Suppose that we have $n\ (\ge 3)$ points in the three dimensional space and that every three of the $n$ points is the vertices of an isosceles triangle. Here, suppose that the vertices of an isosceles triangle do not exist on the same line.
Question : What is the max of $n$?
I have the following conjecture.
Conjecture : The max of $n$ is $7$.
We can see that $n=7$ is possible.
(Example 1) $$A_i\left(\cos\left(\frac{2\pi}{5}i\right),\sin\left(\frac{2\pi}{5}i\right),0\right)\ (i=1,2,3,4,5),A_6(0,0,0),A_7(0,0,1)$$
where $A_1A_2A_3A_4A_5$ is a ragular pentagon.
(Example 2) $$A_1(1,0,0),A_2(-1,0,0),A_3(0,\sqrt 3,0),A_4(1,0,2),A_5(-1,0,2),A_6(0,\sqrt 3,2),A_7\left(0,\frac{1}{\sqrt 3},1\right)$$ where $A_1A_2A_3A_4A_5A_6$ is a regular triangular prism.
(Example 3) $$A_1(1,0,0),A_2(-1,0,0),A_3(0,\sqrt 3,0)$$ $$A_4\left(\frac{1+\sqrt 5}{2},\frac{1-\sqrt 5}{2\sqrt 3},\frac{1+\sqrt 5}{\sqrt 3}\right),A_5\left(-\frac{1+\sqrt 5}{2},\frac{1-\sqrt 5}{2\sqrt 3},\frac{1+\sqrt 5}{\sqrt 3}\right)$$ $$A_6\left(0,\frac{2+\sqrt 5}{\sqrt 3},\frac{1+\sqrt 5}{\sqrt 3}\right),A_7\left(0,\frac{1}{\sqrt 3},\frac{3+\sqrt{5}}{2\sqrt 3}\right)$$ where $A_1A_2A_3A_4A_5A_6$ is a regular triangular frustum.
However, I'm facing difficulty in proving the conjecture. Can anyone help?
It seems that, if we relax the restriction that the points must not be colinear (a rather artificial restriction IMO), the maximum is 8; with 9 it has been proven impossible. See references here.
The construction for the 8-points set seems to be as follows: 5 points equidistant over a circle or radius $1$ (say, over the $x,y$ plane, with the center at the origin), plus an additional point in the origin, plus two additional points on $(0,0,\pm 1)$