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.

- Joined
- 29 Oct '09
- Moves
- 1421

- Joined
- 22 Apr '05
- Moves
- 521304

Linkenheim25 Feb '15 15:03

Form a square from four matches and lay the other two crosswise into the square.*Originally posted by WanderingKing***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 ðŸ™‚- Joined
- 29 Oct '09
- Moves
- 1421

- Joined
- 22 Apr '05
- Moves
- 521304

Linkenheim25 Feb '15 20:38

If I had read properly I would have realized right away ðŸ™‚ 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.*Originally posted by WanderingKing***It's an almost-solution. ðŸ™‚ There's the one problem that the diagonal matches will be too short to reach both vertices.**- Joined
- 29 Oct '09
- Moves
- 1421

- Joined
- 09 Jun '07
- Moves
- 45641

In the den- Joined
- 30 Sep '12
- Moves
- 731

26 Feb '15 21:44

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*Originally posted by WanderingKing***It's an almost-solution. ðŸ™‚ There's the one problem that the diagonal matches will be too short to reach both vertices.****#**, 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.- Joined
- 26 Apr '03
- Moves
- 25805

- Joined
- 29 Oct '09
- Moves
- 1421

27 Feb '15 12:35

That does it too! I didn't think of this solution. ðŸ™‚*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]- Joined
- 29 Oct '09
- Moves
- 1421

- Joined
- 29 Oct '09
- Moves
- 1421

- Joined
- 09 Jun '07
- Moves
- 45641

In the den- Joined
- 29 Oct '09
- Moves
- 1421

- Joined
- 23 Aug '04
- Moves
- 24791

tinyurl.com/y9ls7wbl