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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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[*]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
Inequality
Amin12   8
N an hour ago by A.H.H
Source:  Iran 3rd round-2017-Algebra final exam-P3
Let $a,b$ and $c$ be positive real numbers. Prove that
$$\sum_{cyc} \frac {a^3b}{(3a+2b)^3} \ge \sum_{cyc} \frac {a^2bc}{(2a+2b+c)^3} $$
8 replies
Amin12
Sep 2, 2017
A.H.H
an hour ago
Not so beautiful
m4thbl3nd3r   0
an hour ago
Let $a, b,c>0$ such that $b+c>a$. Prove that $$2 \sqrt[4]{\frac{a}{b+c-a}}\ge 2 +\frac{2a^2-b^2-c^2}{(a+b)(a+c)}.$$
0 replies
m4thbl3nd3r
an hour ago
0 replies
Inequalities
sqing   1
N 2 hours ago by sqing
Let $ a,b> 0 $ and $2a+2b+ab=5. $ Prove that
$$ \frac{a^4}{b^4}+\frac{1}{a^4}+42ab-a^4\geq  43$$$$ \frac{a^5}{b^5}+\frac{1}{a^5}+64ab-a^5\geq  65$$$$ \frac{a^6}{b^6}+\frac{1}{a^6}+90ab-a^6\geq  91$$$$ \frac{a^7}{b^7}+\frac{1}{a^7}+121ab-a^7\geq  122$$
1 reply
sqing
Yesterday at 2:31 PM
sqing
2 hours ago
Every subset of size k has sum at most N/2
orl   50
N 2 hours ago by de-Kirschbaum
Source: USAMO 2006, Problem 2, proposed by Dick Gibbs
For a given positive integer $k$ find, in terms of $k$, the minimum value of $N$ for which there is a set of $2k + 1$ distinct positive integers that has sum greater than $N$ but every subset of size $k$ has sum at most $\tfrac{N}{2}.$
50 replies
orl
Apr 20, 2006
de-Kirschbaum
2 hours ago
Inspired by a9opsow_
sqing   2
N 2 hours ago by sqing
Source: Own
Let $ a,b > 0  .$ Prove that
$$ \frac{(ka^2 - kab-b)^2 + (kb^2 - kab-a)^2 + (ab-ka-kb )^2}{ (ka+b)^2 + (kb+a)^2+(a - b)^2 }\geq  \frac {1}{(k+1)^2}$$Where $ k\geq 0.37088 .$
$$\frac{(a^2 - ab-b)^2 + (b^2 - ab-a)^2 + ( ab-a-b)^2}{a^2 +b^2+(a - b)^2 } \geq 1$$$$ \frac{(2a^2 - 2ab-b)^2 + (2b^2 - 2ab-a)^2 + (ab-2a-2b )^2}{ (2a+b)^2 + (2b+a)^2+(a - b)^2 }\geq  \frac 19$$
2 replies
sqing
3 hours ago
sqing
2 hours ago
Cute NT Problem
M11100111001Y1R   5
N 3 hours ago by compoly2010
Source: Iran TST 2025 Test 4 Problem 1
A number \( n \) is called lucky if it has at least two distinct prime divisors and can be written in the form:
\[
n = p_1^{\alpha_1} + \cdots + p_k^{\alpha_k}
\]where \( p_1, \dots, p_k \) are distinct prime numbers that divide \( n \). (Note: it is possible that \( n \) has other prime divisors not among \( p_1, \dots, p_k \).) Prove that for every prime number \( p \), there exists a lucky number \( n \) such that \( p \mid n \).
5 replies
M11100111001Y1R
Tuesday at 7:20 AM
compoly2010
3 hours ago
Can anyone help me with this inequality problem
a9opsow_   3
N 3 hours ago by sqing
Let a, b, c be nonnegative real numbers, not all equal. Prove that

\frac{(a - bc)^2 + (b - ca)^2 + (c - ab)^2}{(a - b)^2 + (b - c)^2 + (c - a)^2} \geq \frac{1}{2}.
3 replies
a9opsow_
Yesterday at 11:50 AM
sqing
3 hours ago
3 var inequality
SunnyEvan   13
N 4 hours ago by Nguyenhuyen_AG
Let $ a,b,c \in R $ ,such that $ a^2+b^2+c^2=4(ab+bc+ca)$Prove that :$$ \frac{7-2\sqrt{14}}{48} \leq \frac{a^3b+b^3c+c^3a}{(a^2+b^2+c^2)^2} \leq \frac{7+2\sqrt{14}}{48} $$
13 replies
SunnyEvan
May 17, 2025
Nguyenhuyen_AG
4 hours ago
trigonometric inequality
MATH1945   13
N 4 hours ago by sqing
Source: ?
In triangle $ABC$, prove that $$sin^2(A)+sin^2(B)+sin^2(C) \leq \frac{9}{4}$$
13 replies
MATH1945
May 26, 2016
sqing
4 hours ago
Inequalities
sqing   21
N 4 hours ago by sqing
Let $ a,b,c\geq 0 ,a+b+c\leq 3. $ Prove that
$$a^2+b^2+c^2+ab +2ca+2bc +  abc \leq \frac{251}{27}$$$$ a^2+b^2+c^2+ab+2ca+2bc  + \frac{2}{5}abc  \leq \frac{4861}{540}$$$$ a^2+b^2+c^2+ab+2ca+2bc  + \frac{7}{20}abc  \leq \frac{2381411}{26460}$$
21 replies
sqing
May 21, 2025
sqing
4 hours ago
Iran TST Starter
M11100111001Y1R   2
N 5 hours ago by sami1618
Source: Iran TST 2025 Test 1 Problem 1
Let \( a_n \) be a sequence of positive real numbers such that for every \( n > 2025 \), we have:
\[
a_n = \max_{1 \leq i \leq 2025} a_{n-i} - \min_{1 \leq i \leq 2025} a_{n-i}
\]Prove that there exists a natural number \( M \) such that for all \( n > M \), the following holds:
\[
a_n < \frac{1}{1404}
\]
2 replies
M11100111001Y1R
Tuesday at 7:36 AM
sami1618
5 hours ago
Polar Coordinates
pingpongmerrily   4
N 5 hours ago by K124659
Convert the equation $r=\tan(\theta)$ into rectangular form.
4 replies
pingpongmerrily
5 hours ago
K124659
5 hours ago
Twin Prime Diophantine
awesomeming327.   23
N 5 hours ago by HDavisWashu
Source: CMO 2025
Determine all positive integers $a$, $b$, $c$, $p$, where $p$ and $p+2$ are odd primes and
\[2^ap^b=(p+2)^c-1.\]
23 replies
awesomeming327.
Mar 7, 2025
HDavisWashu
5 hours ago
Troublesome median in a difficult inequality
JG666   2
N 5 hours ago by navid
Source: 2022 Spring NSMO Day 2 Problem 3
Determine the minimum value of $\lambda\in\mathbb{R}$, such that for any positive integer $n$ and non-negative reals $x_1, x_2, \cdots, x_n$, the following inequality always holds:
$$\sum_{i=1}^n(m_i-a_i)^2\leqslant \lambda\cdot\sum_{i=1}^nx_i^2,$$Here $m_i$ and $a_i$ denote the median and arithmetic mean of $x_1, x_2, \cdots, x_i$, respectively.

Duanyang ZHANG, High School Affiliated to Renmin University of China
2 replies
JG666
May 22, 2022
navid
5 hours ago
Another binomial coefficients sum
aether01   8
N Apr 27, 2025 by soryn
Prove that $$\sum_{k=0}^{n}{{\left(-1\right)}^{k}\binom{n}{k} \binom{2n-k}{n-k}} = 1$$
8 replies
aether01
Mar 3, 2022
soryn
Apr 27, 2025
Another binomial coefficients sum
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aether01
5 posts
#1 • 1 Y
Y by HWenslawski
Prove that $$\sum_{k=0}^{n}{{\left(-1\right)}^{k}\binom{n}{k} \binom{2n-k}{n-k}} = 1$$
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alexheinis
10631 posts
#2 • 2 Y
Y by HWenslawski, fungarwai
The sum equals $\sum_0^n (-1)^k {n\choose k} p(x-k)$ where $p(x):={x\choose n}$ and $x=2n$. This sum equals $D^n p(x)$ where $D$ is the difference operator. Since $p$ has degree $n$ it is well-known that $D^np(x)=n! a$ where $a$ is the leading cft of $p$. Since $a=1/n!$, the sum equals 1, as stated.
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BackToSchool
1640 posts
#3
Y by
alexheinis wrote:
The sum equals $\sum_0^n (-1)^k {n\choose k} p(x-k)$ where $p(x):={x\choose n}$ and $x=2n$. This sum equals $D^n p(x)$ where $D$ is the difference operator. Since $p$ has degree $n$ it is well-known that $D^np(x)=n! a$ where $a$ is the leading cft of $p$. Since $a=1/n!$, the sum equals 1, as stated.

I just saw lots of new notations without actual prove process.
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alexheinis
10631 posts
#6 • 1 Y
Y by fungarwai
@above: it's not my fault that you don't know the theory, since it is well-known and basic.
The notations might be new to you but they are standard. So you're wrong when you say that there is no actual proof.
Let $p$ be a polynomial. The difference polynomial $Dp$ is defined with $Dp(x)=p(x)-p(x-1)$.
With induction one shows $D^n(p)(x)=\sum_{k=0}^n (-1)^k {n\choose k} p(x-k)$.
If $\deg(p)=s$ then $\deg(Dp)=s-1$ and the leading cft is multiplied by a factor $s$.
In particular, if $\deg (p)=n$ then $D^n (p)$ is a constant polynomial, and it's value is $n! a$, where $a$ is the leading cft of $p$.
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BackToSchool
1640 posts
#7
Y by
@above,
A well proof should be a friendly one, no matter it is well-known or new to the reader.
Anyway, thanks for explanation.
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hopeless404
588 posts
#8 • 1 Y
Y by fungarwai
Consider functions
$F(x) = \sum_{k\ge0} (-1)^k\binom{n}{k} x^k = (1-x)^n$
and the well-known $G_m(x) = \sum_{k\ge0} \binom{m+k}{k}x^k = \frac{1}{(1-x)^{m+1}}$
Then $$F(x)G_m(x) = \sum_{n\ge0} \sum_{k=0}^n (-1)^k\binom{n}{k} \binom{m+n-k}{n-k} x^n = \frac{(1-x)^n}{(1-x)^{m+1}}$$Equating the coefficients of $x^n$ gives
\begin{align*}
[x^n] F(x)G_n(x) &= [x^n] \frac{1}{1-x}\\
\implies \sum_{k=0}^n (-1)^k\binom{n}{k} \binom{2n-k}{n-k} &= 1
\end{align*}
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P162008
226 posts
#9 • 1 Y
Y by soryn
Cute
This post has been edited 6 times. Last edited by P162008, Apr 28, 2025, 3:21 PM
Reason: Typo
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fungarwai
865 posts
#10 • 3 Y
Y by Taiharward, soryn, P162008
This is a special case of the formula $\boxed{\displaystyle \sum_{i=0}^k (-1)^{k-i}\binom ni \binom {m-1+k-i}{k-i}=\binom {n-m}k}$ introduced in Summation with binomial coefficient

The proof is similar to #8 in this post.

$\displaystyle \sum_{k=0}^n (-1)^{n-k}\binom nk \binom {(n+1)-1+n-k}{n-k}=\binom {-1}n$

$\displaystyle \sum_{k=0}^n (-1)^{k}\binom nk \binom {2n-k}{n-k}=(-1)^n\binom {-1}n=1$

Applying difference operator as #2 is also a clever thought in this case.
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soryn
5348 posts
#11
Y by
Very, very nice!
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N Quick Reply
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