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?

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- January 4th 2013, 10:09 AMWilmer4 equations
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? - January 4th 2013, 03:14 PMzhandeleRe: 4 equations
Did you notice that all the integers are divisible by three? From equation four, we know that r is also divisible by three.

- January 4th 2013, 04:50 PMWilmerRe: 4 equations
- January 5th 2013, 08:18 AMBobPRe: 4 equations
Hi Wilmer, can you give us some clue as to background/origin ?

- January 5th 2013, 08:54 AMWilmerRe: 4 equations
Thanks Bob. Best I can do diagramwise(!):

Code:`C`

D E

U V

M

W

B F A

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.