Mathematical Trivia Question

Trivia question: Solve this system of equations.

A+B+C=6
AB+BC+AC=9
ABC=3

 

Let n=Cos(pi/9), m=Cos(2*pi/9)

A=4(1+n)/(5+6n+sqrt(15+20n+2m))
B=(15+18n+sqrt(9(5+6n)^2-96(1+n)^3))/(8(1+n)^2)
C=2+2n

A=0.46791111376204392960....
B=1.6527036446661393023....
C=3.8793852415718167681....

 

How the heck did you do that? I did it another way but shouldn't have had cosine in there at all.

 

Did you come up with the same numbers for A,B and C?

 

(X^3) + (A+B+C)(X^2) + (AB+BC+AC)(X) + ABC = 0

For roots -A, -B, -C.

I never saw the cosines in the cubic equation... and still don't... So I was wondering how you went about solving it.

 

awww you already posted an answer and i thought i could solve it... jk

 

I started with A=3/BC
Then rewrote the 2nd equation:
(3/BC)B + BC + (3/BC) C = 9

Multiplied that equation by B, and then used the quadratic equation to find an expression for B. Then put that expression for B into (3/BC) + B + C = 6, and then rewrote that expression for C. It will simplify down pretty well, but there will be complex numbers in there. You can do a complex expansion to find the corresponding trig functions. C will then be the expression I wrote above. Then I put C's expression into the expression for B, and then both the expressions for B and C into the expression for A. Viola!

btw, where did you find this?

 

Someone had given it to me, and it took me a while to solve it. My peers couldn't solve it so I was wondering if someone on the internet could solve it.