2001 AMC 12 Problems/Problem 10

Revision as of 20:49, 16 November 2013 by Mathcool2009 (talk | contribs) (Solution)
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

$\text{(A) }50 \qquad \text{(B) }52  \qquad \text{(C) }54 \qquad \text{(D) }56 \qquad \text{(E) }58$

[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]

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]

If the side of the small square is $a$, then the area of the tile is $9a^2$, with $4a^2$ covered by squares and $5a^2$ by pentagons. Hence exactly $5/9$ of any tile are covered by pentagons, and therefore pentagons cover $5/9$ of the plane. When expressed as a percentage, this is $55.\overline{5}\%$, and the closest integer to this value is $\boxed{(\mathrm{D})56}$.

See Also

2001 AMC 12 (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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