Clearly if

a

b

and

c

are the lengths of the sides of a triangle and the triangle is equilateral then

a^{2} + b^{2} + c^{2} = ab + bc + ca .

Is the converse true, and if so can you prove it? That is if $a^{2} + b^{2} + c^{2} = ab + bc + ca$ 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.

$a^{2} + b^{2} + c^{2} - ab - bc - ca = 0$

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

$a^{2} + b^{2} + c^{2} - ab - bc - ca = 0$

as a quadratic equation for $a$ in terms of $b$ and $c$ , namely:

$a^{2} - a (b + c) + (b^{2} + c^{2} - bc) = 0,$

and then the condition for this quadratic equation to have real roots requires that $a = b = c .$