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Difference between revisions of "2022 AMC 12A Problems/Problem 5"

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==Solution 1 (Pick's Theorem)==
 
==Solution 1 (Pick's Theorem)==
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Let <math>P = (x, y)</math>. Since the problem asks for taxicab distances from the origin, we want <math>|x| + |y| \le 20</math>. The graph of all solutions to this equation on the <math>xy</math>-plane is a square with vertices at <math>(0, \pm 20)</math> and <math>(\pm 20, 0)</math> (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.  
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  Let <math>P = (x, y)</math>. Since the problem asks for taxicab distances from the origin, we want <math>|x| + |y| \le 20</math>. The graph of all solutions to this equation on the <math>xy</math>-plane is a square with vertices at <math>(0, \pm 20)</math> and <math>(\pm 20, 0)</math> (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 <math>(0,20)</math> to <math>(20, 0)</math>, we can see that it goes through <math>21</math> points in total. In addition, each of the vertices gets counted twice, so the total number of border points is <math>21\cdot4 - 4 = 80</math>. Also, the area of the square is <math>800</math>, so when we plug this information inside Pick's theorem, we get <math>800 = i + \frac{80}{2} - 1 \implies i = 761</math>. Then our answer is <math>761+80 = \boxed{841}.</math>
−
  Each side of the square goes through an equal number of lattice points, so if we focus on one side going from <math>(0,20)</math> to <math>(20, 0)</math>, we can see that it goes through <math>21</math> points in total. In addition, each of the vertices gets counted twice, so the total number of border points is <math>21\cdot4 - 4 = 80</math>. Also, the area of the square is <math>800</math>, so when we plug this information inside Pick's theorem, we get <math>800 = i + \frac{80}{2} - 1 \implies i = 761</math>. Then our answer is <math>761+80 = \boxed{841}.</math>
 
 
   
 
   
 
~ Oxymoronic15
 
~ Oxymoronic15

Revision as of 12:15, 12 November 2022

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 $761+80 = \boxed{841}.$

~ Oxymoronic15

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