Write about Steiner ellipse here.

Dear Hyacinthians,

I am trying out both a new technique and a new theorem I discovered in  Kepetis' very great book on triangle geometry.

First the theorem. The axes of a circumconic can be constructed as  follows:  Find the 4th intersection of the circumconic with the  circumcircle. Draw the Cevian lines to this point. The bisectors of the  A-Cevian with edge a (and similarly for the other directions) are  parallel to the axes of the circumconic.

I like this theorem. Now the technique. There is a dot product on lines  which has always fascinated me.   It can be seen in this: the distance  from point P1 ~  < x1 : y1 : z1 > and line L1 ~ [l : m : n]  ->  lx +  my + nz = 0, where I am using angle brackets for points and square  brackets for lines) is in barycentrics

S   ( l x1 + m y1 + n z1)/  |L1|

where    |L1|  =   aa ll  +  bb mm + cc nn - 2 SA mn - 2SB nl - 2 SC lm 

,

Since bisectors are equally distant from their defining lines, the  bisectors of the lines L1 and L2 are

L1/ |L1|   +-  L2/|L2|

Now to compute the Steiner axes with this formula. Since I do not know  their equations, I would appreciate someone telling me if the resulting  formulas are correct!

The B-Cevian has form      n z - l x = 0 and the b edge is   y = 0.

Normalized these are    (n z - l x)   /  root(ll aa + nn cc +  2 SB ca)  = 0    and   y / b = 0.

Hence the bisectors are nicely and easily

n z - l x  +- root(ll aa + nn cc +  2 SB ca)/b   y  = 0.

To find the actual axes of the conic, we translate these lines by  adding lambda (x+y+z) and solving for lambda by putting in the  coordinates of the center.

Since the 4th intersection of the of the Steiner circumellipse and the  circumcircle is S, the Steiner point, I have, after centering the axes  to go through G, the center

The Steiner axes

(2aa-bb-cc +- R) x + ( 2bb - aa - cc -+ 2R) y + (2cc-aa-bb +- R) z = 0

Note that this goes through G. 

where    R =  root(sum (aaaa - bbcc) ) 

Can anyone tell me if this is right or wrong?

Steve

From: John Conway <conway@...> 

Date: Mon Jan 24, 2000  11:17 pm 

Subject: The foci of the Steiner ellipse conway@... 

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Clark and Antreas - I thought you might like to be informed about my study of Paul Yiu's formula for the foci of the Steiner (circum)ellipse. 

Paul - I've progressed a little further, as you will see.

First, there's a simple proof that there can be no continuous function f(A,B,C) of a triangle whose value is always one such focus. Namely, we restrict to the set of triangles having a given Steiner ellipse, which I take as xx/aa + yy/bb = 1. The vertices of the general such triangle are then

( a.cos(theta + 2n.pi/3), b.sin(theta + 2n.pi/3) ) (n = 0,1,2).

Now if one continuously increases theta until its new value is pi more than the original one, the triangle gets turned through pi about G, but the chosen focus remains fixed. So, "reciprocally", the triangle can be returned to its original position with the foci interchanged.

But Paul has given formulae for the barycentric coordinates of a focus, namely

2(cc-aa)

( : 1 + ----------------------- : ) = (X:Y:Z), say,

root[ (bb+q)^2 - ccaa ]

where q denotes a square root of aaaa+bbbb+cccc-bbcc-ccaa-aabb.

It behooves us to find the discontinuities in this formula. The square root in the denominator of Y can vanish only when bb+q = +-ca, ie., q = ca - bb (since we suppose our triangle has only positive edgelengths, so only when

aaaa+bbbb+cccc-bbcc-ccaa-aabb = qq = ccaa + bbbb - 2bbca,

ie.,

(aa-cc)^2 - bb(a-c)^2 = 0,

ie.,

(a-c)^2.[(a+c)^2 - bb] = 0

ie.,

(a-c)^2.4sob = 0,

ie.,

precisely when a = c.

But then the numerator vanishes also, so the formula becomes indefinite. So the answer given can be discontinuous only for isosceles triangles.

But we also see that if the value does jump, it can only do so along the line FB. Now for a "squat" isosceles triangle (one in which b > c = a) the two foci aren't in line with B (the line they define is parallel to CA), so for such triangles the value doesn't in fact jump.

However, the argument I gave at the beginning proves that it must jump somewhere, and therefore at some "tall" isosceles triangle (b < c = a), since these are now the only possible places. By symmetry, if it jumps for one such triangle, it must jump for all, and this it therefore does.

Several things still puzzle me, though. There are 4 square roots in Paul's formulae, namely the "little" one that produces q, and the three "big" ones like that in the denominator of Y. Just what signs can be taken in these? I had tacitly supposed we could take them all to be positive, but maybe Paul has not actually proved this? Since there are 16 choices of signs and only four foci (counting the imaginary ones), it doesn't seem that all sign-combinations are possible. Just which ones work? Of course the answer should show us that only two of these roots are rationally independent.

John Conway

Did you also send these to Kimberling? I hope so, but don't see his email address in your list. After your first sentence, I'm not sure if you really meant "Steiner circumellipse" in the last one above, rather than "Steiner inellipse", which would of course only change that big vector by a factor of 2.

I'm very interested to see the form of these, because I seem to have proved this morning that there's no continuous way to distinguish one Bickart point from the other. Namely, if the Steiner (circum)ellipse is xx/aa + yy/bb = 1, we can take the three vertices to be ( acos(theta + 2n.pi/3), bsin(theta + 2n.pi/3) ) (n = 0,1,2). Now, fixing the ellipse, increase theta continuously until it reaches its initial value plus pi. Then the triangle ends up rotated by pi from its original position, while the foci have stayed fixed, so that with respect to the triangle, they get swapped.

This seems to be incompatible with your coordinates. I'm not saying that they're wrong, merely that I don't understand the situation! Another thing that I don't understand is what has happened to the imaginary foci. Yet another is the degree 16 appearance of your coordinates. Since there are only four foci, this must of course be illusory, and I think the best way to get a better understanding is to find the way to express them (which must exist) with fewer square roots

Let me rewrite them so that I can stare at them!

          2(cc-aa)

Focus = ( : 1 + --------------------- : ),

          root[(bb+q)^2 - ccaa]

where I've written q for root[aaaa+bbbb+cccc-bbcc-ccaa-aabb] (a function which, by the way, arises all over the theory of the Brocard triangle). It's easy to see that q vanishes only for equilateral triangles, so provided we avoid those, we can't continuously vary our triangle so as to negate q. 

I suppose that somehow the three "big" roots must be dependent, and so I guess that their product is already a rational function of (aa,bb,cc and) q. Let me try to work out this product:

(bb+q)^2 - ccaa = bbbb-ccaa+aaaa+bbbb+cccc-bbcc-ccaa-aabb + 2bbq

= aaaa + 2bbbb + cccc - bbcc - 2ccaa - aabb + 2bbq

... no - it's obviously going to get too complicated for me!

Let me try another tack - can we continuously vary the triangle so as to negate one of the three "big" roots? (The answer must be "yes" if we believe my result that the two foci can be continuously interchanged.) If we can change the sign of (bb+q)^2 - ccaa, then at some point we must have bb+q = ca, so q = ca-bb, so

aaaa+bbbb+cccc-bbcc-ccaa-aabb = qq = ccaa+bbbb-2bbca

so

aaaa + cccc - 2ccaa - bbcc - bbaa + 2bbca = 0,

i.e.,

(aa-cc)^2 -bb(a-c)^2 = 0,

i.e.,

(a-c)^2.[(a+c)^2 - bb] = 0

i.e.,

(a-c)^2.4sob = 0,

which is precisely when a = c, for ordinary triangles for which so and sb can't vanish.

Forgive me for boring you, but I feel I really do need to understand how these foci vary, and think I might be getting there.

That denominator = (bb+q+ca)(bb+q-ca), and we've proved that

(q+bb-ca)(q-bb+ca) = qq - (bb-ca)^2 = (a-c)^2.4sob,

which extraverts to

(q+bb+ca)(q-bb-ca) = qq - (bb+ca)^2 = -(a+c)^2.4sca.

Curses! I'm so nearly there, but STILL can't see what's going on!

I'll send this off anyway.

John Conway

Congratulations. It was a great pleasure : it is a quite interesting and well documented paper. I remember that, a long time ago, I found a quite elementary method in order to get the values of the remarkable parameters of the outer Steiner ellipse. Let a, b, c be the lengths of the sides of ABC e the excentricity p the semimajor axis - thank you, Bernard, for your lexicon q the semiminor axis

      r = GF = sqrt(p^2 - q^2)

>In usual cartesian coordinates - an horror, I know -, we have

A [ p cos(t), q sin(t)] and for B, C the same with t -> t + 2 Pi/3, t -> t -

2 Pi/3 - if, for a triangle, circumcenter = centroid, the triangle is equilateral -

>

Using dotprod(AB, AC) = Sa, ...

we come to a very simple system to find p^2, e^2, tan (2 t).

If (b^2 - c^2)^2 + (c^2 - a^2)^2 + (a^2 - b^2)^2 = 2 d^4,

I get - if I didn't mistake -

p^2 = 1/9 (a^2 + b^2 + c^2 + 2 d^2)

q^2 = 1/9 (a^2 + b^2 + c^2 - 2 d^2)

r = 2/3 d

tan(2 t) = sqrt(3) (c^2 - b^2)/(b^2 + c^2 - 2 a^2)

and symmetric expressions for tan( 2 t + Pi/3) and tan(2 t - Pi/3). Using the areas of FAB, FBC, FCA we can easily find the barycentric coordinates for each focus

X = 1/2 + e cos(t)

Y = 1/2 - e cos(t - Pi/3)

Z = 1/2 - e cos(t + Pi/3)

The length of the six cevians is 3/2 p

So, I have a question - I don't know the answer - :

WHAT ARE THE VALUES OF t IN TERMS OF THE ANGLES A, B, C?

Sure, the answer would be very interesting for Clark Kimberling!! Sorry for my so bas English language and friendly from France. JPE

Here are the algebraic thoughts I mentioned. Connected with the Steiner ellipse(s) are the "four great trails",

e = GO = GH = GN = GL , the Euler line or trail

k = GK , the symmedian trail

s = GS , the Steiner trail

t = GT , the Tarry trail.

The subordinate and superior maps fix all these trails, while the Brocard minor and major maps swap e and t, k and s (since O is the minor of T, and K the minor of S).

This "Brocard minor map" is the antisimilarity that takes A,B,C to the vertices of the Brocard minor triangle, or "Brocard's first triangle". These vertices are (aa:cc:bb), (cc:bb:aa), (bb:aa:cc).

Now I discovered some years ago that the Steiner or inertial axes are the angle-bisectors of k and s, or equally of e and t. To be more precise, the minor map negates the major axis, and then rescales by what I call "the Brocard factor" f = q/(aa+bb+cc), where

qq = aaaa+bbbb+cccc-bbcc-ccaa-aabb.

So the directions of the axes are those of the vectors

f(S-G) +- (K-G) or f(T-G) +- (O-G)

(according to taste), the signs being + for the minor axis and - for the major one (I think).

So if F is a focus, then F-G is a scalar multiple of one of these vectors, say k times it, where kk must be a rational function of q and the squared edgelengths.

This clarifies the algebraic nature of the foci, which was still unclear to me from Paul's formulae. First of all, the sign of q distinguishes between the real and imaginary foci, q being positive for the former. Then k will have the form root(T + Uq), where T and U are rational functions of aa,bb,cc, and the sign of this second root determines which of two opposite foci one is taking.

I don't know if I'm strong enough to work out the details! But it's now clear that the three "big roots", typically root[(bb+q)^2 - ccaa] of Paul's formulae must all be essentially the same as my single new square root, k, in other words, that they all have the form R + Sq, where R and S are rational functions of aa,bb,cc.

I'd like to see the formulae!

John Conway

      PS -- in the process of this I found this formula for the intersection of a line [ l : m : n ], which is the dual of P = ( l : m : n ) with the Steiner ellipse.

       m(P^2–)  +  (infinite point on ~P)  Z

      where the points represent their coordinates in fully reduced form, and

      Z = root( ll+mm+nn-2lm-2ln-2mn)

      Note:   mP^2-, the medial of the Steiner inverse of P is the point of intersection of GP and the dual line of P.

      [JPE]consider three positive numbers p,q,r and the point P with barycentrics p:q:r.

      If U, V, W are the projections of A,B,C upon a variable line L, then the minimum value of p.AU^2 + q.BV^2 + r.CW^2 is

      (-o(P)/2).(p+q+r)(1-OP*/R) where o(P) is the power of P wrt the circumcircle and P* the isogonal conjugate of P.

      The corresponding line L goes through P and is parallel to an asymptot of the rectangular circumhyperbola going through P.

      A funny case is p=1/SA, q=1/SB, r=1/SC (with an acute angle triangle) : then the minimum is 1 and we get the minimum for any line going through H.

      In the case p = q = r = 1, L is the major axis of the Steiner circumellipse and the minimum value is, with K=symedian point,

      (a^2+b^2+c^2)(1-OK/R)/6 = 3/2 (semiminor_axis)^2

      In fact, I've just noticed, but it is probably known, that the squares of the lengths of the semimajor and semiminor axis of the the Steiner circumellipse are

      z(1+OK/R) and z(1-OK/R) where z=(a^2+b^2+c^2)/9.

      Friendly. Jean-Pierre