Difference between revisions of "1992 AHSME Problems/Problem 27"
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== Solution == | == Solution == | ||
− | 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>. | + | <asy> |
+ | import olympiad; | ||
+ | import cse5; import geometry; size(150); defaultpen(fontsize(10pt)); defaultpen(0.8); dotfactor = 4; path circ = Circle(origin, 1); pair A = dir(degrees(7pi/12)); pair D = dir(degrees(-5pi/12)); pair B = dir(degrees(2pi/12)); pair C = dir(degrees(-2pi/12)); pair P = extension(A, B, C, D); draw(circ); draw(A--P--D); label('$A$', A, N); label('$D$', D, S); label('$C$', C, SE); label('$B$', B, NE); label('$P$', P, E); label('$60^\circ$', P, 2 * (dir(P--A) + dir(P--D))); label('$10$', A--B, S); label('$8$', B--P, NE); label('$7$', C--D, N); | ||
+ | </asy> | ||
+ | |||
+ | 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 <math>30-60-90</math> 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 == |
Latest revision as of 16:12, 7 June 2019
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 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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