Difference between revisions of "1991 AIME Problems/Problem 9"
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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>. | 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>. | ||
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+ | Note: The problem is much easier computed if we consider what sec x is, then find the relationship between sin x and cos x (using tan x = <math>\frac{435}{308}</math>, and then computing csc x + cot x using 1/sin x and then the reciprocal of tan x. | ||
=== Solution 2 === | === Solution 2 === |
Revision as of 20:12, 15 July 2019
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 .
Note: The problem is much easier computed if we consider what sec x is, then find the relationship between sin x and cos x (using tan x = , and then computing csc x + cot x using 1/sin x and then the reciprocal of tan x.
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 .
Solution 6
Assign a right triangle with angle , hypotenuse , adjacent side , and opposite side . Then, through the given information above, we have that..
Hence, because similar right triangles can be scaled up by a factor, we can assume that this particular right triangle is indeed in simplest terms.
Hence, ,
Furthermore, by the Pythagorean Theorem, we have that
Solving for in the first equation and plugging in into the second equation...
Hence,
Now, we want
Plugging in, we find the answer is
Hence, the answer is
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.