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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)!

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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
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
Inequalities
sqing   3
N a minute ago by sqing
Let $ a,b> 0 ,\frac{a}{2b+1}+\frac{b}{3}+\frac{1}{2a+1} \leq 1.$ Prove that
$$  a^2+b^2 -ab\leq 1$$$$ a^2+b^2 +ab \leq3$$Let $ a,b,c> 0 , \frac{a}{2b+1}+\frac{b}{2c+1}+\frac{c}{2a+1} \leq 1.$ Prove that
$$    a +b +c +abc \leq 4$$
3 replies
+1 w
sqing
Today at 3:11 AM
sqing
a minute ago
Inequalities
sqing   1
N 9 minutes ago by sqing
Let $ a,b>0 $ and $ \frac { a}{b} +\frac {4b}{a}=5. $ Prove that
$$ \frac {a^{2}+2}{a+2b} + \frac {b^{2}+\frac{1}{2}}{a} \geq \frac{1}{6}\sqrt {\frac {385}{2} }$$$$ \frac {a^{2}+2}{a+2b} + \frac {b^{2}+\frac{1}{3}}{a} \geq \frac {5\sqrt {7}}{6} $$$$ \frac {a^{2}+2}{a+2b} + \frac {b^{2}+\frac{1}{5}}{a} \geq \frac {\sqrt {161}}{6} $$$$ \frac {a^{2}+2}{a+2b} + \frac {b^{2}+3}{a} \geq \frac {\sqrt {455}}{6} $$$$ \frac {a^{2}+2}{a+2b} + \frac {b^{2}+5}{a} \geq \frac {\sqrt {665}}{6} $$
1 reply
sqing
28 minutes ago
sqing
9 minutes ago
[Sipnayan SHS] Written Round, Average, #4.6
LilKirb   1
N 27 minutes ago by maromex
Define the function
\[f(n) = \sum_{k=0}^{n} \binom{n}{k} a_k, \quad n = 1, 2, 3, \ldots\]where
\[a_k = 
    \begin{cases}
        3^k, & \text{if } k \text{ is even}, \\
        0, & \text{if } k \text{ is odd}.
        \end{cases}
\]Find the remainder when \( f(10^9 + 2) \) is divided by \( 2^{20} + 1 \)
1 reply
LilKirb
an hour ago
maromex
27 minutes ago
Nice "if and only if" function problem
ICE_CNME_4   4
N 28 minutes ago by ICE_CNME_4
Let $f : [0, \infty) \to [0, \infty)$, $f(x) = \dfrac{ax + b}{cx + d}$, with $a, d \in (0, \infty)$, $b, c \in [0, \infty)$. Prove that there exists $n \in \mathbb{N}^*$ such that for every $x \geq 0$
\[
f_n(x) = \frac{x}{1 + nx}, \quad \text{if and only if } f(x) = \frac{x}{1 + x}, \quad \forall x \geq 0.
\](For $n \in \mathbb{N}^*$ and $x \geq 0$, the notation $f_n(x)$ represents $\underbrace{(f \circ f \circ \dots \circ f)}_{n \text{ times}}(x)$. )

Please do it at 9th grade level. Thank you!
4 replies
ICE_CNME_4
Yesterday at 7:23 PM
ICE_CNME_4
28 minutes ago
Reflections and midpoints in triangle
TUAN2k8   1
N 28 minutes ago by Funcshun840
Source: Own
Given an triangle $ABC$ and a line $\ell$ in the plane.Let $A_1,B_1,C_1$ be reflections of $A,B,C$ across the line $\ell$, respectively.Let $D,E,F$ be the midpoints of $B_1C_1,C_1A_1,A_1B_1$, respectively.Let $A_2,B_2,C_2$ be the reflections of $A,B,C$ across $D,E,F$, respectively.Prove that the points $A_2,B_2,C_2$ lie on a line parallel to $\ell$.
1 reply
TUAN2k8
2 hours ago
Funcshun840
28 minutes ago
Nice geometry
gggzul   1
N 32 minutes ago by ehuseyinyigit
Let $ABC$ be a acute triangle with $\angle BAC=60^{\circ}$. $H, O$ are the orthocenter and excenter. Let $D$ be a point on the same side of $OH$ as $A$, such that $HDO$ is equilateral. Let $P$ be a point on the same side of $BD$ as $A$, such that $BDP$ is equilateral. Let $Q$ be a point on the same side of $CD$ as $A$, such that $CDP$ is equilateral. Let $M$ be the midpoint of $AD$. Prove that $P, M, Q$ are collinear.
1 reply
gggzul
2 hours ago
ehuseyinyigit
32 minutes ago
Original Problem Unit 1 - Logarithms
ACalculationError   2
N 35 minutes ago by ACalculationError
Let \( a, b, c \) be positive real numbers such that \( abc = 1 \). Compute the minimum value of \(\log(1 + a^2) + \log(1 + b^2) + \log(1 + c^2).\)

Answer Confirmation
2 replies
ACalculationError
an hour ago
ACalculationError
35 minutes ago
Inspired by RMO 2006
sqing   1
N 38 minutes ago by sqing
Source: Own
Let $ a,b>0 $ and $ \frac { a}{b} +\frac {4b}{a}=5. $ Prove that
$$ \frac {a^{2}+2}{a+2b} +  \frac {b^{2}+k}{a}  \geq \frac { \sqrt{5(21k+28)}}{6}  $$Where $ k\geq 0.138. $
$$ \frac {a^{2}+2}{a+2b} +  \frac {b^{2}+1}{a}  \geq \frac {7\sqrt 5}{6}  $$$$ \frac {a^{2}+2}{a+2b} +  \frac {b^{2}+8}{a}  \geq \frac {7\sqrt 5}{3}  $$$$ \frac {a^{2}+2}{a+2b} +  \frac {b^{2}+\frac{1}{4}}{a}  \geq \frac {\sqrt {665}}{12}  $$
1 reply
1 viewing
sqing
2 hours ago
sqing
38 minutes ago
Square NT FE
Why_rF   13
N 43 minutes ago by MathLuis
Source: own
Let $\mathbb{Z} _{>0}$ be the set of positive integers. Find all functions $f: \mathbb{Z} _{>0}\rightarrow \mathbb{Z} _{>0}$ such that
\[f(m)+n \mid m^2f(m) - f(n) \]for all positive integers $m$ and $n$.
13 replies
Why_rF
Apr 14, 2021
MathLuis
43 minutes ago
Russian Diophantine Equation
LeYohan   0
44 minutes ago
Source: Moscow, 1963
Find all integer solutions to

$\frac{xy}{z} + \frac{xz}{y} + \frac{yz}{x} = 3$.
0 replies
LeYohan
44 minutes ago
0 replies
Nice orthocenter config
Rijul saini   12
N an hour ago by Commander_Anta78
Source: India IMOTC 2024 Day 4 Problem 3
Let $ABC$ be an acute-angled triangle with $AB<AC$, and let $O,H$ be its circumcentre and orthocentre respectively. Points $Z,Y$ lie on segments $AB,AC$ respectively, such that \[\angle ZOB=\angle YOC = 90^{\circ}.\]The perpendicular line from $H$ to line $YZ$ meets lines $BO$ and $CO$ at $Q,R$ respectively. Let the tangents to the circumcircle of $\triangle AYZ$ at points $Y$ and $Z$ meet at point $T$. Prove that $Q, R, O, T$ are concyclic.

Proposed by Kazi Aryan Amin and K.V. Sudharshan
12 replies
Rijul saini
May 31, 2024
Commander_Anta78
an hour ago
Arrange marbles
FunGuy1   3
N an hour ago by FunGuy1
Source: Own?
Anna has $200$ marbles in $25$ colors such that there are exactly $8$ marbles of each color. She wants to arrange them on $50$ shelves, $4$ marbles on each shelf such that for any $2$ colors there is a shelf that has marbles of those colors.
Can Anna achieve her goal?
3 replies
FunGuy1
3 hours ago
FunGuy1
an hour ago
Projective geometry
definite_denny   1
N an hour ago by Funcshun840
Source: IDK
Let ABC be a triangle and let DEF be the tangency point of incircirle with sides BC,CA,AB. Points P,Q are chosen on sides AB,AC such that PQ is parallel to BC and PQ is tangent to the incircle. Let M denote the midpoint of PQ. Let EF intersect BC at T. Prove that TM is tangent to the incircle
1 reply
definite_denny
4 hours ago
Funcshun840
an hour ago
Nice problem of concurrency
deraxenrovalo   1
N an hour ago by Funcshun840
Let $(I)$ be an inscribed circle of $\triangle$$ABC$ and touching $BC$, $CA$, $AB$ at $D$, $E$, $F$ respectively. Let $EE'$ and $FF'$ be diameters of $(I)$. Let $X$ and $Y$ be the pole of $DE'$ and $DF'$ with respect to $(I)$, respectively. $BE$ cuts $(I)$ again at $K$. $CF$ cuts $(I)$ again at $L$. The tangent at $K$ of $(I)$ cuts $AX$ at $M$. The tangent at $L$ of $(I)$ cuts $AY$ at $N$. Let $U$ and $V$ be midpoint of $IM$ and $IN$, respectively.

Show that : $UV$, $E'F'$ and perpendicular bisector of $ID$ are concurrent.
1 reply
deraxenrovalo
Today at 4:39 AM
Funcshun840
an hour ago
Octagon Problem
Shiyul   11
N Apr 26, 2025 by Sid-darth-vater
The vertices of octagon $ABCDEFGH$ lie on the same circle. If $AB = BC = CD = DE = 11$ and $EF = FG = GH = HA = \sqrt2$, what is the area of octagon $ABCDEFGH$?

I approached this problem by noticing that the area of the octagon is the area of the eight isoceles triangles with lengths $r$, $r$, and $sqrt2$ or 11. However, I didn't know how to find the radius. Can anyone give me a hint?
11 replies
Shiyul
Apr 24, 2025
Sid-darth-vater
Apr 26, 2025
Octagon Problem
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Shiyul
22 posts
#1
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The vertices of octagon $ABCDEFGH$ lie on the same circle. If $AB = BC = CD = DE = 11$ and $EF = FG = GH = HA = \sqrt2$, what is the area of octagon $ABCDEFGH$?

I approached this problem by noticing that the area of the octagon is the area of the eight isoceles triangles with lengths $r$, $r$, and $sqrt2$ or 11. However, I didn't know how to find the radius. Can anyone give me a hint?
This post has been edited 1 time. Last edited by Shiyul, Apr 25, 2025, 6:47 PM
Reason: latex error
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Sid-darth-vater
43 posts
#2
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Yeah, so what you want to do here is consider angles. They are extremely useful. Try to see what angles to consider yourself. If you can't find anything, look at these angles: Click to reveal hidden text
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Shiyul
22 posts
#3
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Oh yeah, I forgot to say, but I tried that and I noticed that they add to 90, but I'm not sure what to do with that information.
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Sedro
5851 posts
#4
Y by
Building on @Sid-darth-vater's hint: idea

@below oh yeah, how did I miss that ...
you don't need $\angle AOH$ at all
This post has been edited 1 time. Last edited by Sedro, Apr 25, 2025, 1:52 AM
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Sid-darth-vater
43 posts
#5 • 1 Y
Y by Sedro
Shiyul wrote:
Oh yeah, I forgot to say, but I tried that and I noticed that they add to 90, but I'm not sure what to do with that information.
Adding to @Sedro look at the angles of the octagon itself, you might be able to determine an angle
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Shiyul
22 posts
#6
Y by
Okay. I solved the problem, an I'm pretty sure the answer is $121\sqrt3 + 8\sqrt{10}$.
This post has been edited 1 time. Last edited by Shiyul, Apr 25, 2025, 6:06 PM
Reason: casdfasdfa
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Sid-darth-vater
43 posts
#7
Y by
hmm, I got an answer of Click to reveal hidden text, can I ask what you did?
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vanstraelen
9060 posts
#8
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The answer is $167$, the radius is $\frac{\sqrt{290}}{2}$.
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Shiyul
22 posts
#9
Y by
Can you tell me how you found that?
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vanstraelen
9060 posts
#10
Y by
$\sin \alpha=\frac{\frac{11}{2}}{r}=\frac{11}{2r}$ and $\sin \beta=\frac{\frac{\sqrt{2}}{2}}{r}=\frac{\sqrt{2}}{2r}$.

$4 \cdot (2\alpha+2\beta)=360^{\circ} \Rightarrow \alpha+\beta=45^{\circ}$.
$\sin(\alpha+\beta)=\frac{\sqrt{2}}{2}$,
$\sin \alpha\cos \beta+\sin \beta\cos \alpha=\frac{\sqrt{2}}{2}$,
$\frac{11}{2r} \cdot \sqrt{1-(\frac{\sqrt{2}}{2})^{2}}+\frac{\sqrt{2}}{2r} \cdot \sqrt{1-(\frac{11}{2r})^{2}}=\frac{\sqrt{2}}{2} \quad (1)$.
Squaring, calculating, again squaring, two solutions but after (1) only one solution.
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mathprodigy2011
342 posts
#11
Y by
Shiyul wrote:
Oh yeah, I forgot to say, but I tried that and I noticed that they add to 90, but I'm not sure what to do with that information.

law of cosines could be useful, just my speculation, i havent rly solved the problem yet
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Sid-darth-vater
43 posts
#12 • 1 Y
Y by Sedro
Shiyul wrote:
Can you tell me how you found that?

Sure!fullsol also, I'm fairly new to writing proofs so please tell me if any improvements can be made, I'm trying to do better for next year!! :)
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