「斑馬知識+挑戰」證明「在所有相同周界的圖形之中，圓形的面積就是最大。」

Since the area of a regular polygon is larger than the area of a non-regular polygon of the same perimeter (its proof can be found in the reference URL), we only need to show that the area K of a regular n-gon with fixed perimeter P increases as n increases.

Each side has length P/n, so

K = n*(P/n)2*cot(pi/n)/4
K = (P2/4*pi)*(pi/n)/tan(pi/n)

so you are reduced to showing that x/tan(x) increases as x approaches 0 from above (in fact, its limit is 1.)

The limiting case of a regular n-gon as n increases without bound is a circle, which thus is the solution to the problem.

As a shorthand, denote "Among all planar shapes with the same perimeter the circle has the largest area." as A and "Among all planar shapes with the same area the circle has the shortest perimeter." as B.

Assume Aholds and let us prove B.

Suppose, on the contrary, that B is false.Then for agiven circle C there exists a figure F with the same area but with aperimeter shorter than thatof C. Shrink C into a smaller circle C' whose perimeter equals that ofF. The area of C' willclearly be smaller than that of C and, consequently, it will be smallerthan the area of F. Now, thiscontradicts our assumption that A holds: C' and F have the sameperimeter but the area of the circleC' is less than the area of F. Thus A=>B.

A's proof can be found at http://hk.knowledge.yahoo.com/question/question?qid=7008040900309
參考資料 http://www.cut-the-knot.org/do_you_know/isoperimetric.shtml

我未check Fractal Dimension，但我計到魔鬼的曲線長度= 3/2 !! 結果依然係圓的面積最大????
我又試過諗用 variational 既方法，但係就 formulate 唔到個問題，甚至做出 negative 既結果，實在太鬼異啦。
快的搵人了結哩題啦。

2008-04-17 20:10:13 補充：
From wiki, take f0(x)=x and fk(x) as defined in wiki,
Then, length of f0 := L0 = sqrt(2)
length of f1 := L1 = 0.5 + 2(L0)/3
...
length of f_(k+1) := L_(k+1) = 0.5 + 2( L_k)/3
...
Taking limit, we have : L = 0.5 + 2L/3
and hence
L = 3/2

Ivan: 恕我未能同意將"魔鬼"引入這問題。
據我所知，所有碎形連續曲線都是 infinite length 的，連要比較兩條不同的碎形連續曲線的長度比例都唔容易。
(我以前曾和人就比例此問題講得面紅耳赤，但最後我都搵唔到証明或反例，所以不了了之，如果有請必話我知。)

2008-04-16 16:16:16 補充：
所以，由於 infinite length 在此問題係無意義，所以我們可以排除"魔鬼"存在的可能性。
在 wiki裡所說的 length，我認為是指在 R^1 下對那集合的measure，不是指那條在R^2的碎形連續曲線的長度。
而且那measure=1的集合不是連續的，所以我認為很難代入這問題裡。

2008-04-16 16:16:42 補充：
但是在那碎形連續曲線下的面積是絕對有意義。事實上，被碎形連續曲線所包圍的面積都是有意義(limit)的。
因為績分會把碎形的影響平均化去(average)。我一直想寫一個關於數學上"averaging"這概念的小評論，奈何一直手頭上的例子/論証都乏善可陳。
如果 Ivan 為在下回應以上疑難，那份小評論便發表有期了。

Let me say something.
The proposition is WRONG, if you don't give any assumption. (e.g. when the curve CANNOT be approximate by polygons, like a fractal curve)

咁邪惡的counterexample, 當然要找魔鬼...

http://en.wikipedia.org/wiki/Cantor_function
(又叫Devil's Staircase)

2008-04-14 12:17:01 補充：
這是一條由0 到1的 continuous curve. 但係其長度只是1, 因為是由 Cantor Function 的橫線組成的.而對稱性得出 area under curve = 1/2.

現在將4條 Devil's Staircase 貼成凌形的樣子. 這個凌形的周界=4x1 = 4, 面積= 4x 1/2 = 2. 但周界=4的圓形面積只有 4/pi = 1.273, 被遠遠拋離...

2008-04-14 12:19:09 補充：
問題就是出在這條continuous curve 唔係smooth curve, 即是不可微分. 於是便不可以用 approximation by polygon 去做. 不能define Arc Length 的Curve 也會出問題.

所以大前題, 證明這個定理 (Isoperimetric Inequality), 唔多唔少都要假設周界係可微的, 並要用到Differential Geometry 的技巧. 用Polygon 去Approximate 係一個好的Motivation, 但絕對唔可以當係Formal Proof

2008-04-14 12:20:54 補充：
>>假設周界係可微的
正確D 講係Piecewise Smooth (Piecewise C^1), 即係有限個C^1 curve 的union.

2008-04-17 05:35:25 補充：
點解會係 3/2?
留意這個 Cantor Function 的構造:
首先中間一條 1/3 的橫線.
然後兩條 1/9 的橫線
然後四條 1/27 的橫線...
.....
這些線都是 Cantor Set 的 Complement, 由於 Cantor Set measure = 0, 這些直線的總長度是1:
1/3 + 2/9 + 4/27 + 8/81 + ... = 1/3 (1/(1-2/3)) = 1

2008-04-18 01:46:18 補充：
你計錯了...
畫個 graph 出黎. f1 的長度係 1/3 + 2 sqrt(13/36)
中間的橫線的長度是1/3, 唔係0.5
而頭尾兩條鈄線的長度唔係L0/3, 而要用 Pythagorus Theorem 計的呀~

2008-04-18 04:27:35 補充：
計完......... 魔鬼的長度是2.........
看來我錯了.......
沮喪中 T_T

應否也需考慮 irregular polygon 的 case?