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?

2 Answers
Nov 8, 2016

Answer:

#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#

Nov 8, 2016

Answer:

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,

enter image source here