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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
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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
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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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Regular tetrahedron
vanstraelen   7
N 3 minutes ago by ReticulatedPython
Given the points $O(0,0,0),A(1,0,0),B(\frac{1}{2},\frac{\sqrt{3}}{2},0)$
a) Determine the point $C$, above the xy-plane, such that the pyramid $OABC$ is a regular tetrahedron.
b) Calculate the volume.
c) Calculate the radius of the inscribed sphere and the radius of the circumscribed sphere.
7 replies
vanstraelen
May 4, 2025
ReticulatedPython
3 minutes ago
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Geometry Proof
strongstephen   4
N 7 minutes ago by jb2015007
Proof that choosing four distinct points at random has an equal probability of getting a convex quadrilateral vs a concave one.
not cohesive proof alert!

NOTE: By choosing four distinct points, that means no three points lie on the same line on the Gaussian Plane.

Start by picking three of the four points. Next, graph the regions where the fourth point would make the quadrilateral convex or concave. In diagram 1 below, you can see the regions where the fourth point would be convex or concave. Of course, there is the centre region (the shaded triangle), but in an infinite plane, the probability the fourth point ends up in the finite region approaches 0.

Next, I want to prove to you the area of convex/concave, or rather, the probability a point ends up in each area, is the same. Referring to the second diagram, you can flip each concave region over the line perpendicular to the angle bisector of which the region is defined. (Just look at it and you'll get what it means.) Now, each concave region has an almost perfect 1:1 probability correspondence to another convex region. The only difference is the finite region (the triangle, shaded). Again, however, the actual significance (probability) of this approaches 0.

If I call each of the convex region's probability P(a), P(c), and P(e) and the concave ones P(b), P(d), P(f), assuming areas a and b are on opposite sides (same with c and d, e and f) you can get:
P(a) = P(b)
P(c) = P(d)
P(e) = P(f)

and P(a) + P(c) + P(e) = P(convex)
and P(b) + P(d) + P(f) = P(concave)

therefore:
P(convex) = P(concave)
4 replies
strongstephen
Today at 4:54 AM
jb2015007
7 minutes ago
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(not so) small set of residues generates all of F_p upon applying Q many times
62861   14
N 17 minutes ago by john0512
Source: RMM 2019 Problem 6
Find all pairs of integers $(c, d)$, both greater than 1, such that the following holds:

For any monic polynomial $Q$ of degree $d$ with integer coefficients and for any prime $p > c(2c+1)$, there exists a set $S$ of at most $\big(\tfrac{2c-1}{2c+1}\big)p$ integers, such that
\[\bigcup_{s \in S} \{s,\; Q(s),\; Q(Q(s)),\; Q(Q(Q(s))),\; \dots\}\]contains a complete residue system modulo $p$ (i.e., intersects with every residue class modulo $p$).
14 replies
62861
Feb 24, 2019
john0512
17 minutes ago
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find positive n so that exists prime p with p^n-(p-1)^n$ a power of 3
parmenides51   13
N 17 minutes ago by SimplisticFormulas
Source: JBMO Shortlist 2017 NT5
Find all positive integers $n$ such that there exists a prime number $p$, such that $p^n-(p-1)^n$ is a power of $3$.

Note. A power of $3$ is a number of the form $3^a$ where $a$ is a positive integer.
13 replies
parmenides51
Jul 25, 2018
SimplisticFormulas
17 minutes ago
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Functional equation of nonzero reals
proglote   5
N 19 minutes ago by TheHimMan
Source: Brazil MO 2013, problem #3
Find all injective functions $f\colon \mathbb{R}^* \to \mathbb{R}^* $ from the non-zero reals to the non-zero reals, such that \[f(x+y) \left(f(x) + f(y)\right) = f(xy)\] for all non-zero reals $x, y$ such that $x+y \neq 0$.
5 replies
1 viewing
proglote
Oct 24, 2013
TheHimMan
19 minutes ago
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5-th powers is a no-go - JBMO Shortlist
WakeUp   8
N 22 minutes ago by sansgankrsngupta
Prove that there are are no positive integers $x$ and $y$ such that $x^5+y^5+1=(x+2)^5+(y-3)^5$.

Note
8 replies
WakeUp
Oct 30, 2010
sansgankrsngupta
22 minutes ago
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IMO Genre Predictions
ohiorizzler1434   54
N 25 minutes ago by sansgankrsngupta
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
54 replies
ohiorizzler1434
May 3, 2025
sansgankrsngupta
25 minutes ago
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Chess game challenge
adihaya   20
N 25 minutes ago by cursed_tangent1434
Source: 2014 BAMO-12 #5
A chess tournament took place between $2n+1$ players. Every player played every other player once, with no draws. In addition, each player had a numerical rating before the tournament began, with no two players having equal ratings. It turns out there were exactly $k$ games in which the lower-rated player beat the higher-rated player. Prove that there is some player who won no less than $n-\sqrt{2k}$ and no more than $n+\sqrt{2k}$ games.
20 replies
1 viewing
adihaya
Feb 22, 2016
cursed_tangent1434
25 minutes ago
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Permutations inequality
OronSH   13
N 27 minutes ago by sansgankrsngupta
Source: ISL 2023 A5
Let $a_1,a_2,\dots,a_{2023}$ be positive integers such that
[list=disc]
[*] $a_1,a_2,\dots,a_{2023}$ is a permutation of $1,2,\dots,2023$, and
[*] $|a_1-a_2|,|a_2-a_3|,\dots,|a_{2022}-a_{2023}|$ is a permutation of $1,2,\dots,2022$.
[/list]
Prove that $\max(a_1,a_{2023})\ge 507$.
13 replies
OronSH
Jul 17, 2024
sansgankrsngupta
27 minutes ago
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Find all real numbers
sqing   5
N 28 minutes ago by ytChen
Source: IMOC 2021 A1
Find all real numbers x that satisfies$$\sqrt{\sqrt{x-\frac{1}{x}}+\sqrt{1-\frac{1}{x}}-\frac{1}{\sqrt{x-\frac{1}{x}}+\sqrt{1-\frac{1}{x}}}}+\sqrt{1-\frac{1}{\sqrt{x-\frac{1}{x}}+\sqrt{1-\frac{1}{x}}}}=x.$$2021 IMOC Problems
5 replies
sqing
Aug 11, 2021
ytChen
28 minutes ago
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4-var inequality
sqing   0
32 minutes ago
Source: SXTB (4)2025 Q2837
Let $ a,b,c,d> 0 $. Prove that
$$ \frac{1}{(3a+1)^4}+ \frac{1}{(3b+1)^4}+\frac{1}{(3c+1)^4}+\frac{1}{(3d+1)^4} \geq \frac{1}{16(3abcd+1)}$$
0 replies
sqing
32 minutes ago
0 replies
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Looking for users and developers
derekli   16
N 39 minutes ago by John_Mgr
Guys I've been working on a web app that lets you grind high school lvl math. There's AMCs, AIME, BMT, HMMT, SMT etc. Also, it's infinite practice so you can keep grinding without worrying about finding new problems. Please consider helping me out by testing and also consider joining our developer team! :P :blush:

Link: https://stellarlearning.app/competitive
16 replies
derekli
May 4, 2025
John_Mgr
39 minutes ago
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Inspired by Bet667
sqing   0
an hour ago
Source: Own
Let $ a,b $ be a real numbers such that $a^2+kab+b^2\ge a^3+b^3.$Prove that$$a+b\leq k+2$$Where $ k\geq 0. $
0 replies
sqing
an hour ago
0 replies
J
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This shouldn't be a problem 15
derekli   1
N an hour ago by Shan3t
Hey guys I was practicing AIME and came across this problem which is definitely misplaced. It asks for the surface area of a plane within a cylinder which we can easily find out using a projection that is easy to find. I think this should be placed in problem 10 or below. What do you guys think?
1 reply
derekli
an hour ago
Shan3t
an hour ago
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Right Triangle
phiReKaLk6781   4
N Apr 13, 2025 by lpieleanu
A right triangle has sides of integer length. One side has length 11. What is the area of the triangle?
4 replies
phiReKaLk6781
Mar 16, 2010
lpieleanu
Apr 13, 2025
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Right Triangle
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geometry
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phiReKaLk6781
2074 posts
phiReKaLk6781
#1 h  Mar 16, 2010, 10:30 PM • 4 Y
Y by Adventure10, Kizaruno, and 2 other users
A right triangle has sides of integer length. One side has length 11. What is the area of the triangle?
This post has been edited 1 time. Last edited by phiReKaLk6781, Mar 16, 2010, 10:48 PM
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isabella2296
4829 posts
isabella2296
#2 h  Mar 16, 2010, 10:41 PM • 5 Y
Y by Adventure10, Mango247, Kizaruno, and 2 other users
Suppose 11 is a leg. We have $ 121 + b^2 = c^2 \implies 121 = (c+b)(c-b)$. The factors of 121 are (1,121) and (11,11). For (11,11), there's no way for $ c$ and $ b$ to be integers, so we could use (1,121) instead. This gives $ c = 61$ and $ b = 60$.

Thus, the area is $ 30 \times 11 = 330$.
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cmtappu96
274 posts
cmtappu96
#3 h  Apr 3, 2010, 12:44 PM • 2 Y
Y by Adventure10, Kizaruno
isabella2296 wrote:
For (11,11), there's no way for $ c$ and $ b$ to be integers

Correct solution, but just want to point out a mistake-
You said there is no way for $ c$ and $ b$ to be INTEGERS.
But if $ c = 11$ and $ b = 0$, $ c$ and $ b$ are integers, and $ (c + b, c - b) = (11,11)$
Only that degenerates the triangle.
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Abhishek009
1 post
Abhishek009
#4 h  Apr 6, 2010, 11:42 AM • 2 Y
Y by Adventure10, Kizaruno
You can solve this directly if you are aware of Pythagorean triple rule.

For your reference I am providing you with the link. http://en.wikipedia.org/wiki/Pythagorean_triple
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lpieleanu
2994 posts
lpieleanu
#5 h  Apr 13, 2025, 12:13 AM • 1 Y
Y by Kizaruno
Solution
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