Hyacinthian Correspondences on Geometry

Pre - Hyacinthian Correspondences on Geometry

Floor van Lamoen Paul Yiu Goes, and Boca Raton, Florida, The Netherlands USA

April 1999 to December 1999

Floor van Lamoen, Thursday, April 15. Preprints. Dear Mr. Yiu, I received from Professor Clark Kimberling a couple of preprints from your hand. I have been reading these preprints with a lot of pleasure, and the pleasure will continue, because I haven’t fully studied all material yet. I was especially struck by the beauty of “Those Ubiquitous Archimedean Cirles”. It seems to me that two very simple Archimedean circles, that might be of slight interest, are not mentioned in this manuscript. Maybe they might be a tiny addition. Construct a line from O2 tangent to (O1), and let Wx be the point of tangency. Then, if we let (Wx) be the smallest circle that is tangent to the common tangent CD of (O1)and(O2), we find that (Wx) is an Archimedean circle. This r1r2 canbeeasilyverifiedbythep = r formula and Pythagoras’theorem. InthesamewaywecanconstructalinefromO1 tangent to (O2), and let Wy be the point of tangency. And we find an Archimedean circle (Wy). Please let me know if you like to recieve comments like this. Best regards. Floor van Lamoen, Goes, The Netherlands.

Paul Yiu, April 15, 1999. Dear Mr. van Lamoen, It is indeed a great delight to receive your kind words. Your comments and suggestions on the preprints, or any other communication in plane Euclidean geometry, are very much appreciated. The two Archimedean circles you mentioned are indeed wonderful. I have not realized these before. I shall communicate these to my coauthors. I am sure they will be as delighted as I am. The point Wx, in the notation of the preprint Those Ubiquitous Archimedean Circles,isthepointS1 in Figure 26. As you said, that these are Archimedean

1 vanLamoenandYiu:Pre-HyacinthianGeometry 2

follow easily from Euclid’s proof of Pythagoras’ Theorem: the radius of (Wx) being 2 r1 r1r2 r1 − = . r1 + r2 r1 + r2 Also not mentioned in the same article are two other Archimedean circles discovered by Frank Power, who took a geometry course with me last summer. Let R1 be the highest point of the circle (O1), as in Figure 15 of the preprint. ThecircletangenttoOR1 at this point and also to (O) internally has radius x given by 2 2 2 2 x +(r1 + r2)=(r1 + r2 − x) . From this it is readily seen that r1r2 x = , r1 + r2 showing that the circle is Archimedean. The same equation holds for the radius of the analogous circle at the highest point R2 of the circle (O2). Archimedean circles are indeed ubiquitous! Best regards. Sincerely, Paul Yiu Department of Mathematics Florida Atlantic University Boca Raton, FL 33431-0991 USA

Floor van Lamoen, Monday, April 19, 1999. The radical center of the excircles of a triangle Dear Mr. Yiu, I have read your preprint “The radical center of the excircles of a triangle”. I was surprised to find the construction as in figure 2, which has been presented by John Conway in the Geometry-puzzles mailing list in March, 1998. Very interesting to find it this way! I wondered why you did not give a construction for the circle orthogonal to the incircle and two of the excircles. I am not√ fully sure how you derived that the power of S w.r.t. either excircle 1 2 2 is equal to 2 r + s . However, it seems reasonable to me that, when the incircle is taken for the A−excircle, the power of the A−excenter of triangle DEF 1 2 − 2 w.r.t. the incircle will be 2 ra +[(b + c a)/2] . In that case the construction of the circle of the circle orthogonal to the B− and C−excircles and the incircle is a variation on Figure 2: Leave FA and EA unchanged, but take for FB , DB, DC and EC the alter- native intersection of circle and side. These six points lie on the circle centered 1 2 − 2 at the A-excenter of DEF and having radius 2 ra +[(b + c a)/2] .(Isthere a reasonable version of Woo’s observation and its corollary in this case?) When we have the four circles, each perpendicular to three circles out of the set incircle and excircles, then in fact we have two twin sets of four circles: each circle is perpendicular to three circles of the other set. vanLamoenandYiu:Pre-HyacinthianGeometry 3

What would you think? Best regards. Sincerely, Floor van Lamoen. N.B. Please, call me by my ﬁrst name.

Paul Yiu, Monday, April 19, 1999. Dear Floor, Thanks for your comments on the excircles paper. When I wrote that little paper last summer, I intended it to be short and did not fill out all details, since I aimed at sending it to the Mathematics Magazine. I can find out the calculations for the radius of the orthogonal circles, and details of Woo’s statement. I learnt of the circle of Conway in early March this year, from one of the MAA Online articles by Bogomolny. If I remember correctly, I may have written a statement for the circle orthogonal to the incircle and two excircles. I am between classes now, and shall think more about your comments. By the way, the Mathematics Magazine did not want my article. I am still trying to find a home for it. Three weeks ago, I sent it out to Nieuw voor Archief Wiskunde. It is of course too early to hear anything. Best regards. Paul

Floor van Lamoen, Tuesday, June 8, 1999. Ubiquitous Archimedean Circles Dear Paul, Every now and then I think about the arbelos article I received via Professor Kimberling. Somehow I think it is very attractive. I noticed the following: Let M be the midpoint of O1O2.Let(C1)bethe (semi-)circle with O1O2 as diameter, (C2) the circle with center O1 through O and (C3) the circle with center O2 through O.Letthepointswhere(C2)and (C3) meet (C1) be called Wt and Wu. Then the circles with these points as centers, tangent to the line through O perpendicular to AB, are archimedean. This is not really surprising, since these circles are the reflections about the line through O perpendicular to AB of the two circles I wrote you about in April. But then I noticed that these circles are in some way similar to (W13) and (W14). The consctruction on the points on (O) in figure 9 is similar to that of the points Wt and Wu above. And then I realized a baby-arbelos was born: is an arbelos similar to ABC, but with opposite orientation, and lengths of half the size of ABC. Using this baby-arbelos it is possible to find some “new” archimedean circles. The circles (W4), (W5), (W9)and(W10) have children in the baby- arbelos, where the transforms of the points A, F , D and C serve as midpoints in stead of points on the circumference. There is also a very logical child of (W4), (W5), (W12) in the baby-arbelos. How do you like this idea? Best regards. Floor. vanLamoenandYiu:Pre-HyacinthianGeometry 4

Paul Yiu Wednesday, June 9, 1999. Ubiquitous Archimedean Circles Dear Floor, Thank you for your message yesterday on new Archimedean circles. From the explanation you gave, and the resemblance to W13 and W14, I recall the pictures of Thomas Schoch’s wonderful notes written twenty years ago. His very ﬁrst picture conﬁrms your conclusion. What appeals to me even more, however, is the fact that the centers of these two circles, and of the two you gave in April, are all on the semicircle (with O1O2 as diameter). Now this semicircle has quite a number of Archimedean circles associated with it. (1) This being orthogonal to both semicircles (AC)and(BC), its intersections with these semicircle are the points of contact with the tangent from the center of the other semicircle. This can be seen from Figure 26. Your April message shows that the distance from each of these to the line CD is equal to the Archimedean radius. Hence, your two new circles in April. With the two yesterday, now there are four Archimedean circles with centers on this semicircle. (2) Associated with this semicircle also are the Archimedean circles W17, W18 and W19 which Thomas discovered. (3) After reading your message yesterday, I asked myself this question. There is a chain of circles tangent to the semicircle (O1O2) externally and to (AB) internally. How large is the one whose center is on the line CD? Let a and b be, as usual, the radii of the two semicircles (AC)and(BC). If the radius of this circle is x,then

(a + b − x)2 − (a − b)2 =[(a + b)/2 − x]2 − [(a − b)/2]2.

ab From this, x = a+b , Archimedean!!! Archimedean circles are really ubiquitous. How many Archimedean circles are known now? The Math. Mag. paper lists 29. Last July, Frank Power discovered two, and Thomas immediately pointed out their reﬂections, making four of them. Total = 33. In April, Floor gave two, and now two more, all with centers on the semicircle (O1O2). Total = 37. With the one in (3) above, we have now 38 Archimedean circles. Perhaps your idea of baby arbelos would lead to even more. Sincerely, Paul

Paul Yiu to Peter Woo, Thomas Schoch, Clayton Dodge and Floor van Lamoen, Wednesday, June 9, 1999. Incircle of the arbelos Dear friends, Talking about the shoemaker’s knife, I have a question which always puzzles me. I hope you can enlighten me. It concerns the construction of the incircle of the shoemaker’s knife. One way to perform the construction is to make use vanLamoenandYiu:Pre-HyacinthianGeometry 5

of the Bankoff circle W3 to locate the points of tangency with the two smaller semicircles. Once these two points are located, the center can be easily found, leading to the incircle. Now, Bankoff has given, in his 1954 list of 9 surprises, a very simple alterna- tive to this construction. Let R1 and R2 be the highest points of the semicircles (AC)and(BC) respectively. Bankoff’s surprise number 6 is this: Construct the circle R1(A) [center R1, through A] to intersect the semicircle (BC)atQ and the large semicircle (AB)atR. Construct the circle R2(B) to intersect the semicircle (AC)atP . Then PQR is the incircle of the shoemaker’s knife. I can only prove this by a very long calculation, and have always been wondering if there is any elegant proof of this fact. Best regards. Sincerely, Paul

Floor van Lamoen, Thursday, June 10, 1999. Ubiquitous Archimedean Circles Dear Paul, There is (at least) one more of the Math. Mag. paper Archimedean circles that is associated with the (O1O2) semicircle: (W3). I found that (W3)is internally tangent to (O1O2). To see this, let O be the center of (O1O2). And let a and b be as you deﬁned 1 − ab them. In triangle O CW3 we have O C = 2 (a b), and CW3 = a+b . a2+b2 We ﬁnd using Pythagoras’Theorem that O W3 = 2(a+b) .Thisleavesfrom a+b − a2+b2 ab the radius in (O1O2)beyondW3: 2 2(a+b) = a+b ! This means also that one Archimedean circle can be added: Let a circle be tangent to line AB in O, and internally tangent to (O1O2), then this circle is, of course, Archimedean, too. Archimedean triangles are really ubiquitous, and very special! Best regards. Floor.

Paul Yiu, Thursday, June 10, 1999. Ubiquitous Archimedean Circles Dear Floor, How lovely is this, the tangency of W3 with (O1O2), and also its ”reﬂection” giving the 39th Archimedean circle. I am driving 40 miles to teach a geometry class now. I shall tell my class. Paul

Thomas Schoch, Friday, June 11, 1999. Ubiquitous Archimedean Circles Dear Paul and Floor, The new discoveries are truly wonderful, I particularly like that ”baby arbelos” - I’m just thinking through child arbeloses and grandchild arbeloses ... Who ﬁnds No. 40? vanLamoenandYiu:Pre-HyacinthianGeometry 6

Best regards. Thomas PS: How is the Math. Mag. paper?

Floor van Lamoen, Friday, June 11, 1999. Ubiquitous Archimedean Circles Dear Thomas, Paul and others, No. 40 was already hidden in my note on the baby-arbelos. Let the baby-arbelos be made out of the semicircles (O1O2), (OO1)and (OO2). Note that this baby-arbelos is similar to its father/mother but oriented oppositely. To ease naming let x be the point constructed in the baby-arbelos similarly as X in the arbelos. Then, following naming as in figure 7 of your paper, we have four archimedean circles found in the baby: The circle with center e tangent to cd(= OR in figure 21) is archimedean. The circle with center f tangent to cd is archimedean. The circle with center c(= O) tangent to ef is archimedean. The circle with center d tangent to ef is archimedean. And, following figure 8, we find in the baby: The circle g(d) (center g passing throug d) is archimedean. Best regards. Floor van Lamoen.

Floor van Lamoen, Friday, June 11, 1999. Incircle of the arbelos Dear Paul, I haven’t found a proof of the construction (however, I will be looking for it), but the construction shows that W21 and the incircle of the arbelos are both tangent to (O) in the same point. From ﬁgure 18 it is easiest seen that R1I = R1C = R1A, which shows that I from ﬁgure 18 is R from the construction you describe. Was this known? Best regards. Floor.

Paul Yiu, Saturday, June 12, 1999. Ubiquitous Archimedean Circles Dear Floor, Here is Peter Woo’s wonderful proof of surprise # 6, with a few interesting corollaries. Years ago, Peter has sent me a hard copy of a proof of all the surprises in Bankoﬀ’s note, all done by inversion with respect to the circle A(D). This leads to another easy construction of the incircle, using only ruler to locate the points P , Q, R:simply (1) join A to N to intersect the semicircle (BC)atQ; (2) join B to M to intersect the semicircle (AC)atP ; (3) extend BN to interect CD at R,andthenAR to intersect the semicircle (AB)atR. The shoemaker’s knife is truly wonderful. Sincerely, Paul vanLamoenandYiu:Pre-HyacinthianGeometry 7

Paul Yiu, Saturday, June 12, 1999. Incircle of the arbelos Dear Floor, I have forgotten to mention in my previous message that your observation that the incircle PQR of the arbelos touches the semicircle (AB)atthesame point as (W21) does is very interesting. It makes an even more elegant description of the incircle. I did not know of this fact before. It adds to the credential of W21 being Archimedean. Sincerely, Paul

Thomas Schoch, Thursday, June 17, 1999. Ubiquitous Archimedean Circles Dear friends, Consider a “twin” of the baby-arbelos, which has the same size and position as the baby, but the opposite orientation (the same one as the parent). Denote the corresponding things (circles, points) by: X[parent] = x[baby] = x[twin], referring to Clayt’s paper. Hence: a = a = O1,b = b = O2,andc = C. Then: 1. The first two Lamoen circles are corresponding to (W13)and(W14), having the diameters of (w13)and(w14) as radii. 2. The circle with center c tangent to e f is circle (W8). Its point of tangency with ef lies at the “Schoch line”. 3. The circle with center d tangent to ef and touching ef at g corre- sponds to (W4), with the diameter of (w4) as radius. Points W4 and g lie on a line parallel to CD with distance CK to it (see figure 27). It would be nice to bring Paul’s wonderful new circle into any significant conjunction with the other cricles. Best regards. Thomas

Floor van Lamoen, Friday, June 25, 1999. Preprints. Dear Paul, I have completed, and submitted to Elem. Math., a paper called “Triangle centers associated with rhombi” which I think you might like to read. I could send to you a paper version by mail (if I have your address) or I could send a .dvi as email attachment, if you wish. Just let me know! I suppose that Clark Kimberling already gave you my paper “Bicentric triangles”, which will appear in Nieuw Arch. Wisk. If not, then I could send you a copy of that one too, again if you wish. Best regards. Floor.

Paul Yiu, Friday, June 25, 1999. Dear Floor, Please send me your preprint and some of your reprints. I look forward to reading them. Please send to vanLamoenandYiu:Pre-HyacinthianGeometry 8

Paul Yiu Department of Mathematics Florida Atlantic University Boca Raton, FL 33431-0991 USA

I also want to forward a few items to you, including a reprint of the 4-author paper which just appeared in Math. Magazine last week. Please give me your mailing address. Warm regards. Paul

Floor van Lamoen, Friday, June 25. 1999. Dear Paul., The two papers will be on its way to you! My address is: Floor van Lamoen Statenhof 3 4463 TV Goes The Netherlands

Floor.

Paul Yiu to Clark Kimberlingand Floor van Lamoen. Tuesday, July 6, 1999. Miquel associates and conjugates Dear Clark and Floor, A recent fine gift from Antreas Hatzipolakis has stimulated me to think about problems related to Miquel theorem. Here is part of the thank - you note I have just written to Antreas. The analogous problem of inscribing an equilateral triangle in a given triangle is even more interesting. One starts with the firstisodynamic point (α1 in Figure 52) and draws lines making equal angles with the sides of the triangle. The intersections would form the vertices of an equilateral triangle. If these points are allowed to be on the extensions of the sides, then one can use the second isodynamic point as well. So, associated with the first isodynamic point, there is an infinity of equilateral triangles inscribed in the given triangle ABC. Which one of these is cevian? Is this constructible using ruler and compass? While I still cannot completely resolve these questions, these lead to some interesting theorems about Miquel circles. Miquel’s theorem states that if X, Y ,andZ are points on the sides BC, CA,andAB respectively of triangle ABC, then the circles AY Z, BZX,and CXY intersect at a point M. vanLamoenandYiu:Pre-HyacinthianGeometry 9

Now, starting with a point P, consider the traces X, Y , Z of P on the sides. We call the corresponding point M the Miquel associate of P .Thecircles through X, Y , Z intersects the sides BC, CA, AB again at X, Y ,andZ.

Theorem 1. AX, BY , CZ are concurrent. We call this point Q of concurrency the Miquel conjugate of P .

Theorem 2. The Miquel associates of P and Q are isogonal conjugates. Consider the circles AY Z and AY Z. These two circles intersect at A,and at another point which we label X. Likewise, the points Y and Z are deﬁned by the pairs of circles BZX, BZX and CXY , CXY .

Theorem 3. The lines AX, BY , CZ are concurrent. Here are some examples.

Example 1. Let’s start with the centroid G. The traces of G are the midpoints D, E, F of the sides. The circles AEF , BFD, CDE intersect at the circumcenter O.The circumcenter is therefore the Miquel associate of the centroid. Now, the circle DEF is the nine-point circle of triangle ABC. The “second” intersections with the sides are the perpendicular feet D, E, F , the traces of the orthocenter H. Therefore, the Miquel conjugate of G is H.Thecircles AEF , BFD,andCDE intersect at H. The orthocenter H is its own Miquel associate. The Miquel associates of G and H are O and H. These are isogonal conjugates. Here, the lines AX, BY ,andCZ are all perpendicular to the Euler line.

Example 2. If P = Gergonne point, then the Miquel circles intersect at the incenter I. If X, Y , Z are the traces of the Gergonne point, the circle XYZ is the incircle, and is tangent to the sides of the triangle. The Gergonne point is therefore its own Miquel conjugate.

Example 3. If P is the Nagel point, with homogeneous barycentric barycentric coordinates s−a : s−b : s−c, its Miquel associate is the point with homogeneous coordinates

[a(a3 + a2b − ab2 − b3 + a2c − 2abc + b2c − ac2 + bc2 − c3)] vanLamoenandYiu:Pre-HyacinthianGeometry 10

Here, the second and third components can be obtained from the ﬁrst component by cyclic permutations. This is the point X40 in Kimberling’s list, and is the reﬂection of the incenter I with respect to the circumcenter O.

Example 4. If P is the isotomic conjugate of the orthocenter, then the Miquel associate is the de Longschamps point, the reﬂection of the circumcenter O with respect to the orthocenter H.

Example 5. The Miquel associate and Miquel conjugate of the incenter I also have relatively simple coordinates: a2 Associate = (a3 + a2b − ab2 − b3 + a2c − abc − b2c − ac2 − bc2 − c3) , b + c 1 Conjugate = . a3 + a2b − ab2 − b3 + a2c − abc − b2c − ac2 − bc2 − c3

But I have not been able to identify these with any of the ﬁrst 100 centers in Kimberling’s list. Maybe they are among the points in the longer list, or maybe new. There are many interesting questions concerning the Miquel circles. I hope to be able to write a short on these. Does any of these sound familiar to you? I would like to know your opinion. With warm regards, Sincerely, Paul

Floor van Lamoen, Wednesday, July 7, 1999. Miquel associates and conjugates Dear Paul, Many thanks for the interesting note. I think this is new to me, and that this is a nice way to generalize that two concyclic pedal triangles are pedal triangles of isogonal conjugates. Since I am quite busy I will get back to you when I have studied this more precise, and maybe can think of something nice in addition? Thanks again! Kind regards, Floor van Lamoen.

Floor van Lamoen, Thursday, July 22, 1999. Extensions of Feuer- bach’s theorem Dear Paul, First of all, to disappoint you: I am not a professor at all. I am not even attached to a university. I am graduated as a mathematician in the University vanLamoenandYiu:Pre-HyacinthianGeometry 11 of Amsterdam, but I chose to become a High School Teacher (which I still love to do). So I am a mere amateur! When I studied maths in university I was not very interested in geometry. I must say that neither Dutch High Schools, nor Universities offered much Euclidean or Projective Geometry when I went to them. Luckily in the new (national) High School programs, there is an increase in Geometry! Then my parents gave me a booklet by O. Bottema, of which the title translated into English is (Chapters from) Elementary Euclidean Geometry (my father was in High School with a son of O. Bottema) and I was infected. I have read, full of enthusiasm, your preliminary version of Extensions of Feuerbach’s Theorem .... I am very impressed byyour ability to explain all elements of your paper in such a way that it could serve in a book for, say, sophomores. Really very impressive. Funny is, that I am working at the moment on a paper, that describes some of the conics - grouped in homothetic families - that you are describing too, be it from a different point of view. My starting point is that the circumconic CQ you are describing is the locus of points in the Euclidean plane for which the “HP −pseudopedal triangle” is degenerate, as such being an extension of the Simson/Wallace line. Here the “X−pseudopedal triangle” does not use perpendicular altitudes, but altitudes parallel to the Cevians of X. The rest of the paper will be not half as interseting as yours. But, when I will have ready a version of the paper that I think is good enough to serve as a preliminary version, I will send it to you. Of course I do have two minor questions: In my opinion your observations in 8.3 already prove that the conics are homothetic, because the three radii fix six points in the conic, and thus fix the conic. Of course, the observations of section 13 are much more interesting as a proof! Maybe it would be nice to add to 8.3, or to section 13, that for the lengths A B C da, db, dc and d of the principal axes of HP , HP , HP and HP respectively the identity 1 1 1 1 + + = , da db dc d an extension for the famous identity of inradii. Kind regards, Floor.

Paul Yiu, Thursday, July 22, 1999. Extensions of Feuerbach’s Theo- rem Dear Floor, Geometry lovers, I believe, have a good taste of mathematics. I am sure you are a dedicated teacher, and your works clearly show that you are a true amateur. Being this, you have an advantage of being able to impart enthusiasm of mathematics to students at an early stage, before most are them are lured away by “more practical subjects”. Thank you for your kind words about my preprint. I shall try to ﬁnd a home for her, but it can be diﬃcult. I somehow have the feeling that most vanLamoenandYiu:Pre-HyacinthianGeometry 12 of the results have been known before. They could hardly have escaped the nineteenth century geometers. But it was a great satisfaction doing this on my own. I actually obtained these results during the spring semester while teaching a course of geometry to a small class of undergraduates. Please keep me informed of your good works. May be we shall have occasion to work on something together. Warm regards. Paul

Floor van Lamoen, Tuesday, August 3, 1999. Homothetic conics Dear Paul, Today I have put in the mail a preliminary version of the paper about homothetic conics, that I mentioned earlier to you. Any comments are very welcome! Please let me know when you have received this. Kind regards, Floor.

Paul Yiu, Tuesday, August 3, 1999. Homothetic conics Dear Floor, I look forward to reading this, and shall let you know when it arrives. Thank you. Best regards. Paul

Paul Yiu, Wednesday, August 18, 1999. Homothetic conics Dear Floor, Your preprint Homothetic Conics has just arrived, Thank you very much. I shall study it carefully, and see what we can do together. Best regards Sincerely, Paul

Paul Yiu, Friday, August 20, 1999. Homothetic conics Dear Floor, Your paper on homothetic conics is very lovely indeed. It is very interesting, and I am sure a lot more can be done on section 5. The crucial concept you introduce here is the notion of orthogonality with respect to a point. Theorem 3 at the end is just wonderful. In Section 4 you consider the Bottema conic. Now suppose T varies on the circumcircle. Then the locus of the center of the Bottema conic of T is the radical axis of the circumcircle and the nine-point circle. This has a generalization. Let P be a ﬁxed point, and KP the circumscribing conic with center 3G−2P . [It is the same as C1 in your paper]. If T varies on C1, then the locus of the center of the Bottema conic of T is the radical axis of KP and the Feuerbach conic FP . The notion of P −orthogonality is something worth pondering on. vanLamoenandYiu:Pre-HyacinthianGeometry 13

What a coincidence! This year I began subscribing to Nieuw Archief voor Wiskunde. For long time, I have not received my ﬁrst issue. But it arrived exactly the time with your paper. There, the very ﬁrst paper is your joint paper with Clark on central triangles. Warm regards. Sincerely, Paul

Floor van Lamoen, Thursady, August 26, 1999. Homothetic conics Dear Paul, Yesteray I arrived home from my summer vacation travel. In fact I was lucky, because we planned to travel to Turkey, but the trip was cancelled . . . Thanks for your kind remarks on my paper suggestion. I am a bit unsure how to progress with the paper. First of all, I think section 3 should maybe be dropped, being a bit far fetched. Second, I think it would be interesting to incorporate more of your results about the generalization of the Feuerbach theorem. It might be worth consid- ering to put your and my results together in one joint paper? When we would do so, then there should be some care, because not for all P in my paper, there is a quadruple of inscribed conics. I myself was very delighted when I found the results of section 5, and I am glad to read that you like them, too. Indeed, there should be a load of results that could be ﬁt into the “P −perpendicular” concept. I think your result on the centers of various Bottema conics is really nice!! It connects the Bottema conic in a surprising way to the circum- and ninepoint- conic (Feuerbach conic when there are in-/exconics). If you really carefully read the proof of theorem 3, then you see that it proves a stronger statement: For each point X the line joining X and the center of conic C1 is P −perpendicular to the polar line of X. Well, I have to do my “after vacation” things now. Kind regards, Floor.

Floor van Lamoen, Monday, August 30, 1999. P −perspectivity and somethingcompletely different Dear Paul, I have been thinking about ways to use P −perspectivity. In that thinking I noted that the remarks that I made after theorem 3 have to be slightly adjusted. I stated that the P −circumcenter of any triangle in the plane of a triangle ABC with a perspectivity based upon P is found by taking the point of intersection of the P −perpendicular bisectors. The arguments I used do not cover triangles with two sides that are P −perpendicular to each other, because such a triangle would not fix again the perpendicularity (The new “P −orthocenter” is found on the sides). Indeed, one would then have to find the P −circumcenter of such a P −right triangle as the midpoint of the P −hypothenuse. But when the circumconic derived from the perpendicularity is a hyperbola, then the P −hypothenuse might not be able to be a diameter of such a hyperbola. vanLamoenandYiu:Pre-HyacinthianGeometry 14

Also I found a small result where P −perpendicularity can be used for a generalization: It is known that for a point X in ABC the “third pedal triangle” is similar to ABC.(LetABC be the pedal triangle of X. The second pedal triangle are the perpendicular feet of X on the sides of ABC. Etc.). Since I haven’t found a good extension of similarity in P −perpendicularity yet, that is not good for generalization. But, when X = H (orthocenter), then the second and third pedal triangles of X = H are homothetic to ABC. In fact, the third pedal triangle is the medial triangle of the ﬁrst. And X = H is also again the orthocenter of the third pedal triangle. This can be generalized for X = P and the second and third P −pedal triangle. Then, I have something completely diﬀerent, that I think you might like: I have kindly been sent some reprints of papers by Ludwig Stammler. In one of them (and possibly in more) I found something I suppose you are interested in: 3π Let T = ABC be a triangle with all angles ≤ 4 .Lett be the only real solution of the following equation:

x3 − (a2 + b2 + c2)x − 2abc =0.

Then the point given by trilinears: t2 − a2 : t2 − b2 : t2 − c2 is the midpoint of a circle C, with radius √ √2 t √ a t2 − a2 + b t2 − b2 + c t2 − c2 where = area of triangle ABC, that has the following property: The area of the symmetric difference of T and C is minimal over all possible circles. Reference: Ludwig Stammler, Der Kreis mit minimaler Flächendifferenz zum Dreieck, Elemente der Mathematik 33/6 (1978), 143-152. The circle is one of the examples of Stammler’s Cutting Circles (Schnit- tkreise). I hope you like this! What a coincidence that you received your first issue of Nieuw Archief voor Wiskunde. Well, when you have read Clark’s Triangle Centers and Central Triangles, our joint paper is nothing new really. In the third issue of this year my Bicentric Triangles paper will appear. Unfortunately Clark wrote me that he had received a letter by the editors of Nieuw Archief, that they will for some time not accept papers on triangle geometry, unless something “truly exceptional” comes along. That’s sad. You submitted a paper to Nieuw Archief, didn’t you? Kind regards, Floor. vanLamoenandYiu:Pre-HyacinthianGeometry 15

Paul Yiu, Tuesday, August 31, 1999. P −perspectivity and something completely diﬀerent Dear Floor, Thank you for your message. I shall study the P −perpendicularity more closely. It is really something interesting. Thank you also for the information about Stammler. I have browsed through a recent short paper of his on the cutting circle, but have not pursued it further. Here, our semester has just begun, and things shall be more settled in a few days. In the past year, I have made a few good friends in geometry, and some are very competent geometers, like yourself. I have often complained of the diﬃculty of publishing in plane geometry. While in the 19th century, most famous mathematicians have touched upon Euclidean geometry, in our century, it seems that journal editors, and mathematicians in general, seem to have been biased against Euclidean geometry. I actually have no good idea how to sell a geometry paper, but my general feeling is that it should be written non-technically so that general readers can understand. Euclidean geometry is particularly suitable because the methods used are in general quite elementary, but the results are often elegant and beautiful. But referees todays do not seem to see it this way. Yes, I did send my short note The radical center of the excircles of a triangle to Nieuw Archief. But just the past weekend I received a rejection letter with this report:

This paper treats some properties of the so-called Spieker points of a triangle in the Euclidean plane. The presentation is clear, but not particularly attractive for a large proportion of the membership of Het Wisdundig Genootschap. I therefore propose not to accept this paper for publication in Nieuw Archief voor Wiskunde.

Early this year, I have received a similar report on the same note from the Mathematics Magazine. I shall try to search the MathSciNet to get an idea of what kind of journals publishing papers in this area. Actually, although I am a computer (or technology) illiterate, I often harbor the idea of an Electronic Journal of Euclidean Geometry. Ideally, Clark Kim- berling is the editor, and John Conway the consulting editor. Then we can all send our papers there, and get evaluations from experts really knowledgeable and interested in this area. I hope it is a dream that will come true some day, perhaps in the near future when we launch into the 21st century. What do you think? Sincerely, Paul

Floor van Lamoen, September 2, 1999. Dear Paul, I am also quite surprised, or disappointed, that the mathematical world doesn’t seeem to think that plane, Euclidean (or projective or affine) geometry is really serious. It seems to be associated with either school or recreational vanLamoenandYiu:Pre-HyacinthianGeometry 16 mathematics. Or something for problem sections. On the other hand, I think that a lot of mathematicians, and non mathematicians, see the beauty of plane geometry. So, I like your idea to have a Journal - be it electronic, on paper, or both - on Euclidean Geometry!! I think, ideally, that would be a mix of “sophisti- cated” and new results, and surveys on ”old” subjects, made clear in a very comprehensive way. Yes, that seems very attractive. Should it be trouble to start such a Journal? I think that there should be a group of people like Clark Kimberling and John Conway, who should have to put effort in it. I am not at all afraid whether there should be enough papers! And also, electronic publishing would, as far as I can see, not be too difficult. Would you want to turn this into a serious plan? Kind regards, Floor.

Floor van Lamoen to Paul Yiu, October 29, 1999. Dear Paul, On August 20th you wrote:

In Section 4 you consider the Bottema conic. . . .

Let P be a ﬁxed point, and KP the circumscribing conic with center 3G − 2P . [ItisthesameasC1 in your paper]. If T varies on C1, then the locus of the center of the Bottema conic of T is the radical axis of KP and the Feuerbach conic FP .

I am at the moment in the finishing stages of rewriting my paper. One thing is that I would like to include the observation you made, because it is such a beautiful one. I would much like your consent to do so. The paper has, as I wrote already, been extended quite a bit. To give you an idea of what it has become: Title: Perpendicularity with respect to a point in a triangle. I start introducing the circumconic in a “Simson” way, still in the same way. Then the ninepoint conic comes in. I mention your paper with Feuerbach conics. The parasix configuration is dropped. The Bottema conic section is not really changed, except for your addition. The section where I introduce the perpendicularity with respect to a point has had some corrections, and a couple of additions: P −perpendicular Hy- perbolas circumscribing ABC go through P ; introduction of P-Darboux and P −Lucas conics; P is on the Q−Darboux conic iff Q is on the P −Darboux conic. Introduction of the P −iso conjugacy mapping (analogue to isogonal conjugacy with circle-definition); “De Villiers theorem” generalized from “isogonal” to “P-iso”; P −cyclocevian conjugacy. If you consent with adding your observation, I could send you a copy next Monday. vanLamoenandYiu:Pre-HyacinthianGeometry 17

Although I am quite satisﬁed with the resulting paper - I had really a lot of joy writing it - I am absolutely unsure to which paper I should submit it. Do you have any suggestions? Would Journal of Geometry be something? Apart from Zvonko Cerin’sˇ paper on the Neuberg Cubic I have never seen something from this Journal. I like your idea to start a Journal of Euclidean Geometry more and more!

Paul Yiu to Floor van Lamoen, October 29, 1999. Dear Floor, Please feel free to include my earlier remark(s). I look forward to read the new version of the paper. After the weekend, I shall have time to think about this work. I have to go to teach a class in half an hour, and so can only be brief now. It would be very good if you can buy a copy of Bottema and/or other Dutch geometry books. I would love to have these. Please let me know the expenses and I shall reimburse you. I have been aware of the Dutch geometric tradition from the past issues of the Canadian problem journal Crux Mathematicorum. Names likes Bottema, Veldkamp and others I often came across there, Of course, I do not know Dutch. But geometry books with illustration usually make the content understandable. Outside mathematics, I have been deeply nurtured by the paintings of Rem- brandt and van Gogh, and the writings of Henri Nouwen, who wrote mainly in English in this continent. These are my good Dutch inﬂuences.

Paul Yiu to Geometry.college, November 2, 1999. Dear friends, Talking about cubic curves, I would like to introduce a family of cubic curves arising from the notion of P-perspectivity that Floor formulated in his recent preprint. I came upon this family of cubic curves last night when going through a recent Monthly article, whose exact reference Antreas asked me to check up yesterday: Miguel de Guzman, An Extension of the Wallace - Simson Theorem: Projecting in Arbitrary Directions, Amer. Math. Monthly 106 (1999) 574–580. [June-July issue]. Let P be a point with homogeneous barycentric coordinates u : v : w.For each point X the P −traces of X are the intersections of the side lines of the reference triangle with the lines through X parallel to AP , BP, CP respectively. Suppose X has barycentric coordinates x : y : z.TheP −trace triangle is in perspective with the reference triangle if and only if (v + w − u)x(w(u + v)y2 − v(w + u)z2)=0.

[vL] My revised paper includes this cubic. It consists of the points Q such that Q and its P −iso conjugate (see below) are collinear with the point v + w − u : w + u − v : u + v − w. vanLamoenandYiu:Pre-HyacinthianGeometry 18

This ‘pivot’is the reflection of P through OP ,whereOP is the center of the conic CP of points Q such that the P −trace triangle of Q is degenerate (like the circumcircle in case of perpendicularity). Note the similarity with the De Longchamps point. I call the cubic ”P-Darboux cubic”. The P −iso conjugate of a point X = x : y : z is given by yzu(v+w): xzv(u + w):xyw(u + v). It is defined in a way similar to isogonal conjugacy: let the P −trace triangle of X be circumscribed by a conic homothetic to the conic CP above. This conic intersects the sides of ABC in a second triangle. This triangle is again a P −trace triangle of a point Y .ThepointY is the P −iso conjugate of X. A very notable observation: Q is on the P −Darboux cubic iff P is on the Q−Darboux conic.

This is an ISOGONAL cubic if and only if P = u : v : w is the orthocenter, and the triangle of P −traces of a point is the pedal triangle. [In this case, the cubic in question is associated with the de Longchamps point]. It is an ISOTOMIC cubic if and only if P is the centroid.

[vL]: When P is the centroid, then the cubic degenerates into the three medians of ABC.

For example, if P = a : b : c, the incenter, then the cubic is (b + c − a)(c(a + b)xy2 − b(c + a)xz2)=0.

This cubic curve contains at least the following points: incenter, orthocenter, Nagel point, Spieker center, X65 and X72 in Kimberling’s list.

[vL]: There is something funny about taking a weak point for P ,and thus having a weak point as “pivot”. This gives us four cubics as extraversions. In the below cubic, for instance, the three extraversions must all pass through the orthocenter.

Is this family of cubic curves well known, or studied before? [vL]: Add me to the list of people asking! The related cubic of points X = x : y : z such that their Cevian triangle is a P-trace triangle, is remarkably simple:

vwx(yy − zz)+uwy(zz − xx)+uvz(xx − yy)=0.

It is always self-isotomic, consisting of the points Q such that Q and its isotomic conjugate are collinear with the isotomic conjugate of P . [Y]: How nice! I suspected you knew of these. Before receiving your message, I checked the one you sent two days ago, and found you mentioned conics, and realized sometimes you meant cubics! vanLamoenandYiu:Pre-HyacinthianGeometry 19

By the way, what is the “working group”, and who are in that? I suppose it is under the coordination of Guru Conway. [vL]: The working group consisted of Steve, Guru JHC, Barry Wolk and I, but has grown since. We were trying to prove that the Neuberg cubic is the locus of points Q such that the Brocard axes of ABC, ABQ, AQC and QBC are concurrent. The term “working group” is, as far as I know, introduced by Steve. For me the “working group” consists of everybody who has joined discussion until now. So especially I include you and Antreas.

Paul Yiu to Geometry.college, November 2, 1999. This is a supplement of my earlier posting on the cubic curve of points whose P −traces form a triangle in perspective with the reference triangle. This morning, I gave the example of P = incenter = a : b : c, and a number of points on the corresponding cubic curve. I have forgotten to list X40 = reﬂection of incenter in the circumcenter, among the incenter, orthocenter, Nagel point, Spieker center, X65 and X72 in Kimberling’s list. Here is another interesting observation: The above cubic can be rewritten in the form (y + z − x)u(z(x + y)v2 − y(z + x)w2)=0

This gives the following

Theorem The P −trace triangle of Q are in perspective with ABC if and only if the Q−trace triangle of P are in perspective with ABC. Here are a few more examples of P and points on the cubic curve QP associated with P . It is clear that P lies on QP .

P Points on QP [Kimberling index] II,H,Nagel, Spieker, 40, 65, 72 OO,H,9 − point center, 52, 68 H I,O,H,de Longchamps, 40, 64, 84 9 − point O, 9 − point KK,66, 69 Gergonne Gergonne, Mittenpunkt Nagel I,Nagel Mittenpunkt Gergeonne, Mittenpunkt Spieker I,Spieker Feuerbach Feuerbach, 100

Floor van Lamoen to Paul Yiu, November 4, 1999. Yesterday I have put Bottema’s book in the mail. Together with it you’ll ﬁnd the new version of my P −perpendicular paper. It included already the observation vanLamoenandYiu:Pre-HyacinthianGeometry 20 you Tuesday made that “the P −trace triangle of Q are in perspective with ABC if and only if the Q−trace triangle of P are in perspective with ABC”. I have decided to submit the paper to ”Journal of Geometry”, although I don’t really know the Journal. It appears in good references though (for instance Antreas’list about cubics). [Y]: I look forward to reading the new version of the P −perpendicular paper and the Bottema book. Please let me know how much I owe you for the Bottema book. I thought about the notion of P −perpendicularity yesterday, and it is really a very useful concept. The recent discussions about cubics can be extended to this more context. It is really exciting! [vL]: I send you the book as a gift (besides, it is not very expensive). So you owe me nothing. [Y]: It is very kind of you. Floor. Thank you so much. I really look forward to reading the Bottema book. [vL]: I enjoyed reading about your connection with my home country: you seem to be aware of the tradition of painting since you mention Rembrandt en Van Gogh. Perhaps you are familiar with famous Dutch painters like Vermeer, Escher, Appel, Mondriaan and/or De Kooning (the latter two having ended their careers in the USA) too. [Y]: How could I have forgotten M.C.Escher? His paintings has a very strong geometrical theme. Perhaps I was thinking of spiritual nurture when I wrote of Rembrandt and van Gogh. [vL]: I don’t know the writer Henri Nouwen you mention. My favourite in Dutch writer is Simon Vestdijk. His works have been translated into several languages, but I don’t think they’re sold outside the Netherlands and Belgium anymore, since the books are somewhat oldfashioned, slow and typicly Dutch. [Y]: Henri Nouwen was one of my favorite spiritual writers. He was a Catholic priest who came to the United States in the 60’s, taught theology in several famous universities. He quitted his professorship in Harvard in 1986 to join the Daybreak Community in Toronto, Canada, ministering to the severely handi- capped. He was a popular spiritual writer in this country. One of his famous books, the Return of the Prodigal Son, is on Rembrandt’s famous painting. He died suddenly in 1996, at the age of 64, when he was home in the Netherlands, on his way to Russia, to work on a TV project on Rembrandt’s Prodigal Son painting. His books are all very short, easy reading, and penetrating message of the Gospel of Jesus Christ. I read all of his 40+ books, and they are by my bedside. [vL]: I am glad that you agree that the notion of P −perpendicularity seems to be a useful one. Until now I have been very surprised how much geometry can be generalized using this P −perpendicularity. I am however still looking for a good journal to send my paper. Maybe, when you’ve seen the new version, you can help me with a suggestion? [Y]: The notion of P −perpendicularity is a very useful one, I think. I am sure there are a lot more that can be said. For example, the observation of the equivalence of P −trace triangle of Q being cevian and the Q−trace triangle of P being cevian is very interesting. I think you meant the same thing in your vanLamoenandYiu:Pre-HyacinthianGeometry 21 earlier message about “the P −Darboux cubic and Q−Darboux conic”. You see, Floor, I have slipped into the same confusion between “cubic” and “conic”. The other day, when I asked for a sketch of the cubic of points whose duals and those of their isogonal conjugates are perpendicular, I also wrote “conic” instead. But Steve Sigur sent in a cubic anyway! More seriously, how about starting a more systematic, book project on the geometry of P −perpendicularity.

[vL](11/8): This sounds very interesting. There must be so much that can be done using P −perpendicularity! What exactly do you think of when you mention “more systematic, book project”?

I have to teach a class in 20 minutes, and shall write more about this when I come back from classroom. Here are some data illustrating the theorem that the P −trace triangle of Q is cevian if and only if the Q−trace triangle of P is cevian.

P Q perspector of P − trace perspector of Q − trace triangle of Q triangle of P −1 I Nagel I = X75 G −1 I Spieker X10 G X86 = X10 a I X65 Gergonne X81 =[b+c ] a(b+c−a) I X72 Nagel X21 =[ b+c ] −1 1 O H O =[a2(b2+c2−a2) ] G −1 1 O N N =[a2(b2+c2)−(b2−c2)2 ] G −1 H L = X20 H G 1 K X69 3rd Brocard X76 =[a2 ] G −1 1 X7 X9 G X85 = X9 =[a(b+c−a) ] −1 1 X11 X100 X11 =[(b−c)2(b+c−a) ] G

Have a good weekend.

Paul Yiu to Geometry.college, November 8, 1999. I would like to oﬀer a generalization of the family of cubics that John Conway is talking about here. Let P be a point with homogeneous barycentric coordinates u : v : w.For each point X,theP −traces of A[P,X], B[P,X],andC[P,X] are the intersections of the side lines of the reference triangle with the lines through X parallel to AP , BP, CP respectively. Fix a real number k. On each of the the lines XA[P,X], XB[P,X],andXC[P,X] choose a point A[P,k], B[P,k],andC[P,k], such that XA[P,k] = k · XA[P,X], XB[P,k] = k · XB[P,X],andXC[P,k] = k · XC[P,X]. vanLamoenandYiu:Pre-HyacinthianGeometry 22

The triangles A[P,k]B[P,k]C[P,k] and ABC are in perspective if and only if the point X lies on the cubic (v + w − ku)x(w(u + v)y2 − v(w + u)z2)=0.

The pivot of this cubic is the point v + w − ku : w + u − kv : u + v − kw, which is 3G − (k +1)P normalized. Here, G is the centroid of the reference triangle. If k = 1, this is the cubic I mentioned last week. See http://forum.swarthmore.edu/epigone/geometry- college/swelsumswex/[email protected] For k = 2, this is the cubic for the perspectivity of the triangle of P −reﬂections and the reference triangle, which has pivot the point at inﬁnity of the P −Euler line.

Paul Yiu to Geometry.college, November 8, 1999. This is a commentary on the sextic curve that John [Conway] referred to in his earlier posting. Given a point P = f : g : h in homogeneous barycentric coordinates. We talk about P −perpendicularity. If the point of infinity of a line ell is u:v:w, then the point of infinity of the P −perpendicular(s) of ell is v w w u u v − : − : − . g h h f f g Now we consider a point X = u : v : w.TheP −perpendicular to AX at A is the line v v + w w w + w ( + )y − ( + )z =0. g f h f Similarly, we write down the equation of the P −perpendiculars at B to BX, and at C to CX. Let’s call these P −perpendiculars 2a, 2b,and2c respectively. The intersection A of the lines 2b and 2c is the point v w + v w w + u u u + v w w + u u w + u v u + v ( + )( + ):( + )( + ):( + )( + ). g h h g f h h g f g g h The line joining AA intersects BC at the point u u + v w w + u u w + u v u + v 0:( + )( + ):( + )( + ). f h h g f g g h Similarly, we write down the coordinates of the intersections B of 2c, 2a, and C of 2a, 2b. The line BB intersects CA at the point v u + v w v + w v v + w u u + v ( + )( + ):0:( + )( + ), g h h f g f f h vanLamoenandYiu:Pre-HyacinthianGeometry 23 and the line CC intersects AB at the point w w + u v v + w w v + w u w + u ( + )( + ):( + )( + ):0. h g g f h f f g From these coordinates, we easily determine the condition for the perspectivity of the triangles ABC and ABC to be that X lies on the union of two cubic curves. The first of these is the cubic curve (g + h − f)u(h(f + g)v2 − g(h + f)w2)=0, which is the same as the cubic for which the P −trace triangle is cevian. The second cubic is u(h(f + g)v2 + g(h + f)w2)++2(fg + gh + hf)uvw =0.

[This actually factors into (u + v + w)(g(h + f)wu + f(g + h)vw + h(f + g)uv), and is the union of the line and inﬁnity and the standard P −circumconic].

[vL]: From here I would prefer to go a diﬀerent way. Let me refer to the above three points as A, B and C. Remember that the P −trace or P −pedal triangle of X is given by the matrix:   0 gu +(g + h)vhu+(g + h)w  fv +(f + h)u 0 hv +(f + h)w  fw +(f + g)ugw+(f + g)v 0

Which I will refer to as:   0 m12 m13   m21 0 m23 . m31 m32 0

Then triangle ABC can be written as:   0 gm13m21 hm12m31   fm23m12 0 hm21m32 . fm32m13 gm31m23 0

These points form a Cevian triangle iﬀ:

2 2 fgh((m12m23m31) − (m21m32m13) )=0 fgh(m12m23m31 − m21m32m13)(m12m23m31 + m21m32m13)=0.

So we have that the P −pedal/P −trace triangle is either Cevian or degenerate! vanLamoenandYiu:Pre-HyacinthianGeometry 24

[Y]: Excellent! How could I have missed this? I was actually looking for an interpretation of this second cubic. [vL]: So the locus of X for which the P −pedal triangle is perspective to ABC is the union of the cubic (P −Darboux) for which the pedal triangle is a Cevian triangle, the P −circumconic in the sense of Simson’s theorem, and the line at infinity. Specializing this to P = H we find the union of the circumcircle, the Darboux conic (isogonal, pivot= De Longchamps) and the line at infinity. [Y]: cubic again!

The second aside, we obtain from the first one the following corollary: (1) If the P −trace triangle of a point is cevian, then so is the “P −antipedal” triangle of the same point. Specializing to P = H, the orthocenter, we obtain (2) If the pedal triangle of a point is cevian, then so is the antipedal triangle of the same point. Example. For P = H, the orthocenter, the first cubic is the isogonal cubic of the de Longchamps point. On this cubic is, for example, the point 2O − I, the reflection of the incenter about the circumcenter, X40 in Kimberling’s list. The pedal triangle of this point is cevian. In this example, the two perspectors both lie on the OI−line.

Floor van Lamoen to Paul Yiu, November 8, 1999. Dear Paul, You are redoing my work!

[Y]: That was what I meant last Friday when I suggested we work on something together in the form of a book on P −perpendicularity and some related concepts. How does this sound to you? [vL](11/9): This sounds excellent! And I am sure there will be enough to write about. What would be the best way to coordinate things. [Y]: I have actually written a few pages (about 15) on reflections and includes the basics of P −perpendicularity, together with a generalization of the reflections of line I wrote about last week. Specifically, we talk about the “standard P −circumconic”. [vL](11/9): In my revised paper I call them P −circ conics. Pointing to both ‘circum-’and ‘circle’. The observation on the P −antipedal/prepedal triangles is already in my paper. But I am open to any change in naming. [Y]: As generalizations of what I posted on October 29, Friday, here is vanLamoenandYiu:Pre-HyacinthianGeometry 25

Theorem

The P −reﬂections of a line are concurrent if and only if the line passes through P . In this case, the intersection is on the standard P −circumconic. This is a special case of the following

Theorem

The P −reﬂections of a line in the side lines of the reference bound the cevian triangle of a point on the standard P −circumconic. [vL](11/9): Very nice! [Y](11/8): I have a number of cubic curves arising from various problems on P −perpendicularity. This notion is actually equivalent to the approach in my Feuerbach conic preprint, where the incenter is prescribed instead of the orthocenter as in your case. Each of these has its own advantages, depending on the relative simplicity of the coordinates involved. [vL](11/9): Yes, this is what I immediatly noticed when I read your paper. You were in a way doing exactly the same thing. That is why I thought it was a good idea to send you the preliminary version of my paper. About advantages: It surely is an advantage to prescribe the incenter when you write about inconics. However, prescribing the orthocenter seems more general. Some notes (you will read in the refreshed paper you will recieve soon): Let P = t : u : v. The center of the Bottema conic t(u + v): u(t + v):v(t + u)isaP −version of the symmedian point. The points ± t(u + v):± u(t + v):± v(t + u), if exisiting, are the centers of the inconics. These are the points invariant under the P-iso conjugacy, given by:

t(u + v) u(t + v) v(t + u) x : y : z → : : . x y z

There was one thing about this conjugacy that I wanted to add, and which is not in my paper: Consider the B and C vertices of the pedal triangle of the P −iso conjugate of X = x : y : z.

tuxz + t(u + v)yz :0:uvxz + v(t + u)xy, tvxy + t(u + v)yz : uvxy + u(t + v)xz :0. vanLamoenandYiu:Pre-HyacinthianGeometry 26

The line through these points meets the line at inﬁnity at:

tuz − tvy : −uv(y + z) − tuz : uv(y + z)+tvy

which is exactly the point belonging to lines P −perpendicular to AX. We can conclude: The P −prepedal triangle of a point X is homothetic to the P −trace/pedal triangle of the P −iso conjugate of X. [Y](11/8): I wrote the commentary on Conway’s problem to show that the calculation in the case of P −perpendicularity is actually easier than the standard case when P = H. Please keep the discussion on collaboration between us for a while.

Paul Yiu to Floor van Lamoen, November 9, 1999. Dear Floor, I think virtually the whole of triangle geometry can be (or should be) reworked in the framework of P −perpendicularity. The main advantage, as you pointed out to Steve Sigur, is that it gives more general results with simpler calculations. It is a most geometric way of treating the conics. I got this insight from Theorem 3 of your preprint in August.

[vL]: I hope also some good new insights will come up. Like “Q is on the P −Darboux cubic iﬀ P is on the Q−Darboux cubic”.

Here are two examples.

Theorem 1 Let 2 be a line through P ,andQ a point on 2.DenotebyX, Y , Z the P −reﬂections of Q on the sides BC, CA,andAB. Then the circumconics of AY Z, BZX, CXY homothetic to the standard P −circumconic intersect at a point on the standard P −circumconic whose line of P −reﬂections is 2.

Theorem 2 Let 2 be a line through the center of the standard P −circumconic of triangle ABC, intersecting the side lines at X, Y , Z respectively. The homothetic conics with diameters AX, BY ,andCZ intersect at two points, one on the standard P −circumconic, and the other on the Feuerbach conic. The common chord passes through the point P . As such, it is the line of P −reﬂections of the P −reﬂection in 2 of the common point on the standard P −circumconic. The intersection on the Feuerbach conic is the P −orthopole of the line 2. vanLamoenandYiu:Pre-HyacinthianGeometry 27

Corollaries (1’) [Musselman’s Theorem] Let H be the orthocenter, and P an arbitrary point. For every point Q on the line HP,letX, Y , Z be the reﬂections of Q on the sides BC, CA,andAB. Then the circles AY Z, BZX,andCXY intersect at apointK on the circumcircle whose line of images is HP. (2’) [Blanc’s Theorem] Let 2 be a line through the circumcenter of triangle ABC, intersecting the side lines at X, Y , Z respectively. The circles with diameters AX, BY ,andCZ intersect at two points, one on the circumcircle and one on the nine-point circle. The common chord passes through the orthocenter. As such, it is the line of images of the reﬂection of the common point on the circumcircle in the line 2.

[vL]: Did you already prove these theorems? Or do you ﬁrst state and later prove them ? I had (and have) trouble in thinking how to compute the equation for a P −standard conic with a given diameter. That is what I miss when working with these conics: I can’t treat them as easily as I can treat circles! There is something else I am playing with in my head. There are so many geometric properties that I can’t easily generalize in the P −perpendicular sense. Like ”two angles on the same chord are congruent”. Would that make sense/ is it consistent / can we use it to prove something ?? I thought of the following way to generalize similarity of triangles: Two triangles are similar when the Cevian ratios of their P −orthocenter are equal. I think it would be interesting if we can deal with important notions like these! Not everything can be generalized, as it seems. I have unsuccesfully tried to generalize the famous Taylor circle. Maybe I have made mistakes, but it clearly didn’t work out the way I expected!

Conceptually, P −perpendicularity is indeed much more fundamental. In performing calculations, sometimes, switching to a prescribed incenter may be more convenient. This would be especially convenient when one wants to compare results obtained with standard results in triangle geometry using the orthocenter.

[vL]: Oh yes, sure. I was very delighted when I read your paper on the Feuerbach conics. It was such a clear piece of work. One of the reasons for this clarity was that you worked with a prescribed incenter, which made all very transparant. I don’t think you get the transparancy when you start with an incenter of the form t(u + v):... : ... !!

In his thought-provoking message yesterday, Steve said vanLamoenandYiu:Pre-HyacinthianGeometry 28

A problem with geometry is that there has been a tremendous amount of the ﬁrst and not enough of the second. The discovery and naming phase of geometry is important, but in our time, with our level of mathematical development, we should be able to understand the previous discoveries.

I very much agree with this. I can see Guru Conway behind this. However, as for myself, I am relatively new to triangle geometry, and do not know of much of the great things done in the past, though I am always painfully aware of the reality that in plane geometry there is hardly anything new under the sun. I mainly see myself as catching up with (or learning from) the great geometers in the 19th century. On the other hand, the great geometric past is scattered or buried in a few great libraries inaccessible to most of us, at least myself. Think of the famous book by F.G.-M. Every body says it is a treasure, but it is so diﬃcult to acquire a copy. I have never had a chance to see it.

[vL]: Well, I am certainly a newcomer, too. And I believe Steve is. Allmost everything in geometry is not new, but on the other hand we should have possibilities now to gain more insight. In that way I fully agree with Steve. We have the possibility to work with dynamic software, like GSP (and I use the freeware program WinGeom, which roughly has the same limitations). Also, I think we, as modern world citizens, will naturally have a diﬀerent approach from the old famous Geometers. You are in the happy circumstance (well, it is your job) to have easy access to a university library. I don’t, and I feel that as a student I should have bought the books I could get, and otherwise I should have xeroxed them. I very much enjoy Antreas’contributions in coming up with references and tell what they’re roughly about.

In this country, the richest source is probably the American Mathematical Monthly, and I have indeed borrowed very old issues (up to the 1920’s) from our library here and keep them in my oﬃce.

[vL]: Well, and we have Nieuw Archief voor Wiskunde. I subscribed as a student, but stopped subscription in 1991 when I had to reor- ganize my ﬁnancial situation. Two years ago I resubscribed. In the meantime the Problem Section seems to be dead, and NAvW seems not to like triangle geometry to much anymore. On the other hand, they have published some triangle geometry pieces recently.

In the past years, I have often dreamed of writing a good geometry book, categorizing what I have learned from these old sources. But in what ways can a contemporary geometry book be diﬀerent from the old ones? With the convenience of technology in recent years, one obvious way is the emphasis of elegant geometric constructions. I have actually taught my geometry classes in this spirit. vanLamoenandYiu:Pre-HyacinthianGeometry 29

[vL]: It is the synergy of constructions now and computations then, that I like very much. I don’t really like the approach of letting Maple, or Mathematica, or whatever program, compute that some locus leads to some cubic curve. It is very powerful though. I am sure that Guru Conway is having the same dream. He really comes up with wonderful ideas! I am very curious what his “Triangle Book” will be like.

This is why I particularly welcome the problems posed by Antreas. I always try to view his problems in the light of elegant geometric constructions. The only piece of geometry software I know how to use, namely, the geometer’s sketchpad, is of course very wonderful. But it has some severe limitations. It cannot deal with conics (and cubics) eﬃciently.

[vL]: Antreas is a phenomenal reader and I like his problems very much, too. Only, when I have busy times in school, they are too numerous for me!! And very often I don’t have a good approach in mind. I like it very much that, when you come with a solution, he answers with more questions! His problems are very good, and your solutions have been very interesting quite some times, too. I enjoyed!

The work of Clark Kimberling is very appealing to me, since I love old, elementary algebra. I have often wanted to write a commentary on his book. I mean categorizing geometric properties of the various triangle centers. That was why I started working on the conics. I want to compile a repertoire of conics, and more recently I suddenly realized the tractibility of cubic curves. [I must confess that this awareness was awakened by Antreas’asking me to check the Simson-Wallace line reference last week, prior to my posting last Tuesday. Before, I have always subconsciously avoided cubic and higher order curves. I begin more and more to appreciate the loci problems Antreas posed].

Guru Conway seems to be not very impressed by Clark’s work. I can imagine why; it seems that Clark is going for quantity sometimes. Also, his approach is not always very geometric. But, neverthe- less, I really appreciate his work very much, and think it is a good source. He puts energy in computing all kinds of possible things, like collinearities, parallelisms, etc. He is looking for all kinds of new ways to watch triangle geometry. That’s good.

Here is a very preliminary sketch of something that can be done. Preliminaries Miquel Theory

[vL]: You mean Miquel Theory with circles, or with P −perpendicularity? I have been wondering how to ﬁnd an interpretation for ”pivoting about an angle” can be translated in P −perpendicularity. vanLamoenandYiu:Pre-HyacinthianGeometry 30

Feuerbach conics P −perpendicularity

[vL]: This afternoon I realized: a P −standard conic can not be a P −perpendicular hyperbola. (Then P would be on the hyperbola, which would lead to t + u + v =0).

Reﬂections

Maybe it is interesting to include the ideas I worked out in my paper on Bliss’theorem in this section? ... Conjugacy mappings Generalizations of some famous triangle theorems: the isogonal theorem (already in my paper), Napoleons theorem, Morley??, Stammler (probably very complicated)

...... A Dictionary of conics and cubics. Please tell me what you think.

[vL]: I am empty now, for this day...

Paul Yiu to Floor van Lamoen, November 10, 1999. Dear Floor, This is a P −perpendicularity generalization of the Apollonian circles intersecting at the pair of isodynamic points. Let P = f : g : h be a ﬁxed point. Consider Q = u : v : w. The traces of Q on the sides are the points

X =0:v : w, Y = u :0:w, Z = u : v :0.

Construct points X, Y , Z on the sides such that X, X divided BC harmonically; similarly for the other two pairs. These are the points

X =0:v : −w, Y = −u :0:w, Z = u : −v :0.

Every conic homothetic to the standard P −circumconic is of the form

f(g + h)yz + g(h + f)zx + h(f + g)xy − (x + y + z)(px + qy + rz)=0 vanLamoenandYiu:Pre-HyacinthianGeometry 31 for some p, q, r. The center of this conic is the point r − p p − q p − q q − r q − r r − p g + h + − : h + f + − : f + g + − . g h h f f g

[The expressions I gave previously for the center were not correct, due to careless typing]. From this, it is easy to determine the conic with diameter XX homotetic to the standard P −conic. This is given by

h(f + g)v2 − g(h + f)w2 −f(g + h)w2 f(g + h)v2 (p, q, r)=( , , ). v2 − w2 v2 − w2 v2 − w2 Similarly, the P −conic with YY as diameter is deﬁned by

g(h + f)w2 f(g + h)w2 − h(f + g)u2 −g(h + f)u2 (p, q, r)=( , , ). w2 − u2 w2 − u2 w2 − u2 The “radical axis” of these two P −conics is the line

[g(h + f)w2(u2 − v2)+h(f + g)v2(w2 − u2)]x +[h(f + g)u2(v2 − w2)+f(g + h)w2(u2 − v2)]y +[f(g + h)v2(w2 − u2)+g(h + f)u2(v2 − w2)]z =0.

From this, it is clear that the radical axes of the other two pairs of P −conics are the same line. It follows that the three conics are coaxal, and intersect at two common points which are inverse with respect to the standard P −circumconic.

Proposition These intersections are real if and only if f 2(g + h)2v4w4 − 2g(h + f)h(f + g)u4v2w2 ≥ 0.

This is equivalent to the point

f(g + h) g(h + f) h(f + g) : : u2 v2 w2 being on or inside the inner Steiner ellipse

x2 + y2 + z2 − 2xy − 2yz − 2zx =0

(tangent to the sides of the reference triangle at their midpoints).

Corollary The point Q = u : v : w is such that the three P −conics are tangent to each other at a point on the standard P −circumconic if and only if Q is the “square root” of the P −iso conjugate of a point on the inner Steiner ellipse. vanLamoenandYiu:Pre-HyacinthianGeometry 32

Application Animate a point K on the inner Steiner ellipse. Construct semicircles outside the triangle, with BC, CA, AB as diameters. Through the traces of the isogonal conjugate of K, construct perpendiculars intersecting these semicircles at X0, Y0, Z0 respectively. Construct the bisector of angle BX0C to intersect the side BC at X. Simi- larly, construct Y and Z on the other two sides. [AX, BY , CZ are concurrent]. Construct a point X on BC such that X and X divide BC harmonically. Similarly, construct points Y and Z. The circles with diameters XX, YY,andZZ are tangent to each other at a point on the circumcircle. Best regards. Sincerely, Paul vanLamoenandYiu:Pre-HyacinthianGeometry 33

Floor van Lamoen to Paul Yiu, November 10, 1999. Dear Paul, Thanks for your explanation on how to deal with a P −conic on a given diameter. You are doing it roughly the same way I did, but I think I have mistaken on the center somewhere. Anyhow, I will give it a retry later. I intened to use the P −conics on a given diameter to ﬁnd analogues of equilateral triangles, but as you have seen I found another way. The result is again surprisingly beautiful! And again the Steiner ellipse plays such an important role. Best regards. Floor.

Floor van Lamoen to Paul Yiu, November 10, 1999. Dear Paul, I will type it in TEX-style, so that it can be easily transported. I take P =(α : β : γ). We need the condition that αβγ > 0. I put coordinates of points between () and of lines between []. Consider segment AB of ABC. We will look for points C such that ABC is an P -equi triangle, which means that the P -orthocenter and the centroid of ABC arethesame.Torealizethat,C has to be on the P - perpendicular sidebisector of AB which is given with parameter t by C = (1 + 3tα :1+3tβ : −3t(α + β)). The centroid of ABC is thus C =(1+tα : 1+tβ : −t(α + β)). The point where BC meets L∞ is (1 + tα : −1+tβ : −t(α + β)). The point on L∞ where lines P -perpendicular to BC meet is

X =(αγ − α2βt − αβ2t − αβγt : βγ + α2βt + αβ2t + αβγt : −αγ − αβ).

The point where AC meets L∞ is Y =(−1+3tα :1+3tβ : −3t(α + β)). The condition that Y = X gives

−1+3tα αγ − α2βt − αβ2t − αβγt = , 1+3tβ βγ + α2βt + αβ2t + αβγt which reduces to γ t = ±√ . 3αβγ We ﬁnd two possible points for A. We can ﬁnd points A, A, B and B in similar ways. This gives us the possibility to state the following.

Theorem Consider the triangle of centroids ABC of P -equi triangles erected on the triangle sides, either all taken with positive parameter, or all with negative parameter. Then ABC is a P -equi triangle. vanLamoenandYiu:Pre-HyacinthianGeometry 34

Proof. Coordinates of ABC are given by

−α(β + γ): κ + αβ : κ + αγ, κ + αβ : −β(α + γ):κ + βγ, κ + αγ : κ + βγ : −γ(α + β). √ where κ = ± 3αβγ. The centroid of this triangle is G, the centroid of ABC. With straightforward computations we ﬁnd that lines perpendicular to AB meet L∞ at κ κ 2κ + αγ : + βγ : − − γ(α + β), 3 3 3 which is easily seen to be collinear with G and C. I hope you enjoyed this. I am happy that I have succeeded in thinking of a way to work√ with “equilateral triangle” in P −perpendicularity. And I was surprised by 3 popping up so clearly! Best regards. Floor.

Paul Yiu to Floor van Lamoen, November 10, 1999. Dear Floor, I shall study this later. But I am glad you ﬁnd a way to handle equilateral triangles. This has been puzzling to me, though I have not seriously try to deal with it. Very good indeed! Best regards. Sincerely, Paul

Floor van Lamoen to Paul Yiu, November 11, 1999. Dear Paul, Thanks for your explanation on how to deal with a P-conic on a given diameter. You are doing it roughly the same way I did, but I think I have mistaken on the center somewhere. Anyhow, I will give it a retry later. I intened to use the P-conics on a given diameter to ﬁnd analogues of equilateral triangles, but as you have seen I found another way. The result is again surprisingly beautiful! And again the Steiner ellipse plays such an important role. Best regards. Floor.

Floor van Lamoen to Paul Yiu, Thursday, November 11, 1999. P −similarity Dear Paul, I propose the following deﬁnition of P −similarity: Two triangles are similar iﬀ their P −orthocenters have equal Cevian ratios / equal barycentrics (w.r.t. the triangles themselves). For instance: vanLamoenandYiu:Pre-HyacinthianGeometry 35

The ﬁrst Bottema triangle T1 of P = t : u : v is given by: 0 u(t + v) tv tu 0 v(t + u) t(u + v) uv 0

The P −orthocenter of T1 is tu : uv : tv. When we calculate this to barycentrics w.r.t. T1 we ﬁnd v : t : u (It is easier to evaluate v : t : u in T1 and ﬁnd it is tu : uv : tv). So T1 is P −similar to ABC, although not directly. Best regards. Floor

Paul Yiu to Floor van Lamoen, November 11, 1999. Dear Floor, Excellent. I shall study this. Please brain - storm whatever occurs to you about P −perpendicularity. I really believe that the whole of triangle geometry should be reworked in this context. I am working on your message yesterday, and trying to compare it with other natural alternatives. Since “equi” is a preﬁx, perhaps P-regular is more appropriate. Best regards. Sincerely, Paul

Floor van Lamoen to Paul Yiu, Thursday, November 11, 1999. Dear Paul, I agree to avoid “equi” as a suﬃx and use the word P −regular. IhaveworkedonP −perpendicularisation of things around the Kiepert hyperbola. It looks rather promising, and I hope that I can send some of the results tomorrow. Reworking “the whole” of triangle geometry (the non-projective part) is a good challenge!! Best regards. Floor.

Floor van Lamoen, Thursday, November 11, 1999. Brocard circle Dear Paul, In reply to your message “The Brocard circle as a locus of Miquel point” I send you a message I sent to John Conway 3-30-98. It is roughly the same question! I will also send his reply in a seperate message. Kind regards, Floor.

I have been thinking about the following. Let ABC [be] inscribe[d in] ABC in such a way that AB : BC = CA : AB = BC : CA = 1:t. Construct the in-pivot P belonging to inscribed triangles similar to ABC. P has barycentrics:

a2(−t2c2 + ta2 + b2):b2(−t2a2 + tb2 + c2):c2(−t2b2 + tc2 + a2). vanLamoenandYiu:Pre-HyacinthianGeometry 36

When you construct ABC by changing the ratio 1 : t to t :1 then the in-pivot Q has barycentrics: a2(−t2b2 + ta2 + c2):b2(−t2c2 + tb2 + a2):c2(−t2a2 + tc2 + b2). When t =1thenP = Q =circumcenter; When t = 0 then you get the Brocard points; P and Q are on the Brocard circle. (It seems that P and Q are reﬂected in each other over the Brocard-diameter). This must mean that all ABC (and ABC)musthavethesame Brocard angle as ABC. I conjecture also the following: PCB = 1 QC A = 2 POQ,[O = circumcenter]. Is this known?

Floor van Lamoen, Thursday, November 11, 1999. Dear Paul, This is John Conway’s reply. Some things refer to thoughts he had been sending around then, like the “Global Viewpoint”. If you wish, I can find some stored message about it probably, and forward it to you. Best regards. Floor. I did in fact know most of it, although it’s taken me a few minutes to realise that! Part of it resembles something Alex Ryba was telling me (but isn’t quite the same). The assertion that all these triangles have the same Brocard angle is easy, using the fact that the images of all the equilateral triangles in one plane under a fixed orthogonal projection (or affine map) have the same Brocard angle. Then the rest is a consequence of Schoute’s theorem, that the locus of (in)pivots of triangles with a given Brocard angle is one of the so- called ”Schoute circles”, the Brocard circle being the Schoute circle for the Brocard angle of ABC itself. More below on this. The thing Ryba told me is that if you make the angles AAB, BBC, CCA all equal (to any sufficiently small value), then the triangle that’s left when you chop them off shares the appropriate Brocard point with ABC and (I think) has the same Brocard angle. Schoute’s theorem is one of the things that is seen very naturally in my ”global viewpoint”. On the globe, the triangles having given Brocard angle lie on a latitude circle. Since the pivot-point rela- tion is inversively invariant, the Schoute loci must be the images of these latitude circles under the appropriate stereographic projection, namely a coaxal system of circles in the plane. Since I discovered the global viewpoint, I’ve been calling them Schoute’s ”latitude-circles”. Prominent members are: vanLamoenandYiu:Pre-HyacinthianGeometry 37

• The circumcircle (image of the equator) • The Lemoine axis (image of the circle of latitude containing ∆), which I’m now calling Lemoine’s ”latitude time”. • The Brocard circle (image of the circle of opposite latitude to the above) • The two point-circles around the Isogonic points (images of the poles), which I’m now calling the ”isogonic poles”. The fact that P and Q are images of each other in ”the longitude line” OK (I call it this because it’s the ”longitude circle” of infinite radius) I can’t quite see at this moment, though maybe it will be easy with pencil and paper. Oh - I think I see a ”cheating” proof. The correspondence between P and Q is a 1-1 algebraic correspondence (from the Brocard circle to itself) that fixes O and swaps Oe, Ow. So is the reflection in OK. But such a correspondence is determined by the images of three points! I’ll try to think of a more geometric proof. I found your letter most stimulating, because although it turned out that I ”really” knew most of what you were saying, this wasn’t obvious at first, because you were saying it in a way that was new to me. I am very pleased today for another reason, namely that last night I recovered what I’ve been calling my ”Lost Theorem”, which is the ultimate generalization of the Tucker circles, and it suggests an interesting little research programme. [The theorem, by the way, is almost certainly the same as one due to J.A. Third that I’ve seen references too, but I wanted to rediscover it the way I found it 15 or 20 years ago, so have avoided looking up Third’s paper.] Also, a few days ago, I unified things a bit by introducing the following ”brother-sister” terminology in this area. A Tucker hexagon for ∆ has its sides alternately parallel and antiparallel to the sides of ∆. Let ∆1 and ∆2 be the triangles formed by alternate sides, ∆1 being parallel to ∆. Then the Tucker circle we’re talking about is also a Tucker circle for Delta1 - I call this the ”brother” circle for ∆1.For∆2 it’s what Peter Yff class an ”isoscelizer circle”, and I call it the ”sister” circle for ∆2. So we have families consisting of two brothers and one sister, For the Lemoine parallel circle, the brother is the infinite Tucker circle, and the sister the Adams circle. The Lemoine antiparallel circle is its own brother, and its sister is the infinite isoscelizer circle. The Taylor circle’s sister is the “Conway circle” I found a few weeks ago, and its brother is an interesting Tucker circle that doesn’t seem to vanLamoenandYiu:Pre-HyacinthianGeometry 38

have been noticed before. The circumcircle is its own brother, and its sister is the incircle.

Floor van Lamoen, November 12, 1999. Kiepert hyperbola Dear Paul, Here are my ﬁndings on the Kiepert hyperbola. The Kiepert hyperbola in P −perspecivity. Let P = f : g : h, f + g + h =1,fgh > 0. We have found that the vertices of P −regular triangles on the side BC are are given by fgh fgh −f(g + h):fg ± : fh± . 3 3 Therefore the P-Fermat points are found as 1 ···: : ··· fgh hf ± 3

It is iso−P (’iso’ is a preﬁx, so let’s use it that way) conjugates are the P −isodynamic centers: fgh ···: g(f + h)(hf ± ):···. 3 These P −isodynamic centers are the two intersection points of the Appolonius P −conics of the P −incenter. Now let’s generalize the P −Fermat points in the following way 1 ···: : ···. hf ± t These points form the P −Kiepert hyperbola, given by: f(h − g) g(f − h) h(g − f) + + =0. x y z On this hyperbola we also ﬁnd G and P .SinceP is on the hyperbola, P has P −perpendicular asymptotes. (Generalizing the P = H case, it should be that these asymptotes are the Simson lines of the points where the P −Brocard axis meets the P −circumconic. I haven’t checked.) The center is of the P −Kiepert hyperbola is (no mistakes?)

f 2(h − g)2 : g2(f − h)2 : h2(g − f)2, a point on the Feuerbach P −conic. The iso−P conjugates of points on the P −Kiepert hyperbola form line OP KP , the isotomic conjugates the line G − KP . Let 1 V = ···: : ···, n hf + v vanLamoenandYiu:Pre-HyacinthianGeometry 39

1 V = ···: : ··· s hf − v be points on the P −Kiepert hyperbola, and

Wn = ···: wg(f + h)+f + h : ···, Ws = ···: −wg(f + h)+f + h : ··· be iso−P conjugates of such points. Then wVn + vWs = −wVs − vWn = wvv ∗ G + wfgh ∗ P + v ∗ OP ,where G =1:1:1,P = f : g : h, OP = g + h : f + h : f + g. So the lines VnWs and VsWn intersect in a point on the P −Euler line. Have a nice weekend! Best regards. Floor.

Paul Yiu, Friday, November 12, 1999. Dear Floor, I have many things to thank you. First of all, your P −perpendicularity paper and the Bottema book have just arrived this morning. I shall study the paper later. After the last week, you can see that I am now more familiar with P −perpendicularity. This is just a wonderful concept!! I am sure a lot of good things will be coming up. And indeed, they already have, in your messages this past few days. The Kiepert hyperbola message has just popped up on my computer screen. I have here a master’s student defending her thesis next Tuesday, and I am one of the readers of her thesis. After that I can start seriously working on the geometry book. The messages you communicated in the past few days have further convinced me the worth and viability of “reworking triangle geometry”. The Bottema book is just a lovely gem. A lot of wonderful geometry can be done in just 100 pages! I like the fact that Bottema punched into Ceva’s theorem right at the beginning and used barycentric coordinates.

[vL]: I am glad you appreciate Bottema’s book. It is a gem indeed! I have wondered many times how such a small booklet can contain so much. And I reread/browse it quite often.

Chapter 7, on the distances from a point to the vertices of a triangle, is on my favorite “Bottema equation”. I ﬁrst learned of it from Bottema’s short paper in Crux Mathematicorum in the early 1980’s.

[vL]: I suppose you point to equation (13) for the Euler line on page 38 (chapter 8)? That one I used in the newsgroup to show that if the Euler lines of three of the triangles ABC, PBC, AP C and ABP concur, so does the fourth. It is a piece of cake then! [Y]: I meant equation (3) on p.34, the more general one. I also love elementary number theory, speciﬁcally Diophantine equations. vanLamoenandYiu:Pre-HyacinthianGeometry 40

I can also see the cubics of Darboux and Lucas on p.18. The diagram on p.120 is very impressive. Chapter 26 is on the failure of the analogue of the Lehmus Theorem for external bisectors.

[vL]: Actually, sections 1-4 are about diﬀerent proofs of the Lehmus- Steiner theorem. I like the proof in 4.

Next spring, I am teaching undergraduate geometry again. I shall see if it is possible to model the class on this book or not. It is just a gem. Thank you! Also for your response to my Brocard circle posting yesterday. Let me tell you the truth. I posted this with an intention of luring Guru Conway to say something. [I did not know why I got this sense]. And he did! But his explanation was too vague for me. Then I began looking into Johnson’s Advanced Euclidean Geometry and things began to make sense. Then your forwarded message came. It was very eye opening indeed. I like to perform algebraic computation to obtain results, and then interpret (or explain) them elegantly. I am of course interested to know of the “global viewpoint” that Conway was talking about. But there is no hurry. We have enough of geometrical things these days to keep our heartbeats up at very high level!

[vL]: I will send it some time later. Yes, there are many geometrical things to do, and I feel very well about it. However, I will have to focus on a lot of other things too, in the coming weeks. We will have exam week next week in the ﬁnal years of our school, and that is always a busy time. But probably I will ﬁnd some time to think and type. [Y]: There are many good things in front of us. Perhaps we need to learn to relax a little bit, to have the tranquillity to do things well.

Best regards. Sincerely, Paul

Floor van Lamoen, Monday, November 22, 1999. Dear Paul, Much to my surprise today I found your work “Euclidean Geometry - preliminary version” in the mail. Many thanks! I have only had the opportunity to roughly browse through the book, but I already found some things I didn’t know. And I am quite impressed by the nice approach and good pictures you have put in your work. Of course I was happy to ﬁnd my favourite Steiner-Lehmus proof, the one by Descube, on page 17. I was also very impressed by exc. 4 on page 22/23, which I didn’t know. I also very much like the idea of mixtilinear in- and excircles. Page 59 is nice, and I am still surprised by your result in 8.4. Someway I would expect that the three mixtilinear excircles would yield diﬀerent points. Now, in fact, the triangle of points of tangency of the circumcircle and the three mixtilinear excircles is perspective to ABC with the internal center of similitude of the in- vanLamoenandYiu:Pre-HyacinthianGeometry 41 and circumcircle as perpector. This really is a gem. Have you ever had in mind to consider mixtilinear circumcircles: through two vertices and tangent to the incricle? I also enjoyed the Malfatti construction! Paul, I suppose you use this book for geometry courses for your students? I’d wish I’d been one of them... Kind regards, Floor. PS. I have not been doing much on the P −perpendicularity. The only thing is I have been trying to think of a way to tackle Stammler circles and the Stammler hyperbola. Unsuccesfully. By the way: The ‘circular’Stammler hyperbola is given by: b2c2(b2 − c2)x2 + c2a2(c2 − a2)y2 + a2b2(a2 − b2)z2 =0 which is a rectangular hyperbola, passing through the four in-/excenters and the circumcenter. It contains the centers of the ’cutting circles’.

Paul Yiu, November 24, 1999. Dear Floor, Thank you for your kind words about the geometry notes. The other day (Monday), I could not find a copy of the notes in my office, and so could not immediately respond to your message. Then yesterday, I spent a whole day rearranging my office, clearing papers and putting back books on to the shelves. Only then a copy of the notes surfaced. I have actually written two follow-up short notes on mixtilinear incircles and excircles. But since it is so difficult to publish, I just leave them in my file cabinet and am perhaps waiting for better results. Antreas has actually made a lot of interesting suggestions and questions on the mixtilinear incircles and excircles. He has also suggested the mixtilinear circumcircles that you mentioned. I actually want to write something on the theme of three circles. I shall review my notes with Antreas to see what has actually be done. There are just so many interesting things in geometry! And of course the P −perpendicularity. Last week I gave a 15-minute on triangle geometry in a local MAA meeting here on campus. The title was very ambitious: Triangle Geometry on the Web, some recent threads in Math. Forum. But of course I could only talk about one or two things. So I narrowed it down to “Reflections, Euler line, and the Neuberg cubic”. I am converting the transparencies into pdf file. Please see http://www.math.fau.edu/yiu/goldcoast.pdf I like your reflection theorem very much. What puzzles me is that in Adam Bliss’case, he reflected the sides in lines through midpoints in a fixed direction. In your case, you reflected the lines in a fixed direction in the side lines. I have actually done some calculations and strengthened Adam’s result in the following way: vanLamoenandYiu:Pre-HyacinthianGeometry 42

Let P be a point with traces X, Y , Z on the side lines of a reference triangle. If for every direction 2, the reﬂections of the side lines in lines parallel to 2 through X, Y , Z are concurrent, then P must be the centroid. I have not yet worked on your case to see if the orthocenter is the only such point. The Neuberg cubic I mentioned about the reﬂection triangles, the Euler lines at the end of the talk were worked out in the P −perpendicularity context. Of course, I did not mention P −perpendicularity in the talk. I really believe that it is advantageous to see Euclidean geometry with the eyes of P −perpendicularity. Best regards. Sincerely, Paul

Floor van Lamoen, Wednesday, November 24, 1999. Dear Paul, I read your ”Goldcoast” sheets. That has been quite a bunch for a 15 minute talk!! But I suppose your audience liked it. In your sheets you mention Konsita’s theorem. I am not sure if you’re aware of the Neuberg’s original (according to Zvonko Cerin)ˇ locus property of his cubic: The locus of all points P in the plane, such that the lines AOa, BOb and COc are concurrent, where Oa, Ob and Oc denote the circumcentres of the triangles BCP, ACP and ABP . Best regards. Floor.

Paul Yiu, November 24, 1999. Dear Floor, I think I have worked it out myself. Actually, I wanted to ask you about Cerin.ˇ Some time ago, I found in MathSciNet a paper in Journal of Geometry on cubic curve, perhaps by this author. We do not have the journal here. I requested a copy from InterLibrary loan, but for a few month I never got it. What are the main results in this paper? Best regards. Paul

Floor van Lamoen, Wednesday, November 24, 1999. Dear Paul, I asked the author himself to send me a copy, which he kindly did. The main results are an enormous list of locus properties, which he proved by computer calculus (using complex number representation). Here are some of them: First he starts with 7 “classic” properties:

1. The original locus problem by J. Neuberg: Cubic + circumcircle = P such that AOa, BOb and COc concur, where Oa is the circumcenter of PCB, etc. The point of concurrence lies on OP. (J. Neuberg, Sur la parabole vanLamoenandYiu:Pre-HyacinthianGeometry 43

de Kiepert, Ann. de la Soc. sci. de Bruxelles, (1909-1910), 1-11] (I will only accidently give one of the many references) 2. Let x = |AP |, y = |BP| and z = |CP|, then Cubic = P such that   1 x2 + a2 x2a2 det  1 y2 + b2 y2b2  =0. 1 z2 + c2 z2c2

3. . Cubic = P such that the product of powers of P w.r.t. circles [A, AB], [B,BC], and [C, CA] is equal to the product of the powers of P w.r.t. circle [A, AC], [B,BA], [C, CB]. 4. Cubic = P such that the line through P and coP is parallel to the Euler line. [F. Morley, Note on Neuberg cubic curve, Amer. Math. Monthly, 32(1925),407-411.] 5. Cubic = P such that the lines joining A, B, C with the reﬂections of P through the sides are concurrenct. 6. Cubic + circumcircle = P such that the Euler lines of ABP , AP C and PBC concur. 7. Cubic = P such that P lies on the Neuberg cubic of its pedal triangle. Then Cerinˇ proves many locus properties, some of them are quite complicated. Let me give just a couple of the easier ones: • Cubic + circumcircle - sidelines = P , not on sidelines, such that antipedal triangle PaPbPc is perspective to the triangle of the circumcenters of PPbPc, PaPPc, PaPbP . • Cubic - circumcircle = P not on circumcircle such that ABC is perspective of the isogonal conjugates of A w.r.t. PBC, B w.r.t. AP C and C w.r.t. ABP . The perspector is on the line PP∗. • Cubic = P such that ABC is perspective to the triangle formed by the centers of three equally oriented equilateral triangles on the sides of the pedal triangle of P . • Cubic + circumcircle - sidelines = P such that the circumcenter of OaObOc (as in 1.) lies on the Euler line.

• Cubic = P such that triangles XbYcZa and XcYaZb have equal areas, where Ya and Za are the images of B and C under inversion in circle [A, AP ], Zb, Xb images of C and A under inversion in [B,BP]andXc and Yc images of A and B under inversion in [C, CP ]. I can mail you a photocopy of the reprint I’ve got, if you wish. [Y]: Yes, please. It is interesting. Best regards. Floor. vanLamoenandYiu:Pre-HyacinthianGeometry 44

Floor van Lamoen, November 29, 1999 10:28. P −perpendicular project: Stammler conics Dear Paul, Last Friday I have brought a photocopy of ”Locus properties of the Neuberg cubic” to the mail-office. I expect that it will reach Boca Raton by the beginning of next week. Last few days I have spent my free time to translate much of Stammler’s work on his cutting circles into P −perpendicularity, in view of our ‘project’. I am not quite happy, since I had to fly into Cartesian coordinates (just like Stammler does in his paper)... Please enjoy reading the stuff, and see if you have something to add, or to comment. Best regards. Floor A conic is called a proportionally cutting conic, if it cuts off chords A1A2 off BC, B1B2 off AC and C1C2 off AB such that |A1A2| : |B1B2| : |C1C2| = a : b : c.If|A1A2|/a = |B1B2|/b = |C1C2|/c = µ then the conic is called a µ-cutting conic. A conic tritangent to ABC is accepted as a proportionally cutting conic with µ =0, being a limit case. The terms ‘proportionally cutting conic’and ‘ µ−cutting conic’are chosen after the names for similar circles that Ludwig Stammler defines in [1]. We will prove that the centers of proportionally cutting P -conics lie on a conic, the P -Stammler conic. First we will have to prove the following Lemma: Let AB and AC be two distinct lines in the Euclidean plane, and let t1 and t2 be two nonnegative numbers. Let a central conic C be given. The centers of all conics homothetic to the given conic, cutting chords of lengths 2t off AB and 2u off AC,forma conic. Proof: Let AB and AC form the two axes of a grid for the Euclidean plane, with A as origin, AB as x-axis and AC as y-axis. In this grid we can describe C with the equation (x − a)2 + c(x − a)(y − b)+d(y − b)2 = e. If in this equation c, d and e are changed, but a and b are left unchanged, then the conic described is homothetic to C. The center of this conic is (c, d) and the magnitude is fixed by e. Let us consider a general conic homothetic to C for a fixed magnitude: 2 2 (x − x0) + c(x − x0)(y − y0)+d(y − y0) = D . When we intersect this conic with y = 0 we find 2 2 x = x0 + cy0/2 ± D +(cy0/2) − dy0 , vanLamoenandYiu:Pre-HyacinthianGeometry 45

describing two points of intersection with the x-axis. The condition that these points must be 2t apart, gives us

2 2 b 2 t =( − d)y0 + D. 4 In the same way the condition that the two poits of intersection with the y-axis must be 2u apart, gives us

2 2 b 2 du =( − 1)x0 + D. 4c The two equations could be used to calculate the four, or less, possible centers (x0,y0) of conics with a magnitude ﬁxed by D serving the conditions of this lemma. Eliminating D from these equations, we ﬁnd that for all D these centers are bound by the equation

2 2 b 2 b 2 2 2 ( − 1)x0 +( − d)y0 = t − du , 4c 4 which describes that the centers form a conic with A as center. Now, consider in the reference triangle the centers of the P -conics cutting off chords of lengths b off AC and c off AB. By the lemma these are on a conic CA. In a similar way we find conics CB and CC . Naturally the points of intersection of two of these conics are also on the third, and form the centers of the 1-cutting P -conics. Centers of the µ-cutting conics for any other positive µ are found by first multiplying of CA with factor 1/µ w.r.t. A and similarly multiplying CB and CC , resulting in conics CA(µ), CB(µ)andCC (µ), and then find the points of intersection of the three resulting conics. The tritangent centers are found as the limit when we let µ tend to zero. We find that either the limit of CA is a single point (when CA is an ellipse) or the union of two lines (when CA is a hyperbola). We can have two cases: 1. The tritangent P -conics do not exist. In that case at least two of the conics CA, CB and CC are ellipses. 2. The tritangent P -conics exist. In that case CA(µ), CB(µ)and CC (µ)areP -perpendicular hyperbolas, since the tritangent centers form an P -orthecentric system. As a consequence, if there are four points of intersection, the µ-cutting P -conics form an P -orthocentric system (and if there are three points of intersection, the µ-cutting P - conics form a P -right triangle). The P -circumconic of ABC is the Feuerbach P -conic of the triangle IAIBIC formed by the tritangent excenters. Let T be the midpoint of IIA and U the midpoint of IB IC , then TU is a diameter of the P -circumconic of ABC and thus OP is the midpoint of TU. Also the line TU is the P -perpendicular bisector of BC.SinceT and U are each on an asymptote of CA,it follows that the TU is tangent to CA in OP . By symmetry similar results are found for CB and CC . vanLamoenandYiu:Pre-HyacinthianGeometry 46

From this (the tangents to CA, CB and CC in OP have unequal directions), and from the location of the diverse asymptotes, we can derive that the branches of CA, CB and CC passing through OP must meet in a second point. Locations of the other branches and their asymptotes show that there must be two more points of intersection. Hence, there are four 1-cutting P -conics. Open question: In the circular case the three 1-cutting P -exconics form an equilateral triangle circumscribing the circumcircle. Gener- alization to P -perpendicularity? The proof that the centers of proportionally cutting P -conics are on a conic can be ﬁnished with quoting the following theorem from [1].

Theorem (L. Stammler)

Let T1 and T2 be two distinct nondegenerate conics with centers M1 and M2 respectively. For each µ>=0letT (µ)j (j =1, 2) be the image of Tj after multiplication with factor µ w.r.t. Mj (For an ellipse, resp. hyperbola, for µ = 0 the center, resp. the pair of asymptotes, is understood to be T (0)j as limitcase). Then there exists a conic H, that encloses all intersections T (µ)1 ∩ T (µ)2 (µ>=0).

Proof (L. Stammler): In an x, y-grid Tj is given by the equation fj(x, y)=1,where 2 2 fj(x, y)=a(x − xj ) + b(x − xj )(y − yj)+c(y − yj) .

It follows that T (µ)j has the equation 1 f (x, y)=1. µ2 j Each point (x,y) for which there exists a µ>0 such that (x, y) ∈ T (µ)1 ∩ T (µ)2, must fullﬁll the equation

f1(x, y)=f2(x, y), which is an equation of a conic H.Bycontinuity,H encloses also T (0)1 ∩ T (0)2. And the existance of the Stammler P -conic is proven. When P = f : g : h, then the Stammler P -conic is given by: g − h h − f f − g x2 + y2 + z2 =0. g + h h + f f + g

When I = t : u : v is a prescribed incenter, then the P -Stammler conic is given by: x2b2c2(b2 − c2)+y2a2c2(c2 − a2)+z2a2b2(a2 + b2)=0. vanLamoenandYiu:Pre-HyacinthianGeometry 47

Consequently, the P -Stammler conic exists of the square roots of points in the intersection of the line OP KP and the closure of the interior of ABC, together with their preCevian triangles of these square roots. Not all points on the P -Stammler conic are centers of proportionally cutting P -conics. For instance, the point KP is on the conic, but is not the center of a (real) proportionally cutting conic. Problem: Let A B C be the preCevian triangle of OP . Show that the conics with centers A, B and C, passing through A, B and C respectively, are proportionally cutting P -conics. [1] Ludwig Stammler, Dreiecks-Proportionalschnittkreise, ihre Mit- tenhyperbel und eind Pendant zum Satz von Morley, Elem. Math., 47 158-168 (1992).

Paul Yiu, Monday, November 29, 1999. Dear Floor, Thank you for your many messages, and I look forward to seeing the cubic curve paper. We had Thanksgiving holidays here since last Thursday. Wednes- day afternoon, before I left oﬃce, I read your translation of the paper of Cerin’s,ˇ and made a copy to bring home. But I left it on my desk and forgot to bring it with me. It was only this morning that I reread it again. It is very interesting indeed, and I am sure the paper is very inspiring. Some of the results your out- lined I have already known. It is good that the author used complex numbers instead of homogeneous coordinates.

[vL]: It apparently helped Zvonko Cerinˇ to do his computer calculations. I must say, that the number of, quite complicated, results he gets, is impressive. A lot of the calculations I wouldn’t try to do by hand...

The geometry book project is pretty much on my mind these days. Let’s make this a worthwhile project and produce a geometry book surveying old results and bringing in new ones, working in the context of generalized perpendicularity as much as possible. It may sound ambitious, Floor, but I believe it is worth the eﬀort. During the holidays, I have browsed through, and made records from, the Canadian problem journal Crux Mathematicorum, which I have a complete set, beginning with 1975. There are lots of good geometry problems dealing with the triangle, using homogeneous coordinates. [During the 1980’s, several Dutch contributors were particularly prominent. Some familiar names are J.T.Groenman, D.J.Smeenk, P.Penning, and, of course, G.R.Veldkamp and O.Bottema]. Some of these problems can be weaved into uniﬁed themes. I am also accessible to a long run of old issues of the American Mathematical Monthly, from 1940 to the present. The early issues had a lot of interesting short notes and problems in triangle geometry too. So I am going through these with an eye of reworking them in the context of P −perpendicularity. Perhaps vanLamoenandYiu:Pre-HyacinthianGeometry 48 in a week or two, things will take better shape. You mentioned earlier that you subscribed to Nieuw Archief voor Wiskunde. I am sure before the 1990’s there must be lot of good geometry materials in the problem sections. It is worthwhile to do similar things for this journal.

[vL]: I do not have many back copies, but I think I know where to ﬁnd the problems from older issues on the WWW. It seems a good idea when I work them through, and see if I can ﬁt them into something nice. I would expect that the name “J. van IJzeren” would appear many times, too.

Here is what I have in mind about the tentative structure of the book. 1. Preliminaries. [Homogeneous coordinates, Distance formula, and other basic material] 2. Common Triangle Centers [Euler lines, Feuerbach Theorem..] This chapter is an annotation of properties of the more common centers Kimberling’s list. This by itself can be a very interesting place of putting things scattered in various problem journals together. Perhaps it should be divided into two or even three chapters.

[vL]: Sounds very good! This would probably also the place where projective things like “Desmic relationships” (or would you prefer to leave this to Guru Conway) and diverse conjugacies appear?

3. Conics. I don’t know whether this should precede the repertorie of triangle centers. The Monthly in the 40’s has a lot of things here. Unfortunately the several book on analytic conics cited in Kimberling’s works is not easily accessible. But I think there are enough interesting things. This is also the place one should introduce generalized perpendicularity. One hopes that by developing enough of the basic properties, it is possible to do P −perpendicularity synthetically. 4. Reﬂections.

[vL]: For this chapter: did you understand chapter XVI of Hoofd- stukken uit de Elementaire Meetkunde? It is about the reflection triangle (A reflected through BC, etc.): The condition for a triangle that its reflection triangle is degenerate is equivalent with the ninepointcenter being on the circumcircle!

5. Miquel Theory. 6. Brocard Geometry. 7. Other Locus Problems. vanLamoenandYiu:Pre-HyacinthianGeometry 49

8. A Dictionary of Cubic Curves. 9. High order curves (??) This of course is tentative.

[vL]: This seems excellent as a set-up.

Your working with the Stammler conics, I am sure, will ﬁnd a good place here. Please tell me what you think, and keep on brain storming with things in triangle geometry. I shall organize some of the materials better and write a few sample sections in a few days.

[vL]: I am looking forward to it.

Let me give one example from a group of problems posed by George Tsintsi- fas (of Athens) in Crux Math. in the late 1980’s. These problems are on inscribed triangles. There have been mistakes, corrections, and unsolved problems. But obviously they lead to locus problems. What kind of curves does a speciﬁc triangle center of an inscribed triangle trace out if the vertices of the inscribed triangle divide the side lines of the reference triangle equally?

1. Let ABC be an equilateral triangle inscribed in a triangle ABC,sothat A ∈ BC, B ∈ CA, C ∈ AB.If BA CB AC = = , AC BA CB prove that triangle ABC is equilateral. 1 2. Let ABC be an equilateral triangle inscribed in triangle ABC,sothat A ∈ BC,etc.DenotebyG, G the centroid, by O, O the circumcenters, by I, I the incenters, and by H, H the orthocenters of triangles ABC and ABC respectively. Prove that in each of the four cases (a) G = G, (b) O = O, (c) I = I, [not correct; Seimiya gave a counterexample]. (d) H = H, ABC must be equilateral. 2 3. Let ABC be a triangle inscribed in triangle ABC,sothatA ∈ BC,etc. Suppose that BA CB AC = = =1 , AC BA CB 1Tsintsifas, G., Dou, J., Pedoe, D. and Klamkin, M.S. [1990]: Problem 1437 and solution, Crux Math., 16, 187 – 190. 2Tsintsifas, G., Janous, W., Seimiya, T. [1990]: Problem 1446 and solution, Crux Math., 16, 217 – 219. vanLamoenandYiu:Pre-HyacinthianGeometry 50

and that triangle ABC is similar to triangle ABC. Prove that the triangles are equilateral. 3 4. Let ABC be a triangle inscribed in triangle ABC,sothatA ∈ BC,etc. 4

(a) Prove that BA CB AC = = AC BA CB if and only if the centroids G, G of the two triangles coincide. (b) Prove that if (1) holds, and either the circumcenters O, O or the orthocenters H, H of the triangles coincide, then triangle ABC is equilateral. (C)* If (1) holds and the incenters I and I of the triangle coincide, characterize triangle ABC. 5. Let ABC be an equilateral triangle inscribed in a triangle ABC,sothat A ∈ BC, B ∈ CA, C ∈ AB,andsothatABC and ABC are directly similar. If BA = CB = AC, prove that the triangle are equilateral. [Not correct!] 5 6. Let ABC be an equilateral triangle inscribed in a triangle ABC,sothat A ∈ BC, B ∈ CA, C ∈ AB,andsothatABC and ABC are directly similar. (a) Show that, if the centroids G and G of the triangles coincide, then either the triangles are equilateral or A, B,C are the midpoints of the sides of triangle ABC. (b) Show that if either the circumcenters O, O or the incenters I, I of the triangles coincide, then the triangles are equilateral. 6 7. Let ABC be an equilateral triangle inscribed in a triangle ABC,sothat A ∈ BC, B ∈ CA, C ∈ AB. Suppose also that BA = CB = AC. (a) If either the centroids G, G or the circumcenters O, O of the triangles coincide, prove that triangle ABC is equilateral. (b)* If either the incenters I, I or the orthocenters H, H of the triangles coincide, characterize triangle ABC. 7

Warm regards. Sincerely, Paul

3Tsintsifas, G. and Klamkin, M.S. [1990]: Problem 1455 and solution, Crux Math., 17, 249 – 250. 4Tsintsifas, G., and Klamkin, M.S. [1990]: Problem 1464 and partial solution, Crux Math., 16, 282 – 284. 5Tsintsifas, G. and Seimiya, T. [1990]: Problem 1471 and corrected solution, Crux Math., 16, 304 – 305. 6Tsintsifas, G. and Dou, J. [1991]: Problem 1483 and solution, Crux Math., 17, 22 – 23. 7Tsintsifas, G. and Klamkin, M.S., Problem 1492 and partial solution, Crux Math.,17,50 – 52. vanLamoenandYiu:Pre-HyacinthianGeometry 51

Floor van Lamoen, Wednesday, December 1, 1999. Dear Paul, I just sent out a locus problem to geometry.college.Hereisarelatedone, which is related to my earlier post on Stammler conics: The Stammler P −conic is the locus of points Q not on the sidelines of ABC, such that Q,itsiso−P conjugate and its iso−P # conjugate (where P # is the isotomic conjugate of P ) are collinear. Hence, the Stammler P −conic, and the Stammler P #-conic are one and the same. I have collected some problems from Nieuw Arch. Wisk. from the WWW (solutions are not included, but that doesn’t matter to much). I noticed the following problem (proposed by J.T. Groenman): Let ABC be a triangle in the plane. The Spieker circle of ABC is the inscribed circle of A1B1C1,whereA1, B1 and C1 are the midpoints of BC, CA and AB respectively. Show that the center of the Spieker circle coincides with the radical center of three escribed circles. A third note/question will come via Geometry.college. Best regards. Floor.

Floor van Lamoen to Geometry.college, Wednesday, December 1, 1999. Here’salocusproblemforyou: In the plane of triangle ABC, consider points P not on the sidelines of the triangle. Show that the locus of P such that the three points P , its isogonal conjugate/conjugal P ∗ and its isotomic conjugate/isotome P # are collinear is a conic. What is the center of this conic? [Y]: In homogeneous barycentric coordinates, the locus of P which lies on the line joining its isogonal and isotomic conjugates is the conic (b2 − c2)x2 +(c2 − a2)y2 +(a2 − b2)z2 =0. [vL]: The conic is a rectangular hyperbola. It passes through the excenters, and the orthocenter of the excentral triangle, hence the incenter. Its center is on the nine-point circle of the excentral triangle, hence the circumcircle. [Y]: This is the same situation with the Stammler conic. Am I right? I shall think about the rest of the message later. [Y]: The center of this conic is the Steiner point 1 1 1 : : , b2 − c2 c2 − a2 a2 − b2 which is the fourth intersection of the circumcircle and the Steiner ellipse of the reference triangle. This appears as point X99 in Kim- berling’s list. vanLamoenandYiu:Pre-HyacinthianGeometry 52

[vL]: Yes, this is why I posted it. It was my version of a problem proposed by J.T. Groenman:

Let ABC be a triangle with incenter I and excenters Ia, Ib and Ic. Furthermore, let G be the median point [nice name for the centroid, FvL]. We denote by S the Steiner point, i.e. the fourth intersection point of the circumcircle and the Steiner ellipse E, which is the ellipse passing through A, B, C and having G as center. (A, B, C are the other three intersection points).

Show that the conic passing through ﬁve points I, Ia, Ib, Ic and G has center S.

Problem and solution (by R.H. Eddy) published in Nieuw Archief voor Wiskunde, 4-9 (1991) 223-224. Let’s generalize.

Let ABC be a triangle with incenter I and excenters Ia, Ib, Ic.ConsiderX.WedenotebyS the fourth intersection point of the circumcircle and the ellipse E passing through A, B, C and having X as center. (A, B, C are the other three intersection points).

Is it true that the conic passing through the ﬁve points Ia, Ib, Ic and X has center S?

[vL](12/2/99): This generalization is false. The following generalization is true:

Let ABC be a triangle with incenter I and excenters Ia, Ib and Ic.ConsiderX.WedenotebyS the fourth intersection point of the circumcircle and the ellipse E passing through A, B, C and homothetic to the conic tritangent to the sidelines of ABC with X as its center. (A, B, C are the other three intersection points).

The conic passing through the ﬁve points I, Ia, Ib, Ic and X has center S. Another type of generalization: What is the locus of X (not on the sidelines of ABC) such that X is collinear with pro−X and the conjugal/isogonal conjugate of X? What is the locus of X (not on the sidelines of ABC) such that X is collinear with pro−X and the isotome/isotomic conjugate of X? These give quartic curves. Enjoy!

Best regards. Floor van Lamoen. vanLamoenandYiu:Pre-HyacinthianGeometry 53

Floor van Lamoen to geometry-college, Wednesday, December 1, 1999. Isotropic points in a triangle. Does anyone know what ‘isotropic points’in the context of a triangle are supposed to be? Let P be a point in the plane of a given triangle ABC, P the isogonal conjugate of P with respect to ABC, L1 the line PP , L2 the trilinear polar (or harmonical) of P with respect to ABC. Show that the locus of points P such that L1 and L2 are perpendicular is a quintic curve, with nodes at the vertices of ABC, passing through the isotropic points, through the incenter and the three excenters, through the centroid, through the orthocenter and through the vertices of the pedal triangle of the orthocenter. Source: Problem in Nieuw Archief voor Wiskunde, 3-24 (1976), by O. Bot- tema and M.C. van Hoorn:

[APH]: Another Source: A focal curve is defined as a plane cubic curve c passing through the isotropic points J1, J2 and such that the intersection of the tangents at J1 and J2 (the principal focus of c)isonc.LetP1 and P2 be two points, isogonally conjugate with respect to a given triangle such that their midpoint M is on a given 8 line l. Prove that the locus of P1 and P2 is a focal curve. And one more: Toelke, Juergen: Zu den Winkelgegenpunkten der isotropen Dreiecks- geometrie. (On opposite angle points in isotropic triangle geomet- rics). Math. Pannonica 6, No.2, 155-161 (1995). [ISSN 865-2090] Gegeben sei ein Dreieck ∆ := ∆A1A2A3 in einer reellen affinen Ebene A. Fuer jede Wahl einer festen, zu keiner Seite von ∆ paral- lelen Richtung r in A wird A zu einer isotropen Ebene A(r)mit r als isotroper Richtung. Spiegelt man im Sinne der zugehoeri- gen isotropen Metrik fuer i =1, 2, 3 jeweils die Verbindungsgeraden PAi eines Punktes P ∈ A(r) an der durch Ai gehenden Winkel- halbierenden von ∆, so gehen die drei gespiegelten Geraden durch einen Punkt, den “Winkelgegenpunkt” P von P ; umgekehrt ist P dann der Winkelgegenpunkt von P . Bekanntlich sind die Brocard- Punkte erster Art B1(r) und 2. Art B2(r) sowie der Schwerpunkt S und der Lemoine-Punkt L(r)=S(r) von ∆ jeweils Paare von Winkelgegenpunkten. Bei variablem r durchlaufen die Mittelpunkte der Sehnen B1(r)B2(r) eine zur Steinerschen Hypozykloide affin aequivalente Kurve (Satz 1) bzw. die Lemoine-Punkte L(r)eine Steiner-Ellipse (Satz 2). Ferner werden die Paare {B1,B2} bzw. {S, L} nach Herleitung geeigneter metrischer Beziehungen unter den Paaren von Winkelgegenpunkten gekennzeichnet. So gilt z.B. (Satz 6): Seien P , P isotrope Winkelgegenpunkte des zulaessigen Dreiecks ∆=∆A1A2A3;dannistP genau dann der Lemoine-Punkt (und P

8Problem in Nieuw Archief voor Wiskunde 23(1975) 242, #415, by O. Bottema. vanLamoenandYiu:Pre-HyacinthianGeometry 54

der Schwerpunkt) von ∆, wenn d1 : d2 : d3 = s1 : s2 : s3 gilt, wobei si (i =1, 2, 3) die isotrope Laenge der i-ten Seite von ∆ und di den isotropen Lotabstand von P zu dieser Dreiecksseite bezeichnet. R.Koch (Muenchen) (from ZfM) I found this deﬁnition on the Net: Isotropic or cyclic points and lines With respect to orthonormal axes we deﬁne the points I(1,i,0) and J(1, −i, 0) as cyclic or isotropic points. These points are ideal and conjugate imaginary. Each line that contains such point is called an isotropic line

http://www.ping.be/math/imag.htm#Isotropic-or-cyclic-

[Conway]: This is one of the two traditional names for a concept I’m sure you know about, the other name being “the circular points at infinity”. Just in case you don’t, they are the two points on the line at infinity that are common to all circles. One customarily extends the Euclidean plane whose points are (x, y) to the projective one whose points are (x, y, z)=(kx, ky, kz)(k = 0) by the conventions that if z is not 0, then (x, y, z) is the new name for the ”old” point (x/z, y/z) of the Euclidean plane, while the points (x, y, 0) are the “new” points at infinity. The typical circle x2 + y2 = r2 becomes x2 + y2 = r2z2 in the new coordinates, and from this equation we can see that the two points (1,i,0) and (1, −i, 0) lie on it. I don’t know of another meaning that has special reference to a triangle, so I’m sure this is the intended one.

Best regards. Floor van Lamoen.

Paul Yiu, Wednesday, December 1, 1999. Dear Floor, I have, in the past 2 days, done some calculations on the Stammler conic. Perhaps later today I can ﬁnd time to write more about that. I tried to work with homogeneous coordinates from the beginning, and it is very interesting. The Stammler conic, in the classical case, is the [rectangular] hyperbola through the tritangent centers and the circumcenters. Yesterday, I also went to the MathPro problem source to ﬁnd geometry problems from NavW and other sources. There are about 70 geometry problems and I copied them down. So, last night, I realized, as you mentioned, that Groenman had posed as a problem the radical center of the excircles. [I have, for the third time, sent out the Spieker center note to another journal some time last month to try my luck. I shall put this reference in when I receive a referee report]. vanLamoenandYiu:Pre-HyacinthianGeometry 55

I also wonder what the isotropic points are, as you posed to the geometry discussion group. We have both done our homework! Best regards. Paul

Paul Yiu, December 1, 1999. Dear Floor, You wrote

The conic is a rectangular hyperbola. It passes through the excenters, and the orthocenter of the excentral triangle, hence the incenter. Its center is on the nine-point circle of the excentral triangle, hence the circumcircle.

This is the same situation with the Stammler conic. Am I right? I shall think about the rest of the message later. Best regards. Paul

Paul Yiu, December 1, 1999. Dear Floor, Please ignore my earlier remark. It was not right. The Stammler conic passes through the nine-point center of the excentral triangle (=circumcenter). It is not the same as the present one.

[vL]: But the center of the Stammler conic should be on the circumcircle, equal to the conic I presented. By the way, the generalization I gave is false (intentionally, I hope some people try something). Tomorrow I will post the correct one (which heavily depends on the homothetic conics stuﬀ we are working with). I am looking forward to see what you did on the Stammler conics! There are still some thoughts playing around in my head. For instance: Suppose some circle with O as its center cuts of chords t, u and v. It seems to me that the locus of t : u : v-proportionally cutting circles is then the same as the Stammler hyperbola? [Y]: Good that you tell me otherwise. I must be have been (and still am) too tired, and get confused. I shall think about it later. Have a good evening.

Apology! Best regards. Paul vanLamoenandYiu:Pre-HyacinthianGeometry 56

Paul Yiu, December 3, 1999. P −perpendicularity generalization of Gossard’s Theorem Dear Clark, John, Floor, and friends, Here is a further generalization of Gossard’s Theorem, in an arbitrary direction. Let P be a point distinct from the centroid G of triangle ABC. (1) Denote by X, Y , Z the intersections of the line GP with the side lines BC, CA, AB respectively. (2) Mark the centroids G1, G2, G3 of triangles AY Z, BZX,andCXY respectively. (3) Construct points P1 such that YP1//BP and ZP1//CP ; P2 such that ZP2//CP and XP2//AP ; P3 such that XP3//AP and YP3//BP . Then the lines G1P1, G2P2,andG3P3 bound a triangle A B C congruent and oppositely homothetic to ABC, the center of homothety being a point on the line GP . In homogeneous coordinates, let P = f : g : h. Then this P −Gossard perspector is the point

···:(h + f − 2g)(h2 + f 2 − g2 − 3hf + fg + gh):···

The corresponding expressions in the case of the Euler line (when P = orthocenter) are a lot more complicated:

··· ::(c4 + a4 − 2b4 − 2c2a2 + a2b2 + b2c2) (−b8 + b4(c2 + a2)+b4(2c4 − 5c2a2 +2a4) − 3b2(c2 − a2)2(c2 + a2)+(c2 − a2)2(c4 +3c2a2 + a4)) : ···.

[vL]: This is a good way to “sell’ P −perpendicularity!!! [Y]: This is indeed a very good example to show the elegance of working with P −perpendicularity. In a separate message, I am including more details of the calculations.

I should perhaps point out that the three lines G1P1, G2P2, G3P3 (and the triangle they bound) depend only on the direction of GP .[ThepointsGi, Pi vary, but the lines GiPi remain invariant]. This is reﬂected in the coordinates of the perpector, which can be rewritten as

···:((h − f) − (f − g))((g − h)2 +(h − f)(f − g)) : ···

John has pointed out that a parallel translation of this line results in a parallel translation of the “Gossard triangle”. Sincerely, Paul Yiu vanLamoenandYiu:Pre-HyacinthianGeometry 57

Paul Yiu, December 3, 1999. P −-Gossard Theorem Dear Floor, Let P = f : g : h.TheP −Euler line has equation

(g − h)x +(h − f)y +(f − g)z =0.

This intersects the sides at the points

X =0:g − f : h − f, Y = f − g :0:h − g, Z = f − h : g − h :0.

The triangle AY Z has two P −altitudes

g(g − h)x +(f 2 − fg − g2 + hf)y +(f − g)gz =0, h(g − h)x +(h − f)hy +(h2 + hf − f 2 − fg)z =0.

These two lines intersect at the point

f(f 2 − g2 − h2 +3gh − fg − hf):g(h − f)(g − h):h(f − g)(g − h), which is the P −orthocenter of triangle AY Z. The centroid of this triangle is the point

3(f 2 − g2 − h2 +3gh − fg − hf):(g − h)(h + f − 2g):−(g − h)(f + g − 2h).

The P −Euler line of this triangle is therefore the line

(g − h)2x − (g2 + h2 − f 2 − 3gh + hf + fg)(y + z)=0.

This is a line parallel to BC. Similarly, the P −Euler lines of the triangles BZX and CXY are respectively

(h − f)2y − (h2 + f 2 − g2 − 3hf + fg + gh)(z + x)=0, and (f − g)2z − (f 2 + g2 − h2 − 3fg + gh + hf)(x + y)=0. These two latter lines intersect at

A = −(g − h)2(g + h − 2f) :(h + f − 2g)(h2 + f 2 − g2 − 3hf + fg + gh) :(f + g − h)(f 2 + g2 − h2 − 3fg + gh + hf).

Note that A depends only on the direction of the P −Euler line. Similarly, we write down the coordinates of B and C. It is clear that AA, BB, CC intersect at the point

(g + h − 2f)(g2 + h2 − f 2 − 3gh + hf + fg) vanLamoenandYiu:Pre-HyacinthianGeometry 58

:(h + f − 2g)(h2 + f 2 − g2 − 3hf + fg + gh) :(f + g − 2h)(f 2 + g2 − h2 − 3fg + gh + hf).

The centroid of the triangle ABC is the point

(g + h − 2f)((f − g)2 +(h − f)2 − 2(g − h)2) :(h + f − 2g)((g − h)2 +(f − g)2 − 2(h − f)2) :(f + g − 2h)((h − f)2 +(g − h)2 − 2(f − g)2).

Since the midpoint of the two centroids is the perspector, the two triangles are indeed congruent. Since the perspector lies on the P −Euler line GP ,theP −Euler lines of the two triangles ABC and ABC coincide. The perspector in the case of the Euler line is the point ··· :(c4 + a4 − 2b4 − 2c2a2 + a2b2 + b2c2) (−b8 + b4(c2 + a2)+b4(2c4 − 5c2a2 +2a4) −3b2(c2 − a2)2(c2 + a2)+(c2 − a2)2(c4 +3c2a2 + a4)) : ···.

Conway has pointed out that a parallel translation of this line results in a parallel translation of the “Gossard triangle” vanLamoenandYiu:Pre-HyacinthianGeometry 59

Paul Yiu, Friday, December 3, 1999. Dear Floor, Many thanks for the copy of Cerin’sˇ cubic curve paper. It is a very interesting paper indeed. Apart from the solution of the 7 problems, the author was able to include quite a number of interesting and related facts. Problem 7 is quite intriguing. I do not particularly like the use of complex numbers, though I am aware of such a tradition in the early part of the century. I think this paper can give a lot of interesting inspiration. Warm regards. Sincerely, Paul

Floor van Lamoen, Monday, December 6, 1999. Problems Dear Paul, Here are some problems. I start with three problems from NAvW, which are in the one interesting back issue I still appear to have (apparently the other ones are lost):

840 (O.Bottema) The semi-perimeter and the radii of the circumcircle and the inscribed circle of a triangle are denoted by s, R and r, respectively. It is well known that s2 ≥ 16rR − 5r2 holds for all triangles. Prove that for acute triangles we have √ √ s2 ≥ 4R2 − 2( 2 − 1)Rr +(7+4 2)r2.

841 (G.W. Decnop) Let P and Q denote the foci of a conic, tangent to the three sides of a triangle A1A2A3 in the real Euclidean plane. If the conic is tangent to each side at its midpoint, then:

a1 · PA1 · QA1 = a2 · PA2 · QA2 = a3 · PA3 · QA3 = ka1a2a3,

where ai is the length of the side opposite to Ai,(i =1, 2, 3). Prove this, determine k, and examine the validity of the converse.

844 (J.T. Groenman) Let A1A2A3 be a triangle, G its median point, and G1, G2, G3 the midpoints of the sides A2A3, A1A3, A2A3.Furthermore let I be the incenter, and I1, I2, I3 be the excenters in the usual way.

Prove that the lines GkIk fork =1, 2, 3 are concurrent. Let S denote their common point. Show that the line IS passes through the Lemoine point, i.e. the point, isogonally conjugate to the median point.

Then, here is a problem I submitted last week to Amer. Math. Monthly:

Given triangle ABC,letABC be its medial triangle and G its centroid. In [1, 9.7] Clark Kimberling states that the circumcenters of the six triangles BAG, CAG, ABG, CBG, ACG and BCG are circumscribable by a conic. Show that this conic is a circle. Reference: vanLamoenandYiu:Pre-HyacinthianGeometry 60

1. Clark Kimberling, Triangle centers and central triangles, Con- gressus Numerantium, 129 (1998).

I included a synthetic solution. I think it might be P −generalizable. Finally, I started discussion last week in geometry.college on the conic

(b2 − c2)x2 +(c2 − a2)y2 +(a2 − b2)z2 =0.

Steve made an interesting remark that this conic is self-polar (this is also true for the Stammler conic). I found another, a bit far fetched, locus property for this conic: The conic is the locus of X such that X,pro−X and the centroid of the Cevian triangle of X are collinear. Related to this locus property is the following locus property of the P −Jerabek hyperbola (the P −iso-conugates of points on the P −Euler line): The P −Jerabek hyperbolaisthelocusofpointsX, such that X, P and the centroid of the P −trace triangle of X are collinear. Best regards. Floor.

Floor van Lamoen, December 8, 1999. Dear Paul, It seems that your mail address was left out somewhere in this thread. I copy to you this message of Antreas’. Barry Wolk’s statement is a bit hard to read when you don’t realize that ABC is a Cevian triangle, and A”B”C”the points of intersection of the sides of ABC and ABC that form the axis of perspective. The reason that we should consider this, is of course that this might be P −generalized. I came into a complete mess when I tried to P −generalize Barry’s version, but the simpler version Antreas quotes (with four lines) is promising. I will try soon. Kind regards, Floor.

Michael Keyton to Steve Sigur, December 7, 1999. During the last ten years, I have had several students who have discovered in an elementary Euclidean geometry course at the 9th grade various conic sections. Since my knowledge of algebraic geometry is virtually non-existent, I hope someone can help direct me to sources to prove some (all) of the following: (1) [Discoverer: Edward McCullough; I can prove using Pascal’s theorem and point at inﬁnity] The six centroids of the triangles formed by the medians of a triangle lie on an ellipse. There are at least four other ellipses closely related to this. [Treat the six points as the vertices of a hexagon. The opposite sides of the hexagon are parallel, thus the points lie on a conic (ellipse).] (2) [My discovery, I think I have a proof using a result about lines forming constant angles.] The locus of the intersection of the Wallace lines (Simson) for two triangles inscribed in the same circle is an ellipse. vanLamoenandYiu:Pre-HyacinthianGeometry 61

(3) [Discoverer: Travis Waddington; proven by Richard Newcomb of Cister- cian School, Irving, TX] The locus of the orthopole for all lines through a given point for a given triangle is an ellipse. (4) [Discoverer: Aubrey Clayton, equivalent to Kiepert] Construct isosceles triangles on the sides of a triangle, so that their altitudes are proportional to the sides of the triangle, the lines through the vertex of an isosceles triangle to the opposite vertex of the original triangle are concurrent. The locus of the point of concurrence is a hyperbola.

[APH]: Yes it is equivalent to Kiepert’s, since the angle in the base of the a−isosceles triangle is cot−1 a . Similarly for the b−isosceles: 2ha cot−1 b , and for the c−isosceles: cot−1 c , and, because of the 2hb 2hc proportionality of altitudes to sides these angles are equal. So we have similar isosceles triangles on the reference triangle (Kiepert).

(5) [Discoverer: me] Use the construction of an ellipse (hyperbola) as given by Clark Kimberling on Historia Mathematica recently, a variation on Pascal’s theorem, except put E on a circle (the constructed conic will sometimes be a hyperbola. Then the locus of the center of the conics is a hyperbola. Michael Keyton

Conway. [Conway]: I’m jumping in on this without having read the prior messages, because it sounds interesting and I’m sure to have something to say: Barry writes here:

Transforms the cevian points of I (namely ABC) into the cevian points of either Ia or Ib or Ic. Sothe3newtrianglesareA B”C” or A”BC”orA”B”C. ...And the result of extraversionis that the orthocenters of these four triangles all lie on OK,theBrocard axis of ABC. .... should not lie randomnly on the OK.Itwouldbe interesting to know the ratio.

Were A, B, C really the Cevian feet of I, or rather (as I suspect) the pedal feet of I (which I think form the triangle that’s usually called ”the incentral triangle”)? Whichever they are, I take it that the assertion is that the orthocenter of this triangle lies on the Brocard meridian m = OK ? If this is so, then we do indeed get an extraversion quartet of points on OK, and the obvious one is either the coSpiekers or the isoSpiekers - I forget which. Let me think. THE isoSpieker is

(: (b + c)(a + b):)=(:(bc + ca + ab)+bb :) so is on GK rather than OK, so it must be the coSpiekers. I surprised Steve by being right with this guess for another such problem, so I’ll make the same bet again. vanLamoenandYiu:Pre-HyacinthianGeometry 62

I’m trying desperately hard to get back up to date with the rest of my email, so won’t check just now, but hope to do so later this evening.

Antreas P Hatzipolakis, December 9, 1999. Barry Wolk wrote: Extensive computation has just proved some amazing generalizations of these results. For any point P =(x : y : z), let h(P )be the orthocenter of the triangle whose vertices are the cevian points (0 : y : z), (x :0:z), (x : y :0)ofP.LetP1 =(−x : y : z), P2 =(x : −y : z), P3 =(x : y : −z) be the pre-Cevians of P . Antreas wrote about the case where P is the incenter I, so the orthocenters of his four triangles ABC, AB”C”, A”BC”, A”B”C are the points h(I), h(I1), h(I2), h(I3) in this notation. In this special case, I1, I2, I3 are the extraversions of I, usually written Ia, Ib, Ic. Theorem 1. For every point P , the four points h(P ), h(P1), h(P2), h(P3) are collinear. Theorem 2. h(P ) lies on the Brocard axis OK of ABC iﬀ either P is on the circumcircle of ABC,orthepoint

a4 b4 c4 : : x2 y2 z2 lies on the Brocard axis. I have moved most of the calculation to the computer, so handling long polynomials is no longer a problem. The formula I got for h(P ) is

···:(SC/x + bb/y + SA/z)(2SB/xz + aa/xx − bb/yy + cc/zz):···

However, that formula is unlikely to be useful for hand calculations. Theorem 1 is so general that it just can’t be a new result. However, the calculation was such a mess that perhaps it really is new after all. Any references? While we are on Hs: I was searching for references for the Theorems that Michael Keyton sent the other day, and came across a theorem on Hs, as an exercise in a Projective Geometry book: 9 The Orthocenters f the triangles deﬁned by four lines are collinear. Whose is this theorem? Euler’s perhaps?

9Panagiotis Ladopoulos: Projective Geometry [in Greek], p. 293, #95. vanLamoenandYiu:Pre-HyacinthianGeometry 63

[Sigur](12/9/99): This is also in the Penguin Book of Curious and Interesting Geometry. [Wolk](12/10/99): Thanks. That theorem is exactly the same as my result (Theorem 1), after you untangle my deﬁnition of h(P1), etc. And this generalizes the original conjecture of Antreas that started this thread, except that the generalization just says ”these four points are collinear”, without saying anything about which line contains these points. In the original statement, which started with the special point P = incenter, that line was the Brocard axis. [Y](12/8/99): How about this? The circumcenters of four triangles bounded by four lines are concyclic. I found this by drawing a picture, and I am quite sure it is true. Is this something well known? [vL](12/9/99): The circle you describe is known as ‘Steiner’s circle’. Also, this can be generalized from 4-lines to n−lines in a nice way, known as ‘Cliﬀord’s chain’. I didn’t know about this until I asked about it in this forum last January. I then found (grasping from a list produced by Antreas) the instructive page - by Alexander Bogomolny:

http://www.cut-the-knot.com/triangle/Morley/CenterCircle.html

[Y]: Thank you for pointing this out. This result should have a P −perpendicularity generalization:

Let X, Y , Z be the intercepts of a transversal on the side lines of triangle ABC.TheP −circumcenters of the four triangles formed by four generic lines lies on a P -conic (homothetic to the standard P-circumconic).

Yes, please forward the messages. [vL]: I am sure there will be such a generalization to P-conics, and I will (try to) work that out. The idea of ‘Cliﬀord’s chain’ is also very attractive. I found the following page from Eric Weisstein’s CRC Concise encyclopedia of mathematics:

http://www.treasure-troves.com/math/CliﬀordsCircleTheorem.html

Is very nice Euclidean geometry!!

Antreas.

Paul Yiu to Floor van Lamoen, December 8, 1999. Dear Floor, vanLamoenandYiu:Pre-HyacinthianGeometry 64

Thank you very much for forwarding this. For a few days I have no connection to the outside world since our technician somehow inadvertently disconnect my computer from our department network. But now things are back to normal. The P −perpendicular generalization of the theorem Antreas quoted is correct: If the line px + qy + rz = 0 intersects the side lines of the reference triangle at X, Y , Z respectively, the P −orthocenters of the four triangles ABC, AY Z, BZX,andCXY all lie on the line

ghp(q − r)x + hfq(r − p)y + fgr(p − q)z =0.

This line is the P −perpendicular from P to the line x y z + + =0. p q r Best regards. Sincerely, Paul

Floor van Lamoen, December 9, 1999. Dear Paul, Thanks for computing the P −perpendicular generalization, which I didn’t x y z have time for yesterday. A nice result!! Isn’t p + q + r =0.the“isotomic conjugate” of px + qy + rz =0? I saw, after Antreas asked, that you have missed quite a couple of messages from the “My Conjecture” thread. Do you want (most of) them (I didn’t save all)? Then I can forward them to you, just like I did with the one message yesterday. Kind regards, Floor.

Antreas Hatzipolakis to Conway, December 1, 1999. I asked Michael Keyton to verify by Cabri (I have not that program) my conjecture (namely: The Orthocenter of the internal bisectors Triangle lies on the Brocard Axis of the reference Triangle) Here is his reply: How about the Orthocenter of the External Bisectors Triangle?

Paul Yiu to Antreas Hatzipolakis, December 1, 1999. Dear Antreas, Your conjecture is indeed true. To show that the reﬂections of the Brocard axis in the side lines of the incentral triangle, it is enough to show that this line (OK) contains the orthocenter of the incentral triangle. See Kimberling, TCCT, p.148, or my posting on October 25,

http://forum.swarthmore.edu/epigone/geometry- college/glelclalbrimp/[email protected] vanLamoenandYiu:Pre-HyacinthianGeometry 65 where I wrote: More generally, Section 5.6 (p.148) of Kimberling’s book treats concurrent reﬂected lines. There, Clark quotes the following theorem. I have changed the notation slightly.

Theorem: For any line 2,let2a, 2b, 2c be the reﬂections of ell about the sidelines BC, CA, AB respectively. 2a, 2b, 2c concur at a point P if and only if 2 passes through the orthocenter, and in that case, P lies on the circumcircle. Now, it is routine to (i) determine the orthocenter of the incentral triangle; It is the point

a2(b2 − bc + c2 − a2)(a2(b + c)+2abc − (b + c)(b − c)2):···: ···;

(ii) verify that this point lies on the OK-line.

(b2 − c2)x (c2 − a2)y (a2 − b2)z + + =0. a2 b2 c2 The intersection is a point on the circumcircle of the incentral triangle. Best regards. Sincerely, Paul [Barry Wolk]: There should be an extraversion, since wherever we have an IN- result should be expected an EX- one as well. So, the orthocenter of the IN-central triangle lies on OK. The orthocenter of the EX-central triangle [APH]: If ABC, ABC are the the IN- and EX- central triangles, then we can combine their vertices to get “mixed” triangles, and ask for their orthocenters ...... [Wolk]: There is no EX-central triangle, since ABC are collinear. Ex- traversion Writing automatically again :-) [APH]: The EX-line should intersect the OK line at some point. Then this point along with the orthocenters of the triangle ABC and the ones Barry writes here:

transforms the cevian points of I (namely A’B’C’) into the cevian points of either Ia or Ib or Ic. Sothe3newtrianglesareA B C or ABC or ABC. ... And the result of extraversion is that the orthocenters of these four triangles all lie on OK,theBrocardaxisofABC.

...shouldnotlierandomnlyontheOK. It would be interesting to know the ratia. vanLamoenandYiu:Pre-HyacinthianGeometry 66

Floor van Lamoen, Thursday, December 9, 1999. Dear Paul, I found my notes of January, which included a synthetic proof I made of ‘Steiner’s circle:

In ABC with circumcenter O and circumcircle (O), let 2 be a line, intersecting BC, AC and AB in A, B and C respectively. Let A, B and C be the circumcenters of ABC, ABC and ABC respectively, and let (A), B and C be the respective circumcircles. Since ABC is a degenerate triangle inscribed in ABC,its Miquel point M is on (O). Let A1B1C1 be the pedal triangle of, then, by Miquel theory, also MA A1 = MB B1 = MC C1,say α. From this we conclude that MAA , MBB and MCC are similar and equally oriented isosceles triangles with top-angle 2α. This means that ABC can be constructed out of ABC by ﬁrst rotating about M through π/2 − α and then shrinking from M with factor csc(α)/2. Consequently M is on the circle (O)with center O circumscribing ABC. From the fact that by construction MOO is similar to MAA (etc.), and thus isosceles, we see that O is on (O), too.

Maybe we should wait with generalizing this theorem, until we can use generalized notions as used in this proof? Kind regards, Floor.

Paul Yiu, Friday, December 10, 1999. Dear Floor, Thank you for this. AA, BB, C meet on the circumcircle of ABC. What is this point? Best regard. Paul

Paul Yiu, Friday, December 10, 1999. P −perpendicularity generalization of Steiner circle Dear Floor, Here is another triumph of P −perpendicularity, about the Steiner circle. If a line 2 intersects the side lines of ABC at X, Y , Z. The circumcenters of AY Z, BZX,and CXY form a triangle in perspective with ABC, the perspector lying on the circumcircle, as I pointed out in my earlier message. Now, the calculations and resulting expressions are much simpler in the P −perpendicular case. The P −circumcenters of AY Z, BZX,andCXY form a triangle in perspective with ABC, the perspector lying on the standard P −circumconic. Specializing this to the standard case, we choose a line and work out this perspector. vanLamoenandYiu:Pre-HyacinthianGeometry 67

Euler line X110 [This is the same as the point of concurrency of the reﬂections of the Euler line in the side lines]. OI−line X100 van Aubel line HK X112

[vL](12/16/99): Did you calculare coordinates for the P −circumcenters of AY Z, BXZ and CXY ? I would be interested to see them, to check whether the P −orthocenter of these P −circumcenters is on 2, which I think is the case. If you haven’t, I can calculate them myself of course. [Y]: In a separate message, I am sending the tex file containing the coordinates of these P −circumcenters. I need to learn how to organize my computer files. They are so disorganized that it was very difficult to find. Fortunately, this one I name by the date.

A line L : px + qy + rz = 0 intersects the side lines at X =0:−r : q, Y = r :0:−p,andZ = −q : p :0 respectively. The P −circumcenter of triangle AY Z is the point

f(−(g + h)p2 +(f +2g + h)pq +(f + g +2h)pr − 2(f + g + h)qr) A = (h + f)p(f(r − p) − g(q − r)) . (f + g)p(h(q − r) − f(p − q))

Similarly, we obtain the P −circumcenters of BZX and CXY . The triangle ABC is perspective, with perspector g + h h + f f + g : : . g(p − q) − h(r − p) h(q − r) − f(p − q) f(r − p) − g(q − r)

This is clearly a point on the standard P −circumconic:

f(g(p−q)−h(r−p))+g(h(q−r)−f(p−q))+h(f(r−p)−g(q−r)) = 0.

... The four P −circumcenters of AY Z, BZX, CXY ,and ABC lie on a P −conic.

[Y](12/16/99): How wonderful! Your conjecture is indeed correct. Here is the proof, continuing with the notation from the ﬁle I just sent.

Denote by X, Y , Z these P −circumcenters. By working out the P −perpendicular from X to YZ, whose equation vanLamoenandYiu:Pre-HyacinthianGeometry 68

is too long to write down, and ﬁnding the intersection with 2,wehavethepoint

(q − r)(fpq + gpq + fpr+ hpr − 2fqr − gqr − hqr) :(r − p)(fpq + gpq − fpr− 2gpr − hpr + gqr + hqr) :(p − q)(−fpq − gpq − 2hpq + fpr + hpr + gqr + hqr).

This can be written as 1 1 1 f g h (q − r)((f + g + h)(− + + ) − ( + + )) : ···: ···. p q r p q r

The symmetry clearly shows that the same point is the intersection of 2 with the other two P −perpendiculars, from Y to ZX,andfromZ to XY. This proves that the P −orthocenter of XYZ does indeed lies on the line 2.

In a separate message, I shall talk about reﬂections. Best regards. Paul

Paul Yiu, December 10, 1999. Reflections and self P −perpendicular lines Dear Floor, Recently, I am working on the details of results on reflection lines. More specifically, (1) the reflections of parallel lines through the perpendicular feet in the side line, (your communication) (2) the reflections of the side lines through the parallel lines drawn through the midpoints (Adam Bliss). I have always been interested in seeing if the orthocenter is the only point working in (1) and if the centroid the only point working in (2). Now, they are, almost always. In doing the calculations in P −perpendicularity, I ran into self P −perpendicular lines. Does that worry you? Subconsciously I had always tried to avoid that. But now I see that there is no need. You see, in thinking about P −perpendicularity, I tend to see P as an interior point. In fact, if P is in one of the “vertically opposite” regions of the triangle, then one can replace A, B, C, P by P , B, C, A, like the standard orthocenter case. But if P is in one of the regions containing the excenters, [I have better terminology in dealing with these region distinctions] there one cannot assume P an interior point. This is the case when self P −perpendicular lines occur. In fact, in such a case, the standard P −circumconic is a hyperbola. In this case, line parallel to the asymptotes are self P −perpendicular. Back to the reflection problems. vanLamoenandYiu:Pre-HyacinthianGeometry 69

(1) Let Q = u : v : w and 2 : px + qy + rz = 0. Consider the parallels to 2 through the traces of Q.TheP −reflections of these parallels in the respective sidelines are concurrent (i) for all Q if 2 is self P −perpendicular; (ii) for all 2 if Q = P , F+, F−.HereF+ and F− are real points if and only if fg + gh + hf < 0. [Equivalently, P lies outside the Steiner ellipse]. There is a good description (construction) of these points F+ and F−.I hope to be able to write out the details neatly very soon. (2) Let Q = u : v : w and 2 : px + qy + rz = 0. Consider the parallels to 2 through the traces of Q.TheP −reflections of the side lines in these respective parallels are concurrent (i) for all Q if 2 is self P −perpendicular; (ii) for all 2 if Q = centroid. The phenomenon of self P −perpendicularity reminds me of the distinction of what you and other people make between strong and week points. I don’t quite understand the enthusiasm about extraversions. But these results on reflections may warn us that in getting four theorems in the price of one by extraversion, we may have missed out something! Best regards. Sincerely, Paul

Floor van Lamoen, Sunday, December 12, 1999. Reflections and self P −perpendicular lines Dear Paul, This sounds very promising!! The notion of self P −perpendicular lines do not really worry me, but I had not realized that they existed. But somehow, subconsciously, I think I have. And, when I think of their existance, it seems more or less natural that they are there, since we do more than just affinely shrink or expand. Also, I feel it is like conjugacies having fixed points. Your results on the type of reflections by Bliss’and me are very beautiful, and very strong results. Adam Bliss and I must have been very lucky guys to find the right reflections.... What I was wondering, is whether there is a way of generalizing these results. We might reflect a side through 2 and reflect the result through the side again, through 2 again, through the side again,.... Something like that. Could such a thing lead to an interesting concurrence? Until now I haven’t really found anything. I don’t really see the connection between ’extraversion’ and self P −perpendicularity. Can you explain what you are aiming at? The excitement is about getting more pointsthathave-moreorless-thesameproperties.Butaswehaveal- ready seen when working with tritangent P −conics, in P −perpendicularity this doesn’t always equally work out. Four proofs for the price of one is not exciting by itself, in my opinion (Although a Dutchman is supposed to always go for cheap opportunities...). Kind regards, Floor. vanLamoenandYiu:Pre-HyacinthianGeometry 70

Paul Yiu to Antreas Hatzipolakis and Floor van Lamoen, December 9,1999. Dear Antreas and Floor, Take a point P on the circumcircle of triangle ABC, and construct the Simson line. Let it intersect the side lines at X, Y , Z. The circumcenters of AY Z, BZX,andCXY lies on a circle tangent to the circumcircle of ABC at the point P ! Isn’t it wonderful? Is this known? Sincerely, Paul

Antreas Hatzipolakis to Paul Yiu, December 10, 1999. Dear Paul, Wonderful, indeed! It is new - at least to me. (I will search for the keywords “simson line” or “pedal line” and “circumcenters”)

[Y](12/11/99): Indeed, more is true. The circumcenters of AY Z, BZX,andCXY form a triangle homothetic to ABC, the center of 1 homothety being the point P , and the ratio of homothety 2 . In the general case, if we start with an arbitrary line 2, the perspector of the circumcenter triangle lies on the circumcircle. The Simson line of this perspector is parallel to 2. All these extend to P −perpendicularity.

By the way, have you ever heard about the ”Radical Circle”? (it is the brother of the “Radical Axis”).

[Y]: Before I ventured a guess, I tried to look up Court’s College Geometry, 2nd ed. 1952. Here is, on p.215,

484 Theorem. The locus of a point M whose powers with respect to two given circles are equal in magnitude and opposite in sign is a circle coaxal with the given circles. 485 Deﬁnition. The locus of M (Art.484) is called the radical circle of the two given circles.

Also, about the “Angle of two Curves” and particularly, two conics?

[Y]: I think this means the angle between the tangents to the curves at an intersection.

I came across these notions recently (in Greek sources), and probably were new (original). Greetings Antreas vanLamoenandYiu:Pre-HyacinthianGeometry 71

Floor van Lamoen, December 12, 1999. Dear Paul and Antreas, Paul, thank you for these results, which are unknown by me. A couple of notes: The triangle of circumcenters as below (say ABC) is similar to ABC, which can be seen easily from the synthetic proof of the Steiner circle I sent earlier. It is also easily seen that, if we let line XYZ be a Simson line of a point P , then α = π and triangle ABC is just triangle ABC multiplied about P with 1 2 . And thus the circumcircles of ABC and A B C are tangent at P . I suppose this plain corollary must have been known in Steiner’s time. When we consider the general case again, M is the Miquel point of XYZ, and we see from my synthetic proof that MAA = MBB = MCC (base angles of the isosceles triangels I mentioned) and thus AA, BB and CC meet in a point on the circumcircle of ABC. In the same way from the fact that MAA = MBB = MCC we see that AA, BB and CC meet in a point on the circumcircle of ABC, i.e. the Steiner circle. Consequently the perspector of ABC and ABC is the second point of intersection of the Steiner circle and the circumcircle, the ﬁrst being M. This is a rich constellation!! Kind regards, Floor.

Floor van Lamoen to Antreas Hatzipolakis and Paul Yiu, December 13 ,1999. Dear Paul and Antreas, Paul Yiu wrote:

In the general case, if we start with an arbitrary line 2, the perspector of the circumcenter triangle lies on the circumcircle. The Simson line of this perspector is parallel to 2.

The perspector can be found in a diﬀerent way, too. See the following problem (NAvW 578) by J.T. Groenman:

Let ABC be a triangle and γ its circumcircle. The line l cuts BC in A1, CA in B1 and AB in C1. A perpendicular to BC is erected at A1,toCA at B1andtoAB at C1. The perpendiculars enclose a triangle A2B2C2 (where A2 is the point of intersection of the perpendiculars through B1 and C1,etc.).(a)ProvethatAA2, BB2, CC2 are concurrent at a point S of γ. (b) Prove that the Simson line of S with respect to the triangle ABC is parallel to l. vanLamoenandYiu:Pre-HyacinthianGeometry 72

What is remarkable is that the triangle A2B2C2 is similar to ABC, as well as the circumcenter triangle A B C ! It is clear that the segment AA2 is a diameter of the circle (A), or, in other words, A is its midpoint, so indeed this problem gives the same perspector as Paul discovered! Best regards. Floor.

Paul Yiu to Antreas Hatzipolakis and Floor van Lamoen, December 14, 1999. Dear Floor and Antreas, Excellent! I see also other messages from you two. But let me ﬁrst respond to this one. During the weekend I was looking for things like this! What I am interested in is this: given a line 2 through the orthocenter H of ABC, how do we construct the line 2 such that the orthocenters of AY Z, BZX,andCXY all lie on ell. [Here, X, Y , Z are the intercepts of 2 on the side lines of ABC]. Related to this is the construction of the point with given line of reﬂections, which necessarily passes through the orthocenter, and is also called the line of images.

Theorem [Musselman, Amer. Math. Monthly, 1938]: Let 2 be a line through the orthocenter H of triangle ABC. For every point Q on 2,letA, B, C be the reﬂections of Q in the side lines BC, CA, AB respectively. vanLamoenandYiu:Pre-HyacinthianGeometry 73

(1) The circumcircles of ABC, BCA,andCAB intersect at a point M on the circumcircle of ABC. (2) The line of reflections of M is the given line 2. Floor, I think I have communicated this to you on November 9, written in a more general setting of P −perpendicularity. While I am still working on the line of orthocenters problem, I find the following interesting fact about reflections:

Theorem: Let 2 be a line through the orthocenter H of triangle ABC, intersecting the circumcircle at two points Q and Q.DenotebyX, Y , Z the reﬂections of Q in the side lines of ABC, and likewise, X, Y , Z those of Q in the same side lines. (1) The lines XX, YY,andZZ intersect at a point M on the circumcircle. (2) The line of reﬂections of M is the given line 2.

Whether this is new or not, I don’t know. It is just wonderful! Needless to say, this can be extended to the case of P −perpendicularity. Best regards. Sincerely, Paul vanLamoenandYiu:Pre-HyacinthianGeometry 74

Paul Yiu to Antreas Hatzipolakis and Floor van Lamoen, December 14, 1999. Dear Antreas and Floor, I did not point out in my earlier message. But I just realized I used the same notation for the points in the two theorems. They are indeed the same point [for the same line 2]! Best regards. Paul

Paul Yiu to Antreas Hatzipolakis and Floor van Lamoen, December 14, 1999. Dear Floor and Antreas, The theorem I communicated this morning is trivial in light of this observation: Let 2 be a line through the orthocenter H of triangle ABC. For every point Q on 2, the reflections of Q in the side lines clearly lie on the respective reflections of 2 in the same side lines. It follows that any two such reflection triangles, including the degenerate ones when Q lies on the circumcircle, are perspective. The perspector, being the intersection of the reflection lines of 2, is a point on the circumcircle. Best regards. Sincerely, Paul

Floor van Lamoen, December 13, 1999. Dear all, Paul Yiu wrote:

If a line 2 intersects the side lines of triangle ABC at X, Y , Z respectively, then the circumcenters of triangles ABC, AY Z, BZX,and CXY are concyclic. Floor has commented that this circle containing these circumcenters is called the Steiner circle, and has included a nice synthetic proof. Now, the circumcenters A, B, C of AY Z, BZX,andCXY form a triangle in perspective to ABC, the perspector lying on the circumcircle.

If 2 is the Euler line, this perspector is the point X110,whichisalso the point of concurrency of the reﬂections of the Euler line of ABC in its side lines.

If 2 is the OI−line, this perspector is the point X100.

If 2 is the van Aubel line HK, this perspector is the point X112.

Here are some related things for you:

1. Show that the perspector noted by Paul above is the second intersection of the circumcircle of ABC and the Steiner circle induced by 2. vanLamoenandYiu:Pre-HyacinthianGeometry 75

2. (J.T. Groenman, Nieuw Archief voor Wiskunde, 28 202(1980) - in part) Consider the lines through X perpendicular to BC, through Y perpendicular to AC and through Z perpendicular to AB. They form a triangle ABC. Show that AA, BB and CC concur in the same perspector as Paul mentioned. 3. The circumcircle of ABC, together with the circumcircle of ABC and the Steiner circle (circumcircle of ABC) are coaxial. Extending more (and a hint for 2.), let At =(1− t)A + tA , Bt =(1− t)B + tB ,and Ct =(1− t)C + tC .

(a) AtBtCt is similar to and perspective to ABC.

(b) The circumcircle of AtBtCt is coaxial to the circumcircle of ABC and the Steiner circle.

Best regards. Floor.

Floor van Lamoen, December 14, 1999. Dear Paul, Thank you for sending the theorem by Musselman, and your own theorem, of which I didn’t see its triviality before you explained it! This becomes a nicer subject by each day! By accident (Wingeom drawing - dynamic geometry program that I use more than the demo version of GSP that I have) I noted the following: Let ABC be the triangle of circumcenters. The orthocenter of ABC lies on 2 (I haven’t computed yet, but it seems quite clear)! When 2 passes through H then 2 is the locus of all orthocenters of AtBtCt from my last message I sent to the ‘working group’. Kind regards, Floor.

Floor van Lamoen, December 15, 1999. Dear Paul, I found the following intriguing problem on MathPro: TIMES 17840 by R.F. Davis: ABC is a triangle having O, H for its circum- and ortho-centers; E, F the extremities of the circumdiameter OH; U the Euler-point on the on the circum-circle, whose Simson-line is parallel to OH. Prove that the rectangular hyperbola, having its center at U, asymptotes UE and UF and touching EOHF at O, will pass through the in- and ex-centers of ABC. Alsothatanytwopoint which are isogonal conjugates with respect to the triangle ABC will be conjugate points with respect to this rectangular hyperbola. Journal: Mathematical Questions and Solutions from the Educational Times Publisher: Francis Hodgson volume(year)page references: Proposal: 28(1915)92 by R. F. Davis vanLamoenandYiu:Pre-HyacinthianGeometry 76

What would Davis mean by “conjugate points w.r.t. this rectangular hyperbola”?? Might that be a generalization of inversion of a point in a circle? A construction for the mentioned Euler point seems to be revealed by the following problem: TIMES 17969 by R.F. Davis: If φ be the point on the circum-circle of the triangle ABC whose Simson line is parallel to OH,provethatφG passes through the middle point of OK. Journal: Mathematical Questions and Solutions from the Educational Times Publisher: Francis Hodgson volume(year)page references: Proposal: 28(1915)92 by R. F. Davis The rectangular hyperbola of the ﬁrst problem is apparently the Stammler hyperbola, since it passes through O, I, Ia, Ib and Ic.Itisnewformethat, for instance, the Stammler hyperbola is tangent to the Euler line and that the asymptotes are quite easy constructible. Kind regards, Floor.

Floor van Lamoen, December 15, 1999. Dear Paul, I forgot to mention that the point U in the ﬁrst and φ in the second problem is Kimberling’s X110. Kind regards, Floor.

Paul Yiu, December 15, 1999. Dear Floor, What a coincidence! Yesterday, I visited the same site, and copied the 2500+ problems from “Mathematical Questions and Solutions from the Educational Times”. Last night, I tried to go through the geometry problems. But there are so many. On the other hand, I found that most of the problems posed by a Neuberg, (whom I hope was the triangle geometry guru in the last century), have “not been entered into the database yet”. I hope they will be. This is indeed a treasure.

[vL](12/16/99): Yes, it is!! I didn’t copy all the 2500+ problems, but I did select certain keywords in the problem text. And there appear to be quite some very interesting things. How about the following problem on the Steiner constellation (which needs a P −perpendicular generalization, needless to say): TIMES 15237 by W.H. Salmon If the circum-circles of the four triangles determined by any for straight lines meet in P , show that the isogonal conjugates of the six lines drawn from P to the points of intersection of the original four lines are parallel. vanLamoenandYiu:Pre-HyacinthianGeometry 77

(Apparently the isogonal conjugate here is the reﬂection through an angle bisector of the two lines meeting in a point of intersection). [Y]: I shall think about this later. [vL]: And there is so much more! Had you ever heard of this Jour- nal/collection? [Y]: I have been aware of this wonderful 19th century collection for over 20 years, but have never seen one. I imagine it is not easy to ﬁnd this outside Britain. I think it is the 19th century counterpart of the Problem Sections of American Mathematical Monthly, perhaps even more abundant. It is indeed a true treasure. I hope the Neuberg whose many problems were not entered was indeed J. Neuberg. Then it would be ever more eye opening when somebody typed up these geometry problems!

Our geometry book is going to be “encyclopaedic”.

[vL](12/16/99): I hope so!! It should be encyclopedic, but also exciting to read. It will.

I shall say more about reflections later. But I cannot resist telling the excitement I found by reading Chapter 16 of Bottema, which you reminded me of before. The problem is to find triangles ABC so that the reflections of A, B, C on the opposite sides are collinear. My instinct was, of course, to write down the equations in homogeneous coordinates. Bottema very cleverly used trigonometric identities, and the observation that the nine-point center should be on the circumcircle is just wonderful. I just love it!

[vL](12/16/99): I am glad you share the enjoyment of this, kinda weird, chapter. I think these are the kinds of results where many people say: mathematicians are nuts. But I love it, too!

I tried to make a dynamic picture, beginning with the circumcircle and an interior point as the incenter, trying to construct such a triangle. I can construct the incircle, the nine-point center (on the circumcircle), and the orthocenter. But the construction of a triangle from given (or known) I, O, H cannot be done by ruler and compass; it involves the solution of a cubic equation. I shall try to compare with what Bottema has to say.

[vL](12/16/99): He composes possible solutions from equation (9), which doesn’t appear ‘constructable’. [Y]: Bottema’s construction is very wonderful indeed. Equation (9) on p.76 gives the range of p and the cosines of the three angles in terms of square roots, and therefore constructible. The signs in the ﬁrst two lines of (9), by the way, should NOT be there. It is merely a typo.√ But I still have to sit down and work out why he limits 1 p< 4 6. vanLamoenandYiu:Pre-HyacinthianGeometry 78

What I tried yesterday was to follow the suggestion in the next paragraph, as you pointed out earlier, starting with the circumcircle andaninteriorpointforI. In this setup, one can only determine the incircle and the nine-point circle. But it is not possible to construct the triangle by ruler and compass. The other day, I looked at Chapter 15 on the construction of the “Bottema triangles”. I would never have dreamed of such a method. My thinking of plane geometry tend to be quite static. But Old Bottema solved an intriguing problem by starting with something very natural and in a few steps of very elementary calculation arrived at the conclusion. This is very masterful indeed. Best regards. Sincerely, Paul

Floor van Lamoen, December 17, 1999. Dear Paul, Thanks for doing the computations for the P −orthocenter being on 2.To- day I realized that it is more or less trivial: We can derive the fact that the P −orthocenter of XYZ lies on 2 from the ‘line-reflection theorem’: Consider a line 2 : px + qy + rz = 0 intersecting the sidelines of ABC at X =0:−r : q, Y = r :0:−p,andZ = −q : p : 0 respectively. Denote by A, B and C the P −circumcenters of AY Z, XBZ and AY C respectively. We have seen that the P −circumconics (A), (B)and(C) of these three triangles intersect in point M. Note that ZM is the radical axis of (A)and(B). Consequently AB is a perpendicular bisector of segment ZM, and the reflection of 2 through AB passes through M. By symmetry the reflection of 2 through the other sides of ABC pass through M, too. Hence, by the (converse of) the ‘line-reflection theorem’we see that 2 should pass through the P −orthocenter of ABC. If 2 passes through P then we can extend this in the following way: let At = A + t · v(AA )(wherev(AA ) denotes the vector from A to A ), and let Bt and Ct be defined likewise. Triangle A2B2C2 is “Groenman’s triangle”, enclosed by the lines through X, Y , Z and P −perpendicular to BC, AC, AB. Clearly Groenman’s triangle is P −similar to ABC. Consequently all triangles Tt = AtBtCt are similar to ABC and thus the P-orthocenter Ht of Tt moves linearly with t. And from the fact that H0 ∈ 2 and H1 ∈ 2 we see that 2 is the locus of Ht. I hope you like the way one of your favourite theorems, which I call line- reflection theorem here, comes into the picture! Kind regards, Floor.

Paul Yiu, December 17, 1999. Dear Floor, Thanks for this. I am writing a short note to Clark on points arising from the Miquel theory. vanLamoenandYiu:Pre-HyacinthianGeometry 79

[vL]: I suppose you noticed that Clark in TCCT (page 245, p. 8) wrote a nice problem relating to Miquel theory. I try to devote a short section to the case when the points X, Y, Z are intercepts of a line ell. This case is naturally what we are talking about here. The four circumcenters O, Oa, Ob,andOc of ABC, AY Z, BZX,andCXY are concyclic, as I pointed out the other day, and you told me that this is called the Steiner circle. What is the source for this? [vL]: I found this name in the “Cut the knott” webpage by Alexan- der Bogomolsky. It appears in a bunch of columns about Morley’s triangle; Bogomolsky describes that Morley was studying this circle and related results. The URL is: http://www.maa.org/editorial/knot/CenterCircle.html When I posed this problem last January to geometry.college,Dave Wilson replied telling that in the booklet “Some triangle geometry” by Dick Tahta this circle is described. This morning I was browsing through Johnson’s chapter on Miquel theory. There he has an exercise [Art. 197] stating that this circle also passes through the Miquel point, the intersection of three circles AY Z, BZX, CXY . Then I observed this: the perspector of the circumcenter triangle OaObOc, (which is always similar to ABC,evenwhenX, Y , Z are not collinear) is the second intersection of the circumcircle and the Steiner circle! [vL]: Which is, for the circular case, immediatly apparent from the synthetic proof and follow-ups I gave. In fact, all circumcircles of AtBtCt (see my last letter) pass throught this Miquel point and this perspector. Hence triangle AtBtCt induces a family of coaxial circumcircles.

Your remark that OaObOc is always similar to ABC, was also mentioned by Guru Conway, in reply to my January question. He wrote: [Conway]: Yes, I know this, and in a generalized form. I call the generalization “Hallowell’s Theorem”, since it appears in his geometry book of 1871. In the most general form it says that if one puts triangles of apex angles α, β, γ on the sides BC, CA, AB,then provided that α + β + γ = π, their circumcentres will form a triangle of apex angles alpha, beta, gamma. Your theorem is the case when the ABC of Hallowell’s theorem is a straight line. An interesting theorem! This Steiner circle stuﬀ is more exciting than I imagined. Allthese,ofcourse,haveP −perpendicularity generalizations. Best regards. Sincerely, Paul vanLamoenandYiu:Pre-HyacinthianGeometry 80

Paul Yiu, December 17, 1999. Dear Floor, I have forgotten to mention that this orthocenter of OaObOc (on the line 2) is equidistant from the Miquel point (of the three circles) and the orthocenter H of ABC. Best regards. Sincerely, Paul

Paul Yiu, December 17, 1999. Dear Paul, Thanks for this observation. It seems that there must be more! Best regards. Floor.

Paul Yiu, December 20, 1999. Dear friends, While revising my earlier notes on Miquel associates and Miquel conjugates (July 6, 1999), I have found the following interesting theorem. Let P be a point with traces X, Y , Z on the side lines of triangle ABC. The circle XYZ intersects each of the sidelines at a second point, denoted X, Y , Z respectively. These second intersections are the traces of a point P , which we call the Miquel conjugate of P . In [TCCT], this is called the cyclocevian conjugate of P . By Miquel’s theorem, the circles AY Z, BZX,and CXY intersect at a point Q, which we call the Miquel associate of P . Likewise, the circles AY Z, BZX,andCXY intersect at the Miquel associate of Q, which we denote by Q.

Theorem (a) The points Q and Q are isogonal conjugates. (b) The lines PP and QQ are parallel. (c) The four points P , P , Q, Q are collinear if and only if P lies on the Lucas cubic* (b2 + c2 − a2)u(v2 − w2)=0, cyclic which is an isotomic cubic with the isotomic conjugate of the orthocenter as pivot. * I follow Floor in using the name Lucas cubic. Peter Yﬀ called this the Staggall cubic. It is the cubic consisting of points whose cevian triangles are also pedal. (d) The “second intersections” of the pairs of circles AY Z, AY Z; BZX, BZX;andCXY , CXY form a triangle perspective with ABC. (e) The “Miquel perspector” in (d) is the intersection of the duals (trilinear polars) of P and P with respect to triangle ABC. vanLamoenandYiu:Pre-HyacinthianGeometry 81

(f) The Miquel perspector is inﬁnite point, i.e., the common chords of the three pairs of circles are parallel if and only if P lies on the same Lucas cubic in (c) above. (g) If P lies on the Lucas cubic, the line containing P, P’, Q, Q’ also contains the second intersections of the pairs of circles (in d), and is perpendicular to these common chords. I hope to write out the details in a few days.

Floor van Lamoen, December 21, 1999. Dear Paul, The ideas around the Steiner circle conﬁgurations grow and grow. My thoughts of the weekend and yesterday are typed down here: Consider in ABC a line 2 intersecting the triangle sides in X, Y and Z.We have shown that the circumcenters O, OA, OB and OC of ABC, AY Z, BZX, and CXY are concyclic. Also we know that the orthocenter H of OAOB OC lies on 2. By symmetry this means that the orthocenters HAHBHBof OOB OC , OAOOC and OAOB O respectively, form a triangle inscribed in ABC. We can apply the following theorem.

Theorem Let ABC be a triangle with perpendicularity based on P = f : g : h. Sup- pose that D = x : y : z is on the P −circumconic of ABC.LetHd be the P −orthocenter of ABD.ThenHdPCD is a parallelogram. Proof: We ﬁnd for Hd the coordinates

f[h(f + g)xy + fhyz − g(f + g)xz − fgz2] : g[h(f + g)xy − f(f + g)yz + ghxz − fgz2] : h[−(f + g)2xy − f(f + g)yz − g(f + g)xz − fgz2].

Using this, the line HdP has line-coordinates

[ghx(gz +(f + g)y):−fhy((f + g)x + fz):fgz(fy− gx)].

∞ The point of intersection of HdP and L is found as:

−f 2(g + h)yz + fg2xz − fh(f + g)xy : f 2gyz − g2(f + h)xz − gh(f + g)xy : f 2hyz + g2hxz + h(f + g)2xy

Using that f(g + h)yz + g(f + h)xz + h(f + g)xy =0,sinceD lies on the P −circumconic of ABC, this point can be rewritten as x : y : −x − y.Which proves that CD and HdP are parallel. Since clearly also DHd and PC are parallel, both being P −perpendicular to AB, the theorem is proven. vanLamoenandYiu:Pre-HyacinthianGeometry 82

Applying this theorem on the concyclic quadruple OOAOBOC we see that HAHBHC , inscribed in ABC, is homothetic and congruent to OAOB OC and thus P −similar to ABC. From this we can derive the following:

1. K(OAOB OC ) ≥ K(ABC)/4(K denotes area); 2. O is the orthocenter of HAHBHC ; 3. O is the Miquel point of HAHBHC ; 4. There are quite a few congruent circles involved (like the ones that deﬁnd the Miquel point);

The Steiner circle has grown into something quite large! Kind regards, Floor.