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Segments of a polygon

  1. May 12, 2008 #1
    Hi all,

    I have this math problem where an equilateral triangle has each of its sides divided into ratio 2:1, and a smaller equilateral drawn within it from the intersecting lines. There is also a similar problem, but relating to a square. My objective is to figure out a relationship between the ratio of the areas of the two triangles to "n" (the number of divisions of each side)

    In the document attached, the example given is when divided into 1:2 parts. The ratio of the areas for the triangle there is 7.0.

    Has anyone heard of this problem before and knows how to do it?

    I would greatly appreciate any help given.


    Attached Files:

  2. jcsd
  3. May 12, 2008 #2


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    For the square, it is fairly easy, isn't it, to use the Pythagorean theorem to calculate the length of the sides of the inner square, relative to the sides of the outer square.
  4. May 12, 2008 #3
    And the Law of Cosines can do a similar job for the Triangle, since you have two sides and the angle between them (60 degrees).
  5. May 12, 2008 #4
    so far for the triangle i've gotten the rule ((n-1)^2+n)/((n-2)^2) and it works. However, I'm asked to somehow prove this. Anybody have any idea how to?
  6. May 12, 2008 #5
    Your attachment is now available, and now I understand why you get such a high ratio (7) for the triangle: you are doing a different thing than for the square. (Note that the square is not joining point B-E with a segment, for example.)

    I would have expected the triangle to be formed by joining the newly formed points at 2:1 (as you did for the square), rather than throwing lines from the original vertices. For example, with a ratio of 1:1, joining the midpoints with lines splits the triangle into 4 identical ones (and thus the area ratio is 4), while using your construction, midpoints at 1:1 will produce no inner triangle, but only a single point where the three lines intersect.

    Are you sure that the triangle is not meant to be split as you did for the square, instead?
  7. May 12, 2008 #6
    Yeah thanks for that... i got that wrong.. the triangle is right, but the square is wrong. The lines join the segments to the vertices of the polygon. Anyone has any idea how to prove the rule i found?
  8. May 12, 2008 #7
    Well, the ideas have already been mentioned. Your segments will produce a lot of inner triangles; work your way with them to calculate the sides of the inner triangle or square, and thus its area.

    In your case, you'd typically use either Pitagoras' rule (for the square) of the Law of Cosines (for the triangle) to calculate new sides, and then the Law of Sines to calculate new angles, so that you can continue calculating new sides on other inner triangles.
  9. Jun 14, 2008 #8
    Hi Issac

    Hi Issac,

    thanks for posting this. Are u doing I.B maths. I got the same thing for my maths portfolio.
  10. Jun 16, 2008 #9
    Can anybody please explain me in detail about how to use the Law of cosines and what is 'n' in
    the rule ((n-1)^2+n)/((n-2)^2)

    Thank you
  11. Jun 16, 2008 #10
    Can anybody please explain me in detail about how to use the Law of cosines and what is 'n' in
    the rule ((n-1)^2+n)/((n-2)^2) and explain me this rule.

    Thank you
  12. Nov 29, 2008 #11
    omg i just go this portfolio as well...how do you prove that rule? and i don't really see a relationship between the ratios of the sides with the ratios of the triangles. Ive looked at all the measurements but cant see any kind of pattern...please help.

  13. Feb 9, 2009 #12
    can you help me with this portfolio??????????????
  14. Feb 9, 2009 #13

    can you help me with this portfolio??????????????
  15. May 10, 2009 #14
    im also doing this portfolio!!!

    hmmm anyone know how to get to the above formula?
    im kind of stuck after drawing the triangles for 1:2, 1:3, 1:4, 1:5 as i can't seem to find any relation between the side ratios and the area ratios...

    a similar example or a detailed guide on hw to reach the formula above would be nice :D
    hehe ^_^
    thx in advance :D
  16. May 12, 2009 #15
    I am also doing this portfolio right now, and I don't know how to proceed, so what if we work together? All I have found is that the ratio of the lengths squared (the ratio of the "real" lengths if your draw it) equals the ratio of the areas. But that is a very simple thing, there's gotta be more to it, there's gotta be a connection between that and the ratio the lines are cut. As you probably understand i could really use some help. :P
  17. May 14, 2009 #16
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  18. Sep 8, 2009 #17
    the equation that i got is (n^2+n+1)/(n-1)^2
    the length of the inner triangle is (n^2-1)/(n^2+n+1)^(1/2)
    and the area for the inner triangle is ((3)^(1/2)/4) x ((n^2-1)^2/n^2+n+1))
    from this you could get the above equation.
    the area for the big triangle is (3)^(1/2)/4 x (n+1)^2
  19. Sep 25, 2009 #18
    Thanks barom for all of your help.

    I would like to request one more thing. Could you please explain briefly, if possible, on how you got the equations and especially on how you got (n^2+n+1) for the equations.

  20. Oct 27, 2009 #19
    HL maths coursework is so depressing :(
  21. Apr 23, 2010 #20
    I'm doing the math portfolio as well and I've gotton all the ratios but I'm stuck on how to get the equation. How did you derive the equation (n^2+n+1)/(n-1)^2?
  22. Sep 10, 2010 #21
    hello im doing this portfolio too, for a pre-IA task. not formal, only for a practice. and i dunno how to get the conjecture, could any of you help me please? I'm really stuck and dunno what to do
  23. Sep 11, 2010 #22
    I got the conjecture and proved it already, but my teacher said I need to find the formula for the length of the side of the inner triangle to prove it once more time because at first I used approximation for the values of the ratios. can anybody help me to find the formula for the length of the side of the inner triangle? I have found a formula using sine but that aint accurate enough :( how can I arrive to Barom's formula?
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