Isosceles triangle

  • #1

Homework Statement



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.

Homework Equations




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
 
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Answers and Replies

  • #2
761
13
Do you know what a geometric sum is?
 
  • #3
Yes, I write the geometric sums above in the first post. Please see them, now.
 
  • #4
tiny-tim
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Hi! After a tiring excursion, finally I am back... :smile:

strange … I didn't find it tiring! :wink:
[tex]P + \frac{P}{4} + \frac{P}{8} + ... + \frac{P}{2^n}[/tex] for the area of the triangle,

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)
 
  • #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
 
  • #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?
 
  • #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)
 
  • #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]
 
  • #9
tiny-tim
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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]

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:
 
  • #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 ?
 
  • #11
tiny-tim
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It will work also with "k". I sow on wikipedia.
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?
[tex]
\sum_{k=0}^n \left(\frac{1}{x}\right)^k = 1 + \frac{1}{x} + \frac{1}{x^2} + ... + \frac{1}{x^n}
[/tex]

Yes, obviously :rolleyes: … but what is that equal to? :smile:
 
  • #12
[tex]\frac{\frac{1}{x}-\frac{1}{{x}^{n+1}}}{1-\frac{1}{x}}}[/tex]
 
  • #13
tiny-tim
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[tex]\frac{\frac{1}{x}-\frac{1}{{x}^{n+1}}}{1-\frac{1}{x}}}[/tex]

hmm … how about [tex]\frac{1-\frac{1}{{x}^{n+1}}}{1-\frac{1}{x}}}[/tex] ? :smile:
 
  • #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]


?
 
  • #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.
 
  • #16
if d=1 then also down there it will be 1..



[tex]
S=\frac{1-d^n^+^1}{1-1}
[/tex]
 
  • #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.
 
  • #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?
 
  • #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?
 
  • #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.
 
  • #21
tiny-tim
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Physicsissuef, that took ages!.

Look, you will lose marks, and time, in your exams if you keep using formulas without really understanding them.

Multiply (1 - x) by (1 + x + + x^2 + … + x^n) using "long multiplication" to understand why (1 - x^n+1)/(1 - x) = (1 + x + + x^2 + … + x^n). :smile:
 
  • #22
Yes, I know that. :smile:
Just I want to ask you, why on other forum told me that the answer is [itex]/frac{4P}{3}[/itex]?

Here is one post:

4P/3 is indeed the correct answer, since;
a / (1 - r) is the sum to infinity of a G.P.
where |r| < 1, a is the first term and r is the common ratio
so;
P / (1 - 1/4) = 4P/3
 
  • #23
tiny-tim
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Just I want to ask you, why on other forum told me that the answer is [itex]\frac{4P}{3}[/itex]?

Because P(1 - 0)/(1 - 1/4) = 4P/3 is the answer for n = ∞

… and looking all the way back at post #1, even though you kept doing the sums for n = N, that is what your question actually asked for:
[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...
 
  • #24
But why 1-0 ? If the answer is P(1-0)/(1-1/4)=4P/3 then

[itex]\frac{1}{{4}^{n+1}}=0[/itex]

because of [tex]P(1-\frac{1}{{4}^{n+1}})/(1-1/4)[/tex].


How is possible [tex]{4}^{n+1}=0[/tex]?
 
  • #25
tiny-tim
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How is possible [tex]{4}^{n+1}=0[/tex]?

If n —> ∞, then 4^{n+1} —> ∞,

and so (1/4)^{n+1} —> 0. :smile:
 

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