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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.


 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 

   2      2          2
  R  -   r   =  (D/2)
so the area is

 =  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 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

 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. 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 

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:


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
Associated Topics:
High School Logic
High School Probability
High School Puzzles

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