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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
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
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
USAMO Medals
YauYauFilter   17
N 5 minutes ago by rchang25
+1 w
YauYauFilter
Apr 24, 2025
rchang25
5 minutes ago
purple comet discussion
ConfidentKoala4   61
N 13 minutes ago by aopslover08
when can we discuss purple comet
61 replies
+1 w
ConfidentKoala4
May 2, 2025
aopslover08
13 minutes ago
ARMl Local 2025 Final Results
PaulDreyer   28
N 14 minutes ago by lovematch13
Results, problems, and solutions are here. Congratulations to SFBA ARML / AlphaStar: Foxes and Friends and Leading Aces Academy for placing 1st and 2nd overall and to Liam Reddy (Utah Rookies) for their perfect score on the individual round and for being the only student with a perfect score to answer the tiebreaker correctly.
28 replies
PaulDreyer
May 3, 2025
lovematch13
14 minutes ago
NT FE from Taiwan TST
Kitayama_Yuji   13
N 18 minutes ago by bin_sherlo
Source: 2024 Taiwan TST Round 2 Mock P3
Let $\mathbb{N}$ be the set of all positive integers. Find all functions $f\colon \mathbb{N}\to \mathbb{N}$ such that $mf(m)+(f(f(m))+n)^2$ divides $4m^4+n^2f(f(n))^2$ for all positive integers $m$ and $n$.
13 replies
+1 w
Kitayama_Yuji
Mar 29, 2024
bin_sherlo
18 minutes ago
Yet another domino problem
juckter   15
N 21 minutes ago by lksb
Source: EGMO 2019 Problem 2
Let $n$ be a positive integer. Dominoes are placed on a $2n \times 2n$ board in such a way that every cell of the board is adjacent to exactly one cell covered by a domino. For each $n$, determine the largest number of dominoes that can be placed in this way.
(A domino is a tile of size $2 \times 1$ or $1 \times 2$. Dominoes are placed on the board in such a way that each domino covers exactly two cells of the board, and dominoes do not overlap. Two cells are said to be adjacent if they are different and share a common side.)
15 replies
juckter
Apr 9, 2019
lksb
21 minutes ago
Difference of counts of any 2 colors in any interesting rectangle is at most 1
NO_SQUARES   0
22 minutes ago
Source: 239 MO 2025 10-11 p8
Positive integer numbers $n$ and $k > 1$ are given. Losyash likes some of the cells of the $n \times n$ checkerboard. In addition, he is interested in any checkered rectangle with a perimeter of $2n + 2$, the upper-left corner of which coincides with the upper-left corner of the board (there are $n$ such rectangles in total). Given $n$ and $k$, determine whether Losyash can color each cell he likes in one of $k$ colors so that in any rectangle of interest to him the number of cells of any two colors differ by no more than $1$.
0 replies
NO_SQUARES
22 minutes ago
0 replies
Reflection of H about O and SA + SB + SC + AM < AB + BC + CA if US=UM.
NO_SQUARES   0
27 minutes ago
Source: 239 MO 2025 10-11 p7
Point $M$ is the midpoint of side $BC$ of an acute—angled triangle $ABC$. The point $U$ is symmetric to the orthocenter $ABC$ relative to its circumcenter. The point $S$ inside triangle $ABC$ is such that $US = UM$. Prove that $SA + SB + SC + AM < AB + BC + CA$.
0 replies
NO_SQUARES
27 minutes ago
0 replies
If {a^r}={a^s}={a^t}=k, then k=0
NO_SQUARES   0
31 minutes ago
Source: 239 MO 2025 10-11 p6
The real number $a>1$ is given. Suppose that $r$, $s$ and $t$ are different positive integer numbers such that $\{a^r\}=\{a^s\}=\{a^t\}$. Prove that $\{a^r\}=\{a^s\}=\{a^t\}=0$.
0 replies
1 viewing
NO_SQUARES
31 minutes ago
0 replies
MathILy 2025 Decisions Thread
mysterynotfound   32
N 34 minutes ago by tikachaudhuri
Discuss your decisions here!
also share any relevant details about your decisions if you want
32 replies
mysterynotfound
Apr 21, 2025
tikachaudhuri
34 minutes ago
4 wise men and 100 hats. 3 must guess their numbers
NO_SQUARES   0
35 minutes ago
Source: 239 MO 2025 10-11 p5
There are four wise men in a row, each sees only those following him in the row, i.e. the $1$st sees the other three, the $2$nd sees the $3$rd and $4$th, and the $3$rd sees only the $4$th. The devil has $100$ hats, numbered from $1$ to $100$, he puts one hat on each wise man, and hides the extra $96$ hats. After that, each wise man (in turn: first the first, then the second, etc.) loudly calls a number, trying to guess the number of his hat. The numbers mentioned should not be repeated. When all the wise men have spoken, they take off their hats and check which one of them has guessed. Can the sages to act in such a way that at least three of them knowingly guessed?
0 replies
NO_SQUARES
35 minutes ago
0 replies
Flips and flops
NO_SQUARES   0
38 minutes ago
Source: 239 MO 2025 10-11 p4
The numbers from $1$ to $2025$ are arranged in some order in the cells of the $1 \times 2025$ strip. Let's call a flip an operation that takes two arbitrary cells of a strip and swaps the numbers written in them, but only if the larger of these numbers is located to the left of the smaller one. A flop is a set of several flips that do not contain common cells that are executed simultaneously. (For example, a simultaneous flip between the 2nd and 8th cells and a flip between the 5th and 101st cells.) Prove that there exists a sequence of $66$ flops such that for any initial arrangement, applying this sequence of flops to it will result in the numbers being ordered from left to right in ascending order.
0 replies
NO_SQUARES
38 minutes ago
0 replies
Geometry with fix circle
falantrng   32
N 40 minutes ago by bjump
Source: RMM 2018 Problem 6
Fix a circle $\Gamma$, a line $\ell$ to tangent $\Gamma$, and another circle $\Omega$ disjoint from $\ell$ such that $\Gamma$ and $\Omega$ lie on opposite sides of $\ell$. The tangents to $\Gamma$ from a variable point $X$ on $\Omega$ meet $\ell$ at $Y$ and $Z$. Prove that, as $X$ varies over $\Omega$, the circumcircle of $XYZ$ is tangent to two fixed circles.
32 replies
falantrng
Feb 25, 2018
bjump
40 minutes ago
Problem 6
SlovEcience   4
N 44 minutes ago by ZeltaQN2008
Given two points A and B on the unit circle. The tangents to the circle at A and B intersect at point P. Then:
\[ p = \frac{2ab}{a + b} \], \[ p, a, b \in \mathbb{C} \]
4 replies
SlovEcience
May 3, 2025
ZeltaQN2008
44 minutes ago
Perpendicular passes from the intersection of diagonals, \angle AEB = \angle CED
NO_SQUARES   0
44 minutes ago
Source: 239 MO 2025 10-11 p3
Inside of convex quadrilateral $ABCD$ point $E$ was chosen such that $\angle DAE = \angle CAB$ and $\angle ADE = \angle CDB$. Prove that if perpendicular from $E$ to $AD$ passes from the intersection of diagonals of $ABCD$, then $\angle AEB = \angle CED$.
0 replies
NO_SQUARES
44 minutes ago
0 replies
Let's bash some Euler Lines
ABCDE   48
N Sep 2, 2023 by Infinity_Integral
Source: 2014 USAJMO Problem 2
Let $\triangle{ABC}$ be a non-equilateral, acute triangle with $\angle A=60^\circ$, and let $O$ and $H$ denote the circumcenter and orthocenter of $\triangle{ABC}$, respectively.

(a) Prove that line $OH$ intersects both segments $AB$ and $AC$.

(b) Line $OH$ intersects segments $AB$ and $AC$ at $P$ and $Q$, respectively. Denote by $s$ and $t$ the respective areas of triangle $APQ$ and quadrilateral $BPQC$. Determine the range of possible values for $s/t$.
48 replies
ABCDE
Apr 29, 2014
Infinity_Integral
Sep 2, 2023
Let's bash some Euler Lines
G H J
G H BBookmark kLocked kLocked NReply
Source: 2014 USAJMO Problem 2
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ABCDE
1963 posts
#1 • 7 Y
Y by samrocksnature, megarnie, son7, HWenslawski, Adventure10, Mango247, and 1 other user
Let $\triangle{ABC}$ be a non-equilateral, acute triangle with $\angle A=60^\circ$, and let $O$ and $H$ denote the circumcenter and orthocenter of $\triangle{ABC}$, respectively.

(a) Prove that line $OH$ intersects both segments $AB$ and $AC$.

(b) Line $OH$ intersects segments $AB$ and $AC$ at $P$ and $Q$, respectively. Denote by $s$ and $t$ the respective areas of triangle $APQ$ and quadrilateral $BPQC$. Determine the range of possible values for $s/t$.
This post has been edited 1 time. Last edited by djmathman, Apr 6, 2015, 6:43 PM
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AkshajK
4820 posts
#2 • 3 Y
Y by samrocksnature, HWenslawski, Adventure10
Outline
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mymathboy
163 posts
#3 • 3 Y
Y by samrocksnature, HWenslawski, Adventure10
Here is a graph.
Click to reveal hidden text
Attachments:
This post has been edited 6 times. Last edited by mymathboy, Apr 30, 2014, 3:45 AM
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ABCDE
1963 posts
#4 • 4 Y
Y by samrocksnature, HWenslawski, Adventure10, Mango247
2a - The diameter can't go through arc BC because then it doesn't pass through the orthocenter. But wait, it's acute so you can't fit it in one arc so it passes through both and you're done.

2b - I don't think you can do this without trig, but if D is the midpoint of arc BC you get APDQ is a rhombus with one diagonal being 2Rcos something and height of ABC is something R cos something so then everything cancels out nicely and you get a rational linear function which is very easy to find the range of so hoo rah.

Anyone got a "nice" bary, complex, or calc solution?
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DaChickenInc
418 posts
#5 • 4 Y
Y by samrocksnature, HWenslawski, Adventure10, Mango247
Synthetic Solution Sketch
This is a condensation of my fairly rigorous proof.
EDIT: Oops, then I wasted the whole JMO. I was just thinking, "$\angle BOC=120^\circ$ doesn't uniquely define $O$" but thanks to your explanation it turned out to be sufficient to solve the problem. Hopefully I will be more productive on my birthday.
This post has been edited 2 times. Last edited by DaChickenInc, Apr 30, 2014, 12:06 AM
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ABCDE
1963 posts
#6 • 4 Y
Y by samrocksnature, HWenslawski, Adventure10, Mango247
uhh how $BOC=BHC=120$ since $A=60$...?
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numbersandnumbers
258 posts
#7 • 4 Y
Y by samrocksnature, HWenslawski, Adventure10, Mango247
Maybe Nice Solution
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zacchro
179 posts
#8 • 4 Y
Y by samrocksnature, HWenslawski, Adventure10, Mango247
ABCDE wrote:
2b - I don't think you can do this without trig
I got pretty close by continuing coordinate bashing
set point A to be (0,0), C=(1,0), and b=(b,b*sqrt3).
solve for slope of OH=-sqrt3 so APQ is equilateral with area s=(2b+1)^2*sqrt3/36. s/t is then (2b+1)^2/(18b-(2b+1)^2). I couldn't prove that it had range (4/5,1), unfortunately, although I think I could've calculus bashed it to find b=1/2 is the only local minimum.
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mymathboy
163 posts
#9 • 4 Y
Y by samrocksnature, HWenslawski, Adventure10, Mango247
ABCDE wrote:
uhh how $BOC=BHC=120$ since $A=60$...?

$\angle BOC = \angle OBA + \angle BAO + \angle OCA + \angle OAC = 2(\angle BAO + \angle OAC) $
$= 2A = 120^o$

$\angle BHC = 180^o - A = 120^o$
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mathtastic
3258 posts
#10 • 4 Y
Y by samrocksnature, HWenslawski, Adventure10, Mango247
It should say $60^{\circ}$

How can you prove 2b? I got the answer, but how??
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ABCDE
1963 posts
#11 • 4 Y
Y by samrocksnature, HWenslawski, Adventure10, Mango247
You can WLOG it and scale some length to 1 for convenience and pick some angle or length $l$ to vary (some work out better than others), but in the end $s/t$ becomes some function $f(l)$ and you know what domain you need so you get the range.

I personally set the circumradius as 1 and varied $60-B$ and it worked out very nicely.
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tastymath75025
3223 posts
#12 • 4 Y
Y by samrocksnature, HWenslawski, Adventure10, Mango247
So for part a I coordinate bashed.
Then just now, I read Geo Revisited to look at special properties of Euler lines and I see this:
If ABC has the special property that the Euler line is parallel to side BC, then $\tan B \tan C = 3$.

... Great timing -_-

It basically instant-kills the problem.
One angle is 60, $\tan 60 = \sqrt{3}$, so therefore the tangent of the other angle is $\sqrt{3}$, so it is 60 degrees.
However, the triangle is not equilateral, so contradiction.
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qwerty137
3575 posts
#13 • 4 Y
Y by samrocksnature, HWenslawski, Adventure10, Mango247
Somewhat bad way for 2b: (close to) equilateral, then (close to) 30-60-90 right, then assume those are the boundaries
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matholympiad25
426 posts
#14 • 4 Y
Y by samrocksnature, HWenslawski, Adventure10, Mango247
Is it OK if I proved that, according to mathboy's diagram, $\angle AOH < \angle AOB$ and $180^{\circ} - \angle AOH < \angle AOC$ for part 2a?
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ABCDE
1963 posts
#15 • 4 Y
Y by samrocksnature, HWenslawski, Adventure10, Mango247
vincenthuang75025 wrote:
So for part a I coordinate bashed.
Then just now, I read Geo Revisited to look at special properties of Euler lines and I see this:
If ABC has the special property that the Euler line is parallel to side BC, then $\tan B \tan C = 3$.

... Great timing -_-

It basically instant-kills the problem.
One angle is 60, $\tan 60 = \sqrt{3}$, so therefore the tangent of the other angle is $\sqrt{3}$, so it is 60 degrees.
However, the triangle is not equilateral, so contradiction.


Does it? It's segments AB and AC not lines.
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