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2022 AMC 12A Problems/Problem 5

Revision as of 12:23, 12 November 2022 by Jamesl123456 (talk | contribs)

Problem

The $\textit{taxicab distance}$ between points $(x_1, y_1)$ and $(x_2, y_2)$ in the coordinate plane is given by $|x_1 - x_2| + |y_1 - y_2|$. For how many points $P$ with integer coordinates is the taxicab distance between $P$ and the origin less than or equal to $20$?

$\textbf{(A)} \, 441 \qquad\textbf{(B)} \, 761 \qquad\textbf{(C)} \, 841 \qquad\textbf{(D)} \, 921 \qquad\textbf{(E)} \, 924$

Solution 1 (Pick's Theorem)

Let $P = (x, y)$. Since the problem asks for taxicab distances from the origin, we want $|x| + |y| \le 20$. The graph of all solutions to this equation on the $xy$-plane is a square with vertices at $(0, \pm 20)$ and $(\pm 20, 0)$ (In order to prove this, one can divide the sections of this graph into casework on the four quadrants, and tie together the resulting branches.) We want the number of lattice points on the border of the square and inside the square. Each side of the square goes through an equal number of lattice points, so if we focus on one side going from $(0,20)$ to $(20, 0)$, we can see that it goes through $21$ points in total. In addition, each of the vertices gets counted twice, so the total number of border points is $21\cdot4 - 4 = 80$. Also, the area of the square is $800$, so when we plug this information inside Pick's theorem, we get $800 = i + \frac{80}{2} - 1 \implies i = 761$. Then our answer is $761+80 = \boxed{\textbf{(C)} \, 841}.$

~ Oxymoronic15

Solution 2 (Stars and Bars)

Instead of considering all points with integer coordinates, first consider points with nonnegative coordinates only. Then, we want $x + y \le 20$. Note that we can introduce a third variable, say $z$, such that $z = 20 - (x + y)$. Then, counting the ways to have $x + y + z = 20$ is the same as counting the ways to have $x + y \le 20$. Therefore, there are $\binom{20 + 3 - 1}{3 - 1} = 231$ solutions with nonnegative integer coordinates. Then, we can copy our solutions over to the other four quadrants. First, so as not to overcount, we remove all points on the axes. There are $20 + 20 + 1 = 41$ such points. Then we multiply the $190$ remaining points by $4$ to get $760$ points that are not on the axes. Then, we can add the $81$ points on the axes to get $760 + 81 = \boxed{\textbf{(C)} \, 841}.$

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