8. Arrange the numbers $1,2, \cdots, 10$ in a row in some order, such that the sum of any three consecutive numbers does not exceed $n$. Answer the following:
(1) When $n=10$, can it be arranged? Please explain your reasoning;
(2) When it can be arranged, what is the minimum value of $n$?
8. (1) Suppose $n=10$ is already arranged. Then the sum of the last nine numbers is less than or equal to 30. Thus, the first number is not less than 25, which is a contradiction. Therefore, it cannot be arranged.
(2) With the same consideration as (1).
When $n=11,12,13,14$, none can be arranged.
When $n=15$, the sum of the first nine numbers is less than or equal to 45, leading to the tenth number being 10; also, the sum of the last nine numbers is less than or equal to 45, leading to the first number being 10. However, there is only one 10, so it cannot be arranged.
When $n=16$, it can be arranged. For example,
$$
10,5,1,7,6,2,8,3,4,9
$$

or $9,4,3,7,2,6,8,1,5,10$.
From this, we know that the minimum value of $n$ for which it can be arranged is 16.
$$
x_{a}=\frac{2 S}{a+h_{a}}, x_{b}=\frac{2 S}{b+h_{b}}, x_{c}=\frac{2 S}{c+h_{c}} \text {. }
$$

Since $a+h_{a}-\left(b+h_{b}\right)=a-b+h_{a}-h_{b}$
$$
\begin{array}{l}
=a-b+\frac{b h_{b}}{a}-h_{b}=a-b+\frac{b-a}{a} h_{b} \\
=(a-b)\left(1-\frac{h_{b}}{a}\right)>0
\end{array}
$$
(Given $a>b, h_{b}$ is the height on $b$, so $h_{b}>0$.
By analogy, we have $x_{a}<x_{b}<x_{c}$.
Therefore, when a square has two vertices on the shorter side, the area of the square is maximized.