Difference between revisions of "2007 AMC 8 Problems/Problem 11"

(Created page with '== Problem == Tiles <math>I, II, III</math> and <math>IV</math> are translated so one tile coincides with each of the rectangles <math>A, B, C</math> and <math>D</math>. In the …')
 
(Solution)
 
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Tiles <math>I, II, III</math> and <math>IV</math> are translated so one tile coincides with each of the rectangles <math>A, B, C</math> and <math>D</math>. In the final arrangement, the two numbers on any side common to two adjacent tiles must be the same. Which of the tiles is translated to Rectangle <math>C</math>?
 
Tiles <math>I, II, III</math> and <math>IV</math> are translated so one tile coincides with each of the rectangles <math>A, B, C</math> and <math>D</math>. In the final arrangement, the two numbers on any side common to two adjacent tiles must be the same. Which of the tiles is translated to Rectangle <math>C</math>?
  
<center>[[Image:AMC8_2007_11.png]]</center>
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<asy>
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size(400);
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defaultpen(linewidth(0.8));
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path p=origin--(8,0)--(8,6)--(0,6)--cycle;
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draw(p^^shift(8.5,0)*p^^shift(8.5,10)*p^^shift(0,10)*p);
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draw(shift(20,2)*p^^shift(28,2)*p^^shift(20,8)*p^^shift(28,8)*p);
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label("8", (4,6+10), S);
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label("6", (4+8.5,6+10), S);
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label("7", (4,6), S);
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label("2", (4+8.5,6), S);
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label("I", (4,6+10), N);
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label("II", (4+8.5,6+10), N);
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label("III", (4,6), N);
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label("IV", (4+8.5,6), N);
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label("3", (0,3+10), E);
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label("4", (0+8.5,3+10), E);
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label("1", (0,3), E);
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label("9", (0+8.5,3), E);
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label("7", (4,10), N);
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label("2", (4+8.5,10), N);
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label("0", (4,0), N);
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label("6", (4+8.5,0), N);
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label("9", (8,3+10), W);
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label("3", (8+8.5,3+10), W);
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label("5", (8,3), W);
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label("1", (8+8.5,3), W);
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label("A", (24,10), N);
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label("B", (32,10), N);
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label("C", (24,4), N);
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label("D", (32,4), N);</asy>
  
 
<math>\mathrm{(A)}\ I \qquad \mathrm{(B)}\ II \qquad \mathrm{(C)}\ III \qquad \mathrm{(D)}\ IV \qquad \mathrm{(E)}</math> cannot be determined
 
<math>\mathrm{(A)}\ I \qquad \mathrm{(B)}\ II \qquad \mathrm{(C)}\ III \qquad \mathrm{(D)}\ IV \qquad \mathrm{(E)}</math> cannot be determined
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== Solution ==
 
== Solution ==
  
We first notice that tile III has a <math>0</math> on the bottom and a <math>5</math> on the right side. Since no other tile has a <math>0</math> or a <math>5</math>, Tile III must be in rectangle <math>D</math>. Tile III also has a <math>1</math> on the left, so Tile IV must be in Rectangle <math>C</math>.
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We first notice that tile III has a 0 on the bottom and a 5 on the right side. Since no other tile has a 0 or a 5 Tile III must be in rectangle D Tile III also has a 1 on the left, so Tile IV must be in Rectangle C
  
The answer is <math>\boxed{D}</math>
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The answer is D
  
<center>[[Image:AMC8_2007_11S.png]]</center>
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==Video Solution by Why Math==
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https://youtu.be/WobCYII7TRg
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 +
~savannahsolver
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==See Also==
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{{AMC8 box|year=2007|num-b=10|num-a=12}}
 +
{{MAA Notice}}

Latest revision as of 19:09, 1 August 2021

Problem

Tiles $I, II, III$ and $IV$ are translated so one tile coincides with each of the rectangles $A, B, C$ and $D$. In the final arrangement, the two numbers on any side common to two adjacent tiles must be the same. Which of the tiles is translated to Rectangle $C$?

[asy] size(400); defaultpen(linewidth(0.8)); path p=origin--(8,0)--(8,6)--(0,6)--cycle; draw(p^^shift(8.5,0)*p^^shift(8.5,10)*p^^shift(0,10)*p); draw(shift(20,2)*p^^shift(28,2)*p^^shift(20,8)*p^^shift(28,8)*p); label("8", (4,6+10), S); label("6", (4+8.5,6+10), S); label("7", (4,6), S); label("2", (4+8.5,6), S); label("I", (4,6+10), N); label("II", (4+8.5,6+10), N); label("III", (4,6), N); label("IV", (4+8.5,6), N); label("3", (0,3+10), E); label("4", (0+8.5,3+10), E); label("1", (0,3), E); label("9", (0+8.5,3), E); label("7", (4,10), N); label("2", (4+8.5,10), N); label("0", (4,0), N); label("6", (4+8.5,0), N); label("9", (8,3+10), W); label("3", (8+8.5,3+10), W); label("5", (8,3), W); label("1", (8+8.5,3), W); label("A", (24,10), N); label("B", (32,10), N); label("C", (24,4), N); label("D", (32,4), N);[/asy]

$\mathrm{(A)}\ I \qquad \mathrm{(B)}\ II \qquad \mathrm{(C)}\ III \qquad \mathrm{(D)}\ IV \qquad \mathrm{(E)}$ cannot be determined

Solution

We first notice that tile III has a 0 on the bottom and a 5 on the right side. Since no other tile has a 0 or a 5 Tile III must be in rectangle D Tile III also has a 1 on the left, so Tile IV must be in Rectangle C

The answer is D

Video Solution by Why Math

https://youtu.be/WobCYII7TRg

~savannahsolver

See Also

2007 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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 AJHSME/AMC 8 Problems and Solutions

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