Difference between revisions of "1986 AIME Problems/Problem 15"

(Solution)
(Solution)
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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>. <math>/square</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 21:50, 19 December 2009

Problem

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

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 $400$.

See also

1986 AIME (ProblemsAnswer KeyResources)
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