Given a language $\ell$, consider triplets of people $x, \{y, z\}$ such that $x$ talks with both $y$ and $z$ in $\ell$. We consider $x, \{y, z\}$ and $x, \{z, y\}$ to be the same triplet. Suppose there are $T$ such triplets. Because no three members all use the same language, the people in any three-member set generate either zero or one such triplets. Furthermore, three-member sets that generate zero triplets are precisely those counted in the problem statement, so \[ A + T = \binom{2k+1}{3}. \] Now, for any $x$, the number of triplets $x, \{y, z\}$ is at least $k$, seen by discrete smoothing and attained iff $x$ talks to exactly two people using each language. This minimum is attainable, and thus generates a maximum for $A$: label the members from $0$ to $2k$, the languages from $1$ to $k$, and having member $m$ talk with people $m + \ell, m - \ell \bmod{2k+1}$ in language $\ell$. Thus in the maximal example, $T = k(2k+1)$ and \[A = \binom{2k+1}{3} - T = \frac{4}{3}k^3 - 2k^2 - \frac{4}{3}k.\]

math_explorer wrote: This minimum is attainable, and thus generates a maximum for $A$: label the members from $0$ to $2k$, the languages from $1$ to $k$, and having member $m$ talk with people $m + \ell, m - \ell \bmod{2k+1}$ in language $\ell$. But what if $k=4$, $\ell = 3$? Then you get a monochromatic triangle.

This problem was modified from my proposal. Originally there were $n$ people and 3 languages, and the statement was just to prove an inequality.

Represent each member by a point in the plane, and if two members $x$ and $y$ speak to each other in language in $z$, connect points $x$ and $y$ with a segment of color $z$. Call a triangle among such a set of points diverse if none of its edges are the same color, and undiverse if this is not true. There are $\binom{2k+1}{3}$ total triangles in this arrangement. We can count the number of undiverse triangles as follows. Let $F(p_i,c_j)$ be the number of edges of color $j$ that have an endpoint at point $p_i$. Because no triangles are monochrome, each undiverse triangle can be uniquely counted by the one vertex it has that joins sides of the same color. Thus we can count the number of undiverse triangles at point $p_i$ as $\sum_{j=1}^{k}\binom{F(p_i,c_j)}{2}$, since every pair of edges of the same of color represents exactly one undiverse triangle. Since $\sum_{j=1}^{k}F(p_i,c_j)=2k$, the number of undiverse triangles at $p_i$ is minimized if $F(p_i,c_j)=2$ for all j, and thus the total number of undiverse triangles is minimized if $F(p_i,c_j)=2$ for all $i$ and $j$. The number of undiverse triangles is at least $(2k+1)(k)=\binom{2k+1}{2}$, and so the number of diverse triangles is at most $\binom{2k+1}{3}-\binom{2k+1}{2}$. We show that such an arrangement is always attainable. Color all edges which connect points $p_i$ and $p_j$ with $i+j\equiv0\bmod{2k+1}$ a different color. Suppose the edge connecting points $p_i$ and $p_j$ has color $c_k$. Then if $m+n\equiv2i$ or $2j\bmod{2k+1}$, color the edge connecting $p_m$ and $p_n$ with color $c_k$ as well. Edit: if $2k+1$ is divisible by $3$, then connect all points with $i+j\equiv1\bmod{2k+1}$ with a different color and proceed as before. The necessary condition here is that $3i\equiv1\bmod{2k+1}$ has no solution, whereas before it was that $3i\equiv0\bmod{2k+1}$ has no solution for positive integers $i$ less than $2k+1$.

I claim that the answer is $\frac{2k(k-2)(2k+1)}{3}$ except for $k = 1$, where the problem just doesn't work. I will consider a graph with $2k+1$ vertices that represent people and the edge in between each person is colored a language. Consider a triple of people, $a, b, c$. Either $ab$, $bc$, and $ca$ have two languages the same and one different, or all three are distinct. Call the first type bad and the second type good. The total number of triples (without order), is $\binom{2k+1}{3}$. We now have to minimize how many bad triples there are. Notice that each person has degree $2k$. Suppose $x_i$ is the number of edges that are the $i$-th language. Notice that the number of triples that are bad coming from a certain person is $$\sum_{i = 1}^k \binom{x_i}{2} = \sum_{i=1}^k \frac{1}{2} (x_i^2 - x_i) = \left(\sum_{i=1}^k \frac{1}{2} (x_i^2)\right) - \frac{1}{2} \cdot 2k$$From Cauchy Schwarz, we have $$\left(\sum_{i=1}^k \frac{1}{2} (x_i^2)\right) - \frac{1}{2} \cdot 2k \ge \frac{(2k)^2}{2k} - k = k$$The minimum number of bad triples is thus at least $(2k+1)k = 2k^2 + k$. Therefore, the maximum number of good triples is at most $$\binom{2k+1}{3} - 2k^2 - k = \frac{2(k-2)k(2k+1)}{3}$$ We now provide a construction to show this bound is tight. Number the vertices from $0$ to $2k$. Then, for language $i$, have the following edges speak language $i$ (which ranges from $0$ to $k - 1$): $$i \to (i + 1) \to (i -1) \to (i + 2) \to (i - 2) \to \cdots \to (i + k) \to (i - k) \to i$$where all of these indices are taken modulo $2k + 1$. Given $2k + 1 > 3$, we have no monochromatic triangles. (I have edited the construction as the previous one I had here was faulty).

tim9099xxzz wrote: We show that such an arrangement is always attainable. Color all edges which connect points $p_i$ and $p_j$ with $i+j\equiv0\bmod{2k+1}$ a different color. Suppose the edge connecting points $p_i$ and $p_j$ has color $c_k$. Then if $m+n\equiv2i$ or $2j\bmod{2k+1}$, color the edge connecting $p_m$ and $p_n$ with color $c_k$ as well. Edit: if $2k+1$ is divisible by $3$, then connect all points with $i+j\equiv1\bmod{2k+1}$ with a different color and proceed as before. The necessary condition here is that $3i\equiv1\bmod{2k+1}$ has no solution, whereas before it was that $3i\equiv0\bmod{2k+1}$ has no solution for positive integers $i$ less than $2k+1$. When $k=4$ this appears to give a monochromatic triangle $(2,5,8)$ unless I'm mis-interpreting. fclvbfm934 wrote: Label each vertex a number, starting from $0$ to $2k$. Then, for language $i$, person $x$ speaks language $i$ with $x+i$ modulo $2k+1$. This construction clearly works. When $k=4$, or indeed whenever $2k+1 \equiv 0 \pmod 3$, the language $i=(2k+1)/3$ consists entirely of monochromatic equilateral triangles.

The problem with the earlier solutions is their construction for the equality case, fortunately there still exists a nice construction. Take $2k$ of the members and place them into a $2k$-gon with vertices $1,2,...,2k$ counter clockwise. For language $1$, have $1$ speak to $2$ speak to $2k$ speak to $3$ speak to $2k-1$ etc., zig-zagging through the $2k$-gon to create a Hamiltonian path (excluding $2k+1$). Then we can rotate this to form Hamiltonian paths for each language, and complete them into Hamiltonian cycles by connecting their endpoints to $2k+1$. For example, $1$ speaks to $2k+1$ speaks to $k+1$ in language $1$. This is known as a Hamiltonian Decomposition.

Attachments:

Sorry for the revive, but how do we check that the above construction works? I was able to get the bound for this problem but could not find a good construction.

Hope this is correct. ;-; Let $A_1 \dots A_{2k+1}$ be the $2k+1$ people. By pigeonhole, for every $A_i,$ there is at least one language where at least two other people speak it with $A_i$. Given that there are no three people that all speak the same language with each other, if $A_i$ speaks the same language with $A_j$ and $A_k$, then it must be the case that $A_j$ and $A_k$ don't speak the same language as $A_i$ and $A_k/A_j$ do. Now, I claim that the answer is maximized when, for (all) $A_i,$ and any given language, there are exactly two others who speak that same language as $A_i$ (this is easy to prove by smoothing). Let $N$ be the number of such triplets. We know that $A = {2k+1 \choose 3}-N$ by complementary counting, and $N = (2k+1)\frac{(2k+1)-1}{2} = k(2k+1)$ by easy counting. \[A = {2k+1 \choose 3}-k(2k+1) = \boxed{\frac{2}{3} (k-2)k(2k+1)}\]

We claim that the answer is $\frac{2}{3}k(2k + 1)(k - 2)$. Rephrase the problem in terms of graph theory in the obvious ways (using edge colorings of a $K_{2k + 1}$). Call a triangle \textit{rainbow} if its three sides are of distinct colors. We instead find the minimal number of triangles which are \textit{not} rainbow. Note that any non-rainbow triangle contains $2$ edges of the same color, and one vertex of a different color; call the vertex adjacent to the two edges of the same color the \textit{apex} of the triangle. Let $v$ be a fixed vertex, and suppose $v$ is adjacent to $t_c$ edges with color $c$. The number of non-rainbow triangles which have apex $v$ is \begin{align*} \sum_{i = 1}^{k} \binom{t_i}{2} &\geq k\binom{\frac{\sum_{i = 1}^k t_i}{k}}{2} \\ &\geq k\binom{2}{2} \\ &= k, \end{align*}where the first inequality follows by Jensen. Summing over all vertices $v$, we find that there are at least $k(2k + 1)$ non-rainbow triangles (since each non-rainbow triangle has a distinct apex). Now, note that the graph contains $\binom{2k + 1}{3}$ total triangles. It follows that there are at most $\binom{2k + 1}{3} - k(2k + 1) = \frac{2}{3}k(2k + 1)(k - 2)$ rainbow triangles, as claimed. Equality is achieved by having each vertex touch exactly $2$ edges of each color. $\Box$

Solved with porkemon2. Assume $k > 1$, as when $k=1$ the problem scenario can't hold. For any triple of people, there are two scenarios: all three pairs speak in different languages, or $X$ speaks to $Y$ and $C$ in the same language, and $Y$ and $Z$ talk in a different language. We compute the number of triples of the latter type (and call them wacky, with $X$ being ridonkulous), and subtract this from $\binom{2k+1}{3}$. To count the number of wacky triples, we do casework on the ridonkulous person. Label the languages $0$ through $k-1$, and say $X$ talks to $a_i$ people in language $i$. Then there are $$\sum_{i=0}^{k-1} \binom{a_i}{2}$$wacky triples with $X$ being ridonkulous. Since $\sum_{i=0}^{k-1}a_i = 2k$, a smoothing argument shows that our expression is minimized when $a_i=2$ for all $i$. Then there are $k$ wacky triples with $X$ being ridonkulous. Summing over all people, there are $k(2k+1)$ wacky triples, so $A=\boxed{\binom{2k+1}{3}-k(2k+1)}.$ We are left to show that equality can hold. Equality holds when each person speaks to exactly $2$ others in each language, and no three people all speak in the same language. Label the people $0,1, \dots 2k$, and the languages $0,1, \dots k-1$. Let $$f(a,b)=\begin{cases}\left(\left\lfloor \frac{a+b}{2} \right\rfloor \pmod{k}\right) \text{ if } a \neq 2k \text{ and } b \neq 2k, \text{ and} \\ \left(\min(a,b) \pmod{k}\right) \text{ otherwise.} \end{cases}$$Then we claim that if $a$ and $b$ talk in language $f(a,b)$, then the equality conditions are satisfied. First we verify that each person speaks to exactly $2$ people in each of the $k$ languages. For a person $a \neq 2k$, look at people $0,1, \dots a-1,a+1, \dots 2k-1$. When we add them to $a$ and divide by $2$, we get an arithmetic sequence starting at $\frac{a}{2}$ with difference $\frac{1}{2}$, except we delete the term $a$. When we take the floor and take it mod $k$, it is clear that each language besides $a \pmod{k}$ is used exactly twice, while this one is used once. Since this language is used again with $2k$, this works. It is clear that $2k$ speaks to exactly $2$ people in each of the $k$ languages, so we have shown the first equality condition. We now show the second one. Say we have people $a,b,$ and $c$. If $c=2k$, then unless $a \equiv b\pmod{k}$, they will not all speak the same language. If this is true, then WLOG $b=a+k$. Then $$a \equiv \left\lfloor a+\frac{k}{2} \right\rfloor \pmod{k}$$must be true, which is easy to verify is not. Otherwise, we do cases on the parity distribution of $a,b,$ and $c$. If they are all even, let $a=2a_1,b=2b_1,$ and $c=2c_1$. Then the languages spoken (mod $k$ is assumed here) are $$a_1+b_1,b_1+c_1,c_1+a_1.$$However, if these are all the same, then $a_1 \equiv b_1 \equiv c_1 \pmod{k}$, so $a_1=b_1=c_1$, so $a=b=c$, a contradiction. If two of them are even, WLOG $a=2a_1,b=2b_1,c=2c_1+1$. Then the languages spoken (mod $k$ is assumed here) are $$a_1+b_1,b_1+c_1,c_1+a_1.$$However, if these are all the same, then $a_1 \equiv b_1\pmod{k}$, so $a_1=b_1$, so $a=b$, a contradiction. If one of them is even, WLOG $a=2a_1,b=2b_1+1,c=2c_1+1$. Then the languages spoken (mod $k$ is assumed here) are $$a_1+b_1,b_1+c_1+1,c_1+a_1.$$However, if these are all the same, then $b_1 \equiv c_1\pmod{k}$, so $b_1=c_1$, so $b=c$, a contradiction. If none of them are even, let $a=2a_1+1,b=2b_1+1,c=2c_1+1$. Then the languages spoken (mod $k$ is assumed here) are $$a_1+b_1+1,b_1+c_1+1,c_1+a_1+1.$$However, if these are all the same, then $a_1 \equiv b_1 \equiv c_1\pmod{k}$, so $a_1=b_1=c_1$, so $a=b=c$, a contradiction. Thus, all cases are exhausted, so our construction is an equality case.

Observe that the only possible triangles now are where the three languages are different, and when two of the edges are the same. Define $x_{i, j}$ with $i$ varying from $1$ to $2k + 1$ and $j$ varying from $1$ to $k$ as the number of times person $i$ speaks language $j$ with another person. Notice that $x_{i, 1} + x_{i, 2} + \cdots + x_{i, k} = 2k$. We have that \begin{align*} \sum_{i} \sum_j \binom{x_{i,j}}{2} = \text{number of isosceles triangles} \geq (2k + 1) \cdot k, \end{align*}thus \begin{align*} \text{number of scalene triangles} \leq \binom{2k + 1}{3} - k(2k + 1). \end{align*} And now for the construction. Observe equality holds when each person speaks each language with exactly two other people per language; in other words, we want each language to form a Hamiltonian cycle. But it's well known that $K_{2k + 1}$ can always be decomposed into Hamiltonian cycles, thus done.

I claim $A \leq \binom{2k+1}{3}-k(2k+1)$. Note that there are only three classes of triplets. All same color, two colors, and all different colors. Since "all same color" has been precluded, if we let $B$ be the number of two color triples, then $A+B = $total number of triplets $ =\binom{2k+1}{3}$. Thus, it suffices to show $B\geq k(2k+1)$. Clearly we can bound this below. For each of the $2k+1$ vertices, let $c_i$ be the number of edges of color $i$. Then, the number of $B$ triplets centered here is \[\sum_{i=1}^{k} \binom{c_i}{2} \geq k\cdot \binom{\frac{\sum c_i}{k}}{2} = k \cdot \binom{2}{2} = k\]All $2k+1$ edges have the same bound, so $B\geq (2k+1)\cdot k$. Next, for the construction. We number all $2k+1$ points, then all vertices of the form $P_mP_{m+i}$(mod 2k+1) are colored with color $i$, which clearly satisfies both our no monochromatic triples condition, as well as $B=(2k+1)*k$. It is well known that $K_{n+1}$ can be partioned into hamiltonian paths, which clearly satisfies $B=(2k+1)\cdot k$ and we are done. $\blacksquare$

Solved with Félix Walecki. The answer is $\frac{2k(k-2)(2k+1)}{3}$. Assume $k>1$, otherwise the first condition doesn't hold. Now we can view this problem in terms of graph colorings. We have a complete graph on $2k+1$ vertices, and we wish to color its $k(2k+1)$ edges using $k$ colors such that no monochromatic triangles exist, and wish to maximize the number of triangles with three differently-colored edges. Since no monochromatic triangles exist, every triangle which doesn't have differently-colored edges must have exactly one vertex $v$ where the edges of the triangle connected to that vertex have the same color, hence we can simply minimize the number of triangles with two same-colored edges. Now number the vertices $1,2,\ldots,2k+1$, and also number the colors $1,2,\ldots,k$. Let $f(i,j)$ denote the number of edges with color $j$ that have an endpoint in vertex $i$. Then the number of same-colored edges is: $$\sum_{j=1}^k\sum_{i=1}^{2k+1} \binom{f(i,j)}{2}.$$Since $\binom{x}{2}$ is convex on $[0,\infty)$, by Jensen's this quantity is at least: $$k(2k+1)\binom{\left(\sum_{j=1}^k\sum_{i=1}^{2k+1} f(i,j)\right)/\left(k(2k+1)\right)}{2}.$$Now note that: $$(\sum_{j=1}^k\sum_{i=1}^{2k+1} f(i,j)$$counts every edge twice, so it's equal to $2k(2k+1)$. Hence the big binomial expression is just: $$k(2k+1)\binom{2}{2}=k(2k+1).$$Since there are $\binom{2k+1}{3}$ triangles total, after a bit of algebra we find that there at most $\frac{2k(k-2)(2k+1)}{3}$ triangles with differently-colored edges. Equality for our Jensen's application occurs when $f(i,j)$ is the same for all $i,j$. It is easy to see that this common value must be $2$, so it remains to provide a coloring of $K_{2k+1}$ without any monochromatic triangles such that every vertex has two edges of any color. One can easily see that coloring edges based on a Hamiltonian Decomposition (a decomposition of the graph into disjoint hamiltonian cycles) will work. It is actually well-known that such a decomposition exists for complete graph on an odd number of vertices, which was shown by Walecki in the 1890s, so the bound is sharp and we're done. $\blacksquare$ Here is an example (from Wikipedia), where $k=4$ (coloring a $K_9$): Remarks: the bounding part is very similar to that Canada problem in Evan's global (I think?) unit, I think 2009/5 or something. it's 2006/4 AwesomeYRY wrote: Next, for the construction. We number all $2k+1$ points, then all vertices of the form $P_mP_{m+i}$(mod 2k+1) are colored with color $i$, which clearly satisfies both our no monochromatic triples condition, as well as $B=(2k+1)*k$. $\blacksquare$ Unless I'm understanding this incorrectly, this should fail when $k=4$ (in which case $2k+1=9$), since $i=3$ gives you three monochromatic triangles, for instance $P_1P_4P_7$.

Basically, $k$-color all edges of $K_{2k+1}$ such that no monochromatic triangle. Find maximum number of rainbow triangles. There are two types of triplets - isosceles colored triplets or rainbow colored triplets, where there are $B$ of the former and $A$ of the latter. We want $\max(A)$, and we know $A + B = \tbinom{2k+1}{3}$. It suffices to minimize $B$. We count isosceles triplets by casework on their apex points, of which there are $2k+1$ possibilities. Given an apex $v$ which has degree $2k$, if there are $c_i$ edges of color $i$ coming out from it, then the number of isosceles triples with apex $v$ is\[\sum_{i = 1}^k \binom{c_i}{2} \geq k\]by Jensen's, since $c_1 + \ldots + c_k = 2k$. Multiplying over all vertices as possibilities for the apex, there are at least $k(2k+1)$ isosceles triples. Therefore, $A \leq \tbinom{2k+1}{3} - k(2k+1)$, establishing the bound. Equality can be achieved by having each language form a Hamiltonian cycle, which is clearly possible since it is well known that $K_{2k+1}$ can be split into the needed number of Hamiltonian cycles.

The answer is $\frac 23 (k-2)k(2k+1) = {2k+1 \choose 3} - k(2k+1)$. Instead of counting polychromatic triangles, we count the number of triangles with two edges of the same color, which uniquely corresponds to a vertex from which two identically colored edges emanate. Consider the edges emanating from any vertex $k$. Suppose that $a_i$ edges of color $i$ emanate from this vertex for $1 \leq i \leq k$. The number of such identically colored pairs of edges is $$\sum_{i=1}^n {a_i \choose 2} = \frac 12\left[\sum_{i=1}^n a_i^2 - 2k\right] \geq \frac{4k-2k}2 = k$$by Cauchy-Schwarz. As a result, the total number of such pairs across all vertices is $k(2k+1)$, and this upper bounds the number of non-polychromatic triangles. For the construction, it suffices to split $K_{2n+1}$ into $n$ distinct Hamiltonian cycles of length $2n+1$ and assign one color to each cycle. The existence of such a decomposition turns out to be well known.

The answer is $\binom{2k+1}{3} - k(2k+1)$. Interpret the problems as a problem on coloring $K_{2k+1}$ with $k$ colors, with no monochromatic triangle - then we wish to find the number of triangles with all three sides different colors. Call a triangle isosceles if it has exactly two distinct edge colors, and scalene otherwise (yes i know these are not politically correct but idc this is what i used when solving). Then the problem asks for the maximal number of scalene triangles, or equivalently the minimal number of isosceles triangles. First we prove the bound. Fix any vertex $v$ and suppose the number of edges of each color are $a_1$, $a_2$, $a_3$ $\dots$, $a_k$. Now any two edges protruding from $v$ form an isosceles triangle iff they are the same color, which occurs in $\binom{a_i}{2}$ pairs of edges for each color $a_i$ - summing over all $i$ pairs of edges and using Jensen, we note that there are at least $k$ isosceles triangles with vertex $v$. Summing over all possible choices of $v$ (and noting we have no overcount due to no monochromatic triangle) tells us there are at most $k(2k+1)$ isosceles triangles, which gives the desired bound. To prove sufficiency, note that this is equivalent to splitting $K_{2k+1}$ into $k$ distinct Hamiltonian cycles, which is well-known to be possible.

pikapika007 wrote: To prove sufficiency, note that this is equivalent to splitting $K_{2k+1}$ into $k$ distinct Hamiltonian cycles, which is well-known to be possible. welp

asdf334 wrote: pikapika007 wrote: To prove sufficiency, note that this is equivalent to splitting $K_{2k+1}$ into $k$ distinct Hamiltonian cycles, which is well-known to be possible. welp had the exact same reaction.

Take the graph interpretation and color edges with $k$ colors and let a triangle be scalene and isosceles if all edges are differently colored and if two edges are the same respectively. Then let $a_i$ be the number of edges stemming from some vertex with color $i \in \{1, 2, \dots, k\}$. Then the number of triangles with two edges of one color stemming from this vertex can be written as \[\sum_{i=1}^{n} \binom{a_i}{2}\]which by Jensen's is greater than or equal to $k\binom{\frac{2k}{k}}{2} = k$. Summing over all vertices gives us that the total number of isosceles triangles is $k \cdot (2k + 1)$. So our maximum number of scalene triangles is $\binom{2k+3}{3} - k(2k+1)$. The construction for the maximum is when each vertex speaks in language $i$ to exactly two other people. This happens if we partition the $(2k+1)$-clique into $k$ Hamiltonian paths colored with $i = 1, 2, \dots, k$ which is well-known to be possible.

For each vertex, count the number of pairs of edges of the same color, which is the sum of $\tbinom i2$ for $k$ values of $i$ summing to $2k$. This is minimized at $i=2$ by QM-AM for a value of $k$. Over all $2k+1$ vertices, this counts $k(2k+1)$ total triangles with two colors, and thus a maximum of $\binom{2k+1}3-k(2k+1)=\frac{2k(2k+1)(k-2)}3$. To construct, when $3\nmid 2k+1$ then color vertices by their minimum distance. When $3\mid 2k+1$ then color all distances except $1$ and $\tfrac{2k+1}3$. Then for one color, start at some vertex and go around in the equilateral triangle until you run out of vertices, then move one vertex counterclockwise, and repeat, choosing the direction of the last triangle to return to the initial vertex with length $1$. This works since the index of the current vertex $\pmod{\tfrac{2k+1}3}$ is monotonic. Then the uncolored edges may be given the last color, since among the excluded edges no triangle exists since three edges of length $1$ does not loop around unless $k=1$ and the problem is vacuous, three edges of length $\tfrac{2k+1}3$ do not form a triangle by the previous construction, and the others fail $\pmod{\tfrac{2k+1}3}$.

Let $X$, $Y$, and $Z$ denote the sets of triplets with 1, 2, and 3 distinct languages. Then \[A = |Z| = \binom{2k+1}{3} - |X| - |Y| = \binom{2k+1}{3} - |Y|.\] Consider a member $m$ who speaks to $a_i$ other members in language $i$, and thus $a_1 + \ldots + a_k$. This gives \begin{align*} A &= \binom{2k+1}{3} - \sum_{m} \left(\binom{a_1}{2} + \ldots \binom{a_k}{2}\right) \\ &\ge \binom{2k+1}{3} - \sum_{m} k \cdot \binom{2k/k}{2} \\ &= \boxed{\binom{2k+1}{3} - k(2k+1)}. \end{align*} For the construction, we place the members on the vertices of a regular $(2k+1)$-gon and assign their mutual "language" as the distance between them if $3 \nmid (2k+1)$. Note this makes each $a_i=2$, which is the equality case in our bound. If $3 \mid (2k+1)$, we simply swap labels of some of the $\left(\frac{2k+1}{3}\right)$-shortest lengths with the shortest lengths to handle the equilateral triangles. $\blacksquare$