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Difference between revisions of "2021 AMC 12B Problems/Problem 22"

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<math>\textbf{(A) }(6,1,1) \qquad \textbf{(B) }(6,2,1) \qquad \textbf{(C) }(6,2,2)\qquad \textbf{(D) }(6,3,1) \qquad \textbf{(E) }(6,3,2)</math>
 
<math>\textbf{(A) }(6,1,1) \qquad \textbf{(B) }(6,2,1) \qquad \textbf{(C) }(6,2,2)\qquad \textbf{(D) }(6,3,1) \qquad \textbf{(E) }(6,3,2)</math>
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== Solution ==
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First we note that symmetrical positions are losing for the player to move. Then we start checking small positions. <math>(n)</math> is always winning for the first player. Furthermore, <math>(3, 2, 1)</math> is losing and so is <math>(4, 1).</math> We look at all the positions created from <math>(6, 2, 1),</math> as <math>(6, 1, 1)</math> is obviously winning by playing <math>(2, 2, 1, 1).</math> There are several different positions that can be played by the first player from <math>(6, 2, 1).</math> They are <math>(2, 2, 2, 1), (1, 3, 2, 1), (4, 2, 1), (6, 1), (5, 2, 1), (4, 1, 2, 1), (3, 2, 2, 1).</math> Now we list refutations for each of these moves:
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<math>(2, 2, 2, 1) - (2, 1, 2, 1)</math>
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<math>(1, 3, 2, 1) - (3, 2, 1)</math>
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<math>(4, 2, 1) - (4, 1)</math>
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<math>(6, 1) - (4, 1)</math>
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<math>(5, 2, 1) - (3, 2, 1)</math>
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<math>(4, 1, 2, 1) - (2, 1, 2, 1)</math>
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<math>(3, 2, 2, 1) - (1, 2, 2, 1)</math>
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This proves that <math>(6, 2, 1)</math> is losing for the first player.
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-Note: In general, this game is very complicated. For example <math>(8, 7, 5, 3, 2)</math> is winning for the first player but good luck showing that.
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== Solution 2 (Process of Elimination)==
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<math>(6,1,1)</math> can be turned into <math>(2,2,1,1)</math> by Arjun, which is symmetric, so Beth will lose.
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<math>(6,3,1)</math> can be turned into <math>(3,1,3,1)</math> by Arjun, which is symmetric, so Beth will lose.
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<math>(6,2,2)</math> can be turned into <math>(2,2,2,2)</math> by Arjun, which is symmetric, so Beth will lose.
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<math>(6,3,2)</math> can be turned into <math>(3,2,3,2)</math> by Arjun, which is symmetric, so Beth will lose.
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That leaves <math>(6,2,1)</math> or <math>\boxed{\textbf{(B)}}</math>.
  
 
== Video Solution by OmegaLearn (Game Theory) ==
 
== Video Solution by OmegaLearn (Game Theory) ==
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==See Also==
 
==See Also==
 
{{AMC12 box|year=2021|ab=B|num-b=21|num-a=23}}
 
{{AMC12 box|year=2021|ab=B|num-b=21|num-a=23}}
 +
{{AMC10 box|year=2021|ab=B|num-b=23|num-a=25}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 01:01, 13 February 2021

Problem

Arjun and Beth play a game in which they take turns removing one brick or two adjacent bricks from one "wall" among a set of several walls of bricks, with gaps possibly creating new walls. The walls are one brick tall. For example, a set of walls of sizes $4$ and $2$ can be changed into any of the following by one move: $(3,2),(2,1,2),(4),(4,1),(2,2),$ or $(1,1,2).$

[asy] unitsize(4mm); real[] boxes = {0,1,2,3,5,6,13,14,15,17,18,21,22,24,26,27,30,31,32,33}; for(real i:boxes){ draw(box((i,0),(i+1,3))); } draw((8,1.5)--(12,1.5),Arrow()); defaultpen(fontsize(20pt)); label(",",(20,0)); label(",",(29,0)); label(",...",(35.5,0)); [/asy]

Arjun plays first, and the player who removes the last brick wins. For which starting configuration is there a strategy that guarantees a win for Beth?

$\textbf{(A) }(6,1,1) \qquad \textbf{(B) }(6,2,1) \qquad \textbf{(C) }(6,2,2)\qquad \textbf{(D) }(6,3,1) \qquad \textbf{(E) }(6,3,2)$

Solution

First we note that symmetrical positions are losing for the player to move. Then we start checking small positions. $(n)$ is always winning for the first player. Furthermore, $(3, 2, 1)$ is losing and so is $(4, 1).$ We look at all the positions created from $(6, 2, 1),$ as $(6, 1, 1)$ is obviously winning by playing $(2, 2, 1, 1).$ There are several different positions that can be played by the first player from $(6, 2, 1).$ They are $(2, 2, 2, 1), (1, 3, 2, 1), (4, 2, 1), (6, 1), (5, 2, 1), (4, 1, 2, 1), (3, 2, 2, 1).$ Now we list refutations for each of these moves:


$(2, 2, 2, 1) - (2, 1, 2, 1)$


$(1, 3, 2, 1) - (3, 2, 1)$


$(4, 2, 1) - (4, 1)$


$(6, 1) - (4, 1)$


$(5, 2, 1) - (3, 2, 1)$


$(4, 1, 2, 1) - (2, 1, 2, 1)$


$(3, 2, 2, 1) - (1, 2, 2, 1)$


This proves that $(6, 2, 1)$ is losing for the first player.

-Note: In general, this game is very complicated. For example $(8, 7, 5, 3, 2)$ is winning for the first player but good luck showing that.

Solution 2 (Process of Elimination)

$(6,1,1)$ can be turned into $(2,2,1,1)$ by Arjun, which is symmetric, so Beth will lose.

$(6,3,1)$ can be turned into $(3,1,3,1)$ by Arjun, which is symmetric, so Beth will lose.

$(6,2,2)$ can be turned into $(2,2,2,2)$ by Arjun, which is symmetric, so Beth will lose.

$(6,3,2)$ can be turned into $(3,2,3,2)$ by Arjun, which is symmetric, so Beth will lose.

That leaves $(6,2,1)$ or $\boxed{\textbf{(B)}}$.

Video Solution by OmegaLearn (Game Theory)

https://youtu.be/zkSBMVAfYLo

~ pi_is_3.14

See Also

2021 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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 12 Problems and Solutions
2021 AMC 10B (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

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