Question

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 $a + b + c + d?$

$2016$

$2024$

$2032$

$2040$

$2048$

Solution:

Place $P_0 = 0$ and let $z = e^{\pi i/6},$ so each step of length $k$ in direction $z^{k-1}$ gives $P_{2015} = \sum_{k=1}^{2015} k z^{k-1}.$ Summing this (a differentiated geometric series) leads to $P_{2015} = \dfrac{1}{(z - 1)^2}\bigl(2015 z^{2016} - 2016 z^{2015} + 1\bigr).$

Since $z^{12} = 1,$ we have $z^{2016} = 1$ and $z^{2015} = \tfrac1z,$ so $P_{2015} = \dfrac{2016}{z(z - 1)}.$ Using $|z - 1|^2 = 2 - \sqrt3 = \dfrac{(\sqrt3 - 1)^2}{2}$ and $|z| = 1,$ the distance is $\dfrac{2016}{|z - 1|} = 1008\sqrt6 + 1008\sqrt2.$

Hence $a + b + c + d = 1008 + 6 + 1008 + 2 = 2024.$

Thus, the correct answer is B.

2015 AMC 12B Problem 25

Solution:

Problem 25 in Other Years