Non-geometric way to calculate expected value of breaks?"A bar is broken at random in two...

Step 1
Here's a sketch. Pick points $0\le x\le y\le 1$. Now let's find the expected value of x given that the interval [0,x] is shortest. That is, we look at the region satisfying $\begin{array}{rl}x& \le y\\ x& \le y-x\\ x& \le 1-y\end{array}$
You can draw your own pictures. We then get a triangle with $0\le x\le 1/3$, $2x\le y\le 1-x$. This triangle has area 1/6 (by inspection) or by double-integration, and ${\int }_0^{1/3}{\int }_{2x}^{1-x}xdydx=\frac{1}{54},$ so the expected value of x is ${\displaystyle \frac{\frac{1}{54}}{\frac{1}{6}}}={\displaystyle \frac{1}{9}}$.
Step 2
Next case. Suppose the interval [0,x] has middle length. This corresponds to two regions:
$\begin{array}{rl}x& \le y\\ y-x& \le x\\ x& \le 1-y\end{array}$
or $\begin{array}{rl}x& \le y\\ x& \le y-x\\ 1-y& \le x\end{array}$
Interestingly, these both have area 1/12 and for each of these we get ${\displaystyle \iint xdA={\displaystyle \frac{5}{216}}}$. So the expected value of x in the middle case is ${\displaystyle \frac{\frac{5}{216}}{\frac{1}{12}}}={\displaystyle \frac{5}{18}}$

Step 1
For the $k^{th}$ smallest piece indicated by $S_{(k)}$, the expected length is given by
$\begin{array}{rl}\mathbb{E}\left(n{S}_{(k)}\right)& =\mathbb{E}\left(X_{(k)}\right)\\ & =\sum _{i=0}^{k-1}\frac{1}{n-i}\end{array}$
which simplifies to the formula given in the book of 50 challenging problems.
Step 2
For the question asked, $n=3$ and calculate the sums for $k=1,2,3$, which we find to be
$\begin{array}{rl}\mathbb{E}\left(S_{(1)}\right)& =\frac{1}{9}\\ \mathbb{E}\left(S_{(2)}\right)& =\frac{5}{18}\\ \mathbb{E}\left(S_{(3)}\right)& =\frac{11}{18}\end{array}$