Polygons and a circle

We will prove that if a polygon's perimeter is $1$ then we can cover it with a circle which has a radius of $\frac{1}{4}$ .

Let $A$ be a random point at the perimeter of the polygon. Imagine we are walking around the polygon's perimeter form $A$ , and we walk exactly $\frac{1}{2}$ then we stop. Let $B$ the poin where we stopped. Furthermore, let $O$ be the midpoint of $AB$ . Write a circle which has a radius of $\frac{1}{4}$ around $O$ . If M is another point (not $A$ and $B$ ) at the perimeter of the polygon, and $K$ is the $M'$ s mirror image of $O$ , then

$MK=2OM\leq AM+AK =AM+MB\leq \frac{1}{2} \\ OM \leq \frac{1}{4}$ (because $AKBM$ is a paralelogram).

So the circle completely covers the polygon.

Suppose we have any polygon inside the circle. choose any two points on the perimeter of the polygon and draw a line between them. Let's assume that the line's length is more than 0.5 (for example 0.5001) in this case the polygon wouldn't be enclosed completely inside the circle because the circle has a diameter of 0.5. Now choose any of the two points. Because a polygon is a closed shape, we know that there must be two different paths between the two points and the sum of the lengths of these two paths is the perimeter of the polygon. We know that the shortest distance between two points is a straight line (which in this case is 0.5001), therefore, the length of any of the two paths must be greater than/or equal to 0.5001 which means that the perimeter must be greater than 1.

It is known that increasing the sides of a polygon by keeping the perimeter fixed will increase the area of the polygon. Starting from a triangle, a square... the cicrle is the maximum we can go to (with infinite number of point sides.) If by keeping the perimeter fixed at 1 the circle resulting will have an area bigger than the area of preceding polygons, then the given circle of perimeter P=2 pi/4 = 2× 3.14/4= 1.57 > 1, will definitely have a bigger area than the circle of perimeter 1, which already encompasses the polygon of perimeter 1, hence the given circle will encompass the polygon de facto.

Circle radius = $\frac{1}{4}$ = .25 <space> Circle diameter = $\frac{1}{2}$ = .50 Circle Circumference =2 π r = 2 π .25 = 1.57

Polygon (triangle) = perimeter 1 Side A = 0.3 Side B = 0.3 Side C = 0.4

Because the polygon has a perimeter of 1 (divided between at least 3 sides) it will always fit with in a circumference with a radius of at least .25 or $\frac{1}{4}$

We can also rewrite the question as what would be the longest side of a polygon whose perimeter is 1. The lesser number the sides, the longer can the sides be with a fixed perimeter value. Hence we check for a triangle.

Triangle longest side should be <= 0.5 for the circle to enclose it.

Let's take a triangle with the side 0.5 . The other two sides would both be 0.25. They would meet the 0.5 side in the middle creating a line rather than a triangle. The only way a triangle of perimeter 1 would work is if it's longest side is less than 0.5. we have already proved that this can be the longest side of any polygon with perimeter 1. Hence the circle can enclose it.

Consider how you would go about making a polygon as wide as possible for a perimeter of 1. The simplest approach is to take a rectangle and make it as wide as possible i.e. with width close to 1/2 and height close to 0. Clearly this can still be enclosed by the circle radius 1/4.

If the longest line you can draw is 1/2, then you cannot return to your starting point w/o drawing 2 lines with a sum greater than 1/2.

The semantics of "given any polygon" and "does there exist a polygon" is irrelevant when considering the trivial case:

Two parallel lines with length 1/2, perimeter 1, will still be enclosed in the circle of diameter 1/2. Any shape with 2 or more sides will therefore have a smaller effective diameter than 1/2. So if any shape with more than 1 side is enclosed in the circle, isn't the general case for "any polygon" pointless?

I believe this is what Scott Rodham was claiming.

Even a trivial polygon "along" a diameter of 1/2 is contained.

To make this polygon as long and flat as possible, we want to give it the fewest number of sides. So, we make it a triangle. Since the perimeter is 1, we know that the sum of the three side lengths is 1, but we also know that the sum of the two shortest sides has to be greater than the third side. Those two conditions imply that the longest side length of this triangle has to be strictly less than 1/2. Since the diameter of this circle equals 1/2, it is impossible to construct a polygon that cannot be contained by this circle.