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

Revision as of 18:20, 27 February 2018

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

@@ Line 18: / Line 18: @@
 </cmath>
-The postive value of <math>h = 11</math>, so the area is <math>\frac{1}{2}(17 + 3)\cdot 11 = \boxed{110}</math>.
+The positive value of <math>h</math> is <math>11</math>, so the area is <math>\frac{1}{2}(17 + 3)\cdot 11 = \boxed{110}</math>.
 == See also ==

1988 AIME (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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Difference between revisions of "1988 AIME Problems/Problem 7"

Revision as of 18:20, 27 February 2018

Problem

Solution

See also