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

Revision as of 11:05, 14 June 2023

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)

$[asy] size(8cm); pair O, A, B, C, D, F, G, H, I, P, X; O = (0, 0); A = (5, 0); X = (8, 0); P = (5, 5 / sqrt(3)); B = rotate(-120, A) * ((O + A) / 2); C = rotate(-120, B) * ((A + B) / 2); D = rotate(-120, C) * ((B + C) / 2); F = rotate(-120, D) * ((C + D) / 2); G = rotate(-120, F) * ((D + F) / 2); H = rotate(-120, G) * ((F + G) / 2); I = rotate(-120, H) * ((G + H) / 2); draw(O -- A -- B -- C -- D -- F -- G -- H -- I); draw(A -- X, dashed); markscalefactor = 0.05; path angle = anglemark(X, A, B); draw(angle); dot(P); dot(O); dot(A); dot(B); dot(C); dot(D); label("$O$", O, W); label("$P$", P, E); label("$A$", A, S); label("$B$", B, E); label("$C$", C, E); label("$D$", D, W); 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 $\boxed{103}$ .

Remark: To prove that the cycle repeats every six moves, we extend $AB$ to meet $CD$ at $\dotp E$ (Error compiling LaTeX. Unknown error_msg). With angle chasing, we can find that $\angle DEA$ = $60$ degrees; therefore, $OA$ and $CD$ are parallel. This means that $CD$ and the two moves after are basically the reverse of the moves from $\dotp O$ (Error compiling LaTeX. Unknown error_msg) to $\dotp C$ (Error compiling LaTeX. Unknown error_msg), but scaled down. Thus, after two reverses, or six moves, we have a cycle.

-molocyxu -Remarks by mathMagicOPS

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

Solution 4 (Official MAA 1)

Suppose that the bug starts at the origin $(0,0)$ and travels a distance of $a$ units due east on the first day, and that there is a real number $r$ with $0<r < 1$ such that each day after the first, the bug walks $r$ times as far as the previous day. On day $n$ , the bug travels along the vector $\pmb v_{n}$ that has magnitude $ar^{n-1}$ and direction $\langle\cos(n\cdot 60^\circ),\sin(n\cdot 60^\circ)\rangle$ . Then $P$ is the terminal point of the infinite sum of the vectors $\pmb v_{1}+\pmb v_{2}+\pmb v_3+\cdots$ . The $x$ -coordinate of this sum is $a\big(\!\cos0^\circ+r\cos60^\circ + r^{2}\cos120^\circ+r^{3}\cos180^\circ+r^{4}\cos240^\circ$ $+r^{5}\cos300^\circ+r^{6}\cos360^\circ+\cdots\big).$ Because the angles repeat after 6 terms, this sum is equal to $aS(1+r^{6}+r^{12}+r^{18}+\cdots)=\frac{aS}{1-r^{6}},$ where $S=\cos0^\circ+r\cos60^\circ + r^{2}\cos120^\circ+r^{3}\cos180^\circ+r^{4}\cos240^\circ+ r^{5}\cos300^\circ.$ Similarly, the $y$ -coordinate of $P$ will be ${\frac{aT}{1-r^{6}}}$ , where $T=\sin0^\circ+r\sin60^\circ + r^{2}\sin120^\circ+r^{3}\sin180^\circ+r^{4}\sin240^\circ+ r^{5}\sin300^\circ.$ In this case $r=\frac12$ and $a = 5$ , so $S=1+\frac14-\frac18-\frac18-\frac1{32}+\frac1{64}=\frac{63}{64},$ $T=0+\frac{\sqrt3}4+\frac{\sqrt3}8+0-\frac{\sqrt3}{32}-\frac{\sqrt3}{64}=\frac{21\sqrt3}{64},$ and the coordinates of $P$ are $\left(\frac{5S}{1-\frac1{64}}, \frac{5T}{1-\frac1{64}}\right)=\left(5,\frac{5\sqrt3}{3}\right).$ Thus the square of the distance from the origin to $P$ is $25+\frac{25}3=\frac{100}3$ . The requested sum is $100+3=103$ .

Solution 5 (Official MAA 2)

Let point $O$ be the origin in the complex plane. Point $P$ is the complex sum $5(1+z+z^2+\cdots) = \frac{5}{1-z}$ , where $z=\frac{1+i\sqrt3}4$ . The distance squared is ${OP}^2=\left|\frac5{1-\frac{1+i\sqrt3}4}\right|^{2}= \frac{(4\cdot5)^2}{\left|4-(1+i\sqrt3)\right|^2}=\frac{400}{9+3}=\frac{100}3.$

Solution 6 (No coordinates, but basically using the same idea as Solution 1)

The bug goes forward and backward in three directions: straight east, northeast, and northwest. It travels $5-\frac{5}{8}+\frac{5}{64}-\cdots=\frac{40}{9}$ units east. Thus, it goes northeast $\frac{\frac{40}{9}}{2}=\frac{20}{9}$ units northeast and $\frac{10}{9}$ units northwest. Now, the bug travels a total of $\frac{40}{9}+\frac{10}{9}-\frac{5}{9}=\frac{45}{9}=5$ units east and a total of $\frac{10\sqrt{3}}{9}+\frac{5\sqrt{3}}{9}=\frac{15\sqrt{3}}{9}=\frac{5\sqrt{3}}{3}$ units north because of the 30-60-90 right triangles formed. Now, $OP^2=5^2+\frac{5^2}{3}=\frac{100}{3}$ by the Pythagorean Theorem, and the answer is $\boxed{103}$ .

-integralarefun

Video Solution with Motion in Python

https://youtu.be/RrMsMw_ZrBU

Moving Bug in AIME solution with python turtle scripts

Video Solution

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

-avn

@@ Line 56: / Line 56: @@
 Therefore, the coordinates of point <math>P</math> are <math>(5,\frac{5\sqrt{3}}{3})</math>, so using the Pythagorean Theorem, <math>OP^2 = \frac{100}{3}</math>, for an answer of <math>\boxed{103}</math>.
-Remark: To prove that the cycle repeats every six moves, we extend AB to meet CD at point E. With angle chasing, we can find that <math>\angle DEA</math> = 60 degrees; therefore, OA and CD are parallel. This means that CD and the two moves after are basically the reverse of the moves from point O to point C. Therefore, the cycle repeats every six moves.
+Remark: To prove that the cycle repeats every six moves, we extend <math>AB</math> to meet <math>CD</math> at <math>\dotp E</math>. With angle chasing, we can find that <math>\angle DEA</math> = <math>60</math> degrees; therefore, <math>OA</math> and <math>CD</math> are parallel. This means that <math>CD</math> and the two moves after are basically the reverse of the moves from <math>\dotp O</math> to <math>\dotp C</math>, but scaled down. Thus, after two reverses, or six moves, we have a cycle.
 -molocyxu
 -Remarks by mathMagicOPS

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