Mathematical Assoc. of America
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June 13, 2014
12:16 a.m.
Inheritancerelations1.tex
Inheritance relations of hexagons and
ellipses
Mahesh Agarwal and Narasimhamurthi Natarajan
Recessive Gene
Inside each convex hexagon ABCDEF is a diagonal hexagon, U V W XY Z its child
(See figure 1). This establishes a parent-child relationship. This concept can be extended 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
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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)? Recall that the cross product of two vectors is defined as (u1 , u2 , u3 ) × (v1 , v2 , v3 ) =
(u2 v3 − u3 v2 , u3 v1 − u1 v3 , u1 v2 − u2 v1 ). 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,
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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
⇔ [P1 P2 P3 ] := (P1 × P2 ) · P3 = 0
Three lines L1 , L2 and L3 are concurrent ⇔ [L1 L2 L3 ] := (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 F A 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
|
{z
} |
{z
} |
{z
}
AB ∩ DE
BC ∩ EF
CD ∩ F A
(1)
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.
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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 coordinates corresponding to the vertices of the parent hexagon and U1 , U2 , U3 , U4 , U5 , U6
be defined as follows
U1 = P 3 × P 5
U4 = P 6 × P 2
U2 = P 4 × P 6
U5 = P1 × P3
U3 = P5 × P1
U6 = P2 × P4 .
U1 , U2 , U3 , U4 , U5 , U6 form the child hexagon. Using (2) the hexagon P1 P2 P3 P4 P5 P6
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) = [A B D]C − [A B C]D
one can verify that
[(P1 × P4 ), (P2 × P5 ), (P3 × P6 )] = [P1 P4 P6 ][P2 P3 P5 ] − [P1 P3 P4 ][P2 P5 P6 ]
and
[U1 × U4 , U2 × U5 , U3 × U6 ] = [P2 P4 P5 ]{[P1 P3 P6 ][P1 P4 P5 ][P2 P3 P5 ]
− [P1 P3 P4 ][P1 P2 P5 ][P3 P5 P6 ]}
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
([P1 P3 P6 ][P1 P4 P5 ][P2 P3 P5 ] − [P1 P3 P4 ][P1 P2 P5 ][P3 P5 P6 ])
= w[P1 P3 P5 ][P1 P4 P5 ][P2 P3 P5 ] − u[P1 P3 P4 ][P1 P2 P5 ][P3 P5 P1 ]
= [P1 P3 P5 ](w[P1 P4 P5 ][P2 P3 P5 ] − u[P1 P3 P4 ][P1 P2 P5 ])
and
[P1 P4 P6 ][P2 P3 P5 ] − [P1 P3 P4 ][P2 P5 P6 ]
= v[P1 P4 P3 ][P2 P3 P5 ] + w[P1 P4 P5 ][P2 P3 P5 ]
− u[P1 P3 P4 ][P2 P5 P1 ] − v[P1 P3 P4 ][P2 P5 P3 ]
= w[P1 P4 P5 ][P2 P3 P5 ] − u[P1 P3 P4 ][P2 P5 P1 ].
Putting these together we get
[U1 × U4 , U2 × U5 , U3 × U6 ] = [P2 P4 P5 ][P1 P3 P5 ][(P1 × P4 ), (P2 × P5 ), (P3 × P6 )]
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College Mathematics Journal 45:1
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Inheritancerelations1.tex
Figure 4. The whole lineage shares the same trait
In a convex hexagon no three vertices can be collinear hence [P1 P3 P5 ] 6= 0 and
[P2 P4 P6 ] 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
P4 = U6 × U2
P2 = U4 × U6
P5 = U1 × U3
P3 = U5 × U1
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.
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Theorem 5. Given a hexagon ABCDE F̂ , 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 U V W XY Z its child. Then the point of intersection of diagonals V Y 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))
V Y = ((B × D) × (C × A)) × ((D × F ) × (A × E)).
Using a computer algebra system, such as Mathematica, it can be verified that
[ZW × V Y, 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 ∩ V Y lies on the line CF .
Lemma (Common Point Lemma). Let ABCDEF have an inscribed ellipse then
the child, U V W XY Z 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 , V Y intersect on CF . The diagonals V Y and U X intersect on BE and the diagonals W Z and U X intersect on AD.
The major diagonals of the child U V XY ZW 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)
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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
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2. Silverman, Joseph H. and Tate, John, Rational points on elliptic curves, Undergraduate Texts in Mathematics,
Springer-Verlag, 1992
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