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

1. estudier Group Title

|dw:1350233869190:dw|

2. estudier Group Title

Base 20, and sides 26

3. estudier Group Title

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

4. yomamabf Group Title

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

5. estudier Group Title

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

6. yomamabf Group Title

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

7. estudier Group Title

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

8. yomamabf Group Title

oh ok

9. estudier Group Title

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

10. experimentX Group Title

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

11. estudier Group Title

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

12. estudier Group Title

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

13. klimenkov Group Title

No. I will solve it.

14. klimenkov Group Title

15. estudier Group Title

Yes.

16. estudier Group Title

Wow, this a lot of typing....

17. estudier Group Title

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

18. klimenkov Group Title

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 Group Title

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

20. estudier Group Title

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

21. CliffSedge Group Title

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

22. estudier Group Title

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

23. estudier Group Title

Right, I will close this one....