Thursday, September 21, 2017

An Expected Value Problem III


In An Expected Value Problem, we found the expected length of a line segment drawn by picking a random point in a rectangle and choosing a random angle, and extending it until it meets the edge of a rectangle. We ask the same question here for an arbitrary acute angled triangle in this post.

Let $ABC$ be a triangle with $AB=c$, $BC=a$, and $CA=b$. Let $D$, $E$ and $F$ denote the feet of the altitudes from $A$, $B$ and $C$ receptively such that $AD=h_a$, $BE=h_b$ and $CF=h_c$. I considered this question in the same post above, but I was not able to solve it then.

But recently I collaborated with someone who gave me a very valuable hint which reduced the difficulty of the problem drastically. The idea pretty much solved the problem. In summary, if we denote the length of line segment as $L$, we have,

$
\begin{align}
\displaystyle\mathbb{E}(L)&=\sum_{x \in \{a,b,c\}}\frac{h_x}{3\pi}\text{csch}^{-1}\left(\frac{p}{x}\frac{s-x}{p-x}\right)\\
&=\frac{h_a}{3\pi}\text{csch}^{-1}\left(\frac{p}{a}\frac{s-a}{p-a}\right)+\frac{h_b}{3\pi}\text{csch}^{-1}\left(\frac{p}{b}\frac{s-b}{p-b}\right)+\frac{h_c}{3\pi}\text{csch}^{-1}\left(\frac{p}{c}\frac{s-c}{p-c}\right)
\end{align}
$

where $p$ and $s$ denote the perimeter and the semi-perimeter respectively.

Let's calculate the expected values for a few special cases.

$\mathbb{E}(1,1,1)\approx 0.30284835$ and $\mathbb{E}(3,4,5)\approx 1.100147755$

The interest in this question seems very low in the mathematical literature and I was not able to find any material in the internet to verify these values. However, a friend of mine wrote a simulation for the Pythagorean case and the result matched with the value given by the formulae.

Two things are to be mentioned here. First, even though we dealt with the same question for a triangle and a rectangle, note that the formulas are drastically different. Some may find it obvious that they are different but I had an intuition that they'll be pretty 'close' which turned out to be wrong.

Second, the individual terms in the above formulae can themselves are respective expected values for a different question. I'll follow up with the what those questions are and the proof of this formulae in the next few posts.


Until then,
Yours Aye
Me

1 comment:

  1. Hi there. This is really nice work. I have also considered variations on the same problem. If you haven’t published this already then we could combine our methods and results Let me know if you’re interested in this project, and I can share what I’ve written up already. You can teach me at
    David.treeby at Monash dot edu

    ReplyDelete