1988 AIME Problems/Problem 7

Revision as of 23:20, 20 November 2023 by Arcticturn (talk | contribs) (Solution)

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

AIME 1988 Solution 07.png

Call $\angle BAD$ $\alpha$ and $\angle CAD$ $\beta$. So, $\tan \alpha = \frac {17}{h}$ and $\tan \beta = \frac {3}{h}$. Using the tangent addition formula $\tan (\alpha + \beta) = \frac {\tan \alpha + \tan \beta}{1 - \tan \alpha \cdot \tan \ beta}$, we get $\dfrac {\frac {20}{h}}{\frac {x^2 - 51}{x^2}}$.

See also

1988 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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