2003 AMC 12A Problem 17

Question

Square

ABCD

has sides of length

4,

and

M

is the midpoint of

\overline{CD}.

A circle with radius

2

and center

M

intersects a circle with radius

4

and center

A

at points

P

and

D.

What is the distance from

P

to

\overline{AD}?

3

\dfrac{16}{5}

\dfrac{13}{4}

2\sqrt{3}

\dfrac{7}{2}

Answer: B

Concepts:coordinate geometry circle

Difficulty rating: 1730

Small hint:

Put

D

at the origin with

A

on the

y

-axis and

C

on the

x

-axis

Big hint:

The distance from

P

to

\overline{AD}

is the

x

-coordinate of

P

Solution:

Place

D=(0,0),

C=(4,0),

and

A=(0,4).

The circle centered at

M=(2,0)

is

(x-2)^2+y^2=4,

and the circle centered at

A

is

x^2+(y-4)^2=16.

Solving these equations gives the intersection

P=\left(\dfrac{16}{5},\dfrac85\right).

Since

\overline{AD}

lies on the

y

-axis, the distance from

P

to

\overline{AD}

is its

x

-coordinate,

\dfrac{16}{5}.

Thus, the correct answer is B.