1997 AIME Problems/Problem 11
Problem 11
Let . What is the greatest integer that does not exceed ?
Contents
Solution
Solution 1
Note that by the cofunction identities.(We could have also written it as .)
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
Solution 7
We can pair the terms of the summations as below.
From here, we use the cosine and sine subtraction formulas as shown.
\begin{align*} &\dfrac{(\cos(45-44) + \cos(45-1)) + (\cos(45-43) + \cos(45-2))+ \cdots (\cos(45-23) + \cos(45-22))}{(\sin(45-44) + \sin(45-1)) + (\sin(45-43) + \sin(45-2))+ \cdots (\sin(45-23) + \sin(45-22))} \\ &= \dfrac{(\cos{45}\cos{44} +\sin{45}\sin{44} + \cos{45}\cos{1} + \sin{45}\sin{1})+ \cdots + (\cos{45}\cos{23} + \sin{45}\sin{23} + \cos{45}\cos{22} + \sin{45}\sin{22})}{(\sin{45}\cos{44}-\cos{45}\sin{44} + \sin{45}\cos{1} - \cos{45}\sin{1}) + \cdots + (\sin{45}\cos{23} -\cos{45}\sin{23} + \sin{45}\cos{22} - \cos{45}\sin{22})} \\ &= \dfrac{\sqrt{2}/2(\cos{44} + \sin{44} + \cos{1}+\sin{1} +\cos{43} + \sin{43} + \cos{2} + \sin{2} + \cdots + \cos{23} + \sin{23} + \cos{22} + \sin{22})}{\sqrt{2}/2(\cos{44} -\sin{44} +\cos{1} - \sin{1} + \cos{43}-\sin{43} + \cos{2} -\sin{2} +\cdots + \cos{23}-\sin{23} + \cos{22} - \sin{22})} \\ &=\dfrac{\sum\limits_{n=1}^{44} \cos n^\circ + \sum\limits_{n=1}^{44} \sin n^\circ}{\sum\limits_{n=1}^{44} \cos n^\circ - \sum\limits_{n=1}^{44} \sin n^\circ}. \end{align*}
Since all of these operations have been performed to ensure equality, we can set the final line of the above mess equal to our original expression.
For the sake of clarity, let and . Then, we have
Finishing, we have . Adding to both sides gives , or . Taking the positive case gives . Finally,
~NTfish
See also
Video Solution by Osman Nal: https://www.youtube.com/watch?v=o6udScV0F_o
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.