Difference between revisions of "1992 AHSME Problems/Problem 27"
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== Solution == | == Solution == | ||
| − | <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
![[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]](http://latex.artofproblemsolving.com/f/e/8/fe8fc9260e9126d00f90b8d19c6ca9ec9c459dd4.png)
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 | |
| 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. 
