Mathematics and Origami

Circumference tangential equation:^222

1

r

u +v = (6)

,, Cartesian ,, ,, x^2 +y^2 =r^2

In Fig. 4 we can see the circumference of center O and radius OP represented by several
tangents (ui,vi).
(6) can be verified in Fig. 3:

OA

u

− 1

= ;

OB

v

− 1

=

2 2 2

1 1 1

OA OB OP

+ =

()

2 2

(^221)
OA OB OP
OA OB

×

• ; AB^2 ×OP^2 =()OA×OB^2
  AB×OP=OA×OB ;
  OP
  OB
  AO
  AB
  = which is true after the similarity of ∆AOB and BPO.
  In case one would want to make similar checks with the rest of the conics, here we have
  their Plückerian equations with respect to the origin, while keeping their characteristic pa-
  rameters:
  Ellipse: 1a^2 u^2 +b^2 v^2 =
  Parabola: pv^2 = 2 u
  Hyperbola: a^2 u^2 −b^2 v^2 = 1
  O A
  B
  v Y
  t
  X
  u
  P
  4
  (u 1 ,v 1 )
  (u 3 ,v 3 ) (u 2 ,v 2 )
  Interlude