timz From United States of America, joined Sep 1999, 6902 posts, RR: 7 Posted (2 years 2 weeks 5 days 11 hours ago) and read 1754 times:

From one of those Litton ads in the 1960s:

All the faces on a certain polyhedron are triangular; the polyhedron has nine vertices. At six of the vertices, six triangles meet; at the other three vertices, four triangles meet. How many faces does the polyhedron have?

vikkyvik From United States of America, joined Jul 2003, 10340 posts, RR: 26
Reply 4, posted (2 years 2 weeks 4 days 18 hours ago) and read 1513 times:

timz From United States of America, joined Sep 1999, 6902 posts, RR: 7
Reply 5, posted (2 years 2 weeks 4 days 16 hours ago) and read 1478 times:

Last night I decided the answer was 16 faces. Still can't decide whether a polyhedron meeting the given conditions can exist-- anybody know how to figure that?

NoWorries From United States of America, joined Oct 2006, 539 posts, RR: 1
Reply 6, posted (2 years 2 weeks 4 days 15 hours ago) and read 1451 times:

I can't visualize how that would work. Four triangles meeting at a vertex gives us a (four-sided) pyramid with a square base. In order to "hide" the square base, the only thing that I can think of is to join the pyramids at the opposite edges of their bases. That gives three vertices with four triangles and six vertices where five triangles meet (if I'm visualizing correctly) -- essentially three four-sided pyramids glued to the three rectangular faces of a triangular prism -- that's as close as I can get and that gives 14 sides.

timz From United States of America, joined Sep 1999, 6902 posts, RR: 7
Reply 7, posted (2 years 2 weeks 4 days 14 hours ago) and read 1440 times:

Maybe the polyhedron that I described doesn't exist-- I remembered the problem wrong. The correct version is: nine vertices, at six of which four triangles meet, and at the other three vertices six triangles meet. And the correct answer for that is 14 faces.

vikkyvik From United States of America, joined Jul 2003, 10340 posts, RR: 26
Reply 8, posted (2 years 2 weeks 4 days 14 hours ago) and read 1426 times:

Quoting timz (Reply 7): Maybe the polyhedron that I described doesn't exist-- I remembered the problem wrong. The correct version is: nine vertices, at six of which four triangles meet, and at the other three vertices six triangles meet. And the correct answer for that is 14 faces.

You have a picture of that? A septagonal bipyramid is almost that (it has 14 faces and 9 vertices), except that at 7 vertices, 4 triangles meet, and at 2 vertices, 7 triangles meet.

Or a Triaugmented triangular prism, which has 9 vertices and 14 faces, but at 3 vertices, 4 triangles meet, and at 6 vertices, 5 triangles meet.