2011 AMC 12B Problem 23

A lattice point $X=(x,y)$ lies on some path exactly when $$\begin{aligned} d&=|x-3|+|x+3| \\ &\quad {}+|y-2|+|y+2|\le20. \end{aligned}$$ This expression is unchanged when $x\to-x$ or $y\to-y,$ so we count points with $x\ge0,$ $y\ge0,$ multiply by $4,$ and correct for the axes. If $0\le x\le3$ and $0\le y\le2,$ all $4\cdot3=12$ points work. If $0\le x\le3$ and $y\ge3,$ then $y\le7,$ giving $4\cdot5=20$ points. If $x\ge4$ and $0\le y\le2,$ then $x\le8,$ giving $5\cdot3=15$ points. Finally, for $x\ge4$ and $y\ge3,$ the condition is $x+y\le10,$ giving $4+3+2+1=10$ points. Thus there are $57$ in the first quadrant, including $15$ on the nonnegative axes. By symmetry the total is $$4\cdot57-2\cdot15-3=195.$$ Thus, the correct answer is C.