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 P_{1}, P_{2}, ..., P_{n}
(For n=3, this has been shown separately.) In general, we thus have
Theorem 1
If P, P_{1}, P_{2}, ... P_{n}
(1) | (s_{1}·s_{2}· ... ·s_{n})^{2} = d_{1}·d_{2}· ... ·d_{n}·d^{n} |
where, for
Adding some verbiage, (1) reads:
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 PP_{1}, PP_{2}, ..., PP_{n} 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:
Of course some polygons will share some sides as well, not only the vertices. We look into a special example with
Consider the following three polygons with vertices P_{1}, P_{2}, P_{3}, P_{4}:
P_{1}P_{2}P_{3}P_{4}
- sides: P_{1}P_{2}, P_{2}P_{3}, P_{3}P_{4}, P_{4}P_{1}
- distances from P to these sides: d_{1}_{2}, d_{2}_{3}, d_{3}_{4}, d_{4}_{1}.
P_{1}P_{2}P_{4}P_{3}
- sides: P_{1}P_{2}, P_{2}P_{4}, P_{4}P_{3}, P_{3}P_{1}
- distances from P to these sides: d_{1}2, d_{2}4, d_{4}_{3}, d_{3}_{1}
P_{2}P_{3}P_{1}P_{4}
- sides: P_{2}P_{3}, P_{3}P_{1}, P_{1}P_{4}, P_{4}P_{2}
- distances from P to these sides: d_{2}_{3}, d_{3}_{1}, d_{1}_{4}, d_{4}_{2}.
Obviously in our notations
By our theorems,
From these we can have following other equal products of distances:
This result can be formulated as
Theorem 3
Under the assumptions of Theorem 1, for
|Contact| |Front page| |Contents| |Geometry|
Copyright © 1996-2018 Alexander Bogomolny64658856 |