# Difference between revisions of "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

Let $D$ be the intersection of the altitude with $\overline{BC}$, 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{align*} \tan CAB &= \tan (DAB + CAD)\\ \frac{22}{7} &= \frac{\tan DAB + \tan CAD}{1 - \tan DAB \cdot \tan CAD} \\ &=\frac{\frac{17}{h} + \frac{3}{h}}{1 - \left(\frac{17}{h}\right)\left(\frac{3}{h}\right)} \\ \frac{22}{7} &= \frac{20h}{h^2 - 51}\\ 0 &= 22h^2 - 140h - 22 \cdot 51\\ 0 &= (11h + 51)(h - 11) \end{align*}

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