Let ${\mathrm{A}}_{\mathrm{i}}, \mathrm{i}=1,2,3,\dots \dots 21$ be the vertices of a 21-sided regular polygon inscribed in a circle with centre O. Triangle are formed by joining the vertices of the 21-sided polygon

i) A triangle ${\mathrm{A}}_{\mathrm{i}}, {\mathrm{A}}_{\mathrm{j}}, {\mathrm{A}}_{\mathrm{k}}$ (vertices) is equilateral if ${\mathrm{A}}_{\mathrm{i}}, {\mathrm{A}}_{\mathrm{j}}, {\mathrm{A}}_{\mathrm{k}}$ are equally spaced. Out of ${\mathrm{A}}_1, {\mathrm{A}}_2, \dots \dots , {\mathrm{A}}_{21}$ we have only 7 such triplets.

  ${\mathrm{A}}_1{\mathrm{A}}_8{\mathrm{A}}_{15}, {\mathrm{A}}_2{\mathrm{A}}_9{\mathrm{A}}_{16}, \dots \dots \dots , {\mathrm{A}}_7{\mathrm{A}}_{14}{\mathrm{A}}_{21}$

  Therefore there are only 7 equilateral triangles

ii) Consider the diameter A₁ OB where B is the point where A₁O meets the circle. If we have an isosceles triangle A₁ as its vertex then A₁B is the altitude and the base is bisected by A₁B. This means that the other two vertices, A_i and A_k ,are equally spaced from B. We have 10 such pairs, so we have 10 isosceles triangle with A₁ as vertex of which is equilateral. Because proper isosceles triangle with A₁ as vertex (non-equilateral) are 9, with each vertex A_i, i = 1,2,…….,21, we have such isosceles triangles

So, total number of isosceles but non-equilateral triangle are 9 x 21 = 189

But the 7 equilateral triangles are also to be considered as isosceles

Hence total number of isosceles triangle are 196