MHB Is an Equilateral Triangle Possible with Distinct Points in n-Dimensional Space?

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In n-dimensional space with n ≥ 3, a set of more than 2^(n+1)/n distinct points, each with coordinates of the form (±1, ±1, ..., ±1), guarantees the existence of three distinct points that form the vertices of an equilateral triangle. This problem was featured as Problem B-6 in the 2000 William Lowell Putnam Mathematical Competition. Despite its mathematical significance, no responses were provided for this week's Problem of the Week. The solution is credited to Kiran Kedlaya and his team. The discussion emphasizes the intriguing properties of geometric configurations in higher dimensions.
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Here is this week's POTW:

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Let $B$ be a set of more than $2^{n+1}/n$ distinct points with coordinates of the form $(\pm 1,\pm 1,\ldots,\pm 1)$ in $n$-dimensional space with $n\geq 3$. Show that there are three distinct points in $B$ which are the vertices of an equilateral triangle.

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Re: Problem Of The Week # 270 - Jul 03, 2017

This was Problem B-6 in the 2000 William Lowell Putnam Mathematical Competition.

No one answered this week's POTW. The solution, attributed to Kiran Kedlaya and his associates, follows:

For each point $P$ in $B$, let $S_P$ be the set of points with all coordinates equal to $\pm 1$ which differ from $P$ in exactly one coordinate. Since there are more than $2^{n+1}/n$ points in $B$, and each $S_P$ has $n$ elements, the cardinalities of the sets $S_P$ add up to more than $2^{n+1}$, which is to say, more than twice the total number of points. By the pigeonhole principle, there must be a point in three of the sets, say $S_P, S_Q, S_R$. But then any two of $P, Q, R$ differ in exactly two coordinates, so $PQR$ is an equilateral triangle, as desired.
 

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