Difference between revisions of "1991 AIME Problems/Problem 9"
Mathcool2009 (talk | contribs) m (→Solution 3) |
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so <math>m + n = 044</math>. | so <math>m + n = 044</math>. | ||
− | === Solution 3 === | + | === Solution 3 (least computation)=== |
− | + | By the given, | |
<math>\frac {1}{\cos x} + \frac {\sin x}{\cos x} = \frac {22}{7}</math> and | <math>\frac {1}{\cos x} + \frac {\sin x}{\cos x} = \frac {22}{7}</math> and | ||
<math>\frac {1}{\sin x} + \frac {\cos x}{\sin x} = k</math>. | <math>\frac {1}{\sin x} + \frac {\cos x}{\sin x} = k</math>. | ||
Line 46: | Line 46: | ||
<cmath>1 = \frac {22}{7}k - \frac {22}{7} - k.</cmath> | <cmath>1 = \frac {22}{7}k - \frac {22}{7} - k.</cmath> | ||
− | Solving yields <math>k = \frac {29}{15}</math>, and <math>m+n = 044</math> | + | Solving yields <math>k = \frac {29}{15}</math>, and <math>m+n = 044</math> |
=== Solution 4 === | === Solution 4 === |
Revision as of 14:27, 23 June 2014
Problem
Suppose that and that where is in lowest terms. Find
Contents
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.