Problem:
The parallelogram bounded by the lines y=ax+c,y=ax+d,y=bx+c, and y=bx+d has area 18. The parallelogram bounded by the lines y=ax+c, y=ax−d,y=bx+c, and y=bx−d has area 72. Given that a,b,c, and d are positive integers, what is the smallest possible value of a+b+c+d ?
Answer Choices:
A. 13
B. 14
C. 15
D. 16
E. 17
Solution:
Two vertices of the first parallelogram are at (0,c) and (0,d). The x-coordinates of the other two vertices satisfy ax+c=bx+d and ax+d= bx+c, so the x-coordinates are ±(c−d)/(b−a). Thus the parallelogram is composed of two triangles, each of which has area
9=21⋅∣c−d∣⋅∣∣∣∣∣b−ac−d∣∣∣∣∣.
It follows that (c−d)2=18∣b−a∣. By a similar argument using the second parallelogram, (c+d)2=72∣b−a∣. Subtracting the first equation from the second yields 4cd=54∣b−a∣, so 2cd=27∣b−a∣. Thus ∣b−a∣ is even, and a+b is minimized when {a,b}={1,3}. Also, cd is a multiple of 27 , and c+d is minimized when {c,d}={3,9}. Hence the smallest possible value of a+b+c+d is 1+3+3+9=16. Note that the required conditions are satisfied when (a,b,c,d)=(1,3,3,9).
The problems on this page are the property of the MAA's American Mathematics Competitions