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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
D1038 : A generalization of Jensen
Dattier   2
N 43 minutes ago by Ss3qt
Source: les dattes à Dattier
Let $f \in C^1([0,1]), g \in C^2(f([0;1]))$.

Is it true that

$$\min(|g''|)\times \min(|f'|^2) \leq 24 \times\left|\int_0^1g(f(x)) \text{d}x- g(\int_0^1 f(x) \text{d}x) \right| \leq \max(|g''|)\times \max(|f'|^2)$$?
2 replies
Dattier
Today at 12:15 PM
Ss3qt
43 minutes ago
Can't be power of 2
shobber   31
N an hour ago by LeYohan
Source: APMO 1998
Show that for any positive integers $a$ and $b$, $(36a+b)(a+36b)$ cannot be a power of $2$.
31 replies
shobber
Mar 17, 2006
LeYohan
an hour ago
Brilliant Problem
M11100111001Y1R   4
N an hour ago by IAmTheHazard
Source: Iran TST 2025 Test 3 Problem 3
Find all sequences \( (a_n) \) of natural numbers such that for every pair of natural numbers \( r \) and \( s \), the following inequality holds:
\[
\frac{1}{2} < \frac{\gcd(a_r, a_s)}{\gcd(r, s)} < 2
\]
4 replies
M11100111001Y1R
Yesterday at 7:28 AM
IAmTheHazard
an hour ago
Own made functional equation
Primeniyazidayi   1
N an hour ago by Primeniyazidayi
Source: own(probably)
Find all functions $f:R \rightarrow R$ such that $xf(x^2+2f(y)-yf(x))=f(x)^3-f(y)(f(x^2)-2f(x))$ for all $x,y \in \mathbb{R}$
1 reply
Primeniyazidayi
May 26, 2025
Primeniyazidayi
an hour ago
not fun equation
DottedCaculator   13
N 2 hours ago by Adywastaken
Source: USA TST 2024/6
Find all functions $f\colon\mathbb R\to\mathbb R$ such that for all real numbers $x$ and $y$,
\[f(xf(y))+f(y)=f(x+y)+f(xy).\]
Milan Haiman
13 replies
DottedCaculator
Jan 15, 2024
Adywastaken
2 hours ago
Serbian selection contest for the IMO 2025 - P6
OgnjenTesic   12
N 3 hours ago by atdaotlohbh
Source: Serbian selection contest for the IMO 2025
For an $n \times n$ table filled with natural numbers, we say it is a divisor table if:
- the numbers in the $i$-th row are exactly all the divisors of some natural number $r_i$,
- the numbers in the $j$-th column are exactly all the divisors of some natural number $c_j$,
- $r_i \ne r_j$ for every $i \ne j$.

A prime number $p$ is given. Determine the smallest natural number $n$, divisible by $p$, such that there exists an $n \times n$ divisor table, or prove that such $n$ does not exist.

Proposed by Pavle Martinović
12 replies
OgnjenTesic
May 22, 2025
atdaotlohbh
3 hours ago
Geometry with fix circle
falantrng   33
N 3 hours ago by zuat.e
Source: RMM 2018 Problem 6
Fix a circle $\Gamma$, a line $\ell$ to tangent $\Gamma$, and another circle $\Omega$ disjoint from $\ell$ such that $\Gamma$ and $\Omega$ lie on opposite sides of $\ell$. The tangents to $\Gamma$ from a variable point $X$ on $\Omega$ meet $\ell$ at $Y$ and $Z$. Prove that, as $X$ varies over $\Omega$, the circumcircle of $XYZ$ is tangent to two fixed circles.
33 replies
falantrng
Feb 25, 2018
zuat.e
3 hours ago
USAMO 2001 Problem 2
MithsApprentice   54
N 3 hours ago by lpieleanu
Let $ABC$ be a triangle and let $\omega$ be its incircle. Denote by $D_1$ and $E_1$ the points where $\omega$ is tangent to sides $BC$ and $AC$, respectively. Denote by $D_2$ and $E_2$ the points on sides $BC$ and $AC$, respectively, such that $CD_2=BD_1$ and $CE_2=AE_1$, and denote by $P$ the point of intersection of segments $AD_2$ and $BE_2$. Circle $\omega$ intersects segment $AD_2$ at two points, the closer of which to the vertex $A$ is denoted by $Q$. Prove that $AQ=D_2P$.
54 replies
MithsApprentice
Sep 30, 2005
lpieleanu
3 hours ago
German-Style System of Equations
Primeniyazidayi   1
N 4 hours ago by Primeniyazidayi
Source: German MO 2025 11/12 Day 1 P1
Solve the system of equations in $\mathbb{R}$

\begin{align*}
\frac{a}{c} &= b-\sqrt{b}+c \\
\sqrt{\frac{a}{c}} &= \sqrt{b}+1 \\
\sqrt[4]{\frac{a}{c}} &=\sqrt[3]{b}-1
\end{align*}
1 reply
Primeniyazidayi
4 hours ago
Primeniyazidayi
4 hours ago
gcd nt from switzerland
AshAuktober   5
N 4 hours ago by Siddharthmaybe
Source: Swiss 2025 Second Round
Let $a, b$ be positive integers. Prove that the expression
\[\frac{\gcd(a+b,ab)}{\gcd(a,b)}\]is always a positive integer, and determine all possible values it can take.
5 replies
AshAuktober
5 hours ago
Siddharthmaybe
4 hours ago
Shortlist 2017/G1
fastlikearabbit   92
N 4 hours ago by Ilikeminecraft
Source: Shortlist 2017
Let $ABCDE$ be a convex pentagon such that $AB=BC=CD$, $\angle{EAB}=\angle{BCD}$, and $\angle{EDC}=\angle{CBA}$. Prove that the perpendicular line from $E$ to $BC$ and the line segments $AC$ and $BD$ are concurrent.
92 replies
fastlikearabbit
Jul 10, 2018
Ilikeminecraft
4 hours ago
Problem 2, Grade 12th RMO Shortlist - Year 2002
sticknycu   5
N 5 hours ago by P_Fazioli
Let $A \in M_2(C), A \neq O_2, A \neq I_2, n \in \mathbb{N}^*$ and $S_n = \{ X \in M_2(C) | X^n = A \}$.
Show:
a) $S_n$ with multiplication of matrixes operation is making an isomorphic-group structure with $U_n$.
b) $A^2 = A$.

Marian Andronache
5 replies
sticknycu
Jan 3, 2020
P_Fazioli
5 hours ago
3rd AKhIMO for University Students, P1
UzbekMathematician   1
N 6 hours ago by KAME06
Source: AKhIMO 2025, P1
There are two circles in the $xy-$plane centered on the $y-$axis which are tangent to both the parabola $y=x^2$ and the line $y=2025$. Determine the lengths of the circles' diameters.
1 reply
UzbekMathematician
Today at 1:44 PM
KAME06
6 hours ago
3rd AKhIMO for university students, p4
UzbekMathematician   1
N Today at 2:29 PM by grupyorum
Source: AKhIMO 2025, P4
Define a sequence $a_1, a_2, a_3, ... $ by $a_1=2, a_2=5$ and $a_{n+2}=f(a_n, a_{n+1})$ for all $n \ge 1$, where $$f(x,y)=5(x+y)+2\sqrt{6x^2+15xy+6y^2}.$$Show that $a_n$ is an integer for all $n\ge 1$.
1 reply
UzbekMathematician
Today at 2:03 PM
grupyorum
Today at 2:29 PM
Jordan form and canonical base of a matrix
And1viper   3
N Apr 19, 2025 by Suan_16
Find the Jordan form and a canonical basis of the following matrix $A$ over the field $Z_5$:
$$A = \begin{bmatrix}
    2 & 1 & 2 & 0 & 0 \\
    0 & 4 & 0 & 3 & 4 \\
    0 & 0 & 2 & 1 & 2 \\
    0 & 0 & 0 & 4 & 1 \\
    0 & 0 & 0 & 0 & 2
  \end{bmatrix}
$$
3 replies
And1viper
Feb 26, 2023
Suan_16
Apr 19, 2025
Jordan form and canonical base of a matrix
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And1viper
13 posts
#1
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Find the Jordan form and a canonical basis of the following matrix $A$ over the field $Z_5$:
$$A = \begin{bmatrix}
    2 & 1 & 2 & 0 & 0 \\
    0 & 4 & 0 & 3 & 4 \\
    0 & 0 & 2 & 1 & 2 \\
    0 & 0 & 0 & 4 & 1 \\
    0 & 0 & 0 & 0 & 2
  \end{bmatrix}
$$
This post has been edited 1 time. Last edited by And1viper, Feb 26, 2023, 1:51 PM
Reason: Edited Latex
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Suan_16
67 posts
#2
Y by
bro $Z$ is not a field, the problem is faulty
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rchokler
2975 posts
#3 • 1 Y
Y by aidan0626
Suan_16 wrote:
bro $Z$ is not a field, the problem is faulty

Actually it is a field. $\mathbb{Z}_p$ is a finite field for all primes $p$.

The eigenvalues are $\lambda_1=\lambda_2=\lambda_3=2$ and $\lambda_4=\lambda_5=4$.

$A-2I_5=\begin{bmatrix}
0 & 1 & 2 & 0 & 0 \\
0 & 2 & 0 & 3 & 4 \\
0 & 0 & 0 & 1 & 2 \\
0 & 0 & 0 & 2 & 1 \\
0 & 0 & 0 & 0 & 0
\end{bmatrix}$

Observe from rows 3 and 4, we must take none of columns 4 and 5. Then row 2 forces us to take none of column 2. Then row 1 means we must take none of column 3. So $E_2=\text{span}(\mathbf{e}_1)$. The geometric multiplicity is 1.

$(A-2I_5)^3=\begin{bmatrix}
0 & 4 & 0 & 1 & 3 \\
0 & 3 & 0 & 1 & 3 \\
0 & 0 & 0 & 4 & 2 \\
0 & 0 & 0 & 3 & 4 \\
0 & 0 & 0 & 0 & 0
\end{bmatrix}$

If we take none of columns 4 and 5, then we must take none of column 2. So the kernel here contains $\mathbf{e}_1$ and $\mathbf{e}_3$. If we take instead 1 of column 4 and 3 of column 5, which means we have to take none of column 2. So another vector in this kernel is $\mathbf{e}_4+3\mathbf{e}_5$.

$A-4I_5=\begin{bmatrix}
3 & 1 & 2 & 0 & 0 \\
0 & 0 & 0 & 3 & 4 \\
0 & 0 & 3 & 1 & 2 \\
0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 3
\end{bmatrix}$

From row 4 or 5, we must take none of column 5. Then from row 2, we must take none of column 4. Then from row 3, we must take none of column 3. So the geometric multiplicity is again 1.

$(A-4I_5)^2=\begin{bmatrix}
4 & 3 & 2 & 0 & 3 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 4 & 3 & 3 \\
0 & 0 & 0 & 0 & 3 \\
0 & 0 & 0 & 0 & 4
\end{bmatrix}$

We still must take none of column 5.
If we take none of columns 3 and 4, then we can take 1 of column 1 and 2 of column 2. So the kernel here contains $\mathbf{e}_1+2\mathbf{e}_2$. We can take instead 1 of column 3 and 2 of column 4. In that case, we can take 1 of column 1 and 3 of column 2. This gives us $\mathbf{e}_1+3\mathbf{e}_2+\mathbf{e}_3+2\mathbf{e}_4$.

So the generalized eigenvectors are $v_1=\begin{bmatrix}1\\0\\0\\0\\0\end{bmatrix}$, $v_2=\begin{bmatrix}0\\0\\1\\0\\0\end{bmatrix}$, $v_3=\begin{bmatrix}0\\0\\0\\1\\3\end{bmatrix}$, $v_4=\begin{bmatrix}1\\2\\0\\0\\0\end{bmatrix}$, and $v_5=\begin{bmatrix}1\\3\\1\\2\\0\end{bmatrix}$

So $A=PJP^{-1}$ where:

$P=\begin{bmatrix}
1 & 0 & 0 & 1 & 1 \\
0 & 0 & 0 & 2 & 3 \\
0 & 1 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 & 2 \\
0 & 0 & 3 & 0 & 0
\end{bmatrix}$, $J=\begin{bmatrix}
2 & 2 & 0 & 0 & 0 \\
0 & 2 & 2 & 0 & 0 \\
0 & 0 & 2 & 0 & 0 \\
0 & 0 & 0 & 4 & 3 \\
0 & 0 & 0 & 0 & 4
\end{bmatrix}$, $P^{-1}=\begin{bmatrix}
1 & 2 & 0 & 4 & 2 \\
0 & 0 & 1 & 2 & 1 \\
0 & 0 & 0 & 0 & 2 \\
0 & 3 & 0 & 3 & 4 \\
0 & 0 & 0 & 3 & 4
\end{bmatrix}$

Note that the three positions where we would normally have 1s are instead 2s and a 3. These are chosen so the product comes out correct.
This post has been edited 2 times. Last edited by rchokler, Apr 19, 2025, 3:10 AM
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Suan_16
67 posts
#4
Y by
Oh sorry I forgot that it is $Z_5={0,1,2,3,4}$ instead of $Z^5$
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