1973 AMC 12 Problem 11

Attempt Problem 11 of the 1973 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1973 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

11.

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}
none of these
Answer: B
Concepts:inequalitycoordinate geometrycirclesquare (geometry)
Small hint:
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|)
Big hint:
The inequality reverses the order of containment of the corresponding sublevel sets
Solution:
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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