Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution: solution by altheman)
(→‎Solution)
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>\displaystyle  3600 = 2a^2 + 5b^2 - 6ab</math>  (1)</div>
+
<math> 3600 = 2a^2 + 5b^2 - 6ab</math>  (1)</div>
  
 
AC and BC are [[perpendicular]], so the product of their [[slope]]s is -1, giving
 
AC and BC are [[perpendicular]], so the product of their [[slope]]s is -1, 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>\displaystyle 2a^2 + 5b^2 = - \frac {15}{2}ab \displaystyle </math>  (2)</div>
+
<math>2a^2 + 5b^2 = - \frac {15}{2}ab</math>  (2)</div>
  
 
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>.
+
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>
  
 
== See also ==
 
== See also ==

Revision as of 20:49, 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 $400$. $/square$

See also

1986 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Last Question
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1986_AIME_Problems/Problem_15&oldid=33064"