1986 AIME Problems/Problem 15

Problem

Let triangle $ABC$ be a right triangle in the xy-plane with a right angle at $C_{}$ . Given that the length of the hypotenuse $AB$ is $60$ , and that the medians through $A$ and $B$ lie along the lines $y=x+3$ and $y=2x+4$ respectively, find the area of triangle $ABC$ .

Solution

Translate so the medians are $y = x$ , and $y = 2x$ , then model the points $A: (a,a)$ and $B: (b,2b)$ . $(0,0)$ is the centroid, and is the average of the vertices, so $C: (- a - b, - a - 2b)$
$AB = 60$ so

$3600 = (a - b)^2 + (2b - a)^2$
$3600 = 2a^2 + 5b^2 - 6ab \ \ \ \ (1)$

$AC$ and $BC$ are perpendicular, so the product of their slopes is $-1$ , giving

$\left(\frac {2a + 2b}{2a + b}\right)\left(\frac {a + 4b}{a + 2b}\right) = - 1$
$2a^2 + 5b^2 = - \frac {15}{2}ab \ \ \ \ (2)$

Combining $(1)$ and $(2)$ , we get $ab = - \frac {800}{3}$

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 $\left|\frac {3}{2}ab\right|$ , so we get the answer to be \boxed{400}$

1986 AIME (Problems • Answer Key • Resources)
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1986 AIME Problems/Problem 15

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