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Subject: "Math Layout"     Previous Topic | Next Topic
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Conferences The CTK Exchange Middle school Topic #109
Reading Topic #109
AEROX
guest
Jan-31-05, 08:04 PM (EST)
 
"Math Layout"
 
   I am having some trouble figuring out a way to make a layout of a cube I have to draw different ways to amke a cube and do not understand?


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alexb
Charter Member
1905 posts
Jan-31-05, 08:08 PM (EST)
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1. "RE: Math Layout"
In response to message #0
 
   How many square sides has a cube? Cut as many as you need out of paper. All must be equal in size. Put two of the squares so that they share an edge. Attach one to the other by a sticky tape. Place this two on a cube. Think where you can attach a third square to the first two so that you'd be able to flatten out all the three. Repeat for the fourth square and as needed.


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Layout
guest
Feb-04-05, 11:36 AM (EST)
 
2. "RE: Math Layout"
In response to message #0
 
   Well you could draw a square then about in the middle and up draw another. Connect the squares together with lines. Hope this helps, really simple sorry if you already knew that and it didnt help much!


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GEB
guest
Mar-02-05, 06:00 PM (EST)
 
3. "RE: Math Layout"
In response to message #2
 
   If you know there are 6 squares that make up the cube, how many unique ways could you arrange these 6 squares so they would still fold up into a cube?
Having played for a while, can you start thinking of a system that would help you check?


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alexb
Charter Member
1905 posts
Mar-02-05, 11:46 PM (EST)
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4. "RE: Math Layout"
In response to message #3
 
   >If you know there are 6 squares that make up the cube, how
>many unique ways could you arrange these 6 squares so they
>would still fold up into a cube?
>Having played for a while, can you start thinking of a
>system that would help you check?

Of the top of my head:

Think of the cube as a graph with 8 vertices, 12 edges and 6 faces. Its dual graph is the octahedron, a graph with 6 vertices and 12 edges, each vertex being of degree 4. What you are looking for is the number of spanning trees on the octahedron (as a graph). Any such spanning tree has 5 = 6 - 1 edges. Thus the upper bound on your number is C(5, 12) - 12 choose 5. This is not the exact number as there are spanning subgraphs that contain cycles.


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Silver_dragon
guest
Oct-23-06, 10:08 PM (EST)
 
5. "RE: Math Layout"
In response to message #0
 
   I am having the same question, but I know that there are 11 ways total. IF ANYONE KNOWS THE WAYS, PLEASE SHOW ME!!!


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