2007 AMC 10A Problem 19

Attempt Problem 19 of the 2007 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2007 AMC 10A solutions, or check the answer key.

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19.

A paint brush is swept along both diagonals of a square to produce the symmetric painted area, as shown. Half the area of the square is painted. What is the ratio of the side length of the square to the brush width?

22+12\sqrt{2} + 1

323\sqrt{2}

22+22\sqrt{2} + 2

32+13\sqrt{2} + 1

32+23\sqrt{2} + 2

Answer: C
Concepts:area decompositionspecial right trianglerationalizing denominator
Difficulty rating: 1820
Solution:

Let ss be the side, ww the brush width, and xx the leg of one unpainted isosceles right triangle. Each triangle has area 18s2,\tfrac18 s^2, so 12x2=18s2\tfrac12 x^2 = \tfrac18 s^2 and x=s2.x = \tfrac{s}{2}.

The leg plus the brush width is half the diagonal: x+w=22s.x + w = \tfrac{\sqrt2}{2} s. Thus w=22ss2.w = \tfrac{\sqrt2}{2} s - \tfrac{s}{2}.

Therefore sw=221=22+2. \dfrac{s}{w} = \dfrac{2}{\sqrt2 - 1} = 2\sqrt2 + 2.

Thus, the correct answer is C.

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