Birds on a Wire

By Moshe Eliner

Let's pass through all birds, starting, say, from the left, stopping at every fourth bird. I'll call this segments 4b. Notice a delicate point:

Denote the shortest connecting segments sc (i.e. yellow lines). Each 4b has 3 inner connecting segments (average 2).

Since:

1. the states of sc=1 and sc=3 force inner point positions on the neighboring 4b's.

2. on a uniform wire birds tend to have unfiorm distribution, hence inner point positions cannot be predefined.

3. so there cannot be sc=1 or 3 more than average. This implies that each 4b has on average sc=2.

Now it's enough to calulate the density when sc=2:

There are 3 possible sc=2's , but since one can rearrange segments inside the 4b, all have the same density, let's choose one of them, e.g., the one with two shortest at first:

Integrate ydxdy:

1. y=1/2..2/3, x=2y-1..1-y

2. y=0..1/2, x=0..y.

This yields 7/108, but since the total integration area is 1/6, the result density is: 7/18.

Copyright © 1996-2009 Alexander Bogomolny