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

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<center>[[Image:AIME_1988_Solution_07.png]]</center>
 
<center>[[Image:AIME_1988_Solution_07.png]]</center>
  
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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]],
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Call <math>\angle BAD</math> <math>\alpha</math> and <math>\angle CAD</math> <math>\beta</math>. So, <math>\tan \alpha = \frac {17}{h}</math> and <math>\tan \beta = \frac {3}{h}</math>. Using the tangent addition formula <math>\tan (\alpha + \beta) = \dfrac {\tan \alpha + \tan \beta}{1 - \tan \alpha \cdot \tan \beta}</math>, we get <math>\tan (\alpha + \beta) = \dfrac {\frac {20}{h}}{\frac {h^2 - 51}{h^2}} = \frac {22}{7}</math>.  
  
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<cmath>
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Simplifying, we get <math>\frac {20h}{h^2 - 51} = \frac {22}{7}</math>. Cross-multiplying and simplifying, we get <math>11h^2-70h-561 = 0</math>. Factoring, we get <math>(11h+51)(h-11) = 0</math>, so we take the positive positive solution, which is <math>h = 11</math>. Therefore, the answer is <math>\frac {20 \cdot 11}{2} = 110</math>, so the answer is <math>\boxed {110}</math>.
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\begin{align*}
 
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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{\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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0 &= 22h^2 - 140h - 22 \cdot 51\\
 
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0 &= (11h + 51)(h - 11)
 
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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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~Arcticturn
  
 
== See also ==
 
== See also ==

Latest revision as of 22:25, 20 November 2023

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

Call $\angle BAD$ $\alpha$ and $\angle CAD$ $\beta$. So, $\tan \alpha = \frac {17}{h}$ and $\tan \beta = \frac {3}{h}$. Using the tangent addition formula $\tan (\alpha + \beta) = \dfrac {\tan \alpha + \tan \beta}{1 - \tan \alpha \cdot \tan \beta}$, we get $\tan (\alpha + \beta) = \dfrac {\frac {20}{h}}{\frac {h^2 - 51}{h^2}} = \frac {22}{7}$.

Simplifying, we get $\frac {20h}{h^2 - 51} = \frac {22}{7}$. Cross-multiplying and simplifying, we get $11h^2-70h-561 = 0$. Factoring, we get $(11h+51)(h-11) = 0$, so we take the positive positive solution, which is $h = 11$. Therefore, the answer is $\frac {20 \cdot 11}{2} = 110$, so the answer is $\boxed {110}$.

~Arcticturn

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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