1976 AMC 12 Problem 18

Attempt Problem 18 of the 1976 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 1976 AMC 12 solutions, or check the answer key.

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18.

In the adjoining figure,

AB

is tangent at

A

to the circle with center

O;

point

D

is interior to the circle; and

DB

intersects the circle at

C.

BC=DC=3,

OD=2

and

AB=6,

then the radius of the circle is

3+\sqrt3

\frac{15}{\pi}

\frac92

2\sqrt6

\sqrt{22}

Extend

BD

through

D

to meet the circle again at

E

Use

BC\cdot BE=AB^2

first, then use the power of interior point

D

Extend

BD

through

D

to meet the circle at

E.

Since

BC=3

and

BD=6,

the tangent-secant theorem gives

3\cdot BE=AB^2=36,

BE=12

and

DE=6.

The power of

D

gives

\begin{aligned} DC\cdot DE&=(R-OD)(R+OD)\\ &=R^2-OD^2, \end{aligned}

hence

3\cdot6=(R-2)(R+2)=R^2-4.

Thus

R^2=22

and

R=\sqrt{22}.

Therefore, the correct answer is E.

Problem 18 in other years