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Area sum of inscribed pentagons

  1. Aug 15, 2011 #1
    1. The problem statement, all variables and given/known data

    Ok, I wanted to calculate the sums of the areas of inscribed petagons into an initial pentagon (the smaller pentagon's vertices touch the larger pentagon's midpoints). However, I wanted all of the even pentagons to create negative space, and then the odd # pentagon after that to create positive space.

    The initial pentagon is regular with side length of 1. To calculate the area of the smaller pentagons, i assumed that the area would be proportional to the next bigger pentagon by a constant factor. apothem/radius

    i.e. area of small pentagon = area of big pentagon *(apothem/radius)

    Total area = area1 - area2 + area3 - area4 + area5 - ...

    So to do this I calculated a sum (see picture). Can someone verify that my sum is correct?

    My question: How does my calculator take an infinite sum and convert it into a nice and clean polynomial? Can this always be done? What conditions need to be met to do so (obviously the sum need to converge, but are there any other requirements?)

    2. Relevant equations

    http://img.photobucket.com/albums/v298/Swiffer/Math.jpg [Broken] Sorry for the blur, my phone can only do 3 megapixel resolution.

    3. The attempt at a solution

    See picture.

    Ty in advance!
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Aug 16, 2011 #2


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    Homework Helper

    I believe your sum is an infinite geometric series, no? If you know the area of one of the outermost triangles, five times that is the first term of the series, a . If you have found the ratio of the area of the next "positive area" triangles to the initial ones, you should find that ratio is a constant, r . The sum of the infinite series of all the positive area triangles is just a / ( 1 - r ) . You can just ignore the "empty space" triangles, since it appears that you only want the total area of all those concentric "pentagonal rings" of isosceles triangles.
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