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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
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0 replies
jlacosta
Apr 2, 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
Diophantine equation !
ComplexPhi   5
N 11 minutes ago by aops.c.c.
Source: Romania JBMO TST 2015 Day 1 Problem 4
Solve in nonnegative integers the following equation :
$$21^x+4^y=z^2$$
5 replies
ComplexPhi
May 14, 2015
aops.c.c.
11 minutes ago
Combo problem
soryn   0
37 minutes ago
The school A has m1 boys and m2 girls, and ,the school B has n1 boys and n2 girls. Each school is represented by one team formed by p students,boys and girls. If f(k) is the number of cases for which,the twice schools has,togheter k girls, fund f(k) and the valute of k, for which f(k) is maximum.
0 replies
soryn
37 minutes ago
0 replies
Parity and sets
betongblander   7
N 39 minutes ago by ihategeo_1969
Source: Brazil National Olympiad 2020 5 Level 3
Let $n$ and $k$ be positive integers with $k$ $\le$ $n$. In a group of $n$ people, each one or always
speak the truth or always lie. Arnaldo can ask questions for any of these people
provided these questions are of the type: “In set $A$, what is the parity of people who speak to
true? ”, where $A$ is a subset of size $ k$ of the set of $n$ people. The answer can only
be $even$ or $odd$.
a) For which values of $n$ and $k$ is it possible to determine which people speak the truth and
which people always lie?
b) What is the minimum number of questions required to determine which people
speak the truth and which people always lie, when that number is finite?
7 replies
betongblander
Mar 18, 2021
ihategeo_1969
39 minutes ago
Mount Inequality erupts on a sequence :o
GrantStar   88
N 44 minutes ago by Nari_Tom
Source: 2023 IMO P4
Let $x_1,x_2,\dots,x_{2023}$ be pairwise different positive real numbers such that
\[a_n=\sqrt{(x_1+x_2+\dots+x_n)\left(\frac{1}{x_1}+\frac{1}{x_2}+\dots+\frac{1}{x_n}\right)}\]is an integer for every $n=1,2,\dots,2023.$ Prove that $a_{2023} \geqslant 3034.$
88 replies
+1 w
GrantStar
Jul 9, 2023
Nari_Tom
44 minutes ago
JBMO Shortlist 2022 N1
Lukaluce   8
N an hour ago by godchunguus
Source: JBMO Shortlist 2022
Determine all pairs $(k, n)$ of positive integers that satisfy
$$1! + 2! + ... + k! = 1 + 2 + ... + n.$$
8 replies
Lukaluce
Jun 26, 2023
godchunguus
an hour ago
P(x) | P(x^2-2)
GreenTea2593   4
N an hour ago by GreenTea2593
Source: Valentio Iverson
Let $P(x)$ be a monic polynomial with complex coefficients such that there exist a polynomial $Q(x)$ with complex coefficients for which \[P(x^2-2)=P(x)Q(x).\]Determine all complex numbers that could be the root of $P(x)$.

Proposed by Valentio Iverson, Indonesia
4 replies
GreenTea2593
4 hours ago
GreenTea2593
an hour ago
USEMO P6 (Idk what to say here)
franzliszt   16
N an hour ago by MathLuis
Source: USEMO 2020/6
Prove that for every odd integer $n > 1$, there exist integers $a, b > 0$ such that, if we let $Q(x) = (x + a)^
2 + b$, then the following conditions hold:
$\bullet$ we have $\gcd(a, n) = gcd(b, n) = 1$;
$\bullet$ the number $Q(0)$ is divisible by $n$; and
$\bullet$ the numbers $Q(1), Q(2), Q(3), \dots$ each have a prime factor not dividing $n$.
16 replies
franzliszt
Oct 25, 2020
MathLuis
an hour ago
Prove that the fraction (21n + 4)/(14n + 3) is irreducible
DPopov   110
N 2 hours ago by Shenhax
Source: IMO 1959 #1
Prove that the fraction $ \dfrac{21n + 4}{14n + 3}$ is irreducible for every natural number $ n$.
110 replies
DPopov
Oct 5, 2005
Shenhax
2 hours ago
Let \( a, b, c \) be positive real numbers satisfying \[ a^2 + c^2 = b(a + c). \
Jackson0423   3
N 2 hours ago by Mathzeus1024
Let \( a, b, c \) be positive real numbers satisfying
\[
a^2 + c^2 = b(a + c).
\]Let
\[
m = \min \left( \frac{a^2 + ab + b^2}{ab + bc + ca} \right).
\]Find the value of \( 2024m \).
3 replies
Jackson0423
Apr 16, 2025
Mathzeus1024
2 hours ago
real+ FE
pomodor_ap   3
N 2 hours ago by MathLuis
Source: Own, PDC001-P7
Let $f : \mathbb{R}^+ \to \mathbb{R}^+$ be a function such that
$$f(x)f(x^2 + y f(y)) = f(x)f(y^2) + x^3$$for all $x, y \in \mathbb{R}^+$. Determine all such functions $f$.
3 replies
pomodor_ap
Yesterday at 11:24 AM
MathLuis
2 hours ago
Inspired by hlminh
sqing   1
N 2 hours ago by sqing
Source: Own
Let $ a,b,c $ be real numbers such that $ a^2+b^2+c^2=1. $ Prove that $$ |a-kb|+|b-kc|+|c-ka|\leq \sqrt{3k^2+2k+3}$$Where $ k\geq 0 . $
1 reply
sqing
2 hours ago
sqing
2 hours ago
Is this FE solvable?
ItzsleepyXD   3
N 3 hours ago by jasperE3
Source: Original
Let $c_1,c_2 \in \mathbb{R^+}$. Find all $f : \mathbb{R^+} \rightarrow \mathbb{R^+}$ such that for all $x,y \in \mathbb{R^+}$ $$f(x+c_1f(y))=f(x)+c_2f(y)$$
3 replies
ItzsleepyXD
Yesterday at 3:02 AM
jasperE3
3 hours ago
PQ bisects AC if <BCD=90^o, A, B,C,D concyclic
parmenides51   2
N 3 hours ago by venhancefan777
Source: Mathematics Regional Olympiad of Mexico Northeast 2020 P2
Let $A$, $B$, $C$ and $D$ be points on the same circumference with $\angle BCD=90^\circ$. Let $P$ and $Q$ be the projections of $A$ onto $BD$ and $CD$, respectively. Prove that $PQ$ cuts the segment $AC$ into equal parts.
2 replies
parmenides51
Sep 7, 2022
venhancefan777
3 hours ago
Inequality with three conditions
oVlad   3
N 3 hours ago by sqing
Source: Romania EGMO TST 2019 Day 1 P3
Let $a,b,c$ be non-negative real numbers such that \[b+c\leqslant a+1,\quad c+a\leqslant b+1,\quad a+b\leqslant c+1.\]Prove that $a^2+b^2+c^2\leqslant 2abc+1.$
3 replies
oVlad
Yesterday at 1:48 PM
sqing
3 hours ago
50 points in plane
pohoatza   12
N Apr 6, 2025 by de-Kirschbaum
Source: JBMO 2007, Bulgaria, problem 3
Given are $50$ points in the plane, no three of them belonging to a same line. Each of these points is colored using one of four given colors. Prove that there is a color and at least $130$ scalene triangles with vertices of that color.
12 replies
pohoatza
Jun 28, 2007
de-Kirschbaum
Apr 6, 2025
50 points in plane
G H J
Source: JBMO 2007, Bulgaria, problem 3
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pohoatza
1145 posts
#1 • 13 Y
Y by Adventure10, Mathlover_1, OronSH, aidan0626, Blue_banana4, PikaPika999, and 7 other users
Given are $50$ points in the plane, no three of them belonging to a same line. Each of these points is colored using one of four given colors. Prove that there is a color and at least $130$ scalene triangles with vertices of that color.
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bodan
267 posts
#2 • 24 Y
Y by amirmath1995, Catalanfury, Nguyenhuyhoang, silouan, raven_, DepressedCubic, Adventure10, Mathlover_1, Mango247, aidan0626, Blue_banana4, Funcshun840, PikaPika999, kiyoras_2001, and 10 other users
Lemma. Among $n$ points in a plane positioned generally (no three collinear) we have at least $\frac{n(n-1)(n-8)}{6}$ scalene triangles.
Proof. Suppose that $n$ points are fixed and the number of isosceles triangles is $\alpha$. Set $\beta$ to be the number of bases of these triangles (we count three bases for each equilateral and one for each nonequilateral isosceles triangle). Clearly $\beta\geq \alpha$. Then each segment connecting a pair of points can be a base of at most two triangles, as if this is not the case three points will lie on this pairs' perpendicular bisector. Thus there are at most $2{n\choose 2}$ isosceles triangles, yielding that there are at least ${n\choose 3}-2{n\choose 2}=\frac{n(n-1)(n-8)}{6}$ scalene triangles. $\square$


In the problem there are at least $13$ points of the same color and we apply the lemma.
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reveryu
218 posts
#3 • 3 Y
Y by Adventure10, Mango247, PikaPika999
why the fact "there are at most $2{n\choose 2}$ isosceles triangles"
does not imply "#scalene triangle + #acute triangle(except isosceles and equilateral triangle) is at least ${n\choose 3}-2{n\choose 2}=\frac{n(n-1)(n-8)}{6}$ " ??
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huashiliao2020
1292 posts
#4 • 1 Y
Y by PikaPika999
bodan wrote:
Lemma. Among $n$ points in a plane positioned generally (no three collinear) we have at least $\frac{n(n-1)(n-8)}{6}$ scalene triangles.
Proof. Suppose that $n$ points are fixed and the number of isosceles triangles is $\alpha$. Set $\beta$ to be the number of bases of these triangles (we count three bases for each equilateral and one for each nonequilateral isosceles triangle). Clearly $\beta\geq \alpha$. Then each segment connecting a pair of points can be a base of at most two triangles, as if this is not the case three points will lie on this pairs' perpendicular bisector. Thus there are at most $2{n\choose 2}$ isosceles triangles, yielding that there are at least ${n\choose 3}-2{n\choose 2}=\frac{n(n-1)(n-8)}{6}$ scalene triangles. $\square$


In the problem there are at least $13$ points of the same color and we apply the lemma.


Thanks. I had almost the exact same solution, except one question. Why do you care about “ we count three bases for each equilateral and one for each nonequilateral isosceles triangle”? It seems that already 2 bases of triangles at most x n choose 2 pairs of points to form an isosceles triangle is enough.
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megarnie
5587 posts
#5 • 2 Y
Y by OronSH, PikaPika999
We solve the following problem first:

Given are $13$ points in the plane, no three of them belonging to a same line. Prove that there are at least $130$ scalene triangles with vertices in the plane.

Notice that any edge between two points in the plane can be a base of at most two isosceles triangles (because otherwise we would have $3$ points on the perpendicular bisector). Hence there are at most $\binom{13}{2} \cdot 2 = 256$ isosceles triangles, so at lesat $\binom{13}{3} - 256 = 130$ scalene triangles.



This solves the original problem because there must exist a color with at least $\left\lceil \frac{50}{4} \right\rceil = 13$ points of that color.
This post has been edited 3 times. Last edited by megarnie, Aug 9, 2023, 9:11 PM
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asdf334
7586 posts
#6 • 1 Y
Y by PikaPika999
We prove that given 13 points in standard position, there are 130 scalene triangles. This clearly solves the original question.

Note that every base (i.e. two points) gives at most two isosceles triangles, else we have three points along the perpendicular bisector.

Therefore there are at most $\binom{13}{2}\cdot 2=156$ isosceles triangles. At the same time there are $\binom{13}{3}=286$ triangles in total, making 130 non-isosceles AKA scalene triangles. $\blacksquare$
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dolphinday
1320 posts
#8 • 1 Y
Y by PikaPika999
By pigeonhole, there is a color that has at least $13$ points of the same color. This means there are $\binom{13}{3} = 286$ triangles that can be formed.
Let $I$ be the number of isosceles triangles in the $13$ points.
Then each pair of two points can contribute at most $2$ to $I$(due to the collinear condition), so $\binom{13}{2} \cdot 2 = 156$ is the maximum number of isosceles triangles. Then the minimum number of scalene triangles is $130$.
This post has been edited 1 time. Last edited by dolphinday, Dec 9, 2023, 3:09 PM
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mannshah1211
651 posts
#9 • 1 Y
Y by PikaPika999
By Pigeonhole, some color has at least $13$ points of that color. So, assume there are exactly $13$ points, we'll show that we can find at least $130$ scalene triangles whose vertices all belong to those $13$. First, the total number of triangles is $\binom{13}{3} = 286$. Also, note that for each pair of points $(A, B)$, there are at most two isosceles triangles $ABC$ which have $\overline{AB}$ as a base, since in order for that to happen, $C$ must lie on the perpendicular bisector of $\overline{AB}$, but if three or more $C$ exist, this is a contradiction to the fact that no three points are collinear. So, there are at most $2 \cdot \binom{13}{2} = 156$ isosceles triangles, thus at least $130$ scalene triangles among those $13$ points (basically at least $130$ of that color), so done.
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InterLoop
274 posts
#10 • 1 Y
Y by PikaPika999
smol
solution
This post has been edited 1 time. Last edited by InterLoop, May 1, 2024, 5:00 AM
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RedFireTruck
4221 posts
#11 • 1 Y
Y by PikaPika999
https://cdn.aops.com/images/b/8/2/b824a1d2347b72751984bb68d01bd220d96d5d63.png
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kotmhn
58 posts
#12 • 1 Y
Y by PikaPika999
On each base at most 2 isosceles can lie else the collinearity condition is violated.
By PHP 13 dots of same color exist. then at most $2{13 \choose 2} = 156$ isosceles triangles and total triangles equal ${13 \choose 3}=286$ so at least $286-156=50$ triangles.
done
This post has been edited 1 time. Last edited by kotmhn, Aug 13, 2024, 11:14 AM
Reason: arithmetic skill issue
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ihategeo_1969
205 posts
#13 • 1 Y
Y by PikaPika999
Start with a claim.

Claim: For any $x$ points in the plane (no three collinear), there are atleast \[\binom{x}{3}-2\binom{x}2\]scalene triangles formed by it.
Proof: The maximum number of isosceles triangles by choosing a fixed base is atmost $2$ or else there will be $3$ points on the perpendicular bisector of the base.$\blacksquare$.

And hence choose the colour with atleast $\left \lceil \frac{50}4 \right \rceil=13$ points and the number of scalene triangles it form are atleast \[\binom{13}3-2\binom{13}2=130\]as desired.
This post has been edited 1 time. Last edited by ihategeo_1969, Sep 11, 2024, 8:20 AM
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de-Kirschbaum
196 posts
#14 • 1 Y
Y by PikaPika999
Note that by pigeonhole there exists a color that repeats at least $13$ times. Then, since there are no three points that are colinear, each line segment drawn in these $13$ points can be the base for at most $2$ isoceles triangles, thus there are at least $\binom{13}{3}-\binom{13}{2}2=130$ scalene triangles with same colored vertices.
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