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Math Help - pentagon

  1. #1
    Member Veronica1999's Avatar
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    pentagon

    Let A=(0,0), B=(7,2), C=(3,4), D=(3,7) and E=(-1,5). Cameron walks the polygonal path ABCDEA, writing down the number of degrees turned at each corner. What is the sum of these five numbers? Notice that ABCDE in not a convex pentagon.

    I got 491.(using a protractor) Could someone take a look at my attached work?
    oops, on my last line of attached work i meant if angle c was convex
    Attached Thumbnails Attached Thumbnails pentagon-pentagon.pdf  
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  2. #2
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    Why does convexity make any difference? Don't you have to end up where you started and facing the direction you started?
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  3. #3
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    Hi Veronica 1999,
    I calculated the angle of the first turn at B to be 137.5. At C 116.6and 116.6 at D.Finish and let me know if you get the right answer


    bjh
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  4. #4
    Super Member TheChaz's Avatar
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    TKHunny is suggesting a more theoretical approach, with which I agree. Are we explicitly calculating each angle as a means of verification?
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  5. #5
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    Hello, Veronica1999!

    \text{Let: }A=(0,0),\;B=(7,2),\;C=(3,4),\;D=(3,7),\;E=(\text{-}1,5).

    \text{Cameron walks the polygonal path }ABCDEA,

    \text{writing down the number of degrees turned at each corner.}

    \text}What is the sum of these five numbers?}

    \text{Note that }ABCDE\text{ is not a convex pentagon.}

    \text{I got 491 (using a protractor).}

    While sketching, I found the problem more complex than anticipated.

    Code:
          |         D
          |         o
          |   Y     *
          |     . β *
          |       . *
          |         o                 X
          |        C  *           .
          |             *  α  .
          |               o
          |           *    B
          |       *
          |   *
        A o - - - - - - - - - - - - -
          |

    When he walks from \,A to \,B, he is walking in the direction \overrightarrow{AX}.

    To walk to \,C, he turns through \text{angle }\alpha = \angle XBC . . . not \angle ABC.


    Then he walks from \,B to \,C, in the direction \overrightarrow{BY}.

    To walk to \,D, he turns through \text{angle }\beta = \angle YCD


    And so on . . .

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  6. #6
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    pentagon

    I had the right idea but made a mistake in the turn at C I recalculated all the turns with the following results 137,5,63.5,116.5,74.5, 85.35.Is this correct?
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