1970 AMC 12 第 26 题

先试着解答 1970 AMC 12 第 26 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 1970 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。

26.

在 xyxy 平面内,考虑图形 (x+y−5)(2x−3y+5)=0 (x+y-5)(2x-3y+5)=0 和图形 (x−y+1)(3x+2y−12)=0。 (x-y+1)(3x+2y-12)=0\text{。}这两个图形有多少个不同的公共点?
The number of distinct points in the xyxy-plane common to the graphs of (x+y−5)(2x−3y+5)=0 (x+y-5)(2x-3y+5)=0 and (x−y+1)(3x+2y−12)=0 (x-y+1)(3x+2y-12)=0 is:
00
11
22
33
44
无穷多个
infinite
答案:B
知识点:方程组零积性质交点计数
小提示:
每个乘积方程都表示两条直线的并集
Each product equation represents a union of two lines
大提示:
先检查每组两条直线的交点,再考虑四种配对
Check the intersection of the two lines in each pair before considering all four pairings
解答:
第一个图形是以下两条直线的并集:x+y=5,2x−3y=−5, x+y=5,\qquad 2x-3y=-5\text{,}这两条直线交于 (2,3)(2,3)。第二个图形是以下两条直线的并集:x−y=−1,3x+2y=12, x-y=-1,\qquad 3x+2y=12\text{,}它们也交于 (2,3)(2,3)。事实上,将 (2,3)(2,3) 代入可知,它满足四条直线的方程。由于四条直线的斜率各不相同,不会有其他点同时属于两个图形。因此恰有一个公共点。
因此,正确答案是 B。
The first graph is the union of x+y=5,2x−3y=−5, x+y=5,\qquad 2x-3y=-5, and those two lines meet at (2,3).(2,3). The second graph is the union of x−y=−1,3x+2y=12, x-y=-1,\qquad 3x+2y=12, and these also meet at (2,3).(2,3). In fact, substituting (2,3)(2,3) satisfies all four line equations. Since the four lines have distinct slopes, no other point can belong to a line from each graph. There is exactly one common point.
Therefore, the correct answer is B.
LIVE by Po-Shen Loh
LIVE项目

LIVE直播课程

Self-Paced录播课程

启明星Star计划

奖学金项目

关于我们

罗博深是谁?

邮件联系

订阅英文简报

加入我们

申请明星导师

For a more Thought Full world.

© 2026 Expii, Inc.