foci of conic

If ABC is a triangle, P ( x : y : z ) is a point in barycentrics and P* is the isogonal conjugate of P then the inconic (The conic tangent to the sidelines of ABC) with focii (must be isogonal conjugate) P and P* has perspector the point Q that is the isotomic conjugate of the point in barycentrics ( x(ccyy+bbzz+2yzSA) : . . : . . ) There is an easy construction of this perspector that I call Q = In_perspector(P). Let P1 be the reflection of P in BC. The line P*P1 meets the line BC at A1 that is the contact point of the inconic with BC. Similarly construct the points B1, C1 and the lines AA1, BB1, CC1 are concurrent at Q.

For the ETC points I think we have the following list very interesting in the cases 13, 14, 15, 16.

In_perspector(1) = 7 incircle

In_perspector(2) = 598 1/( cc+aa-2bb)

In_perspector(3) = 264 tO MacBeath

In_perspector(4) = 264

In_perspector(6) = 598

In_perspector(13) = 13 Fermat

In_perspector(14) = 14

In_perspector(15) = 13 isodynamic Simmons

In_perspector(16) = 14

In_perspector(74) = 1494 g einf

In_perspector(98) = 290 gX

In_perspector(99) = 670 tripole of tK

In_perspector(100) = 668

In_perspector(101) = 190 bb/(c-a) 1/(c-a)

In_perspector(105) = 2481

In_perspector(106) = 903

In_perspector(109) = 664

In_perspector(110) = 99

In_perspector(111) = 671

In_perspector(112) = 648

In_perspector(691) = 892

In_perspector(898) = 889

In_perspector(919) = 666

In_perspector(2291) = 1121

In_perspector(2715) = 2966

I have not test all these cases since I worked approximately.

Nikolaos

{Inconic perspector -> {focus,focus}},{ETC points on inconic}

{7, {1,1}}, {11,1314,1315,1317,1354,1355,1356,1357,1358,1359,1360,1361,1362,1363,1364,1365,1366,1367,2446,2447,3021,3022,3023,3024,3025,3026,3027,3028}

{13, {13, 15}}, {}

{14, {14, 16}}, {}

{99, {110, 523}}, {523,669,1649,2528}

{190, {101, 514}}, {514,649}

{264, {3, 4}}, {339,1312,1313,2967,2968,2969,2970,2971,2972,2973,2974}

{290, {98, 511}}, {511}

{598, {2, 6}}, {}

{648, {112, 525}}, {525,2501}

{664, {109, 522}}, {522}

{666, {918, 919}}, {918,2284}

{668, {100, 513}}, {513,693}

{670, {99, 512}}, {512,850}

{671, {111, 524}}, {524,1648},

{886, {888, g X(888)}}, {888}

{889, {891, 898}}, {891}

{892, {690, 691}}, {}

{903, {106, 519}}, {519,1647}

{1121, {527, 2291}}, {527}

{1494, {30, 74}}, {30,1650}

{2481, {105, 518}}, {518}

{2966, {2715, 2799}}, {2395,2799}

g X(888) = g t X(886) = {99,669} /\ {111,385} and on the circumcircle.

Given an inconic focus P{p,q,r}, {1 / (p (b^2 r^2 + c^2 q^2 + 2 q r SA)) ::} is the inconic perspector.

Nikolaos has given a nice relation that, given a focus of an inconic, gives all information about the inconic. This works because, for an inconic, the knowledege of one focus gives the other, which then gives the center, which then gives the perspector, and so on.

I am puzzled why this does not seem to work for a circumconic.

Given knowledge of one focus, can the other be uniquely found, and if now what is the locua of the other focus.

I like these general results on conics. I am going to put Nikolaos' results in a form I am used to and generalize to circumconics.

If P and gP (g = isogonal conjugate) are foci of an inconic, then its center is the midpoint of the two.

Q ~ P + gP

Since the relation between perspector and center is

center = t d perspector,

where t is the isotomic conjugate and d the dilated or anticomplementary operation.

The relation between the center and perspector for a circumconic is

center = m t d perspector, where m is the medial or complementary operation.

Given knowledge of the two foci one can then find the center as the midpoint and then find the perspector and thus most features of the conic. The formulas will be uglier because midpoint formulas are typically ugly.

Steve

Steve Sigur asked, of circumconics:

Given knowledge of one focus, can the other be uniquely found, and if

now what is the locua of the other focus.

No. In general there are four circumconics with a given focus F. (Let C(F) be a circle about F; let A'B'C' be the polar reciprocal of the reference triangle with respect to C(F). The polar reciprocals of the incircles of A'B'C' are circumconics of the reference triangle with a focus at F.)

I've fiddled with the problem of locating the other foci, but I'm away from the office and can't check. The case of the circumcenter is interesting; the directrices of the noncircular circumconics are the edges of the medial triangle and, if I recall correctly, the other foci turn out to lie on the cevians of K.

Jim Parish

Following Jim's idea, we can notice that the inverses wrt C(F) of the incircle and of the excircles of A'B'C' are the auxiliary circles of the 4 circumconics with focus F (I'm not sure of the English word "auxiliary circle" : I mean the circle with diameter the vertices of the conic lying on the focal axis) Of course, this gives a very easy construction of the center and of the other focus.

We can notice too the following :

If U,V,W is the contact triangle of an in-excircle of A'B'C', then

the polar lines of U, V, W are the tangents at A,B,C to the corresponding circumconic with focus F.

Friendly. Jean-Pierre

is that the construction of the circumconic with focus F is equivalent to the following construction. It is sufficient to construct the triangle A'B'C' that is tangent to the circumconic at A, B, C.

1. Construct the point F* isogonal conjugate of F wrt ABC

2. Construct the pedal triangle A*B*C* of F* wrt ABC.

3. Construct the intouch triangle A1B1C1 of A*B*C*

4. Construct the line B'C' from A perpendicular to the line F*A1 and the lines C'A', A'B' from B, C and perpendicular to the lines F*B1, F*C1 respectively. The other focus F' of the circumconic is the isogonal conjugate of F wrt A'B'C' and the rest of the construction as in inconics.

The other 3 circumconics if A1B1C1 is an extouch triangle.

Best regards

Nikos Dergiades

Lets see.

If the perspector is (l:m:n) then this triangle has three edges like

s/l + y/m = 0 and vertices ( ±l : ±m : ±n ) (if funny symbols on your browser, they are plus/minus symbols)

So write the focus interms of these vertices then take the isogonal conjugate. There will be a unique formula for the other focus.

This makes sense (and I will do this as soon as I finish grading final exams), but this contradicts Jim's statement that there should be four solutions.

Well some nice things to think about.

Steve

No. In general there are four circumconics with a given focus F. (Let

C(F) be a circle about F; let A'B'C' be the polar reciprocal of the

reference triangle with respect to C(F). The polar reciprocals of the

incircles of A'B'C' are circumconics of the reference triangle with a

focus at F.)

jim,

But how do we know that the new circumconics have a focus at F?

Steve

I'm reconstructing the argument from memory; there are parts of it that I don't fully understand. I'm drawing on Veblen and Young and on a book by Cremona on projective geometry.

1. Let P1, P2 be the circular points, and let c be a conic. Let l11, l12 be the lines through P1 tangent to c, and define l21, l22 similarly. In general, of the four intersections l1i.l2j, two are real, and they are the foci of c. (Special things happen if c is a circle or a parabola.) [This is the part that I don't quite follow.]

2. In particular, if c is a circle with center F, the lines F-P1 and F-P2 are tangent to c. That is, the polars of P1, P2 with respect to c are the lines F-P1 and F-P2.

3. Now let c1 be another circle, and let d be its polar reciprocal with respect to c. Since P1, P2 lie on c1, their polars with respect to c (which is to say, the lines F-P1 and F-P2) are tangent to d. Since they intersect at F and F is real, F is one of the foci of d by 1).

Jim Parish

Here is a possible construction (there exists a better one - Paul mailed me one but I can't find it -) Suppose that your conic touches the lines AC and AB at V and W (we can get V and W using the fact that the Brianchon's point of the conic is the isotomic conjugate of the anticomplement of the center O of the conic)

Let T and C(T) be the circumcenter and the circumcircle of OVW. The parallels through O to AC and AB intersect again C(T) respectively at V' and W'. U = VV' inter WW'. The line TU intersects C(T) at M, M'. Then OM and OM' are the axis of the conic. If AC and the perpendicular at V to AC intersect one axis at P and Q, this axis will be the focal axis if O is outside the segment [PQ]. Now, the focii lie on the circle with center O orthogonal to the circle with diameter [PQ].

Friendly. Jean-Pierre

Dear Steve

Foci are indeed a big problem to compute.

Here in France since scores of years, we had trained our teachers to classify quadratic forms on any fields but most of them are unable to identify a conic on its equation not to say to find their foci even in an orthonormal frame with origin at the center of the conic. I think the best to compute foci in cartesian , trilinear or barycentic coordinates is to use their Plucker definition. Even in orthonormal coordinates, calculus is not easy not to say in barycentric ones with maybe one exception, inscribed parabola where focus and directrix are easy to get on the equation.

I think here in France, we have lost foci for ever except the sun in the sky for those who still remember Newton's (line?) name.

About my question on the construction of 2 isogonal points F and F' (wrt ABC), given their middle O, of course it's equivalent to find foci of the inscribed conic with center O.

It's already a good problem to draw this conic using only affine tools and then after drawing it to trace foci but I don't have a direct construction of F and F' not using that conic. Maybe somebody in Hyacinthos can help me? Thanks in advance

François

I think there's certainly quite a bit of interesting material in those areas. For example: let F be a point not on any sideline of ABC, and suppose that there is a conic with focus F with respect to which ABC is self- polar. (IIRC, such a conic exists iff, of the directed angles AFB, BFC, CFA, at least one is obtuse and at least one acute.) Then the directrix of this conic, the directrix of the inconic with focus F, and the tripolar of F are concurrent.

The proof is not hard; if A'B'C' is the polar reciprocal of ABC with respect to some circle about F, then the three lines above are the polars, with respect to that circle, of the orthocenter, circumcenter, and centroid of A'B'C'.

I'm currently working on a paper which applies polar reciprocals to a number of problems, and foci and directrices appear very naturally in that context.

---

Your constuction is very nice.

I try to understand it! I see 2 steps:

1°Draw axis of the inscribed conic {Gamma}.

2°Draw foci.

For the moment, I only understand the second step which uses a very known property of tangent and normal in a point of {Gamma} wrt its axis. I have to think more on the first.

My own construction is too intricate but it works even in the case of a parabola when the perspector of the contact triangle is on the Steiner circum-ellipse and "center" of {Gamma} is on line of infinity. Of course in this case, there are more simple construction of the focus which is on the circumcircle. Besides in this case, a funny thing arises in my construction, the direct circular map I call g having one fixed point at infinity is a direct similarity and we can get the other fixed point very easily by the usual euclidian construction.

Now I must think again on your first step. Friendly