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Difference between pages "2024 AMC 10A Problems/Problem 25" and "2024 AMC 10A Problems/Problem 24"

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{{duplicate|[[2024 AMC 10A Problems/Problem 25|2024 AMC 10A #25]] and [[2024 AMC 12A Problems/Problem 22|2024 AMC 12A #22]]}}
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== Problem ==
  
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==Problem==
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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?
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The figure below shows a dotted grid <math>8</math> cells wide and <math>3</math> cells tall consisting of <math>1''\times1''</math> squares. Carl places <math>1</math>-inch toothpicks along some of the sides of the squares to create a closed loop that does not intersect itself. The numbers in the cells indicate the number of sides of that square that are to be covered by toothpicks, and any number of toothpicks are allowed if no number is written. In how many ways can Carl place the toothpicks?
 
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<asy>
 
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size(6cm);
 
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for (int i=0; i<9; ++i) {
 
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  draw((i,0)--(i,3),dotted);
 
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}
 
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for (int i=0; i<4; ++i){
 
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  draw((0,i)--(8,i),dotted);
 
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}
 
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for (int i=0; i<8; ++i) {
 
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  for (int j=0; j<3; ++j) {
 
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    if (j==1) {
 
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      label("1",(i+0.5,1.5));
 
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}}}
 
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</asy>
 
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<math>\textbf{(A) }130\qquad\textbf{(B) }144\qquad\textbf{(C) }146\qquad\textbf{(D) }162\qquad\textbf{(E) }196</math>
 
  
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==(Best) Solution 1==
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<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>
  
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===Diagrams===
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Diagrams have been moved to the bottom of the solutions.
  
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Observations:
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==Solution 1 (Simplest)==
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We start by imagining the three dimensional plane.
  
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1. You can not have a vertical line in any place other than the first and second columns and the last and second-to-last columns.
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[[File:Pic1.png | center | 650px]]
  
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2. You can place a box around the top row or along the bottom row, otherwise all the solutions have a vertical line at the first and second columns and the last and second-to-last columns.
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Notice how the three dimensional plane is split into 8 different regions. There exists 8 cubes with one vertex on the point <math>(0,0,0)</math>. We need to consider each of these cases.
  
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Thus, Using casework, we can split this problem into 4 cases. However, we can focus on only the first one for right now.  
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We arbitrarily take a cube from one region of the three dimensional plane.
  
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For case 1, we assume that the green lines shown below are given (always have toothpicks on them)
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Below is a sample drawing.
  
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[[Image:Scrasdfasd.png|600px]]
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[[File:Q241.png | center | 650px]]
  
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[[Image:Screenshot 2024-11-08 192200.png|600px]]
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Roll 1: From <math>(0,0,0)</math>, there are 3 favorable outcomes that can occur. They are listed below.
  
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And the only toothpicks we can place that will connect to the red lines are to go horizontally inward:
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[[File:Q242.png | center | 650 px]]
  
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[[Image:Screenshot 2024-11-08 192516.png|600px]]
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Roll 2: WLOG, say the bee moved up one, there are then 2 ways the bee can go.
  
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Now, concentrate on the first row of squares. A toothpick can be placed on either the bottom or top and connected to a continuous squiggle by adding vertical toothpicks, for example,
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[[File:Q243.png | center | 650px]]
  
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[[Image:Screenshot 2024-11-08 193401.png|600px]]
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Roll 3: WLOG, say the bee moved forward one. From this point, there is again 2 ways the bee can go.
  
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How many squiggles are possible?
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[[File:Q244.png | center | 650px]]
  
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[[Image:Screenshot 2024-11-08 194804.png|600px]]
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Roll 4: Finally, WLOG, say the bee moves to the left one. From that point, there are still 2 ways the bee can go.
  
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We can summarize this by giving a high squiggle position a 1 and a low position a 0, thus we have a 6-digit binary sequence. Thus, we can have <math>2^6=64</math> ways to make this squiggle.  
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[[File:Q245.png | center | 650px]]
  
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Case 2: We can also, as the observations state, pull in one of the sides, thus we can have a squiggle with 5 binary digits. only utilizing the first 7 columns:
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The total probability for one cube is <math>\frac{3 \cdot 2 \cdot 2 \cdot 2}{6^4} \Rightarrow \frac{1}{54} </math>. We have 8 of these cubes, which gives us a total probability of <math> \frac{8}{54} </math>.
  
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[[Image:Screenshot 2024-11-08 195135.png|600px]]
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However, we have overcounted. Notice that if the bee moved down one instead of left one at roll 3, it would only have one way. This is because the bee has a choice between either the origin or a unique point. This will always occur on the third move, and because we have 8 cubes, the probability of this happening is <math> \frac{3 \cdot 8}{6^4} = \frac{1}{54} </math>.
  
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Here, we only have 5 binary digits to work with, so there are <math>2^5=32</math> ways to make this squiggle.  
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Our probability is <math> \frac{8}{54} - \frac{1}{54} = </math> <math>\boxed{\textbf{(B) }\frac{7}{54}}</math>.
  
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Case 3: Similarly, we can utilize the last 7 columns.
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~Pinotation
  
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[[Image:Screenshot 2024-11-08 195553.png|600px]]
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== Solution 2 ==
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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.
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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.
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The first case is if the bee moves to the corner of a cube farthest away from the origin.
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Here, there is a <math>\frac{2}{6}</math> chance of this happening and a <math>\frac{2}{6}</math> chance of the fourth move remaining on one of the cubes.
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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.
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Now, multiply and sum for the answer.
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<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>
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Evaluating this gives you the answer of <math>\boxed{\textbf{(B) }\frac{7}{54}}</math>.
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Solution by [[User:Juwushu|juwushu]].
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Minor edits by andliu766
  
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Again, we only have 5 binary digits to work with, so there are <math>2^5=32</math> ways to make this squiggle.  
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==Solution 3==
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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.
  
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Case 4: We can use an even smaller section. Using only the middle 6 columns gives us a 4-wide squiggle:
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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.
  
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[[Image:Screenshot 2024-11-08 195945.png|600px]]
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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>
  
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Thus, there are <math>2^4=16</math> ways to make this squiggle.  
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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.
  
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Adding up all our cases: <math>64+32+32+16=144</math>
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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>  
  
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However, there are two more ways to draw a qualifying shape:
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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>
  
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[[Image:Screenshot 2024-11-08 200416.png|600px]]
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~mathboy282
  
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We can draw a rectangle like that in the first row or third row. Thus, we have a grand total of
 
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<math>144+2=\boxed{\textbf{(C) }146}</math> ways.
 
  
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== Video Solution by Power Solve ==
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https://www.youtube.com/watch?v=2uTWPiWzfB0
  
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A note to (potential) editors:
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== Video Solution by Pi Academy (Fast and Easy ⚡️🚀) ==
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This answer was not made to be concise or especially professional. It was made to explicitly explain this problem in a way so that it is easy to understand and follow.
 
  
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~hermanboxcar5
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https://youtu.be/3Dk-vmdxbrw?si=G6TcyLN2Rfwd2YYJ
  
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Notes:
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==Video Solution by Innovative Minds==
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Remember these are the ONLY possible cases. It is impossible to cross through the rows of boxes of ones to connect the loop around the bottom since then the loop would intersect itself if you are to put only one toothpick on each box with a one (a more clear definition of Observation 1).
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https://youtu.be/CxLGUPhYYDI
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You are not undercounting by only counting the binary digits on the top, because all of the digits on one side will have corresponding opposite digits on the other.
 
  
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~juwushu
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~i_am_suk_at_math_2
  
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==Solution 2 (Cheese)==
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==Video Solution by SpreadTheMathLove==
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Notice that for any case where the closed loop does not connect from the top side of the ones and bottom side of the ones, there are two of these cases. A cheese solution can be found from this; noting that B and C are the only two options to each other, and, being two apart, with people likely to forget this case, <math>\boxed{\textbf{(C) }146}</math> is likely to be the correct answer.
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https://youtu.be/ryqz2nVuG34?si=T3IU_aF8TJD1qmhT
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Cheese solution done by [[User:Juwushu|juwushu]].
 
  
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==Solution 3 (Observation)==
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== Video solution by TheNeuralMathAcademy ==
 +
https://www.youtube.com/watch?v=4b_YLnyegtw&t=5535s
  
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We have <math>2</math> cases where the loop does not go through the middle.  
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== Diagrams ==
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[[Image:Diagrams.jpeg|200px]]
  
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If the loop goes through the middle, we must have a full column on <math>(0,8),(0,7),(1,8),(1,7).</math> Then we have <math>6,5,5,4</math> empty middle squares. For each one we can have one on top or one on bottom, so <math>2^6+2^5+2^5+2^4=144.</math> Notice that for each case of fixed toothpicks, there is only one way to form the loop. Then we just add <math>2+144=\boxed{\textbf{(C) } 146.}</math>
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[[File:Bee total.png|600px|right]]
  
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~nevergonnagiveup
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~ Diagram by PaperMath
  
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== Video Solution by Power Solve ==
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[[Image:2024amc10a24solution.jpg|300px]]
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https://www.youtube.com/watch?v=FRNbJ5wIGRo
 
  
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==Video Solution By SpreadTheMathLove==
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~ Solution diagram by juwushu
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https://www.youtube.com/watch?v=huMQ9J7rIj0&t=77s
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Up down left right of course
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== Video solution by TheNeuralMathAcademy ==
 
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https://www.youtube.com/watch?v=4b_YLnyegtw&t=5850s
 
  
 
==See Also==
 
==See Also==

Revision as of 19:08, 6 October 2025

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 have been moved to the bottom of the solutions.

Solution 1 (Simplest)

We start by imagining the three dimensional plane.

Pic1.png

Notice how the three dimensional plane is split into 8 different regions. There exists 8 cubes with one vertex on the point $(0,0,0)$. We need to consider each of these cases.

We arbitrarily take a cube from one region of the three dimensional plane.

Below is a sample drawing.

Q241.png

Roll 1: From $(0,0,0)$, there are 3 favorable outcomes that can occur. They are listed below.

Q242.png

Roll 2: WLOG, say the bee moved up one, there are then 2 ways the bee can go.

Q243.png

Roll 3: WLOG, say the bee moved forward one. From this point, there is again 2 ways the bee can go.

Q244.png

Roll 4: Finally, WLOG, say the bee moves to the left one. From that point, there are still 2 ways the bee can go.

Q245.png

The total probability for one cube is $\frac{3 \cdot 2 \cdot 2 \cdot 2}{6^4} \Rightarrow \frac{1}{54}$. We have 8 of these cubes, which gives us a total probability of $\frac{8}{54}$.

However, we have overcounted. Notice that if the bee moved down one instead of left one at roll 3, it would only have one way. This is because the bee has a choice between either the origin or a unique point. This will always occur on the third move, and because we have 8 cubes, the probability of this happening is $\frac{3 \cdot 8}{6^4} = \frac{1}{54}$.

Our probability is $\frac{8}{54} - \frac{1}{54} =$ $\boxed{\textbf{(B) }\frac{7}{54}}$.

~Pinotation

Solution 2

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 3

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://youtu.be/ryqz2nVuG34?si=T3IU_aF8TJD1qmhT

Video solution by TheNeuralMathAcademy

https://www.youtube.com/watch?v=4b_YLnyegtw&t=5535s

Diagrams

Diagrams.jpeg

Bee total.png

~ Diagram by PaperMath

2024amc10a24solution.jpg

~ Solution diagram by juwushu Up down left right of course

See Also

2024 AMC 10A (Problems • Answer Key • Resources)
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2023 AMC 10B Problems 		Followed by
2024 AMC 10B Problems
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All AMC 10 Problems and Solutions

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