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

MS2.SA = (OS2 + QS2).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 OS2 = CS.SA . The Pythagorean theorem applied to OSA gives OA2 = OS2 + SA2, so
OA2 = CS.SA + SA2 = CA.SA,
yielding MS.SQ = OA2 = OS2 + SQ2.

Since AH = CA it follows that
AH/SA = CA/SA = MS/SQ (substituting equals for equals) = MS2/MS.SQ = MS2/(OS2 + SQ2). Cross-multipling gives the desired equation.

© copyright 1999, American Mathematical Society.