Question

If $P(x,y,z)$ is the point nearest to the origin lying on the intersection of the surfaces $z=xy+5$ and $x+y+z=1$, then the distance (in integer) between the origin and $P$ is __________.

Hint:

On intersections of smooth surfaces, minimize $\|x\|^2$ with Lagrange multipliers. Symmetry checks can quickly rule out candidate cases.

Answer 1

Correct Answer: 3

Minimize $f(x,y,z)=x^2+y^2+z^2$ subject to \[ g_1(x,y,z)=z-xy-5=0, \qquad g_2(x,y,z)=x+y+z-1=0. \] Use Lagrange multipliers: $\nabla f=\lambda \nabla g_1+\mu \nabla g_2$. \[ (2x,2y,2z)=\lambda(-y,-x,1)+\mu(1,1,1). \] From the first two coordinates, \[ 2x+\lambda y=\mu,\qquad 2y+\lambda x=\mu \Rightarrow (2-\lambda)(x-y)=0. \] If $x=y$, then from constraints $z=1-2x$ and $z=x^2+5$ give $x^2+2x+4=0$, impossible; hence $\lambda=2$. Then from the third coordinate $2z=\lambda+\mu=2+\mu\Rightarrow \mu=2z-2$. Using $2x+2y=\mu$ gives $x+y=z-1$. With $x+y+z=1$ we get $z=1$ and $x+y=0$. Using $z=xy+5$ yields $xy=-4$; together with $y=-x$ gives $x=\pm2$, $y=\mp2$. Thus $P$ is $(2,-2,1)$ or $(-2,2,1)$ and \[ \|OP\|=\sqrt{2^2+(-2)^2+1^2}=\sqrt{9}= \boxed{3}. \]