![\[\tan^2 \frac {\pi}{16} \cdot \tan^2 \frac {3\pi}{16} + \tan^2 \frac {\pi}{16} \cdot \tan^2 \frac {5\pi}{16}+\tan^2 \frac {3\pi}{16} \cdot \tan^2 \frac {7\pi}{16}+\tan^2 \frac {5\pi}{16} \cdot \tan^2 \frac {7\pi}{16}?\]](http://latex.artofproblemsolving.com/2/d/2/2d248ce9756f0f09265dd538de3b16b22a944c46.png)
Y by
Let 
be real numbers such that 
Prove that 
Where 
be real numbers such that 
Prove that 
Where 
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Metamorphosis of Medial and Contact Triangles
djmathman 102
N
by Mathandski
Source: 2014 USAJMO Problem 6
Let 
be a triangle with incenter 
, incircle 
and circumcircle 
. Let 
be the midpoints of sides 
, 
, 
and let 
be the tangency points of with and , respectively. Let 
be the intersections of line 
with line 
and line 
, respectively, and let 
be the midpoint of arc 
of .
(a) Prove that lies on ray 
.
(b) Prove that line
bisects 
.
be a triangle with incenter 
, incircle 
and circumcircle 
. Let 
be the midpoints of sides 
, 
, 
and let 
be the tangency points of with and , respectively. Let 
be the intersections of line 
with line 
and line 
, respectively, and let 
be the midpoint of arc 
of .(a) Prove that
lies on ray 
.(b) Prove that line
bisects 
.
102 replies
high tech FE as J1?!
imagien_bad 62
N
by jasperE3
Source: USAJMO 2025/1
Let 
be the set of integers, and let 
be a function. Prove that there are infinitely many integers 
such that the function 
defined by 
is not bijective.
Note: A function is bijective if for every integer 
, there exists exactly one integer 
such that 
.
be the set of integers, and let 
be a function. Prove that there are infinitely many integers 
such that the function 
defined by 
is not bijective.Note: A function
is bijective if for every integer 
, there exists exactly one integer 
such that 
.
62 replies
1 viewing
Goals for 2025-2026
Airbus320-214 141
N
by ZMB038
Please write down your goal/goals for competitions here for 2025-2026.
141 replies
[MAIN ROUND STARTS MAY 17] OMMC Year 5
DottedCaculator 62
N
by Craftybutterfly
Hello to all creative problem solvers,
Do you want to work on a fun, untimed team math competition with amazing questions by MOPpers and IMO & EGMO medalists?
Do you want to have a chance to win thousands in cash and raffle prizes (no matter your skill level)?
Check out the fifth annual iteration of the
Online Monmouth Math Competition!
Online Monmouth Math Competition, or OMMC, is a 501c3 accredited nonprofit organization managed by adults, college students, and high schoolers which aims to give talented high school and middle school students an exciting way to develop their skills in mathematics.
Our website: https://www.ommcofficial.org/
Our Discord (6000+ members): https://tinyurl.com/joinommc
Test portal: https://ommc-test-portal-2025.vercel.app/
This is not a local competition; any student 18 or younger anywhere in the world can attend. We have changed some elements of our contest format, so read carefully and thoroughly. Join our Discord or monitor this thread for updates and test releases.
We plan to raffle out a TON of prizes over all competitors regardless of performance. So just submit: a few minutes of your time will give you a great chance to win amazing prizes!
You can check out our past problems and sample problems here:
https://www.ommcofficial.org/sample
https://www.ommcofficial.org/2022-documents
https://www.ommcofficial.org/2023-documents
https://www.ommcofficial.org/ommc-amc
Timeline:
Main Round: May 17th - May 24th
Test Portal Released. The Main Round of the contest is held. The Main Round consists of 25 questions that each have a numerical answer. Teams will have the entire time interval to work on the questions. They can submit any time during the interval. Teams are free to edit their submissions before the period ends, even after they submit.
Final Round: May 26th - May 28th
The top placing teams will qualify for this invitational round (5-10 questions). The final round consists of 5-10 proof questions. Teams again will have the entire time interval to work on these questions and can submit their proofs any time during this interval. Teams are free to edit their submissions before the period ends, even after they submit.
Conclusion of Competition: Early June
Solutions will be released, winners announced, and prizes sent out to winners.
We hope for your participation, and good luck!
OMMC staff
OMMC’S 2025 EVENTS ARE SPONSORED BY:
[list]
[*]Nontrivial Fellowship
[*]Citadel
[*]SPARC
[*]Jane Street
[*]And counting!
[/list]
Do you want to work on a fun, untimed team math competition with amazing questions by MOPpers and IMO & EGMO medalists?
Do you want to have a chance to win thousands in cash and raffle prizes (no matter your skill level)?
Check out the fifth annual iteration of the
Online Monmouth Math Competition!
Online Monmouth Math Competition, or OMMC, is a 501c3 accredited nonprofit organization managed by adults, college students, and high schoolers which aims to give talented high school and middle school students an exciting way to develop their skills in mathematics.
Our website: https://www.ommcofficial.org/
Our Discord (6000+ members): https://tinyurl.com/joinommc
Test portal: https://ommc-test-portal-2025.vercel.app/
This is not a local competition; any student 18 or younger anywhere in the world can attend. We have changed some elements of our contest format, so read carefully and thoroughly. Join our Discord or monitor this thread for updates and test releases.
We plan to raffle out a TON of prizes over all competitors regardless of performance. So just submit: a few minutes of your time will give you a great chance to win amazing prizes!
You can check out our past problems and sample problems here:
https://www.ommcofficial.org/sample
https://www.ommcofficial.org/2022-documents
https://www.ommcofficial.org/2023-documents
https://www.ommcofficial.org/ommc-amc
Timeline:
Main Round: May 17th - May 24th
Test Portal Released. The Main Round of the contest is held. The Main Round consists of 25 questions that each have a numerical answer. Teams will have the entire time interval to work on the questions. They can submit any time during the interval. Teams are free to edit their submissions before the period ends, even after they submit.
Final Round: May 26th - May 28th
The top placing teams will qualify for this invitational round (5-10 questions). The final round consists of 5-10 proof questions. Teams again will have the entire time interval to work on these questions and can submit their proofs any time during this interval. Teams are free to edit their submissions before the period ends, even after they submit.
Conclusion of Competition: Early June
Solutions will be released, winners announced, and prizes sent out to winners.
We hope for your participation, and good luck!
OMMC staff
OMMC’S 2025 EVENTS ARE SPONSORED BY:
[list]
[*]Nontrivial Fellowship
[*]Citadel
[*]SPARC
[*]Jane Street
[*]And counting!
[/list]
62 replies
2v2 (bob lost the game)
GoodMorning 85
N
by maromex
Source: 2023 USAJMO Problem 5/USAMO Problem 4
A positive integer is selected, and some positive integers are written on a board. Alice and Bob play the following game. On Alice's turn, she must replace some integer 
on the board with 
, and on Bob's turn he must replace some even integer on the board with 
. Alice goes first and they alternate turns. If on his turn Bob has no valid moves, the game ends.
After analyzing the integers on the board, Bob realizes that, regardless of what moves Alice makes, he will be able to force the game to end eventually. Show that, in fact, for this value of and these integers on the board, the game is guaranteed to end regardless of Alice's or Bob's moves.
is selected, and some positive integers are written on a board. Alice and Bob play the following game. On Alice's turn, she must replace some integer 
on the board with 
, and on Bob's turn he must replace some even integer on the board with 
. Alice goes first and they alternate turns. If on his turn Bob has no valid moves, the game ends.After analyzing the integers on the board, Bob realizes that, regardless of what moves Alice makes, he will be able to force the game to end eventually. Show that, in fact, for this value of
and these integers on the board, the game is guaranteed to end regardless of Alice's or Bob's moves.
85 replies
Suggestions for preparing for AMC 12
peppermint_cat 3
N
by Konigsberg
So, I have decided to attempt taking the AMC 12 this fall. I don't have any experience with math competitions, and I thought that here might be a good place to see if anyone who has taken the AMC 12 (or done any other math competitions) has any suggestions on what to expect, how to prepare, etc. Thank you!
3 replies
Harmonic Mean
Happytycho 4
N
by elizhang101412
Source: Problem #2 2016 AMC 12B
The harmonic mean of two numbers can be calculated as twice their product divided by their sum. The harmonic mean of 
and 
is closest to which integer?
and 
is closest to which integer?
4 replies
Jane street swag package? USA(J)MO
arfekete 31
N
by vsarg
Hey! People are starting to get their swag packages from Jane Street for qualifying for USA(J)MO, and after some initial discussion on what we got, people are getting different things. Out of curiosity, I was wondering how they decide who gets what.
Please enter the following info:
- USAMO or USAJMO
- Grade
- Score
- Award/Medal/HM
- MOP (yes or no, if yes then color)
- List of items you got in your package
I will reply with my info as an example.
Please enter the following info:
- USAMO or USAJMO
- Grade
- Score
- Award/Medal/HM
- MOP (yes or no, if yes then color)
- List of items you got in your package
I will reply with my info as an example.
31 replies
Inspired by hlminh
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Y by
Here's the proof of the first problem:
By the triangle inequality, we have
|a-kb|+|b-kc|+|c-ka| ≤ |a|+|b|+|c| +k(|a|+|b|+|c|)=(1+k)(|a|+|b|+|c|).
Moreover, if a^2 < b^2, then a<b, given that both a and b are positive. Hence, all we need to show is that the inequality holds for the squares of the given expressions. Thus, we have
(1+k)^2(|a|+|b|+|c|)^2=(1+k)^2(a^2+b^2+c^2+|ab|+|bc|+|ca|)
We are given that a^2+b^2+c^2=1. Hence,
ab+bc+ca ≤ (a^2+b^2)/2 + (b^2+c^2)/2 + (c^2+a^2)/2 = a^2+b^2+c^2=1,
And since this holds for all a, b, c, it surely holds if we substitute a by |a|, b by |b|, and c by |c|.
Thus,
(1+k)^2(a^2+b^2+c^2+|ab|+|bc|+|ca|) ≤ 2(1+k)^2=2k^2+4k+2.
Furthermore, we have
(sqrt(3k^2 + 2k + 3))^2=3k^2 + 2k + 3.
Lastly, 2k^2+4k+2 ≤ 3k^2+2k+3 for all k ≥ 0.
Therefore, the inequality
|a-kb|+|b-kc|+|c-ka| ≤ sqrt(3k^2+2k+3) holds for all k ≥ 0, as required.
By the triangle inequality, we have
|a-kb|+|b-kc|+|c-ka| ≤ |a|+|b|+|c| +k(|a|+|b|+|c|)=(1+k)(|a|+|b|+|c|).
Moreover, if a^2 < b^2, then a<b, given that both a and b are positive. Hence, all we need to show is that the inequality holds for the squares of the given expressions. Thus, we have
(1+k)^2(|a|+|b|+|c|)^2=(1+k)^2(a^2+b^2+c^2+|ab|+|bc|+|ca|)
We are given that a^2+b^2+c^2=1. Hence,
ab+bc+ca ≤ (a^2+b^2)/2 + (b^2+c^2)/2 + (c^2+a^2)/2 = a^2+b^2+c^2=1,
And since this holds for all a, b, c, it surely holds if we substitute a by |a|, b by |b|, and c by |c|.
Thus,
(1+k)^2(a^2+b^2+c^2+|ab|+|bc|+|ca|) ≤ 2(1+k)^2=2k^2+4k+2.
Furthermore, we have
(sqrt(3k^2 + 2k + 3))^2=3k^2 + 2k + 3.
Lastly, 2k^2+4k+2 ≤ 3k^2+2k+3 for all k ≥ 0.
Therefore, the inequality
|a-kb|+|b-kc|+|c-ka| ≤ sqrt(3k^2+2k+3) holds for all k ≥ 0, as required.
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