Interesting inequality

by sqing, May 31, 2025, 2:54 AM

ai+aj is the multiple of n

by Jackson0423, May 30, 2025, 12:41 AM

Consider an strictly increasing sequence of integers \( a_n \).
For every positive integer \( n \), there exist indices \( 1 \leq i < j \leq n \) such that \( a_i + a_j \) is divisible by \( n \).
Given that \( a_1 \geq 1 \), find the minimum possible value of \( a_{100} \).
This post has been edited 1 time. Last edited by Jackson0423, 3 hours ago
Reason: Sf

Central sequences

by EeEeRUT, Apr 16, 2025, 1:37 AM

An infinite increasing sequence $a_1 < a_2 < a_3 < \cdots$ of positive integers is called central if for every positive integer $n$ , the arithmetic mean of the first $a_n$ terms of the sequence is equal to $a_n$.

Show that there exists an infinite sequence $b_1, b_2, b_3, \dots$ of positive integers such that for every central sequence $a_1, a_2, a_3, \dots, $ there are infinitely many positive integers $n$ with $a_n = b_n$.
This post has been edited 3 times. Last edited by EeEeRUT, May 11, 2025, 11:47 AM

Gcd of N and its coprime pair sum

by EeEeRUT, Apr 16, 2025, 1:33 AM

For a positive integer $N$, let $c_1 < c_2 < \cdots < c_m$ be all positive integers smaller than $N$ that are coprime to $N$. Find all $N \geqslant 3$ such that $$\gcd( N, c_i + c_{i+1}) \neq 1$$for all $1 \leqslant i \leqslant m-1$

Here $\gcd(a, b)$ is the largest positive integer that divides both $a$ and $b$. Integers $a$ and $b$ are coprime if $\gcd(a, b) = 1$.

Proposed by Paulius Aleknavičius, Lithuania
This post has been edited 3 times. Last edited by EeEeRUT, May 11, 2025, 11:49 AM

Parallelograms and concyclicity

by Lukaluce, Apr 14, 2025, 10:59 AM

Let $ABC$ be an acute triangle with incentre $I$ and $AB \neq AC$. Let lines $BI$ and $CI$ intersect the circumcircle of $ABC$ at $P \neq B$ and $Q \neq C$, respectively. Consider points $R$ and $S$ such that $AQRB$ and $ACSP$ are parallelograms (with $AQ \parallel RB, AB \parallel QR, AC \parallel SP$, and $AP \parallel CS$). Let $T$ be the point of intersection of lines $RB$ and $SC$. Prove that points $R, S, T$, and $I$ are concyclic.

Easy right-angled triangle problem

by gghx, Aug 3, 2024, 2:17 AM

In triangle $ABC$, $\angle B=90^\circ$, $AB>BC$, and $P$ is the point such that $BP=BC$ and $\angle APB=90^\circ$, where $P$ and $C$ lie on the same side of $AB$. Let $Q$ be the point on $AB$ such that $AP=AQ$, and let $M$ be the midpoint of $QC$. Prove that the line through $M$ parallel to $AP$ passes through the midpoint of $AB$.
This post has been edited 3 times. Last edited by gghx, Aug 3, 2024, 3:02 AM

<BAC = 2 <ABC wanted, AC + AI = BC given , incenter I

by parmenides51, Nov 21, 2020, 9:53 PM

In acute-angled triangle $ABC, I$ is the center of the inscribed circle and holds $| AC | + | AI | = | BC |$. Prove that $\angle BAC = 2 \angle ABC$.

Inequality with abc=1

by tenplusten, May 15, 2016, 7:48 AM

$\boxed{\text{A7}}$ Let $a,b,c$ be positive reals such that $abc=1$.Prove the inequality $\sum\frac{2a^2+\frac{1}{a}}{b+\frac{1}{a}+1}\geq 3$
This post has been edited 1 time. Last edited by tenplusten, May 15, 2016, 7:49 AM

IMO Shortlist 2014 C7

by hajimbrak, Jul 11, 2015, 8:38 AM

Let $M$ be a set of $n \ge 4$ points in the plane, no three of which are collinear. Initially these points are connected with $n$ segments so that each point in $M$ is the endpoint of exactly two segments. Then, at each step, one may choose two segments $AB$ and $CD$ sharing a common interior point and replace them by the segments $AC$ and $BD$ if none of them is present at this moment. Prove that it is impossible to perform $n^3 /4$ or more such moves.

Proposed by Vladislav Volkov, Russia
This post has been edited 1 time. Last edited by hajimbrak, Jul 23, 2015, 10:39 AM
Reason: Added proposer

China South East Mathematical Olympiad 2014 Q3B

by sqing, Aug 17, 2014, 12:46 AM

Let $p$ be a primes ,$x,y,z $ be positive integers such that $x<y<z<p$ and $\{\frac{x^3}{p}\}=\{\frac{y^3}{p}\}=\{\frac{z^3}{p}\}$.
Prove that $(x+y+z)|(x^5+y^5+z^5).$

"Do not worry too much about your difficulties in mathematics, I can assure you that mine are still greater." - Albert Einstein

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