Xrfgtjtk

Given the system
$$begin{aligned}
x^2+y^2+frac{sqrt 3}{2}xy&=32\
x^2+z^2+frac{1}{2}xz&=16\
y^2+z^2&=16,end{aligned}$$
find the value of $(xy+sqrt{3}xz+2yz).$

The answer is $32,$ so I think there will be a really nice solution to this.

I tried completing the square for each pair $x$ and $y$ but I could not find anything, I tried adding all three equations but I got nothing.

I guess the two factors (1/2) shouldn't be there, which mean we have instead
begin{align}
x^2+y^2+sqrt{3}xy&=32tag{A}\
x^2+z^2+xz&=16tag{B}\
y^2+z^2&=16tag{C}.
end{align}
The rationale being that it gives a much nicer set of coefficients when completing the square
begin{align}
(x+frac{sqrt3}2y)^2+frac14y^2&=32\
(x+frac12z)^2+frac34z^2&=16\
end{align}
and we have accidental cancellation when eliminating $y,z$, giving the biquartic $37x^4-896x^2+4096=0$, and hence a nice form of the four real solutions
$$
(x,y,z)=left(8 sqrt{frac{7-2sqrt{3}}{37}},
4(3 + sqrt3)sqrt{frac{7-2sqrt3}{111}},
4sqrt{frac{21 - 6sqrt3}{37}}right),text{ and similar}
$$
and it is easy to check $(xy+sqrt{3}xz+2yz)^2=2^{10}$. Indeed, it follows immediately from noting the LHS of
$$
mathrm{(A)^2+(B)^2+(C)^2-2times(A)times(B)-2times(B)times(C)-2times(C)times(A)}
$$
is $-(xy+sqrt{3}xz+2yz)^2$, which I suppose is the only way to solve this within the 120 seconds limit.

Possible hint:

Let's introduce spherical coordinates:

$$x=r cos a cos b \ y=r sin a cos b \ z= r sin b$$

Then we have:

$$r^2 cos^2 b left(1+frac{sqrt{3}}{2}sin a cos a right)=32 \ r^2 left(cos^2 a cos^2 b+sin^2 b+frac{1}{2}cos a sin b cos b right)=16 \ r^2 (sin^2 a cos^2 b+sin^2 b)=16$$

And we need to find:

$$r^2 left(sin a cos a cos^2 b+(sqrt{3} cos a +2 sin a)cos b sin b right)$$

Let's rewrite the last equation as: $$r^2 cos^2 b (sin^2 a-1)=16-r^2 \ cos^2 a cos^2 b=frac{r^2-16}{r^2}$$

So we need to have $r^2 >16$ for a non-trivial and real solution.

The trivial solution $r^2=16$ doesn't work, because then we have either $cos a=0$ or $cos b=0$, which, after substitution, don't satisfy the whole system of equations.

Edit

I have used Newton's method to solve the system numerically. With the initial guess $(1,2,3)$, I get:

$$x=3.12392804494377 \ y=3.55350700146376 \ z=1.83646072393286$$

And:

$$xy+sqrt{3} xz+2yz=34.0893777894208 neq 32$$

With another initial guess $(-1,2,3)$, I get:

$$x=-3.70737480877305 \ y=-2.95892185457043 \ z=2.69161317030243$$

And:

$$xy+sqrt{3} xz+2yz=-22.2425349920001 neq 32$$

I think those are the only real solutions, because for the polynomial of Dr. Sonnhard Graubner $$9,{x}^{8}-735,{x}^{6}+21328,{x}^{4}-253952,{x}^{2}+1048576=0$$

we have only $4$ real roots:

$$x_{1,2}= pm 3.12392804494377 \ x_{3,4}=pm 3.70737480877305$$

Here's the R code I used (you would need the matrixcalc package to find the inverse Jacobian matrix):

library(matrixcalc);

f1 <- function(x,y,z){y^2+z^2-16};

f2 <- function(x,y,z){x^2+z^2+x*z/2-16};

f3 <- function(x,y,z){x^2+y^2+sqrt(3)/2*x*y-32};

J <- function(x,y,z){matrix(c(0, 2*y, 2*z, 2*x+z/2, 0, 2*z+x/2, 2*x+sqrt(3)*y/2, 2*y+sqrt(3)*x/2, 0), nrow=3, byrow=TRUE)}

Nm <- 18;

r0 <- c(1,2,3);

n <- 0;

while(n < Nm){n <- n+1;

        r0 <- r0 - matrix.inverse(J(r0[1],r0[2],r0[3])) %*% c(f1(r0[1],r0[2],r0[3]),f2(r0[1],r0[2],r0[3]),f3(r0[1],r0[2],r0[3]))};

x <- r0[1];

y <- r0[2];

z <- r0[3];

f <- function(x,y,z){x*y+sqrt(3)*x*z+2*y*z};

paste(r0)

paste(f(x,y,z))

f1(x,y,z)

f2(x,y,z)

f3(x,y,z)