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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.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

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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
How many numbers
brokendiamond   0
4 minutes ago
How many 5-digit numbers can be formed using the digits 1, 3, 5, 7, 9 such that the smaller digits are not positioned between two larger digits?
0 replies
brokendiamond
4 minutes ago
0 replies
Sum of squares in 1865
Twoisaprime   2
N 10 minutes 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
10 minutes ago
USA 97 [1/(b^3+c^3+abc) + ... >= 1/(abc)]
Maverick   47
N 10 minutes 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
10 minutes ago
Inspired by old results
sqing   1
N 22 minutes 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
40 minutes ago
sqing
22 minutes ago
Orthocenter
jayme   7
N 23 minutes 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
23 minutes ago
something...
SunnyEvan   1
N 42 minutes 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
3 hours ago
SunnyEvan
42 minutes ago
2-var inequality
sqing   3
N an hour ago by sqing
Source: Own
Let $ a,b\geq 0    $. Prove that
$$ \frac{a }{a^2+2b^2+1}+ \frac{b }{b^2+2a^2+1}\leq \frac{1}{\sqrt{3}} $$$$   \frac{a }{2a^2+ b^2+2ab+1}+ \frac{b }{2b^2+ a^2+2ab+1}  \leq \frac{1}{\sqrt{5}} $$$$ \frac{a }{2a^2+ b^2+ ab+1}+ \frac{b }{2b^2+ a^2+ ab+1} \leq \frac{1}{2} $$$$\frac{a }{a^2+2b^2+2ab+1}+ \frac{b }{b^2+2a^2+2ab+1}\leq \frac{1}{2} $$
3 replies
1 viewing
sqing
2 hours ago
sqing
an hour ago
HK bisect QS
lssl   24
N an hour 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
an hour ago
Points in general position
AshAuktober   3
N 2 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
2 hours ago
Interesting inequalities
sqing   2
N 2 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
2 hours ago
IMO 2014 Problem 1
Amir Hossein   132
N 2 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
2 hours ago
IMO Genre Predictions
ohiorizzler1434   31
N 2 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
2 hours ago
weird FE
tobiSALT   10
N 2 hours ago by NicoN9
Source: Pan American Girls' Mathematical Olympiad 2024, P5
Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that
$f(f(x+y) - f(x)) + f(x)f(y) = f(x^2) - f(x+y),$

for all real numbers $x, y$.
10 replies
tobiSALT
Nov 27, 2024
NicoN9
2 hours ago
2015 solutions for quotient function!
raxu   49
N 3 hours ago by blueprimes
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
49 replies
raxu
Jun 26, 2015
blueprimes
3 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
1321 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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