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Homework Help: 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 http://pic.mkd.net/images/404616untitled.JPG" [Broken]

    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 by a moderator: May 3, 2017
  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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