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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.

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[*]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
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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
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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
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nice fe with non-linear function being the answer
jjkim0336   2
N 9 minutes ago by jjkim0336
Source: own
f:R+ -> R+

f(xf(y)+y) = y f(y^2 +x)
2 replies
jjkim0336
Apr 8, 2025
jjkim0336
9 minutes ago
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Prove that expression is always even.
shivangjindal   20
N 13 minutes ago by EVKV
Source: INMO 2014- Problem 2
Let $n$ be a natural number. Prove that,
\[ \left\lfloor \frac{n}{1} \right\rfloor+ \left\lfloor \frac{n}{2} \right\rfloor + \cdots + \left\lfloor \frac{n}{n} \right\rfloor + \left\lfloor \sqrt{n} \right\rfloor \]
is even.
20 replies
shivangjindal
Feb 2, 2014
EVKV
13 minutes ago
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Romanian National Olympiad 1997 - Grade 9 - Problem 4
Filipjack   1
N 28 minutes ago by navier3072
Source: Romanian National Olympiad 1997 - Grade 9 - Problem 4
Consider the numbers $a,b, \alpha, \beta \in \mathbb{R}$ and the sets $$A=\left \{x \in \mathbb{R} : x^2+a|x|+b=0 \right \},$$$$B=\left \{ x \in \mathbb{R} : \lfloor x \rfloor^2 + \alpha \lfloor x \rfloor + \beta = 0\right \}.$$If $A \cap B$ has exactly three elements, prove that $a$ cannot be an integer.
1 reply
Filipjack
Apr 6, 2025
navier3072
28 minutes ago
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I need help with this problem
VIATON   0
an hour ago
Let $x,y$ satisfy:
$\frac{4}{(x-1)^2 + (y-2)^2 +4} + \frac{9}{(x-2)^2 + (y-4)^2 +9} = 1$
$[(x-2)^2+(y-4)^2][ (x-1)^2 + (y-2)^2] = 36$
Find Max of :$x+y$
0 replies
VIATON
an hour ago
0 replies
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BMO Shortlist 2021 A5
Lukaluce   17
N an hour ago by jasperE3
Source: BMO Shortlist 2021
Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ such that
$$f(xf(x + y)) = yf(x) + 1$$holds for all $x, y \in \mathbb{R}^{+}$.

Proposed by Nikola Velov, North Macedonia
17 replies
Lukaluce
May 8, 2022
jasperE3
an hour ago
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Function equation
luci1337   3
N an hour ago by jasperE3
find all function $f:R \rightarrow R$ such that:
$2f(x)f(x+y)-f(x^2)=\frac{x}{2}(f(2x)+f(f(y)))$ with all $x,y$ is real number
3 replies
luci1337
Yesterday at 3:01 PM
jasperE3
an hour ago
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Circumcenter of reflection of collinear points over sides
a1267ab   27
N an hour ago by Giant_PT
Source: USA TST 2025
Let $ABC$ be a triangle, and let $X$, $Y$, and $Z$ be collinear points such that $AY=AZ$, $BZ=BX$, and $CX=CY$. Points $X'$, $Y'$, and $Z'$ are the reflections of $X$, $Y$, and $Z$ over $BC$, $CA$, and $AB$, respectively. Prove that if $X'Y'Z'$ is a nondegenerate triangle, then its circumcenter lies on the circumcircle of $ABC$.

Michael Ren
27 replies
a1267ab
Jan 11, 2025
Giant_PT
an hour ago
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a+b+c=abc
KhuongTrang   1
N an hour ago by KhuongTrang
Source: own
Problem. Let $a,b,c$ be three positive real numbers satisfying $a+b+c=abc.$ Prove that$$\sqrt{a^2+b^2+3}+\sqrt{b^2+c^2+3}+\sqrt{c^2+a^2+3}\ge4\cdot \frac{a^2b^2c^2-3}{ab+bc+ca-3}-7.$$There is a very elegant proof :-D Could anyone think of it?
1 reply
KhuongTrang
Wednesday at 11:51 AM
KhuongTrang
an hour ago
J
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multiple of 15-15 positive factors
britishprobe17   0
2 hours ago
Source: KTOM Maret 2025
Find the sum of all natural numbers $n$ such that $n$ is a multiple of $15$ and has exactly $15$ positive factors.
0 replies
britishprobe17
2 hours ago
0 replies
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general form
pennypc123456789   0
2 hours ago
If $a,b,c$ are positive real numbers, $k \ge 3$ then
$$ \frac{a + b}{a + kb + c} + \dfrac{b + c}{b + kc + a}+\dfrac{c + a}{c + ka + b} \geq \dfrac{6}{k+2}$$
0 replies
pennypc123456789
2 hours ago
0 replies
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Multi-equation
giangtruong13   2
N 2 hours ago by cazanova19921
Solve equations: $$\begin{cases} x^4+x^3y+x^2y^2=7x+9 \\ x(y-x+1)=3 \end{cases} $$
2 replies
giangtruong13
Yesterday at 12:30 PM
cazanova19921
2 hours ago
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Right-angled triangle if circumcentre is on circle
liberator   77
N 2 hours ago by Ihatecombin
Source: IMO 2013 Problem 3
Let the excircle of triangle $ABC$ opposite the vertex $A$ be tangent to the side $BC$ at the point $A_1$. Define the points $B_1$ on $CA$ and $C_1$ on $AB$ analogously, using the excircles opposite $B$ and $C$, respectively. Suppose that the circumcentre of triangle $A_1B_1C_1$ lies on the circumcircle of triangle $ABC$. Prove that triangle $ABC$ is right-angled.

Proposed by Alexander A. Polyansky, Russia
77 replies
liberator
Jan 4, 2016
Ihatecombin
2 hours ago
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Beautiful geometry
m4thbl3nd3r   2
N 2 hours ago by Captainscrubz
Let $\omega$ be the circumcircle of triangle $ABC$, $M$ is the midpoint of $BC$ and $E$ be the second intersection of $AM$ and $\omega$. Tangent line of $\omega$ at $E$ intersects $BC$ at $P$, let $PKL$ be a transversal of $\omega$ and $X,Y$ be intersections of $AK,AL$ with $BC$. Let $PF$ be a tangent line of $\omega$. Prove that $LYFP$ is cyclic
2 replies
m4thbl3nd3r
Yesterday at 4:41 PM
Captainscrubz
2 hours ago
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Maximum with positive integers
SMOJ   3
N 2 hours ago by lightsynth123
Source: 2018 Singapore Mathematical Olympiad Senior Q4
Let $a,b,c,d$ be positive integers such that $a+c=20$ and $\frac{a}{b}+\frac{c}{d}<1$. Find the maximum possible value of $\frac{a}{b}+\frac{c}{d}$.
3 replies
SMOJ
Mar 31, 2020
lightsynth123
2 hours ago
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Quadrilateral
sqing   2
N Apr 11, 2019 by ririri
Source: Nordic 2019, P3
The quadrilateral $ABCD$ satisfies $\angle ACD = 2\angle CAB, \angle ACB = 2\angle CAD $ and $CB = CD.$ Show that $$\angle CAB=\angle CAD.$$
2 replies
sqing
Apr 7, 2019
ririri
Apr 11, 2019
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Quadrilateral
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Reveal topic tags
geometry
quadrilateral
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Source: Nordic 2019, P3
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sqing
41677 posts
sqing
#1 h  Apr 7, 2019, 5:39 AM • 1 Y
Y by Adventure10
The quadrilateral $ABCD$ satisfies $\angle ACD = 2\angle CAB, \angle ACB = 2\angle CAD $ and $CB = CD.$ Show that $$\angle CAB=\angle CAD.$$
This post has been edited 1 time. Last edited by sqing, Apr 7, 2019, 6:17 AM
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MathInfinite
187 posts
MathInfinite
#2 h  Apr 7, 2019, 1:12 PM • 3 Y
Y by ktddtk1977, Adventure10, Mango247
sqing wrote:
The quadrilateral $ABCD$ satisfies $\angle ACD = 2\angle CAB, \angle ACB = 2\angle CAD $ and $CB = CD.$ Show that $$\angle CAB=\angle CAD.$$

Let $X = AB \cap CD$ and $Y = AD \cap BC$. Note that $\angle CAX = \angle AXC$. So, $AC = CX$. Similarly, $AC = CY$. So, $C$ is the centre of $AXY$. Note that $\triangle BCX \cong \triangle CDY$. Thus, $\angle CAB = \angle CAD$. $\blacksquare$
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ririri
24 posts
ririri
#3 h  Apr 11, 2019, 10:19 AM • 1 Y
Y by Adventure10
This is almost exactly Canada 2000/4.
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