Difference between revisions of "2009 AMC 12A Problems/Problem 14"
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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>. Using Vieta's formulas, we find that the sum of the roots is <math>\boxed{(B) | + | 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>. |
Revision as of 22:57, 22 December 2020
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
. Using Vieta's formulas, we find that the sum of the roots is $\boxed{\textbf{(B)} - \!\frac {1}{6}$ (Error compiling LaTeX. Unknown error_msg).
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.