Home > Shapes
| 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)
|
|
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:
Forums (0)
Newest Problems
Random Problem
FAQ |
About This Site
Site Statistics
New Comments (9)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On
Chatterbox:
|
