# I Triangle problem

1. Apr 8, 2016

### joshmccraney

Hi PF!

A friend asked me to look at a problem, said it was on his mind. Rather than restate it here (since a picture helps) please check out the attachment. I drew a picture in tikZ and gave my attempt at a solution, but something in my gut feels the answer is infinite. I'd love to talk with someone about it after you look over my work.

My question is, is the sum of all the triangles really infinite? If so, what's wrong with my reasoning on the attachment?

Thanks!

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2. Apr 8, 2016

### Samy_A

$\int_0^1 f(x) dx$ is not equal to the sum of all values f(x) for x ranging from 0 to 1.

3. Apr 8, 2016

### joshmccraney

I know, but if $A(\theta)$ is the area of a triangle, then $A(\theta) d\theta$ can't also be that area.

4. Apr 8, 2016

### joshmccraney

I am assuming you are referring to why I took $\int ds$ rather than $\int A d\theta$.

5. Apr 8, 2016

### Samy_A

No. In general, when you have a real function like your A(θ), the sum of all these values will not be given by the integral of A(θ) over the relevant interval, or by the arclength of that function.

It is quite obvious that the sum of all the areas of your triangles is infinite.

To give a trivial example, take the function f: [0,1] → ℝ defined by f(x) = 1.
The sum of all these values is infinite. The arclenght is 1.

6. Apr 8, 2016

### joshmccraney

I totally agree, especially since the harmonic sequence $1/n \in [0,1]$, whose series diverges. I definitely see what the issue is now that you mention it: I wasn't adding up the heights for all $\theta$; I was adding up the length of the heights.

My initial intuition was that the sum of all the triangles would diverge since the area of each triangle doesn't diminish as $\theta$ changes values at smaller step-sizes.

Does this sound correct?

7. Apr 8, 2016

### Samy_A

The number of triangles is uncountable. Any (however small) interval where the area is not 0 will already give you a divergent sum.

Last edited: Apr 8, 2016
8. Apr 8, 2016

### joshmccraney

Awesome! This is definitely what I was thinking, though I couldn't say it so precise.

Thanks Samy_A!