MarioNintendo Geometry/Algebra problem about pizza

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  • MarioNintendo
    Expect delays.
    FFR Simfile Author
    FFR Music Producer
    • Mar 2008
    • 4177

    #1

    MarioNintendo Geometry/Algebra problem about pizza

    Hey guys, I need help from you majors in math and stuff. While I was eating pizza today (with fork and knife, otherwise this problem wouldn't exist) I was wondering at what height we'd have to cut a slice in order to end up with two pieces of the same area. Kinda silly, I know, but I'm beginning University tomorrow after 6 months of not doing any kind of maths (I was more physics oriented) and I wanted to refresh my memory.

    Here's how far I got. I got a bit lazy in the end, but I was more curious at how you guys thought this problem could be solved. Seems fairly easy and definitely possible, just need the right tools to work it.




    (I pretended the sides of the pizza were all the same size so I could have 60° angles)

    Thanks! Hope I didn't screw up at some part... D:
  • dag12
    FFR Simfile Author
    FFR Simfile Author
    • Dec 2004
    • 468

    #2
    Re: MarioNintendo Geometry/Algebra problem about pizza

    Use the r:r^2:r^3 theorem.

    For two similar 2D shapes, if the lengths are in the proportion of r:1, then the areas are in the proportion of r^2:1

    So that just means just cut it so that one of the two pieces is an equilateral triangle of side length x/sqrt(2), where x is the length of one of the sides of the original pizza.

    Edit: So that means from the top, using a 30-60-90 triangle relation, you would need to cut it at the point x*sqrt(3)/2sqrt(2) (from the top of the pizza).
    Last edited by dag12; 09-30-2012, 08:40 PM.

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    • MarioNintendo
      Expect delays.
      FFR Simfile Author
      FFR Music Producer
      • Mar 2008
      • 4177

      #3
      Re: MarioNintendo Geometry/Algebra problem about pizza

      Originally posted by dag12
      Use the r:r^2:r^3 theorem.

      For two similar 2D shapes, if the lengths are in the proportion of r:1, then the areas are in the proportion of r^2:1

      So that just means just cut it so that one of the two pieces is an equilateral triangle of side length x/sqrt(2), where x is the length of one of the sides of the original pizza.

      Edit: So that means from the top, using a 30-60-90 triangle relation, you would need to cut it at the point x*sqrt(3)/2sqrt(2) (from the top of the pizza).
      Whoa, I don't recall that theorem, but somehow it makes so much sense, no wonder I couldn't think of it...!

      I can't figure out how you came to those answers, though, I'd need to re-do it on paper XD

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      • LongGone
        -
        FFR Simfile Author
        • Jul 2008
        • 1679

        #4
        Re: MarioNintendo Geometry/Algebra problem about pizza

        Are you eating hexagonal pizzas?
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        • MarioNintendo
          Expect delays.
          FFR Simfile Author
          FFR Music Producer
          • Mar 2008
          • 4177

          #5
          Re: MarioNintendo Geometry/Algebra problem about pizza

          Originally posted by LongGone
          Are you eating hexagonal pizzas?
          :s you found my weakness

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          • Reincarnate
            x'); DROP TABLE FFR;--
            • Nov 2010
            • 6332

            #6
            Re: MarioNintendo Geometry/Algebra problem about pizza

            I would just do

            Area of slice: BH/2
            Thus
            Area of minitriangle: b*(H-h)/2
            Area of trapezoid: h*(b+B)/2

            Set equal and solve for h

            b*(H-h)/2 = h*(b+B)/2
            bH-bh = hb+hB
            bH = hb+hB+bh
            bH = h(b+B+b)
            h = bH/(2b+B)

            But we don't want to care about what b is. We just want to write this as a function of h.

            Note that, due to similar triangles:
            (H-h)/(b/2) = H/(B/2)
            (B/2)*(H-h) = H*(b/2)
            b = 2*(B/2)*(H-h)/H

            So:

            h = bH/(2b+B) from earlier, substitute in the b expression
            h = (2*(B/2)*(H-h)/H)*H/(2*(2*(B/2)*(H-h)/H)+B)
            h in this case reduces to

            h = H*(2-sqrt(2))/2
            or
            h = 0.2928932188134H

            In other words, it's a tad under 30% of the way up the pizza slice (from the bottom)
            Last edited by Reincarnate; 10-1-2012, 07:57 AM.

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            • Reincarnate
              x'); DROP TABLE FFR;--
              • Nov 2010
              • 6332

              #7
              Re: MarioNintendo Geometry/Algebra problem about pizza

              Actually let me double check that with dag's results

              He says x*sqrt(3)/(2*sqrt(2)) from the tip of the pizza works
              In this case 2*H/sqrt(3) = x

              So
              (2*H/sqrt(3))*sqrt(3)/(2*sqrt(2)) + H*(2-sqrt(2))/2 should equal H
              And indeed it does

              Because
              (2*H/sqrt(3))*sqrt(3)/(2*sqrt(2)) reduces to H/sqrt(2)
              and
              H/sqrt(2) + H*(2-sqrt(2))/2 = H(1/sqrt(2) + (2-sqrt(2))/2) = H(sqrt(2)/2 + (2-sqrt(2))/2) = H(2/2) = H
              Last edited by Reincarnate; 09-30-2012, 10:21 PM.

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              • qqwref
                stepmania archaeologist
                FFR Simfile Author
                • Aug 2005
                • 4092

                #8
                Re: MarioNintendo Geometry/Algebra problem about pizza

                I'm bored, so... what happens if the pizza is circular, with radius x, and you cut a 60 degree slice?

                The area of the section is πx²/6, and we want the equilateral triangle on top to have an area of half of that or πx²/12. If it has side length w, its area is w²√(3/4), and then we have
                πx²/12 = w²√(3/4) => w² = πx²/6√3 => w = x√(π/6√3).

                Now what's h? It's the radius of the circle (or x) minus the height of the equilateral triangle (or w√(3/4)), so:
                h = x - x√(π/8√3)
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