Difference between revisions of "1988 AIME Problems/Problem 7"
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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>. | + | 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>. |
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>. | 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>. |
Latest revision as of 23:25, 20 November 2023
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)
Call
and
. So,
and
. Using the tangent addition formula
, we get
.
Simplifying, we get . Cross-multiplying and simplifying, we get
. Factoring, we get
, so we take the positive positive solution, which is
. Therefore, the answer is
, so the answer is
.
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.