1992 AHSME Problems/Problem 27

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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

[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 $P$, 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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