Given $A(0,0,1),B(2,1,2)$ be rwo points in a 3_D space and $P(a,b),a,b>0$ be a point on the xy plane such that $\angle APB$ is maximum then find $a,b$,(A,P,B are non collenear)
This question was given in a mock exam I was writing. It is clear that the plane passing through $A,P,B$ will be perpendicular to the $x-y$ plane.So after some work we can conclude the plane is $x-2y=0$ thus $a=2b$.Also the cosine of the angle $<APB$ is
$$\cos (\angle APB)=\frac{2b(2b-2)+b(b-1)+2}{\sqrt{(5b^2+1)(5{(b^2-2b+1)+4)}}}=L$$ so it remains to minimize $L$.But that will be very tedious task especially in an entrance exam mock test.So what is the best way to solve this problem?
If you draw a circumcircle of $ABP$ with the centre $O$, one can see that $\angle APB$ is maximal when $OP\perp xy$.
If $P$ has coordinates $(2y,y,0)$, then $O$ has coordinates $(2y,y,z)$ and we can write $AO^2=OP^2$ and $BO^2=OP^2$:
$$4y^2+y^2+(z-1)^2 = z^2, \\ 4(y-1)^2+(y-1)^2+(z-2)^2=z^2$$
It's easy to eliminate $z$, solve a quadratic equation on $y$ and choose the root $0<y<1$.