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]], | ||
− | < | + | <cmath> |
− | + | \begin{align*} | |
− | \tan CAB &= | + | \tan CAB &= \tan (DAB + CAD)\\ |
− | \frac{22}{7} &= | + | \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{22}{7} &= \frac{20h}{h^2 - 51}\\ |
− | 0 &= | + | 0 &= 22h^2 - 140h - 22 \cdot 51\\ |
− | 0 &= | + | 0 &= (11h + 51)(h - 11) |
− | \end{ | + | \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 ,
, and the altitude from
divides
into segments of length 3 and 17. What is the area of triangle
?
Solution
![AIME 1988 Solution 07.png](https://wiki-images.artofproblemsolving.com//d/d6/AIME_1988_Solution_07.png)
Let be the intersection of the altitude with
, and
be the length of the altitude. Without loss of generality, let
and
. Then
and
. Using the tangent sum formula,
The postive value of , so the area is
.
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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