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 $\overline{AD}?$

$3$

$\dfrac{16}{5}$

$\dfrac{13}{4}$

$2\sqrt{3}$

$\dfrac{7}{2}$

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.

2003 AMC 12A Problem 17

Solution:

Problem 17 in Other Years