AMC 10 Step by Step

2001 AMC 10 Problem 18

P16-20~3 minPrint

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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Key insight

The pattern repeats in 3 by 3 blocks. The four corner squares of a block cover 4 of its 9 units of area, so pentagons cover 5 out of 9.

Solution

  1. The plane goes on forever, so we cannot measure all of it. But the picture is one block repeated over and over, so the fraction covered by pentagons in one block is the fraction for the whole plane.
  2. Use the side of a small square as the unit of length. One block is a 3×33 \times 3 square, so its area is 99.
  3. Inside one block, there is a small 1×11 \times 1 square in each of the four corners. Together they have area 44.
  4. Everything else in the block is pentagon. (Each pentagon is the 1×11 \times 1 square in the middle of one edge, plus a triangle that is one quarter of the center square.) So the pentagons cover 9−4=59 - 4 = 5.
  5. The pentagon fraction is 59\frac{5}{9}. As a decimal, 5÷9=0.5555…5 \div 9 = 0.5555\ldots, which is about 55.6%55.6\%.
  6. The closest choice is 5656.

The answer is (D) 56\boxed{\textbf{(D)}\ 56}.

Why this works

For a pattern that repeats forever, "what fraction of the plane" is the same as "what fraction of one repeating block." Pick the block, then count the easy pieces (here, the squares) and subtract, rather than working out the area of the odd-shaped pieces directly. The answer choices are only 22 apart, so get an exact fraction before rounding.

Alternative approach

Find the area of one pentagon directly. It is a 1×11 \times 1 square (area 11) plus a quarter of the center square (area 14\frac14), so its area is 54\frac54. Four pentagons give 4×54=54 \times \frac54 = 5 out of a block of area 99, the same 59\frac59.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

The tiling repeats: the picture is made of identical 3×33 \times 3 blocks (measured in units of the small square's side), so the fraction covered by pentagons in one block is the fraction for the whole plane.

Take one block, area 3⋅3=93 \cdot 3 = 9. It contains four unit squares, one in each corner, with total area 44. Everything else in the block belongs to the four pentagons (each pentagon is a unit square along an edge plus a quarter of the central unit square, area 1+141 + \frac14), so the pentagons cover 9−4=59 - 4 = 5.

The pentagon share is 59=0.555…\frac{5}{9} = 0.555\ldots, i.e. about 55.6%55.6\%, and the nearest choice is 5656.

The answer is (D) 56\boxed{\textbf{(D)}\ 56}.

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site. Also appeared as 2001 12 #10.