Problem 1

[Towelie]

If an equilateral triangle and a square have the same area, what is the ratio of their perimeters?

by the way, my math teacher is having us email a bum and he sends us problems, if we get it right 1% of extra credit added to our grade. I have 99% in that class so... whatever , I'll pass the problems onto you guys as they come. This is problem #1!

[while1]

seriously a bum?

[Towelie]

Yeah, pretty much. He doesn't have a job and lives in an RV, but he has money from relatives to spend etc. To get internet he jacks it from other people's wireless connection.

[dark_drake]

1:(2/3)*3^(1/4)

Well, if x is the length of one side of the triangle, it's:
3x:2x*3^(1/4)

That's a wild guess from what I remember from geometry. 

[Sidoh]

I get this: , but I have a creeping suspicion that it's supposed to be the golden ratio...

[Towelie]

by plugging in numbers I got 1.139753528:1 , but I can't replicate that algebraically

[dark_drake]

Ok, here goes how I did it.

If x is the length of the side of the triangle, the height is sqrt(3)*x/2.  So, the area is now: 1/2*sqrt(3)*x/2*x=(sqrt(3)*x^2)/4.

All right, so now that the areas have to be equal, let y be the length of the side of a square.  y^2=(sqrt(3)*x^2)/4, so the length of the side is: y=(3^(1/4)*x)/2

Ok, so now we have the perimeters.  3x for the triangle, and 4y or 4*(3^(1/4)*x)/2 or 2x*3^(1/4) for the square.

3x:2x*3^(1/4).
 
edit:

by plugging in numbers I got 1.139753528:1 , but I can't replicate that algebraically

If you put a number into mine and simplify, you will get 1:0.877383, and 1/0.877383=what you got.

[Sidoh]

Yep, that's right.  Working out the problem after reading it correctly (and realizing I found the height of the triangle incorrectly... lol), I got the same thing.

I probably should have eaten something before I tried that, even as trivial is it is... lol