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Math Help - Cyclic Hexagon?

  1. #1
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    Can CaptainBlack help me with this?

    Right I have been given the following problem and cannot resolve it. I have had an attempt but without much success. Could anyone help me with this exercise, please?

    A cyclic hexagon is a hexagon whose vertices all lie on the circumference of a circle.

    The vertices of a cyclic hexagon are labelled in order A to F. Prove that the sum of the interior angles at A, C and E is equal to the sum of the interior angles at B, D and F.

    Generalise (concisely) to other cyclic polygons?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Natasha
    Right I have been given the following problem and cannot resolve it. I have had an attempt but without much success. Could anyone help me with this exercise, please?

    A cyclic hexagon is a hexagon whose vertices all lie on the circumference of a circle.

    The vertices of a cyclic hexagon are labelled in order A to F. Prove that the sum of the interior angles at A, C and E is equal to the sum of the interior angles at B, D and F.

    Generalise (concisely) to other cyclic polygons?
    1. Proving that the sum of the interior angles at verices A, C, and E
    equals the sum of the interior angles at vertices B, D, F is equivalent
    to proving that a-b+c-d+e-f = 0 where a, b, c, d, e, f are the interior
    angles at the vertices A, B, C, D, E, F respectivly.

    2. Observe that this property (which I will call the alternating angle
    sum property) holds for a regular hexagon, since all of these angles
    are equal in a regular hexagon.

    3. Observe that for any cyclic hexagon that the alternating angle
    sum does not change if we move one vertex along the arc connecting
    its neighbours, as long as we leave the vertices in the same realtive order
    (sketch of a proof of this is given below).

    4. Observe that any cyclic hexagon can be transformed into a regular
    hexagon by a sequence of transformations like that described in
    para 3 above.

    5. Hence conclude that the alternating angle sum property holds
    for any cyclic hexagon.
    Attached Thumbnails Attached Thumbnails Cyclic Hexagon?-hexagon.gif  
    Last edited by CaptainBlack; November 14th 2005 at 05:18 AM. Reason: Hopefully to add clarity and spellig errors
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  3. #3
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    Thanks ever so much Captain. That's so clear!
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  4. #4
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    How do you proove that

    1. We want to prove that a-b+c-d+e-f = 0 where a, b, c, d, e, f
    are the interior angles at the vertices A, B, C, D, E, F respectivly.

    Do you say that a+b+c+d+e+f = 360 and then what do you do?
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by Natasha
    How do you proove that

    1. We want to prove that a-b+c-d+e-f = 0 where a, b, c, d, e, f
    are the interior angles at the vertices A, B, C, D, E, F respectivly.

    Do you say that a+b+c+d+e+f = 360 and then what do you do?

    You are asked to prove that the sum of the interior angles at
    vertices A, C and E is equal to the sum of the interior angles
    at vertices B, D and F. Using a, b, c, d, e, and f to denote the
    interior angles at the corresponding vertices this becomes:

    Prove that:

    a + c + e = b + d + f,

    rearranging, you need to prove that:

    a - b + c - d + e - f = 0.

    RonL
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  6. #6
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    Captain I still don't get this?

    I mean the only way I can do this is by saying that:

    a+f+b+c+e+d = a+f+b+c+e+d

    where angle A is formed of angles f and a, B of a and b, C, b and c, D of c and d, E of d and e and F of e and f.

    But I just drew the hexagon and just wrote down the above but don't really prove it???

    Could you tell me exactly how you get your first point of

    1. We want to prove that a-b+c-d+e-f = 0 where a, b, c, d, e, f
    are the interior angles at the vertices A, B, C, D, E, F respectivly.

    Don't get it sorry...
    Last edited by Natasha; November 13th 2005 at 07:09 AM.
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  7. #7
    Grand Panjandrum
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    Smile

    Quote Originally Posted by Natasha
    Captain I still don't get this?

    I mean the only way I can do this is by saying that:

    a+f+b+c+e+d = a+f+b+c+e+d

    where angle A is formed of angles f and a, B of a and b, C, b and c, D of c and d, E of d and e and F of e and f.

    But I just drew the hexagon and just wrote down the above but don't really prove it???

    Could you tell me exactly how you get your first point of

    1. We want to prove that a-b+c-d+e-f = 0 where a, b, c, d, e, f
    are the interior angles at the vertices A, B, C, D, E, F respectivly.

    Don't get it sorry...

    The statement:

    "We want to prove that a-b+c-d+e-f = 0 where a, b, c, d, e, f are
    the interior angles at the vertices A, B, C, D, E, F respectivly",

    is just another way of writting:

    "The vertices of a cyclic hexagon are labelled in order A to F.
    Prove that the sum of the interior angles at A, C and E is equal
    to the sum of the interior angles at B, D and F."

    Rewriting:

    a-b+c-d+e-f = 0

    we have

    a+c+e = b+d+f,

    which is just another way of saying that "the sum of the interior
    angles at A, C and E is equal to the sum of the interior angles
    at B, D and F".

    This is because a+c+e is the sum of the interior angles at A, C and E,
    and b+d+f is the sum of the interior angles at B, D and F.

    Hope this helps explain.

    RonL
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  8. #8
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    The problem here Captain is that you don't seem to be proving anything are you?

    I need to use angle sizes I think, no? To state that I have 6 isosceles triangles, no?
    Last edited by Natasha; November 13th 2005 at 01:48 PM.
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  9. #9
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    It's actually cyclic polygons in the question not cyclic hexagons (at the end).
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  10. #10
    Grand Panjandrum
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    Quote Originally Posted by Natasha
    It's actually cyclic polygons in the question not cyclic hexagons (at the end).
    The only property of the hexagon used in the proof was that
    it was cyclic and had an even number of sides (it's not obvious
    but it is true).

    So the conclusion for a even sided cyclic polygon with vertices
    V1, V2, .. Vn is that the sum of the interior angles at V1, V3,..
    is equal to the sum of the interior angles at vertices V2, V4, ..

    Hope that helps

    RonL
    Last edited by CaptainBlack; November 16th 2005 at 08:14 AM.
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  11. #11
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    Thanks captain! but what about an odd sided polygon like a triangle?
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  12. #12
    Grand Panjandrum
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    I received the page proofs for the paper based on this question today

    Its to be published in September 2007

    RonL
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  13. #13
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    Hello, Natasha!

    Some basic geometry (plus a little algebra) is required.


    A cyclic hexagon is a hexagon whose vertices all lie on the circumference of a circle.

    The vertices of a cyclic hexagon are labelled in order A to F.
    Prove that the sum of the interior angles at A, C and E
    is equal to the sum of the interior angles at B, D and F.
    Code:
                    A
                  * o *
              * o       o *
          F o               o B
           *o               o*
            o               o
          * o               o *
          * o       *       o *
          * o               o *
            o               o
           *o               o*
          E o               o C
              * o       o *
                  * o *
                    D

    An inscribed angle is measured by one-half its intercepted arc.

    . . \angle A \;=\;\frac{1}{2}\left(\widehat{BC} + \widehat{CD} + \widehat{DE} + \widehat{EF}\right)

    . . \angle C \;=\;\frac{1}{2}\left(\widehat{DE} + \widehat{EF} + \widehat{FA} + AB\right)

    . . \angle E \;=\;\frac{1}{2}\left(\widehat{FA} + \widehat{AB} + \widehat{BC} + \widehat{CD}\right)


    Then: . \angle A + \angle C + \angle E \;=\;\frac{1}{2}\left(2\!\cdot\!\widehat{AB} + 2\!\cdot\!\widehat{BC} + 2\!\cdot\!\widehat{CD} + 2\!\cdot\!\widehat{DE} + 2\!\cdot\!\widehat{EF} + 2\!\cdot\!\widehat{FA}\right)

    Hence: . \angle A + \angle C + \angle E \;=\;AB + BC + CD + DE + EF + FA \;=\;360^o


    In a similar fashion, it can be shown that: . \angle B + \angle D + \angle F \;=\;360^o

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  14. #14
    Grand Panjandrum
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    Quote Originally Posted by Soroban View Post
    Hello, Natasha!
    I doubt Natasha will be reading this given how old the thread is (though it
    would be nice to be proved wrong).

    RonL
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  15. #15
    Grand Panjandrum
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    Quote Originally Posted by Soroban View Post
    Hello, Natasha!

    Some basic geometry (plus a little algebra) is required.

    Code:
                    A
                  * o *
              * o       o *
          F o               o B
           *o               o*
            o               o
          * o               o *
          * o       *       o *
          * o               o *
            o               o
           *o               o*
          E o               o C
              * o       o *
                  * o *
                    D
    An inscribed angle is measured by one-half its intercepted arc.

    . . \angle A \;=\;\frac{1}{2}\left(\widehat{BC} + \widehat{CD} + \widehat{DE} + \widehat{EF}\right)

    . . \angle C \;=\;\frac{1}{2}\left(\widehat{DE} + \widehat{EF} + \widehat{FA} + AB\right)

    . . \angle E \;=\;\frac{1}{2}\left(\widehat{FA} + \widehat{AB} + \widehat{BC} + \widehat{CD}\right)


    Then: . \angle A + \angle C + \angle E \;=\;\frac{1}{2}\left(2\!\cdot\!\widehat{AB} + 2\!\cdot\!\widehat{BC} + 2\!\cdot\!\widehat{CD} + 2\!\cdot\!\widehat{DE} + 2\!\cdot\!\widehat{EF} + 2\!\cdot\!\widehat{FA}\right)

    Hence: . \angle A + \angle C + \angle E \;=\;AB + BC + CD + DE + EF + FA \;=\;360^o


    In a similar fashion, it can be shown that: . \angle B + \angle D + \angle F \;=\;360^o
    The generalisation to a 2n cyclic polygon looks a bit fiddly, though not undoable.

    RonL
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