Yet another maths question

I got inspired by superchargednut and decided to propose a question too. This one isn't as easy, though.



Find a formula that relates r (the radius of the circle) and L (the side of the bigger pentagon). Both pentagons and the circle are as regular as they can be.

Whoever gets it right receives a basket of cookies.


PS: No, the answer isn't 42.

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Um dois três quatro cinco seis sete oito nove dez
Eins zwei drei vier fünf sechs sieben acht neun zehn
One two three four five six seven eight nine ten
Один два три четыре пять шесть семь восемь девять десять



Edited 1 time(s). Last edit at 03/09/2009 01:30AM by Lex.

L=3r

my guess:



this forum needs a LaTeX function lol

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used to be GPGSL's Nick Heidfeld

15. :D just a guess i have no qualifactions in maths :D

@Gui: Almost.

@n00binio: If we limit the answer to five decimal digits (like 0,xxxxx), you've got it! You could have used sec 36º instead of 1/cos 36º, as to make your formula look more difficult.

There you go:




Though I must say, I've found a formula that, although means the same thing, looks much simpler than yours:

r = L.(cotg 36º - tg 36º)/2

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Um dois três quatro cinco seis sete oito nove dez
Eins zwei drei vier fünf sechs sieben acht neun zehn
One two three four five six seven eight nine ten
Один два три четыре пять шесть семь восемь девять десять

wohooo, thanks.

i solved that last night (2am) i didn't have the motivation to simplify the whole thing. i read gui's anwer and had a good laugh, then i put mine in a calculator and it was something like r=0,32...*L

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used to be GPGSL's Nick Heidfeld

What amazes me is that Gui is almost right.

But how did you solve it? I connected the center of the circle to the middle point of a L side, and then connected one of the angles of the bigger pentagon to the center of the circle, in order to create a right-angled triangle with catheti L/2 and (r+x). x and r+x can be found (based on L/2) with tg 36º and cotg 36º(in this order). You find the formula I posted above doing (r+x)-(x) = (cotg 36º . L/2) - (tg 36º . L/2).

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Um dois três quatro cinco seis sete oito nove dez
Eins zwei drei vier fünf sechs sieben acht neun zehn
One two three four five six seven eight nine ten
Один два три четыре пять шесть семь восемь девять десять

first i had a look at the relation between the lenght of a pentagon side and the radius of the circumscribed circle (a), then the lenght of the diagonal lines, out of that you can calculate the lenght of a side of the small pentagon in the middle. put that in (a) and you're done (not a beautiful solution but it works).

maybe there's some interest in another problem:

an easy one:



find DS:SB

and for the nerds:


(if anyone tries and does not succeed i post some hints)

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used to be GPGSL's Nick Heidfeld

I did the second one last year in maths and I have no idea how to do it anymore! :S

me too, not easy when you don't know where to begin

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used to be GPGSL's Nick Heidfeld

negative sine square over 2, noobinio? It's been almost 6 years since I did that :P.

Guimengo schrieb:
-------------------------------------------------------
> negative sine square over 2, noobinio? I

nope

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used to be GPGSL's Nick Heidfeld

secant or cosine :P

still nope. it's not that easy

and don't forget that the limits are given. i want a number

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used to be GPGSL's Nick Heidfeld

Fine, ln x :P


Edit: Oops.... 1



Edited 1 time(s). Last edit at 03/10/2009 12:41AM by Guimengo.

nope, and why 1? ln(0) is not defined and ln(infinity)=infinity hehe.

1st hint: substitution

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used to be GPGSL's Nick Heidfeld

Wait...for sin functions the "extreme points", for want of a better term, are at pi/2...In this the upper limit increases to infinity...so we assume that the limit is pi/2...Actually that sounds too simple, but meh, that's my guess.

wohoo, we have a winner. pi/2 is correct. elegant solution:

only instuctions (if somebody wants i can post the complete solution tomorrow)

-make use of the substitution i posted
- fubini -> 2 times partial integration for x
- a bit of messing around with it and you get

\int arctan't)dt

insert the limits and it's done

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used to be GPGSL's Nick Heidfeld

I HATE MATHS!!!!!!! Ahem sorry.

hehehe, noobinio can't you tell I had not been following any pattern at all? It was very fun trying to remember high school and what some integrals are, I wanted to say more absurd things but I at least stopped myself