Mathematics VII: matchsticks

[⛄]

How many triangles can you build by using six matches?



Edit : The vertices of the triangles, of course, have to be the endpoints of the matchsticks.

[H.E. VOLODYMYR ZELENKSYY]

A theoretically infinite amount, although if you want us to do the maximum amount of triangles in one matchstick combination I believe the answer would be to make a Star of David, thus creating eight triangles.

[True Federalist (진정한 연방 주의자)]

At least eight, assuming that the vertices of the triangles do not have to be the endpoints of the matchsticks. Use the matchsticks to form a star of David.  If you do require the vertices to be endpoints, then at least five. Use four to build a tetrahedron with four triangles and then use the remaining two to form a fifth triangle using one of the tetrahedron edges as the third side of the fifth triangle. I don't know why one is listed as an option.  Even someone using the most naive method of forming triangles should be able make at least two.

[H.E. VOLODYMYR ZELENKSYY]

At least eight, assuming that the vertices of the triangles do not have to be the endpoints of the matchsticks. Use the matchsticks to form a star of David.  If you do require the vertices to be endpoints, then at least five. Use four to build a tetrahedron with four triangles and then use the remaining two to form a fifth triangle using one of the tetrahedron edges as the third side of the fifth triangle. I don't know why one is listed as an option.  Even someone using the most naive method of forming triangles should be able make at least two.

I'm sure there's someone out there who thinks you'd make one triangle with sides two matchsticks in length and throw up your hands in despair.

[⛄]

At least eight, assuming that the vertices of the triangles do not have to be the endpoints of the matchsticks. Use the matchsticks to form a star of David.  If you do require the vertices to be endpoints, then at least five. Use four to build a tetrahedron with four triangles and then use the remaining two to form a fifth triangle using one of the tetrahedron edges as the third side of the fifth triangle. I don't know why one is listed as an option.  Even someone using the most naive method of forming triangles should be able make at least two.

Since I didn't include the "eight" option, your second answer is correct. Congrats!

P.S.: I listed the 1 as an option to make it look more aesthetic.

[SUSAN CRUSHBONE]

If you do require the vertices to be endpoints, then at least five. Use four to build a tetrahedron with four triangles and then use the remaining two to form a fifth triangle using one of the tetrahedron edges as the third side of the fifth triangle.

i would've thought you need all six matches to form a tetrahedron...

[muon2]

If you do require the vertices to be endpoints, then at least five. Use four to build a tetrahedron with four triangles and then use the remaining two to form a fifth triangle using one of the tetrahedron edges as the third side of the fifth triangle.

i would've thought you need all six matches to form a tetrahedron...

Yes, a tetrahedron has six edges of equal length giving four faces. In general Euler's formula for non-intersecting polyhedrons is that faces + vertices - edges = 2.

[True Federalist (진정한 연방 주의자)]

If you do require the vertices to be endpoints, then at least five. Use four to build a tetrahedron with four triangles and then use the remaining two to form a fifth triangle using one of the tetrahedron edges as the third side of the fifth triangle.

i would've thought you need all six matches to form a tetrahedron...

Yes, a tetrahedron has six edges of equal length giving four faces. In general Euler's formula for non-intersecting polyhedrons is that faces + vertices - edges = 2.

Yup.  Brainfart.  I was thinking of the four one needs to build a caltrop.  So yeah, Four is the answer with a tetrahedron.  It even makes more sense, since usually with this sort of problem the author doesn't intend a solution that has dangly bits like that extra triangle I got by using two fewer sticks for the core pyramid.