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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
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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
xf(x + xy) = xf(x) + f(x^2)f(y)
orl   15
N 3 minutes ago by MathIQ.
Source: MEMO 2008, Team, Problem 5
Determine all functions $ f: \mathbb{R} \mapsto \mathbb{R}$ such that
\[ x f(x + xy) = x f(x) + f \left( x^2 \right) f(y) \quad  \forall  x,y \in \mathbb{R}.\]
15 replies
orl
Sep 10, 2008
MathIQ.
3 minutes ago
points on sides of a triangle, intersections, extensions, ratio of areas wanted
parmenides51   2
N 21 minutes ago by MathIQ.
Source: Mexican Mathematical Olympiad 1997 OMM P5
Let $P,Q,R$ be points on the sides $BC,CA,AB$ respectively of a triangle $ABC$. Suppose that $BQ$ and $CR$ meet at $A', AP$ and $CR$ meet at $B'$, and $AP$ and $BQ$ meet at $C'$, such that $AB' = B'C', BC' =C'A'$, and $CA'= A'B'$. Compute the ratio of the area of $\triangle PQR$ to the area of $\triangle ABC$.
2 replies
parmenides51
Jul 28, 2018
MathIQ.
21 minutes ago
integer functional equation
ABCDE   156
N 35 minutes ago by MathIQ.
Source: 2015 IMO Shortlist A2
Determine all functions $f:\mathbb{Z}\rightarrow\mathbb{Z}$ with the property that \[f(x-f(y))=f(f(x))-f(y)-1\]holds for all $x,y\in\mathbb{Z}$.
156 replies
ABCDE
Jul 7, 2016
MathIQ.
35 minutes ago
Three numbers cannot be squares simultaneously
WakeUp   38
N 38 minutes ago by SomeonecoolLovesMaths
Source: APMO 2011
Let $a,b,c$ be positive integers. Prove that it is impossible to have all of the three numbers $a^2+b+c,b^2+c+a,c^2+a+b$ to be perfect squares.
38 replies
WakeUp
May 18, 2011
SomeonecoolLovesMaths
38 minutes ago
How can I know the sequences's convergence value?
Madunglecha   3
N Today at 1:39 PM by Figaro
What is the convergence value of the sequence??
(n^2)*ln(n+1/n)-n
3 replies
Madunglecha
Today at 6:56 AM
Figaro
Today at 1:39 PM
Reduction coefficient
zolfmark   1
N Today at 1:26 PM by Mathzeus1024

find Reduction coefficient of x^10

in(1+x-x^2)^9
1 reply
zolfmark
Jul 17, 2016
Mathzeus1024
Today at 1:26 PM
Metric space
wiseman   3
N Today at 10:33 AM by alinazarboland
Source: IMS 2014 - Day1 - Problem4
Let $(X,d)$ be a metric space and $f:X \to X$ be a function such that $\forall x,y\in X : d(f(x),f(y))=d(x,y)$.
$\text{a})$ Prove that for all $x \in X$, $\lim_{n \rightarrow +\infty} \frac{d(x,f^n(x))}{n}$ exists, where $f^n(x)$ is $\underbrace{f(f(\cdots f(x)}_{n \text{times}} \cdots ))$.
$\text{b})$ Prove that the amount of the limit does not depend on choosing $x$.
3 replies
wiseman
Oct 2, 2014
alinazarboland
Today at 10:33 AM
Double integration
Tricky123   2
N Today at 9:34 AM by Mathzeus1024
Q)
\[\iint_{R} \sin(xy) \,dx\,dy, \quad R = \left[0, \frac{\pi}{2}\right] \times \left[0, \frac{\pi}{2}\right]\]
How to solve the problem like this I am using the substitution method but its seems like very complicated in the last
Please help me
2 replies
Tricky123
May 18, 2025
Mathzeus1024
Today at 9:34 AM
Unsolving differential equation
Madunglecha   3
N Today at 7:12 AM by solyaris
For parameter t
I made a differential equation :
y"=y*(x')^2
for here, '&" is derivate and second order derivate for t
could anyone tell me what is equation between y&x?
3 replies
Madunglecha
May 18, 2025
solyaris
Today at 7:12 AM
Prove the statement
Butterfly   11
N Today at 6:58 AM by solyaris
Given an infinite sequence $\{x_n\} \subseteq  [0,1]$, there exists some constant $C$, for any $r>0$, among the sequence $x_n$ and $x_m$ could be chosen to satisfy $|n-m|\ge r $ and $|x_n-x_m|<\frac{C}{|n-m|}$.
11 replies
Butterfly
May 7, 2025
solyaris
Today at 6:58 AM
External Direct Sum
We2592   0
Today at 2:45 AM
Q) 1. Let $V$ be external direct sum of vector spaces $U$ and $W$ over a field $\mathbb{F}$.let $\hat{U}={\{(u,0):u\in U\}}$ and $\hat{W}={\{(0,w):w\in W\}}$
show that
i) $\hat{U}$ and $\hat{W}$ is subspaces.
ii)$V=\hat{U}\oplus\hat{W}$

Q)2. Suppose $V=U+W$. Let $\hat{V}$ be the external direct sum of $U$ and $W$. show that $V$ is isomorphic to $\hat{V}$ under the correspondence $v=u+w\leftrightarrow(u,w)$

I face some trouble to solve this problems help me for understanding.
thank you.

0 replies
We2592
Today at 2:45 AM
0 replies
Definite integration
girishpimoli   2
N Yesterday at 11:59 PM by Amkan2022
If $\displaystyle g(t)=\int^{t^{2}}_{2t}\cot^{-1}\bigg|\frac{1+x}{(1+t)^2-x}\bigg|dx.$ Then $\displaystyle \frac{g(5)}{g(3)}$ is
2 replies
girishpimoli
Apr 6, 2025
Amkan2022
Yesterday at 11:59 PM
Putnam 1968 A6
sqrtX   11
N Yesterday at 11:47 PM by ohiorizzler1434
Source: Putnam 1968
Find all polynomials whose coefficients are all $\pm1$ and whose roots are all real.
11 replies
sqrtX
Feb 19, 2022
ohiorizzler1434
Yesterday at 11:47 PM
Affine variety
YamoSky   1
N Yesterday at 9:01 PM by amplreneo
Let $A=\left\{z\in\mathbb{C}|Im(z)\geq0\right\}$. Is it possible to equip $A$ with a finitely generated k-algebra with one generator such that make $A$ be an affine variety?
1 reply
YamoSky
Jan 9, 2020
amplreneo
Yesterday at 9:01 PM
Outcome related combinatorics problem
egxa   1
N Apr 29, 2025 by iliya8788
Source: All Russian 2025 10.7
A competition consists of $25$ sports, each awarding one gold medal to a winner. $25$ athletes participate, each in all $25$ sports. There are also $25$ experts, each of whom must predict the number of gold medals each athlete will win. In each prediction, the medal counts must be non-negative integers summing to $25$. An expert is called competent if they correctly guess the number of gold medals for at least one athlete. What is the maximum number \( k \) such that the experts can make their predictions so that at least \( k \) of them are guaranteed to be competent regardless of the outcome?
1 reply
egxa
Apr 18, 2025
iliya8788
Apr 29, 2025
Outcome related combinatorics problem
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G H BBookmark kLocked kLocked NReply
Source: All Russian 2025 10.7
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egxa
211 posts
#1
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A competition consists of $25$ sports, each awarding one gold medal to a winner. $25$ athletes participate, each in all $25$ sports. There are also $25$ experts, each of whom must predict the number of gold medals each athlete will win. In each prediction, the medal counts must be non-negative integers summing to $25$. An expert is called competent if they correctly guess the number of gold medals for at least one athlete. What is the maximum number \( k \) such that the experts can make their predictions so that at least \( k \) of them are guaranteed to be competent regardless of the outcome?
This post has been edited 1 time. Last edited by egxa, Apr 18, 2025, 5:21 PM
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iliya8788
8 posts
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We claim that the maximum $k$ is equal to $24$.
We correspond each athlete to a number from $1$ to $25$.
Define Podium as the $n$-tuple $(a_{1},a_{2},...,a_{25})$ where $a_{i}$ is the amount of medals won by athlete $i$.
First notice that none of the experts can predict the $n$-tuple $(1,1,...,1)$ because then if the podium is equal to $(25,0,0,...,0)$ then the most possible amount of judges predicting correctly is equal to $24$. From the previous statement it follows that one of the judges predicts a $n$-tuple which has a $0$. So we proceed with the following algorithm. Pick one of the prediction $n$-tuples and convert it with the following algorithm to a new $n$-tuple: turn anything that isn't equal to $0$ to $0$ and then spread the 25 medals among the zero's.
Now if the podium $n$-tuple is the same as this $n$-tuple then at least one of the experts isn't competent and so maximum $k$ is equal to $24$.
Now we will prove that no matter what the podium looks like we can achieve $k=24$.
So we propose the following predictions: $(24,1,0,0,...,0),(24,0,1,0,0,...,0),(24,0,0,1,0,0,...0),...,(24,0,0,...,1),(1,1,...,1,1,1)$.
Define the count of zeros of the podium as $C$.
Now 3 cases might happen:
$C>2$: then it's easy to notice that every expert except the last one is competent.
$C=2$: an expert isn't competent if and only if both zeros are in the same place as the $1$ and $24$. notice that no $2$ of the predictions have these 2 numbers in the same spot so worst case possible $24$ of the experts are competent.
$C=1$: notice that the podium $n$-tuple must have $23$, $1$'s and a single $2$ and a single $0$. an expert isn't competent if and only if one of the following 2 cases occur:
First case: the $1$ in the prediction is in the same place as the $0$ of the podium.
Second case: the $24$ in the prediction is in the same place as the $0$ of the podium and the $1$ in the prediction is in the same place as the $2$ of the podium.
notice that the position of the $1$ differs among the predictions so for a fixed podium only one expert can be not competent because of the first case. The same applies for the second case as well. Since the union of the positions of $1$ and $24$ differs among the predictions there can't be 2 experts that are incompetent because of the second case. Also both cases can't make experts incompetent simultaneously since in case of both cases happening the $1$ and $24$ must basically switch places but in each of the first $24$ predictions $24$ is not in the first spot and all of the $1$'s are in the first spot. So only one of these $24$ experts can be incompetent and the last expert is competent so at least $24$ of the experts are competent. $\blacksquare$
This post has been edited 5 times. Last edited by iliya8788, Apr 29, 2025, 12:11 PM
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