1991 AMC 12 第 10 题

先试着解答 1991 AMC 12 第 10 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 1991 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。

10.

点 PP 到某圆圆心的距离为 99,该圆的半径为 1515。经过 PP 且长度为整数的不同弦共有多少条?
Point PP is 99 units from the center of a circle of radius 15.15. How many different chords of the circle contain PP and have integer lengths?
1111
1212
1313
1414
2929
答案:B
知识点:circle chorddistancecounting
小提示:
求经过 PP 的最短弦和最长弦
Find the shortest and longest chord through PP
大提示:
除两个极值长度外,每个可取的弦长都对应两个方向
Except at the extreme lengths, each attainable chord length occurs in two directions
解答:
经过 PP 的最长弦是长为 3030 的直径。最短弦垂直于经过 PP 的半径,其长度为 2152−92=24。 2\sqrt{15^2-9^2}=24\text{。}当弦在这两个位置之间转动时,其长度连续地从 2424 变到 3030。两个端点长度各出现一次,而中间的五个整数长度 25,26,27,28,2925,26,27,28,29 各出现两次。因此总数为 1+2(5)+1=121+2(5)+1=12。
因此正确答案为 B。
The longest chord through PP is a diameter of length 30.30. The shortest is perpendicular to the radius through P,P, and has length 2152−92=24. 2\sqrt{15^2-9^2}=24. As the chord rotates between these positions, its length varies continuously from 2424 to 30.30. The endpoint lengths each occur once, while each of the five interior integer lengths 25,26,27,28,2925,26,27,28,29 occurs twice. Thus the count is 1+2(5)+1=12.1+2(5)+1=12.
Thus the correct answer is B.
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