# Given the system {(x+y+z=a),(x^2+y^2+z^2=b^2),(xy=z^2):} determine the conditions over a,b such that x,y,z are distinct positive numbers?

Nov 8, 2016

$c \le \frac{a}{\sqrt{3}} \le b$, for the third surface taken as $x y = {c}^{2}$. Instead, for (c-absent) $x y = {z}^{2}$, the answer is $\frac{a}{\sqrt{3}} \le b$

#### Explanation:

In the first octant (${O}_{1}$), the plane $x + y + z = a$ is just an equilateral

triangular area, with vertices (a, 0, 0), (0, a, 0) and (0, 0, a). The

section of this area by the sphere ${x}^{2} + {y}^{2} + {z}^{2} = {b}^{2}$ exists, if and

only if the length of the perpendicular from the origin on this plane

$\frac{a}{\sqrt{3}} \le b$.

Consider the third as $x y = {c}^{2}$. It represents a rectangular

hyperbolic cylinder. The part of this in ${O}_{1}$ exists from and

beyond the plane $x + y = \sqrt{2} c$.

For the hyperboloid to meet the section of the other two,

the length of the perpendicular from the origin O on this plane ( that

touches the hyperboloid ).,

$c \le \frac{a}{\sqrt{3}} \le b$.

Instead, for (c-absent) $x y = {z}^{2}$, the answer is $\frac{a}{\sqrt{3}} \le b$

Under this condition, $z \ge 0$ is like the parameter c for $x y = {c}^{2}$.

Despite that $x y = {z}^{2}$ is a single surface and $x y = {c}^{2}$ is a family of

surfaces, it is relevant that the section of $x y = {z}^{2}$ by the plane

z=c is the RH $x y = {c}^{2}$

Nov 8, 2016

See below.

#### Explanation:

Substituting $x y = {z}^{2}$ into ${x}^{2} + {y}^{2} + {z}^{2} = {b}^{2}$ we get at

${x}^{2} + {y}^{2} + x y = {b}^{2}$

$x + y = a - z \to {\left(x + y\right)}^{2} = {\left(a - z\right)}^{2}$ so

${x}^{2} + {y}^{2} + 2 x y = {z}^{2} - 2 a z + {a}^{2}$ or substituting $x y = {z}^{2}$

${x}^{2} + {y}^{2} + x y = {a}^{2} - 2 a z$ so we have obtained

${b}^{2} = {a}^{2} - 2 a z$ and $z = \frac{{a}^{2} - {b}^{2}}{2 a}$

so

$x + y = a - \frac{{a}^{2} - {b}^{2}}{2 a} = \frac{{a}^{2} + {b}^{2}}{2 a}$

but

$x y = {z}^{2} = {\left({a}^{2} - {b}^{2}\right)}^{2} / \left(4 {a}^{2}\right)$

Solving the polynomial

${\gamma}^{2} - \left(x + y\right) \gamma + x y = 0$ or equivalently

${\gamma}^{2} - \frac{{a}^{2} + {b}^{2}}{2 a} \gamma + {\left({a}^{2} - {b}^{2}\right)}^{2} / \left(4 {a}^{2}\right) = 0$

we have

$\gamma = \frac{{a}^{2} + {b}^{2}}{4 a} \pm \frac{\sqrt{\left(3 {b}^{2} - {a}^{2}\right) \left(3 {a}^{2} - {b}^{2}\right)}}{4 a}$

$\gamma$ represents here the two solutions $x , y$ so the conditions for realness are:

$\left(3 {b}^{2} - {a}^{2}\right) \left(3 {a}^{2} - {b}^{2}\right) \ge 0$. We let to the reader as an exercise, to work out those conditions.

Attached a figure showing the surfaces,