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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.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 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
Nice Number Theory Question (Inspired by AIME)
MathRook7817   2
N 10 minutes ago by MathRook7817
Let $n$ be a positive integer. Find the smallest value of $n$ such that:

$2^n + 3^n - n$ is divisible by $216$.

2 replies
MathRook7817
an hour ago
MathRook7817
10 minutes ago
Diodes and usamons
v_Enhance   46
N an hour ago by HamstPan38825
Source: USA December TST for the 56th IMO, by Linus Hamilton
A physicist encounters $2015$ atoms called usamons. Each usamon either has one electron or zero electrons, and the physicist can't tell the difference. The physicist's only tool is a diode. The physicist may connect the diode from any usamon $A$ to any other usamon $B$. (This connection is directed.) When she does so, if usamon $A$ has an electron and usamon $B$ does not, then the electron jumps from $A$ to $B$. In any other case, nothing happens. In addition, the physicist cannot tell whether an electron jumps during any given step. The physicist's goal is to isolate two usamons that she is sure are currently in the same state. Is there any series of diode usage that makes this possible?

Proposed by Linus Hamilton
46 replies
v_Enhance
Dec 17, 2014
HamstPan38825
an hour ago
China South East Mathematical Olympiad 2013 problem 2
s372102   3
N an hour ago by AGCN
$\triangle ABC$, $AB>AC$. the incircle $I$ of $\triangle ABC$ meet $BC$ at point $D$, $AD$ meet $I$ again at $E$. $EP$ is a tangent of $I$, and $EP$ meet the extension line of $BC$ at $P$. $CF\parallel PE$, $CF\cap AD=F$. the line $BF$ meet $I$ at $M,N$, point $M$ is on the line segment $BF$, the line segment $PM$ meet $I$ again at $Q$. Show that $\angle ENP=\angle ENQ$
3 replies
s372102
Aug 10, 2013
AGCN
an hour ago
f(x+y) = max(f(x), y) + min(f(y), x)
Zhero   50
N an hour ago by lpieleanu
Source: ELMO Shortlist 2010, A3
Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that $f(x+y) = \max(f(x),y) + \min(f(y),x)$.

George Xing.
50 replies
Zhero
Jul 5, 2012
lpieleanu
an hour ago
Similar to iran 1996
GreekIdiot   1
N an hour ago by Lufin
Let $f: \mathbb R \to \mathbb R$ be a function such that $f(f(x)+y)=f(f(x)-y)+4f(x)y \: \forall x,y \: \in \: \mathbb R$. Find all such $f$.
1 reply
GreekIdiot
Apr 26, 2025
Lufin
an hour ago
Prove that 4p-3 is a square - Iran NMO 2005 - Problem1
sororak   24
N an hour ago by EmersonSoriano
Let $n,p>1$ be positive integers and $p$ be prime. We know that $n|p-1$ and $p|n^3-1$. Prove that $4p-3$ is a perfect square.
24 replies
sororak
Sep 21, 2010
EmersonSoriano
an hour ago
IMO ShortList 1999, combinatorics problem 4
orl   27
N 2 hours ago by cursed_tangent1434
Source: IMO ShortList 1999, combinatorics problem 4
Let $A$ be a set of $N$ residues $\pmod{N^{2}}$. Prove that there exists a set $B$ of of $N$ residues $\pmod{N^{2}}$ such that $A + B = \{a+b|a \in A, b \in B\}$ contains at least half of all the residues $\pmod{N^{2}}$.
27 replies
orl
Nov 14, 2004
cursed_tangent1434
2 hours ago
Geometry
Lukariman   2
N 2 hours ago by Lukariman
Given circle (O) and point P outside (O). From P draw tangents PA and PB to (O) with contact points A, B. On the opposite ray of ray BP, take point M. The circle circumscribing triangle APM intersects (O) at the second point D. Let H be the projection of B on AM. Prove that <HDM = 2∠AMP.
2 replies
Lukariman
Yesterday at 12:43 PM
Lukariman
2 hours ago
IMO 2010 Problem 1
canada   119
N 2 hours ago by lpieleanu
Find all function $f:\mathbb{R}\rightarrow\mathbb{R}$ such that for all $x,y\in\mathbb{R}$ the following equality holds \[
f(\left\lfloor x\right\rfloor y)=f(x)\left\lfloor f(y)\right\rfloor \] where $\left\lfloor a\right\rfloor $ is greatest integer not greater than $a.$

Proposed by Pierre Bornsztein, France
119 replies
canada
Jul 7, 2010
lpieleanu
2 hours ago
f(n) <= f(a(G)) + f(b(G))
dangerousliri   5
N 2 hours ago by awesomeming327.
Source: FEOO, Problem 2, Shortlist C2
Given a group $G$ of people, we define $a(G)$ to be the least number of tables needed such that each person from the group sits in one of them with no two friends sitting on the same table, and $b(G)$ to be the least number of tables needed such that each person from the group sits in one of them with no two enemies sitting on the same table.
Consider all functions $f:\mathbb{N} \to \mathbb{N}$ such that for each group $G$ of $n$ people we have
\[ f(n) \leqslant f(a(G)) + f(b(G)) \quad \text{and} \quad f(2) = 1.\]Find all possible values of $f(2020)$.

Note: We assume that for every pair of people in a group, they are either both friends with each other or both enemies with each other. Also, $\mathbb{N}$ is the set of all positive integers.

Proposed by Demetres Christofides, Cyprus
5 replies
dangerousliri
May 30, 2020
awesomeming327.
2 hours ago
A Collection of Good Problems from my end
SomeonecoolLovesMaths   24
N 2 hours ago by ReticulatedPython
This is a collection of good problems and my respective attempts to solve them. I would like to encourage everyone to post their solutions to these problems, if any. This will not only help others verify theirs but also perhaps bring forward a different approach to the problem. I will constantly try to update the pool of questions.

The difficulty level of these questions vary from AMC 10 to AIME. (Although the main pool of questions were prepared as a mock test for IOQM over the years)

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5
24 replies
SomeonecoolLovesMaths
May 4, 2025
ReticulatedPython
2 hours ago
f(w^2+x^2+y^2+z^2)=f(w^2+x^2)+f(y^2+z^2)
dangerousliri   23
N 3 hours ago by awesomeming327.
Source: FEOO, Problem 1, Shortlist N1
Find all functions $f:\mathbb{N}_0\rightarrow\mathbb{R}$ such that,
$$f(w^2+x^2+y^2+z^2)=f(w^2+x^2)+f(y^2+z^2)$$for all non-negative integers $w,x,y$ and $z$.

Note: $\mathbb{N}_0$ is the set of all non-negative integers and $\mathbb{R}$ is the set of all real numbers.

Proposed by Dorlir Ahmeti, Kosovo
23 replies
dangerousliri
May 30, 2020
awesomeming327.
3 hours ago
Geometry Proof
strongstephen   9
N 3 hours ago by strongstephen
Proof that choosing four distinct points at random has an equal probability of getting a convex quadrilateral vs a concave one.
not cohesive proof alert!

NOTE: By choosing four distinct points, that means no three points lie on the same line on the Gaussian Plane.
NOTE: The probability of each point getting chosen don’t need to be uniform (as long as it is symmetric about the origin), you just need a way to choose points in the infinite plane (such as a normal distribution)

Start by picking three of the four points. Next, graph the regions where the fourth point would make the quadrilateral convex or concave. In diagram 1 below, you can see the regions where the fourth point would be convex or concave. Of course, there is the centre region (the shaded triangle), but in an infinite plane, the probability the fourth point ends up in the finite region approaches 0.

Next, I want to prove to you the area of convex/concave, or rather, the probability a point ends up in each area, is the same. Referring to the second diagram, you can flip each concave region over the line perpendicular to the angle bisector of which the region is defined. (Just look at it and you'll get what it means.) Now, each concave region has an almost perfect 1:1 probability correspondence to another convex region. The only difference is the finite region (the triangle, shaded). Again, however, the actual significance (probability) of this approaches 0.

If I call each of the convex region's probability P(a), P(c), and P(e) and the concave ones P(b), P(d), P(f), assuming areas a and b are on opposite sides (same with c and d, e and f) you can get:
P(a) = P(b)
P(c) = P(d)
P(e) = P(f)

and P(a) + P(c) + P(e) = P(convex)
and P(b) + P(d) + P(f) = P(concave)

therefore:
P(convex) = P(concave)
9 replies
strongstephen
Yesterday at 4:54 AM
strongstephen
3 hours ago
n and n+100 have odd number of divisors (1995 Belarus MO Category D P2)
jasperE3   4
N 5 hours ago by KTYC
Find all positive integers $n$ so that both $n$ and $n + 100$ have odd numbers of divisors.
4 replies
jasperE3
Apr 6, 2021
KTYC
5 hours ago
Simiplifying a Complicated Expression
phiReKaLk6781   6
N Apr 23, 2025 by lbh_qys
Simplify: $ \frac{a^3}{(a-b)(a-c)}+\frac{b^3}{(b-a)(b-c)}+\frac{c^3}{(c-a)(c-b)}$
6 replies
phiReKaLk6781
Mar 15, 2010
lbh_qys
Apr 23, 2025
Simiplifying a Complicated Expression
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phiReKaLk6781
2074 posts
#1 • 2 Y
Y by Adventure10 and 1 other user
Simplify: $ \frac{a^3}{(a-b)(a-c)}+\frac{b^3}{(b-a)(b-c)}+\frac{c^3}{(c-a)(c-b)}$
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varunrocks
1134 posts
#2 • 2 Y
Y by Adventure10, Mango247
There are too many steps but I am getting a+b+c.
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Truffles
1459 posts
#3 • 2 Y
Y by Adventure10, Mango247
You could use the fact that the fractions in the expression are symmetric?
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Vo Duc Dien
341 posts
#4 • 2 Y
Y by Adventure10 and 1 other user
After expanding the expression, just divide the numerator by the denominator and we will get a + b + c.
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dgreenb801
1896 posts
#5 • 1 Y
Y by Adventure10
If we combine it into one fraction we get $ \frac{a^3b-ab^3+b^3c-bc^3+c^3a-ca^3}{(a-b)(a-c)(b-c)}$. But in the numerator if $ a=b$ then the numerator would would $ 0$, thus $ a-b$ must be a factor of the numerator, similarly $ a-c$ and $ b-c$ must be also, dividing them out we get $ a+b+c$.
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P162008
182 posts
#6
Y by
Let $$\frac{a^3}{(a - b)(a - c)} + \frac{b^3}{(b - c)(b - a)} + \frac{c^3}{(c - a)(c - b)} = t$$
$$t = -\frac{a^3(b - c) + b^3(c - a) + c^3(a - b)}{(a - b)(b - c)(c - a)}$$
Now, $$ N_r = a^3(b - c) + b^3(c - a) + c^3(a - b)$$
$$= a^3 b - ab^3 + b^3c - ca^3 + c^3a - bc^3$$
$$= ab(a^2 - b^2) - c(a^3 - b^3) + c^3(a - b)$$
$$= (a - b)[ab(a + b) - c(a^2 + ab + b^2) + c^3]$$
$$= (a - b)[a²b + ab² - ca² - abc - b²c + c³]$$
$$= (a - b)[a²(b - c) + ab(b - c) - c(b² - c²)]$$
$$= (a - b)(b - c)[a² + ab - c(b + c)]$$
$$= (a - b)(b - c)[a² + ab - bc - c²)]$$
$$= (a - b)(b - c)[-b(c - a) - (c² - a²)]$$
$$= -(a - b)(b - c)(c - a)(a + b + c)$$
Therefore, $$t = \frac{(a - b)(b - c)(c - a)(a + b + c)}{(a - b)(b - c)(c - a)} = \boxed{a + b + c}$$
This post has been edited 1 time. Last edited by P162008, Apr 22, 2025, 12:49 AM
Reason: Typo
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lbh_qys
558 posts
#7
Y by
This is the coefficient of the \(x^2\) term in \(x^3 - (x-a)(x-b)(x-c)\) by Lagrange interpolation.
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