Let the letter that denotes a circle also denote its center. Let the radius of
circle O be r and R the common radius of the three circles. Consider the
triangles AA'P and AOP:
|PA'|2+|A'A|2 = |PA|2 = |PO|2+|OP|2-2|PO||OP|cos(∠AOP)
or
a2+R2 = r2+(r+R)2-2r(r+R)cos(∠AOP)
or
a2 = 2r(r+R)[1-cos(∠AOP)]
or
a2 = 4r(r+R)sin(∠AOP/2)2
WOLOG assume 4r(r+R) = 1. Then
a = ±sin(∠AOP/2)
A similar argument gives
b = ±sin(∠BOP/2)
and
c = ±sin(∠COP/2)
Where all angles are directed angles (CCW is positive).
Letting t = ∠AOP and with ∠AOB = ∠COA = 120° we get
a(t) = ±sin(t/2)
b(t) = ±sin(t/2-60°)
c(t) = ±sin(t/2+60°)
With t∈[0°,120°] and a(t),b(t),c(t) ≥ 0; we get
a(t) = sin(t/2)
b(t) = -sin(t/2-60°) = cos(t/2)sin(60°) - sin(t/2)cos(60°)
c(t) = sin(t/2+60°) = sin(t/2)cos(60°) + cos(t/2)sin(60°)
Therefore,
a(t)+b(t)-c(t) = sin(t/2)[1-2*cos(60°] = 0
It is easily checked that
b(t)+c(t)-a(t) = 0 if t∈[120°,240°]
and
c(t)+a(t)-b(t) = 0 if t∈[240°,360°]
Therefore,
[a(t)+b(t)-c(t)]*[b(t)+c(t)-a(t)]*[c(t)+a(t)-b(t)] = 0
if t∈[0°,360°).
QED
|
