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184 geometry/hike.p




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This article is from the Puzzles FAQ, by Chris Cole chris@questrel.questrel.com and Matthew Daly mwdaly@pobox.com with numerous contributions by others.

184 geometry/hike.p


You are hiking in a half-planar woods, exactly 1 mile from the edge,
when you suddenly trip and lose your sense of direction. What's the
shortest path that's guaranteed to take you out of the woods? Assume
that you can navigate perfectly relative to your current location and
(unknown) heading.

geometry/hike.s

Go 2/sqrt(3) away from the starting point, turn 120 degrees and head
1/sqrt(3) along a tangent to the unit circle, then traverse an arc of
length 7*pi/6 along this circle, then head off on a tangent 1 mile.

This gives a minimum of sqrt(3) + 7*pi/6 + 1 = 6.397...

It remains to prove this is the optimal answer.

geometry/hole.in.sphere.p

Old Boniface he took his cheer,
Then he bored a hole through a solid sphere,
Clear through the center, straight and strong,
And the hole was just six inches long.

Now tell me, when the end was gained,
What volume in the sphere remained?
Sounds like I haven't told enough,
But I have, and the answer isn't tough!

geometry/hole.in.sphere.s

The volume of the leftover material is equal to the volume of a 6" sphere.

First, lets look at the 2 dimensional equivalent of this problem. Two
concentric circles where the chord of the outer circle that is tangent
to the inner circle has length D. What is the annular area between the
circles?

It is pi * (D/2)^2. The same area as a circle with that diameter.
Proof:

	big circle radius is R
	little circle radius is r
 
			      2		 2
	area of donut = pi * R   - pi * r
 
			       2    2
	=		pi * (R	 - r )
 
 
Draw a right triangle and apply the Pythagorean Theorem to see that
		 2      2	   2
		R  -   r   =  (D/2)
so the area is
			          2
	=		pi * (D/2)

Start with a sphere of radius R (where R > 6"), drill out the 6"
high hole. We will now place this large "ring" on a plane. Next to it
place a 6" high sphere. By Archemedes' theorem, it suffices
to show that for any plane parallel to the base plane, the cross-
sectional area of these two solids is the same.

Take a general plane at height h above (or below) the center
of the solids. The radius of the circle of intersection on the sphere is

	radius = srqt(3^2 - h^2)
 
so the area is 	
 
	pi * ( 3^2  - h^2 ) 

For the ring, once again we are looking at the area between two concentric
circles. The outer circle has radius sqrt(R^2 - h^2),
The area of the outer circle is therefore

pi (R^2 - h^2)

The inner circle has
radius sqrt(R^2 - 3^2). So the area of the inner circle is

	pi * ( R^2  - 3^2 ) 
 
the area of the doughnut is therefore
 
		pi(R^2 - h^2)  - pi( R^2  - 3^2 ) 
 
	=	pi (R^2 - h^2 - R^2 + 3^2)
 
	=	pi (3^2  - h^2)

Therefore the areas are the same for every plane intersecting the solids.
Therefore their volumes are the same.
QED

There also is a meta-theoretic answer to this puzzle. Assume the puzzle
can be solved. Then it must be solvable with a hole of any diameter, even
zero. But if you drill a hole of zero diameter that is six inches long,
you leave behind the volume of a six inch diameter sphere.

 

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