Random Points On A Square

The $x$ - and $y$ - coordinates of the two points are each uniformly distributed on $[0,1]$ , and all four coordinates are independent of each other. Thus the expected length is $E \; = \; \int_0^1 \int_0^1 \int_0^1 \int_0^1 \sqrt{(a-b)^2 + (c-d)^2}\,da\,db\,dc\,dd$ Using the substitution $u = a+b$ , $v = a-b$ , we notice that $\int_0^1 \int_0^1 f(a-b)\,da\,db \; = \; \tfrac12\int_{-1}^1 dv \int_{|v|}^{2-|v|} f(v)\,du \; = \; \int_{-1}^1 (1 - |v|)f(v)\,dv$ for any function $f$ , and hence $E \; = \; \int_{-1}^1 \int_{-1}^1 (1-|u|)(1-|w|)\sqrt{u^2 + w^2}\,du\,dw \; = \; 4\int_0^1 \int_0^1 (1-u)(1-w)\sqrt{u^2 + w^2}\,du\,dw$ But $\int_0^1 (1-u)\sqrt{u^2+w^2}\,du \; = \; \tfrac16\left[ 2w^3 + \sqrt{1+w^2} - 2w^2\sqrt{1+w^2}+ 3w^2 \ln\left(\frac{1 + \sqrt{1+w^2}}{w}\right)\right]$ for any $w > 0$ , and hence, integrating again, $E \; = \; \frac{2 + \sqrt{2} + 5\sinh^{-1}1}{15} \; = \; \frac{2 + \sqrt{2} + 5\ln\big(1 + \sqrt{2}\big)}{15}$ making the answer $2 + 2 + 5 + 1 + 2 + 15 = \boxed{27}$ .

The distance between two randomly selected points $(x_1,y_1),(x_2,y_2)$ is $d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$ . Now, $x_1,x_2,y_1,y_2$ are i.i.d. $\sim \mathcal{U}[0,1]$ , which implies that $X:=x_1-x_2,Y:=y_1-y_2$ are i.i.d. traingular distributed on the interval $[-1,1]$ with the density given by $f(x)=\left\{\begin{array}{rl} 1+x, & x\in[-1,0]\\ 1-x, & x\in(0,1] \end{array}\right.$ Thus the average distance is $L=\int_{-1}^1\int_{-1}^1 \sqrt{x^2+y^2}f_X(x)f_Y(y)dx dy$ Since the densities of $X,Y$ are even functions, we can simplify the integral as $L=4\int_{0}^1\int_0^1\sqrt{x^2+y^2}(1-x)(1-y)dxdy$ To evaluate the above integral, we use transformation to polar coordinates $x=r\cos \theta,y=r\sin \theta$ , and we will use the symmetry of the two parts of the square region of integration, above and below the diagonal through the origin, to get $L=8\int_0^{\pi/4}\int_0^{\sec\theta}r\cdot r(1-r\cos\theta)(1-r\sin\theta)dr d\theta\\=8\int_0^{\pi/4}\int_0^{\sec\theta} r^2(1-r(\cos\theta+\sin\theta)+r^2\sin\theta\cos\theta)drd\theta\\=8\int_0^{\pi/4}\left(\frac{\sec^3\theta}{3}-\frac{\sec^4\theta}{4}(\cos\theta+\sin\theta)+\frac{\sec^5\theta}{5}\sin\theta\cos\theta\right)d\theta\\=\frac{2}{3}\int_0^{\pi/4}\sec^3\theta d\theta-\frac{2}{5}\int_0^{\pi/4}\sin\theta \sec^4\theta d\theta\\=\frac{2}{3}I_1-\frac{2}{5}I_2$ where $I_1=\int_0^{\pi/4}\sec^3\theta d\theta\\=\int_0^1\sqrt{1+x^2} dx=\frac{1}{2}\left(\sqrt{2}+\ln(1+\sqrt{2})\right)$ $I_2=\int_0^{\pi/4}\sin\theta\sec^4\theta d\theta\\=\int_0^1 x\sqrt{1+x^2}dx\\=\frac{1}{3}(2\sqrt{2}-1)$ Thus, $I=\frac{2+\sqrt{2}+5\ln(1+\sqrt{2})}{15}$ Giving the answer $\boxed{27}$ .