I have a toy train set with lots of curved pieces. If you put any 8 of these pieces together you can make a circle with radius r.
Ignoring the width of the track, find the area (in terms of r) enclosed by each of the following configurations:
1) As the train goes around the track clockwise it goes 5 turns right then 1 turn left. This pattern continues twice.
2) 6 right 2 left, twice.
3) 5 right 1 left 1 right 1 left, twice.
4) 3 right 1 left, four times.
5) 4 right 1 left 2 right 1 left, twice.
6) 5 right 1 left 2 right 1 left 5 right 2 left.
Bonus: Are there any other configurations using 16 or fewer of these pieces?
In a circular setup, each section of track subtends a sector of area pi*r^2 / 8.
1) There are two lobes of area 5*pi*r^2 / 8 plus a "waist" area connecting them. This waist area consists of a rhombus of side 2*r with two of the abovementioned sectors cut out. The vertices of the rhombus are the centers of the convex arcs (5/8 circles) and the centers of the concave arcs (1/8 circles). The base of the rhombus is 2*r and its height is 2*r*cos(45°) = r*sqrt(2).
The total area in case 1 is therefore:
10*pi*r^2/8 + 2*sqrt(2)*r^2  2*pi*r^2 / 8
= (pi + 2*sqrt(2))*r^2
2) This time there are two 3/4 circles plus a square of side 2*r minus two quartercircles:
6*pi*r^2 / 4 + 4*r^2  pi*r^2 / 2
= (pi + 4) * r^2
Interesting aside: case 1 has an area smaller than two circles, while case 2 has an area larger than two circles. If fractional curved pieces could be used, there'd be some point at which the area would be that of exactly 2 circles. What would be the shape of such a track?

Posted by Charlie
on 20110801 14:00:06 