3 solved out of 3 shown

SOLVED

If $n$ distinct points in $\mathbb{R}^2$ form a convex polygon then they determine at least $\lfloor \frac{n+1}{2}\rfloor$ distinct distances.

Solved by Altman [Al63]. The stronger variant that says there is one point which determines at least $\lfloor \frac{n+1}{2}\rfloor$ distinct distances is still open. Fishburn in fact conjectures that if $R(x)$ counts the number of distinct distances from $x$ then
\[\sum_{x\in A}R(x) \geq \binom{n}{2}.\]

Szemerédi conjectured (see [Er97e]) that this stronger variant remains true if we only assume that no three points are on a line, and proved this with the weaker bound of $n/3$.

See also [660].

SOLVED

Suppose $n$ points in $\mathbb{R}^2$ determine a convex polygon and the set of distances between them is $\{u_1,\ldots,u_t\}$. Suppose $u_i$ appears as the distance between $f(u_i)$ many pairs of points. Then
\[\sum_i f(u_i)^2 \ll n^3.\]

SOLVED - $500

Let $x_1,\ldots,x_n\in\mathbb{R}^2$ determine the set of distances $\{u_1,\ldots,u_t\}$. Suppose $u_i$ appears as the distance between $f(u_i)$ many pairs of points. Then for all $\epsilon>0$
\[\sum_i f(u_i)^2 \ll_\epsilon n^{3+\epsilon}.\]