1. Joined
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    24 Feb '15 21:181 edit
    You've got six matches. Can you make four triangles out of them? You can't break the matches.
  2. SubscriberPonderable
    chemist
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    25 Feb '15 15:03
    Originally posted by WanderingKing
    You've got six matches. Can you make four triangles out of them? You can't break the matches.
    Form a square from four matches and lay the other two crosswise into the square.

    My though cane from the fact that I need three matches for one triangle two more for the next, the have one in common. if you now add the remaining match perpenticular to the one use by both you get the figure 🙂
  3. Joined
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    25 Feb '15 20:34
    It's an almost-solution. 🙂 There's the one problem that the diagonal matches will be too short to reach both vertices.
  4. SubscriberPonderable
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    25 Feb '15 20:38
    Originally posted by WanderingKing
    It's an almost-solution. 🙂 There's the one problem that the diagonal matches will be too short to reach both vertices.
    If I had read properly I would have realized right away 🙂 I used four to make four..in fact I should have built a tetrahedron and have made four triangles. a triangle on the floor and three to make a three-cornered pyramide.
  5. Joined
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    26 Feb '15 00:39
    Indeed!
  6. Standard memberwolfgang59
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    26 Feb '15 03:14
    Originally posted by WanderingKing
    You've got six matches. Can you make four triangles out of them? You can't break the matches.
    Make a star of David ... and you've got a bonus two triangles!
  7. Joined
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    26 Feb '15 21:44
    Originally posted by WanderingKing
    It's an almost-solution. 🙂 There's the one problem that the diagonal matches will be too short to reach both vertices.
    In defense of P's first solution, nothing was stated about the ends of the matches having to be the points of intersection. I am picturing four matches laid down like an orthogonal #, with the remaining two matches lying diagonally across the hollow core of that figure, those two matches possibly (but not of necessity) having their ends lying on the four points of intersection of the first four matches.
  8. Joined
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    27 Feb '15 04:18
    Easily, not at the same time though

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  9. Joined
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    27 Feb '15 12:35
    Originally posted by Paul Dirac II
    In defense of P's first solution, nothing was stated about the ends of the matches having to be the points of intersection. I am picturing four matches laid down like an orthogonal [b]#, with the remaining two matches lying diagonally across the hollow core of that figure, those two matches possibly (but not of necessity) having their ends lying on the four points of intersection of the first four matches.[/b]
    That does it too! I didn't think of this solution. 🙂
  10. Joined
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    27 Feb '15 12:35
    Originally posted by iamatiger
    Easily, not at the same time though

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    I don't get it. Can you explain?
  11. Joined
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    27 Feb '15 12:37
    Originally posted by wolfgang59
    Make a star of David ... and you've got a bonus two triangles!
    You do!
  12. Standard memberwolfgang59
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    28 Feb '15 08:16
    Originally posted by WanderingKing
    I don't get it. Can you explain?
    1, 3, 6 & 10 mare triangle numbers. FOUR triangle numbers.
  13. Joined
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    01 Mar '15 04:42
    Originally posted by wolfgang59
    1, 3, 6 & 10 mare triangle numbers. FOUR triangle numbers.
    Oh, right. 🙂 But then it's not hard to make four actual triangles out of six matches if you don't need to make them at the same time 😉
  14. SubscriberAThousandYoung
    All My Soldiers...
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    16 Mar '15 04:54
    Tetrahedron
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