 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Find Points, Maximise Ratio (Posted on 2015-06-05) 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.

 No Solution Yet Submitted by K Sengupta Rating: 4.0000 (1 votes) Comments: ( Back to comment list | You must be logged in to post comments.) Some analytical thoughts Comment 6 of 6 | I agree with Daniel & Charlie that triangle MNR is isosceles:

ratio = (NM + MR)/NR = (sin/R + sin/N)/sin/M

= (sin/R + sin/N)/(sin(/R + /N)

This symmetric function in /R and /N will be maximum

when /R = /N, giving NM = MR, so that R lies on the circle, C,

in X, with centre at M and radius MR.

Using the x-y plane as X, with the origin, O, at the projection
of N on to this plane, let N have coordinates (0, 0, n) and let
M be the point (m, 0, 0). Then R lies on the circle with centre
M and radius r = sqrt(n^2 + m^2).

Since NM and therefore MR are fixed lengths, the ratio
(NM + MR)/NR will be maximum when NR is minimum, i.e.
when R lies on the line MO produced, at which point

OR = r – MO = r – m      where r = sqrt(n2 + m2)

and the ratio is then equal to  2r/sqrt(n2 + (r – m)2)

which simplifies to:         ratio = sqrt(2/(1 – m/r))

With Charlie’s values of n = 3 and m = 5, these formulae

give      OR = sqrt(34) – 5  ~= 0.830952

and       ratio = sqrt(2(34 + 5*sqrt(34)))/3 ~= 3.74625

as computed by Charlie.

 Posted by Harry on 2015-06-06 16:31:27 Please log in:

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