G
Topic
First Poster
Last Poster
Classic complex number geo
Ciobi_   1
N an hour ago by TestX01
Source: Romania NMO 2025 10.1
Let $M$ be a point in the plane, distinct from the vertices of $\triangle ABC$. Consider $N,P,Q$ the reflections of $M$ with respect to lines $AB, BC$ and $CA$, in this order.
a) Prove that $N, P ,Q$ are collinear if and only if $M$ lies on the circumcircle of $\triangle ABC$.
b) If $M$ does not lie on the circumcircle of $\triangle ABC$ and the centroids of triangles $\triangle ABC$ and $\triangle NPQ$ coincide, prove that $\triangle ABC$ is equilateral.
1 reply
Ciobi_
Today at 12:56 PM
TestX01
an hour ago
The greatest length of a sequence that satisfies a special condition
EmersonSoriano   0
an hour ago
Source: 2018 Peru TST Cono Sur P9
Find the largest possible value of the positive integer $N$ given that there exist positive integers $a_1, a_2, \dots, a_N$ satisfying
$$ a_n = \sqrt{(a_{n-1})^2 + 2018 \, a_{n-2}}\:, \quad \text{for } n = 3,4,\dots,N. $$
0 replies
EmersonSoriano
an hour ago
0 replies
Olympiad Geometry problem-second time posting
kjhgyuio   5
N an hour ago by kjhgyuio
Source: smo problem
In trapezium ABCD,AD is parallel to BC and points E and F are midpoints of AB and DC respectively. If
Area of AEFD/Area of EBCF =√3 + 1/3-√3 and the area of triangle ABD is √3 .find the area of trapezium ABCD
5 replies
kjhgyuio
Today at 1:03 AM
kjhgyuio
an hour ago
Summing the GCD of a number and the divisors of another.
EmersonSoriano   0
an hour ago
Source: 2018 Peru TST Cono Sur P8
For each pair of positive integers $m$ and $n$, we define $f_m(n)$ as follows:
$$ f_m(n) = \gcd(n, d_1) + \gcd(n, d_2) + \cdots + \gcd(n, d_k), $$where $1 = d_1 < d_2 < \cdots < d_k = m$ are all the positive divisors of $m$. For example,
$f_4(6) = \gcd(6,1) + \gcd(6,2) + \gcd(6,4) = 5$.

$a)\:$ Find all positive integers $n$ such that $f_{2017}(n) = f_n(2017)$.

$b)\:$ Find all positive integers $n$ such that $f_6(n) = f_n(6)$.
0 replies
EmersonSoriano
an hour ago
0 replies
Sum of whose elements is divisible by p
nntrkien   42
N an hour ago by cubres
Source: IMO 1995, Problem 6, Day 2, IMO Shortlist 1995, N6
Let $ p$ be an odd prime number. How many $ p$-element subsets $ A$ of $ \{1,2,\dots,2p\}$ are there, the sum of whose elements is divisible by $ p$?
42 replies
nntrkien
Aug 8, 2004
cubres
an hour ago
kind of well known?
dotscom26   3
N an hour ago by Svenskerhaor
Source: MBL
Let $ y_1, y_2, ..., y_{2025}$ be real numbers satisfying
$
y_1^2 + y_2^2 + \cdots + y_{2025}^2 = 1.
$
Find the maximum value of
$
|y_1 - y_2| + |y_2 - y_3| + \cdots + |y_{2025} - y_1|.
$

I have seen many problems with the same structure, Id really appreciate if someone could explain which approach is suitable here
3 replies
dotscom26
Yesterday at 4:11 AM
Svenskerhaor
an hour ago
Locus of a point on the side of a square
EmersonSoriano   0
an hour ago
Source: 2018 Peru TST Cono Sur P7
Let $ABCD$ be a fixed square and $K$ a variable point on segment $AD$. The square $KLMN$ is constructed such that $B$ is on segment $LM$ and $C$ is on segment $MN$. Let $T$ be the intersection point of lines $LA$ and $ND$. Find the locus of $T$ as $K$ varies along segment $AD$.
0 replies
EmersonSoriano
an hour ago
0 replies
Chess queens on a cylindrical board
EmersonSoriano   0
an hour ago
Source: 2018 Peru TST Cono Sur P6
Let $n$ be a positive integer. In an $n \times n$ board, two opposite sides have been joined, forming a cylinder. Determine whether it is possible to place $n$ queens on the board such that no two threaten each other when:

$a)\:$ $n=14$.

$b)\:$ $n=15$.
0 replies
EmersonSoriano
an hour ago
0 replies
2015 solutions for quotient function!
raxu   48
N 2 hours ago by zuat.e
Source: TSTST 2015 Problem 5
Let $\varphi(n)$ denote the number of positive integers less than $n$ that are relatively prime to $n$. Prove that there exists a positive integer $m$ for which the equation $\varphi(n)=m$ has at least $2015$ solutions in $n$.

Proposed by Iurie Boreico
48 replies
raxu
Jun 26, 2015
zuat.e
2 hours ago
GCD of x^2-y, y^2-z and z^2-x
EmersonSoriano   0
2 hours ago
Source: 2018 Peru TST Cono Sur P5
Find all positive integers $d$ that can be written in the form
$$ d = \gcd(|x^2 - y| , |y^2 - z| , |z^2 - x|), $$where $x, y, z$ are pairwise coprime positive integers such that $x^2 \neq y$, $y^2 \neq z$, and $z^2 \neq x$.
0 replies
EmersonSoriano
2 hours ago
0 replies
Two permutations
Nima Ahmadi Pour   11
N Jan 30, 2024 by Bataw
Source: Iran prepration exam
Suppose that $ a_1$, $ a_2$, $ \ldots$, $ a_n$ are integers such that $ n\mid a_1 + a_2 + \ldots + a_n$.
Prove that there exist two permutations $ \left(b_1,b_2,\ldots,b_n\right)$ and $ \left(c_1,c_2,\ldots,c_n\right)$ of $ \left(1,2,\ldots,n\right)$ such that for each integer $ i$ with $ 1\leq i\leq n$, we have
\[ n\mid a_i - b_i - c_i
\]

Proposed by Ricky Liu & Zuming Feng, USA
11 replies
Nima Ahmadi Pour
Apr 24, 2006
Bataw
Jan 30, 2024
Two permutations
G H J
Source: Iran prepration exam
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Nima Ahmadi Pour
160 posts
#1 • 2 Y
Y by Adventure10 and 1 other user
Suppose that $ a_1$, $ a_2$, $ \ldots$, $ a_n$ are integers such that $ n\mid a_1 + a_2 + \ldots + a_n$.
Prove that there exist two permutations $ \left(b_1,b_2,\ldots,b_n\right)$ and $ \left(c_1,c_2,\ldots,c_n\right)$ of $ \left(1,2,\ldots,n\right)$ such that for each integer $ i$ with $ 1\leq i\leq n$, we have
\[ n\mid a_i - b_i - c_i
\]

Proposed by Ricky Liu & Zuming Feng, USA
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ZetaX
7579 posts
#2 • 3 Y
Y by Adventure10, Mango247, and 1 other user
This problem and the following generalisation appeared 1979 in Ars Combinatoria (thanks to Darij who found it):

Let $ (G, + )$ be a finite abelian group of order $ n$.
Let also $ a_1,a_2,...,a_{n - 1} \in G$ be arbitrary.
Then there exist pairwise distinct $ b_1,b_2,...,b_{n - 1} \in G$ and pairwise distinct $ c_1,c_2,...,c_{n - 1} \in G$ such that $ a_k = b_k + c_k$ for $ k = 1,2,...,n - 1$.

[Moderator edit: The Ars Combinatoria paper is:
F. Salzborn, G. Szekeres, A problem in Combinatorial Group Theory, Ars Combinatoria 7 (1979), pp. 3-5.]
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
epitomy01
240 posts
#3 • 3 Y
Y by Dan37kosothangnao, Adventure10, Mango247
so could someone post a proof of either the problem or its generalization?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
spdf
136 posts
#4 • 1 Y
Y by Adventure10
you can find the proof in file shortlist 2005 which has been posted by orl
The main idea is given two sequence $a_{1}...a_{n}$ and $b_{1}...b_{n}$ s.t $\sum a_{i}\equiv 0(mod n)$ and $\sum b_{i}\equiv 0(mod n)$ and there are exactly two i;j s.t $a_{i}\neq\ b_{i}(modn)$ and $a_{j}\neq\ b_{j}(modn)$.Then if we know two permutation good for the sequence (a_1...a_n) then we can build two permuttionm good for (b_1...b_n)
i will come back with detail if you need
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ZetaX
7579 posts
#5 • 1 Y
Y by Adventure10
Well, I will post the solution from Ars Combinatoria if a re-find that two sheets of paper...
It's a bit different from the ISL one.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
keira_khtn
485 posts
#6 • 2 Y
Y by Adventure10, Mango247
I think you didnt keep promise,Zetax :lol: Please post it here and now!
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
bodom
123 posts
#7 • 2 Y
Y by Adventure10, Mango247
to spdf: that was also my idea when i first saw the problem but i can't find a good way to contruct those 2 permutations for $ b_j$.you said you can post details.please do so :)
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ZetaX
7579 posts
#8 • 2 Y
Y by Adventure10, Mango247
Sorry for not repsonding (I merely forgot...). But I just saw that problem again: In a slightly different manner (but being equivalent to this one here) it is solved in "The Mathematics of Juggling", called the "Converse of the Average Theorem".

Main ideas:
You show that this property (being a sum of two permutations) is invariant under the operations $ a_{i,j,d}$ that add $ d$ to $ a_i$ and subtract $ d$ from $ a_j$.
For this, you need to do it algorithmically (but describing it is a bit hard without that graphics...).
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
arandomperson123
430 posts
#9 • 2 Y
Y by Adventure10, Mango247
ZetaX wrote:
Sorry for not repsonding (I merely forgot...). But I just saw that problem again: In a slightly different manner (but being equivalent to this one here) it is solved in "The Mathematics of Juggling", called the "Converse of the Average Theorem".

Main ideas:
You show that this property (being a sum of two permutations) is invariant under the operations $ a_{i,j,d}$ that add $ d$ to $ a_i$ and subtract $ d$ from $ a_j$.
For this, you need to do it algorithmically (but describing it is a bit hard without that graphics...).

that is what I tried to do, but I can not prove that we can do it for the general case... can someone please help?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ThE-dArK-lOrD
4071 posts
#10 • 16 Y
Y by Lam.DL.01, Mosquitall, nmd27082001, Arc_archer, MathbugAOPS, iceillusion, Aryan-23, magicarrow, k12byda5h, gabrupro, Mop2018, Adventure10, Mango247, bhan2025, CyclicISLscelesTrapezoid, winniep008hfi
Since it's almost twelve years without complete solution, here's the official solution:

Suppose there exists permutations $\sigma$ and $\tau$ of $[n]$ for some sequence $\{ a_i\}_{i\in [n]}$ so that $a_i\equiv_n \sigma (i)+\tau (i)$ for all $i\in [n]$.
Given a sequence $\{ b_i\}_{i\in [n]}$ with sum divisible by $n$ that differ, in modulo $n$, from $\{ a_i\}_{i\in [n]}$ in only two positions, say $i_1$ and $i_2$.
We want to construct permutations $\sigma'$ and $\tau'$ of $[n]$ so that $b_i\equiv_n \sigma' (i) +\tau' (i)$ for all $i\in [n]$.
Recall that $b_i\equiv a_i\pmod{n}$ for all $i\in [n]$ that $i\neq i_1,i_2$.
Construct a sequence $i_1,i_2,i_3,...$ by, for each integer $k\geq 2$, define $i_{k+1}\in [n]$ to be the unique integer satisfy $\sigma (i_{k-1})+\tau (i_{k+1})\equiv_n b_{i_k}$.
Let (clearly exists) $p<q$ are the indices that $i_p=i_q$ with minimal $p$, and then minimal $q$.

If $p>2$. This means $i_j\not\in \{ i_1,i_2\} \implies \sigma (i_j) +\tau (i_j) \equiv_n b_{i_j}$ for all $j\in \{ p,p+1,...,q\}$.
Summing the equation $\sigma (i_{k-1})+\tau (i_{k+1})\equiv_n b_{i_k}$ for $k\in \{ p,p+1,...,q-1\}$ gives us
$$\sum_{j=p-1}^{q-2}{\sigma (i_j) } +\sum_{j=p+1}^{q}{\tau (i_j)} \equiv_n\sum_{j=p}^{q-1}{b_{i_j}} \implies \sigma (i_{p-1}) +\sigma (i_p) +\tau (i_{q-1}) +\tau (i_q) \equiv_nb_{i_p}+b_{i_{q-1}}.$$Plugging $i_p=i_q$ and use $\sigma (i_p) +\tau (i_p)\equiv_n b_{i_p}$ gives us $\sigma (i_{p-1}) +\tau (i_{q-1})\equiv_n b_{i_{q-1}} \equiv_n \sigma (i_{q-1})+\tau (i_{q-1})$.
Hence, $\sigma (i_{p-1}) \equiv_n \sigma (i_{q-1})\implies i_{p-1}=i_{q-1}$, contradiction to the definition of $p,q$.

So, we've $p\in \{ 1,2\}$. Let $p'=3-p$. Define the desired permutations $\sigma'$ and $\tau'$ as follows:
$$\sigma' (i_l)=\begin{cases} 
\sigma (i_{l-1}), & \text{ if } l\in \{ 2,3,...,q-1\} \\
\sigma (i_{q-1}), & \text{ if } l=1
\end{cases} ,\tau' (i_l)= \begin{cases} 
\tau (i_{l+1}), & \text{ if } l\in \{ 2,3,...,q-1\} \\
\tau (i_{p'}), & \text{ if } l=1
\end{cases}  $$and $\sigma' (i) =\sigma (i),\tau' (i)=\tau (i)$ for the rest $i\in [n]$ that $i\not\in \{ i_1,i_2,...,i_{q-1}\}$.
Note that the reason we choose $\tau (i_{p'})$ is just to not use $\tau (i_p)=\tau (i_{(q-1)+1})$ more than one time.
This construction gives us $\sigma' (i)+\tau' (i)\equiv_n b_i$ for all $i\in [n]$ except when $i=i_1$.
But since both $\sigma'$ and $\tau'$ are permutations of $[n]$, we've $\sum_{i\in [n]}{(\sigma' (i)+\tau' (i))} \equiv_n 2\times \frac{n(n-1)}{2}\equiv_n 0\equiv_n \sum_{i\in [n]}{b_i}$.
This guarantee that $\sigma' (i) +\tau' (i)\equiv_n b_i$ when $i=i_1$ too. This prove the validity of permutations we constructed.
This post has been edited 3 times. Last edited by ThE-dArK-lOrD, Jan 16, 2018, 3:07 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
mathleticguyyy
3217 posts
#11
Y by
The case with $n$ prime is also resolved in this paper
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Bataw
43 posts
#13
Y by
any other solutions ?
Z K Y
N Quick Reply
G
H
=
a