# Isosceles triangle

1. Jun 20, 2008

### 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 $A_1,A_2,A_3$ 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" [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:

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

Last edited by a moderator: May 3, 2017
2. Jun 20, 2008

### dirk_mec1

Do you know what a geometric sum is?

3. Jun 20, 2008

### Physicsissuef

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

4. Jun 20, 2008

### 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)

5. Jun 21, 2008

### 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

6. Jun 21, 2008

### 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?

7. Jun 21, 2008

### Redbelly98

Staff Emeritus
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. Jun 21, 2008

### 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:

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

9. Jun 21, 2008

### tiny-tim

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

ok, now what is:

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

10. Jun 21, 2008

### 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 ?

11. Jun 21, 2008

### tiny-tim

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?
Yes, obviously … but what is that equal to?

12. Jun 21, 2008

### Physicsissuef

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

13. Jun 21, 2008

### tiny-tim

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

14. Jun 21, 2008

### Physicsissuef

But isn't something like this:

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

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

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

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

?

15. Jun 21, 2008

### Defennder

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

16. Jun 21, 2008

### Physicsissuef

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

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

17. Jun 21, 2008

### 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.

18. Jun 21, 2008

### 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?

19. Jun 21, 2008

### 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?

20. Jun 22, 2008

### Physicsissuef

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