INTERNATIONAL JOURNAL OF GEOMETRY Vol. 3 (2014) No. 2 74-80 A SYNTHETIC PROOF OF A. MYAKISHEV S GENERALIZATION OF VAN LAMOEN CIRCLE THEOREM AND AN APPLICATION OAI THANH DAO Abstract. In this article we give a synthetic proof of A.Myakishev s generalization of van Lamoen s circle theorem and introduce a family six circumcenters lie on circle of a triangle associated with Kiepert s con guration. 1. Introduction The famous van Lamoen circle theorem states that: If a triangle is divided by its three medians into 6 smaller triangles then the circumcenters of these smaller triangles lie on a circle. The Van Lamoen circle be introduced in Floor van Lamoen Problem 10830 American Mathematical Monthly 107 (2000) 863 (see [5]); solution by the editors 109 (2002) 396-397. The proof of van Lamoen circle theorem can be found in many texts see [3] [4] [6] or [8]. In 2002 A.Myakishev s generalization of van Lamoen circle theorem as follows: Theorem 1.1 (A. Myakishev-[7]). Let two triangles ABC and A 1 B 1 C 1 perpective and its have same the centroid if the perpector of two triangles is D then circumcenter of six triangles ADB 1 B 1 DC CDA 1 A 1 DB BDC 1 C 1 DA lie on a circle. The rst proof of A. Myakishev s theorem by Darij Grinberg see [2]. In the paper we give another synthetic proof A. Myakishev s theorem and in the application we show that exist a family six circumcenters lie on a circle of a triangle associated with Kiepert s con guration. Keywords and phrases: van Lamoen circle Triangle six circumcenter lie on a circle (2010)Mathematics Subject Classi cation: 51P99 60A99 Received: 25.06.2014. In revised form: 08.09.2014. Accepted: 30.09.2014.
A Synthetic proof of a Myakishev s generalization of van Lamoen circle theorem 75 2. A PROOF OF A.MYAKISHEV THEOREM We omit the proof of a easy lemma following: Lemma 2.1. Let four circles (O 1 ); (O 2 ); (O 3 ); (O 4 ) concurrent at D. The circle (O i ) meets (O i+1 ) again at D i(i+1) with i = 1; 2; 3; 4 and (O 5 ) (O 1 ). Then O 1 ; O 2 ; O 3 ; O 4 lie on a circle if only if \D 23 DD 12 = \D 34 DD 41 (mod): Figure 1 Lemma 2.2. Let ACA 1 C 1 be a quadrilateral AA 1 meets CC 1 at D. Let N; N 1 be the midpoint of AC; A 1 C 1 respectively. NN 1 meets AD at F. The circle (ADC 1 ) meets the circle (CDA 1 ) again at Q. Then \AF N=\QDC (mod ) Figure 2 Proof. Since \C 1 AQ = \C 1 DQ = \CA 1 Q (mod ) and \AC 1 Q = \A 1 DQ = \A 1 CQ (mod ) hence two triangles AQC 1 ; A 1 QC are similar ) C 1Q AC 1 = CQ A 1 C.
76 Oai Thanh Dao (1)! Let E be the midpoint of AA 1. We obtain AC 1 = 2EN 1 A 1 C = 2 EN! ) C 1 Q = CQ EN 1 EN On the other hand: \C 1 QC = \C 1 QD+\DQC (mod ). But \C 1 QD = \C 1 DA (mod ) = \( AC 1 ; DA) and \DQC = \DA 1 C (mod ) = \( A 1 D; A 1 C). Thus: \C 1 QC = \( AC 1 ; DA) + \( A 1 D; A 1 C) = (2) \( AC 1 ; A 1 C) = \( EN 1 ; EN) = \N 1 EN( mod ) Since (1) and (2) we get that two triangles C 1 QC; N 1 EN are similar since (3) \ENN 1 = \QCC 1 We obtain: \AF N = \F EN + \ENF (mod ). On the other hand: \F EN = \DA 1 C (mod ). And since (3) we obtain \ENF = \QCD = \QA 1 D (mod ). Therefore \AF N = \QA 1 D + \DA 1 C = \QA 1 C = \QDC (mod ). The completes the proof of Lemma 2.2. Proof of A. Myakishev theorem: Figure 3 Let A c ; B c ; B a ; C a ; C b ; A b are circumcenter of six circles (ADC 1 ) (C 1 DB) (BDA 1 ) (A 1 DC) (CDB 1 ) (B 1 DA) respectively. Let N; N 1 be the midpoint of AC; A 1 C 1 respectively and NN 1 meets AA 1 at F. The circle (ADC 1 ) meets (CDA 1 ) again at Q. Easily we deduce that BB 1 NN 1 are parallel therefor \ADB 1 = \AF N (mod ). By Lemma 2.2 we obtain \AF N = \QDC. Thus \ADB 1 = \QDC. By Lemma 2.1 we get that four circumcenters A c ; A b ; C b ; C a lie on a circle. Similarly four circumcenters A b ; C b ; C a ; B a lie on a circle and C b ; C a ; B a ; A b lie on a circles. Hence six circumcenters A c ; B c ; B a ; C a ; C b ; A b lie on a circle. This completes the proof of Myakishev s theorem.
A Synthetic proof of a Myakishev s generalization of van Lamoen circle theorem 77 3. AN APPLICATION OF A.MYAKISHEV THEOREM Theorem 3.1 (Dao-[1]). Let ABC be a triangle G is the centroid. Constructed three similar isosceles triangles AC 0 B; BA 0 C; CB 0 A (either all outward or all inward). Let A 1 ; B 1 ; C 1 lie on AA 0 ; BB 0 ; CC 0 respectively such that: (4) (5) AA 1 AA 0 = BB 1 BB 0 = CC 1 CC 0 = k 1 Let A 2 ; B 2 ; C 2 lie on GA 1 ; GB 1 ; GC 1 respectively such that: GA 2 GA 1 = GB 2 GB 1 = GC 2 GC 1 = k 2 Then AA 2 ; BB 2 ; CC 2 are concurrent at a point K lie on Kiepert hyperbola. And circumcenter of six triangles AKB 2 B 2 KC CKA 2 A 2 KB BKC 2 C 2 KA lie on a circle. - When A 0 ; B 0 ; C 0 at midpoint of BC; CA; AB respectively and k 1 = k 2 = 1 this circle is called as the van Lamoen circle. - When A 0 ; B 0 ; C 0 at midpoint of BC; CA; AB respectively and k 2 = 0 and k 1 is any real number this circle is called as the Dao six point circle [4]. Proof. By (4) we get Figure 4 (6) AA 0 + A 0 A 1 AA 0 = BB 0 + B 0 B 1 BB 0 = CC 0 + C 0 C 1 CC 0
78 Oai Thanh Dao (7) A 0 A = B 0B = C 0C A 0 A 1 B 0 B 1 C 0 C 1 (8) A 0 A 1 + A 1 A A 0 A 1 = B 0B 1 + B 1 B B 0 B 1 = C 0C 1 + C 1 C C 0 C 1 (9) A 1 A 0 A 1 A = B 1B 0 B 1 B = C 1C 0 C 1 C Figure 5 Denote A 3 ; B 3 ; C 3 are midpoint of BC; CA; AB repectively. Let AA 2 ; BB 2 ; CC 2 meet A 3 A 0 ; B 3 B 0 ; C 3 C 0 at A 4 ; B 4 ; C 4 respectively. Let GA 1 ; GB 1 ; GC 1 meet A 3 A 0 ; B 3 B 0 ; C 3 C 0 at A 5 ; B 5 ; C 5 respectively. By Menelaus theorem for 4AA 3 A 0 cut by A 1 A 5 G we obtain: (10) Similarly we obtain: A 5 A 0 = GA A 1A 0 A 5 A 3 GA 3 A 1 A = 2A 1A 0 A 1 A (11) B 5 B 0 = 2 B 1B 0 B 5 B 3 B 1 B (12) C 5 C 0 = 2 C 1C 0 C 5 C 3 C 1 C Since (10)(11)(12) and (9) we obtain:
A Synthetic proof of a Myakishev s generalization of van Lamoen circle theorem 79 (13) A 5 A 0 A 5 A 3 = B 5B 0 B 5 B 3 = C 5C 0 C 5 C 3 By Menelaus theorem for 4AGA 1 cut by A 0 A 5 A 3 we obtain: A 5 G (14) = A 3G A 5 A 1 A 3 A A 0A = 1 A 0 A 1 3 A 0A A 0 A 1 Similarly we obtain: (15) B 5 G B 5 B 1 = 1 3 B 0B B 0 B 1 C 5 G (16) = 1 C 5 C 1 3 C 0C C 0 C 1 Since (14)(15)(16) and (7) we get: (17) A 1 A 5 GA 5 = B 1B 5 GB 5 = C 1C 5 GC 5 (18) A 1 G + GA 5 GA 5 = B 1G + GB 5 GB 5 = C 1G + GC 5 GC 5 GA 1 (19) = GB 1 = GC 1 GA 5 GB 5 GC 5 Since (19) we have: A 1 B 1 k A 5 B 5 ; B 1 C 1 k B 5 C 5 ; C 1 A 1 k C 5 A 5. And since (5) we have: A 1 B 1 k A 2 B 2 ; B 1 C 1 k B 2 C 2 ; C 1 A 1 k C 2 A 2. on the other hand G; A 1 ; A 2 ; A 5 are collinear G; B 1 ; B 2 ; B 5 are collinear; G; C 1 ; C 2 ; C 5 are collinear. Now by Thales theorem we obtain: (20) A 2 A 5 A 2 A 1 = B 2B 5 B 2 B 1 = C 2C 5 C 2 C 1 By Menelaus theorem for 4A 5 A 0 A 1 cut by AA 2 A 4 we obtain: (21) Similarly we get: A 4 A 5 A 4 A 0 = AA 1 AA 0 A 2A 5 A 2 A 1 (22) (23) B 4 B 5 B 4 B 0 = BB 1 BB 0 B 2B 5 B 2 B 1 C 4 C 5 C 4 C 0 = CC 1 CC 0 C 2C 5 C 2 C 1
80 Oai Thanh Dao Since (4)(20)(21) (22) and (23) we have: A 4 A 5 (24) = B 4B 5 = C 4C 5 A 4 A 0 B 4 B 0 C 4 C 0 Three triangles BC 0 A AB 0 C and CA 0 B are three similar isosceles triangle either all outward or all inward on the sides 4ABC. Since (13) we get that three triangles BA 5 C; CB 5 A; AC 5 B are similar isosceles triangle. Since (25) we get that three triangles BC 4 A AB 4 C CA 4 B are similar isosceles triangle on the sides. By famous Kiepert theorem we have the lines AA 4 BB 4 CC 4 concurrent on Kiepert hyperbola so AA 2 ; BB 2 ; CC 2 concurrent on Kiepert hyperbola. It is well-known that two triangles ABC A 0 B 0 C 0 have same the centroid. Since (4) we can show that two triangles A 0 B 0 C 0 and A 1 B 1 C 1 have same the centroid. Since (5) we can show that two triangles A 1 B 1 C 1 and A 2 B 2 C 2 have same the centroid. Therefore two triangles ABC and A 2 B 2 C 2 have same the centroid. By Myakishev theorem we get that the circumcenter of six triangles AKB 2 B 2 KC CKA 2 A 2 KB BKC 2 C 2 KA lie on a circle. This completes the proof of Theorem 3.1. References [1] Dao T. Dao Thanh Oai problem Art of Problem Solving Oct. 24 2013. available at http://www.artofproblemsolving.com/forum/viewtopic.php?f=47t=559546 [2] Grinberg D. Hyacinthos message 7351 Jul 13 2003. [3] Ha N. M. Another Proof of van Lamoen s Theorem and Its Converse Forum Geom. 5(2005) 127-132. [4] Kimberling C. X(5569) Center of the Dao six point circle Nov. 3 2013. Encyclopedia of Triangle Centers available at http://faculty.evansville.edu/ck6/encyclopedia/etc.html. [5] Lamoen F. M. Problem 10830 Amer. Math. Monthly 107(2000) 893. [6] Li K. Concyclic problem Mathematical Excalibur 6(1)(2001) 1-2. [7] Myakishev A. and Wolk B. Hyacinthos message 5612 May 31 2002. [8] Myakishev A. and Woo P. On the Circumcenters of Cevasix Con guration Forum Geom. 3(2003) 57-63. SON LA HYDROPOWER COMPANY VIETNAM E-mail address: daothanhoai@hotmail.com