Did you notice that all the integers are divisible by three? From equation four, we know that r is also divisible by three.
Thanks Bob. Best I can do diagramwise(!):
Acute triangle ABC: M is circumcenter.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.