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Triangle proof help

  1. May 24, 2006 #1
    I was wondering whether someone could come up with the proof of this exercise I set myself:

    We have four different types of triangles to play with (unlimited amounts of each):

    one triangle which has 3 sides of length 1
    one triangle which has 2 sides of length 1 and one of length square root 2
    one triangle which has 2 sides of length square root 2 and one of side length 1
    one triangle which has 3 sides of length square root 2

    When I assumble 4 triangles together, what is the maximum amout of irregular tetrahedron (3-D shape) I can form. And what is the proof to make sure I have found the maximum?

    My answer so far is:

    Using:

    4 triangles the same shape ----- you can make 2 irregular tetrahedrons.
    3 triangles the same shape----- you can make 2 irregular tetrahedrons.
    2 the same + 2 same (but different to the 2 others) ---- you can make 1 irregular tetrahedron.
    2 the same + 2 different --- you can make 3 irregular tetrahedrons (I'm not sure on this one)
    0 different ---- obviously 0 irregular tetrahedron.

    Does this make any sense?
     
    Last edited: May 24, 2006
  2. jcsd
  3. May 24, 2006 #2

    AKG

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    Well your "0 different" and "4 triangles the same shape" should be the same category, but for one you say you can make 2 irregular tetrahedrons and for the other you say you can make none.

    I think with 4 the same shape, you can make 2.
    With 3 the same shape, I think you can make 2.
    With 2+2 you can make 3.
    With 2+1+1 you can make 2.

    Let's give the different types of triangles labels:

    A : one triangle which has 3 sides of length 1
    B : one triangle which has 2 sides of length 1 and one of length square root 2
    C : one triangle which has 2 sides of length square root 2 and one of side length 1
    D : one triangle which has 3 sides of length square root 2

    So for 4 triangles the same, you can make BBBB and CCCC
    For 3+1 you can make ACCC and BBBD
    For 2+2 you can amke AABB, BBCC, and CCDD
    For 2+1+1 you can make ABBC and BCCD
     
  4. May 25, 2006 #3
    That's great thanks. But is this a proof? I mean does this support 200% the fact that 9 is the maximum amount of irregular tetrahedrons and garanties there are no other possible ones?
     
  5. May 25, 2006 #4

    AKG

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    No, it's not a proof at all. I just tried to break down all the possibilities into easily manageable cases, and count the possibilities.

    Case 1: has an A piece (w.l.o.g. we can make the base an A piece)
    Subcase 1a: has at least 2 A pieces
    Subcase 1b: has only 1 A piece

    Case 2: has a B piece but no A piece (w.l.o.g. let the base be a B piece)
    Subcase 2a: has at least 2 B pieces
    Subcase 2b: has only 1 B piece

    Case 3: has a C piece, but no A or B
    Subcases:... (you can figure this out)

    Case 4:... (figure it out)

    Once you've figured out your subcases, there aren't that many possibilities for tetrahedrons in each subcase, so you can work out the possibilities. You also have to check that they work out geometrically. For example, in the attached drawing, you have three faces of a shape, and you see sides of length 1, 1, and 21/2, but looking at it you'll see that you can't put a triangle of those side lengths to join those three faces. Just because the numbers match up doesn't mean that the shapes will. I'm not sure if the same problems will occur with tetrahedrons (in my image, there's a rectangle because it's easier to illustrate the problem with a rectangle), you'll have to check.
     

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  6. May 25, 2006 #5

    AKG,

    In your earlier post you say that:


    A - SMALL EQUI [3 * 1]
    B - RIGHT ANGLE [1 * 1] +[ 2 * ROOT 2]
    C - BIG ISOSC [2 * ROOT 2] +[ 1]
    D - BIG EQUI [3 * ROOT 2]

    You say we can make BBBB. If you join two sets of root 2 sides together (which you HAVE to do! - where else would they go?), you'll be left with basically two squares (imagine diamonds with bisecting lines down the middle). Now if you join these two diamonds together at one edge (any edge), you can't match up the others. Not without using a naked flame, anyway.
     
  7. May 25, 2006 #6
    I've just emailed my lecturer with this problem who told me there are actually 10 possibilities and not 9. I must say I'm puzzled at this stage!
     
  8. May 25, 2006 #7

    AKG

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    I should have mentioned that I wasn't sure my stuff was right. You're right, BBBB doesn't really work. It gives a "flat tetrahedron", which is why you have to check that the possibilites work out geometrically, i.e. it's not enough to check that the lengths of the sides match up.
     
  9. May 25, 2006 #8

    matt grime

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    Why? Just count the cases. They are a very small number.
     
  10. May 25, 2006 #9

    AKG

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    If BBBB counts as a (very) irregular tetrahedron, then the most I can see possible is 9. Certainly not 10. I broke it up into the same cases you did:

    All the same: 1 one possibility CCCC (+ possibly BBBB)
    3+1: CCCA, BBBD
    2+2: AABB, DDCC, [BBCC - didn't check if this one works geometrically]
    2+1+1: [ABBC, DCCB - didn't check if these work geometrically]

    If you work it out, you should see that these are the only possibilities. I really don't think there are any others.
     
  11. May 25, 2006 #10

    matt grime

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    Not having looked too closely, can I ask if you are, or need to, take symmetries into account?

    For instance, taking line segments of lengths 3,4,5 I can make either one or two different triangles. I.e. what are you classifying them 'up to'?
     
  12. May 25, 2006 #11
    No no Matt Grime, the triangles have fixed lengths as mentioned in first post. Nice thinking though!
     
  13. May 25, 2006 #12

    matt grime

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    No, you didn't understand what I wrote. It was 'an analogy' that allows you to visualize the point I was making, nothing to do with 'changing the lengths'. It is a perfectly valid question: if you construct two tetrahedra from the same triangles differing by some reflection are they the same or different?
     
  14. May 25, 2006 #13
    Ok sorry I got it now... they are the same two tetrahedra.
     
  15. May 25, 2006 #14
    After speaking to my lecture he told me to "try marrying side lengths" to make it to 10!

    Can anyone see what he actually means?
     
  16. May 27, 2006 #15
    For every tetrahedron you will always have one of the triangles as a 'base'. Working from that, there are only 4 possible bases. Now it is just a matter of going through the possible arrangements for each one - bearing in mind adjacent sides must always match in the 'standing' triangles
     
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