25 Feb 15
Originally posted by WanderingKingForm a square from four matches and lay the other two crosswise into the square.
You've got six matches. Can you make four triangles out of them? You can't break the matches.
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 🙂
25 Feb 15
Originally posted by WanderingKingIf 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.
It's an almost-solution. 🙂 There's the one problem that the diagonal matches will be too short to reach both vertices.
26 Feb 15
Originally posted by WanderingKingIn 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.
It's an almost-solution. 🙂 There's the one problem that the diagonal matches will be too short to reach both vertices.
27 Feb 15
Originally posted by Paul Dirac IIThat does it too! I didn't think of this solution. 🙂
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]
