1992 AMC 12 Problem 27

Attempt Problem 27 of the 1992 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1992 AMC 12 solutions, or check the answer key.

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27.

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 the circle. If

\angle APD=60^\circ

and

BP=8,

then

r^2=

70

71

72

73

74

Answer: D

Concepts:power of a point secants right triangle circle diameter

Difficulty rating: 2330

Small hint:

Apply the secant-secant theorem to find

PC

and

PD

Big hint:

Once

PA=18,

PC=9,

and

\angle APC=60^\circ,

identify a

30^\circ

-

60^\circ

-

90^\circ

triangle

Solution:

Power of

P

gives

PA\cdot PB=PC\cdot PD.

Here

PA=18,

PB=8,

and

PD=PC+7,

so

PC(PC+7)=144.

Thus

PC=9

and

PD=16.

Since

PA=2PC

and

\angle APC=60^\circ,

triangle

APC

is right at

C,

with

AC=9\sqrt3.

Therefore

AD

is a diameter, and

\begin{aligned} (2r)^2&=AD^2=AC^2+CD^2\\ &=243+49=292. \end{aligned}

Hence

r^2=73.

Thus the correct answer is D.