Medallions
Problem
I keep three circular medallions in a rectangular box in which they just fit with each one touching the other two. The smallest one has radius $4$ cm and touches one side of the box, the middle sized one has radius $9$ cm and touches two sides of the box and the largest one touches three sides of the box. What is the radius of the largest one?
This problem by R B J T Allenby appeared in the Sunday Times on 12.11.00.
Student Solutions
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
$$ (2\sqrt{x})^2 + (2x-13)^2 = 13^2 $$ $$ 4x + 4x^2 - 52x + 169 =
169 $$ $$ 4x^2 - 48x = 0 $$ $$ 4x(x - 12) = 0 $$ $$ x = 12 $$.