1986 AIME Problems/Problem 15: Difference between revisions
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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> 3600 = 2a^2 + 5b^2 - 6ab</math> | <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> | <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 
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
and 
are perpendicular, so the product of their slopes is 
, giving
Combining 
and 
, 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 | ||
- AIME Problems and Solutions
- American Invitational Mathematics Examination
- Mathematics competition resources
These problems are copyrighted © by the Mathematical Association of America. 
