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+1​21​​=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