Problem:
A triangle has vertices (0,0),(1,1), and (6m,0), and the line y=mx divides the triangle into two triangles of equal area. What is the sum of all possible values of m?
Answer Choices:
A. −31
B. −61
C. 61
D. 31
E. 21
Solution:
The line must contain the midpoint of the segment joining (1,1) and (6m,0), which is (26m+1,21). Thus
m=26m+121=6m+11
from which 0=6m2+m−1=(3m−1)(2m+1). The two possible values of m are −21 and 31, and their sum is −61.
If m=−21 then the triangle with vertices (0,0),(1,1), and (−3,0) is bisected by the line passing through the origin and (−1,21). Similarly, when m=31 the triangle with vertices (0,0),(1,1), and (2,0) is bisected by the line passing through the origin and (23,21).
The problems on this page are the property of the MAA's American Mathematics Competitions