114 桃園市陽明高中

[thepiano]

請參考附件

[thepiano]

計算第 1 題
x，y，z > 0，x + y + z = 1

令 x = tan(A/2)tan(B/2)，y = tan(B/2)tan(C/2)，z = tan(C/2)tan(A/2)

(x - yz)/(x + yz) + (y - zx)/(y + zx) + (z - xy)/(z + xy)
= (1 - yz/x)/(1 + yz/x) + (1 - zx/y)/(1 + zx/y) + (1 - xy/z)/(1 + xy/z)
= {1 - [tan(C/2)]^2}/{1 + [tan(C/2)]^2} + {1 - [tan(A/2)]^2}/{1 + [tan(A/2)]^2} + {1 - [tan(B/2)]^2}/{1 + [tan(B/2)]^2}
= 1 - 2[tan(C/2)]^2/{1 + [tan(C/2)]^2} + 1 - 2[tan(A/2)]^2/{1 + [tan(A/2)]^2} + 1 - 2[tan(B/2)]^2/{1 + [tan(B/2)]^2}
= 1 - 2[sin(C/2)]^2 + 1 - 2[sin(A/2)]^2 + 1 - 2[sin(B/2)]^2
= cosC + cosA + cosB
≦ 3/2

[thepiano]

填充題第 12 題
不能碰到圖上的兩條紅線
所求 = 729 / 2^14

[thepiano]

計算第 2 題
令 m = logx (以 2 為底)，n = logy (以 3 為底)
P = (m - 1)n^2 - (6m + 2a)n + (m + 1)

(1) a = 0，1 ≦ x ≦ 2，0 ≦ m ≦ 1
P = (m - 1)n^2 - 6mn + (m + 1) = (n^2 - 6n + 1)m - (n^2 - 1) > 0
m = 0 代入，- (n^2 - 1) > 0，-1 < n < 1
m = 1 代入，(n^2 - 6n + 1) - (n^2 - 1) > 0，n < 1/3
取 -1 < n < 1/3
1/3 < y < 3^(1/3)
官方答案給錯了

(2) x ≠ 2，m ≠ 1
P = (m - 1)n^2 - (6m + 2a)n + (m + 1)
對 x > 0 且 x ≠ 2，均有 y 使得 P = 0
故 [-(6m + 2a)]^2 - 4(m - 1)(m + 1) ≧ 0
8m^2 + 6am + (a^2 + 1) ≧ 0
(6a)^2 - 4 * 8 * (a^2 + 1) ≦ 0
-2√2 ≦ a ≦ 2√2