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### A Geometry and a Logic Problem

```
Date: 11 Mar 1995 15:53:26 -0500
From: Anonymous
Subject: Math Questions

Problem 1:

A cylindrical hole six inches long is drilled straight through
the center of a solid sphere. What is the volume remaining in
the sphere?

Problem 2: (This Is More Of A Logic Problem)

You are a contestant on a game show and you are asked to choose
from among three boxes - A,B, and C. One of the boxes contains
a diamond ring and two of them contain pop-out snakes. After you
have made a choice, the master of ceremonies (who knows which box
contains the ring) opens one of the boxes with the pop out snakes.
You are offered a choice to either stay with the box you originally
chose, or switch to the other unopened box. Are your chances better
if you stay with your original choice or if you switch, or does it
matter?  Why?

Thanks for your time and help.
```

```
Date: 11 Mar 1995 17:36:05 -0500
From: Elizabeth Weber
Subject: Re: Math Questions

Hello there!

Problem 2 (the famous "stay-or switch" problem)...

Your chances of winning are actually better if you switch. When you
pick a box in the beginning, the chance that you've picked a winner
is 1/3, right? So the chances that the winning box is really one of
the other boxes is 2/3. By giving you more information, the master
of ceremonies doesn't change this fact: *the probability that the
box you have chosen is the wrong one is STILL 2/3*.  So, after the
master of ceremonies opens one of the other boxes, you should switch,
since the probability that the other box is the winning one is 2/3.

Thanks for writing to Dr. Math!
Elizabeth, a math doctor
```

```
Date: 15 Mar 1995 10:59:36 -0500
From: Stephen Weimar
Subject: Re: Math Questions

These puzzles are fun and challenging, aren't they?  Both of these
problems are famous and any such puzzles have been well-discussed
on the Internet in the rec.puzzles newsgroup. You can read the
newsgroup archives by going to:

Elizabeth responded to the Monty Hall problem.  Here's what the
archive has to say about the bored hole in a sphere problem:

==  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, let's 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:

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)

6" high hole.  We will now place this large "ring" on a plane.
Next to it place a 6" high sphere. By Archimedes' 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

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. Q.E.D.

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.
```

```
Date: 09/28/98 at 15:34:42
From: Sarah Cook
Subject: Problem on question on the sphere

In order to know the remaining volume of the sphere you have to not
only know that the length of the hole is 6 inches long, but what the
radius of the hole is. What is it?
```

```
Date: 09/28/98 at 17:07:15
From: Doctor Peterson
Subject: Re: Problem on question on the sphere

Hi, Sarah. We also discuss this problem in:

http://mathforum.org/dr.math/problems/klein12.30.96.html

If you read carefully, you will see (though you probably still won't
believe) that it doesn't matter how big the sphere itself is. If you
drill a hole in any sphere of diameter greater than 6 inches, and
make sure that the diameter of the hole is just right to make the
hole 6 inches long, then what is left of the sphere will always have
the same volume. So we don't know, and can't even find out, how big
the hole is, yet we can tell what volume is left.

This sort of thing happens in many interesting problems: you assume
some unknown value for a variable (such as the radius of the hole)
and when you're done calculating, you find that the variable has
disappeared, and it doesn't have any effect on the result.

So what is happening? Try to picture a small sphere, maybe with
diameter 7 inches. When you drill through it with a small drill, you
will be cutting out a narrow cylinder of height 6 inches, plus two
shallow caps:

*****
**+-------+**
*XX|       |XX*
*XXX|       |XXX*
*XXXX|       |XXXX*
*XXXX|       |XXXX*
*XXX|       |XXX*
*XX|       |XX*
**+-------+**
*****

Now picture a huge sphere, maybe the size of the earth. When you drill
through this with a drill almost as wide as the earth, you will be
cutting out a very wide cylinder of height 6 inches, plus two caps
that take out most of the remaining volume from the sphere:

***********
******           ******
***                       ***
***                             ***
*                                   *
*                                     *
**                                       **
*                                           *
*+-------------------------------------------+*
*|                                           |*
*X|                                           |X*
*X|                                           |X*
*X|                                           |X*
*X|                                           |X*
*|                                           |*
*+-------------------------------------------+*
*                                           *
**                                       **
*                                     *
*                                   *
***                             ***
***                       ***
*******         *******
*********

It just happens that even though the sphere is so much larger, the
drill has to take out just as much more in order to make the height of
the hole the same, so that the volume left doesn't depend on the size
of the sphere itself, or of the hole itself - only on their relation,
which is forced by requiring the hole to be 6 inches long.

It's still surprising, but maybe it makes a little sense now.

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Logic
High School Probability
High School Puzzles

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