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Difference between revisions of "2009 AMC 12A Problems/Problem 14"

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==Solution 2==
 
==Solution 2==
  
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The line must pass through the triangle's centroid, whose coordinates are found by averaging those of the vertices. The slope of the line from the origin through the centroid is thus <math>\frac{\frac{1}{3}}{\frac{1+6m}{3}}</math>, which is equal to <math>m</math>. Solving this equation, we arrive at the answer: <math>\boxed{-\frac{1}{6}}</math>.
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The line must pass through the triangle's centroid, whose coordinates are found by averaging those of the vertices. The slope of the line from the origin through the centroid is thus <math>\frac{\frac{1}{3}}{\frac{1+6m}{3}}</math>, which is equal to <math>m</math>. This becomes a quadratic, and using Viete's Formulas, we get our answer, <math>\boxed{-\frac 1 6}</math>
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC12 box|year=2009|ab=A|num-b=13|num-a=15}}
 
{{AMC12 box|year=2009|ab=A|num-b=13|num-a=15}}

Revision as of 23:24, 8 February 2012

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

[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, whose coordinates 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 $m$. This becomes a quadratic, and using Viete's Formulas, we get our answer, $\boxed{-\frac 1 6}$

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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