Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2024 AMC 10A Problems/Problem 24"

(→‎Solution 1)
(→‎Diagram by PaperMath)
 
(23 intermediate revisions by 12 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
  
−
A bee is moving in three-dimensional space. A fair six-sided die with faces labeled <math>A^+, A^-, B^+, B^-, C^+,</math> and <math>C^-</math> is rolled. Suppose the bee occupies the point <math>(a,b,c).</math> If the die shows <math>A^+</math>, then the bee moves to the point <math>(a+1,b,c)</math> and if the die shows <math>A^-,</math> then the bee moves to the point <math>(a+1,b,c).</math> Analogous moves are made with the other four outcomes. Suppose the bee starts at the point <math>(0,0,0)</math> and the die is rolled four times. What is the probability that the bee traverses four distinct edges of some unit cube?
+
A bee is moving in three-dimensional space. A fair six-sided die with faces labeled <math>A^+, A^-, B^+, B^-, C^+,</math> and <math>C^-</math> is rolled. Suppose the bee occupies the point <math>(a,b,c).</math> If the die shows <math>A^+</math>, then the bee moves to the point <math>(a+1,b,c)</math> and if the die shows <math>A^-,</math> then the bee moves to the point <math>(a-1,b,c).</math> Analogous moves are made with the other four outcomes. Suppose the bee starts at the point <math>(0,0,0)</math> and the die is rolled four times. What is the probability that the bee traverses four distinct edges of some unit cube?
  
 
<math>\textbf{(A) }\frac{1}{54}\qquad\textbf{(B) }\frac{7}{54}\qquad\textbf{(C) }\frac{1}{6}\qquad\textbf{(D) }\frac{5}{18}\qquad\textbf{(E) }\frac{2}{5}</math>
 
<math>\textbf{(A) }\frac{1}{54}\qquad\textbf{(B) }\frac{7}{54}\qquad\textbf{(C) }\frac{1}{6}\qquad\textbf{(D) }\frac{5}{18}\qquad\textbf{(E) }\frac{2}{5}</math>
 +
 +
== Diagrams ==
 +
[[Image:Diagrams.jpeg|200px]]
 +
[[File:Bee total.png|600px|right]]
 +
~ Diagram by PaperMath
 +
 +
[[Image:2024amc10a24solution.jpg|300px]]
 +
 +
~ Solution diagram by juwushu
  
 
== Solution 1 ==
 
== Solution 1 ==
−
WLOG, assume that the first two moves are equal for all possible combinations, since the direction does not matter. The first move has a <math>\frac{6}{6}</math> probability of being along one of the 8 unit cubes around the origin, and the second move has a <math>\frac{4}{6}</math> chance.
+
WLOG, assume that the first two moves are equal for all possible combinations, since the direction does not matter. The first move has a <math>\frac{6}{6}</math> probability of being along one of the <math>8</math> unit cubes around the origin, and the second move has a <math>\frac{4}{6}</math> chance.
 
Now, there are two cases. We are currently on one of the points of the <math>2</math> by <math>2</math> squares that are aligned with the axes.
 
Now, there are two cases. We are currently on one of the points of the <math>2</math> by <math>2</math> squares that are aligned with the axes.
 
The first case is if the bee moves to the corner of a cube farthest away from the origin.
 
The first case is if the bee moves to the corner of a cube farthest away from the origin.
Line 12: Line 21:
 
The second case is if the bee moves along the same plane of the <math>2</math> by <math>2</math> squares previously, ending up on a point 1 away from the origin. There is a <math>\frac{1}{6}</math> chance of this happening and a <math>\frac{3}{6}</math> chance of remaining on one of the cubes.
 
The second case is if the bee moves along the same plane of the <math>2</math> by <math>2</math> squares previously, ending up on a point 1 away from the origin. There is a <math>\frac{1}{6}</math> chance of this happening and a <math>\frac{3}{6}</math> chance of remaining on one of the cubes.
 
Now, multiply and sum for the answer.
 
Now, multiply and sum for the answer.
−
<cmath>\frac{2/3}\cdot(\frac{1}{3}\cdot\frac{1}{3}+\frac{1}{6}\cdot\frac{1}{2})=\frac{2}{3}\cdot(\frac{1}{9}+\frac{1}{12})</cmath>
+
<cmath>\frac{2}{3}\cdot(\frac{1}{3}\cdot\frac{1}{3}+\frac{1}{6}\cdot\frac{1}{2})=\frac{2}{3}\cdot(\frac{1}{9}+\frac{1}{12})</cmath>
−
Evaluating this gives you the answer of <math>\box{\textbf{(B) }\frac{7}{54}}</math>.
+
Evaluating this gives you the answer of <math>\boxed{\textbf{(B) }\frac{7}{54}}</math>.
 
Solution by [[User:Juwushu|juwushu]].
 
Solution by [[User:Juwushu|juwushu]].
 +
Minor edits by andliu766
 +
 +
==Solution 2==
 +
Remove the dice. It is not necessary for this problem. It is the same problem if you instead view this problem as the bee moving one up or down in the <math>x, y,</math> and <math>z</math> directions.
 +
 +
Count the total number of ways for one cube of the 8 total ones. It is not hard to see that after the first step (which you can travel to <math>(1,0,0), (0,1,0),</math> or <math>(0,0,1)</math> -  3 ways), you have a total of 8 moves to traverse a face of that 1x1 cube. Hence, for each cube, you would have <math>8 \cdot 3 = 24</math> total moves.
 +
 +
There are a total of 8 cubes you could have traversed, though, so multiplying this by <math>8</math> yields <math>8 \cdot 8 \cdot 3.</math>
 +
 +
We have to account for overcounting. For example, when we traversed the face in the <math>xz</math> plane <math>(0,0,0), (0,0,1),(1,0,0),(1,0,1),</math> the number of ways that we traverse that particular face got counted twice, once from the cube in the far upper right and once from the cube in the closer upper right.
 +
 +
For a given face, you can go clockwise or counterclockwise, hence yielding <math>2</math> ways. There are a total of <math>12</math> shared faces, each one shared by exactly two faces so no additonal PIE is needed. We simply subtract <math>2\cdot12</math> from our original count now, yielding <math>8 \cdot 3 \cdot 8 - 2 \cdot 12 = 168.</math>
 +
 +
With no restrictions, there are a total of <math>6^4</math> ways to move four steps. Your probability is <math>\frac{168}{6^4}=\frac{7}{54}.</math>
 +
 +
~mathboy282
 +
 +
 +
== Video Solution by Power Solve ==
 +
https://www.youtube.com/watch?v=2uTWPiWzfB0
 +
 +
== Video Solution by Pi Academy (Fast and Easy ⚡️🚀) ==
 +
 +
https://youtu.be/3Dk-vmdxbrw?si=G6TcyLN2Rfwd2YYJ
 +
 +
==Video Solution by Innovative Minds==
 +
https://youtu.be/CxLGUPhYYDI
 +
 +
~i_am_suk_at_math_2
 +
 +
==Video Solution by SpreadTheMathLove==
 +
https://www.youtube.com/watch?v=6SQ74nt3ynw
 +
 +
==See also==
 +
{{AMC10 box|year=2024|ab=A|num-b=23|num-a=25}}
 +
{{MAA Notice}}

Latest revision as of 12:24, 21 November 2024

Problem

A bee is moving in three-dimensional space. A fair six-sided die with faces labeled $A^+, A^-, B^+, B^-, C^+,$ and $C^-$ is rolled. Suppose the bee occupies the point $(a,b,c).$ If the die shows $A^+$, then the bee moves to the point $(a+1,b,c)$ and if the die shows $A^-,$ then the bee moves to the point $(a-1,b,c).$ Analogous moves are made with the other four outcomes. Suppose the bee starts at the point $(0,0,0)$ and the die is rolled four times. What is the probability that the bee traverses four distinct edges of some unit cube?

$\textbf{(A) }\frac{1}{54}\qquad\textbf{(B) }\frac{7}{54}\qquad\textbf{(C) }\frac{1}{6}\qquad\textbf{(D) }\frac{5}{18}\qquad\textbf{(E) }\frac{2}{5}$

Diagrams

Diagrams.jpeg

Bee total.png

~ Diagram by PaperMath

2024amc10a24solution.jpg

~ Solution diagram by juwushu

Solution 1

WLOG, assume that the first two moves are equal for all possible combinations, since the direction does not matter. The first move has a $\frac{6}{6}$ probability of being along one of the $8$ unit cubes around the origin, and the second move has a $\frac{4}{6}$ chance. Now, there are two cases. We are currently on one of the points of the $2$ by $2$ squares that are aligned with the axes. The first case is if the bee moves to the corner of a cube farthest away from the origin. Here, there is a $\frac{2}{6}$ chance of this happening and a $\frac{2}{6}$ chance of the fourth move remaining on one of the cubes. The second case is if the bee moves along the same plane of the $2$ by $2$ squares previously, ending up on a point 1 away from the origin. There is a $\frac{1}{6}$ chance of this happening and a $\frac{3}{6}$ chance of remaining on one of the cubes. Now, multiply and sum for the answer. \[\frac{2}{3}\cdot(\frac{1}{3}\cdot\frac{1}{3}+\frac{1}{6}\cdot\frac{1}{2})=\frac{2}{3}\cdot(\frac{1}{9}+\frac{1}{12})\] Evaluating this gives you the answer of $\boxed{\textbf{(B) }\frac{7}{54}}$. Solution by juwushu. Minor edits by andliu766

Solution 2

Remove the dice. It is not necessary for this problem. It is the same problem if you instead view this problem as the bee moving one up or down in the $x, y,$ and $z$ directions.

Count the total number of ways for one cube of the 8 total ones. It is not hard to see that after the first step (which you can travel to $(1,0,0), (0,1,0),$ or $(0,0,1)$ - 3 ways), you have a total of 8 moves to traverse a face of that 1x1 cube. Hence, for each cube, you would have $8 \cdot 3 = 24$ total moves.

There are a total of 8 cubes you could have traversed, though, so multiplying this by $8$ yields $8 \cdot 8 \cdot 3.$

We have to account for overcounting. For example, when we traversed the face in the $xz$ plane $(0,0,0), (0,0,1),(1,0,0),(1,0,1),$ the number of ways that we traverse that particular face got counted twice, once from the cube in the far upper right and once from the cube in the closer upper right.

For a given face, you can go clockwise or counterclockwise, hence yielding $2$ ways. There are a total of $12$ shared faces, each one shared by exactly two faces so no additonal PIE is needed. We simply subtract $2\cdot12$ from our original count now, yielding $8 \cdot 3 \cdot 8 - 2 \cdot 12 = 168.$

With no restrictions, there are a total of $6^4$ ways to move four steps. Your probability is $\frac{168}{6^4}=\frac{7}{54}.$

~mathboy282


Video Solution by Power Solve

https://www.youtube.com/watch?v=2uTWPiWzfB0

Video Solution by Pi Academy (Fast and Easy ⚡️🚀)

https://youtu.be/3Dk-vmdxbrw?si=G6TcyLN2Rfwd2YYJ

Video Solution by Innovative Minds

https://youtu.be/CxLGUPhYYDI

~i_am_suk_at_math_2

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=6SQ74nt3ynw

See also

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_10A_Problems/Problem_24&oldid=235240"