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Euler's Formula and Poncelet Porism

Euler's formula that relates the circumradius, the inradius and the distance between the circumcenter and the incenter of a triangle serves the basis for the Poncelet porism for triangles.

The circumradius (R), the inradius (r) and the distance between the circumcenter and the incenter (d) of a triangle stand in an elegant relationship

  1/(R - d) + 1/(R + d) = 1/r,

which we shall prove in the form

(1) 2Rr = R2 - d2.

Let ABC be a triangle with circumcenter O and incenter I. Extend AI to meet the circumcircle the second time at K. Extend OK and OI to obtain two diameters UV and KK' of the circumcricle.

 

By the Intersecting Chords Theorem,

(2) AI·IK = IU·IV = (R + d)(R - d) = R2 - d2.

Next, in ΔICK, CIK is external to ΔACI. Thus, if angles of ΔABC are denoted α, β, and γ, we see that

  CIK = (α + γ)/2.

On the other hand,

 
ICK= BCI + BCK
 = γ/2 + BAK
 = γ/2 + α/2.

It follows that CIK = ICK making ΔICK isosceles so that

(3) IK = KC.

Let Z be the point of tangency of AB and the incircle. ΔAIZ is right with angle at A equal α/2. ΔCKK' is also right with angle at K' equal CAK = α/2. The two triangles are similar, from which we get a proportion:

  AI / IZ = KK' / KC.

This is the same as

(4) AI·KC = IZ·KK' = r·2R

The combination of (2), (3), and (4) yields (1). (Elsewhere there is a different proof of the formula.)

Note that a similar result holds for excircles, for example,

(5) 2Rra = da2 - R2,

where ra is the A-excircle and da the distance between the circumcenter and the A-excenter.

 

The proof must be slightly modified. For example, instead of the Intersecting Chords Theorem, we make use of the Intersecting Secants Theorem.

 
(da + R)(da - R)= IaU · IaV
 = IaA · IaK
 = IaA · KC
 = (IaZ · KK'/KC) · KC
 = ra · 2R.

The identity can be written in an elegant way:

(6) 1/(d - R) - 1/(d + R) = 1/r.

In addition,

 
IIa2= 4R(ra - r), and
IbIc2= 4R(rb + rc).

To see that, apply the formula for the length of a median to triangles OIIa and OIbIb:

 
OI2 + OIa2= 2 OK2 + IIa2/2, and
OIb2 + OIc2= 2 OK'2 + IbIb2/2

and use the just derived formulas for OI2, OIa2, and similar.

By reversing the argument, we can establish Poncelet porism for triangles.

Poncelet Porism

If the distance d between the centers of two circles O(R) and I(r) satisfies (1), an infinite number of triangles may be circumscribed about the circle I(r) which also be inscribed in the circle O(R).

Indeed, from any point A on O(R) draw two chords AB and AC tangent to the circle I(r). The line AI bisects angle BAC, and it follows from (1) that

  AI · IK = 2Rr.

From similar triangles AIZ and CKK',

  AI · CK = 2Rr,

and therefore IK = KC making I the incenter of ΔABC.

Note: in a similar manner, (6) insures that a porism exists also in case where the incircle is replaced by an excircle.

References

  1. N. Altshiller-Court, College Geometry, Dover, 1980, pp. 85-87
  2. J. Casey, A Sequel to Euclid, Scholarly Publishing Office, University of Michigan Library (December 20, 2005), reprint of the 1888 edition, pp. 107-110
  3. J. L. Coolidge, A Treatise On the Circle and the Sphere, AMS - Chelsea Publishing, 1971, p. 45
  4. F. G.-M., Exercices de Géométrie, Éditions Jacques Gabay, sixiéme édition, 1991, pp. 837-839
  5. R. A. Johnson, Advanced Euclidean Geometry (Modern Geometry), Dover, 1960, p. 172
  6. T. Lalesco, La Géométrie du Triangle, Éditions Jacques Gabay, sixiéme édition, 1987

Poncelet Porism

Copyright © 1996-2008 Alexander Bogomolny

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