The problem provides a table relating a figure number to the number of outer lines of unit length (which gives the perimeter of the figure). The task is to complete the missing rows in the table and then to prove that no figure in the sequence can have a perimeter of 100.

1. Problem Description

The problem provides a table relating a figure number to the number of outer lines of unit length (which gives the perimeter of the figure). The task is to complete the missing rows in the table and then to prove that no figure in the sequence can have a perimeter of

1

0

0.

2. Solution Steps

First, we need to identify the pattern in the table.

The figure number increases by 1 in the first column.

The perimeter increases.

The perimeter for figure number 1 is

5. The perimeter for figure number 2 is

8. The perimeter for figure number 3 is

1

1. The differences between consecutive perimeters are: $8 - 5 = 3$, and $11 - 8 = 3$.

Therefore, the perimeter increases by 3 for each increment of 1 in the figure number.

Also, we can deduce the formula of the number of lines of unit length based on the figure number $n$:

Number of lines = $n + 2 + 2(n-1) = n + 2 + 2n - 2 = 3n$.

Now we can use this information to fill in the missing rows.

(i)

Given the figure number is

6. The number of outer lines is $6+2+2(6-1) = 6 + 2 + 10 = 18$.

The perimeter is $3 \times 6 + 2 = 18 + 2 = 20$. This row is already provided so, we can use it to find the other values.

(ii)

Given the perimeter is 65, we can find the figure number $n$ using the formula for the perimeter which can also be expressed as perimeter = $3n+2$. Thus, $3n + 2 =65$, so $3n = 63$, and $n = 21$.

The number of outer lines will be $3n=3(21)=63$. Also, we can derive this by using our $n + 2 + 2(n-1)$ formula by substituting $n=21$: so we have $21 + 2 + 2(20) = 21 + 2 + 40 = 63$.

(iii)

The general case for figure $n$:

The number of outer lines of unit length is $n + 2 + 2(n-1) = n + 2 + 2n - 2 = 3n$.

The perimeter is $3n + 2$.

To prove that no figure can have a perimeter of 100, we can set up the equation $3n + 2 = 100$. Solving for $n$:

$3n = 100 - 2$

$3n = 98$

$n = \frac{98}{3} = 32.666...$

Since $n$ must be an integer for a figure number, and $32.666...$ is not an integer, there is no figure with a perimeter of

1

0

0.

3. Final Answer

(i) Figure number = 6, Number of lines of unit length = 18, Perimeter = 20 (given).

(ii) Figure number = 21, Number of lines of unit length = 63, Perimeter = 65 (given).

(iii) Figure number = $n$, Number of lines of unit length = $3n$, Perimeter = $3n+2$.

Proof that no figure can have a perimeter of 100:

$3n + 2 = 100$

$3n = 98$

$n = \frac{98}{3} \approx 32.67$

Since $n$ is not an integer, no figure has a perimeter of

1

0

0.