Dog on a Leash

My National Review Online "Diary" column for February 2003 included a brainteaser about a dog on a leash.  Here it is..

Dog on a Leash

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A lighthouse has a circular floor plan, 20 feet in diameter.  On the outside wall of the lighthouse is fixed a hook.  A leash is tied to the hook, with a dog at the other end.  The leash is 10 pi feet (that is to say, a tad more than 31.4159265358 9793238462 6433832795 0288419716 9399375105 8209749445 9230781640 6286208998 6280348253 4211706798 feet) long.  Assuming the dog cannot enter the lighthouse, and the surface he can travel is flat and unobstructed, what area can he cover?

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Solution.  This one needs a diagram.  The inner circle represents the lighthouse.  The hook is at the right on the lighthouse wall.  At full stretch, the dog can be 10 pi feet away from the hook;  that's the right-most point of the diagram.

If the dog heads "north" from this point (assuming, with no loss of generality, that the diagram is oriented in the usual north-south-east-west way), and still keeps the leash taut, he can walk out a quarter-circle, ending at the top of the diagram.

If he keeps on walking — he'd be heading directly due west at first — the leash starts to wrap round the wall of the lighthouse.  So the radius of his path is no longer constant.  It's lessening all the time, as more and more leash wraps round the lighthouse wall...  until the dog himself crashes into the wall.  Since the leash is 10 pi feet long, and that is precisely half the circumference of the lighthouse (circumference of a circle is pi times its diameter, remember), the dog crashes into the lighthouse wall at precisely the west-most point of the wall, with all his leash wrapped round the north half of the lighthouse.

The same applies in the south half of the diagram, of course.  We can just solve the problem for the north half, then double the answer.

So what is the area the dog can cover in the north half?  Well, it's that quarter of a circle, plus that non-circular bit the dog traces out when his leash is wrapping round the wall.

The quarter circle is easy.  Area of a full circle is pi times square of radius, in this case pi times square of 10 pi.  That makes 100 times pi cubed.  We only need a quarter of that:  25 times pi cubed.

To get the non-circle area, I believe you have to use calculus.  (Though if there is a non-calculus solution, I'd be glad to hear it.)  Consider the situation, leash at full stretch, when the leash has already wrapped round an arc of the wall that makes an angle t at the center.  The angle t is of course measured in radians.  Length of this arc is 10t, by elementary geometry (length of an arc of a circle).  The remainder of the leash — the part that's straight — is therefore 10 pi minus 10t, or 10 (pi - t).  Check:  When the angle t is zero, the straight part is 10 pi.  When the angle t is pi radians (that is, 180 degrees), the straight length is zero.  Yep.

Now increase t by a tiny amount dt.  The tangent advances correspondingly; and the angle between the advanced tangent and the original one, again by elementary geometry, is dt.  The area of the shaded triangle between the two tangents, to a first-order approximation, is 50 times (pi - t) squared times dt, again by elementary geometry (area of a sector of a circle).  If you integrate that from t equals zero to t equals pi, you get 50/3 times pi cubed.

The top half of the required area is therefore (25 times pi cubed) plus (50/3 times pi cubed).  This is 125/3 times pi cubed.

The entire area is twice this (north plus south):  250/3 times pi cubed.

Solution:  250/3 times pi cubed, or about 2583.856390 0249850146 2302625559 1782933519 square feet.