Challenging UKMT geometry question I was attempting this problem and am struggled to find an approach. Could someone please show how to solve it. If you could also give me any advice on any thoughts to consider when seeing a question like this that would be very helpful (although I appreciate that this is a vague request).

 A: Adopt Cartesian coordinates where $O$ is the bottom left vertex of the square, and the sides meeting there are the positive $x$- and $y$-axes. The $i$th rightmost line passes through $(0,\,\frac{i}{n})$ and $(\frac{n-i+1}{n},\,0)$, so has equation $x=\frac{n-i+1}{n}-\frac{n-i+1}{i}y$. This intersects line $i+1$ at $y=\frac{i(i+1)}{n(n+1)}$.
The lines create $n$ triangles each of base $\frac1n$. Moving from right to left, the shaded area is$$\sum_{i=1}^n\frac12\frac1n\frac{i(i+1)}{n(n+1)}=\frac{n+2}{6n}.$$For $n=3$, this is $\frac{5}{18}$; for $n=10$, it's $\frac15$.
With a bit of work you can show the large-$n$ case approximates the curve $y=(1-\sqrt{x})^2$, bounding a grey area of $\frac16$.
A: You might like to use Euclidean Geometry, alternatively. Here's, for example, how.
Label the segments in which the square sides are divided as shown below, and let $1$ be the square side.

By Menelaus's Theorem on $\triangle A_nA_0B_1$ cut by $B_2A_1$ we have
\begin{eqnarray}\frac{\overline{B_2B_1}}{\overline{B_2A_n}}\cdot\frac{\overline{A_nA_1}}{\overline{A_1A_0}}\cdot\frac{\overline{A_0X_1}}{\overline{X_1B_1}}&=&1\\
\frac{1}{2}\cdot\frac{n-1}{1}\cdot\frac{\overline{A_0X_1}}{\overline{X_1B_1}}&=&1 
\end{eqnarray}
which gives
\begin{eqnarray}
\frac{\overline{A_0X_1}}{\overline{X_1B_1}} &=& \frac{2}{n-1}\\
\frac{\overline{A_0X_1}}{\overline{A_0B_1}} &=& \frac2{n+1}.
\end{eqnarray}
By Thales Theorem you obtain the altitude of $\triangle A_1A_0X_1$ with respect to the side $A_1A_0$
$$h_1=\frac{\overline{A_0X_1}}{\overline{A_0B_1}}\cdot \overline{A_nB_1}=\frac{2}{n(n+1)}.$$

Generalizing this approach, for $i=1,2,\dots,n-1$, you apply Menealus's Theorem to $\triangle A_nA_{i-1}B_i$, cut by line $B_{i+1}A_i$ to get
\begin{eqnarray}\frac{\overline{B_{i+1}B_i}}{\overline{B_{i+1}A_n}}\cdot\frac{\overline{A_nA_i}}{\overline{A_iA_{i-1}}}\cdot\frac{\overline{A_{i-1}X_i}}{\overline{X_iB_i}}&=&1\\
\frac{1}{i+1}\cdot\frac{n-i}{1}\cdot\frac{\overline{A_{i-1}X_i}}{\overline{X_iB_i}}&=&1, 
\end{eqnarray}
yielding
\begin{eqnarray}
\frac{\overline{A_{i-1}X_i}}{\overline{X_iB_i}} &=& \frac{i+1}{n-i}\\
\frac{\overline{A_{i-1}X_i}}{\overline{A_{i-1}B_i}} &=& \frac{i+1}{n+1},
\end{eqnarray}
and, finally, the altitude of $\triangle A_iA_{i-1}X_i$
$$h_i=\frac{\overline{A_{i-1}X_i}}{\overline{A_{i-1}B_i}}\cdot \overline{A_nB_i}=\frac{i(i+1)}{n(n+1)}.$$

From here, as in J.G.'s answer, the shaded area is obtained as
$$\mathcal A= \frac{1}{2n^2(n+1)} \cdot \sum_{i=1}^ni(i+1)=\frac{n+2}{6n}.$$
