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2009 AMC 12A Problems/Problem 14

Revision as of 21:56, 22 December 2020 by Smileymittens (talk | contribs) (→‎Solution)

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 $m$?

$\textbf{A} - \!\frac {1}{3} \qquad \textbf{(B)} - \!\frac {1}{6} \qquad \textbf{(C)}\ \frac {1}{6} \qquad \textbf{(D)}\ \frac {1}{3} \qquad \textbf{(E)}\ \frac {1}{2}$

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 $6m^2 + m - 1 = 0$. Using Vieta's formulas, we find that the sum of the roots is $boxed{(B) \text{ }-\frac 16}$

This is a quadratic equation with roots $m=\frac 13$ and $m=-\frac 12$. Both roots represent valid solutions, and their sum is $\frac 13 - \frac 12 = \boxed{-\frac 16}$. You could also use Vieta's formulas to find this.

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

[asy] unitsize(1cm); defaultpen(0.8); real m=1.5; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-0.5)*(1,m)) -- ((1.5)*(1,m)), dashed ); label("$A$",(0,0),NW); label("$B$",(1,1),NE); label("$C$",(6*m,0),E); [/asy]

[asy] unitsize(2cm); defaultpen(0.8); real m=0.5; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-1)*(1,m)) -- (3*(1,m)), dashed ); label("$A$",(0,0),NW); label("$B$",(1,1),N); label("$C$",(6*m,0),E); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label("$D$",D,S); [/asy]

[asy] unitsize(2cm); defaultpen(0.8); real m=1/3; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-1)*(1,m)) -- (3*(1,m)), dashed ); label("$A$",(0,0),NW); label("$B$",(1,1),N); label("$C$",(6*m,0),E); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label("$D$",D,S); [/asy]


[asy] unitsize(2cm); defaultpen(0.8); real m=-1/2; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-2)*(1,m)) -- (1*(1,m)), dashed ); label("$A$",(0,0),S); label("$B$",(1,1),NE); label("$C$",(6*m,0),W); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label("$D$",D,S); [/asy]

Solution 2

The line must pass through the triangle's centroid, since the line divides the triangle in half. The coordinates of the centroid are found by averaging those of the vertices. The slope of the line from the origin through the centroid is thus $\frac{\frac{1}{3}}{\frac{1+6m}{3}}$, which is equal to $m$.

$\frac{\frac{1}{3}}{\frac{1+6m}{3}}=m$

$\frac{1}{3}=m\left(\frac{1+6m}{3}\right)$

$1=m(1+6m)$

$6m^2+m-1=0$

Using Vieta's Formulas, the sum of the possible values of $m$ is $\boxed{\textbf{(B)}\; -\frac{1}{6}}$

See Also

2009 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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