Difference between revisions of "1996 AIME Problems/Problem 13"
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<cmath>\dfrac{Area(\triangle ADB)}{Area(\triangle ABC)}</cmath> | <cmath>\dfrac{Area(\triangle ADB)}{Area(\triangle ABC)}</cmath> | ||
| − | can be written in the form <math>\dfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math> | + | can be written in the form <math>\dfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. |
== Solution == | == Solution == | ||
Revision as of 12:58, 12 November 2007
Problem
In triangle 
, 
, 
, and 
. There is a point 
for which 
bisects 
, and 
is a right angle. The ratio
![\[\dfrac{Area(\triangle ADB)}{Area(\triangle ABC)}\]](http://latex.artofproblemsolving.com/d/e/8/de8053c594f285593dc7e5ba0d32da9388b4dcf1.png)
can be written in the form 
, where 
and 
are relatively prime positive integers. Find 
.
Solution
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See also
| 1996 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
