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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
Finding all possible $n$ on a strange division condition!!
MathLuis   10
N 13 minutes ago by justaguy_69
Source: Bolivian Cono Sur Pre-TST 2021 P1
Find the sum of all positive integers $n$ such that
$$\frac{n+11}{\sqrt{n-1}}$$is an integer.
10 replies
MathLuis
Nov 12, 2021
justaguy_69
13 minutes ago
IMO 2012 P5
mathmdmb   123
N 21 minutes ago by SimplisticFormulas
Source: IMO 2012 P5
Let $ABC$ be a triangle with $\angle BCA=90^{\circ}$, and let $D$ be the foot of the altitude from $C$. Let $X$ be a point in the interior of the segment $CD$. Let $K$ be the point on the segment $AX$ such that $BK=BC$. Similarly, let $L$ be the point on the segment $BX$ such that $AL=AC$. Let $M$ be the point of intersection of $AL$ and $BK$.

Show that $MK=ML$.

Proposed by Josef Tkadlec, Czech Republic
123 replies
mathmdmb
Jul 11, 2012
SimplisticFormulas
21 minutes ago
Fixed line
TheUltimate123   14
N 23 minutes ago by amirhsz
Source: ELMO Shortlist 2023 G4
Let \(D\) be a point on segment \(PQ\). Let \(\omega\) be a fixed circle passing through \(D\), and let \(A\) be a variable point on \(\omega\). Let \(X\) be the intersection of the tangent to the circumcircle of \(\triangle ADP\) at \(P\) and the tangent to the circumcircle of \(\triangle ADQ\) at \(Q\). Show that as \(A\) varies, \(X\) lies on a fixed line.

Proposed by Elliott Liu and Anthony Wang
14 replies
TheUltimate123
Jun 29, 2023
amirhsz
23 minutes ago
Computing functions
BBNoDollar   7
N 37 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 $\underbrace{(f \circ f \circ \dots \circ f)}_{n \text{ times}}(x)$. )
7 replies
BBNoDollar
May 18, 2025
ICE_CNME_4
37 minutes ago
RMO 2024 Q2
SomeonecoolLovesMaths   14
N 44 minutes ago by Adywastaken
Source: RMO 2024 Q2
For a positive integer $n$, let $R(n)$ be the sum of the remainders when $n$ is divided by $1,2, \cdots , n$. For example, $R(4) = 0 + 0 + 1 + 0 = 1,$ $R(7) = 0 + 1 + 1 + 3 + 2 + 1 + 0 = 8$. Find all positive integers such that $R(n) = n-1$.
14 replies
SomeonecoolLovesMaths
Nov 3, 2024
Adywastaken
44 minutes ago
Decimal functions in binary
Pranav1056   3
N an hour ago by ihategeo_1969
Source: India TST 2023 Day 3 P1
Let $\mathbb{N}$ be the set of all positive integers. Find all functions $f : \mathbb{N} \rightarrow \mathbb{N}$ such that $f(x) + y$ and $f(y) + x$ have the same number of $1$'s in their binary representations, for any $x,y \in \mathbb{N}$.
3 replies
Pranav1056
Jul 9, 2023
ihategeo_1969
an hour ago
Beautiful numbers in base b
v_Enhance   21
N an hour ago by Martin2001
Source: USEMO 2023, problem 1
A positive integer $n$ is called beautiful if, for every integer $4 \le b \le 10000$, the base-$b$ representation of $n$ contains the consecutive digits $2$, $0$, $2$, $3$ (in this order, from left to right). Determine whether the set of all beautiful integers is finite.

Oleg Kryzhanovsky
21 replies
v_Enhance
Oct 21, 2023
Martin2001
an hour ago
Polynomial method of moving points
MathHorse   6
N an hour ago by Potyka17
Two Hungarian math olympians achieved significant breakthroughs in the field of polynomial moving points. Their main results are summarised in the attached pdf. Check it out!
6 replies
MathHorse
Jun 30, 2023
Potyka17
an hour ago
Intertwined numbers
miiirz30   2
N an hour ago by Gausikaci
Source: 2025 Euler Olympiad, Round 2
Let a pair of positive integers $(n, m)$ that are relatively prime be called intertwined if among any two divisors of $n$ greater than $1$, there exists a divisor of $m$ and among any two divisors of $m$ greater than $1$, there exists a divisor of $n$. For example, pair $(63, 64)$ is intertwined.

a) Find the largest integer $k$ for which there exists an intertwined pair $(n, m)$ such that the product $nm$ is equal to the product of the first $k$ prime numbers.
b) Prove that there does not exist an intertwined pair $(n, m)$ such that the product $nm$ is the product of $2025$ distinct prime numbers.
c) Prove that there exists an intertwined pair $(n, m)$ such that the number of divisors of $n$ is greater than $2025$.

Proposed by Stijn Cambie, Belgium
2 replies
miiirz30
Yesterday at 10:12 AM
Gausikaci
an hour ago
Geometry
shactal   0
an hour ago
Two intersecting circles $C_1$ and $C_2$ have a common tangent that meets $C_1$ in $P$ and $C_2$ in $Q$. The two circles intersect at $M$ and $N$ where $N$ is closer to $PQ$ than $M$ . Line $PN$ meets circle $C_2$ a second time in $R$. Prove that $MQ$ bisects angle $\widehat{PMR}$.
0 replies
shactal
an hour ago
0 replies
Inspired by 2025 KMO
sqing   1
N 2 hours ago by sqing
Source: Own
Let $ a,b,c,d  $ be real numbers satisfying $ a+b+c+d=0 $ and $ a^2+b^2+c^2+d^2= 6 .$ Prove that $$ -\frac{3}{4} \leq abcd\leq\frac{9}{4}$$Let $ a,b,c,d  $ be real numbers satisfying $ a+b+c+d=6 $ and $ a^2+b^2+c^2+d^2= 18 .$ Prove that $$ -\frac{9(2\sqrt{3}+3)}{4} \leq abcd\leq\frac{9(2\sqrt{3}-3)}{4}$$
1 reply
sqing
2 hours ago
sqing
2 hours ago
RMO 2024 Q1
SomeonecoolLovesMaths   25
N 2 hours ago by Adywastaken
Source: RMO 2024 Q1
Let $n>1$ be a positive integer. Call a rearrangement $a_1,a_2, \cdots , a_n$ of $1,2, \cdots , n$ nice if for every $k = 2,3, \cdots , n$, we have that $a_1 + a_2 + \cdots + a_k$ is not divisible by $k$.
(a) If $n>1$ is odd, prove that there is no nice arrangement of $1,2, \cdots , n$.
(b) If $n$ is even, find a nice arrangement of $1,2, \cdots , n$.
25 replies
SomeonecoolLovesMaths
Nov 3, 2024
Adywastaken
2 hours ago
4 variables
Nguyenhuyen_AG   10
N 2 hours ago by Butterfly
Let $a,\,b,\,c,\,d$ are non-negative real numbers and $0 \leqslant k \leqslant \frac{2}{\sqrt{3}}.$ Prove that
$$a^2+b^2+c^2+d^2+kabcd \geqslant k+4+(k+2)(a+b+c+d-4).$$hide
10 replies
Nguyenhuyen_AG
Dec 21, 2020
Butterfly
2 hours ago
2025 KMO Inequality
Jackson0423   3
N 2 hours ago by sqing
Source: 2025 KMO Round 1 Problem 20

Let \(x_1, x_2, \ldots, x_6\) be real numbers satisfying
\[
x_1 + x_2 + \cdots + x_6 = 6,
\]\[
x_1^2 + x_2^2 + \cdots + x_6^2 = 18.
\]Find the maximum possible value of the product
\[
x_1 x_2 x_3 x_4 x_5 x_6.
\]
3 replies
Jackson0423
Yesterday at 4:32 PM
sqing
2 hours ago
Cevian Triangle with Perpendiculars
v_Enhance   4
N Jul 7, 2020 by amar_04
Source: ELMO 2014 Shortlist G10, by Sammy Luo
We are given triangles $ABC$ and $DEF$ such that $D\in BC, E\in CA, F\in AB$, $AD\perp EF, BE\perp FD, CF\perp DE$. Let the circumcenter of $DEF$ be $O$, and let the circumcircle of $DEF$ intersect $BC,CA,AB$ again at $R,S,T$ respectively. Prove that the perpendiculars to $BC,CA,AB$ through $D,E,F$ respectively intersect at a point $X$, and the lines $AR,BS,CT$ intersect at a point $Y$, such that $O,X,Y$ are collinear.

Proposed by Sammy Luo
4 replies
v_Enhance
Jul 24, 2014
amar_04
Jul 7, 2020
Cevian Triangle with Perpendiculars
G H J
G H BBookmark kLocked kLocked NReply
Source: ELMO 2014 Shortlist G10, by Sammy Luo
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v_Enhance
6877 posts
#1 • 5 Y
Y by narutomath96, swamih, Gaussian_cyber, Adventure10, swynca
We are given triangles $ABC$ and $DEF$ such that $D\in BC, E\in CA, F\in AB$, $AD\perp EF, BE\perp FD, CF\perp DE$. Let the circumcenter of $DEF$ be $O$, and let the circumcircle of $DEF$ intersect $BC,CA,AB$ again at $R,S,T$ respectively. Prove that the perpendiculars to $BC,CA,AB$ through $D,E,F$ respectively intersect at a point $X$, and the lines $AR,BS,CT$ intersect at a point $Y$, such that $O,X,Y$ are collinear.

Proposed by Sammy Luo
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XmL
552 posts
#2 • 4 Y
Y by narutomath96, Adventure10, Mango247, and 1 other user
We know that $AD,BE,CF$ concur at the orthocenter of $DEF$, denoted by $H$. Let $X'$ denote the isogonal conjugate of $H$ wrt $ABC$, If $DEF$ isn't the pedal triangle of $X'$, then its pedal triangle is homothetic with $DEF$ which is impossible since the lines that connect the corresponding points of the two homothetic triangles don't concur. Hence $X'\equiv X$. It's a well known property of isogonal conjugates that $O,X,H$ are collinear and $RST$ is the pedal triangle of $H$.

We will prove that $BS\cap CT=Y'$ lies on line $OXH$, in which it's clear that the concurrence at $Y$ follows. Apply Pappus' theorem on $CES,BFT$ and we have $H,Y',ET\cap FS$ are collinear. Let $HS,HT\cap (O)=S',T'$ again and since $SS'\cap TT'=O$, by Pascal theorem we have $O,H,ET\cap FS$ are collinear and hence $O,X,H,ET\cap FS,Y'$ are collinear and we are done.
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TelvCohl
2312 posts
#3 • 5 Y
Y by Gaussian_cyber, Ru83n05, Adventure10, Mango247, and 1 other user
My solution:

Since the lines through $ A, B, C $ and perpendicular to $ EF, FD, DE $ are concurrent at the orthocenter of $\triangle DEF $ ,
so $ \triangle DEF $ and $ \triangle ABC $ are orthologic $ \Longrightarrow $ the perpendiculars through $ D, E, F $ to $ BC, CA, AB $ are concurrent at $ X $ .

Since $ AD, BE, CF $ are concurrent ,
so by Terquem theorem we get $ AR, BS, CT $ are concurrent at $ Y $ .

Since $ \triangle RST $ is the pedal triangle of $ X' $ ($ X' $ is the isogonal conjugate of $ X $ ) ,
so the line passing $ A, B, C $ and perpendicular to $ ST, TR, RS $ are concurrent at $ X $ ,
hence by Sondat theorem (for $\triangle ABC $ and $\triangle RST $ ) we get $ Y, X', X $ are collinear.
ie. $ X, O, Y $ are collinear

Q.E.D

Remark:

(1) Easy to see $ X' = AD \cap BE \cap BF $ which is the orthocenter of $\triangle DEF $

(2) Another interesting property in this configuration :
Denote $ O' $ as the circumcenter of $ \triangle ABC $ , then the isogonal conjugate of $ Y $ lie on $ O'H $ .
(see http://www.artofproblemsolving.com/community/c6h366335 )
This post has been edited 1 time. Last edited by TelvCohl, Apr 13, 2015, 12:53 AM
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JuanOrtiz
366 posts
#4 • 2 Y
Y by Adventure10, Mango247
Here is my solution.

Let $H$ be the orthocenter of $DEF$, which clearly lies on $AD,BE,CF$. By Desargues on $EDF,CH_DB$, where $H_D$ is the foot of the perpendicular from $D$ to $EF$ (define $H_E,H_F$ similarly), we get that $H_EH_F,EF,CB$ are concurrent and therefore $D(T,E,H,F)$ is a harmonic pencil, and so $TEH_DF$ is a harmonic quadrilateral. Thus if $H_D'$ is the reflection of $H_D$ across the perpendicular bisector of $EF$ (which is also clearly the point diametrically opposite to $D$), since $TH_D$ is symmedian of $TEF$ we get that $TH_D'$ is a median of $TEF$. Thus $T$ lies on $MH_D'$, where $M$ is the midpoint of $EF$, but $H$ also lies here, and so these 4 points are collinear. Therefore, $\boxed{\angle HTF=90}$ (since $DH_D'$ is a diameter).

Reflecting $\ell$, the line perpendicular to $CB$ that passes through $D$, through $O$, it is easily seen that it transforms into line $TH$. The analogous happens for $E$ and $F$. Thus the three original lines are concurrent at the reflection of $H$ through $O$. Therefore, $X$ is the reflection of $H$ through $O$. We must now prove $Y$ exists and lies on the Euler line of $\triangle DEF$.

To this end, we invert through $H$ with any radius. Let $Z'$ denote the inverse of $Z$ for any point $Z \neq H$. Lines $HD,HE,HF$ invert to themselves. Notice that the circle $HEH_DH_FS$ has a diameter $HE$, and hence its inverse is a line perpendicular to $HE$ (the analogous applies to $HD,HF$). From this we gather that $D',E',F'$ are the feet of the altitudes in a triangle, and that $S,R,T$ lie on the sides. Since $DEFSRT$ are concyclic, so are their inverses and so $S', R', T'$ are the midpoints of said triangle, and their circumcenter is the 9-point circle. Finally, since $\angle HSA=\angle HTA=90$, we gather that $STAH$ are cyclic, so $S',T',A'$ are collinear. From this, $A'=T'S' \cap HD'$. The analogous occurs for $B',C'$. From now on, for convenience, let $Z$ denote what we previously denoted by $Z'$, for any letter $Z$. Let the big triangle be $XYZ$.

The question is now to prove that the circumcenters of $HTC,HRA,HSB$ have the Euler line of $XYZ$ as a common radical axis (since $O$ inverted to a point on this line, because it also contains the circumcenter of $DEF$, and $H$ and $O$ are collinear with this circumcenter by symmetry). Let $G$ be the barycenter of $XYZ$, let's prove that $G$ has the same power of a point to all 3 circles.

Let $Z_0$ be the intersection of $ZT$ and the 9-point circle. Then $(ZZ_0)(ZT)=(ZS)(ZE)=(ZC)(ZH)$ since $HCSE$ is cyclic because $\angle HCS= \angle HES=90$. Then $THCZ_0$ are concyclic. But $ZT$ is the median of $XYZ$, and so $G$ lies on it. Therefore $G$ has the same power of a point to the circumcircle of $THCZ_0$ than to the 9-point circle, and by analogy we are done. $\blacksquare \text{ Q.E.D}$.
This post has been edited 2 times. Last edited by JuanOrtiz, Apr 13, 2015, 12:09 AM
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amar_04
1916 posts
#5 • 6 Y
Y by GeoMetrix, Bumblebee60, Gaussian_cyber, Ru83n05, Mango247, Mango247
ELMOSL 2014 G10 wrote:
We are given triangles $ABC$ and $DEF$ such that $D\in BC, E\in CA, F\in AB$, $AD\perp EF, BE\perp FD, CF\perp DE$. Let the circumcenter of $DEF$ be $O$, and let the circumcircle of $DEF$ intersect $BC,CA,AB$ again at $R,S,T$ respectively. Prove that the perpendiculars to $BC,CA,AB$ through $D,E,F$ respectively intersect at a point $X$, and the lines $AR,BS,CT$ intersect at a point $Y$, such that $O,X,Y$ are collinear.

Proposed by Sammy Luo

Clearly $AD,BE,CF$ are concurrent at the Orthocenter $U$ of $\Delta DEF$. So, $AR,BS,CT$ are concurrent at the Cyclocevian Conjugate $(Y)$ of $U$ WRT $\Delta ABC$ $(\bigstar)$. Also Clearly $\{\Delta ABC,\Delta DEF\}$ are Orthologic, so the perpendiculars to $BC,CA,AB$ through $D,E,F$ concurs at $X$. Now notice that the Perpendiculars from $R,S,T$ to $BC,CA,AB$ respectively concurs at a point $X^*$ which is the Isogonal Conjugate of $X$. Also the perpendiculars from $A,B,C$ to $ST,RT,SR$ are concurrent at $X$. Hence, $\{\Delta ABC,\Delta RST\}$ are Orthologic with centers of Orthology $\{X,X^*\}$. Also from $(\bigstar)$ we get that $\{\Delta ABC,\Delta RST\}$ are perspective. So, by Sondat's Theorem $Y\in\overline{XX^*}$ but $O\in\overline{XX^*}$. Hence. $\overline{O-X-Y}$. $\blacksquare$
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