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

m (→‎Solution 3 (Solution 1 faster))
(→‎Solution 1 (Coordinates))
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== Solution 1 (Coordinates) ==
 
== 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 <math>O</math> as the origin. We will keep a tally of the x-coordinate and y-coordinate separately.
 
We plot this on the coordinate grid with point <math>O</math> as the origin. We will keep a tally of the x-coordinate and y-coordinate separately.
  

Revision as of 13:43, 13 March 2020

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 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 $(1/2)^3=1/8$ 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

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

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