$\triangle ABC$ has sides of length $45,60$ and $75$. Place a point $D$ randomly and uniformly inside the triangle. What is the expected value of the sum of perpendicular distances from point $D$ to the triangle's three sides?
Geometrically it should be $47$ but I cannot rigorously prove it. Any help please?
Best Answer
Let's use $AB=60$ along the $x$ axis, $AC=45$ along the $y$ axis. $A=(0,0)$, $B=(60,0)$, $C=(0,45)$. $BC=75$, and we will need to first calculate the equation of the line. $$\begin{align}y&=ax+b\\0&=60a+b\\45&=0a+b\end{align}$$ This yield $b=45$ and $a=-\frac34$. The distance from a point$(x,y)$ inside the triangle to the $AB$ side is $y$, to the $AC$ side is $x$, and to the $BC$ side is (from wikipedia)$$d(x,y)=\frac{|45x+60y-45\cdot 60|}{75}=36-\frac35x-\frac45y$$ The sum of the distances will be $$S=x+y+d(x,y)=36+\frac25x+\frac15y$$. With uniform distribution, $$\begin{align}\bar S&=\frac{\iint_\triangle S dxdy}{\iint_\triangle dxdy}\\&=\frac{\int_0^{60}dx\int_0^{45-\frac34x}dy (36+\frac25 x+\frac 15 y)}{\int_0^{60}dx\int_0^{45-\frac34x}dy}\end{align}$$