2015 AMC 12B Problem 25

Problem 25 of 25HarderAlgebra

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?

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Solution

Place

P_0 = 0

and let

z = e^{\frac{\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

\begin{aligned} &P_{2015} = \dfrac{1}{(z - 1)^2} \\ &\quad {}\cdot \bigl(2015 z^{2016} - 2016 z^{2015} + 1\bigr). \end{aligned}

Since

z^{12} = 1,

we have

z^{2016} = 1

and

z^{2015} = \tfrac1z,

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.