Let the $ k$-clique be $ (a_1,b_1),(a_2,b_2),\ldots,(a_{201},b_{201})$. For all $ 1\le i,j\le201$ there exist integers $ x,y$ such that $ \pm 2k=(a_ib_j-a_jb_i)+(a_j-a_i)y+(b_i-b_j)x$. By Pigeonhole Principle, there exist $ (a_i,b_i)$ and $ (a_j,b_j)$ which are congruent modulo 1,2,3,4,5,6,7,8,9,10,11,12,13. So $ 2k$ is divisible by $ lcm(1,2,3,4,5,6,7,8,9,10,11,12,13)$, which implies $ 540540|k$. I'm not sure how to continue this. I think $ k=540540$ is the answer and we should construct a $ k$-clique such that $ gcd(a_j-a_i,b_j-b_i)=1$ except when $ (a_i,b_i)\equiv(a_j,b_j)$ modulo something between 2 and 13.

The construction part is easy; take a $ 15 \times 15$ square.

I didn't notice that. I was thinking the number theory part too much. What do you think, if this problem were selected in IMO 2008 which problem would it be? 1? 2? 4? 5?

By the way, you miscalculated; it should be $ k=180180$.

I'd say only a tiny bit harder than 1/4, but for combinatorics I'm apparently biased toward thinking they're much easier than they are. (I think this is true of a lot of people in the US for various reasons.)

Peruvian trainer $ Jorge$ $ Tipe$ (nickname $ Tipe$ at mathlinks) proposed this problem. Last year and for the first time, two peruvian problems were selected for the IMO shortlist (these are C3 and G3)

Smallest possible $k$ is $180180$. Consider two points with coordinates $(a_1, a_2)$ and $(b_1, b_2)$. Then they are $k$-friends iff there exists a third point $(c_1, c_2)$ such that $$\frac{1}{2}\begin{vmatrix} a_1 & a_2 & 1\\ b_1 & b_2 & 1\\ c_1 & c_2 &1 \end{vmatrix} = k $$. So \[a_1(b_2-c_2)-a_2(b_1-c_1)+(b_1c_2-c_1b_20 = \pm 2k\]\[\Rightarrow (b_1-a_1)c_2+(a_2-b_2)c_1 = \pm 2k+a_2b_1-a_1b_2\]If say $b_1-a_1 = 0$ then we see that $a_2-b_2 | 2k$. Otherwise if $b_1 \neq a_1$ and $a_2 \neq b_2$, by taking mod $b_1-a_1$ we obtain \[2k+(a_1-c_1)(a_2-b_1) \equiv 0 \pmod{b_1-a_1}\]This equation has solution iff $gcd(b_1-a_1, a_2-b_1) |2k$. We see that even if one of $b_1-a_1$, $b_2-a_2$ is $0$, this is still a necessary and sufficient condition for the two points to be $k$-friends. Now suppose $2k$ is not divisible for some positive integer $i$ where $i < 15$. Then we consider all points mod $i$, there are $i^2$ possible distinct pairs so that if there are at least $i^2+1$ distinct points there exist two such that the gcd of their difference in $x$ coordinates and difference in $y$ coordinates is divisible by $i$ and it would be impossible. Then there would be at most $i^2 \le 196$ points. Otherwise, we indeed see that the smallest possible $k$ such that $2k$ is divisible by all positive integers less than $15$ is $lcm(1,2,3,4,5,6,7,8,9,10,11,12,13,14)/2 = 180180$. We see that $k = 180180$ can have a clique with more than $200$ elements by considering a lattice points on a square with side length $14$.

The NT is harder than the combo in this combo problem.... $k$ is $lcm(1,2,3,4,5,6,7,8,9,10,12,13)/2=180180$ Let us look at any 2 points that are k-friends, we can shift the graph so that they are only on the 2 axis wlog. Let the third point that makes the k-area triangle be $(a,0)$ and $(0.b)$. Let a third point be $(a+x, b+y)$. It is easy to see that the area of this triangle is $\frac{ab+bx+ay}{2}$. Thus $ay+bx=2k-ab$. Obviously since $gcd(a,b)$ divides $ay,bx,ab$, it must also divide $2k$ also. If it does, then $\frac{a}{gcd(a,b)}*y+\frac{b}{gcd(a,b)}*x=\frac{2k}{gcd(a,b)}-\frac{ab}{gcd(a,b)}$. Since $\frac{a}{gcd(a,b)}$ and $\frac{b}{gcd(a,b)}$ are coprime, there must exist integers $x$ and $y$ that satisfy the conditions. Thus the points $(x_1,y_1)$ and $(x_2,y_2)$ are k-friends iff $gcd(x_1-x_2,y_1-y_2)$ divides $2k$. Let us look at the coordinates mod $m$ for $0<m<14$. There are $m^2$ choices. Since there are more than $m^2$ points, there exists 2 points that have the same modulus for both their coordinates. Thus $m$ divides $2k$. Thus $lcm(1,2,3,4,5,6,7,8,9,10,11,12,13)/2=180180$ divides $k$. We see this minimum is satisfied by a 14x14 square with 5 more points on its border.(too lazy to include graph but you get the idea)

I misread problem i thought that for two points $A$ and $B$ to be $k-friends$ there needs to be point $K$ in $T$ s.t. area is $k$.What should be answer to this question ?

Lemma: $(a,b)$ and $(c,d)$ are $k$-friends if and only if $\gcd(c-a,d-b)\mid 2k$. Proof of Lemma: Note that $(a,b)$ and $(c,d)$ are $k$-friends if and only if there exists an integer pair $(x,y)$ such that \[\begin{vmatrix}1 & 0 & 0 \\ 1 & c-a & d-b \\ 1 & x & y\end{vmatrix} = \pm 2k,\]or \[(c-a)y-(d-b)x = \pm 2k.\]This is equivalent to $\gcd(c-a,d-b)\mid 2k$ by Bezout's lemma. $\blacksquare$ We claim that in order to get a $k$-clique of size at least $200$, we must have \[\mathrm{lcm}(1,2,\ldots,14)\mid 2k.\]To see this, note that for each $1\le n\le 14$, we have $n^2+1\le 200$, so there exist two points in the clique with matching residue classes mod $n$ for each coordinate, call the points $(a,b)$ and $(c,d)$. Thus, $n\mid \gcd(c-a,d-b)\mid 2k$, so $n\mid 2k$ for each $1\le n\le 14$, as desired. We claim that $k=\boxed{\frac{1}{2}\mathrm{lcm}(1,2,\ldots,14)}$ is in fact attainable. To do this, place $225$ points in a $15\times 15$ grid formation, and arbitrarily remove $25$ points. For each pair of points $(a,b)$ and $(c,d)$, we have $0\le|c-a|,|d-b|\le 14$, so certainly $\gcd(c-a,d-b)\mid \mathrm{lcm}(1,2,\ldots,14)$, as desired. $\blacksquare$

Why has this thread been bumped? From 2009-2017? Then 2018-2019? Really? I always find it weird when a 2009 thread is here ten years later.....

dragonrider1 wrote: Why has this thread been bumped? From 2009-2017? Then 2018-2019? Really? I always find it weird when a 2009 thread is here ten years later..... This question has been asked and answered before. The primary way people reach this particular thread is through the IMO Shortlist contest collection, and certainly not by searching through HSO. I was browsing through ISL looking for a problem to do, this one caught my eye, I spent about 20 minutes with it, and then wrote up and posted my solution.

Claim: $A,B$ are $k$-friends iff $(x_A-x_B,y_A-y_B)|2k$. Proof: WLOG $A=(0,0)$, $B=(m,n)$. Then, if $C=(x,y)$, $$2[ABC]=\begin{vmatrix}1 & 0 & 0\\1 & m & n\\ 1 & x & y\end{vmatrix}=my-nx$$If we want $my-nx=2k$, then we need $\gcd(m,n)|2k$. For the reverse, just note that $x,y$ are guaranteed to exist by Bezout. Now, I claim that if I have a set of 200 points, then $\forall x\le 14$, there exists two points $(a,b)$, $(c,d)$ such that $x|\gcd(a-c,b-d)$, which implies that $\text{lcm}(1,2,\ldots,14)=360360|2k$. Of course, there exist only $x^2$ elements in $(\mathbb{Z}/ x\mathbb{Z})^2$, so by pigeon-hole, there exists two points as desired. Hence, $360360|2k\implies 180180|k$. To show that $k=180180$ is sufficient, consider an arbitrary subset of $201$ points in the set $\{(x,y)|0\le x,y\le 14\}$.

Solved with Th3Numb3rThr33. First, we characterize all \(k\)-friends: Claim: \((a_1,b_1)\), \((a_2,b_2)\) are \(k\)-friends if and only if \(\gcd(a_2-a_1,b_2-b_1)\mid2k\). Proof. By translation, it is equivalent to show \((0,0)\), \((a,b)\) are \(k\)-friends if and only if \(\gcd(a,b)\mid2k\). By Shoelace formula, the area of the triangle with vertices \((0,0)\), \((a,b)\), \((x,y)\) should be \(\pm2k=bx-ay\), so the claim follows from B\'ezout. \(\blacksquare\) What remains is as follows: select 200 points \((a_1,b_1)\), \ldots, \((a_{200},b_{200})\), and minimize \[2k:=\mathop{\operatorname{lcm}}_{i,j}\big\{\gcd(a_i-a_j,b_i-b_j)\big\}.\]By Pigeonhole, for each \(n=1,\ldots,14\) we have \(n\mid\gcd(a_i-a_j,b_i-b_j)\) for some \(i\), \(j\); thus \[k\ge\frac{\operatorname{lcm}(1,\ldots,14)}2=180180.\]Finally, 180180 is attainable by taking \(\{(i,j):0\le i,j\le14\}\), so we are done.

Nice one April wrote: In the coordinate plane consider the set $ S$ of all points with integer coordinates. For a positive integer $ k$, two distinct points $A$, $ B\in S$ will be called $ k$-friends if there is a point $ C\in S$ such that the area of the triangle $ ABC$ is equal to $ k$. A set $ T\subset S$ will be called $ k$-clique if every two points in $ T$ are $ k$-friends. Find the least positive integer $ k$ for which there exits a $ k$-clique with more than 200 elements. Proposed by Jorge Tipe, Peru Solution: Claim: Points $\mathcal{A}\ (x,y)$ and $\mathcal{B}\ (w,z)$ are $k$-friends if and only if $\gcd(w-x, z-y) | 2k.$ Proof. We observe that $\mathcal{A}\ (x,y)$ and $\mathcal{B}\ (w,z)$ are $k$-friends if and only if $\gcd(w-x, z-y) | 2k$ if and only if there exists an integer pair $\mathcal{C}\ (u,v)$ such that (by Shoelace formula) $$\dfrac 12 \begin{vmatrix}1 & x & y \\ 1 & w & z \\ 1 & u & v\end{vmatrix} = k$$$$\implies \begin{vmatrix}1 & 0 & 0 \\ 1 & w-x & z-y \\ 1 & u & v\end{vmatrix} = 2k$$$$\implies v(w-x) + u(z-y) = \pm 2k.$$Now, the above is a Diophantine equation in variables $v, u$, hence by Bezout's lemma we get $\gcd(w-x, z-y) | 2k. \quad \blacksquare$ Next, we note that $n^2+1 = n\cdot n + 1 < 200$ for all $1 \leqslant n \leqslant 14.$ Now, we consider all coordinates modulo $n$. Hence, by PHP there exist $2$ pairs $(x,y)$ and $(w, z)$ in the clique which have the same residue classes modulo $n$. Thus, $n | \gcd(w-x,z-y)$. So, $n | 2k$ for each $1 \leqslant n \leqslant 14$ and therefore, $\mathrm{lcm} (1, 2, . . . , 14)|2k.$ For least positive integer $k$, we equate them and thus, $$k = \frac{\mathrm{lcm} (1, 2, . . . , 14)}{2} = \boxed{180180}.$$In the view of attainability, we consider the set of points $\mathcal{S} \overset{\text{Def}}{:=} \{(m,n): 0 \leqslant m, n \leqslant 14\}$, and an arbitrary subset of $201$ points from that $\mathcal{S}.\quad \square$

That's a beautiful problem, mixing combinatorial and number-theoretic ideas. Fresh and original. It is a pity that it did not appear on the IMO paper.

For three points to have area $k$, shift $A$ to $(0,0)$, let $B = (x,y)$ and $C = (a,b)$, by shoelace, we want $ax - by = 2k$, and since $ax-by$ can take on every multiple of the gcd, we have that gcd$(x,y) | 2k$, so letting all points have coordinates $(x_1, y_1), (x_2, y_2), \cdots, (x_{201}, y_{201})$, we want gcd$(x_i - x_j, y_i - y_j) | 2k$ always. Note that for any number $n$ with $n^2 < 201$, we must have two points that have the same coordinates mod $n$, so $n | 2k$, since we can do this for all $n = 1,2, \cdots, 14$, we must have that lcm$(1,2,\cdots, 14) | 2k \implies 360360 | 2k \implies 180180 | k$, so $k \ge 180180$. For a construction, consider a $15 \times 15$ square in the lattice grid. (You dont need to ignore any points because problem says $> 200$, not exactly any number). $\blacksquare$

The answer is $k = \frac{1}{2} \operatorname{lcm}(1, 2, \dots, 14) = 180180$. First, we describe the relation. Claim: Given two points $A,B \in {\mathbb Z}^2$, let $A-B = (x,y)$. Then $A$ and $B$ are $k$-friends if and only if $\gcd(x,y)$ divides $2k$. Proof. [Sketch of proof] If $\gcd(x,y) = 1$, then use Bezout to find a triangle with area $\frac{1}{2}$. In this way we can construct triangles whose area is a multiple of $\frac{1}{2} k$. Conversely, if $\gcd(x,y) = n$ then any triangle using $AB$ as a side can be dissected by cevians into $n$ lattice triangles of equal area, and any lattice triangle has area a multiple of $\frac{1}{2}$. $\blacksquare$ We claim that if $q$ is a prime power with $q^2 \le 200$, then $q$ must divide $2k$. Indeed, this just follows by the pigeonhole principle as two of the points have both the same $x$ and $y$ coordinates mod $q$. Thus $2k$ must be divisible by $\operatorname{lcm}(1, \dots, 14)$. Conversely, if one takes a $15 \times 15$ square, this gives a valid construction.

We claim the minimal $k$ is 180180 = $\frac{1}{2}lcm(1, 2, ..., 14)$. Claim 1: $P_i = (x_i, y_i)$ and $P_j = (x_j, y_j)$ are $k$-friends if and only if $d = \gcd(x_j - x_i, y_j - y_i) | 2k$. Proof: Let $x_j - x_i = da$ and $y_j - y_i = db$. Note that if $(x_i + \alpha, y_i + \beta) \in P_iP_j$, then by similar triangles $\frac{\beta}{\alpha} = \frac{db}{da} = \frac{b}{a}$ where $\gcd(a, b) = 1$. Hence, as integers $\alpha = ca$, $\beta = cb$, and since $\alpha \in [0, da]$, $c \in [0, d]$. Thus, the segment $P_iP_j$ contains $d+1$ equally-spaced points. Let $P_{i+1}$ satisfy $\overrightarrow{P_iP_{i+1}} = \frac{1}{d} \overrightarrow{P_iP_j}$. Note that $P_{i+1} = (x_i + a, y_i + b)$ is a lattice point. Proof of if direction: Consider the square bound by $P_i, P_j, (x_i + db, y_i - da), (x_j + db, y_j - da)$. It is a finite area and hence contains finitely many points. Consider the lattice point $X$ in the square (excluding $P_iP_j$ but including the other two corners) with the shortest distance $h$ to $P_iP_j$. By minimality, no lattice point can lie on $P_iX$ or $P_jX$, or inside of $\triangle P_iP_jX$. Furthermore $h$ is non-zero so the triangle isn't degenerate. Hence by Pick's theorem, its area is $0 + \frac{d+1 + 1}{2} - 1 = \frac{d}{2}$. Now consider the point $Y$ satisfying $\overrightarrow{P_iY} = \frac{2k}{d} \cdot \overrightarrow{P_iX}$. Note that $\frac{2k}{d}$ is an integer so $Y$ is a lattice point. Furthermore we get that the area of $\triangle P_iP_jY$ is $\frac{2k}{d} \cdot \frac{d}{2} = k$, since height is $\frac{2k}{d} \cdot h$ by similar triangles but base is invariant. Proof of only if direction: Let $\triangle P_iP_jZ$ have area $k$. Then $\triangle P_iP_{i+1}Z$ has area $\frac{1}{d}k$ since the height from the line $P_iP_j$ is invariant but the base length $|P_iP_{i+1}| = \frac{1}{d} |P_iP_j|$. Note that by Pick's theorem, all lattice polygons have area of the form $\frac{n}{2}$ where $n$ is an integer, and hence $\frac{k}{d} = \frac{n}{2} \Rightarrow \frac{2k}{d} = n \Rightarrow d | 2k$. $\square$ Proof of Bound Note that for $m = 1, 2, ..., 14$, we have $200 \ge m^2 + 1$. Since there are only $m^2$ pairs of residues $(x, y) \pmod{m}$, for each $m$ there exist two points $(x_m, y_m)$, $(x_m', y_m')$ such that $m | \gcd(x_m' - x_m, y_m' - y_m)$. Hence by Claim 1, $m | 2k$. Thus $lcm(1, 2, ..., 14) | 2k$, and $k \ge \frac{1}{2} lcm(1, 2, ...., 14)$. Construction Consider a $15 \times 15$ grid of points, with $225 \ge 200$ points. Let $k = \frac{1}{2} lcm(1, 2, ..., 14)$. $|x_i - x_j| \le 14$ and similarly for $y$; hence $\gcd(x_j - x_i, y_j - y_i) \le 14$ over all pairs of points. Hence $\frac{d}{2} \le \frac{14}{2} \Rightarrow k = n \cdot \frac{d}{2}$, and by Claim 1, all pairs of points are $k$-friends.

Remark

Let $(x_1,y_1)$ and $(x_2,y_2)$ be two points. We show that these two are $k$-friends if and only if $\gcd(x_1-x_2,y_1-y_2)\mid 2k$. Note that we can shift until $(x_2,y_2)$ is on the origin, then the area of the triangle formed by the origin, $(a_1,b_1)$, and $(a_2,b_2)$ is $\tfrac12 |a_1b_2-a_2b_1|$ which by Bezout's can be all the multiples of $\gcd(a_1,b_1)$ but no other values. Thus, our claim is proved. Now, given $200$ points we claim that there is either $15$ distinct $y$-values or $15$ distinct $x$-values. Otherwise, there are only $196$ points. WLOG, let it be the distinct $y$-values. Now, for each integer $n\le 14$, by pigeonhole principle on the residues $\pmod n$ there will be two points with $y$-values that are different but differ by a multiple of $n$. Thus, \[k\mid \tfrac{1}{2} \text{lcm}(1,2,\dots,14)=180180\]and note that $k=180180$ works with any subset of $15\times 15$ square.

pasted from overleaf so here you go \begin{soln} \begin{claim*} Given two points $A,B$ with usual definitions, let $A-B = (x,y)$. Then A and B are k-friends iff $\gcd(x,y)\mid2k$. \end{claim*} \begin{proof} By shifting to 0,0 and a,b, Shoelace implies $|2k|=bx-ay$, and this is Bezout. \end{proof} Next, notice by Pigeonhole that any $x:x^2\le 200$ makes it s.t. two points have the same coordinate mod x, so subtstituting into the lemma implies that cx|2k, which readily implies k=lcm(1,2,...,14)=180180. \end{soln}

The area of a triangle $(0,0),(a,b),(c,d)$ is $\frac{|ad-bc|}{2}$, so we want $ad-bc=\pm 2k$. By Bezout, there exists $c,d$ that satisfy this if and only if $\gcd(a,b)\mid 2k$. Thus, $(0,0)$ and $(a,b)$ are $k$-friendly iff $\gcd(a,b)\mid 2k$. We claim that the answer is $$2k=lcm(1,2,3\cdots 14)\rightarrow k=180180.$$ Consider any positive integer $n\leq 14$. Partition the lattice points into $n^2$ sets based on the x-coordinate mod $n$ and the y-coordinate mod $n$. By pigeonhole, there must be some two of the 200 points in the same set. Thus, the gcd of their difference must be a multiple of $n$, hence $n\mid 2k$. This shows the lower bound. To show that $180180$ works, take a 15 by 15 square grid, and pick any subset of 200 points. The coordinate differences between any two points are at most 14, so their gcd will clearly divide $2k,$ hence done.

We claim the answer is $\frac 12 \text{lcm} (1, 2, 3, \ldots, 14) = 180180$. Construction: Consider the 225 points of a 15 by 15 array of lattice points, and denote $A = (a_1, a_2)$. The distance from $B$ to either $x = a_1$ or $y = a_2$, whichever is nonzero, must be an integer from 1 to 14 inclusive, meaning we can construct $C$ on one of the lines such that $[ABC] = \frac 12 \text{lcm} (1, 2, 3, \ldots, 14)$. ${\color{blue} \blacksquare}$ Minimality: Translate an arbitrary triangle $\triangle ABC$ such that $A = (0, 0)$, $B = (m, n)$, and $C = (x, y)$. Then \[[ABC] = \frac{|xn - ym|}{2}.\]From Bezout's, we know the least possible positive integer value of $xn - ym$ is $\gcd(m, n)$, so $\gcd{m, n} \mid 2k$. For any integer $\ell$ between 1 and 14 inclusive, we can group our points into $\ell^2 < 200$ sets based on $\ell$ possible $x$-coordinate residues and $\ell$ possible $y$-coordinate residues. By Pigeonhole, we find two points must have their $x$- and $y$-coordinates differ by a multiple of $\ell$, so \[\ell \mid \gcd(m, n) \mid 2k,\]which proves $2k \ge \text{lcm} (1, 2, 3, \ldots, 14)$. ${\color{blue} \blacksquare}$ Thus our answer is $\boxed{180180}$. $\blacksquare$

The answer is $\frac{1}{2} \operatorname{lcm}(1,2,3\dots, 14) = \boxed{180180}$. Notice that this is possible by arranging the points in a $15 \times 15$ square since the distances between each square horizontally and vertically lie in the set $\{1,2,\dots, 14 \}$. Now, we will prove that this is the minimum. Begin by translating two points $A$ and $B$ so that point $A$ is the origin and point $B$ is $(a,b)$. In order for $A$ and $B$ to be $k$-friends, there must exist a point $C =(c,d)$ such that the area of $\triangle ABC$ equals $k$. The Shoelace Theorem states that the area of $\triangle ABC$ is \[\frac{|ad-bc|}{2},\] so we would like to have $ad-bc=\pm 2k$. By Bezout's Identity, there exist such $c$ and $d$ if and only if $\gcd(a,b) \mid 2k$. Now, for each value $\ell=1,2,\dots,14$, we have $\ell^2 \le 200$, so by Pigeonhole, we must have at least two points $C$ and $D$ whose $x$ and $y$-coordinates differ by a multiple of $\ell$. Translate them such that $C$ is the origin and $D=(x,y)$. We must have \[\ell \mid \gcd(x,y) \mid 2k\] Thus, $k \ge \frac{1}{2}\operatorname{lcm} (1,2,\dots, 14)$, and we are done.

(solved w/ 70% hint on ARCH) We claim that the answer is $k=\tfrac12 \text{lcm}(1, 2, 3,..., 14)$. Let two points $A=(x_A, y_A)$ and $B=(x_B, y_B)$ be $k$-friends. Translate $B$ to the origin such that $A=(x_A-x_B, y_A-y_B)$ and $B=(0,0)$. Then, let point $C=(x_C, y_C)$ be the point such that $[ABC]=k$. By the Shoelace Theorem, note that \[[ABC]=\tfrac12|y_C(x_B-x_A)+x_C(y_B-y_A)|=k\]\[y_C(x_B-x_A)+x_C(y_B-y_A)=2k\](we do not need to worry about the signs). Since $x_B-x_A$ and $y_B-y_A$ are constants, by Bezout's Theorem, this only holds true when $\gcd(x_B-x_A, y_B-y_A)=2k$. Now, consider the coordinates of the points mod $q$. By the Pigeonhole Principle, note that if there is a prime power $q$ where $q^2 \le 200$, then $q \mid 2k$. From here it follows that $\text{lcm}(1, 2, 3,..., 14) \mid 2k$, or $\tfrac12 \text{lcm}(1, 2, 3,..., 14) \mid k$, as desired. (edit: forgot construction oops)

The answer is $k=\frac12\operatorname{lcm}(1,2,3,4,\dots,14)$, achieved by a $15$ by $15$ square. By the Shoelace Formula, we can use Bezout's Lemma to achieve everything(I will elaborate more on this later in the proof). To prove that $k$ is minimal, we will show that for all $n\le 14$, we must have $2k\mid n$. By the pigeonhole principle, two of the $200$ points will have the same coordinates mod $n$. Then let them be $(a,b)$ and $(a+cn,b+dn)$. Using the Shoelace Formula, the third point $(x,y)$ must satisfy \[ \left|\pm xdn\pm ycn\pm(ad-bc)n\right|=2k, \]where I am too lazy to figure out the signs. We are done. $\blacksquare$

The answer is 180180

Let $K$ be the clique with size $\ge 200$ mentioned in the statement. We claim the answer is $k=\frac{1}{2}\times \mathrm{lcm}(1,2,\ldots,14)=180180$. The proof is heavily based on this one observation : Claim : Two points $A=(x_A,y_A)$, $B=(x_B,y_B)$ in $S$ are $k$-friends if and only if $\gcd(\Delta x,\Delta y)$ divides $2k$. Proof : Note that a point $C\in S$ satisfies $\mathcal{A}(ABC)=k$ if and only if there exists $x_C$, $y_C$ $\in \mathbb Z$ with \[k=\frac{1}{2}\times \sum x_A(y_B-y_C)\iff 2k=x_Ay_B-x_By_A-x_C\cdot \Delta y+y_C\cdot \Delta x\]The result then follows from Bezout's lemma since $\gcd(\Delta x,\Delta y)$ divides $-y_A\Delta x+x_A\Delta y=x_Ay_B-x_By_A$. $\square$ In particular, if $m$ denotes the least positive integer that doesn't divide $2k$, then pairs $(x_v,y_v)$ where $v\in K$ never take the same value twice modulo $m$. Since there are $m^2$ such pairs we must have $m^2\ge \vert K\vert \ge 200$, from which it follows that $m\ge 15$. Therefore, we must have $k\ge 180180$. Conversely, the set $\{(x,y)\mid x,y\in \{0,1,\ldots,14\}\}$ is a $180180$-clique of size $225$. Indeed, for points $A$ and $B$ in this set we will always have $\gcd(\Delta x,\Delta y)\le 14$ dividing $\mathrm{lcm}(1,2,\ldots,14)$, as desired. $\square$

The answer is $180180$. Before going any further in the solution we need the following lemma which characterize all $k$-friends: Lemma: $(x_1,y_1)$ and $(x_2,y_2)$ are $k$-friends if and only if $(x_1-x_2,y_1-y_2)|2k$ Proof: we need $(x_3,y_3)\in \mathbb Z^2$ such that $x_3(y_1-y_2)+y_3(x_2-x_1)=2k+x_2y_1-x_1y_2$ from which the conclusion follows by Bezout. Now we have $(x_1,y_1),(x_2,y_2),\dots (x_{201},y_{201})$ such that $(x_i-x_j,y_i-y_j)|2k$ for every $1\le i<j\le 201$ Let's fix $1\le t\le 14$. Considering $(x_i,y_i)$ modulo $t$ we have $t^2$ possible values but at least $t^2+1$ pairs so by pigeonhole we have $(x_i,y_i)\equiv (x_j,y_j)\pmod t$ so $t|2k$. Thus $lcm(1,2,..,14)|2k$ from which we get $k\ge 180180$ The construction part is easy: choose $(i,j)$ where $1\le i,j \le 14$ and $(15,1);(15,2);(15,3);(15,4);(15,5)$. Now observe that chosing $2$ pairs they differ by at most $14$ in both coordinates so the gcd is at most $14$ as wanted

Answer is $\mathrm{lcm}(1,2,3,\dots,14)/2$ with the construction being a $15\times 15$ square. Associate each pair of points with a vector $[a,b]$ that is the difference of the points. The claim is that these two points are $k$ friends for $\gcd(a,b)\mid 2k$, which is true by shoelace and Bezout. Consider a point and the $N>200$ vectors connected to it. $2$ of these vectors must be equivalent modulo $n$ for $1\leq n \leq 14$ by pigeonhole, implying all such $n$ divide $2k$.

My first $C3$ $\textbf{Answer:}$ $180180$ $\textbf{Solution:}$ Let $A(a_1,a_2)$ , $B(b_1,b_2)$. So for $A,B$ to be $k$-friends there exists a point $C(c_1,c_2)$ such that the area of $\triangle ABC$ is $k$. Now by Shoelace formula we have: $k=\dfrac{1}{2} \cdot \begin{vmatrix}a_1 & a_2 & 1 \\ b_1 & b_2 & 1 \\ c_1 & c_2 & 1 \end{vmatrix} \implies 2k= \begin{vmatrix}a_1 & a_2 & 1 \\ b_1 & b_2 & 1 \\ c_1 & c_2 & 1 \end{vmatrix} \implies 2k=\begin{vmatrix}0 & 0 & 1 \\ b_1-a_1 & b_2-a_2 & 1 \\ c_1 & c_2 & 1 \end{vmatrix} \implies$ $$2k=c_2(b_1-a_1)+c_1(b_2-a_2) $$which by Bezout's identity there exists such $c_1$ and $c_2$ only iff $$gcd(b_1-a_1,b_2-a_2) \mid 2k ...(*)$$Let $(x_1,y_1),(x_2,y_2) \dots (x_{201},y_{201}) \dots $ be the cordinates of the points in the set $T$. Also note that iff we pick $n \leq 14$ then $n^2+1 \leq 14^2+1=197 \leq 201$. So we know that $x_i \pmod{n}$ has $n$ possible values, simmilarly $y_i \pmod{n}$ $n$ possible values. So $(x_i,y_i) \pmod{n}$ has $n^2$ possible values $\forall i: 1 \leq i \leq 201$. But we know that we have at least $201$ points but since $n^2+1 \leq 201$ we can say that we have at least $n^2+1$ points. So we are having at least $n^2+1$ points and $n^2$ remainders$\pmod{n}$ now by pigeonhole principle we have that there are at least $2$ points that their cordiantes have the same remainder $\pmod{n} \iff$ $$(x_i,y_i) \equiv (x_j,y_j) \pmod{n} \forall i,j: 1\leq i,j \leq 201$$ $(x_i,y_i) \equiv (x_j,y_j) \pmod{n} \implies (x_i,y_i) - (x_j,y_j) \equiv 0 \pmod{n}\implies (x_i-x_j,y_i-y_j) \equiv 0 \pmod{n} \implies$ $ n \mid (x_i-x_j,y_i-y_j) \implies $ $$ n \mid \text{gcd}(x_i-x_j,y_i-y_j) ...(@)$$ Combining $(*)$ and $(@)$ we get the following: $$ n \mid \text{gcd}(x_i-x_j,y_i-y_j) \mid 2k \implies n \mid 2k$$ Since $1 \leq n \leq 14$ then we have: $\text{lcm}(1,2, \dots , 14) \mid 2k \implies 216216 \mid 2k \implies 216216 \leq 2k \implies 108108 \leq k$ which means that $\boxed{k=108108}$ $\blacksquare$.