Processing math: 27%

2024年11月27日 星期三

[107指考數甲,圓] 圓內兩線段互相垂直,求¯CD長度

==問題== 

ABCD為圓上的相異四點。已知圓的半徑為72¯AB=5,兩線段¯AC¯BD互相垂直,如圖所示(此為示意圖,非依實際比例)。則¯CD的長度為何?(化成最簡根式)


[107,指考,數學甲]

==答案==

26

==解析==

¯AC¯BD的交點為O。然後再假設BAC=θ,於是BC之弧度為2 \theta。由於\angle BDC所對的弧也是\overparen{BC},因此\angle BDC = \theta

接著座標化,取\overline{AC}x軸,\overrightarrow{AC}之方向為正向;取\overline{BD}y軸,\overrightarrow{BD}之方向為正向。於是O點座標為(0, 0)

再假設\overline{CD} = x,於是\overline{OC} = x \sin \theta, \overline{OD} = x \cos \theta, \overline{OA} = 5 \cos \theta, \overline{OB} = 5 \sin \theta,從而可得各點座標A = (-5 \cos \theta, 0), B = (0, -5 \sin \theta), C = (x \sin \theta, 0), D = (0, x \cos \theta)

再設圓心為K,注意到K點的x座標與\overline{AC}中點的x座標相同,y座標與\overline{BD}中點的y座標相同,因此可得K點的座標為\left( \frac{x \sin \theta - 5 \cos \theta}{2}, \frac{x \cos \theta - 5 \sin \theta}{2} \right)

由題目條件「圓的半徑為\frac{7}{2}」可知

\overline{KC} = \frac{7}{2},

\sqrt{ \left( \frac{x \sin \theta - 5 \cos \theta}{2} - x \sin \theta \right)^2 + \left( \frac{x \cos \theta - 5 \sin \theta}{2} - 0 \right)^2 } = \frac{7}{2},

等號兩邊左右同時平方,展開得

\frac{x^2 \sin^2 \theta + 10 x \sin \theta \cos \theta + 25 \cos^2 \theta + x^2 \cos^2 \theta - 10 x \sin \theta \cos \theta + 25 \sin^2 \theta}{4} = \frac{49}{4},

於是

\frac{(x^2 \sin^2 \theta + x^2 \cos^2 \theta) + (25 \cos^2 \theta + 25 \sin^2 \theta)}{4} = \frac{49}{4},

\frac{ x^2(\sin^2 \theta + \cos^2 \theta) + 25( \cos^2 \theta + \sin^2 \theta)}{4} = \frac{49}{4},

\frac{ x^2 \cdot 1 + 25 \cdot 1}{4} = \frac{49}{4},

x^2 + 25 = 49,

x^2 = 24,

x = \pm \sqrt{24} = \pm 2 \sqrt{6} \quad {\text{負不合}},

x = 2 \sqrt{6}.

因此所求\overline{CD} = 2 \sqrt{6}

(解答終了)