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1988 AIME Problems/Problem 7

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Problem

In triangle ABC, \tan \angle CAB = 22/7, and the altitude from A divides BC into segments of length 3 and 17. What is the area of triangle ABC?

Solution

Image:AIME_1988_Solution_07.png

Let D be the intersection of the altitude and \overline{BC}, and h be the length of the altitude. Without loss of generality, let BD = 17 and CD = 3. Then \tan \angle DAB = \frac{17}{h} and \tan \angle CAD = \frac{3}{h}. Using the tangent sum formula,

\begin{eqnarray*}\tan CAB &=& \tan (DAB + CAD)\\\frac{22}{7} &=& \frac{\tan DAB + \tan CAD}{1 - \tan DAB \cdo...

The postive value of h = 11, so the area is \frac{1}{2}(17 + 3)\cdot 11 = 110.

See also

1988 AIME (ProblemsResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Looking for a challenging geometry text? Preparing for MATHCOUNTS or the AMC exams? Check out Art of Problem Solving's Introduction to Geometry by Richard Rusczyk.
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