**The Method of Archimedes**

## 4. The plane geometry behind the argument

The areas of the three discs in question are proportional to
the square of their radii. With points labeled as in this diagram
(adapted from T. L. Heath, *The Works of Archimedes with
The Method of Archimedes*, Dover, New York), the equation
m d = M D from the previous page
becomes

MS^{2}.SA = (OS^{2} + QS^{2}).AH .

Since `MS = CA` and `SQ = SA`, we have `MS.SQ = CA.SA` .
Now `OS` is the altitude of the right triangle `OSA`
and therefore `OS`^{2} = CS.SA . The Pythagorean
theorem applied to `OSA` gives `OA`^{2} = OS^{2} + SA^{2},
so

`OA`^{2} = CS.SA + SA^{2} = CA.SA,
yielding
`MS.SQ = OA`^{2} = OS^{2} + SQ^{2}.
Since `AH = CA` it follows that

`AH/SA = CA/SA = MS/SQ` (substituting
equals for equals) `= MS`^{2}/MS.SQ = MS^{2}/(OS^{2} + SQ^{2}). Cross-multipling gives the desired equation.

© copyright 1999, American Mathematical Society.