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Isosceles triangle

  1. Jun 20, 2008 #1
    1. The problem statement, all variables and given/known data

    Hi! After a tiring excursion, finally I am back... :smile:

    Here is one for you:

    In isosceles triangle ABC (triangle with all sides equal size) with side a, is drawn another isosceles triangle [itex]A_1,A_2,A_3[/itex] which points [itex]A_1,A_2,A_3[/itex] are in AB/2, BC/2, AC/2. Again third triangle with same attributes is drawn, fourth, fifth.... infinite...

    This looks like on this picture.

    Find the sum of the perimeter and calculate the sum of the areas of the triangles.

    2. Relevant equations


    3. The attempt at a solution

    I think it is something like this:

    [tex]P + \frac{P}{4} + \frac{P}{8} + ... + \frac{P}{2^n}[/tex] for the area of the triangle, and
    [tex]L + \frac{L}{2} + \frac{L}{4} + ... + \frac{L}{2^n^-^1}[/tex]
    for the perimeter.

    I think also, that I can write them as:
    [tex]
    P + \sum_{n=2}^n \frac{P}{2^n} = P + \frac{P}{4} + \frac{P}{8} + ... + \frac{P}{2^n}
    [/tex]

    [tex]
    L + \sum_{n=2}^n \frac{L}{{2}^{n-1}} = L + \frac{L}{2} + \frac{L}{4} + ... + \frac{L}{{2}^{n-1}}
    [/tex]

    [tex]n \in \mathbb{N}[/tex]

    [tex]n\geq 2[/tex]

    n - number of triangles
     
    Last edited: Jun 20, 2008
  2. jcsd
  3. Jun 20, 2008 #2
    Do you know what a geometric sum is?
     
  4. Jun 20, 2008 #3
    Yes, I write the geometric sums above in the first post. Please see them, now.
     
  5. Jun 20, 2008 #4

    tiny-tim

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    strange … I didn't find it tiring! :wink:
    No. :frown:

    Hint: if the lengths are halved each time, what happens to the area?

    And what is ∑x^n? :smile:

    (oh, and they're actually equilateral … isoceles means two sides equal)
     
  6. Jun 21, 2008 #5
    tiny-tim I was on excursion, Skopje - Belgrade - Bratislava - Prague - Vienna (approximately 3000 km in both ways):smile:
    Is this correct:

    [tex]

    P + \sum_{k=2}^n \frac{P}{2^k} = P + \frac{P}{4} + \frac{P}{8} + ... + \frac{P}{2^k}

    [/tex]


    [tex]

    L + \sum_{k=2}^n \frac{L}{{2}^{k-1}} = L + \frac{L}{2} + \frac{L}{4} + ... + \frac{L}{{2}^{k-1}}

    [/tex]

    [tex]
    n \in \mathbb{N}
    [/tex]

    [tex]
    n\geq 2
    [/tex]

    n - number of triangles
     
  7. Jun 21, 2008 #6

    tiny-tim

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    Hi Physicsissuef! :smile:

    The L equation is correct, the P equation isn't.

    Hint, repeated: if the lengths are halved each time, what happens to the area?
     
  8. Jun 21, 2008 #7

    Redbelly98

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    The area equation is incorrect, starting with the term P/8. P and P/4 are correct.

    Hint: think carefully, what is the area of the 3rd triangle? (It's not P/8)
     
  9. Jun 21, 2008 #8
    Ahh... I understand. Maybe this is better:

    [tex]

    \sum_{k=0}^n \frac{P}{4^k} = P + \frac{P}{4} + \frac{P}{16} + ... + \frac{P}{4^k}

    [/tex]

    [tex]n \in \mathbb{N}[/tex]

    [tex]n \geq 2[/tex]

    n- number of triangles.

    Also for L:

    [tex]


    \sum_{k=0}^n \frac{L}{2^k} = L + \frac{L}{2} + \frac{L}{4} + ... + \frac{L}{2^k}


    [/tex]
     
  10. Jun 21, 2008 #9

    tiny-tim

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    Yes (with n at the end instead of k, of course :wink:)

    ok, now what is:

    [tex]\sum_{k=0}^n \left(\frac{1}{x}\right)^k[/tex] ? :smile:
     
  11. Jun 21, 2008 #10
    It will work also with "k". I sow on wikipedia.

    [tex]
    \sum_{k=0}^n \left(\frac{1}{x}\right)^k = 1 + \frac{1}{x} + \frac{1}{x^2} + ... + \frac{1}{x^n}
    [/tex]

    Why 1/x ?
     
  12. Jun 21, 2008 #11

    tiny-tim

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    No … it has to be … [tex] \sum_{k=0}^n \frac{P}{4^k} = P + \frac{P}{4} + \frac{P}{16} + ... + \frac{P}{4^n}[/tex] … doesn't it?
    Yes, obviously :rolleyes: … but what is that equal to? :smile:
     
  13. Jun 21, 2008 #12
    [tex]\frac{\frac{1}{x}-\frac{1}{{x}^{n+1}}}{1-\frac{1}{x}}}[/tex]
     
  14. Jun 21, 2008 #13

    tiny-tim

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    hmm … how about [tex]\frac{1-\frac{1}{{x}^{n+1}}}{1-\frac{1}{x}}}[/tex] ? :smile:
     
  15. Jun 21, 2008 #14
    But isn't something like this:

    [tex]S=d+d^2+...+d^n[/tex]

    [tex]Sd=d^2+d^3+...+d^n^+^1[/tex]

    [tex]S-Sd=S(1-d)=d-d^n^+^1[/tex]

    [tex]S=\frac{d-d^n^+^1}{1-d}[/tex]


    ?
     
  16. Jun 21, 2008 #15

    Defennder

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    Your first term is 1, not 1/x. The formula you derived needs to be modified slightly to suit your problem.
     
  17. Jun 21, 2008 #16
    if d=1 then also down there it will be 1..



    [tex]
    S=\frac{1-d^n^+^1}{1-1}
    [/tex]
     
  18. Jun 21, 2008 #17

    Defennder

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    No, in your derivation, the first term is equivalent to r, the ratio of the n+1 term to the nth term. But in [tex]\sum_{k=0}^n \left(\frac{1}{x}\right)^k = 1 + \frac{1}{x} + \frac{1}{x^2} + ... + \frac{1}{x^n}[/tex], the first term is not equivalent to the ratio. Which is why the formula needs to be modifed in order to suit your problem.
     
  19. Jun 21, 2008 #18
    [tex]S = 1 + \frac{1}{x} + \frac{1}{x^2} + ... + \frac{1}{x^n} [/tex]

    [tex]S\frac{1}{x}=\frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} + ...+ \frac{1}{{x}^{n+1}}[/tex]

    [tex]S-S\frac{1}{x}=S(1-\frac{1}{x})=1-\frac{1}{{x}^{n+1}}[/tex]

    [tex]S=\frac{1-\frac{1}{{x}^{n+1}}}{1-\frac{1}{x}}[/tex]

    Sorry, you're right.

    But I need to do the same for P and L, right?
     
  20. Jun 21, 2008 #19
    [tex]S = P + \frac{P}{4} + \frac{P}{16} + ... + \frac{P}{4^n}[/tex]

    [tex]SP=P^2+\frac{P^2}{4}+\frac{P^2}{16}+ ... + \frac{P^2}{4^n}[/tex]

    [tex]S-SP=S(1-P)=P - \frac{P^2}{4^n}[/tex]

    [tex]S=\frac{P - \frac{P^2}{4^n}}{1-P}[/tex]

    is this correct?
     
  21. Jun 22, 2008 #20
    Ohh... Again I made mistake.
    It should be

    [tex]S(1-\frac{1}{4})=P-\frac{P}{{4}^{n+1}}[/tex]

    [tex]S=\frac{P-\frac{P}{{4}^{n+1}}}{1-\frac{1}{4}}[/tex]

    Is the final answer.
     
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