2012 AMC 12B Problem 17

Problem 17 of 25IntermediateAlgebra Geometry

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?

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Solution

Let

\theta

be the acute angle line

PQ

makes with the

x

-axis. Sides

SR=PQ

span the segment from

(3,0)

(5,0)

2\cos\theta,

while

SP=QR

span the segment from

(7,0)

(13,0)

6\sin\theta.

Since the square has equal sides,

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

\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.

Tagged: coordinate geometry · slope · system of equations