平行四邊形周界一問

設圓心為 O ,
延長AC交圓於E,連DE ,則△AOB ~ △CDE
設AB = x , 則AC = OE + 1 = 2x
OE = 2x - 1 , EC = 1 - OE = 1 - (2x - 1) = 2 - 2x
有 : AB / OA = EC / CD
x / 1 = (2 - 2x )/ x
x^2 + 2x - 2 = 0
x =
圖片參考：http://www.edp.ust.hk/previous/math/history/7/root_4.jpg
3 - 1(只取正根)
周界 = 6x
= 6
圖片參考：http://www.edp.ust.hk/previous/math/history/7/root_4.jpg
3 - 1 cm


2009-09-23 22:30:50 補充：
更正答案 :

6√3 - 6 cm

= 4.392304845cm

2009-09-23 22:35:27 補充：
OE 改為 OC

OA= OB = OD = radius = 1.
Let AB = a and let angle BAC = x.
Applying cosine rule to triangle BOA, we get
OB^2 = AB^2 + OA^2 - 2(AB)(OA)cos x
1 = a^2 + 1 - 2a cos x
cos x = a/2......(1)
AC = 2AB = 2a, so OC = AC - OA = 2a - 1.
Angle OCD = 180 - x
Applying cosine rule to triangle ODC we get
OD^2 = DC^2 + OC^2 - 2(DC)(OC)cos (180 - x)
1 = a^2 + (2a - 1)^2 + 2a(2a - 1) cos x
1 = a^2 + 4a^2 + 1 - 4a + 2a(2a - 1)(a/2)
0 = 5a^2 - 4a + a(2a - 1)
0 = 5a - 4 + a(2a - 1)
2a^2 + 4a - 4 = 0
a^2 + 2a - 2 = 0
a = sqrt 3 - 1
so perimeter = 6a = 6(sqrt 3 - 1).