Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I want your critique of the following derivation of the mean chord length in a convex polygon.

Q: Let $\mathbf{P}= P_1P_2..P_n$ be a convex polygon. Pick two of its edges at random, and further pick a random point on each of these sides. What is the expected length of the chord joining them?

Answer: Let $e_i$ be the edge $P_iP_{i+1}$ (with $P_{n+1}=P_{1}$) and let $\hat{l}_{ij}$ denote the mean chord length from $e_i$ to $e_j$. Then our final answer is $\hat{l}(\mathbf{P}) = \frac{\displaystyle\sum_{1 \leq i\leq n} \displaystyle\sum_{1\leq j < i} \hat{l}_{ij}}{{n \choose 2}}$.

To compute $\hat{l}_{ij}$, suppose without losing generality that $e_i$ and $e_j$ are nonadjacent. If either of them is vertical, rotate the polygon suitably until they cease to be so. Suppose now that $P_i=(x_i,y_i)$ and similarly for $P_{i+1},P_j$ and $P_{j+1}$. Since the edges are both non-vertical, we can write them as $e_i : y=a_i x+b_i , x \in [x_i, x_{i+1}]$ and similarly for $e_{j}$.

Now consider a test point $p=(x',a_ix'+b_i)$ on $e_i$ . A line segment from $p$ to any point on $e_j$ will lie entirely within $\mathbf{P}$, as $\mathbf{P}$ is convex. So the mean distance from $p$ to points in $e_j$ is just

$ \hat{l}_{p,j} = \frac{ \int_{x_j} ^{x_{j+1}} \sqrt { (x- x')^2 + (a_{j}x +b_{j} - (a_ix' +b_i))^2 } dx} { ||e_j|| }$

This gives the following value for the mean distance between $e_i $ and $e_j$

$\hat{l}_{ij} = \frac{ \int_{x_i} ^{x_{i+1}} \int_{x_j} ^{x_{j+1}} \sqrt { (x- x')^2 + (a_{j}x +b_{j} - (a_ix' +b_i))^2 } dxdx'} {||e_i||||e_j||} $

share|improve this question
    
What is the question? –  Douglas Zare Apr 6 '11 at 6:30
    
Is the derivation right? –  Ganesh Apr 6 '11 at 6:52
    
Why are you interested in this particular average of chord lengths? It looks to me like there are other much more natural averages, and that you have built in a bad definition. However, if you are sure you want this definition of mean chord length, then perhaps that isn't a flaw. –  Douglas Zare Apr 6 '11 at 9:42
    
This definition naturally arose in an application. Can you tell me about the other, more natural averages? –  Ganesh Apr 6 '11 at 16:44
    
You could select the points uniformly with respect to arc length. Then breaking a side in two or cutting off a vertex would not cause a discontinuous change in the mean chord length. As part of this, you need to be able to choose two points on the same edge. –  Douglas Zare Apr 11 '11 at 16:38

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.