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

Mathematics
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|dw:1350233869190:dw|
Base 20, and sides 26
Sum of perimeters of all the circles, I mean...

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Other answers:

wait r u trying to teach or is this something you really need help on? cuz ur at level 99
Think of it as a tutorial (or not):-)
so circles dont matter so i would add 20 to 26?
I already corrected that above, it is the perimeter of all the circles
oh ok
Hint This is just Pythagorus and a splash of imagination....
|dw:1350234945917:dw| is something missing ... like side of triangle or ... radius of circle of first and second??
Maybe you didn't see it, sorry.. I posted above "Base 20, and sides 26"
Maybe I should close it and repost it with out the error and including the figures :-(
No. I will solve it.
Do you have an answer?
Yes.
Wow, this a lot of typing....
Anyone just want to write out the infinite sum...?
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 \(\frac 12\cdot\sqrt{26^2-10^2}=12\)
Sorry, forgot to multiply by \(2\pi\). Answer is \(24\pi\).
That's right, pi*(d1 + d2 +.....) = pi * altitude of triangle
I think I've seen this done before using a geometric progression.
@CliffSedge Yup, u could do that, lot more work though....
Right, I will close this one....

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