Bang's Theorem

If all the faces of a tetrahedron have the same perimeter then show that they are all congruent.

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

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