Finding a solution for non-linear simultaneous equations

Hi

I am not sure if the title accurately reflects my issue, apologies if this is not the case.

I am an academic engineer, I have encountered a problem where i would like to equate a circle to a triangle to simplfy a problem..... My criteria is to transform a circle to a 120, 30, 30 triangle with equivalent area and perimeter, when i quickly derived the simultaeous equations i get the following two expressions

y1 = b+2*h

y2=(b*h)/2

where:

y1=the circles perimeter

y2=the circles area

b= base of triangle

h= height of triangle

I thought i would ask as i do not want to waste time if it is not possible to solve and satisfy the 120, 30, 30 condition......... Are there any methods that can be used to tackle this issue?

Re: Finding a solution for non-linear simultaneous equations

if b,h are given, $\displaystyle y_1$ and $\displaystyle y_2$ may simply be calculated. so i assume that $\displaystyle y_1, y_2$ are given, and you wish to find b and h.

write: $\displaystyle b = y_1 - 2h = \frac{2y_2}{h}$

from $\displaystyle y_1 - 2h = \frac{2y_2}{h}$ we get:

$\displaystyle 2h^2 - y_1h - 2y_2 = 0$, which is a quadratic in h.

the discriminant is $\displaystyle y_1^2 - 8y_2$

however, $\displaystyle y_1,y_2$ aren't "independent" they both depend on the radius of the circle, r:

$\displaystyle y_1 = 2\pi r$

$\displaystyle y_2 = \pi r^2 $

so $\displaystyle y_1^2 - 8y_2 = (2\pi r)^2 - 8\pi r^2 = -4\pi r^2 < 0$, therefore: no real solutions for h.

however, in reviewing your original equations, it appears that your formula for either the area or the perimeter of the isoceles triangle is incorrect, if h is the height, then the perimeter length should be:

$\displaystyle b + 2(\sqrt{h^2 + \frac{b^2}{4}})$.

you might want to review that, and confirm (the calcuations get a lot messier if this is true).

Re: Finding a solution for non-linear simultaneous equations

Thanks, I did make a couple of mistakes actually, sorry about that, the equations should be

y1 = b+2*(sqrt(b^2+h^2))

y2=(b*h)/2

where:

y1=the circle z's perimeter (GIVEN)

y2=the circle z's area (GIVEN)

b= base of triangle

h= height of triangle

I want to calculate values for b, and h for a 120, 30, 30 triangle, my main question is there a solution? i.e. can you take a circle and construct a 120, 30, 30 triangle with an identicle Area and perimeter?