### 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?

### Square Mean

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

# Medallions

##### Stage: 4 Challenge Level:

This solution was sent in by Liwei Deng, aged 17, of The Latymer School in London.

First of all, join up the centres of the three circles. This will form a triangle with side lengths $x+9$, $x+4$ and $13$.
Then draw vertical and horizontal lines as shown to create three right-angled triangles (shaded on diagram).
Taking the unknown radius to be $x$, we can mark on the lengths shown in green.

We can now use Pythagoras' Theorem on the two yellow right-angled triangles to find the missing sides. The upper one is $6 \sqrt{x}$,

and the lower one is $4 \sqrt{x}$ .

Now we move to the pale blue shaded triangle.
The horizontal side can be found by subtracting:

$$6\sqrt{x} - 4\sqrt{x} = 2\sqrt{x}.$$

The vertical side can also be found by subtracting: as we know that the height of the rectangle is $2x$, this side must be $2x-9-4 = 2x-13$.
Using Pythagoras' Theorem on this triangle, we find that

$$(2\sqrt{x})^2 + (2x-13)^2 = 13^2$$ $$4x + 4x^2 - 52x + 169 = 169$$ $$4x^2 - 48x = 0$$ $$4x(x - 12) = 0$$ $$x = 12$$.