Dear Willy, I was intrigued by your findings and have asked around about the result.John Sylvester of King's College, London has identified it - here is his reply. This is a known result, Frégier's theorem, and S is the Frégier point of P w.r.t. the conic. (It doesn't have to be an ellipse.) Proof probably easiest by complex projective geometry. The orthogonal lines through P are line-pairs of an involution on the pencil at P, and cut an involution on the conic. S is the vertex of this involution. If you want to fill in the gaps, take the conic as y2 = xz, P as (t2, t, 1), and the lines PX, PZ as the united lines of the involution (X = (1,0,0), Z = (0,0,1)). With a bit of work, S comes out as Y (0,1,0), unless I made a mistake. It appears as an exercise in Semple & Kneebone, Algebraic Projective Geometry, p.153, ex. 16. (No proof, of course.) You can also find a statement of the theorem at http://mathworld.wolfram.com/FregiersTheorem.html (but not a proof), and in Wells' book, referenced there (but no proof, again). Interesting extension: the locus of S as P moves around the conic is a second conic, concentric and homothetic to the first. There is something about this at http://tech.groups.yahoo.com/group/Hyacinthos/message/21195 All the best, John So, many thanks indeed Willy for sending this interesting result around, but sorry that you cannot name it Leenders Theorem! Best wishes, Peter Peter RansomPresident Designate, The Mathematical Association From: [email protected] Subject: I found out (graphic) a theorem Date: Sun, 3 Feb 2013 23:11:15 +0100 To: [email protected]
In his book "Die Sonnenuhr und ihre Theorie" (The sundial and his theory) Jörg Meyer writes on page 200 (my English translation): I found the following remarkable theorem in the book of Heinz Schilt. 'Ebene Sonnenuhren' (Plane sundials) Through the point P where all the hour lines of any sundial come together, a circle is drawn.The center M and the radius of the circle are irrelevant.The hour lines whose hour angle differ from each 6 hours or 90 ° were grouped into pairs.The points where the hour lines of a pair intersect the circle, are connected with a chord (a straight line joining the ends of an arc )Then is applicable: all these chords pass through a common point Q. In the study of this case and assuming that the projection of a circle is an ellipse and vice versa, I found out (graphic) the theorem: Of all right triangles, inscribed in an ellipse, of which the right angle point (point P) is common, the hypotenuses intersect in the same point (point S). See drawing. I found that theorem never formulated, certainly no proof. Who can prove this theorem? Willy LeendersHasselt in Flanders (Belgium) Visit my website about the sundials in the province of Limburg (Flanders) with a section 'worth knowing about sundials' (mostly in Dutch): http://www.wijzerweb.be --------------------------------------------------- https://lists.uni-koeln.de/mailman/listinfo/sundial
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