1988 AIME Problems/Problem 7
Problem
In triangle , , and the altitude from divides into segments of length 3 and 17. What is the area of triangle ?
Solution
Let be the intersection of the altitude with , and be the length of the altitude. Without loss of generality, let and . Then and . Using the tangent sum formula,
$\begin{eqnarray*} \tan CAB &=& \tan (DAB + CAD)\\ \frac{22}{7} &=& \frac{\tan DAB + \tan CAD}{1 - \tan DAB \cdot \tan CAD} \\ &=& \frac{\frac{17}{h} + \frac{3}{h}}{1 - \left(\frac{17}{h}\right)\left(\frac{3}{h}\right)} \\ \frac{22}{7} &=& \frac{20h}{h^2 - 51}\\ 0 &=& 22h^2 - 140h - 22 \cdot 51\\ 0 &=& (11h + 51)(h - 11)
\end{eqnarray*}$ (Error compiling LaTeX. ! Missing \endgroup inserted.)The postive value of , so the area is .
See also
1988 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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