1988 AIME Problems/Problem 7

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 $\tan (\alpha + \beta) = \dfrac {\tan \alpha + \tan \beta}{1 - \tan \alpha \cdot \tan \beta}$, we get $\tan (\alpha + \beta) = \dfrac {\frac {20}{h}}{\frac {h^2 - 51}{h^2}} = \frac {22}{7}$.

Simplifying, we get $\frac {20h}{h^2 - 51} = \frac {22}{7}$. Cross-multiplying and simplifying, we get $11h^2-70h-561 = 0$. Factoring, we get $(11h+51)(h-11) = 0$, so we take the positive positive solution, which is $h = 11$. Therefore, the answer is $\frac {20 \cdot 11}{2} = 110$, so the answer is $\boxed {110}$.

~Arcticturn

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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