AMC12 2015 12B, Problem 25

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 $a+b+c+d$ ?

Solution: 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 arithmetico-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 $1008+2+1008+6=\fbox{2024}$.

Answer: 2024