Ok, here is a test for the engineers, I think. I would love to see if a student gets it first. Try it out and email me at [email protected] if your able to figure it out. Thanks
c:\my documents\math.jpg
I dont think u can attach a file that way…u either have to use the “attach file” thing at the bottom or find some webspace, etc…until then we wont be able to see your math problemo…
I am so confused…I totally forget geometry and, well, just about all math outside of basic arthimatic…yeah…I’m gonna die in precalc next semester…lol
I got the answer, but I won’t tell how to get it until someone else do so…
So, here it is:
R = S / ( ( 8^(1/2) ) + 6*sin(120) )
hehe, so just plug 10 into S and get the answer.
I don’t think I am really a student anymore.
For all you know, this could be a fake answer! 
Was Ken right??? did he win???
Just wondering…plus this forum seemed kinda dead
Originally posted by Clark Gilbert *
Was Ken right??? did he win
Well, I can confirm that Ken was right as far as he went. I got
R = S/(2(2^0.5) + 3*(3^0.5))
which (if I typed it correctly) is equivalent to Ken’s
R = S / ( ( 8^(1/2) ) + 6*sin(120) )
But he only gets partial credit. He didn’t do the second part of the problem: solving for BD. This part gets a little hairier. I get
BD = ((15+4*(6^0.5))/(35+12*(6^0.5)))^0.5*S
So if S=10, BD = 6.2056272
I agree with Ken’s answer (although i got it in a slightly different form)
S = (3sqrt(3) + 2sqrt(2))R = Ken’s answer since 3sqrt(3) = 6Sin120
R = S / (3sqrt(3) + 2sqrt(2))
and BD = S * sqrt( (237 - 40*sqrt(6)) / 361)
The solution for finding R in terms of S basically boils down to finding the sides of 3 right triangles for those who are looking for a hint. (Remember the special stuff about tangents to circles? yeah…) And to find BD, cosine law 
Soo… do I win a prize too?
Anthony.