# isosceles triangle

by Physicsissuef
Tags: isosceles, triangle
 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 $A_1,A_2,A_3$ 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 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: $$P + \frac{P}{4} + \frac{P}{8} + ... + \frac{P}{2^n}$$ 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
 P: 664 Do you know what a geometric sum is?
 P: 909 Yes, I write the geometric sums above in the first post. Please see them, now.
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P: 26,167

## isosceles triangle

 Quote by Physicsissuef 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)
 P: 909 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
 Sci Advisor HW Helper Thanks P: 26,167 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?
 Mentor P: 11,984 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)
 P: 909 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: $$\sum_{k=0}^n \frac{L}{2^k} = L + \frac{L}{2} + \frac{L}{4} + ... + \frac{L}{2^k}$$
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P: 26,167
 Quote by Physicsissuef Ahh... I understand. Maybe this is better: $$\sum_{k=0}^n \frac{P}{4^k} = P + \frac{P}{4} + \frac{P}{16} + ... + \frac{P}{4^k}$$
Yes (with n at the end instead of k, of course )

ok, now what is:

$$\sum_{k=0}^n \left(\frac{1}{x}\right)^k$$ ?
 P: 909 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 ?
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P: 26,167
 Quote by Physicsissuef 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?
 P: 909 $$\frac{\frac{1}{x}-\frac{1}{{x}^{n+1}}}{1-\frac{1}{x}}}$$
 Quote by Physicsissuef $$\frac{\frac{1}{x}-\frac{1}{{x}^{n+1}}}{1-\frac{1}{x}}}$$
hmm … how about $$\frac{1-\frac{1}{{x}^{n+1}}}{1-\frac{1}{x}}}$$ ?
 P: 909 But isn't something like this: $$S=d+d^2+...+d^n$$ $$Sd=d^2+d^3+...+d^n^+^1$$ $$S-Sd=S(1-d)=d-d^n^+^1$$ $$S=\frac{d-d^n^+^1}{1-d}$$ ?
 P: 909 if d=1 then also down there it will be 1.. $$S=\frac{1-d^n^+^1}{1-1}$$
 HW Helper P: 2,618 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.
 P: 909 $$S = 1 + \frac{1}{x} + \frac{1}{x^2} + ... + \frac{1}{x^n}$$ $$S\frac{1}{x}=\frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} + ...+ \frac{1}{{x}^{n+1}}$$ $$S-S\frac{1}{x}=S(1-\frac{1}{x})=1-\frac{1}{{x}^{n+1}}$$ $$S=\frac{1-\frac{1}{{x}^{n+1}}}{1-\frac{1}{x}}$$ Sorry, you're right. But I need to do the same for P and L, right?