1973 AMC 12 第 11 题

先试着解答 1973 AMC 12 第 11 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 1973 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。

11.

在直角坐标系中,原点为 OO,正 xx 轴与正 yy 轴分别为 OXOX 与 OYOY。以下每幅图 I\mathrm{I} 至 IV\mathrm{IV} 都画出了一个圆及其外接正方形和内接正方形,三者均以原点为中心。
不等式 ∣x∣+∣y∣≤2(x2+y2)≤2Max⁡(∣x∣,∣y∣) \begin{aligned} |x|+|y| &\le \sqrt{2(x^2+y^2)}\\ &\le 2\operatorname{Max}(|x|,|y|) \end{aligned} 的几何表示对应编号为下列哪一幅图:
几何表示:若不等式 f(x,y)≤g(x,y)f(x,y)\le g(x,y) 对所有 xx、yy 成立,则其几何表示是一幅对一个典型实数 aa 展示下列包含关系的图:{(x,y):g(x,y)≤a}⊂{(x,y):f(x,y)≤a}。 \begin{aligned} &\{(x,y):g(x,y)\le a\}\\ &\quad\subset \{(x,y):f(x,y)\le a\} \end{aligned}\text{。}
A circle with a circumscribed and an inscribed square centered at the origin OO of a rectangular coordinate system with positive xx- and yy-axes OXOX and OYOY is shown in each figure I\mathrm{I} to IV\mathrm{IV} below.
The inequalities ∣x∣+∣y∣≤2(x2+y2)≤2Max⁡(∣x∣,∣y∣). \begin{aligned} |x|+|y| &\le \sqrt{2(x^2+y^2)}\\ &\le 2\operatorname{Max}(|x|,|y|). \end{aligned} are represented geometrically by the figure numbered
Geometric representation: An inequality f(x,y)≤g(x,y)f(x,y)\le g(x,y) for all x,x, yy is represented by a figure showing, for a typical real number a,a, the containment {(x,y):g(x,y)≤a}⊂{(x,y):f(x,y)≤a}. \begin{aligned} &\{(x,y):g(x,y)\le a\}\\ &\quad\subset \{(x,y):f(x,y)\le a\}. \end{aligned}
I\mathrm{I}
II\mathrm{II}
III\mathrm{III}
IV\mathrm{IV}
以上都不是
none of these
答案:B
知识点:不等式坐标几何圆正方形(几何)
小提示:
确定 ∣x∣+∣y∣|x|+|y|、2(x2+y2)\sqrt{2(x^2+y^2)} 和 2Max⁡(∣x∣,∣y∣)2\operatorname{Max}(|x|,|y|) 的次水平集形状
Identify the shapes of the sublevel sets for ∣x∣+∣y∣,|x|+|y|, 2(x2+y2),\sqrt{2(x^2+y^2)}, and 2Max⁡(∣x∣,∣y∣)2\operatorname{Max}(|x|,|y|)
大提示:
不等式会使对应次水平集的包含顺序反向
The inequality reverses the order of containment of the corresponding sublevel sets
解答:
对固定的正数 aa,集合 2Max⁡(∣x∣,∣y∣)≤a 2\operatorname{Max}(|x|,|y|)\le a 是边与坐标轴平行的正方形。集合 2(x2+y2)≤a \sqrt{2(x^2+y^2)}\le a 是它的外接圆,而 ∣x∣+∣y∣≤a |x|+|y|\le a 是该圆的外接菱形。因此所需的嵌套顺序是内部的轴向正方形、圆、外部菱形,对应图 II\mathrm{II}。
所以正确答案是 B。
For a fixed positive a,a, the set 2Max⁡(∣x∣,∣y∣)≤a 2\operatorname{Max}(|x|,|y|)\le a is an axis-aligned square. The set 2(x2+y2)≤a \sqrt{2(x^2+y^2)}\le a is its circumscribed circle, and ∣x∣+∣y∣≤a |x|+|y|\le a is the diamond circumscribed about that circle. Thus the required nesting is an inner axis-aligned square, then a circle, then an outer diamond, which is figure II.\mathrm{II}.
Therefore, the correct answer is B.
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