Absolute Value Sum Inequality

$|x|+|y| \leq 10$ represent a square with its vertices $(10,0),(0,10),(-10,0)$ and $(0,-10)$ . The area of the square is $200$ units.

From Pick's Theorem , we have

$\displaystyle A=i+\frac{b}{2}-1$

where $A$ is the area of polygon (in this case, square), $i$ is the number of points lying inside the square and $b$ is the number of points lying on the edge of square.

From manual counting, the number of points on the edge are 40 i.e $b=40$ .

We have to find $N=i+b$ .

$\displaystyle A=i+b-\frac{b}{2}-1 \Rightarrow 200=N-21$

$\displaystyle \fbox{N=221}$

First, we solve for the number of ordered pairs of positive integer solutions (where $x \neq 0, y \neq 0$ ). When $x=1$ , $y=\{1,2,...,9\}$ , so there are $n(y)=9$ solutions; when $x=2$ , $n(y)=8$ ; . . . when $x=9$ , $n(y)=1$ . This gives us a total of 45. However, considering there are positive and negative x's and y's, so there are $45\times 4=180$ solutions.

Now it's the time to count when either x or y is 0. when $x=0$ , $y={ -10,-9,...9,10}$ , $n(y)=21$ similarly, when $y=0$ , $n(x)=21$ . However, we have repeated the extreme case, $(x,y) = (0,0)$ so altogether there are 180+21+21-1=221 ordered pairs :)

The same problem without the absolute value signs is equivalent to choosing 2 balls out of 12 balls in a row: x would be the number of balls before the first chosen ball, and y would be the number of balls between the first and second chosen balls. The total number of pairs of integer values for x and y would then be $\dbinom{12}{2}=66$ .

With the absolute value signs, for each possible pairs of integers (x,y) in the above problem, x and y can now be either positive or negative. This means, for all combinations when x and y are not zero, there are now four combinations: x and y are positive, x is positive but y is negative, x is negative but y is positive, or x and y are negative.

When only x is zero, there are two combinations: y is positive or y is negative. When only y is zero, there are, similarly, two combinations. When both x and y are zero, there is only one combination.

There is one combination when x and y are zero, and 10 combinations each for when only one of x or y is zero. This leaves $66-1-10-10=45$ combinations where neither x nor y is zero.

Thus, the total number of possible combinations is $45 \times 4+2 \times (10 \times 2) + 1=221$ .

Case 1 .when zero is included then For x=0,y has 21 choises For y=0,x has 21 choises But (0,0) counted two times so net solution in case 1 are 21+21-1=41 Case 2.when x and y are non zeroFor x>0, y>0 We have x+y from 2 to 10. Thus we get [1+2+3+4+5+6+7+8+9]=45 sol Now we have four types of solution as (x,y),(x,-y),(-x,y),(-x,-y) SO net sol in case 2 =45*4=180 Final ans =case1+case2=41+180=221

Recall that absolute value is a piecewise-defined function, $-y=x\\ y=x$ .Our inequality can be written in the form $|y|\leq-|x|+10$ . If we use our above definition of absolute value, we can split this further into two inequalities, $y\leq-|x|+10\\ y\geq|x|-10,$ remembering to flip the direction of the inequality in the second inequality to account for multiplying $y$ by $-1$ . Now we can graph these two absolute value functions separately by applying transformations to the function $y=|x|$ . When we graph them, we see the region they enclose is a square with the axes as its diagonals, and vertices at the points $(10,0),(-10,0),(0,10),(0,-10).$ The number of ordered pairs $(x,y)$ that satisfy our original inequality is the number of lattice points (points with integer coordinates) on or enclosed by this region. The "successful" ordered pairs form four "triangles", one in each quadrant. The number of points in each "triangle" is the sum of all the numbers from $1$ to $10$ , which is $1+2+3+\cdots+10=\frac{10(10+1)}{2}=55.$ Since we have four of these "triangles", there are a total of $4\times55=220$ points. Finally, we must account for the origin, which is not counted in any of the four "triangles." This gives us a total of $220+1=\boxed{221}$ ordered pairs.

We have the easiest solution $x=0 ,y=0$ Then all the solutions of the form $x=0, 1 \leq y \leq 10$ so 10 of these but also $y=0, 1 \leq x \leq 10$ which gives another 10. $10+10=20$ . But for each of these pairs of $x,y$ the value which is greater than $0$ can be replaced with its negative counterpart since we wish to satisfy $|x| + |y| \leq 10$ so really we can double this to get $2 \times 20=40$ Remembering $x=0, y=0$ was a solution we have $41$ pairs of x and y with that satisfy the inequality with one or both of $x$ or $y$ being $0$

Now for the easy part where we don't have to consider as many split cases in order to accommodate for the fact that $-0=0$ . We're not interested in $0$ anymore because we've found all solutions with zeroes.

We have:

$x=1, 1 \leq y \leq 9$

$9 cases$

$x=2, 1 \leq y \leq 8$

$8 cases$

                            Continue up to x=9

$x=9, y=1$

$1 case$

A total of $9+8+7+6+5+4+3+2+1$ = $\frac{(9)(10)}{2}=45$

Since we can change the sign of $EACH$ number of $EACH$ pair we have $4\times 45=180$

Giving a total of $180+41=\boxed{221}$

Well when I first saw the problem, I thought of Polya's comment about generalizing a problem using a smaller version of it, so I drew the graph of $|x|+|y|\leq2$ and of $|x|+|y|\leq3$ which is an trivial one with the help of graph transformations. The graph was symmetrical and a square, I could figure that for $|x|+|y|\leq2$ the number of integral solutions are $2\cdot (2(2(2-1)+1+2(2-2)+1)+2\cdot 2 +1=13$ and for $|x|+|y|\leq3$ it was $2(2(3-1)+1+2(3-2)+1+2(3-3)+1)+2\cdot 3 +1$ . So we generalize the formula into $P_n=2(2(n-1)+1+2(n-2)+1+ \cdots +2(n-n)+1)+2\cdot n +1=2n^2+2n+1$ , which can be proved by induction, which we do so, Let $P_n=2(n(n-1)+1+n(n-2)+1+ \cdots+ n(n-n)+1)+2 \cdot n +1=2n^2+2n+1$ be our induction basis, since $P_1=5$ is true, we can assume $P_k$ to be true, then $P_{k+1}$ should also be true, and we verify it. $P_{k+1}=2(2(k+1-1)+1+2(k+1-2)+1+\cdots+2(k+1-(k+1))+2\cdot (k+1)+1=2(2(k+(k-1)+(k-2)+ \cdots + 1)+1+1+ \cdots + \text {k+1 times})+2(k+1)+1$ $\implies$ $P_{k+1}=2(2(\dfrac{k(k+1)}{2})+k)+2(k+1)+1=2(k+1)(k+1)+2(k+1)+1=2(k+1)^2+2(k+1)+1$ , thus $P_{k+1}$ is true, hence $P_n$ is true.

So $P_{10}=2(10)^2+2(10)+1=221$ . $_\square$

This is not a solution. How would you generalize this problem? Consider two related problems:

How many ordered pairs of integers are there that satisfy $|x| + |y| \leq n?$

How many ordered $m$ -tuples of integers are there that satisfy $|x_1| + |x_2| + \cdots |x_m| \leq 10?$

These are both generalizations of the original problem. Can you solve these? What if we combine them both to get:

How many ordered $m$ -tuples of integers are there that satisfy $(|x_1| + |x_2| + \cdots |x_m| \leq n?$

Can you find a closed formula in terms of $m$ and $n?$

|x|+|y|<=10 means the length of the diagonal =20 and it's a square with two sides parallel to the line y=x.

Any square whose two sides parallel to x=y satisfies/has the property that the number of integral points it covers(including the points that lie on the sides/border ) will be equal to ((diagonal/2)+1)^2 +((diagonal/2))^2. so here it will be 11^2+10^2. Simple! works for any bigger sided squares.

We have $1$ pair of integers that equals to $0$ , $4$ pairs that equals to $1$ , $8$ pairs that equals to $2$ , $12$ pairs that equals to $3$ .....

So clearly we have the pattern $1, 4, 8, 12, ..., 40$ .

Why $40$ ? Expressing it mathematically, we get from the second element a pattern that can be expressed as $4(n + 1)$ (beginning with $n=0$ ), and as the inequality says $\leq{10}$ , we go for the $10th$ element of the pattern. If we make $1 + 4 + 8 + ••• + 40$ it is equal to $1 + 44*5$ and this is $\boxed{221}$ .

First we consider when $x=10$ . In this case there is only $1$ solution ( $y=0$ ). When $x=9$ , there are $3$ possible values for $y$ ( $y = \pm 1$ or $y=0$ ). Following this pattern, we notice that there is an arithematic series, with common difference $2$ and $n^{th}$ term, $U_{n} = 2(10 - n) + 1$ We want to sum this arithematic progression for values of $n$ between $1$ and $10$ , for a total of $10$ terms. Thus we have the sum, $S_{10} = \frac{10}{2} \cdot [2 \cdot 1 + (10-1) \cdot 2] = 100$ Now we multiply this by $2$ to account for when $x < 0$ to get $200$ . Finally, we consider the case when $x=0$ . Following the arithmetic progression, we get the value $U_0 = 20 + 1 = 21$ We do not have to consider the negative equivalent of $x$ in this case, so the final answer is, $200 + 21 = \fbox{221}$

reform the inequality to be

$|y| \leq 10-|x| \rightarrow -(10-|x|) \leq y \leq 10-|x|$

that means to for each $x$ the variable $y$ can be $2(10-|x|)+1$ values and $x$ can be $21$ values that is $-10,-9,...,10$ so the number of $(x,y)$ is

$\sum_{x=-10}^{10}[2(10-|x|)+1]=[2(10-|0|)+1]+2\sum_{x=1}^{10}[2(10-|x|)+1]=221$

Calling $F_n$ the number of integer solutions to $|x| + |y| <= n$ we have $F_0=1$ (just the point $(0,0)$ )

$F_1=F_0+4$ because it's the point $(0,0)$ plus the points where $|x|+|y|=1$ (that's a rotated square on the $xy$ plane)

$F_2=F_1+8$ because it's $F_1$ plus the points where $|x|+|y|=2$ that's a square larger where we have 1 more point on each side

using this idea for induction we have $F_n = F_n-1 + 4n$

now it's easy to find $F_{10}=221$