isosceles triangle


by Physicsissuef
Tags: isosceles, triangle
Physicsissuef
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#1
Jun20-08, 03:36 PM
P: 909
1. The problem statement, all variables and given/known data

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

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
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dirk_mec1
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#2
Jun20-08, 03:48 PM
P: 664
Do you know what a geometric sum is?
Physicsissuef
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#3
Jun20-08, 03:55 PM
P: 909
Yes, I write the geometric sums above in the first post. Please see them, now.

tiny-tim
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#4
Jun20-08, 05:44 PM
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isosceles triangle


Quote Quote by Physicsissuef View Post
Hi! After a tiring excursion, finally I am back...
strange I didn't find it tiring!
[tex]P + \frac{P}{4} + \frac{P}{8} + ... + \frac{P}{2^n}[/tex] for the area of the triangle,
No.

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

And what is ∑x^n?

(oh, and they're actually equilateral isoceles means two sides equal)
Physicsissuef
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#5
Jun21-08, 03:14 AM
P: 909
tiny-tim I was on excursion, Skopje - Belgrade - Bratislava - Prague - Vienna (approximately 3000 km in both ways)
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
tiny-tim
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#6
Jun21-08, 05:52 AM
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Hi Physicsissuef!

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

Hint, repeated: if the lengths are halved each time, what happens to the area?
Redbelly98
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#7
Jun21-08, 06:01 AM
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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)
Physicsissuef
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#8
Jun21-08, 06:25 AM
P: 909
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]
tiny-tim
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#9
Jun21-08, 06:34 AM
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Quote Quote by Physicsissuef View Post
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 )

ok, now what is:

[tex]\sum_{k=0}^n \left(\frac{1}{x}\right)^k[/tex] ?
Physicsissuef
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#10
Jun21-08, 08:50 AM
P: 909
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 ?
tiny-tim
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#11
Jun21-08, 10:10 AM
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Quote Quote by Physicsissuef View Post
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 but what is that equal to?
Physicsissuef
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#12
Jun21-08, 11:02 AM
P: 909
[tex]\frac{\frac{1}{x}-\frac{1}{{x}^{n+1}}}{1-\frac{1}{x}}}[/tex]
tiny-tim
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#13
Jun21-08, 11:06 AM
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Quote Quote by Physicsissuef View Post
[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] ?
Physicsissuef
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#14
Jun21-08, 12:31 PM
P: 909
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]


?
Defennder
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#15
Jun21-08, 12:57 PM
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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.
Physicsissuef
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#16
Jun21-08, 01:22 PM
P: 909
if d=1 then also down there it will be 1..



[tex]
S=\frac{1-d^n^+^1}{1-1}
[/tex]
Defennder
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#17
Jun21-08, 01:26 PM
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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.
Physicsissuef
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#18
Jun21-08, 01:27 PM
P: 909
[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?


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