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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 $a = b = c.$
Converse
Age 14 to 16
Challenge Level 
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 $a = b = c.$
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