1986 AIME Problems/Problem 15
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
![$3600 = (a - b)^2 + (2b - a)^2$](http://latex.artofproblemsolving.com/1/8/a/18a313e3956fe0bcd1660ecabc9ac19b9aa56b2a.png)
![$3600 = 2a^2 + 5b^2 - 6ab \ \ \ \ (1)$](http://latex.artofproblemsolving.com/e/d/b/edb50bd2bb2dbca072addb484ed9837d12d12621.png)
and
are perpendicular, so the product of their slopes is
, giving
![$\left(\frac {2a + 2b}{2a + b}\right)\left(\frac {a + 4b}{a + 2b}\right) = - 1$](http://latex.artofproblemsolving.com/e/7/2/e72bb70cf41d0dccad96abfa6cd34625f432fe32.png)
![$2a^2 + 5b^2 = - \frac {15}{2}ab \ \ \ \ (2)$](http://latex.artofproblemsolving.com/d/3/9/d39bb21c6df6873302b21d1001209d1fa0a9c0a6.png)
Combining and
, 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 | ||
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