1998 AMC 12 第 25 题

先试着解答 1998 AMC 12 第 25 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 1998 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。

25.

把一张方格纸折叠一次,使 (0,2)(0,2) 与 (4,0)(4,0) 重合,并使 (7,3)(7,3) 与 (m,n)(m,n) 重合。求 m+nm+n。
A piece of graph paper is folded once so that (0,2)(0,2) is matched with (4,0),(4,0), and (7,3)(7,3) is matched with (m,n).(m,n). Find m+n.m+n.
6.76.7
6.86.8
6.96.9
7.07.0
8.08.0
答案:B
知识点:reflections坐标几何
小提示:
折痕是连接 (0,2)(0,2) 与 (4,0)(4,0) 的线段的垂直平分线
The crease is the perpendicular bisector of the segment joining (0,2)(0,2) and (4,0)(4,0)
大提示:
先求出直线 y=2x−3y=2x-3,再把 (7,3)(7,3) 关于它反射
Find the line y=2x−3y=2x-3, then reflect (7,3)(7,3) across it
解答:
折痕经过 (2,1)(2,1),斜率为 22,所以其方程为 y=2x−3y=2x-3。过 (7,3)(7,3) 的垂线斜率为 −12-\frac{1}{2},与折痕交于 (195,235)(\frac{19}{5},\frac{23}{5})。该交点是 (7,3)(7,3) 与 (m,n)(m,n) 的中点,所以 m=35, n=315m=\frac{3}{5},\ n=\frac{31}{5}。因此 m+n=345=6.8m+n=\frac{34}{5}=6.8,正确答案是 B。
The crease passes through (2,1)(2,1) with slope 2,2, so it is y=2x−3.y=2x-3. The perpendicular through (7,3)(7,3) has slope −12-\frac{1}{2} and meets the crease at (195,235).(\frac{19}{5},\frac{23}{5}). This intersection is the midpoint of (7,3)(7,3) and (m,n),(m,n), giving m=35, n=315.m=\frac{3}{5},\ n=\frac{31}{5}. Hence m+n=345=6.8,m+n=\frac{34}{5}=6.8, so B is correct.
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