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Math Help - Area of Octagon

  1. #1
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    Area of Octagon

    An octagon is inscribed in a circle. The octagon has vertices P1,P2, ... P8 around the circumference of the circle.

    Vertices P1,P3,P5,P7 form a square of area 5 and sides \sqrt{5}

    Vertices P2,P4,P6,P8 form a rectangle of area 4 and sides of length \sqrt{2} and 2\sqrt{2}

    How do I find the maximum area of the octagon? Thanks!
    Last edited by RubyRed; September 9th 2008 at 10:48 AM.
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  2. #2
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    Quote Originally Posted by RubyRed View Post
    An octagon is inscribed in a circle. The octagon has vertices P1,P2, ... P8 around the circumference of the circle.

    Vertices P1,P3,P5,P7 form a square of area 5 and sides \sqrt{5}

    Vertices P2,P4,P6,P8 form a rectangle of area 4 and sides of length \sqrt{2} and 2\sqrt{2}

    How do I find the area of the octagon? Thanks!
    The rarea of the octagon is the area of the square plus the areas of the 4 congruent triangles that are outside of the square.

    One long way to do this is:
    Get the equations of the lines P1P3, P2P8, and P2P4.
    Find their intersection points so that you can have the coordinates of the 3 corners of one of the mentioned 4 congruent triangles.
    With the coordinates known, get the lengths of the 3 sides.
    Use the Heron's formula to get the area of the said triangle.

    It's a long way. I don't have the energy to show it all here.
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  3. #3
    Member courteous's Avatar
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    Isn't RubyRed referring to irregular octagon? Would statement of 4 congruent triangles still be valid?
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  4. #4
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    Yes it is irregular so I can't use that method at least I think. I've made an extremely crude drawing in Paint below:


    I think I can use symmetry to a certain degree. For instance: the triangle P2P3P4 is congruent to P6P7P8 so I only need to find the area of one and then multiple it by 2. The same holds for P4P5P6 and P8P1P2.

    I just don't know how to find the area of these triangles.







    Last edited by RubyRed; September 9th 2008 at 06:21 AM. Reason: Formatting
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  5. #5
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    Hello, everyone!

    I'm still working on this one . . . so I'll just babble . . .


    I having trouble controlling the variable.
    You see, the rectangle can be "tipped" at a variety of angles.

    Code:
                  * * *    P2  
              *           o
          P1*---------o-----*P3
           *|     o        o|*
            | o             |
        P8o |               |o*
          * |               | *
          *o|               | oP4
            |             o |
           *|o        o     |*
          P7*-----o---------*P5
              o           *
            P6    * * *

    The area of the octagon would be:

    . . \text{area of: }\:(\text{square }P_1P_3P_5P_7) + \Delta P_1P_2P_3 + \Delta P_3P_4P_5 + \Delta P_5P_6P_7 + \Delta P_7P_8P_1



    Edit: You're right, RubyRed . . . I overlooked the word "maximum".


    Yes, the top and bottom triangles are congruent,
    . . as are the left and right triangles.


    If my diagram is hard to read . . .

    Points P_1, P_3, P_5, P_7 are the vertices of the square,
    . . which is clearly outlined.

    Points P_2, P_4, P_6, P_8 are the vertices of the rectangle,
    . . outlined with o's (harder to see).

    Last edited by Soroban; September 9th 2008 at 11:20 AM.
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  6. #6
    Member courteous's Avatar
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    Post Congruent triangles?

    I still don't see congruency. Can you point it out for me

    This is my add to common effort:
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  7. #7
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    The problem should read find the maximum area of the octagon.

    courteous look at your picture (much better than mine by the way ), the two triangles on opposite sides of each other have equal area (ie top and bottom and left and right). I'm having trouble reading some of your labeling so I can't give you the vertex names.
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  8. #8
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    Quote Originally Posted by courteous View Post
    I still don't see congruency. Can you point it out for me

    This is my add to common effort:
    That's the figure, except that I "centered" the rectangle to the diameter P3P7. So all of the four triangles outside of the square are congruent.
    I "misread" the question that I thought the rectangle could be positioned any which way as long as the P1, P2,P3,P4,....are consecutive or in order.

    That the four triangles outside of the square are congruent does not make the octagon regular. It is still irregular.

    Now for the intent of the question to find the maximum area of the octagon, maybe my solution about is not the maximum. Maybe it is. I don't know yet. It can be shown by using Calculus in positioning the rectangle such that the 4 triangles outside of the square are not congruent...but there'd be two pairs of congruent triangles instead of 4 congruent triangles.

    When I have time, I see what I can do. I need another approach.
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  9. #9
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    I have "solved" this problem although you have to make an assumption I don't entirely agree with.


    The area of the octagon is: P2P4P6P8 + 2 * P2P3P4 + 2 * P4P5P6

    You then must assume that P3 lies on the midpoint of the arc between P2P4 and also P5 lies on the midpoint of the arc between P4P6. If this is the case then you can simply use the radius of the circle to solve the problem.

    You get area of P2P3P4 as
    \frac{1}{2}(\sqrt{2})(\frac{\sqrt{10}}{2}-\sqrt{2}) = \frac{\sqrt{5}}{2} - 1


    You get area of P4P5P6 as
    \frac{1}{2}(2\sqrt{2})
    (\frac{\sqrt{10}}{2}-\frac{\sqrt{2}}{2} ) = \sqrt{5} - 1

    And then 4+2(\frac{\sqrt{5}}{2} - 1)+2(\sqrt{5} - 1) = 3\sqrt{5}

    which is the answer my teacher told us today.

    I'm still not sure how you can assume that P3 and P5 are the midpoints to give maximum area.
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