1960 AMC 12 Problem 32

Question

In this figure the center of the circle is

O.

\overline{AB}\perp\overline{BC},

ADOE

is a straight line,

\overline{AP}=\overline{AD},

and

\overline{AB}

has a length twice the radius. Then:

\overline{AP}^{\,2}=\overline{PB}\cdot\overline{AB}

\overline{AP}\cdot\overline{DO}=\overline{PB}\cdot\overline{AD}

\overline{AB}^{\,2}=\overline{AD}\cdot\overline{DE}

\overline{AB}\cdot\overline{AD}=\overline{OB}\cdot\overline{AO}

none of these

Let the radius be

r

and write

AD

in terms of

AO

and

r

Use the tangent relation

AB^2=AD\cdot AE

together with

AB=2r

Let the radius be

r.

Since

AB

is tangent at

B,

AB^2=AD\cdot AE.

Write

AD=t.

Because the secant passes through the center,

AE=t+2r,

while

AB=2r.

Hence

t(t+2r)=4r^2.

Also

AP=AD=t

and

PB=AB-AP=2r-t.

The displayed equation rearranges to

t^2=2r(2r-t)=PB\cdot AB.

Therefore

AP^2=PB\cdot AB.

Thus, the correct answer is A.

Problem 32 in other years