Difference between revisions of "1991 AIME Problems/Problem 9"
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<cmath>\csc^2 x = (y - \cot^2 x)^2 \Longrightarrow 1 = y^2 - 2y\cot x.</cmath> | <cmath>\csc^2 x = (y - \cot^2 x)^2 \Longrightarrow 1 = y^2 - 2y\cot x.</cmath> | ||
− | Substituting <math>\cot x = \frac{1}{\tan x} = \frac{308}{435}</math> yields a [[quadratic equation]]: <math>0 = 435y^2 - | + | Substituting <math>\cot x = \frac{1}{\tan x} = \frac{308}{435}</math> yields a [[quadratic equation]]: <math>0 = 435y^2 - 616y - 435 = 15y - 29)(29y + 15)</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 = \boxed{044}</math>. |
=== Solution 2 === | === Solution 2 === |
Revision as of 22:34, 18 March 2012
Problem
Suppose that and that where is in lowest terms. Find
Contents
[hide]Solution
Solution 1
Use the two trigonometric Pythagorean identities and .
If we square the given , we find that
This yields .
Let . Then squaring,
Substituting yields a quadratic equation: . It turns out that only the positive root will work, so the value of and .
Solution 2
Recall that , from which we find that . Adding the equations
together and dividing by 2 gives , and subtracting the equations and dividing by 2 gives . Hence, and . Thus, and . Finally,
so .
Solution 3
(least computation) By the given, and .
Multiplying the two, we have
Subtracting both of the two given equations from this, and simpliyfing with the identity , we get
Solving yields , and
Solution 4
Make the substitution (a substitution commonly used in calculus). , so . Now note the following:
Plugging these into our equality gives:
This simplifies to , and solving for gives , and . Finally, .
Solution 5
We are given that , or equivalently, . Note that what we want is just .
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 |