delegat wrote: Given is a convex polygon $ P$ with $ n$ vertices. Triangle whose vertices lie on vertices of $ P$ is called good if all its sides are equal in length. Prove that there are at most $ \frac {2n}{3}$ good triangles. That's a demanding, but well-known problem. A solution can be found for instance in: ---Janos Pach, Rom Pinchasi ---How many unit equilateral triangles can be generated by N points in convex position? ---American Mathematical Monthly 110(5): 400-406 (2003)

Hmmm, I think it is also true that there are at most $ \frac{n}{3}$ squares among the vertices of a convex $ n$-gon.

What is the maximum number of times that the unit distance can occur among the distances between n points in the plane?

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Nice problem!! Here is my solution: To each vertex $V$, assign at most two good triangles in the following way: Consider all the good triangles which have a vertex $V$, assign to $V$ the leftmost clockwise and the rightmost clockwise (this is possible because all good triangles lie on a half-plane determined by some line through $V$). Now take $ABC$ a good triangle (those vertices are in clockwise order). I prove that it is either assigned to $B$ as the leftmost clockwise triangle, or assigned to $C$ as the rightmost clockwise triangle. To see this, assume not, then there are good triangles $BB'B''$ and $CC'C''$ more to the right clockwise and more to the left clockwise than $ABC$, respectively, such that $B'$ and $C$ are on different sides of $AB$, and $C'$ and $B$ are on different sides of $AC$. Then $\angle B'AC' > 180$ but $\angle BAC < 60$, contradiction since the polygon is convex. From this, we see that every good triangle is assigned at least 3 times. Since each vertex is assigned at most 2 triangles, we easily finish and therefore there are at most $2n/3$ good triangles. $\blacksquare$

Consider triples $(T,v,\varepsilon)$ where $T$ is a good triangle, $v$ is one of its vertices, and $\varepsilon=1$ if there is no other good triangle with vertex $v$ that is $T$ rotated counterclockwise by some angle less than $\pi$, and $\varepsilon=-1$ if the previous statement holds with clockwise. Note that for a good triangle $T$ and one of its vertices $v$, there may not always exist an appropriate $\varepsilon$. Call such triples extremal. We first claim that there exactly $2n$ extremal triples. Indeed, for each vertex $v$ of $P$, there is exactly one ``most clockwise'' good triangle, and one ``most counterclockwise'' good triangle. Note that these may be the same, but still count towards different triples since $\varepsilon$ is different. Thus, for each $v$, there are exactly $2$ extremal triples, so we have total $2n$ such triples. The key claim is the following. Claim: Consider a good triangle $T=abc$. Then, at least three of the following triples are extremal: \[(T,a,1),(T,a,-1),(T,b,1),(T,b,-1),(T,c,1),(T,c,-1)\] Proof: Suppose WLOG that $abc$ is oriented counterclockwise. We claim that one of $(T,b,1)$ and $(T,c,-1)$ is extremal, and by symmetry thus one of $(T,c,1)$ and $(T,a,-1)$, and one of $(T,a,1)$ and $(T,b,-1)$ are extremal. This would complete the proof. Suppose for the sake of contradiction that neither $(T,b,1)$ nor $(T,c,-1)$ are extremal. Let $\ell$ be the line through $a$ parallel to $bc$. It is easy to see then that one of the vertices of a good triangle with vertex $b$ which is $T$ rotated counterclockwise is above or on $\ell$, and similarly one of the vertices of a good triangle with vertex $c$ which is $T$ rotated clockwise is above or on $\ell$, so two vertices of $P$ are above or on $\ell$. Furthermore, these two are on opposite sides of the perpendicular line to $\ell$ through $a$, so $a$ is actually inside or on the convex hull of these two vertices along with $b$ and $c$, which violates the convexity of $P$. This completes the proof of the claim. $\blacksquare$ This now finishes, as we have at least $3X$ extremal triples, where $X$ is the number of good triangles, so $3X\le 2n$, or $X\le\frac{2n}{3}$, as desired.