Is Archimedes’ proof correct?

Question

I began studying Archimedes’ work "on the sphere and cylinder". I currently believe that proof for proposition 3 is not good. I attached the pages for reference. Now, I don't understand why he believes there's some king of a relationship between the angles and the ratios he states (bottom of p.6). It seems to me that he bases his demonstration solely on the angles' size when he does the correlation between the elements of the circle and the initial magnitudes. In other words, I don't see any relationship between the circle and his previous construct (the right triangle) to justify the ratio relationships (i.e. MK/LK > OC/OH). The only link between the two seems to be the size of their angles, but I don't see how this justifies the relationships he mentions further.

What do I get wrong?

Answer 1

From $\angle LKM>\angle HOC$ it follows we can construct point $M'$ on side $LM$ such that $\angle LKM'=\angle HOC$. Then triangles $LKM'$, $COH$ are similar.

In triangle $KMM'$ we have $\angle M'> 90° > \angle M$, hence $KM>KM'$ (Euclid I-19) and: $$ {KM\over KL}>{KM'\over KL}={OC\over OH}. $$

Answer 2

Let me address the issue of why the following implication is true: $$\angle HOC < \angle LKM \implies MK : LK > OC : OH $$ Although it is of course impossible to know what Archimedes was thinking, it is my impression that an understanding of trigonometry was well established in the ancient Mediterranean. I suspect that Archimedes knew his trigonometry very, very well; this snippet of his works that you have provided in your post is evidence for that suspicion.

The inequality in question can be translated, in modern terms, into a fact about trigonometry: When the secant function is restricted to the interval $[0,\pi/2)$ the resulting function $$\sec : [0,\pi/2) \to [1,\infty) $$ is strictly increasing: $$\theta < \phi \implies \sec\theta < \sec\phi $$ One can observe this fact by inspecting a graph of the secant function. This can also be proved rigorously using calculus (and Archimedes knew his calculus pretty well too).

We can apply this implication to derive the inequality in question, as follows:

• letting $\theta = \angle HOC$ it follows that $\sec \theta = OC / OH$.
• letting $\phi = \angle LKM$ it follows that $\sec \phi = MK / LK$.

In a bit more detail, in any right triangle the secant of one of the acute angles is equal to the length of the hypotenuse divided by the length of the leg of the triangle that is adjacent to the given angle. So for example in the right triangle $OHC$, the secant of the angle $\theta = \angle HOC$ is equal to the length of the hypotenuse $OC$ divided by the length of the adjacent leg $OH$.

So Archimedes was right all along.