Viviani by Inversion

Ptolemy's theorem can be proved by inversion from a simple identity AB + BC = AC, where point B is between A and C and all three are collinear. By exactly same reasoning we can prove another non-trivial statement:

Let point M lie on the arc between vertices A1An of the circumcircle of a regular n-gon A1A2...An. Let dj denote the distance from M to Aj, j = 1, ..., n. Then

(1)
1

d1d2
+
1

d2d3
+
1

d3d4
+ ... +
1

dn-1dn
=
1

d1dn

Solution

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Copyright © 1996-2017 Alexander Bogomolny

(1)
1

d1d2
+
1

d2d3
+
1

d3d4
+ ... +
1

dn-1dn
=
1

d1dn

Solution

Perform inversion f with center M and radius 1. Let f(Aj) = Bj, j = 1, ..., n. Then by the distance formula:

BjBj+1 = AjAj+1/MAjMAj+1 = a/djdj+1,

where a is the side length of the given n-gon.

Points B are all collinear stretching successively from B1 to Bn, implying that

B1B2 + B2B3 + ... + Bn-1Bn = B1Bn.

Substitution now gives the desired identity (1).

Note: For n = 3, we may multiply by the product d1d2d3 to obtain

d3 + d1 = d2,

which is Viviani's Theorem.

References

  1. I. M. Yaglom, Geometric Transformations IV, MAA, 2009

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  • Equilateral Triangle on Parallel Lines
  • Equilateral Triangle on Parallel Lines II
  • When a Triangle is Equilateral?
  • Viviani's Theorem
  • Viviani's Theorem (PWW)
  • Tony Foster's Proof of Viviani's Theorem
  • Viviani in Isosceles Triangle
  • Viviani by Vectors
  • Slanted Viviani
  • Slanted Viviani, PWW
  • Morley's Miracle
  • Triangle Classification
  • Napoleon's Theorem
  • Sum of Squares in Equilateral Triangle
  • A Property of Equiangular Polygons
  • Fixed Point in Isosceles and Equilateral Triangles
  • Parallels through the Vertices of Equilateral Triangle
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    Copyright © 1996-2017 Alexander Bogomolny

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