Question

Three vertices are chosen randomly from the nine vertices of a regular 9-sided polygon. The probability that they form the vertices of an isosceles triangle, is

• A. \( \frac{4}{7} \)
• B. \( \frac{3}{7} \)
• C. \( \frac{2}{7} \)
• D. \( \frac{5}{7} \)

Hint:

For regular polygons, symmetry can be used to calculate probabilities involving isosceles or equilateral triangles by considering the number of symmetrical vertex pairs and the total combinations.

Answer 1

The Correct Option is B

A regular 9-sided polygon has 9 vertices. To form an isosceles triangle, we need two vertices that are symmetrically placed about the center of the polygon. 
When we select the first vertex, we have a fixed reference point. Now, for the other two vertices to form an isosceles triangle, they must be symmetrically placed relative to this first vertex. 
The number of ways this can happen depends on selecting vertices that are separated by the same number of steps (which is how symmetry works in a regular polygon). 
We can visualize this as picking any vertex and ensuring the next two vertices are equidistant from the first one. Out of the 9 vertices, there are 3 possible ways to choose these pairs of equidistant vertices that form isosceles triangles. 
The total number of ways to choose any three vertices from the 9 vertices is \( \binom{9}{3} = 84 \). 
The favorable cases, where the three vertices form an isosceles triangle, are 36 (since we have 3 such possible sets of vertices as discussed earlier). Thus, the probability is: \[ P = \frac{36}{84} = \frac{3}{7} \] 
Thus, the probability that the chosen vertices form an isosceles triangle is \( \frac{3}{7} \).