Find the limit of the sum of the areas of all these triangles

[tellmesomething]

Homework Statement

Starting from an equilateral triangle of side a, a new triangle is constructed from the three heights of the first triangle, and so on η times ; find the limit of the sum of the areas of all the triangles as η - > oo.

Relevant Equations

None

Firstly im not sure if the new triangle that we make of height 3 times the original height of an equilateral triangle of side a, will be an equilateral triangle as well or not.

I assumed it would be, please let me know if I interpreted the question wrongly.

Continuing this train of thought I did the easy part:

The height of an equilateral triangle of side a would be $$ √(a²-\frac{a²} {4})$$ (from pythogoras theorm)

==> $$\frac{√3a} {2}$$

Height of the new triangle would be
$$3(\frac{√3a} {2})$$
Assuming this to be an equilateral triangle as well we know that the side of this new triangle would be x, therefore applying pythogoras theorem once again we get x as 3a

We see a pattern forming if we multiply the new height of this new triangle with 3 and do the same
The general formula for the height would be
$$\frac{3^r√3 a} {2}$$
The general formula for the base would be $$(3^r a)$$

Where r represents the number of triangles we have used to reach the current triangle.

So if we were to write the general term for the summation of its area it would Be

Summation from r=0 to n ($$\frac{(3)^{2r}√3a²} {2}$$)

Taking the constants out leaves us with summation $$3^{2r}$$ inside which I believe is a geometric series

We get
$$\frac{√3a²} {2} (\frac{1(1-(3²)^n)} {1-3²}$$)
If we tend n to ∞ we get this sum as infinite which is not the Answer.


Im obviously very wrong somewhere. Please consider giving a hint. Thankyou :-)

 

[FactChecker]

IMO, the homework statement is very vague. (For instance, how are there 3 heights associated with an equilateral triangle?) Is that the exact wording of the problem statement?

 

[tellmesomething]

IMO, the homework statement is very vague. (For instance, how are there 3 heights associated with an equilateral triangle?) Is that the exact wording of the problem statement?

 

[tellmesomething]

IMO, the homework statement is very vague. (For instance, how are there 3 heights associated with an equilateral triangle?) Is that the exact wording of the problem statement?

Just realised it most probably means that the heights become the sides of the new triangle...ughhhh

 

[Mark44]

Just realised it most probably means that the heights become the sides of the new triangle...ughhhh

Yes. It took me a little while to come to this conclusion as well. The gist of the problem is to calculate the sum of an infinite series. In fact I believe this is a geometric series, for which formulas are well known.

 

[tellmesomething]

Yes. It took me a little while to come to this conclusion as well. The gist of the problem is to calculate the sum of an infinite series. In fact I believe this is a geometric series, for which formulas are well known.

:-( took you a little while, took me 4 hours..

On that note I had a question on the sum of the infinite converging series..
How is the formula for the said derived? Can we use the normal geometric series summation formula and then tend the variable n to infinity?
Would that give us the same answer as the formula
$\frac{a} {1-r}$ ?

 

[FactChecker]

View attachment 348180

IMO, that is vague and incomprehensible. Don't waste your time on it.

 

[Mark44]

Can we use the normal geometric series summation formula and then tend the variable n to infinity?

Yes, if I'm correct in thinking that each successive triangle that is created is the same fraction of the previous triangle. I haven't worked the problem, so my assumption might not be correct.

Would that give us the same answer as the formula $\frac{a} {1-r}$ ?

Yes.

Heights above what?

Heights AKA altitudes. The three heights/altitudes of the first triangle are all the same size, but smaller. The idea is to form a new triangle from these three lengths. You don't need to use any coordinates.
Then do the same with the triangle that was formed to get the third triangle. And so on.