Difference between revisions of "1986 AIME Problems/Problem 15"
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<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> | + | <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> | + | <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 21:49, 19 December 2009
Problem
Let triangle be a right triangle in the xy-plane with a right angle at
. Given that the length of the hypotenuse
is
, and that the medians through
and
lie along the lines
and
respectively, find the area of triangle
.
Solution
Translate so the medians are , and
, then model the points
and
.
is the centroid, and is the average of the vertices, so
so
![$3600 = (a - b)^2 + (2b - a)^2$](http://latex.artofproblemsolving.com/1/8/a/18a313e3956fe0bcd1660ecabc9ac19b9aa56b2a.png)
![$3600 = 2a^2 + 5b^2 - 6ab$](http://latex.artofproblemsolving.com/0/8/6/0869465d44effb909c77b496764859a1cfbd6685.png)
AC and BC are perpendicular, so the product of their slopes is -1, giving
![$\left(\frac {2a + 2b}{2a + b}\right)\left(\frac {a + 4b}{a + 2b}\right) = - 1$](http://latex.artofproblemsolving.com/e/7/2/e72bb70cf41d0dccad96abfa6cd34625f432fe32.png)
![$2a^2 + 5b^2 = - \frac {15}{2}ab$](http://latex.artofproblemsolving.com/0/8/2/0827ab0c201a3b6c4e47bd90cf6c6bdfc6418086.png)
Combining (1) and (2), we get
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 , so we get the answer to be
.
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 |