You may also like

problem icon

Building Tetrahedra

Can you make a tetrahedron whose faces all have the same perimeter?

problem icon

Rudolff's Problem

A group of 20 people pay a total of £20 to see an exhibition. The admission price is £3 for men, £2 for women and 50p for children. How many men, women and children are there in the group?

problem icon

Square Mean

Is the mean of the squares of two numbers greater than, or less than, the square of their means?


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 $a = b = c.$