Puzzle time: Complex polyhedrons
-
Oh, I think I got it. Good old Dirichlet's pigeonhole principle comes to the rescue.
:::
No. Let N be the maximum number of sides of any face. Consider a face with N sides. It has N neighbor faces. Each of these neighbors must have between 3 and N sides. Hence there must be at least one pair of neighbors that has the same number of sides.
:::
-
:::
If you order the sides by the number of edges, the number of open (unconnected) edges, as the polyhedron is constructed side by side, necessarily increases as more sides are added, in that order.
Or you could start with the highest-edge side with N edges, and realize that the total number of other sides must be at most N-3, since you can't have one-edged or two-edged sides. N-3 other sides aren't enough to connect to all N edges of your highest-edge side.
:::
-
@klaus said in Puzzle time: Complex polyhedrons:
Each of these neighbors must have between 3 and N sides.
Why?
@jon-nyc said in Puzzle time: Complex polyhedrons:
@klaus said in Puzzle time: Complex polyhedrons:
Each of these neighbors must have between 3 and N sides.
Why?
Because every polygon has at least three sides and the maximum number of sides is N. No?
-
@jon-nyc said in Puzzle time: Complex polyhedrons:
@klaus said in Puzzle time: Complex polyhedrons:
Each of these neighbors must have between 3 and N sides.
Why?
Because every polygon has at least three sides and the maximum number of sides is N. No?
-
Bonus question:
Solve the problem using Euler's formula V+F-E=2
“Don’t make me repeat myself”
- history
-
@klaus said in Puzzle time: Complex polyhedrons:
Why the restriction to convex polyhedra?
From today’s solution:
Convexity is necessary for this proof to work, because without it, there may be two faces that meet at more than one edge; therefore, the faces touching the E edges of our polygon might not all be different. For example, cut a tetrahedral notch out of one edge of a cube; then the two square faces that met at that edge become heptagons that now meet at two edges.
Those of you that used Euler's formula V - E + F = 2 (as envisioned by the puzzle's composer, for the 1973 Moscow Mathematical Olympiad) to solve this puzzle didn't need to assume that no two faces meet in more than one edge, but did (perhaps implicitly) use the fact that convex polyhedra are simply connected. If the polygon has holes all the way through it, like a faceted doughnut, Euler's formula changes. So I don't actually know whether there's a solid with polygonal faces each of which has a different number of sides!
