timz
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A Math/Logic/Geometry Problem

Sun Dec 02, 2012 11:28 pm

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?
 
smittyone
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RE: A Math/Logic/Geometry Problem

Mon Dec 03, 2012 12:52 pm

Thanks Tim, my head just exploded like JFK in the Zapruder video.
 
yooyoo
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RE: A Math/Logic/Geometry Problem

Mon Dec 03, 2012 2:56 pm

the answer is ...... ice cream.
I am so smart, i am so smart... S-M-R-T... i mean S-M-A-R-T
 
Birdwatching
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RE: A Math/Logic/Geometry Problem

Mon Dec 03, 2012 4:13 pm

The answer is 42

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All the things you probably hate about travelling are warm reminders that I'm home
 
vikkyvik
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RE: A Math/Logic/Geometry Problem

Mon Dec 03, 2012 4:45 pm

Quoting Birdwatching (Reply 3):
The answer is 42

Hey, and now we finally know the question too!

I shall go make myself a jynnan tonnyx in celebration.
I'm watching Jeopardy. The category is worst Madonna songs. "This one from 1987 is terrible".
 
timz
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RE: A Math/Logic/Geometry Problem

Mon Dec 03, 2012 7:06 pm

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
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RE: A Math/Logic/Geometry Problem

Mon Dec 03, 2012 8:12 pm

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
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RE: A Math/Logic/Geometry Problem

Mon Dec 03, 2012 8:37 pm

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
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RE: A Math/Logic/Geometry Problem

Mon Dec 03, 2012 8:52 pm

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.
I'm watching Jeopardy. The category is worst Madonna songs. "This one from 1987 is terrible".

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