Difference between revisions of "1992 AHSME Problems/Problem 27"
(Created page with "== Problem == A circle of radius <math>r</math> has chords <math>\overline{AB}</math> of length <math>10</math> and <math>\overline{CD}</math> of length 7. When <math>\overline{...") |
(→Solution) |
||
Line 10: | Line 10: | ||
== 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 09:51, 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 | |
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 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.