A Geometry and a Logic ProblemDate: 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: ftp://rtfm.mit.edu/pub/usenet/news.answers/puzzles/archive/ 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: 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 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 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/ |
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