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Hyacinthian Correspondences on Geometry

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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 tan- gency. 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 incir- cle 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: <O2O1O>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 intersec- tions 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 semicir- cle. (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.
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