Difference between revisions of "2009 AMC 12A Problems/Problem 14"
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==Solution 2== | ==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 <math>\frac{\frac{1}{3}}{\frac{1+6m}{3}}</math>, which is equal to <math>m</math>. | + | 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 22:24, 8 February 2012
Contents
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 . This is a quadratic equation with roots and . Both roots represent valid solutions, and their sum is .
For illustration, below are pictures of the situation for , , , and .
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 , which is equal to . This becomes a quadratic, and using Viete's Formulas, we get our answer,
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 |