2011 AMC 12B Problem 23

Attempt Problem 23 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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

23.

A bug travels in the coordinate plane, moving only along the lines that are parallel to the xx-axis or yy-axis. Let A=(−3,2)A=(-3, 2) and B=(3,−2).B=(3, -2). Consider all possible paths of the bug from AA to BB of length at most 20.20. How many points with integer coordinates lie on at least one of these paths?
161161
185185
195195
227227
255255
Answer: C
Concepts:lattice pointcaseworksymmetry
Small hint:
A point XX lies on such a path iff its taxicab distances satisfy d(A,X)+d(X,B)≤20d(A,X)+d(X,B)\le20
Big hint:
This is ∣x−3∣+∣x+3∣|x-3|+|x+3| +∣y−2∣+∣y+2∣≤20;+|y-2|+|y+2|\le20; count by symmetry across the axes
Solution:
A lattice point X=(x,y)X=(x,y) lies on some path exactly when d=∣x−3∣+∣x+3∣+∣y−2∣+∣y+2∣≤20. \begin{aligned} d&=|x-3|+|x+3| \\ &\quad {}+|y-2|+|y+2|\le20. \end{aligned} This expression is unchanged when x→−xx\to-x or y→−y,y\to-y, so we count points with x≥0,x\ge0, y≥0,y\ge0, multiply by 4,4, and correct for the axes.
If 0≤x≤30\le x\le3 and 0≤y≤2,0\le y\le2, all 4⋅3=124\cdot3=12 points work. If 0≤x≤30\le x\le3 and y≥3,y\ge3, then y≤7,y\le7, giving 4⋅5=204\cdot5=20 points. If x≥4x\ge4 and 0≤y≤2,0\le y\le2, then x≤8,x\le8, giving 5⋅3=155\cdot3=15 points. Finally, for x≥4x\ge4 and y≥3,y\ge3, the condition is x+y≤10,x+y\le10, giving 4+3+2+1=104+3+2+1=10 points. Thus there are 5757 in the first quadrant, including 1515 on the nonnegative axes. By symmetry the total is 4⋅57−2⋅15−3=195. 4\cdot57-2\cdot15-3=195.
Thus, the correct answer is C.
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