2012 AMC 12B Problem 17

Question

Square

PQRS

lies in the first quadrant. Points

(3, 0),

(5, 0),

(7, 0),

and

(13, 0)

lie on lines

SP,

RQ,

PQ,

and

SR,

respectively. What is the sum of the coordinates of the center of the square

PQRS?

6

6.2

6.4

6.6

6.8

Answer: C

Concepts:coordinate geometry slope system of equations

Difficulty rating: 1910

Small hint:

Let

\theta

be the angle line

PQ

makes with the

x

-axis; the square condition relates the projections of the given segments

Big hint:

From

2\cos\theta=6\sin\theta

get

\tan\theta=\tfrac13,

then intersect the two lines through the midpoints

(4,0)

and

(10,0)

Solution:

Let

\theta

be the acute angle line

PQ

makes with the

x

-axis. Sides

SR=PQ

span the segment from

(3,0)

to

(5,0)

as

2\cos\theta,

while

SP=QR

span the segment from

(7,0)

to

(13,0)

as

6\sin\theta.

Since the square has equal sides,

2\cos\theta=6\sin\theta,

so

\tan\theta=\tfrac13.

Thus lines

SP,RQ

have slope

3

and lines

SR,PQ

have slope

-\tfrac13.

The center lies on the line through

(4,0)

with slope

3

and the line through

(10,0)

with slope

-\tfrac13:

y=3(x-4),

y=-\tfrac13(x-10).

These meet at

(4.6,1.8).

The sum of the coordinates is

4.6+1.8=6.4.

Thus, the correct answer is C.