The Simmons inconics
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Nikolaos Dergiades recently posted a result showing that there was an inconic for which the perspector was a focus. Read about this here.
This is the Simmons inconic for which the Fermat point Fn and isodynamic point In are the foci and Fn the perspector. This unusual circumstance immediately got me interested. In addition this conic is defined by a point of some algebraic complexity, which interests me as well.
Bernard Gibert has written an excellent article on the Simmons conics; and a second one. The first is prior two this page and covers much of the same material. The second is about the relation of this conic to other conics and cubics and shares almost nothing with these pages.
This conic has two versions. I have generally avoided weak conics because one has to consider all the versions to properly analyze them. But I have found this pair interesting to contemplate. (note written a year later: I now love weak conics; it is all the versions that makes them interesting!)
The Fermat point contains a square root that the general run of points does not have and is thus algebraically isolated from most points. It seems that every point on these conics inherits this square root (or has worse ones), so that the points on this conic must by their very nature be isolated from most of the geometry that we know. They are a lost tribe of points, with strange customs, isolated in the land of the Simmons conics. [Note: Peter Moses has since verified that none of these points are in ETC.]
Here is a picture of the Simmons conics in their ellipse form. Inconics are ellipses if their perspector is inside the Steiner circumellipse.
Figure: The two Simmons conics are red filled with green. Strong points are blue and fissile (2-fold) points are green. "of X" refers to the tripolar line of point X. ~X is the dual line of point X. The 4th tangent is ~S which is perpendicular to G—T, the Euler line of the Brocard image triangle. The duals of the perspectors go through the centers of the conics.
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Details about the Simmons inconics
In the following n and s indicate the "normal" and "switched" versions of an object that comes in 2 versions. Often we take this to mean the version inside the triangle and the one outside the triangle, but since the inside can be continuously varied to the outside, this is not a valid distinction.
Perspectors:
The Fermat points Fn and Fs = (: 1/(SA±Sπ/3) :)
These conics can both be ellipses, as seen in the illustration; both can be hyperbolas, or one of each. These conics cannot be parabolae as they the transition between ellipse and hyperbola occurs as the Fermat point is at a vertex, so the parabola is degenerate (see movie below).
Equations:
√((SA±Sπ/3)x) + √((SB±Sπ/3)y) + √((SC±Sπ/3)z) = 0.
Note: square roots are always assumed to have two possibilities (we live in an "absolute-value" free zone).
where Sπ/3 = S cot π/3 = S/√3 is an irrational factor that gives these points a unique algebraic flavor while insuring that they do not interact well with the great mass of strong and weak points, which have a different algebraic nature.
Algebraic uniqueness:
This is a new consideration, perhaps the most powerful one, in Geometry. It tells us the circumstances under which geometric objects can relate to eachother. Let us say that we have a triangle ABC and on BC we have constructed an equilateral triangle and an equilateral pentagon. Let D be the equilateral triangle vertex and E one of the pentagon vertices. Now D must include a √3 in its coordinates and E a √5 in its. Now imagine that we draw AD. This line must include √3 in its coefficients. A point whose coordinates have √5 essentially cannot be on this line or any line whose coefficients do not include √5. The points with √3 and √5 are algebraically isolated from the great mass of points not having these factors (or equivalently not requiring a construction that gives it). Similarly the Euler line cannot go though this point, nor the Brocard line, nor the Nagel line.
Here is a picture that uses coloring to show points and lines with different algebraic natures. The two starting points exist in some computational field which we indicate by coloring blue. The equal circles intersect, introducing a factor of √3 (red). Since the root can be positive or negative, there are two versions of them. Lines drawn through the red point acquire the √3 in their coefficients and so they are red. Each of the shown red lines has another version not shown. The sole line through the red points that is not red is the one through them both. If √3 changes sign, the red points change places but this line is unchanged, so √3 is not in its coefficients. The midpoint, being the intersection of blue lines, is blue; i.e., of the same algebraic nature as the starting point.
Foci:
n version of Simmons inconic: Fn, In = g Fn where g indicates the isogonal conjugate.
s version of Simmons inconic: Fs, Is = g Fn
Centers:
The centers are mtFn and mtFs where m is the medial operation and t the isotomic conjugate. They are on the line G—K.
[from Peter Moses] The centers of the Simmons conics are X(396) and X(395)
Focal axes
are the In—Fn line and Is—Fs, both parallel to the Euler line. The InIs, FnFs, and center to center lines all go though K, so that there is a natural mapping between points on the two focal axes.
Since the focal axis of the first conic went though the Euler infinity point, a strong point, so must that of the second. Hence both conics must have an axis parallel to the Euler line.
Non-focal axes:
through the center parallel to the dual of K and perpendicular to the Euler line. Each non-focal axis is the dual of the opposite perspector. The non-focal axes are the generating lines of the opposite conic in that the tripolar of points on the non-focal axis are also tangent to the opposite conic.
4th tangent: The 4th tangent is the line joining J, the center of the Jerabek hyperbola; mS, the medial Steiner point, the center of the Kiepert hyperbola; and the fourth intersection of the Jerabek hyperbola with the Steiner ellipse. This line is the dual of the Steiner point and meets the conics at Sn = :(c2–a2)2(SB+Sπ/3): and its algebraic cousin Ss = :(c2–a2)2(SB–Sπ/3):
note: the dual of a line is the tripolar of the isotomic conjugate. The dual of P is indicated as ~P. Unlike the tripolar it is s proper dual. The relation of point to dual is affine invariant.
The intersections of the Lemoine axis with the non-focal axes form a fissile pair whose coordinates are of the form JP ± l•∞ Z, where JP is the Jerabek perspector, l•∞ is the endpoint of the Lemoine line, and Z = 2S/√3.
The equation of the line at infinity is x + y + z = 0, and that of an inconic with perspector (:m:) is ...+√y/m + ... = 0. So that a point on the inconic is (: m y2 :) where (:y:) is at infinity. This means that (:y2/(SB±Sπ/3):) are on the Simmons inconic. With our knowledge of points at infinity we can easily generate points on the first Simmons conic. As explained here the points on a conic can each be considered to originate from one of the multitude of triangle centers. We list the points in groups that correspond to a point in the plane of ABC.
The Fermat point contains a square root that the general run of points does not have and is thus algebraically isolated from most points. It seems that every point on these conics inherits this square root (or has worse ones), so that the points on this conic must by their very nature be isolated from most of the geometry that we know. They are a lost tribe of points, with strange customs, isolated in the land of the Simmons conics. [Note: Peter Moses has verified that none of these points are in ETC.]
All the points in the following chart generated from weak points such as the incenter and Gergonne points lead to 8-fold points, four on each conic. The strong points lead to points with two versions.
I did these points just because I wanted to see what they were like. Well...they are what they are. If I came across these in ETC, I would term them "silly points" because the almost certainly do not connect with much known triangle geometry. Inconics seem to be resistant to containing points we have discovered. Perhaps here we can see why.
In the next picture the octile points in rows 1 and 2 of the table below are plotted. There are four per conic. Hence octile points to a pair of conics are as weak (quartile) points are to strong conics. The table below uses points at infinity to find points on the Simmons conics. These points are constructed from the equivalent points on the Steiner ellipse which are shown in the picture. From there it is an exercise in modern construction, which is not done with a ruler and compass (although in principle it can be done that way). I have analyzed another octile set of points here.
From the points on the Steiner ellipse, take the barycentric square (Paul Yiu has shown us this construction), then the isotomic conjugate, and finally take the barycentric product with either Fermat point to create the points on the Simmons conics.
The four weak points on the Steiner ellipse will become 8 weaker points on the Simmons conics, four per conic.
Figure: Octile points on the Simmons conics. Points labeled 1xy (x = n,s and y = o,a,b,c) are the 8 versions of point 1 listed below. Sx and nSx originate with the Steiner point and its reflection in the centroid.
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Line at infinity
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Simmons inconics
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(: y :)
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(:y2/(SB±Sπ/3):)
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From the Incenter
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514 c–a
∞•~Io |
point 1 (c–a)2/(SB±Sπ/3)
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519 c+a–2b
∞•(G—Io) |
point 2 (c+a–2b)2/(SB±Sπ/3)
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513 b(c–a)
twS = ∞•~tIo |
? b2(c–a)2/(SB±Sπ/3)
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900 (c-a)(c+a-2b)
(∞•~190o) |
? (c–a)2(c+a–2b)2/(SB±Sπ/3)
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812 (c–a)(b2–ca)
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? (c–a)2(b2–ca)2/(SB±Sπ/3)
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740 (c+a)(b2–ca)
∞•(Io—tIo) |
? (c+a)2(b2–ca)2/(SB±Sπ/3)
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518 b(c2+a2–ab–bc)
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? b2(c2+a2–ab–bc)2/(SB±Sπ/3)
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g934 b(c-a)sbb,
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? b2(c–a)2sbb2/(SB±Sπ/3)
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From the Gergonne point
also the Mittenpunkt |
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522 = g109 (c–a)sb
∞•~Go |
? (c–a)2sbb/(SB±Sπ/3)
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527 csc+asa–2bsb
∞•(G—Go) |
? (csc+asa–2bsb)2/(SB±Sπ/3)
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From the Symmedian point
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523 c2–a2
tS = ∞•~K |
? (c2–a2)2/(SB±Sπ/3)
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524 c2+a2–2b2
∞•(G—K) |
? ( c2+a2–2b2)2/(SB±Sπ/3)
t(∞•(G—K)) |
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512 b2(c2-a2)
gS = ∞•~tK |
? b4(c2–a2)2/(SB±Sπ/3)
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g98 b2SB2–
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? b4(SB2–)2/(SB±Sπ/3)
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From the Orthocenter
also the Circumcenter |
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525 (c2–a2)SB
∞•~H = ∞•~O |
? (c2–a2)2SB2/(SB±Sπ/3)
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30 SBC+SAB–2SCA
∞•(G—K) infinite point on Euler line |
? (SBC+SAB–2SCA)2/(SB±Sπ/3)
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520 b2(c2–a2)SBB
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? b4(c2–a2)2SBBBB/(SB±Sπ/3)
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From the Fermat point
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? (c2–a2)(SB±Sπ/3)
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? (c2–a2)2(SB±Sπ/3)
Sn, Ss These are the meets of the ~S, the 4th tangent, with each conic. |
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Table of points on the Simmons conics. These points were generated by using the main points on the line at infinity and moving them down to this conic. The colors indicate the origin of the generating points. The red mark to the right indicated that the point is not in ETC.
This movie (2 Megs) shows both conics in red. It is formatted for the video
iPod. The triangle varies showing the various shapes of the pair
of conics. A
larger version (20 M) is here.
An inconic is an ellipse if its perspector is inside the Steiner ellipse. The Simmon's conics cover all types. It degenerates and changes from ellipse to hyperbola when a Fermat point moves onto a vertex, which happens when an angle is ±π/3. Both conics can simultaneously be ellipses or hyperbolas.
Duals and tripolars
The duals of all points on G—P are parallel to ~P. We have three conspicuous lines through G in the Simmons configuration. First we consider G—In—Fs and G—Is—Fn. This means that the duals of In and Fs are parallel as are those of Is and Fn.
The next line is tFn—G— In+Fn —K— Is+Fs —tFs. The duals of all these points are parallel to ~K which is perpendicular to the Euler line. Since the dual of tP is the tripolar of P, we conclude that the tripolars of Fn and Fs are perpendicular to the Euler line, as are the duals of the conic centers.
Note: In+Fn is the affine notation for the midpoint of In and Fn.
Since Fn, K, mS, Fs are collinear., their duals are concurrent. Since In, K, Is are collinear., their duals are concurrent. The coordinates of these two concurrence points have the form JP ± l•∞ Z, where JP is the Jerabek perspector, l•∞ is the endpoint of the Lemoine line, and Z = 2S/√3.
The tripolars of the perspectors Fn and Fs are each the non-focal axis of the other ellipse. The tripolars of In and Is are parallel. These four lines form a parallelogram for which the Lemoine line (the tripolar of K) is a diagonal.
The duals of all points on G—K are parallel to ~K and therefore parallel to the non-focal axis. Therefore the duals of the centers go through Fn and Fs and are parallel to the non-focal axis; i.e., perpendicular to the Euler line.
Tangent Lines
Each non-focal axis (a tripolar of a Fermat point) is the axis of perpective of the other conic. Hence the tripolar of any point on it is tangent to one of the Simmons conics. Special points on this, the not focal axis seem are:
The intersection with Kiepert hyperbola with a perspectrix yields a tangent parallel to non-focal axis at a vertex. This is because the tripolars of any point on the Kiepert hyperbola is perpendicular to the Euler line. There are 4 of these lines.
The dual circumconics (under construction)
This pair has perspectors tFn and tFs centers mFn and mFs.
From Peter Moses
The dual of both conics pass through X(99) .. The Steiner point.
The perspectors of the duals are t X(13) = X(298) and t X(14) = X(299)
The center of the dual of an inconic with perspector P is the G-Ceva conjugate of t P.
In the dual of the Simmons case, the centers become ...
(a2 + S /√3 – SA) (SA + S /√3) :: &
(a2 – S /√3 – SA) (SA – S /√3) ::
