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Sum of squares in 1865
Twoisaprime   2
N 2 hours ago by EmptyMachine
Source: 2024 CWMO P1
For positive integer $n$, note $S_n=1^{2024}+2^{2024}+ \cdots +n^{2024}$.
Prove that there exists infinitely many positive integers $n$, such that $S_n$ isn’t divisible by $1865$ but $S_{n+1}$ is divisible by $1865$
2 replies
Twoisaprime
Aug 6, 2024
EmptyMachine
2 hours ago
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USA 97 [1/(b^3+c^3+abc) + ... >= 1/(abc)]
Maverick   47
N 2 hours ago by justaguy_69
Source: USAMO 1997/5; also: ineq E2.37 in Book: Inegalitati; Authors:L.Panaitopol,V. Bandila,M.Lascu
Prove that, for all positive real numbers $ a$, $ b$, $ c$, the inequality
\[ \frac {1}{a^3 + b^3 + abc} + \frac {1}{b^3 + c^3 + abc} + \frac {1}{c^3 + a^3 + abc} \leq \frac {1}{abc} \]
holds.
47 replies
Maverick
Sep 12, 2003
justaguy_69
2 hours ago
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Inspired by old results
sqing   1
N 2 hours ago by sqing
Source: Own
Let $ a,b>0 $ and $ \frac{a}{b^2}+\frac{2\sqrt{3}}{2a^2+b^2}\leq 3\sqrt 3. $ Prove that$$a^2+2b^2\geq 1$$
1 reply
sqing
2 hours ago
sqing
2 hours ago
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Orthocenter
jayme   7
N 2 hours ago by Ianis
Dear Mathlinkers,

1. ABC an acuatangle triangle
2. H the orthcenter of ABC
3. DEF the orthic triangle of ABC
4. A* the midpoint of AH
5. X the point of intersection of AH and EF.

Prove : X is the orthocenter of A*BC.

Sincerely
Jean-Louis
7 replies
jayme
Mar 25, 2015
Ianis
2 hours ago
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something...
SunnyEvan   1
N 2 hours ago by SunnyEvan
Source: unknown
Try to prove : $$ \sum csc^{20} \frac{2^{i} \pi}{7} csc^{23} \frac{2^{j}\pi }{7} csc^{2023} \frac{2^{k} \pi}{7} $$is a rational number.
Where $ (i,j,k)=(1,2,3) $ and other permutations.
1 reply
SunnyEvan
4 hours ago
SunnyEvan
2 hours ago
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HK bisect QS
lssl   24
N 3 hours ago by LeYohan
Source: 1998 HK
In a concyclic quadrilateral $PQRS$,$\angle PSR=\frac{\pi}{2}$ , $H,K$ are perpendicular foot from $Q$ to sides $PR,RS$ , prove that $HK$ bisect segment$SQ$.
24 replies
lssl
Jan 5, 2012
LeYohan
3 hours ago
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Points in general position
AshAuktober   3
N 4 hours ago by blackbluecar
Source: 2025 Nepal ptst p1 of 4
Shining tells Prajit a positive integer $n \ge 2025$. Prajit then tries to place n points such that no four points are concyclic and no $3$ points are collinear in Euclidean plane, such that Shining cannot find a group of three points such that their circumcircle contains none of the other remaining points. Is he always able to do so?

(Prajit Adhikari, Nepal and Shining Sun, USA)
3 replies
AshAuktober
Mar 15, 2025
blackbluecar
4 hours ago
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Interesting inequalities
sqing   2
N 4 hours ago by sqing
Source: Own
Let $ a,b,c>0 $ and $ a+b\leq 16abc. $ Prove that
$$ a+b+kc\geq\sqrt{k}$$$$ a+b+kc^2\geq\frac{3\sqrt[3]{k}}{4}$$Where $ k>0. $
$$ a+b+c\geq1$$$$ a+b+4c\geq2$$$$ a+b+c^2\geq\frac{3}{4}$$$$ a+b+8c^2\geq\frac{3}{2}$$
2 replies
sqing
Yesterday at 12:23 PM
sqing
4 hours ago
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IMO 2014 Problem 1
Amir Hossein   132
N 4 hours ago by maromex
Let $a_0 < a_1 < a_2 < \dots$ be an infinite sequence of positive integers. Prove that there exists a unique integer $n\geq 1$ such that
\[a_n < \frac{a_0+a_1+a_2+\cdots+a_n}{n} \leq a_{n+1}.\]
Proposed by Gerhard Wöginger, Austria.
132 replies
Amir Hossein
Jul 8, 2014
maromex
4 hours ago
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IMO Genre Predictions
ohiorizzler1434   31
N 4 hours ago by ohiorizzler1434
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
31 replies
ohiorizzler1434
Saturday at 6:51 AM
ohiorizzler1434
4 hours ago
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Putnam 1947 B1
sqrtX   3
N Yesterday at 5:38 PM by Levieee
Source: Putnam 1947
Let $f(x)$ be a function such that $f(1)=1$ and for $x \geq 1$
$$f'(x)= \frac{1}{x^2 +f(x)^{2}}.$$Prove that
$$\lim_{x\to \infty} f(x)$$exists and is less than $1+ \frac{\pi}{4}.$
3 replies
sqrtX
Apr 3, 2022
Levieee
Yesterday at 5:38 PM
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Equivalent definition for C^1 functions
Ciobi_   1
N Apr 3, 2025 by KAME06
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.
1 reply
Ciobi_
Apr 2, 2025
KAME06
Apr 3, 2025
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Equivalent definition for C^1 functions
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Source: Romania NMO 2025 11.3
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Ciobi_
25 posts
Ciobi_
#1 h  Apr 2, 2025, 1:54 PM
Y by
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.
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KAME06
158 posts
KAME06
#2 h  Apr 3, 2025, 2:17 AM
Y by
Let $(x_n)_{n\geq 1},(y_n)_{n\geq 1}$ sequences that converge to $a$ and $x_n \neq y_n$ for any $n$.
First, note that $(x_n)$ or $(y_n)$ mustn't have an $n$ such that $x_n=a$ or $y_n=a$ (because $x_n \neq y_n$ and they are convergent to $a$), so WLOG consider $y_n \neq a$ for any $n$.
(a)$\rightarrow$(b):
If $f$ is differentiable, with continuous first derivative, then the function $\lim_{b \rightarrow x} \frac{f(x)-f(b)}{x-b}$ is continuous.
That implies that: $\lim_{x \rightarrow a} \frac{f(a)-f(b)}{a-b}$=$\lim_{n \rightarrow \infty}\left(\lim_{b \rightarrow x_n} \frac{f(x_n)-f(b)}{x_n-b}\right)$.
As $(x_n)$ and $(y_n)$ converge, for large enough $n$, $|x_n-a|<\frac{\epsilon}{2}$, $|y_n-a|<\frac{\epsilon}{2}$, so $|x_n-y_n|<\epsilon$.
That implies (and the fact that that $y_n \neq x_n$ for any $n$) if $n \rightarrow \infty$, then $y_n \rightarrow x_n$, so:
$\lim_{n \rightarrow \infty}\left(\lim_{b \rightarrow x_n} \frac{f(x_n)-f(b)}{x_n-b}\right)=\lim_{n \rightarrow \infty} \frac{f(x_n)-f(y_n)}{x_n-y_n}$. The limit exists.
We conclude that $\left(\frac{f(x_n)-f(y_n)}{x_n-y_n}\right)_{n\geq 1}$ is convergent.
(b) $\rightarrow$ (a):
As above: $\lim_{n \rightarrow \infty} \frac{f(x_n)-f(y_n)}{x_n-y_n}=\lim_{n \rightarrow \infty}\left(\lim_{b \rightarrow x_n} \frac{f(x_n)-f(b)}{x_n-b}\right)=\lim_{n \rightarrow \infty}\left(\lim_{b \rightarrow a} \frac{f(x_n)-f(b)}{a-b}\right)$.
$\lim_{n \rightarrow \infty}f(x_n)$ must be defined because the limit exist. If $f(x_n)$ doesn't converge to $f(a)$, notice that $b \rightarrow a$ and $a-b$ converges to $0$ and it can be as little as we want, so $\lim_{b \rightarrow a} \frac{f(x_n)-f(b)}{a-b}$ is $\infty$ or $-\infty$, but $\lim_{n \rightarrow \infty} \frac{f(x_n)-f(y_n)}{x_n-y_n}$ converges. Contradiction.
Then $f(x_n)$ converges to $f(a)$ and $b \rightarrow a$, so $\lim_{n \rightarrow \infty}\left(\lim_{b \rightarrow x_n} \frac{f(x_n)-f(b)}{x_n-b}\right)=\lim_{b \rightarrow a} \frac{f(a)-f(b)}{a-b}$.
We conclude that $f$ is differentiable and that $\lim_{b \rightarrow x} \frac{f(x)-f(b)}{x-b}$ is continuous.
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