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- anonymous

Circles, tangent everywhere, all the way up the triangle.
Sum of perimeters of all the triangles
(Bit difficult to draw this, wait a minute)

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- anonymous

- katieb

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- anonymous

|dw:1350233869190:dw|

- anonymous

Base 20, and sides 26

- anonymous

Sum of perimeters of all the circles, I mean...

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- anonymous

wait r u trying to teach or is this something you really need help on? cuz ur at level 99

- anonymous

Think of it as a tutorial (or not):-)

- anonymous

so circles dont matter so i would add 20 to 26?

- anonymous

I already corrected that above, it is the perimeter of all the circles

- anonymous

oh ok

- anonymous

Hint This is just Pythagorus and a splash of imagination....

- experimentX

|dw:1350234945917:dw| is something missing ... like side of triangle or ... radius of circle of first and second??

- anonymous

Maybe you didn't see it, sorry..
I posted above "Base 20, and sides 26"

- anonymous

Maybe I should close it and repost it with out the error and including the figures :-(

- klimenkov

No. I will solve it.

- klimenkov

Do you have an answer?

- anonymous

Yes.

- anonymous

Wow, this a lot of typing....

- anonymous

Anyone just want to write out the infinite sum...?

- klimenkov

Try to check my solution. Lets sign the radiuses of the circles \(r_1, r_2,...\). Then their length will be \(2\pi r_1+2\pi r_2+...=2\pi(r_1+r_2+...)\). Now watch the picture.|dw:1350235644893:dw|From the picture you can see that radiuses take a half of the height of the triangle. So I conclude that the whole perimeter of the circumferences is \(\frac 12\cdot\sqrt{26^2-10^2}=12\)

- klimenkov

Sorry, forgot to multiply by \(2\pi\). Answer is \(24\pi\).

- anonymous

That's right, pi*(d1 + d2 +.....) = pi * altitude of triangle

- anonymous

I think I've seen this done before using a geometric progression.

- anonymous

@CliffSedge Yup, u could do that, lot more work though....

- anonymous

Right, I will close this one....

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