Difference between revisions of "1992 AHSME Problems/Problem 27"
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== Solution == | == Solution == | ||
− | <math>\fbox{D}</math> | + | Applying Power of a Point on <math>P</math>, we find that <math>PC=9</math> and thus <math>PD=16</math>. Observing that <math>PD=2BP</math> and that <math>\angle BPD=60^{\circ}</math>, we conclude that <math>BPD</math> is a 30-60-90 right triangle with right angle at <math>B</math>. Thus, <math>BD=8\sqrt{3}</math> and triangle <math>ABD</math> is also right. Using that fact that the circumcircle of a right triangle has its diameter equal to the hypotenuse, we compute using the Pythagorean Theorem <math>AD=2r=2\sqrt{73}</math>. From here we see that <math>r^2=73</math>. The answer is thus <math>\fbox{D}</math>. |
== See also == | == See also == |
Revision as of 08:51, 1 August 2016
Problem
A circle of radius has chords of length and of length 7. When and are extended through and , respectively, they intersect at , which is outside of the circle. If and , then
Solution
Applying Power of a Point on , we find that and thus . Observing that and that , we conclude that is a 30-60-90 right triangle with right angle at . Thus, and triangle is also right. Using that fact that the circumcircle of a right triangle has its diameter equal to the hypotenuse, we compute using the Pythagorean Theorem . From here we see that . The answer is thus .
See also
1992 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 26 |
Followed by Problem 28 | |
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All AHSME Problems and Solutions |
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