Difference between revisions of "1995 AIME Problems/Problem 7"
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== Solution 3 == | == Solution 3 == | ||
− | We want <math>1+\sin | + | We want <math>1+\sin t \cos t-\sin t-\cos t</math>. However, note that we only need to find <math>\sin t+\cos t</math>. |
− | Let <math>y = \sin | + | Let <math>y = \sin t+\cos t \rightarrow y^2 = \sin^2 t + \cos^2 t + 2\sin t \cos t = 1 + 2\sin t \cos x</math> |
− | From this we have <math>\sin | + | From this we have <math>\sin t \cos t = \frac{y^2-1}{2}</math> and <math>\sin t + \cos t = y</math> |
Substituting, we have <math>2y^2+4y-3=0 \rightarrow y = \frac {-2 \pm \sqrt{10}}{2}</math> | Substituting, we have <math>2y^2+4y-3=0 \rightarrow y = \frac {-2 \pm \sqrt{10}}{2}</math> |
Revision as of 19:44, 9 October 2020
Problem
Given that and
where and are positive integers with and relatively prime, find
Solution
From the givens, , and adding to both sides gives . Completing the square on the left in the variable gives . Since , we have . Subtracting twice this from our original equation gives , so the answer is .
Solution 2
Let . Multiplying with the given equation, , and . Simplifying and rearranging the given equation, . Notice that , and substituting, . Rearranging and squaring, , so , and , but clearly, . Therefore, , and the answer is .
Solution 3
We want . However, note that we only need to find .
Let
From this we have and
Substituting, we have
.
See also
1995 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.