**Schröder
Points Database / Darij Grinberg**

This page contains some results and questions about the Schröder points of a triangle and configuration related to these. The dates are the dates of last changes.

The idea of this page was to collect all known results about the Schröder points of a triangle from different sources (Hyacinthos messages, MathLinks discussions). Feel free to mail me (see main site for address) if you have something to add.

[Schroeder2]The Schröder points,Darij Grinberg, 4 Jul 2004

[Schroeder3]Incentral Triangle Question,Darij Grinberg, 27 Feb 2004

[Schroeder4]Re: Incentral Triangle Question,Eric Danneels, 3 May 2003

[Schroeder5]Poristically fixed points,Darij Grinberg, 25 Mar 2004

[Schroeder6]Re: Incentral Triangle Question,Darij Grinberg, 27 Feb 2004

[Schroeder7]Some newer results from MathLinks,Darij Grinberg, 4 Jul 2004

[Schroeder2] The Schröder pointsDarij Grinberg, 4 Jul 2004

This message contains all properties and generalizations of the Schröder points known hitherto.1. The Schröder point by inversionIn Hyacinthos message #6319, I defined the Schröder point as follows: Let ABC be a triangle with the incenter I. The incircle of triangle ABC touches BC, CA, AB at X, Y, Z. The triangle XYZ is called the Gergonne triangle (= intouch triangle = contact triangle = pedal triangle of I) of triangle ABC. Then the circles AIX, BIY and CIZ are coaxal and concur at two points. The first point of concurrence is obviously I; the second one is a point Sc which I callSchröder pointof triangle ABC. My naming refers to the article Heinz Schröder:Die Inversion und ihre Anwendung im Unterricht der Oberstufe, Der Mathematikunterricht 1/1957, pages 59 - 80. In this article, the concurrence is proven using inversion. One can verify that the vertices of the tangential triangle of an arbitrary triangle ABC are the inverses of the midpoints of ABC's sides in the circumcircle. Applying this fact to triangle XYZ, we easily see that the circles AIX, BIY and CIZ are the inverses of the medians of triangle XYZ in the circumcircle of triangle XYZ, i. e. in the incircle of triangle ABC. Hence, the Schröder point Sc is the inverse of the centroid of XYZ in the incircle of ABC. In other words, Sc is the Far-out point of the Gergonne triangle XYZ. From this we get that Sc lies on the Euler line of XYZ, i. e. on the line OI, where O is the circumcenter and I is the incenter of ABC. (This follows from the theorem that the Euler line of XYZ is the line OI, which can be shown using the homothety of triangle XYZ with the excentral triangle of ABC.) The common chord of the circles AIX, BIY and CIZ is the line OI.2. Trilinears of the Schröder pointHomogeneous trilinears for the Schröder point Sc found by Jean-Pierre Ehrmann are ( (b-c)^{2}+ a(b+c-2a) : (c-a)^{2}+ b(c+a-2b) : (a-b)^{2}+ c(a+b-2c) ). Later I proved another trilinear representation: ( cos B + cos C - 2 cos A : cos C + cos A - 2 cos B : cos A + cos B - 2 cos C ). Clark Kimberling has included the Schröder point Sc as X(1155) in the ETC (Encyclopedia of Triangle Centers). I also propose to call Sc theGergonne-Schröder pointof triangle ABC, since we will later find some analogues related to the Nagel and Bevan points. Clark Kimberling introduces the Schröder point as follows:X(1155) = SCHRÖDER POINTLet XYZ be the intouch triangle of ABC; i.e., the pedal triangle of the incenter, I. The circles AIX, BIY, CIZ concur in two points. One of them is I; the other is X(1155). This result is obtain by inversion inHeinz Schröder,"Die Inversion und ihre Anwendung im Unterricht der Oberstufe,"Der Mathematikunterricht1 (1957) 59-80. Each vertex of the tangential triangle of any triangle T is the inverse-in-the-circumcircle-of-T of the midpoints of the sides of T. Applying this to triangle XYZ shows that X(1155) is the inverse-in-the-incircle of the centroid of XYZ; i.e., X(1155) is X(23)-of-the-intouch-triangle. (Darij Grinberg, #6319, 1/11/03; coordinates by Jean-Pierre Ehrmann, #6320, 1/11/03) In Clark Kimberling's ETC, the isogonal conjugate of the Schröder point Sc is the point X(1156); this point lies on the line GF, where G is the Gergonne point and F the Feuerbach point of triangle ABC.3. Jean-Pierre Ehrmann's approachThe key to the trilinear coordinates of Sc is another proof of the fact that the circles AIX, BIY and CIZ concur at two points. In fact, if the external bisectors of A, B, C in triangle ABC meet the opposite sides in A", B", C", then the points A", B", C" lie on one line h, the so-called tripolar of the incenter I of triangle ABC. Since A", B", C" lie on one line h, the midpoints of segments IA", IB", IC" lie on the image of this line in the homothety with center I and factor 1/2. Call this image h'. Now, the circle AIX is the circle with diameter IA", because the angles IAA" and IXA" are 90°. Analogously, the circles BIY and CIZ are the circles with diameters IB" and IC". Hence, the centers of the circles AIX, BIY and CIZ lie on the line h'. Consequently, they pass through the reflection of I in h'. This proves again that they have two common points, but this also gives more: In fact, this reflection is easily seen to be the pedal of I on h. This yields that the Schröder point lies on h and that ISc is orthogonal to h. But we know that the Schröder point Sc lies on OI. Thus, we get the following result: The tripolar h of the incenter of a triangle ABC is orthogonal to the line joining its circumcenter O with its incenter I, and these two lines meet at the Schröder point Sc of triangle ABC. From this result, we easily get the trilinears of Sc given above. These nice observations are due to Jean-Pierre Ehrmann (Hyacinthos messages #6326 and #6327) - many thanks!4. The Schröder point as inversive image of X(55)The Schröder point Sc of triangle ABC is the inversive image of the centroid S' of the Gergonne triangle XYZ in theincircleof ABC. Note that S' is X(354) in Kimberling's ETC, known as theWeill pointof triangle ABC. On the other hand, in Kimberling's ETC I read that the Schröder point Sc is the inversive image of X(55) in thecircumcircleof ABC. The point X(55) is the internal center of similtude of the circumcircle and the incircle of triangle ABC, also known as the isogonal conjugate of the Gergonne point. I have found a proof that Sc is the inverse of X(55), using some ratios between points on the line OI. This proof is not easy enough to describe it here, but it is pretty straightforward, using the relation IS' r --- = --, OI 3R where r is the inradius and R is the circumradius of triangle ABC, and the Euler formula OI^{2}= R^{2}- 2Rr.5. The Nagel-Schröder pointIn Hyacinthos message #6544, I have defined three analogues of the Schröder point. While the Schröder point is the second intersection of the circles AIX, BIY and CIZ, where I is the incenter and X, Y, Z are the vertices of the Gergonne triangle, other points can be obtained by replacing the incenter by the excenters or the Gergonne triangle by the Nagel triangle, for example. Draw the points X', Y', Z' where the sides BC, CA, AB of triangle ABC meet therespectiveexcircles. The triangle X'Y'Z' is called the Nagel triangle (= extouch triangle = pedal triangle of X(40)) of triangle ABC. Now I conjectured that the circles AIX', BIY' and CIZ' are coaxal and concur at two points. The first point is I, and the second point, denoted by Sn, is called theNagel-Schröder pointof triangle ABC. Unlike the Schröder point Sc, for whose existence we have two rather simple synthetic proofs, the first elementary proof for the existence of Sn was found more than a year after the discovery of the point itself, and this proof is very long. The homogeneous trilinears of Sn are also much more difficult: / (b+c-2a)(2bc(b+c-a)-(a+b+c)S_{A}) \ ( ------------------------------ : ... ) \ b+c-3a / (the two other coordinates follow by symmetry), where b^{2}+c^{2}-a^{2}S_{A}= --------, and analogously S_{B}and S_{C}. 2 These trilinears were found by Jean-Pierre Ehrmann (and I have verified them with a dynamic sketch). In Clark Kimberling's ETC, the Nagel-Schröder point is X(1339).6. The Stutensee pointOne of my attempts to prove the existence of the Nagel-Schröder point Sn synthetically was an inversion argument like that I used for Sc. In fact, the inverses of the circles AIX', BIY' and CIZ' in the incircle of triangle ABC are the lines M_{x}X", M_{y}Y" and M_{z}Z", where M_{x}, M_{y}and M_{z}are the midpoints of the sides of triangle XYZ, and X", Y" and Z" are the inverses of X', Y' and Z' in the incircle of ABC. By inversion, the concurrence of the lines M_{x}X", M_{y}Y" and M_{z}Z" is equivalent to the coaxality of the circles AIX', BIY' and CIZ'. I had no success with this method but it produced a new triangle center. Namely, the intersection of the lines M_{x}X", M_{y}Y" and M_{z}Z", or equivalently the inverse of the Nagel-Schröder point Sn in the incircle of triangle ABC, is a triangle center not in Kimberling's ETC; I call it theStutensee pointof triangle ABC. [My naming is voluntary and has no reasons.] Thanks to Edward Brisse for trilinears of the Stutensee point. Unfortunately, these trilinears are too long to be shown here.7. The Mitten-Schröder pointsA further variation of the Schröder point is not only replacing the Gergonne triangle by the Nagel triangle, but also the incenter I by the excenters I_{a}, I_{b}, I_{c}. In fact, in Hyacinthos message #6544, I showed: The circles AI_{a}X', BI_{b}Y', CI_{c}Z' are coaxal. They intersect at two points. The common chord passes through I.My proofwas similar to Jean-Pierre Ehrmann's proof for the Schröder point Sc. The circle AI_{a}X' has diameter I_{a}A", where A" is the intersection of the external angle bisector of A with BC, since the angles I_{a}X'A" and I_{a}AA" are both 90°. So we have to prove that the circles having diameters I_{a}A", I_{b}B", I_{c}C" are coaxal, and that I lies on the radical axis. But from the Bodenmiller and Steiner theorems [see Hyacinthos #6124, §4, or the equivalent geometry-college messages], applied to the complete quadrilateral by the lines I_{b}I_{c}A", I_{c}I_{a}B", I_{a}I_{b}C" and A"B"C", this circles are coaxal, and the orthocenter of triangle I_{a}I_{b}I_{c}, i. e. the incenter I of ABC, lies on the radical axis. Now it remains to prove that in our case, the circles have common points (in fact, there could be also the case that they don't have common points). But this is easy: the incenter I lies on the chords AI_{a}, BI_{b}, CI_{c}of each of these circles; therefore, it is an inner point of all three circles, and they must have common points. One could think that this completely closes the problem. But Jean-Pierre Ehrmann added that the common chord of the three circles also passes through the symmedian point K of triangle ABC. I. e., the common chord is the line IK, and after a well-known theorem it also passes through the Mitten point of triangle ABC. The two points at which the circles AI_{a}X', BI_{b}Y', CI_{c}Z' intersect can be called theMitten-Schröder pointsof triangle ABC. I don't know their trilinears. I am also very interested in a synthetic proof that the common chord of the circles AI_{a}X', BI_{b}Y', CI_{c}Z' is the line IK.8. The Bevan-Schröder pointNow we are going to consider the last and most fruitful analogon of the Schröder point. TheBevan pointW of triangle ABC is defined as the circumcenter of the excentral triangle I_{a}I_{b}I_{c}; it is known to be the reflection of the incenter I of ABC in the circumcenter O of ABC, and it is also the intersection of the lines I_{a}X', I_{b}Y', I_{c}Z'. In fact, X', Y' and Z' are the pedals of the Bevan point W on BC, CA and AB. The Bevan point is X(40) in Clark Kimberling's ETC. Now I found that the circles AWX', BWY' and CWZ' are coaxal and intersect at two points. Their first intersection is W; the other intersection is a point which I callBevan-Schröder pointof triangle ABC and denote by Sb. The common chord of the circles AWX', BWY' and CWZ' is the line OI; the Bevan-Schröder point Sb lies on OI.9. Trilinears of the Bevan-Schröder pointHomogeneous trilinears for the Bevan-Schröder point Sb were found by Jean-Pierre Ehrmann: / b+c-2a c+a-2b a+b-2c \ ( ------ : ------ : ------ ). \ b+c-a c+a-b a+b-c / In Clark Kimberling's ETC, the Bevan-Schröder point Sb occurs as X(1319). The isogonal conjugate of the Bevan-Schröder point Sb is the point X(1320); this point lies on the line NF, where N is the Nagel point and F the Feuerbach point of triangle ABC.10. Some synthetic conjectures about the Bevan-Schröder pointAgain, I am missing a synthetic proof of the existence of the Bevan-Schröder point. This time, we have two results about Sb which remind on similar results for Sc. In fact, while the Schröder point Sc is the inverse of the centroid of triangle XYZ in the incircle of ABC, the Bevan-Schröder point Sb is the inverse of the orthocenter of triangle XYZ in the incircle of ABC. And while the Schröder point Sc is the inverse of X(55), the internal center of similtude of circumcircle and incircle, in the circumcircle of ABC, the Bevan-Schröder point Sb is the inverse of X(56), the external center of similtude of circumcircle and incircle, in the circumcircle of ABC. The centroid of triangle XYZ is X(354) in Kimberling's ETC, while the orthocenter of triangle XYZ is X(65) in Kimberling's ETC. So we can state the two results on Sb as follows: The Bevan-Schröder point Sb is the inverse of X(65) in the incircle of triangle ABC and the inverse of X(56) in the circumcircle of triangle ABC. I can't prove either of these two facts. However, it is not hard to show that the inverse of X(65) in the incircle coincides with the inverse of X(56) in the circumcircle, i. e. it is sufficient to establish one of the two results. Remark that X(56) is the isogonal conjugate of the Nagel point.11. Poristically fixed pointsA triangle center of a triangle is calledporistically fixedif every triangle having the same circumcircle and the same incircle has the same corresponding triangle center. For example, the circumcenter is obviously poristically fixed. The incenter is also poristically fixed. The midpoint of OI is poristically fixed (where O is the circumcenter and I is the incenter). It can be also shown that the isogonal Mitten point (i. e. the isogonal conjugate of the Mitten point) is poristically fixed, the orthocenter X(65) of the Gergonne triangle XYZ is poristically fixed, and the centroid X(354) of the Gergonne triangle XYZ is poristically fixed. The internal and external centers of similtude of the circumcircle and the incircle are poristically fixed. By inversion in the circumcircle, this yields that the Schröder point and the Bevan-Schröder point are poristically fixed (what means that all triangles which share the same circumcircle and the same incircle have the same Schröder point and the same Bevan-Schröder point). The Nagel-Schröder point is not poristically fixed. The examples suggest the following theorem:Fundamental Poristic Theorem.All poristically fixed triangle centers lie on the line OI.NOTE.This theorem must be understood with a slight restriction: For example, constructing equilateral triangles OX_{1}I and OX_{2}I on the segment OI yields two points X_{1}and X_{2}, which are no triangle centers yet, but can be made triangle centers by defining X_{1}as the point for which triangle OX_{1}I has the same orientation as the original triangle ABC and X_{2}as the point for which triangle OX_{2}I has opposite orientation to triangle ABC. But the orientation of triangle ABC is not poristically fixed, i. e. the points X_{1}and X_{2}can be interchanged if we get over to another triangle with the same circumcircle and the same incircle. A proof of the Fundamental Poristic Theorem was given by Barry Wolk (I cite Hyacinthos message #6895): If you reflect ABC about a line through O, the image triangle has the same circumcircle as ABC. Similarly, if you reflect ABC about a line through I, the image has the same incircle as ABC. So the reflection of ABC about its OI line is poristic with ABC. This easily shows that any poristically fixed triangle center must lie on the line OI.12. A generalization of the Schröder and Bevan-Schröder pointsleading to the Darboux cubicNow we are going to treat two generalizations of the Schröder points and their variations. The Schröder point is the second intersection of the circles AIX, BIY and CIZ, where I is the incenter of triangle ABC and XYZ is the pedal triangle of I. The Bevan-Schröder point is the second intersection of the circles AWX', BWY' and CWZ', where W is the Bevan point of triangle ABC and X'Y'Z' is the pedal triangle of W. This challenges the following generalizing question (Floor van Lamoen, Hyacinthos message #6321): For which points P with pedal triangle P_{a}P_{b}P_{c}do the circles APP_{a}, BPP_{b}and CPP_{c}have two common points? In Hyacinthos message #6329, Floor van Lamoen proved that they have two common points if and only if P lies on the Darboux cubic of triangle ABC. I cite Floor van Lamoen (notations changed): The perpendicular bisectors of AP, BP, CP, P_{a}P, P_{b}P and P_{c}P are the homothetics through P with factor 1/2 of the sidelines of ABC and the prepedal triangle of P. The circumcenters are the intersections of the corresponding sides of these smaller triangles. So the circumcenters are collinear iff ABC and prepedal of P are lineperspective iff P lies on Darboux. Of course the second point of intersection of the three circles lies on the perspectrix of ABC and the prepedal triangle of P. Twonotes: The prepedal triangle of a point P (also called antipedal triangle) is the triangle whose the sides are the perpendiculars to AP, BP, CP at A, B, C. The idea of Floor van Lamoen is a generalization of the proof that Jean-Pierre Ehrmann gave for the Schröder point (see 3.). Since the Darboux cubic is also the locus of all P whose pedal triangle P_{a}P_{b}P_{c}is perspective with ABC, we can rewrite our result as follows: The circles APP_{a}, BPP_{b}and CPP_{c}have two common points if and only if the lines AP_{a}, BP_{b}and CP_{c}concur. Here are some points P on the Darboux cubic and the corresponding second intersections Q of the circles APP_{a}, BPP_{b}and CPP_{c}:·If P is the incenter X(1) of triangle ABC, then Q is the Schröder point X(1155) and has trilinears ( (b-c)^{2}+ a(b+c-2a) : (c-a)^{2}+ b(c+a-2b) : (a-b)^{2}+ c(a+b-2c) ).·If P is the circumcenter X(3) of triangle ABC, then Q is the triangle center X(187), defined as the inversive image of the symmedian point in the circumcircle, and known as theSchoute pointof triangle ABC. This point has trilinears ( a(b^{2}+c^{2}-2a^{2}) : b(c^{2}+a^{2}-2b^{2}) : c(a^{2}+b^{2}-2c^{2}) ), and lies on the Brocard axis and on the Lemoine axis of ABC.·If P is the orthocenter X(4) of triangle ABC, then Q is the infinite point of the inversive plane. The circles APP_{a}, BPP_{b}and CPP_{c}degenerate to lines.·If P is the Longchamps point X(20) of triangle ABC, then Q is X(468), having trilinears / b^{2}+c^{2}-2a^{2}c^{2}+a^{2}-2b^{2}a^{2}+b^{2}-2c^{2}\ ( --------- : --------- : --------- ). \ b^{2}+c^{2}-a^{2}c^{2}+a^{2}-b^{2}a^{2}+b^{2}-c^{2}/ Isn't it a good idea to call X(468) theLongchamps-Schröder pointof triangle ABC ?·If P is the Bevan point X(40) of triangle ABC, then Q is the Bevan-Schröder point X(1319) and has trilinears / b+c-2a c+a-2b a+b-2c \ ( ------ : ------ : ------ ). \ b+c-a c+a-b a+b-c / The trilinears of X(468) were found by Floor van Lamoen in Hyacinthos message #6352. In Hyacinthos messages #6385 and #6390, Floor van Lamoen and Bernard Gibert discussed a variation of this generalized Schröder points.13. A generalization of the Schröder and Nagel-Schröder pointsand the Feuerbach hyperbolaIn Hyacinthos message #6764, I suggested another generalization of the Schröder point. While the previous generalization covered the Schröder and Bevan-Schröder point, this one contains the Schröder and Nagel-Schröder points. Here is how I introduced the latter generalization: If ABC is a triangle with incenter I, and XYZ is the cevian triangle of the Gergonne point (called Gergonne triangle, intouch triangle, contact triangle), then the circles AIX, BIY and CIZ are coaxal. Their second point of concurrence is called the Schröder point or the 1st Schröder point of triangle ABC. If X'Y'Z' is the cevian triangle of the Nagel point (called Nagel triangle, extouch triangle), then the circles AIX', BIY' and CIZ' are also coaxal. Their second point of concurrence is called the Nagel-Schröder point of triangle ABC.Generalization:If P is a point with cevian triangle A'B'C', then I conjecture that the circles AIA', BIB', CIC' are coaxal if and only if P lies on the Feuerbach hyperbola of triangle ABC. By the way, what curve is drawn by the second points of concurrence of the circles? In Hyacinthos message #6785, Barry Wolk verified my conjecture with slight adjustment: The circles are coaxal if and only if P lies on the the Feuerbach hyperbola or on the line at infinity. Barry Wolk also generalized the generalization (I cite Hyacinthos message #6785, with little changes): I tried generalizing, from using the incenter I to using an arbitrary point Q. Given Q, find all P such that the circles AQA', BQB' and CQC' have collinear centers, where A'B'C' is the cevian triangle of P. The answer in general is a cubic, with a complicated equation. That cubic factors into (line at infinity)·(a conic), only when Q=I or Q=an excenter. And when Q=I the conic is indeed the Feuerbach hyperbola. A few other choices for Q didn't give anything interesting. When Q=G, the cubic is isotomic, with pivot (a^{4}+ b^{2}c^{2}- b^{4}- c^{4}: : ) in barycentrics. The latter pivot has a name: It is the so-calledDroussent pivotof triangle ABC, the point X(316) in Kimberling's ETC. Another nice choice for Q is the circumcenter of triangle ABC. Then the locus of P is the union of the Euler line and the circumcircle of triangle ABC, as Paul Yiu and Jean-Pierre Ehrmann found in some later Hyacinthos messages.Darij Grinberg

[Schroeder3] Incentral Triangle QuestionDarij Grinberg, 27 Feb 2004

I post this problem here, since it has to do with Schröder points. Here are two of my Hyacinthos messages (slightly edited):

Hyacinthos message #6542 Subject:Incentral Triangle questionFrom:Darij Grinberg Dear friends, Let the internal bisectors of a triangle ABC intersect the opposite sides at A', B', C'. To prove that the distance of I to B'C' is rR / OI_{a}and B'C' orthogonal OI_{a}, where O is the circumcenter of ABC, I is the incenter, I_{a}is the a-excenter, r is the inradius and R is the circumradius. (The orthogonality relation I have proven, but how to manage the equation?) Note that if we also take the points A", B", C" where the external bisectors intersect the opposite sides, we get the collinear triples of points A"B'C', A'B"C', A'B'C" and A"B"C", and then the lines A"B'C', A'B"C', A'B'C", A"B"C" are orthogonal to OI_{a}, OI_{b}, OI_{c}, OI respectively, where I_{a}, I_{b}, I_{c}are the excenters of triangle ABC and I is the incenter. Darij GrinbergHyacinthos message #6627 Subject:Re: Incentral Triangle questionFrom:Darij Grinberg In message #6542, I wrote: >> Let the internal bisectors of a triangle ABC >> intersect the opposite sides at A', B', C'. >> To prove that >> >> the distance of I to B'C' is rR / OI_{a}and >> B'C' orthogonal OI_{a}, >> >> where O is the circumcenter of ABC, I is the >> incenter, I_{a}is the a-excenter, r is the >> inradius and R is the circumradius. >> >> (The orthogonality relation I have proven, but >> how to manage the equation?) >> >> Note that if we also take the points A", B", C" >> where the external bisectors intersect the >> opposite sides, we get the collinear triples >> of points A"B'C', A'B"C', A'B'C" and A"B"C", >> and then the lines A"B'C', A'B"C', A'B'C", >> A"B"C" are orthogonal to OI_{a}, OI_{b}, OI_{c}, OI >> respectively, where I_{a}, I_{b}, I_{c}are the >> excenters of triangle ABC and I is the >> incenter. Here is a little idea: Call M the midpoint of I_{b}I_{c}, then OM = R. Let L_{a}be the pedal of I on B'C', and X be the pedal of I on BC, then IX = r. To prove that the distance of I to B'C' is rR / OI_{a}, we have to show that triangles L_{a}IX and MOI_{a}are similar. I have tested this by computer drawing, but still don't have a glimmer how to prove it. (Angles L_{a}IX and MOI_{a}are equal, but we need more.) Darij Grinberg

Eric Danneels has found a trigonometric proof - see [Schroeder4]. Later, I found a synthetic proof - see [Schroeder6].Darij Grinberg

[Schroeder4] Re: Incentral Triangle QuestionEric Danneels, 3 May 2003 (edited by Darij Grinberg)

Dear Darij, in Forum Geometricorum Volume I pages 121-124 Lev Emelyanov and Tatiana Emelyanova proved that B'C' = abc·sqrt(R·(R + 2·r_{a})) / [R·(a + b)·(a + c)] Application of the sinus law in triangles AIB' and AIC' leads to IB' = b·c·sin(A/2) / [(a + c)·cos(C/2)] and IC' = b·c·sin(A/2) / [(a + b)·cos(B/2)] So the surface of triangle IB'C' becomes 1/2·IB'·IC'·cos(A/2) with sin^{2}(A/2) = (s-b)·(s-c)/bc and cos^{2}(A/2) = s·(s-a)/bc etc... this becomes S(IB'C') = a·b·c·r / [2·(a + b)·(a + c)] and therefore IL_{a}= 2·S(IB'C') / B'C' = R·r / sqrt(R·(R + 2·r_{a})) since sqrt(R·(R + 2·r_{a})) = OI_{a}(Euler) we have IL_{a}= R·r / OI_{a}I hope this can be of some help Kind regardsEric Danneels

[Schroeder5] Poristically fixed pointsDarij Grinberg, 25 Mar 2004

In [Schroeder2], 11. I have discussed "poristically fixed" points. With the theoretical assistance of Barry Wolk and computational help of Paul Yiu - many thanks - I have made a list of all triangle centers on the line OI showing which of them are poristically fixed. See also Hyacinthos messages #6873, #6895, #6897 and #6900. The following list contains all points on the line OI from X(1) up to X(2445).Triangle centers X(i) in Clark Kimberling's ETC lying on OI, where I = X(1) and O = X(3).----------------------------------------------------------- X(1) = incenter; PORISTICALLY FIXED X(3) = circumcenter; PORISTICALLY FIXED X(35) = harmonical conjugate of X(36) with respect to OI; PORISTICALLY FIXED X(36) = inverse of the incenter in the circumcircle; PORISTICALLY FIXED X(40) = Bevan point = reflection of I in O; PORISTICALLY FIXED X(46) = reflection of I in X(56); PORISTICALLY FIXED X(55) = internal center of similtude of circumcircle and incircle; PORISTICALLY FIXED X(56) = external center of similtude of circumcircle and incircle; PORISTICALLY FIXED X(57) = isogonal Mitten point; PORISTICALLY FIXED X(65) = orthocenter of intouch triangle; PORISTICALLY FIXED X(165) = centroid of excentral triangle; PORISTICALLY FIXED X(171) = isogonal conjugate of 1st Sharygin point; NOT FIXED X(241) = intersection of OI line and Gergonne axis; NOT FIXED X(260) = isogonal conjugate of 1st mid-arc point; NOT FIXED X(354) = Weill point; PORISTICALLY FIXED X(484) = (1st) Evans perspector = reflection of I in X(36); PORISTICALLY FIXED X(517) = intersection of OI and line at infinity; PORISTICALLY FIXED X(559) ; a fissile and quartile (--> octile) point; NOT FIXED X(940) = intersection of OI and GK, where G centroid and K symmedian point; NOT FIXED X(942) = midpoint of I and X(65); PORISTICALLY FIXED X(980) ; NOT FIXED X(982) ; NOT FIXED X(986) ; NOT FIXED X(988) ; NOT FIXED X(999) = midpoint of I and X(57); PORISTICALLY FIXED X(1038) ; NOT FIXED X(1040) ; NOT FIXED X(1060) ; NOT FIXED X(1062) ; NOT FIXED X(1082) ; a fissile and quartile (--> octile) point; NOT FIXED X(1155) = Schröder point; PORISTICALLY FIXED X(1159) = Greenhill point; PORISTICALLY FIXED X(1214) ; NOT FIXED X(1319) = Bevan-Schröder point; PORISTICALLY FIXED X(1381) = 1st intercept of line OI and circumcircle; PORISTICALLY FIXED X(1382) = 2nd intercept of line OI and circumcircle; PORISTICALLY FIXED X(1385) = midpoint of OI; PORISTICALLY FIXED X(1388) = midpoint of I and X(36); PORISTICALLY FIXED X(1402) ; NOT FIXED X(1403) ; NOT FIXED X(1420) ; PORISTICALLY FIXED X(1429) ; NOT FIXED X(1454) ; PORISTICALLY FIXED X(1460) ; NOT FIXED X(1466) ; PORISTICALLY FIXED X(1467) ; PORISTICALLY FIXED X(1470) ; PORISTICALLY FIXED X(1482) = reflection of O in I; PORISTICALLY FIXED X(1617) ; PORISTICALLY FIXED X(1622) ; NOT FIXED X(1697) = internal center of similtude of incircle of triangle ABC and circumcircle of excentral triangle; PORISTICALLY FIXED X(1715) ; NOT FIXED X(1735) ; NOT FIXED X(1754) ; NOT FIXED X(1758) ; NOT FIXED X(1771) ; NOT FIXED X(1936) ; NOT FIXED X(2061) ; NOT FIXED X(2077) = reflection of X(36) in O; PORISTICALLY FIXED X(2078) = inverse of X(57) in the circumcircle; PORISTICALLY FIXED X(2093) = reflection of I in X(57); PORISTICALLY FIXED X(2095) = reflection of O in X(57); PORISTICALLY FIXED X(2098) = reflection of X(56) in I; PORISTICALLY FIXED X(2099) = reflection of X(55) in I; PORISTICALLY FIXED X(2223) ; NOT FIXED X(2283) ; NOT FIXED X(2352) ; NOT FIXEDPS.Many thanks to Edward Brisse and to Paul Yiu for determining the poristic behaviour of many ETC centers.Darij Grinberg

[Schroeder6] Re: Incentral Triangle QuestionDarij Grinberg, 27 Feb 2004

In [Schroeder3], I wrote: >> Let the internal bisectors of a triangle ABC intersect the opposite >> sides at A', B', C'. To prove that >> >> the distance of I to B'C' is rR / OI_{a}and >> B'C' orthogonal OI_{a}, >> >> where O is the circumcenter of ABC, I is the incenter, I_{a}is the >> a-excenter, r is the inradius and R is the circumradius. Here is a synthetic proof. I am going to use directed angles modulo 180°. I will write "<" for "angle". We will use the following lemmata:

Lemma 1.Let H be the orthocenter of a triangle ABC, and X, Y, Z bethe feet of the altitudes from A, B, C. The sidelines YZ, ZX, XY of triangle XYZ intersect the sidelines BC, CA, AB of triangle ABC at the points O, P, Q, respectively. Then the points O, P, Q lie on one line - the so-calledorthic axisof triangle ABC -, and this line is perpendicular to the Euler line of triangle ABC.Proof.Letobe the circumcircle andnthe nine-point circle of triangle ABC. The circumcircleopasses through A, B, C, and the nine-point circlenpasses through X, Y, Z. Since the points Z and X lie on the circle with diameter CA, we have PZ * PX = PC * PA, and hence the point P has equal powers with respect to the circlesnando(in fact, PZ * PX is the power of P with respect ton, and PC * PA is the power of P with respect too). Hence, P lies on the radical axis ofnando. Similarly, O and Q also lie on this radical axis; altogether, the points O, P, Q lie on one line, namely the radical axis ofnando. Moreover, this line is perpendicular to the Euler line of triangle ABC, since the radical axis of two circles is perpendicular to the central line, while the centers ofnandolie on the Euler line of triangle ABC. Lemma 1 is proven.Lemma 2.Let H be the orthocenter of a triangle ABC, let X and Y be the feet of the altitudes from A and B, and C' the midpoint of the side AB. Finally, call Q the intersection of the lines XY and AB. Then, the line HQ is perpendicular to the median CC'.Proof(from: M. Volchkevich,Reshenie zadachi M1724, Kvant).Let N be the foot of the perpendicular from H to CQ. The points X, Y and N lying on the circle with diameter CH, we have < CNX = < CYX. The points X and Y lying on the circle with diameter AB, we get < ABX = < AYX, i. e. < ABX = < CYX = < CNX. In other words, < QBX = < QNX. Consequently, the points Q, N, X and B lie on a circle, and < NBQ = < NXQ. But, again, the circle with diameter CH passing through X, Y and N, we obtain < NXY = < NCY, and thus < NBA = < NBQ = < NXQ = < NCY = < NCA. Hence, the points N, B, A, C lie on one circle, i. e., the point N lies on the circumcircle of triangle ABC. Let the line NH meet the circumcircle again at the point R (apart from N). Since < RNC = 90°, the segment RC is a diameter of the circumcircle, so that < RAC = 90°, and RA is perpendicular to CA. Therefore, RA is parallel to the altitude BH. Similarly, RB || AH. Hence, the quadrilateral AHBR is a parallelogram, and the diagonal HR passes through the midpoint C' of the diagonal AB. In other words, the line C'H is perpendicular to CQ. On the other hand, the line CH is perpendicular to the line C'Q (since CH is perpendicular to AB). Hence, the point H lies on two altitudes of triangle CC'Q; consequently, it also lies on the third altitude, i. e. the line HQ is perpendicular to CC', qed..

Okay, now we go on to the proof. Let the external angle bisector of A meet the sideline BC at A". The excenters of triangle ABC will be called I_{a}, I_{b}, I_{c}. The points A, B, C are the feet of the altitudes of triangle I_{a}I_{b}I_{c}, and the point I is the orthocenter of this triangle. On the other hand, we may consider the triangle II_{c}I_{b}; the feet of the altitudes of this triangle are A, B, C, again, and the orthocenter of this triangle is I_{a}. Now, the sidelines BC, CA, AB of triangle ABC intersect the sidelines I_{c}I_{b}, I_{b}I, II_{c}of triangle II_{c}I_{b}in the points A", B', C', respectively; hence, after Lemma 1, the points A", B', C' lie on one line perpendicular to the Euler line of triangle II_{c}I_{b}. However, the Euler line of this triangle is the line OI_{a}(because O is the nine-point center and I_{a}is the orthocenter of this triangle); hence, the points A", B', C' lie on one line perpendicular to OI_{a}. This proves the second part of our problem. Now to the first part. Let X be the orthogonal projection of I on the line BC, i. e. the point where the incircle of triangle ABC touches BC. Then, IX = r. Since A, B, C are the feet of the altitudes of triangle I_{a}I_{b}I_{c}, the circumcircle of triangle ABC is the nine-point circle of triangle I_{a}I_{b}I_{c}; hence, it also passes through the midpoints of the sides of triangle I_{a}I_{b}I_{c}. For instance, it passes through the midpoint M of the side I_{b}I_{c}. Hence, OM = R. As the points B and C lie on the circle with diameter I_{b}I_{c}(remember < I_{b}BI_{c}= 90° and < I_{b}CI_{c}= 90°), the perpendicular bisector of BC passes through the center of this circle, i. e. through the midpoint M of I_{b}I_{c}. On the other hand, this perpendicular bisector obviously passes through the circumcenter O of triangle ABC. Consequently, MO is perpendicular to BC. Together with IX perpendicular to BC (this is trivial), we obtain MO || IX. We have shown before that the line A"B'C' is perpendicular to OI_{a}. On the other hand, if L_{a}is the orthogonal projection of I on the line A"B'C', the line IL_{a}is perpendicular to the line A"B'C'. Hence, the lines IL_{a}and OI_{a}are parallel. Since MO || IX and OI_{a}|| IL_{a}, we have < (IX; IL_{a}) = < (MO; OI_{a}) = - < (OI_{a}; MO), i. e. < XIL_{a}= - < I_{a}OM. Now apply Lemma 2 to the triangle I_{a}I_{b}I_{c}with I as orthocenter, A and B as feet of the altitudes, and A" as intersection of BC with I_{b}I_{c}. This yields that the line IA" is perpendicular to the median I_{a}M of triangle I_{a}I_{b}I_{c}. The points X and L_{a}lie on the circle with diameter IA"; thus, < IXL_{a}= < IA"L_{a}. But < IA"L_{a}= < (IA"; B'C'). Now, on one hand, IA" is perpendicular to I_{a}M; on the other hand, B'C' is perpendicular to OI_{a}. Hence, < IA"L_{a}= < (IA"; B'C') = < (IA"; I_{a}M) + < (I_{a}M; OI_{a}) + < (OI_{a}; B'C') = 90° + < MI_{a}O + 90° = 180° + < MI_{a}O = < MI_{a}O = - < OI_{a}M. Therefore, < IXL_{a}= - < OI_{a}M. On the other hand, < XIL_{a}= - < I_{a}OM, as we have seen before. Hence, triangles IXL_{a}and OI_{a}M are oppositely similar, and IL_{a}OM --- = ---, IX OI_{a}thus IX * OM rR IL_{a}= -------- = ---. OI_{a}OI_{a}In other words, the distance of I to the line B'C' is rR / OI_{a}, proving the second assertion. Sometimes geometry is nontrivial...Darij Grinberg

[Schroeder7] Some newer results from MathLinksDarij Grinberg, 4 Jul 2004

1. Treegoner's TheoremIn [Schroeder2], I wrote: >> Let ABC be a triangle with the incenter I. [...] >> Draw the points X', Y', Z' where the sides BC, CA, AB of triangle >> ABC meet therespectiveexcircles. [...] Now I conjectured that >> the circles AIX', BIY' and CIZ' are coaxal and concur at two >> points. The first point is I, and the second point, denoted by >> Sn, is called theNagel-Schröder pointof triangle ABC. Lately, the existence of the Nagel-Schröder point was proven elementarily by Treegoner on the MathLinks forum. See the thread http://www.mathlinks.ro/viewtopic.php?p=22210 started by Treegoner. We begin with the following theorem he has discovered:Theorem 1.Let ABC be a triangle, let A', B', C' be the midpoints of its sides BC, CA, AB, and let X, Y, Z be the feet of its altitudes. These altitudes AX, BY, CZ concur at the orthocenter H of triangle ABC. Finally, let (W) be the nine-point circle of triangle ABC, passing through the points A', B', C', X, Y, Z; and let X', Y', Z' be the second intersections of the lines A'H, B'H, C'H with (W). Then,(a)The lines XX', YY', ZZ' meet at one point P, which lies on the Euler line of triangle ABC.(b)The lines AX', BY', CZ' meet at one point Q. Here are myproofsof(a)and(b). Another proof of(a), using inversion with respect to the polar circle of triangle ABC, was given by Grobber on MathLinks. Let W be the center of the nine-point circle (W). Also, the midpoints A_{1}, B_{1}, C_{1}of the segments AH, BH, CH lie on the nine-point circle. The segments A_{1}A', B_{1}B', C_{1}C' are diameters of the nine-point circle (this is clear from < A_{1}XA' = 90°, < B_{1}YB' = 90°, < C_{1}ZC' = 90°); hence, the center W of the nine-point circle must lie on these segments and bisect them, and the distances WA_{1}, WA', WB_{1}, WB', WC_{1}, WC' must all be equal to the radius of the nine-point circle. Hence, WA' * WA_{1}= WB' * WB_{1}= WC' * WC_{1}. In other words, the powers of the point W with respect to the circles HA_{1}A', HB_{1}B', HC_{1}C' are equal. But the point H must also have equal powers with respect to these three circles (since it actually lies on these circles). Hence, we have found two points - H and W - with equal powers with respect to the circles HA_{1}A', HB_{1}B', HC_{1}C'. Therefore, these circles are coaxal, and their common radical axis is the line HW, i. e. the Euler line of triangle ABC. Since our circles meet at H, they must also meet at another point R on the Euler line. We state this fact:(c)The circles HA_{1}A', HB_{1}B', HC_{1}C' meet at H and at another point R on the Euler line of triangle ABC. Another proof of(c)is obtained quickly by defining the point R as the point on the Euler line of triangle ABC satisfying the equation WH * WR = -t^{2}, where t is the radius of the nine-point circle of triangle ABC. Then, since the segments WA' and WA_{1}both equal t but with opposite signs, we have WH * WR = -t^{2}= WA' * WA_{1}, so that the point R lies on the circle HA_{1}A', and similarly R lies on the circles HB_{1}B' and HC_{1}C'. Thus we have not only proved(c)again, but also shown that(d)We have WH * WR = -t^{2}, where t is the radius of the nine-point circle of triangle ABC. (In other words, if we invert the point H with respect to the nine-point circle of triangle ABC, and then reflect the resulting inversive image in W, then we get the point R.) Now let the line ZZ' meet the Euler line at a point P. Then < PZH = < Z'ZC_{1}= < Z'C'C_{1}= < HC'C_{1}= < HRC_{1}= < C_{1}RH. Also, evidently, < PHZ = < C_{1}HR. Hence, the triangles PZH and C_{1}RH are similar, so that HP : HZ = HC_{1}: HR, and HP * HR = HZ * HC_{1}. But HZ * HC_{1}= p, where p is the power of the point H with respect to the nine-point circle of triangle ABC. Hence, HP * HR = p. This is a symmetric term, so the point P does not really depend on the choice of the line ZZ', but could be obtained for the lines XX' and YY' as well. Hence, the point P lies on the lines XX', YY', ZZ' and the Euler line of triangle ABC. This proves Theorem 1(a). Now, in an arbitrary circle, the ratio of two chords equals the ratio of the sines of their chordal angles. Hence, in the nine-point circle, XZ' sin < XZZ' --- = ----------. Z'Y sin < Z'ZY Similarly, we can find YX' / X'Z and ZY' / Y'X, and see that XZ' YX' ZY' sin < XZZ' sin < YXX' sin < ZYY' --- * --- * --- = ---------- * ---------- * ---------- = 1, Z'Y X'Z Y'X sin < Z'ZY sin < X'XZ sin < Y'YX the latter following from the trigonometric version of Ceva's theorem, applied to the concurrent lines XX', YY', ZZ' in triangle XYZ. Now, using the sine law in triangles XCZ' and YCZ', we have sin < BCZ' sin < XCZ' XZ' * sin < CXZ' : CZ' ---------- = ---------- = ---------------------- sin < Z'CA sin < Z'CY Z'Y * sin < CYZ' : CZ' XZ' sin < CXZ' XZ' sin < A'XZ' XZ' sin < A'C'Z' = --- * ---------- = --- * ----------- = --- * ------------ Z'Y sin < CYZ' Z'Y sin < B'YZ' Z'Y sin < B'C'Z' XZ' sin < A'C'H = --- * -----------, Z'Y sin < B'C'H and similarly sin < CAX' YX' sin < B'A'H ---------- = --- * -----------; sin < X'AB X'Z sin < C'A'H sin < ABY' ZY' sin < C'B'H ---------- = --- * -----------. sin < Y'BC Y'X sin < A'B'H Hence, sin < BCZ' sin < CAX' sin < ABY' ---------- * ---------- * ---------- sin < Z'CA sin < X'AB sin < Y'BC / XZ' sin < A'C'H \ / YX' sin < B'A'H \ / ZY' sin < C'B'H \ = ( --- * ----------- ) * ( --- * ----------- ) * ( --- * ----------- ) \ Z'Y sin < B'C'H / \ X'Z sin < C'A'H / \ Y'X sin < A'B'H / / XZ' YX' ZY' \ / sin < A'C'H sin < B'A'H sin < C'B'H \ = ( --- * --- * --- ) * ( ----------- * ----------- * ----------- ) \ Z'Y X'Z Y'X / \ sin < B'C'H sin < C'A'H sin < A'B'H / sin < A'C'H sin < B'A'H sin < C'B'H = ----------- * ----------- * ----------- sin < B'C'H sin < C'A'H sin < A'B'H / XZ' YX' ZY' \ ( since --- * --- * --- = 1 ) \ Z'Y X'Z Y'X / = 1; this time, the last step was a consequence of the trigonometric version of Ceva's theorem applied to the concurrent lines A'H, B'H, C'H in triangle A'B'C'. Hence, the trigonometric version of Ceva's theorem, applied to triangle ABC, yields that the lines AX', BY', CZ' are concurrent, proving Theorem 1(b).2. More circlesIn the following, theantipodeof a point P on a circle k will mean the point lying diametrically opposite to the point P on k. Now consider the reflections A_{0}, B_{0}, C_{0}of the point H in the points A', B', C'. Since A_{0}is the reflection of H in A', the point A' is the midpoint of the segment HA_{0}. But remember that we have defined the point A' as the midpoint of the segment BC. Hence, the midpoints of the segments HA_{0}and BC coincide, and the quadrilateral BHCA_{0}is a parallelogram. Therefore, BA_{0}|| CH, or, in other words, BA_{0}is perpendicular to AB, so that < ABA_{0}= 90°. Similarly, < ACA_{0}= 90°. Therefore, the points B and C lie on the circle with diameter AA_{0}. This is equivalent to saying that the point A_{0}is the antipode of A on the circumcircle of triangle ABC. Similar facts will hold for B_{0}and C_{0}, and thus we can summarize that the points A_{0}, B_{0}, C_{0}are the antipodes of the points A, B, C on the circumcircle of triangle ABC. A nearly trivial observation is that the circles AHA_{0}, BHB_{0}, CHC_{0}concur at one point L (apart from H); this point L is the reflection of the inverse of H with respect to the circumcircle of triangle ABC in the circumcenter of triangle ABC. Now we are going to prove:Theorem 2.(a)The circles HXA_{0}, HYB_{0}, HZC_{0}concur at one point on the line HQ (apart from the point H).(b)These circles pass through the orthogonal projections A_{2}, B_{2}, C_{2}of the points A, B, C onto the lines YZ, ZX, XY, respectively.Proof.Let S be the point on the line HQ satisfying the equation HQ * HS = 2p, where p is the power of the point H with respect to the nine-point circle of triangle ABC. (Of course, the segments are directed.) We have p = HA_{1}* HX, so that 2p = 2 HA_{1}* HX = HA * HX (remember that A_{1}is the midpoint of AH). Also, since the point A_{0}is the reflection of H in A', we have HA_{0}= 2 HA', so that 2p = 2 HX' * HA' = HX' * (2 HA') = HX' * HA_{0}. Altogether we see that 2p = HQ * HS = HA * HX = HX' * HA_{0}. From HQ * HS = HX' * HA_{0}, we conclude HQ / HX' = HA_{0}/ HS; this, together with < QHX' = < A_{0}HS, shows that the triangles QHX' and A_{0}HS are similar. Therefore, < HQX' = < HA_{0}S. On the other hand, from HQ * HS = HA * HX, it follows that HQ / HA = HX / HS, and this, combined with < QHA = < XHS, leads to the similarity of the triangles QHA and XHS. Thus, < HQA = < HXS. Therefore, < HA_{0}S = < HQX' = < HQA = < HXS. It follows that the point S lie on the circle HXA_{0}. Similarly, the same point S lies on the circles HYB_{0}and HZC_{0}. This proves Theorem 2(a). Now, we have to show that the orthogonal projection A_{2}of the point A onto the line YZ lies on the circle HXA_{0}. Here is a proof of this: We will work with directed angles modulo 180°. The triangles AYZ and ABC are indirectly similar. In the triangle AYZ, the point A_{2}is the foot of the A-altitude, and the point H is the antipode of A on the circumcircle of triangle AYZ. In the triangle ABC, the point X is the foot of the A-altitude, and the point A_{0}is the antipode of A on the circumcircle of triangle ABC. Hence, the points A_{2}and X are corresponding points in the triangles ABC and AYZ, and so are H and A_{0}. Since corresponding points in indirectly similar triangles form oppositely equal angles, we get < A_{2}AH = - < XAA_{0}and < AHA_{2}= - < AA_{0}X. At first, from < A_{2}AH = - < XAA_{0}, it follows that < A_{2}AH = - < XAA_{0}= < A_{0}AH, so that the point A_{2}lies on the line AA_{0}. Then, from < AHA_{2}= - < AA_{0}X, we imply that < XHA_{2}= < AHA_{2}= - < AA_{0}X = < XA_{0}A_{2}, so that the points H, X, A_{2}and A_{0}lie on one circle, i. e. the point A_{2}lies on the circle HXA_{0}. Similarly, we can show that the points B_{2}and C_{2}lie on the circles HYB_{0}and HZC_{0}, respectively. This finally proves Theorem 2(b).3. The Nagel-Schröder pointTheorem 2(a)states that the circles HXA_{0}, HYB_{0}, HZC_{0}concur at one point (apart from H). But after Theorem 2(b),these circles coincide with the circles HXA_{2}, HYB_{2}, HZC_{2}, respectively. Therefore, we can state:Corollary 3.The circles HXA_{2}, HYB_{2}, HZC_{2}concur at one point, apart from H. If we now consider a triangle ABC with its incenter I and its excenters I_{a}, I_{b}, I_{c}, then it is well-known that the points A, B, C are the feet of the altitudes of triangle I_{a}I_{b}I_{c}, and the point I is the orthocenter of triangle I_{a}I_{b}I_{c}. The orthogonal projections of the points I_{a}, I_{b}, I_{c}onto the lines BC, CA, AB are the points X', Y', Z' where the A-, B-, C-excircles of triangle ABC touch the sides BC, CA, AB. Hence, from Corollary 3, the circles IAX', IBY', ICZ' concur at one point, apart from H. The existence of the Nagel-Schröder point is proven!4. ETC referenceHere are the numbers under which the points defined above occur in Clark Kimberling's ETC: P = triangle center X(235) = the midpoint between the orthocenter H = X(4) and the triangle center X(24). X(24) = a very interesting triangle center. Here are three constructions of this point: - If you take the orthic triangle XYZ of triangle ABC, and the orthic triangle X_{1}Y_{1}Z_{1}of triangle XYZ, then the lines AX_{1}, BY_{1}, CZ_{1}concur at X(24). - If you take the inverses X_{v}, Y_{v}, Z_{v}, of the points X, Y, Z in the circumcircle of triangle ABC, then the lines AX_{v}, BY_{v}, CZ_{v}concur at X(24). - If you call O the circumcenter of triangle ABC, and X_{2}, Y_{2}, Z_{2}the circumcenters of triangles BOC, COA, AOB, then the lines XX_{2}, YY_{2}, ZZ_{2}concur at X(24). Q : not in the ETC. R : not in the ETC, but: R = the reflection of X(403) in X(5). X(403) = the inverse of the orthocenter H in the nine-point circle of triangle ABC. Two properties of X(403): - If O is the circumcenter of triangle ABC, then the circles AOX, BOY, COZ pass through X(403). - The point X(403) is the inverse of X(24) in the circumcircle of triangle ABC. L : not in the ETC, but: L = the reflection of X(186) in X(3). X(186) = "threefold angle point" = the inverse of H with respect to the circumcircle of triangle ABC. S : not in the ETC.5. Alternative proof of Theorem 1 (a) by TreegonerAt first, we define the notion ofcircumcevian triangles: If ABC is a triangle, and P is a point, then thecircumcevian triangleof the point P with respect to the triangle ABC is the triangle A'B'C', where A', B', C' are the points of intersection of the lines AP, BP, CP with the cirucmcircle of triangle ABC (different from A, B, C, of course). An alternative proof of Theorem 1(a)found by Treegoner used the following lemma:Lemma 4.Let M and N be two points in the plane of a triangle ABC. Let A_{1}B_{1}C_{1}be the circumcevian triangle of the point M with respect to triangle ABC. Let A_{3}B_{3}C_{3}be the circumcevian triangle of the point N with respect to triangle A_{1}B_{1}C_{1}. Let A_{4}B_{4}C_{4}be the circumcevian triangle of the point M with respect to triangle A_{3}B_{3}C_{3}. Then the lines AA_{4}, BB_{4}, CC_{4}are concurrent at a point which lies on the line MN. Now apply this lemma to triangle XYZ with its M = H and N = W, and you get a new proof of Theorem 1(a). Theproof of Lemma 4is very simple: In the following, I will denote by g /\ h the point of intersection of two given lines g and h. After Pascal's theorem, applied to the cyclic hexagon A_{1}A_{3}BB_{1}B_{3}A, the points A_{1}A_{3}/\ B_{1}B_{3}(= N), A_{3}B /\ B_{3}A and BB_{1}/\ AA_{1}(= M) lie on one line. Hence, the point A_{3}B /\ B_{3}A lies on the line MN. On the other hand, after Pascal's theorem, applied to the cyclic hexagon A_{4}A_{3}BB_{4}B_{3}A, the points A_{4}A_{3}/\ B_{4}B_{3}(= M), A_{3}B /\ B_{3}A (lying on MN) and BB_{4}/\ AA_{4}are collinear. Hence, the point BB_{4}/\ AA_{4}lies on the line MN, i. e. the lines AA_{4}, BB_{4}and MN concur. Similarly, the lines BB_{4}, CC_{4}and MN concur. Hence, the four lines AA_{4}, BB_{4}and CC_{4}concur at one point, and Lemma 4 is proven.Darij Grinberg

Schröder Points Database

*Darij Grinberg*