MHB Four Non-Linear Simultaneous Equations

[Wilmer]

4 equations, 4 unknowns:

\[\frac{u(r^2 - u^2)}{r^2 + u^2}=156~~~~~~~~~~(1)\]
\[\frac{v(r^2 - v^2)}{r^2 + v^2} = 96~~~~~~~~~(2)\]
\[\frac{w(r^2 - w^2)}{r^2 + w^2} = 63~~~~~~~~~~(3)\]
\[\frac{315uvw + 24336vw + 9216uw + 3969uv}{2r} = 943488~~~~~~~~~(4)\]

Who can solve that mess?

 

[Sudharaka]

4 equations, 4 unknowns:

u(r^2 - u^2) / (r^2 + u^2) = 156 [1]
v(r^2 - v^2) / (r^2 + v^2) = 96 [2]
w(r^2 - w^2) / (r^2 + w^2) = 63 [3]
(315uvw + 24336vw + 9216uw + 3969uv) / (2r) = 943488 [4]

Who can solve that mess?

Hi Wilmer, :)

If you are only concerned about real roots these are the solutions Maxima gives. Note that the first one is only an approximate.\[u=-166.2623906705539,v=-110.8238636363636,r=-29.66972878390201,w=-81.98732171156894\]

\[u=260,v=104,r=520,w=65\]

Kind Regards,
Sudharaka.

 

[Wilmer]

\[u=260,v=104,r=520,w=65\]

Hi Suds!
Yes, that's correct: I had this one solved,
but by brute force.

All I'm trying to do is find a way to solve
this "by hand".

 

[Opalg]

Hi Suds!
Yes, that's correct: I had this one solved,
but by brute force.

All I'm trying to do is find a way to solve
this "by hand".

I think that I may have been part way towards finding Sudharaka's solution "by hand". I was looking at the function $f(x) = \dfrac{x(1-x^2)}{1+x^2}.$ For positive values of $x$, this has a maximum value of 3/10, which occurs when $x=1/2.$

Your equation (1) says that $f(u/r) = 156/r.$ This tells you that $156/r\leqslant 3/10$, or $r\geqslant 520.$ Also, the value $r=520$ can only occur if $u=520/2=260$. I was going to explore this further, to see if there were values of $v$ and $w$ compatible with those values of $r$ and $u$, but Sudharaka got there first.

 

[Wilmer]

Background info, in case useful:

              C
 
 
        D                     E
              U          V                    
                   M
                   
                   W
                   
B                  F                        A

Acute triangle ABC: M is circumcenter.
U, V and W are the incenters of triangles BCM, ACM and ABM respectively:
and DM, EM and FM are the perpendicular heights.

NOT GIVENS: a = BC = 624, b = AC = 960, c = AB = 1008, r = 520 = AM=BM=CM.
NOT GIVENS: u = UM = 260, v = VM = 104, w = WM = 65.
GIVENS: d = DU = 156, e = EV = 96, f = FW = 63.

Work to set up the 4 equations:
a = 2dr / u , b = 2er / b , c = 2fr / w

from triangleBCM: u(r^2 - u^2) / (r^2 + u^2) = d [1]
from triangleACM: v(r^2 - v^2) / (r^2 + v^2) = e [2]
from triangleABM: w(r^2 - w^2) / (r^2 + w^2) = f [3]

area(BCM + ACM + ABM) = areaABC; leads to :
[uvw(d + e + f) + vwd^2 + uwe^2 + uvf^2] / (2r) = def [4]

Inserting the givens gives us:
u(r^2 - u^2) / (r^2 + u^2) = 156 [1]
v(r^2 - v^2) / (r^2 + v^2) = 96 [2]
w(r^2 - w^2) / (r^2 + w^2) = 63 [3]
(315uvw + 24336vw + 9216uw + 3969uv) / (2r) = 943488 [4]

I'm simply curious as to the possibility of solving these 4 simultaneous equations.