archimedes4## The Method of Archimedes

## 4. The plane geometry behind the argument

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

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