Difference between revisions of "1991 AIME Problems/Problem 9"
(solution, though I'm sure there is a more elegant manner of going about this) |
m (→Solution: fmt) |
||
| Line 5: | Line 5: | ||
Use the two [[Trigonometric identities#Pythagorean Identities|trigonometric Pythagorean identities]] <math>\displaystyle 1 + \tan^2 x = \sec^2 x</math> and <math>\displaystyle 1 + \cot^2 x = \csc^2 x</math>. | Use the two [[Trigonometric identities#Pythagorean Identities|trigonometric Pythagorean identities]] <math>\displaystyle 1 + \tan^2 x = \sec^2 x</math> and <math>\displaystyle 1 + \cot^2 x = \csc^2 x</math>. | ||
| − | If we square <math>\sec x = \frac{22}{7} - \tan x</math>, we find that <math>\sec^2 x = (\frac{22}7)^2 - 2(\frac{22}7)\tan x + \tan^2 x</math>, so <math>1 = (\frac{22}7)^2 - \frac{44}7 \tan x</math>. Solving shows that <math>\tan x = \frac{435}{308}</math>. | + | If we square <math>\sec x = \frac{22}{7} - \tan x</math>, we find that <math>\sec^2 x = \left(\frac{22}7\right)^2 - 2\left(\frac{22}7\right)\tan x + \tan^2 x</math>, so <math>1 = \left(\frac{22}7\right)^2 - \frac{44}7 \tan x</math>. Solving shows that <math>\tan x = \frac{435}{308}</math>. |
| − | Call <math>y = \frac mn</math>. Rewrite the second equation in a similar fashion: <math>\displaystyle 1 = y^2 - 2y\cot x</math>. Substitute in <math>\cot x = \frac{1}{\tan x} = \frac{308}{435}</math> to get a [[quadratic equation|quadratic]]: <math>0 = y^2 - \frac{616}{435} - 1</math>. The quadratic is [[factor]]able (though somewhat ugly); <math>(15y - 29)(29y + 15) = 0</math>. It turns out that only the [[positive]] root will work, so the value of <math>y = \frac{29}{15}</math> and <math>m + n = 044</math>. | + | Call <math>y = \frac mn</math>. Rewrite the second equation in a similar fashion: <math>\displaystyle 1 = y^2 - 2y\cot x</math>. Substitute in <math>\cot x = \frac{1}{\tan x} = \frac{308}{435}</math> to get a [[quadratic equation|quadratic]]: <math>0 = y^2 - \frac{616}{435} - 1</math>. The quadratic is [[factor]]able (though somewhat ugly); <math>(15y - 29)(29y + 15) = 0</math>. It turns out that only the [[positive]] root will work, so the value of <math>y = \frac{29}{15}</math> and <math>m + n = 044</math>. |
== See also == | == See also == | ||
Revision as of 20:11, 11 March 2007
Problem
Suppose that 
and that 
where 
is in lowest terms. Find 
Solution
Use the two trigonometric Pythagorean identities 
and 
.
If we square 
, we find that 
, so 
. Solving shows that 
.
Call 
. Rewrite the second equation in a similar fashion: 
. Substitute in 
to get a quadratic: 
. The quadratic is factorable (though somewhat ugly); 
. It turns out that only the positive root will work, so the value of 
and 
.
See also
| 1991 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
