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

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== Solution ==
 
== Solution ==
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{{solution}}
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Translate so the [[median]]s are <math>y = x</math>, and <math>y = 2x</math>, then model the [[point]]s <math>A: (a,a)</math> and <math>B: (b,2b)</math>. <math>(0,0)</math> is the [[centroid]], and is the average of the vertices, so <math>C: (- a - b, - a - 2b)</math><br />
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<math>AB = 60</math> so
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<div style="text-align:center;"><math>3600 = (a - b)^2 + (2b - a)^2</math><br />
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<math>\displaystyle  3600 = 2a^2 + 5b^2 - 6ab</math>  (1)</div>
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AC and BC are [[perpendicular]], so the product of their [[slope]]s is -1, giving
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<div style="text-align:center;"><math>\left(\frac {2a + 2b}{2a + b}\right)\left(\frac {a + 4b}{a + 2b}\right) = - 1</math><br />
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<math>\displaystyle 2a^2 + 5b^2 = - \frac {15}{2}ab \displaystyle </math>  (2)</div>
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Combining (1) and (2), we get <math>ab = - \frac {800}{3}</math>
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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 16:46, 14 September 2007

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

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

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

$\left(\frac {2a + 2b}{2a + b}\right)\left(\frac {a + 4b}{a + 2b}\right) = - 1$
$\displaystyle 2a^2 + 5b^2 = - \frac {15}{2}ab \displaystyle$ (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 (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Last Question
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All AIME Problems and Solutions
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