Determinants, Area, and Barycentric Coordinates
Let the vertices of ΔABC have coordinates
which can also be written as
(1) |
Determinants are introduced in Linear Algebra and are mostly used for solving linear equations. No one likes Cramer's rule, and there are much simpler ways to solve a system of linear equations. However, there is nothing like determinants to compress long expressions into a very concise form. I'll need just the linearity property of determinants. Using boldface to denote whole columns of a determinant, linearity means that
(2) | | ka + mb c d | = k | a c d | + m | b c d | |
A determinant with two equal columns is zero which is only a very particular case of a much more general statement. Returning to the triangle ABC, let there be three points K_{1}, K_{2}, and K_{3} in the interior of ΔABC. I want to express the area of ΔK_{1}K_{2}K_{3} in terms of area(ΔABC).
Let the barycentric coordinates of the three points (u_{1}, v_{1}, w_{1}), (u_{2}, v_{2}, w_{2}), and (u_{3}, v_{3}, w_{3}) normalized so that, for
(*) | u_{i} + v_{i} + w_{i} = 1. |
Then, for
K_{i} = (u_{i}x_{A} + v_{i}x_{B} + w_{i}x_{C}, u_{i}y_{A} + v_{i}y_{B} + w_{i}y_{C})
Use now (1) to express the area of ΔK_{1}K_{2}K_{3} as a determinant. Apply repeatedly the properties of determinants to obtain the following nice formula
(3) |
Although (3) has been derived under the condition (*), the determinant in (3) is zero or not for all equivalent representations of the three points in the barycentric coordinates. The area of the triangle K_{1}K_{2}K_{3} is 0 iff the three points are collinear, i.e., all lie on the same straight line. For a generic point
(4) |
which is an equation of a straight line through two points in barycentric coordinates, that need not satisfy (*). Thus a straight line in the barycentric coordinates is given by a linear homogeneous equation
With the tools we now possess we can solve an interesting problem:
Given a point K inside a ΔABC, let E, D, F be the feet of the cevains through K on BC, AC, AB. Determine K such that the area of triangle DEF is as large as possible. (Solution)
Note that (4) supplies an algebraic proof of Menelaus' theorem. Indeed, assume there are three points A', B', C' on the sides of triangle ABC. In the barycentric coordinates the points can be written as
(5) |
A' = (0, v_{1}, w_{1}), B' = (u_{2}, 0, w_{2}), and C' = (u_{3}, v_{3}, 0). |
The corresponding determinant evaluates to
u_{3}· v_{1}· w_{2} + u_{2}· v_{3}· w_{1}. |
The three points are collinear iff the determinant is 0, which is equivalent to
u_{3}/u_{2}· v_{1}/v_{3}· w_{2}/w_{1} = -1. |
We can also prove a generalization of Ceva's theorem due to E. J. Routh (1896).
For simplicity, in (5), I'll use a somewhat different form of the barycentric coordinates:
(5') |
A' = (0, 1, λ), B' = (μ, 0, 1), and C' = (1, ν, 0). |
This is always possible. Note that the points A', B', C' are located on the side lines BC, AC and AB, respectively. In general, the three lines AA', BB' and CC' form a triangle whose area is established by Routh's theorem. The area of that triangle vanishes iff the three lines are concurrent which, as we shall see shortly, leads to Ceva's theorem.
In the barycentric coordinates (u, v, w) the lines AA', BB' and CC' have equations (see (4)):
AA': λv = w,
BB': μw = u
CC': νu = v.
Further, the lines AA' and BB' intersect in point
Theorem
If three points A'(0, 1, λ), B'(μ, 0, 1), C'(1, ν, 0) are selected on the sides BC, AC, and AB of triangle ABC of unit area. Then the area of the triangle formed by the three lines AA', BB', CC' is given by
(6) | (λμν - 1)^{2} / (μ + μν + 1)(1 + ν + λν)(λμ + 1 + λ). |
Ceva's theorem is an immediate consequence of (6).
Formula (6) helps us verify a more specialized result where
In passing, the area of triangle A'B'C' relative to that of ABC is given by
(7) | (λμν + 1) / (λ + 1)(μ + 1)(ν + 1). |
The latter formula implies Menelaus' theorem.
References
- G. Chang and Th. W. Sederberg, Over And Over Again, MAA 1997
- H. S. M. Coxeter, Introduction to Geometry, John Wiley & Sons, 1961
Barycenter and Barycentric Coordinates
- 3D Quadrilateral - a Coffin Problem
- Barycentric Coordinates
- Barycentric Coordinates: a Tool
- Barycentric Coordinates and Geometric Probability
- Ceva's Theorem
- Determinants, Area, and Barycentric Coordinates
- Maxwell Theorem via the Center of Gravity
- Bimedians in a Quadrilateral
- Simultaneous Generalization of the Theorems of Ceva and Menelaus
- Three glasses puzzle
- Van Obel Theorem and Barycentric Coordinates
- 1961 IMO, Problem 4. An exercise in barycentric coordinates
- Centroids in Polygon
- Center of Gravity and Motion of Material Points
- Isotomic Reciprocity
- An Affine Property of Barycenter
- Problem in Direct Similarity
- Circles in Barycentric Coordinates
- Barycenter of Cevian Triangle
- Concurrent Chords in a Circle, Equally Inclined
|Contact| |Front page| |Contents| |Geometry|
Copyright © 1996-2018 Alexander Bogomolny
Solution
From the definition, points D, E, and F have the following barycentric coordinates:
D(0, r_{a}/(r_{a}+1), 1/(r_{a}+1)),
E(1/(r_{b}+1), 0, r_{b}/(r_{b}+1)), and
F(r_{c}/(r_{c}+1), 1/(r_{c}+1), 0).
The determinant in (3) appears as
(11) |
which we should maximize subject to r_{a}r_{b}r_{c}=1. This is the same as minimizing
|Contact| |Front page| |Contents| |Geometry| |Up|
Copyright © 1996-2018 Alexander Bogomolny
64222880 |