Question

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?

Answer 1

`c<=a/sqrt 3<=b`, for the third surface taken as `xy=c^2`. Instead, for (c-absent) `xy = z^2`, the answer is `a/sqrt 3<=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

`a/sqrt 3<=b`.

Consider the third as `xy=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<=a/sqrt 3<=b`.

Instead, for (c-absent) `xy = z^2`, the answer is `a/sqrt 3<=b`

Under this condition, `z>=0` is like the parameter c for `xy=c^2`.

Despite that `xy=z^2` is a single surface and `xy=c^2` is a family of

surfaces, it is relevant that the section of `xy=z^2` by the plane

z=c is the RH `xy=c^2`

Answer 2

See below.

Explanation:

Substituting `xy=z^2` into `x^2+y^2+z^2=b^2` we get at

`x^2+y^2+xy=b^2`

`x+y=a-z->(x+y)^2=(a-z)^2` so

`x^2+y^2+2xy=z^2-2az+a^2` or substituting `xy=z^2`

`x^2+y^2+xy=a^2-2az` so we have obtained

`b^2 = a^2-2az` and `z = (a^2-b^2)/(2a)`

so

`x+y=a- (a^2-b^2)/(2a)=(a^2+b^2)/(2a)`

but

`xy=z^2= (a^2-b^2)^2/(4a^2)`

Solving the polynomial

`gamma^2-(x+y)gamma+xy=0` or equivalently

`gamma^2-(a^2+b^2)/(2a)gamma+(a^2-b^2)^2/(4a^2)=0`

we have

`gamma = (a^2+b^2)/(4a)pmsqrt((3 b^2-a^2) (3 a^2 - b^2))/(4a)`

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

`(3 b^2-a^2) (3 a^2 - b^2)ge 0`. We let to the reader as an exercise, to work out those conditions.

Attached a figure showing the surfaces,