Difference between revisions of "2009 AMC 12A Problems/Problem 14"
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A triangle has vertices <math>(0,0)</math>, <math>(1,1)</math>, and <math>(6m,0)</math>, and the line <math>y = mx</math> divides the triangle into two triangles of equal area. What is the sum of all possible values of <math>m</math>? | A triangle has vertices <math>(0,0)</math>, <math>(1,1)</math>, and <math>(6m,0)</math>, and the line <math>y = mx</math> divides the triangle into two triangles of equal area. What is the sum of all possible values of <math>m</math>? | ||
− | <math>\textbf{ | + | <math>\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}</math> |
== Solution == | == Solution == | ||
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Hence they have equal areas if and only if <math>D</math> is the midpoint of <math>BC</math>. | Hence they have equal areas if and only if <math>D</math> is the midpoint of <math>BC</math>. | ||
− | The midpoint of the segment <math>BC</math> has coordinates <math>\left( \frac{6m+1}2, \frac 12 \right)</math>. This point lies on the line <math>y=mx</math> if and only if <math>\frac 12 = m \cdot \frac{6m+1}2</math>. This simplifies to <math>6m^2 + m - 1 = 0</math>. | + | The midpoint of the segment <math>BC</math> has coordinates <math>\left( \frac{6m+1}2, \frac 12 \right)</math>. This point lies on the line <math>y=mx</math> if and only if <math>\frac 12 = m \cdot \frac{6m+1}2</math>. This simplifies to <math>6m^2 + m - 1 = 0</math>. Using Vieta's formulas, we find that the sum of the roots is <math>\boxed{\textbf{(B)} - \!\frac {1}{6}}</math>. |
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For illustration, below are pictures of the situation for <math>m=1.5</math>, <math>m=0.5</math>, <math>m=1/3</math>, and <math>m=-1/2</math>. | For illustration, below are pictures of the situation for <math>m=1.5</math>, <math>m=0.5</math>, <math>m=1/3</math>, and <math>m=-1/2</math>. | ||
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<math>\frac{\frac{1}{3}}{\frac{1+6m}{3}}=m</math> | <math>\frac{\frac{1}{3}}{\frac{1+6m}{3}}=m</math> | ||
− | <math>\frac{1}{3}=m(\frac{1+6m}{3})</math> | + | <math>\frac{1}{3}=m\left(\frac{1+6m}{3}\right)</math> |
<math>1=m(1+6m)</math> | <math>1=m(1+6m)</math> | ||
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<math>6m^2+m-1=0</math> | <math>6m^2+m-1=0</math> | ||
− | Using Vieta's Formulas, the sum of the possible values of <math>m</math> is <math>\boxed{\textbf{(B | + | Using Vieta's Formulas, the sum of the possible values of <math>m</math> is <math>\boxed{\textbf{(B)}\; -\frac{1}{6}}</math> |
== See Also == | == See Also == |
Latest revision as of 21:58, 22 December 2020
Contents
[hide]Problem
A triangle has vertices , , and , and the line divides the triangle into two triangles of equal area. What is the sum of all possible values of ?
Solution
Let's label the three points as , , and .
Clearly, whenever the line intersects the inside of the triangle, it will intersect the side . Let be the point of intersection.
The triangles and have the same height, which is the distance between the point and the line . Hence they have equal areas if and only if is the midpoint of .
The midpoint of the segment has coordinates . This point lies on the line if and only if . This simplifies to . Using Vieta's formulas, we find that the sum of the roots is .
For illustration, below are pictures of the situation for , , , and .
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 , which is equal to .
Using Vieta's Formulas, the sum of the possible values of is
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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