
As outlined in this (locked) post, the forum/blog creation message is out of date. It has been over a year since that post, and it still has not changed. I don't know if this got lost somewhere in the "passing this along" chain or if it was determined to be irrelevant, but it seems like a relatively simple fix ;)
4 replies
ostriches88
Mar 10, 2025
jlacosta
Mar 14, 2025
Pressing &#039;go down button&#039; always creates a gray box on the last post
Craftybutterfly   13
N Mar 14, 2025 by Craftybutterfly
Summary of the problem: Pressing go down to last post button always creates a gray box overlapping last post
Page URL: any forum
Steps to reproduce:
1. Go to any topic in a forum
2. The gray box at the bottom overlaps part of the first post
Expected behavior: Should not show a gray box
Frequency: 100% of the time
Operating system(s): Linux HP EliteBook 835 G8 Notebook PC
Browser(s), including version: Chrome 133.0.6943.142 (Official Build) (64-bit) (cohort: Stable)
Additional information: It works on any other device, on my iPhone XR, a MacOS, and my iPad. Took the screenshot a month ago. The gray box still appears
13 replies
Craftybutterfly
Mar 12, 2025
Craftybutterfly
Mar 14, 2025
k Somehow I broke the unique contest collection naming...
Equinox8   1
N Mar 14, 2025 by jlacosta
Somehow I broke the unique contest collection naming scheme; for what it's worth, I also ran into an AJAX timeout error while trying to make this particular post collection for the first time, and pressed "create" twice.

Not sure if there's anything to be done here, but are there potentially unforeseen consequences here?

https://artofproblemsolving.com/community/c4258765_2024_puerto_rico_team_selection_test (this is the one I'm working on; it's also not finished yet)
https://artofproblemsolving.com/community/c4258764_2024_puerto_rico_team_selection_test
1 reply
Equinox8
Mar 12, 2025
jlacosta
Mar 14, 2025
k New Forums Duplicates
k1glaucus   3
N Mar 14, 2025 by jlacosta
Summary of the problem: In the New Forums collection, many of the forums are duplicated, although some have slightly different information (likely due to the forum being edited by its admin(s)). Typically only one of the two has threads, and I believe this is the only way to access
Page URL: https://artofproblemsolving.com/community/c74_new_forums
Steps to reproduce:
1. Go to the link
2. You should see duplicates of some forums
Expected behavior: Each forum appears once
Frequency: Every time
Additional information: Typically only one of the two has threads, and from creating forums in the past I believe this is the only way to access the duplicate. Also not all of the forums are duplicated. Another reason to suspect these are duplicates are the forums with similar names have the same admin.
3 replies
k1glaucus
Mar 12, 2025
jlacosta
Mar 14, 2025
k Pi Day!!
SomeonecoolLovesMaths   66
N Mar 14, 2025 by LawofCosine
Happy $\pi$ day everyone!
66 replies
SomeonecoolLovesMaths
Mar 14, 2025
LawofCosine
Mar 14, 2025
k Help with beast academy
tyrantfire4   1
N Mar 14, 2025 by tyrantfire4
So my siblings do BA and it won't work if anyone can fix it that would be great!
All browers won't work
Apple and fire
Open BA account
1 reply
tyrantfire4
Mar 14, 2025
tyrantfire4
Mar 14, 2025
Search function in private messages not working
WisteriaV   2
N Mar 12, 2025 by mathlearner2357
For the past while, the search function in private messages hasn’t been working. Whenever I search for anything, it says, “No topics here!” after trying to load for a while. I’ve tried different devices (laptop and ipad) and browsers (chrome on both devices, safari on ipad, and microsoft edge on laptop), and the results are the same. I’ve also had friends say the same happens for them.
2 replies
WisteriaV
Mar 12, 2025
mathlearner2357
Mar 12, 2025
Viewing next classes
lilorocks11   2
N Mar 11, 2025 by jlacosta
Hello,

I'm wondering how to look at my next classes. I'm not entirely sure if I'm registered for a class. I tried looking around on the My AoPS page, but didn't find anything that had to do with the course I wanted.
2 replies
lilorocks11
Mar 8, 2025
jlacosta
Mar 11, 2025
Error when abandoning quest
dragonborn56   3
N Mar 11, 2025 by pb0975
I attempted to abandon the Alcumus quest Hit The Gym: Accuracy. I clicked on it and selected abandon, but the popup disappeared and the quest was still there. There were no changes to the log, and my daily abandon was used, as when I clicked the quest again it said that all of my abandons were used.
3 replies
dragonborn56
Mar 10, 2025
pb0975
Mar 11, 2025
What????
LawofCosine   21
N Mar 10, 2025 by LawofCosine
Umm....what is happening? The $-\infty$ seems to be squashed but the $\infty$ is normal. This is from the AoPS Calculus ebook.
21 replies
LawofCosine
Mar 7, 2025
LawofCosine
Mar 10, 2025
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
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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
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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.
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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
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#51 • 1 Y
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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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