2009 AMC 12A Problems/Problem 14

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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 $m$ ?

$\textbf{(A)} - \!\frac {1}{3} \qquad \textbf{(B)} - \!\frac {1}{6} \qquad \textbf{(C)}\ \frac {1}{6} \qquad \textbf{(D)}\ \fr...$

Solution

Let's label the three points as $A=(0,0)$ , $B=(1,1)$ , and $C=(6m,0)$ .

Clearly, whenever the line $y=mx$ intersects the inside of the triangle, it will intersect the side $BC$ . Let $D$ be the point of intersection.

The triangles $ABD$ and $ACD$ have the same height, which is the distance between the point $A$ and the line $BC$ . Hence they have equal areas if and only if $D$ is the midpoint of $BC$ .

The midpoint of the segment $BC$ has coordinates $\left( \frac{6m+1}2, \frac 12 \right)$ . This point lies on the line $y=mx$ if and only if $\frac 12 = m \cdot \frac{6m+1}2$ . This simplifies to $6m^2 + m - 1 = 0$ . This is a quadratic equation with roots $m=\frac 13$ and $m=-\frac 12$ . Both roots represent valid solutions, and their sum is $\frac 13 - \frac 12 = \boxed{-\frac 16}$ .

For illustration, below are pictures of the situation for $m=1.5$ , $m=0.5$ , $m=1/3$ , and $m=-1/2$ .

2009 AMC 12A (Problems • Resources)
Preceded by Problem 13	Followed by Problem 15
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2009 AMC 12A Problems/Problem 14

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Problem

Solution

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