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The New Coffee Room

  1. TNCR
  2. General Discussion
  3. Puzzle time - Silo dog

Puzzle time - Silo dog

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  • jon-nycJ Offline
    jon-nycJ Offline
    jon-nyc
    wrote on last edited by
    #1

    A farm has a grain silo with a circular floor plan, 20 meters in diameter. On the outside of the silo is a hook. A dog is tied up to that hook on a leash of length 10*pi meters.

    Assuming the dog can’t enter the silo, what area can he cover?

    You were warned.

    1 Reply Last reply
    • KlausK Offline
      KlausK Offline
      Klaus
      wrote on last edited by
      #2

      silently meditating over the complexity of the formula for circle circle intersection

      1 Reply Last reply
      • jon-nycJ Offline
        jon-nycJ Offline
        jon-nyc
        wrote on last edited by jon-nyc
        #3

        Do keep in mind the perimeter traced out by the leash is not a circle, since the leash wraps around the silo when he goes behind it.

        You were warned.

        1 Reply Last reply
        • KlausK Offline
          KlausK Offline
          Klaus
          wrote on last edited by
          #4

          Uh oh...

          1 Reply Last reply
          • jon-nycJ Offline
            jon-nycJ Offline
            jon-nyc
            wrote on last edited by
            #5

            This isn’t a ‘find the trick and do the answer in your head’ problem. It’s a pencil-to-paper problem.

            You were warned.

            1 Reply Last reply
            • bachophileB Offline
              bachophileB Offline
              bachophile
              wrote on last edited by
              #6

              I figure the dog wanting to mark its territory simply pees all over the place.

              1 Reply Last reply
              • KlausK Offline
                KlausK Offline
                Klaus
                wrote on last edited by
                #7

                Game plan would be:

                1. Find function that computes max height/distance from any point on silo, i.e., the shape of the "wrap around" part.

                2. Presuming the formula is complicated, ask computer algebra system to integrate the function to find its area.

                3. Multiply by two and add area of the semi-circle.

                1 Reply Last reply
                • AxtremusA Away
                  AxtremusA Away
                  Axtremus
                  wrote on last edited by
                  #8

                  Yeah, getting the the “game plan” is easy, actually getting step 1 is the harder part.

                  1 Reply Last reply
                  • jon-nycJ Offline
                    jon-nycJ Offline
                    jon-nyc
                    wrote on last edited by jon-nyc
                    #9

                    Actually integrating the function is straightforward.

                    The trick, if you want to call it that, is just figuring out the function. Totally doable by both of you.

                    You were warned.

                    1 Reply Last reply
                    • bachophileB Offline
                      bachophileB Offline
                      bachophile
                      wrote on last edited by
                      #10

                      I glad he didn’t say the three of you

                      1 Reply Last reply
                      • jon-nycJ Offline
                        jon-nycJ Offline
                        jon-nyc
                        wrote on last edited by
                        #11

                        lol.

                        You were warned.

                        1 Reply Last reply
                        • AxtremusA Away
                          AxtremusA Away
                          Axtremus
                          wrote on last edited by
                          #12

                          ||Cycloid is key.
                          Imagine the silo "rolls over" by 180º, the arc traced out by the point opposite the hook is that of a cycloid, and the area under the cycloid is given by the formula 3πr^2 where r is the silo's radius.
                          Solution is (area of cycloid traced out by the silo)-(area of silo)+(half the area of circle with radius that is the leash's length).
                          Area covered by the dog = 3π(10)^2 - π(10)^2 + (π(10π)^2)/2
                          ||

                          KlausK 1 Reply Last reply
                          • AxtremusA Axtremus

                            ||Cycloid is key.
                            Imagine the silo "rolls over" by 180º, the arc traced out by the point opposite the hook is that of a cycloid, and the area under the cycloid is given by the formula 3πr^2 where r is the silo's radius.
                            Solution is (area of cycloid traced out by the silo)-(area of silo)+(half the area of circle with radius that is the leash's length).
                            Area covered by the dog = 3π(10)^2 - π(10)^2 + (π(10π)^2)/2
                            ||

                            KlausK Offline
                            KlausK Offline
                            Klaus
                            wrote on last edited by
                            #13

                            @Axtremus said in Puzzle time - Silo dog:

                            Imagine the silo "rolls over" by 180º, the arc traced out by the point opposite the hook is that of a cycloid,

                            But how does that arc relate to where the dog can go? Your solution - which may well be correct - is missing an argument here.

                            1 Reply Last reply
                            • KlausK Offline
                              KlausK Offline
                              Klaus
                              wrote on last edited by Klaus
                              #14

                              Wouldn't the resulting shape be a (part of a) cardoid and not a cycloid?

                              alt text

                              1 Reply Last reply
                              • jon-nycJ Offline
                                jon-nycJ Offline
                                jon-nyc
                                wrote on last edited by jon-nyc
                                #15

                                Yes. This is the shape. Note the 10pi leash length is just perfectly half the circumference of the silo.

                                609F2B11-E3C8-4B7D-8EF5-10F86F0519CE.jpeg

                                You were warned.

                                1 Reply Last reply
                                • AxtremusA Away
                                  AxtremusA Away
                                  Axtremus
                                  wrote on last edited by
                                  #16

                                  Right, cardioid rather than cycloid.

                                  1 Reply Last reply
                                  • jon-nycJ Offline
                                    jon-nycJ Offline
                                    jon-nyc
                                    wrote on last edited by jon-nyc
                                    #17

                                    Maybe you can google up a formula if you can’t figure out the integral.

                                    You were warned.

                                    1 Reply Last reply
                                    • KlausK Offline
                                      KlausK Offline
                                      Klaus
                                      wrote on last edited by
                                      #18

                                      I haven't looked at formulas yet but it looks like one needs only a part of the cardoid area since it overlaps with the semi circle.

                                      1 Reply Last reply
                                      • jon-nycJ Offline
                                        jon-nycJ Offline
                                        jon-nyc
                                        wrote on last edited by
                                        #19

                                        Yep. It’s the quarter circle plus that bit of cartioid x 2

                                        You were warned.

                                        1 Reply Last reply
                                        • KlausK Offline
                                          KlausK Offline
                                          Klaus
                                          wrote on last edited by
                                          #20

                                          The area formula is:

                                          1c7fca0a-26a5-4be8-935f-b300870327a7-image.png

                                          but the integral would not have to be taken to pi but to the angle at which the y value is maximal, would be the case at the angle 3/4 * sqrt(3).

                                          But this gets rather complicated hence I assume something's wrong here...

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