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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
JBMO TST Bosnia and Herzegovina 2024 P4
FishkoBiH   0
a few seconds ago
Source: JBMO TST Bosnia and Herzegovina 2024 P4
Let $m$ and $n$ be natural numbers. Every one of the $m*n$ squares of the $m*n$ board is colored either black or white, so that no 2 neighbouring squares are the same color(the board is colored like in chess").In one step we can pick 2 neighbouring squares and change their colors like this:
- a white square becomes black;
-a black square becomes blue;
-a blue square becomes white.
For which $m$ and $n$ can we ,in a finite sequence of these steps, switch the starting colors from white to black and vice versa.
0 replies
FishkoBiH
a few seconds ago
0 replies
JBMO TST Bosnia and Herzegovina 2023 P1
FishkoBiH   1
N 8 minutes ago by clarkculus
Source: JBMO TST Bosnia and Herzegovina 2023 P1
Determine all real numbers $a, b, c, d$ for which

$ab+cd=6$
$ac+bd=3$
$ad+bc=2$
$a+b+c+d=6$
1 reply
FishkoBiH
an hour ago
clarkculus
8 minutes ago
JBMO TST Bosnia and Herzegovina 2024 P3
FishkoBiH   0
8 minutes ago
Source: JBMO TST Bosnia and Herzegovina 2024 P3
Let $ABC$ be a right-angled triangle where $ACB$=90°.Let $CD$ be an altitude of that triangle and points $M$ and $N$ be the midpoints of $CD$ and $BC$, respectively.If $S$ is the circumcenter of the triangle $AMN$, prove that $AS$ and $BC$ are paralel.
0 replies
FishkoBiH
8 minutes ago
0 replies
JBMO TST Bosnia and Herzegovina 2024 P2
FishkoBiH   0
15 minutes ago
Source: JBMO TST Bosnia and Herzegovina 2024 P2
Determine all $x$, $y$, $k$ and $n$ positive integers such that:

$10^x$ + $10^y$ + $n!$ = $2024^k$

0 replies
FishkoBiH
15 minutes ago
0 replies
Inspired by 2025 Beijing
sqing   7
N 31 minutes ago by ytChen
Source: Own
Let $ a,b,c,d >0  $ and $ (a^2+b^2+c^2)(b^2+c^2+d^2)=36. $ Prove that
$$ab^2c^2d \leq 8$$$$a^2bcd^2 \leq 16$$$$ ab^3c^3d \leq \frac{2187}{128}$$$$ a^3bcd^3 \leq \frac{2187}{32}$$
7 replies
sqing
Yesterday at 4:56 PM
ytChen
31 minutes ago
Inequality em981
oldbeginner   18
N 36 minutes ago by xzlbq
Source: Own
Let $a, b, c>0, a+b+c=3$. Prove that
\[\sqrt{a+\frac{9}{b+2c}}+\sqrt{b+\frac{9}{c+2a}}+\sqrt{c+\frac{9}{a+2b}}+\frac{2(ab+bc+ca)}{9}\ge\frac{20}{3}\]
18 replies
oldbeginner
Sep 22, 2016
xzlbq
36 minutes ago
JBMO TST Bosnia and Herzegovina 2023 P2
FishkoBiH   1
N 36 minutes ago by clarkculus
Source: JBMO TST Bosnia and Herzegovina 2023 P2
Determine all non negative integers $x$ and $y$ such that $6^x$ + $2^y$ + 2 is a perfect square.
1 reply
FishkoBiH
an hour ago
clarkculus
36 minutes ago
Divisiblity...
TUAN2k8   1
N 39 minutes ago by Natrium
Source: Own
Let $m$ and $n$ be two positive integer numbers such that $m \le n$.Prove that $\binom{n}{m}$ divides $lcm(1,2,...,n)$
1 reply
TUAN2k8
Today at 6:13 AM
Natrium
39 minutes ago
JBMO TST Bosnia and Herzegovina 2023 P4
FishkoBiH   0
an hour ago
Source: JBMO TST Bosnia and Herzegovina 2023 P4
Let $n$ be a positive integer. A board with a format $n*n$ is divided in $n*n$ equal squares.Determine all integers $n$3 such that the board can be covered in $2*1$ (or $1*2$) pieces so that there is exactly one empty square in each row and each column.
0 replies
FishkoBiH
an hour ago
0 replies
JBMO TST Bosnia and Herzegovina 2023 P3
FishkoBiH   0
an hour ago
Source: JBMO TST Bosnia and Herzegovina 2023 P3
Let ABC be an acute triangle with an incenter $I$.The Incircle touches sides $AC$ and $AB$ in $E$ and $F$ ,respectively. Lines CI and EF intersect at $S$. The point $T$$I$ is on the line AI so that $EI$=$ET$.If $K$ is the foot of the altitude from $C$ in triangle $ABC$,prove that points $K$,$S$ and $T$ are colinear.
0 replies
FishkoBiH
an hour ago
0 replies
Locus of points P in triangle ABC
v_Enhance   25
N an hour ago by alexanderchew
Source: USA January TST for IMO 2016, Problem 3
Let $ABC$ be an acute scalene triangle and let $P$ be a point in its interior. Let $A_1$, $B_1$, $C_1$ be projections of $P$ onto triangle sides $BC$, $CA$, $AB$, respectively. Find the locus of points $P$ such that $AA_1$, $BB_1$, $CC_1$ are concurrent and $\angle PAB + \angle PBC + \angle PCA = 90^{\circ}$.
25 replies
v_Enhance
May 17, 2016
alexanderchew
an hour ago
number theory diophantic with factorials and primes
skellyrah   4
N 2 hours ago by skellyrah
Source: by me
find all triplets of non negative integers (a,b,p) where p is prime such that $$ a! + b! + 7ab = p^2 $$
4 replies
skellyrah
Feb 16, 2025
skellyrah
2 hours ago
primes,exponentials,factorials
skellyrah   6
N 2 hours ago by skellyrah
find all primes p,q such that $$ \frac{p^q+q^p-p-q}{p!-q!} $$is a prime number
6 replies
skellyrah
Apr 30, 2025
skellyrah
2 hours ago
Serbian selection contest for the IMO 2025 - P5
OgnjenTesic   2
N 2 hours ago by GreenTea2593
Source: Serbian selection contest for the IMO 2025
Determine the smallest positive real number $\alpha$ such that there exists a sequence of positive real numbers $(a_n)$, $n \in \mathbb{N}$, with the property that for every $n \in \mathbb{N}$ it holds that:
\[
        a_1 + \cdots + a_{n+1} < \alpha \cdot a_n.
    \]Proposed by Pavle Martinović
2 replies
OgnjenTesic
May 22, 2025
GreenTea2593
2 hours ago
A+b+c=0
Xixas   18
N Apr 17, 2025 by sqing
Source: Lithuanian Mathematical Olympiad 2006
Show that if $a+b+c=0$ then $(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b})(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c})=9$.
18 replies
Xixas
Apr 12, 2006
sqing
Apr 17, 2025
A+b+c=0
G H J
G H BBookmark kLocked kLocked NReply
Source: Lithuanian Mathematical Olympiad 2006
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Xixas
297 posts
#1 • 2 Y
Y by Adventure10, Mango247
Show that if $a+b+c=0$ then $(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b})(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c})=9$.
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Michael
99 posts
#2 • 3 Y
Y by Mathskidd, Adventure10, Mango247
Xixas wrote:
Show that if $a+b+c=0$ then $(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b})(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c})=9$.

The following identity is easily verified:
\[ {{y-z}\over{x}}+{{z-x}\over{y}}+{{x-y}\over{z}}=-{{(y-z)(z-x)(x-y)}\over{xyz}}\quad\hbox{for}\quad x,y,z\neq 0.\leqno (1) \]
Set $x=b-c,y=c-a,z=a-b$ in (1). In view of $a+b+c=0$ we have $y-z=b+c-2a=-3a,z-x=-3b,x-y=-3c$, and (1) becomes
\[ {{-3a}\over{b-c}}+{{-3b}\over{c-a}}+{{-3c}\over{a-b}}=-{{(-3a)(-3b)(-3c)}\over{(b-c)(c-a)(a-b)}}, \]
i.e.,
\[ {{a}\over{b-c}}+{{b}\over{c-a}}+{{c}\over{a-b}}=-{{9abc}\over{(b-c)(c-a)(a-b)}}. \]
Now set $x=a,y=b,z=c$ in (1):
\[ {{b-c}\over{a}}+{{c-a}\over{b}}+{{a-b}\over{c}}=-{{(b-c)(c-a)(a-b)}\over{abc}}. \]
Multiplying the last two equalities we get the claimed one.
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stergiu
1648 posts
#3 • 2 Y
Y by Adventure10, Mango247
Nice work !

The problem comes from the Austrian- Polish competition.I have recently read a very similar - almost the same - solution in an older issue of Crux.

Babis
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indybar
398 posts
#4 • 2 Y
Y by Adventure10, Mango247
Indeed :D
http://www.kalva.demon.co.uk/aus-pol/ap85.html
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silouan
3952 posts
#5 • 2 Y
Y by Adventure10, Mango247
The problem indeed is classic and quite well-known (at least in Greece ).
Another solution is . Let $S$ the given product

Notice that $\frac{a}{b-c}(\frac{c-a}{b}+\frac{a-b}{c})=\frac{2a^2}{bc}$

Similar for the other two .

We expand the given product and we use he above .so
$S=3+\frac{2(a^3+b^3+c^3)}{abc}$ .But $a^3+b^3+c^3=3abc$ when $a+b+c=0$

so $S=9$ and we are done
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ArcticMonkey
46 posts
#6 • 2 Y
Y by Adventure10, Mango247
Another solution..

Click to reveal hidden text
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Xixas
297 posts
#7 • 3 Y
Y by Kunihiko_Chikaya, Adventure10, Mango247
Ghm :wacko: Can we apply AM-HM to potentialy negative $x, y, z$?
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ArcticMonkey
46 posts
#8 • 2 Y
Y by Adventure10, Mango247
Hm I have to admit I really didn't consider that restriction. Isn't it only for the inequalities with geometric mean? :? Actually it is also possible to proof this inequality using Cauchy-Schwarz: (Though it seems that this time we really have a restriction because of those d$a$mn sqrt{} things. :))

$(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq (\sqrt{x}\cdot\sqrt{\frac{1}{x}} +\sqrt{y}\cdot\sqrt{\frac{1}{y}}+ \sqrt{z}\cdot\sqrt{\frac{1}{z}})^2=9$

[EDIT]

Ok, we just have to try $(-10, 1, 1)$ and see that my solution sucks. :oops:
This post has been edited 1 time. Last edited by ArcticMonkey, Apr 22, 2006, 4:23 PM
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Xixas
297 posts
#9 • 2 Y
Y by Adventure10, Mango247
That's the reason that prevented me from using Cauchy-Schwarz on the contest.
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dgreenb801
1896 posts
#10 • 2 Y
Y by Adventure10, Mango247
Since $ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$
It follows that if $ a + b + c = 0$, then $ a^3 + b^3 + c^3 = 3abc$
The numerator of the left factor is
$ - \sum (a^3 - a^2b - a^2c + abc) = - \sum (a^3 - a^2(b + c) + abc) = - \sum (2a^3 + abc)$
Since $ 3abc = a^3 + b^3 + c^3$, the left factor is
$ \frac { - 3 \sum a^3}{(a - b)(b - c)(c - a)}$
The right factor is
$ \frac {b^2c - bc^2 + c^2a - ca^2 + a^2b - ab^2}{abc}$
But the denominator, $ abc = \frac {1}{3} \sum a^3$. Everything cancels, and we're left with 9.
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CrazyBoy
1 post
#11 • 2 Y
Y by Adventure10, Mango247
silouan wrote:
Notice that $\frac{a}{b-c}(\frac{c-a}{b}+\frac{a-b}{c})=\frac{2a^2}{bc}$
How did you come up with that ?
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sqing
42398 posts
#12 • 3 Y
Y by boss_1998, Adventure10, Mango247
Xixas wrote:
Show that if $a+b+c=0$ then $(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b})(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c})=9$.
$\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=-\frac{9abc}{(a-b)(b-c)(c-a)},$

$\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}=-\frac{(a-b)(b-c)(c-a)}{abc}.$
here
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Kunihiko_Chikaya
14514 posts
#13 • 2 Y
Y by Adventure10, Mango247
The problem is very classic. In Japanese univesity entrance exam, Nara University entrance exam/Medicine has posed te same problem in early 1960 or later 1950.

I would like to know the original problem or articles regarding the problem.

Are there anyone having the information ?

Any information would be appreciated.

Thanks in advance.

kunny
This post has been edited 2 times. Last edited by Kunihiko_Chikaya, Aug 3, 2015, 8:17 AM
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arqady
30258 posts
#14 • 3 Y
Y by Kunihiko_Chikaya, Adventure10, Mango247
I have found this problem in the following book.
В.Кречмар, "Задачник по алгебре", (1937 year, page 15, problem 18).
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soupynoodles
1 post
#15 • 2 Y
Y by Adventure10, Mango247
Since a+b+c=0, c=-a-b
Therefore, substituting this in you get the attached image.
Take out 9 from the numerator, cancel 2a^4 + 5(a^3)b - 5a(b^3) - 2b^4 from the top and bottom... and you're done!
Attachments:
This post has been edited 1 time. Last edited by soupynoodles, Aug 6, 2015, 4:36 PM
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sqing
42398 posts
#16 • 3 Y
Y by Adventure10, Mango247, alice211
Xixas wrote:
Show that if $a+b+c=0$ then $(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b})(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c})=9$.
Kyrgyzstan national 2016
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sqing
42398 posts
#17 • 3 Y
Y by Adventure10, Mango247, AlexCenteno2007
Let $a,b,c$ be real numbers such that $a+b+c=0$ .Prove that$$\frac{a^2+b^2+c^2}{2}\cdot \frac{a^5+b^5+c^5}{5}=\frac{a^7+b^7+c^7}{7}$$
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AlexCenteno2007
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#18
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sqing wrote:
Let $a,b,c$ be real numbers such that $a+b+c=0$ .Prove that$$\frac{a^2+b^2+c^2}{2}\cdot \frac{a^5+b^5+c^5}{5}=\frac{a^7+b^7+c^7}{7}$$

Solution:
By elementary symmetric polynomials we have S1=$a+b+c=0$
S2=-2$\sigma(2)$ and
S3=3$\sigma(3)$
By substituting in a direct way we have to
$$\frac{S2}{2}\cdot \frac{S2}{2}\cdot\frac{S3}{3}=\frac{a^7+b^7+c^7}{7}$$But I know that S7=$7\sigma(3)\cdot \sigma(2)^2$
From this the result is immediate.
This post has been edited 3 times. Last edited by AlexCenteno2007, Apr 17, 2025, 12:08 AM
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sqing
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#19
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Thanks.
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