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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 9th (Webinar), 4:00pm PT/7:00pm ET, Learn about Video-based Summer Camps at the Virtual Campus
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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
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
Stylish Numbers
pedronis   3
N 5 minutes ago by pedronis
A positive even integer $n$ is called stylish if the set $\{1, 2, \ldots, n\}$ can be partitioned into $\frac{n}{2}$ pairs such that the sum of the elements in each pair is a power of $3$. For example, $6$ is stylish because the set $\{1, 2, 3, 4, 5, 6\}$ can be partitioned as $\{1,2\}, \{3,6\}, \{4,5\}$, with sums $3$, $9$, and $9$ respectively. Determine the number of stylish numbers less than $3^{2025}$.
3 replies
pedronis
Apr 13, 2025
pedronis
5 minutes ago
Cool functional equation
Rayanelba   4
N an hour ago by ATM_
Source: Own
Find all functions $f:\mathbb{Z}_{>0}\to \mathbb{Z}_{>0}$ that verify the following equation for all $x,y\in \mathbb{Z}_{>0}$:
$max(f^{f(y)}(x),f^{f(y)}(y))|min(x,y)$
4 replies
Rayanelba
3 hours ago
ATM_
an hour ago
primes,exponentials,factorials
skellyrah   3
N 2 hours ago by skellyrah
find all primes p,q such that $$ \frac{p^q+q^p-p-q}{p!-q!} $$is a prime number
3 replies
skellyrah
5 hours ago
skellyrah
2 hours ago
Transformation of a cross product when multiplied by matrix A
Math-lover1   2
N 2 hours ago by Math-lover1
I was working through AoPS Volume 2 and this statement from Chapter 11: Cross Products/Determinants confused me.
[quote=AoPS Volume 2]A quick comparison of $|\underline{A}|$ to the cross product $(\underline{A}\vec{i}) \times (\underline{A}\vec{j})$ reveals that a negative determinant [of $\underline{A}$] corresponds to a matrix which reverses the direction of the cross product of two vectors.[/quote]
I understand that this is true for the unit vectors $\vec{i} = (1 \ 0)$ and $\vec{j} = (0 \ 1)$, but am confused on how to prove this statement for general vectors $\vec{v}$ and $\vec{w}$ although its supposed to be a quick comparison.

How do I prove this statement easily with any two 2D vectors?
2 replies
Math-lover1
Yesterday at 10:29 PM
Math-lover1
2 hours ago
af(a)+bf(b)+2ab=x^2 for all natural a, b - show that f(a)=a
shoki   26
N 2 hours ago by MathLuis
Source: Iran TST 2011 - Day 4 - Problem 3
Suppose that $f : \mathbb{N} \rightarrow \mathbb{N}$ is a function for which the expression $af(a)+bf(b)+2ab$ for all $a,b \in \mathbb{N}$ is always a perfect square. Prove that $f(a)=a$ for all $a \in \mathbb{N}$.
26 replies
shoki
May 14, 2011
MathLuis
2 hours ago
Very easy NT
GreekIdiot   8
N 2 hours ago by vsamc
Prove that there exists no natural number $n>1$ such that $n \mid 2^n-1$.
8 replies
GreekIdiot
Today at 2:49 PM
vsamc
2 hours ago
Another quadrilateral in a circle
v_Enhance   110
N 2 hours ago by Marco22
Source: APMO 2013, Problem 5
Let $ABCD$ be a quadrilateral inscribed in a circle $\omega$, and let $P$ be a point on the extension of $AC$ such that $PB$ and $PD$ are tangent to $\omega$. The tangent at $C$ intersects $PD$ at $Q$ and the line $AD$ at $R$. Let $E$ be the second point of intersection between $AQ$ and $\omega$. Prove that $B$, $E$, $R$ are collinear.
110 replies
v_Enhance
May 3, 2013
Marco22
2 hours ago
Rectangle EFGH in incircle, prove that QIM = 90
v_Enhance   64
N 2 hours ago by lpieleanu
Source: Taiwan 2014 TST1, Problem 3
Let $ABC$ be a triangle with incenter $I$, and suppose the incircle is tangent to $CA$ and $AB$ at $E$ and $F$. Denote by $G$ and $H$ the reflections of $E$ and $F$ over $I$. Let $Q$ be the intersection of $BC$ with $GH$, and let $M$ be the midpoint of $BC$. Prove that $IQ$ and $IM$ are perpendicular.
64 replies
v_Enhance
Jul 18, 2014
lpieleanu
2 hours ago
unfair coin, points winning 2024 TMC AIME Mock #9
parmenides51   5
N 2 hours ago by Math-lover1
Krithik has an unfair coin with a $\frac13$ chance of landing heads when flipped. Krithik is playing a game where he starts with $1$ point. Every turn, he flips the coin, and if it lands heads, he gains $1$ point, and if it lands tails, he loses $1$ point. However, after the turn, if he has a negative number of points, his point counter resets to $1$. Krithik wins when he earns $8$ points. Find the expected number of turns until Krithik wins.
5 replies
parmenides51
Apr 26, 2025
Math-lover1
2 hours ago
Queue geo
vincentwant   2
N 2 hours ago by MathLuis
Let $ABC$ be an acute scalene triangle with circumcenter $O$. Let $Y, Z$ be the feet of the altitudes from $B, C$ to $AC, AB$ respectively. Let $D$ be the midpoint of $BC$. Let $\omega_1$ be the circle with diameter $AD$. Let $Q\neq A$ be the intersection of $(ABC)$ and $\omega$. Let $H$ be the orthocenter of $ABC$. Let $K$ be the intersection of $AQ$ and $BC$. Let $l_1,l_2$ be the lines through $Q$ tangent to $\omega,(AYZ)$ respectively. Let $I$ be the intersection of $l_1$ and $KH$. Let $P$ be the intersection of $l_2$ and $YZ$. Let $l$ be the line through $I$ parallel to $HD$ and let $O'$ be the reflection of $O$ across $l$. Prove that $O'P$ is tangent to $(KPQ)$.
2 replies
vincentwant
Today at 3:54 PM
MathLuis
2 hours ago
Functional Geometry
GreekIdiot   2
N 3 hours ago by Double07
Source: BMO 2024 SL G7
Let $f: \pi \to \mathbb R$ be a function from the Euclidean plane to the real numbers such that $f(A)+f(B)+f(C)=f(O)+f(G)+f(H)$ for any acute triangle $\Delta ABC$ with circumcenter $O$, centroid $G$ and orthocenter $H$. Prove that $f$ is constant.
2 replies
GreekIdiot
Apr 27, 2025
Double07
3 hours ago
Right-angled triangle if circumcentre is on circle
liberator   78
N 3 hours ago by bin_sherlo
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
78 replies
liberator
Jan 4, 2016
bin_sherlo
3 hours ago
Can you construct the incenter of a triangle ABC?
PennyLane_31   3
N 3 hours ago by cj13609517288
Source: 2023 Girls in Mathematics Tournament- Level B, Problem 4
Given points $P$ and $Q$, Jaqueline has a ruler that allows tracing the line $PQ$. Jaqueline also has a special object that allows the construction of a circle of diameter $PQ$. Also, always when two circles (or a circle and a line, or two lines) intersect, she can mark the points of the intersection with a pencil and trace more lines and circles using these dispositives by the points marked. Initially, she has an acute scalene triangle $ABC$. Show that Jaqueline can construct the incenter of $ABC$.
3 replies
PennyLane_31
Oct 29, 2023
cj13609517288
3 hours ago
BrUMO 2025 Team Round Problem 15
lpieleanu   1
N 3 hours ago by vanstraelen
Let $\triangle{ABC}$ be an isosceles triangle with $AB=AC.$ Let $D$ be a point on the circumcircle of $\triangle{ABC}$ on minor arc $AB.$ Let $\overline{AD}$ intersect the extension of $\overline{BC}$ at $E.$ Let $F$ be the midpoint of segment $AC,$ and let $G$ be the intersection of $\overline{EF}$ and $\overline{AB}.$ Let the extension of $\overline{DG}$ intersect $\overline{AC}$ and the circumcircle of $\triangle{ABC}$ at $H$ and $I,$ respectively. Given that $DG=3, GH=5,$ and $HI=1,$ compute the length of $\overline{AE}.$
1 reply
lpieleanu
Apr 27, 2025
vanstraelen
3 hours ago
IOQM 2022 P1
L567   7
N Aug 19, 2024 by stonycapybara
Three parallel lines $L_1, L_2, L_2$ are drawn in the plane such that the perpendicular distance between $L_1$ and $L_2$ is $3$ and the perpendicular distance between lines $L_2$ and $L_3$ is also $3$. A square $ABCD$ is constructed such that $A$ lies on $L_1$, $B$ lies on $L_3$ and $C$ lies on $L_2$. Find the area of the square.
7 replies
L567
Mar 8, 2022
stonycapybara
Aug 19, 2024
IOQM 2022 P1
G H J
G H BBookmark kLocked kLocked NReply
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L567
1184 posts
#1 • 3 Y
Y by Master_of_Aops, HWenslawski, son7
Three parallel lines $L_1, L_2, L_2$ are drawn in the plane such that the perpendicular distance between $L_1$ and $L_2$ is $3$ and the perpendicular distance between lines $L_2$ and $L_3$ is also $3$. A square $ABCD$ is constructed such that $A$ lies on $L_1$, $B$ lies on $L_3$ and $C$ lies on $L_2$. Find the area of the square.
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Nuterrow
254 posts
#2 • 2 Y
Y by HWenslawski, stonycapybara
Say the side length is $x$, now say $\angle (L_3, AB)=\theta$, this means $\angle (BC, L_3)=90^{\circ}- \theta$, now you have $x\cos{\theta}=3$ and $x\sin{\theta}=6$, squaring and adding them you get $x^2=45$.
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Armonista
10 posts
#3 • 2 Y
Y by HWenslawski, Prabh2005
18 gang :(
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mathlearner2357
391 posts
#4 • 1 Y
Y by HWenslawski
$\frac{x^2}{2}=3\times\frac{15}{2}$ but i somehow calculated $x^2$ as $30$ instead of $45$ :wallbash:
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NTistrulove
183 posts
#5 • 1 Y
Y by HWenslawski
Take a line $L_0$ above $L_1$ which is again at a distance of $3$. We will have $D\in L_0$, then,

\[ \text{Area}=a^2=6^2+3^2+45 \]
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thrayambhakesh1111
18 posts
#6 • 1 Y
Y by HWenslawski
Redacted
This post has been edited 1 time. Last edited by thrayambhakesh1111, Jul 19, 2022, 4:57 AM
Reason: .
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SatisfiedMagma
458 posts
#7
Y by
Uh oh... That $90^\circ$ angle between $A,B,C$ screams for complex come on! For sure unnecessary, but if first post can be trig, then this is fine as well.

Solution: Toss this onto the complex number with $B$ as origin. By the problem condition, assume $c = m + 3i$ for some real $m$. Then $A$ is precisely $90^\circ$ anti-clockwise rotation of $C$ at $B$. Thus, we can get $a = i \cdot c = -3 + mi$. But now since $a \in L_1$, we get $m = 6$. Apply Pythagoras once to get that side length of square is $\sqrt{6^2 + 3^2}$. Thus the area shall be $\boxed{45}$ which is the desired answer. $\blacksquare$
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stonycapybara
5 posts
#8
Y by
I think this method is easier and more intuitive than every method above:

Drop a perpendicular from A to a point M on L3 and another from C to a point N on L3.
The triangles AMB and NCB can be proved congruent using the ASA congruency criterion.
Hence, in triangle NCB,
BN = AM = 6
CN = 3
BC^2 = BN^2 + CN^2 = 36+9 = 45
As BC is a side of the square ABCD, BC^2 is just the area of the square and thus 45 is the answer.
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