Art of Problem Solving

1996 AIME Problems/Problem 13: Difference between revisions

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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 $ABC$, $AB=\sqrt{30}$, $AC=\sqrt{6}$, and $BC=\sqrt{15}$. There is a point $D$ for which $\overline{AD}$ bisects $\overline{BC}$, and $\angle ADB$ is a right angle. The ratio

\[\dfrac{Area(\triangle ADB)}{Area(\triangle ABC)}\]

can be written in the form $\dfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

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See also

1996 AIME (ProblemsAnswer KeyResources)
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