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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

WOOT early bird pricing is in effect, don’t miss out! If you took MathWOOT Level 2 last year, no worries, it is all new problems this year! Our Worldwide Online Olympiad Training program is for high school level competitors. AoPS designed these courses to help our top students get the deep focus they need to succeed in their specific competition goals. Check out the details at this link for all our WOOT programs in math, computer science, chemistry, and physics.

Looking for summer camps in math and language arts? Be sure to check out the video-based summer camps offered at the Virtual Campus that are 2- to 4-weeks in duration. 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 events:
[list][*]April 3rd (Webinar), 4pm PT/7:00pm ET, Learning with AoPS: Perspectives from a Parent, Math Camp Instructor, and University Professor
[*]April 8th (Math Jam), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS State Discussion
April 9th (Webinar), 4:00pm PT/7:00pm ET, Learn about Video-based Summer Camps at the Virtual Campus
[*]April 10th (Math Jam), 4:30pm PT/7:30pm ET, 2025 MathILy and MathILy-Er Math Jam: Multibackwards Numbers
[*]April 22nd (Webinar), 4:00pm PT/7:00pm ET, Competitive Programming at AoPS (USACO).[/list]
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0 replies
jlacosta
Apr 2, 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
Inspired by SXTX (4)2025 Q712
sqing   1
N 17 minutes ago by sqing
Source: Own
Let $ a ,b,c>0 $ and $ (a+b)^2+2(b+c)^2+(c+a)^2=12. $ Prove that$$  abc(a+b+c) \leq  \frac{9}{5} $$Let $ a ,b,c>0 $ and $ 2(a+b)^2+ (b+c)^2+2(c+a)^2=12. $ Prove that$$  abc(a+b+c) \leq  \frac{9}{8} $$
1 reply
1 viewing
sqing
Yesterday at 11:59 AM
sqing
17 minutes ago
2^x+3^x = yx^2
truongphatt2668   4
N 25 minutes ago by Jackson0423
Prove that the following equation has infinite integer solutions:
$$2^x+3^x = yx^2$$
4 replies
truongphatt2668
Apr 22, 2025
Jackson0423
25 minutes ago
Operations on Pebbles
MarkBcc168   23
N 26 minutes ago by quantam13
Source: ISL 2022 C6
Let $n$ be a positive integer. We start with $n$ piles of pebbles, each initially containing a single pebble. One can perform moves of the following form: choose two piles, take an equal number of pebbles from each pile and form a new pile out of these pebbles. Find (in terms of $n$) the smallest number of nonempty piles that one can obtain by performing a finite sequence of moves of this form.
23 replies
MarkBcc168
Jul 9, 2023
quantam13
26 minutes ago
Polynomials
P162008   1
N 34 minutes ago by thehound
Define a family of polynomials by $P_{0}(x) = x - 2$ and $P_{k}(x) = \left(P_{k - 1} (x)\right)^2 - 2$ if $k \geq 1$ then find the coefficient of $x^2$ in $P_{k}(x)$ in terms of $k.$
1 reply
P162008
Today at 2:05 AM
thehound
34 minutes ago
Irrational equation
giangtruong13   4
N 43 minutes ago by giangtruong13
Solve the equation : $$(\sqrt{x}+1)[2-(x-6)\sqrt{x-3}]=x+8$$
4 replies
giangtruong13
Yesterday at 1:44 PM
giangtruong13
43 minutes ago
\frac{1}{9}+\frac{1}{\sqrt{3}}\geq a^2+\sqrt{a+ b^2} \geq \frac{1}{4}
sqing   3
N an hour ago by sqing
Source: Own
Let $a,b\geq  0 $ and $3a+4b =1 .$ Prove that
$$\frac{2}{3}\geq a +\sqrt{a^2+ 4b^2}\geq  \frac{6}{13}$$$$\frac{1}{9}+\frac{1}{\sqrt{3}}\geq a^2+\sqrt{a+ b^2} \geq  \frac{1}{4}$$$$2\geq a+\sqrt{a^2+16b} \geq  \frac{2}{3}\geq  a+\sqrt{a^2+16b^3} \geq  \frac{2(725-8\sqrt{259})}{729}$$
3 replies
sqing
Oct 3, 2023
sqing
an hour ago
Find all possible values of BT/BM
va2010   54
N an hour ago by ja.
Source: 2015 ISL G4
Let $ABC$ be an acute triangle and let $M$ be the midpoint of $AC$. A circle $\omega$ passing through $B$ and $M$ meets the sides $AB$ and $BC$ at points $P$ and $Q$ respectively. Let $T$ be the point such that $BPTQ$ is a parallelogram. Suppose that $T$ lies on the circumcircle of $ABC$. Determine all possible values of $\frac{BT}{BM}$.
54 replies
va2010
Jul 7, 2016
ja.
an hour ago
Functions
Potla   23
N an hour ago by Ilikeminecraft
Source: 0
Find all functions $ f: \mathbb{R}\longrightarrow \mathbb{R}$ such that
\[f(x+y)+f(y+z)+f(z+x)\ge 3f(x+2y+3z)\]
for all $x, y, z \in \mathbb R$.
23 replies
Potla
Feb 21, 2009
Ilikeminecraft
an hour ago
Functional equation over the integers
Jutaro   32
N an hour ago by Ilikeminecraft
Source: Centroamerican 2020, problem 3
Find all the functions $f: \mathbb{Z}\to \mathbb{Z}$ satisfying the following property: if $a$, $b$ and $c$ are integers such that $a+b+c=0$, then

$$f(a)+f(b)+f(c)=a^2+b^2+c^2.$$
32 replies
Jutaro
Oct 28, 2020
Ilikeminecraft
an hour ago
f(x+y) = max(f(x), y) + min(f(y), x)
Zhero   49
N an hour ago by Ilikeminecraft
Source: ELMO Shortlist 2010, A3
Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that $f(x+y) = \max(f(x),y) + \min(f(y),x)$.

George Xing.
49 replies
Zhero
Jul 5, 2012
Ilikeminecraft
an hour ago
Identical or Periodic?
L567   12
N an hour ago by Ilikeminecraft
Source: India EGMO TST 2023/4
Let $f, g$ be functions $\mathbb{R} \rightarrow \mathbb{R}$ such that for all reals $x,y$, $$f(g(x) + y) = g(x + y)$$Prove that either $f$ is the identity function or $g$ is periodic.

Proposed by Pranjal Srivastava
12 replies
L567
Dec 10, 2022
Ilikeminecraft
an hour ago
f(2) = 7, find all integer functions [Taiwan 2014 Quizzes]
v_Enhance   58
N an hour ago by Ilikeminecraft
Find all increasing functions $f$ from the nonnegative integers to the integers satisfying $f(2)=7$ and \[ f(mn) = f(m) + f(n) + f(m)f(n) \] for all nonnegative integers $m$ and $n$.
58 replies
v_Enhance
Jul 18, 2014
Ilikeminecraft
an hour ago
USAMO 2002 Problem 4
MithsApprentice   90
N an hour ago by Ilikeminecraft
Let $\mathbb{R}$ be the set of real numbers. Determine all functions $f: \mathbb{R} \to \mathbb{R}$ such that \[ f(x^2 - y^2) = x f(x) - y f(y)  \] for all pairs of real numbers $x$ and $y$.
90 replies
MithsApprentice
Sep 30, 2005
Ilikeminecraft
an hour ago
IMO ShortList 2002, algebra problem 1
orl   130
N an hour ago by Ilikeminecraft
Source: IMO ShortList 2002, algebra problem 1
Find all functions $f$ from the reals to the reals such that

\[f\left(f(x)+y\right)=2x+f\left(f(y)-x\right)\]

for all real $x,y$.
130 replies
orl
Sep 28, 2004
Ilikeminecraft
an hour ago
Show that two angles are equal
crazyfehmy   5
N Feb 16, 2025 by ehuseyinyigit
Source: Junior Turkish Mathematical Olympiad 2010 Problem 1
A circle that passes through the vertex $A$ of a rectangle $ABCD$ intersects the side $AB$ at a second point $E$ different from $B.$ A line passing through $B$ is tangent to this circle at a point $T,$ and the circle with center $B$ and passing through $T$ intersects the side $BC$ at the point $F.$ Show that if $\angle CDF= \angle BFE,$ then $\angle EDF=\angle CDF.$
5 replies
crazyfehmy
Jul 24, 2011
ehuseyinyigit
Feb 16, 2025
Show that two angles are equal
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Source: Junior Turkish Mathematical Olympiad 2010 Problem 1
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crazyfehmy
1345 posts
#1 • 2 Y
Y by Adventure10, Mango247
A circle that passes through the vertex $A$ of a rectangle $ABCD$ intersects the side $AB$ at a second point $E$ different from $B.$ A line passing through $B$ is tangent to this circle at a point $T,$ and the circle with center $B$ and passing through $T$ intersects the side $BC$ at the point $F.$ Show that if $\angle CDF= \angle BFE,$ then $\angle EDF=\angle CDF.$
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sunken rock
4384 posts
#2 • 2 Y
Y by Adventure10, Mango247
Hint: See that $F$ is the midpoint of $BC$.

Best regards,
sunken rock
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professordad
4549 posts
#3 • 3 Y
Y by Skidish, Adventure10, Mango247
Wow it took me this long...I really fail
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Skidish
2 posts
#4 • 2 Y
Y by Adventure10, Mango247
We find that BF^2=BE*BA.Then BF is the tangent to the circumcircle of the triangle EFA.That means angle EAF=angle BFE.Since angle CDF=angle BFE,then angle DFE=90.So , the quadrilateral AEFD is cyclic.In that we get angle EDF = angle EAF .So , angle EDF = angle EAF = angle BFE = angle CDF.That`s all.
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Jalil_Huseynov
439 posts
#5
Y by
BT = BF and BT² = BE * BA ―> BF² = BE * BA. It gives us that <BAF = <EFB.
Also <DFB = <FDC + 90° = <EFD + <EFB. Then,<EFD = 90°. And it gives us that AEFD is cyclic.
Then we are done, because <EAF = <EDF = <FDC.
Note: After posting my solution I saw that it is same with @above. Anyway I want to post it.
This post has been edited 2 times. Last edited by Jalil_Huseynov, Dec 19, 2021, 5:14 PM
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ehuseyinyigit
810 posts
#6
Y by
SIMPLE. $BE.BA=BT^2=BF^2$ gives $(AEF)$ is tangent to $BF$. On the other hand, equal angles imply $DF\perp EF\Rightarrow AEFD$ is cyclic. Hence, $\angle CDF=\angle BFE=\angle FDE$.
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