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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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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
Inspired by SXJX (12)2022 Q1167
sqing   2
N 10 minutes ago by pooh123
Source: Own
Let $ a,b,c>0 $. Prove that$$\frac{kabc-1} {abc(a+b+c+8(2k-1))}\leq \frac{1}{16 }$$Where $ k>\frac{1}{2}.$
2 replies
sqing
Yesterday at 4:01 AM
pooh123
10 minutes ago
Interesting inequalities
sqing   1
N 14 minutes ago by pooh123
Source: Own
Let $ a,b>0 $. Prove that
$$\frac{ab-1} {ab(a+b+2)} \leq \frac{1} {8}$$$$\frac{2ab-1} {ab(a+b+1)} \leq 6\sqrt 3-10$$
1 reply
sqing
Today at 4:24 AM
pooh123
14 minutes ago
Inequality with rational function
MathMystic33   4
N 15 minutes ago by ytChen
Source: Macedonian Mathematical Olympiad 2025 Problem 2
Let \( n > 2 \) be an integer, \( k > 1 \) a real number, and \( x_1, x_2, \ldots, x_n \) be positive real numbers such that \( x_1 \cdot x_2 \cdots x_n = 1 \). Prove that:

\[
\frac{1 + x_1^k}{1 + x_2} + \frac{1 + x_2^k}{1 + x_3} + \cdots + \frac{1 + x_n^k}{1 + x_1} \geq n.
\]
When does equality hold?
4 replies
MathMystic33
May 13, 2025
ytChen
15 minutes ago
Combinatorics: Easy or tough?
AlexLewandowski   3
N 22 minutes ago by FrancoGiosefAG
Source: Mexican MO 2000
A board $n$×$n$ is coloured black and white like a chessboard. The following steps are permitted: Choose a rectangle inside the board (consisting of entire cells)whose side lengths are both odd or both even, but not both equal to $1$, and invert the colours of all cells inside the rectangle. Determine the values of $n$ for which it is possible to make all the cells have the same colour in a finite number of such steps.

3 replies
AlexLewandowski
Feb 6, 2017
FrancoGiosefAG
22 minutes ago
D1033 : A problem of probability for dominoes 3*1
Dattier   1
N 27 minutes ago by Dattier
Source: les dattes à Dattier
Let $G$ a grid of 9*9, we choose a little square in $G$ of this grid three times, we can choose three times the same.

What the probability of cover with 3*1 dominoes this grid removed by theses little squares (one, two or three) ?
1 reply
Dattier
May 15, 2025
Dattier
27 minutes ago
inequality with roots
luci1337   2
N 39 minutes ago by Nguyenhuyen_AG
Source: an exam in vietnam
let $a,b,c\geq0: a+b+c=1$
prove that $\Sigma\sqrt{a+(b-c)^2}\geq\sqrt{3}$
2 replies
luci1337
May 18, 2025
Nguyenhuyen_AG
39 minutes ago
Very hard
steven_zhang123   0
an hour ago
Source: An article
Given a positive integer \(n\) and a positive real number \(m\), non-negative real numbers \(a_1, a_2, \cdots, a_n\) satisfy \(\sum_{i=1}^n a_i = m\). Define \(N\) as the number of elements in the following collection of subsets:

$$
\left\{ I \subseteq \{1, 2, \cdots, n\} : \prod_{i \in I} a_i \geq 1 \right\}.
$$
Find the maximum possible value of \(N\).
0 replies
steven_zhang123
an hour ago
0 replies
Interesting Inequality
lbh_qys   1
N an hour ago by nexu
Let $ a, b, c$ be real numbers such that $ (3a-2b-c)(3b-2c-a)(3c-2a-b)\neq 0 $ and $ a + b + c = 3 . $ Prove that
$$ \left( \frac{1}{3a - 2b - c} + \frac{1}{3b - 2c - a} + \frac{1}{3c - 2a - b} \right)^2 + a^2 + b^2 + c^2 \geq 3 + 3\sqrt{\frac 27}$$
1 reply
lbh_qys
3 hours ago
nexu
an hour ago
Goofy geometry
giangtruong13   1
N an hour ago by nabodorbuco2
Source: A Specialized School's Math Entrance Exam
Given the circle $(O)$, from $A$ outside the circle, draw tangents $AE,AF$ ($E,F$ are tangential points) and secant $ABC$ ($B,C$ lie on circle $O$, $B$ is between $A$ and $C$). $OA$ intersects $EF$ at $H$; $I$ is midpoint of $BC$. The line crossing $I$, paralleling with $CE$, intersects $EF$ at $D$. $CD$ intersects $AE$ at $K$. Let $N$ lie inside the triangle $FBC$ such that: $AF$=$AN$. From $N$ draw chords $BQ$, $RC$, $FP$ on circle $(O)$. Prove that: $PRQ$ is a isosceles triangle
1 reply
giangtruong13
Yesterday at 4:23 PM
nabodorbuco2
an hour ago
Rainbow vertices
goodar2006   3
N 2 hours ago by Sina_Sa
Source: Iran 3rd round 2012-Combinatorics exam-P1
We've colored edges of $K_n$ with $n-1$ colors. We call a vertex rainbow if it's connected to all of the colors. At most how many rainbows can exist?

Proposed by Morteza Saghafian
3 replies
goodar2006
Sep 20, 2012
Sina_Sa
2 hours ago
Geometry hard problem.
noneofyou34   1
N 2 hours ago by User21837561
In a circle of radius R, three chords of length R are given. Their ends are joined with segments to
obtain a hexagon inscribed in the circle. Show that the midpoints of the new chords are the vertices of
an equilateral triang
1 reply
noneofyou34
3 hours ago
User21837561
2 hours ago
Prove that two different boards can be obtained
hectorleo123   2
N 2 hours ago by alpha31415
Source: 2014 Peru Ibero TST P2
Let $n\ge 4$ be an integer. You have two $n\times n$ boards. Each board contains the numbers $1$ to $n^2$ inclusive, one number per square, arbitrarily arranged on each board. A move consists of exchanging two rows or two columns on the first board (no moves can be made on the second board). Show that it is possible to make a sequence of moves such that for all $1 \le i \le n$ and $1 \le j \le n$, the number that is in the $i-th$ row and $j-th$ column of the first board is different from the number that is in the $i-th$ row and $j-th$ column of the second board.
2 replies
hectorleo123
Sep 15, 2023
alpha31415
2 hours ago
Inspired by lbh_qys
sqing   2
N 3 hours ago by JARP091
Source: Own
Let $ a, b $ be real numbers such that $ (a-3)(b-3)(a-b)\neq 0 $ and $ a + b =6 . $ Prove that
$$ \left( \frac{a}{b - 3} + \frac{b}{3 - a} + \frac{3}{a - b} \right)^2 + 2(a^2 + b^2 )\geq54$$Equality holds when $ (a,b)=\left(\frac{3}{2},\frac{9}{2}\right)$
$$\left( \frac{a+1}{b - 3} + \frac{b+1}{3 - a} + \frac{4}{a - b} \right)^2 + 2(a^2 + b^2 )\geq 60$$Equality holds when $ (a,b)=\left(3-\sqrt 3,3+\sqrt 3\right)$
$$  \left( \frac{a+3}{b - 3} + \frac{b+3}{3 - a} + \frac{6}{a - b} \right)^2 + 2\left(a^2 + b^2\right)\geq 72$$Equality holds when $ (a,b)=\left(3-\frac{3}{\sqrt 2},3+\frac{3}{\sqrt 2}\right)$
2 replies
sqing
3 hours ago
JARP091
3 hours ago
Interesting Inequality
lbh_qys   4
N 3 hours ago by lbh_qys
Given that \( a, b, c \) are pairwise distinct $\mathbf{real}$ numbers and \( a + b + c = 9 \), find the minimum value of

\[
\left( \frac{a}{b - c} + \frac{b}{c - a} + \frac{c}{a - b} \right)^2  + a^2 + b^2 + c^2.
\]
4 replies
1 viewing
lbh_qys
Today at 5:42 AM
lbh_qys
3 hours ago
Subsets of points lying in disks
oVlad   4
N Apr 11, 2025 by Andyexists
Source: Romania TST 2023 Day 2 P4
Fix a positive integer $n.{}$ Consider an $n{}$-point set $S{}$ in the plane. An eligible set is a non-empty set of the form $S\cap D,{}$ where $D$ is a closed disk in the plane. In terms of $n,$ determine the smallest possible number of eligible subsets $S{}$ may contain.

Proposed by Cristi Săvescu
4 replies
oVlad
Apr 7, 2024
Andyexists
Apr 11, 2025
Subsets of points lying in disks
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G H BBookmark kLocked kLocked NReply
Source: Romania TST 2023 Day 2 P4
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oVlad
1746 posts
#1
Y by
Fix a positive integer $n.{}$ Consider an $n{}$-point set $S{}$ in the plane. An eligible set is a non-empty set of the form $S\cap D,{}$ where $D$ is a closed disk in the plane. In terms of $n,$ determine the smallest possible number of eligible subsets $S{}$ may contain.

Proposed by Cristi Săvescu
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Randomization
40 posts
#2
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Maybe problem is near to trivialize by BAMO 2020/5 problem....
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YaoAOPS
1541 posts
#3 • 1 Y
Y by FairyBlade
Interesting

We claim that there are at minimum $\binom{n+1}{2}$ such subsets.

Claim: This can be constructed.
Proof. A construction follows by taking all of $S$ on a line, which gives $\binom{n}{2}$ subsets when there are more than two elements in it, and $n$ more.
This gives a total of $n + \binom{n}{2} = \binom{n+1}{2}$. $\blacksquare$

Claim: For a set $S$, we can generate $n+1$ distinct pairs of eligible sets $(A, A')$ where $A' = A \cup \{O\}$.
Proof. Let one point $O$ of $S$ have minimal $x$ coordinate (rotate the plane until its uniquely minimal).
Take a line that goes through $O$.
By taking a sufficiently large circle, it can approximate a half plane bordering the line. First have this half plane containg on elements.
Now rotate the line with pivot $O$. Suppose that when sorted by when they hit the ray, the elements can be ordered as sets $T_1, T_2, \dots$ where $T_i$ is a set of point(s) collinear with $O$. Then if $T_i = \{P\}$ has size one, we get a pair by considering when point $P$ is added to the set, and when the ray is perturbed such that $O$ isn't in the half plane.
In the case where $T_i$ has size more than one, then we get $|T_i|$ pairs by considering half planes that contain the $k$th elements closest to $O$ for $1 \le k \le |T_i|$. We can once again guarentee such a perturbation to not have $O$ in the subsets. $\blacksquare$

Claim: This is minimal.
Proof. We prove this inductively, the base case of $n = 2$ is obvious.
Now suppose that we have a set $S$ of $n + 1$ points. Fix $O$ as before. Then $S \setminus \{O\}$ inductively has at least $\binom{n+1}{2}$ eligible sets.
For each such eligible set, take a representation of it. This creates a matching from eligible sets on $S \setminus \{O\}$ to $S$ which agree on all elements but perhaps $O$.
For each pair, then take the element of the pair which disagrees on $O$. This gives that $S$ has at least $\binom{n+1}{2} + n+1 = \binom{n+2}{2}$. $\blacksquare$
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matematica007
17 posts
#4
Y by
This is my solution.
The minimum is $\binom{n+1}{2}$ .
We will initially show that for any initial configuration of $S$ we have at least $\binom{n+1}{2}$ eligible subsets.
Claim : For every point in S there is a circle that passes through it and does not pass through the other $n-1$ points.

Let $S={ A_1,A_2,\ldots,A_n }$.We will prove the claim for $A_1$. We consider $l$ a straight line that passes through $A_1$ and does not pass through the other n-1 points.We consider $O$ a point on $l$ such that $O$ is not on the perpendicular bisector of $A_1A_i$ for every i from 2 to n.
Now the circle with the center $O$ that passes through $A_1$ respects the condition in the claim .So the claim is proved.

Analogously, we will prove that for any pair of 2 points in $S$ there is a circle that passes through them and does not pass through the other $n-2$ points.So we have at least $n+\binom{n}{2}=\binom{n+1}{2}$ eligible subsets of $S$.

Now we will build an example in which the minimum is reached.

Thus we consider $n$ distinct collinear points .So the configuration reaches the minimum.
This post has been edited 1 time. Last edited by matematica007, Apr 9, 2024, 4:31 PM
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Andyexists
7 posts
#5
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Well

There's at least $\binom{n+1}{2}$ eligible subsets. For the intuition, we will first generate the example.

Take $n$ points all in a line, numbered $1, 2, \dots, n$. If an eligible subset $D$ has "lowest" point $a$ and "greatest" point $b$, then segment $ab$ will be contained within the disc $D$ that represents the subest, from which all the points $a, a+1, a+2, \dots, b$ are contained within $D$. This means all of the subsets can be represented by a pair $(a, b)$, with $1 \leq a \leq b \leq n$, which can be picked in $\binom{n+1}{2}$ ways.

Now to prove this is the minimum: using the intuition from the example, we "order" the $n$ points from left to right, and number them $1, 2, \dots, n$.

Explanation and argumentation: pick a line $l$ that is not parallel to any of the lines determined by any pair of points, and place that line outside of the convex hull of the $n$ points. Sort the points from left to right by sorting them in increasing order from $l$. No two points are at the same distance, because their line can't be parallel to $l$

We need to prove we can construct at least $\binom{n+1}{2}$ sets. For each point $A$, Start creating a circle passing through $A$ with center $O$ such that $AO \perp l$, and $A$ is the closest point to $l$ on the circle; we gradually increase the radius of the disc, adding more points into the set until we have every point "greater' than $A$ in our disc. We start only with $A$, and as we gradually increase the disc we keep adding exactly one* other point greater than $A$ into the disc, we end up with $n - a$ new sets (here $a$ is the order number of $A$).

There's an asterisk there, because this logic breaks if at one step we add two or more points. This however happens only if we have three or more points on the same circle and one of the lines through the center and a point in the circle is perpendicular to $l$. There's only a finite number of such problematic lines, and $l$ takes an infinity of orientations, so just pick an orientation of $l$ for which this does not happen.

The reason why this process only adds points greater than $A$ into the disc is the following: as we gradually increase the disc through $A$, each such generated disc contains any disc that came before it. As the radius approaches infinity, the disc becomes the half-plane found to the right of the line through $A$ parallel to $l$, which doesn't contain any point to the left of $A$.

This means there's at least $n + (n-1) + (n-2) + \dots + 1 = \binom{n+1}{2}$ eligible subsets.
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