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 $PQRS?$

$6$

$6.2$

$6.4$

$6.6$

$6.8$

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 $-\tfrac13:$ $y=3(x-4),\qquad y=-\tfrac13(x-10).$ 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.

2012 AMC 12B Problem 17

Solution:

Problem 17 in Other Years