1960 AHSME Problems/Problem 32
Problem
In this figure the center of the circle is . , is a straight line, , and has a length twice the radius. Then:
Solution 1
We claim that is the right answer.
Let the radius of circle be , and let the length of . Since , is a tangent to circle . Thus, by the tangent-secant theorem, we have , or, . Through some algebraic manipulation, we find .
Since , , and , we see that is identical to , hence our answer is
Solution 2 (guess and check)
Let be the radius of the circle, so . By the Pythagorean Theorem, . That means, , so .
Substitute values for each answer choice to determine which one is correct for all .
For option A, substitution results in
For option B, substitution results in
For option C, substitution results in
For option D, substitution results in
From each option, only option A has both sides equaling each other, so the answer is .
See Also
1960 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 31 |
Followed by Problem 33 | |
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