Relative to the triangle, the circle moves three places counterclockwise in each move. A simple count, keeping the triangle fixed, shows that four moves are required.

When the circle moves two places anti-clockwise at each stage and the triangle moves one place clockwise at each stage they always meet at vertex 4 in every four stages.

The movements can be represented using modular arithmetic as follows: $(+1) x\equiv0\mod(5) \ , \ (-2) y \equiv 2 \mod(5)$ Where the numbers in the brackets represent the movements after each step. If ${a}$ represents the number of movements such that the triangle and circle end up in the same position then: $y - 2a \equiv\ (x+a) mod(5) \Rightarrow\ 2 \equiv 3a \Rightarrow\ 12 \equiv 3a \Rightarrow\ 4 \equiv a$ Testing this out find that they do indeed end up at the same vertex after 4 moves.

relative velocity between the triangle and the circle = 3 places/stage Distance between the shapes = 2 , 7, 12, 17, 22, ... (each loop around the pentagon would consist of 5 places, so adding 5 places per loop)

12 is the minimum distance that is divisible by the velocity i.e., 3

Therefore, number of stages required = 12/3 = 4

Its just imagination

Suppose, at present triangle is at corner 1 & circle is at corner 3 ( named the corners 1, 2, 3, 4 & 5 clockwise direction ). At 1st stage triangle comes to at corner 2 and in this way at 4th stage triangle comes to at corner 5, likewise at 1st stage circle comes to at corner 1 (move anticlockwise from corner 3 to on 1) and in this way at 4th stage it get co-inside with triangle at corner 4 ( Ans.)