### Building Tetrahedra

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

### 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?

### Medallions

Three circular medallions fit in a rectangular box. Can you find the radius of the largest one?

# Converse

##### Stage: 4 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 $a = b = c.$