2003 AMC 12A Problem 17

Problem 17 of 25IntermediateGeometry

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}?

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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.

More practice

Concepts: coordinate geometry · circle

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.