A watchdog is tied to the outside wall of a round building 20 feet in diameter. If the dog's chain is long enough to wind half way around the building, how large an area can the watchdog patrol?

(In reply to re: Pi Dog Night by Charlie)

Charlie's comment has made me rethink.  I believe my model to 'find' all the area was not optimal, and has left some area unaccounted for.   I should have made the model with rope wrapped tightly around the building to a given point, and then tangent.  I was making the rope point away from the building radially (in line with the tether point, not in line with the center of the building).  Either way, by simply allowing the rope to be not stuck to the building and make a sharp angle, the dog could go a little farther.

Given my assumption, it is a linear relationship, though.  The rope not tight against the building is the same as the arc length of building without rope up against it (ignoring the southern half of the building).  The angle of that arc length measured from the center of the building is 2a ( or 2 theta).  ie double the angle made by 2 chords going from the 2 ends of that same arc to meet somewhere on the far side of the circle (in this case the horizontal axis and the chord that is angle a (or theta) away from the horizontal.    Anyway, the angle is 2a, the radius is 10, so the length of rope is 20 a.    It is different for a rope running tangent to the building.

My mistake was trying to fit the reality of the problem into what I thought would be easier math.

ps:  I expect the correct area to be a few percent larger than what I found.

Posted by Larry on 2004-10-29 23:49:19