In the seminal "Isocubics Book", Ehrmann and Gibert point out that each cubic nK meets the
Circumcircle at A,B,C and at three other points whose Simson Lines pass through a point ω.
Indeed, the cubic can be defined using pedal circles and a circle centred at ω . Here, we
study the point ω and its relation to the cubic. Further details
appear in Footnote 1.
Some useful information on Simson Lines is collected in Footnote 2.
In Hyacinthos, Jean-Pierre Ehrmann has proposed two methods of finding the three points on
the Circumcircle whose Simson Lines pass through a given point. Maple calculation confirms
these results. Indeed, Jean-Pierre has just publshed a paper in Forum Geometricum which
mentions some of these. These are summarized in Footnote 4.
The general case is considered in Footnote 5.
The Simson Line of a point P p:q:r on the Circumcircle equation given by
(a) The Gibert-Simson Transform.
The first coordinate of the tripole is given by (b^{2}(c^{2}+a^{2}-b^{2})/q-c^{2}(a^{2}+b^{2}-c^{2})/r)p/a^{2}.
Using this, it is quite easy to prove that this point lies on the Simson Cubic cK(#G,X69) when
the point P is on the Circumcircle. It is not at all clear that there are just three points whose
Simson Lines pass through a given point X.
(b) First variation.
For a point P on the Circumcircle, a^{2}/p may be replaced by -(b^{2}/q+c^{2}/r).
Writing down the equation and clearing fractions, we see that the points for which this line
contains the fixed point X lie on a (complicated) circumcubic. The points whose Simson Lines
contain X are then the three intersections of this with the Circumcircle, other than A,B,C.
Since we are interested in these three, we may replace the cubic by any other in the pencil
which also contains the degenerate cubics consisting of a line and the Circumcircle. Some
tedious calculations lead to the cubics we find below, but no other isogonal nK.
(c) The Casey-Simson Transformation.
This is named for John Casey, who gave another formula for the Simson Line (webpage).
Here, the coefficient of x in an equation of the Simson Line is given by
p((2c^{2}q+(b^{2}+c^{2}-a^{2})r)(2b^{2}r+(b^{2}+c^{2}-a^{2})q))
In fact, this line is the radical axis of the Pedal Circle of P = p:q:r with the Circumcircle. When the
point P lies on the Circumcircle, the Pedal Circle becomes the Simson Line.
Now it is quite clear that the points whose Simson Lines pass through X lie on a circumcubic
nK(K,X&X69,?), where & denotes the barycentric product. X69 is the isotomic conjugate of H.
We shall refer to this as the Casey Cubic for the point X. Unfortunately, although the relationship
between X and the root of the circumcubic is simple, the cubic passes through no obvious points.
This makes it hard to identify.
Let C0 be the degenerate isogonal nK consisting of the Circumcircle and the Line at Infinity.
Then it is clear that the pencil of cubics containing the points whose Simson Lines pass through X
is generated by the Casey Cubic and C0. In each case, we shall select the member of the pencil
which contains O, and hence also H. Maple calculation proves
Theorem 1
Given a point X, the isogonal nK through O and the points with Simson Lines through X has
root R which is the anticomplement of the barycentric product of X and X(69).
Conversely, the cubic nK(K,R,O) meets the Circumcircle at points whose Simson Lines pass
though the barycentric product of the complement of R with H.
The maps f taking X to R, and its inverse g taking R to X are clearly projective. They can be
easily implemented in Cabri using my macro "pqoverr" which maps a triple P,Q,R to the point
which is the barycentric product of P,Q and the isotomic conjugate of R.
The maps have the following geometric properties :
f maps ΔABC to the anticomplementary triangle, and H to G.
g maps ΔABC to the orthic triangle, and G to H.
As an immediate consequence, f maps a line through H to a line through G. Since f also maps
K to H, it maps lines through K to lines through H.
These maps f and g are related to some transformations mentioned in ETC :
f(X) is the Zamosa transform of the barycentric product of X and I.
g(X) is the barycentric product of X(75) and the Mimosa transform of the isogonal conjugate of X.
Related Structures :
We write
C(X) for the circumconic of ΔABC with perspector X,
O(X) for the circumconic of the Orthic Triangle, with perspector X,
A(X) for the circumconic of the Anticomplementary Triangle, with perspector X
X on | R on | Comments |
Euler Line | Line GK | |
Line HK | Euler Line | |
Orthic Axis | Line at Infinity | |
Line at Infinity | Tripolar of X(264) | X(264) is the isotomic conjugate of O |
Nine-point Circle | C(O) | X(53) is X(6) of the Orthic Triangle, C(O) is the Macbeath Conic |
O(K) | C(H) | O(K) has centre H, C(H) is the Orthic Conic |
O(X(427)) | C(K) | |
C(K) | A(H) | H is O of Anticomplementary Triangle, so A(H) is its Macbeath Conic |
C(H) | A(G) | A(G) is the Steiner Ellipse of the Anticomplementary Triangle |
Theorem 2
Let nK(K,R,O) have associated Simson Lines meeting in X. The following are equivalent :
(a) X lies on the Line at Infinity,
(b) R lies on the line with tripolar X(264),
(c) nK(K,R,O) is circular.
This uses Fact 2.1. If X is at infinity, then the only points with Simson Line through X are the
infinite circular points and one other. Thus the cubic must be circular. The other facts are clear
from the above table.
Theorem 3
The circumcubic nK(K,R,O) contains the points whose Simson lInes contain X.
nK(K,R,O) degenerates if and only if either
(a) R lies on the Macbeath Conic C(O), i.e. X lies on the Nine Point Circle, or
(b) R lies on the Orthic Conic C(H), i.e. X lies on the circumconic of the Orthic Triangle, centre H.
Footnote 3 provides an outline of the proof, and some information about the circumconics. We
now discuss these degenerate cases. In the first case, much is well known.
Example 1. Theorem 3(a) : R lies on the Macbeath Conic, X on the Nine Point Circle.
Suppose that we know X. Then
(1) One of the points on the Circumcircle and the cubic is P, the reflection of H in X.
(2) Let PH meet the Circumcircle again at P*. Then P*X(110) meets C(O) again at R.
(3) The other two points on the Circumcircle are the intersections with the tripolar of R -
this is a diameter. The Simson Line of P is perpendicular to this.
Suppose that we know R. Then
(4) Let RX(110) meet the Circumcircle again at P*. Then P as above is the second
intersection of P*H with the Circumcircle.
(5) X is the mid-point of segment PH.
(6) The other points on the Circumcircle are the intersections with the tripolar of R.
Example 2. Theorem 3(b) : R lies on the Orthic Conic, X on the conic O(K) - see Table.
Suppose that we know R. Then
(1) Let R* be the other intersection of RX(107) with the Circumcircle. Then OR* meets the
Circumcircle again at a point P0 with Simson Line through X.
(2) The other two points P1, P2 on the Circumcircle are the intersections with the tripolar of R.
(3) The Simson Line of Pi is the line throgh the mid-point of the segment HPi perpendicular to
the line PjPk. The three Simson Lines determine X.
Suppose that we know X. Then
(4) Let X* denote the other intersection of XX(125) with the Nine-point Circle. Then HX* meets
the Circumcircle twice. One point is the reflection of H in X*. We label the other as R*. Then
R*X(107) meets C(H) again at the root R.
(5) We can now determine the Pi as above.
Some sample values :
X | R | notes |
G | X(193) | |
H | G | cubic K074 |
X(5) | X(1993) | cubic K105 |
K | H | |
X(11) | X(651) | |
X(19) | X(8) | |
X(25) | X(69) | |
X(33) | X(7) | |
X(53) | O | |
X(115) | X(110) | |
X(125) | X(648) | |
X(427) | X(6) | X(427) is on Euler Line |
X(647) | X(520) | |
X(523) | X(525) | |
X(647) | X(520) | |
X(1562) | X(107) | |
X(1824) | X(75) | |
X(1826) | X(1) | |
X(1839) | X(10) | |
X(2489) | X(512) | |
X(2501) | X(523) | cubic K164 |
Footnote 1.
In the Isocubics Book, it is noted that a cubic nK can be defined as a locus using a fixed circle.
Let the tripolar of the root of the cubic meet BC, CA, AB at U, V, W respectively. The centre Ω
lies on the radical axis of the circles on diameters AU, BV, CW. The following were observed
using Cabri sketches, they can be proved by Maple. Most appear in the Isocubic Book.
Fact 1.1
(1) The radical axis of the above circles passes through H.
(2) If the conic is isogonal and passes through O and H, then
(a) the circle centre Ω passes through the complement of the root,
(b) ω is the mid-point of Ω and H.
Footnote 2.
These are stated without proof.
Fact 2.1
The Simson Line of either infinite circular point is the Line at Infinity.
Fact 2.2
The Simson Line of a point P bisects the segment PH.
Fact 2.3
If P,Q,R lie on The Circumcircle and have Simson Lines passing through X, then
(1) P is the isogonal conjugate of the infinite point on QR, and similarly for Q,R.
(2) The Simson Line of P is perpendicular to QR.
Footnote 3.
If a circumcubic degenerates, then it must consist of a line, and the conic which is its isoconjugate.
Some simple algebra indicates that, if nK(P,R.?) degenerates, then the linear factor must be the
tripolar of the root R. Now, nK(P,R,?) contains the intersections of the tripolar of R with the sidelines
of ΔABC, so we get degeneracy precisely when this contains at least one more point of the cubic.
In our case, we have P = K, and we know that the cubic contains O and H. This is equivalent to the
root R lying on C(O) or C(H).
C(O) meets the Circumcircle at X(110),
C(H) meets the Circumcircle at X(107),
C(O) meets C(H) at X(648).
Footnote 4.
Here are two ways of finding the points on the Circumcircle whose Simson Lines pass through a
given point X. Each has been mentioned in Hyacinthos by Jean-Pierre Ehrmann.
Theorem 4
Suppose that X is a fixed point.
(1) Let C be the circumconic through X and H. In other words, C is the rectangular circumconic
through the point X. Let u be the vector HX. Then the translation of C by vector u cuts the
Circumcircle at its intersection with C, and at the three points whose Simson Lines contain X.
(2) Let D be the (diagonal) conic through the incentre, the excentres and P', the anticomplement
of P. Let P* be the point on OP' such that PP*:P*O = 4:-1. Then the image of D under the
homothety H(P*,-1/2) is a conic which meets the Circumcircle at the antipode of the centre of D,
and at the three points whose Simson Lines contain X.
(3) If X* is the orthocentre of the triangle whose vertices are the points on the Circumcircle whose
Simson Lines pass through X, then X is the mid-point of the segment HX*.
This can be verified by Maple. The conics mentioned do meet the Circumcircle at the required points.
Both contain the required point X*. Since the conics are rectangular and are circumconics of the
triangle containing the three points, they meet only in its orthocentre.
Note
The description in (2) is a shorter version of that proposed by Jean-Pierre Ehrmann.
Footnote 5.
The special case of isocubics through O yields simple algebra relating the point ω and the root R.
In general, each isogonal nK can be described in the same fashion, as in the Isocubics Book, but the
algebra becomes more complicated.
In the Book, each isogonal nK is described as the locus of points for which the Pedal Circle is orthogonal
to a fixed circle with centre ω. This is equivalent to saying that, for any two non-isogonal points on the
cubic, the radical axis of their Pedal Circles passes through ω. If the root R is known, then the isocubic
contains the intersections of the sidelines with the tripolar of R. The Pedal Circles are those mentioned in
Footnote 1. We then have an easy general result, and one that is useful for us here.
Fact 5.1
(a) The Pedal Circles of a point and of its isogonal conjugate are identical.
(b) The centre of the Pedal Circle of P is the mid-point of P and its isogonal conjugate.
Fact 5.2
For R ≠ G, the Pedal Circles above have radical axis the line through H and the barycentric product of R
and H.
Note that one further point U on the isocubic determines ω by looking at the radical axis of the Pedal
Circle of P and any one of the Pedal Circles above.
Note on calculations
If we begin with the point ω and the radius r of the circle orthogonal to the Pedal Circles, then we can
write down the equation of the locus. This is an isogonal nK. The root appears to be quadratic in the
coordinates of ω. If we now assume that the locus contains a point U not on any sideline, or on the
Circumcircle or Line at Infinity, then we can determine the radius r. Then the equation reduces to one
with the root linear in the coordinates of ω. This allows us to resolve the case R = G.
Fact 5.3
The isogonal nK locus has root G if and only if ω = H. These isocubics form the class CL020 on Bernard
Gibert's webpages.
The linearity mentioned above as further consequences. We have a projective map from ω to R. Then
the inverse is also projective. This allows us to describe the cases of degeneracy. We begin with the
elementary
Fact 5.4
The cubic nK(K,R,U) is degenerate if and only if R is on one of the circumconics with perspector U or U*,
the isogonal conjugate of U.
This arises since the cubic must degenerate into the tripolar of R and its isogonal conjugate. Thus we have
degeneracy if and only if U is on one of these. Looking at this from the point of view of R, we get our result.
Fact 5.5
The locus defined by ω and the point U is degenerate if and only if ω lies on one of two conics determined
by the point U.
The equations of these conics are complicated. They are not in general "nine-point conics", unlike the case
were U is O (or H).
As above, a non-pivotal isocubic can be described as the locus of points P such that the circle with diameter
PP*, where P* is the appropriate isoconjugate of P, is orthogonal to a fixed circle with centre Ω. We then
have an analogue of Fact 1.1 (already noted in the Isocubics Book).
Fact 5.6
For an isogonal nK, ω is the mid-point of Ω and H.