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解法一(代數法) :x² + ⅓ y² + xy = 25
⅓ y² + z² = 9
z² + x² + zx = 16
利用前式等如後兩式之和得
⅓ y² + z² + z² + x² + zx = 9 + 16 = 25 = x² + ⅓ y² + xy
2z² + zx = xy
y = 2z²/x + z , 代入 ⅓ y² + z² = 9 :
⅓ (2z²/x + z)² + z² = 9
4z⁴/x² + 4z³/x + 4z² = 27
4z² (z² + zx + x²) = 27x²
4z² (16) = 27x²
z = 3√3/8 x , 代入 z² + x² + zx = 16 :
27/64 x² + x² + 3√3/8 x² = 16
(27 + 64 + 24√3)x² = 1024
x² = 1024 / (91 + 24√3)
故 xy + 2zy + 3zx
= x(2z²/x + z) + 2(3√3/8 x)(2z²/x + z) + 3(3√3/8 x)x
= 2z² + zx + 3√3/2 z² + 3√3/4 zx + 9√3/8 x²
= (4 + 3√3)/2 z² + (4 + 3√3)/4 zx + 9√3/8 x²
= (4 + 3√3)/2 (27/64 x²) + (4 + 3√3)/4 3√3/8 x² + 9√3/8 x²
= ( (108 + 81√3)/128 + (12√3 + 27)/32 + 9√3/8 ) x²
= ( (108 + 81√3 + 48√3 + 108 + 144√3)/128 ) 1024 / (91 + 24√3)
= 8 (273√3 + 216) / (91 + 24√3)
= 24 (91√3 + 72) / (91 + 24√3)
= 24√3
解法二(幾何法) :
Ｂ
y/√3
150° ３
Ｐ 90°
x 120° z
Ａ ─────────────────── Ｃ
４
觀察條件形式,聯想直角三角形及餘弦定理。如圖,邊長3 , 4 , 5的直角三角形內之一點P使得∠APB = 150° , ∠BPC = 90° , ∠CPA = 120° ; PA = x , PB = y/√3 , PC = z。
由餘弦定理得
x² + y²/3 - 2xy/√3 cos150° = 5²
y²/3 + z² - 2zy/√3 cos90° = 3²
z² + x² - 2zx cos120° = 4²
⇔
x² + y²/3 + xy = 25
y²/3 + z² = 9
z² + x² + zx = 16
恰為題設條件。最後由 △APB + △BPC + △CPA = △ABC 得½ x y/√3 sin150° + ½ z y/√3 + ½ zx sin120° = ½ × 4 × 3
xy/√3 (½) + zy/√3 + zx√3/2 = 12
xy + 2zy + 3zx = 12 × 2√3 = 24√3。
2015-06-24 05:46:40 補充：
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