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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
Yesterday at 11:16 PM
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
Yesterday at 11:16 PM
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
existence of a circle tangent to AB and AC
NicoN9   0
2 minutes ago
Source: Japan Junior MO Preliminary 2020 P10
Let $ABC$ be a triangle with integer side lengths. Let $D, E$ be points on segment $BC$ such that $B, D, E,C$ are in this order, $BD=4$, and $EC=7$.
Suppose that there exists a circle which is tangent to sides $AB$ and $AC$, passes through $D, E$. Find the minimum of the perimeter of triangle $ABC$.
0 replies
+1 w
NicoN9
2 minutes ago
0 replies
filling tiles again?
NicoN9   0
4 minutes ago
Source: Japan Junior MO Preliminary 2020 P9
There is a board with regular hexagon shape with side length $1$. As shown below, we dessert the board into $24$ of equilateral triangle, with side length $1/2$. We call the $19$ points of $\circ$ is good in the figure.

IMAGE
There are $12$ of tiles with side length $\frac{1}{2}$, $\frac{\sqrt{3}}{2}$, $1$ (thus the tile is right-angled). How many ways are there to fill the board with these tiles such that
$\bullet$ Each vertex of the tiles are on good points, and
$\bullet$ There doesn't exist $2$ tiles, such that it forms a equilateral triangle of side length $1$.
0 replies
NicoN9
4 minutes ago
0 replies
3 variables NT
NicoN9   0
8 minutes ago
Source: Japan Junior MO Preliminary 2020 P8
Find all triples $(l, m, n)$ such that \[
l^2+mn=m^2+ln,\quad  n^2+lm=2020,\quad  l\le m\le n.
\]
0 replies
NicoN9
8 minutes ago
0 replies
Filling with tiles
NicoN9   0
10 minutes ago
Source: Japan Junior MO Preliminary 2020 P7
Consider the following tiles, created by using three and five unitsquare, respectively.
IMAGE
There are twelve of L, and four of X. We fill the following gray region created by $56$ unitsquare, using L and X.

IMAGE
Find the number of ways to do so.
0 replies
NicoN9
10 minutes ago
0 replies
3D combo puzzle
NicoN9   0
12 minutes ago
Source: Japan Junior MO Preliminary 2020 P6
As shown below, there is a figure $Q$ created by removing the unitcube at the cornor of the cube with side length $5$. Also, there are infinitely many figure $L$ created with four unitcube, and infinitely many unitcubes.

IMAGE
We paste together $L$ and unitcubes to create $Q$.
What is the maximum possible number of $L$ that we can use?
0 replies
NicoN9
12 minutes ago
0 replies
quadrilateral ABCD and mid of AC, BD
NicoN9   0
14 minutes ago
Source: Japan Junior MO Preliminary 2020 P5
Suppose $ABCD$ is a convex quadrilateral such that $AB=CD=7$, $DA=6$, $\angle B=72^\circ$, and $\angle C=48^\circ$. Let $P$, and $Q$ be the midpoint of segment $AC$, and $BD$, respectively. Find the value of $PQ$.
0 replies
NicoN9
14 minutes ago
0 replies
isosceles right triangle with four squares inscribed
NicoN9   0
16 minutes ago
Source: Japan Junior MO Preliminary 2020 P4
Four squares with side length $1$ are inscribed in isosceles right triangle, as shown below. Find the area of the isosceles right triangle.
IMAGE
0 replies
NicoN9
16 minutes ago
0 replies
number of palindromes divisible by 11
NicoN9   0
18 minutes ago
Source: Japan Junior MO Preliminary 2020 P3
We call an integer palindrome if it's the same value when read it backwards, and the unit digit is nonzero. Find the number of positive integers less than or equal to $10000$ such that it is a palindrome, and divisible by $11$.
0 replies
NicoN9
18 minutes ago
0 replies
Consecutive sum of integers sum up to 2020
NicoN9   0
20 minutes ago
Source: Japan Junior MO Preliminary 2020 P2
Let $a$ and $b$ be positive integers. Suppose that the sum of integers between $a$ and $b$, including $a$ and $b$, are equal to $2020$.
All among those pairs $(a, b)$, find the pair such that $a$ achieves the minimum.
0 replies
NicoN9
20 minutes ago
0 replies
3 right-angled triangle area
NicoN9   0
21 minutes ago
Source: Japan Junior MO Preliminary 2020 P1
Right angled triangle $ABC$, and a square are drawn as shown below. Three numbers written below implies each of the area of shaded small right angled triangle. Find the value of $AB/AC$.

IMAGE
0 replies
NicoN9
21 minutes ago
0 replies
Function equation
LeDuonggg   3
N 24 minutes ago by luutrongphuc
Find all functions $f: \mathbb{R^+} \rightarrow \mathbb{R^+}$ , such that for all $x,y>0$:
\[ f(x+f(y))=\dfrac{f(x)}{1+f(xy)}\]
3 replies
LeDuonggg
Yesterday at 2:59 PM
luutrongphuc
24 minutes ago
A sequence containing every natural number exactly once
Pomegranat   4
N 33 minutes ago by Pomegranat
Source: Own
Does there exist a sequence \( \{a_n\}_{n=1}^{\infty} \), which is a permutation of the natural numbers (that is, each natural number appears exactly once), such that for every \( n \in \mathbb{N} \), the sum of the first \( n \) terms is divisible by \( n \)?
4 replies
Pomegranat
3 hours ago
Pomegranat
33 minutes ago
hard square root problem
kjhgyuio   0
43 minutes ago
........
0 replies
kjhgyuio
43 minutes ago
0 replies
Queue geo
vincentwant   5
N an hour ago by Ilikeminecraft
Let $ABC$ be an acute scalene triangle with circumcenter $O$. Let $Y, Z$ be the feet of the altitudes from $B, C$ to $AC, AB$ respectively. Let $D$ be the midpoint of $BC$. Let $\omega_1$ be the circle with diameter $AD$. Let $Q\neq A$ be the intersection of $(ABC)$ and $\omega$. Let $H$ be the orthocenter of $ABC$. Let $K$ be the intersection of $AQ$ and $BC$. Let $l_1,l_2$ be the lines through $Q$ tangent to $\omega,(AYZ)$ respectively. Let $I$ be the intersection of $l_1$ and $KH$. Let $P$ be the intersection of $l_2$ and $YZ$. Let $l$ be the line through $I$ parallel to $HD$ and let $O'$ be the reflection of $O$ across $l$. Prove that $O'P$ is tangent to $(KPQ)$.
5 replies
vincentwant
Wednesday at 3:54 PM
Ilikeminecraft
an hour 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
1742 posts
#1
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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
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Maybe problem is near to trivialize by BAMO 2020/5 problem....
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YaoAOPS
1535 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
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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
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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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