Hi again, everyone! Perhaps you’ve seen my previous post discussing complex numbers in geometry, and I want to return back to that topic with another interesting problem from the 2020 AIME I:
A bug walks all day and sleeps all night. On the first day, it starts at point O, faces east, and walks a distance of 5 units due east. Each night, the bug rotates 60º counterclockwise. Each day, it walks in this new direction half as far as it walked the previous day. The bug gets arbitrarily close to the point P. Then OP2 = m/n, where m and n are relatively prime positive integers. Find m + n.
This was placed #8, which I believe to be a little too high. There are multiple ways to do this problem without complex numbers, such as using geometric series to represent the change in x and y coordinates, but using complex numbers is what I believe to be the most straightforward solution.
The idea is to represent each day’s walk as adding a complex number, and doing this infinitely will represent the entire journey. Due to the unique properties of complex numbers, this will be easy to do.
On the first day, the bug walks 5 units in the positive x direction, which can be represented as 5cis(0º). On the second day, the bug walks 5/2 units in the 60º direction, which can be represented as (5/2)cis(60º). The third day can be represented as (5/4)cis(120º), and the pattern is clear. Adding all the terms together will give us a complex number representing the point the bug will approach, point P.
But how do you find the sum 5cis(0º) + (5/2)cis(60º) + (5/4)cis(120º) + (5/8)cis(180º)…?
Well, it can be represented as a geometric series. Each term is the previous term multiplied by (½)*cis(60º), so the sum can be found using the infinite geometric series sum formula a/(1-r) where a is the first term and r is the common ratio.
5/(1-(½)cis(60º)) = 5 + (5√3/3)i. Thus, the ant will approach the point (5, 5√3/3). The square of the distance from this point to the origin is 25 + 25/3 = 100/3, so m+n = 103, which is the answer.
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