114 桃園市陽明高中
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114 桃園市陽明高中
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Re: 114 桃園市陽明高中
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
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