2011 AMC 12B Problem 19

Attempt Problem 19 of the 2011 AMC 12B 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 2011 AMC 12B solutions, or check the answer key.

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19.

A lattice point in an xyxy-coordinate system is any point (x,y)(x, y) where both xx and yy are integers. The graph of y=mx+2y=mx+2 passes through no lattice point with 0<x1000 \lt x \le 100 for all mm such that 12<m<a.\dfrac{1}{2} \lt m \lt a. What is the maximum possible value of a?a?

51101\dfrac{51}{101}

5099\dfrac{50}{99}

51100\dfrac{51}{100}

52101\dfrac{52}{101}

1325\dfrac{13}{25}

Answer: B
Concepts:lattice pointslope
Difficulty rating: 2090
Solution:

For 0<x100,0\lt x\le100, the nearest lattice point above the line y=12x+2y=\tfrac12x+2 is (x,12x+3)\left(x,\tfrac12x+3\right) if xx is even and (x,12x+52)\left(x,\tfrac12x+\tfrac52\right) if xx is odd.

The slope from (0,2)(0,2) to that point is 12+1x\dfrac12+\dfrac1x for even xx and 12+12x\dfrac12+\dfrac{1}{2x} for odd x.x. The minimum such slope is 51100\dfrac{51}{100} for even xx and 5099\dfrac{50}{99} for odd x.x.

Since 5099<51100,\dfrac{50}{99}\lt\dfrac{51}{100}, the line avoids all these lattice points exactly when 12<m<5099,\dfrac12\lt m\lt\dfrac{50}{99}, so the maximum is a=5099.a=\dfrac{50}{99}.

Thus, the correct answer is B.

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