Relations in a Cyclic Polygon

The results below are due to Bui Quang Tuan, a contemporary Vietnamese mathematician.

Starting with one of his theorems concerning two cyclic n-gons with a common circumcircle we establish a special case where one of the polygons shrinks into a point.

Given point P and n points P1, P2, ..., Pn (n > 1) on the circle (O) with diameter d. s1, s2, ..., sn are lengths of segments PP1, PP2, ..., PPn respectively while d1, d2, ..., dn are distances from P to segments P1P2, P2P3, ..., PnP1 respectively. Then

(s1·s2· ... ·sn)2 = d1·d2· ... ·dn·dn.

(For n=3, this has been shown separately.) In general, we thus have

Theorem 1

If P, P1, P2, ... Pn (n > 1) are concyclic on a circle with diameter d then

(1) (s1·s2· ... ·sn)2 = d1·d2· ... ·dn·dn

where, for i = 1, ..., n, si is the length of the segment from PPi, di is the distance from P to the side Pi i+1, where indices are counted cyclically 1, ..., n, n+1 = 1.

Adding some verbiage, (1) reads:

Assume a point and vertices of an n-gon are concyclic. The square of the product of all the segments from the point to all the vertices of the polygon is equal to the product of all the distances from the point to all sides of the polygon multiplied by the nth power of the diameter of the circle.

Of course, there are many polygons with the same set of vertices, and the distances from a point to the sides of a polygon vary from one polygon to another. However, obviously, all such polygons share the same set of segments PP1, PP2, ..., PPn and hence the product of such segments from P to the n vertices in the left-hand side of (1) does not depend on the selection of the polygon. This leads to

Theorem 2

Under the assumptions of Theorem 1, the product of all distances from a point to the sides of a polygon with a given set of vertices is independent of the polygon:

d1·d2· ... ·dn = constant.

Of course some polygons will share some sides as well, not only the vertices. We look into a special example with n = 4.

Consider the following three polygons with vertices P1, P2, P3, P4:

  1. P1P2P3P4

    • sides: P1P2, P2P3, P3P4, P4P1
    • distances from P to these sides: d12, d23, d34, d41.

  2. P1P2P4P3

    • sides: P1P2, P2P4, P4P3, P3P1
    • distances from P to these sides: d12, d24, d43, d31

  3. P2P3P1P4

    • sides: P2P3, P3P1, P1P4, P4P2
    • distances from P to these sides: d23, d31, d14, d42.

Obviously in our notations di j = dj i.

By our theorems,

d12·d23·d34·d41 = d12·d24·d43·d31 = d23·d31·d14·d42.

From these we can have following other equal products of distances:

d12·d34 = d14·d23 = d13·d24.

This result can be formulated as

Theorem 3

Under the assumptions of Theorem 1, for n = 4, the products of the distances from a point on the circumcircle of a cyclic quadrilateral to a pair of opposite sides is the same as for the other pair and for the diagonals.

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