We denote by $a,b,c$ the sides of a triangle whose vertices were chosen from the vertices of a regular pentagon. Calculate $(a-b)(b-c)(c-a)$.

Question

PROBLEM

We denote by $a,b,c$ the lengths of the sides of a triangle whose vertices were randomly chosen from among the vertices of a regular pentagon. Calculate $(a-b)(b-c)(c-a)$.

WHAT I THOUGHT OF

First of all: the drawing

I don't really understand the requirment. If we chose the 3 vertices to be $A,B,C$, as $AB=BC$ (because ABCDE is a regular pentagon) we will have the product $0$, becuase one of the terms will be $AB-BC=0$.

Does i understand the requirment wrong or something? Hope one of you can help me!

Answer 1

You are on correct path.

Steps :

1. Show that all diagonals of a regular pentagon are of same length.
2. Only two types of triangles can be formed. If each side has length $x$ and each diagonal has length $d$, the two triangles have side-lengths $$x,\, x,\,d \quad \quad or \quad \quad d,\, d,\, x$$

Thus any triangle formed is isosceles and $(a-b)(b-c)(c-a) = \ldots \,?$