## anonymous 4 years ago Circles, tangent everywhere, all the way up the triangle. Sum of perimeters of all the triangles (Bit difficult to draw this, wait a minute)

1. anonymous

|dw:1350233869190:dw|

2. anonymous

Base 20, and sides 26

3. anonymous

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

4. anonymous

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

5. anonymous

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

6. anonymous

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

7. anonymous

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

8. anonymous

oh ok

9. anonymous

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

10. experimentX

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

11. anonymous

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

12. anonymous

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

13. klimenkov

No. I will solve it.

14. klimenkov

15. anonymous

Yes.

16. anonymous

Wow, this a lot of typing....

17. anonymous

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

18. 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 $$\frac 12\cdot\sqrt{26^2-10^2}=12$$

19. klimenkov

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

20. anonymous

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

21. anonymous

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

22. anonymous

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

23. anonymous

Right, I will close this one....