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 …') |
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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>? | ||
− | < | + | <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> | ||
<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 | ||
Line 11: | Line 40: | ||
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>. | 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>. | ||
− | The answer is <math>\boxed{D}</math> | + | The answer is <math>\boxed{\textbf{(D)}}</math> |
<center>[[Image:AMC8_2007_11S.png]]</center> | <center>[[Image:AMC8_2007_11S.png]]</center> | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=66MVYKyetGc | ||
+ | |||
+ | |||
+ | ==Video Solution by Why Math== | ||
+ | https://youtu.be/WobCYII7TRg | ||
+ | |||
+ | ~savannahsolver | ||
+ | ==Video Solution by SpreadTheMathLove== | ||
+ | https://www.youtube.com/watch?v=omFpSGMWhFc | ||
+ | ==See Also== | ||
+ | {{AMC8 box|year=2007|num-b=10|num-a=12}} | ||
+ | {{MAA Notice}} |
Latest revision as of 14:53, 2 July 2024
Contents
[hide]Problem
Tiles and are translated so one tile coincides with each of the rectangles and . 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 ?
cannot be determined
Solution
We first notice that tile III has a on the bottom and a on the right side. Since no other tile has a or a , Tile III must be in rectangle . Tile III also has a on the left, so Tile IV must be in Rectangle .
The answer is
Video Solution
https://www.youtube.com/watch?v=66MVYKyetGc
Video Solution by Why Math
~savannahsolver
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=omFpSGMWhFc
See Also
2007 AMC 8 (Problems • Answer Key • Resources) | ||
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.