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Difference between revisions of "1986 AIME Problems/Problem 15"
m (Fixed LaTeX)
Line 6: Line 6:
 
<math>AB = 60</math> so  
 
<math>AB = 60</math> so  
 
<div style="text-align:center;"><math>3600 = (a - b)^2 + (2b - a)^2</math><br />
 
<div style="text-align:center;"><math>3600 = (a - b)^2 + (2b - a)^2</math><br />
−
<math> 3600 = 2a^2 + 5b^2 - 6ab</math> (1)</div>
+
<math> 3600 = 2a^2 + 5b^2 - 6ab \ \ \ \ (1)</math></div>
  
−
AC and BC are [[perpendicular]], so the product of their [[slope]]s is -1, giving
+
<math>AC</math> and <math>BC</math> are [[perpendicular]], so the product of their [[slope]]s is <math>-1</math>, giving
 
<div style="text-align:center;"><math>\left(\frac {2a + 2b}{2a + b}\right)\left(\frac {a + 4b}{a + 2b}\right) = - 1</math><br />
 
<div style="text-align:center;"><math>\left(\frac {2a + 2b}{2a + b}\right)\left(\frac {a + 4b}{a + 2b}\right) = - 1</math><br />
−
<math>2a^2 + 5b^2 = - \frac {15}{2}ab</math> (2)</div>
+
<math>2a^2 + 5b^2 = - \frac {15}{2}ab \ \ \ \  (2)</math></div>
  
−
Combining (1) and (2), we get <math>ab = - \frac {800}{3}</math>
+
Combining <math>(1)</math> and <math>(2)</math>, 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>.
 
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>.

Revision as of 22:01, 2 April 2018

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)
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Problem 14 		Followed by
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All AIME Problems and Solutions

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