1. Assume the contrary: Suppose there are 51 points chosen from the 101 vertices of the regular polygon such that no three of these points form an isosceles triangle.

2. Coloring the vertices: Let the vertices of the 101-regular polygon be labeled as $A_0, A_1, A_2, \ldots, A_{100}$. Color the chosen 51 points red and the remaining 50 points blue.

3. Symmetry argument: For each red point, consider the line passing through this point and the center of the polygon. This line divides the polygon into two halves, each containing 50 points. By our assumption, each pair of points that are symmetric with respect to this line can have at most one red point.

4. Counting pairs: Since there are 51 red points, each symmetric pair must contain exactly one red point and one blue point. This is because if any pair had two red points, it would contradict our assumption that no three red points form an isosceles triangle.

5. Contradiction with consecutive points: Let's find two consecutive red points. Without loss of generality, assume $A_0$ and $A_1$ are red points.

- By the symmetry argument, the point symmetric to $A_0$ (which is $A_{50}$) must be blue.
- Similarly, the point symmetric to $A_1$ (which is $A_{51}$) must be blue.

6. Propagation of colors:
- Since $A_0$ is red, $A_{50}$ is blue.
- Since $A_1$ is red, $A_{51}$ is blue.
- By the symmetry argument at $A_1$, $A_2$ must be blue.
- By the symmetry argument at $A_0$, $A_{-1}$ must be blue.
- By the symmetry argument at $A_1$, $A_3$ must be red.
- By the symmetry argument at $A_0$, $A_{-2}$ must be red.

7. Contradiction: Now, consider the point $A_2$. By the symmetry argument at $A_0$, $A_2$ must be blue. However, by the symmetry argument at $A_1$, $A_2$ must be blue. This leads to a contradiction because $A_2$ cannot be both red and blue.

8. Conclusion: The assumption that there are 51 points such that no three form an isosceles triangle leads to a contradiction. Therefore, among any 51 vertices of the 101-regular polygon, there must be three that form an isosceles triangle.

$\blacksquare$