Problem:
Let points A=(0,0),B=(1,2),C=(3,3), and D=(4,0). Quadrilateral ABCD is cut into equal area pieces by a line passing through A. This line intersects CD at point (qp​,sr​), where these fractions are in lowest terms. What is p+q+r+s ?
Answer Choices:
A. 54
B. 58
C. 62
D. 70
E. 75
Solution:
Let line AG be the required line, with G on CD. Divide ABCD into triangle ABF, trapezoid BCEF, and triangle CDE, as shown. Their areas are 1,5 , and 23​, respectively. Hence the area of ABCD=215​, and the area of triangle ADG=415​. Because AD=4, it follows that GH=815​=sr​. The equation of CD is y=−3(x−4), so when y=815​,x=qp​=827​. Therefore p+q+r+s=58​.
The problems on this page are the property of the MAA's American Mathematics Competitions
