Difference between revisions of "1997 AIME Problems/Problem 11"
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We write <math>x =\frac{\sum_{n=46}^{89} \sin n^{\circ}}{\sum_{n=1}^{44} \sin n^{\circ}}</math> since <math>\cos x = \sin (90^{\circ}-x).</math> Now we by the sine angle sum we know that <math>\sin (x+45^{\circ}) = \sin 45^{\circ}(\sin x + \cos x).</math> So the expression simplifies to <math>\sin 45^{\circ}\left(\frac{\sum_{n=1}^{44} (\sin n^{\circ}+\cos n^{\circ})}{\sum_{n=1}^{44} \sin n^{\circ}}\right) = \sin 45^{\circ}\left(1+\frac{\sum_{n=1}^{44} \cos n^{\circ}}{\sum_{n=1}^{44} \sin n^{\circ}}\right)=\sin 45^{\circ}(1+x).</math> Therefore we have the equation <math>x = \sin 45^{\circ}(1+x) \implies x = \sqrt{2}+1.</math> Finishing, we have <math>\lfloor 100x \rfloor = \boxed{241}.</math> | We write <math>x =\frac{\sum_{n=46}^{89} \sin n^{\circ}}{\sum_{n=1}^{44} \sin n^{\circ}}</math> since <math>\cos x = \sin (90^{\circ}-x).</math> Now we by the sine angle sum we know that <math>\sin (x+45^{\circ}) = \sin 45^{\circ}(\sin x + \cos x).</math> So the expression simplifies to <math>\sin 45^{\circ}\left(\frac{\sum_{n=1}^{44} (\sin n^{\circ}+\cos n^{\circ})}{\sum_{n=1}^{44} \sin n^{\circ}}\right) = \sin 45^{\circ}\left(1+\frac{\sum_{n=1}^{44} \cos n^{\circ}}{\sum_{n=1}^{44} \sin n^{\circ}}\right)=\sin 45^{\circ}(1+x).</math> Therefore we have the equation <math>x = \sin 45^{\circ}(1+x) \implies x = \sqrt{2}+1.</math> Finishing, we have <math>\lfloor 100x \rfloor = \boxed{241}.</math> |
Revision as of 15:46, 4 September 2020
Problem 11
Let . What is the greatest integer that does not exceed ?
Contents
Solution
Solution 1
Note that
Now use the sum-product formula We want to pair up , , , etc. from the numerator and , , etc. from the denominator. Then we get:
To calculate this number, use the half angle formula. Since , then our number becomes: in which we drop the negative roots (as it is clear cosine of and are positive). We can easily simplify this:
And hence our answer is
Solution 2
Using the identity , that summation reduces to
This fraction is equivalent to . Therefore,
Solution 3
A slight variant of the above solution, note that
This is the ratio we are looking for. reduces to , and .
Solution 4
Consider the sum . The fraction is given by the real part divided by the imaginary part.
The sum can be written (by De Moivre's Theorem with geometric series)
(after multiplying by complex conjugate)
Using the tangent half-angle formula, this becomes .
Dividing the two parts and multiplying each part by 4, the fraction is .
Although an exact value for in terms of radicals will be difficult, this is easily known: it is really large!
So treat it as though it were . The fraction is approximated by .
Solution 5
Consider the sum . The fraction is given by the real part divided by the imaginary part.
The sum can be written as . Consider the rhombus on the complex plane such that is the origin, represents , represents and represents . Simple geometry shows that , so the angle that makes with the real axis is simply . So is the sum of collinear complex numbers, so the angle the sum makes with the real axis is . So our answer is .
Note that the can be shown easily through half-angle formula.
Solution 6
We write since Now we by the sine angle sum we know that So the expression simplifies to Therefore we have the equation Finishing, we have
See also
1997 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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.