Clearly if a, b and c are the lengths of the sides of a
triangle and the triangle is equilateral then
a2 + b2 + c2 = ab + bc + ca.
Is the converse true, and if so can you prove it?
That is if a2 + b2 + c2 = ab + bc + ca is the triangle with
side lengths a, b and c necessarily equilateral?
Again you don't require much mathematical knowledge to do this, just the
ability to use elementary algebra. Here is a very neat solution from Koopa Koo,
Boston College, USA.
a2 + b2 + c2 −ab − bc − ca = 0
implies that
(1/2)[(a − b)2 + (b − c)2 + (c − a)2] = 0
which implies a = b = c. So, the converse is also true.
Another, rather clumsier, method is to consider the expression
a2 + b2 + c2 −ab − bc − ca = 0
as a quadratic equation for a in terms of b and c, namely:
a2 − a(b + c) + (b2 + c2 − bc) = 0,
and then the condition for this quadratic equation to have real roots
requires that a = b = c.