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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
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Inequalities
sqing   0
20 minutes ago
Let $ a,b,c\geq 2 . $ Prove that
$$(a-1)(b^2-2)(c-1) - \frac{1}{2}abc\geq -2$$$$(a^2-2)(b-1)(c^2-2) - \frac{3}{2}abc\geq -6$$
0 replies
1 viewing
sqing
20 minutes ago
0 replies
J
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one intersting
teomihai   3
N 24 minutes ago by hexuhdecimal
Let $n $ it is fixed positiv number.
Find $card\{(x,y) | |x|+|y|<n \}$, where $x,y$ are integer numbers .
3 replies
teomihai
Mar 15, 2025
hexuhdecimal
24 minutes ago
J
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Integer equation in 3 variables
Kimchiks926   2
N 33 minutes ago by MuradSafarli
Source: Latvian TST for Baltic Way 2019 Problem 15
Determine all tuples of integers $(a,b,c)$ such that:
$$(a-b)^3(a+b)^2 = c^2 + 2(a-b) + 1$$
2 replies
Kimchiks926
May 29, 2020
MuradSafarli
33 minutes ago
J
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Interesting inequality
sqing   2
N 36 minutes ago by SunnyEvan
Source: Own
Let $ a,b,c\geq 2 . $ Prove that
$$(a^2-1)(b-1)(c^2-1) -\frac{9}{4}abc\geq -9$$$$(a^2-1)(b-1)(c^2-1) -\frac{11}{5}abc\geq -\frac{43}{5}$$$$(a^2-1)(b-1)(c^2-1) -2abc\geq -7$$$$(a-1)(b^2-1)(c-1) -\frac{3}{4}abc\geq -3$$$$(a-1)(b^2-1)(c-1) -\frac{3}{5}abc\geq -\frac{9}{5}$$$$(a-1)(b^2-1)(c-1) -\frac{1}{2}abc\geq -1$$
2 replies
sqing
an hour ago
SunnyEvan
36 minutes ago
J
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Interesting inequality
sqing   1
N 39 minutes ago by sqing
Source: Own
Let $ a,b,c\geq 2 . $ Prove that
$$(a^2-1)(b-\frac{3}{2})(c^2-1) - \frac{9}{4}abc\geq -15$$$$(a^2-1)(b-\frac{3}{2})(c^2-1) - 2abc\geq -\frac{73}{6}$$$$(a^2-1)(b-\frac{3}{2})(c^2-1) - abc\geq -\frac{7}{2}$$$$(a^2-2)(b-\frac{3}{2})(c^2-2) - abc\geq -6$$
1 reply
sqing
an hour ago
sqing
39 minutes ago
J
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Hard problem involving circumcenter and concurrent lines
GeoMetrix   6
N 40 minutes ago by bin_sherlo
Source: AQGO 2020 Problem 3
Let $\triangle{ABC}$ be a triangle with circumcenter $O$. Let $M,N$ be the midpoints of $\overline{AB}$ and $\overline{AC}$ respectively and let $T$ be the projection of $O$ on $\overline{MN}$. Let $D$ be the projection of $A$ on $\overline{BC}$. Let $\overline{TD}$ intersect $\odot(BOC)$ at points $U$ and $V$. Let $\odot(AUV)$ intersct $\overline{MN}$ at points $X,Y$. Let $\overline{AY}$ intersect $\odot(AMN)$ at $R$ and $\overline{AX}$ intersect $\odot(AMN)$ at $S$. Then show that $\overline{AO},\overline{RS},\overline{MN}$ are concurrent.

Proposed by GeoMetrix
6 replies
GeoMetrix
Jun 20, 2020
bin_sherlo
40 minutes ago
J
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Oi! These lines concur
Rg230403   17
N an hour ago by L13832
Source: LMAO 2021 P5, LMAOSL G3(simplified)
Let $I, O$ and $\Gamma$ respectively be the incentre, circumcentre and circumcircle of triangle $ABC$. Points $A_1, A_2$ are chosen on $\Gamma$, such that $AA_1 = AI = AA_2$, and point $A'$ is the foot of the altitude from $I$ to $A_1A_2$. If $B', C'$ are similarly defined, prove that lines $AA', BB'$ and $CC'$ concurr on $OI$.
Original Version from SL
Proposed by Mahavir Gandhi
17 replies
Rg230403
May 10, 2021
L13832
an hour ago
J
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Find the period
Anto0110   1
N an hour ago by Anto0110
Let $a_1, a_2, ..., a_k, ...$ be a sequence that consists of an initial block of $p$ positive distinct integers that then repeat periodically. This means that $\{a_1, a_2, \dots, a_p\}$ are $p$ distinct positive integers and $a_{n+p}=a_n$ for every positive integer $n$. The terms of the sequence are not known and the goal is to find the period $p$. To do this, at each move it possible to reveal the value of a term of the sequence at your choice.
If $p$ is one of the first $k$ prime numbers, find for which values of $k$ there exist a strategy that allows to find $p$ revealing at most $8$ terms of the sequence.
1 reply
Anto0110
Yesterday at 7:37 PM
Anto0110
an hour ago
J
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Discrete problem
ThanhNg   1
N an hour ago by RPCX
Prove that : From any 161 natural numbers, we can always find 81 numbers whose sum is divisible by 81.
1 reply
ThanhNg
Yesterday at 4:23 PM
RPCX
an hour ago
J
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inequality
senku23   2
N an hour ago by User21837561
Let x,y,z in R+ prove that 8(x^3+y^3+z^3)2≥9(x^2+yz)(y^2+zx)(z^2+xy).
2 replies
senku23
3 hours ago
User21837561
an hour ago
J
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Differentiable functional
bakerbakura   2
N an hour ago by Gryphos
Find all differentiable functions $ f;\mathbb{R}\to\mathbb{R}$ such that, for all real numbers $ a,b,t$ with $ 0<t<1$, $ t^2f(a)+(1-t^2)f(b)\geq f(ta+(1-t)b)$
2 replies
bakerbakura
Jan 11, 2010
Gryphos
an hour ago
J
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function equation
cipher703516247   2
N an hour ago by luutrongphuc

Find all the functions $f: \mathbb R^{+} \to \mathbb R^{+}$such that:
$$f(xy^n +f(y)^{2n}) )=f(x)f(y)^n +(yf(y))^n $$
2 replies
cipher703516247
Feb 14, 2025
luutrongphuc
an hour ago
J
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gcd (a^n+b,b^n+a) is constant
EthanWYX2009   78
N 2 hours ago by SimplisticFormulas
Source: 2024 IMO P2
Determine all pairs $(a,b)$ of positive integers for which there exist positive integers $g$ and $N$ such that
$$\gcd (a^n+b,b^n+a)=g$$holds for all integers $n\geqslant N.$ (Note that $\gcd(x, y)$ denotes the greatest common divisor of integers $x$ and $y.$)

Proposed by Valentio Iverson, Indonesia
78 replies
EthanWYX2009
Jul 16, 2024
SimplisticFormulas
2 hours ago
J
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Does there exist a function $f:\mathbb N^* \rightarrow \mathbb Z$ other than a p
kyotaro   0
3 hours ago
Does there exist a function $f:\mathbb N^* \rightarrow \mathbb Z$ other than a polynomial satisfying
$$f(a)-f(b) \mid a-b$$
0 replies
kyotaro
3 hours ago
0 replies
J
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J
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Cool one
MTA_2024   6
N Mar 16, 2025 by MTA_2024
Prove that for all real numbers $a$ and $b$ verifying $a>b>0$ . $$(n+1) \cdot b^n \leq \frac{a^{n+1}-b^{n+1}}{a-b} \leq (n+1) \cdot a^n $$
6 replies
MTA_2024
Mar 15, 2025
MTA_2024
Mar 16, 2025
J
Cool one
G H J
Reveal topic tags
algebra
powers
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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 $a>b>0$ . $$(n+1) \cdot b^n \leq \frac{a^{n+1}-b^{n+1}}{a-b} \leq (n+1) \cdot a^n $$
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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
1106 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
317 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
709 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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