Given a plane X, a point M in the plane and a point N not in the plane.
Find all points R in X such that the ratio (NM + MR)/NR is the maximum.
Let a be then angle MRN
b the angle MNR
and c the angle NMR
then from law of sines we have
sin(a)/MN=sin(b)/MR=sin(c)/NR
then (MN+MR)/NR = (MN/NR)+(MR/NR)
= [sin(a)+sin(b)]/sin(c)
this is maximized when a=b thus we need
MR=NR
Now rotate the plane in question so it coincides with the xy plane. And flip if needed so M is above the plane.
Now let the points be at the following coordinates:
N (a,b,0)
M (p,q,r)
R (x,y,0)
then we need
MR=NR
MR^2=NR^2
(x-p)^2+(y-q)^2+r^2=(p-a)^2+(q-b)^2+r^2
(x-p)^2+(y-q)^2=(p-a)^2+(q-b)^2
this is the same as saying that R needs to be equidistant from N as a point at (p,q,0)
(p,q,0) is simply the point M projected onto the chosen plane.
Thus the locus of R is simply a circle on the plane with center at N and radius equal to the distance from N to the projection of M onto the plane.
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Posted by Daniel
on 2015-06-05 13:02:15 |
solution
