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. Unknown error_msg)The postive value of , so the area is
.
See also
1988 AIME (Problems • Answer Key • Resources) | ||
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.