AMC12 2015 B

AMC12 2015 B · Q25

AMC12 2015 B · Q25. It mainly tests Complex numbers (rare), Trigonometry (basic).

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 square-free. What is $a + b + c + d$?

一只蜜蜂从点$P_0$开始飞行。她向正东飞1英寸到点$P_1$。对于$j \ge 1$，到达点$P_j$后，她逆时针转$30^\circ$，然后直飞$j+1$英寸到点$P_{j+1}$。当蜜蜂到达$P_{2015}$时，她离$P_0$正好$a\sqrt{b} + c\sqrt{d}$英寸，其中$a, b, c,$和$d$是正整数且$b$和$d$是无平方因子。$a + b + c + d$是多少？

(A) 2016 2016

(B) 2024 2024

(D) 2040 2040

(E) 2048 2048

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Modeling the bee’s path with complex numbers, set $P_0=0$ and $z=e^{\pi i/6}$. It follows that for $j\ge 1$, $$ P_j=\sum_{k=1}^{j}kz^{k-1}. $$ Thus $$ P_{2015}=\sum_{k=0}^{2015}kz^{k-1}=\sum_{k=0}^{2014}(k+1)z^k=\sum_{k=0}^{2014}\sum_{j=0}^{k}z^k. $$ Interchanging the order of summation and summing the geometric series gives $$ P_{2015}=\sum_{j=0}^{2014}\sum_{k=j}^{2014}z^k=\sum_{j=0}^{2014}z^j\sum_{k=0}^{2014-j}z^k $$ $$ =\sum_{j=0}^{2014}\frac{z^j\left(z^{2015-j}-1\right)}{z-1} =\sum_{j=0}^{2014}\frac{z^{2015}-z^j}{z-1} =\frac{1}{z-1}\sum_{j=0}^{2014}\left(z^{2015}-z^j\right) $$ $$ =\frac{1}{z-1}\left(2015z^{2015}-\sum_{j=0}^{2014}z^j\right) =\frac{1}{z-1}\left(2015z^{2015}-\frac{z^{2015}-1}{z-1}\right) $$ $$ =\frac{1}{(z-1)^2}\left(2015z^{2015}(z-1)-z^{2015}+1\right) =\frac{1}{(z-1)^2}\left(2015z^{2016}-2016z^{2015}+1\right). $$ Note that $z^{12}=1$ and thus $z^{2016}=(z^{12})^{168}=1$ and $z^{2015}=\frac{1}{z}$. It follows that $$ P_{2015}=\frac{2016}{(z-1)^2}\left(1-\frac{1}{z}\right)=\frac{2016}{z(z-1)}. $$ Finally, $$ |z-1|^2=\left|\cos\left(\frac{\pi}{6}\right)-1+i\sin\left(\frac{\pi}{6}\right)\right|^2 =\left|\frac{\sqrt3}{2}-1+i\frac12\right|^2 =2-\sqrt3=\frac{(\sqrt3-1)^2}{2}, $$ and thus $$ |P_{2015}|=\left|\frac{2016}{z(z-1)}\right|=\frac{2016}{|z-1|} =\frac{2016\sqrt2}{\sqrt3-1}=1008\sqrt2(\sqrt3+1) =1008\sqrt6+1008\sqrt2. $$ The requested sum is $1008+6+1008+2=2024$.

答案（B）：用复数来刻画蜜蜂的路径，令 $P_0=0$ 且 $z=e^{\pi i/6}$。则对 $j\ge 1$， $$ P_j=\sum_{k=1}^{j}kz^{k-1}. $$ 因此 $$ P_{2015}=\sum_{k=0}^{2015}kz^{k-1}=\sum_{k=0}^{2014}(k+1)z^k=\sum_{k=0}^{2014}\sum_{j=0}^{k}z^k. $$ 交换求和次序并对等比数列求和，得到 $$ P_{2015}=\sum_{j=0}^{2014}\sum_{k=j}^{2014}z^k=\sum_{j=0}^{2014}z^j\sum_{k=0}^{2014-j}z^k $$ $$ =\sum_{j=0}^{2014}\frac{z^j\left(z^{2015-j}-1\right)}{z-1} =\sum_{j=0}^{2014}\frac{z^{2015}-z^j}{z-1} =\frac{1}{z-1}\sum_{j=0}^{2014}\left(z^{2015}-z^j\right) $$ $$ =\frac{1}{z-1}\left(2015z^{2015}-\sum_{j=0}^{2014}z^j\right) =\frac{1}{z-1}\left(2015z^{2015}-\frac{z^{2015}-1}{z-1}\right) $$ $$ =\frac{1}{(z-1)^2}\left(2015z^{2015}(z-1)-z^{2015}+1\right) =\frac{1}{(z-1)^2}\left(2015z^{2016}-2016z^{2015}+1\right). $$ 注意 $z^{12}=1$，因此 $z^{2016}=(z^{12})^{168}=1$，且 $z^{2015}=\frac{1}{z}$。于是 $$ P_{2015}=\frac{2016}{(z-1)^2}\left(1-\frac{1}{z}\right)=\frac{2016}{z(z-1)}. $$ 最后， $$ |z-1|^2=\left|\cos\left(\frac{\pi}{6}\right)-1+i\sin\left(\frac{\pi}{6}\right)\right|^2 =\left|\frac{\sqrt3}{2}-1+i\frac12\right|^2 =2-\sqrt3=\frac{(\sqrt3-1)^2}{2}, $$ 因此 $$ |P_{2015}|=\left|\frac{2016}{z(z-1)}\right|=\frac{2016}{|z-1|} =\frac{2016\sqrt2}{\sqrt3-1}=1008\sqrt2(\sqrt3+1) =1008\sqrt6+1008\sqrt2. $$ 所求的和为 $1008+6+1008+2=2024$。

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