Let consider a surface $z=1/(xy)$ and a sphere defined by $(x-1.5)^2+(y-1.5)^2+(z-1.5)^2=3/4$. The sphere touches the surface at (1,1,1). Is it possible to prove that point (1,1,1) is the only intersection based on the fact that this surface is convex and sphere's radius is not large as given?
$\begingroup$
$\endgroup$
20
-
1$\begingroup$ Decent question, wrong forum. $\endgroup$– Lev BorisovCommented Apr 6, 2016 at 13:36
-
$\begingroup$ Consider the polynomial of degree three with roots $x,y,z$. The information given implies that it is equal to $f(t)=t^3-st^2 +(s^2/2 -3s/2 + 3)t -1$. Three roots are real iff the discriminant is nonnegative. This gives you a polynomial inequality on $s$, which you presumably can check. $\endgroup$– Lev BorisovCommented Apr 6, 2016 at 13:56
-
$\begingroup$ I am getting discriminant to be $(1/4)(s-3)^2(-s^4 + 6s^3 - 24s^2 +56s - 60)$. The remaining polynomial of degree $4$ is negative for all $s$. $\endgroup$– Lev BorisovCommented Apr 6, 2016 at 14:21
-
$\begingroup$ Somebody of the usual set "put on hold" may prove it? Will be thankful... $\endgroup$– SergeiCommented Apr 6, 2016 at 14:40
-
$\begingroup$ Certainly, computer gives accurate enough coordinates of four roots to know that they are complex. If that's not enough for you, try to write it as opposite of sum of squares of quadratic polynomials. $\endgroup$– Lev BorisovCommented Apr 6, 2016 at 14:45
|
Show 15 more comments