Let A, B, and C be circles with equal radii and tangent externally
pairwise. Let O be the circle tangent externally to A, B, and C and
P any point on O. For X∈{A,B,C}, let X' be one of the two points
on X such that line PX' is tangent to X and let x = |PX'|.

Prove that (a+b-c)(b+c-a)(c+a-b) = 0.
  

Let the centres of circles A and O be points D and E, and
their radii be r and R respectively. Denote angle PED by Q_A.

The cosine rule in triangle PED and Pythagoras in PAD

give:   a² = |PD|² – r²

             =  R² + (R – r)² - 2R(R – r)cos(Q_A) – r²

            =  2R(R – r)(1 – cos(Q_A))

            =  4R(R – r) sin²(Q_A/2)

Thus     a = k sin(Q_A/2)               where k² = 4R(R – r)

By similar reasoning, b = k sin(Q_B/2) and c = k sin(Q_C/2)

where Q_A, Q_B and Q_C are angles subtended at the centre of O

by arcs extending from P to the tangent points of O with A, B
and C respectively.
If P lies on the minor arc between contacts with A and B

then    Q_B = 120^o – Q_A   and   Q_C = 240^o – Q_A   which give  

   c - b   =  k(sin(120^o – Q_A/2) - sin(60^o – Q_A/2))

            = 2k(cos(90^o – Q_A/2)sin(30^o))

            = k sin(Q_A/2)  =  a,   giving  a + b – c = 0 .

By cycling a, b, c, it follows that whichever arc P lies on, one
of the three factors will be zero giving

            ((a + b – c)(b + c – a)(c + a – b) = 0

By now the reader will be as convinced as I am that geometry
would provide a better proof! We look to Bractals for closure.

Posted by Harry on 2014-06-07 12:32:40