A 3-D Object that Fits 3 Holes
Date: 06/03/99 at 20:25:08 From: Gretchen Subject: Geometry Dr. Math, I have three shapes: a triangle with the same bisector length from each vertex, a square with the same length as the bisector of the triangle, and a circle with a diameter the same length as a side of the square and the length as the bisector. I have to come up with an object that can fit in the three different holes blocking all light and sliding all the way through without forcing it. I have come up with the coke can squash, but it doesn't seem to work. What I did was create a cylinder and cut the two faces so that it lay like this: /\ /__\ thus creating the triangle from the front, the cone from the faces and the square from the bottom, only the square fit only halfway through. Is there any other way to the problem, or is my venture correct? Thank you for your time. Hope you have as much fun with this problem as I have.
Date: 06/04/99 at 11:49:42 From: Doctor Rick Subject: Re: Geometry Hi, Gretchen. This is an interesting challenge! In your description of the problem, is it possible that you mean for the triangle to have a height equal to its base? The way you describe it, it would be an equilateral triangle, with height equal to sqrt(3)/2 times the base, but in this case either the square would have to be a rectangle or the circle would be an ellipse. They couldn't both have the same height and width. If I understand your description of your proposed solution correctly, the axis of the cone is horizontal in your triangle diagram. The two tilted sides are ellipses where the cylinder was cut off. I don't know quite what you mean when you say "the square fit only halfway through," but you do have a problem in this direction. Another way to think of the problem is in terms of the shadows cast by a light directly above, in front of, or to the side of the object. The shadow cast by the overhead light with your object will be a rectangle with half ellipses stuck on the sides. This is what it looks like from the top: ************* * : * * : * * : * * : * * : * : * * : * : * * : * : * * : * : * * : * : * * : * : * * : * * : * * : * * : * ************* This is obviously not a square, though its maximum width and its height are correct so it will fit through the hole; it just won't fill it so as to block the light. There is a way to do it correctly. In fact, the shape I have in mind came up recently in the answer to another question about why the volume of a cone isn't half the base area times the height. Stand your cylinder up vertically. It should have a height equal to the diameter of the base circle. You will see a circle from the top and a square from front and side. Only one view must be changed. Draw a diameter of the top circle, front to back. Can you see what parts have to be removed to give the correct front view without spoiling the other views? There is more than one way to do this. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/
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