# Archimedes: Inscribed regular polygon smaller perimeter than a circle?

When Archimedes found the upper and lower boundary for the value of pi, he used an $$\color{red}{\textrm{inscribed regular polygon}}$$ and a $$\color{blue}{\textrm{encapsulating outer regular polygon}}$$. He then let the number of edges increase, as depicted here:

I would love to know if it was proven or known at the time, that the $$\color{red}{\textrm{inscribed regular polygon}}$$ had a smaller perimeter than the $$\color{green}{\textrm{circle}}$$? I am not a mathematician but I feel like this should be proven before continuing finding the bounds of pi.

I would arguer the outer object, has to have a greater perimeter, but the inner object could be constructed not to be smaller but actually have a greater perimeter, like this star-like-shape:

I would love to hear your input on this, and if you have any idea how to prove an inscribed regular polygon has to have a perimeter smaller than that of a outer encapsulating circle?

• What do you mean by the "circumference of the inner polygon"? Do you mean it's perimeter ? Jul 20 at 20:51
• Of course your counterexample is not inscribed in the circle. But if it were, that its perimeter must be less than the circumference of the circle would follow from the fact that a straight line is the shortest path between two points in the plane: Just consider the line and the circular arc between any two successive vertices. Jul 20 at 21:12
• @AdamRubinson yes you are right, I mean perimeter. I updated the post Jul 21 at 6:38
• @nammerkage The key missing word here is convex: "As Archimedes already knew, if two convex curves have the same endpoint, and one of the two curves lies between the other and the line through their endpoints, then the inner curve is shorter than the outer one." (more references in my answer here). Both circles and regular polygons are convex curves.
– dxiv
Jul 21 at 6:46
• Note the proof summarized by @DanAsimov doesn't require a regular polygon, only that all the polygon's vertices are on the circle, and the polygon and circle visit them in the same order (or, the polygon is not self-intersecting). Jul 21 at 7:14

## 1 Answer

Let $$AB$$ be the side of a regular $$n$$-gon inscribed in a circle. Joining $$C$$, the midpoint of $$\overset{\frown}{AB}$$, to $$A$$ and $$B$$ produces triangle $$ABC$$. Archimedes knew, as did Euclid before him (Elements I, 20), that in any triangle two sides taken together are greater than the remaining one. Therefore$$AC+CB>AB$$and hence$$\overset{\frown}{AB}>AB$$And generally, any arc of a circle is greater than the chord that subtends it.

Thus $$AC$$, $$CB$$, the sides of a $$2n$$-gon, sum to less than $$\overset{\frown}{AC}+\overset{\frown}{CB}$$.

Similarly, if $$D$$ is the midpoint of $$\overset{\frown}{CB}$$, sides $$CD$$, $$DB$$ of a $$4n$$-gon sum to less than less than $$\overset{\frown}{CD}+\overset{\frown}{DB}$$.

And so on: no matter how many times Archimedes doubles the number of sides of the inscribed regular polygon, he knows its perimeter remains less than the circumference of the circle.