Difference between revisions of "2020 AIME I Problems/Problem 8"
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+ | ==Problem== | ||
+ | A bug walks all day and sleeps all night. On the first day, it starts at point <math>O</math>, faces east, and walks a distance of <math>5</math> units due east. Each night the bug rotates <math>60^\circ</math> 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 <math>P</math>. Then <math>OP^2=\tfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | ||
− | + | ==Solution 1 (Coordinates)== | |
− | |||
− | |||
− | == Solution 1 (Coordinates) == | ||
<asy> | <asy> | ||
size(8cm); | size(8cm); | ||
− | pair O, A, B, C, D, F, G, P, X; | + | pair O, A, B, C, D, F, G, H, I, P, X; |
O = (0, 0); | O = (0, 0); | ||
A = (5, 0); | A = (5, 0); | ||
Line 16: | Line 15: | ||
F = rotate(-120, D) * ((C + D) / 2); | F = rotate(-120, D) * ((C + D) / 2); | ||
G = rotate(-120, F) * ((D + F) / 2); | G = rotate(-120, F) * ((D + F) / 2); | ||
− | draw(O -- A -- B -- C -- D -- F -- G); | + | 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); | draw(A -- X, dashed); | ||
markscalefactor = 0.05; | markscalefactor = 0.05; | ||
Line 24: | Line 25: | ||
dot(P); | dot(P); | ||
dot(O); | dot(O); | ||
+ | dot(A); | ||
+ | dot(B); | ||
+ | dot(C); | ||
+ | dot(D); | ||
label("$O$", O, W); | label("$O$", O, W); | ||
label("$P$", P, E); | 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); | label("$60^\circ$", angle, ENE*3); | ||
</asy> | </asy> | ||
− | We plot | + | We notice that the moves cycle every 6 moves, so we plot out the first 6 moves 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. Then, we will combine them and account for the cycling using the formula for an infinite geometric series. |
First move: The ant moves right <math>5</math>. | First move: The ant moves right <math>5</math>. | ||
Line 44: | Line 53: | ||
After this cycle of six moves, all moves repeat with a factor of <math>(\frac{1}{2})^6 = \frac{1}{64}</math>. Using the formula for a geometric series, multiplying each sequence by <math>\frac{1}{1-\frac{1}{64}} = \frac{64}{63}</math> will give us the point <math>P</math>. | After this cycle of six moves, all moves repeat with a factor of <math>(\frac{1}{2})^6 = \frac{1}{64}</math>. Using the formula for a geometric series, multiplying each sequence by <math>\frac{1}{1-\frac{1}{64}} = \frac{64}{63}</math> will give us the point <math>P</math>. | ||
− | <math>\frac{315}{64} \cdot \frac{64}{63} = 5</math>, <math>\frac{105\sqrt{3}}{64} \cdot \frac{64}{63} = \frac{5\sqrt{3}}{3}</math>. | + | Now, knowing the initial <math>x</math> and <math>y,</math> we plug this into the geometric series formula (<math>\frac{a}{1-r}</math>), and we get <math>\frac{315}{64} \cdot \frac{64}{63} = 5</math>, <math>\frac{105\sqrt{3}}{64} \cdot \frac{64}{63} = \frac{5\sqrt{3}}{3}</math>. |
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>. | 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>. | ||
+ | |||
-molocyxu | -molocyxu | ||
− | == Solution 2 (Complex) == | + | ==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 | 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 | ||
<cmath>5\left(1 + \frac{e^{\frac{i\pi}{3}}}{2} + \left(\frac{e^{\frac{i\pi}{3}}}{2}\right)^2 + \cdots\right)</cmath> | <cmath>5\left(1 + \frac{e^{\frac{i\pi}{3}}}{2} + \left(\frac{e^{\frac{i\pi}{3}}}{2}\right)^2 + \cdots\right)</cmath> | ||
Line 58: | Line 68: | ||
so the answer is <math>100 + 3 = \boxed{103}</math>. | so the answer is <math>100 + 3 = \boxed{103}</math>. | ||
− | == Solution 3 (Solution 1 | + | ==Solution 3 (Solution 1 Faster)== |
The ant goes in the opposite direction every <math>3</math> moves, going <math>(1/2)^3=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> . | The ant goes in the opposite direction every <math>3</math> moves, going <math>(1/2)^3=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 | ~Lcz | ||
− | == Video Solution == | + | ==Solution 4 (Official MAA 1)== |
+ | Suppose that the bug starts at the origin <math>(0,0)</math> and travels a distance of <math>a</math> units due east on the first day, and that there is a real number <math>r</math> with <math>0<r < 1</math> such that each day after the first, the bug walks <math>r</math> times as far as the previous day. On day <math>n</math>, the bug travels along the vector <math>\pmb v_{n}</math> that has magnitude <math>ar^{n-1}</math> and direction <math>\langle\cos(n\cdot 60^\circ),\sin(n\cdot 60^\circ)\rangle</math>. Then <math>P</math> is the terminal point of the infinite sum of the vectors <math>\pmb v_{1}+\pmb v_{2}+\pmb v_3+\cdots</math>. The <math>x</math>-coordinate of this sum is | ||
+ | <cmath>a\big(\!\cos0^\circ+r\cos60^\circ + r^{2}\cos120^\circ+r^{3}\cos180^\circ+r^{4}\cos240^\circ</cmath> | ||
+ | <cmath>+r^{5}\cos300^\circ+r^{6}\cos360^\circ+\cdots\big).</cmath> | ||
+ | Because the angles repeat after 6 terms, this sum is equal to | ||
+ | <cmath>aS(1+r^{6}+r^{12}+r^{18}+\cdots)=\frac{aS}{1-r^{6}},</cmath> | ||
+ | where | ||
+ | <cmath>S=\cos0^\circ+r\cos60^\circ + r^{2}\cos120^\circ+r^{3}\cos180^\circ+r^{4}\cos240^\circ+ r^{5}\cos300^\circ.</cmath> | ||
+ | Similarly, the <math>y</math>-coordinate of <math>P</math> will be <math>{\frac{aT}{1-r^{6}}}</math>, where | ||
+ | <cmath>T=\sin0^\circ+r\sin60^\circ + r^{2}\sin120^\circ+r^{3}\sin180^\circ+r^{4}\sin240^\circ+ r^{5}\sin300^\circ.</cmath>In this case <math>r=\frac12</math> and <math>a = 5</math>, so | ||
+ | <cmath>S=1+\frac14-\frac18-\frac18-\frac1{32}+\frac1{64}=\frac{63}{64},</cmath> | ||
+ | <cmath>T=0+\frac{\sqrt3}4+\frac{\sqrt3}8+0-\frac{\sqrt3}{32}-\frac{\sqrt3}{64}=\frac{21\sqrt3}{64},</cmath>and the coordinates of <math>P</math> are | ||
+ | <cmath>\left(\frac{5S}{1-\frac1{64}}, \frac{5T}{1-\frac1{64}}\right)=\left(5,\frac{5\sqrt3}{3}\right).</cmath>Thus the square of the distance from the origin to <math>P</math> is <math>25+\frac{25}3=\frac{100}3</math>. The requested sum is <math>100+3=103</math>. | ||
+ | |||
+ | ==Solution 5 (Official MAA 2)== | ||
+ | Let point <math>O</math> be the origin in the complex plane. Point <math>P</math> is the complex sum <math>5(1+z+z^2+\cdots) = \frac{5}{1-z}</math>, where <math>z=\frac{1+i\sqrt3}4</math>. The distance squared is<cmath> {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.</cmath> | ||
+ | |||
+ | ==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 <math>5-\frac{5}{8}+\frac{5}{64}-\cdots=\frac{40}{9}</math> units east. Thus, it goes northeast <math>\frac{\frac{40}{9}}{2}=\frac{20}{9}</math> units northeast and <math>\frac{10}{9}</math> units northwest. Now, the bug travels a total of <math>\frac{40}{9}+\frac{10}{9}-\frac{5}{9}=\frac{45}{9}=5</math> units east and a total of <math>\frac{10\sqrt{3}}{9}+\frac{5\sqrt{3}}{9}=\frac{15\sqrt{3}}{9}=\frac{5\sqrt{3}}{3}</math> units north because of the 30-60-90 right triangles formed. Now, <math>OP^2=5^2+\frac{5^2}{3}=\frac{100}{3}</math> by the Pythagorean Theorem, and the answer is <math>\boxed{103}</math>. | ||
+ | |||
+ | -integralarefun | ||
+ | |||
+ | ==Solution 7== | ||
+ | The bug's bearings on each traversal are <math>0^\circ, 60^\circ, 120^\circ,</math> and so on; in general, the <math>n-</math>th traversal has length <math>5\cdot (1/2)^{n-1}</math> and bearing <math>60(n-1).</math> This means that the <math>x</math> and <math>y</math> displacements for the <math>n-</math>th traversal are | ||
+ | |||
+ | <cmath>(\Delta x_n, \Delta y_n)=(5\cdot (1/2)^{n-1}\cos (60(n-1))^\circ,5\cdot (1/2)^{n-1}\sin (60(n-1))^\circ).</cmath> | ||
+ | |||
+ | Summing this over all the displacements, we get | ||
+ | |||
+ | <cmath>x_P=\sum_{n=1}^{\infty} 5\cdot (1/2)^{n-1}\cos (60(n-1))^\circ, y_P=\sum_{n=1}^{\infty} 5\cdot (1/2)^{n-1}\sin (60(n-1))^\circ.</cmath> | ||
+ | |||
+ | We then have | ||
+ | |||
+ | <cmath> | ||
+ | \begin{align} | ||
+ | OP^2 &= x_P^2+y_P^2 \\ | ||
+ | &=\sum_{n=1}^{\infty} (5\cdot (1/2)^{n-1}\cos ^2(60(n-1))^\circ) + (5\cdot (1/2)^{n-1}\sin ^2(60(n-1))^\circ) \\ | ||
+ | &= \sum_{n=1}^{\infty} (5\cdot (1/2)^{n-1})^2(\cos ^2 (60(n-1))^\circ+\sin ^2 (60(n-1))^\circ) \\ | ||
+ | &= \sum_{n=1}^{\infty} (25\cdot (1/4)^{n-1}) \\ | ||
+ | &= \dfrac{25}{1-1/4} \\ | ||
+ | &= 100/3. | ||
+ | \end{align} | ||
+ | </cmath> | ||
+ | |||
+ | Thus, the answer is <math>100+3=\boxed{103}.</math> --MenuThreeOne | ||
+ | |||
+ | ==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 | https://www.youtube.com/watch?v=BtMBSoZ3cMQ |
Latest revision as of 16:12, 2 January 2024
Contents
- 1 Problem
- 2 Solution 1 (Coordinates)
- 3 Solution 2 (Complex)
- 4 Solution 3 (Solution 1 Faster)
- 5 Solution 4 (Official MAA 1)
- 6 Solution 5 (Official MAA 2)
- 7 Solution 6 (No coordinates, but basically using the same idea as Solution 1)
- 8 Solution 7
- 9 Video Solution with Motion in Python
- 10 Video Solution
- 11 See Also
Problem
A bug walks all day and sleeps all night. On the first day, it starts at point , faces east, and walks a distance of
units due east. Each night the bug rotates
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
. Then
, where
and
are relatively prime positive integers. Find
.
Solution 1 (Coordinates)
We notice that the moves cycle every 6 moves, so we plot out the first 6 moves on the coordinate grid with point as the origin. We will keep a tally of the x-coordinate and y-coordinate separately. Then, we will combine them and account for the cycling using the formula for an infinite geometric series.
First move: The ant moves right .
Second move: We use properties of a
triangle to get
right,
up.
Third move:
left,
up.
Fourth move:
left.
Fifth move:
left,
down.
Sixth move:
right,
down.
Total of x-coordinate: .
Total of y-coordinate:
.
After this cycle of six moves, all moves repeat with a factor of . Using the formula for a geometric series, multiplying each sequence by
will give us the point
.
Now, knowing the initial and
we plug this into the geometric series formula (
), and we get
,
.
Therefore, the coordinates of point
are
, so using the Pythagorean Theorem,
, for an answer of
.
-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
Using the formula for an infinite geometric series, this is equal to
We are looking for the square of the modulus of this value:
so the answer is
.
Solution 3 (Solution 1 Faster)
The ant goes in the opposite direction every moves, going
the distance backwards. Using geometric series, he travels
the distance of the first three moves over infinity moves. Now, we use coordinates meaning
or
. Multiplying these by
, we get
.
~Lcz
Solution 4 (Official MAA 1)
Suppose that the bug starts at the origin and travels a distance of
units due east on the first day, and that there is a real number
with
such that each day after the first, the bug walks
times as far as the previous day. On day
, the bug travels along the vector
that has magnitude
and direction
. Then
is the terminal point of the infinite sum of the vectors
. The
-coordinate of this sum is
Because the angles repeat after 6 terms, this sum is equal to
where
Similarly, the
-coordinate of
will be
, where
In this case
and
, so
and the coordinates of
are
Thus the square of the distance from the origin to
is
. The requested sum is
.
Solution 5 (Official MAA 2)
Let point be the origin in the complex plane. Point
is the complex sum
, where
. The distance squared is
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 units east. Thus, it goes northeast
units northeast and
units northwest. Now, the bug travels a total of
units east and a total of
units north because of the 30-60-90 right triangles formed. Now,
by the Pythagorean Theorem, and the answer is
.
-integralarefun
Solution 7
The bug's bearings on each traversal are and so on; in general, the
th traversal has length
and bearing
This means that the
and
displacements for the
th traversal are
Summing this over all the displacements, we get
We then have
Thus, the answer is --MenuThreeOne
Video Solution with Motion in Python
Moving Bug in AIME solution with python turtle scripts
Video Solution
https://www.youtube.com/watch?v=BtMBSoZ3cMQ
-avn
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.