Difference between revisions of "1986 AIME Problems/Problem 15"
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== Problem == | == Problem == | ||
− | Let [[triangle]] <math> | + | Let [[triangle]] <math>ABC</math> be a [[right triangle]] in the xy-plane with a right angle at <math>C_{}</math>. Given that the length of the [[hypotenuse]] <math>AB</math> is <math>60</math>, and that the [[median]]s through <math>A</math> and <math>B</math> lie along the lines <math>y=x+3</math> and <math>y=2x+4</math> respectively, find the area of triangle <math>ABC</math>. |
== Solution == | == Solution == | ||
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[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 19:05, 4 July 2013
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 |
- AIME Problems and Solutions
- American Invitational Mathematics Examination
- Mathematics competition resources
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.