Practice · Three reflections make a rectangle sized by absolute values · Sensim Math

Variant 1 easy

For a point $P(a, b)$ in the coordinate plane, point $A$ is symmetric to $P$ with respect to the origin, point $B$ is symmetric to $P$ with respect to the $x$ -axis, and point $C$ is symmetric to $P$ with respect to the $y$ -axis. How many ordered pairs $(a, b)$ of integers make the quadrilateral with vertices $P$ , $A$ , $B$ , and $C$ have a perimeter of $12$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Reflect a lattice point $P(a, b)$ through the origin and across both axes. The four points form a quadrilateral; count the integer pairs $(a, b)$ that give it perimeter $12$ .

Givens

$A$ is the reflection of $P(a, b)$ through the origin.
$B$ is the reflection of $P$ across the $x$ -axis.
$C$ is the reflection of $P$ across the $y$ -axis.
The quadrilateral with vertices $P$ , $A$ , $B$ , $C$ has perimeter $12$ , and $a$ , $b$ are integers.

Unknowns

The number of ordered pairs $(a, b)$ .

Constraints

The four points must form a real quadrilateral, so neither $a$ nor $b$ can be $0$ (otherwise the points pile up on an axis).

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #13 Convert to Algebra#2 Make a Systematic List

Draw the four points: they are the corners of a rectangle centered at the origin with sides parallel to the axes. Its side lengths depend only on $|a|$ and $|b|$ , so the perimeter condition becomes one short equation in $|a|$ and $|b|$ . Then list its solutions and attach signs.

3 · Execute5 carry out the plan

1Write the reflected points

#1 Draw a Diagram 6.NS.C.6

Reflecting through the origin flips both signs, across the

x

-axis flips the

y

-sign, and across the

y

-axis flips the

x

-sign.

A(-a, -b), \quad B(a, -b), \quad C(-a, b)

Each reflection just changes signs.

2See the rectangle

#1 Draw a Diagram 7.NS.A.1

The points

(\pm a, \pm b)

are the corners of a rectangle with horizontal sides from

-a

to

a

and vertical sides from

-b

to

b

. The quadrilateral exists only if

a \ne 0

and

b \ne 0

.

|a-(-a)| = 2|a|, \quad |b-(-b)| = 2|b|

The distance between a number and its opposite is twice its absolute value.

3Turn the perimeter into an equation

#13 Convert to Algebra 7.NS.A.1

Two widths and two heights add to

12

, so

4(|a| + |b|) = 12

.

2(2|a| + 2|b|) = 12 \Rightarrow |a| + |b| = 3

Divide both sides by

4

.

4List the absolute values

#2 Make a Systematic List 6.NS.C.7

With

|a|

and

|b|

both positive integers adding to

3

, the only choices are

1+2

and

2+1

.

(|a|, |b|) = (1, 2) \text{ or } (2, 1)

Neither part may be

0

.

5Attach the signs

#2 Make a Systematic List 6.NS.C.7

Each absolute-value pair allows

a

to be positive or negative and

b

to be positive or negative:

4

sign choices. With

2

absolute-value pairs that is

8

ordered pairs.

2 \times 4 = 8

Every nonzero absolute value comes from two numbers.

Answer:

8

4 · Reviewdoes it hold up?

The $8$ pairs are $(\pm 1, \pm 2)$ and $(\pm 2, \pm 1)$ . For example $(1, 2)$ gives a $2 \times 4$ rectangle with perimeter $2 \times (2+4) = 12$ .

Another way: Use symmetry: count only $P$ in Quadrant I, where $a+b=3$ with $a, b \ge 1$ gives $(1, 2)$ and $(2, 1)$ . Each of these has a matching point in each of the four quadrants, so $2 \times 4 = 8$ .

Standardsmin grade 7

6.NS.C.6 Rational numbers on the number line and coordinate plane — Reading the coordinates of reflections across the axes and through the origin as sign changes.
6.NS.C.7 Order and absolute value of rational numbers — Using absolute values and counting the numbers with a given absolute value.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Finding side lengths as distances between signed numbers.

💡Takeaway. Reflections only flip signs -- so the size of the shape lives in the absolute values.

Variant 2 easy

For a point $P(a, b)$ in the coordinate plane, point $A$ is symmetric to $P$ with respect to the origin, point $B$ is symmetric to $P$ with respect to the $x$ -axis, and point $C$ is symmetric to $P$ with respect to the $y$ -axis. How many ordered pairs $(a, b)$ of integers make the quadrilateral with vertices $P$ , $A$ , $B$ , and $C$ have a perimeter of $16$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Reflect a lattice point $P(a, b)$ through the origin and across both axes. The four points form a quadrilateral; count the integer pairs $(a, b)$ that give it perimeter $16$ .

Givens

$A$ is the reflection of $P(a, b)$ through the origin.
$B$ is the reflection of $P$ across the $x$ -axis.
$C$ is the reflection of $P$ across the $y$ -axis.
The quadrilateral with vertices $P$ , $A$ , $B$ , $C$ has perimeter $16$ , and $a$ , $b$ are integers.

Unknowns

The number of ordered pairs $(a, b)$ .

Constraints

The four points must form a real quadrilateral, so neither $a$ nor $b$ can be $0$ (otherwise the points pile up on an axis).

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #13 Convert to Algebra#2 Make a Systematic List

Draw the four points: they are the corners of a rectangle centered at the origin with sides parallel to the axes. Its side lengths depend only on $|a|$ and $|b|$ , so the perimeter condition becomes one short equation in $|a|$ and $|b|$ . Then list its solutions and attach signs.

3 · Execute5 carry out the plan

1Write the reflected points

#1 Draw a Diagram 6.NS.C.6

Reflecting through the origin flips both signs, across the

x

-axis flips the

y

-sign, and across the

y

-axis flips the

x

-sign.

A(-a, -b), \quad B(a, -b), \quad C(-a, b)

Each reflection just changes signs.

2See the rectangle

#1 Draw a Diagram 7.NS.A.1

The points

(\pm a, \pm b)

are the corners of a rectangle with horizontal sides from

-a

to

a

and vertical sides from

-b

to

b

. The quadrilateral exists only if

a \ne 0

and

b \ne 0

.

|a-(-a)| = 2|a|, \quad |b-(-b)| = 2|b|

The distance between a number and its opposite is twice its absolute value.

3Turn the perimeter into an equation

#13 Convert to Algebra 7.NS.A.1

Two widths and two heights add to

16

, so

4(|a| + |b|) = 16

.

2(2|a| + 2|b|) = 16 \Rightarrow |a| + |b| = 4

Divide both sides by

4

.

4List the absolute values

#2 Make a Systematic List 6.NS.C.7

With

|a|

and

|b|

both positive integers adding to

4

, the only choices are

1+3

,

2+2

, and

3+1

.

(|a|, |b|) = (1, 3) \text{ or } (2, 2) \text{ or } (3, 1)

Neither part may be

0

.

5Attach the signs

#2 Make a Systematic List 6.NS.C.7

Each absolute-value pair allows

a

to be positive or negative and

b

to be positive or negative:

4

sign choices. With

3

absolute-value pairs that is

12

ordered pairs.

3 \times 4 = 12

Every nonzero absolute value comes from two numbers.

Answer:

12

4 · Reviewdoes it hold up?

The $12$ pairs are $(\pm 1, \pm 3)$ , $(\pm 2, \pm 2)$ , and $(\pm 3, \pm 1)$ . For example $(1, 3)$ gives a $2 \times 6$ rectangle with perimeter $2 \times (2+6) = 16$ .

Another way: Use symmetry: count only $P$ in Quadrant I, where $a+b=4$ with $a, b \ge 1$ gives $(1, 3)$ , $(2, 2)$ , and $(3, 1)$ . Each of these has a matching point in each of the four quadrants, so $3 \times 4 = 12$ .

Standardsmin grade 7

6.NS.C.6 Rational numbers on the number line and coordinate plane — Reading the coordinates of reflections across the axes and through the origin as sign changes.
6.NS.C.7 Order and absolute value of rational numbers — Using absolute values and counting the numbers with a given absolute value.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Finding side lengths as distances between signed numbers.

💡Takeaway. Reflections only flip signs -- so the size of the shape lives in the absolute values.

Variant 3 easy

For a point $P(a, b)$ in the coordinate plane, point $A$ is symmetric to $P$ with respect to the origin, point $B$ is symmetric to $P$ with respect to the $x$ -axis, and point $C$ is symmetric to $P$ with respect to the $y$ -axis. How many ordered pairs $(a, b)$ of integers make the quadrilateral with vertices $P$ , $A$ , $B$ , and $C$ have a perimeter of $20$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Reflect a lattice point $P(a, b)$ through the origin and across both axes. The four points form a quadrilateral; count the integer pairs $(a, b)$ that give it perimeter $20$ .

Givens

$A$ is the reflection of $P(a, b)$ through the origin.
$B$ is the reflection of $P$ across the $x$ -axis.
$C$ is the reflection of $P$ across the $y$ -axis.
The quadrilateral with vertices $P$ , $A$ , $B$ , $C$ has perimeter $20$ , and $a$ , $b$ are integers.

Unknowns

The number of ordered pairs $(a, b)$ .

Constraints

The four points must form a real quadrilateral, so neither $a$ nor $b$ can be $0$ (otherwise the points pile up on an axis).

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #13 Convert to Algebra#2 Make a Systematic List

Draw the four points: they are the corners of a rectangle centered at the origin with sides parallel to the axes. Its side lengths depend only on $|a|$ and $|b|$ , so the perimeter condition becomes one short equation in $|a|$ and $|b|$ . Then list its solutions and attach signs.

3 · Execute5 carry out the plan

1Write the reflected points

#1 Draw a Diagram 6.NS.C.6

Reflecting through the origin flips both signs, across the

x

-axis flips the

y

-sign, and across the

y

-axis flips the

x

-sign.

A(-a, -b), \quad B(a, -b), \quad C(-a, b)

Each reflection just changes signs.

2See the rectangle

#1 Draw a Diagram 7.NS.A.1

The points

(\pm a, \pm b)

are the corners of a rectangle with horizontal sides from

-a

to

a

and vertical sides from

-b

to

b

. The quadrilateral exists only if

a \ne 0

and

b \ne 0

.

|a-(-a)| = 2|a|, \quad |b-(-b)| = 2|b|

The distance between a number and its opposite is twice its absolute value.

3Turn the perimeter into an equation

#13 Convert to Algebra 7.NS.A.1

Two widths and two heights add to

20

, so

4(|a| + |b|) = 20

.

2(2|a| + 2|b|) = 20 \Rightarrow |a| + |b| = 5

Divide both sides by

4

.

4List the absolute values

#2 Make a Systematic List 6.NS.C.7

With

|a|

and

|b|

both positive integers adding to

5

, the only choices are

1+4

,

2+3

,

3+2

, and

4+1

.

(|a|, |b|) = (1, 4) \text{ or } (2, 3) \text{ or } (3, 2) \text{ or } (4, 1)

Neither part may be

0

.

5Attach the signs

#2 Make a Systematic List 6.NS.C.7

Each absolute-value pair allows

a

to be positive or negative and

b

to be positive or negative:

4

sign choices. With

4

absolute-value pairs that is

16

ordered pairs.

4 \times 4 = 16

Every nonzero absolute value comes from two numbers.

Answer:

16

4 · Reviewdoes it hold up?

The $16$ pairs are $(\pm 1, \pm 4)$ , $(\pm 2, \pm 3)$ , $(\pm 3, \pm 2)$ , and $(\pm 4, \pm 1)$ . For example $(1, 4)$ gives a $2 \times 8$ rectangle with perimeter $2 \times (2+8) = 20$ .

Another way: Use symmetry: count only $P$ in Quadrant I, where $a+b=5$ with $a, b \ge 1$ gives $(1, 4)$ , $(2, 3)$ , $(3, 2)$ , and $(4, 1)$ . Each of these has a matching point in each of the four quadrants, so $4 \times 4 = 16$ .

Standardsmin grade 7

6.NS.C.6 Rational numbers on the number line and coordinate plane — Reading the coordinates of reflections across the axes and through the origin as sign changes.
6.NS.C.7 Order and absolute value of rational numbers — Using absolute values and counting the numbers with a given absolute value.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Finding side lengths as distances between signed numbers.

💡Takeaway. Reflections only flip signs -- so the size of the shape lives in the absolute values.

Variant 4 easy

For a point $P(a, b)$ in the coordinate plane, point $A$ is symmetric to $P$ with respect to the origin, point $B$ is symmetric to $P$ with respect to the $x$ -axis, and point $C$ is symmetric to $P$ with respect to the $y$ -axis. How many ordered pairs $(a, b)$ of integers make the quadrilateral with vertices $P$ , $A$ , $B$ , and $C$ have a perimeter of $24$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Reflect a lattice point $P(a, b)$ through the origin and across both axes. The four points form a quadrilateral; count the integer pairs $(a, b)$ that give it perimeter $24$ .

Givens

$A$ is the reflection of $P(a, b)$ through the origin.
$B$ is the reflection of $P$ across the $x$ -axis.
$C$ is the reflection of $P$ across the $y$ -axis.
The quadrilateral with vertices $P$ , $A$ , $B$ , $C$ has perimeter $24$ , and $a$ , $b$ are integers.

Unknowns

The number of ordered pairs $(a, b)$ .

Constraints

The four points must form a real quadrilateral, so neither $a$ nor $b$ can be $0$ (otherwise the points pile up on an axis).

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #13 Convert to Algebra#2 Make a Systematic List

Draw the four points: they are the corners of a rectangle centered at the origin with sides parallel to the axes. Its side lengths depend only on $|a|$ and $|b|$ , so the perimeter condition becomes one short equation in $|a|$ and $|b|$ . Then list its solutions and attach signs.

3 · Execute5 carry out the plan

1Write the reflected points

#1 Draw a Diagram 6.NS.C.6

Reflecting through the origin flips both signs, across the

x

-axis flips the

y

-sign, and across the

y

-axis flips the

x

-sign.

A(-a, -b), \quad B(a, -b), \quad C(-a, b)

Each reflection just changes signs.

2See the rectangle

#1 Draw a Diagram 7.NS.A.1

The points

(\pm a, \pm b)

are the corners of a rectangle with horizontal sides from

-a

to

a

and vertical sides from

-b

to

b

. The quadrilateral exists only if

a \ne 0

and

b \ne 0

.

|a-(-a)| = 2|a|, \quad |b-(-b)| = 2|b|

The distance between a number and its opposite is twice its absolute value.

3Turn the perimeter into an equation

#13 Convert to Algebra 7.NS.A.1

Two widths and two heights add to

24

, so

4(|a| + |b|) = 24

.

2(2|a| + 2|b|) = 24 \Rightarrow |a| + |b| = 6

Divide both sides by

4

.

4List the absolute values

#2 Make a Systematic List 6.NS.C.7

With

|a|

and

|b|

both positive integers adding to

6

, the choices run

1+5

,

2+4

, and so on up to

5+1

, which is

5

choices.

(|a|, |b|) = (1, 5), (2, 4), \dots, (5, 1)

Neither part may be

0

.

5Attach the signs

#2 Make a Systematic List 6.NS.C.7

Each absolute-value pair allows

a

to be positive or negative and

b

to be positive or negative:

4

sign choices. With

5

absolute-value pairs that is

20

ordered pairs.

5 \times 4 = 20

Every nonzero absolute value comes from two numbers.

Answer:

20

4 · Reviewdoes it hold up?

The $20$ pairs are $(\pm 1, \pm 5)$ , $(\pm 2, \pm 4)$ , and so on up to $(\pm 5, \pm 1)$ . For example $(1, 5)$ gives a $2 \times 10$ rectangle with perimeter $2 \times (2+10) = 24$ .

Another way: Use symmetry: count only $P$ in Quadrant I, where $a+b=6$ with $a, b \ge 1$ gives $(1, 5)$ through $(5, 1)$ , $5$ points. Each of these has a matching point in each of the four quadrants, so $5 \times 4 = 20$ .

Standardsmin grade 7

6.NS.C.6 Rational numbers on the number line and coordinate plane — Reading the coordinates of reflections across the axes and through the origin as sign changes.
6.NS.C.7 Order and absolute value of rational numbers — Using absolute values and counting the numbers with a given absolute value.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Finding side lengths as distances between signed numbers.

💡Takeaway. Reflections only flip signs -- so the size of the shape lives in the absolute values.

Variant 5 medium

For a point $P(a, b)$ in the coordinate plane, point $A$ is symmetric to $P$ with respect to the origin, point $B$ is symmetric to $P$ with respect to the $x$ -axis, and point $C$ is symmetric to $P$ with respect to the $y$ -axis. How many ordered pairs $(a, b)$ of integers make the quadrilateral with vertices $P$ , $A$ , $B$ , and $C$ have a perimeter of $28$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Reflect a lattice point $P(a, b)$ through the origin and across both axes. The four points form a quadrilateral; count the integer pairs $(a, b)$ that give it perimeter $28$ .

Givens

$A$ is the reflection of $P(a, b)$ through the origin.
$B$ is the reflection of $P$ across the $x$ -axis.
$C$ is the reflection of $P$ across the $y$ -axis.
The quadrilateral with vertices $P$ , $A$ , $B$ , $C$ has perimeter $28$ , and $a$ , $b$ are integers.

Unknowns

The number of ordered pairs $(a, b)$ .

Constraints

The four points must form a real quadrilateral, so neither $a$ nor $b$ can be $0$ (otherwise the points pile up on an axis).

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #13 Convert to Algebra#2 Make a Systematic List

Draw the four points: they are the corners of a rectangle centered at the origin with sides parallel to the axes. Its side lengths depend only on $|a|$ and $|b|$ , so the perimeter condition becomes one short equation in $|a|$ and $|b|$ . Then list its solutions and attach signs.

3 · Execute5 carry out the plan

1Write the reflected points

#1 Draw a Diagram 6.NS.C.6

Reflecting through the origin flips both signs, across the

x

-axis flips the

y

-sign, and across the

y

-axis flips the

x

-sign.

A(-a, -b), \quad B(a, -b), \quad C(-a, b)

Each reflection just changes signs.

2See the rectangle

#1 Draw a Diagram 7.NS.A.1

The points

(\pm a, \pm b)

are the corners of a rectangle with horizontal sides from

-a

to

a

and vertical sides from

-b

to

b

. The quadrilateral exists only if

a \ne 0

and

b \ne 0

.

|a-(-a)| = 2|a|, \quad |b-(-b)| = 2|b|

The distance between a number and its opposite is twice its absolute value.

3Turn the perimeter into an equation

#13 Convert to Algebra 7.NS.A.1

Two widths and two heights add to

28

, so

4(|a| + |b|) = 28

.

2(2|a| + 2|b|) = 28 \Rightarrow |a| + |b| = 7

Divide both sides by

4

.

4List the absolute values

#2 Make a Systematic List 6.NS.C.7

With

|a|

and

|b|

both positive integers adding to

7

, the choices run

1+6

,

2+5

, and so on up to

6+1

, which is

6

choices.

(|a|, |b|) = (1, 6), (2, 5), \dots, (6, 1)

Neither part may be

0

.

5Attach the signs

#2 Make a Systematic List 6.NS.C.7

Each absolute-value pair allows

a

to be positive or negative and

b

to be positive or negative:

4

sign choices. With

6

absolute-value pairs that is

24

ordered pairs.

6 \times 4 = 24

Every nonzero absolute value comes from two numbers.

Answer:

24

4 · Reviewdoes it hold up?

The $24$ pairs are $(\pm 1, \pm 6)$ , $(\pm 2, \pm 5)$ , and so on up to $(\pm 6, \pm 1)$ . For example $(1, 6)$ gives a $2 \times 12$ rectangle with perimeter $2 \times (2+12) = 28$ .

Another way: Use symmetry: count only $P$ in Quadrant I, where $a+b=7$ with $a, b \ge 1$ gives $(1, 6)$ through $(6, 1)$ , $6$ points. Each of these has a matching point in each of the four quadrants, so $6 \times 4 = 24$ .

Standardsmin grade 7

6.NS.C.6 Rational numbers on the number line and coordinate plane — Reading the coordinates of reflections across the axes and through the origin as sign changes.
6.NS.C.7 Order and absolute value of rational numbers — Using absolute values and counting the numbers with a given absolute value.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Finding side lengths as distances between signed numbers.

💡Takeaway. Reflections only flip signs -- so the size of the shape lives in the absolute values.

Variant 6 medium

For a point $P(a, b)$ in the coordinate plane, point $A$ is symmetric to $P$ with respect to the origin, point $B$ is symmetric to $P$ with respect to the $x$ -axis, and point $C$ is symmetric to $P$ with respect to the $y$ -axis. How many ordered pairs $(a, b)$ of integers make the quadrilateral with vertices $P$ , $A$ , $B$ , and $C$ have a perimeter of $32$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Reflect a lattice point $P(a, b)$ through the origin and across both axes. The four points form a quadrilateral; count the integer pairs $(a, b)$ that give it perimeter $32$ .

Givens

$A$ is the reflection of $P(a, b)$ through the origin.
$B$ is the reflection of $P$ across the $x$ -axis.
$C$ is the reflection of $P$ across the $y$ -axis.
The quadrilateral with vertices $P$ , $A$ , $B$ , $C$ has perimeter $32$ , and $a$ , $b$ are integers.

Unknowns

The number of ordered pairs $(a, b)$ .

Constraints

The four points must form a real quadrilateral, so neither $a$ nor $b$ can be $0$ (otherwise the points pile up on an axis).

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #13 Convert to Algebra#2 Make a Systematic List

Draw the four points: they are the corners of a rectangle centered at the origin with sides parallel to the axes. Its side lengths depend only on $|a|$ and $|b|$ , so the perimeter condition becomes one short equation in $|a|$ and $|b|$ . Then list its solutions and attach signs.

3 · Execute5 carry out the plan

1Write the reflected points

#1 Draw a Diagram 6.NS.C.6

Reflecting through the origin flips both signs, across the

x

-axis flips the

y

-sign, and across the

y

-axis flips the

x

-sign.

A(-a, -b), \quad B(a, -b), \quad C(-a, b)

Each reflection just changes signs.

2See the rectangle

#1 Draw a Diagram 7.NS.A.1

The points

(\pm a, \pm b)

are the corners of a rectangle with horizontal sides from

-a

to

a

and vertical sides from

-b

to

b

. The quadrilateral exists only if

a \ne 0

and

b \ne 0

.

|a-(-a)| = 2|a|, \quad |b-(-b)| = 2|b|

The distance between a number and its opposite is twice its absolute value.

3Turn the perimeter into an equation

#13 Convert to Algebra 7.NS.A.1

Two widths and two heights add to

32

, so

4(|a| + |b|) = 32

.

2(2|a| + 2|b|) = 32 \Rightarrow |a| + |b| = 8

Divide both sides by

4

.

4List the absolute values

#2 Make a Systematic List 6.NS.C.7

With

|a|

and

|b|

both positive integers adding to

8

, the choices run

1+7

,

2+6

, and so on up to

7+1

, which is

7

choices.

(|a|, |b|) = (1, 7), (2, 6), \dots, (7, 1)

Neither part may be

0

.

5Attach the signs

#2 Make a Systematic List 6.NS.C.7

Each absolute-value pair allows

a

to be positive or negative and

b

to be positive or negative:

4

sign choices. With

7

absolute-value pairs that is

28

ordered pairs.

7 \times 4 = 28

Every nonzero absolute value comes from two numbers.

Answer:

28

4 · Reviewdoes it hold up?

The $28$ pairs are $(\pm 1, \pm 7)$ , $(\pm 2, \pm 6)$ , and so on up to $(\pm 7, \pm 1)$ . For example $(1, 7)$ gives a $2 \times 14$ rectangle with perimeter $2 \times (2+14) = 32$ .

Another way: Use symmetry: count only $P$ in Quadrant I, where $a+b=8$ with $a, b \ge 1$ gives $(1, 7)$ through $(7, 1)$ , $7$ points. Each of these has a matching point in each of the four quadrants, so $7 \times 4 = 28$ .

Standardsmin grade 7

6.NS.C.6 Rational numbers on the number line and coordinate plane — Reading the coordinates of reflections across the axes and through the origin as sign changes.
6.NS.C.7 Order and absolute value of rational numbers — Using absolute values and counting the numbers with a given absolute value.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Finding side lengths as distances between signed numbers.

💡Takeaway. Reflections only flip signs -- so the size of the shape lives in the absolute values.

Variant 7 medium

For a point $P(a, b)$ in the coordinate plane, point $A$ is symmetric to $P$ with respect to the origin, point $B$ is symmetric to $P$ with respect to the $x$ -axis, and point $C$ is symmetric to $P$ with respect to the $y$ -axis. How many ordered pairs $(a, b)$ of integers make the quadrilateral with vertices $P$ , $A$ , $B$ , and $C$ have a perimeter of $36$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Reflect a lattice point $P(a, b)$ through the origin and across both axes. The four points form a quadrilateral; count the integer pairs $(a, b)$ that give it perimeter $36$ .

Givens

$A$ is the reflection of $P(a, b)$ through the origin.
$B$ is the reflection of $P$ across the $x$ -axis.
$C$ is the reflection of $P$ across the $y$ -axis.
The quadrilateral with vertices $P$ , $A$ , $B$ , $C$ has perimeter $36$ , and $a$ , $b$ are integers.

Unknowns

The number of ordered pairs $(a, b)$ .

Constraints

The four points must form a real quadrilateral, so neither $a$ nor $b$ can be $0$ (otherwise the points pile up on an axis).

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #13 Convert to Algebra#2 Make a Systematic List

Draw the four points: they are the corners of a rectangle centered at the origin with sides parallel to the axes. Its side lengths depend only on $|a|$ and $|b|$ , so the perimeter condition becomes one short equation in $|a|$ and $|b|$ . Then list its solutions and attach signs.

3 · Execute5 carry out the plan

1Write the reflected points

#1 Draw a Diagram 6.NS.C.6

Reflecting through the origin flips both signs, across the

x

-axis flips the

y

-sign, and across the

y

-axis flips the

x

-sign.

A(-a, -b), \quad B(a, -b), \quad C(-a, b)

Each reflection just changes signs.

2See the rectangle

#1 Draw a Diagram 7.NS.A.1

The points

(\pm a, \pm b)

are the corners of a rectangle with horizontal sides from

-a

to

a

and vertical sides from

-b

to

b

. The quadrilateral exists only if

a \ne 0

and

b \ne 0

.

|a-(-a)| = 2|a|, \quad |b-(-b)| = 2|b|

The distance between a number and its opposite is twice its absolute value.

3Turn the perimeter into an equation

#13 Convert to Algebra 7.NS.A.1

Two widths and two heights add to

36

, so

4(|a| + |b|) = 36

.

2(2|a| + 2|b|) = 36 \Rightarrow |a| + |b| = 9

Divide both sides by

4

.

4List the absolute values

#2 Make a Systematic List 6.NS.C.7

With

|a|

and

|b|

both positive integers adding to

9

, the choices run

1+8

,

2+7

, and so on up to

8+1

, which is

8

choices.

(|a|, |b|) = (1, 8), (2, 7), \dots, (8, 1)

Neither part may be

0

.

5Attach the signs

#2 Make a Systematic List 6.NS.C.7

Each absolute-value pair allows

a

to be positive or negative and

b

to be positive or negative:

4

sign choices. With

8

absolute-value pairs that is

32

ordered pairs.

8 \times 4 = 32

Every nonzero absolute value comes from two numbers.

Answer:

32

4 · Reviewdoes it hold up?

The $32$ pairs are $(\pm 1, \pm 8)$ , $(\pm 2, \pm 7)$ , and so on up to $(\pm 8, \pm 1)$ . For example $(1, 8)$ gives a $2 \times 16$ rectangle with perimeter $2 \times (2+16) = 36$ .

Another way: Use symmetry: count only $P$ in Quadrant I, where $a+b=9$ with $a, b \ge 1$ gives $(1, 8)$ through $(8, 1)$ , $8$ points. Each of these has a matching point in each of the four quadrants, so $8 \times 4 = 32$ .

Standardsmin grade 7

6.NS.C.6 Rational numbers on the number line and coordinate plane — Reading the coordinates of reflections across the axes and through the origin as sign changes.
6.NS.C.7 Order and absolute value of rational numbers — Using absolute values and counting the numbers with a given absolute value.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Finding side lengths as distances between signed numbers.

💡Takeaway. Reflections only flip signs -- so the size of the shape lives in the absolute values.

Variant 8 medium

For a point $P(a, b)$ in the coordinate plane, point $A$ is symmetric to $P$ with respect to the origin, point $B$ is symmetric to $P$ with respect to the $x$ -axis, and point $C$ is symmetric to $P$ with respect to the $y$ -axis. How many ordered pairs $(a, b)$ of integers make the quadrilateral with vertices $P$ , $A$ , $B$ , and $C$ have a perimeter of $40$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Reflect a lattice point $P(a, b)$ through the origin and across both axes. The four points form a quadrilateral; count the integer pairs $(a, b)$ that give it perimeter $40$ .

Givens

$A$ is the reflection of $P(a, b)$ through the origin.
$B$ is the reflection of $P$ across the $x$ -axis.
$C$ is the reflection of $P$ across the $y$ -axis.
The quadrilateral with vertices $P$ , $A$ , $B$ , $C$ has perimeter $40$ , and $a$ , $b$ are integers.

Unknowns

The number of ordered pairs $(a, b)$ .

Constraints

The four points must form a real quadrilateral, so neither $a$ nor $b$ can be $0$ (otherwise the points pile up on an axis).

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #13 Convert to Algebra#2 Make a Systematic List

Draw the four points: they are the corners of a rectangle centered at the origin with sides parallel to the axes. Its side lengths depend only on $|a|$ and $|b|$ , so the perimeter condition becomes one short equation in $|a|$ and $|b|$ . Then list its solutions and attach signs.

3 · Execute5 carry out the plan

1Write the reflected points

#1 Draw a Diagram 6.NS.C.6

Reflecting through the origin flips both signs, across the

x

-axis flips the

y

-sign, and across the

y

-axis flips the

x

-sign.

A(-a, -b), \quad B(a, -b), \quad C(-a, b)

Each reflection just changes signs.

2See the rectangle

#1 Draw a Diagram 7.NS.A.1

The points

(\pm a, \pm b)

are the corners of a rectangle with horizontal sides from

-a

to

a

and vertical sides from

-b

to

b

. The quadrilateral exists only if

a \ne 0

and

b \ne 0

.

|a-(-a)| = 2|a|, \quad |b-(-b)| = 2|b|

The distance between a number and its opposite is twice its absolute value.

3Turn the perimeter into an equation

#13 Convert to Algebra 7.NS.A.1

Two widths and two heights add to

40

, so

4(|a| + |b|) = 40

.

2(2|a| + 2|b|) = 40 \Rightarrow |a| + |b| = 10

Divide both sides by

4

.

4List the absolute values

#2 Make a Systematic List 6.NS.C.7

With

|a|

and

|b|

both positive integers adding to

10

, the choices run

1+9

,

2+8

, and so on up to

9+1

, which is

9

choices.

(|a|, |b|) = (1, 9), (2, 8), \dots, (9, 1)

Neither part may be

0

.

5Attach the signs

#2 Make a Systematic List 6.NS.C.7

Each absolute-value pair allows

a

to be positive or negative and

b

to be positive or negative:

4

sign choices. With

9

absolute-value pairs that is

36

ordered pairs.

9 \times 4 = 36

Every nonzero absolute value comes from two numbers.

Answer:

36

4 · Reviewdoes it hold up?

The $36$ pairs are $(\pm 1, \pm 9)$ , $(\pm 2, \pm 8)$ , and so on up to $(\pm 9, \pm 1)$ . For example $(1, 9)$ gives a $2 \times 18$ rectangle with perimeter $2 \times (2+18) = 40$ .

Another way: Use symmetry: count only $P$ in Quadrant I, where $a+b=10$ with $a, b \ge 1$ gives $(1, 9)$ through $(9, 1)$ , $9$ points. Each of these has a matching point in each of the four quadrants, so $9 \times 4 = 36$ .

Standardsmin grade 7

6.NS.C.6 Rational numbers on the number line and coordinate plane — Reading the coordinates of reflections across the axes and through the origin as sign changes.
6.NS.C.7 Order and absolute value of rational numbers — Using absolute values and counting the numbers with a given absolute value.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Finding side lengths as distances between signed numbers.

💡Takeaway. Reflections only flip signs -- so the size of the shape lives in the absolute values.

Variant 9 hard

For a point $P(a, b)$ in the coordinate plane, point $A$ is symmetric to $P$ with respect to the origin, point $B$ is symmetric to $P$ with respect to the $x$ -axis, and point $C$ is symmetric to $P$ with respect to the $y$ -axis. How many ordered pairs $(a, b)$ of integers make the quadrilateral with vertices $P$ , $A$ , $B$ , and $C$ have a perimeter of $44$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Reflect a lattice point $P(a, b)$ through the origin and across both axes. The four points form a quadrilateral; count the integer pairs $(a, b)$ that give it perimeter $44$ .

Givens

$A$ is the reflection of $P(a, b)$ through the origin.
$B$ is the reflection of $P$ across the $x$ -axis.
$C$ is the reflection of $P$ across the $y$ -axis.
The quadrilateral with vertices $P$ , $A$ , $B$ , $C$ has perimeter $44$ , and $a$ , $b$ are integers.

Unknowns

The number of ordered pairs $(a, b)$ .

Constraints

The four points must form a real quadrilateral, so neither $a$ nor $b$ can be $0$ (otherwise the points pile up on an axis).

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #13 Convert to Algebra#2 Make a Systematic List

Draw the four points: they are the corners of a rectangle centered at the origin with sides parallel to the axes. Its side lengths depend only on $|a|$ and $|b|$ , so the perimeter condition becomes one short equation in $|a|$ and $|b|$ . Then list its solutions and attach signs.

3 · Execute5 carry out the plan

1Write the reflected points

#1 Draw a Diagram 6.NS.C.6

Reflecting through the origin flips both signs, across the

x

-axis flips the

y

-sign, and across the

y

-axis flips the

x

-sign.

A(-a, -b), \quad B(a, -b), \quad C(-a, b)

Each reflection just changes signs.

2See the rectangle

#1 Draw a Diagram 7.NS.A.1

The points

(\pm a, \pm b)

are the corners of a rectangle with horizontal sides from

-a

to

a

and vertical sides from

-b

to

b

. The quadrilateral exists only if

a \ne 0

and

b \ne 0

.

|a-(-a)| = 2|a|, \quad |b-(-b)| = 2|b|

The distance between a number and its opposite is twice its absolute value.

3Turn the perimeter into an equation

#13 Convert to Algebra 7.NS.A.1

Two widths and two heights add to

44

, so

4(|a| + |b|) = 44

.

2(2|a| + 2|b|) = 44 \Rightarrow |a| + |b| = 11

Divide both sides by

4

.

4List the absolute values

#2 Make a Systematic List 6.NS.C.7

With

|a|

and

|b|

both positive integers adding to

11

, the choices run

1+10

,

2+9

, and so on up to

10+1

, which is

10

choices.

(|a|, |b|) = (1, 10), (2, 9), \dots, (10, 1)

Neither part may be

0

.

5Attach the signs

#2 Make a Systematic List 6.NS.C.7

Each absolute-value pair allows

a

to be positive or negative and

b

to be positive or negative:

4

sign choices. With

10

absolute-value pairs that is

40

ordered pairs.

10 \times 4 = 40

Every nonzero absolute value comes from two numbers.

Answer:

40

4 · Reviewdoes it hold up?

The $40$ pairs are $(\pm 1, \pm 10)$ , $(\pm 2, \pm 9)$ , and so on up to $(\pm 10, \pm 1)$ . For example $(1, 10)$ gives a $2 \times 20$ rectangle with perimeter $2 \times (2+20) = 44$ .

Another way: Use symmetry: count only $P$ in Quadrant I, where $a+b=11$ with $a, b \ge 1$ gives $(1, 10)$ through $(10, 1)$ , $10$ points. Each of these has a matching point in each of the four quadrants, so $10 \times 4 = 40$ .

Standardsmin grade 7

6.NS.C.6 Rational numbers on the number line and coordinate plane — Reading the coordinates of reflections across the axes and through the origin as sign changes.
6.NS.C.7 Order and absolute value of rational numbers — Using absolute values and counting the numbers with a given absolute value.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Finding side lengths as distances between signed numbers.

💡Takeaway. Reflections only flip signs -- so the size of the shape lives in the absolute values.

Variant 10 hard

For a point $P(a, b)$ in the coordinate plane, point $A$ is symmetric to $P$ with respect to the origin, point $B$ is symmetric to $P$ with respect to the $x$ -axis, and point $C$ is symmetric to $P$ with respect to the $y$ -axis. How many ordered pairs $(a, b)$ of integers make the quadrilateral with vertices $P$ , $A$ , $B$ , and $C$ have a perimeter of $48$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Reflect a lattice point $P(a, b)$ through the origin and across both axes. The four points form a quadrilateral; count the integer pairs $(a, b)$ that give it perimeter $48$ .

Givens

$A$ is the reflection of $P(a, b)$ through the origin.
$B$ is the reflection of $P$ across the $x$ -axis.
$C$ is the reflection of $P$ across the $y$ -axis.
The quadrilateral with vertices $P$ , $A$ , $B$ , $C$ has perimeter $48$ , and $a$ , $b$ are integers.

Unknowns

The number of ordered pairs $(a, b)$ .

Constraints

The four points must form a real quadrilateral, so neither $a$ nor $b$ can be $0$ (otherwise the points pile up on an axis).

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #13 Convert to Algebra#2 Make a Systematic List

Draw the four points: they are the corners of a rectangle centered at the origin with sides parallel to the axes. Its side lengths depend only on $|a|$ and $|b|$ , so the perimeter condition becomes one short equation in $|a|$ and $|b|$ . Then list its solutions and attach signs.

3 · Execute5 carry out the plan

1Write the reflected points

#1 Draw a Diagram 6.NS.C.6

Reflecting through the origin flips both signs, across the

x

-axis flips the

y

-sign, and across the

y

-axis flips the

x

-sign.

A(-a, -b), \quad B(a, -b), \quad C(-a, b)

Each reflection just changes signs.

2See the rectangle

#1 Draw a Diagram 7.NS.A.1

The points

(\pm a, \pm b)

are the corners of a rectangle with horizontal sides from

-a

to

a

and vertical sides from

-b

to

b

. The quadrilateral exists only if

a \ne 0

and

b \ne 0

.

|a-(-a)| = 2|a|, \quad |b-(-b)| = 2|b|

The distance between a number and its opposite is twice its absolute value.

3Turn the perimeter into an equation

#13 Convert to Algebra 7.NS.A.1

Two widths and two heights add to

48

, so

4(|a| + |b|) = 48

.

2(2|a| + 2|b|) = 48 \Rightarrow |a| + |b| = 12

Divide both sides by

4

.

4List the absolute values

#2 Make a Systematic List 6.NS.C.7

With

|a|

and

|b|

both positive integers adding to

12

, the choices run

1+11

,

2+10

, and so on up to

11+1

, which is

11

choices.

(|a|, |b|) = (1, 11), (2, 10), \dots, (11, 1)

Neither part may be

0

.

5Attach the signs

#2 Make a Systematic List 6.NS.C.7

Each absolute-value pair allows

a

to be positive or negative and

b

to be positive or negative:

4

sign choices. With

11

absolute-value pairs that is

44

ordered pairs.

11 \times 4 = 44

Every nonzero absolute value comes from two numbers.

Answer:

44

4 · Reviewdoes it hold up?

The $44$ pairs are $(\pm 1, \pm 11)$ , $(\pm 2, \pm 10)$ , and so on up to $(\pm 11, \pm 1)$ . For example $(1, 11)$ gives a $2 \times 22$ rectangle with perimeter $2 \times (2+22) = 48$ .

Another way: Use symmetry: count only $P$ in Quadrant I, where $a+b=12$ with $a, b \ge 1$ gives $(1, 11)$ through $(11, 1)$ , $11$ points. Each of these has a matching point in each of the four quadrants, so $11 \times 4 = 44$ .

Standardsmin grade 7

6.NS.C.6 Rational numbers on the number line and coordinate plane — Reading the coordinates of reflections across the axes and through the origin as sign changes.
6.NS.C.7 Order and absolute value of rational numbers — Using absolute values and counting the numbers with a given absolute value.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Finding side lengths as distances between signed numbers.

💡Takeaway. Reflections only flip signs -- so the size of the shape lives in the absolute values.

Variant 11 hard

For a point $P(a, b)$ in the coordinate plane, point $A$ is symmetric to $P$ with respect to the origin, point $B$ is symmetric to $P$ with respect to the $x$ -axis, and point $C$ is symmetric to $P$ with respect to the $y$ -axis. How many ordered pairs $(a, b)$ of integers make the quadrilateral with vertices $P$ , $A$ , $B$ , and $C$ have a perimeter of $52$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Reflect a lattice point $P(a, b)$ through the origin and across both axes. The four points form a quadrilateral; count the integer pairs $(a, b)$ that give it perimeter $52$ .

Givens

$A$ is the reflection of $P(a, b)$ through the origin.
$B$ is the reflection of $P$ across the $x$ -axis.
$C$ is the reflection of $P$ across the $y$ -axis.
The quadrilateral with vertices $P$ , $A$ , $B$ , $C$ has perimeter $52$ , and $a$ , $b$ are integers.

Unknowns

The number of ordered pairs $(a, b)$ .

Constraints

The four points must form a real quadrilateral, so neither $a$ nor $b$ can be $0$ (otherwise the points pile up on an axis).

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #13 Convert to Algebra#2 Make a Systematic List

Draw the four points: they are the corners of a rectangle centered at the origin with sides parallel to the axes. Its side lengths depend only on $|a|$ and $|b|$ , so the perimeter condition becomes one short equation in $|a|$ and $|b|$ . Then list its solutions and attach signs.

3 · Execute5 carry out the plan

1Write the reflected points

#1 Draw a Diagram 6.NS.C.6

Reflecting through the origin flips both signs, across the

x

-axis flips the

y

-sign, and across the

y

-axis flips the

x

-sign.

A(-a, -b), \quad B(a, -b), \quad C(-a, b)

Each reflection just changes signs.

2See the rectangle

#1 Draw a Diagram 7.NS.A.1

The points

(\pm a, \pm b)

are the corners of a rectangle with horizontal sides from

-a

to

a

and vertical sides from

-b

to

b

. The quadrilateral exists only if

a \ne 0

and

b \ne 0

.

|a-(-a)| = 2|a|, \quad |b-(-b)| = 2|b|

The distance between a number and its opposite is twice its absolute value.

3Turn the perimeter into an equation

#13 Convert to Algebra 7.NS.A.1

Two widths and two heights add to

52

, so

4(|a| + |b|) = 52

.

2(2|a| + 2|b|) = 52 \Rightarrow |a| + |b| = 13

Divide both sides by

4

.

4List the absolute values

#2 Make a Systematic List 6.NS.C.7

With

|a|

and

|b|

both positive integers adding to

13

, the choices run

1+12

,

2+11

, and so on up to

12+1

, which is

12

choices.

(|a|, |b|) = (1, 12), (2, 11), \dots, (12, 1)

Neither part may be

0

.

5Attach the signs

#2 Make a Systematic List 6.NS.C.7

Each absolute-value pair allows

a

to be positive or negative and

b

to be positive or negative:

4

sign choices. With

12

absolute-value pairs that is

48

ordered pairs.

12 \times 4 = 48

Every nonzero absolute value comes from two numbers.

Answer:

48

4 · Reviewdoes it hold up?

The $48$ pairs are $(\pm 1, \pm 12)$ , $(\pm 2, \pm 11)$ , and so on up to $(\pm 12, \pm 1)$ . For example $(1, 12)$ gives a $2 \times 24$ rectangle with perimeter $2 \times (2+24) = 52$ .

Another way: Use symmetry: count only $P$ in Quadrant I, where $a+b=13$ with $a, b \ge 1$ gives $(1, 12)$ through $(12, 1)$ , $12$ points. Each of these has a matching point in each of the four quadrants, so $12 \times 4 = 48$ .

Standardsmin grade 7

6.NS.C.6 Rational numbers on the number line and coordinate plane — Reading the coordinates of reflections across the axes and through the origin as sign changes.
6.NS.C.7 Order and absolute value of rational numbers — Using absolute values and counting the numbers with a given absolute value.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Finding side lengths as distances between signed numbers.

💡Takeaway. Reflections only flip signs -- so the size of the shape lives in the absolute values.

Variant 12 hard

For a point $P(a, b)$ in the coordinate plane, point $A$ is symmetric to $P$ with respect to the origin, point $B$ is symmetric to $P$ with respect to the $x$ -axis, and point $C$ is symmetric to $P$ with respect to the $y$ -axis. How many ordered pairs $(a, b)$ of integers make the quadrilateral with vertices $P$ , $A$ , $B$ , and $C$ have a perimeter of $60$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

Reflect a lattice point $P(a, b)$ through the origin and across both axes. The four points form a quadrilateral; count the integer pairs $(a, b)$ that give it perimeter $60$ .

Givens

$A$ is the reflection of $P(a, b)$ through the origin.
$B$ is the reflection of $P$ across the $x$ -axis.
$C$ is the reflection of $P$ across the $y$ -axis.
The quadrilateral with vertices $P$ , $A$ , $B$ , $C$ has perimeter $60$ , and $a$ , $b$ are integers.

Unknowns

The number of ordered pairs $(a, b)$ .

Constraints

The four points must form a real quadrilateral, so neither $a$ nor $b$ can be $0$ (otherwise the points pile up on an axis).

2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #13 Convert to Algebra#2 Make a Systematic List

Draw the four points: they are the corners of a rectangle centered at the origin with sides parallel to the axes. Its side lengths depend only on $|a|$ and $|b|$ , so the perimeter condition becomes one short equation in $|a|$ and $|b|$ . Then list its solutions and attach signs.

3 · Execute5 carry out the plan

1Write the reflected points

#1 Draw a Diagram 6.NS.C.6

Reflecting through the origin flips both signs, across the

x

-axis flips the

y

-sign, and across the

y

-axis flips the

x

-sign.

A(-a, -b), \quad B(a, -b), \quad C(-a, b)

Each reflection just changes signs.

2See the rectangle

#1 Draw a Diagram 7.NS.A.1

The points

(\pm a, \pm b)

are the corners of a rectangle with horizontal sides from

-a

to

a

and vertical sides from

-b

to

b

. The quadrilateral exists only if

a \ne 0

and

b \ne 0

.

|a-(-a)| = 2|a|, \quad |b-(-b)| = 2|b|

The distance between a number and its opposite is twice its absolute value.

3Turn the perimeter into an equation

#13 Convert to Algebra 7.NS.A.1

Two widths and two heights add to

60

, so

4(|a| + |b|) = 60

.

2(2|a| + 2|b|) = 60 \Rightarrow |a| + |b| = 15

Divide both sides by

4

.

4List the absolute values

#2 Make a Systematic List 6.NS.C.7

With

|a|

and

|b|

both positive integers adding to

15

, the choices run

1+14

,

2+13

, and so on up to

14+1

, which is

14

choices.

(|a|, |b|) = (1, 14), (2, 13), \dots, (14, 1)

Neither part may be

0

.

5Attach the signs

#2 Make a Systematic List 6.NS.C.7

Each absolute-value pair allows

a

to be positive or negative and

b

to be positive or negative:

4

sign choices. With

14

absolute-value pairs that is

56

ordered pairs.

14 \times 4 = 56

Every nonzero absolute value comes from two numbers.

Answer:

56

4 · Reviewdoes it hold up?

The $56$ pairs are $(\pm 1, \pm 14)$ , $(\pm 2, \pm 13)$ , and so on up to $(\pm 14, \pm 1)$ . For example $(1, 14)$ gives a $2 \times 28$ rectangle with perimeter $2 \times (2+28) = 60$ .

Another way: Use symmetry: count only $P$ in Quadrant I, where $a+b=15$ with $a, b \ge 1$ gives $(1, 14)$ through $(14, 1)$ , $14$ points. Each of these has a matching point in each of the four quadrants, so $14 \times 4 = 56$ .

Standardsmin grade 7

6.NS.C.6 Rational numbers on the number line and coordinate plane — Reading the coordinates of reflections across the axes and through the origin as sign changes.
6.NS.C.7 Order and absolute value of rational numbers — Using absolute values and counting the numbers with a given absolute value.
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers — Finding side lengths as distances between signed numbers.

💡Takeaway. Reflections only flip signs -- so the size of the shape lives in the absolute values.

Three reflections make a rectangle sized by absolute values · 12 practice problems

Generated variants — 12

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Where next?