Inheritance Relations of Hexagons and Ellipses Mahesh Agarwal and Narasimhamurthi Natarajan

Home , Diagonal

Mathematical Assoc. of America College Mathematics Journal 45:1 June 13, 2014 12:16 a.m. Inheritancerelations1.tex page 1

Inheritance relations of hexagons and ellipses Mahesh Agarwal and Narasimhamurthi Natarajan

Recessive Gene Inside each convex hexagon ABCDEF is a diagonal hexagon, UVWXYZ its child (See ﬁgure 1). This establishes a parent-child relationship. This concept can be ex- tended to grandchildren and grandparents, etc. Brianchon’s theorem gives a criterion for inscribing an ellipse in a hexagon and Pascal’s theorem gives a criterion for circumscribing an ellipse around a hexagon. So it is natural to ask, if we know that a hexagon inscribes in an ellipse, will its child or grandchild inherit this trait? Similarly, if a hexagon is circumscribed by an ellipse, will its child or grandchild inherit this trait? These are the questions we explore here, and the answer leads us to a recessive trait. We show that a hexagon can be circumscribed by an ellipse if and only if its child has an ellipse inscribed inside it. This parent-child relationship is symmetric in that a hexagon has an ellipse inscribed inside it if and only if the child can be circumscribed by an ellipse. In short, Circumscribed begets inscribed and inscribed begets circumscribed. This shows that inscribing and circumscribing are recessive traits, in the sense that it can skip a generation but certainly resurfaces in the next.

The language of projective geometry The conditions for circumscribing a hexagon by an ellipse was answered by Pascal in 1640 when he showed that for such hexagons, the points of intersections of the

Figure 1. Parent-child-grandchild

VOL. 45, NO. 1, JANUARY 2014 THE COLLEGE MATHEMATICS JOURNAL 1 Mathematical Assoc. of America College Mathematics Journal 45:1 June 13, 2014 12:16 a.m. Inheritancerelations1.tex page 2

Figure 2. Parent and grandchild are both circumscribed by ellipses and child has an inscribed ellipse

opposite sides must lie on a straight line (see Figure 3). The problem of inscribing an ellipse was fully answered by Brianchon [1] in early 1800’s. He showed that an ellipse can be inscribed inside a hexagon if and only if the main diagonals intersect at a point (see Figure 3). We will make extensive use of these two results. The conditions of Pascal and Brianchon deal with finding the point of intersection of straight lines, and drawing lines between points. If we use standard coordinate geometry, the formulas for these become complicated. However, the language of projective geometry provides an elegant way to handle such problems using the dot product and cross product. In projective geometry, the point (x, y) in the plane is identified with a homogeneous point P = (X,Y,Z) in space where x = X/Z and y = Y/Z. Note that since only ratios are important, (4, 6, 2) and (14, 21, 7) are two different representations of the point (2, 3). In the plane, equation of a line is ax + by + c = 0. If we move to the projective space, this becomes aX/Z + bY/Z + c = 0 or aX + bY + cZ = 0. In the projective space, a line is identified by a vector L = (a, b, c). The dot product gives us a characterization: a point P lies on a line L if and only if P · L = 0. The cross product is useful in the following characterization: the point of intersection of two lines L1 and L2 is L1 × L2, and the line connecting two points P1 and P2 is P1 × P2. We illustrate these ideas using examples. Example. What is the equation of the line connecting A = (7, 8), B = (4, 2)? Re- call that the cross product of two vectors is defined as (u1, u2, u3) × (v1, v2, v3) = (u2v3 − u3v2, u3v1 − u1v3, u1v2 − u2v1). Since (7, 8, 1) × (4, 2, 1) = (6, −3, −18) the equation of the line connecting P1 and P2 is 6X − 3Y − 18Z = 0. Dividing by 3Z we get 2x − y − 6 = 0. Example. Given two lines 4x + 5y = 14 and 7x + 10y = 27, find the point of intersection. Using projective geometry this is easily calculated as (4, 5, −14) × (7, 10, −27) = (5, 10, 5) which corresponds to the point (1, 2). Now we are in a position to characterize the conditions of Pascal and Brianchon,

2  THE MATHEMATICAL ASSOCIATION OF AMERICA Mathematical Assoc. of America College Mathematics Journal 45:1 June 13, 2014 12:16 a.m. Inheritancerelations1.tex page 3

Figure 3. Pascal’s condition and Brianchon’s condition

that is, co-linearity of points and concurrency of lines.

Three points P1, P2 and P3 are collinear ⇔ [P1P2P3] := (P1 × P2) · P3 = 0

Three lines L1, L2 and L3 are concurrent ⇔ [L1L2L3] := (L1 × L2) · L3 = 0

The similarity between points and lines is striking in projective spaces. Lines and points have similar representations. The intersection of two lines is a point and the intersection of two points in a line. In general, in any statement which is true for the projective space, one can reword the statement, replacing points by lines and lines by points, still preserving its truth. This is the principle of duality in projective spaces. We now give a statement of Pascal and Brianchon conditions using the language of projective spaces. Theorem 1 (Pascal’s Theorem). A convex hexagon ABCDEF circumscribed by an ellipse if and only if the point of intersection of the opposite sides AB and DE, BC and EF and CD and FA are collinear. In the language of projective spaces, a convex hexagon ABCDEF can be circumscribed by an ellipse if and only if

[(A × B) × (D × E), (B × C) × (E × F ), (C × D) × (F × A)] = 0 (1) | {z } | {z } | {z } AB ∩ DE BC ∩ EF CD ∩ FA

Theorem 2 (Brianchon’s Theorem). A convex hexagon ABCDEF has an inscribed ellipse if and only if the the major diagonals AD, BE and CF are concurrent. In the language of projective spaces, a convex hexagon ABCDEF has an inscribed ellipse if and only if

[(A × D), (B × E), (C × F )] = 0 (2)

Projective geometry has a lot more to offer. For readers interested in learning more about projective geometry, Appendix A in Silverman-Tate [2] is an excellent source.

Hereditary property of Hexagons We are now ready to answer the central questions of our work. We will explore how hereditary properties are passed on in hexagonal families.

VOL. 45, NO. 1, JANUARY 2014 THE COLLEGE MATHEMATICS JOURNAL 3 Mathematical Assoc. of America College Mathematics Journal 45:1 June 13, 2014 12:16 a.m. Inheritancerelations1.tex page 4

Theorem 3 (Inheritance theorem). A parent hexagon has an inscribed ellipse if and only if the child hexagon is circumscribed by an ellipse; also a parent hexagon can be circumscribed by an ellipse if and only if the child hexagon has an inscribed ellipse.

Proof. We prove the ﬁrst statement. Let P1,P2, ··· ,P6 denote the homogeneous co- ordinates corresponding to the vertices of the parent hexagon and U1,U2,U3,U4,U5,U6 be deﬁned as follows

U1 = P3 × P5 U2 = P4 × P6 U3 = P5 × P1 U4 = P6 × P2 U5 = P1 × P3 U6 = P2 × P4.

U1,U2,U3,U4,U5,U6 form the child hexagon. Using (2) the hexagon P1P2P3P4P5P6 has an inscribed ellipse if and only if

[(P1 × P4), (P2 × P5), (P3 × P6)] = 0 (3)

and the child will have a circumscribed ellipse if and only if

[U1 × U4,U2 × U5,U3 × U6] = 0. (4)

Using standard the vector identity (A × B) × (C × D) = [ABD]C − [ABC]D

one can verify that

[(P1 × P4), (P2 × P5), (P3 × P6)] = [P1P4P6][P2P3P5] − [P1P3P4][P2P5P6]

and

[U1 × U4,U2 × U5,U3 × U6] = [P2P4P5]{[P1P3P6][P1P4P5][P2P3P5]

− [P1P3P4][P1P2P5][P3P5P6]}

Due to the non-collinearity of P1,P3,P4 there exist u, v and w such that P6 = uP1 + vP3 + wP5. On expressing P6 in terms of P1,P3 and P5 and using the fact that [AAB] = 0, we get

([P1P3P6][P1P4P5][P2P3P5] − [P1P3P4][P1P2P5][P3P5P6])

= w[P1P3P5][P1P4P5][P2P3P5] − u[P1P3P4][P1P2P5][P3P5P1]

= [P1P3P5](w[P1P4P5][P2P3P5] − u[P1P3P4][P1P2P5])

and

[P1P4P6][P2P3P5] − [P1P3P4][P2P5P6]

= v[P1P4P3][P2P3P5] + w[P1P4P5][P2P3P5]

− u[P1P3P4][P2P5P1] − v[P1P3P4][P2P5P3]

= w[P1P4P5][P2P3P5] − u[P1P3P4][P2P5P1].

Putting these together we get

[U1 × U4,U2 × U5,U3 × U6] = [P2P4P5][P1P3P5][(P1 × P4), (P2 × P5), (P3 × P6)]

4  THE MATHEMATICAL ASSOCIATION OF AMERICA Mathematical Assoc. of America College Mathematics Journal 45:1 June 13, 2014 12:16 a.m. Inheritancerelations1.tex page 5

Figure 4. The whole lineage shares the same trait

In a convex hexagon no three vertices can be collinear hence [P1P3P5] 6= 0 and [P2P4P6] 6= 0. From (3) and (4) it follows that a parent hexagon has an inscribed ellipse if and only if the child hexagon is circumscribed by an ellipse. To show that a child hexagon has an inscribed ellipse if and only if the parent hexagon is circumscribed by an ellipse notice that

P1 = U3 × U5 P2 = U4 × U6 P3 = U5 × U1 P4 = U6 × U2 P5 = U1 × U3 P6 = U2 × U4

Comparing this with the deﬁnition of Ui, ··· ,U6 and the structure of (3) and (4), it is clear that result follows from the same vector identity. Remark. The theorem and its proof are an example of the principle of duality where the roles of lines and points can be interchanged. Putting together the two statements that constitute the previous theorem, we get a theorem that completely characterizes the heredity trait of hexagon vis-a-vis inscribed and circumscribed ellipses (see Figure 2). Theorem 4 (Recessive trait theorem). A hexagon has an inscribed ellipse if and only if its grandparent has an inscribed ellipse; also a hexagon has a circumscribed ellipse if and only if its grandparent has a circumscribed ellipse.

Lineage property By the discussion above, if a hexagon can be circumscribed by an ellipse and an ellipse can be inscribed in it, then this property is shared by its entire lineage. Regular hexagons generate such lineages. We now show how to construct irregular hexagons that have both these properties. In fact, starting with any a hexagon we can perturb any one vertex so that the resulting hexagon generates such a lineage.

VOL. 45, NO. 1, JANUARY 2014 THE COLLEGE MATHEMATICS JOURNAL 5 Mathematical Assoc. of America College Mathematics Journal 45:1 June 13, 2014 12:16 a.m. Inheritancerelations1.tex page 6

Theorem 5. Given a hexagon ABCDEFˆ, one can construct a unique hexagon ABCDEF so that the hexagon ABCDEF can be both circumscribed by an ellipse and has an inscribed ellipse. There are at least two approaches one can take to constructing these hexagons. The ﬁrst approach is to draw an ellipse passing through the ﬁve points A, B, C, D and E. This can certainly be done, as there exists a unique ellipse passing through any given 5 points, no three of which are collinear. Then one can determine the sixth point, as the point of intersection of the ellipse with a line passing through C and the point of intersection of the diagonals AD and BE. We note that this construction requires one to draw the ellipse. We focus on the second approach which provides a straight edge construction for such hexagons. Our construction relies on the following two Lemmas which are of interest in their own right. Lemma (Diagonal Lemma). Let ABCDEF be a hexagon with an inscribed ellipse, and UVWXYZ its child. Then the point of intersection of diagonals VY and ZW of the child lies on the diagonal of CF of the parent hexagon. Proof. Note

ZW = ((A × E) × (B × F )) × ((C × E) × (B × D))

VY = ((B × D) × (C × A)) × ((D × F ) × (A × E)).

Using a computer algebra system, such as Mathematica, it can be veriﬁed that

[ZW × VY,C,F ] = [BCD][ACE][AEF ][BDF ][A × D,B × E,C × F ]

Since the lines AD, BE and CF are concurrent by the Brianchon criterion [A × D,B × E,C × F ] = 0. Hence ZW ∩ VY lies on the line CF . Lemma (Common Point Lemma). Let ABCDEF have an inscribed ellipse then the child, UVWXYZ has an inscribed ellipse if and only if the major diagonals of the parent and the child intersect at the same point. Proof. By the Diagonal Lemma, the diagonals ZW , VY intersect on CF . The diagonals VY and UX intersect on BE and the diagonals WZ and UX intersect on AD. The major diagonals of the child UVXYZW are concurrent if and only if the point of concurrency lies on all the three major diagonals of the parent. Hence the result follows. We now provide the construction for the Lineage Theorem. Construction. Let O be the intersection point of the major diagonals of AD and BF . So O is given by

O := (A × D) × (B × E)

Let V be the vertex of the child at the intersection of AC and BD.

V := (A × C) × (B × D)

6  THE MATHEMATICAL ASSOCIATION OF AMERICA Mathematical Assoc. of America College Mathematics Journal 45:1 June 13, 2014 12:16 a.m. Inheritancerelations1.tex page 7

Figure 5. Construction showing hexagon with both inscribed and circumscribed ellipses

By the Diagonal Lemma, the vertex Y opposite V of the child, must lie on the line joining V and O as well as the line AE. Hence

Y := (V × O) × (A × E)

For the parent to have an inscribed ellipse, the desired point F must be on the intersection of CO and DY .

F := (D × Y ) × (C × O)

The outer hexagon ABCDEF is Brianchon since the three major diagonals AD, BE and CF are concurrent at O. By application of Common Point Lemma, the inner major diagonals must also pass through O. Hence both the parent and the child have inscribed ellipses. So by the Inheritance Theorem, the hexagon ABCDEF has both an inscribed and circumscribed ellipse. The point F is unique by construction.

Acknowledgment. The authors wish to acknowledge the support of **********

Summary. We show that a hexagon can be circumscribed by an ellipse if and only if the child hexagon (diagonal hexagon) has an ellipse inscribed inside it. This parent-child relationship is symmetric in that a hexagon has an ellipse inscribed inside it if and only if the child can be circumscribed by an ellipse. In short, Circumscribed begets inscribed and inscribed begets circumscribed. This shows that inscribing and circumscribing are recessive traits, in the sense that it can skip a generation but certainly resurfaces in the next.

References

1. Coxeter, H. S. M. Projective geometry, Second edition, University of Toronto Press, 1974

VOL. 45, NO. 1, JANUARY 2014 THE COLLEGE MATHEMATICS JOURNAL 7 Mathematical Assoc. of America College Mathematics Journal 45:1 June 13, 2014 12:16 a.m. Inheritancerelations1.tex page 8

2. Silverman, Joseph H. and Tate, John, Rational points on elliptic curves, Undergraduate Texts in Mathematics, Springer-Verlag, 1992

8  THE MATHEMATICAL ASSOCIATION OF AMERICA