Difference between revisions of "2020 AIME I Problems/Problem 8"
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label("$P$", P, E); | label("$P$", P, E); | ||
label("$A$", A, S); | label("$A$", A, S); | ||
− | label("$C$, C, | + | label("$B$", B, E); |
− | label("$D$", D, | + | 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 52: | 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)== | ||
Line 95: | Line 95: | ||
-integralarefun | -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== | ==Video Solution with Motion in Python== |
Latest revision as of 15:12, 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.