Suppose there are some 51 points which there are no three points that form an isosceles triangle.
Let the vertex of the 101-regular polygon $A_0 A_1 A_2 \cdots A_{100}$ .
Lets mark red color to the chosen 51 points, and blue color to the others.
For each red points, we can cut the polygon in half with the diameter passing that point.
this line divides points into 50 points each.
By the supposition, each pair of two points which is in symmetrical places has at most 1 red points.
There are 51 red points, so each pair should have exactly one red point and one blue point. $ \cdots (*)$
We can find two consecutive red points, WLOG $A_0 , A_1$ are red points.
by $(*)$ at $A_0$ , $A_{-1}$ is a blue point. $\rightarrow$by $(*)$ at $A_1$ , $A_{3}$ is a red point.
by $(*)$ at $A_1$ , $A_{2}$ is a blue point.$\rightarrow$ by $(*)$ at $A_0$ , $A_{-2}$ is a red point.$\rightarrow$ by $(*)$ at $A_0$ , $A_{2}$ is a blue point. & by $(*)$ at $A_1$ , $A_{4}$ is a blue point.
This is contradiction if we use $(*)$ at $A_3$, and we are done.