1988 AIME Problems/Problem 7: Difference between revisions

Revision as of 19:12, 15 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 $ABC$ ?

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 == Solution ==
 <center>[[Image:AIME_1988_Solution_07.png]]</center>
-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]],
-<cmath>
-\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}\\
-&= 22h^2 - 140h - 22 \cdot 51\\
-&= (11h + 51)(h - 11)
-\end{align*}
-</cmath>
-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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1988 AIME Problems/Problem 7: Difference between revisions

Revision as of 19:12, 15 November 2023

Problem

Solution

See also