Difference between revisions of "2020 AIME I Problems/Problem 8"

Revision as of 16:34, 12 March 2020

Note: Please do not post problems here until after the AIME.

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)

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 put the ant in the complex plane, with its first move going in the positive real direction. Take $|\sum_{k=0}^{\infty} (5\frac{e^{k\pi i / 3}}{2^k})|^2$ and this is an infinite geometric series. Summing using $\frac{a}{1-r}$ gives $\boxed{103}.$ ~awang11

Solution 3 (Solution 1 faster)

The ant goes in the opposite direction every $3$ moves, going $1/2^=1/8$ 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

@@ Line 29: / Line 29: @@
 <cmath>|\sum_{k=0}^{\infty} (5\frac{e^{k\pi i / 3}}{2^k})|^2</cmath>
 and this is an infinite geometric series. Summing using <math>\frac{a}{1-r}</math> gives <math>\boxed{103}.</math> ~awang11
+== Solution 3 (Solution 1 faster) ==
+The ant goes in the opposite direction every <math>3</math> moves, going <math>1/2^=1/8</math> the distance backwards. Using geometric series, he travels <math>1-1/8+1/64-1/512...=(7/8)(1+1/64+1/4096...)=(7/8)(64/63)=8/9</math> the distance of the first three moves over infinity moves. Now, we use coordinates meaning <math>(5+5/4-5/8, 0+5\sqrt3/4+5\sqrt3/8)</math> or <math>(45/8, 15\sqrt3/8)</math>. Multiplying these by <math>8/9</math>, we get <math>(5, 5\sqrt3/3)</math> <math>\implies</math> <math>\boxed{103}</math> .
+~Lcz
 ==See Also==

2020 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search