2. A pattern of Figures is shown below. Figure 1 is a regular pentagon with side length 1. Figure 2 is a regular pentagon of side length 2 drawn around Figure 1 so that the two shapes share the top vertex, $T$, and the sides on either side of $T$ overlap. The pattern continues so that for each $n>1$, Figure $n$ is a regular pentagon of side length $n$ drawn around the previous Figure so that the two shapes share the top vertex, $T$, and the sides on either side of $T$ overlap. The ink length of each Figure is the sum of the lengths of all of the line segments in the Figure.
(a) Determine the ink length of Figure 4.
(b) Determine the difference between the ink length of Figure 9 and the ink length of Figure 8.
(c) Determine the ink length of Figure 100.
2. (a) Figure 4 is shown below. Its ink length can be found in a number of ways, including by labeling the various segment lengths as shown:

The total ink length is thus $11 \cdot 1+3 \cdot 2+3 \cdot 3+3 \cdot 4=38$.
(We could instead use a similar method to the methods from (b) or (c) to calculate this.)
(b) We calculate the difference between these ink lengths by mentally starting with Figure 8 and determining what needs to be added to Figure 8 to produce Figure 9 .
Figure 9 is a copy of Figure 8 with a regular pentagon of side length 9 added and overlapping segments removed.
A regular pentagon with side length 9 has an ink length of $5 \cdot 9=45$.
The overlapping segments that must be removed are the two edges coming out of $T$ in Figure 8, which each have length 8 .
Therefore, the difference in ink length is $5 \cdot 9-2 \cdot 8=45-16=29$.
(c) Solution 1
Figure 100 consists of 100 regular pentagons of side lengths 1 to 100 , inclusive, each sharing a common vertex at $T$ and with edges on either side of $T$ overlapping.
Figure 100 can be constructed using the bottom three edges of each of these pentagons and the top two edges of the pentagon of side length 100 .
Its ink length is thus
$$
\begin{aligned}
3(1)+3 & (2)+\cdots+3(99)+3(100)+2(100) \\
& =3(1+2+\cdots+99+100)+2(100) \\
& =3\left(\frac{1}{2}(100)(101)\right)+200 \\
& =150(101)+200 \\
& =15150+200 \\
& =15350
\end{aligned}
$$

Therefore, the ink length of Figure 100 is 15350.
Solution 2
The ink length of Figure 100 can be calculated by starting with Figure 99 and determining how much ink length needs to be added.
Similarly, the ink length of Figure 99 can be calculated by starting with Figure 98 and determining how much ink length needs to be added.
In general, the ink length of Figure $k$ can be calculated by starting with Figure $(k-1)$ and determining how much ink length needs to be added.
Using this process, we can start with Figure 1, add to this to get Figure 2, add to this to get Figure 3, and so on, all of the way up to Figure 100.
To obtain the difference in ink length between Figure $k$ and Figure $(k-1)$, we can model what we did in (b):
We calculate the difference between these ink lengths by mentally starting with Figure $(k-1)$ and determining what needs to be added to Figure $(k-1)$ to produce Figure $k$.
Figure $k$ is a copy of Figure $(k-1)$ with a regular pentagon of side length $k$ added and overlapping segments removed.
A regular pentagon with side length $k$ has an ink length of $5 k$.
The overlap segments that must be removed are the two edges coming out of $T$ in Figure $(k-1)$, which each have length $(k-1)$.
Therefore, the difference in ink length is $5 k-2(k-1)=3 k+2$.

Therefore, to get from the ink length of Figure 1 (which is 5 ) to the ink length of Figure 100 we need to add each of the differences $3 k+2$ (calculated above) from $k=2$ to $k=100$.

Therefore, the ink length of Figure 100 is
$$
\begin{aligned}
5+(3(2) & +2)+(3(3)+2)+\cdots+(3(99)+2)+(3(100)+2) \\
= & (3(1)+2)+(3(2)+2)+(3(3)+2)+\cdots+(3(99)+2)+(3(100)+2) \\
= & 3(1+2+3+\cdots+99+100)+100(2) \\
= & 3\left(\frac{1}{2}(100)(101)\right)+200 \\
= & 150(101)+200 \\
= & 15150+200 \\
= & 15350
\end{aligned}
$$

Therefore, the ink length of Figure 100 is 15350.