2020 AIME I Problems/Problem 8

Problem

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 $60^\circ$ 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 $OP^2=\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution 1 (Coordinates)

$[asy] size(8cm); pair O, A, B, C, D, F, G, P, X; O = (0, 0); A = (5, 0); X = (8, 0); P = (5, 5 / sqrt(3)); B = rotate(-120, A) * ((O + A) / 2); C = rotate(-120, B) * ((A + B) / 2); D = rotate(-120, C) * ((B + C) / 2); F = rotate(-120, D) * ((C + D) / 2); G = rotate(-120, F) * ((D + F) / 2); draw(O -- A -- B -- C -- D -- F -- G); draw(A -- X, dashed); markscalefactor = 0.05; path angle = anglemark(X, A, B); draw(angle); dot(P); dot(O); label("$O$", O, W); label("$P$", P, E); label("$60^\circ$", angle, ENE*3); [/asy]$

We plot this on the coordinate grid with point $O$ as the origin. We will keep a tally of the x-coordinate and y-coordinate separately.

First move: The ant moves right $5$ . Second move: We use properties of a $30-60-90$ triangle to get $\frac{5}{4}$ right, $\frac{5\sqrt{3}}{4}$ up. Third move: $\frac{5}{8}$ left, $\frac{5\sqrt{3}}{8}$ up. Fourth move: $\frac{5}{8}$ left. Fifth move: $\frac{5}{32}$ left, $\frac{5\sqrt{3}}{32}$ down. Sixth move: $\frac{5}{64}$ right, $\frac{5\sqrt{3}}{64}$ down.

Total of x-coordinate: $5 + \frac{5}{4} - \frac{5}{8} - \frac{5}{8} - \frac{5}{32} + \frac{5}{64} = \frac{315}{64}$ . Total of y-coordinate: $0 + \frac{5\sqrt{3}}{4} + \frac{5\sqrt{3}}{8} + 0 - \frac{5\sqrt{3}}{32} - \frac{5\sqrt{3}}{64} = \frac{105\sqrt{3}}{64}$ .

After this cycle of six moves, all moves repeat with a factor of $(\frac{1}{2})^6 = \frac{1}{64}$ . Using the formula for a geometric series, multiplying each sequence by $\frac{1}{1-\frac{1}{64}} = \frac{64}{63}$ will give us the point $P$ .

$\frac{315}{64} \cdot \frac{64}{63} = 5$ , $\frac{105\sqrt{3}}{64} \cdot \frac{64}{63} = \frac{5\sqrt{3}}{3}$ . Therefore, the coordinates of point $P$ are $(5,\frac{5\sqrt{3}}{3})$ , so using the Pythagorean Theorem, $OP^2 = \frac{100}{3}$ , for an answer of $\boxed{103}$ .

-molocyxu

Solution 2 (Complex)

We place the ant at the origin of the complex plane with its first move being in the positive real direction. Then the ant's journey can be represented as the infinite series $5\left(1 + \frac{e^{\frac{i\pi}{3}}}{2} + \left(\frac{e^{\frac{i\pi}{3}}}{2}\right)^2 + \cdots\right)$ Using the formula for an infinite geometric series, this is equal to $\frac{5}{1 - \frac12e^{\frac{i\pi}{3}}} = \frac{5}{1 - \frac{1 + i\sqrt{3}}{4}} = \frac{20}{3 - i\sqrt{3}} = 5 + \frac{5i\sqrt{3}}{3}$ We are looking for the square of the modulus of this value: $\left|\frac{5 + 5i\sqrt{3}}{3}\right|^2 = 25 + \frac{25}{3} = \frac{100}{3}$ so the answer is $100 + 3 = \boxed{103}$ .

Solution 3 (Solution 1 faster)

The ant goes in the opposite direction every $3$ moves, going $(1/2)^3=1/8$ the distance backwards. Using geometric series, he travels $1-1/8+1/64-1/512...=(7/8)(1+1/64+1/4096...)=(7/8)(64/63)=8/9$ the distance of the first three moves over infinity moves. Now, we use coordinates meaning $(5+5/4-5/8, 0+5\sqrt3/4+5\sqrt3/8)$ or $(45/8, 15\sqrt3/8)$ . Multiplying these by $8/9$ , we get $(5, 5\sqrt3/3)$ $\implies$ $\boxed{103}$ .

~Lcz

Video Solution

https://www.youtube.com/watch?v=BtMBSoZ3cMQ

-avn

2020 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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