1952 AMC 12 第 29 题

先试着解答 1952 AMC 12 第 29 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 1952 AMC 12 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

29.

在半径为 55 个单位的圆中,CDCD 与 ABAB 是互相垂直的直径。弦 CHCH 与 ABAB 相交于 KK,且长 88 个单位。直径 ABAB 被分成两段,其长度分别为:
In a circle of radius 55 units, CDCD and ABAB are perpendicular diameters. A chord CHCH cutting ABAB at KK is 88 units long. The diameter ABAB is divided into two segments whose dimensions are:
1.251.25, 8.758.75
1.25,1.25, 8.758.75
2.752.75, 7.257.25
2.75,2.75, 7.257.25
22, 88
2,2, 88
44, 66
4,4, 66
以上答案均不正确
None of these
答案:A
知识点:圆弦坐标几何
小提示:
将圆心置于原点,取 C=(0,5)C=(0,5),并用坐标表示弦的另一端点 HH
Put the center at the origin, take C=(0,5)C=(0,5), and represent the other endpoint HH of the chord by coordinates
大提示:
利用 CH=8CH=8 和 OH=5OH=5,再求直线 CHCH 与水平直径的交点
Use CH=8CH=8 and OH=5OH=5, then find where line CHCH crosses the horizontal diameter
解答:
将圆心置于原点,取 C=(0,5)C=(0,5),并设 H=(x,y)H=(x,y)。方程组 x2+y2=25,x2+(y−5)2=64 \begin{aligned} x^2+y^2&=25,\\ x^2+(y-5)^2&=64 \end{aligned} 给出 y=−75y=-\frac{7}{5} 和 x=±245x=\pm\frac{24}{5}。取 x=245x=\frac{24}{5} 不影响两段的长度。
从 C=(0,5)C=(0,5) 到 H=(245,−75)H=(\frac{24}{5},-\frac{7}{5}) 的直线与 y=0y=0 相交于 x=154x=\frac{15}{4} 处。因此,该交点到端点 (−5,0)(-5,0) 和 (5,0)(5,0) 的距离如下;这两个端点构成直径 ABAB:5+154=354=8.75,5−154=54=1.25。 \begin{aligned} 5+\frac{15}{4}&=\frac{35}{4}=8.75,\\ 5-\frac{15}{4}&=\frac54=1.25 \end{aligned}\text{。}
因此,正确答案是 A。
Put the circle at the origin with C=(0,5)C=(0,5) and let H=(x,y).H=(x,y). The equations x2+y2=25,x2+(y−5)2=64 \begin{aligned} x^2+y^2&=25,\\ x^2+(y-5)^2&=64 \end{aligned} give y=−75y=-\frac{7}{5} and x=±245.x=\pm\frac{24}{5}. Taking x=245x=\frac{24}{5} does not change the segment lengths.
The line from C=(0,5)C=(0,5) to H=(245,−75)H=(\frac{24}{5},-\frac{7}{5}) crosses y=0y=0 at x=154.x=\frac{15}{4}. Its distances to the endpoints (−5,0)(-5,0) and (5,0)(5,0) of diameter ABAB are therefore 5+154=354=8.75,5−154=54=1.25. \begin{aligned} 5+\frac{15}{4}&=\frac{35}{4}=8.75,\\ 5-\frac{15}{4}&=\frac54=1.25. \end{aligned}
Thus, the correct answer is A.
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