Difference between revisions of "1995 AIME Problems/Problem 7"
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Let <math>(1 - \sin t)(1 - \cos t) = x</math>. Multiplying <math>x</math> with the given equation, <math>\frac{5x}{4} = (1 - \sin^2 t)(1 - \cos^2 t) = \sin^2 t \cos ^2 t</math>, and <math>\frac{\sqrt{5x}}{2} = \sin t \cos t</math>. Simplifying and rearranging the given equation, <math>\sin t + \cos t = \frac{5}{4} - (\sin^2 t + \cos^2 t) - \sin t \cos t = \frac{1}{4} - \frac{\sqrt{5x}}{2}</math>. Notice that <math>(1 + \sin t)(1 + \cos t) - 2(\sin t + \cos t) = x</math>, and substituting, <math>x = \frac{5}{4} - 2( \frac{1}{4} - \frac{\sqrt{5x}}{2}) = \frac{3}{4} + \sqrt{5x}</math>. Rearranging and squaring, <math>5x = x^2 - \frac{3}{2} x + \frac{9}{16}</math>, so <math>x^2 - \frac{13}{2} x + \frac{9}{16} = 0</math>, and <math>x = \frac{13}{4} \pm \sqrt{10}</math>, but clearly, <math>0 \leq x < 4</math>. Therefore, <math>x = \frac{13}{4} - \sqrt{10}</math>, and the answer is <math> 13 + 4 + 10 = \boxed{027}</math>. | Let <math>(1 - \sin t)(1 - \cos t) = x</math>. Multiplying <math>x</math> with the given equation, <math>\frac{5x}{4} = (1 - \sin^2 t)(1 - \cos^2 t) = \sin^2 t \cos ^2 t</math>, and <math>\frac{\sqrt{5x}}{2} = \sin t \cos t</math>. Simplifying and rearranging the given equation, <math>\sin t + \cos t = \frac{5}{4} - (\sin^2 t + \cos^2 t) - \sin t \cos t = \frac{1}{4} - \frac{\sqrt{5x}}{2}</math>. Notice that <math>(1 + \sin t)(1 + \cos t) - 2(\sin t + \cos t) = x</math>, and substituting, <math>x = \frac{5}{4} - 2( \frac{1}{4} - \frac{\sqrt{5x}}{2}) = \frac{3}{4} + \sqrt{5x}</math>. Rearranging and squaring, <math>5x = x^2 - \frac{3}{2} x + \frac{9}{16}</math>, so <math>x^2 - \frac{13}{2} x + \frac{9}{16} = 0</math>, and <math>x = \frac{13}{4} \pm \sqrt{10}</math>, but clearly, <math>0 \leq x < 4</math>. Therefore, <math>x = \frac{13}{4} - \sqrt{10}</math>, and the answer is <math> 13 + 4 + 10 = \boxed{027}</math>. | ||
− | ==Solution 3== (-synergy) | + | == Solution 3 == (-synergy) |
We have <math>1+\sin x\cos x+\sin x+\cos x = \frac{5}{4}</math> | We have <math>1+\sin x\cos x+\sin x+\cos x = \frac{5}{4}</math> |
Revision as of 20:35, 3 January 2019
Contents
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 == (-synergy)
We have
We want to find
If we find , we will be done with the problem.
Let
Squaring, we have
From this we have and
Substituting this into the first equation 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.