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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
1 viewing
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
problemo
hashbrown2009   2
N 22 minutes ago by Pin123
if x/(3^3+4^3) + y/(3^3+6^3) =1

and

x/(5^3+4^3) + y/(5^3+6^3) =1

find the 2 values of x and y.
2 replies
hashbrown2009
Mar 30, 2025
Pin123
22 minutes ago
Doubt on a math problem
AVY2024   15
N an hour ago by giratina3
Solve for x and y given that xy=923, x+y=84
15 replies
AVY2024
Apr 8, 2025
giratina3
an hour ago
Geo problem
MathWinner121   6
N an hour ago by giratina3
Using analytical geometry, prove that the sum of the squares of a parallelograms diagonals equal the sum of the squares of the side lengths.
6 replies
MathWinner121
Yesterday at 10:15 PM
giratina3
an hour ago
IMO Shortlist 2009 - Problem C5
April   38
N an hour ago by MathematicalArceus
Five identical empty buckets of $2$-liter capacity stand at the vertices of a regular pentagon. Cinderella and her wicked Stepmother go through a sequence of rounds: At the beginning of every round, the Stepmother takes one liter of water from the nearby river and distributes it arbitrarily over the five buckets. Then Cinderella chooses a pair of neighbouring buckets, empties them to the river and puts them back. Then the next round begins. The Stepmother goal's is to make one of these buckets overflow. Cinderella's goal is to prevent this. Can the wicked Stepmother enforce a bucket overflow?

Proposed by Gerhard Woeginger, Netherlands
38 replies
April
Jul 5, 2010
MathematicalArceus
an hour ago
1500th Post!
PikaPika999   124
N an hour ago by PikaPika999
I hit my 1500th post!!
This is just gonna be a 24 game thread then ig :)
You can't use concatenation
To start off, ig make 24 using the following numbers:

3,3,8,8

EDIT: NOTE THAT I DIDN'T MEAN TO GET THIS FROM THE TEAM CONTEST THING! I GOT THIS FROM MY FRIENDS, SO PLEASE STOP BUGGING ME ABOUT THIS
124 replies
PikaPika999
May 1, 2025
PikaPika999
an hour ago
Inspired by Austria 2025
sqing   4
N an hour ago by Tkn
Source: Own
Let $ a,b\geq 0 ,a,b\neq 1$ and $  a^2+b^2=1. $ Prove that$$   (a + b ) \left( \frac{a}{(b -1)^2} + \frac{b}{(a - 1)^2} \right) \geq 12+8\sqrt 2$$
4 replies
sqing
Today at 2:01 AM
Tkn
an hour ago
Erasing the difference of two numbers
BR1F1SZ   1
N 2 hours ago by BR1F1SZ
Source: Austria National MO Part 1 Problem 3
Consider the following game for a positive integer $n$. Initially, the numbers $1, 2, \ldots, n$ are written on a board. In each move, two numbers are selected such that their difference is also present on the board. This difference is then erased from the board. (For example, if the numbers $3,6,11$ and $17$ are on the board, then $3$ can be erased as $6 - 3=3$, or $6$ as $17 - 11=6$, or $11$ as $17 - 6=11$.)

For which values of $n$ is it possible to end with only one number remaining on the board?

(Michael Reitmeir)
1 reply
BR1F1SZ
Yesterday at 9:48 PM
BR1F1SZ
2 hours ago
3-var inequality
sqing   1
N 2 hours ago by Natrium
Source: Own
Let $ a,b,c\geq 0 ,a+b+c =1. $ Prove that
$$\frac{ab}{2c+1} +\frac{bc}{2a+1} +\frac{ca}{2b+1}+\frac{27}{20} abc\leq \frac{1}{4} $$
1 reply
sqing
May 3, 2025
Natrium
2 hours ago
mathemetics
Pangbowen   0
2 hours ago
Let a,b,c≥0 and a+b+c=7. Prove that : a/b+b/c+c/a+abc≥ab+bc+ca-2
0 replies
Pangbowen
2 hours ago
0 replies
Property of a function
Ritangshu   1
N 3 hours ago by Natrium
Let \( f(x, y) = xy \), where \( x \geq 0 \) and \( y \geq 0 \).
Prove that the function \( f \) satisfies the following property:

\[
f\left( \lambda x + (1 - \lambda)x',\; \lambda y + (1 - \lambda)y' \right) > \min\{f(x, y),\; f(x', y')\}
\]
for all \( (x, y) \ne (x', y') \) and for all \( \lambda \in (0, 1) \).

1 reply
Ritangshu
May 3, 2025
Natrium
3 hours ago
max value
Bet667   2
N 3 hours ago by Natrium
Let $a,b$ be a real numbers such that $a^2+ab+b^2\ge a^3+b^3.$Then find maximum value of $a+b$
2 replies
Bet667
4 hours ago
Natrium
3 hours ago
Geometry
gggzul   2
N 3 hours ago by gggzul
In trapezoid $ABCD$ segments $AB$ and $CD$ are parallel. Angle bisectors of $\angle A$ and $\angle C$ meet at $P$. Angle bisectors of $\angle B$ and $\angle D$ meet at $Q$. Prove that $ABPQ$ is cyclic
2 replies
gggzul
4 hours ago
gggzul
3 hours ago
thank you !
Piwbo   2
N 3 hours ago by Piwbo
Given positive integers $a,b$ such that $a$ is even , $b$ is odd and $ab(a+b)^{2023}$ is divisible by $a^{2024}+b^{2024}$ .Prove that there exists a prime number $p$ such that $a^{2024}+b^{2024}$ is divisible by $p^{2025}$
2 replies
+1 w
Piwbo
4 hours ago
Piwbo
3 hours ago
Inequality involving square root cube root and 8th root
bamboozled   2
N 3 hours ago by bamboozled
If $a,b,c,d,e,f,g,h,k\in R^+$ and $a+b+c=d+e+f=g+h+k=8$, then find the maximum value of $\sqrt{ad^3 g^4} +\sqrt[3]{be^3 h^4} + \sqrt[8]{cf^3 k^4}$
2 replies
bamboozled
Today at 4:46 AM
bamboozled
3 hours ago
two solutions
τρικλινο   10
N Apr 14, 2025 by Safal
in a book:CORE MATHS for A-LEVEL ON PAGE 41 i found the following


1st solution


$x^2-5x=0$



$ x(x-5)=0$



hence x=0 or x=5



2nd solution



$x^2-5x=0$

$x-5=0$ dividing by x



hence the solution x=0 has been lost



is the above correct?
10 replies
τρικλινο
Apr 12, 2025
Safal
Apr 14, 2025
two solutions
G H J
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τρικλινο
502 posts
#1
Y by
in a book:CORE MATHS for A-LEVEL ON PAGE 41 i found the following


1st solution


$x^2-5x=0$



$ x(x-5)=0$



hence x=0 or x=5



2nd solution



$x^2-5x=0$

$x-5=0$ dividing by x



hence the solution x=0 has been lost



is the above correct?
Z K Y
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Safal
169 posts
#2
Y by
2nd Solution is basically wrong. Why? Here is the explanation.

$$x(x-5)=0$$Then there are two cases either $x=0$ or $x\neq 0$.When we are admitting the case $x=0$ we cannot divide by $0$. So, in the case we apply divison by $x$ then $x\neq 0$ is a solid prerequisite to do so.Thus, $x-5=0$ from $x(x-5)=0$ we must take the assumption in hand that $x\neq 0$. For example take the extension of the same problem in $\mathbb{F}_5$ then the same problem reads $$x^2=0$$,implying only one solution with optimistic repetation of root $0$, two times that is the multiplicity of $0$ in $x^2$. Thankfully, we are lucky enough that we are in the field of $\text{char}$ $0$.The reason is that, the book you mention was a book for below 10std (as far as I remember it is below 10th std) students, where prerequisite assumption is that ,we should work on field of $\text{char}$ $0$.
This post has been edited 10 times. Last edited by Safal, Apr 12, 2025, 7:52 PM
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τρικλινο
502 posts
#3
Y by
Safal wrote:
2nd Solution is basically wrong. Why? Here is the explanation.

$$x(x-5)=0$$Then there are two cases either $x=0$ or $x\neq 0$.When we are admitting the case $x=0$ we cannot divide by $0$. So, in the case we apply divison by $x$ then $x\neq 0$ is a solid prerequisite to do so.Thus, $x-5=0$ from $x(x-5)=0$ we must take the assumption in hand that $x\neq 0$. For example take the extension of the same problem in $\mathbb{F}_5$ then the same problem reads $$x^2=0$$,implying only one solution with optimistic repetation of root $0$, two times that is the multiplicity of $0$ in $x^2$. Thankfully, we are lucky enough that we are in the field of $\text{char}$ $0$.The reason is that, the book you mention was a book for below 10std (as far as I remember it is below 10th std) students, where prerequisite assumption is that ,we should work on field of $\text{char}$ $0$.

so how do we get x=0 or x=5 ,since we assumed $x\neq 0$.
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maxamc
566 posts
#4
Y by
Move this to MSM, reported
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Safal
169 posts
#5
Y by
τρικλινο wrote:
Safal wrote:
2nd Solution is basically wrong. Why? Here is the explanation.

$$x(x-5)=0$$Then there are two cases either $x=0$ or $x\neq 0$.When we are admitting the case $x=0$ we cannot divide by $0$. So, in the case we apply divison by $x$ then $x\neq 0$ is a solid prerequisite to do so.Thus, $x-5=0$ from $x(x-5)=0$ we must take the assumption in hand that $x\neq 0$. For example take the extension of the same problem in $\mathbb{F}_5$ then the same problem reads $$x^2=0$$,implying only one solution with optimistic repetation of root $0$, two times that is the multiplicity of $0$ in $x^2$. Thankfully, we are lucky enough that we are in the field of $\text{char}$ $0$.The reason is that, the book you mention was a book for below 10std (as far as I remember it is below 10th std) students, where prerequisite assumption is that ,we should work on field of $\text{char}$ $0$.

so how do we get x=0 or x=5 ,since we assumed $x\neq 0$.

If you read carefully I haven't said that we cannot get $x=0$.The assumption whenever $x\neq 0$ we get $x=5$ else we get the case $x=0$.I can explain you more but the fact is I cannot use argument of field theory to explain it in total details.The reason why it's actually the case lies in field theory logics.
This post has been edited 1 time. Last edited by Safal, Apr 13, 2025, 6:23 AM
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τρικλινο
502 posts
#6
Y by
what is field theory logic.
IS the logic that suports the development of field theory?
THIS post should not be moved to Middle School Math
Because in the 2nd solution we have the answer : x different than zero this implies x=5
And according to logic this is equivelant to x=0 or x=5.Hence no solution is lost as the book claims
There for it should be removed back to at least college algebra although i doupt if even there anyone knew of that solution
WEmake use of the law of propositional calculus: ¬p implies q this is equivelant to p or q
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Safal
169 posts
#7
Y by
"If you judge a fish because it cannot climb a tree , it will be foolish"-Unknown.

I am not commenting further in this post thank you.

Thanks to aops for moving it to MSM and I support it.
This post has been edited 1 time. Last edited by Safal, Apr 13, 2025, 4:51 PM
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SpeedCuber7
1833 posts
#8
Y by
@triklino dude that's an awesome username i didn't even know greek letters were allowed lol
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sadas123
1261 posts
#9
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Proof: $x^2-5x=0$ Which means that the roots of this equation have to be real so we can use an method that is lost in the darkness called factoring. We can factor out the $x$ from each of the terms on the left hand side and get $x(x-5)=0$ which with more logic we can find that the possible outcomes is that if the Parantheses are 0 or the x=0 so first we can subsitute a value of x into that to make the value 0 so we get that x=5 and we finally get the solutions of $x=5$ and $x=0$ and to wrap up our proof we can prove that factoring is the best way to go because with quadratics you would only find 2 possibiliteis or 1 depending on the plus minus. And the other thing is that if you divide by x and just solve it with algebra then you will only get the solution of 5. Thus, proving that factoring is the best method out of all of them. We can use the remainder theorem to prove this which can be done easily. $\blacksquare$
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τρικλινο
502 posts
#10
Y by
please read the previous posts


The question here is not which is the best method to solve the problem,but if we lose a solution if we solve the problem by dividing the equation by a non zero x
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Safal
169 posts
#13
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Dear Triklino , I am high yesterday I am sorry for being rude. Can you please explain your Question properly that is what the thing you exactly want to know. According to what I understand, you wanted to know why in 2nd Solution $x=0$ is lost? right. Well forget about field theory and all that, Let me explain it in layman's term what is actually happening. In second solution , the solution $x=0$ is not actually lost. The reason we are getting $x=5$ but not $x=0$ is beacuse when we are dividing by $x$ we making an assumption that $x\neq 0$ and since we are making this assumption the solution $x=0$ is lost. For example when we divide by $x-5$ the solution $x=5$ is lost why $x-5=y(say)$ and we are assuming $y\neq 0$ which is equivalent to $x\neq 5$. Now divison by zero is not possible which is not at all very easy to explain. Now $x=5$ and $x=0$ is not possible at the same time. Thus either $x=0$ or $x=5$.

Now why I am talking about field beacuse $0$ and $5$ can be same when we are in a field of $\text{char} 5$. If you are avoid knowing what is field that's perfectly fine to learn later, but just in layman's term note that $0=5$ is possible in finite fields of charecteristic $5$.

Well I like sour grapes but fox will be happy if he clear your doubt thanks.

I hope it is clear now. If it is not then text me in dm.
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