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Math Help - A problem with a regular hexagon

  1. #1
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    A problem with a regular hexagon

    Dear forum members, I am currently doing a problem related to hexagons.


    We have a regular hexagon. By joining the midpoints of the sides we get another hexagon inside of the original, and by joining the midpoints of the sides of the smaller hexagon we get another one and so forth. I am told to calculate the first hexagon, whose area is less than one millionth of the original. I know how to do this, but I have trouble in formulating the equation.

    The apothem of the original hexagon would be given by \frac{\sqrt{3}}{2}*a, right?

    Then comes something I can't get through my head. the book answer tells me that when I connect the two radii of the circumcircle of the original hexagon, I get an equilateral triangle, yet the book does not prove it. Is this really true?

    And then the book tells me that the apothem of the orinal hexagon, equals the lenght of the side of the hexagon drawn inside the original hexagon. Can someone verify this or prove it, please. My thick head refuses to accept it.





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  2. #2
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    Quote Originally Posted by Coach View Post
    ...

    Then comes something I can't get through my head. the book answer tells me that when I connect the two radii of the circumcircle of the original hexagon, I get an equilateral triangle, yet the book does not prove it. Is this really true?
    ...
    With a regular hexagon you have 6 isosceles triangles because the legs of the triangle are the radius of the circumcircle.

    With a regular hexagon the central angle must be \alpha = \frac{360^\circ}{6}= 60^\circ

    An isosceles triangle with one interior angle of 60 must be an equilateral triangle because all interior angles are 60 and therefore all sides must have equal length.

    And that's the reason why the circumference of a regular 6gon is

    c = 6r
    Attached Thumbnails Attached Thumbnails A problem with a regular hexagon-regular_6gon.gif  
    Last edited by earboth; March 24th 2008 at 01:32 AM.
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  3. #3
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    Oh, yes...

    stupid me


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  4. #4
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    But I still don't understand the second part, how can the height of the isosceles/equilateral triangle equal, the lenght of the side of the smaller triangle inside the smaller hexagon?
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  5. #5
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    Quote Originally Posted by Coach View Post
    But I still don't understand the second part, how can the height of the isosceles/equilateral triangle equal, the lenght of the side of the smaller triangle inside the smaller hexagon?
    All apothem = heights = angle bisectors have the same length. Therefore the red colored triangles are isosceles triangles.

    Since the height in the original equilateral triangle bisect the angle of 60 two heights include an angle of 2 * 30 = 60. I've typed this angle in question in red.

    And therefore the red triangles are equilateral too.
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