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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.

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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
Another integral limit
RobertRogo   1
N an hour ago by alexheinis
Source: "Traian Lalescu" student contest 2025, Section A, Problem 3
Let $f \colon [0, \infty) \to \mathbb{R}$ be a function differentiable at 0 with $f(0) = 0$. Find
$$\lim_{n \to \infty} \frac{1}{n} \int_{2^n}^{2^{n+1}} f\left(\frac{\ln x}{x}\right) dx$$
1 reply
RobertRogo
Today at 2:28 PM
alexheinis
an hour ago
AB=BA if A-nilpotent
KevinDB17   3
N 2 hours ago by loup blanc
Let A,B 2 complex n*n matrices such that AB+I=A+B+BA
If A is nilpotent prove that AB=BA
3 replies
KevinDB17
Mar 30, 2025
loup blanc
2 hours ago
Interesting functional equation with geometry
User21837561   3
N 3 hours ago by Double07
Source: BMOSL 2024 G7
For an acute triangle $ABC$, let $O$ be the circumcentre, $H$ be the orthocentre, and $G$ be the centroid.
Let $f:\pi\rightarrow\mathbb R$ satisfy the following condition:
$f(A)+f(B)+f(C)=f(O)+f(G)+f(H)$
Prove that $f$ is constant.
3 replies
User21837561
Today at 8:14 AM
Double07
3 hours ago
Equilateral triangle formed by circle and Fermat point
Mimii08   2
N 4 hours ago by Mimii08
Source: Heard from a friend
Hi! I found this interesting geometry problem and I would really appreciate help with the proof.

Let ABC be an acute triangle, and let T be the Fermat (Torricelli) point of triangle ABC. Let A1, B1, and C1 be the feet of the perpendiculars from T to the sides BC, AC, and AB, respectively. Let ω be the circle passing through points A1, B1, and C1. Let A2, B2, and C2 be the second points where ω intersects the sides BC, AC, and AB, respectively (different from A1, B1, C1).

Prove that triangle A2B2C2 is equilateral.

2 replies
Mimii08
Yesterday at 10:36 PM
Mimii08
4 hours ago
two circumcenters and one orthocenter, vertices of parallelogram
parmenides51   4
N 4 hours ago by AylyGayypow009
Source: Greece JBMO TST 2015 p2
Let $ABC$ be an acute triangle inscribed in a circle of center $O$. If the altitudes $BD,CE$ intersect at $H$ and the circumcenter of $\triangle BHC$ is $O_1$, prove that $AHO_1O$ is a parallelogram.
4 replies
parmenides51
Apr 29, 2019
AylyGayypow009
4 hours ago
Very nice equivalence in matrix equations
RobertRogo   3
N 4 hours ago by Etkan
Source: "Traian Lalescu" student contest 2025, Section A, Problem 4
Let $A, B \in \mathcal{M}_n(\mathbb{C})$ Show that the following statements are equivalent:

i) For every $C \in \mathcal{M}_n(\mathbb{C})$ there exist $X, Y \in \mathcal{M}_n(\mathbb{C})$ such that $AX + YB = C$
ii) For every $C \in \mathcal{M}_n(\mathbb{C})$ there exist $U, V \in \mathcal{M}_n(\mathbb{C})$ such that $A^2 U + V B^2 = C$

3 replies
RobertRogo
Today at 2:34 PM
Etkan
4 hours ago
Group Theory
Stephen123980   0
4 hours ago
Let G be a group of order $45.$ If G has a normal subgroup of order $9,$ then prove that $G$ is abelian without using Sylow Theorems.
0 replies
Stephen123980
4 hours ago
0 replies
Find the angle alpha [Iran Second Round 1994]
Amir Hossein   4
N Today at 1:36 PM by Mysteriouxxx
In the following diagram, $O$ is the center of the circle. If three angles $\alpha, \beta$ and $\gamma$ be equal, find $\alpha.$
IMAGE
4 replies
Amir Hossein
Nov 26, 2010
Mysteriouxxx
Today at 1:36 PM
official solution of IGO
ABCD1728   7
N Today at 12:32 PM by ABCD1728
Source: IGO official website
Where can I get the official solution of IGO for 2023 and 2024, there are some inhttps://imogeometry.blogspot.com/p/iranian-geometry-olympiad.html, but where can I find them on the official website, thanks :)
7 replies
ABCD1728
May 4, 2025
ABCD1728
Today at 12:32 PM
Combo geo with circles
a_507_bc   10
N Today at 12:30 PM by EthanWYX2009
Source: 239 MO 2024 S8
There are $2n$ points on the plane. No three of them lie on the same straight line and no four lie on the same circle. Prove that it is possible to split these points into $n$ pairs and cover each pair of points with a circle containing no other points.
10 replies
a_507_bc
May 22, 2024
EthanWYX2009
Today at 12:30 PM
Really fun geometry problem
Sadigly   6
N Today at 11:45 AM by farhad.fritl
Source: Azerbaijan Senior MO 2025 P6
In an acute triangle $ABC$ with $AB<AC$, the foot of altitudes from $A,B,C$ to the sides $BC,CA,AB$ are $D,E,F$, respectively. $H$ is the orthocenter. $M$ is the midpoint of segment $BC$. Lines $MH$ and $EF$ intersect at $K$. Let the tangents drawn to circumcircle $(ABC)$ from $B$ and $C$ intersect at $T$. Prove that $T;D;K$ are colinear
6 replies
Sadigly
Yesterday at 4:29 PM
farhad.fritl
Today at 11:45 AM
find angle
TBazar   5
N Today at 9:56 AM by TBazar
Given $ABC$ triangle with $AC>BC$. We take $M$, $N$ point on AC, AB respectively such that $AM=BC$, $CM=BN$. $BM$, $AN$ lines intersect at point $K$. If $2\angle AKM=\angle ACB$, find $\angle ACB$
5 replies
TBazar
Yesterday at 6:57 AM
TBazar
Today at 9:56 AM
Grouping angles in a pentagon with bisectors
Assassino9931   0
Today at 9:28 AM
Source: Al-Khwarizmi International Junior Olympiad 2025 P2
Let $ABCD$ be a convex quadrilateral with \[\angle ADC = 90^\circ, \ \ \angle BCD = \angle ABC > 90^\circ, \mbox{ and } AB = 2CD.\]The line through \(C\), parallel to \(AD\), intersects the external angle bisector of \(\angle ABC\) at point \(T\). Prove that the angles $\angle ATB$, $\angle TBC$, $\angle BCD$, $\angle CDA$, $\angle DAT$ can be divided into two groups, so that the angles in each group have a sum of $270^{\circ}$.

Miroslav Marinov, Bulgaria
0 replies
Assassino9931
Today at 9:28 AM
0 replies
Tangent circles
Sadigly   1
N Today at 9:19 AM by lbh_qys
Source: Azerbaijan Junior MO 2025 P6
Let $T$ be a point outside circle $\omega$ centered at $O$. Tangents from $T$ to $\omega$ touch $\omega$ at $A;B$. Line $TO$ intersects bigger $AB$ arc at $C$.The line drawn from $T$ parallel to $AC$ intersects $CB$ at $E$. Ray $TE$ intersects small $BC$ arc at $F$. Prove that the circumcircle of $OEF$ is tangent to $\omega$.
1 reply
Sadigly
Today at 8:06 AM
lbh_qys
Today at 9:19 AM
MVT question
mqoi_KOLA   10
N Apr 15, 2025 by mqoi_KOLA
Let \( f \) be a function which is continuous on \( [0,1] \) and differentiable on \( (0,1) \), with \( f(0) = f(1) = 0 \). Assume that there is some \( c \in (0,1) \) such that \( f(c) = 1 \). Prove that there exists some \( x_0 \in (0,1) \) such that \( |f'(x_0)| > 2 \).
10 replies
mqoi_KOLA
Apr 10, 2025
mqoi_KOLA
Apr 15, 2025
MVT question
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mqoi_KOLA
91 posts
#1
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Let \( f \) be a function which is continuous on \( [0,1] \) and differentiable on \( (0,1) \), with \( f(0) = f(1) = 0 \). Assume that there is some \( c \in (0,1) \) such that \( f(c) = 1 \). Prove that there exists some \( x_0 \in (0,1) \) such that \( |f'(x_0)| > 2 \).
This post has been edited 3 times. Last edited by mqoi_KOLA, Apr 10, 2025, 9:51 PM
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KAME06
159 posts
#2 • 1 Y
Y by mqoi_KOLA
Case 1: $c > 0.5$
Then, using Mean Value Theorem, there exist an $x_0$ such that $f'(x_0)=\frac{f(c)-f(0)}{c-0}=\frac{1-0}{c}=\frac{1}{c}>2 \Rightarrow |f'(x_0)|>2$.
Case 2: $c < 0.5$
That implies that $c-1 > -0.5$ then using Mean Value Theorem, there exist an $x_0$ such that $f'(x_0)=\frac{f(c)-f(1)}{c-1}=\frac{1-0}{c-1}=\frac{1}{c-1}<-2 \Rightarrow |f'(x_0)|>2$
Case 3: $c=0.5$
Here idk for \( |f'(x_0)| > 2 \)
This post has been edited 3 times. Last edited by KAME06, May 3, 2025, 3:14 AM
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mqoi_KOLA
91 posts
#3 • 1 Y
Y by KAME06
u left the case which i wanted the proof for.. :noo: :wallbash_red:
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ddot1
24748 posts
#4 • 2 Y
Y by mqoi_KOLA, KAME06
To handle the $c=1/2$ case, suppose $f(1/2)=1$ and assume $|f'(x)|\le 2$ on the whole interval $(0,1)$. Then by the mean value theorem, there is some $x_0$ such that $$\frac{f(1/2)-f(0)}{1/2-0}=f'(x_0),$$so $f'(x_0)=2$. That by itself isn't a contradiction, but we can do something similar to get a contradiction. The intuitive idea is that since we're right on the "edge" of a contradiction, we have no room to move the graph of $f$. If $|f'|\le 2$ and $f(1/2)=1$, that forces $f(x)=2x$ on the interval $[0,1/2]$. Moving the graph up or down at any point makes the slope larger than $2$ somewhere. Similarly, $f(x)=2-2x$ on the interval $[1/2,1]$.

We first prove that $f(x)=2x$ for every $x\in [0,1/2]$. On any interval $[0,a]$ with $a\le 1/2$, we have $$\frac{f(a)-f(0)}{a-0}=f'(x_a)$$for some $x_a$, depending on $a$. Since we're assuming $f'$ is always bounded by $2$, this means that $\dfrac{f(a)}{a}\le 2,$ so $f(a)\le 2a$ for all $a\in[0,1/2]$.

This inequality can never be strict, either. If we had $f(a)<2a$, then we could use the mean value theorem on the interval $[a,1/2]$ to get \begin{align*}\frac{f(1/2)-f(a)}{1/2-a}&=f'(x_a)\\
\frac{1-f(a)}{1/2-a}&=f'(x_a).
\end{align*}But the left side is larger than $2$, and the right side is at most $2$, a contradiction.

The same reasoning forces $f(x)=2-2x$ on the interval $[1/2,1]$, so $$f(x)=\begin{cases} 2x,\, &0\le x\le 1/2\\ 2-2x, &1/2<x\le 1.\end{cases}$$However, this function is not differentiable at $1/2$, so we are done.
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Alphaamss
247 posts
#5
Y by
Proof from MSE https://math.stackexchange.com/questions/1752763/suppose-f0-f1-0-and-fx-0-1-show-that-there-is-rho-with-lve?noredirect=1
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mqoi_KOLA
91 posts
#6
Y by
as a novice, this qns was good . thanks @alphaamss and @ddot1
This post has been edited 1 time. Last edited by mqoi_KOLA, Apr 11, 2025, 4:06 AM
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MS_asdfgzxcvb
70 posts
#7 • 1 Y
Y by mqoi_KOLA
\(c=\frac 12\):
\(\emph{Proof.}\) Assume towards a contradiction that \(\forall 0<\xi<1:\big|f'(\xi)\big|\le 2\).
LMVT on \(0<x<\frac 12\) and \(\frac 12\):
\[\usepackage{mathtools}
\tfrac {1-f(x)}{\frac 12-x}\le 2\xRightarrow{\hspace{40pt}}\left(\forall 0<x<\tfrac 12\right):f(x)\ \ge\ 2x\tag{1}\]LMVT on \(\frac 12\) and \(\frac 12<x<1\):
\[\usepackage{mathtools}
\tfrac {f(x)-1}{x-\frac 12}\ge -2\xRightarrow{\hspace{40pt}}\left(\forall \tfrac 12<x< 1\right):f(x)\ \ge\ 2-2x\tag{2}\]Differentiability at \(x=\frac 12\) implies \(f\equiv\begin{cases} 2x &0\le x\le 1/2\\ 2-2x &1/2<x\le 1\end{cases}\) is not possible.
Thus, using \((1)\) and \((2)\), reflecting about the line \(x=\frac 12\) if necessary, we may assume \(\exists 0<\alpha<\frac 12, \exists\eta>0:f(\alpha)=2\alpha+\eta\).
LMVT on \(0<x<\alpha\) and \(\alpha\):\[\usepackage{mathtools}
\tfrac {2\alpha+\eta-f(x)}{\alpha-x}\le 2\xRightarrow{\hspace{40pt}}\left(\forall 0<x<\alpha\right):f(x)\ \ge\ 2x+\eta\tag{3}\]\(\epsilon\big/\delta\) continuity at \(x=0\):\[(\forall \epsilon>0)(\exists\delta>0):0<x<\delta\xRightarrow{\hspace{40pt}}\big|f(x)\big|<\epsilon\tag{4} \]\((4)\) contradicts \((3)\) for \(0<\epsilon<\eta\).\(\usepackage{amsthm}
\qed\)
This post has been edited 1 time. Last edited by MS_asdfgzxcvb, Apr 11, 2025, 5:16 AM
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mqoi_KOLA
91 posts
#8
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thanks @MS_asdfgzxcvb :)
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Rohit-2006
242 posts
#9 • 1 Y
Y by mqoi_KOLA
Easy peasy....vai toone mujhe nehi bola.....ye le soln
$f$ is differentiable on $(0,1)$. For $c>0.5$ and $c<0.5$ you can hopefully do!!
For $c=0.5$ put the value of $c$ in the two equations,
$$f(x)=\begin{cases} 2x,\, &0\le x\le 1/2\\ 2-2x, &1/2<x\le 1.\end{cases}$$So clearly visible that not differentiable on $c=0.5$....
$LHD=2$ and $RHD=-2$
This post has been edited 1 time. Last edited by Rohit-2006, Apr 15, 2025, 4:27 PM
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mqoi_KOLA
91 posts
#11
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Rohit-2006 wrote:
Easy peasy....vai toone mujhe nehi bola.....ye le soln
$f$ is differentiable on $(0,1)$. For $c>0.5$ and $c<0.5$ you can hopefully do!!
For $c=0.5$ put the value of $c$ in the two equations,
$$f(x)=\begin{cases} 2x,\, &0\le x\le 1/2\\ 2-2x, &1/2<x\le 1.\end{cases}$$So clearly visible that not differentiable on $c=0.5$....
$LHD=2$ and $RHD=-2$

orz rohit ka question kal wale UGb mock mein aya tha (u solved that romanain grade 11 problem orz)
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mqoi_KOLA
91 posts
#12
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Rohit-2006 wrote:
Easy peasy....vai toone mujhe nehi bola.....ye le soln
$f$ is differentiable on $(0,1)$. For $c>0.5$ and $c<0.5$ you can hopefully do!!
For $c=0.5$ put the value of $c$ in the two equations,
$$f(x)=\begin{cases} 2x,\, &0\le x\le 1/2\\ 2-2x, &1/2<x\le 1.\end{cases}$$So clearly visible that not differentiable on $c=0.5$....
$LHD=2$ and $RHD=-2$

bro u only told to forget you :( :noo:
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