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

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== Problem ==
 
== Problem ==
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Let [[triangle]] <math>\displaystyle ABC</math> be a [[right triangle]] in the xy-plane with a right angle at <math>\displaystyle C_{}</math>. Given that the length of the [[hypotenuse]] <math>\displaystyle AB</math> is <math>\displaystyle 60</math>, and that the [[median]]s through <math>\displaystyle A</math> and <math>\displaystyle B</math> lie along the lines <math>\displaystyle y=x+3</math> and <math>\displaystyle y=2x+4</math> respectively, find the area of triangle <math>\displaystyle ABC</math>.
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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]]
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Revision as of 18:05, 4 July 2013

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 $400$.

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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