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

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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]],
  
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<div style="text-align:center;">
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<cmath>
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<math>\begin{eqnarray*}
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\begin{align*}
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\tan CAB &=& \tan (DAB + CAD)\\
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\tan CAB &= \tan (DAB + CAD)\\
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\frac{22}{7} &=& \frac{\tan DAB + \tan CAD}{1 - \tan DAB \cdot \tan CAD} \\
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\frac{22}{7} &= \frac{\tan DAB + \tan CAD}{1 - \tan DAB \cdot \tan CAD} \\
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&=& \frac{\frac{17}{h} + \frac{3}{h}}{1 - \left(\frac{17}{h}\right)\left(\frac{3}{h}\right)} \\
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&=\frac{\frac{17}{h} + \frac{3}{h}}{1 - \left(\frac{17}{h}\right)\left(\frac{3}{h}\right)} \\
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\frac{22}{7} &=& \frac{20h}{h^2 - 51}\\
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\frac{22}{7} &= \frac{20h}{h^2 - 51}\\
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0 &=& 22h^2 - 140h - 22 \cdot 51\\
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0 &= 22h^2 - 140h - 22 \cdot 51\\
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0 &=& (11h + 51)(h - 11)
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0 &= (11h + 51)(h - 11)
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\end{eqnarray*}</math></div>
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\end{align*}
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</cmath>
  
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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 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 ==
 
== See also ==

Revision as of 19:20, 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

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

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

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