Calculating Perimeter and Area Sums in Infinite Isosceles Triangles

[Physicsissuef]

Homework Statement

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 $A_1,A_2,A_3$ which points $A_1,A_2,A_3$ 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"

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:

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

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

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

$$n \in \mathbb{N}$$

$$n\geq 2$$

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]

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

strange … I didn't find it tiring!

$$P + \frac{P}{4} + \frac{P}{8} + ... + \frac{P}{2^n}$$ 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]

tiny-tim I was on excursion, Skopje - Belgrade - Bratislava - Prague - Vienna (approximately 3000 km in both ways)
Is this correct:

$$P + \sum_{k=2}^n \frac{P}{2^k} = P + \frac{P}{4} + \frac{P}{8} + ... + \frac{P}{2^k}$$ $$L + \sum_{k=2}^n \frac{L}{{2}^{k-1}} = L + \frac{L}{2} + \frac{L}{4} + ... + \frac{L}{{2}^{k-1}}$$

$$n \in \mathbb{N}$$

$$n\geq 2$$

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:

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

$$n \in \mathbb{N}$$

$$n \geq 2$$

n- number of triangles.

Also for L:

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

 

[tiny-tim]

Ahh... I understand. Maybe this is better:

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

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

ok, now what is:

$$\sum_{k=0}^n \left(\frac{1}{x}\right)^k$$ ?

 

[Physicsissuef]

It will work also with "k". I sow on wikipedia.

$$\sum_{k=0}^n \left(\frac{1}{x}\right)^k = 1 + \frac{1}{x} + \frac{1}{x^2} + ... + \frac{1}{x^n}$$

Why 1/x ?

 

[tiny-tim]

It will work also with "k". I sow on wikipedia.

No … it has to be … $$\sum_{k=0}^n \frac{P}{4^k} = P + \frac{P}{4} + \frac{P}{16} + ... + \frac{P}{4^n}$$ … doesn't it?

$$\sum_{k=0}^n \left(\frac{1}{x}\right)^k = 1 + \frac{1}{x} + \frac{1}{x^2} + ... + \frac{1}{x^n}$$

Yes, obviously … but what is that equal to?

 

[Physicsissuef]

$$\frac{\frac{1}{x}-\frac{1}{{x}^{n+1}}}{1-\frac{1}{x}}}$$

 

[tiny-tim]

$$\frac{\frac{1}{x}-\frac{1}{{x}^{n+1}}}{1-\frac{1}{x}}}$$

hmm … how about $$\frac{1-\frac{1}{{x}^{n+1}}}{1-\frac{1}{x}}}$$ ?

 

[Physicsissuef]

But isn't something like this:

$$S=d+d^2+...+d^n$$

$$Sd=d^2+d^3+...+d^n^+^1$$

$$S-Sd=S(1-d)=d-d^n^+^1$$

$$S=\frac{d-d^n^+^1}{1-d}$$?

 

[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..
$$S=\frac{1-d^n^+^1}{1-1}$$

 

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

$$S = 1 + \frac{1}{x} + \frac{1}{x^2} + ... + \frac{1}{x^n}$$

$$S\frac{1}{x}=\frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} + ...+ \frac{1}{{x}^{n+1}}$$

$$S-S\frac{1}{x}=S(1-\frac{1}{x})=1-\frac{1}{{x}^{n+1}}$$

$$S=\frac{1-\frac{1}{{x}^{n+1}}}{1-\frac{1}{x}}$$

Sorry, you're right.

But I need to do the same for P and L, right?

 

[Physicsissuef]

$$S = P + \frac{P}{4} + \frac{P}{16} + ... + \frac{P}{4^n}$$

$$SP=P^2+\frac{P^2}{4}+\frac{P^2}{16}+ ... + \frac{P^2}{4^n}$$

$$S-SP=S(1-P)=P - \frac{P^2}{4^n}$$

$$S=\frac{P - \frac{P^2}{4^n}}{1-P}$$

is this correct?

 

[Physicsissuef]

Ohh... Again I made mistake.
It should be

$$S(1-\frac{1}{4})=P-\frac{P}{{4}^{n+1}}$$

$$S=\frac{P-\frac{P}{{4}^{n+1}}}{1-\frac{1}{4}}$$

Is the final answer.

 

[tiny-tim]

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).

 

[Physicsissuef]

Yes, I know that.
Just I want to ask you, why on other forum told me that the answer is $/frac{4P}{3}$?

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

 

[tiny-tim]

Just I want to ask you, why on other forum told me that the answer is $\frac{4P}{3}$?

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 $A_1,A_2,A_3$ which points $A_1,A_2,A_3$ are in AB/2, BC/2, AC/2. Again third triangle with same attributes is drawn, fourth, fifth... infinite...

 

[Physicsissuef]

But why 1-0 ? If the answer is P(1-0)/(1-1/4)=4P/3 then

$\frac{1}{{4}^{n+1}}=0$

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


How is possible $${4}^{n+1}=0$$?

 

[tiny-tim]

How is possible $${4}^{n+1}=0$$?

If n —> ∞, then 4^{n+1} —> ∞,

and so (1/4)^{n+1} —> 0.

 

[Physicsissuef]

Strange... So what is the right answer now?

 

[HallsofIvy]

But why 1-0 ? If the answer is P(1-0)/(1-1/4)=4P/3 then

$\frac{1}{{4}^{n+1}}=0$

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


How is possible $${4}^{n+1}=0$$?

It isn't. Nor is $4^{-(n+1)}= 0$ for any fixed n. However, the limit $4^{-(n+1)}= 1/4^{n+1}= 0$ as n goes to infinity.

 

[tiny-tim]

Strange... So what is the right answer now?

You tell us! ​