1966 AHSME Problems/Problem 4

Problem

Circle I is circumscribed about a given square and circle II is inscribed in the given square. If $r$ is the ratio of the area of circle I to that of circle II, then $r$ equals:

$\text{(A) } \sqrt{2} \quad \text{(B) } 2 \quad \text{(C) } \sqrt{3} \quad \text{(D) } 2\sqrt{2} \quad \text{(E) } 2\sqrt{3}$

Solution

Make half of the square's side $x$ . Now the radius of the smaller circle is $x$ , so it's area is $\pi x^2$ . Now find the diameter of the bigger circle. Since half of the square's side is $x$ , the full side is $2x$ . Using the Pythagorean theorem, you get the diagonal to be $2\sqrt{2}x$ . Half of that is the radius, or $x\sqrt{2}$ . Using the same equation as before, you get the area of the larger circle to be $2x^2 \pi$ . Putting one over the other and dividing, you get two as the answer: or $\boxed{(B)}$ .

1966 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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1966 AHSME Problems/Problem 4

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