2003 AMC 12B Problem 22

Problem 22 of 25HarderAlgebra Geometry

Let

ABCD

be a rhombus with

AC = 16

and

BD = 30.

Let

N

be a point on

\overline{AB},

and let

P

and

Q

be the feet of the perpendiculars from

N

\overline{AC}

and

\overline{BD},

respectively. Which of the following is closest to the minimum possible value of

PQ?

Show solution

Solution

Let

O

be the intersection of the diagonals. Then

\triangle AOB

is right-angled at

O

with legs

OA = 8

and

OB = 15.

Quadrilateral

OPNQ

has right angles at

O,

P,

and

Q,

so it is a rectangle and

PQ = ON.

The minimum of

ON

is the altitude from

O

\overline{AB}

\triangle AOB.

Since

AB = \sqrt{8^2 + 15^2} = 17,

equating the two area expressions gives

\begin{aligned} ON &= \frac{OA \cdot OB}{AB} \\ &= \frac{8 \cdot 15}{17} \\ &= \frac{120}{17} \approx 7.06. \end{aligned}

This is closest to

7.

Thus, the correct answer is C.