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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.

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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
Only consecutive terms are coprime
socrates   37
N 7 minutes ago by blueprimes
Source: 7th RMM 2015, Problem 1
Does there exist an infinite sequence of positive integers $a_1, a_2, a_3, . . .$ such that $a_m$ and $a_n$ are coprime if and only if $|m - n| = 1$?
37 replies
socrates
Feb 28, 2015
blueprimes
7 minutes ago
Cyclic roots are not real, they can't hurt you
anantmudgal09   21
N 15 minutes ago by bjump
Source: INMO 2023 P2
Suppose $a_0,\ldots, a_{100}$ are positive reals. Consider the following polynomial for each $k$ in $\{0,1,\ldots, 100\}$:
$$a_{100+k}x^{100}+100a_{99+k}x^{99}+a_{98+k}x^{98}+a_{97+k}x^{97}+\dots+a_{2+k}x^2+a_{1+k}x+a_k,$$where indices are taken modulo $101$, i.e., $a_{100+i}=a_{i-1}$ for any $i$ in $\{1,2,\dots, 100\}$. Show that it is impossible that each of these $101$ polynomials has all its roots real.

Proposed by Prithwijit De
21 replies
anantmudgal09
Jan 15, 2023
bjump
15 minutes ago
Erasing the difference of two numbers
BR1F1SZ   2
N 30 minutes ago by sami1618
Source: Austria National MO Part 1 Problem 3
Consider the following game for a positive integer $n$. Initially, the numbers $1, 2, \ldots, n$ are written on a board. In each move, two numbers are selected such that their difference is also present on the board. This difference is then erased from the board. (For example, if the numbers $3,6,11$ and $17$ are on the board, then $3$ can be erased as $6 - 3=3$, or $6$ as $17 - 11=6$, or $11$ as $17 - 6=11$.)

For which values of $n$ is it possible to end with only one number remaining on the board?

(Michael Reitmeir)
2 replies
BR1F1SZ
Monday at 9:48 PM
sami1618
30 minutes ago
GCD of terms in a sequence
BBNoDollar   0
2 hours ago
Determine the general term of the sequence of non-zero natural numbers (a_n)n≥1, with the property that gcd(a_m, a_n, a_p) = gcd(m^2 ,n^2 ,p^2), for any distinct non-zero natural numbers m, n, p.

⁡Note that gcd(a,b,c) denotes the greatest common divisor of the natural numbers a,b,c .
0 replies
BBNoDollar
2 hours ago
0 replies
Number Theory
fasttrust_12-mn   13
N 2 hours ago by KTYC
Source: Pan African Mathematics Olympiad P1
Find all positive intgers $a,b$ and $c$ such that $\frac{a+b}{a+c}=\frac{b+c}{b+a}$ and $ab+bc+ca$ is a prime number
13 replies
fasttrust_12-mn
Aug 15, 2024
KTYC
2 hours ago
GCD of terms in a sequence
BBNoDollar   0
2 hours ago
Determine the general term of the sequence of non-zero natural numbers (a_n)n≥1, with the property that gcd(a_m, a_n, a_p) = gcd(m^2 ,n^2 ,p^2), for any distinct non-zero natural numbers m, n, p.

⁡Note that gcd(a,b,c) denotes the greatest common divisor of the natural numbers a,b,c .
0 replies
BBNoDollar
2 hours ago
0 replies
Aime type Geo
ehuseyinyigit   3
N 2 hours ago by sami1618
Source: Turkish First Round 2024
In a scalene triangle $ABC$, let $M$ be the midpoint of side $BC$. Let the line perpendicular to $AC$ at point $C$ intersect $AM$ at $N$. If $(BMN)$ is tangent to $AB$ at $B$, find $AB/MA$.
3 replies
ehuseyinyigit
Monday at 9:04 PM
sami1618
2 hours ago
minimizing sum
gggzul   1
N 3 hours ago by RedFireTruck
Let $x, y, z$ be real numbers such that $x^2+y^2+z^2=1$. Find
$$min\{12x-4y-3z\}.$$
1 reply
gggzul
4 hours ago
RedFireTruck
3 hours ago
Equilateral Triangle inside Equilateral Triangles.
abhisruta03   2
N 3 hours ago by Reacheddreams
Source: ISI 2021 P6
If a given equilateral triangle $\Delta$ of side length $a$ lies in the union of five equilateral triangles of side length $b$, show that there exist four equilateral triangles of side length $b$ whose union contains $\Delta$.
2 replies
abhisruta03
Jul 18, 2021
Reacheddreams
3 hours ago
USAMO 1984 Problem 5 - Polynomial of degree 3n
Binomial-theorem   8
N 3 hours ago by Assassino9931
Source: USAMO 1984 Problem 5
$P(x)$ is a polynomial of degree $3n$ such that

\begin{eqnarray*}
P(0) = P(3) = \cdots &=& P(3n) = 2, \\
P(1) = P(4) = \cdots &=& P(3n-2) = 1, \\
P(2) = P(5) = \cdots &=& P(3n-1) = 0, \quad\text{ and }\\
&& P(3n+1) = 730.\end{eqnarray*}

Determine $n$.
8 replies
Binomial-theorem
Aug 16, 2011
Assassino9931
3 hours ago
Finding positive integers with good divisors
nAalniaOMliO   2
N 3 hours ago by KTYC
Source: Belarusian National Olympiad 2025
For every positive integer $n$ write all its divisors in increasing order: $1=d_1<d_2<\ldots<d_k=n$.
Find all $n$ such that $2025 \cdot n=d_{20} \cdot d_{25}$.
2 replies
nAalniaOMliO
Mar 28, 2025
KTYC
3 hours ago
Balkan MO 2025 p1
Mamadi   1
N 3 hours ago by KevinYang2.71
Source: Balkan MO 2025
An integer \( n > 1 \) is called good if there exists a permutation \( a_1, a_2, \dots, a_n \) of the numbers \( 1, 2, 3, \dots, n \), such that:

\( a_i \) and \( a_{i+1} \) have different parities for every \( 1 \le i \le n - 1 \)

the sum \( a_1 + a_2 + \dots + a_k \) is a quadratic residue modulo \( n \) for every \( 1 \le k \le n \)

Prove that there exist infinitely many good numbers, as well as infinitely many positive integers which are not good.

Remark: Here an integer \( z \) is considered a quadratic residue modulo \( n \) if there exists an integer \( y \) such that \( y^2 \equiv z \pmod{n} \).
1 reply
Mamadi
5 hours ago
KevinYang2.71
3 hours ago
Random Points = Problem
kingu   4
N 4 hours ago by zuat.e
Source: Chinese Geometry Handout
Let $ABC$ be a triangle. Let $\omega$ be a circle passing through $B$ intersecting $AB$ at $D$ and $BC$ at $F$. Let $G$ be the intersection of $AF$ and $\omega$. Further, let $M$ and $N$ be the intersections of $FD$ and $DG$ with the tangent to $(ABC)$ at $A$. Now, let $L$ be the second intersection of $MC$ and $(ABC)$. Then, prove that $M$ , $L$ , $D$ , $E$ and $N$ are concyclic.
4 replies
kingu
Apr 27, 2024
zuat.e
4 hours ago
CooL geo
Pomegranat   2
N 4 hours ago by Curious_Droid
Source: Idk

In triangle \( ABC \), \( D \) is the midpoint of \( BC \). \( E \) is an arbitrary point on \( AC \). Let \( S \) be the intersection of \( AD \) and \( BE \). The line \( CS \) intersects with the circumcircle of \( ACD \), for the second time at \( K \). \( P \) is the circumcenter of triangle \( ABE \). Prove that \( PK \perp CK \).
2 replies
Pomegranat
Monday at 5:57 AM
Curious_Droid
4 hours ago
A game optimization on a graph
Assassino9931   3
N Apr 28, 2025 by dgrozev
Source: Bulgaria National Olympiad 2025, Day 2, Problem 6
Let \( X_0, X_1, \dots, X_{n-1} \) be \( n \geq 2 \) given points in the plane, and let \( r > 0 \) be a real number. Alice and Bob play the following game. Firstly, Alice constructs a connected graph with vertices at the points \( X_0, X_1, \dots, X_{n-1} \), i.e., she connects some of the points with edges so that from any point you can reach any other point by moving along the edges.Then, Alice assigns to each vertex \( X_i \) a non-negative real number \( r_i \), for \( i = 0, 1, \dots, n-1 \), such that $\sum_{i=0}^{n-1} r_i = 1$. Bob then selects a sequence of distinct vertices \( X_{i_0} = X_0, X_{i_1}, \dots, X_{i_k} \) such that \( X_{i_j} \) and \( X_{i_{j+1}} \) are connected by an edge for every \( j = 0, 1, \dots, k-1 \). (Note that the length $k \geq 0$ is not fixed and the first selected vertex always has to be $X_0$.) Bob wins if
\[
  \frac{1}{k+1} \sum_{j=0}^{k} r_{i_j} \geq r;
  \]otherwise, Alice wins. Depending on \( n \), determine the largest possible value of \( r \) for which Bob has a winning strategy.
3 replies
Assassino9931
Apr 8, 2025
dgrozev
Apr 28, 2025
A game optimization on a graph
G H J
G H BBookmark kLocked kLocked NReply
Source: Bulgaria National Olympiad 2025, Day 2, Problem 6
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Assassino9931
1324 posts
#1 • 1 Y
Y by cubres
Let \( X_0, X_1, \dots, X_{n-1} \) be \( n \geq 2 \) given points in the plane, and let \( r > 0 \) be a real number. Alice and Bob play the following game. Firstly, Alice constructs a connected graph with vertices at the points \( X_0, X_1, \dots, X_{n-1} \), i.e., she connects some of the points with edges so that from any point you can reach any other point by moving along the edges.Then, Alice assigns to each vertex \( X_i \) a non-negative real number \( r_i \), for \( i = 0, 1, \dots, n-1 \), such that $\sum_{i=0}^{n-1} r_i = 1$. Bob then selects a sequence of distinct vertices \( X_{i_0} = X_0, X_{i_1}, \dots, X_{i_k} \) such that \( X_{i_j} \) and \( X_{i_{j+1}} \) are connected by an edge for every \( j = 0, 1, \dots, k-1 \). (Note that the length $k \geq 0$ is not fixed and the first selected vertex always has to be $X_0$.) Bob wins if
\[
  \frac{1}{k+1} \sum_{j=0}^{k} r_{i_j} \geq r;
  \]otherwise, Alice wins. Depending on \( n \), determine the largest possible value of \( r \) for which Bob has a winning strategy.
This post has been edited 1 time. Last edited by Assassino9931, Apr 23, 2025, 4:03 PM
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ayeen_izady
32 posts
#2 • 3 Y
Y by GoodGuy2008, sami1618, dgrozev
Hopefully this is correct! I claim that the answer is: $\begin{cases}n=2k\implies r= \frac{1}{k(k+1)}\\n=2k+1\implies r=\frac{1}{(k+1)^2}\end{cases} $
WLOG assume that Alice makes a tree $T$ (a connected graph without cycles). Now use BFS starting from $X_0$. Let $S$ be the set of vertices of $T$ that are not connected to any vertex below them. I claim that in the best strategy for Alice(worst case scenario for Bob) any vertex $X_i\not\in S$ should be labeled with $r_i=0$. The proof of this claim is quite easy since when we have a vertex $X_i\not\in S$ such that $r_i=c>0$ we can turn $r_i$ into $0$ and add $c$ to vertices below $X_i$(maybe more than one layer) that are in $S$. Note that by doing this algorithm largest possible value for $r$ will be decreasing(not necessarily strictly decreasing). Hence we may assume that in the best strategy for Alice, our claim holds. Now I claim that the distance between any vertex in $S$ and $X_0$ should be a constant(every two vertices in $S$ should have equal distance to $X_0$). In order to prove this, assume that we have vertices $X_i,X_j\in S$ such that $d(X_i,X_0)>d(X_j,X_0)$. in this case by deleting $X_i$ and putting it between $X_0$ and the second layer of our BFS Tree we will decrease $r$. Thus by doing this process again and again we may assume that in the worst case scenario for Bob, all vertices in $S$ have equal distance to $X_0$. Here is an example for this process:
[asy] 
unitsize(2cm);
pair A,B,C,D,E,F,G; A=(0,0); B=(0.5,-1); C=(-0.5,-1); D=(-1,-2); E=(0,-2); F=(1,-2); G=(1,-3);
dot(A); label("$X_0$",A,dir(100)); dot(B); dot(C); dot(D); dot(E); dot(F); dot(G); draw(G--F--B--A);draw(A--C--D); draw(A--C--E); label("$X_i$",G,dir(270));
[/asy]
The above tree turns into:
[asy] 
unitsize(2cm);
pair A,B,C,D,E,F,G; A=(0,0); B=(0.5,-1); C=(-0.5,-1); D=(-1,-2); E=(0,-2); F=(1,-2); G=(0,1);
dot(A); label("$X_0$",G,dir(100)); dot(B); dot(C); dot(D); dot(E); dot(F); dot(G); draw(F--B--A--G);draw(A--C--D); draw(A--C--E); label("$X_i$",A,dir(180));
[/asy]
So we finally have that the distance between any two vertices of $S$ and $X_0$ is equal. Since they are equidistant we may assume that their labeling numbers are equal. Thus $r_{i_1}=r_{i_2}=\ldots r_{i_{|S|}}=\frac{1}{|S|}$. So the least value that Alice can bound is: $$\frac{1}{|S|}\times\frac{1}{d}\ge\frac{1}{|S|}\times\frac{1}{n+1-|S|}\ge\begin{cases}n=2k\implies r\ge \frac{1}{k(k+1)}\\n=2k+1\implies r\ge\frac{1}{(k+1)^2}\end{cases}$$Which we used AM-GM for the last equality. And the structure is that the graph will be a path of length $\lfloor\frac{n}{2}\rfloor-1$ which the last vertex is connected to $\lceil\frac{n}{2}\rceil$ other vertices. Here is an example for $n=8$:
[asy] 
unitsize(1.5cm);
pair A,B,C,D,E,F,G,H; pair A=(0,0);B=(0,-1);C=(0,-2);D=(0,-3);E=(1.5,-4);F=(0.5,-4);G=(-0.5,-4);H=(-1.5,-4);
dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);dot(G);dot(H); draw(A--B--C--D--F); draw(D--G); draw(D--H);draw(D--E);label("$r_1=0$",B,dir(180));label("$r_2=0$",C,dir(180));label("$r_3=0$",D,dir(180));label("$r_0=0$",A,dir(90));label("$r_4=\frac{1}{4}$",E,dir(270));
label("$r_5=\frac{1}{4}$",F,dir(270));label("$r_6=\frac{1}{4}$",G,dir(270));label("$r_7=\frac{1}{4}$",H,dir(270));
[/asy]
Which gives $r=\frac{1}{20}$ for $n=8$. Q.E.D $\blacksquare$
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dgrozev
2463 posts
#3 • 1 Y
Y by ayeen_izady
@above: It's ok, but why do you need a BFS tree? The job can be done using any spanning tree! Btw, your set $S$ is usually called "leaves".
This post has been edited 2 times. Last edited by dgrozev, Apr 28, 2025, 11:17 AM
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dgrozev
2463 posts
#4
Y by
(Ilya Bogdanov) The required maximum is
$$
\frac1{\bigl\lfloor\frac{n+1}2\bigr\rfloor\cdot\bigl\lceil\frac{n+1}2\bigr\rceil} = 
\begin{cases}
	\frac{4}{(n+1)^2} & \text{ if } n \text{ is odd,}\\
	\frac{4}{n(n+2)} & \text{ if } n \text{ is even,}
\end{cases}
$$and is achieved by the tree described at the end of the solution.

Consider the number $r_i$ Alice assigns to vertex $X_i$. If she replaces her graph by some of its spanning trees, that makes Bob's job just harder, so we assume she draws a tree.
Now, assume that she has drawn a graph and assigned some numbers $r_i$ to vertices. We show how to modify those numbers to make Bob's job not easier.
Consider every leaf $X_i$ with $i>0$, and assign it the sum of the numbers on the (unique) path from $X_0$ to $X_i$; all other numbers are replaced by zeroes. Then Bob's sum on every path does not increase. On the other hand, every number at a vertex is accounted for at least one leaf, so the sum of the numbers does not decrease. Now Alice may decrease the numbers at the leaves so as to fulfil the condition on the sum.

The problem now reads: Consider a tree on $n$ vertices rooted at $X_0$. Let $L_1,\dots,L_k$ be the leaves of this tree different from the root, and let $d_i$ be the number of vertices of the path from $X_0$ to $L_i$. We are to choose non-negative numbers $s_1,\dots,s_k$ adding up to $1$ so as to minimize the quantity
$$
 r=\max_{1\leq i\leq k} \frac{s_i}{d_i}.
 $$Let $d=\max_i d_i$; without loss of generality, let $d=d_1$. Then the path from $X_0$ to $L_1$ has $d-1$ vertices distinct from the $L_i$, so $k\leq n-(d-1)$. Hence
$$
 r\geq \frac1k \sum_{i=1}^k \frac{s_i}{d_i}\geq \frac 1{dk}\sum_{i=1}^k s_i=\frac1{dk}\geq \frac1{d(n-d+1)}\geq \frac1{\bigl\lfloor\frac{n+1}2\bigr\rfloor\cdot\bigl\lceil\frac{n+1}2\bigr\rceil}.
 $$Equality is achieved, if, say, $d=\left\lceil\frac{n+1}2\right\rceil$, and the graph consists of a path of length $d-1$, one of whose endpoints is $X_0$, and to the other $n-d+1=\left\lfloor\frac{n+1}2\right\rfloor$ leaves are attached. Each of those leaves should be assigned the number $1/(n-d+1)$, while all other vertices are assigned zeroes.

Remark. When I proposed this problem, I hadn't expected it would be selected as p6. One can see the originally proposed solution as well as some further comments in my blog.
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