Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎See also)
(solution)
Line 1: Line 1:
−
{{empty}}
 
−
 
 
== Problem ==
 
== Problem ==
 +
In [[triangle]] <math>ABC</math>, <math>\tan \angle CAB = 22/7</math>, and the [[altitude]] from <math>A</math> divides <math>BC</math> into [[segment]]s of length 3 and 17.  What is the area of triangle <math>ABC</math>?
  
 
== Solution ==
 
== Solution ==
 +
{{image}}
 +
 +
Let <math>D</math> be the intersection of the [[altitude]] and <math>\overline{BC}</math>, and <math>h</math> be the length of the altitude. [[Without loss of generality]], let <math>BD = 17</math> and <math>CD = 3</math>. Then <math>\tan \angle DAB = \frac{17}{h}</math> and <math>\tan \angle CAD = \frac{3}{h}</math>. Using the [[tangent sum formula]],
 +
 +
<div style="text-align:center;">
 +
<math>\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*}</math></div>
 +
 +
The postive value of <math>h = 11</math>, so the area is <math>\frac{1}{2}(17 + 3)\cdot 11 = 110</math>.
  
 
== See also ==
 
== See also ==
−
* [[1988 AIME Problems]]
+
{{AIME box|year=1988|num-b=6|num-a=8}}
  
−
{{AIME box|year=1988|num-b=6|num-a=8}}
+
[[Category:Intermediate Trigonometry Problems]]

Revision as of 20:32, 26 September 2007

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

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

Let $D$ be the intersection of the altitude and $\overline{BC}$, 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{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 $h = 11$, so the area is $\frac{1}{2}(17 + 3)\cdot 11 = 110$.

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1988_AIME_Problems/Problem_7&oldid=17026"