Difference between revisions of "2001 AMC 12 Problems/Problem 10"
(New 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 <math> \text{(A) }50 ...) |
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| + | {{duplicate|[[2001 AMC 12 Problems|2001 AMC 12 #10]] and [[2001 AMC 10 Problems|2001 AMC 10 #18]]}} | ||
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== Problem == | == Problem == | ||
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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 | 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 | ||
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<asy> | <asy> | ||
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</asy> | </asy> | ||
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| + | <math> \textbf{(A) }50 \qquad \textbf{(B) }52 \qquad \textbf{(C) }54 \qquad \textbf{(D) }56 \qquad \textbf{(E) }58 </math> | ||
== Solution == | == Solution == | ||
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Consider any single tile: | Consider any single tile: | ||
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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{56}</math>. | + | 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>. |
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| + | == Video Solution == | ||
| + | https://www.youtube.com/watch?v=P43NYTbZH4k | ||
== See Also == | == See Also == | ||
{{AMC12 box|year=2001|num-b=9|num-a=11}} | {{AMC12 box|year=2001|num-b=9|num-a=11}} | ||
| + | {{AMC10 box|year=2001|num-b=17|num-a=19}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 12:26, 13 September 2024
- The following problem is from both the 2001 AMC 12 #10 and 2001 AMC 10 #18, so both problems redirect to this page.
Contents
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
![[asy] unitsize(3mm); defaultpen(linewidth(0.8pt)); path p1=(0,0)--(3,0)--(3,3)--(0,3)--(0,0); path p2=(0,1)--(1,1)--(1,0); path p3=(2,0)--(2,1)--(3,1); path p4=(3,2)--(2,2)--(2,3); path p5=(1,3)--(1,2)--(0,2); path p6=(1,1)--(2,2); path p7=(2,1)--(1,2); path[] p=p1^^p2^^p3^^p4^^p5^^p6^^p7; for(int i=0; i<3; ++i) { for(int j=0; j<3; ++j) { draw(shift(3*i,3*j)*p); } } [/asy]](http://latex.artofproblemsolving.com/2/c/8/2c8743395e4a086d2ae9a30a2d161157e7dc3077.png)
Solution
Consider any single tile:
![[asy] unitsize(1cm); defaultpen(linewidth(0.8pt)); path p1=(0,0)--(3,0)--(3,3)--(0,3)--(0,0); path p2=(0,1)--(1,1)--(1,0); path p3=(2,0)--(2,1)--(3,1); path p4=(3,2)--(2,2)--(2,3); path p5=(1,3)--(1,2)--(0,2); path p6=(1,1)--(2,2); path p7=(2,1)--(1,2); path[] p=p1^^p2^^p3^^p4^^p5^^p6^^p7; draw(p); [/asy]](http://latex.artofproblemsolving.com/3/6/f/36f7a20120732a8955c3f77ea0d1e81fc34cc6ea.png)
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 
.
Video Solution
https://www.youtube.com/watch?v=P43NYTbZH4k
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. 
