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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
6 hours ago
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

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0 replies
jlacosta
6 hours ago
0 replies
k a My Retirement & New Leadership at AoPS
rrusczyk   1571
N Mar 26, 2025 by SmartGroot
I write today to announce my retirement as CEO from Art of Problem Solving. When I founded AoPS 22 years ago, I never imagined that we would reach so many students and families, or that we would find so many channels through which we discover, inspire, and train the great problem solvers of the next generation. I am very proud of all we have accomplished and I’m thankful for the many supporters who provided inspiration and encouragement along the way. I'm particularly grateful to all of the wonderful members of the AoPS Community!

I’m delighted to introduce our new leaders - Ben Kornell and Andrew Sutherland. Ben has extensive experience in education and edtech prior to joining AoPS as my successor as CEO, including starting like I did as a classroom teacher. He has a deep understanding of the value of our work because he’s an AoPS parent! Meanwhile, Andrew and I have common roots as founders of education companies; he launched Quizlet at age 15! His journey from founder to MIT to technology and product leader as our Chief Product Officer traces a pathway many of our students will follow in the years to come.

Thank you again for your support for Art of Problem Solving and we look forward to working with millions more wonderful problem solvers in the years to come.

And special thanks to all of the amazing AoPS team members who have helped build AoPS. We’ve come a long way from here:IMAGE
1571 replies
1 viewing
rrusczyk
Mar 24, 2025
SmartGroot
Mar 26, 2025
egmo 2018 p4
microsoft_office_word   28
N 7 minutes ago by akliu
Source: EGMO 2018 P4
A domino is a $ 1 \times 2 $ or $ 2 \times 1 $ tile.
Let $n \ge 3 $ be an integer. Dominoes are placed on an $n \times n$ board in such a way that each domino covers exactly two cells of the board, and dominoes do not overlap. The value of a row or column is the number of dominoes that cover at least one cell of this row or column. The configuration is called balanced if there exists some $k \ge 1 $ such that each row and each column has a value of $k$. Prove that a balanced configuration exists for every $n \ge 3 $, and find the minimum number of dominoes needed in such a configuration.
28 replies
microsoft_office_word
Apr 12, 2018
akliu
7 minutes ago
Polynomial
EtacticToe   3
N 12 minutes ago by EmersonSoriano
Source: Own
Let $f(x)$ be a monic polynomial with integer coefficient. And suppose there exist 4 distinct integer $a,b,c,d$ such that $f(a)=…=f(d)=5$.

Find all $k$ such that $f(k)=8$
3 replies
EtacticToe
Dec 14, 2024
EmersonSoriano
12 minutes ago
inequalities hard
Cobedangiu   5
N 27 minutes ago by Primeniyazidayi
problem
5 replies
Cobedangiu
Mar 31, 2025
Primeniyazidayi
27 minutes ago
Geo Final but hard to solve with Conics...
Seungjun_Lee   5
N 28 minutes ago by L13832
Source: 2025 Korea Winter Program Practice Test P4
Let $\omega$ be the circumcircle of triangle $ABC$ with center $O$, and the $A$ inmixtilinear circle is tangent to $AB, AC, \omega$ at $D,E,T$ respectively. $P$ is the intersection of $TO$ and $DE$ and $X$ is the intersection of $AP$ and $\omega$. Prove that the isogonal conjugate of $P$ lies on the line passing through the midpoint of $BC$ and $X$.
5 replies
1 viewing
Seungjun_Lee
Jan 18, 2025
L13832
28 minutes ago
Integral inequality with differentiable function
Ciobi_   1
N 5 hours ago by MS_asdfgzxcvb
Source: Romania NMO 2025 12.2
Let $f \colon [0,1] \to \mathbb{R} $ be a differentiable function such that its derivative is an integrable function on $[0,1]$, and $f(1)=0$. Prove that \[ \int_0^1 (xf'(x))^2 dx \geq 12 \cdot \left( \int_0^1 xf(x) dx\right)^2 \]
1 reply
Ciobi_
Today at 2:29 PM
MS_asdfgzxcvb
5 hours ago
On coefficients of a polynomial over a finite field
Ciobi_   0
6 hours ago
Source: Romania NMO 2025 12.4
Let $p$ be an odd prime number, and $k$ be an odd number not divisible by $p$. Consider a field $K$ be a field with $kp+1$ elements, and $A = \{x_1,x_2, \dots, x_t\}$ be the set of elements of $K^*$, whose order is not $k$ in the multiplicative group $(K^*,\cdot)$. Prove that the polynomial $P(X)=(X+x_1)(X+x_2)\dots(X+x_t)$ has at least $p$ coefficients equal to $1$.
0 replies
Ciobi_
6 hours ago
0 replies
On non-negativeness of continuous and polynomial functions
Ciobi_   0
Today at 2:51 PM
Source: Romania NMO 2025 12.3
a) Let $a\in \mathbb{R}$ and $f \colon \mathbb{R} \to \mathbb{R}$ be a continuous function for which there exists an antiderivative $F \colon \mathbb{R} \to \mathbb{R} $, such that $F(x)+a\cdot f(x) \geq 0$, for any $x \in \mathbb{R}$, and$ \lim_{|x| \to \infty} \frac{F(x)}{e^{|\alpha \cdot x|}}=0$ holds for any $\alpha \in \mathbb{R}^*$. Prove that $F(x) \geq 0$ for all $x \in \mathbb{R}$.
b) Let $n\geq 2$ be a positive integer, $g \in \mathbb{R}[X]$, $g = X^n + a_1X^{n-1}+ \dots + a_{n-1}X+a_n$ be a polynomial with all of its roots being real, and $f \colon \mathbb{R} \to \mathbb{R}$ a polynomial function such that $f(x)+a_1\cdot f'(x)+a_2\cdot f^{(2)}(x)+\dots+a_n\cdot f^{(n)}(x) \geq 0$ for any $x \in \mathbb{R}$. Prove that $f(x) \geq 0$ for all $x \in \mathbb{R}$.
0 replies
Ciobi_
Today at 2:51 PM
0 replies
Proving AB-BA is singular from given conditions
Ciobi_   0
Today at 2:04 PM
Source: Romania NMO 2025 11.4
Let $A,B \in \mathcal{M}_n(\mathbb{C})$ be two matrices such that $A+B=AB+BA$. Prove that:
a) if $n$ is odd, then $\det(AB-BA)=0$;
b) if $\text{tr}(A)\neq \text{tr}(B)$, then $\det(AB-BA)=0$.
0 replies
Ciobi_
Today at 2:04 PM
0 replies
Equivalent definition for C^1 functions
Ciobi_   0
Today at 1:54 PM
Source: Romania NMO 2025 11.3
Prove that, for a function $f \colon \mathbb{R} \to \mathbb{R}$, the following $2$ statements are equivalent:
a) $f$ is differentiable, with continuous first derivative.
b) For any $a\in\mathbb{R}$ and for any two sequences $(x_n)_{n\geq 1},(y_n)_{n\geq 1}$, convergent to $a$, such that $x_n \neq y_n$ for any positive integer $n$, the sequence $\left(\frac{f(x_n)-f(y_n)}{x_n-y_n}\right)_{n\geq 1}$ is convergent.
0 replies
Ciobi_
Today at 1:54 PM
0 replies
RREF of some matrices
tommy2007   3
N Today at 1:51 PM by tommy2007
for $\forall n \in \mathbb{N},$
what is the maximum integer that appears in one of the Reduced Row Echelon Forms of $n \times n$ matrices which has only $-1$ and $1$ for their entries?
3 replies
tommy2007
Today at 6:57 AM
tommy2007
Today at 1:51 PM
Finding pairs of functions of class C^2 with a certain property
Ciobi_   0
Today at 1:31 PM
Source: Romania NMO 2025 11.1
Find all pairs of twice differentiable functions $f,g \colon \mathbb{R} \to \mathbb{R}$, with their second derivative being continuous, such that the following holds for all $x,y \in \mathbb{R}$: \[(f(x)-g(y))(f'(x)-g'(y))(f''(x)-g''(y))=0\]
0 replies
Ciobi_
Today at 1:31 PM
0 replies
Inverse of absolute value function
MetaphysicalWukong   2
N Today at 12:32 PM by paxtonw
how does the function have an inverse for k= 101, 203, 305, 509, 611 and 713?

how do we deduce this without graphing software?
2 replies
MetaphysicalWukong
Today at 7:21 AM
paxtonw
Today at 12:32 PM
Unique global minimum points
chirita.andrei   0
Today at 11:06 AM
Source: Own. Proposed for Romanian National Olympiad 2025.
Let $f\colon[0,1]\rightarrow \mathbb{R}$ be a continuous function. Suppose that for each $t\in(0,1)$, the function \[f_t\colon[0,1-t]\rightarrow\mathbb{R}, f_t(x)=f(x+t)-f(x)\]has an unique global minimum point, which we will denote by $g(t)$. Prove that if $\lim\limits_{t\to 0}g(t)=0$, then $g$ is constant zero.
0 replies
chirita.andrei
Today at 11:06 AM
0 replies
2024 Putnam B1
KevinYang2.71   7
N Today at 9:27 AM by anudeep
Let $n$ and $k$ be positive integers. The square in the $i$th row and $j$th column of an $n$-by-$n$ grid contains the number $i+j-k$. For which $n$ and $k$ is it possible to select $n$ squares from the grid, no two in the same row or column, such that the numbers contained in the selected squares are exactly $1,\,2,\,\ldots,\,n$?
7 replies
KevinYang2.71
Dec 10, 2024
anudeep
Today at 9:27 AM
Inequality with 3 variables
bel.jad5   11
N Dec 7, 2020 by LeonhardEuler0
Source: alijadallah belabess
Let $a$, $b$ and $c$ positive real numbers such that: $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$. Prove that:
\[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3 \geq 2(a+b+c)\]
11 replies
bel.jad5
Nov 4, 2018
LeonhardEuler0
Dec 7, 2020
Inequality with 3 variables
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Source: alijadallah belabess
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bel.jad5
3750 posts
#1 • 2 Y
Y by nguyendangkhoa17112003, Adventure10
Let $a$, $b$ and $c$ positive real numbers such that: $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$. Prove that:
\[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3 \geq 2(a+b+c)\]
This post has been edited 1 time. Last edited by bel.jad5, Nov 4, 2018, 3:03 PM
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luofangxiang
4613 posts
#2 • 3 Y
Y by ywq233, Adventure10, Mango247
[(ab(a+c)+bc(b+a)+ca(c+b)]^2>=4(ab+bc+ca)[(ab)(bc)+(bc)(ca)+(ca)(ab)]
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bel.jad5
3750 posts
#3 • 2 Y
Y by Adventure10, Mango247
luofangxiang wrote:
[(ab(a+c)+bc(b+a)+ca(c+b)]^2>=4(ab+bc+ca)[(ab)(bc)+(bc)(ca)+(ca)(ab)]

And?
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luofangxiang
4613 posts
#4 • 2 Y
Y by ywq233, Adventure10
[a(x+y)+b(y+z)+c(z+x)]^2>=4(ab+bc+ca)(xy+yz+zx)
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anhduy98
1080 posts
#5 • 3 Y
Y by bel.jad5, luofangxiang, Adventure10
bel.jad5 wrote:
Let $a$, $b$ and $c$ positive real numbers such that: $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$. Prove that:
\[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3 \geq 2(a+b+c)\]

Use $$AX-BY\ge \sqrt{A^2-B^2}.\sqrt{X^2-Y^2},(*),(A\ge B \ge 0, X\ge Y\ge 0)$$Because $$(*) \Leftrightarrow (AY-BX)^2\ge 0 , (true)$$$$LHS=\frac{ab^2+bc^2+ca^2+3abc}{abc}=\frac{(a+b+c)(ab+bc+ca)-(a^2b+b^2c+c^2a)}{abc}$$$$LHS \ge \frac{(a+b+c)(ab+bc+ca)-\sqrt{a^2+b^2+c^2}.\sqrt{a^2b^2+b^2c^2+c^2a^2}}{abc}$$$$LHS \ge \frac{\sqrt{(a+b+c)^2-(\sqrt{a^2+b^2+c^2})^2}.\sqrt{(ab+bc+ca)^2-(\sqrt{a^2b^2+b^2c^2+c^2a^2})^2}}{abc}$$$$LHS \ge \frac{\sqrt{4abc(a+b+c)(ab+bc+ca)}}{abc}=\frac{2(ab+bc+ca)}{abc}=RHS$$
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bel.jad5
3750 posts
#6 • 2 Y
Y by Adventure10, Mango247
anhduy98 wrote:
bel.jad5 wrote:
Let $a$, $b$ and $c$ positive real numbers such that: $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$. Prove that:
\[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3 \geq 2(a+b+c)\]

Use $$AX-BY\ge \sqrt{A^2-B^2}.\sqrt{X^2-Y^2},(*),(A\ge B \ge 0, X\ge Y\ge 0)$$Because $$(*) \Leftrightarrow (AY-BX)^2\ge 0 , (true)$$$$LHS=\frac{ab^2+bc^2+ca^2+3abc}{abc}=\frac{(a+b+c)(ab+bc+ca)-(a^2b+b^2c+c^2a)}{abc}$$$$LHS \ge \frac{(a+b+c)(ab+bc+ca)-\sqrt{a^2+b^2+c^2}.\sqrt{a^2b^2+b^2c^2+c^2a^2}}{abc}$$$$LHS \ge \frac{\sqrt{(a+b+c)^2-(\sqrt{a^2+b^2+c^2})^2}.\sqrt{(ab+bc+ca)^2-(\sqrt{a^2b^2+b^2c^2+c^2a^2})^2}}{abc}$$$$LHS \ge \frac{\sqrt{4abc(a+b+c)(ab+bc+ca)}}{abc}=\frac{2(ab+bc+ca)}{abc}=RHS$$

Brilliant!!!
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sqing
41383 posts
#7
Y by
Let $a$, $b$ and $c$ positive real numbers such that $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$. Prove that
\[2(a^2+b^2+c^2)\geq  a\sqrt{a^2+3}+b\sqrt{b^2+3}+c\sqrt{c^2+3} \]Let $a,b,c$ be positive real numbers, such that: $a+b+c \ge \frac {1}{a}+\frac {1}{b}+\frac{1}{c}$. Prove that:
$$a+b+c \ge \frac{3}{a+b+c}+\frac {2}{abc}$$Let $a$, $b$ and $c$ positive real numbers such that: $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$. Prove that:
$$ a+b+c+abc \geq 4$$$$\frac 1{a+2}+\frac 1{b+2}+\frac 1{c+2}\leq 1$$\[\frac{1}{(2a+b+c)^2}+\frac{1}{(a+2b+c)^2}+\frac{1}{(a+b+2c)^2}\leq \frac{3}{16} \]\[3(a+b+c)\geq\sqrt{8a^{2}+1}+\sqrt{8b^{2}+1}+\sqrt{8c^{2}+1}.\]\[(a+b+c)^{2}+3\geq 4(a+b+c)\sqrt[3]{abc}\]$$(ab+bc+ca)(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})^2 \geq 27$$https://artofproblemsolving.com/community/c6h2896381p25793992
https://artofproblemsolving.com/community/c4h2821329p25787158
https://artofproblemsolving.com/community/c4h1813778p20534203
https://artofproblemsolving.com/community/c6h355781p1932917
https://artofproblemsolving.com/community/c6h1529847p9194503
https://artofproblemsolving.com/community/c6h1739555p11304368
https://artofproblemsolving.com/community/c6h1659898p10533543
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Auditor
108 posts
#8
Y by
nice

Stronger .
Let $a, b, c, $ be positive real numbers such that $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$. Prove that $$a+b+c \ge abc + \frac {2}{abc} .$$
This post has been edited 1 time. Last edited by Auditor, Nov 26, 2020, 7:52 PM
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Auditor
108 posts
#9
Y by
Auditor wrote:
nice

Stronger .
Let $a, b, c, $ be positive real numbers such that $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$. Prove that $$a+b+c \ge abc + \frac {2}{abc} .$$

Let $a, b, c, $ be non-zero real numbers such $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$. Prove that $$a^2 + b^2 + c^2 + 6 \ge 3 (a+b+c) .$$
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denery
180 posts
#10 • 3 Y
Y by Mango247, Mango247, Mango247
@auditor im pretty sure that equality is occuring at a=b=c=1
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Universes
96 posts
#11 • 1 Y
Y by Mango247
denery wrote:
@auditor im pretty sure that equality is occuring at a=b=c=1

And so ?
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LeonhardEuler0
103 posts
#12 • 1 Y
Y by TerenceTao11235
bel.jad5 wrote:
Let $a$, $b$ and $c$ positive real numbers such that: $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$. Prove that:
\[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3 \geq 2(a+b+c)\]

First,Note that $ab*(ac+bc)=ab*(1-c^2)$ so,there is no solution for $c^2>1$ (a,b,c are positive) so $0<c=<1$ and because of the symmetry $0<a=<1$ and $0<b=<1$
Thus, $0<2(a+b+c)=<6$.
Then use AM-GM:
$a/b+b/c+c/a+3>=6$
Henceforth,
$a/b+b/c+c/a+3>=6>=2(a+b+c)$
Moreover if you use AM-HM,
$9=<(1/a+1/b+1/c)^2=(a+b+c)^2$
$3=<(a+b+c) , 6=<2(a+b+c)=<6 $
So $a+b+c=3$ and because of the nature of AM-GM-HM inequalities, $(a,b,c)$ is $(1,1,1)$.
This post has been edited 1 time. Last edited by LeonhardEuler0, Dec 7, 2020, 11:27 AM
Reason: Typo
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