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Difference between revisions of "2021 AMC 10B Problems/Problem 24"

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{{AMC10 box|year=2021|ab=B|num-b=23|num-a=25}}

Revision as of 23:54, 11 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] /* CREDITS */ /* Made by samrocksnature */ /* Modified commas an periods by forester2015 */ /* Import and Set variables */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -20.98617651190462, xmax = 71.97119514540032, ymin = -24.802885633158763, ymax = 28.83570218998556; /* image dimensions */ /* draw figures */ draw((14,4)--(13.010050506338834,3.0100505063388336), linewidth(1)); draw((14,4)--(13.010050506338834,4.989949493661166), linewidth(1)); draw((10,4)--(14,4), linewidth(1)); draw((4,6)--(8,6), linewidth(1)); draw((4,2)--(8,2), linewidth(1)); draw((8,2)--(8,6), linewidth(1)); draw((4,6)--(4,2), linewidth(1)); draw((6,6)--(6,2), linewidth(1)); draw((-6,6)--(-6,2), linewidth(1)); draw((-6,6)--(2,6), linewidth(1)); draw((2,6)--(2,2), linewidth(1)); draw((2,2)--(-6,2), linewidth(1)); draw((-4,2)--(-4,6), linewidth(1)); draw((-2,6)--(-2,2), linewidth(1)); draw((0,2)--(0,6), linewidth(1)); draw((50,6)--(50,2), linewidth(1)); draw((50,2)--(58,2), linewidth(1)); draw((58,2)--(58,6), linewidth(1)); draw((58,6)--(50,6), linewidth(1)); draw((52,6)--(52,2), linewidth(1)); draw((54,6)--(54,2), linewidth(1)); draw((56,6)--(56,2), linewidth(1)); draw((32,6)--(32,2), linewidth(1)); draw((46,2)--(46,6), linewidth(1)); draw((34,6)--(34,2), linewidth(1)); draw((36,2)--(36,6), linewidth(1)); draw((38,6)--(38,2), linewidth(1)); draw((40,2)--(40,6), linewidth(1)); draw((42,6)--(42,2), linewidth(1)); draw((44,2)--(44,6), linewidth(1)); draw((16,6)--(16,2), linewidth(1)); draw((28,2)--(28,6), linewidth(1)); draw((18,6)--(18,2), linewidth(1)); draw((20,6)--(20,2), linewidth(1)); draw((22,6)--(22,2), linewidth(1)); draw((24,6)--(24,2), linewidth(1)); draw((26,6)--(26,2), linewidth(1)); draw((16,6)--(22,6), linewidth(1)); draw((24,6)--(28,6), linewidth(1)); draw((16,2)--(22,2), linewidth(1)); draw((24,2)--(28,2), linewidth(1)); draw((32,6)--(36,6), linewidth(1)); draw((32,2)--(36,2), linewidth(1)); draw((38,6)--(40,6), linewidth(1)); draw((38,2)--(40,2), linewidth(1)); draw((42,6)--(46,6), linewidth(1)); draw((42,2)--(46,2), linewidth(1)); /* dots and labels */ label(",",(59,2)); label(".",(60,2)); label(".",(61,2)); label(".",(62,2)); label(",",(29,2)); label(",",(47,2)); [/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

2021 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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All AMC 10 Problems and Solutions
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