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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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0 replies
jlacosta
May 1, 2025
0 replies
Inspired by Philippine 2025
sqing   0
3 minutes ago
Source: Own
Let $ a,b,c,d $ be real numbers . Prove that
$$\frac{(a-1)(b-3)(c-3)(d-1)}{  (a^2+3)(b^2+3)(c^2+3)(d^2+3)} \ge -\frac{7+4\sqrt 3}{144}$$$$\frac{(a-1)(b-2)(c-2)(d-1)}{  (a^2+3)(b^2+3)(c^2+3)(d^2+3)} \ge -\frac{11+4\sqrt 7}{432}$$


0 replies
+1 w
sqing
3 minutes ago
0 replies
Find all functions $f$: \(\mathbb{R}\) \(\rightarrow\) \(\mathbb{R}\) such : $(f
guramuta   1
N 15 minutes ago by jasperE3
Find all functions $f$: \(\mathbb{R}\) \(\rightarrow\) \(\mathbb{R}\) such :
$(f(x-y))^2= (f(x))^2 - 2f(xy) + (f(y))^2$
1 reply
1 viewing
guramuta
an hour ago
jasperE3
15 minutes ago
Quadrangle, nine-point conic, Steiner line
kosmonauten3114   0
15 minutes ago
Source: My own
Let $P_1P_2P_3P_4$ be a general quadrangle which does not form an orthocentric system. Let $P$, $I$, $M$, $T$ be the Euler-Poncelet point ($\text{QA-P2}$), isogonal center ($\text{QA-P4}$), midray homothetic center ($\text{QA-P8}$), inscribed square axes crosspoint ($\text{QA-P23}$) of $P_1P_2P_3P_4$, respectively.
Let $H_1$ be the orthocenter of $\triangle{P_2P_3P_4}$, and define $H_2$, $H_3$, $H_4$ cyclically.
Let $A_{ij}=P_iP_j \cap H_iH_j$ ($\{i, j\} \in \{1, 2, 3, 4\}, i<j$).
Let $B_{ij}=P_iP_j \cap H_kH_l$ ($\{i, j, k, l\} \in \{1, 2, 3, 4\}, i<j$).
Then, the 12 points $A_{12}$, $A_{13}$, $A_{14}$, $A_{23}$, $A_{24}$, $A_{34}$, $B_{12}$, $B_{13}$, $B_{14}$, $B_{23}$, $B_{24}$, $B_{34}$ lie on the same conic, here denoted by $\mathcal{C}_1$.
Let $\mathcal{C}_2$ be the nine-point conic of $P_1P_2P_3P_4$.
Suppose that $\mathcal{C}_1$ and $\mathcal{C}_2$ have 4 distinct intersection points, and let $U$, $V$, $W$ be the intersections, other than $P$, of $\mathcal{C}_1$ and $\mathcal{C}_2$.

Prove that the Steiner line of $P$ with respect to $\triangle{UVW}$ passes through $I$ and $M$, and show that the center of $\mathcal{C}_1$ and the orthocenter of $\triangle{UVW}$ coincide with $T$.
0 replies
kosmonauten3114
15 minutes ago
0 replies
Inspired by Philippine 2025
sqing   1
N 30 minutes ago by sqing
Source: Own
Let $ a,b,c $ be real numbers . Prove that
$$\frac{(a-1)(b-1)(c-1)}{(a^2+1)(b^2+1)(c^2+1)} \ge -\frac{7+5\sqrt 2}{8}$$$$\frac{(a-1)(b-1)(c-1)}{(a^2+2)(b^2+2)(c^2+2)} \ge -\frac{5+3\sqrt 3}{32}$$
1 reply
1 viewing
sqing
38 minutes ago
sqing
30 minutes ago
Metric space
wiseman   3
N May 21, 2025 by alinazarboland
Source: IMS 2014 - Day1 - Problem4
Let $(X,d)$ be a metric space and $f:X \to X$ be a function such that $\forall x,y\in X : d(f(x),f(y))=d(x,y)$.
$\text{a})$ Prove that for all $x \in X$, $\lim_{n \rightarrow +\infty} \frac{d(x,f^n(x))}{n}$ exists, where $f^n(x)$ is $\underbrace{f(f(\cdots f(x)}_{n \text{times}} \cdots ))$.
$\text{b})$ Prove that the amount of the limit does not depend on choosing $x$.
3 replies
wiseman
Oct 2, 2014
alinazarboland
May 21, 2025
Double integration
Tricky123   2
N May 21, 2025 by Mathzeus1024
Q)
\[\iint_{R} \sin(xy) \,dx\,dy, \quad R = \left[0, \frac{\pi}{2}\right] \times \left[0, \frac{\pi}{2}\right]\]
How to solve the problem like this I am using the substitution method but its seems like very complicated in the last
Please help me
2 replies
Tricky123
May 18, 2025
Mathzeus1024
May 21, 2025
Find solution of IVP
neerajbhauryal   2
N May 15, 2025 by Mathzeus1024
Show that the initial value problem \[y''+by'+cy=g(t)\] with $y(t_o)=0=y'(t_o)$, where $b,c$ are constants has the form \[y(t)=\int^{t}_{t_0}K(t-s)g(s)ds\,\]

What I did
2 replies
neerajbhauryal
Sep 23, 2014
Mathzeus1024
May 15, 2025
fourier series?
keroro902   2
N May 15, 2025 by Mathzeus1024
f(x)=$\sum _{n=0}^{\infty } \text{cos}(nx)/2^{n}$
f(x) = ?
thanks
2 replies
keroro902
May 14, 2010
Mathzeus1024
May 15, 2025
uniformly continuous of multivariable function
keroro902   1
N May 14, 2025 by Mathzeus1024
How can I determine which of the following functions are uniformly continuous on the given domain A?

$f \left( x, y \right) = \frac{x^3 + y^2}{x^2 + y}$ , $A = \left\{ \left( x, y
\right) \in \mathbb m{R}^2 \left|  \right.  \left| y \right| \leq \frac{x^2}{2}
%Error. "nocomma" is a bad command.
, x^2 + y^2 < 1 \right\}$

$g \left( x, y \right) = \frac{y^2 + 4 x^2}{y^2 - 4 x^2 - 1}$, $A = \left\{
\left( x, y \right) \in \mathbb m{R}^2 \left| 0 \leq x^2 - y^2 \leqslant 1
\right\} \right.$
1 reply
keroro902
Nov 2, 2012
Mathzeus1024
May 14, 2025
Miklos Schweitzer 1971_5
ehsan2004   2
N May 9, 2025 by pi_quadrat_sechstel
Let $ \lambda_1 \leq \lambda_2 \leq...$ be a positive sequence and let $ K$ be a constant such that \[  \sum_{k=1}^{n-1} \lambda^2_k < K \lambda^2_n \;(n=1,2,...).\] Prove that there exists a constant $ K'$ such that \[  \sum_{k=1}^{n-1} \lambda_k < K' \lambda_n \;(n=1,2,...).\]

L. Leindler
2 replies
ehsan2004
Oct 29, 2008
pi_quadrat_sechstel
May 9, 2025
Miklos Schweitzer 1968_9
ehsan2004   1
N May 8, 2025 by pi_quadrat_sechstel
Let $ f(x)$ be a real function such that
\[ \lim_{x \rightarrow +\infty} \frac{f(x)}{e^x}=1\]
and $ |f''(x)|\leq c|f'(x)|$ for all sufficiently large $ x$. Prove that \[ \lim_{x \rightarrow +\infty} \frac{f'(x)}{e^x}=1.\]

P. Erdos
1 reply
ehsan2004
Oct 8, 2008
pi_quadrat_sechstel
May 8, 2025
Range of 2 parameters and Convergency of Improper Integral
Kunihiko_Chikaya   3
N May 1, 2025 by Mathzeus1024
Source: 2012 Kyoto University Master Course in Mathematics
Let $\alpha,\ \beta$ be real numbers. Find the ranges of $\alpha,\ \beta$ such that the improper integral $\int_1^{\infty} \frac{x^{\alpha}\ln x}{(1+x)^{\beta}}$ converges.
3 replies
Kunihiko_Chikaya
Aug 21, 2012
Mathzeus1024
May 1, 2025
f'(1)>1 implies f has a fixed point in (0,1)
Sayan   13
N Apr 27, 2025 by Apple_maths60
Source: ISI(BS) 2010 #4
A real valued function $f$ is defined on the interval $(-1,2)$. A point $x_0$ is said to be a fixed point of $f$ if $f(x_0)=x_0$. Suppose that $f$ is a differentiable function such that $f(0)>0$ and $f(1)=1$. Show that if $f'(1)>1$, then $f$ has a fixed point in the interval $(0,1)$.
13 replies
Sayan
May 17, 2012
Apple_maths60
Apr 27, 2025
Maximum value of ∫_0^1 e^x logf(x) dx when ∫_0^1 f(x) dx =1
tom-nowy   2
N Apr 26, 2025 by Dattier
Source: 1982 Niigata University entrance exam
Let $\mathcal{F}$ be the set of continuous functions $f: [0,1] \to (0, \infty )$ such that $ \int_0^1 f(x) \, \mathrm dx =1 $. For $f \in \mathcal{F}$, let $$I(f)=\int_0^1 e^x \log f(x) \, \mathrm dx .$$Determine $\max_{f \in \mathcal{F}}I(f)$.
2 replies
tom-nowy
Jul 7, 2013
Dattier
Apr 26, 2025
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.
1729 posts
#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
4190 posts
#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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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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