#40 THE MICE PROBLEM

Six mice of negligible size are sitting in the corners of a hexagonal room (see picture). The length of each wall is three meters. Each mouse focuses on its right neighbor and at the same time all mice start running with the same velocity. Each mouse always runs straight to the current position of the focused neighbor.

What distance does each mouse run until they meet?

Correct answer: 3m

Explanation:



If we look at the mice at the bottom we can consider the following running directions for these mice for the very first moment:

m1=(x=1,y=0)
m2=(x=0.5,y=sqrt(3)/2)

After a infinitesimal small step of size e the distance decreases from A to A*.
According to Pythgoras' theorem we will get A*=sqrt((A-0.5*e)²+(sqrt(3)/2)²*e²)
A*=sqrt(A²-A*e+e²)
For every small step of size e the mice take the distance will decrease from A to A*. After this small step all six mice still form a hexagon but the the distance is now A*.
The lengths of the distances form a decreasing geometrical series. Thus the length of the path a mouse will travel until they meet is:
e/[1-A*/A]=e/[1-sqrt((A²-A*e+e²)/A²)]=e/[1-sqrt(1-e/A+e²/A²)]

This is according to the formula that the sum of an infinite decreasing geometric series is a1/(1-q) where a1 is the value of its first item and q is the proportion between two consecutive items.

Now we need to limit e to 0: 

                e
lim   --------------------- =
e->0   1-sqrt(1-e/A+e²/A²)


                e              1+sqrt(1-e/A+e²/A²)
lim   --------------------- * --------------------- =
e->0   1-sqrt(1-e/A+e²/A²)     1+sqrt(1-e/A+e²/A²)


        e * (1+sqrt(1-e/A+e²/A²))
lim   ---------------------------- =
e->0        1-(1-e/A+e²/A²)



       e * (1+sqrt(1-e/A+e²/A²))
lim   --------------------------- =
e->0         e/A-e²/A²



       (1+sqrt(1-e/A+e²/A²))
lim   ----------------------- =
e->0         1/A-e/A²



   (1+sqrt(1-0/A+0²/A²))
  ----------------------- =
       1/A-0/A²

   2
 ----- = 2*A = 2*3m = 6m
  1/A

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