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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.
[*]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!
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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
Yet another circle!
Rushil   7
N 12 minutes ago by FireMaths
Source: INMO 1999 Problem 4
Let $\Gamma$ and $\Gamma'$ be two concentric circles. Let $ABC$ and $A'B'C'$ be any two equilateral triangles inscribed in $\Gamma$ and $\Gamma'$ respectively. If $P$ and $P'$ are any two points on $\Gamma$ and $\Gamma'$ respectively, show that \[ P'A^2 + P'B^2 + P'C^2 = A'P^2 + B'P^2 + C'P^2. \]
7 replies
Rushil
Oct 7, 2005
FireMaths
12 minutes ago
Binary Operator from AMC 10
pinetree1   36
N 24 minutes ago by Ilikeminecraft
Source: USA TSTST 2019 Problem 1
Find all binary operations $\diamondsuit: \mathbb R_{>0}\times \mathbb R_{>0}\to \mathbb R_{>0}$ (meaning $\diamondsuit$ takes pairs of positive real numbers to positive real numbers) such that for any real numbers $a, b, c > 0$,
[list]
[*] the equation $a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c$ holds; and
[*] if $a\ge 1$ then $a\,\diamondsuit\, a\ge 1$.
[/list]
Evan Chen
36 replies
+1 w
pinetree1
Jun 25, 2019
Ilikeminecraft
24 minutes ago
Interesting inequality
sqing   4
N 36 minutes ago by lbh_qys
Source: Own
Let $ a,b\geq 2  . $ Prove that
$$(a^2-1)(b^2-1) -6ab\geq-15$$$$(a^2-1)(b^2-1)  -7ab\geq  -\frac{58}{3}$$$$(a^3-1)(b^3-1)  -\frac{21}{4}a^2b^2\geq -35$$$$(a^3-1)(b^3-1)  -6a^2b^2\geq-\frac{2391}{49}$$
4 replies
sqing
an hour ago
lbh_qys
36 minutes ago
Incredible vanilla geometry again
anantmudgal09   47
N 41 minutes ago by kes0716
Source: INMO 2021 Problem 5
In a convex quadrilateral $ABCD$, $\angle ABD=30^\circ$, $\angle BCA=75^\circ$, $\angle ACD=25^\circ$ and $CD=CB$. Extend $CB$ to meet the circumcircle of triangle $DAC$ at $E$. Prove that $CE=BD$.

Proposed by BJ Venkatachala
47 replies
anantmudgal09
Mar 7, 2021
kes0716
41 minutes ago
USAMO 1985 #4
Mrdavid445   5
N an hour ago by RedFireTruck
There are $n$ people at a party. Prove that there are two people such that, of the remaining $n-2$ people, there are at least $\left\lfloor\frac{n}{2}\right\rfloor-1$ of them, each of whom either knows both or else knows neither of the two. Assume that knowing is a symmetric relation, and that $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$.
5 replies
Mrdavid445
Jul 26, 2011
RedFireTruck
an hour ago
surprisingly not trivial for a P2
iStud   1
N an hour ago by MathLuis
Source: Monthly Contest KTOM March 2025 P2 Essay
Find all natural numbers $(m,n)$ such that
\[2^{n!}+1\mid 2^{m!}+19\]
Hint
1 reply
iStud
2 hours ago
MathLuis
an hour ago
Eventually constant sequence with condition
PerfectPlayer   0
an hour ago
Source: Turkey TST 2025 Day 3 P8
A positive real number sequence $a_1, a_2, a_3,\dots $ and a positive integer \(s\) is given.
Let $f_n(0) = \frac{a_n+\dots+a_1}{n}$ and for each $0<k<n$
\[f_n(k)=\frac{a_n+\dots+a_{k+1}}{n-k}-\frac{a_k+\dots+a_1}{k}\]Then for every integer $n\geq s,$ the condition
\[a_{n+1}=\max_{0\leq k<n}(f_n(k))\]is satisfied. Prove that this sequence must be eventually constant.
0 replies
PerfectPlayer
an hour ago
0 replies
Number of modular sequences with different residues
PerfectPlayer   0
an hour ago
Source: Turkey TST 2025 Day 3 P9
Let \(n\) be a positive integer. For every positive integer $1 \leq k \leq n$ the sequence ${\displaystyle {\{ a_{i}+ki\}}_{i=1}^{n }}$ is defined, where $a_1,a_2, \dots ,a_n$ are integers. Among these \(n\) sequences, for at most how many of them does all the elements of the sequence give different remainders when divided by \(n\)?
0 replies
PerfectPlayer
an hour ago
0 replies
Find max no. of gangsters
sk2005   4
N an hour ago by flower417477
In Chicago, there are 36 criminal gangs, some of which are at war with
each other. Each gangster belongs to several gangs and every pair of gangsters
belongs to a different set of gangs. It is known that no gangster is a member of
two gangs which are at war with each other. Furthermore, each gang that some
gangster does not belong to is at war with some gang he does belong to. What is
the largest possible number of gangsters in Chicago?
4 replies
sk2005
Sep 13, 2021
flower417477
an hour ago
super duper ez radax problem
iStud   1
N an hour ago by MathLuis
Source: Monthly Contest KTOM March 2025 P1 Essay
Given an acute triangle $ABC$ with $BC<AB<AC$. Points $D$ and $E$ are on $AB$ and $AC$ respectively such that $DB=BC=CE$. Lines $CD$ and $BE$ meet at $F$. $I$ is the incenter of $\triangle{ABC}$ and $H$ is the orthocenter of $\triangle{DEF}$. $\omega_b$ and $\omega_c$ are circles with diameter $BD$ and $CE$, respectively, intersecting each other at points $X$ and $Y$. Prove that $I$ and $H$ lie on $XY$.

Hint
1 reply
iStud
2 hours ago
MathLuis
an hour ago
how do we find a construction?
iStud   0
2 hours ago
Source: Monthly Contest KTOM March 2025 P4 Essay
Given a chess board $n\times n$ with $n>3$ with all the unit squares are initially white coloured. Every move, we can turn the color (from white to black or otherwise) from the 5 unit squares that form this T-pentomino which can be rotated or reflexed (see the image below). Determine all natural numbers $n$ such that all unit squares on the board can be made into all black after a finite number of moves.
0 replies
iStud
2 hours ago
0 replies
unnecessary wrapped FE on Q
iStud   0
2 hours ago
Source: Monthly Contest KTOM March 2025 P3 Essay
Find all functions $f:\mathbb{Q}\to\mathbb{Q}$ such that
\[f(f(f(\frac{x+y}{2}))+x+y)=f(x)+f(y)+f(\frac{x+y}{2})\]for all rational numbers $x,y$.

Hint
0 replies
iStud
2 hours ago
0 replies
Inspired by Mihaela Berindeanu
sqing   2
N 2 hours ago by sqing
Source: Own
Let $ a,b,c>0  . $ Prove that
$$(4a+b+c)(a+b)(b+c)(c+a)\geq (ab+bc+ca)(2a+b+c)^2$$
2 replies
sqing
Yesterday at 2:08 PM
sqing
2 hours ago
The Sums of Elements in Subsets
bobaboby1   1
N 2 hours ago by bobaboby1
Given a finite set \( X = \{x_1, x_2, \ldots, x_n\} \), and the pairwise comparison of the sums of elements of all its subsets (with the empty set defined as having a sum of 0), which amounts to \( \binom{2}{2^n} \) inequalities, these given comparisons satisfy the following three constraints:

1. The sum of elements of any non-empty subset is greater than 0.
2. For any two subsets, removing or adding the same elements does not change their comparison of the sums of elements.
3. For any two disjoint subsets \( A \) and \( B \), if the sums of elements of \( A \) and \( B \) are greater than those of subsets \( C \) and \( D \) respectively, then the sum of elements of the union \( A \cup B \) is greater than that of \( C \cup D \).

The question is: Does there necessarily exist a positive solution \( (x_1, x_2, \ldots, x_n) \) that satisfies all these conditions?
1 reply
bobaboby1
Mar 12, 2025
bobaboby1
2 hours ago
Sharygin 2025 CR P2
Gengar_in_Galar   4
N Yesterday at 1:31 PM 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
Yesterday at 1:31 PM
Sharygin 2025 CR P2
G H J
G H BBookmark kLocked kLocked NReply
Source: Sharygin 2025
The post below has been deleted. Click to close.
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Gengar_in_Galar
29 posts
#1 • 1 Y
Y by kiyoras_2001
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
This post has been edited 1 time. Last edited by Gengar_in_Galar, Mar 11, 2025, 9:49 AM
Z K Y
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Siddharthmaybe
106 posts
#2
Y by
used up a lot of time on this to no avail, this is prolly a troll :(
Z K Y
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YaoAOPS
1486 posts
#3 • 1 Y
Y by MS_asdfgzxcvb
Let $M_{IJ}$ be the midpoint of $I$, $J$. Let $P = (M_{AD}M_{BD}M_{CD}) \cap (M_{AB}M_{BD}M_{BC})$ so
\[
\measuredangle M_{AD}E - \measuredangle EM_{AB} = \measuredangle M_{AD}E - \measuredangle EM_{BD} + \measuredangle EM_{BD} - \measuredangle EM_{AB} = \measuredangle M_{AD}M_{CD} - \measuredangle M_{CD}M_{BD} + \measuredangle M_{BD}M_{BC} - \measuredangle M_{AB}M_{BC} = \measuredangle AC - \measuredangle BC + \measuredangle CD - \measuredangle AC = \measuredangle DCB = \measuredangle M_{AD}M_{AC}M_{AB}
\]so $P$ lies on $(M_{AD}M_{AC}M_{AB})$ and we are done by symmetry.
This post has been edited 1 time. Last edited by YaoAOPS, Mar 10, 2025, 6:58 PM
Z K Y
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ND_
11 posts
#4
Y by
Z is simply the Gergonne-Steiner Point.
Z K Y
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FKcosX
1 post
#5
Y by
Suppose the four given points are
\(
A,\; B,\; C,\; D,
\)
with no three collinear and not all four concyclic. (Thus, for each vertex the circle through the other three is well–defined.) We shall now define for each vertex \(X\) the “vertex–circle”
\(
\omega_X=(\text{three points other than }X);
\)
in other words,
\[
\omega_A=(BCD),\quad \omega_B=(ACD),\quad \omega_C=(ABD),\quad \omega_D=(ABC).
\]Next, for each vertex \(X\) we perform the homothety (dilation) \(h_X\) with center \(X\) and ratio \(1/2\). (A homothety with ratio \(1/2\) sends every point \(Y\) to the midpoint of \(XY\).) Denote
\[
\omega'_X = h_X(\omega_X).
\]Thus, for example,
\[
\omega'_A \text{ is the circle through } M_{AB},\; M_{AC},\; M_{AD},
\]where \(M_{AB}\) is the midpoint of \(AB\), etc. (Similarly, \(\omega'_B\) passes through \(M_{BA},M_{BC},M_{BD}\); note that \(M_{BA}=M_{AB}\).)



Observe that if \(X\) and \(Y\) are two distinct vertices then by construction the circle \(\omega'_X\) contains the midpoint \(M_{XY}\) (since \(M_{XY}=h_X(Y)\)) and similarly \(\omega'_Y\) contains \(M_{XY}\) (since \(M_{XY}=h_Y(X)\)). Hence for any two vertices \(X\) and \(Y\) the circles \(\omega'_X\) and \(\omega'_Y\) have the point \(M_{XY}\) in common.

For each vertex \(X\), note that the homothety \(h_X\) with center \(X\) and ratio 2 sends \(\omega'_X\) back to the circumcircle \(\omega_X\). (Indeed, if a point \(P\) lies on \(\omega'_X\) then by definition \(P=h_X(Q)\) for some \(Q\in\omega_X\); applying the inverse homothety of ratio 2 shows that \(2P-X=Q\in\omega_X\).) In particular, if a point \(Z\) lies on \(\omega'_X\) then the point
\[
X' = 2Z-X
\]lies on \(\omega_X\). (This is the same as saying that reflecting \(X\) in \(Z\) lands on \(\omega_X\).)

Thus, if we can show that there exists a point \(Z\) lying simultaneously on all four circles \(\omega'_A,\omega'_B,\omega'_C,\omega'_D\) then for each vertex \(X\) the reflection of \(X\) in \(Z\) lies on \(\omega_X\) (that is, on the circle through the three vertices other than \(X\)). This is exactly the property desired in the original problem.



We now prove that the four circles \(\omega'_A,\omega'_B,\omega'_C,\omega'_D\) have a common point. The proof will use two key facts:

1. Common “side–midpoints”: As we have shown earlier, for any two vertices \(X\) and \(Y\) the circles \(\omega'_X\) and \(\omega'_Y\) meet at the midpoint \(M_{XY}\).

2. The Miquel configuration for the complete quadrilateral:
In any quadrilateral
\(ABCD\)
the four circles
\(\omega_A,\omega_B,\omega_C,\omega_D\)
(i.e. the circumcircles of the triangles determined by three of the four vertices) have a well‐known associated configuration: The three circles determined by the triangles
\(ABC\), \(BCD\),
and
\(CDA\)
have a common “Miquel point”. (In our situation the four points are not concyclic so these circles are distinct and their “Miquel point” is defined by three of them.)

Now, fix two vertices, say \(A\) and \(B\). The circles \(\omega'_A\) and \(\omega'_B\) meet in \(M_{AB}=H\) and in a second point, say \(Z_{AB}\). (Because two distinct circles have two intersections unless they are tangent, and in our configuration tangency is avoided by the general‐position hypothesis.) Next, consider the circles \(\omega'_A\) and \(\omega'_C\); they meet in \(M_{AC}=I\) and in a second point \(Z_{AC}\). One now shows by using the homothetic relationships
\[
h_A(\omega'_A)=\omega_A,\quad h_B(\omega'_B)=\omega_B,\quad h_C(\omega'_C)=\omega_C,
\]and the standard fact about Miquel points in the complete quadrilateral \(ABCD\) (namely, that the circles \(\omega_A,\omega_B,\omega_C,\omega_D\) “rotate” about a unique Miquel point when three are taken at a time) that the second intersections \(Z_{AB}\) and \(Z_{AC}\) must in fact coincide $=Z.$ (One way to see this is to “lift” the configuration via the homothety of ratio 2 from each vertex. For example, applying the homothety with center \(A\) sends the pair \(\omega'_A\) and \(\omega'_B\) to \(\omega_A\) and a circle through the image of \(M_{AB}=H\); the unique intersection point of the corresponding circumcircles coming from the Miquel configuration forces the corresponding “half‐points” to agree.)

A similar argument shows that the second intersection of \(\omega'_B\) and \(\omega'_C\) agrees with these, and then by symmetry the same point \(Z\) lies on all four circles.

Thus, we obtain a unique point \(Z\) such that
\[
Z\in\omega'_A\cap\omega'_B\cap\omega'_C\cap\omega'_D.
\]


Because the homothety with center \(X\) (and ratio 2) sends \(\omega'_X\) to \(\omega_X\), the fact that
\[
Z\in\omega'_X
\]implies that the point
\[
X' = 2Z - X,
\]i.e. the reflection of \(X\) in \(Z\), lies on \(\omega_X\). Since this holds for every vertex \(X\in\{A,B,C,D\}\), the point \(Z\) is exactly the desired point with the property that reflecting any one of \(A\), \(B\), \(C\), \(D\) in \(Z\) lands on the circle determined by the three remaining vertices.

This completes the proof of the concurrency of the four circles.

Hence, such a unique point $Z$ that satisfies the question conditions surely exists and is constructed as said in the solution.
Z K Y
N Quick Reply
G
H
=
a