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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

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0 replies
jlacosta
Mar 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
Local-global with Fibonacci numbers
MarkBcc168   26
N 4 minutes ago by pi271828
Source: ELMO 2020 P2
Define the Fibonacci numbers by $F_1 = F_2 = 1$ and $F_n = F_{n-1} + F_{n-2}$ for $n\geq 3$. Let $k$ be a positive integer. Suppose that for every positive integer $m$ there exists a positive integer $n$ such that $m \mid F_n-k$. Must $k$ be a Fibonacci number?

Proposed by Fedir Yudin.
26 replies
MarkBcc168
Jul 28, 2020
pi271828
4 minutes ago
Cauchy functional equations
syk0526   10
N 6 minutes ago by imagien_bad
Source: FKMO 2013 #2
Find all functions $ f : \mathbb{R}\to\mathbb{R}$ satisfying following conditions.
(a) $ f(x) \ge 0 $ for all $ x \in \mathbb{R} $.
(b) For $ a, b, c, d \in \mathbb{R} $ with $ ab + bc + cd = 0 $, equality $ f(a-b) + f(c-d) = f(a) + f(b+c) + f(d) $ holds.
10 replies
syk0526
Mar 24, 2013
imagien_bad
6 minutes ago
Three circles are concurrent
Twoisaprime   21
N 16 minutes ago by L13832
Source: RMM 2025 P5
Let triangle $ABC$ be an acute triangle with $AB<AC$ and let $H$ and $O$ be its orthocenter and circumcenter, respectively. Let $\Gamma$ be the circle $BOC$. The line $AO$ and the circle of radius $AO$ centered at $A$ cross $\Gamma$ at $A’$ and $F$, respectively. Prove that $\Gamma$ , the circle on diameter $AA’$ and circle $AFH$ are concurrent.
Proposed by Romania, Radu-Andrew Lecoiu
21 replies
+1 w
Twoisaprime
Feb 13, 2025
L13832
16 minutes ago
IMO Shortlist 2011, Algebra 3
orl   45
N 23 minutes ago by Ilikeminecraft
Source: IMO Shortlist 2011, Algebra 3
Determine all pairs $(f,g)$ of functions from the set of real numbers to itself that satisfy \[g(f(x+y)) = f(x) + (2x + y)g(y)\] for all real numbers $x$ and $y$.

Proposed by Japan
45 replies
1 viewing
orl
Jul 11, 2012
Ilikeminecraft
23 minutes ago
Hard FE with positive reals
egxa   8
N 33 minutes ago by megarnie
Source: Turkey Olympic Revenge 2023 Shortlist A4
Find all functions $f:\mathbb{R^+}\to \mathbb{R^+}$ such that for all $x,y\in \mathbb{R^+}$
$f(xf(y)+y)=f(f(y))+yf(x)$
Proposed by Şevket Onur Yılmaz
8 replies
egxa
Jan 22, 2024
megarnie
33 minutes ago
Like Father Like Son... (or Like Grandson?)
AlperenINAN   1
N 38 minutes ago by hakN
Source: Turkey TST 2025 P4
Let $a,b,c$ be given pairwise coprime positive integers where $a>bc$. Let $m<n$ be positive integers. We call $m$ to be a grandson of $n$ if and only if, for all possible piles of stones whose total mass adds up to $n$ and consist of stones with masses $a,b,c$, it's possible to take some of the stones out from this pile in a way that in the end, we can obtain a new pile of stones with total mass of $m$. Find the greatest possible number that doesn't have any grandsons.
1 reply
AlperenINAN
Today at 6:09 AM
hakN
38 minutes ago
Crazy number theory
MTA_2024   5
N 38 minutes ago by bjump
Find all couple $(p;q)$ of primes (greater than 5) such that : $$pq \mid (5^q-3^q)(5^p-3^p)$$
5 replies
MTA_2024
3 hours ago
bjump
38 minutes ago
hard number theory problem
Zavyk09   0
an hour ago
Source: forgotten
Find all couple $(x, y)$ of positive integers such that:
$$2^n + 3^n \mid x^n + y^n, \forall n \in \mathbb{N}^*$$
0 replies
Zavyk09
an hour ago
0 replies
The return of a legend inequality
giangtruong13   2
N an hour ago by polishedhardwoodtable
Source: Legacy
Given that $0<a,b,c,d<1$.Prove that: $$ (1-a)(1-b)(1-c)(1-d) > 1-a-b-c-d $$
2 replies
giangtruong13
2 hours ago
polishedhardwoodtable
an hour ago
Slightly weird points which are not so weird
Pranav1056   9
N an hour ago by Retemoeg
Source: India TST 2023 Day 4 P1
Suppose an acute scalene triangle $ABC$ has incentre $I$ and incircle touching $BC$ at $D$. Let $Z$ be the antipode of $A$ in the circumcircle of $ABC$. Point $L$ is chosen on the internal angle bisector of $\angle BZC$ such that $AL = LI$. Let $M$ be the midpoint of arc $BZC$, and let $V$ be the midpoint of $ID$. Prove that $\angle IML = \angle DVM$
9 replies
Pranav1056
Jul 9, 2023
Retemoeg
an hour ago
2023 factors and perfect cube
proxima1681   4
N an hour ago by anudeep
Source: Indian Statistical Institute (ISI) UGB 2023 P4
Let $n_1, n_2, \cdots , n_{51}$ be distinct natural numbers each of which has exactly $2023$ positive integer factors. For instance, $2^{2022}$ has exactly $2023$ positive integer factors $1,2, 2^{2}, 2^{3}, \cdots 2^{2021}, 2^{2022}$. Assume that no prime larger than $11$ divides any of the $n_{i}$'s. Show that there must be some perfect cube among the $n_{i}$'s.
4 replies
proxima1681
May 14, 2023
anudeep
an hour ago
circle geometry showing perpendicularity
Kyj9981   1
N an hour ago by Retemoeg
Two circles $\omega_1$ and $\omega_2$ intersect at points $A$ and $B$. A line through $B$ intersects $\omega_1$ and $\omega_2$ at points $C$ and $D$, respectively. Line $AD$ intersects $\omega_1$ at point $E \neq A$, and line $AC$ intersects $\omega_2$ at point $F \neq A$. If $O$ is the circumcenter of $\triangle AEF$, prove that $OB \perp CD$.
1 reply
Kyj9981
Today at 11:53 AM
Retemoeg
an hour ago
Hard problem
Tendo_Jakarta   5
N an hour ago by Tendo_Jakarta
Let the sequence \(x_{n}\) be such that
\[u_{1} = 1; \quad u_{n+1} = \dfrac{u_{1} + u_{2} +...+u_{n}}{n}+n-1 \quad \forall n \in \mathbb{N^{*}}\]and \(y_{n} =\dfrac{1}{u_{1}u_{2}} + \dfrac{1}{u_{3}u_{4}} + ... + \dfrac{1}{u_{2n-1}u_{2n}}  \quad \forall n \geq 1\). Find \(\lim_{n\rightarrow\infty}{y_{n}}\).
5 replies
Tendo_Jakarta
2 hours ago
Tendo_Jakarta
an hour ago
Oh no! Inequality again?
mathisreaI   108
N an hour ago by Maximilian113
Source: IMO 2022 Problem 2
Let $\mathbb{R}^+$ denote the set of positive real numbers. Find all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ such that for each $x \in \mathbb{R}^+$, there is exactly one $y \in \mathbb{R}^+$ satisfying $$xf(y)+yf(x) \leq 2$$
108 replies
mathisreaI
Jul 13, 2022
Maximilian113
an hour ago
Immaculate Geometry
anantmudgal09   6
N Oct 1, 2024 by Pyramix
Source: The 1st India-Iran Friendly Competition Problem 6
Let $ABC$ be a triangle with midpoint $M$ of $BC$. A point $X$ is called immaculate if the perpendicular line from $X$ to line $MX$ intersects lines $AB$ and $AC$ at two points that are equidistant from $M$. Suppose $U, V, W$ are three immaculate points on the circumcircle of triangle $ABC$. Prove that $M$ is the incentre of $\triangle UVW$.


Proposed by Pranjal Srivastava and Rohan Goyal
6 replies
anantmudgal09
Jun 13, 2024
Pyramix
Oct 1, 2024
Immaculate Geometry
G H J
G H BBookmark kLocked kLocked NReply
Source: The 1st India-Iran Friendly Competition Problem 6
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anantmudgal09
1979 posts
#1 • 4 Y
Y by GeoKing, Supercali, Rg230403, kiyoras_2001
Let $ABC$ be a triangle with midpoint $M$ of $BC$. A point $X$ is called immaculate if the perpendicular line from $X$ to line $MX$ intersects lines $AB$ and $AC$ at two points that are equidistant from $M$. Suppose $U, V, W$ are three immaculate points on the circumcircle of triangle $ABC$. Prove that $M$ is the incentre of $\triangle UVW$.


Proposed by Pranjal Srivastava and Rohan Goyal
This post has been edited 1 time. Last edited by anantmudgal09, Jun 13, 2024, 3:42 PM
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anantmudgal09
1979 posts
#2 • 3 Y
Y by GeoKing, Rijul saini, Supercali
anantmudgal09 wrote:
Let $ABC$ be a triangle with midpoint $M$ of $BC$. A point $X$ is called immaculate if the perpendicular line from $X$ to line $MX$ intersects lines $AB$ and $AC$ at two points that are equidistant from $M$. Suppose $U, V, W$ are three immaculate points on the circumcircle of triangle $ABC$. Prove that $M$ is the incentre of $\triangle UVW$.


Proposed by Pranjal Srivastava and Rohan Goyal

Let $X$ be an immaculate point and $P, Q$ be the points where lines $AB, AC$ meet the perpendicular to line $MX$ at $X$. Since $XP=XQ$, we get $$-1 = A(PQ; X \infty) = (AB, AC; AX, \ell)$$where $\ell$ is the perpendicular to $MX$ from $A$. Let $D$ be the antipode of $A$ in $BC$. Look at lines perpendicular to lines of this pencil passing through $D$, we get $(DB, DC; DX, DK) = -1$ where $K$ lies on the line through $D$ parallel to line $XM$. Let tangents to $B$ and $C$ meet at $T$; then $T, K, X$ are collinear so we redefine $K$ as such and compute the conditions on $X$ with $XM \parallel DK$. We now use complex numbers with $(ABC)$ as the unit circle. Let $X$ be an immaculate point with coordinate $x^2$, with an appropriate sign to be chosen later.


Suppose $M$ has coordinate $m \in \mathbb{R}$ by aligning the diameter line through $M$ with the real axis (using a rotation if necessary). Then $m = \overline{m}$. So $T$ has coordinate $\frac{1}{m}$. Now $$K \in \overline{TX} \iff \frac{k-x^2}{x^2-\frac{1}{m}} = \frac{\frac{1}{k}-\frac{1}{x^2}}{\frac{1}{x^2}-\frac{1}{m}} \iff k = \frac{x^2-\frac{1}{m}}{x^2\left(\frac{x^2-m}{mx^2}\right)} = \frac{mx^2-1}{x^2-m}$$and $$XM \parallel DK \iff  \frac{k-d}{x^2-m} = \frac{\frac{1}{k}-\frac{1}{d}}{\frac{1}{x^2}-m} \iff kd = \frac{x^2(x^2-m)}{mx^2-1} \iff d = \frac{x^2(x^2-m)^2}{(mx^2-1)^2}.$$Fix a square root $\lambda$ of $d$. Choosing the value of $x$ such that $$\frac{x(x^2-m)}{(mx^2-1)} = \lambda \iff x^3-\lambda mx^2-mx+\lambda = 0.$$Thus, if $X, Y, Z$ are immaculate points, there exists complex numbers $x, y, z$ such that $X=x^2, Y=y^2, Z=z^2$, and $-m = xy+yz+zx$ by Vieta's formula, hence $m=-(xy+yz+zx)$. Thus, $M$ is either the incentre or one of the three excentres of $\triangle ABC$. Since $M$ lies in the interior of $(XYZ)$, we are done.


Anecdote
This post has been edited 1 time. Last edited by anantmudgal09, Jun 13, 2024, 5:51 PM
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mueller.25
86 posts
#3 • 2 Y
Y by GeoKing, anantmudgal09
Let $U'$ denote the point on $(ABC)$ such that $(UU';BC)=-1$. Define $V'$ and $W'$ similarly. Projecting $(UU';BC)=-1$ from $A$ onto the line through $U$ perpendicular to $MU$, we get that $AU' \perp UM$.

Note that we have the cyclic quadrilaterals $UU'OM$, $VV'OM$, $WW'OM$ where $O$ is the circumcentre of $ABC$. As $V'A \perp MV$ and $W'A \perp MW$,
\[\measuredangle V'AW'=\measuredangle VMW\]\begin{align*}
	2 \measuredangle V'AW'
	&=\measuredangle V'OW'\\
    &= \measuredangle V'OM +\measuredangle MOW'\\
    &= \measuredangle V'VM + \measuredangle MWW'\\
    &= \measuredangle V'VU+\measuredangle UVM+ \measuredangle MWU+ \measuredangle UWW'\\
	&= \measuredangle V'WU+\measuredangle UWW'+\measuredangle UVM+ \measuredangle MWU\\
	&= \measuredangle V'AW'+ \measuredangle UVM+ \measuredangle MWU\\
\end{align*}\[\measuredangle VMW=\measuredangle UVM+ \measuredangle MWU\implies 2 \measuredangle VMW= \measuredangle VUW\]The centre $N_1$ of $(VMW)$ lies on $(UVW)$. Similarly, the centres $N_2,N_3$ of $(UMW)$ and $UMV$ lie on $(UVW)$. If any two of $N_1,N_2,N_3$ are the midpoints of the corresponding arcs not containing the opposite vertex, then $M$ is the incentre. If not, then $N_1,N_2,N_3$ are the arc midpoints of $\widehat{VUW},\widehat{UVW},\widehat{UWV}$. It follows that $M$ is the reflection of $U$ in $N_2N_3$, but then $UM$ internally bisects $\angle U$ and similarly $VM,WM$ are also internal angle bisectors which implies that $M$ must be the incentre.
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math_comb01
659 posts
#4 • 1 Y
Y by GeoKing
Nice Problem!
Let $BB \cap CC = T$ and $TU,TV,TW \cap (ABC) = U^{*},V^{*},V^{*}$.
Claim 1: $AU^{*} \perp MU$
Proof
Claim 2: If $O$ is the circumcenter of $ABC$, then $UU^{*}OM$ is cyclic.
Proof
Claim 3: The center of $(MUV)$ lies on $(ABC)$.
Proof
Claim 3 and cyclic variants imply $M$ is incenter of $UVW$.
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mathscrazy
113 posts
#5 • 2 Y
Y by GeoKing, Om245
Relabel the immaculate points on $(ABC)$ as $X_1,X_2,X_3$. Let $\omega=(ABC)$ with center $O$. Let $Y_i$ be the point on $\omega$ such that $(X_iY_i;BC)=-1$ for $i=1,2,3$.

Claim 1 : $AY_i \perp MX_i$ and $X_iY_iOM$ is cyclic for $i=1,2,3$.
Proof : Projecting the cross ratio $(X_iY_i;BC)=-1$, from $A$ onto the line through $X_i$ perpendicular to $MX_i$, we infer that $AY_i \perp X_iM$.
If $BB\cap CC=T$, then $TX_i\cdot TY_i = TB^2 = TO\cdot TM$ which implies that $X_iY_iOM$ is cyclic. $\blacksquare$

Claim 2 : $M$ is orthocenter of $Y_1Y_2Y_3$.
Proof : Angel chase $\measuredangle Y_1Y_2Y_3 = \measuredangle Y_1AY_3 = \measuredangle X_1MX_3 = \measuredangle Y_3MY_1$.
The second equality follows since $AY_i \perp MX_i$; and the last equality follows by reflection over line $OM$, since it is angle bisector of $\angle X_iMY_i$.
Similarly doing the angle chase for $\measuredangle Y_1Y_3Y_2$ implies the claim. $\blacksquare$

Claim 3 : $M$ lies on angle bisector $X_1X_3X_2$.
Proof : Look at quadrilateral $X_3Y_2Y_3X_2$ with center $O$. Since $M = (X_3Y_3O) \cap (X_2Y_2O)$, $M$ is the spiral center of the quadrilateral.
Hence $\measuredangle MX_3X_2 = \measuredangle MY_2Y_3 = 90 - \measuredangle Y_2Y_3Y_1$ by claim 2.
Hence, $90 - \measuredangle MX_3X_2 =  \measuredangle Y_2Y_3Y_1 = - \measuredangle Y_1Y_3Y_2 = -90 + \measuredangle MX_3X_1$, which implies the claim. $\blacksquare$

Hence $M$ is incentre or excentre of $X_1X_2X_3$. Since $M$ lies inside $(X_1X_2X_3)$, it must be the incentre and we are done!
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sami1618
873 posts
#6 • 1 Y
Y by GeoKing
Seems too simple, but I couldn't find a mistake :)

Claim: $\measuredangle AWM+\measuredangle BMW=90^{\circ}$
Intersect the perpendicular to $MW$ through $W$ with $AB$ and $AC$ at $B_W$ and $C_W$. Let $W'$ be the point such that $AB_WW'C_W$ is a parallelogram. Then triangle $AB_W W'$ and $CWB$ are similar with opposite orientation so $\measuredangle WMB= \measuredangle C_W W A$ and the result follows.

Cyclic variants also hold. Thus $$\angle WMV=-\measuredangle BMW-\measuredangle VMC=\measuredangle AWM+\measuredangle MVA=\measuredangle WAV-\measuredangle  WMV\iff 2\measuredangle WMV=\measuredangle WUV$$Cyclic variants imply $M$ is the incenter of $UVW$ as desired. remark
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Pyramix
419 posts
#7
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We make the use of complex numbers. Toss $ABC$ onto the unit circle. Let $X$ denote an immaculate point and let $X_1,X_2$ be points of $AB,AC$ such that $X_1,X_2,X$ are collinear and that line is perpendicular to $MX$.

First note that since $MX_1=MX_2$ and $MX\perp X_1X_2$, we must have $X$ to be mid-point of $\overline{X_1X_2}$. So, $2X-X_1$ must lie on line $AC$. We have,
\[x_1+ab\overline{x_1}=a+b, \ \ \ x_2+ac\overline{x_2}=a+c, \ \ \ x_1+x_2=2x\]This systems gives
\[x_1=\frac{2bc+ab+ac-\frac{2abc}{x}-2bx}{c-b},x_2=\frac{2bc+ab+ac-\frac{2abc}{x}-2cx}{b-c}\]Hence, $\frac{x_2-x_1}{\overline{x_2-x_1}}=\frac{ax\left(b+c-\frac{2bc}{x}\right)}{b+c-2x}$
Since $MX\perp X_1X_2$, we want
\[ax\left(\frac{m-\frac{bc}{x}}{m-x}\right)=ax\left(\frac{b+c-\frac{2bc}{x}}{b+c-2x}\right)=\frac{x_2-x_1}{\overline{x_2-x_1}}=\frac{x-m}{\overline{m}-\frac1x}\]\[\Longrightarrow x(x-m)^2+a(mx-bc)(\overline{m}x-1)=0.\]The roots of the cubic are $x_1,x_2,x_3$, then we need to compute $-\left(\sqrt{x_1x_2}+\sqrt{x_2x_3}+\sqrt{x_3x_1}\right)$
The rest follows by Vieta's relations.
This post has been edited 1 time. Last edited by Pyramix, Oct 1, 2024, 2:24 PM
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