Difference between revisions of "1997 AIME Problems/Problem 11"

Latest revision as of 16:05, 8 September 2024

Problem

Let $x=\frac{\sum\limits_{n=1}^{44} \cos n^\circ}{\sum\limits_{n=1}^{44} \sin n^\circ}$ . What is the greatest integer that does not exceed $100x$ ?

Solution 1

Note that $\frac{\sum_{n=1}^{44} \cos n}{\sum_{n=1}^{44} \sin n} = \frac{\sum_{n=1}^{44} \cos n}{\sum_{n=46}^{89} \cos n} = \frac {\cos 1 + \cos 2 + \dots + \cos 44}{\cos 89 + \cos 88 + \dots + \cos 46}$ by the cofunction identities.(We could have also written it as $\frac{\sum_{n=1}^{44} \cos n}{\sum_{n=1}^{44} \sin n} = \frac{\sum_{n=46}^{89} \sin n}{\sum_{n=1}^{44} \sin n} = \frac {\sin 89 + \sin 88 + \dots + \sin 46}{\sin 1 + \sin 2 + \dots + \sin 44}$ .)

Now use the sum-product formula $\cos x + \cos y = 2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)$ We want to pair up $[1, 44]$ , $[2, 43]$ , $[3, 42]$ , etc. from the numerator and $[46, 89]$ , $[47, 88]$ , $[48, 87]$ etc. from the denominator. Then we get: $\frac{\sum_{n=1}^{44} \cos n}{\sum_{n=1}^{44} \sin n} = \frac{\sum_{n=1}^{44} \cos n}{\sum_{n=46}^{89} \cos n} = \frac{2\cos(\frac{45}{2})[\cos(\frac{43}{2})+\cos(\frac{41}{2})+\dots+\cos(\frac{1}{2})]}{2\cos(\frac{135}{2})[\cos(\frac{43}{2})+\cos(\frac{41}{2})+\dots+\cos(\frac{1}{2})]} \Rightarrow \frac{\cos(\frac{45}{2})}{\cos(\frac{135}{2})}$

To calculate this number, use the half angle formula. Since $\cos\left(\frac{x}{2}\right) = \pm \sqrt{\frac{\cos x + 1}{2}}$ , then our number becomes: $\frac{\sqrt{\frac{\frac{\sqrt{2}}{2} + 1}{2}}}{\sqrt{\frac{\frac{-\sqrt{2}}{2} + 1}{2}}}$ in which we drop the negative roots (as it is clear cosine of $22.5$ and $67.5$ are positive). We can easily simplify this:

$\begin{eqnarray*} \frac{\sqrt{\frac{\frac{\sqrt{2}}{2} + 1}{2}}}{\sqrt{\frac{\frac{-\sqrt{2}}{2} + 1}{2}}} &=& \sqrt{\frac{\frac{2+\sqrt{2}}{4}}{\frac{2-\sqrt{2}}{4}}} \\ &=& \sqrt{\frac{2+\sqrt{2}}{2-\sqrt{2}}} \cdot \sqrt{\frac{2+\sqrt{2}}{2+\sqrt{2}}} \\ &=& \sqrt{\frac{(2+\sqrt{2})^2}{2}} \\ &=& \frac{2+\sqrt{2}}{\sqrt{2}} \\ &=& \sqrt{2}+1 \end{eqnarray*}$

And hence our answer is $\lfloor 100x \rfloor = \lfloor 100(1 + \sqrt {2}) \rfloor = \boxed{241}$ .

Solution 2

$\begin{eqnarray*} x &=& \frac {\sum_{n = 1}^{44} \cos n^\circ}{\sum_{n = 1}^{44} \sin n^\circ} = \frac {\cos 1 + \cos 2 + \dots + \cos 44}{\sin 1 + \sin 2 + \dots + \sin 44}\\ &=& \frac {\cos (45 - 1) + \cos(45 - 2) + \dots + \cos(45 - 44)}{\sin 1 + \sin 2 + \dots + \sin 44} \end{eqnarray*}$

Using the identity $\sin a + \sin b = 2\sin \frac{a+b}2 \cos \frac{a-b}{2}$ $\Longrightarrow \sin x + \cos x$ $= \sin x + \sin (90-x)$ $= 2 \sin 45 \cos (45-x)$ $= \sqrt{2} \cos (45-x)$ , that summation reduces to

$\begin{eqnarray*}x &=& \left(\frac {1}{\sqrt {2}}\right)\left(\frac {(\cos 1 + \cos2 + \dots + \cos44) + (\sin1 + \sin2 + \dots + \sin44)}{\sin1 + \sin2 + \dots + \sin44}\right)\\ &=& \left(\frac {1}{\sqrt {2}}\right)\left(1 + \frac {\cos 1 + \cos 2 + \dots + \cos 44}{\sin 1 + \sin 2 + \dots + \sin 44}\right) \end{eqnarray*}$

This fraction is equivalent to $x$ . Therefore, $\begin{eqnarray*} x &=& \left(\frac {1}{\sqrt {2}}\right)\left(1 + x\right)\\ \frac {1}{\sqrt {2}} &=& x\left(\frac {\sqrt {2} - 1}{\sqrt {2}}\right)\\ x &=& \frac {1}{\sqrt {2} - 1} = 1 + \sqrt {2}\\ \lfloor 100x \rfloor &=& \lfloor 100(1 + \sqrt {2}) \rfloor = \boxed{241}\\ \end{eqnarray*}$

Solution 3

A slight variant of the above solution, note that

$\begin{eqnarray*} \sum_{n=1}^{44} \cos n + \sum_{n=1}^{44} \sin n &=& \sum_{n=1}^{44} \sin n + \sin(90-n)\\ &=& \sqrt{2}\sum_{n=1}^{44} \cos(45-n) = \sqrt{2}\sum_{n=1}^{44} \cos n\\ \sum_{n=1}^{44} \sin n &=& (\sqrt{2}-1)\sum_{n=1}^{44} \cos n \end{eqnarray*}$

This is the ratio we are looking for. $x$ reduces to $\frac{1}{\sqrt{2} - 1} = \sqrt{2} + 1$ , and $\lfloor 100(\sqrt{2} + 1)\rfloor = \boxed{241}$ .

Solution 4

Consider the sum $\sum_{n = 1}^{44} \text{cis } n^\circ$ . The fraction is given by the real part divided by the imaginary part.

The sum can be written $- 1 + \sum_{n = 0}^{44} \text{cis } n^\circ = - 1 + \frac {\text{cis } 45^\circ - 1}{\text{cis } 1^\circ - 1}$ (by De Moivre's Theorem with geometric series)

$= - 1 + \frac {\frac {\sqrt {2}}{2} - 1 + \frac {i \sqrt {2}}{2}}{\text{cis } 1^\circ - 1} = - 1 + \frac {\left( \frac {\sqrt {2}}{2} - 1 + \frac {i \sqrt {2}}{2} \right) (\text{cis } ( - 1^\circ) - 1)}{(\cos 1^\circ - 1)^2 + \sin^2 1^\circ}$ (after multiplying by complex conjugate)

$= - 1 + \frac {\left( \frac {\sqrt {2}}{2} - 1 \right) (\cos 1^\circ - 1) + \frac {\sqrt {2}}{2}\sin 1^\circ + i\left( \left(1 - \frac {\sqrt {2}}{2} \right) \sin 1^\circ + \frac {\sqrt {2}}{2} (\cos 1^\circ - 1)\right)}{2(1 - \cos 1^\circ)}$

$= - \frac {1}{2} - \frac {\sqrt {2}}{4} - \frac {i\sqrt {2}}{4} + \frac {\sin 1^\circ \left( \frac {\sqrt {2}}{2} + i\left( 1 - \frac {\sqrt {2}}{2} \right) \right)}{2(1 - \cos 1^\circ)}$

Using the tangent half-angle formula, this becomes $\left( - \frac {1}{2} + \frac {\sqrt {2}}{4}[\cot (1/2^\circ) - 1] \right) + i\left( \frac {1}{2}\cot (1/2^\circ) - \frac {\sqrt {2}}{4}[\cot (1/2^\circ) + 1] \right)$ .

Dividing the two parts and multiplying each part by 4, the fraction is $\frac { - 2 + \sqrt {2}[\cot (1/2^\circ) - 1]}{2\cot (1/2^\circ) - \sqrt {2}[\cot (1/2^\circ) + 1]}$ .

Although an exact value for $\cot (1/2^\circ)$ in terms of radicals will be difficult, this is easily known: it is really large!

So treat it as though it were $\infty$ . The fraction is approximated by $\frac {\sqrt {2}}{2 - \sqrt {2}} = \frac {\sqrt {2}(2 + \sqrt {2})}{2} = 1 + \sqrt {2}\Rightarrow \lfloor 100(1+\sqrt2)\rfloor=\boxed{241}$ .

Solution 5

Consider the sum $\sum_{n = 1}^{44} \text{cis } n^\circ$ . The fraction is given by the real part divided by the imaginary part.

The sum can be written as $\sum_{n=1}^{22} (\text{cis } n^\circ + \text{cis } 45-n^\circ)$ . Consider the rhombus $OABC$ on the complex plane such that $O$ is the origin, $A$ represents $\text{cis } n^\circ$ , $B$ represents $\text{cis } n^\circ + \text{cis } 45-n^\circ$ and $C$ represents $\text{cis } n^\circ$ . Simple geometry shows that $\angle BOA = 22.5-k^\circ$ , so the angle that $\text{cis } n^\circ + \text{cis } 45-n^\circ$ makes with the real axis is simply $22.5^\circ$ . So $\sum_{n=1}^{22} (\text{cis } n^\circ + \text{cis } 45-n^\circ)$ is the sum of collinear complex numbers, so the angle the sum makes with the real axis is $22.5^\circ$ . So our answer is $\lfloor 100 \cot(22.5^\circ) \rfloor = \boxed{241}$ .

Note that the $\cot(22.5^\circ) = \sqrt2 + 1$ can be shown easily through half-angle formula.

Solution 6

We write $x =\frac{\sum_{n=46}^{89} \sin n^{\circ}}{\sum_{n=1}^{44} \sin n^{\circ}}$ since $\cos x = \sin (90^{\circ}-x).$ Now we by the sine angle sum we know that $\sin (x+45^{\circ}) = \sin 45^{\circ}(\sin x + \cos x).$ So the expression simplifies to $\sin 45^{\circ}\left(\frac{\sum_{n=1}^{44} (\sin n^{\circ}+\cos n^{\circ})}{\sum_{n=1}^{44} \sin n^{\circ}}\right) = \sin 45^{\circ}\left(1+\frac{\sum_{n=1}^{44} \cos n^{\circ}}{\sum_{n=1}^{44} \sin n^{\circ}}\right)=\sin 45^{\circ}(1+x).$ Therefore we have the equation $x = \sin 45^{\circ}(1+x) \implies x = \sqrt{2}+1.$ Finishing, we have $\lfloor 100x \rfloor = \boxed{241}.$

Solution 7

We can pair the terms of the summations as below.

$\dfrac{(\cos{1} + \cos{44}) + (\cos{2} + \cos{43}) + (\cos{3} + \cos{42}) + \cdots + (\cos{22} + \cos{23})}{(\sin{1} + \sin{44}) + (\sin{2} + \sin{43}) + (\sin{3} + \sin{42}) + \cdots + (\sin{22} + \sin{23})}.$

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.

$\dfrac{\sum\limits_{n=1}^{44} \cos n^\circ + \sum\limits_{n=1}^{44} \sin n^\circ}{\sum\limits_{n=1}^{44} \cos n^\circ - \sum\limits_{n=1}^{44} \sin n^\circ} = \frac{\sum\limits_{n=1}^{44} \cos n^\circ}{\sum\limits_{n=1}^{44} \sin n^\circ}.$

For the sake of clarity, let $\sum\limits_{n=1}^{44} \cos n^\circ = C$ and $\sum\limits_{n=1}^{44} \sin n^\circ = S$ . Then, we have

$\dfrac{C+S}{C-S} = \dfrac{C}{S} \implies CS+S^2 = C^2-CS.$

Finishing, we have $S^2+2CS=C^2$ . Adding $C^2$ to both sides gives $(C+S)^2 = 2C^2$ , or $C+S = \pm C\sqrt{2}$ . Taking the positive case gives $S= C(\sqrt{2}-1)$ . Finally,

$x=\frac{\sum\limits_{n=1}^{44} \cos n^\circ}{\sum\limits_{n=1}^{44} \sin n^\circ} = \dfrac{C}{S} = \dfrac{1}{\sqrt{2}-1} = \sqrt{2} +1 \implies \lfloor{100x\rfloor} = \boxed{241}.$

~NTfish

Solution 8 (Integral Calculus, not rigorous)

Because the numerator and denominator are both sums, we can replace each with $44$ times the average (i.e. arithmetic mean) of the terms of the respective sums. These $44$ s will cancel out, leaving simply the ratio of the averages of the respective sums.

Given that the sums span from $1^{\circ}\approx 0$ radians to $44^{\circ}\approx \tfrac\pi4$ radians, we would expect the average heights of cosine and sine on $[0,\tfrac\pi4]$ to be very close if not equal to the given ratio. We can use definite integrals to calculate these averages.

The average height of $\cos x$ on $[0,\tfrac\pi4]$ is $\frac1{\pi/4}\int^{\pi/4}_0(\cos x) dx = \frac4\pi(\sin\frac\pi4-\sin0) = \frac4\pi\cdot\frac{\sqrt2}2$ .

Similarly, the average height of $\sin x$ on $[0,\tfrac\pi4]$ is $\frac1{\pi/4}\int^{\pi/4}_0(\sin x) dx = \frac4\pi(-\cos\frac\pi4+\cos0) = \frac4\pi\cdot(1-\frac{\sqrt2}2)$ .

Taking the ratio of these two values yields $\frac{\sqrt2/2}{1-\sqrt2/2}=\frac1{\sqrt2-1}\cdot\frac{\sqrt2+1}{\sqrt2+1}=\frac{\sqrt2+1}{2-1}=\sqrt2+1 \approx 1.41+1 = 2.41$ . Thus, our answer is $\boxed{241}$ .

@@ Line 114: / Line 114: @@
 ~NTfish
-== Solution 8 ==
+== Solution 8 (Integral Calculus, not rigorous) ==
+Because the numerator and denominator are both sums, we can replace each with <math>44</math> times the average (i.e. [[arithmetic mean]]) of the terms of the respective sums. These <math>44</math>s will cancel out, leaving simply the ratio of the averages of the respective sums.
+Given that the sums span from <math>1^{\circ}\approx 0</math> radians to <math>44^{\circ}\approx \tfrac\pi4</math> radians, we would expect the average heights of cosine and sine on <math>[0,\tfrac\pi4]</math> to be very close if not equal to the given ratio. We can use definite [[integrals]] to calculate these averages.
+The average height of <math>\cos x</math> on <math>[0,\tfrac\pi4]</math> is <math>\frac1{\pi/4}\int^{\pi/4}_0(\cos x) dx = \frac4\pi(\sin\frac\pi4-\sin0) = \frac4\pi\cdot\frac{\sqrt2}2</math>.
+Similarly, the average height of <math>\sin x</math> on <math>[0,\tfrac\pi4]</math> is <math>\frac1{\pi/4}\int^{\pi/4}_0(\sin x) dx = \frac4\pi(-\cos\frac\pi4+\cos0) = \frac4\pi\cdot(1-\frac{\sqrt2}2)</math>.
+Taking the ratio of these two values yields <math>\frac{\sqrt2/2}{1-\sqrt2/2}=\frac1{\sqrt2-1}\cdot\frac{\sqrt2+1}{\sqrt2+1}=\frac{\sqrt2+1}{2-1}=\sqrt2+1 \approx 1.41+1 = 2.41</math>. Thus, our answer is <math>\boxed{241}</math>.
 == 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

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