2017 AMC 12A 第 9 题

先试着解答 2017 AMC 12A 第 9 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 2017 AMC 12A 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。

9.

设 SS 是坐标平面中满足以下条件的点 (x,y)(x,y) 的集合:三个量 33,x+2x+2,和 y−4y-4 中有两个相等,而第三个量不大于这个公共值。以下哪一项正确描述了 SS?
Let SS be the set of points (x,y)(x,y) in the coordinate plane such that two of the three quantities 3,3, x+2,x+2, and y−4y-4 are equal and the third of the three quantities is no greater than this common value. Which of the following is a correct description of S?S?
一个点
a single point
两条相交直线
two intersecting lines
三条直线,且两两交点是三个不同的点
three lines whose pairwise intersections are three distinct points
一个三角形
a triangle
三条有共同端点的射线
three rays with a common endpoint
答案:E
知识点:坐标几何分类讨论
小提示:
令三个量中的两个相等,并要求第三个量至多等于这个公共值
Set two of the three quantities equal and require the third to be at most that common value
大提示:
三种配对各产生一条射线,且都从同一点 (1,7)(1,7) 出发
Each of the three pairings produces a ray, and all three start at the same point (1,7)(1,7)
解答:
考虑 33,x+2x+2,y−4y-4 中哪两个是相等的较大值。
若 3=x+2≥y−43=x+2\ge y-4,则 x=1x=1 且 y≤7y\le7,得到一条从 (1,7)(1,7) 向下的射线。若 3=y−4≥x+23=y-4\ge x+2,则 y=7y=7 且 x≤1x\le1,得到一条从 (1,7)(1,7) 向左的射线。若 x+2=y−4≥3x+2=y-4\ge3,则 y=x+6y=x+6 且 x≥1x\ge1,得到一条从 (1,7)(1,7) 向右上方的射线。
三条射线共享端点 (1,7)(1,7),所以 SS 是三条有共同端点的射线。
所以正确答案是 E。
Consider which two of 3,3, x+2,x+2, y−4y-4 are the (equal) larger pair.
If 3=x+2≥y−4:3=x+2\ge y-4: then x=1x=1 and y≤7,y\le7, a downward ray from (1,7).(1,7). If 3=y−4≥x+2:3=y-4\ge x+2: then y=7y=7 and x≤1,x\le1, a leftward ray from (1,7).(1,7). If x+2=y−4≥3:x+2=y-4\ge3: then y=x+6y=x+6 and x≥1,x\ge1, a ray from (1,7)(1,7) going up and to the right.
All three rays share the endpoint (1,7),(1,7), so SS is three rays with a common endpoint.
Thus, the correct answer is E.
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