Difference between revisions of "1988 AIME Problems/Problem 7"

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The postive value of <math>h = 11</math>, so the area is <math>\frac{1}{2}(17 + 3)\cdot 11 = 110</math>.
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The postive value of <math>h = 11</math>, so the area is <math>\frac{1}{2}(17 + 3)\cdot 11 = \boxed{110}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 18:40, 23 April 2017

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

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 $\frac{1}{2}(17 + 3)\cdot 11 = \boxed{110}$.

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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