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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
IMO Shortlist 2008, Geometry problem 2
April   42
N a minute ago by s27_SaparbekovUmar
Source: IMO Shortlist 2008, Geometry problem 2, German TST 2, P1, 2009
Given trapezoid $ ABCD$ with parallel sides $ AB$ and $ CD$, assume that there exist points $ E$ on line $ BC$ outside segment $ BC$, and $ F$ inside segment $ AD$ such that $ \angle DAE = \angle CBF$. Denote by $ I$ the point of intersection of $ CD$ and $ EF$, and by $ J$ the point of intersection of $ AB$ and $ EF$. Let $ K$ be the midpoint of segment $ EF$, assume it does not lie on line $ AB$. Prove that $ I$ belongs to the circumcircle of $ ABK$ if and only if $ K$ belongs to the circumcircle of $ CDJ$.

Proposed by Charles Leytem, Luxembourg
42 replies
April
Jul 9, 2009
s27_SaparbekovUmar
a minute ago
3^n + 61 is a square
VideoCake   25
N 7 minutes ago by endless_abyss
Source: 2025 German MO, Round 4, Grade 11/12, P6
Determine all positive integers \(n\) such that \(3^n + 61\) is the square of an integer.
25 replies
VideoCake
Monday at 5:14 PM
endless_abyss
7 minutes ago
Problem 5
blug   3
N 13 minutes ago by Jt.-.
Source: Czech-Polish-Slovak Junior Match 2025 Problem 5
For every integer $n\geq 1$ prove that
$$\frac{1}{n+1}-\frac{2}{n+2}+\frac{3}{n+3}-\frac{4}{n+4}+...+\frac{2n-1}{3n-1}>\frac{1}{3}.$$
3 replies
blug
May 19, 2025
Jt.-.
13 minutes ago
Inequality with xy+yz+zx=1
Kimchiks926   14
N 38 minutes ago by math-olympiad-clown
Source: Baltic Way 2022, Problem 4
The positive real numbers $x,y,z$ satisfy $xy+yz+zx=1$. Prove that:
$$ 2(x^2+y^2+z^2)+\frac{4}{3}\bigg (\frac{1}{x^2+1}+\frac{1}{y^2+1}+\frac{1}{z^2+1}\bigg) \ge 5 $$
14 replies
Kimchiks926
Nov 12, 2022
math-olympiad-clown
38 minutes ago
Problem 7
SlovEcience   7
N 43 minutes ago by GreekIdiot
Consider the sequence \((u_n)\) defined by \(u_0 = 5\) and
\[
u_{n+1} = \frac{1}{2}u_n^2 - 4 \quad \text{for all } n \in \mathbb{N}.
\]a) Prove that there exist infinitely many positive integers \(n\) such that \(u_n > 2020n\).

b) Compute
\[
\lim_{n \to \infty} \frac{2u_{n+1}}{u_0u_1\cdots u_n}.
\]
7 replies
SlovEcience
May 14, 2025
GreekIdiot
43 minutes ago
D1033 : A problem of probability for dominoes 3*1
Dattier   1
N an hour ago by Dattier
Source: les dattes à Dattier
Let $G$ a grid of 9*9, we choose a little square in $G$ of this grid three times, we can choose three times the same.

What the probability of cover with 3*1 dominoes this grid removed by theses little squares (one, two or three) ?
1 reply
Dattier
May 15, 2025
Dattier
an hour ago
Inequality
lgx57   1
N an hour ago by sqing
Source: Own
$a,b,c \in \mathbb{R}^{+}$,$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1$. Prove that
$$a^abc+b^bac+c^cab \ge 27(ab+bc+ca)$$
1 reply
lgx57
2 hours ago
sqing
an hour ago
Easy but Nice 12
TelvCohl   2
N an hour ago by AuroralMoss
Source: Own
Given a $ \triangle ABC $ with orthocenter $ H $ and a point $ P $ lying on the Euler line of $ \triangle ABC. $ Prove that the midpoint of $ PH $ lies on the Thomson cubic of the pedal triangle of $ P $ WRT $ \triangle ABC. $
2 replies
TelvCohl
Mar 8, 2025
AuroralMoss
an hour ago
Serbian selection contest for the IMO 2025 - P6
OgnjenTesic   8
N an hour ago by atdaotlohbh
Source: Serbian selection contest for the IMO 2025
For an $n \times n$ table filled with natural numbers, we say it is a divisor table if:
- the numbers in the $i$-th row are exactly all the divisors of some natural number $r_i$,
- the numbers in the $j$-th column are exactly all the divisors of some natural number $c_j$,
- $r_i \ne r_j$ for every $i \ne j$.

A prime number $p$ is given. Determine the smallest natural number $n$, divisible by $p$, such that there exists an $n \times n$ divisor table, or prove that such $n$ does not exist.

Proposed by Pavle Martinović
8 replies
OgnjenTesic
May 22, 2025
atdaotlohbh
an hour ago
Balkan Mathematical Olympiad
ABCD1728   1
N 2 hours ago by ABCD1728
Can anyone provide the PDF version of the book "Balkan Mathematical Olympiads" by Mircea Becheanu and Bogdan Enescu (published by XYZ press in 2014), thanks!
1 reply
ABCD1728
May 24, 2025
ABCD1728
2 hours ago
Unexpecredly Quick-Solve Inequality
Primeniyazidayi   2
N 2 hours ago by sqing
Source: German MO 2025,Round 4,Grade 11/12 Day 2 P1
If $a, b, c>0$, prove that $$\frac{a^5}{b^2}+\frac{b}{c}+\frac{c^3}{a^2}>2a$$
2 replies
Primeniyazidayi
4 hours ago
sqing
2 hours ago
cute geo
Royal_mhyasd   0
2 hours ago
Source: own(?)
Let $\triangle ABC$ be an acute triangle and $I$ it's incenter. Let $A'$, $B'$ and $C'$ be the projections of $I$ onto $BC$, $AC$ and $AB$ respectively. $BC \cap B'C' = \{K\}$ and $Y$ is the projection of $A'$ onto $KI$. Let $M$ be the middle of the arc $BC$ not containing $A$ and $T$ the second intersection of $A'M$ and the circumcircle of $ABC$. If $N$ is the midpoint of $AI$, $TY \cap IA' = \{P\}$, $BN \cap PC' = \{D\}$ and $CN \cap PB' =\{E\}$, prove that $NEPD$ is cyclic.
PS i'm not sure if this problem is actually original so if it isn't someone please tell me so i can change the source (if that's possible)
0 replies
Royal_mhyasd
2 hours ago
0 replies
Nice inequality
TUAN2k8   2
N 2 hours ago by TUAN2k8
Source: Own
Let $n \ge 2$ be an even integer and let $x_1,x_2,...,x_n$ be real numbers satisfying $x_1^2+x_2^2+...+x_n^2=n$.
Prove that
$\sum_{1 \le i < j \le n} \frac{x_ix_j}{x_i^2+x_j^2+1} \ge \frac{-n}{6}$
2 replies
TUAN2k8
Today at 2:03 AM
TUAN2k8
2 hours ago
An NT for a break
reni_wee   1
N 2 hours ago by reni_wee
Source: ONTCP 2.4.1
Prove that there are no positive integers $x,k$ and $n \geq 2$ such that $x^2+1 = k(2^n -1)$.
1 reply
reni_wee
2 hours ago
reni_wee
2 hours ago
Vertices of a convex polygon if and only if m(S) = f(n)
orl   12
N Apr 24, 2025 by Maximilian113
Source: IMO Shortlist 2000, C3
Let $ n \geq 4$ be a fixed positive integer. Given a set $ S = \{P_1, P_2, \ldots, P_n\}$ of $ n$ points in the plane such that no three are collinear and no four concyclic, let $ a_t,$ $ 1 \leq t \leq n,$ be the number of circles $ P_iP_jP_k$ that contain $ P_t$ in their interior, and let \[m(S)=a_1+a_2+\cdots + a_n.\]Prove that there exists a positive integer $ f(n),$ depending only on $ n,$ such that the points of $ S$ are the vertices of a convex polygon if and only if $ m(S) = f(n).$
12 replies
orl
Aug 10, 2008
Maximilian113
Apr 24, 2025
Vertices of a convex polygon if and only if m(S) = f(n)
G H J
Source: IMO Shortlist 2000, C3
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orl
3647 posts
#1 • 6 Y
Y by Adventure10, junioragd, and 4 other users
Let $ n \geq 4$ be a fixed positive integer. Given a set $ S = \{P_1, P_2, \ldots, P_n\}$ of $ n$ points in the plane such that no three are collinear and no four concyclic, let $ a_t,$ $ 1 \leq t \leq n,$ be the number of circles $ P_iP_jP_k$ that contain $ P_t$ in their interior, and let \[m(S)=a_1+a_2+\cdots + a_n.\]Prove that there exists a positive integer $ f(n),$ depending only on $ n,$ such that the points of $ S$ are the vertices of a convex polygon if and only if $ m(S) = f(n).$
This post has been edited 1 time. Last edited by djmathman, Oct 3, 2016, 3:25 AM
Reason: changed formatting to match imo compendium
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Thjch Ph4 Trjnh
205 posts
#2 • 4 Y
Y by Adventure10, Mango247, and 2 other users
$ f(n) = 2.(_4^n)$
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Myth
4464 posts
#3 • 4 Y
Y by Ali3085, Adventure10, Mango247, and 1 other user
It is strange to see such an easy and evident problem in IMO SL.
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SnowEverywhere
801 posts
#4 • 12 Y
Y by Catalanfury, Vaijan_Mama, k12byda5h, DCMaths, Adventure10, Mango247, Stuffybear, winniep008hfi, and 4 other users
We claim that the function $f(n)=2 \binom{n}{4}$ satisfies the requirements.

Let the score $s(a,b,c,d)$ of the four points $P_a$, $P_b$, $P_c$ and $P_d$ be the number of points $P_i$ where $i \in \{a,b,c,d \}$ such that $P_i$ is properly contained in the circle passing through the remaining three points. Observe that

\[m(S) = \sum_{1 \le a<b<c<d \le n} s(a,b,c,d)\]

First we prove the following lemma.

Lemma 1. The score of a convex quadrilateral $2$.

Proof. Let the vertices of the convex quadrilateral be denoted as $A, B, C$ and $D$. We have that point $D$ lies within the circumcircle of $\triangle{ABC}$ if and only if

\[\angle{ABC} > 180 - \angle{ADC} \quad \Leftrightarrow \quad \angle{ABC} + \angle{ADC} > 180\]

Therefore if $D$ lies within the circumcircle of $\triangle{ABC}$, it also follows by symmetry that $B$ lies within the circumcircle of $\triangle{ADC}$. If not, then since the sum of the interior angles of $ABCD$ is $360$,

\[\angle{ABC} + \angle{ADC} < 180 \quad \Rightarrow \quad \angle{BAD} + \angle{ACD} > 180\]

Therefore $A$ lies within the circumcircle of $\triangle{BCD}$ and $C$ lies within the circumcircle of $\triangle{ABD}$. In both cases, the score of $ABCD$ is equal to $2$.

Lemma 2. The score of a concave quadrilateral is $1$.

Proof. Let $ABCD$ denote the concave quadrilateral. Without the loss of generality, let $A$ be such that interior angle $\angle{BAC} > 180$. It follows that $A$ lies in the interior of triangle $\triangle{BCD}$. Therefore, the circumcircle of $\triangle{BCD}$ contains $A$. However, none of the remaining three circles passing through three of the points $A, B, C$ and $D$ contain the remaining point. Hence the score of $ABCD$ is equal to $1$.

If Direction. If the vertices of $S$ form a convex $n$-gon, then each quadrilateral formed by four distinct points in $S$ is a convex quadrilateral and therefore

\[m(S) = \sum_{1 \le a<b<c<d \le n} s(a,b,c,d) = 2 \binom{n}{4}\]

Only If Direction. If the vertices of $S$ form a concave $n$-gon, then at least one of the quadrilaterals formed by four distinct points in $S$ is a concave quadrilateral and therefore

\[m(S) = \sum_{1 \le a<b<c<d \le n} s(a,b,c,d) < 2 \binom{n}{4}\]

The function $f(n)=2 \binom{n}{4}$ therefore satisfies that $m(S)=f(n)$ if and only if the points in $S$ are the vertices of a convex polygon.
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JackXD
151 posts
#5 • 1 Y
Y by Adventure10
Lemma Every quadrilateral $S$ satisfies $m(S) \le 2$ with equality if and only if it is convex,

Proof:If a quadrilateral $ABCD$ is convex then one out of $\angle{ABC}+\angle{ADC}$ and $\angle{DAB}+\angle{DCB}$ is greater than $\pi$ and the other is less than $\pi$.This implies $m(S)=2$.One the other hand if it is concave then clearly $m(S)=1$


Back to our main problem.If S forms a convex polygon then every quadrilateral $P_{i}P_{j}P_{k}P_{l}$ contributes $1$ to two of $a_{i},a_{j},a_{k}$ and $a_{l}$ (from the lemma) and thus contributes two to $m(S)$.This immediately implies $f(n)=2\binom{n}{4}$

Now let $m(S)=f(n)=2\binom{n}{4}$.Each quadrilateral $P_{i}P_{j}P_{k}P_{l}$ contributes atmost $2$ to $f(n)$ and thus $m(S) \le 2\binom{n}{4}$.As there is equality,we have that each quadrilateral contributes exactly $2$ to $f(n)$,hence from the lemma every quadrilateral is convex,implying that $S$ forms a convex polygon.
This post has been edited 3 times. Last edited by JackXD, Jan 7, 2016, 4:22 PM
Reason: xx
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william122
1576 posts
#6 • 2 Y
Y by Adventure10, Mango247
Note that given any 4 points, they add 2 to the count if their convex hull is a quadrilateral, and 1 otherwise. So, $f(n)=2\binom{n}{2}$, with equality achieved iff all quadruplets of points form convex quadrilaterals. However, if there exists a point $P_i$ inside the convex hull $Q_1,Q_2,\ldots,Q_k$, then it must be in one of the triangles $Q_1Q_2Q_3$, $Q_1Q_3Q_4,\ldots Q_1Q_{k-1}Q_k$, which cover the convex hull. So, there exist 4 points whose convex hull is a triangle. Therefore, the only way equality is reached is if all points are on the convex hull, as desired.
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niyu
830 posts
#7
Y by
The key idea is to consider four-tuples.

Lemma: If $ABCD$ is a non-cyclic convex quadrilateral, $m(ABCD) = 2$.

Proof: Suppose $ABCD$ is convex and non-cyclic. Note that $\angle ABC + \angle ADC \neq \angle BCD + \angle BAD \neq 180^\circ$. Hence, exactly one of these two sums is greater than $180^\circ$. WLOG, suppose $\angle ABC + \angle ADC > 180^\circ$ and $\angle BCD + \angle BAD < 180^\circ$. Since $\angle ABC > 180^\circ - \angle ADC$ it follows that $B$ lies within $(ADC)$. Similarly, $D$ lies within $(ABC)$. Meanwhile, since $\angle BCD < 180^\circ - \angle BAD$, it follows that $A$ does not lie within $(BCD)$, and similarly, $C$ does not lie within $(ABD)$. This implies that $m(ABCD) = 2$, proving the lemma. $\blacksquare$

Lemma: If $ABCD$ is a concave quadrilateral, $m(ABCD) = 1$.

Proof: Say $\angle BAD > 180^\circ$. Since $A$ and $C$ both lie on the same side of $\overline{BD}$, and $\angle BCD < \angle BAD$, it follows that $C$ does not lie within $(ABD)$, while $A$ lies within $(BCD)$. Meanwhile, since $\angle ABC + \angle ADC < 180^\circ$, by the same argument as in the previous lemma we find that $B$ does not lie within $(ACD)$ and that $D$ does not lie within $(ABC)$. Hence, $m(ABCD) = 1$. $\blacksquare$

We now return to the given problem. We claim that $m(S) \leq 2\binom{n}{4}$, and that equality holds iff the points of $S$ form a convex quadrilateral. Indeed, as no four points in $S$ are concyclic, we have
\begin{align*}
	m(S) &= \sum_{1 \leq w < x < y < z \leq n} m(P_wP_xP_yP_z) \\
	&\leq \sum_{1 \leq w < x < y < z \leq n} 2 \\
	&\leq 2\binom{n}{4}.
\end{align*}Equality holds here iff each of the quadrilaterals $P_wP_xP_yP_z$ is convex, which occurs iff the points of $S$ form a cyclic quadrilateral. This completes the proof. $\Box$
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bluelinfish
1449 posts
#8
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Rather easy for a C3.

All angles are in degrees.

Claim: Suppose $Q$ consists of four points $A,B,C,D$. Then $m(Q)$ is $2$ if $A,B,C,D$ form a convex quadrilateral and $1$ otherwise.
Proof. Notice that if $A,B,C,D$ form a convex quadrilateral, then $D$ appears inside the circumcircle of $ABC$, $\angle D$ must be greater than $180-\angle B$, which is equivalent to $\angle B + \angle D >180$. Since exactly one opposite pair of angles sum to greater than $180$ degrees, there will be exactly two points that are contained in the circumcircle of the other three.

If $A,B,C,D$ are not convex, WLOG let $A,B,C$ be the convex hull. Then it is clear that the only point that is contained in the circumcircle of the other three is $D$. $\blacksquare$

The key step is to notice that $$m(S)=\sum_{1\le a<b<c<d\le n} m\left(\{P_a,P_b,P_c,P_d\}\right)$$because both quantities count the amount of ordered pairs containing a single point of $S$ and a set of three points in $S$, with all four points distinct, such that the single point is inside the circumcircle of the three points.

Using our claim, $m(S)$ must be equal to twice the number of four-point subsets of $S$ that consist of points forming a convex quadrilateral plus the four-point subsets of $S$ that do not. Moreover, every subset of four points form a convex quadrilateral iff $S$ does, so $m(S)=2\binom{n}{4}$ iff $S$ forms a convex quadrilateral. We are done.
This post has been edited 1 time. Last edited by bluelinfish, Jan 18, 2022, 10:45 PM
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awesomeming327.
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#9 • 3 Y
Y by Mango247, Mango247, Mango247
Let $m(\{a,b,c,d\})$ be the number $m(S)$ for the quadrilateral $P_aP_bP_cP_d.$ It is easy to see that $m(S)$ is the sum of $m(\{a,b,c,d\})$ for all choices of subset $\{a,b,c,d\} \subseteq S$. Note that $(P_aP_bP_c)$ contains $P_d$ if only $P_d$ lies inside of the angle $P_aP_bP_c$ and $\angle P_b+\angle P_d> 180^\circ.$ Clearly, when $P_aP_bP_cP_d$ then $m(a,b,c,d)=2$ and when it is nonconvex $m(a,b,c,d)=1$. Therefore, $m(S)=2\tbinom{n}4$ if and only if $S$ is convex.
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john0512
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#10
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Note that by swapping the order of summation, $m(S)$ is equal to the number of quadruples of points $(A,B,C,D)$ where $D$ is inside $(ABC)$ and $A,B,C$ are unordered.

Claim: Each set of 4 points contributes $2$ if they are convex, and $1$ if they are not.

If $ABCD$ is convex and non-cyclic, then we either have $\angle A+\angle C>180$ or $\angle B+\angle D>180$ but not both. However, since $A$ is inside $(BCD)$ if and only if $\angle A+\angle C>180$, etc, the convex quadriateral contributes $2$. If $D$ is inside $\triangle ABC$, then $D$ will be inside $(ABC)$ but nothing else works.

Thus, the maximum possible value of $m(S)$ is $2{n\choose 4}$, with equality if and only if each quadrilateral is convex, which is the same as saying the entire set is convex. We are done as $f(n)=2{n\choose 4}$.
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asdf334
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#11
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For any four points $S'=\{P_a,P_b,P_c,P_d\}$ count the value $m(S')$. Notice this is either $2$ if the convex hull is a quadrilateral (i.e. the points form a convex polygon) and $1$ if the convex hull is a triangle (i.e. there is an interior point).
Clearly $f(n)$ is the sum of $m(S')$ over all such $S'$. Hence the maximum occurs if every convex hull of four points is a quadrilateral. If the points of $S$ are the vertices of a convex polygon this occurs. If the points of $S$ are not the vertices of a convex polygon then triangulate the convex hull. Any interior point is contained in a triangle and for these four points we have $m(S')=1$. $\blacksquare$
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onyqz
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#12
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storage
solution
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Maximilian113
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#13
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Observe that in any $4$ points $A, B, C, D,$ if they form a convex polygon there are $2$ pairs of the form $(i, \omega)$ where $i$ is a point from $A, B, C, D$ and $\omega$ is the circumcircle of the other points. However if they form a non-convex polygon there is $1$ only.

Clearly, if we consider all quadruples of points from our $n$ points, and sum up the number of valid pairs, every point, along with a circumcircle it lies in, is counted. In addition, the point and circumcircle pair uniquely determines which quadruple it was counted in, meaning that this count yields a injection and surjection to $m(S),$ so there is a bijection.

Therefore if $x$ quadruples are convex and $y$ are concave, $$m(S) = 2x+y.$$But $x+y=\binom{n}{4}$ so $$m(S)=\binom{n}{4}+x,$$and the desired result follows. QED
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