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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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[*]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
Interesting Inequality
lbh_qys   1
N 21 minutes 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

\[
\left( \frac{a}{b - c} + \frac{b}{c - a} + \frac{c}{a - b} \right)^2  + a^2 + b^2 + c^2.
\]
1 reply
lbh_qys
2 hours ago
lbh_qys
21 minutes ago
Consecutive QR
FireBreathers   1
N 37 minutes ago by GreekIdiot
Is it true that the longest chain of consecutive quadratic residues in modulo $p$ $\leq 2\sqrt{p}$?
1 reply
FireBreathers
Sunday at 2:34 PM
GreekIdiot
37 minutes ago
Problem you can do in angle chasing but I chose to do it with Pascal
prosciutto   1
N 42 minutes ago by Matteo_Degli_Apostoli
Source: ITAMO 2025 p5
Let $ABC$ be a triangle and call $D$ the intersection of the internal angle bisector from $A$ with the side $BC$. The perpendicular bisector of $AD$ intersects the circumcircle of $ABC$ in $E$ and $F$, with $E$ and $B$ on two opposite sides of the line $AD$. Let $G$ be the intersection of lines $BE$ and $DF$, and let $H$ be the intersection of lines $CF$ and $DE$.
Prove that $GH$ and $BC$ are parallel.
1 reply
prosciutto
an hour ago
Matteo_Degli_Apostoli
42 minutes ago
Good prime...
Jackson0423   3
N an hour ago by GreekIdiot
A prime number \( p \) is called a good prime if there exists a positive integer \( n \) such that
\[
n^2 + 1 \text{ is divisible by } p^{2025}.
\]Prove that there are infinitely many good primes.
3 replies
Jackson0423
Yesterday at 3:46 PM
GreekIdiot
an hour ago
Inspired by 2007 Bulgarian
sqing   1
N 2 hours ago by cazanova19921
Source: Own
Let $a$, $b$, $c$ be real numbers such that $a+b+c=0$ and $a^2+b^2+c^2+a^4+b^4+c^4=2$. Prove that $$ab+bc+ca=\frac{1-\sqrt 5}{2}$$Let $a$, $b$, $c$ be real numbers such that $a+b+c=0$ and $ab+bc+ca+a^2+b^2+c^2+a^4+b^4+c^4=2$. Prove that $$ab+bc+ca=\frac{1-\sqrt{17}}{4}$$
1 reply
sqing
4 hours ago
cazanova19921
2 hours ago
Hard Inequality
danilorj   1
N 2 hours ago by Phat_23000245
Let $a, b, c > 0$ with $a + b + c = 1$. Prove that:
\[
\sqrt{a + (b - c)^2} + \sqrt{b + (c - a)^2} + \sqrt{c + (a - b)^2} \geq \sqrt{3},
\]with equality if and only if $a = b = c = \frac{1}{3}$.
1 reply
danilorj
2 hours ago
Phat_23000245
2 hours ago
Orthocenter lies on circumcircle
whatshisbucket   89
N 3 hours ago by Mathandski
Source: 2017 ELMO #2
Let $ABC$ be a triangle with orthocenter $H,$ and let $M$ be the midpoint of $\overline{BC}.$ Suppose that $P$ and $Q$ are distinct points on the circle with diameter $\overline{AH},$ different from $A,$ such that $M$ lies on line $PQ.$ Prove that the orthocenter of $\triangle APQ$ lies on the circumcircle of $\triangle ABC.$

Proposed by Michael Ren
89 replies
whatshisbucket
Jun 26, 2017
Mathandski
3 hours ago
Hard math inequality
noneofyou34   5
N 3 hours ago by JARP091
If a,b,c are positive real numbers, such that a+b+c=1. Prove that:
(b+c)(a+c)/(a+b)+ (b+a)(a+c)/(c+b)+(b+c)(a+b)/(a+c)>= Sqrt.(6(a(a+c)+b(a+b)+c(b+c)) +3
5 replies
noneofyou34
Sunday at 2:00 PM
JARP091
3 hours ago
Interesting inequalities
sqing   0
3 hours ago
Source: Own
Let $ a,b>0 $. Prove that
$$\frac{ab-1} {ab(a+b+2)} \leq \frac{1} {8}$$$$\frac{2ab-1} {ab(a+b+1)} \leq 6\sqrt 3-10$$
0 replies
sqing
3 hours ago
0 replies
Inspired by SXJX (12)2022 Q1167
sqing   1
N 4 hours ago by sqing
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}.$
1 reply
sqing
Yesterday at 4:01 AM
sqing
4 hours ago
Algebra manipulation excercise
Marinchoo   3
N 4 hours ago by compoly2010
Source: 2007 Bulgarian Autumn Math Competition, Problem 9.2
Let $a$, $b$, $c$ be real numbers, such that $a+b+c=0$ and $a^4+b^4+c^4=50$. Determine the value of $ab+bc+ca$.
3 replies
Marinchoo
Mar 17, 2022
compoly2010
4 hours ago
Numbers on a circle
navi_09220114   2
N 4 hours ago by ja.
Source: TASIMO 2025 Day 1 Problem 1
For a given positive integer $n$, determine the smallest integer $k$, such that it is possible to place numbers $1,2,3,\dots, 2n$ around a circle so that the sum of every $n$ consecutive numbers takes one of at most $k$ values.
2 replies
navi_09220114
Yesterday at 11:35 AM
ja.
4 hours ago
Gives typical russian combinatorics vibes
Sadigly   4
N 6 hours ago by lbd4203
Source: Azerbaijan Senior MO 2025 P3
You are given a positive integer $n$. $n^2$ amount of people stand on coordinates $(x;y)$ where $x,y\in\{0;1;2;...;n-1\}$. Every person got a water cup and two people are considered to be neighbour if the distance between them is $1$. At the first minute, the person standing on coordinates $(0;0)$ got $1$ litres of water, and the other $n^2-1$ people's water cup is empty. Every minute, two neighbouring people are chosen that does not have the same amount of water in their water cups, and they equalize the amount of water in their water cups.

Prove that, no matter what, the person standing on the coordinates $(x;y)$ will not have more than $\frac1{x+y+1}$ litres of water.
4 replies
Sadigly
May 8, 2025
lbd4203
6 hours ago
Product of Sum
shobber   4
N 6 hours ago by alexanderchew
Source: CGMO 2006
Given that $x_{i}>0$, $i = 1, 2, \cdots, n$, $k \geq 1$. Show that: \[\sum_{i=1}^{n}\frac{1}{1+x_{i}}\cdot \sum_{i=1}^{n}x_{i}\leq \sum_{i=1}^{n}\frac{x_{i}^{k+1}}{1+x_{i}}\cdot \sum_{i=1}^{n}\frac{1}{x_{i}^{k}}\]
4 replies
1 viewing
shobber
Aug 9, 2006
alexanderchew
6 hours ago
Sum of First, Second, and Third Powers
Brut3Forc3   48
N Apr 25, 2025 by Ilikeminecraft
Source: 1973 USAMO Problem 4
Determine all roots, real or complex, of the system of simultaneous equations
\begin{align*} x+y+z &= 3, \\
x^2+y^2+z^2 &= 3, \\
x^3+y^3+z^3 &= 3.\end{align*}
48 replies
Brut3Forc3
Mar 7, 2010
Ilikeminecraft
Apr 25, 2025
Sum of First, Second, and Third Powers
G H J
Source: 1973 USAMO Problem 4
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AlanLG
241 posts
#36 • 1 Y
Y by cubres
Let $P(t)=t^3+a_2t^2+a_1t+a_0$ the monic pollynomial with roots $x,y,z$; by Newton Sums
\begin{align}
0
&= 1 \cdot (x+y+z)+a_2\Rightarrow a_2=-3\nonumber
\\
&=1\cdot (x^2+y^2+z^2)+a_2(x+y+z)+a_1\cdot 2\Rightarrow a_1=3 \nonumber
\\
&=1\cdot(x^3+y^3+z^3)+a_2(x^2+y^2+z^2)+a_1(x+y+z)+3a_0\Rightarrow a_0=-1\nonumber
\end{align}Therefore $P(t)=t^3-3t^2+3t-1=(t-1)^3\Rightarrow x=y=z=1$
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mathmax12
6051 posts
#37 • 1 Y
Y by cubres
Let, $q=x+y+z, r=xy+xz+yz, s=xyz,$ then $3=q, 3=q^2-2r, 3=q^3-qr+3s.$ Note, that $r=\frac{q^2-3}{2}=\frac{9-3}{2}=3.$ And, we find that $s=1.$ Now, by Vieta's this is the cubic polynomial, $x^3-3x^2+3x-1=(x-1)^3$, so we must have $x=y=z=1.$
This post has been edited 1 time. Last edited by mathmax12, Sep 4, 2023, 9:32 PM
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joshualiu315
2534 posts
#38 • 1 Y
Y by cubres
We will use standard SQP notation:

\begin{align*} 
s &= x+y+z \\
q &= xy+yz+zx \\
p &= xyz
\end{align*}
From the first equation, we immediately know $s=3$. Then, we have $x^2+y^2+z^2=s^2-2q$, so we can solve that $q=3$. To apply the last equation, notice that

\[x^3+y^3+z^3=3xyz+(x+y+z)(x^2+y^2+z^2-xy-yz-xz)\]
Simplifying this and solving, we get $p=1$.

By Vieta's, we see that $x$, $y$, and $z$ are the roots of polynomial $P(t)=t^3-3t^2+3t-1=(t-1)^3$. Thus, the only triple $(x,y,z)$ is $\boxed{(1,1,1)}$.
This post has been edited 1 time. Last edited by joshualiu315, Oct 13, 2023, 4:49 PM
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Cusofay
85 posts
#39 • 1 Y
Y by cubres
By using Newton's sum, we can find that $x$,$y$ and $z$ are roots of the polynomial $P(t)=(t-1)^3$. This implies that $x=y=z=1$ is the only solution to this system of equations.
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peppapig_
280 posts
#40 • 1 Y
Y by cubres
I claim that the only possible solution is $(1,1,1)$.

Note that from the second equation, we get that
\[xy+yz+zx=\frac{(x+y+z)^2-(x^2+y^2+z^2)}{2}=\frac{9-3}{2}=3,\]and using this, along with the third equation, we get that
\[3-3xyz=(x^3+y^3+z^3)-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)=3(0)=0 \iff xyz=1,\]which gives us the system of equations
\[x+y+z=3,\]\[xy+yz+zx=3,\]\[xyz=1,\]which implies that $x$, $y$, $z$ should all be roots of $f(a)=a^3-3a^2+3a-1=(a-1)^3$ by Vieta's. This gives us that the only solution to $(x,y,z)$ is $(1,1,1)$, finishing the problem.
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David_Kim_0202
384 posts
#41 • 1 Y
Y by cubres
since, $x+y+z=3, x²+y²+z²=3$, we can say $xy+yz+zx = 3$.
then, by subtracting $3xyz$ by both sides by the third equation,
$$x³+y³+z³-3xyz=(x+y+z)(x²+y²+z²-xy-yz-zx)=0$$also, $x³+y³+z³=3$ so $xyz=1$.
therefore we know all of the sum, pair sum, product of $x, y, x$, we can know these 3 variables are the roots of a polynomial $t³-3t²+3t-1=0$
therefore the answer is $\boxed{x=y=z=1}$
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blueberryfaygo_55
340 posts
#42 • 1 Y
Y by cubres
From equation $(1)$ and equation $(2)$, we obtain $xy+yz+zx=3$. It follows that $x^2+y^2+z^2=xy+yz+zx$, and multiplying both sides by $2$ and rearranging to factor into binomials yields$$(x-y)^2+(y-z)^2+(z-x)^2 = 0$$so we must have $x=y=z=1$ as the only solution.
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AshAuktober
1008 posts
#43 • 1 Y
Y by cubres
No way is this USAMO.
From standard identities for $(a+b+c)^2$ and $a^3 + b^3 + c^3 -3abc$, we can obtain $$xy + yz + zx = 3, xyz = 1,$$so $x, y, z$ are the zeroes of $$t^3 - 3t^2 + 3t - 1 = (t-1)^3,$$and we get $$x = y = z = 1.$$Since it works, we're done.$\square$
This post has been edited 1 time. Last edited by AshAuktober, Aug 10, 2024, 4:20 AM
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AshAuktober
1008 posts
#44 • 1 Y
Y by cubres
blueberryfaygo_55 wrote:
From equation $(1)$ and equation $(2)$, we obtain $xy+yz+zx=3$. It follows that $x^2+y^2+z^2=xy+yz+zx$, and multiplying both sides by $2$ and rearranging to factor into binomials yields$$(x-y)^2+(y-z)^2+(z-x)^2 = 0$$so we must have $x=y=z=1$ as the only solution.

How do you get the last step when $x, y, z$ need not be real?
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Mr.Sharkman
501 posts
#45 • 1 Y
Y by cubres
Solution
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Markas
150 posts
#46 • 1 Y
Y by cubres
Lets assume that x, y, z are the roots of a cubic polynomial $ax^3 + bx^2 + cx + d$. By the condition we have $p_1 = 3$, $p_2 = 3$, $p_3 = 3$ $\Rightarrow$ by Newton's sums we have that 0 = 3a + b = 3a + 3b + 2c = 3a + 3b + 3c + 3d. Also we can assume a = 1 $\Rightarrow$ from this we get b = -3, c = 3, d = -1 $\Rightarrow$ the polynomial which roots are x, y and z is $x^3 - 3x^2 + 3x - 1$. Also $x^3 - 3x^2 + 3x - 1 = (x - 1)^3$ $\Rightarrow$

x = y = z = 1.
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eg4334
637 posts
#47 • 1 Y
Y by cubres
Consider the polynomial $(t-x)(t-y)(t-z) = 0$. The $t^2$ coefficient is $-3$ here by vietas and the $t$ coefficient is $\frac{3^2-3}{2} = 3$ by vietas as well. Also $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)$ so we can find that $xyz=1$. Thus the polynomial is $x^3-3x^2+3x-1 = (x-1)^3$ so $\boxed{x=y=z=1}$ is the only solution.
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megahertz13
3183 posts
#48 • 1 Y
Y by cubres
The answer is $(1,1,1)$ only. Let $$a=x+y+z$$$$b=xy+yz+zx$$$$c=xyz.$$Note that $a=3$, $a^2-2b=3$, and $$3-3c=a(a^2-3b)\implies (a,b,c)=(3,3,1).$$By Vieta's, we find that $x$, $y$, and $z$ are the roots of the cubic $$a^3-3a^2+3a-1=(a-1)^3=0,$$so $(x,y,z)=(1,1,1)$ is the only solution.
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cubres
119 posts
#49
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Solution
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Ilikeminecraft
658 posts
#50
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I claim that $\boxed{(1, 1, 1)}$ is the only solution. Let $x, y, z$ be the roots to the polynomial $\alpha^3 + a\alpha^2 + b\alpha + c.$ By newton's sums, we have that
\begin{align*}
    P_1 + a = 0 & \implies P_1 = -a \\
    P_2 + a P_1 + 2 b = 0 & \implies P_2 = a^2 - 2b \\
    P_3 + a P_2 + b P_1 + 3c = 0 & \implies P_3 = -a^3 + 3ab - 3c
\end{align*}Using the fact that $P_1 = P_2 = P_3 = 3,$ we get that $a = -3, b = 3, c = -1.$ Thus, $(\alpha - 1)^3 = 0,$ and thus we are done.
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