huyên đề về đường tròn Hagge

Hagge circles revisited Nguyen Van Linh 24/12/2011 Abstract In 1907, Karl Hagge wrote an article on the construction of circles that always pass through the orthocenter of a given triangle. The purpose of his work is to find an extension of the Wallace-Simson theorem when the generating point is not on the circumcircle. Then they was named Hagge circles. In this paper, we present the new insight into Hagge circles with many corollaries. Furthermore, we also introduce three problems which are similar to Hagge circles. 1 Hagge circles and their corollaries Problem 1 (The construction of Hagge circles). Given triangle ABC with its circumcircle (O) and orthocenter H. Choose P an arbitrary point in the plane. The cevian lines AP, BP, CP meets (O) again at A 1 , B 1 , C 1 . Denote A 2 , B 2 , C 2 the reflections of A 1 , B 1 , C 1 with respect to BC, CA, AB, respectively. Then H, A 1 , B 1 , C 1 are concyclic. The circle (H, A 1 , B 1 , C 1 ) is called P-Hagge circle. Here we give two proofs for problem 1. Both of them start from the problem which appeared in China Team Selection Test 2006. Problem 2 (China TST 2006). Given triangle ABC with its circumcircle (O) and orthocenter H. Let P be an arbitrary point in the plane. The cevian lines AP, BP, CP meets (O) again at A 1 , B 1 , C 1 . Denote A 2 , B 2 , C 2 the reflections of A 1 , B 1 , C 1 across the midpoints of BC, CA, AB, respectively. Then H, A 1 , B 1 , C 1 are concyclic. Proof. We introduce a lemma: Lemma 1. (B 2 C 2 , B 2 A 2 ) ≡ (PC, PA)(modπ) Proof. 1 B' C' A' M N B 2 C 2 A 2 X Z Y C 1 B 1 A 1 O P A B C Denote M, N the reflections of A 2 , C 2 across the midpoint of AC then ∆MB 1 N is the reflection of ∆A 2 B 2 C 2 across the midpoint of AC. Since AM  CA 2 , AM = CA 2 and BA 1  CA 2 , BA 1 = CA 2 we get AMA 1 B is a parallelogram. Similarly, BC 1 NC is a parallelogram too. Let A  , B  , C  be the midpoints of AA 1 , BB 1 , CC 1 then A  , C  are the midpoints of BM, BN , respectively. The homothety H 1 2 B : M → A  , B 1 → B  , N → C  therefore ∆MB 1 N → ∆A  B  C  . This means (B 1 M, B 1 N) ≡ (B  A  , B  C  )(modπ). On the other side, OA  , OB  , OC  are perpendicular to AA 1 , BB 1 , CC 1 hence O, A  , B  , C  , P are concyclic, which follows that: (B 2 A 2 , B 2 C 2 ) ≡ (B 1 M, B 1 N) ≡ (B  A  , B  C  ) ≡ (PA  , PC  ) ≡ (PA, PC)(modπ). We are done. Back to problem 2. C 3 B 3 A 3 B 4 C 4 A 4 H A 2 B 2 C 2 Z Y X C 1 B 1 A 1 O P A B C 2 Construct three lines through A 1 , B 1 , C 1 and perpendicular to AA 1 , BB 1 , CC 1 , respectively. They intersect each other and form triangle A 3 B 3 C 3 , intersect (O) again at A 4 , B 4 , C 4 , respectively. Note that AA 4 , BB 4 , CC 4 are diameters of (O) therefore A 4 , C 4 are the reflections of H across the midpoints of BC, AB. We get HA 2 A 4 A 1 , HC 2 C 4 C 1 are parallelograms, which follows that HA 2  A 1 A 4 , HC 2  C 1 C 4 . But P C 1 B 3 A 1 is a cyclic quadrilateral and applying lemma 1 we have (HC 2 , HA 2 ) ≡ (B 3 C 1 , B 3 A 1 ) ≡ (P C 1 , PA 1 ) ≡ (PC, PA) ≡ (B 2 C 2 , B 2 A 2 )(modπ) or H, A 2 , B 2 , C 2 are concyclic. Our proof is completed. Remark 1: Now come back to problem 1. Let A 3 , B 3 , C 3 be the reflections of A 2 , B 2 , C 2 across the midpoints of BC, CA, AB, respectively. Since (BA 2 C) is the reflections of (ABC) wrt BC then A 3 ∈ (ABC). It is easy to see that BC is the midline of the triangle A 1 A 2 A 3 hence A 1 A 3  BC. We claim that AA 3 , BB 3 , CC 3 concur at the isogonal conjugate Q of P wrt ∆ABC. Then according to problem 2 we obtain H, A 2 , B 2 , C 2 are concyclic. Another proof (Nsato) [9] We introduce a lemma. Lemma 2. Given triangle ABC and its circumcircle (O). Let D be an arbitrary point on (O). A line d through D and d ⊥ BC.d ∩ (O) = {G} then AG is parallel to the Simson line of point D. Proof. F E G A B C D Denote d ∩ BC = {E}, F the projection of D on AC then EF is the Simson line of D. Since DEF C is a cyclic quadrilateral we get ∠GAC = ∠GDC = ∠EF A. Therefore AG  EF . Back to problem 1. 3 C 4 B 4 Y A 1 L Z K A 4 X Q A 3 H A 2 P A B C Let Q be the isogonal conjugate of P wrt ∆ABC; X, Y, Z be the reflections of A, B, C wrt Q; A 3 B 3 C 3 be the circumcevian triangle of Q; A 4 B 4 C 4 be the median triangle of ∆ABC. The symmetry with center Q takes A to X, B to Y, C to Z then ∆ABC → XY Z. But A 4 B 4 C 4 is the median triangle of ∆ABC hence there exist a point K which is the center of the homothety with ratio 1 2 . This transformation maps ∆XY Z to ∆A 4 B 4 C 4 . On the other hand, let L be the second intersection of A 1 A 2 and (O). Applying lemma 2, AL  HA 2 . Moreover, LA 1 is perpendicular to BC and A 3 A 1 is parallel to BC therefore A 3 A 1 ⊥ LA 1 , which follows that AA 3 ⊥ AL or AA 3 ⊥ HA 2 . From remark 1, A 4 is the midpoint of A 2 A 3 hence XA 2 KA 3 is a parallelogram. This means KA 2  AA 3 . So KA 2 ⊥ HA 2 or A 2 lies on the circle with diamenter KH. Similarly we are done. Problem 3. Given triangle ABC with its circumcircle (O) and orthocenter H. Let P be an arbitrary point in the plane. The cevian lines AP, BP, CP meets (O) again at A 1 , B 1 , C 1 . Denote A 2 , B 2 , C 2 the reflections of A 1 , B 1 , C 1 with respect to BC, CA, AB, respectively. Then ∆A 2 B 2 C 2 ∼ ∆A 1 B 1 C 1 . Proof. Firstly we introduce a lemma: Lemma 3. Given triangle ABC and its circumcircle (O). Let E, F be two arbitrary points on (O). Then the angle between the Simson lines of two points E and F is half the measure of the arc EF . Proof. 4 L K G H I J O A B C F E See on the figure, IJ, HG are the Simson lines of two points E and F , respectively. IJ ∩ HG = {K} Denote L the projection of K on BC. We have KL  F G and F HGC is a cyclic quadrilateral thus (KG, KL) ≡ (GK, GF) ≡ (CA, CF )(modπ)(1) Similarly, (KL, KJ) ≡ (BE, BA)(modπ)(2) From (1) and (2) we are done. Back to problem 3. C 2 B 2 A 2 C 1 B 1 A 1 H O P A B C According to Hagge circle, we have H, A 2 , B 2 , C 2 are concyclic. Since A 2 is the reflections of A 1 wrt BC we get HA 2 is the Steiner line of A 1 , which is parallel to it’s Simson line. Analogously, HB 2 is the Steiner line of B 1 . Then applying lemma 3, (C 2 A 2 , C 2 B 2 ) ≡ (HA 2 , HB 2 ) ≡ (C 1 A 1 , C 1 B 1 )(modπ). Do similarly with other direct angles we refer to the similarity of two tri- angles A 1 B 1 C 1 and A 2 B 2 C 2 . Problem 4. Using the same notations as problem 1, let A 3 , B 3 , C 3 be the second intersections of AH, BH, CH and (A 2 B 2 C 2 ), respectively. Then A 2 A 3 , B 2 B 3 , C 2 C 3 concur at P . Proof. 5 A 3 C 3 B 3 H B 2 B 1 A 2 A 1 C 2 C 1 P A B C (A 2 C 2 , A 2 A 3 ) ≡ (HC 2 , HA 3 ) ≡ (HC 2 , BC)+(BC, HA) ≡ π/2+(HC 2 , BC)(modπ). Let L, Z be the projections of C 1 on BC, AB, respectively then LZ is the Simson line of C 1 . We get HC 2  LZ Therefore (HC 2 , BC) ≡ (LZ, LC) ≡ (C 1 Z, C 1 B)(modπ) Hence (A 2 C 2 , A 2 A 3 ) ≡ π/2 + (C 1 Z, C 1 B) ≡ (BA, BC 1 ) ≡ (CA, CC 1 )(modπ) Since H, A 3 , B 3 , C 3 are concyclic then (C 3 B 3 , C 3 A 3 ) ≡ (HB 3 , HA 3 ) ≡ (CA, CB)(modπ) . Sim- ilarly we get ∆A 3 B 3 C 3 ∼ ∆ABC. But from problem 3, ∆A 2 B 2 C 2 ∼ ∆A 1 B 1 C 1 . Thus there exist a similarity transformation which takes A 3 C 2 B 3 A 2 C 3 to AC 1 BA 1 CB 1 and it follows that A 2 A 3 , B 2 B 3 , C 2 C 3 concur at P  . Moreover, AP A 1 P = A 3 P  A 2 P  hence P P   AH. Similarly, P P   BH, CH. This means P ≡ P  . Our proof is completed then. Problem 5. Let I be the circumcenter of triangle A 2 B 2 C 2 . K is the reflection of H across I. AK, BK, CK cut (I) again at A 4 , B 4 , C 4 . Then A 3 A 4 , B 3 B 4 , C 3 C 4 are concurrent. Proof. A 4 B 4 C 4 K A 3 B 3 C 3 I H O P A B C 6 Since KA 3 ⊥ AH we have (C 3 C 4 , C 3 A 3 ) ≡ (KC 4 , KA 3 ) ≡ (CK, CB)(modπ). Similarly, (C 3 C 4 , C 3 B 3 ) ≡ (CK, CA)(modπ). Do the same with other angles then applying Ceva-sine theorem for triangle A 3 B 3 C 3 we are done. Problem 6. Given triangle ABC with its circumcircle (O). Let P and Q be two isogonal conjugate points, O 1 , O 2 be the circumcenters of P-Hagge and Q-Hagge circles, respectively. Then P Q  O 1 O 2 and P Q = O 1 O 2 . Proof. Lemma 4. Given triangle ABC with its orthocenter H, circumcenter O and let P be an arbitrary point in the plane. AP, BP, CP intersect (O) again at A 1 , B 1 , C 1 , respectively. Denote A 2 , B 2 , C 2 the reflections of A 1 , B 1 , C 1 across the midpoints of BC, CA, AB, respectively, O 2 the center of (A 2 B 2 C 2 ). Then OO 2 HP is a parallelogram. Proof. H' U Z Y X O' 2 A' C' H O 2 B 2 C 2 A 2 C 1 B 1 A 1 O P A B C Denote A  , C  the reflections of A 2 , C 2 across the midpoint of AC, respectively. Then ∆A  B 1 C  is the reflections of ∆A 2 B 2 C 2 across the midpoint of AC. Since AA   A 2 C, AA  = A 2 C, A 2 C  BA 1 , A 2 C = BA 1 we get AA  A 1 B is a parallelogram. (1) Similarly, BCC  C 1 is a parallelogram. (2) Let X, Y, Z be the midpoints of AA 1 , BB 1 , CC 1 , respectively then X, Y, Z lies on (OP ). But from (1) and (2), X, Z are the midpoints of BA  , BC  . A homothety H 1 2 B : A  → X, B 1 → Y, C  → Z therefore ∆A  B 1 C  → ∆XY Z A symmetry with center is the midpoint of BC: S : B → B  , H → H  the orthocenter of triangle AB  C, O 2 → O  2 ⇒ O 2 H → O  2 H  It is easy to show that B, O, H  are collinear. Hence the homothety H 1 2 B : H  → O, O  2 → U the circumcenter of triangle XY Z. We conclude that OP = O  2 H = O 2 H. Therefore HPOO 2 is a parallelogram. We are done. Back to our problem. From the lemma above, OO 1 = QH, OO 2 = P H thus P Q = O 1 O 2 7 Our proof is completed. Remark 2. The radius of P-Hagge circle is equal to the distance of Q and O. Problem 7. Given triangle ABC, its Nine-point circle (E) and an arbitrary point P on (E). Let Q be the isogonal conjugate point of P wrt ∆ABC. Then the center I of Q-Hagge circle lies on (E) too. Moreover, I is the reflection of P across point E. Proof. The conclusion follows immediately from remark 2. Note that the center of Q-Hagge circle is always the reflection of P wrt E. Problem 8. Given triangle ABC and its circumcenter O. P and Q are isogonal conjugate wrt triangle ABC such that O, P, Q are collinear. Then P-Hagge circle and Q-Hagge circle are tangent. Proof. This problem is also a corollary of remark 2. We will leave the proof for the readers. Problem 9. Given triangle ABC and its circumcircle (O, R). A circle which has center O and radius r < R meets BC, CA, AB at A 1 , A 2 , B 1 , B 2 , C 1 , C 2 , respectively. Let E, F be the Miquel points of triangle ABC wrt (A 1 , B 1 , C 1 ) and (A 2 , B 2 , C 2 ) respectively. Then E-Hagge circle and F-Hagge circle are congruent. Proof. We will introduce a lemma. Lemma 5. Given triangle ABC. A circle (O) intersects BC, CA, AB at 6 points A 1 , A 2 , B 1 , B 2 , C 1 , C 2 , respectively. Let P, Q be the Miquel points of ∆ABC wrt (A 1 , B 1 , C 1 ) and (A 2 , B 2 , C 2 ). Then P and Q are isogonal conjugate and OP = OQ. Proof. Y Z X Q P C 2 C 1 B 2 B 1 A 2 A B C O A 1 Let X, Y, Z be the second intersections of A 1 P, B 1 P, C 1 P and (O), respectively. We have (ZY, ZX) ≡ (A 1 Y, A 1 X) ≡ (A 1 Y, PY ) + (P Y, A 1 X) ≡ (A 2 C, A 2 B 1 ) + (CA 2 , CB 1 ) ≡ (B 1 B 2 , B 1 A 2 )(modπ) This means XY = A 2 B 2 . Similarly, Y Z = B 2 C 2 , XZ = A 2 C 2 . Therefore ∆XY Z = ∆A 2 B 2 C 2 . On the other side, (P X, P Y ) ≡ (QA 2 , QB 2 )(modπ), (PY, P Z) ≡ (QB 2 , QC 2 )(modπ) then there exist a rotation with center O which takes ∆XY Z to ∆A 2 B 2 C 2 and P to Q. We conclude that OP = OQ. Moreover, (AP, AB) ≡ (B 1 P, B 1 C 1 ) ≡ (ZY, ZP ) ≡ (C 2 B 2 , C 2 Q) ≡ (AB 2 , AQ)(modπ). Similarly we get P and Q are isogonal conjugate wrt ∆ABC. Back to our problem. According to lemma 5, P and Q are isogonal conjugate wrt ∆ABC and OP = OQ. Then applying remark 2 the conclusion follows. 8 Problem 10. Given triangle ABC with its orthocenter H. Let P, Q be two isogonal conjugate points wrt ∆ABC. Construct the diameter HK of P-Hagge circle. Let G be the centroid of the triangle ABC. Then Q, G, K are collinear and QG = 1 2 GK. Proof. X Q G M a K H C 2 C 1 B 2 B 1 A 2 A 1 P A B C Let X be the reflection of A wrt Q, M a be the midpoint of BC then from proof 2 of problem 1, we obtain M a is the midpoint of XK. Then applying Menelaus theorem for the triangle AXM a with the line (Q, G,K) we have Q, G, K are collinear. Applying Menelaus theorem again for the triangle QXK with the line (A, G, M a ) then QG = 1 2 . Problem 11. The centroid G lies on L-Hagge circle where L is the Symmedian point. Proof. C 1 B 3 C 3 A 3 C 2 A 2 B 2 B 1 A 1 G A B C Let A 1 B 1 C 1 , A 2 B 2 C 2 be the circumcevian triangles of G and L, respectively; A 3 , B 3 , C 3 be the reflections of A 2 , B 2 , C 2 wrt BC, CA, AB, respectively. It is easy to see that A 3 , B 3 , C 3 lie on AG, BG, CG, respectively. From problem 3, ∆A 3 B 3 C 3 ∼ ∆A 2 B 2 C 2 . Then (A 2 B 2 , A 2 C 2 ) ≡ (A 3 C 3 , A 3 B 3 )(modπ) On the other side, (GC 3 , GB 3 ) ≡ 1 2 (  CB 1 +  C 1 B) ≡ 1 2 (  B 2 A +  AC 2 ) ≡ (A 2 B 2 , A 2 C 2 ) ≡ (A 3 C 3 , A 3 B 3 )(modπ). 9 This means G, A 3 , B 3 , C 3 are concyclic. We are done. More results. 1. When P ≡ I the incenter of triangle ABC, P-Hagge circle is called Fuhrmann circle of ∆ABC, after the 19th-century German geometer Wilhelm Fuhrmann. The line segment con- necting the orthocenter H and the Nagel point N is the diameter of Fuhrmann circle. The solution for this result can be found in [8]. Moreover, denote O, I the circumcenter and the incenter of triangle ABC then HN = 2OI. 2. When P lies on the circumcircle of triangle ABC, P-Hagge circle becomes the Steiner line of P . 3. When Q lies on the circumcircle of triangle ABC, we have a problem: Given triangle ABC and an arbitrary point P on its circumcircle (O). Denote A  , B  , C  the reflections of P across the midpoints of BC, CA, AB. Then H, A  , B  , C  are concyclic. 4. When P lies at the infinity, problem 1 can be re-written as below: Given triangle ABC with its circumcircle (O), orthocenter H. Denote A 1 , B 1 , C 1 points on (O) such that AA 1  BB 1  CC 1 , A 2 , B 2 , C 2 the reflections of A 1 , B 1 , C 1 across the lines BC, CA, AB, respectively. Then H, A 2 , B 2 , C 2 are concylic. 5. When P ≡ H the orthocenter of ∆ABC, P-Hagge circle becomes point H. 6. When P ≡ O the circumcenter of ∆ABC, the center of P-Hagge circle coincides with O. 2 Results from the triads of congruent circles Three year ago, Quang Tuan Bui, the Vietnamese geometer, wrote an article about ”two triads of congruent circles from reflections” in the magazine Forum Geometricorum. In that note, he constructed two triads of congruent circles through the vertices, one associated with reflections in the altitudes, and the other reflections in the angle bisectors. In 2011, Tran Quang Hung, the teacher at High School for Gifted Student, Hanoi University of Science, also constructed another triad which associated with reflections in the lines joining the vertices and the Nine-point center. His work was published in a abook called ”Exploration and Creativity”. Here we introduce the involvement of Hagge circles in these problems. Problem 12. Given triangle ABC with its orthocenter H, circumcenter O. Let B a , C a be the reflections of B and C across the line AH. Similarly, consider the reflections C b , A b of C, A, respectively across the line BH, and A c , B c of A, B across the line CH. Then the circumcircles of three triangles AC b B c , BA c C a , CA b B a and O-Hagge circle are congruent. 10 [...]... conjugate of E) 2 The orthocenter of triangle Ia Ib Ic is the reflection of H wrt the center of N -Hagge circle (Ia , Ib , Ic are the circumcenters of three triangles ACb Bc , BAc Ca , CAb Ba , respectively) 3 N is the center of K -Hagge circle 3 Near Hagge circles Finally, we give 3 interesting problems which involve in Hagge circles Problem 15 Given triangle ABC, its orthocenter H and its circumcircle (O) Denote... triangle.) http://www.artofproblemsolving.com/Forum/viewtopic.php?t=345850 [13] Nguyen Van Linh, Hagge circle http://nguyenvanlinh.wordpress.com [14] Nguyen Van Linh, A concyclic problem, Hyacinthos message 18801 http://tech.groups.yahoo.com/group/Hyacinthos/message/18801 [15] Jean-Louis Ayme, Le P-Cercle De Hagge, Une preuve purement synthetique http://perso.orange.fr/jl.ayme Nguyen Van Linh, Foreign... Analogously, AHB2 Cb is also a isosceles trapezoid We claim (ABc Cb ) is the image of (HC2 A2 ) across the symmetry whose the axis is the perpendicular bisector of AH This means two circles (ABc Cb ) and E -Hagge circle are congruent Similarly we are done Remark 3 Other synthetic proofs of the congruent of three circles (ACb Bc ), (BAc Ca ), (CAb Ba ) can be found in [6] or [7] We have some properties: 1...Ca A C1 B1 Ba C2 O H Cb Ac B C Bc A1 Ab Note In [2], Quang Tuan Bui proved that three circles (ACb Bc ), (BAc Ca ), (CAb Ba ) are congruent with the circle (H, HO) By remark 2, HO is the radius of O -Hagge circle We are done Problem 13 Let I be the incenter of triangle ABC Consider the reflections of the vertices in the angles bisectors: Ba , Ca of B, C in AI, Ab , Cb of A, C in BI, Ac , Bc of A, B... Applying lemma 6 we get (HA3 , HB3 ) ≡ (C3 A3 , C3 B3 )(modπ) or H, A3 , B3 , C3 are concyclic Our proof is completed then 16 References [1] Christopher J Bradley and Geoff C.Smith, On a Construction of Hagge, Forum Geom., 7 (2007) 231-247 [2] Quang Tuan Bui, Two triads of congruent circles from reflections, Forum Geom., 8 (2008) 7-12 [3] Nguyen Van Linh, Two similar geometry problems http://nguyenvanlinh.wordpress.com... AE Similarly, consider the reflections Cb , Ab of C, A, respectively across the line BE, and Ac , Bc of A, B across the line CE Then the circumcircles of three triangles ACb Bc , BAc Ca , CAb Ba and E -Hagge circle are congruent Proof 11 Cb B1 A Bc Oc A2 C1 J C2 E O Q H C B B2 A1 Denote A1 B1 C1 the circumcevian triangle of E A2 , B2 , C2 the reflections of A1 , B1 , C1 across BC, CA, AB, respectively . named Hagge circles. In this paper, we present the new insight into Hagge circles with many corollaries. Furthermore, we also introduce three problems which are similar to Hagge circles. 1 Hagge. the center of N -Hagge circle (I a , I b , I c are the circumcenters of three triangles AC b B c , BA c C a , CA b B a , respectively). 3. N is the center of K -Hagge circle. 3 Near Hagge circles Finally,. circumcircle (O). Let P and Q be two isogonal conjugate points, O 1 , O 2 be the circumcenters of P -Hagge and Q -Hagge circles, respectively. Then P Q  O 1 O 2 and P Q = O 1 O 2 . Proof. Lemma 4. Given