Problem:
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 ?
Answer Choices:
A. 6
B. 6.2
C. 6.4
D. 6.6
E. 6.8
Solution:
Let A=(3,0),B=(5,0),C=(7,0),D=(13,0), and ΞΈ be the acute angle formed by the line PQ and the x-axis. Then SR=PQ= ABcosΞΈ=2cosΞΈ, and SP=QR=CDsinΞΈ=6sinΞΈ. Because PQRS is a square, it follows that 2cosΞΈ=6sinΞΈ and tanΞΈ=31β. Therefore lines SP and RQ have slope 3, and lines SR and PQ have slope β31β. Let the points M=(4,0) and N=(10,0) be the respective midpoints of segments AB and CD. Let β1β be the line through M parallel to line SP. Let β2β be the line through N parallel to line SR. Lines β1β and β2β intersect at the center of the square PQRS. Line β1β satisfies the equation y=3(xβ4), and line β2β satisfies the equation y=β31β(xβ10). Thus the lines β1β and β2β intersect at the point (4.6,1.8), and the required sum of coordinates is 6.4β .
The problems on this page are the property of the MAA's American Mathematics Competitions
