Kiepert hyperbola

Conway

... but remark that I prefer to use "extraversion" inclusively, so that anything is the trivial extraversion of itself. The line joining a nubile pair of points is indeed strong. (It's the only strong line through either of those points.)

Now, about isogonics, isodynamics, meridian axis, etc. I'll be writing from memory as usual, but hope I won't get anything wrong.

Lurking around here is a generalization of the notion of pre-pivot point. Namely, let tau = alpha,beta,gamma be any triple of angles. Then erect triangles on the three sides of ABC, with base-angles of alpha (at A), beta (at B), gamma (at C). Their apices I call A^tau, B^tau, C^tau:

A^tau-----B-----C^tau

\ / \ /

C-------A

\ /

B^tau

It's a theorem that this triangle of apices is in perspective with ABC - the perspector I call P^tau. In the special case when alpha+beta+gamma = pi, P^tau is the prepivot for triangles of shape tau. However, what we're really concerned with here is another special case, when the three angles are equal, to theta, say, in which case I write theta for tau.

Pursuing the general case for a moment, I'll give coordinates - we have:

B^tau = ( cotC + cot(gamma) : -cotC-cotA : cotA + cot(alpha) )

and so

P^tau = ( : 1/(cotB + cot(beta)) : ).

whose isotome and conjugal are

(P^tau)| = ( : cotB + cot(beta) : )

(P^tau)* = ( : bb(cotB + cot(beta) : ).

[I've changed my sign-convention a few times, and am a bit worried that perhaps alpha, beta, gamma should be negated in the above? The exact convention won't affect much of what I say, so I won't stop to check now.]

Now let's put alpha=beta=gamma=theta. Then we see that the loci of the last two points mentioned are both straight lines, namely the isotome, which I'll abbreviate to Ptheta|, is on the line joining

K+ = ( :cotB: ) = ( : cc+aa-bb : )

G = ( :cot(theta): ) = (1:1:1),

which we see is the symmedian trail GK, while the conjugal Ptheta*, joining

O = ( :bb(cc+aa-bb): )

K = ( : bb.cot(theta): ) = (aa:bb:cc),

is on the meridian axis OK.

What's the locus of Ptheta itself? It's easiest to see this as the conjugal of OK, which is a conic passings through A,B,C and the conjugals

O* = H of O and K* = G of K.

Since this passes through the orthocentric set {A,B,C,H} it's a rectangular hyperbola, called the Kiepert hyperbola.

latitude

Now let's take particular values for theta: |

Ow |

| \ |

| \ _

meridian------------O----Is-O=--J+--K---K2----------In---K-----axis

| / |

Oe |

| | | | | | | |

pi pi 2w w O -w -pi

-- -- ---

2 3 3

I'm not sure if I've put Is on the correct side of the Brocard center O= ; we should soon see. (In fact I did!)

Let's still further abbreviate (P^theta)* to just

(theta) = ( : bb(cotB + cot(theta) : ).

Then since cot0 = infinity and cot(pi/2) = 0 we have

(0) = (:bb:) = K, (pi/2) = (:bbcotB:) = O.

I should say how I think, which is usuing the formulae

4DELTA.cotB = cc+aa-bb (etc.)

4DELTA.cot(omega) = aa+bb+cc

which show that the cots of various angles are proportional to simple functions of aa,bb,cc.

Now put theta = -omega:

(-omega) = ( : bb( cotB - cot(omega) : )

= ( :bb( cc+aa-bb -aa-bb-cc): ) = ( :bbbb: )

a point I must admit to calling K^2, abbreviated to K2 above.

Similarly

(omega) = ( : bb(cc+aa-bb + aa+bb+cc) : ) = (:bb(cc+aa):) = J+

and (this one I'll leave to you):

(2omega) = O=.

I've just gone back to insert all these angles in the Figure, writing w for omega. In fact theta is the angle

O (Ow or Oe) (theta).

Some other interesting values of theta are

+- pi/6 - the perspectors of the Napoleon triangles

( w - pi/2 ) = the point at infinity on the meridian axis.

My "meridian axis" OK has been called "the Brocard axis", so its full name is "Brocard's meridian axis". Similarly, my "latitude axis", which is the perpendicular to OK through the inverse of K in the circumcircle, has been called "Lemoine's axis", so its full name is "Lemoine's latitude axis". [There'll be a piece in the book explaining that the nomenclature "Somebody's something" does refer to a "something" that was discovered by Somebody, but does not imply that Somebody actually used that name for it.]

Now my convention is that (P^tau)* = P_tau (which, when the angles of tau add to pi, is the pivot for tau-shaped triangles). Taking tau = +- pi/3 (which I'll call "60") in this, we get

the (normal) Fermat point Fn = P_60

and

the switched Fermat point Fs = P(-60),

which we see lie on the Kiepert hyperbola, as do the conjugals of all the other points on OK.

I think I'd better stop there for now, leaving you to disentangle whatever misapprehension led you astray. I note that since the Fermat points are a nubile pair, the line joining them is a strong one (which I don't think I've studied), and their midpoint a strong point (ditto). Perhaps these would repay further investigation.

Steve - further investigation of the quadrangle Fs Is Fn In led to the following figure, that contains quite a few properties of the Kiepert hyperbola (h).

T-----------*-----------U cuts h at T = Tarry, and U

Fs----------Is-------> tgt to h at Fs

|\ /

| \ /

| mS-vK--/-----> center of h is mS

| K /

| \ /

In----Fn-------> tgt to h at Fn

| / \

|/ \

O-----------G-----N-----R-----------H---> e cuts h at G,H

The horizontal lines are all parallel to the Euler line e, and the chords they cut off from h are all bisected by the Fermat line f. For the horizontal lines through Fn, Fs these chords shrink to those points; ie. these lines are tangents to h. K is the pole of e with respect to h, so that the tangents at G and H pass through K.

The center of h is mS, and so this is the midpoint of Fn--Fs, H--T, G--U. Also, since the segments Fs--Is and Fn--In are horizontal, and vK is the midpoint of Is--In, the segment mS--vK must also be horizontal (meaning: parallel to the Euler line). All three of thse segments are bisected by the symmedian track k = GK.

Lots of points on the figure are simply related to the Neuberg cubic - for instance Fn,Fs,In,Is,H,O,%e are on it, %e being its pivot. This should get more study!

We've met quite a few of these properties at various times, but never all at once like this. Plan for TB: put practically all of them into this figure and its caption, so that we say a lot in a little space. Perhaps there should be another figure showing the Spiekers on h?

JHC

One great advantage from the great upheaval is that we can now afford to use "iso" as an abbreviation for "isosceles". So I plan to speak of "isoNapoleons" (= isosceles Napoleons), whose perspectors are the "isoperspectors" P_theta = P_(theta,theta,theta), isogonal to the corresponding "isopivots" P^theta = P^(theta,...,theta).

(These are particular cases of the generalized perspectors and pivots P^tau P_tau, where tau = al,be,ga is any triple of angles, not necessarily adding to pi).

Kiepert's great theorems are that every isoperspector lies on K's hyperbola, while every isoperspectrix (of course we have those too) touches his hyperbola. Correspondingly, every isopivot lies on the Brocard meridian isogonal to the K H.

We should mention notable particular iso-p-things, namely

isoperspectors A,B,C, H, Fn, Fs, sM, rG, Spiekers,...

isopivots O, In, Is, M, pK,iO, symSpiekers,...

isoperspectrices a,b,c e, ... .

I rather like saying, for instance, that "The Spieker points are isoperspectors" rather than saying that they're on K's H. By making the full name of the curve be "Kiepert's Isoperspector Hyperbola", we can effectively make the two forms equivalent, and at the same time help our readers to remember one of his theorems; a beautiful application of our "descriptive name" policy.

Similar comments apply to "Kiepert's Isoperspectrix Parabola".

At the time I was in Chicago, Larry Evans et al were finding some generalizations of some facts about In,Is,Fn,Fs that we noticed about a year ago. I didn't take much notice at the time, but have recovered them since, and will decribe them now.

Fs------Is----------------------> %e

\ /|

\ K |

\ / |

In---Fn----------------------> %e

/ \ |

O------G-----R-----------------> %e

The original properties are summarized in the above diagram. They are that Fs--In and Fn--Is meet in G, that Fs--Is and Fn--In are parallel to the Euler line, and that Fs--Fn passes through K and the Ringcenter R. The assertion that "the diagonal triangle of Fs Is Fn In is G K %e" contains

most of them.

All these generalize to an arbitrary quadrangle of the form P^(+-theta), P_(+-theta), with the replacement of R and %e by (1/(3-2ss))e and (1/(3-4ss))e, where s = sin(theta).

I'll draw a little table of particular cases:

theta 1/(3-2ss) 1/(3-4ss) pers pivs Euler points

0 1/3 1/3 G K G, G

w (1+tt)/(3+tt) (1+tt)/(3-tt) sM,rG M,pK

pi/6 2/7 1/2 Nn,Ns (2/7)e, N

pi/4 1/2 1 Vn,Vs N, H

pi/3 2/3 % Fn,Fs In,Is R, %e

w' (1+tt)/(1+3tt) (1+tt)/(1-3tt) T, %m,iO

pi/2 1 -1 H O H, L

The columns "pers" and "pivs" give the names of the points P^(+-theta), P_(+-theta), while the two "Euler points" are those with parameters 1/(3-2ss) and 1/(3-4ss) that generalize from R and %e in the "Fermat/Isodynamic" case.

One of the recent "minidiscoveries" was that P_(+-theta) are mutual inverses in the Brocard circle. I've used Vn,Vs in the above for Vectin's Pythagorean perspectors, but still think it might be better in TB to call them Pn,Ps. We don't seem to have named the isogonals of these or the Napoleonic perspectors.

In the case theta = omega' it's one of the perspectors that's unnamed, namely the isogonal of the center of the Brocard circle, which we now see is also the inverse in that circle of the Tarry point T. The fact that it's the only unnamed perspector in the above table suggests that it should have some nice geometrical properties.

We should give some form of this table in TB, in the "isothings" section, and expand it to include the Spiekers. That section should also say something about the isotomic transforms of all these perspectors, which lie on the symmedial track (the isotome of K's H).

This "isothings" section should, with luck, be hardly longer than one that just stated Kiepert's theorems, but manage by including this table to say lots of things about a host of interesting points.

JHC

I sent you a message on the KH, in which there appeared the point U that completes the parallelogram TGHU. Let me draw a figure:

T------U

|\ \

| \ mS \

| \ \

O---G------H

The four vertices T,U,G,H of this parallelogram all lie on the KH, whose center is mS. As you'll see, this implies that U is the reflection of S in G, and since S lies on the Steiner ellipse, whos ecenter is G, we see that U must also lie on the Steiner ellipse. Now S is traditionally defined to be the 4th intersection (after A,B,C) of the Steiner ellipse and the circumcircle, while the Tarry point T is the other end of the diameter through S. So T is the 4th intersection of KH with the circumcircle, and U its fourth intersection with the Steiner ellipse.

[PY]

Dear John and friends,

[continuing]

It is interesting to find the intersections of the Fermat line with the orthocentroidal circle. I found that

(b^2-c^2)^2+ t a^2 : (c^2-a^2)^2 + t b^2 : (a^2-b^2)^2 + t c^2

lies on the circle GH if and only if

t^2 = a^4+b^4+c^4-a^2b^2-b^2c^2-c^2a^2.

This means that these two intersections divide harmonically K and the Kiepert center. In other words, the symmedian point and the Kiepert center are inverses

with respect to the orthocentroidal circle!

On Fri, 18 Aug 2000, Lang wrote:

> About inflection points:

> 1) O is an inflection point of the Darboux cubic, with inflectional tangent

> equal to OK (K = Lemoine point).

Oh yes; of course I knew THAT one, which is obvious from the symmetry. I believe I proved once that the other 8 require the solution of an octic that's irreducible in general.

> 2) The isogonal inverse of OK is Kiepert hyperbola, so the cubic of Darboux

> is tangent to this hyperbola with triple contact

[at H, of course]

the other 3 points of intersection are A, B and C.

3.4. X671 . The point P = X671 = :1/(SC +SA−2SB) is the antipode of the Steiner point on the Steiner circum-ellipse. It is also on the Kiepert hyperbola, with Kiepert parameter −arccot( 1 /3 cot ω), where ω is the Brocard angle. In this case, the circumcenters are on the altitudes.