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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.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]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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interesting functional
Pomegranat   2
N 23 minutes ago by Pomegranat
Source: I don't know sorry
Find all functions \( f : \mathbb{R}^+ \to \mathbb{R}^+ \) such that for all positive real numbers \( x \) and \( y \), the following equation holds:
\[ \frac{x + f(y)}{x f(y)} = f\left( \frac{1}{y} + f\left( \frac{1}{x} \right) \right) \]
2 replies
Pomegranat
2 hours ago
Pomegranat
23 minutes ago
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Function equation algebra
TUAN2k8   1
N 25 minutes ago by TUAN2k8
Source: Balkan MO 2025
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x \in \mathbb{R}$ and $y \in \mathbb{R}$,
\begin{align} f(x+yf(x))+y=xy+f(x+y). \end{align}
1 reply
TUAN2k8
40 minutes ago
TUAN2k8
25 minutes ago
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Functional equation with a twist (it's number theory)
Davdav1232   1
N 25 minutes ago by NO_SQUARES
Source: Israel TST 8 2025 p2
Prove that for all primes \( p \) such that \( p \equiv 3 \pmod{4} \) or \( p \equiv 5 \pmod{8} \), there exist integers
\[ 1 \leq a_1 < a_2 < \cdots < a_{(p-1)/2} < p \]such that
\[ \prod_{\substack{1 \leq i < j \leq (p-1)/2}} (a_i + a_j)^2 \equiv 1 \pmod{p}. \]
1 reply
Davdav1232
Thursday at 8:32 PM
NO_SQUARES
25 minutes ago
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Coolabra
Titibuuu   3
N an hour ago by sqing
Let \( a, b, c \) be distinct real numbers such that
\[ a + b + c + \frac{1}{abc} = \frac{19}{2} \]Find the maximum possible value of \( a \).
3 replies
Titibuuu
Today at 2:21 AM
sqing
an hour ago
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HMMT Masters Round
rrusczyk   0
Feb 16, 2011
For those of you college students who miss the camaraderie and challenge of on-site math competitions, the Harvard-MIT Math Tournament offers a college competition. Here's a note from Maria Monks and Rishi Gupta, AoPSers who are running the competition:

[quote="Maria Monks and Rishi Gupta"]
The HMMT (Harvard-MIT Math Tournament) Masters Round is an annual math contest for undergraduates, written by former HMMT directors and problem czars. The contest aims to bring the spirit of competition and the art of problem solving into higher mathematics, with challenging problems in abstract algebra, analysis, topology, combinatorics, calculus, linear algebra, and number theory at the undergraduate level. The Masters Round will consist of a 4 hour, proof-based test with 10 questions of varying difficulty.

Any student currently enrolled in an undergraduate institution is eligible to compete, and awards are given to the top 7 undergraduates. Additionally, there will be prizes awarded for particularly clever or elegant solutions. College graduates may also compete, but they will not be eligible for awards.

The team scoring is as follows. Any undergraduate institution with more than five participating undergraduates automatically becomes a team, and that team comprises all the undergraduates from that school. A school's team score is the sum of the ranks of the top 5 individuals from that school, after all non-team participants are removed from the results. Ties are broken by the 6th place individuals from the respective schools.

Problem submissions from college graduates are welcome. Please send any ideas you may have to hmmt-masters@mit.edu. College graduates are also invited to help grade after the contest on April 2, including those who are competing unofficially, and free dinner will be provided for volunteers.

The second annual Masters Round will be held at Harvard University in the Science Center, Hall A, on April 2, 2011. For details and to sign up, please visit http://hmmt.mit.edu/masters. Hope to see you there![/quote]
0 replies
rrusczyk
Feb 16, 2011
0 replies
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AMC changes
DPatrick   3
N Dec 2, 2010 by dragon96
Today the American Mathematics Competitions announced changes to the AMC10/12 - AIME - USA(J)MO series of contests for high school students.

There are two major changes:

1. The cutoff score to advance from the AMC10 to the AIME is still 120, but in the event that fewer than 2.5% of all students score 120 or higher, they will lower the cutoff score so that the top 2.5% of students advance. (In previous years this percentage was 1%.) (Also, no changes for advancement from AMC12 to AIME: it's still 100 or the top 5%.)

2. Students can qualify for the USAMO only by taking the AMC12. Students can qualify for the USAJMO only by taking the AMC10. If a student takes both the AMC10 and AMC12 and qualifies for both olympiads, he or she must take the USAMO.

Official rules are on the AMC's website. Discussion continues on our AMC forum.
3 replies
DPatrick
Dec 1, 2010
dragon96
Dec 2, 2010
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Math Prize for Girls
DPatrick   1
N Nov 15, 2010 by fprosk
Congratulations to all the winners and participants in the 2010 Math Prize for Girls, which was held this past weekend in New York.

Ravi Boppana, the director of the contest, has very graciously posted all of the problems on our forum.
1 reply
DPatrick
Nov 15, 2010
fprosk
Nov 15, 2010
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USAMTS 2010-11 has started!
DPatrick   0
Oct 8, 2010
The 2010-11 edition of the USA Mathematical Talent Search is now underway. The Round 1 problems have been posted on http://www.usamts.org, and the deadline to submit solutions is November 22.

The USAMTS is different from most math contests: you get several weeks to think about the problems. You can use any resources you like to help you explore the problems (other than asking other people). Then you write up full solutions showing all your work. You'll not only receive scores for your work, but you'll also receive individual feedback on your solutions.

For those who have participated in the past, you'll notice a new format for 2010-11: we're doing 2 rounds of 6 problems each this year. There are many, many more activities in the math contest universe now as compared to when the USAMTS was founded in the late 1980s, and as such we felt that a more streamlined USAMTS would give students more opportunity to participate in both the USAMTS and in other mathematical activities.
0 replies
DPatrick
Oct 8, 2010
0 replies
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International Math Olympiad Problems Now Available
rrusczyk   1
N Jul 9, 2010 by phiReKaLk6781
The problems from the 2010 International Math Olympiad are now available here (in many languages). You can join discussion of the problems on AoPS here. Best of luck to all our community members at the IMO!
1 reply
rrusczyk
Jul 8, 2010
phiReKaLk6781
Jul 9, 2010
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No more topics!
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Inequality with three conditions
oVlad   3
N Apr 22, 2025 by sqing
Source: Romania EGMO TST 2019 Day 1 P3
Let $a,b,c$ be non-negative real numbers such that \[b+c\leqslant a+1,\quad c+a\leqslant b+1,\quad a+b\leqslant c+1.\]Prove that $a^2+b^2+c^2\leqslant 2abc+1.$
3 replies
oVlad
Apr 21, 2025
sqing
Apr 22, 2025
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Inequality with three conditions
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Reveal topic tags
algebra
inequalities
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Source: Romania EGMO TST 2019 Day 1 P3
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oVlad
1746 posts
oVlad
#1 h  Apr 21, 2025, 1:48 PM
Y by
Let $a,b,c$ be non-negative real numbers such that \[b+c\leqslant a+1,\quad c+a\leqslant b+1,\quad a+b\leqslant c+1.\]Prove that $a^2+b^2+c^2\leqslant 2abc+1.$
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Haris1
77 posts
Haris1
#2 h  Apr 21, 2025, 6:50 PM
Y by
Nice ineq,
$3(a+1)(b+1)(c+1)\geq (a+1)(b+1)(a+b)+(a+1)(a+c)(c+1)+(b+c)(b+1)(c+1)$
and using $(a+1)(b+1)(c+1)\geq (a+b)(b+c)(c+a)$ completes it.
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Quantum-Phantom
272 posts
Quantum-Phantom
#3 h  Apr 21, 2025, 11:24 PM
Y by
Because
\begin{align*}8abc+4-4\sum_{\rm cyc}a^2=&\left(1+\sum_{\rm cyc}a\right)\sum_{\rm cyc}(a+1-b-c)(b+1-c-a)\\&+\prod_{\rm cyc}(a+1-b-c)\ge0.\end{align*}
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sqing
42092 posts
sqing
#4 h  Apr 22, 2025, 4:24 AM
Y by
oVlad wrote:
Let $a,b,c$ be non-negative real numbers such that \[b+c\leqslant a+1,\quad c+a\leqslant b+1,\quad a+b\leqslant c+1.\]Prove that $a^2+b^2+c^2\leqslant 2abc+1.$
Indian 2007
https://artofproblemsolving.com/community/c6h1754566p11450296
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