Difference between revisions of "2001 AMC 12 Problems/Problem 10"
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If the side of the small square is <math>a</math>, then the area of the tile is <math>9a^2</math>, with <math>4a^2</math> covered by squares and <math>5a^2</math> by pentagons. | If the side of the small square is <math>a</math>, then the area of the tile is <math>9a^2</math>, with <math>4a^2</math> covered by squares and <math>5a^2</math> by pentagons. | ||
− | Hence exactly <math>5/9</math> of any tile are covered by pentagons, and therefore pentagons cover <math>5/9</math> of the plane. When expressed as a percentage, this is <math>55.\overline{5}\%</math>, and the closest integer to this value is <math>\boxed{ | + | Hence exactly <math>5/9</math> of any tile are covered by pentagons, and therefore pentagons cover <math>5/9</math> of the plane. When expressed as a percentage, this is <math>55.\overline{5}\%</math>, and the closest integer to this value is <math>\boxed{\textbf{(D) }56}</math>. |
== See Also == | == See Also == |
Latest revision as of 00:11, 15 August 2022
- The following problem is from both the 2001 AMC 12 #10 and 2001 AMC 10 #18, so both problems redirect to this page.
Problem
The plane is tiled by congruent squares and congruent pentagons as indicated. The percent of the plane that is enclosed by the pentagons is closest to
Solution
Consider any single tile:
If the side of the small square is , then the area of the tile is
, with
covered by squares and
by pentagons.
Hence exactly
of any tile are covered by pentagons, and therefore pentagons cover
of the plane. When expressed as a percentage, this is
, and the closest integer to this value is
.
See Also
2001 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 9 |
Followed by Problem 11 |
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 |
2001 AMC 10 (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.