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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.

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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
Classic Diophantine
Adywastaken   4
N 39 minutes ago by mrtheory
Source: NMTC 2024/6
Find all natural number solutions to $3^x-5^y=z^2$.
4 replies
Adywastaken
Today at 3:39 PM
mrtheory
39 minutes ago
Where are the Circles?
luminescent   43
N an hour ago by Amkan2022
Source: EGMO 2022/1
Let $ABC$ be an acute-angled triangle in which $BC<AB$ and $BC<CA$. Let point $P$ lie on segment $AB$ and point $Q$ lie on segment $AC$ such that $P \neq B$, $Q \neq C$ and $BQ = BC = CP$. Let $T$ be the circumcenter of triangle $APQ$, $H$ the orthocenter of triangle $ABC$, and $S$ the point of intersection of the lines $BQ$ and $CP$. Prove that $T$, $H$, and $S$ are collinear.
43 replies
luminescent
Apr 9, 2022
Amkan2022
an hour ago
Divisibilty...
Sadigly   0
2 hours ago
Source: Azerbaijan Junior NMO 2025 P2
Find all $4$ consecutive even numbers, such that the square of their product divides the sum of their squares.
0 replies
Sadigly
2 hours ago
0 replies
Quadratic system
juckter   35
N 3 hours ago by shendrew7
Source: Mexico National Olympiad 2011 Problem 3
Let $n$ be a positive integer. Find all real solutions $(a_1, a_2, \dots, a_n)$ to the system:

\[a_1^2 + a_1 - 1 = a_2\]\[ a_2^2 + a_2 - 1 = a_3\]\[\hspace*{3.3em} \vdots \]\[a_{n}^2 + a_n - 1 = a_1\]
35 replies
juckter
Jun 22, 2014
shendrew7
3 hours ago
IMO Shortlist 2012, Geometry 3
lyukhson   75
N 4 hours ago by numbertheory97
Source: IMO Shortlist 2012, Geometry 3
In an acute triangle $ABC$ the points $D,E$ and $F$ are the feet of the altitudes through $A,B$ and $C$ respectively. The incenters of the triangles $AEF$ and $BDF$ are $I_1$ and $I_2$ respectively; the circumcenters of the triangles $ACI_1$ and $BCI_2$ are $O_1$ and $O_2$ respectively. Prove that $I_1I_2$ and $O_1O_2$ are parallel.
75 replies
lyukhson
Jul 29, 2013
numbertheory97
4 hours ago
Diophantine
TheUltimate123   31
N 4 hours ago by SomeonecoolLovesMaths
Source: CJMO 2023/1 (https://aops.com/community/c594864h3031323p27271877)
Find all triples of positive integers \((a,b,p)\) with \(p\) prime and \[a^p+b^p=p!.\]
Proposed by IndoMathXdZ
31 replies
TheUltimate123
Mar 29, 2023
SomeonecoolLovesMaths
4 hours ago
Cyclic ine
m4thbl3nd3r   1
N 4 hours ago by arqady
Let $a,b,c>0$ such that $a^2+b^2+c^2=3$. Prove that $$\sum \frac{a^2}{b}+abc \ge 4$$
1 reply
m4thbl3nd3r
Today at 3:34 PM
arqady
4 hours ago
Non-homogenous Inequality
Adywastaken   7
N 4 hours ago by ehuseyinyigit
Source: NMTC 2024/7
$a, b, c\in \mathbb{R_{+}}$ such that $ab+bc+ca=3abc$. Show that $a^2b+b^2c+c^2a \ge 2(a+b+c)-3$. When will equality hold?
7 replies
Adywastaken
Today at 3:42 PM
ehuseyinyigit
4 hours ago
FE with devisibility
fadhool   2
N 4 hours ago by ATM_
if when i solve an fe that is defined in the set of positive integer i found m|f(m) can i set f(m) =km such that k is not constant and of course it depends on m but after some work i find k=c st c is constant is this correct
2 replies
fadhool
Today at 4:25 PM
ATM_
4 hours ago
Japan MO Finals 2023
parkjungmin   2
N 5 hours ago by parkjungmin
It's hard. Help me
2 replies
parkjungmin
Yesterday at 2:35 PM
parkjungmin
5 hours ago
Iranian geometry configuration
Assassino9931   2
N 5 hours ago by Captainscrubz
Source: Al-Khwarizmi Junior International Olympiad 2025 P7
Let $ABCD$ be a cyclic quadrilateral with circumcenter $O$, such that $CD$ is not a diameter of its circumcircle. The lines $AD$ and $BC$ intersect at point $P$, so that $A$ lies between $D$ and $P$, and $B$ lies between $C$ and $P$. Suppose triangle $PCD$ is acute and let $H$ be its orthocenter. The points $E$ and $F$ on the lines $BC$ and $AD$, respectively, are such that $BD \parallel HE$ and $AC\parallel HF$. The line through $E$, perpendicular to $BC$, intersects $AD$ at $L$, and the line through $F$, perpendicular to $AD$, intersects $BC$ at $K$. Prove that the points $K$, $L$, $O$ are collinear.

Amir Parsa Hosseini Nayeri, Iran
2 replies
Assassino9931
Today at 9:39 AM
Captainscrubz
5 hours ago
f(m + n) >= f(m) + f(f(n)) - 1
orl   30
N 6 hours ago by ezpotd
Source: IMO Shortlist 2007, A2, AIMO 2008, TST 2, P1, Ukrainian TST 2008 Problem 8
Consider those functions $ f: \mathbb{N} \mapsto \mathbb{N}$ which satisfy the condition
\[ f(m + n) \geq f(m) + f(f(n)) - 1
\]
for all $ m,n \in \mathbb{N}.$ Find all possible values of $ f(2007).$

Author: Nikolai Nikolov, Bulgaria
30 replies
orl
Jul 13, 2008
ezpotd
6 hours ago
Add d or Divide by a
MarkBcc168   25
N 6 hours ago by Entei
Source: ISL 2022 N3
Let $a > 1$ be a positive integer and $d > 1$ be a positive integer coprime to $a$. Let $x_1=1$, and for $k\geq 1$, define
$$x_{k+1} = \begin{cases}
x_k + d &\text{if } a \text{ does not divide } x_k \\
x_k/a & \text{if } a \text{ divides } x_k
\end{cases}$$Find, in terms of $a$ and $d$, the greatest positive integer $n$ for which there exists an index $k$ such that $x_k$ is divisible by $a^n$.
25 replies
MarkBcc168
Jul 9, 2023
Entei
6 hours ago
Alice and Bob play, 8x8 table, white red black, minimum n for victory
parmenides51   14
N 6 hours ago by Ilikeminecraft
Source: JBMO Shortlist 2018 C3
The cells of a $8 \times 8$ table are initially white. Alice and Bob play a game. First Alice paints $n$ of the fields in red. Then Bob chooses $4$ rows and $4$ columns from the table and paints all fields in them in black. Alice wins if there is at least one red field left. Find the least value of $n$ such that Alice can win the game no matter how Bob plays.
14 replies
parmenides51
Jul 22, 2019
Ilikeminecraft
6 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
Y by
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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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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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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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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solution
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Maximilian113
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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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