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

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

1988 AIME (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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1988 AIME Problems/Problem 7

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