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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
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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2-var inequality
sqing   8
N 21 minutes ago by sqing
Source: Own
Let $ a,b\geq 0 $. Prove that
$$ \frac{a }{a^2+2b^2+1}+ \frac{b }{b^2+2a^2+1}\leq \frac{1}{\sqrt{3}} $$$$ \frac{a }{2a^2+ b^2+2ab+1}+ \frac{b }{2b^2+ a^2+2ab+1} \leq \frac{1}{\sqrt{5}} $$$$ \frac{a }{2a^2+ b^2+ ab+1}+ \frac{b }{2b^2+ a^2+ ab+1} \leq \frac{1}{2} $$$$\frac{a }{a^2+2b^2+2ab+1}+ \frac{b }{b^2+2a^2+2ab+1}\leq \frac{1}{2} $$
8 replies
sqing
Today at 1:39 AM
sqing
21 minutes ago
J
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Solution
KTYC   0
25 minutes ago





(a + b)/(a + c) = (b + c)/(b + a)


Cross-multiply,

(a + b)² = (a + c)(b + c)

(a + b)² = ab + bc + ac + c²

ab + bc + ac = (a + b)² - c²

ab + bc + ac = (a + b - c)(a + b + c)
—-Eqn(1)

—————————-

Since a, b, c ∈ ℕ,

(a + b + c) ∈ ℕ

(ab + bc + ac) ∈ ℕ


From Eqn(1),

(a + b - c) ∈ ℕ

—————————-

Since (ab + bc + ac) is prime,

The only positive factors of
(ab + bc + ac) are
1 and (ab + bc + ac)


Since c ∈ ℕ,

(a + b - c) < (a + b + c)


Hence,


a + b - c = 1
—-Eqn(2)

a + b + c = ab + bc + ac
—-Eqn(3)


From Eqn(2),

a + b = c + 1
—-Eqn(4)


From Eqn(3),

a + b + c = c(a + b) + ac
—-Eqn(5)


Sub. Eqn(4) into Eqn(5):

c + 1 + c = c(c + 1) + ac

2c + 1 = c² + c + ac

c² + c + ac - 2c - 1 = 0

c² + ac - c - 1 = 0

c² + c(a - 1) - 1 = 0
—-Eqn(6)

——————————

Eqn(6) is a quadratic equation about c


Hence,

Discriminant of Eqn(6)
= (a - 1)² - 4(1)(-1)
= (a - 1)² + 4


Since c ∈ ℕ,
Discriminant of Eqn(6) is a perfect square


Hence,

[(a - 1)² + 4] is a perfect square


Let
(a - 1)² + 4 = k²
where k ∈ ℕ


Hence,

k² - (a - 1)² = 4

[k - (a - 1)][k + (a - 1)] = 4

(k - a + 1)(k + a - 1) = 4
—-Eqn(7)

——————————

Since a, k ∈ ℕ,

a ≥ 1

k ≥ 1


Hence,

(k + a - 1) ≥ (1 + 1 - 1)

(k + a - 1) ≥ 1


Hence,

(k + a - 1) ∈ ℕ


4 ∈ ℕ


Hence,

From Eqn(7),

(k - a + 1) ∈ ℕ


Hence,

4 is a factor of two positive integers in Eqn(7)

——————————

(k - a + 1) + (k + a - 1)
= 2k,
which is even


Hence,

(k - a + 1) and (k + a - 1) have the same parity


Since 4 is even,
(k - a + 1) and (k + a - 1) are even

——————————

Express 4 as a product of two even positive integers,

4 = 2 x 2


Hence,

k - a + 1 = 2
—-Eqn(8)

k + a - 1 = 2
—-Eqn(9)


From Eqn(8) and Eqn(9),

k - a + 1 = k + a - 1

-a + 1 = a - 1

-a - a = -1 - 1

-2a = -2

a = 1
—-Eqn(10)


Sub. Eqn(10) into Eqn(6):

c² + c(1 - 1) - 1 = 0

c² = 1


Hence,

c = 1 —-Eqn(11)
Or
c = -1 (rej. as c > 0)


Sub. Eqn(10) and Eqn(11) into Eqn(4):

1 + b = 1 + 1

b = 1

——————————————————

Therefore,


Ans:

(a, b, c)
= (1, 1, 1)
0 replies
KTYC
25 minutes ago
0 replies
J
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Inequality
MathsII-enjoy   4
N 30 minutes ago by ehuseyinyigit
A interesting problem generalized :-D
4 replies
MathsII-enjoy
Saturday at 1:59 PM
ehuseyinyigit
30 minutes ago
J
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old one but good one
Sunjee   0
35 minutes ago
If $x_1,x_2,...,x_n $ are positive numbers, then prove that
$$\frac{x_1}{1+x_1^2}+\frac{x_2}{1+x_1^2+x_2^2}+\cdots+ \frac{x_n}{1+x_1^2+\cdots+x_n^2}\geq \sqrt{n}$$
0 replies
Sunjee
35 minutes ago
0 replies
J
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Insects walk
Giahuytls2326   1
N an hour ago by removablesingularity
Source: somewhere in the internet
A 100 × 100 chessboard is divided into unit squares. Each square has an arrow pointing up, down, left, or right. The board square is surrounded by a wall, except to the right of the top right corner square. An insect is placed in one of the squares.

Every second, the insect moves one unit in the direction of the arrow in its square. As the insect moves, the arrow of the square it just left rotates 90° clockwise.

If the specified movement cannot be performed, then the insect will not move for that second, but the arrow in the square it is standing on will still rotate. Is it possible that the insect never leaves the board?
1 reply
Giahuytls2326
May 3, 2025
removablesingularity
an hour ago
J
H
something...
SunnyEvan   1
N an hour ago by SunnyEvan
Source: unknown
Try to prove : $$ \sum csc^{20} \frac{2^{i} \pi}{7} csc^{23} \frac{2^{j}\pi }{7} csc^{2023} \frac{2^{k} \pi}{7} $$is a rational number.
Where $ (i,j,k)=(1,2,3) $ and other permutations.
1 reply
SunnyEvan
Today at 1:13 AM
SunnyEvan
an hour ago
J
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something interesting...
SunnyEvan   1
N an hour ago by SunnyEvan
Source: old result
Let $x$, $y$, $z$ be non-negative real numbers, no two of which are zero. Such that $ x+y+z=3.$ Prove that :
$$ \sum \frac{16(9-xyz)}{9(z+x)^2(y+3)^2} \geq \frac{2xyz}{\sum x^2} + \frac{\sum (z^2-x^2)(z^2-y^2)(z+x)(z+y)}{\sum(x+y)^3(y+z)^3} $$
1 reply
SunnyEvan
3 hours ago
SunnyEvan
an hour ago
J
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we can find one pair of a boy and a girl
orl   17
N an hour ago by raghu7
Source: Vietnam TST 2001 for the 42th IMO, problem 3
Some club has 42 members. It’s known that among 31 arbitrary club members, we can find one pair of a boy and a girl that they know each other. Show that from club members we can choose 12 pairs of knowing each other boys and girls.
17 replies
orl
Jun 26, 2005
raghu7
an hour ago
J
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IMO Genre Predictions
ohiorizzler1434   37
N an hour ago by flower417477
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
37 replies
ohiorizzler1434
May 3, 2025
flower417477
an hour ago
J
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Inequality
lgx57   6
N 2 hours ago by MihaiT
Source: Own
$a,b,c>0,ab+bc+ca=1$. Prove that

$$\sum \sqrt{8ab+1} \ge 5$$
(I don't know whether the equality holds)
6 replies
lgx57
Saturday at 3:14 PM
MihaiT
2 hours ago
J
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purple comet discussion
ConfidentKoala4   59
N 2 hours ago by hzbrl
when can we discuss purple comet
59 replies
ConfidentKoala4
May 2, 2025
hzbrl
2 hours ago
J
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USAJMO problem 3: Inequality
BOGTRO   104
N Today at 4:00 AM by justaguy_69
Let $a,b,c$ be positive real numbers. Prove that $\frac{a^3+3b^3}{5a+b}+\frac{b^3+3c^3}{5b+c}+\frac{c^3+3a^3}{5c+a} \geq \frac{2}{3}(a^2+b^2+c^2)$.
104 replies
BOGTRO
Apr 24, 2012
justaguy_69
Today at 4:00 AM
J
H
2025 Math and AI 4 Girls Competition: Win Up To $1,000!!!
audio-on   68
N Today at 3:51 AM by RainbowSquirrel53B
Join the 2025 Math and AI 4 Girls Competition for a chance to win up to $1,000!

Hey Everyone, I'm pleased to announce the dates for the 2025 MA4G Competition are set!
Applications will open on March 22nd, 2025, and they will close on April 26th, 2025 (@ 11:59pm PST).

Applicants will have one month to fill out an application with prizes for the top 50 contestants & cash prizes for the top 20 contestants (including $1,000 for the winner!). More details below!

Eligibility:
The competition is free to enter, and open to middle school female students living in the US (5th-8th grade).
Award recipients are selected based on their aptitude, activities and aspirations in STEM.

Event dates:
Applications will open on March 22nd, 2025, and they will close on April 26th, 2025 (by 11:59pm PST)
Winners will be announced on June 28, 2025 during an online award ceremony.

Application requirements:
Complete a 12 question problem set on math and computer science/AI related topics
Write 2 short essays

Prizes:
1st place: $1,000 Cash prize
2nd place: $500 Cash prize
3rd place: $300 Cash prize
4th-10th: $100 Cash prize each
11th-20th: $50 Cash prize each
Top 50 contestants: Over $50 worth of gadgets and stationary

Many thanks to our current and past sponsors and partners: Hudson River Trading, MATHCOUNTS, Hewlett Packard Enterprise, Automation Anywhere, JP Morgan Chase, D.E. Shaw, and AI4ALL.

Math and AI 4 Girls is a nonprofit organization aiming to encourage young girls to develop an interest in math and AI by taking part in STEM competitions and activities at an early age. The organization will be hosting an inaugural Math and AI 4 Girls competition to identify talent and encourage long-term planning of academic and career goals in STEM.

Contact:
mathandAI4girls@yahoo.com

For more information on the competition:
https://www.mathandai4girls.org/math-and-ai-4-girls-competition

More information on how to register will be posted on the website. If you have any questions, please ask here!


68 replies
audio-on
Jan 26, 2025
RainbowSquirrel53B
Today at 3:51 AM
J
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Question about AMC 10
MathNerdRabbit103   15
N Today at 3:02 AM by GallopingUnicorn45
Hi,

Can anybody predict a good score that I can get on the AMC 10 this November by only being good at counting and probability, number theory, and algebra? I know some geometry because I took it in school though, but it isn’t competition math so it probably doesn’t count.

Thanks.
15 replies
MathNerdRabbit103
May 2, 2025
GallopingUnicorn45
Today at 3:02 AM
J
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J
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AMC and JMO qual question
HungryCalculator   4
N Apr 22, 2025 by eyzMath
Say that on the AMC 10, you do better on the A than the B, but you still qualify for AIME thru both. Then after your AIME, it turns out that you didn’t make JMO through the A+AIME index but you did pass the threshold for the B+AIME index.

does MAA consider your B+AIME index over the A+AIME index and consider you a JMO qualifier even tho Your A test score was higher?

4 replies
HungryCalculator
Apr 17, 2025
eyzMath
Apr 22, 2025
J
AMC and JMO qual question
G H J
Reveal topic tags
AMC 10
AMC
AIME
AIME I
L
G H BBookmark kLocked kLocked NReply
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HungryCalculator
539 posts
HungryCalculator
#1 h  Apr 17, 2025, 12:05 AM
Y by
Say that on the AMC 10, you do better on the A than the B, but you still qualify for AIME thru both. Then after your AIME, it turns out that you didn’t make JMO through the A+AIME index but you did pass the threshold for the B+AIME index.

does MAA consider your B+AIME index over the A+AIME index and consider you a JMO qualifier even tho Your A test score was higher?
Z K Y
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bebebe
992 posts
bebebe
#2 h  Apr 17, 2025, 12:19 AM
Y by
i think in this case you'll make jmo
Z K Y
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Bread10
94 posts
Bread10
#3 h  Apr 17, 2025, 12:47 AM
Y by
Yes of course. A more interesting question would be if you DIDN'T qualify for AIME through B, then would you still be able to do it, which I believe the answer would be no.
This post has been edited 1 time. Last edited by Bread10, Apr 17, 2025, 12:47 AM
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mhgelgi
708 posts
mhgelgi
#4 h  Apr 17, 2025, 12:57 AM
Y by
They have four separate cutoffs {10A + A1, 10A + A2, 10B + A1, 10B + A2}
whichever you succeed in doesn't matter.
Z K Y
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eyzMath
9 posts
eyzMath
#5 h  Apr 22, 2025, 12:04 PM
Y by
yeah i think so
Z K Y
N Quick Reply
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a