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The Napoleon Configuration Forum Geometricorum Volume 2 (2002) 39–46. ¡ FORUM GEOM ISSN 1534-1178 The Napoleon Configuration Gilles Boutte Abstract. It is an elementary fact in triangle geometry that the two Napoleon triangles are equilateral and have the same centroid as the reference triangle. We recall some basic properties of the Fermat and Napoleon configurations, and use them to study equilateral triangles bounded by cevians. There are two families of such triangles, the triangles in the two families being oppositely oriented. The locus of the circumcenters of the triangles in each family is one of the two Napoleon circles, and the circumcircles of each family envelope a conchoid of a circle. 1. The Fermat-Napoleon configuration + Consider a reference triangle ABC, with side lengths a, b, c. Let Fa be the + point such that CBFa is equilateral with the same orientation as ABC; similarly + + for Fb and Fc . See Figure 1. + Fc ΓC ΓB A + Fb ΓC F+ ΓB + A B C Nc + Nb F+ B C ΓA N + + a Fa Figure 1. The Fermat configuration Figure 2. The Napoleon configuration + + + The triangle Fa Fb Fc is called the first Fermat triangle. It is an elementary + + + fact that triangles Fa Fb Fc and ABC are perspective at the first Fermat point − − − − F+. We define similarly the second Fermat triangle Fa Fb Fc in which CBFa , − − ACFb and BAFc are equilateral triangles with opposite orientation of ABC. 1 This is perspective with ABC at the second Fermat point F−. Denote by ΓA the + + + + circumcircle of CBFa , and Na its center; similarly for ΓB, ΓC , Nb , Nc . The Publication Date: April 30, 2002. Communicating Editor: Paul Yiu. 1 In [1], these are called the isogonic centers, and are referenced as X13 and X14. 40 G. Boutte + + + triangle Na Nb Nc is called the first Napoleon triangle, and is perspective with ABC at the first Napoleon point N+. Similarly, we define the second Napoleon − − − 2 triangle Na Nb Nc perspective with ABC at the second Napoleon point N−. − + See Figure 2. Note that Na is the antipode of Fa on ΓA. We summarize some of the important properties of the Fermat and Napoleon points. Theorem 1. Let ABC be a triangle with side lengths a, b, c and area . 4 (1) The first Fermat point F+ is the common point to ΓA, ΓB and ΓC. + + + (2) The segments AFa , BFb , CFc have the same length ` given by 1 `2 = (a2 + b2 + c2 + 4p3 ): 2 4 + + + (3) The first Napoleon triangle Na Nb Nc is equilateral with the same orien- ` tation as ABC. Its circumradius is 3 . (4) The Fermat and Napoleon triangles have the same centroid G as ABC. (5) The first Fermat point lies on the circumcircle of the second Napoleon tri- angle. We shall call this circle the second Napoleon circle. + + (6) The lines Nb Nc and AF+ are respectively the line of centers and the common chord of ΓB and ΓC . They are perpendicular. Remarks. (i) Similar statements hold for the second Fermat and Napoleon points F− and N−, with appropriate changes of signs. (ii) (4) is an easy corollary of the following important result: Given a triangle A0B0C0 with ABC0, BCA0, CAB0 positively similar. Thus ABC and A0B0C0 have the same centroid. See, for example, [3, p.462]. F + − c Fa A + − Nc A + Na + Fb Nb F+ N − − a G Nb − − Nb Fb − Nc B − C B C Nc N + a − Fc + Fa Figure 3. The Fermat and Napoleon Figure 4. The Fermat point on the second triangles Napoleon circle 2 In [1], the Napoleon points as X17 and X18. The Napoleon configuration 41 2. Equilateral triangles bounded by cevians Let A1B1C1 be an equilateral triangle, with the same orientation as ABC and whose sides are cevian lines in ABC, i.e. A lies on B1C1, B lies on C1A1, C lies π on A1B1. See Figure 5. Thus, CB is seen from A1 at an angle , i.e., \CA1B = π 3 3 , and A1 lies on ΓA. Similarly B1 lies on ΓB and C1 lies on ΓC . Conversely, let A1 be any point on ΓA. The line A1B intersects ΓC at B and C1, the line A1C intersects ΓB at C and B1. C1 A − Na F+ B1 − O1 Nb B − Nc C A1 Figure 5. Equilateral triangle bounded by cevians π Three of the angles of the quadrilateral A1B1AC1 are ; so A lies on B1C1 and 3 A1B1C1 is equilateral with the same orientation as ABC. We obtain an equilateral triangle bounded by cevians. There is an infinity of such triangles. π Let O1 be the center of A1B1C1. BO1 is seen from A1 at an angle (mod π); − − 6 similarly for BNa . The line A1O1 passes through Na . Similarly the lines B1O1 − − − − and C1O1 pass through Nb and Nc respectively. It follows that Nb Nc and B1C1 2π π are seen from O1 at the same angle = (mod π), and the point O1 lies on − − − 3 − 3 the circumcircle of Na Nb Nc . Thus we have: Theorem 2. The locus of the center of equilateral triangles bounded by cevians, and with the same orientation as ABC, is the second Napoleon circle. Similarly, the locus of the center of equilateral triangles bounded by cevians, and with the opposite orientation of ABC, is the first Napoleon circle. 42 G. Boutte 3. Pedal curves and conchoids We recall the definitions of pedal curves and conchoids from [2]. Definitions. Given a curve and a point O, C (1) the pedal curve of with respect to O is the locus of the orthogonal pro- jections of O on theCtangent lines of ; (2) for a positive number k, the conchoidCof with respect to O and with offset k is the locus of the points P for which thereC exists M on with O, M, P collinear and MP = k. C For the constructions of normal lines, we have Theorem 3. Let O be the pedal curve of with respect to O. For any point M on , if P is the projectionP of O on the tangentCto at M, and Q is such that OP MQ C C is a rectangle, then the line P Q is normal to O at P . P P M PO O C Q Figure 6. The normal to a pedal curve Theorem 4. Let O;k be the conchoid of with respect to O and offset k. For any C C point P on O;k, if M is the intersection of the line OP with , then the mormal C C lines to O;k at P and to at M intersect on the perpendicular to OP at O. C C 4. Envelope of the circumcircles Consider one of the equilateral triangles with the same orientation of ABC. Let 1 be the circumcircle of A1B1C1, R1 its radius. Its center O1 lies on the Napoleon C circle and the vertex A1 lies on the circle ΓA. The latter two circles pass through − \ − \ − F+ and Na . The angles Na A1F+ and Na O1F+ have constant magnitudes. The shape of triangle A1O1F+ remains unchanged when O1 traverses the second O1A1 R1 Napoleon circle . The ratio = remains constant, say, λ. N O1F+ O1F+ The Napoleon configuration 43 Q O P 0 M P C CO Figure 7. The normal to a conchoid C1 C1 A − M Na N 0 F+ B1 O O1 O1 B C Q F F 0 ΓA C0 A1 P Figure 8. Equilateral triangle Figure 9. The pedal of C0 with respect bounded by cevians and its circumcir- to F cle For convenience we denote by the Napoleon circle which is the locus of O1, N 0 F the Fermat point lying on this circle, F the antipode of F on , and 0 the 0 N C particular position of 1 when O1 and F coincide. See Figure 7. Let P be any C 0 0 0 point on , the line P F intersects 0 at Q and Q (F between P and Q), we N C construct the point M such that F P QM is a rectangle. The locus F of M is the P pedal curve of 0 with respect to F and, by Theorem 3, the line MP is the normal C 44 G. Boutte to F at M. The line MP intersects at P and O1 and the circle through M with P N center O1 is tangent to F at M. P 0 0 The triangles F MO1 and F QF are similar since \F MO1 = \F QF and 0 \ \ 0 3 O1M F Q F O1M = F F Q. It follows that = 0 = λ, and O1M = R1. The O1F F F circle through M with center O1 is one in the family of circle for which we search the envelope. Furthermore, the line F M intersects at F and O0, and O0MQF 0 is a rectan- 0 0 N gle. Thus, O M = F Q, the radius of 0. It follows that M lies on the external C branch of the conchoid of with respect to F and the length R = radius of 0. N 0 0 C By the same reasoning for the point Q , we obtain M on F , but on the internal P branch of the conchoid. Each circle 1 touches both branches of the conchoid. C Theorem 5. Let be the family of circumcircles of equilateral triangles bounded by cevians whoseFlocus of centers is the Napoleon circle passing through the Fermat point F . The envelope of this family is the pedalN with respect to F of F the circle 0 of whose center is the antipode of F on , i.e.
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