Difference between revisions of "1986 AIME Problems/Problem 15"
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Combining (1) and (2), we get <math>ab = - \frac {800}{3}</math> | Combining (1) and (2), we get <math>ab = - \frac {800}{3}</math> | ||
− | Using the [[determinant]] product for area of a triangle (this simplifies nicely, add columns 1 and 2, add rows 2 and 3), the area is <math>\left|\frac {3}{2}ab\right|</math>, so we get the answer to be <math>400</math>. | + | Using the [[determinant]] product for area of a triangle (this simplifies nicely, add columns 1 and 2, add rows 2 and 3), the area is <math>\left|\frac {3}{2}ab\right|</math>, so we get the answer to be <math>400</math>. |
== See also == | == See also == |
Revision as of 20:50, 19 December 2009
Problem
Let triangle be a right triangle in the xy-plane with a right angle at . Given that the length of the hypotenuse is , and that the medians through and lie along the lines and respectively, find the area of triangle .
Solution
Translate so the medians are , and , then model the points and . is the centroid, and is the average of the vertices, so
so
(1)
AC and BC are perpendicular, so the product of their slopes is -1, giving
(2)
Combining (1) and (2), we get
Using the determinant product for area of a triangle (this simplifies nicely, add columns 1 and 2, add rows 2 and 3), the area is , so we get the answer to be .
See also
1986 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |