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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

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0 replies
jlacosta
Mar 2, 2025
0 replies
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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
J
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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
J
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One inequality 1
prof.   2
N 20 minutes ago by invisibleman
If $x>0$ prove inequality $$\sqrt{x}\cdot (x+1)+x\cdot (x-4)+1\ge0.$$
2 replies
prof.
3 hours ago
invisibleman
20 minutes ago
J
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Inequality
JK1603JK   1
N 21 minutes ago by CHESSR1DER
Source: unknown
Let a,b,c>=0 and ab+bc+ca>0 then prove \sqrt{a+b}+\sqrt{c+b}+\sqrt{a+c}\ge 2\sqrt[4]{ab+bc+ca}+\sqrt{\frac{a(b-c)^2+b(c-a)^2+c(a-b)^2}{ab+bc+ca}}
1 reply
1 viewing
JK1603JK
Today at 2:58 AM
CHESSR1DER
21 minutes ago
J
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Cool one
MTA_2024   7
N 29 minutes ago by invisibleman
Prove that for all real numbers $a$ and $b$ verifying $a>b>0$ . $$(n+1) \cdot b^n \leq \frac{a^{n+1}-b^{n+1}}{a-b} \leq (n+1) \cdot a^n $$
7 replies
MTA_2024
Mar 15, 2025
invisibleman
29 minutes ago
J
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geometry coordinates
CHESSR1DER   0
29 minutes ago
Source: simplified version of Belarus TST
Points $A, B, C$ with rational coordinates lie on a plane. It turned out that the distance between every pair of points is an integer. Prove that there exist points $D, E, F$ with integer coordinates such that $AB = DE$, $AC = DF$, $BC = EF$
0 replies
CHESSR1DER
29 minutes ago
0 replies
J
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Cutting a big square into smaller squares
nAalniaOMliO   5
N 32 minutes ago by RagvaloD
Source: Belarusian National Olympiad 2020
A $20 \times 20$ checkered board is cut into several squares with integer side length. The size of a square is it's side length.
What is the maximum amount of different sizes this squares can have?
5 replies
nAalniaOMliO
Jan 29, 2025
RagvaloD
32 minutes ago
J
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a/b + b/a never integer ?
MTA_2024   4
N 36 minutes ago by invisibleman
Let $a$ and $b$ be 2 distinct positive integers.
Can $\frac a b +\frac b a $ be in an integer. Prove why ?
4 replies
MTA_2024
Mar 16, 2025
invisibleman
36 minutes ago
J
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Ah yes, very interesting
Quidditch   23
N 40 minutes ago by quantam13
Source: EGMO 2024 P4
For a sequence $a_1<a_2<\cdots<a_n$ of integers, a pair $(a_i,a_j)$ with $1\leq i<j\leq n$ is called interesting if there exists a pair $(a_k,a_l)$ of integers with $1\leq k<l\leq n$ such that $$\frac{a_l-a_k}{a_j-a_i}=2.$$For each $n\geq 3$, find the largest possible number of interesting pairs in a sequence of length $n$.
23 replies
Quidditch
Apr 14, 2024
quantam13
40 minutes ago
J
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Nice FE as the First Day Finale
swynca   3
N 43 minutes ago by PerfectPlayer
Source: 2025 Turkey TST P3
Find all $f: \mathbb{R} \rightarrow \mathbb{R}$ such that, for all $x,y \in \mathbb{R}-\{0\}$,
$$ f(x) \neq 0 \text{ and } \frac{f(x)}{f(y)} + \frac{f(y)}{f(x)} - f \left( \frac{x}{y}-\frac{y}{x} \right) =2 $$
3 replies
swynca
Mar 18, 2025
PerfectPlayer
43 minutes ago
J
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keep one card and discard the other
Scilyse   1
N an hour ago by g0USinsane777
Source: CGMO 2024 P2
There are $8$ cards on which the numbers $1$, $2$, $\dots$, $8$ are written respectively. Alice and Bob play the following game: in each turn, Alice gives two cards to Bob, who must keep one card and discard the other. The game proceeds for four turns in total; in the first two turns, Bob cannot keep both of the cards with the larger numbers, and in the last two turns, Bob also cannot keep both of the cards with the larger numbers. Let $S$ be the sum of the numbers written on the cards that Bob keeps. Find the greatest positive integer $N$ for which Bob can guarantee that $S$ is at least $N$.
1 reply
Scilyse
Jan 28, 2025
g0USinsane777
an hour ago
J
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Simultaneous eqs. with matrix
RenheMiResembleRice   7
N an hour ago by RenheMiResembleRice
Source: Ningyi Hou
Solve the attached with steps
7 replies
RenheMiResembleRice
4 hours ago
RenheMiResembleRice
an hour ago
J
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Find the minimum
sqing   6
N an hour ago by sqing
Source: 2019 China Mathematical Olympiad Hope League Summer Camp
Let $x,y,z $ be positive real number such that $xyz(x+y+z)=4.$ Find the minimum value of $(x+y)^2+2(y+z)^2+3(z+x)^2.$
6 replies
1 viewing
sqing
Aug 10, 2019
sqing
an hour ago
J
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Interesting inequality
sqing   2
N 2 hours ago by sqing
Source: Own
Let $ a,b\geq 0 . $ Prove that
$$ a^4+b^4 +kab\geq\left(\sqrt{k(k+2)}-k\right)ab(a+b+k)$$Where $ k>0 . $
$$ a^4+b^4 +ab\geq (\sqrt 3-1)ab(a+b+1)$$$$ a^4+b^4 +2ab\geq 2(\sqrt 2-1)ab(a+b+2)$$
2 replies
sqing
5 hours ago
sqing
2 hours ago
J
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Does there exist a function $f:\mathbb N^* \rightarrow \mathbb Z$ other than a p
kyotaro   3
N 5 hours ago by quantam13
Does there exist a function $f:\mathbb N^* \rightarrow \mathbb Z$ other than a polynomial satisfying
$$f(a)-f(b) \mid a-b$$
3 replies
kyotaro
Yesterday at 10:51 AM
quantam13
5 hours ago
J
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solve the system of equations
Havu   3
N Today at 2:27 AM by lbh_qys
Solve the system of equations:
\[\begin{cases} 3x^2-2xy+3y^2+\dfrac{2}{x^2-2xy+y^2}=8\\ 2x+\dfrac{1}{x-y}=4 \end{cases}\]
3 replies
Havu
Yesterday at 7:52 AM
lbh_qys
Today at 2:27 AM
J
Cool one
G H J
Reveal topic tags
algebra
powers
L
G H BBookmark kLocked kLocked NReply
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MTA_2024
19 posts
MTA_2024
#1 h  Mar 15, 2025, 9:09 PM • 1 Y
Y by byron-aj-tom
Prove that for all real numbers $a$ and $b$ verifying $a>b>0$ . $$(n+1) \cdot b^n \leq \frac{a^{n+1}-b^{n+1}}{a-b} \leq (n+1) \cdot a^n $$
Z K Y
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vsarg
239 posts
vsarg
#2 h  Mar 15, 2025, 10:11 PM
Y by
Very easy , Just use the geometric Serie Expansion

(a^(n+1) - b^(n+1)) / (a-b) = a^n + b * a^(n-1) + b^2 * a^(n-2) + ... + a * b^(n-1) + b^n

Since a > b, and Both are positive, this is > (n+1) * b^n and < (n+1) * a^n, And we are done ! Easy Peasy Lemon Squeasy as they say in Kentucky xD :blush:
This post has been edited 1 time. Last edited by vsarg, Mar 15, 2025, 10:12 PM
Reason: Many mistake for me. I edit to ffix it
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EaZ_Shadow
1107 posts
EaZ_Shadow
#3 h  Mar 15, 2025, 10:32 PM
Y by
vsarg wrote:
Very easy , Just use the geometric Serie Expansion

(a^(n+1) - b^(n+1)) / (a-b) = a^n + b * a^(n-1) + b^2 * a^(n-2) + ... + a * b^(n-1) + b^n

Since a > b, and Both are positive, this is > (n+1) * b^n and < (n+1) * a^n, And we are done ! Easy Peasy Lemon Squeasy as they say in Kentucky xD :blush:

Not valid you didnt prove.
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MTA_2024
19 posts
MTA_2024
#4 h  Mar 15, 2025, 10:42 PM
Y by
vsarg wrote:
Very easy , Just use the geometric Serie Expansion

(a^(n+1) - b^(n+1)) / (a-b) = a^n + b * a^(n-1) + b^2 * a^(n-2) + ... + a * b^(n-1) + b^n

Since a > b, and Both are positive, this is > (n+1) * b^n and < (n+1) * a^n, And we are done ! Easy Peasy Lemon Squeasy as they say in Kentucky xD :blush:

EXACTLY WHY IS IT GREATER THAT $(n+1) \cdot b^n$ ?
This post has been edited 1 time. Last edited by MTA_2024, Mar 15, 2025, 10:42 PM
Reason: Miswriting
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no_room_for_error
318 posts
no_room_for_error
#5 h  Mar 15, 2025, 11:55 PM
Y by
EaZ_Shadow wrote:
vsarg wrote:
Very easy , Just use the geometric Serie Expansion

(a^(n+1) - b^(n+1)) / (a-b) = a^n + b * a^(n-1) + b^2 * a^(n-2) + ... + a * b^(n-1) + b^n

Since a > b, and Both are positive, this is > (n+1) * b^n and < (n+1) * a^n, And we are done ! Easy Peasy Lemon Squeasy as they say in Kentucky xD :blush:

Not valid you didnt prove.

His proof is valid (and imo explained thoroughly enough).
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ohiorizzler1434
714 posts
ohiorizzler1434
#6 h  Mar 15, 2025, 11:59 PM
Y by
MTA_2024 wrote:
Prove that for all real numbers $a$ and $b$ verifying $a>b>0$ . $$(n+1) \cdot b^n \leq \frac{a^{n+1}-b^{n+1}}{a-b} \leq (n+1) \cdot a^n $$

bro! the middle term is a^n + a^{n-1}b + ... + b^n, so it has 11 terms made up of mixed powers of a and b. Because a>b, then the middle term is larger than (n+1)b^n. Because b<a, the middle term is less than (n+1)a^n. Easy Peasy Lemon Squeezy! Now that's rizz!



I agree! vsarg's proof is thorough and complete! Anyone who says the opposite is just a hater unwilling to embrace the fundamental property of multiplication and factorisations!
This post has been edited 1 time. Last edited by ohiorizzler1434, Mar 15, 2025, 11:59 PM
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MTA_2024
19 posts
MTA_2024
#7 h  Mar 16, 2025, 12:19 AM
Y by
Sorry I'm dumb, found it out myself a bit later.
I was stuck on a problem till I reached right here. Thought I was still a long way through, before realising what is the fudging problem quoting at the very beginning. $a>b$
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invisibleman
13 posts
invisibleman
#8 h  29 minutes ago
Y by
This is the natural solution above! I have an idea that is not only valid for natural numbers $n$! Let's see! Consider the function $f(x)=x^{n+1}$ on the interval $[a;b]$. If we apply Lagrange's finite growth theorem, then there is a $c$ in the interval $(a;b)$ for which $$\frac{f(b)-f(a)}{b-a}=(n+1){{c}^{n}}$$But since $$a<c<b$$we already got the problem. And we didn't have to use the abbreviated calculation formula anywhere, which is only valid for natural numbers $n$!
This post has been edited 1 time. Last edited by invisibleman, 28 minutes ago
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