isosceles triangle

Physicsissuef

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

dirk_mec1

Do you know what a geometric sum is?

Physicsissuef

Yes, I write the geometric sums above in the first post. Please see them, now.

tiny-tim

strange … I didn't find it tiring!
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

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

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

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

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

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

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

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 … but what is that equal to?

Physicsissuef

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

tiny-tim

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

Physicsissuef

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

Your first term is 1, not 1/x. The formula you derived needs to be modified slightly to suit your problem.

Physicsissuef

if d=1 then also down there it will be 1..



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

Defennder

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.