2009 AMC 12A Problems/Problem 14
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 . You could also use Vieta's formulas to find this.
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 |
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All AMC 12 Problems and Solutions |
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