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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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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
Integer equation in 3 variables
Kimchiks926   2
N 3 minutes ago by MuradSafarli
Source: Latvian TST for Baltic Way 2019 Problem 15
Determine all tuples of integers $(a,b,c)$ such that:
$$(a-b)^3(a+b)^2 = c^2 + 2(a-b) + 1$$
2 replies
Kimchiks926
May 29, 2020
MuradSafarli
3 minutes ago
Interesting inequality
sqing   2
N 6 minutes ago by SunnyEvan
Source: Own
Let $ a,b,c\geq 2  . $ Prove that
$$(a^2-1)(b-1)(c^2-1) -\frac{9}{4}abc\geq -9$$$$(a^2-1)(b-1)(c^2-1) -\frac{11}{5}abc\geq -\frac{43}{5}$$$$(a^2-1)(b-1)(c^2-1) -2abc\geq -7$$$$(a-1)(b^2-1)(c-1) -\frac{3}{4}abc\geq -3$$$$(a-1)(b^2-1)(c-1) -\frac{3}{5}abc\geq -\frac{9}{5}$$$$(a-1)(b^2-1)(c-1) -\frac{1}{2}abc\geq -1$$
2 replies
+1 w
sqing
43 minutes ago
SunnyEvan
6 minutes ago
Interesting inequality
sqing   1
N 10 minutes ago by sqing
Source: Own
Let $ a,b,c\geq 2  . $ Prove that
$$(a^2-1)(b-\frac{3}{2})(c^2-1) - \frac{9}{4}abc\geq -15$$$$(a^2-1)(b-\frac{3}{2})(c^2-1) - 2abc\geq -\frac{73}{6}$$$$(a^2-1)(b-\frac{3}{2})(c^2-1) - abc\geq -\frac{7}{2}$$$$(a^2-2)(b-\frac{3}{2})(c^2-2) - abc\geq -6$$
1 reply
1 viewing
sqing
19 minutes ago
sqing
10 minutes ago
Hard problem involving circumcenter and concurrent lines
GeoMetrix   6
N 10 minutes ago by bin_sherlo
Source: AQGO 2020 Problem 3
Let $\triangle{ABC}$ be a triangle with circumcenter $O$. Let $M,N$ be the midpoints of $\overline{AB}$ and $\overline{AC}$ respectively and let $T$ be the projection of $O$ on $\overline{MN}$. Let $D$ be the projection of $A$ on $\overline{BC}$. Let $\overline{TD}$ intersect $\odot(BOC)$ at points $U$ and $V$. Let $\odot(AUV)$ intersct $\overline{MN}$ at points $X,Y$. Let $\overline{AY}$ intersect $\odot(AMN)$ at $R$ and $\overline{AX}$ intersect $\odot(AMN)$ at $S$. Then show that $\overline{AO},\overline{RS},\overline{MN}$ are concurrent.

Proposed by GeoMetrix
6 replies
GeoMetrix
Jun 20, 2020
bin_sherlo
10 minutes ago
Oi! These lines concur
Rg230403   17
N 18 minutes ago by L13832
Source: LMAO 2021 P5, LMAOSL G3(simplified)
Let $I, O$ and $\Gamma$ respectively be the incentre, circumcentre and circumcircle of triangle $ABC$. Points $A_1, A_2$ are chosen on $\Gamma$, such that $AA_1 = AI = AA_2$, and point $A'$ is the foot of the altitude from $I$ to $A_1A_2$. If $B', C'$ are similarly defined, prove that lines $AA', BB'$ and $CC'$ concurr on $OI$.
Original Version from SL
Proposed by Mahavir Gandhi
17 replies
Rg230403
May 10, 2021
L13832
18 minutes ago
Find the period
Anto0110   1
N 22 minutes ago by Anto0110
Let $a_1, a_2, ..., a_k, ...$ be a sequence that consists of an initial block of $p$ positive distinct integers that then repeat periodically. This means that $\{a_1, a_2, \dots, a_p\}$ are $p$ distinct positive integers and $a_{n+p}=a_n$ for every positive integer $n$. The terms of the sequence are not known and the goal is to find the period $p$. To do this, at each move it possible to reveal the value of a term of the sequence at your choice.
If $p$ is one of the first $k$ prime numbers, find for which values of $k$ there exist a strategy that allows to find $p$ revealing at most $8$ terms of the sequence.
1 reply
Anto0110
Yesterday at 7:37 PM
Anto0110
22 minutes ago
inequality
senku23   2
N an hour ago by User21837561
Let x,y,z in R+ prove that 8(x^3+y^3+z^3)2≥9(x^2+yz)(y^2+zx)(z^2+xy).
2 replies
senku23
2 hours ago
User21837561
an hour ago
Differentiable functional
bakerbakura   2
N an hour ago by Gryphos
Find all differentiable functions $ f;\mathbb{R}\to\mathbb{R}$ such that, for all real numbers $ a,b,t$ with $ 0<t<1$, $ t^2f(a)+(1-t^2)f(b)\geq f(ta+(1-t)b)$
2 replies
bakerbakura
Jan 11, 2010
Gryphos
an hour ago
function equation
cipher703516247   2
N an hour ago by luutrongphuc

Find all the functions $f: \mathbb R^{+} \to \mathbb R^{+}$such that:
$$f(xy^n +f(y)^{2n}) )=f(x)f(y)^n +(yf(y))^n $$
2 replies
cipher703516247
Feb 14, 2025
luutrongphuc
an hour ago
gcd (a^n+b,b^n+a) is constant
EthanWYX2009   78
N an hour ago by SimplisticFormulas
Source: 2024 IMO P2
Determine all pairs $(a,b)$ of positive integers for which there exist positive integers $g$ and $N$ such that
$$\gcd (a^n+b,b^n+a)=g$$holds for all integers $n\geqslant N.$ (Note that $\gcd(x, y)$ denotes the greatest common divisor of integers $x$ and $y.$)

Proposed by Valentio Iverson, Indonesia
78 replies
EthanWYX2009
Jul 16, 2024
SimplisticFormulas
an hour ago
Maximum Cost of Identifying Two Digits in a Three-Digit Number
Borisaurus   0
an hour ago
Source: 2025 MBL QQ P7 Variant
You find yourself on a flight to Bosnia and Herzegovina. You are seated next to a strange man wearing a hat—similar to the one worn by Indiana Jones. The man spends most of the flight rambling about some pyramids.

However, as you approach your destination, he tells you the following:

"I have thought of a $three-digit$ number. I will allow you to ask me questions. A question consists of you saying a three-digit number, to which I respond 'yes' if at least two of its digits are correct, or 'no' otherwise. (A digit is correct if it matches the corresponding digit in my number.)

You can ask as many questions as you like, but they come at a price. The game ends when you have correctly identified two digits of my number. If
you guess correctly, I will sell you my precious pyramid. You will have to pay a number of Bosnian marks equal to the number of questions you asked.

What is the maximum number of Bosnian marks you will have to pay using the best strategy?"
0 replies
Borisaurus
an hour ago
0 replies
BMN is equilateral iff rectangle ABCD is square
parmenides51   3
N 2 hours ago by Tsikaloudakis
Source: 2004 Romania NMO SL - Shortlist VII-VIII p8 https://artofproblemsolving.com/community/c3950157_
Consider a point $M$ on the diagonal $BD$ of a given rectangle $ABCD$, such that $\angle AMC = \angle  CMD$. The point $N$ is the intersection point between $AM$ and the parallel line to $CM$ that contains $B$. Prove that the triangle $BMN$ is equilateral if and only if $ABCD$ is a square.

Valentin Vornicu
3 replies
parmenides51
Sep 16, 2024
Tsikaloudakis
2 hours ago
D1015 : A strange EF for polynomials
Dattier   1
N 2 hours ago by Dattier
Source: les dattes à Dattier
Find all $P \in \mathbb R[x,y]$ with $P \not\in \mathbb R[x] \cup \mathbb R[y]$ and $\forall g,f$ homeomorphismes of $\mathbb R$, $P(f,g)$ is an homoemorphisme too.
1 reply
Dattier
Mar 16, 2025
Dattier
2 hours ago
Cutting a big square into smaller squares
nAalniaOMliO   4
N 2 hours ago by anudeep
Source: Belarusian National Olympiad 2020
A $20 \times 20$ checkered board is cut into several squares with integer side length. The size of a square is it's side length.
What is the maximum amount of different sizes this squares can have?
4 replies
nAalniaOMliO
Jan 29, 2025
anudeep
2 hours ago
Variable point on the median
MarkBcc168   47
N Mar 17, 2025 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
Mar 17, 2025
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
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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 $P$.
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DottedCaculator
7304 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
251 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
965 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, Mar 17, 2025, 2:37 AM
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