2009 AMC 12A Problem 14

Question

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?

-\dfrac{1}{3}

-\dfrac{1}{6}

\dfrac{1}{6}

\dfrac{1}{3}

\dfrac{1}{2}

The line passes through the vertex

(0, 0),

so to bisect the area it must hit the midpoint of the opposite side

Set the midpoint of the segment from

(1, 1)

to

(6m, 0)

on the line

y = mx

The line

y = mx

passes through the vertex

(0, 0),

so it bisects the triangle’s area exactly when it passes through the midpoint of the opposite side, joining

(1, 1)

and

(6m, 0).

That midpoint is

\left(\dfrac{6m + 1}{2}, \dfrac{1}{2}\right).

Requiring it to satisfy

y = mx

gives

\frac{1}{2} = m\cdot\frac{6m + 1}{2},

so

6m^2 + m - 1 = 0,

that is

(3m - 1)(2m + 1) = 0.

The possible values are

m = \dfrac{1}{3}

and

m = -\dfrac{1}{2},

whose sum is

-\dfrac{1}{6}.

Thus, the correct answer is B.

Problem 14 in other years