Kiepert Parabola

[JHC] There follows a lot of stuff about Kiepert's parabola, of which I don't think much has got into the book (especially the fair amount that I discovered only in the last few days):

Typical isoperspectrix [:1/(SS+@@ + 2bb@):], where I'm writing @ for the parameter angle theta.

Intersection of two of these <:(C-A)(SS+@@+2bb@)(SS+%%+2bb%):>, and so contact point of one

< : (C-A)(SS+@@+2bb@)^2 : >.

Let's write SS+@@ + 2bb@ = lambda + mu.bb, so that

[ : 1/(lambda + mu.bb) : ] and < : (C-A)(lambda + mu.bb)^2 : >

are the typical isoperspectrix and its contact point, if we're working over the complex numbers. But if we're working over the reals, they are only so if |mu| = < |lambda|. If this inequality is satisfied, all call them "true", otherwise "false".

The way it comes about that not every isoperspectrix is true is that as @ increases, the isoperspectrix starts as the line at infinity (for @ = 0) and then rolls around the parabola until it gets to the outer Pythagorean (or Vecten) one when @ = 45 degrees, after which it rolls back, becoming the line at infinity again at @ = 90, then the inner Pythagorean one at @ = 135, after which it again reverses and rolls back to infinity at @ = 180.

The two Pythagorean cases and their contact points are

[ :1/(bb +- S): ] , < : (C-A)(bb +- S)^2 : >.

Their midpoint and meeting point are <: (C-A)(bbbb +- SS) :>, and the midpoint <: (C-A)bbbb :> of these is the contact point of Lemoine's latitude line l, which is itself an isoperspectrix. These last three points lie on the "vertical" line through P- and (the new) Y, which is the doubly-dilated polar axis, ddp. So the parabola touches l at the intersection l.ddp = <: bbbb(cc-aa) :> !

This gives you a few more things to put on the big picture.

You'll want to know about finality. I really think the strong alphabet is now final, but don't want to make that an absolute promise just in case I suddenly discover some other missing point that, like the Taylor center, MUST be in. But there can't be more than one or two such.

Oh - I forgot to tell you which are true and which false.

The tangent at the vertex V is always true.

The Lemoine line l is always false.

The B-edge is true just if S =< bb, which amounts to saying that the B-height is at most b.

So the picture looks like this:

P-

/|\

----------------Y-------------Euler line

___ |

/ \ |

/ *|

| |

| |*

| | |

where the vertical line is ddp and the two stars next to it are the contact points of the Pythagorean isoperspectrices. The bit of the parabola between them is the "false" part, and l is the tangent where it crosses the vertical line (at <:(C-A)bbbb:>), the mid-point of the two stars (<:(C-A)(bbbb+SS):>) being also on that line, as is the place (<:(C-A)(bbbb-SS):>) where their two tangents meet.

Oh, I never said that the join of the two stars is [:BB:], which is parallel to l. I worked out a whole lot more coordinates, but everything is easily reconstructible from these ones.

JHC

On Wed, 26 Apr 2000, Barry Wolk wrote:

> So the vertex [of the Kiepert isoperspectrix parabola] is

(... :(3SAC-SS)^2/(c^2-a^2) :...)

> which is just a little simpler.

Oh good! Since SS = SBC+SCA+SAB, we have

3SCA - SS = 2SCA-SAB-SBC,

which is the typical coordinate of the "Euler end" (ie., the point at infinity on the Euler line):

%e = (:2SCA-SAB-SBC:),

which might be trying to teach us something.

[I'm glad to see how many people are beginning to use "the S notation". I was decidedly reluctant to circulate it at first, but after I'd used it myself for a few months, found it so convenient that I could hardly do without it. Without it, we'd have to write the above point as

(:(aaaa-2bbbb+cccc-2bbcc+4ccaa-2aabb)^2/(cc-aa):),

for instance.

Let me summarise this notation for new readers, who should understand that letters immediately following an s or S are subscripts, and that w is the Brocard angle omega. I define

so = (a + b + c)/2, sa = so - a, sb = so - b, sc = so -c

Sw = (aa+bb+cc)/2, SA = Sw - aa, SB = Sw - bb, SC = Sw - cc,

and abbreviate

sx.sy. ... .sz to sxy...z, SX.SY. ... .SZ to SXY...Z.

The notations Sw,SA,SB,SC generalise to

Stheta = S.cot(theta),

where S = 2Delta = root(soabc)/2.

These conventions enable us to express the coordinates of many points very succinctly. In particular, there is a simple formula for the coordinates of a point P for which one knows the angles made by two Cevians with the sides - for example in the figure

C-------A

\C' A'/

\ /

the coordinates of P are

( SC + SC' : -bb : SA + SA' )

and the perspector of the triangle formed by 3 such points is

1 1 1

( ------- : ------- : ------- ). ]

SA + SA' SB + SB' SC + SC'

John Conway

I'm not sure whether you noticed that Paul corrected your message on this? He says the focus is the proSteiner point pS = (:bb/(cc-aa):) itself, rather than the slightly more complicated thing that you gave. Do you agree?

Presuming that this is right, it makes me happier, essentially because we don't have to coin a new name, and because it neatly exemplifies the virtues of the present system of prefixes. The conjugation chain looks like

points comments

.....

rS = (:1/bb(cc-aa):) on Steiner ellipse

quo

gS = (:bb(cc-aa):) = %l at infinity (= "Latitude End")

S = (:1/(cc-aa):) on Steiner ellipse & circumcircle

quo

tS = (:(cc-aa):) = %p at infinity (= "Polar End")

pS = (:bb/(cc-aa):) on circumcircle (= KP focus)

quo

.....

and we don't really need any other names. I think it's nice that the center of the Kiepert hyperbola is also simply related to S, being the medial Steiner point

mS = (:(cc-aa)^2:).

The fact that so many geometrically important points now have such simple names seems to indicate that we've managed to pick out the right prefixes. To my mind, the shortness of the resulting names is very important, as is the fact that one can easily translate them into coordinate form. In this case, for example, when we tell the reader that "the focus of the Kiepert parabola is the proSteiner point" he'll be able to know and remember its coordinates (:bb/(cc-aa):) in terms of those of S. I hope you're also beginning to appreciate the virtues of the short notations!

I add to this for completeness Barry's evaluation of the vertex of the KP as:

(2SCA-SAB-SBC)^2 (2SCA-bbSB)^2 (3SCA-SS)^2

(:----------------:) = (:-------------:) = (:-----------:).

cc - aa cc - aa cc - aa

About the Kiepert hyperbola: I think I once showed that the axes of this are parallel to those of the Steiner ellipse, but was unable to reconstruct this the other day, and now am not quite sure that this was the thing I proved. So can you draw them both in the same picture to check? If not, it was probably the axes of some other hyperbola; I think if we can find what's true it should be fairly easy to prove.

Regards, JHC

Steve - I passed this by, thinking that it was about the Kiepert hyperbola, of which I think I have fairly good control. This will be a good opportunity to learn, and I'll make comments as I do so. (near the end I go into calculating mood - sorry!)

> Let A'B'C' be the vertices of "isosceles Napoleons" constructed on > the sides of ABC. The Keipert hyperbola is the locus of the > perspector of this triangle with ABC as the base angle of the > isosceles triangles are varied.

Since we'll probably keep something like the "quo" solution, we'll be able to use the terms "isoperspector" and "(Kiepert's) isoperspector hyperbola"...

> The Kiepert parabola is the envelope of the axes of perspective.

... also "isoperspectrix" and "(Kiepert's) isoperspectrix parabola", in accordance with our general position that descriptive names are best.

> The KP is inscribed in ABC. Its focus is :(cc-aa)/bb: which is

> related to the Steiner point. Its directrix is the Euler line.

Let's think about that relation, with a view to naming things. Apply "pro" twice to the focus F:

>From F = :(cc-aa)/bb: we get

pF = :cc-aa: = %p (ie., the point at infinity perp to e,

so on the polar axis p), then

ppF = :(cc-aa)bb: = %l (ie., the point at infinity on the Lemoine

latitude line, perpendicular to m).

These are in the conjugation chain of the Steiner point S:

... co F quo pS co %p quo S co %l quo rS co...

so the name-two-of-a-chain principle tells us to try to find a short name for &p, say "the Polar point". Then F and %l become the "retroPolar" and "proPolar" points.

> The sides of the triangles are tangent, as is the line at infinity,

> as is the Lemoine axis.

Of course we'll just say that these lines are isoperspectrices.

I note that the line at infinity is tangent at the Polar point %p,

and wonder about the contact point of the Lemoine latitude line.

> Its Brianchon point (the point of concurrence of the contact points

> and ABC) is the Steiner point.

For me this is "the perspector of this conic (and ABC)" because this is a much more general term. Namely, ABC will have a dual triangle with any conic, and by the perspector of the conic (and ABC) we mean the perspector of that triangle (and ABC). Let's recall that there's a unique circumconic with a given perspector (u:v:w), namely u/X + v/Y + w/H, and a unique inconic with that perspector, namely

root(X/u) +- root(Y/v) +- root(Z/w) = 0 (loosely)

or more precisely

XX/uu + YY/vv + ZZ/ww - 2YZ/vw - 2ZX/wu - 2XY/uv = 0,

agreeing with your

> ff XX + gg YY + hh ZZ - 2 fg XY - 2 gh YZ - 2 hf ZX = 0.

Let me first find the contact points of the edges, and then that of the latitude line. Putting Y = 0 gives X/u = Z/w, so the edge b touches at (u:0:w) = ( 1/(bb-cc) : 0 : 1/(aa-bb) ), a Cevian foot of the Steiner point S.

Now for the latitude line l. This joins the versoBrocard points - how lightly that trips off the tongue! - which I find from the general formula vP = (: aabbccY.sigma - 2bbSB.phi :), where for Oe = (1/bb:1/cc:1/aa) we have phi = SUM(aa/ccaa) = sigma, so the formula simplifies to

vOe = (:aabbccY - 2bbSB:) = (:bb(aa-cc-aa+bb):) = (:bb(bb-cc):).

But this (and similarly vOw) obviously lies on (1/aa:1/bb:1/cc|), which must therefore be the equation of the latitude line. I'll rescale it to (bbcc:ccaa:aabb|0) to avoid fractions. Oh, come to think of it, I could have found this from the Schoute formula

(bbcc:ccaa:aabb|Sw+Stheta) by putting theta = -w;

still, it's a good check.

I want this line to be the polar of (u:v:w) with respect to the isospectrix parabola, namely

... + ggvY + ... - fguY - ... - ghwY - ... = 0

where I've displayed only the Y terms. This gives three equations like

fgu - ggv + ghw ~ ccaa

(with the same implied constant in ~), or (changing this constant)

fu - gv + hw ~ hfccaa to which I add

fu + gv - hw ~ fgaabb to get (new constant)

fu ~ faa(cch-bbg), u ~ aa(cch-bbg)

into which I substitute f = bb-cc &c to get

u ~ aa(ccaa-ccbb-bbcc+bbaa).

So the latitude line touches the isoperspectrix parabola at

( : bb(aabb+bbcc-2ccaa) : ),

which is the pro of the point at infinity on the line GJ.

By the way, a day or two ago I redrew this picture of the latitude and meridian lines. I don't know how much of it you're familiar with:

| <-------- Lemoine latitude line l

vOe

/ | \

Oe / | \

/ vKw \

/ | \

-----O----bO-In--K--vK-----Is-------> Brocard meridian line m

\ | /

\ vKe /

Ow \ | /

a Brocard point^^ \ | /

\ | /

vOw <--- a Beltrami versoBrocard point

I couldn't draw Ke and Kw themselves - they are the other places where the circles through In,Is,Oe and In,Is,Ow cut the Brocard circle - but I've drawn their inverses, which are the centers of the equilateral triangles marked (and which therefore form equilateral triangles with either In or Is).

[Notation: on the globe, Pe and Pw denote the points that are 120 degrees due east or west from P.]

Regards, JHC

On Wed, 26 Apr 2000, Paul Yiu quoted Steve:

> The KP is inscribed in ABC. Its focus is :(cc-aa)/bb: which is

> related to the Steiner point. Its directrix is the Euler line.

and wrote:

> The focus of the Kiepert parabola is (:bb/(cc-aa):).

In other words, it is the "proSteiner point".

> It is the antipode of the isogonal conjugate of the infinite point

> on the Euler line, and has many interesting properties, including:

Let me discuss terminology here, since I may have been the one to introduce "antipode" for the point directly opposite to a given one through the circumcenter. Unfortunately, it now has another natural meaning, namely the antipode in the global picture, so some time ago I switched to "negative", with the abbreviating prefix "nega-".

Now the isogonal conjugates of a point P and its negative nP are points at infinity in orthogonal directions, so Paul's assertion is that the focus is isogonal to the point %p at infinity on the polar axis p. ["%" is email for the infinity symbol.]

The fact that (:cc-aa:) is this point is equivalent to the fact that the isotomic conjugates

(bb:cc:aa) and (cc:aa:bb)

of the Brocard points lie on a line perpendicular to the Euler line e. So we have three such lines: this one, the polar axis, and the symmetry axis of the Kiepert parabola.

Let me try to find the vertex of the KP. The equation of the Euler line is

( (bb-cc)SA : (cc-aa)SB : (aa-bb)SC | 0 )

and I'll first find the projection of the focus on this, which has the form

( : bb/(cc-aa) - t(cc-aa) : )

(since (:cc-aa:) is perpendicular to e), giving the equation

t.SUM( SB.(cc-aa)^2 ) = SUM(bbSB) = 2SS.

What is the left-hand sum? I write it as SUM( SB(SA-SC)^2 )

so see that it's

SAAB + ... (to 6 terms) - 6SABC,

and compare that with aabbcc, which equals

(SB+SC)(SC+SA)(SA+SB) = (same sum) + 3SABC

to see that it's aabbcc - 9SABC = SSSw - 8SABC.

So t = 2SS/(aabbcc-9SABC),

and therefore the above point is

( : (aabbcc-9SABC)bb/(cc-aa) - 2SS(cc-aa) : )

which I doubt can be greatly simplified. The vertex of the parabola is the midpoint of this and the focus, which I don't feel inclined to attempt!

> X_(110) is the point of concurrency of the reflections of

> the Euler line in the side lines.

It seems that the focus is by far the most natural of these three points.

------------------------------------------------------------

On another matter, I've been in communication with Steve about the fact that each of the two most recent notations has defects not shared by the other one, and we've agreed on a hybrid system that has the advantages but not the defects, namely

dP = super-P = (:Z+X-Y:) mP = medial-P = (:Z+X:)

gP = co-P = (:bb/Y:) tP = isotomic-P = (:1/Y:)

pP = pro-P = (:Ybb:) rP = retro-P = (:Y/bb:)

bP = Brocard P = (:ccX+bbY+aaZ:) aP = antiBrocard P

vP = verso-P = (:aabbccY-bbSBphi:) nP = nega-P = 2O - P

where in the last line O and P are supposed normalised.

The only new one is "quo", for the isotomic conjugate, or "isotomic quotient", which neatly rhymes with my old "co" for the isogonal conjugate or "conjugal". By changing to this from my old "iso" we enable "i" to be used for "infra", and also can use "iso" to abbreviate "isosceles", which has great advantages. For example, similar isosceles triangles on the edges are "isoNapoleons", and the perspector of their apices is an "isoperspector". So the Kiepert hyperbola is the locus of isoperspectors.

John Conway

On Wed, 26 Apr 2000, Barry Wolk wrote:

> So the vertex [of the Kiepert isoperspectrix parabola] is

(... :(3SAC-SS)^2/(c^2-a^2) :...)

> which is just a little simpler.

Oh good! Since SS = SBC+SCA+SAB, we have

3SCA - SS = 2SCA-SAB-SBC,

which is the typical coordinate of the "Euler end" (ie., the point at infinity on the Euler line):

%e = (:2SCA-SAB-SBC:),

which might be trying to teach us something.

(:(aaaa-2bbbb+cccc-2bbcc+4ccaa-2aabb)^2/(cc-aa):),

for instance.

Let me summarise this notation for new readers, who should understand that letters immediately following an s or S are subscripts, and that w is the Brocard angle omega. I define

so = (a + b + c)/2, sa = so - a, sb = so - b, sc = so -c

Sw = (aa+bb+cc)/2, SA = Sw - aa, SB = Sw - bb, SC = Sw - cc,

and abbreviate

sx.sy. ... .sz to sxy...z, SX.SY. ... .SZ to SXY...Z.

The notations Sw,SA,SB,SC generalise to

Stheta = S.cot(theta),

where S = 2Delta = root(soabc)/2.

Date: 12/01 11:46 PM

Received: 12/02 4:01 AM

From: Steve Sigur, ssigur@netdepot.com

To: John Conway, conway@math.Princeton.EDU

CC: geometry-college@forum.swarthmore.edu

SS> If you take the axes of perspective of the points on the Euler line,

you

SS> get a set of lines whose envelope is a parabola.

JC>My guess is that this is the Kiepert parabola, which is the one that

JC>shows its face most often, so to speak. But I'll have to check.

Date: 12/02 11:23 AM

Received: 12/02 2:31 PM

From: John Conway, conway@math.Princeton.EDU

To: Steve Sigur, ssigur@netdepot.com

CC: geometry-college@forum.swarthmore.edu

This parabola is not the Kiepert parabola, which has the Euler line as

directrix. The axis of this parabola is not perpendicular to the Euler

line.

Like the Kiepert parabola it seems to be tangent to the sides of the

triangle (not proved, just observed in Geometer's Sketchpad).

On Tue, 1 Dec 1998, Steve Sigur wrote:

> SS> If you take the axes of perspective of the points on the Euler line,

> you

> SS> get a set of lines whose envelope is a parabola.

> This parabola is not the Kiepert parabola, which has the Euler line as

> directrix. The axis of this parabola is not perpendicular to the Euler

> line.

OK, so let me look at it. First of all, let me say that I now see

that you used the term "axes of perspective" for trilinear polars; the

tripolar of P is the axis of perspective (or "perspectrix") of ABC

with the Cevian triangle of P.

> Like the Kiepert parabola it seems to be tangent to the sides of the

> triangle (not proved, just observed in Geometer's Sketchpad).

This at least is obvious because of the way the tripolarity relation

collapses at the edges. Let me explain this. If P = (U:V:W) then the

typical vertex B_P of its Cevian triangle is (U:0:W) whose harmonic

conjugate w.r.t. A and C is (-U:0:V); so the tripolar of P is

the line X/U + Y/V + W/Z = 0. [I'M thinking in barycentrics of course,

but you can think in orthogonal trilinears; since this is a projective

construction the formulae will be the same.]

Well, all that is strictly valid only if UVW =/= 0; ie., if P

isn't on an edge. If it is, say if V = 0, the tripolar actually IS

that edge. So tripolarity fails to be 1-to-1; the tripolar of any

(nonvertex) point on an edge is that edge. [The tripolar of a vertex

fails to be uniquely defined - it "is" the set of all lines through

that vertex.]

Everything we've said so far, by the way, applies to any line L

through G, not just to the Euler line. [Reason - in barycentrics -

the tripolar of G = (1:1:1) is the line at infinity X+Y+Z=0, showing

that the conic which is the envelope touches the line at infinity.]

We can find the equation of this conic by taking a point on L

that's near an edge but not on it. I let P = (U:V:W) be the tripole

of L; then R = (-U:0:W) will be where L meets CA. Now let

Q = (u:v:w) be the point at infinity on L. Then (using e for

"epsilon") such a point will be

R + eQ = (-U+eu,ev,W+ew), whose tripolar will be

X/(-U+eu) + Y/ev + Z/(W+ew) = 0

which cuts CA where X/(-U+eu) + Z/(W+ew) = 0, in which I can now

afford to neglect e to see it's at (U:0:V). [Obviously I didn't

in fact need to introduce e ; I thought the argument was going to

go rather differently.]

The argument shows that this parabola is the inconic whose

perspector is P, which I suppose you've already noticed, so

its equation is

+- root(X/U) +- root(Y/U) +- root(Z/W) = 0

or more strictly, what you get by rationalizing this :

(X/U)^2 + (U/V)^2 + (Z/W)^2 = 2( YZ/VW + ZX/WU + XY/UV ).

Now for the general inconic the director circle is, as I said,

the pedal circle of either focus; so in this case, it's the "pedal

circle" of both Q and its conjugal point Q* - in other words

the Simson line of Q*.

As I said, all this holds for a general L through G - the

focus of the corresponding parabola is the point Q* on the

circumcircle that's conjugal to the infinite point on L, and its

directrix is the Simson line of Q*. This is actually the

general parabola inscribed to ABC.

The identification of the particular case when L is the Euler

line will have to wait for a time!