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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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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
Geometric inequality with Fermat point
Assassino9931   1
N a minute ago by Circumcircle
Source: Balkan MO Shortlist 2024 G2
Let $ABC$ be an acute triangle and let $P$ be an interior point for it such that $\angle APB = \angle BPC = \angle CPA$. Prove that
$$ \frac{PA^2 + PB^2 + PC^2}{2S} + \frac{4}{\sqrt{3}} \leq \frac{1}{\sin \alpha} + \frac{1}{\sin \beta} + \frac{1}{\sin \gamma}. $$When does equality hold?
1 reply
Assassino9931
an hour ago
Circumcircle
a minute ago
Involved conditional geo
Assassino9931   1
N 15 minutes ago by hukilau17
Source: Balkan MO 2024 Shortlist G4
Let $ABC$ be an acute-angled triangle with $AB < AC$, orthocenter $H$, circumcircle $\Gamma$ and circumcentre $O$. Let $M$ be the midpoint of $BC$ and let $D$ be a point such that $ADOH$ is a parallellogram. Suppose that there exists a point $X$ on $\Gamma$ and on the opposite side of $DH$ to $A$ such that $\angle DXH + \angle DHA = 90^{\circ}$. Let $Y$ be the midpoint of $OX$. Prove that if $MY = OA$, then $OA = 2OH$.
1 reply
Assassino9931
an hour ago
hukilau17
15 minutes ago
Inversion exercise
Assassino9931   2
N 20 minutes ago by awesomeming327.
Source: Balkan MO Shortlist 2024 G5
Let $ABC$ be an acute scalene triangle $ABC$, $D$ be the orthogonal projection of $A$ on $BC$, $M$ and $N$ are the midpoints of $AB$ and $AC$ respectively. Let $P$ and $Q$ are points on the minor arcs $\widehat{AB}$ and $\widehat{AC}$ of the circumcircle of triangle $ABC$ respectively such that $PQ \parallel BC$. Show that the circumcircles of triangles $DPQ$ and $DMN$ are tangent if and only if $M$ lies on $PQ$.
2 replies
Assassino9931
an hour ago
awesomeming327.
20 minutes ago
all functions satisfying f(x+yf(x))+y = xy + f(x+y)
falantrng   26
N 35 minutes ago by awesomeming327.
Source: Balkan MO 2025 P3
Find all functions $f\colon \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x,y \in \mathbb{R}$,
\[f(x+yf(x))+y = xy + f(x+y).\]
Proposed by Giannis Galamatis, Greece
26 replies
falantrng
Today at 11:52 AM
awesomeming327.
35 minutes ago
One more problem defined only with lines
Assassino9931   0
an hour ago
Source: Balkan MO 2024 Shortlist G6
Let $ABC$ be a triangle and the points $K$ and $L$ on $AB$, $M$ and $N$ on $BC$, and $P$ and $Q$ on $AC$ be such that $AK = LB < \frac{1}{2}AB, BM = NC < \frac{1}{2}BC$ and $CP = QA < \frac{1}{2}AC$. The intersections of $KN$ with $MQ$ and $LP$ are $R$ and $T$ respectively, and the intersections of $NP$ with $LM$ and $KQ$ are $D$ and $E$, respectively. Prove that the lines $DR, BE$ and $CT$ are concurrent.
0 replies
Assassino9931
an hour ago
0 replies
Fixed point in a small configuration
Assassino9931   0
an hour ago
Source: Balkan MO Shortlist 2024 G3
Let $A, B, C, D$ be fixed points on this order on a line. Let $\omega$ be a variable circle through $C$ and $D$ and suppose it meets the perpendicular bisector of $CD$ at the points $X$ and $Y$. Let $Z$ and $T$ be the other points of intersection of $AX$ and $BY$ with $\omega$. Prove that $ZT$ passes through a fixed point independent of $\omega$.
0 replies
Assassino9931
an hour ago
0 replies
Sum of divisors
DinDean   1
N an hour ago by Tintarn
Does there exist $M>0$, such that $\forall m>M$, there exists an integer $n$ satisfying $\sigma(n)=m$?
$\sigma(n)=$ the sum of all positive divisors of $n$.
1 reply
DinDean
Apr 18, 2025
Tintarn
an hour ago
Projections on collections of lines
Assassino9931   0
an hour ago
Source: Balkan MO Shortlist 2024 C6
Let $\mathcal{D}$ be the set of all lines in the plane and $A$ be a set of $17$ points in the plane. For a line $d\in \mathcal{D}$ let $n_d(A)$ be the number of distinct points among the orthogonal projections of the points from $A$ on $d$. Find the maximum possible number of distinct values of $n_d(A)$ (this quantity is computed for any line $d$) as $A$ varies.
0 replies
Assassino9931
an hour ago
0 replies
Interesting polygon game
Assassino9931   0
an hour ago
Source: Balkan MO Shortlist 2024 C5
Let $n\geq 3$ be an integer. Alice and Bob play the following game on the vertices of a regular $n$-gon. Alice places her token on a vertex of the n-gon. Afterwards Bob places his token on another vertex of the n-gon. Then, with Alice playing first, they move their tokens alternately as follows for $2n$ rounds: In Alice’s turn on the $k$-th round, she moves her token $k$ positions clockwise or anticlockwise. In Bob’s turn on the $k$-th round, he moves his token $1$ position clockwise or anticlockwise. If at the end of any person’s turn the two tokens are on the same vertex, then Alice wins the game, otherwise Bob wins. Decide for each value of $n$ which player has a winning strategy.
0 replies
Assassino9931
an hour ago
0 replies
An equation from the past with different coefficients
Assassino9931   13
N an hour ago by grupyorum
Source: Balkan MO Shortlist 2024 N2
Let $n$ be an integer. Prove that $n^4 - 12n^2 + 144$ is not a perfect cube of an integer.
13 replies
Assassino9931
Today at 1:00 PM
grupyorum
an hour ago
Euler Totient optimality - why combinatorics?
Assassino9931   0
an hour ago
Source: Balkan MO Shortlist 2024 C4
Let $k$ be a positive integer. Prove that there exists a positive integer $n$ and distinct primes $p_1,p_2,\ldots,p_k$ such that if $A(n)$ denotes the number of positive integers less than or equal to $n$ and not divisible by any of $p_1,p_2,\ldots,p_k$, then
$$ \left|n\left(1 - \frac{1}{p_1}\right)\left(1 - \frac{1}{p_2}\right)\cdots \left(1-\frac{1}{p_k}\right) - A(n)\right| > 2^{k-3} $$
0 replies
Assassino9931
an hour ago
0 replies
Abstraction function in combinatorics
Assassino9931   0
an hour ago
Source: Balkan MO Shortlist 2024 C2
Let $n\geq 2$ be an integer and denote $S = \{1,2,\ldots,n^2\}$. For a function $f: S \to S$ we denote Im $f = \{b\in S: \exists a\in S, f(a) = b\}$, Fix $f = \{x \in S: f(x) = x\}$ and $f^{-1}(k) = \{a\in S: f(a) = k\}$. Find all possible values of $|$Im $f|$ + $|$Fix $f|$ + $\max_{k\in S} |f^{-1}(k)|$.
0 replies
Assassino9931
an hour ago
0 replies
2016 Kmo Final round
Jackson0423   1
N an hour ago by Tintarn
Source: 2016 FKMO P4
Let \(x,y,z\in\mathbb R\) with \(x^{2}+y^{2}+z^{2}=1\).
Find the maximum value of
\[
(x^{2}-yz)(y^{2}-zx)(z^{2}-xy).
\]
1 reply
Jackson0423
Apr 22, 2025
Tintarn
an hour ago
hard problem
Rename   1
N an hour ago by GeoMorocco
Determine the largest constant $K\geq 0$ so that:
$$\frac{a^a(b^2+c^2)}{(a^a-1)^2}+\frac{b^b(c^2+a^2)}{(b^b-1)^2}+\frac{c^c(a^2+b^2)}{(c^c-1)^2}\geq K\left (\frac{a+b+c}{abc-1}\right)^2$$with all real numbers $a; b; c$ satisfies $ab+bc+ca=abc$

P/s: Do you know which exam question this problem is actually in, at first I remembered but now I forgot
1 reply
Rename
Today at 3:24 PM
GeoMorocco
an hour ago
Shortlist 2017/G8
SHARKYKESA   8
N Dec 16, 2023 by signifance
There are $2017$ mutually external circles drawn on a blackboard, such that no two are tangent and no three share a common tangent. A tangent segment is a line segment that is a common tangent to two circles, starting at one tangent point and ending at the other one. Luciano is drawing tangent segments on the blackboard, one at a time, so that no tangent segment intersects any other circles or previously drawn tangent segments. Luciano keeps drawing tangent segments until no more can be drawn.

Find all possible numbers of tangent segments when Luciano stops drawing.
8 replies
SHARKYKESA
Jul 10, 2018
signifance
Dec 16, 2023
Shortlist 2017/G8
G H J
G H BBookmark kLocked kLocked NReply
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SHARKYKESA
436 posts
#1 • 6 Y
Y by Amir Hossein, ValidName, scimaths, megarnie, Adventure10, Mango247
There are $2017$ mutually external circles drawn on a blackboard, such that no two are tangent and no three share a common tangent. A tangent segment is a line segment that is a common tangent to two circles, starting at one tangent point and ending at the other one. Luciano is drawing tangent segments on the blackboard, one at a time, so that no tangent segment intersects any other circles or previously drawn tangent segments. Luciano keeps drawing tangent segments until no more can be drawn.

Find all possible numbers of tangent segments when Luciano stops drawing.
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tastymath75025
3223 posts
#2 • 22 Y
Y by DVDthe1st, yugrey, rkm0959, coolstuff, StoryScene, ac_math, Amir Hossein, DrMath, AIME12345, trumpeter, Kagebaka, bluephoenix, shankarmath, TheUltimate123, amar_04, Toinfinity, Aryan-23, BVKRB-, megarnie, rayfish, Adventure10, Mango247
Beautiful geometry problem!
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SleepWalker
104 posts
#5 • 2 Y
Y by CANBANKAN, Adventure10
Solution?
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SleepWalker
104 posts
#6 • 3 Y
Y by CANBANKAN, Adventure10, Mango247
ABCDE wrote:
Beautiful gold medalist! Beautiful solution!
Beautiful solution?
:help:
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JuanOrtiz
366 posts
#7 • 3 Y
Y by SleepWalker, Adventure10, Mango247
I believe it's always $3n-3$ (let $n=2017$). Say we have $n$ circles and $m$ segments such that no more can be drawn. Choose a circle and continuously dilate and translate it, together with its segments, while keeping the rest of the picture the same. Do this until either a new segment becomes available or an old segment becomes intersected (wlog assume the latter). If the old segment becomes intersected by a circle, we easily see that at this moment 3 circles must have a common tangent. If the old segment becomes intersected by another segment, assume these 2 segments are tangent to the same circle. (We can assume this wlog if we assume all circles are always $\ge 1$ cm apart). Then again at this moment, 3 circles must have a common tangent.

In any case, the maximality of $m$ can only change when 3 circles share a common tangent. It is easy to check, however, that this situation only affects 3 segments (the ones lying on the common tangent line), and both before and after the event, exactly 2 of these 3 segments can be drawn. Therefore, as long as the circles are always $\ge 1$ cm apart, the number $m$ can never change. Consider the following configuration: a giant circle and little tiny circles very close to it, resembling a regular $n-1$-gon. We can easily check that no segments are drawable between the tiny circles. And the segments drawn between two distinct tiny circles and the giant circle don't affect each other. Between each tiny circle and the giant one we can draw 3 segments. Thus, no matter how he draws the segments, Luciano will always draw $m=3n-3$ segments.

Thus, if we start with any picture, re-scale it so that all circles are at least $\ge G$ cm apart (where $G$ is very big), and continuously deform it circle-by-circle to the configuration previously explained, we see that $m=3n-3$ always. $\square$
This post has been edited 1 time. Last edited by JuanOrtiz, Jul 17, 2018, 9:36 PM
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SleepWalker
104 posts
#8 • 2 Y
Y by Adventure10, Mango247
Thank you! :thumbup:
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Inconsistent
1455 posts
#9 • 2 Y
Y by Awesome_guy, ihatemath123
We can think of the edges as windmills along faces of the graph permitting circular edges between vertices (tangency points): The outer windmill has an angle sum of $2\pi$. It can be shown that if a windmill changes direction, it must only touch two circles and have angle sum $\pi$. If a windmill doesn't change direction, it can be shown by various convexity arguments that it must touch only three circles and by extending to a triangle, the angle sum of the windmill is $\pi$.

By the Euler characteristic, since every vertex in this graph has degree 3, $E + 2 = F + \frac{2}{3}E$, where since the sum of the angle measures of all windmills must allow you to recover all circular arcs of the graph, if $F'$ is the number of non-circular faces then $(F'-1)\pi + 2\pi = n(2\pi)$ so $F = 3n-1$, so $E = 9n-9$. Then notice the number of circular edges is equal to the number of vertices on each circle, so the number of straight edges is $E - V = 3n-3$, and so the only number is $6048$.
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Leo.Euler
577 posts
#10 • 1 Y
Y by The_Great_Learner
Let $2017$ be replaced with arbitrary $n$. I claim the only answer is $3n-3$.

Draw a multigraph $G$ on the circles where two vertices are connected if and only if their corresponding circles have a tangent drawn.

Split the figure into multiples regions, which correspond to the faces of $G$, some of which are bounded. A short proof shows that the sum of the angles induced by the arcs on the boundary of each bounded region $\mathcal{R}$ is $2\pi$; this can be seen by considering the orientation of a directed point moving along the path of the boundary. Now call a tangency point that forms an acute angle with the previous arc on the path a pocket knife, and all other tangency points revolvers. As a consequence of the arc sum result, there are at least $3$ pocket knives.

Consider the line tangent to the boundary of $\mathcal{R}$ as a point moves along it, and refer to these lines as lightsabers. By geometric continuity, there exists a lightsaber from one arc of a circle (not at an endpoint of the arc) that is tangent to another. This inductively proves that $G$ is connected.

Now we show that there are at most $3$ pocket knives. Assume for contradiction that there are $4$ pocket knives, and WLOG $3$ of them are consecutive. Label the consecutive circles $\omega_1, \omega_2, \omega_3$ in that order and the 4th one $\omega_4$. Then we have a contradiction since there is then a lightsaber from $\omega_2$ through both $\omega_4$ and either one of $\omega_1$ and $\omega_3$. Thus there are exactly $3$ pocket knives.

The pocket knives allow us to use a double counting strategy to prove a correlation between $E$ and $F$. Each edge of $G$ corresponds to $2$ pocket knives, and each face of $G$ corresponds to $3$ pocket knives, so counting the total pocket knives we have $2E=3F$. Thus by Euler's formula, \[ V+F-E=2 \rightarrow E = 3n-3, \]as desired.

Remark: This is just a convexity problem where you have to know what you want to prove, which is essentially the small graph theory portion of the problem.
This post has been edited 3 times. Last edited by Leo.Euler, Sep 25, 2023, 10:35 PM
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signifance
140 posts
#11
Y by
this problem took me like 2 hrs to digest and then 1 hr to finish, after given the hint that it's local. i maintain that g6 summit should be g8 summit (this is unrelated, this is from zcx globloc)

Consider a situation where the common external tangent between two consecutive circles ``covers'' the other circles to the left; it's evident that now you can only draw tangents between consec circles which gives 3(n-1) with n circles. We prove this is always the answer by local technique of moving/dilating one circle at a time. The main claim is that a \emph{change} where a tangent is (WLOG) now obstructed (rather than made possible) can only occur when three circles share a common tangent at some moment of moving before then. Indeed, if a tangent line t is blocked by a circle, at some point it must've been tangent so at that moment done; while if t is blocked by a segment t', t,t' must both have a vertex on a common circle (otherwise to intersect you'd have to move the circle through t'), so we observe that at some moment t,t' must be ``tangent'' (in the sense that all points of t are on the same side of t'), but since they're on same circle they must be collinear (tangent at the same point), proved.

and oh wait!! we're done!!!!!!!!!!!! if we perturb by ever so slightly s.t. no other circles/segments obstruct anything, either t12 and t23 are available but t13 is obstructed by C2 (where C1,C2,C3 are the circles), or all are available but then t12 and t23 intersect. hence exactly 2 tangents exist, but remember. we started with t,t' as the 2 tangents, so no. of tangents is invariant!!!
This post has been edited 1 time. Last edited by signifance, Dec 16, 2023, 5:32 AM
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