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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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0 replies
jlacosta
May 1, 2025
0 replies
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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
J
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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
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orthocenter on sus circle
DVDTSB   2
N 2 minutes ago by Diamond-jumper76
Source: Romania TST 2025 Day 2 P1
Let \( ABC \) be an acute triangle with \( AB < AC \), and let \( O \) be the center of its circumcircle. Let \( A' \) be the reflection of \( A \) with respect to \( BC \). The line through \( O \) parallel to \( BC \) intersects \( AC \) at \( F \), and the tangent at \( F \) to the circle \( \odot(BFC) \) intersects the line through \( A' \) parallel to \( BC \) at point \( M \). Let \( K \) be a point on the ray \( AB \), starting at \( A \), such that \( AK = 4AB \).
Show that the orthocenter of triangle \( ABC \) lies on the circle with diameter \( KM \).

Proposed by Radu Lecoiu

2 replies
DVDTSB
Yesterday at 12:18 PM
Diamond-jumper76
2 minutes ago
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problem 5
termas   74
N 15 minutes ago by maromex
Source: IMO 2016
The equation
$$(x-1)(x-2)\cdots(x-2016)=(x-1)(x-2)\cdots (x-2016)$$is written on the board, with $2016$ linear factors on each side. What is the least possible value of $k$ for which it is possible to erase exactly $k$ of these $4032$ linear factors so that at least one factor remains on each side and the resulting equation has no real solutions?
74 replies
termas
Jul 12, 2016
maromex
15 minutes ago
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Inspired by nhathhuyyp5c
sqing   2
N 18 minutes ago by lbh_qys
Source: Own
Let $ x,y>0, x+2y- 3xy\leq 4. $Prove that
$$ \frac{1}{x^2} + 2y^2 + 3x + \frac{4}{y} \geq 3\left(2+\sqrt[3]{\frac{9}{4} }\right)$$
2 replies
sqing
an hour ago
lbh_qys
18 minutes ago
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I think I know why this problem was rejected by IMO PSC several times...
mshtand1   1
N 35 minutes ago by sarjinius
Source: Ukrainian Mathematical Olympiad 2025. Day 2, Problem 11.8
Exactly $102$ country leaders arrived at the IMO. At the final session, the IMO chairperson wants to introduce some changes to the regulations, which the leaders must approve. To pass the changes, the chairperson must gather at least \(\frac{2}{3}\) of the votes "FOR" out of the total number of leaders. Some leaders do not attend such meetings, and it is known that there will be exactly $81$ leaders present. The chairperson must seat them in a square-shaped conference hall of size \(9 \times 9\), where each leader will be seated in a designated \(1 \times 1\) cell. It is known that exactly $28$ of these $81$ leaders will surely support the chairperson, i.e., they will always vote "FOR." All others will vote as follows: At the last second of voting, they will look at how their neighbors voted up to that moment — neighbors are defined as leaders seated in adjacent cells \(1 \times 1\) (sharing a side). If the majority of neighbors voted "FOR," they will also vote "FOR." If there is no such majority, they will vote "AGAINST." For example, a leader seated in a corner of the hall has exactly $2$ neighbors and will vote "FOR" only if both of their neighbors voted "FOR."

(a) Can the IMO chairperson arrange their $28$ supporters so that they vote "FOR" in the first second of voting and thereby secure a "FOR" vote from at least \(\frac{2}{3}\) of all $102$ leaders?

(b) What is the maximum number of "FOR" votes the chairperson can obtain by seating their 28 supporters appropriately?

Proposed by Bogdan Rublov
1 reply
mshtand1
Mar 14, 2025
sarjinius
35 minutes ago
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Quadrilateral with Congruent Diagonals
v_Enhance   38
N an hour ago by Giant_PT
Source: USA TSTST 2012, Problem 2
Let $ABCD$ be a quadrilateral with $AC = BD$. Diagonals $AC$ and $BD$ meet at $P$. Let $\omega_1$ and $O_1$ denote the circumcircle and the circumcenter of triangle $ABP$. Let $\omega_2$ and $O_2$ denote the circumcircle and circumcenter of triangle $CDP$. Segment $BC$ meets $\omega_1$ and $\omega_2$ again at $S$ and $T$ (other than $B$ and $C$), respectively. Let $M$ and $N$ be the midpoints of minor arcs $\widehat {SP}$ (not including $B$) and $\widehat {TP}$ (not including $C$). Prove that $MN \parallel O_1O_2$.
38 replies
v_Enhance
Jul 19, 2012
Giant_PT
an hour ago
J
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Interesting inequality
sealight2107   4
N 2 hours ago by NguyenVanHoa29
Source: Own
Let $a,b,c>0$ such that $a+b+c=3$. Find the minimum value of:
$Q=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{1}{a^3+b^3+abc}+\frac{1}{b^3+c^3+abc}+\frac{1}{c^3+a^3+abc}$
4 replies
sealight2107
May 6, 2025
NguyenVanHoa29
2 hours ago
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Inequality
nguyentlauv   3
N 2 hours ago by NguyenVanHoa29
Source: Own
Let $a,b,c$ be positive real numbers such that $ab+bc+ca=3$ and $k\ge 0$, prove that
$$\frac{\sqrt{a+1}}{b+c+k}+\frac{\sqrt{b+1}}{c+a+k}+\frac{\sqrt{c+1}}{a+b+k} \geq \frac{3\sqrt{2}}{k+2}.$$
3 replies
nguyentlauv
May 6, 2025
NguyenVanHoa29
2 hours ago
J
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schur weighted
Ducksohappi   1
N 2 hours ago by truongngochieu
Schur-weighted:
let a,b,c be positive. Prove that:
$a^3+b^3+c^3+3abc\ge \sum ab\sqrt{a^2+b^2}$
1 reply
Ducksohappi
4 hours ago
truongngochieu
2 hours ago
J
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forced vertices in graphs
Davdav1232   1
N 2 hours ago by CBMaster
Source: Israel TST 7 2025 p2
Let \( G \) be a graph colored using \( k \) colors. We say that a vertex is forced if it has neighbors in all the other \( k - 1 \) colors.

Prove that for any \( 2024 \)-regular graph \( G \), there exists a coloring using \( 2025 \) colors such that at least \( 1013 \) of the colors have a forced vertex of that color.

Note: The graph coloring must be valid, this means no \( 2 \) vertices of the same color may be adjacent.
1 reply
Davdav1232
May 8, 2025
CBMaster
2 hours ago
J
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Cyclic inequality with rational functions
MathMystic33   1
N 3 hours ago by Nguyenhuyen_AG
Source: 2025 Macedonian Team Selection Test P3
Let \(x_1,x_2,x_3,x_4\) be positive real numbers. Prove the inequality
\[ \frac{x_1 + 3x_2}{x_2 + x_3} \;+\; \frac{x_2 + 3x_3}{x_3 + x_4} \;+\; \frac{x_3 + 3x_4}{x_4 + x_1} \;+\; \frac{x_4 + 3x_1}{x_1 + x_2} \;\ge\;8. \]
1 reply
MathMystic33
Yesterday at 6:00 PM
Nguyenhuyen_AG
3 hours ago
J
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I got stuck in this combinatorics
artjustinhere237   2
N 3 hours ago by artjustinhere237
Let $S = \{1, 2, 3, \ldots, k\}$, where $k \geq 4$ is a positive integer.
Prove that there exists a subset of $S$ with exactly $k - 2$ elements such that the sum of its elements is a prime number.
2 replies
artjustinhere237
Yesterday at 4:56 PM
artjustinhere237
3 hours ago
J
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d1-d2 divides n for all divisors d1, d2
a_507_bc   5
N 3 hours ago by Assassino9931
Source: Romania 3rd JBMO TST 2023 P1
Determine all natural numbers $n \geq 2$ with at most four natural divisors, which have the property that for any two distinct proper divisors $d_1$ and $d_2$ of $n$, the positive integer $d_1-d_2$ divides $n$.
5 replies
a_507_bc
May 20, 2023
Assassino9931
3 hours ago
J
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Something weird with this one FE in integers (probably challenging, maybe not)
Gaunter_O_Dim_of_math   2
N 4 hours ago by aaravdodhia
Source: Pang-Cheng-Wu, FE, problem number 52.
During FE problems' solving I found a very specific one:

Find all such f that f: Z -> Z and for all integers a, b, c
f(a^3 + b^3 + c^3) = f(a)^3 + f(b)^3 + f(c)^3.

Everything what I've got is that f is odd, f(n) = n or -n or 0
for all n from 0 to 11 (just bash it), but it is very simple and do not give the main idea.
I actually have spent not so much time on this problem, but definitely have no clue. As far as I see, number theory here or classical FE solving or advanced methods, which I know, do not work at all.
Is here a normal solution (I mean, without bashing and something with a huge number of ugly and weird inequalities)?
Or this is kind of rubbish, which was put just for bash?
2 replies
Gaunter_O_Dim_of_math
Yesterday at 8:10 PM
aaravdodhia
4 hours ago
J
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Bulgaria 8
orl   9
N 4 hours ago by Assassino9931
Source: IMO LongList 1959-1966 Problem 34
Find all pairs of positive integers $\left( x;\;y\right) $ satisfying the equation $2^{x}=3^{y}+5.$
9 replies
orl
Sep 2, 2004
Assassino9931
4 hours ago
J
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J
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AbD+AcD=60
sinajackson   1
N Oct 7, 2006 by yetti
$D$ is a point in triangle $ABC$ and the real numbers $a,b,c,d$ exist such that :
$AB=ab$, $AC=ac$, $AD=ad$, $BC=bc$, $BD=bd$, $CD=cd$
Prove that:
$\angle ABD+\angle ACD=60^{\circ}$
1 reply
sinajackson
Sep 29, 2006
yetti
Oct 7, 2006
J
AbD+AcD=60
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Reveal topic tags
geometry
circumcircle
trigonometry
geometry proposed
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sinajackson
594 posts
sinajackson
#1 h  Sep 29, 2006, 3:45 PM • 2 Y
Y by Adventure10, Mango247
$D$ is a point in triangle $ABC$ and the real numbers $a,b,c,d$ exist such that :
$AB=ab$, $AC=ac$, $AD=ad$, $BC=bc$, $BD=bd$, $CD=cd$
Prove that:
$\angle ABD+\angle ACD=60^{\circ}$
Z K Y
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This post has been deleted. Click here to see post.
yetti
2643 posts
yetti
#2 h  Oct 7, 2006, 4:45 AM • 4 Y
Y by Adventure10, Mango247, Mango247, Mango247
$\frac{AC}{BC}= \frac{ac}{bc}= \frac{ad}{bd}= \frac{AD}{BD}$

$\frac{BA}{CA}= \frac{ba}{ca}= \frac{bd}{cd}= \frac{BD}{CD}$

$\frac{CB}{AB}= \frac{cb}{ab}= \frac{cd}{ad}= \frac{CD}{AD}$

Thus D is the concurrency point of the 3 Apollonius circles of $\triangle ABC$ inside its circumcircle, i.e., the 1st isodynamic point. For D to be inside of $\triangle ABC,$ the largest internal angle must be less than $120^\circ.$ Let X, Y, Z be the feet of normals from D to BC, CA, AB. From the cyclic quadrilaterals CXDY, AYDZ, BZDX with their circumcircle diameters CD, AD, BD, we get $XY = CD \sin C,\ YZ = AD \sin A,\ ZX = BD \sin B$ and

$\frac{XY}{YZ}= \frac{CD}{AD}\cdot \frac{\sin C}{\sin A}= \frac{CB}{AB}\cdot \frac{\sin C}{\sin A}= 1$

i.e., XY = YZ and similarly, YZ = ZX. Therefore, $\triangle XYZ$ is equilateral. From the cyclic quadrilaterals CXDY, BZDX,

$\angle ABD+\angle ACD \equiv \angle ZBD+\angle YCD = \angle ZXD+\angle YXD = \angle ZXY = 60^\circ$
Z K Y
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