# Every three of $n$ points is the vertices of an isosceles triangle. What is the max of $n$?

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?

• I advertise a loosely related paper: We call a metric space X (m,n)-equidistant if, when A⊆X has exactly m points, there are exactly n points in X each of which is equidistant from (the points of) A. We prove that, for k≥2, the Euclidean space ℝ$^k$ contains an (m,1)-equidistant set if and only if k≥m. Although the sphere $S^2$ is (3,2)-equidistant, $S^3$ and ℝ$^4$ contain no (4,2)-equidistant sets. We discuss related results about projective spaces, and state a conjecture about $S^2$ analogous to the Double Midset Conjecture. – Mirko Nov 27 '14 at 13:41
• @user48481MirkoSwirko: Thank you for showing the paper(although I don't think I can get a clue from the paper). – mathlove Nov 29 '14 at 12:17
• Consider cases. Say there were 8 points $a_i$, $i=1..8$. Make a table indicating all non-zero distances $d(a_i,a_j)$. How many distinct distances could there be? (As many as 4,in Ex.1 if you move the two extra points apart.) Do you have the solution for the plane? In $R^3$ the intersection of your sets with a plane could be an isosceles or equilateral triangle (with its centroid), a square, rhombus, a regular pentagon: Could it be anything else? Once you fix the intersection with a plane (considering cases) there should not be many possibilities left how to add more points not in this plane. – Mirko Nov 29 '14 at 13:53
• @user48481MirkoSwirko: I've tried your ideas, but I've been facing difficulty even in finding the solution for the plane. (It looked easy, but it's indeed hard for me) Can you share your idea about the case for the plane if you have? – mathlove Dec 1 '14 at 16:02
• Start with triangle $ABC$. How could we add another point $D$, so that say $D$ is the apex of $ABD$? In the plane you have the perpendicular bisector of $AB$ as the set of points to choose from, in $R^3$ you have a whole plane to choose from. There are restrictions coming from $BC$ (whether $D$ is to be an apex in $BCD$ or not). The more points you add the smaller the set of possible new choices: It will go down from 2D (2 dimensions) to 1D, to 0D, to only finitely many possible choices, eventually to no choices. There will be cases to consider, branching as you make a new choice. Needs work. – Mirko Dec 2 '14 at 19:53

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 $(0,0,\pm 1)$
• So, we can see that the answer to my question is $n=7$. Thank you! (Thanks to @user48481, I found this paper.) – mathlove Dec 3 '14 at 9:55