Expected Area of 4 Lines

Let's place the square in the $xy$ -plane and let be

$A(0,0)$ , $B(1,0)$ , $C(1,1)$ , $D(0,1)$ .

Let's also define

$E(x_e,0)$ , $F(1,y_f)$ , $G(x_g,1)$ , $H(0,y_h)$

and

$P_1(x_1,y_1)$ , $P_2(x_2,y_2)$ , $P_3(x_3,y_3)$ , $P_4(x_4,y_4)$ .

The area of the region we're interested in, $\textbf{A}(P_1P_2P_3P_4)$ , can be written as

$\textbf{A}(P_1P_2P_3P_4)=\textbf{A}(ABCD)-\sum_{s \in S}\textbf{A}(s)+\sum_{t \in T} \textbf{A}(t)$

where $S=\{ABH, BCE, CDF, ADG\}$ and $T=\{BP_2E, FP_3C, DP_4G, AP_1H\}$ .

In other words, the area of the central quadrilateral is the area of the entire square ( $\textbf{A}(ABCD)$ ) from which we remove the area of big triangles adjactent to the sides ( $\sum_{s \in S}\textbf{A}(s)$ ). But, doing so, we remove the area of the smaller cornered triangles ( $\sum_{t \in T} \textbf{A}(t)$ ) twice, so we need to add them.

The expected area of $\textbf{A}(P_1P_2P_3P_4)$ is

$\mathbb{E}[\textbf{A}(P_1P_2P_3P_4)]=\mathbb{E}\big[\textbf{A}(ABCD)-\sum_{s \in S}\textbf{A}(s)+\sum_{t \in T} \textbf{A}(t) \big]=\textbf{A}(ABCD)-\mathbb{E}\big[\sum_{s \in S}\textbf{A}(s) \big]+\mathbb{E}\big[\sum_{t \in T}\textbf{A}(t) \big]$ .

Notice that

$\mathbb{E}[\textbf{A}(ABH)]=\mathbb{E}[\textbf{A}(BCE)]=\mathbb{E}[\textbf{A}(CDF)]=\mathbb{E}[\textbf{A}(ADG)]$

and

$\mathbb{E}[\textbf{A}(BP_2E)]=\mathbb{E}[\textbf{A}(FP_3C)]=\mathbb{E}[\textbf{A}(DP_4G)]=\mathbb{E}[\textbf{A}(AP_1H)]$

due to the symmetry of the problem, i.e. each triangle in $T$ can be obtained from the first one by rotating the square. The same holds for $S$ . For the sake of brevity we'll evaluate $\mathbb{E}[\textbf{A}(ABH)]$ and $\mathbb{E}[\textbf{A}(BP_2E)]$ , that is

$\displaystyle \mathbb{E}[\textbf{A}(ABH)]=\mathbb{E}\bigg(\frac{x_h}{2}\bigg)= \int_{0}^{1} \frac{x_h}{2} dx_h=\frac{1}{4}$

and

$\displaystyle \mathbb{E}[\textbf{A}(BP_2E)]=\mathbb{E}\bigg( \frac{1}{2} \frac{(x_e-1)x_h}{((x_e-1)x_h-1)}(1-x_e)\bigg)^{(1)}=\frac{1}{2} \int_{0}^{1} \int_{0}^{1} \frac{(x_e-1)x_h}{((x_e-1)x_h-1)}(1-x_e) dx_edx_h =\frac{3}{4}-\ln{2}$

Eventually, being $\textbf{A}(ABCD)=1$ , we have

$\displaystyle \mathbb{E}[\textbf{A}(P_1P_2P_3P_4)]=1-4 \cdot \frac{1}{4} + 4 \cdot \bigg(\frac{3}{4}-\ln{2}\bigg) = 3-\ln{2} \approx \boxed{0.2274}$

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$^{(1)}$ : The area for $BP_2E$ has been obtained intersecting $HB$ and $EC$ , in order to find $y_2$ (the height), and multiplying it by $EB=1-x_e$ (the base) divided by $2$