maths, prove area of circle?

Question

Please can you prove that the if all shapes had the same perimeter that the circle would have the largest area

Also prove that Reg. shapes will have larger area than irregular shapes if they each have the same perimeter.

Please don't be afraid to be long winded or refer me to websites or use Pi or Cos/Tan/Sin although keeping it simple would obvo be nice.

Thanks in advance!

Answer 1

1) Any non-convex shape cannot have the maximum area for a given perimeter because a line tangent to the shape enclosing an indentation has a shorter length than the perimeter of the indentation, thereby leading to another shape with greater area and shorter perimeter.
2) The perimeter and area of any shape is unaffected by an arbitrary slice through it, making a mirror image of one of the pieces, and reattaching it. Thus, if any shape can result in a non-convex shape, it also cannot have the maximum area for a given perimeter.
3) Therefore, the area of maximum shape for a given perimeter must have the property that any slice through it cuts the perimeter at the same angle, so that "flipping" one piece will not result in a non-convex shape.
4) Only the circle has this property. QED.
5) Draw any shape (including irregular ones) inscribed inside a circle. Imagine that the sectors of the circle is attached to the edges of the shape as the shape is bent, as if it had hinges at the corners. Any non-circular shape formed cannot have the maximum area for a given perimeter (per #1-#4). Therefore the area inside the bent shape cannot also, as the areas of the sectors of the circle is unchanged. Therefore any shape with the maximum area for given sides can be inscribed inside a circle.
6) Take any two adjacent equal sides of an irregular polygon inscribed inside a circle, and draw the chord between the end points. The triangle with the maximum area for a given sum of sides is the one where the sides are equal. Repeat process with the other sides of the irregular polygon until all sides are equal, thus proving that it is the regular polygon that has the maximum area for its given perimeter.

(#6 is not spot on, it's incomplete. I'll get back to that later)

Addendum: Expanding on #6, first we prove that any given triangle of base length 1 with sides A, B, for a total perimeter of 1 + A + B has a smaller area than a triangle of base length 1 with sides A - x, B + x, if I A - B I > x. This can be done with a lot of tedious algebra which we won't go into here. Thus, a triangle of base length 1 with equal sides (A + B)/2 has the maximum area. Then beginning with any irregular polygon, we can rearrange it by cutting chords and flipping triangles so that we have two adjacent sides such that A > s > B, where s the perimeter length of the polygon divided by the number of sides. We form a new triangle of sides s and B + A - s, with the same chord, hence increasing the area of the polygon without changing the perimeter length. We continue rearranging the polygon in this fashion until all the sides are equal to s, each step at least not decreasing the area. Finally, we inscribe this polygon into a circle, thereby increasing its area to the maximum, again without changing the perimeter length. At this point, we have a regular polygon.

Addendum: ksoileau, you've just described another way of arriving at a circle for the maximum area for a given perimeter. Not bad, interesting approach.

Answer 2

"2) The perimeter and area of any shape is unaffected by an arbitrary slice through it, making a mirror image of one of the pieces, and reattaching it. Thus, if any shape can result in a non-convex shape, it also cannot have the maximum area for a given perimeter."

Do you mean:
Choose an arbitrary point A on the perimeter. Find the unique point B which splits the perimeter into two curves of equal length. Draw the line segment AB. This divides the area into two pieces. Take the piece with larger area, create its mirror image and discard the smaller piece. Attach the two remaining mirror twins together along the AB edges. This shape will have the same perimeter and at least the area of the previous shape. Continue this process until no increase in area is possible. This will be the maximal solution.

(I realize this description is imperfect but time and space are limited.)

Answer 3

Just to add, this can be proven in 2 steps.
1) Take a polygon of n sides. Prove that the A/P ratio is largest when this polygon is regular.
2) Then for regular polygons, prove that the A/P ratio increases with n. As n --> ∞, the polygon tends to a circle and the ratio tends to r/2.

Answer 4

This page gives some good ways of expressing the proofs. Note that the second proof is a step on the way to the first one.

Answer 5

pie x raffle x 2