「斑馬知識+挑戰」幾何挑戰題 2

C = circumscribed
I = inscribed

Ac : Ai = [csc@]^2

@ = [90(n-2)]/n

2008-07-21 18:25:14 補充：
我是不是要列出步驟??

circumscribed circle 果圓心係由n咁多條既 angel bisector 線的相交點
inscribed circle 果圓心係由n咁多條既 perpendicular bisector 伸出來果條線的相交點

苜先證明相交點是在同一位置:
設 L 為circumscribed circle的半徑
設 R 為inscribed circle的半徑
R/L = sin@
@2 = half of the an angle of the regular polygon
180(n-2)/2n = [90(n-2)]/n

2008-07-21 18:25:21 補充：
@1 = @2 = @3 = @4 ..(由於這是一個regular polygon(每隻內角都一樣))

R1/L1 = sin@1
R2/L2 = sin@2
.....
又由於R and L 都是圓的半徑
所以 R1 = R2 = ...
L1 = L2 = ....


Rp/Lp = sin@p
這可行的條件只有一個,
R1 與 L1 的夾點 都在 R2 與 L2 的夾點 都在 Rp 與 Lp 的夾點

2008-07-21 18:28:52 補充：
所以兩圓心點是在同一位置
而 Rp / Lp = Rp : Lp
R1 = R2 = ...
L1 = L2 = ...

R:L = sin@
@ = [90(n-2)]/n

而兩圓是similar ,
兩圓的面積比例是,長度比的2次方:
Ai : Ac = R^2 : L^2
Ai : Ac = [sin@]^2

Ac行先,調轉分子分母
Ac : Ai = [csc@]^2

@ = [90(n-2)]/n

Not difficult...

the required ratio = (radius of C. circle)^2 : (radius of I. circle)^2
= (distance from vertice to centroid)^2 : (distance from mid-pt of edge to centroid)^2
= sec^2 (360/2n)
= sec^2 (180/n)