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Fontene Theorems and Some Corollaries

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Fontene Theorems and Some Corollaries Fontene theorems and some corollaries Linh Nguyen Van 30/04/2010 Abstract In 1905 and 1906, Fontene. G presented three theorems about the generalization of Feuerbach point in the magazine "Nouvelles Annales de Mathmatiques ". In this note, we give their proofs and learn about their corollaries. 1 Theorems Fontene theorem 1. Given triangle ABC. Let P be an arbitrary point in the plane. A1;B1;C1 are the midpoints of BC; CA; AB; A2B2C2 is the pedal triangle of P with respect to triangle ABC. Let X; Y; Z be the intersections of B1C1 and B2C2;A1C1 and A2C2;A1B1 and A2B2. Then A2X; B2Y; C2Z concur at the intersection of (A1B1C1) and (A2B2C2). Proof by Bricard (ibid., 1906). L A Q B2 X C 1 B1 O C 2 F P E O' B A2 A1 C 0 Let E be the center of (A1B1C1);O be the center of (A2B2C2);F be the intersection of OP and the circle with diameter OA; L be the reflection of A2 with respect to B1C1 then AL==BC. We obtain o \ALP = 90 . o Since \AF P = \AB2P = \AC2P = \ALP = 90 we get L; F; B2;C2 are on (AP ). \FC1X = \F AB1 = \B2C2F then FXC1C2 is a cyclic quadrilateral. 0 0 Denote L the intersection of FX and (AP ). We have AL C2F is a cyclic quadrilateral. But 0 0 FXC1C2 is also cyclic therefore AL ==B1C1 or L ≡ L, which follows that L; X; F are collinear. 0 Denote Q the intersection of A2X and (E):F is the reflection of Q with respect to B1C1. 1 0 Consider the Symmetry SB1C1 :(AO) 7! (E), but Q 2 (E) hence F 2 (AO). 0 On the other side, SB1C1 maps A2 to L. Furthermore A2; X; Q are collinear so L; X; F are collinear, which is equivalent to F 0 ≡ F . We deduce that A2LQF is a isosceles trapezoid. This means XQ:XA2 = XL:XF = XB2:XC2. 0 Therefore Q lies on (O ). Similarly B2Y; C2Z also pass through Q. We are done. Fontene theorem 2. If a point P moves on the fixed line d which passes through the circumcenter O of triangle ABC then the pedal circle of P with respect to triangle ABC intersects the Nine-point circle of triangle ABC at a fixed point. Proof. According to the proof of Fontene theorem 1, the point of contact Q of (E) and (O0) is the reflec- tion of a point F which lies on OP with respect to the line B1C1. It is easy to show that O is the orthocenter of triangle A1B1C1 thus Q is the Anti-Steiner point of d. Therefore Q is fixed. Our proof is completed. Fontene theorem 3. Denote the isogonal conjugate of P with respect to triangle ABC as P 0. Then the pedal circle of P is tangent to the Nine-point circle of triangle ABC if and only if O; P; P 0 are collinear. Proof. According to Fontene theorem 2 we can prove that the second intersection Q0 of (O0) and (E) is the Anti-Steiner point of OP 0. This means Q0 ≡ Q if and only if OP ≡ OP 0 or O; P; P 0 are collinear. We are done. Note. Feuerbach point is a corollary of Fontene theorem 3, when P coincides with the incenter or 3 excenters. 2 Some corollaries Corollary 1. O0 is the orthocenter of triangle XYZ. Proof. In fact, applying Fontene theorem 1 we claim A2X; B2Y; C2Z concur at a point Q which lies on (O0), which implies that XZ is the polar of Y with respect to (O0);XY is the polar of Z with respect to (O0). Therefore O0 is the orthocenter of triangle XYZ. 0 Corollary 2. Denote A3B3C3 the pedal triangle of the isogonal conjugate P of P with respect to triangle ABC. Then the Simson line of Q with respect to triangle A3B3C3 is parallel to the Simson line of Q with respect to triangle A1B1C1. Proof. 2 L A Q B2 S X B C1 1 T C3 R B3 O C2 F P' P A1 B A2 A3 C W 0 Let R; S be the projections of Q on B3C3;A3B3. Extend QR to W which lies on (O ), AF \ QW = fT g. Since \QRS = \QB3S = \QW A3 we deduce that RS==A3W . 0 0 Moreover, \P C3B3 = \P AB3 = \P AB hence AP ? B3C3, which implies that AP==QW . This means \TWA3 + \FTW = \QA2A3 + \P AF = \QA2A3 + \P LF = \QA2A3 + \PA2Q = o 90 . So SR==A3W==OP:(1) According to the proof of Fontene theorem 2, Q is the Anti-Steiner point of OP with respect to triangle A1B1C1 so the Simson line of Q with respect to triangle A1B1C1 is parallel to OP:(2) From (1) and (2) we are done. Corollary 3. Given triangle ABC with its circumcircle (O) and its orthocenter H. Let N be the Nagel point of triangle ABC. ON meets (O) at Q. Then the Simson line of Q with respect to triangle ABC is parallel to NH. Proof. First let us introduce a lemma: Lemma 1. Let I; G; N be the incenter, centroid and Nagel point of triangle ABC, respectively. Then I; G; N are collinear and IN = 3IG. Proof. 3 A T Y B' Y' I G N R C B X A' X' K P Ia Let X; Y be the tangencies of (I) with BC; AC. XI cuts (I) at T . We will show that A; T; N are collinear. 0 IT IY IA In fact, let Ia be the A-excenter. (Ia) contacts BC; AC at X ;K. We have 0 = = , IaX IaK IaA which follows that A; T; X0 are collinear or A; T; N are collinear. Let P be a point on AI such that I is the midpoint of AP . Let B0 be the midpoint of AC. IY intersects (I) at R; BR intersects AC at Y 0. Because IB0 is the midline of two triangles Y RY 0 and AP C at the same time hence BN==IB0==P C0. Likewise, CN==BP . Therefore BNCP is a parallelogram. We conclude that A0 is the midpoint of NP . GA Let G0 be the intersection of AA0 and IN. Since IA0 is the midline of triangle AP N then = GA0 GI = 2. Our lemma is solved. GN Back to our problem. 4 C' A B' O I G N H B C P A' Through A; B; C construct three lines which are parallel to opposite side, they intersect each other and make triangle A0B0C0. We have (ABC) is the Nine-point circle of triangle A0B0C0. Consider the homothetic H−2 :(ABC) 7! (A0B0C0);O 7! H; I 7! N. But O is the center of (ABC) so H is the center of (A0B0C0), I is the incenter of triangle ABC so N is the incenter of triangle A0B0C0. We get P is the Feuerbach point of triangle A0B0C0. According to the proof of Fontene theorem 2 , P is the Anti-Steiner point of NH with respect to triangle ABC. This means the Simson line of P with respect to triangle ABC is parallel to NH. Corollary 4. Given triangle ABC with its circumcenter O. Let l be a line which passes through O. l intersects BC; CA; AB at X; Y; Z, repspectively. Then four circles (AX); (BY ); (CZ), the Nine-point circle of triangle ABC are concurrent. Proof. In fact, the result follows immediately from Fontene theorem 2 because X; Y; Z are three special cases of P on the line l. Corollary 5. Given triangle ABC:P is an arbitrary point in the plane. A1B1C1 is the pedal triangle of P with respect to ∆ABC: A2;B2;C2 are the midpoints of BC; CA; AB, respectively. A3;B3;C3 are the reflections of A1;B1;C1 with respect to A2;B2;C2, respectively. Then three cir- cles (A1B1C1); (A2B2C2); (A3B3C3) are concurrent. Proof. 5 A C3 Y B1 C2 B2 Q C1 P O B3 E C B A1 A2 A3 Since A1B1C1 is the pedal triangle of P with respect to triangle ABC then applying Carnot theorem we obtain: 2 2 2 2 2 2 BA1 − CA1 + CB1 − AB1 + AC1 − BC1 = 0. 2 2 2 2 2 2 Hence CA3 − BA3 + BC3 − AC3 + AB3 − CB3 = 0. We deduce that A3B3C3 is the pedal triangle of Q with respect to triangle ABC. Applying Thales theorem, it is easy to see that the perpendicular bisector of the line segment BC passes through the midpoint of PQ. Similarly we get the circumcenter O of triangle ABC is the midpoint of PQ. Denote Y the Anti-Steiner point of OP with respect to triangle A2B2C2 then according to the proof of Fontene theorem 2, Y 2 (A1B1C1). Moreover O; P; Q are collinear so Y is the Anti-Steiner point of OQ with respect to triangle A2B2C2. But A3B3C3 is the pedal triangle of Q with respect to triangle ABC thus according to the proof of Fontene theorem 2 again, Y 2 (A3B3C3). We are done. Corollary 6. Given triangle ABC. Let A1 be the projection of A on BC; A2;B2;C2 be the midpoints of BC; CA; AB, respectively. P is an arbitrary point in the plane, A0B0C0 is the pedal 0 0 0 0 0 triangle of P with respect to ∆ABC:(A B C ) \ (A2B2C2) = fF; F g. A line through A and parallel 00 0 00 0 to AP meets AA1 at A .
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