Is it possible to place 8 points in $\mathbb{R} ^3$ , such that any 3 points determine an isosceles or equilateral triangle?

Note: For this problem, the triangle is allowed to be degenerate, meaning we have 3 points on a line.

Yes
No

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Take the origin, the 5 vertices of the regular pentagon that lies on $x^2 + y^2 = 1$ , and the 2 points $(0, 0, \pm 1 )$ .

Claim:

1. We can verify that this set satisfies the conditions.

2. Every set that satisfies the conditions is isomorphic to this set (by scaling, rotating and translating)

3. There is no 9 points that satisfies the conditions.

This is known as

Isosceles Setsin the literature. Look it up if you're interested.