Areas of Conic Quadrilaterals

A quadrilateral is said to be conic for a given eccentricity if all four vertices lie on a conic locus with that eccentricity. A special case of this is with zero eccentricity, in which case the conic is a circle and the quadrilateral is said to be cyclic. Brahmagupta’s formula gives the area of a cyclic quadrilateral in terms of the usual metrical lengths of the four edges, but that formula is actually just a special case of a formula for the area of any conic quadrilateral in terms of the generalized “lengths” of the four edges, where the metric used to define the “lengths” corresponds to the character of the conic.

Given the Cartesian coordinates (x_n,y_n), n = 1, 2, …, N of the vertices of a N-gon, the signed area of the N-gon is easily computed by the formula

From this can be derived several interesting algebraic relations between the vertex coordinates and the area of the polygon, such as Heron’s formula for the area of a triangle and Brahmagupta’s formula for the area of a quadrilateral, both in terms of the lengths (i.e., the root-sum-squares of the coordinate differences) of the edges, provided the vertices lie on a circle. In the following we explain how those formulas can be generalized to give the areas in terms of generalized edge lengths, showing that the formulas of Heron and Brahmagupta are actually just useful special cases of more general algebraic identities.

Given a set of points P₁, P₂, P₃, … on a plane with some Cartesian coordinate system, let (x_j,y_j) denote the coordinates of the jth point, and define the quantities

Thus s_mn is simply the ordinary metrical distance from P_m to P_n.  In terms of these quantities Heron’s formula for the squared area of the triangle with vertices P₁, P₂, P₃ can be written as

For convenience we can define s = (s₁₂ + s₂₃ + s₃₁)/2, so Heron’s formula can be written in the form

This is just a special case of a more general algebraic identity. For any arbitrary constants a,b,c we can define the generalized “distances”

In terms of these quantities, and again putting σ = (σ₁₂ + σ₂₃ + σ₃₁)/2, the squared area of the triangle with vertices P₁, P₂, P₃ is given by

This equation is identical to Heron’s formula, except for the denominator of the right side, which is the discriminant of the quadratic form used in the definition of the σ_mn values. In the case of Heron’s traditional formula we have a = c = 1 and b = 0, so this discriminant is +1, and hence the factor is unnoticed. On the other hand, if we define the σ_mn with a = c = 0 and b = 1, the discriminant equals −1, and hence the squared area is the negative of the right hand side. In general, we can base the quantities σ_mn on any quadratic function we choose, and compute the area using the generalized version of Heron’s formula given by equation (4).

Now, a natural extension of Heron’s formula, originally due to Brahmagupta, is to consider a quadrilateral with vertices P₁, P₂, P₃, P₄. If we define s = s₁₂ + s₂₃ + s₃₄ + s₄₁, Brahmagupta says the area of the quadrilateral is

However, this formula is understood to have restricted applicability, because it is valid only for quadrilaterals whose vertices all lie on a circle. These are called cyclic quadrilaterals, as illustrated in the figure below.

Naturally Brahmagupta’s formula reduces to Heron’s formula if any two of the vertices coincide, in which case the polygon reduces to a triangle. (Every triangle is cyclic in the sense that it can be inscribed in a circle.)

As we might expect, for any constants a,b,c we can substitute the generalized distances σ_mn in place of the s_mn values, and divide the overall expression by the discriminant, to give the generalized version of Brahmagupta’s formula

As with Brahmagupta’s original formula, this has restricted applicability. Specifically, it applies only to quadrilaterals whose vertices all lie on a locus with the same quadratic character as the function used in the definition of the σ_mn values. For Brahmagupta’s original formula the s values are based on the circular quadratic form, with a = c = 1 and b = 0, so the formula applies only if the four vertices lie on a circle. On the other hand, if we take a = c = 0 and b = 1 we have a discriminant of −1, and the quadratic form is a right hyperbola, so the area formula applies only if the four vertices lie on such a hyperbola.

In general, for any given quadratic form characterized by the constants a,b,c in the definition of the generalized distances σ_mn, the generalized formula of Brahmagupta gives the correct area of a quadrilateral if and only if the vertices satisfy the condition

where A_ijk is the signed area given by equation (1) for the triangle with vertices P_i, P_j, P_k. If the coordinates of three points are given, this equation is a quadratic in the coefficients of the fourth point, and its quadratic character is given by the respective parenthetical expression in (7), because the coordinates of that point appear only linearly in the triangle area factors of the other term (so they merely represent translations of the locus).

To illustrate, consider a quadrilateral inscribed in a right hyperbola, taking a = c = 0 and b = 1. We can begin with any three vertices, so let us take for example the points P₁ = {1,1}, P₂ = {3,5}, and P₃ = {4,2}. The generalized “lengths” are of the right hyperbolic form

Inserting the coordinates of P₁, P₂, and P₃ into (7), we find that the fourth vertex must lie on the right hyperbola described by

As an example we can take P₄ = {7/3, −1/7}, so we have the quadrilateral shown in the figure below.

From equation (1) we know the squared area of this quadrilateral is (155/21)². The generalized “distances” for these points are

Inserting these σ values into the generalized formula (6) we confirm that it yields the same result. (We can also show that, as expected, the original formula of Brahmagupta does not give the correct area for this quadrilateral, because it cannot be inscribed in a circle.)

Note that for the purely circular case (i.e., the original formulas of Heron and Brahmagupta) equation (7) is invariant under rotations, which gives that case its special usefulness, but it also implies that if a quadrilateral is not cyclic in one orientation then it is not cyclic in any orientation. In contrast, for hyperbolic cases we can always find an orientation such that the vertices of any given quadrilateral lie on a real hyperbola of the required form. (We can do the same for elliptical and parabolic cases too, but in those cases the coefficients may be complex.)  To show this, consider again the simple hyperbolic case with a = c = 0 and b = 1, for which equation (7) reduces to

This equation may not be satisfied for the given points in terms of this coordinate system, but the value of the expression on the left side will vary (for a given configuration of points) if we rotate the coordinate system, so we can seek an orientation for which the equation is satisfied. Note that the A_ijk coefficients in this expression are invariant under rotations, so the effect of rotating the coordinate system about the origin through an angle θ is given by simply making the following substitutions for the explicit x and y coordinates

Therefore each x_jy_j in the preceding equation is transformed into

Consequently equation (8) can be written in terms of these rotated coordinates as

Dividing through by cos(2θ) gives an equation in tan(2θ)

which can be solved for θ. Thus given four points with coordinates X_j,Y_j, we can use this formula to compute the angle θ, and then apply the inverse rotation to get back to the x_j,y_j coordinates in terms of which the four points lie on the required hyperbolic locus.

By the way, one might question the generality of (6) for quadrilaterals with verticies on a suitable conic (i.e., satisfying equation (7)), because in general a set of four line segments of  given lengths can be arranged as the edges of a quadrilateral in distinct ways, not necessarily having equal areas. Consider for example the normal Euclidean metric distance function, and four line segments with lengths 1, 1, 2, and 2 respectively. We can arrange these segments as the edges of a quadrilateral inscribed in a circle in three different ways, as depicted below.

The two quadrilaterals on the left have equal areas (and in fact the circumscribed circles have equal radii). In each of these two cases, the area is given by equation (6), taking each of the edge lengths as the positive square roots of the sum of squared coordinate differences. Noting that σ = (1+2+1+2)/2 = 3 and ac − b² = 1 in these two case, the area is given by

and hence A = 2. However, the net area of the (non-convex) right hand quadrilateral is zero, so that area clearly is not given by equation (6) with all positive edge lengths, despite the fact that the verticies satisfy equation (7), i.e., they all lie on a circle. What has gone wrong? The explanation is that there is an ambiguity in the choice of the signs for the edge lengths, which are algebraically the square roots of the sum of squared coordinate differences. The effect of this ambiguity can be see by expanding the product appearing in the numerator of equation (6) using the algebraic identity

In all but the final term, the individual arguments appear to even powers, so the signs have no effect, but the arguments appear to the first power in the final term 8urst. Since the signs of the edge lengths are ambiguous, the sign of the last term is also ambiguous, so we could just as well replace +8urst with –8urst. We can do this by using all positive edge lengths but subtracting 16urst from the product. It follows that the algebraic expression for the signed area (in terms of positive edge lengths) can also have the form

In our example this implies that the area of a quadrilateral with edge lengths 1, 1, 2, 2, inscribed in a circle can also be given by

and hence A = 0, which represents the case with crossed edges.

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