Suppose point $C$ is directly above point $B$ of $\overline{AB}$. $|\overline{AB}| =2$, and the distance from $C$ to $B$ is $1$.
How can I find the average distance from point $C$ to every point on $\overline{AB}$?
To further clarify my question, suppose we start to randomly pick ten points $A,D,E,F,G,H,I,J,K,B$ on $\overline{AB}$ and connect them all to point $C$.
The average distance from point $C$ to points $A,D,E,F,G,H,I,J,K,B$ would then be $$\frac{|\overline{CA}|+|\overline{CD}|+|\overline{CE}|+|\overline{CF}|+|\overline{CG}|+|\overline{CH}|+|\overline{CI}|+|\overline{CJ}|+|\overline{CK}|+|\overline{CB}|}{10}$$Ideally, there should be infinitly many segments, but I cannot draw them all.
Now, return to my question: what is the the average distance from point $C$ to every point on $\overline{AB}$?
I think this question is related to Triangle Line Picking, but I have not been able to figure out the set up for solving this problem. Any hints would be appreciated.
$\endgroup$ 12 Answers
$\begingroup$This is $\frac12\int_0^2 \sqrt{x^2+1}\;dx$, as Aditya Dua explains. To evaluate this integral, make the substitution $x=\sinh u$, so that $x^2+1=\cosh^2u$ and $dx=\cosh u\;du$. We get
$$\begin{align} \frac12\int_0^{\sinh^{-1}2}\cosh^2 u\;du&=\frac14\int_0^{\sinh^{-1}2}(\cosh 2u+1)\;du\\ &=\frac14\left[\frac12\sinh 2u+u\right]_0^{\sinh^{-1}2}\\ &=\frac14\left[\sinh u\cosh u+u\right]_0^{\sinh^{-1}2}\\ &=\frac14(2\sqrt{2^2+1}+\sinh^{-1}2)\\ &=\frac12\sqrt 5+\frac14\sinh^{-1}2 \end{align}$$
which is about $1.47894285754460$. You can find the various hyperbolic identities at this link.
$\endgroup$ $\begingroup$Without loss of generality, let the coordinates of A, B, C be: $(0, 0), (b, 0)$, and $(b, c)$. An arbitrary point on AB is described by the coordinates $(x, 0)$, where $x \in [0, b]$. The distance from C to this point is given by $d(x) = \sqrt{(b-x)^2 + c^2}$. Now you can integrate $d(x)$ w.r.t. to $x$ to get the average distance $\bar{d}$.
$\bar{d} = \frac{1}{b}\int_0^{b} \sqrt{(b-x)^2 + c^2} dx$.
I think a substitution like $(b-x) = c \tan \theta$ might work.
Specifically, in your example, $b=2$ and $c=1$.
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