2015 AMC 12B Problems/Problem 25

Problem

A bee starts flying from point $P_0$ . She flies $1$ inch due east to point $P_1$ . For $j \ge 1$ , once the bee reaches point $P_j$ , she turns $30^{\circ}$ counterclockwise and then flies $j+1$ inches straight to point $P_{j+1}$ . When the bee reaches $P_{2015}$ she is exactly $a \sqrt{b} + c \sqrt{d}$ inches away from $P_0$ , where $a$ , $b$ , $c$ and $d$ are positive integers and $b$ and $d$ are not divisible by the square of any prime. What is $a+b+c+d$ ?

$\textbf{(A)}\; 2016 \qquad\textbf{(B)}\; 2024 \qquad\textbf{(C)}\; 2032 \qquad\textbf{(D)}\; 2040 \qquad\textbf{(E)}\; 2048$

Solution 1

Let $x = e^{i \pi / 6}$ , a $30^\circ$ counterclockwise rotation centered at the origin. Notice that $P_k$ on the complex plane is:

$1 + 2x + 3x^2 + \cdots + (k+1)x^k$

We need to find the magnitude of $P_{2015}$ on the complex plane. This is an arithmetic/geometric series.

$\begin{align*} S &= 1 + 2x + 3x^2 + \cdots + 2015x^{2014} \\ xS &= x + 2x^2 + 3x^3 + \cdots + 2015x^{2015} \\ (1-x)S &= 1 + x + x^2 + \cdots + x^{2014} - 2015x^{2015} \\ S &= \frac{1 - x^{2015} }{(1-x)^2} - \frac{2015x^{2015}}{1-x} \end{align*}$

We want to find $|S|$ . First, note that $x^{2015} = x^{11} = x^{-1}$ because $x^{12} = 1$ . Therefore

$S = \frac{1 - \frac{1}{x}}{(1-x)^2} - \frac{2015}{x(1-x)} = -\frac{1}{x(1-x)} - \frac{2015}{x(1-x)} = -\frac{2016}{x(1-x)}.$

Hence, since $|x|=1$ , we have $|S| = \frac{2016}{|1-x|}.$

Now we just have to find $|1-x|$ . This can just be computed directly:

$1 - x = 1 - \frac{\sqrt{3}}{2} - \frac{1}{2}i$

$|1-x|^2 = \left(1 - \sqrt{3} + \frac{3}{4} \right) + \frac{1}{4} = 2 - \sqrt{3} = {\left( \frac{\sqrt{6}-\sqrt{2}}{2} \right)}^2$

$|1-x| = \frac{\sqrt{6} - \sqrt{2}}{2}.$

Therefore $|S| = 2016 \cdot \frac{2}{\sqrt{6} -\sqrt{2}} = 2016 \left( \frac{\sqrt{6} + \sqrt{2}}{2} \right) = 1008 \sqrt{2} + 1008 \sqrt{6}.$

Thus the answer is $1008 + 1008 + 2 + 6 = \boxed{\textbf{(B)}\; 2024}.$

Solution 2

Here is an alternate solution that does not use complex numbers:

We will calculate the distance from $P_{2015}$ to $P_0$ using the Pythagorean theorem. Assume $P_0$ lies at the origin, so we will calculate the distance to $P_{2015}$ by calculating the distance traveled in the x-direction and the distance traveled in the y-direction. We can calculate this by summing each movement:

$x=1\cos{0}+2\cos {30}+3\cos {60}+4\cos {90}+5\cos {120}+\cdot \cdot \cdot+2011\cos{180}+2012\cos {210}+2013\cos{240}+2014\cos{270}+2015\cos{300}$

A movement of $p$ units at $q$ degrees is the same thing as a movement of $-p$ units at $q-180$ degrees, so we can adjust all the cosines with arguments greater than 180 as follows:

$x=1\cos{0}+2\cos {30}+3\cos {60}+4\cos {90}+5\cos {120}+6\cos{150}-7\cos{0}-8\cos{30}-\cdot \cdot \cdot -2015 \cos{120}$

Now we group terms with like-cosines and factor out the cosines:

$x=(1-7+13-\cdot \cdot \cdot +2005-2011)\cos{0}+(2-8+14- \cdot \cdot \cdot +2006-2012)\cos{30}+\cdot \cdot \cdot +(5-11+17- \cdot \cdot \cdot +2008-2014)\cos{120}+(6-12+18- \cdot \cdot \cdot -2004+2010)\cos{150}$

Each sum in the parentheses has 336 terms (except the very last one, which has 335), so by pairing each term, we can see that there are $\frac {336}{2}$ pairs of $-6$ . So each sum evaluates to $168\cdot -6=-1008$ , except the very last sum, which has 167 pairs of $-6$ and an extra 2010, so it evaluates to $167\cdot -6+2010=1008$ . Plugging in these values:

$x=-1008\cos{0}-1008\cos{30}-1008\cos{60}-1008\cos{90}-1008\cos{120}+1008\cos{150}$ $x=1008(-1-\frac{\sqrt{3}}{2}-\frac{1}{2}-0+\frac{1}{2}-\frac{\sqrt{3}}{2})=-1008(1+\sqrt{3})$

Now that we have how far was traveled in the x-direction, we need to find how far was traveled in the y-direction. Using the same logic as above, we arrive at the sum:

$y=-1008\sin{0}-1008\sin{30}-1008\sin{60}-1008\sin{90}-1008\sin{120}+1008\sin{150}$

$y=1008(0-\frac{1}{2}-\frac{\sqrt{3}}{2}-1-\frac{\sqrt{3}}{2}+\frac{1}{2})=-1008(1+\sqrt{3})$

The last step is to use the Pythagorean to find the distance from $P_0$ . This distance is given by:

$\sqrt{x^2+y^2}=\sqrt{(-1008(1+\sqrt{3}))^2+(-1008(1+\sqrt{3}))^2}=\sqrt{2\cdot 1008^2 \cdot (1+\sqrt{3})^2}=1008(1+\sqrt{3})\sqrt{2}$

Multiplying out, we have $1008\sqrt{2}+1008\sqrt{6}$ , so the answer is $1008+2+1008+6= \boxed {\bold {(B)}\; 2024}$ .

Solution 3

We first notice that if the bee is turning 30 degrees each turn, it will take 12 turns to be looking in the same direction when the bee initially left. This means we simply need to answer the question; how far will the bee be when the bee is facing in the same direction?

First we use the fact that after 3 turns, the bee will be facing in a direction perpendicular to the the initial direction. From here we can draw a perpendicular from $P_2$ to the line $\overline{P_0P_1}$ intersecting a point $C_0$ . We will also place the point $C_1$ at the intersection of $\overline{P_0P_1}$ and $\overline{P_3P_4}$ . In addition, the point $C_2$ is placed at the perpendicular dropped from $P_2$ to the line $\overline{P_3C_1}$ . We will also set the distance $\overline{P_0P_1} = n$ and thus $\overline{P_1P_2} = n+1$ . With this perpendicular we see that the triangle $\triangle{P_1P_2C_0}$ is a 30-60-90 triangle. This means that the length $\overline{P_1C_0} = \frac{(n+1)\sqrt{3}}{2}$ and the length $\overline{C_1C_2} = \frac{n+1}{2}$ . We can also see that the triangle $\triangle{P_2C_1P_3}$ is a 30-60-90 triangle and thus $\overline{C_0C_1} = \frac{n+2}{2}$ and $\overline{C_2P_3} = \frac{(n+2)\sqrt{3}}{2}$ . Now if we continue this across all $P_i$ and set the point $P_0$ to the coordinates $(0, 0)$ . As you can see, we are inherently putting a “box” around the figure. Doing similar calculations for all four “sides” of this spiral we get that the length

$\overline{P_0C_1} = n + \frac{(n+1)\sqrt{3}}{2} + \frac{n+2}{2}$ , $\overline{C_1C_4} = \frac{(n+1)}{2} + \frac{(n+2)\sqrt{3}}{2} + (n+3) + \frac{(n+4)\sqrt{3}}{2} + \frac{n+5}{2}$ , $\overline{C_4C_7} = \frac{(n+4)}{2} + \frac{(n+5)\sqrt{3}}{2} + (n+6) + \frac{(n+7)\sqrt{3}}{2} + \frac{n+8}{2}$ , $\overline{C_7C_{10}} = \frac{(n+7)}{2} + \frac{(n+8)\sqrt{3}}{2} + (n+9) + \frac{(n+10)\sqrt{3}}{2} + \frac{n+11}{2}$ , and finally $\overline{C_{10}P_{12}} = \frac{(n+10)}{2} + \frac{(n+11)\sqrt{3}}{2}$ .

Here the point $C_4$ is defined as the intersection of lines $\overline{P_3P_4}$ and $\overline{P_6P_7}$ . The point $C_7$ is defined as the intersection of lines $\overline{P_6P_7}$ and $\overline{P_9P_{10}}$ . Finally, the point $C_10$ is defined as the intersection of lines $\overline{P_{9}P_{10}}$ and $\overline{P_{12}P_{13}}$ . Note that our spiral stops at $P_{12}$ before the next spiral starts. Calculating the offset from the x and the y direction, we see that the offset, or the new point $P_{12}$ , is $({-6}, {-6}-12 \sqrt{3})$ . This is an interesting property that the points’ coordinate changes by a constant offset no matter what $n$ is. Since the new point’s subscript changes by 12 each time and we see that 2016 is divisible by 12, the point $P_{2016} = ({-168} \cdot {6}, {168} \cdot ({-6} \sqrt{3} {-12}))$ . Using similar 30-60-90 triangle properties, we see that $P_{2015} = ({-6} \cdot 168-1008 \sqrt{3}, 168({-6} \sqrt{3} - 12) + 1008)$ . Using the distance formula, the numbers cancel out nicely (1008 is divisible by 168, so take 168 when using the distance formula) and we see that the final answer is $(1008)(1+\sqrt{3})(\sqrt{2})$ which gives us a final answer of $\boxed{2024}$ .

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2015 AMC 12B Problems/Problem 25

Contents

Problem

Solution 1

Solution 2

Solution 3

See Also