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The New Coffee Room

  1. TNCR
  2. General Discussion
  3. Puzzle time: Complex polyhedrons

Puzzle time: Complex polyhedrons

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  • KlausK Offline
    KlausK Offline
    Klaus
    wrote on last edited by
    #3

    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.

    :::

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    • jon-nycJ Online
      jon-nycJ Online
      jon-nyc
      wrote on last edited by
      #4

      @klaus said in Puzzle time: Complex polyhedrons:

      Each of these neighbors must have between 3 and N sides.

      Why?

      "You never know what worse luck your bad luck has saved you from."
      -Cormac McCarthy

      KlausK 1 Reply Last reply
      • HoraceH Online
        HoraceH Online
        Horace
        wrote on last edited by
        #5

        :::

        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.

        :::

        Education is extremely important.

        1 Reply Last reply
        • jon-nycJ jon-nyc

          @klaus said in Puzzle time: Complex polyhedrons:

          Each of these neighbors must have between 3 and N sides.

          Why?

          KlausK Offline
          KlausK Offline
          Klaus
          wrote on last edited by
          #6

          @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-nycJ 1 Reply Last reply
          • KlausK Klaus

            @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-nycJ Online
            jon-nycJ Online
            jon-nyc
            wrote on last edited by
            #7

            @Klaus Oh of course. You defined N as the maximum.

            "You never know what worse luck your bad luck has saved you from."
            -Cormac McCarthy

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            • jon-nycJ Online
              jon-nycJ Online
              jon-nyc
              wrote on last edited by
              #8

              Bonus question:

              Solve the problem using Euler's formula V+F-E=2

              "You never know what worse luck your bad luck has saved you from."
              -Cormac McCarthy

              HoraceH KlausK 2 Replies Last reply
              • jon-nycJ jon-nyc

                Bonus question:

                Solve the problem using Euler's formula V+F-E=2

                HoraceH Online
                HoraceH Online
                Horace
                wrote on last edited by
                #9

                @jon-nyc said in Puzzle time: Complex polyhedrons:

                V+F-E=2

                V+F-E=2, ergo every face of a convex polyhedron cannot have a different number of sides, QED.

                Education is extremely important.

                1 Reply Last reply
                • jon-nycJ Online
                  jon-nycJ Online
                  jon-nyc
                  wrote on last edited by
                  #10

                  Horace gets it in one!

                  "You never know what worse luck your bad luck has saved you from."
                  -Cormac McCarthy

                  1 Reply Last reply
                  • jon-nycJ jon-nyc

                    Bonus question:

                    Solve the problem using Euler's formula V+F-E=2

                    KlausK Offline
                    KlausK Offline
                    Klaus
                    wrote on last edited by
                    #11

                    @jon-nyc said in Puzzle time: Complex polyhedrons:

                    Bonus question:

                    Solve the problem using Euler's formula V+F-E=2

                    ...Each of these neighbors must have between 3 1+V+F-E and N sides....

                    That was easy.

                    1 Reply Last reply
                    • KlausK Klaus

                      Why the restriction to convex polyhedra?

                      jon-nycJ Online
                      jon-nycJ Online
                      jon-nyc
                      wrote on last edited by
                      #12

                      @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!

                      "You never know what worse luck your bad luck has saved you from."
                      -Cormac McCarthy

                      1 Reply Last reply
                      • jon-nycJ Online
                        jon-nycJ Online
                        jon-nyc
                        wrote on last edited by
                        #13

                        Funny the problem composer didn’t see the simple pigeon hole proof. But I feel better about jumping down the Euler rabbit hole.

                        "You never know what worse luck your bad luck has saved you from."
                        -Cormac McCarthy

                        1 Reply Last reply
                        • jon-nycJ Online
                          jon-nycJ Online
                          jon-nyc
                          wrote on last edited by
                          #14

                          Wish they showed the Euler formula based proof. I have one but it’s clunky. Basically shows a lower and upper bound for V which shows it’s impossible.

                          "You never know what worse luck your bad luck has saved you from."
                          -Cormac McCarthy

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