# 1973 AHSME Problems/Problem 11

## Problem 11

A circle with a circumscribed and an inscribed square centered at the origin of a rectangular coordinate system with positive and axes and is shown in each figure to below. $[asy] size((400)); draw((0,0)--(22,0), EndArrow); draw((10,-10)--(10,12), EndArrow); draw((25,0)--(47,0), EndArrow); draw((35,-10)--(35,12), EndArrow); draw((-25,0)--(-3,0), EndArrow); draw((-15,-10)--(-15,12), EndArrow); draw((-50,0)--(-28,0), EndArrow); draw((-40,-10)--(-40,12), EndArrow); draw(Circle((-40,0),6)); draw(Circle((-15,0),6)); draw(Circle((10,0),6)); draw(Circle((35,0),6)); draw((-34,0)--(-40,6)--(-46,0)--(-40,-6)--(-34,0)--(-34,6)--(-46,6)--(-46,-6)--(-34,-6)--cycle); draw((-6.5,0)--(-15,8.5)--(-23.5,0)--(-15,-8.5)--cycle); draw((-10.8,4.2)--(-19.2,4.2)--(-19.2,-4.2)--(-10.8,-4.2)--cycle); draw((14.2,4.2)--(5.8,4.2)--(5.8,-4.2)--(14.2,-4.2)--cycle); draw((16,6)--(4,6)--(4,-6)--(16,-6)--cycle); draw((41,0)--(35,6)--(29,0)--(35,-6)--cycle); draw((43.5,0)--(35,8.5)--(26.5,0)--(35,-8.5)--cycle); label("I", (-49,9)); label("II", (-24,9)); label("III", (1,9)); label("IV", (26,9)); label("X", (-28,0), S); label("X", (-3,0), S); label("X", (22,0), S); label("X", (47,0), S); label("Y", (-40,12), E); label("Y", (-15,12), E); label("Y", (10,12), E); label("Y", (35,12), E);[/asy]$

The inequalities $$|x|+|y|\leq\sqrt{2(x^{2}+y^{2})}\leq 2\mbox{Max}(|x|, |y|)$$

are represented geometrically* by the figure numbered $\textbf{(A)}\ I\qquad\textbf{(B)}\ II\qquad\textbf{(C)}\ III\qquad\textbf{(D)}\ IV\qquad\textbf{(E)}\ \mbox{none of these}$

* An inequality of the form $f(x, y) \leq g(x, y)$, for all $x$ and $y$ is represented geometrically by a figure showing the containment $\{\mbox{The set of points }(x, y)\mbox{ such that }g(x, y) \leq a\} \subset\\ \{\mbox{The set of points }(x, y)\mbox{ such that }f(x, y) \leq a\}$

for a typical real number $a$.

## Solution

First, note that the following inequality $$|x|+|y|\leq\sqrt{2(x^{2}+y^{2})}\leq 2\mbox{Max}(|x|, |y|)$$ represents the graphs $|x| + |y| = a$, $\sqrt{2(x^{2}+y^{2})} = a$, and $2\mbox{Max}(|x|, |y|) = a$. We don't actually have to worry about the inequality, we just want to find the picture that graphs those $3$ equations for some constant $a$.

WLOG, we can set $a=1$. To make our life easier, we can also focus entirely on the first quadrant, since all $4$ graphs have different shapes in the first quadrant. Therefore, $x \geq 0$ and $y \geq 0$.

First, we have $|x| + |y| = 1$. Since $x \geq 0$ and $y \geq 0$, this becomes $x + y = 1$. After a bit of rearranging, we get $$y = -x + 1$$. $[asy] size((100)); draw((-50,0)--(-28,0), EndArrow); draw((-40,-10)--(-40,12), EndArrow); draw((-34,0)--(-40,6)); label("X", (-28,0), S); label("Y", (-40,12), E); [/asy]$

This eliminates $III$, which does not have a diagonal line in the first quadrant.

Next, we have $\sqrt{2(x^{2}+y^{2})} = 1$. Squaring both sides and dividing by $2$, we get $$x^2+y^2=\frac{\sqrt{2}}{2}^2$$ Note that this describes a circle with the center $(0,0)$ and radius $\frac{\sqrt{2}}{2}$. We can find that $(\frac{1}{2}, \frac{1}{2})$ is on the circle, which is also on the line $y = -x + 1$. Therefore, the circle intersects the line at that point. We can graph this as follows: $[asy] size((100)); draw((-50,0)--(-28,0), EndArrow); draw((-40,-10)--(-40,12), EndArrow); draw((-34,0)--(-40,6)); draw(arc((-40,0),4.25, 0, 90)); label("X", (-28,0), S); label("Y", (-40,12), E); [/asy]$

This eliminates $I$, which does not include the intersection point listed above.

Finally, we have $2\mbox{Max}(|x|, |y|) = 1$. Rearranging and substituting $x$ for $|x|$ and likewise for $y$, we get $$\mbox{Max}(x, y) = \frac{1}{2}$$ Note that this equation means that either $x \leq y = \frac{1}{2}$ or $y \leq x = \frac{1}{2}$. Therefore, from $x = 0$ to $x = \frac{1}{2}$, the graph looks like the line $y = \frac{1}{2}$. Then, from $y = 0$ to $y = \frac{1}{2}$, the graph looks like the line $x = \frac{1}{2}$. Graphing this, we get: $[asy] size((100)); draw((-50,0)--(-28,0), EndArrow); draw((-40,-10)--(-40,12), EndArrow); draw((-34,0)--(-40,6)); draw(arc((-40,0),4.25, 0, 90)); draw((-40,3)--(-37,3)--(-37,0)); label("X", (-28,0), S); label("Y", (-40,12), E); [/asy]$

Thus, we can eliminate $IV$. However, at this point, we don't know if the answer is $II$ or none of the graphs, since this entire time we were only graphing in the first quadrant to simplify things. However, note that no matter what sign $x$ or $y$ is, the equations all surround the variables with absolute value signs or exponents and make the values positive. Therefore, the graphs are symmetric about the $x$ and $y$ axes. As a result, the final graph looks like this: $[asy] size((100)); draw((-25,0)--(-3,0), EndArrow); draw((-15,-10)--(-15,12), EndArrow); draw(Circle((-15,0),6)); draw((-6.5,0)--(-15,8.5)--(-23.5,0)--(-15,-8.5)--cycle); draw((-10.8,4.2)--(-19.2,4.2)--(-19.2,-4.2)--(-10.8,-4.2)--cycle); label("X", (-3,0), S); label("Y", (-15,12), E);[/asy]$

Therefore, we can choose $\boxed{\textbf{B}}$ as our answer.

## See Also

 1973 AHSME (Problems • Answer Key • Resources) Preceded byProblem 10 Followed byProblem 12 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 All AHSME Problems and Solutions
Invalid username
Login to AoPS