1983 AIME 第 4 题

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4.

如图所示,一个机械加工切削工具的形状是带缺口的圆。圆的半径为 50\sqrt{50} 厘米,ABAB 的长度为 66 厘米,BCBC 的长度为 22 厘米,且角 ABCABC 为直角。求 BB 到圆心的距离(单位为厘米)的平方。

A machine-shop cutting tool has the shape of a notched circle, as shown. The radius of the circle is 50\sqrt{50} cm, the length of ABAB is 66 cm, and that of BCBC is 22 cm. The angle ABCABC is a right angle. Find the square of the distance (in centimeters) from BB to the center of the circle.

答案:26
知识点:坐标几何垂直平分线
难度评级:2210
小提示:

B=(0,0)B=(0,0)A=(0,6)A=(0,6)C=(2,0)C=(2,0)

Put B=(0,0),B=(0,0), A=(0,6),A=(0,6), and C=(2,0)C=(2,0)

大提示:

圆心位于线段 ACAC 的垂直平分线上,并且与其中点相距 40\sqrt{40}

The center lies on the perpendicular bisector of ACAC at distance 40\sqrt{40} from its midpoint

解答:

B=(0,0)B=(0,0)A=(0,6)A=(0,6)C=(2,0)C=(2,0)ACAC 的中点为 M=(1,3)M=(1,3),且 AC=40AC=\sqrt{40}。若 OO 为圆心,则 OM=OA2AM2=5010=40 \begin{aligned} OM&=\sqrt{OA^2-AM^2}\\ &=\sqrt{50-10}=\sqrt{40} \end{aligned}\text{。}AC=(2,6)AC=(2,-6) 垂直的向量 (6,2)(6,2) 的长度恰为 40\sqrt{40}。因此两个可能的圆心是 M+(6,2)=(7,5),M(6,2)=(5,1) \begin{aligned} M+(6,2)&=(7,5),\\ M-(6,2)&=(-5,1) \end{aligned}\text{。}图中的圆心位于缺口的另一侧,所以 O=(5,1)O=(-5,1)。因此 BO2=(5)2+12=26BO^2=(-5)^2+1^2=26

Put B=(0,0),B=(0,0), A=(0,6),A=(0,6), and C=(2,0).C=(2,0). The midpoint of ACAC is M=(1,3),M=(1,3), and AC=40.AC=\sqrt{40}. If OO is the center, then OM=OA2AM2=5010=40. \begin{aligned} OM&=\sqrt{OA^2-AM^2}\\ &=\sqrt{50-10}=\sqrt{40}. \end{aligned} A vector perpendicular to AC=(2,6)AC=(2,-6) is (6,2),(6,2), which already has length 40.\sqrt{40}. Thus the two possible centers are M+(6,2)=(7,5),M(6,2)=(5,1). \begin{aligned} M+(6,2)&=(7,5),\\ M-(6,2)&=(-5,1). \end{aligned} The pictured notched circle has its center on the side opposite the notch, so O=(5,1).O=(-5,1). Therefore BO2=(5)2+12=26.BO^2=(-5)^2+1^2=26.

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