1973 AHSME Problems/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.
are represented geometrically* by the figure numbered
* An inequality of the form , for all and is represented geometrically by a figure showing the containment
for a typical real number .
First, note that the following inequality represents the graphs , , and . We don't actually have to worry about the inequality, we just want to find the picture that graphs those equations for some constant .
WLOG, we can set . To make our life easier, we can also focus entirely on the first quadrant, since all graphs have different shapes in the first quadrant. Therefore, and .
First, we have . Since and , this becomes . After a bit of rearranging, we get .
This eliminates , which does not have a diagonal line in the first quadrant.
Next, we have . Squaring both sides and dividing by , we get Note that this describes a circle with the center and radius . We can find that is on the circle, which is also on the line . Therefore, the circle intersects the line at that point. We can graph this as follows:
This eliminates , which does not include the intersection point listed above.
Finally, we have . Rearranging and substituting for and likewise for , we get Note that this equation means that either or . Therefore, from to , the graph looks like the line . Then, from to , the graph looks like the line . Graphing this, we get:
Thus, we can eliminate . However, at this point, we don't know if the answer is 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 or 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 and axes. As a result, the final graph looks like this:
Therefore, we can choose as our answer.
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