Systems of Equations degree 2

Question

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.

Answer 1

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:

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

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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.

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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.

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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:

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

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

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

And:

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

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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:

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

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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)

Answer 2

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&=32\tag{A}\\ x^2+z^2+xz&=16\tag{B}\\ y^2+z^2&=16\tag{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 $$ (x,y,z)=\left(8 \sqrt{\frac{7-2\sqrt{3}}{37}}, 4(3 + \sqrt3)\sqrt{\frac{7-2\sqrt3}{111}}, 4\sqrt{\frac{21 - 6\sqrt3}{37}}\right),\text{ and similar} $$ and it is easy to check $(xy+\sqrt{3}xz+2yz)^2=2^{10}$. Indeed, it follows immediately from noting the LHS of $$ \mathrm{(A)^2+(B)^2+(C)^2-2\times(A)\times(B)-2\times(B)\times(C)-2\times(C)\times(A)} $$ is $-(xy+\sqrt{3}xz+2yz)^2$, which I suppose is the only way to solve this within the 120 seconds limit.