幾何雜題2條

(1) 圖B表示該四邊形各長16的兩條對邊(紅色)及對角線(藍色),並以θ及φ表示對角線和兩對邊的角度.點線為另一對角線.


圖片參考：http://img12.imageshack.us/img12/8849/20oct20093.png

兩三角形的面積和為(1/2)(16)(16)sinΘ+(1/2)(16)(16)sinΦ = 32
=> 4sinΘ + 4sinΦ = 1
圖A顯示當Θ趨於0時,第二對角線迫近下方的紅線, 圖C顯示當Φ趨於0時,第二對角線迫近上方的紅線.所以第二對角線的一個極值是16.當Θ=0過渡至Φ=0時,會經過另一極值點,就是當Θ=Φ,其時8sinΘ = 1 => Θ = 7.18度
花一點功夫算得第二對角線長度為16.249.
由此得知 16 <= 第二對角線長度 <= 16.249
(2) 由於兩三角形為等邊,每個內角 = 60 度

圖片參考：http://img12.imageshack.us/img12/9536/20oct20092.png

另如圖所示,星形每角的三角形,其內角均為60, x 及y, 60 + x + y = 180 => y = 120 - x
利用正弦公式,可得L/sinx = a3/sin60 => L = a3sinx/sin60
H = a1siny / sin60
由此得到大三角形邊長為 a3sinx /sin60 + b1 + a1siny / sin60
= (a3sinx + b1sin60 + a1siny) / sin60 … (1)
同理,可得大三角形邊長為
(b1siny + a1sin60 + b2sinx) / sin60 … (2)
(a1sinx + b2sin60 + a2siny) / sin60 … (3)
(b2siny + a2sin60 + b3sinx) / sin60 … (4)
(a2sinx + b3sin60 + a3siny) / sin60 … (5)
(b3siny + a3sin60 + b1sinx) / sin60 … (6)
大三角形周界 = (1) + (3) + (5) = (2) + (4) + (6)
(a3sinx + b1sin60 + a1siny) / sin60 + (a1sinx + b2sin60 + a2siny) / sin60 + (a2sinx + b3sin60 + a3siny) / sin60 = (b1siny + a1sin60 + b2sinx) / sin60 + (b2siny + a2sin60 + b3sinx) / sin60 + (b3siny + a3sin60 + b1sinx) / sin60
(a1 + a2 + a3)(sinx + siny) + (b1 + b2 + b3)sin60 = (b1 + b2 + b3)(sinx + siny) + (a1 + a2 + a3)sin60
(a1 + a2 + a3)(sinx + siny – sin60) = (b1 + b2 + b3)(sinx + siny – sin60)
=> a1 + a2 + a3 = b1 + b2 + b3或sinx + siny – sin60 = 0
由於 y = 120 – x
考慮sinx + siny – sin60 = 0 => sinx + sin(120 – x) – √3/2 = 0
sinx + √3cosx/2 + sinx/2 = √3/2
3sinx/2 + √3cosx/2 = √3/2
(√3/2)sinx + (1/2)cosx = 1/2
sin60sinx + cos60cos = cos60
cos(60 – x) = cos60 => x = 0 (y = 120)或120 (x = 0)六角形皆不存在,未能貼題.
於是a1 + a2 + a3 = b1 + b2 + b3
(3) 見圖c = bcosA + acosB
c = cosA + acosB
因此 c > cosA
另c = 2Rcos(A – θ)
最大值發生於當 R 及 θ 為最大時
留意 2Rcosθ = 1
由於 R < 1, 1 < 2cosθ => θ < 60 degrees
Max(R) = 1; Max(θ) = 60 度
Max(c) = 2cos(A – 60)
= 2cosAcos60 + 2sinAsin60
= cosA + √3sinA
總結 cosA < c < cosA + √3sinA

圖片參考：http://img260.imageshack.us/img260/9955/20oct20091.png

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