can anyone post solution not using grap?

Please check my solution (this looks a bit different to that of #2)

the distance function to a new and easier to handle distance function $d'(P',Q') = \max \{ |P'_x - Q'_x|, |P'_y - Q'_y| \}$ without changing the structure of the problem. So now through the solution we work with the new distance function. I claim the answer is $\frac{3*100^2}{4*2} = 3750$. To see that this bound is achieved first, put $25$ points each at some small neighborhood of the four points $(\pm \frac{1.0201082102011209}{2}, \pm \frac{1.0201082102011209}{2})$. Now we prove that this is the maximum we can achieve. Create a graph $G$ with $100$ vertices with two vertices joined iff the two points are mutually harmonic. Claim $G$ has no $K_5$ Proof : Firstly note this two facts, both of which are easy to verify by case-bashing: If a coloring of the edges of $K_5$ with two colors doesn't produces a monochromatic triangle, then it must have a monochromatic cycle with lenght $5$ It's impossible to find three real numbers $A,B,C$ for which all the points $(A,0), (B,0), (C,0)$ are mutually harmonic. Now, for each edge $\bar{PQ}$ already in $G$, color the edge $PQ$ red if $\max \{ |P_x - Q_x|, |P_y - Q_y| \} = |P_x - Q_x|$, or blue otherwise. Now suppose for the sake of contraction, there is a $K_5$ in $G$, with the points $A,B,C,D,E$. Then by #2, it has no monochromatic triangle, and so by #1, it has a monochromatic cycle of lenght $5$. WLOG the cycle is of red color, and let the cycle be $A \rightarrow B \cdots \rightarrow E$. Note that now if $\max(A_y,B_y,C_y,D_y,E_y) - \min(A_y,B_y,C_y,D_y,E_y) > 2$, we have a contradiction (then the extreme points would not be harmonic), so $\max(A_y,B_y,C_y,D_y,E_y) - \min(A_y,B_y,C_y,D_y,E_y) \leq 2$. WLOG assume $\min(A_y,B_y,C_y,D_y,E_y) = A_y = 0$, so $\max(A_y,B_y,C_y,D_y,E_y) > 0$. Since $\max(A_y,B_y,C_y,D_y,E_y) \leq 2 + \min(A_y,B_y,C_y,D_y,E_y) = 2$, we have $A_y,B_y,C_y,D_y,E_y \in [0,2]$. So color the vertices with ordinate in the interval $[0,1]$ black and ordinate in the interval $(1,2]$ white. Now we have reached a contradiction - firstly note that by transversing $A \rightarrow B \rightarrow \cdots E$, we change the color of the interval (because no two adjacent vertices in the cycle can't be in the same interval since $|P_y - Q_y| > 1$) each time, so that implies the odd cycle is bipartite, a contradiction ! $\square$ So by Turan's theorem on $G$, we now find that it's the strictest bound possible.

BTW, Let $P = (p_1,p_2,\cdots, p_n)$ and $Q = (q_1,q_2,\cdots, q_n)$. Also define the functions $d_1(P, Q) = |p_1 -q_1| +|p_2 -q_2|+ \cdots + |p_n -q_n|$ and $d_2(P, Q) = \max(|p_1 -q_1|, |p_2 -q_2|, \cdots, |p_n -q_n|) $. Also, let $T(m, n)$ denote the maximum number of edges in an $K_n$ free graph with $m$ vertices.

of a ISL 1983 problem found in PfTB Chapter 22, it's not hard to adopt the above proof to show that maximum number of harmonic pairs among $k$ points is $T(k, 2^n+1)$. I have no idea what's the analogous answer when the first function is used for defining harmonic pairs, but I think it's $T(k, 2n+1)$.

diagram solves the problem.

redacted

@above isn't that usamo 2008 3, not usa tst

We claim that the answer is $3750$. For the construction, put $25$ points near each of $(0.75,0),(0,0.75),(-0.75,0),(0,-0.75)$. Connect two points if they are harmonic. Claim: There cannot be a $5$-clique. Proof: Instead of doing normal stuff, we use the orz ERDOS SZEKERES theorem to show that there must be an increasing or decreasing subsequence of $3$ points. However, since taxicab distance is additive for monotonic subsequences, these $3$ points cannot all be harmonic. By Turan’s theorem, we are done.

We claim that the answer is $3750$; this is achievable by taking $25$ points in a sufficiently small disk centered at each of $(0, \pm \frac{3}{4})$ and $(\pm \frac{3}{4}, 0)$. Construct graph $G$ on 100 points in the plane where we place an edge between two points if they form a harmonic pair. Claim: $G$ has no $5$-cliques. Proof: Assume to the contrary $G$ has a $5$-clique. By Erdos-Szekeres, we can find three vertices $P_1, P_2, P_3$ of this clique where $P_i=(x_i, y_i)$ for $1 \le i \le 3$ such that $x_1 \le x_2 \le x_3$, and $y_1, y_2, y_3$ are in either increasing or decreasing order. However $d(P_1, P_3)=d(P_1, P_2) + d(P_2, P_3)>2$ while $d(P_1, P_3) \le 2$, a contradiction. Thus there is no $5$-clique. $\square$ By Turan's theorem, the number of edges in $G$ is at most $t(100, 4)=100^2 \cdot \frac{4-1}{2 \cdot 4} = 3750$, and since we have a construction of $3750$ edges, the final answer is $3750$, as claimed.