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Archimedes approximated $\pi$ by circumscribing and inscribing polygons around and in circles. It's obvious that the inner polygon has a perimeter inferior to the circle's. But how did he justify that the outer polygon has a longer perimeter? It seems a bit out of reach for the tools available to ancient Greeks, except for appeal to intuition.

Edit: I have since learned that Archimedes used areas and the formula $A=\pi r^2$ to get the upper bound and he used lengths only for the lower bound.

Martin Frenzel
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    Aren't both cases equally obvious and intuitive, but also equally in need of modern tools for rigorous justification? – ziggurism Apr 10 '20 at 15:17
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    What kind of answer are you looking for? "No, it wasn't properly justified"? It seems you're already aware of that. – Najib Idrissi Apr 10 '20 at 15:19
  • @ziggurism The circle connects the vertices of the inscribed polygons via circular arcs, and the polygon does via straight lines. As circular arcs deviate from the straight line, they are longer paths and therefore the polygon has a shorter perimeter. This is much more obvious to me than the claim about the outer polygon, unless I overlook a simple argument. – Martin Frenzel Apr 10 '20 at 15:30
  • @NajibIdrissi An argument that the outer polygon has a longer perimeter than the circle it circumscribes. There could be an argument that I don't know of. – Martin Frenzel Apr 10 '20 at 15:39
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    @AlgebraicSlug Fair enough, I agree that while it may be somewhat intuitive that the circumscribed polygonal path is longer than circular arc, it is far more obvious that the inscribed straight line is shorter than every other path between two points. – ziggurism Apr 10 '20 at 18:58

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You make a monotonically decreasing sequence of polygon perimeters.

Consider a regular $n$-gon circumscribed around a circle. Ler $A,B,C$ be any three consecutive vertices in rotational order. Now, draw the radius from the center of the circle to $B$, which intersects the circle at $B'$. You then draw the tangent at $B'$ which intersects $\overline{AB}$ at $B''$ and $\overline{BC}$ at $B'''$. Then the straight line segment $\overline{B''B'''}$ is shorter than the combination of $\overline{B'B}$ and $\overline{BB''}$ on the original polygon.

Apply this operation tothe entire $n$-gon with each vertex in turn playing the role of $B$. This leads to a circumscribed regular $2n$-gon whose perimeter is shorter than that of the original circumscribed regular $n$-gon. Upon iterating, the decreasing perimeters of the polygons aporoach that of the circle forcing the polygon perimeters to be longer.

ziggurism
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Oscar Lanzi
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