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>.
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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 ==

Revision as of 10:42, 4 December 2017

Problem

A circle of radius $r$ has chords $\overline{AB}$ of length $10$ and $\overline{CD}$ of length 7. When $\overline{AB}$ and $\overline{CD}$ are extended through $B$ and $C$, respectively, they intersect at $P$, which is outside of the circle. If $\angle{APD}=60^\circ$ and $BP=8$, then $r^2=$

$\text{(A) } 70\quad \text{(B) } 71\quad \text{(C) } 72\quad \text{(D) } 73\quad \text{(E) } 74$

Solution

Applying Power of a Point on $P$, we find that $PC=9$ and thus $PD=16$. Observing that $PD=2BP$ and that $\angle BPD=60^{\circ}$, we conclude that $BPD$ is a $30-60-90$ right triangle with right angle at $B$. Thus, $BD=8\sqrt{3}$ and triangle $ABD$ 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 $AD=2r=2\sqrt{73}$. From here we see that $r^2=73$. The answer is thus $\fbox{D}$.

See also

1992 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 26
Followed by
Problem 28
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