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 (qpβ,srβ), 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 23β, respectively. Hence the area of ABCD=215β, and the area of triangle ADG=415β. Because AD=4, it follows that GH=815β=srβ. The equation of CD is y=β3(xβ4), so when y=815β,x=qpβ=827β. Therefore p+q+r+s=58β.
The problems on this page are the property of the MAA's American Mathematics Competitions