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

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m (Solution)
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Let <math>D</math> be the intersection of the [[altitude]] with <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 [[Trigonometric_identities#Angle_Addition.2FSubtraction_Identities|tangent sum formula]],
 
Let <math>D</math> be the intersection of the [[altitude]] with <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 [[Trigonometric_identities#Angle_Addition.2FSubtraction_Identities|tangent sum formula]],
  
<div style="text-align:center;">
+
<cmath>
<math>\begin{eqnarray*}
+
\begin{align*}
\tan CAB &=& \tan (DAB + CAD)\\
+
\tan CAB &= \tan (DAB + CAD)\\
\frac{22}{7} &=& \frac{\tan DAB + \tan CAD}{1 - \tan DAB \cdot \tan 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{\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}\\
+
\frac{22}{7} &= \frac{20h}{h^2 - 51}\\
0 &=& 22h^2 - 140h - 22 \cdot 51\\
+
0 &= 22h^2 - 140h - 22 \cdot 51\\
0 &=& (11h + 51)(h - 11)
+
0 &= (11h + 51)(h - 11)
\end{eqnarray*}</math></div>
+
\end{align*}
 +
</cmath>
  
 
The postive value of <math>h = 11</math>, so the area is <math>\frac{1}{2}(17 + 3)\cdot 11 = 110</math>.
 
The postive value of <math>h = 11</math>, so the area is <math>\frac{1}{2}(17 + 3)\cdot 11 = 110</math>.

Revision as of 18:30, 10 March 2015

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

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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