2022 AMC 12A Problems/Problem 5

1 Problem
2 Diagram
3 Solution 1 (Arithmetic Series)
4 Solution 2 (Arithmetic Series With Symmetry)
5 Solution 3 (Triangular Numbers)
6 Solution 4 (Two Square Arrays)
7 Solution 5 (Pick's Theorem)
8 Solution 6 (Stars and Bars)
9 Video Solution 1 (HOW TO THINK CREATIVELY!!!)
10 Video Solution 2 (First Understand the problem)
11 See Also

Problem

The 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 $20$ ?

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

Diagram

All possible locations of $P$ are lattice points such that $|x|+|y|\leq 20,$ whose graph is shown below: $[asy] /* Made by MRENTHUSIASM */ size(350); int xMin = -21; int xMax = 21; int yMin = -21; int yMax = 21; filldraw((20,0)--(0,20)--(-20,0)--(0,-20)--cycle,yellow,linewidth(1.25)); //Draws the horizontal gridlines void horizontalLines() { for (int i = yMin+1; i < yMax; ++i) { draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4)); } } void verticalLines() { for (int i = xMin+1; i < xMax; ++i) { draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4)); } } void horizontalTicks() { for (int i = yMin+1; i < yMax; ++i) { draw((-3/8,i)--(3/8,i), black+linewidth(1.5)); } } //Draws the vertical ticks void verticalTicks() { for (int i = xMin+1; i < xMax; ++i) { draw((i,-3/8)--(i,3/8), black+linewidth(1.5)); } } horizontalLines(); verticalLines(); horizontalTicks(); verticalTicks(); label("$20$",(0,20),(-2,0),fontsize(10)); label("$20$",(20,0),(0,-2),fontsize(10)); label("${-}20$",(0,-20),(-2,0),fontsize(10)); label("${-}20$",(-20,0),(0,-2),fontsize(10)); draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("$x$",(xMax,0),(2,0)); label("$y$",(0,yMax),(0,2)); for (int y = 20; y >= 0; --y) { for (int x = y-20; x <= 20-y; ++x) { dot((x,y),linewidth(4)); } } for (int y = -1; y >= -20; --y) { for (int x = -y-20; x <= y+20; ++x) { dot((x,y),linewidth(4)); } } [/asy]$

~MRENTHUSIASM

Solution 1 (Arithmetic Series)

Let us consider the number of points for a certain $x$ -coordinate. For any $x$ , the viable points are in the range $[-20 + |x|, 20 - |x|]$ . This means that our total sum is equal to $\begin{align*} 1 + 3 + 5 + \cdots + 41 + 39 + 37 + \cdots + 1 &= (1 + 3 + 5 + \cdots + 39) + (1 + 3 + 5 + \cdots + 41) \\ & = 20^2 + 21^2 \\ & = 29^2 \\ &= \boxed{\textbf{(C)} \, 841}. \end{align*}$ ~mathboy100

Solution 2 (Arithmetic Series With Symmetry)

Since the second point is the origin, this is equivalent to finding all points $(x, y)$ such that $|x| + |y| \leq 20$ . Due to the absolute values, the set of all such points will be symmetric about the origin meaning we can focus on the first quadrant and multiply by $4$ .

To avoid overcounts, ignore points on the axes. This means $x, y > 0$ . If $x = 1$ , there are $19$ solutions for $y$ ( $y = 1, 2, 3, \ldots, 19$ ). If $x = 2$ , there are $18$ solutions. This pattern repeats until $x = 19$ , at which point there is $1$ solution for $y$ .

So we get $19 + 18 + 17 + \cdots + 1 = \frac{19(20)}{2} = 190$ points in the first quadrant. Multiplying by $4$ gives $760$ . Now, the $x$ axis has $y = 0$ which gives $|x| \leq 20$ , meaning there are $41$ solutions. This is the same with the $y$ axis, but we overcounted the origin by $1$ .

Our final answer is $760 + 41 + 41 - 1 = \boxed{\textbf{(C)} \, 841}$ .

Solution 3 (Triangular Numbers)

This solution refers to the Diagram section. $[asy] /* Made by MRENTHUSIASM */ size(350); for (int y = 20; y >= 1; --y) { for (int x = 0; x <= 20-y; ++x) { dot((x,y),green+linewidth(4)); } } for (int y = 0; y >= -20; --y) { for (int x = 1; x <= y+20; ++x) { dot((x,y),blue+linewidth(4)); } } for (int y = 20; y >= 0; --y) { for (int x = y-20; x <= -1; ++x) { dot((x,y),purple+linewidth(4)); } } for (int y = -1; y >= -20; --y) { for (int x = -y-20; x <= 0; ++x) { dot((x,y),red+linewidth(4)); } } dot(origin,black+linewidth(4)); [/asy]$ The problem can be visualized as depicted on the right split equally into four "triangular" parts excluding the origin. The "triangular" parts are identical the ones that would be used in a visual proof of the formula for triangular numbers. Becuase of this the number of points in each part is equal to $\frac{n(n+1)}{2}$ where $n$ is the length of a "leg" of the "triangle" which is $20$ for this problem. Substituting and computing, we get $210.$ Multiplying by $4$ and adding $1$ to account for all parts and the origin, we get $210\cdot4 + 1 = \boxed{\textbf{(C)} \, 841}.$

~Apersoma

Solution 4 (Two Square Arrays)

This solution refers to the Diagram section.

As shown below, the taxicab distance between each red point and the origin is even, and the taxicab distance between each blue point and the origin is odd. $[asy] /* Made by MRENTHUSIASM */ size(350); for (int y = 20; y >= 0; --y) { for (int x = y-20; x <= 20-y; x+=2) { dot((x,y),red+linewidth(4)); } } for (int y = -1; y >= -20; --y) { for (int x = -y-20; x <= y+20; x+=2) { dot((x,y),red+linewidth(4)); } } for (int y = 19; y >= 0; --y) { for (int x = y-19; x <= 19-y; x+=2) { dot((x,y),blue+linewidth(4)); } } for (int y = -1; y >= -19; --y) { for (int x = -y-19; x <= y+19; x+=2) { dot((x,y),blue+linewidth(4)); } } draw((20,0)--(0,20)--(-20,0)--(0,-20)--cycle,red+linewidth(1.25)); draw((19,0)--(0,19)--(-19,0)--(0,-19)--cycle,blue+linewidth(1.25)); [/asy]$ Note that the red array consists of $21^2=441$ points, and the blue array consists of $20^2=400$ points.

Together, the answer is $441+400=\boxed{\textbf{(C)} \, 841}.$

~MRENTHUSIASM

Solution 5 (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 6 (Stars and Bars)

Instead of considering all points with integer coordinates, first consider points with nonnegative coordinates only. Then, we want $x + y \le 20$ where $x$ and $y$ are nonnegative integers. We can introduce a third variable, say $z$ , such that $z = 20 - (x + y)$ . Note that counting the number of ways to have $x + y + z = 20$ is the same as counting the number of ways to have $x + y \le 20$ . Therefore, by stars and bars, there are $\dbinom{20 + 3 - 1}{3 - 1} = 231$ 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 with nonnegative integer coordinates. We multiply the $190$ remaining points by $4$ to get $760$ points that are not on the axes. Then, we can add back the $41$ nonnegative points on the axes, as well as the $40$ other points on the negative axes to get $760 + 41 + 40 = \boxed{\textbf{(C)} \, 841}.$

~ jamesl123456

Video Solution 1 (HOW TO THINK CREATIVELY!!!)

https://youtu.be/zTGRlr7eYKM

~Education, the Study of Everything

Video Solution 2 (First Understand the problem)

https://youtu.be/7yAh4MtJ8a8?si=0CHICdALPhDnrgTx&t=834

~Math-X

2022 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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