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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
inequality
senku23   1
N 8 minutes ago by giangtruong13
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).
1 reply
+1 w
senku23
15 minutes ago
giangtruong13
8 minutes ago
D1015 : A strange EF for polynomials
Dattier   1
N 9 minutes 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
9 minutes ago
BMN is equilateral iff rectangle ABCD is square
parmenides51   2
N 14 minutes 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
2 replies
parmenides51
Sep 16, 2024
Tsikaloudakis
14 minutes ago
Cutting a big square into smaller squares
nAalniaOMliO   4
N 26 minutes 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
26 minutes ago
2015 Taiwan TST Round 1 Quiz 1 Problem 1
wanwan4343   11
N 28 minutes ago by ariopro1387
Source: 2015 Taiwan TST Round 1 Quiz 1 Problem 1
Find all primes $p,q,r$ such that $qr-1$ is divisible by $p$, $pr-1$ is divisible by $q$, $pq-1$ is divisible by $r$.
11 replies
wanwan4343
Jul 12, 2015
ariopro1387
28 minutes ago
D1010 : How it is possible ?
Dattier   13
N 29 minutes ago by Dattier
Source: les dattes à Dattier
Is it true that$$\forall n \in \mathbb N^*, (24^n \times B \mod A) \mod 2 = 0 $$?

A=1728400904217815186787639216753921417860004366580219212750904
024377969478249664644267971025952530803647043121025959018172048
336953969062151534282052863307398281681465366665810775710867856
720572225880311472925624694183944650261079955759251769111321319
421445397848518597584590900951222557860592579005088853698315463
815905425095325508106272375728975

B=2275643401548081847207782760491442295266487354750527085289354
965376765188468052271190172787064418854789322484305145310707614
546573398182642923893780527037224143380886260467760991228567577
953725945090125797351518670892779468968705801340068681556238850
340398780828104506916965606659768601942798676554332768254089685
307970609932846902
13 replies
1 viewing
Dattier
Mar 10, 2025
Dattier
29 minutes ago
Eventually constant sequence with condition
PerfectPlayer   2
N 31 minutes ago by egxa
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.
2 replies
PerfectPlayer
Yesterday at 4:27 AM
egxa
31 minutes ago
JBMO Shortlist 2021 C2
Lukaluce   5
N 40 minutes ago by Frd_19_Hsnzde
Source: JBMO Shortlist 2021
Let $n$ be a positive integer. We are given a $3n \times 3n$ board whose unit squares are colored in black and white in such way that starting with the top left square, every third diagonal is colored in black and the rest of the board is in white. In one move, one can take a $2 \times 2$ square and change the color of all its squares in such way that white squares become orange, orange ones become black and black ones become white. Find all $n$ for which, using a finite number of moves, we can make all the squares which were initially black white, and all squares which were initially white black.

Proposed by Boris Stanković and Marko Dimitrić, Bosnia and Herzegovina
5 replies
Lukaluce
Jul 2, 2022
Frd_19_Hsnzde
40 minutes ago
Flips in triangulation
Oksutok   0
40 minutes ago
Prove that every triangulation of a convex $n$-polygon can be constructed from any other triangulation by at most $2n-10$ flips (if $n>12$).

Note. A flip is an operation that transforms one triangulation to another by removing an edge between two triangles and adding the opposite diagonal to the resulting quadrilateral.
0 replies
Oksutok
40 minutes ago
0 replies
D1016 : A strange result about the palindrom polynomials
Dattier   4
N an hour ago by lbh_qys
Source: les dattes à Dattier
Let $Q\in \{-1,1,0\}[x]$ with $Q(1)=0$.

Is it true that $\exists P \in \mathbb Z[x]$ palindrom with $P | Q$ ?
4 replies
Dattier
Monday at 12:13 PM
lbh_qys
an hour ago
Laura fixing roads with cheap price
CrazyInMath   1
N an hour ago by shanelin-sigma
Source: 2024 CK Summer MSG Mock C3; Shortlist C5
In the nation of Peach Blossom Origin, there are $n$ cities numbered $1$ to $n$. City $a$ and city $b$ has a bidirectional road of lengths $(a+b)^a$ if and only if $a<b$ and $a+b$ is a prime number.
As Typhoon Gaemi strikes, all roads are destroyed. The admin Laura wants to repair some roads such that for each pair of city, you can get from one to another just using the repaired roads. However, each unit length of road costs $1$ dollar to fix, and Laura wants to minimize the cost. Prove that Laura would have one unique way to do this.

Proposed by CrazyInMath.
1 reply
CrazyInMath
Aug 16, 2024
shanelin-sigma
an hour ago
Slightly weird points which are not so weird
Pranav1056   10
N an hour ago by kes0716
Source: India TST 2023 Day 4 P1
Suppose an acute scalene triangle $ABC$ has incentre $I$ and incircle touching $BC$ at $D$. Let $Z$ be the antipode of $A$ in the circumcircle of $ABC$. Point $L$ is chosen on the internal angle bisector of $\angle BZC$ such that $AL = LI$. Let $M$ be the midpoint of arc $BZC$, and let $V$ be the midpoint of $ID$. Prove that $\angle IML = \angle DVM$
10 replies
Pranav1056
Jul 9, 2023
kes0716
an hour ago
classical R+ FE
jasperE3   1
N an hour ago by pco
Source: kent2207, based on 2019 Slovenia TST
wanted to post this problem in its own thread: https://artofproblemsolving.com/community/c6h1784825p34307772
Find all functions $f:\mathbb R^+\to\mathbb R^+$ for which:
$$f(f(x)+f(y))=yf(1+yf(x))$$for all $x,y\in\mathbb R^+$.
1 reply
jasperE3
Yesterday at 3:55 PM
pco
an hour ago
Physics disguised as math
everythingpi3141592   7
N an hour ago by kes0716
Source: India IMOTC 2024 Day 4 Problem 2
There are $n\ge 3$ particles on a circle situated at the vertices of a regular $n$-gon. All these particles move on the circle with the same constant speed. One of the particles moves in the clockwise direction while all others move in the anti-clockwise direction. When particles collide, that is, they are all at the same point, they all reverse the direction of their motion and continue with the same speed as before.

Let $s$ be the smallest number of collisions after which all particles return to their original positions. Find $s$.

Proposed by N.V. Tejaswi
7 replies
everythingpi3141592
May 31, 2024
kes0716
an hour ago
Nonsymmetric inequality for sum of three fractions
steppewolf   5
N Jun 3, 2023 by F10tothepowerof34
Source: Macedonian TST for IMO 2013 - P1 day 2
Let $a>0,b>0,c>0$ and $a+b+c=1$. Show the inequality
$$\frac{a^4+b^4}{a^2+b^2}+\frac{b^3+c^3}{b+c} + \frac{2a^2+b^2+2c^2}{2} \geq \frac{1}{2}$$
5 replies
steppewolf
Mar 28, 2021
F10tothepowerof34
Jun 3, 2023
Nonsymmetric inequality for sum of three fractions
G H J
G H BBookmark kLocked kLocked NReply
Source: Macedonian TST for IMO 2013 - P1 day 2
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steppewolf
351 posts
#1
Y by
Let $a>0,b>0,c>0$ and $a+b+c=1$. Show the inequality
$$\frac{a^4+b^4}{a^2+b^2}+\frac{b^3+c^3}{b+c} + \frac{2a^2+b^2+2c^2}{2} \geq \frac{1}{2}$$
Z K Y
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grupyorum
1400 posts
#2 • 1 Y
Y by steppewolf
By Cauchy, $(a^4+b^4)\ge (a^2+b^2)^2/2$. Hence, the given object is lower bounded by
\[
\frac{3a^2+2b^2+2c^2 +2b^2-2bc+2c^2}{2} = \frac{3a^2+4b^2+4c^2 -2bc}{2}.
\]Now, using (a) $b^2+c^2-2bc\ge 0$; and (b) $3a^2+3b^2+3c^2\ge (a+b+c)^2=1$ valid by the Cauchy-Schwarz, we complete the proof.
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sqing
41095 posts
#3 • 1 Y
Y by steppewolf
steppewolf wrote:
Let $a>0,b>0,c>0$ and $a+b+c=1$. Show the inequality
$$\frac{a^4+b^4}{a^2+b^2}+\frac{b^3+c^3}{b+c} + \frac{2a^2+b^2+2c^2}{2} \geq \frac{1}{2}$$
https://artofproblemsolving.com/community/c6h1195489p5852461
Let $a>0,b>0,c>0$ and $a+b+c=1$. Show the inequality
\[\frac{1}{2}\left(\frac{a^2+b^2+c^2}{ab+bc+ca}\right)^2\geq\frac{a^4+b^4}{a^2+b^2}+\frac{b^3+c^3}{b+c}+\frac{2a^2+b^2+2c^2}{2} \geq \frac{1}{2} \]
Z K Y
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CadmiumDwight
139 posts
#4
Y by
Muirhead on LHS first and second term gives $a^4 + b^4 \ge ab(a^2 + b^2)$ and $b^3 + c^3 \ge bc(b + c)$. Then, AM-GM gives $\frac{a^2 + c^2}{2} \ge ac$ and C-S gives $\frac{a^2 + b^2 + c^2}{2} \ge \frac{1}{6}$.

Thus, $LHS \ge ab + bc + bc + \dfrac{1}{6}$ and since $ab + bc + ac \ge \dfrac{1}{3}$ (which is trivial), $LHS \ge \dfrac{1}{2}$. $\blacksquare$
This post has been edited 1 time. Last edited by CadmiumDwight, Mar 29, 2021, 8:20 AM
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Math00954
405 posts
#5
Y by
CadmiumDwight wrote:
$ab + bc + ac \ge \dfrac{1}{3}$ (which is trivial)

Are you sure?

~~~~~~~~~~~~~~~~~~~~~~~~~
steppewolf wrote:
Let $a>0,b>0,c>0$ and $a+b+c=1$. Show the inequality
$$\frac{a^4+b^4}{a^2+b^2}+\frac{b^3+c^3}{b+c} + \frac{2a^2+b^2+2c^2}{2} \geq \frac{1}{2}$$

We have, by Chebyshev and AM-GM:

$$LHS\ge \frac{a^2+b^2}{2}+\frac{b^2+c^2}{2}+\frac{2a^2+b^2+2c^2}{2}=\frac{3}{2}(a^2+b^2+c^2)\ge \frac{3}{2} \cdot \frac{(a+b+c)^2}{3}=\frac{1}{2}$$
This post has been edited 2 times. Last edited by Math00954, Mar 29, 2021, 11:13 AM
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F10tothepowerof34
195 posts
#6
Y by
Notice that, $\frac{a^4+b^4}{a^2+b^2}\ge\frac{a^2+b^2}{2}\Longrightarrow 2a^4+2b^4\ge(a^2+b^2)^2=a^2+2a^2b^2+b^4\Longrightarrow a^4+b^4\overset{\text{AM-GM}}{\ge}2a^2b^2$
Furthermore, $\frac{b^3+c^3}{b+c}\ge\frac{b^2+c^2}{2b^3+2c^3}\Longrightarrow \ge(b^2+c^2)(b+c)=b^3+b^2c+c^2b+c^3\Longrightarrow b^3+c^3\ge b^2c+c^2b\Longrightarrow (b+c)(b^2+c^2-bc)\ge bc(b+c)=b^2c+c^2b$
Thus by summing the previously stated inequalities, $\sum\frac{a^4+b^4}{a^2+b^2}\ge\frac{3\sum_{cyc}a^2}{2}=\frac{3}{2}\left(\bigg(\sum_{cyc}a\bigg)^2-2\sum_{cyc}ab\right)=\frac{3}{2}\bigg(1-2\sum_{cyc}ab\bigg)$, thus the inequality boils down to proving: $\frac{3}{2}-3\sum_{cyc}ab\ge\frac{1}{2}\Longleftrightarrow 1\ge3\sum_{cyc}ab$ thus after we homogenize the $LHS$, the inequality turns into $\bigg(\sum_{cyc}a\bigg)^2\ge3\sum_{cyc}ab$ which is clearly true $\blacksquare$.
This post has been edited 2 times. Last edited by F10tothepowerof34, Jun 19, 2023, 3:13 AM
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