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

Answer: A

Concepts:tangent line power of a point circle

Difficulty rating: 2000

Small Hint:

Let the radius be $r$ and write $AD$ in terms of $AO$ and $r$

Big Hint:

Use the tangent relation $AB^2=AD\cdot AE$ together with $AB=2r$

Solution:

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.

1960 AMC 12 Problem 32

Answer: A

Difficulty rating: 2000

Small Hint:

Big Hint:

Solution:

Problem 32 in Other Years