AIME 1999 · 第 11 题

Solution 1

Let $s = \sum_{k=1}^{35}\sin 5k = \sin 5 + \sin 10 + \ldots + \sin 175$. We could try to manipulate this sum by wrapping the terms around (since the first half is equal to the second half), but it quickly becomes apparent that this way is difficult to pull off. Instead, we look to telescope the sum. Using the identity $\sin a \sin b = \frac 12(\cos (a-b) - \cos (a+b))$, we can rewrite $s$ as

$$\begin{aligned} s \cdot \sin 5 = \sum_{k=1}^{35} \sin 5k \sin 5 &= \sum_{k=1}^{35} \frac{1}{2}(\cos (5k - 5)- \cos (5k + 5))\\ &= 0.5(\cos 0 - \cos 10 + \cos 5 - \cos 15 + \cos 10 \ldots + \cos 165 - \cos 175+ \cos 170 - \cos 180) \end{aligned}$$ This telescopes to

$$s = \frac{\cos 0 + \cos 5 - \cos 175 - \cos 180}{2 \sin 5} = \frac{1 + \cos 5}{\sin 5}.$$ Manipulating this to use the identity $\tan x = \frac{1 - \cos 2x}{\sin 2x}$, we get

$$s = \frac{1 - \cos 175}{\sin 175} \Longrightarrow s = \tan \frac{175}{2},$$ and our answer is $\boxed{177}$.

Solution 2

This solution will work entirely in degrees. Let

$$z = e^{5^\circ i}.$$ For each integer $k$, consider the unit vector in the complex plane

$$v_k = z^k = e^{5k^\circ i}.$$ Geometrically, $v_k$ is a unit vector making an angle of $5k ^\circ$ with the positive real axis. Now draw each vector $v_k$ for $1 \leq k \leq 35$ tip-to-tail on the complex plane. Begin at $O$, the origin, and draw the vector $v_1$. Then starting at where $v_1$ ended, draw $v_2$. Continue until you draw $v_{35}$, and the sum of these $35$ vectors will end at $B$, as seen in the diagram below. Because each successive vector is obtained by rotating the previous vector by $5^\circ$, we have a portion of a regular polygon with an exterior angle of $5$ degrees, so it has $\frac{360^\circ}{5^\circ} = 72$ sides.

Additionally, notice that $B$ is directly above $A$. This is because the vector sum will have its horizontal components completely cancel out. We realize that $\cos{5} + \cos{175} = 0$, and we can form a similar pair for each of the other complex numbers, except for $v_18 = i$.

The above diagram starts at the origin $O$ and draws the $35$ vectors.

Note that

$$\text{Im}(v_k) = \sin{5k}.$$ Therefore, $\sum_{k=1}^{35}\sin 5k$ is simply the y-coordinate of the vector sum $v_1 + v_2 + \dots v_{35}$. In the diagram, this is the distance of the vertical blue line connecting $O$ to the topmost point.

Let $A$ be the vertex of the $72$-gon that is directly to the left of $O$ and $B$ be the resulting vertex of summing the $35$ vectors. Looking at right triangle $AOB$, we notice that $AO=1$. Remember that our desired sum equals $BO = \tan{\angle BAO}.$ We now just need to find $\angle BAO$.

Notice that $BA$ bisects $A$'s interior angle, so $\angle BAO$ is half of the interior angle of the $72$-gon. So $\angle BAO = \frac{175}{2}$.

Recall that our sum equals $\tan{\frac{m}{n}} = \tan{\frac{175}{2}}$, so our answer is $175 + 2 = \boxed{177}$.

~lprado

Solution 3

We note that $\sin x = \text{Im } e^{ix}\text{*}$. We thus have that

$$\begin{aligned} \sum_{k = 1}^{35} \sin 5k &= \sum_{k = 1}^{35} \text{Im } e^{5ki}\\ &= \text{Im } \sum_{k = 1}^{35} e^{5ki}\\ &= \text{Im } \frac{e^{5i}(1 - e^{180i})}{1 - e^{5i}}\\ &= \text{Im } \frac{2\cos5 + 2i \sin 5}{(1 - \cos 5) - i \sin 5}\\ &= \text{Im } \frac{(2 \cos 5 + 2i \sin 5)[(1 - \cos 5) + i \sin 5]}{(1 - \cos 5)^2 + \sin^2 5}\\ &= \frac{2 \sin 5}{2 - 2 \cos 5}\\ &= \frac{\sin 5}{1 - \cos 5}\\ &= \frac{\sin 175}{1 + \cos 175} \\ &= \tan \frac{175}{2}.\\ \end{aligned}$$ The desired answer is thus $175 + 2 = \boxed{177}$.

• Only if $x$ is in radians, which it is not. However, the solution is still viable, so keep reading.

Solution 4

Let $x=e^{\frac{i\pi}{36}}$. By Euler's Formula, $\sin{5k^\circ}=\frac{x^k-\frac{1}{x^{k}}}{2i}$.

The sum we want is thus $\frac{x-\frac{1}{x}}{2i}+\frac{x^2-\frac{1}{x^{2}}}{2i}+\cdots+\frac{x^{35}-\frac{1}{x^{35}}}{2i}$

We factor the $\frac{1}{2i}$ and split into two geometric series to get $\frac{1}{2i}\left(\frac{-\frac{1}{x^{35}}(x^{35}-1)}{x-1}+\frac{x(x^{35}-1)}{x-1}\right)$

However, we note that $x^{36}=-1$, so $-\frac{1}{x^{35}}=x$, so our two geometric series are actually the same. We combine the terms and simplify to get $\frac{1}{i}\left(\frac{x^{36}-x}{x-1}\right)$

Apply Euler's identity and simplify again to get $\frac{1}{i}\left(\frac{-x-1}{x-1}\right)$

Now, we need to figure out how to express this as the tangent of something. We note that $\tan(5k^\circ)=\frac{\sin(5k^\circ)}{\cos(5k^\circ)}=\frac{\frac{x^k-\frac{1}{x^k}}{2i}}{\frac{x^k+\frac{1}{x^k}}{2}}=\frac{1}{i}\frac{x^{2k}-1}{x^{2k}+1}$.

So, we set the two equal to each other to solve for $k$. Cross multiplying gets $(-x-1)(x^{2k}+1)=(x-1)(x^{2k}-1)$. Expanding yields $-x^{2k+1}-x-x^{2k}-1=x^{2k+1}-x-x^{2k}+1$. Simplifying yields $x^{2k+1}=-1$. Since $2k+1=36$ is the smallest solution, we have $k=\frac{35}{2}$, and the argument of tangent is $5k=\frac{175}{2}$. The requested sum is $175+2=\boxed{177}$.

Solution 5

This solution is inspired by 1997 AIME Problems/Problem 11. First of all, $\sum_{k=1}^{35}\sin 5k=\sum_{k=0}^{35}\sin 5k$ because the difference is $\sin 0 = 0$. Now consider the sum $\sum_{k=0}^{35}\cos 5k$. Since $\cos 5=-\cos 175$, $\cos 10=-\cos 170$ and so on, this simplifies to $\cos 0=1$. Now, consider the ratio

$$\dfrac{\sum\limits_{n=0}^{35}\sin 5k}{\sum\limits_{n=0}^{35}\cos 5k.}$$ Using the sum-to-product identities, the numerator can be written as

$$\sin \frac{175}{2} (\cos \frac{175}{2} + \cos \frac {165}{2}+\cdots + \cos \frac{5}{2}.)$$ Similarly (via sum-to-product identities), the denominator is equivalent to

$$\cos \frac{175}{2} (\cos \frac{175}{2} + \cos \frac {165}{2}+\cdots + \cos \frac{5}{2}.)$$ If we substitute these equivalent expressions into the original ratio, many parts cancel. We are left with

$$\frac{\sin \frac{175}{2}}{\cos \frac{175}{2}} = \tan{\frac{175}{2}}.$$ Aha! Remembering that the original denominator $\sum_{k=0}^{35}\cos 5k$ was equal to 1, we realize that the numerator, $\sum_{k=0}^{35}\sin 5k=\tan{\frac{175}{2}}$. Since $\sum_{k=0}^{35}\sin 5k=\sum_{k=1}^{35}\sin 5k$, we get $\tan \frac{175}{2}$, or $m+n=\boxed{177}$, as our final answer. $\blacksquare$ ~ewei12