Orthopoles, Flanks, and Vecten Points

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1 Forum Geometricorum Volume 17 (2017) FORUM GEOM ISSN Orthopoles, Flanks, and Vecten Points Todor Zaharinov Abstract. Erect a square outwardly (or inwardly) from each side of the triangle. For the flank lines construct orthopoles. The lines from this orthopoles to corresponding vertexes of triangle concur in the Vecten point (or in the Inner Vecten point). This fact is showing together with a number of other interesting results, which include complex coordinates of Vecten points and orthopole. 1. Introduction We will use complex numbers in the proofs. Denote by a lowercase letter the complex coordinate of a point denoted by an uppercase letter. Let ABC be a triangle. With no loss of generality, we can say that the circumcirclec is the union circle. Thena.ā = b. b = c. c = The Outer Vecten point. Theorem 1. Erect a square outwardly from each side of triangle ABC. Let A o B o C o be the triangle formed by the respective centers of the squares. The lines AA o,bb o,cc o concur inx 485. [1] Figure 1. The Outer Vecten point Point X 485 is the Outer Vecten point. Proof. Let the squares are: AC a C b B, BA b A c C, and CB c B a A with centers C o, A o,b o respectively (Figure 1). Publication Date: November 28, Communicating Editor: Paul Yiu.

2 402 T. Zaharinov Then a b = b+(b c)i, b c = c+(c a)i, c a = a+(a b)i, ā b = c+(b c)i bc = a+(c a)i ca, c a = b+(a b)i. The coordinates ofa o,b o,c o and its complex conjugates are bc, a o = 1 2 (a b +c) = 1 ((b+c)+(b c)i) (1) 2 ā o = (b+c)+(b c)i 2bc b o = 1 2 (b c +a) = 1 2 ((c+a)+(c a)i) bo = (c+a)+(c a)i 2ca c o = 1 2 (c a +b) = 1 2 ((a+b)+(a b)i) c o = (a+b)+(a b)i 2 The lineaa o, joining a and a o has equation z a a o 0 = z ā ā o = (ā o ā)z (a o a) z +(a o ā ā o a) = āo ā a o a z z + a oā ā o a a o a LinesBB o and CC o have equations 0 = ( b o b)z (b o b) z +(b o b bo b) = b o b b o b z z + b o b b o b b o b 0 = ( c o c)z (c o c) z +(c o c c o c) = c o c c o c z z + c o c c o c c o c The intersection of the last two lines is given by ( bo b 0 = b o b c ) ( o c bo b bo b z + c ) o c c o c c o c b o b c o c = 2(a b)(b c)(c a) i(a(b c)2 +b(c a) 2 +c(a b) 2 ) z + (a b)(b c)(c a)(a+b+c)+i(a2 (b c) 2 +b 2 (c a) 2 +c 2 (a b) 2 ) z = (a b)(b c)(c a)(a+b+c)+i(a2 (b c) 2 +b 2 (c a) 2 +c 2 (a b) 2 ) 2(a b)(b c)(c a)+i(a(b c) 2 +b(c a) 2 +c(a b) 2 ) Note that this is symmetric ina,b,c. This means that the three linesaa o,bb o, CC o are concurrent. The point of concurrency is the Vecten point X 485. (2) (3) (4)

3 Orthopoles, flanks, and Vecten points 403 Corollary 2. Let the circumcircle of triangle ABC be the unit circle. The Outer Vecten point X 485 has coordinates x 485 = (a b)(b c)(c a)(a+b+c)+i(a2 (b c) 2 +b 2 (c a) 2 +c 2 (a b) 2 ) 2(a b)(b c)(c a)+i(a(b c) 2 +b(c a) 2 +c(a b) 2 ) (5) 1.2. The Inner Vecten point. Theorem 3. Erect a square inwardly from each side of triangleabc. LetA ob oc o be the triangle formed by the respective centers of the squares. The lines AA o, BB o, CC o concur inx 486. [1] Point X 486 is the Inner Vecten point. Figure 2. The Inner Vecten point Proof. (See Figure 2) Let the squares are: AB ab cc,ca ca b B, andbc b C aa with centersb o,a o,c o respectively. Then a b = b (b c)i, b c = c (c a)i, c a = a (a b)i, ā b = c (b c)i b c = a (c a)i ca, c a = b (a b)i. The coordinates ofa o,b o,c o and its complex conjugates are bc, a o = 1 2 (a b +c) = 1 ((b+c) (b c)i) (6) 2 ā o = (b+c) (b c)i 2bc b o = 1 2 (b c +a) = 1 2 ((c+a) (c a)i) b o = (c+a) (c a)i 2ca

4 404 T. Zaharinov c o = 1 2 (c a +b) = 1 2 ((a+b) (a b)i) c o = (a+b) (a b)i. 2 The lineaa o, joining a anda o has equation z a a o 0 = z ā ā o = (ā o ā)z (a o a) z +(a oā ā oa) (7) LinesBB o andcc o have equations 0 = ( b o b)z (b o b) z +(b o b b ob) = b o b b o b z z + b o b b ob b o b 0 = ( c o c)z (c o c) z +(c o c c oc) = c o c c o c z z + c o c c oc c o c The intersection of the last two lines is given by ) ( ( b 0 = o b b o b c o c b c z + o b b o b o c b c o c c ) oc o b c o c = 2(a b)(b c)(c a)+i(a(b c)2 +b(c a) 2 c(a b) 2 ) z + (a b)(b c)(c a)(a+b+c) i(a2 (b c) 2 +b 2 (c a) 2 +c 2 (a b) 2 ) z = (a b)(b c)(c a)(a+b+c) i(a2 (b c) 2 +b 2 (c a) 2 +c 2 (a b) 2 ) 2(a b)(b c)(c a) i(a(b c) 2 +b(c a) 2 +c(a b) 2 ) This is symmetric in a,b,c and means that the three lines AA o,bb o,cc o are concurrent. The point of concurrency is the Inner Vecten point X 486. Corollary 4. Let the circumcircle of triangle ABC be the unit circle. The Inner Vecten point X 486 has coordinates x 486 = (a b)(b c)(c a)(a+b+c) i(a2 (b c) 2 +b 2 (c a) 2 +c 2 (a b) 2 ) 2(a b)(b c)(c a) i(a(b c) 2 +b(c a) 2 +c(a b) 2 ) (10) 1.3. The Orthopole. Lemma 5. Let V and W be points of the plane. The orthogonal projection P 1 of a point P onto the linevw is given by p 1 = ( w v)p+(w v) p (w v wv) 2( w v) (8) (9) (11)

5 Orthopoles, flanks, and Vecten points 405 Proof. The linewv contains a point with coordinates p 1 if and only if p 1 p = w w 1 w v = v v 1 w v p w v wv 1 p 1 + w v P 1 P WV : w v 0 = (p 1 p)( w v)+( p 1 p)(w v) = w v p 1 + p 1 w v p p w v Adding the last two expressions, receive w v 2 w v p w v w v wv 1 p p+ = 0 w v w v Dividing by2 w v w v, we obtainp 1, as given in (11) ove. Remark. [7, Lemma 3] In casev andw be points on the unit circle, p 1 = 1 2 (v +w +p vw p) ; p 1 = 1 (v +w p+vw p) (12) 2vw Lemma 6. Let the circumcircle of triangle ABC be the unit circle. Let EF be a arbitrary line in the plane. LetA 1 be the orthogonal projection ofaonto lineef. The perpendicular froma 1 tobc has equation z bc f) (e f)) z+a(bc(ē + 2bc(e f) bc(ē f) (e f) (ē f) + (bc(ē f)+(e f))(e f ēf) 2bc(e f)(ē f) = 0 (13) Proof. By Lemma 5, the coordinatesa 1 of A 1 and its complex conjugate are 1 ( 1 a 1 = 2(ē f) (ē f)a+(e f) a (e f ēf) ) 1 ( 1 ā 1 = (ē f)a+(e f) 2(e f) a +(e f ēf) ) By (12), (Remark to Lemma 5), the coordinate a 2 of the orthogonal projection A 2 ofa 1 ontobc, together with its conjugate, are a 2 = 1 2 (b+c+a 1 bcā 1 ) ā 2 = 1 2bc (b+c a 1 +bcā 1 ) The linea 1 A 2 contains a point with coordinates z if and only if z a 1 a 2 0 = z ā 1 ā = ā1 ā 2 z z + a 1ā 2 ā 1 a 2 = 1 a 1 a 2 a 1 a 2 bc z z + (a2 (ē f)+(e f))(bc(ē f) (e f))+a(e f ēf)(bc(ē f)+(e f)) (e f)(ē f) (14) This is equal to (13).

6 406 T. Zaharinov Theorem 7 (Orthopole, [2, p 247, Theorem 406]). If perpendiculars are dropped on any line from the vertices of a triangle, the perpendiculars to the opposite sides from their feet are concurrent at a point called the orthopole of the point. Figure 3. The Orthopole Proof. Let EF be a arbitrary line of the plane. Let A 1,B 1,C 1 be the orthogonal projections from vertexesa,b,c respectively to the line EF (Figure 3). We consider the construction in Lemma 6 beginning with all three vertices of triangleabc. This results in the three lines z bc z ca z f) (e f)) z+a(bc(ē + 2bc(e f) f) (e f)) z+b(ca(ē + 2ca(e f) f) (e f)) z+c((ē + 2(e f) The intersection of the last two lines is z(b c) + c c(b c)(ē f) (e f) Multiplying by c b c, we obtain bc(ē f) (e f) (ē f) + (bc(ē f)+(e f))(e f ēf) 2bc(e f)(ē f) = 0 ca(ē f) (e f) (ē f) + (ca(ē f)+(e f))(e f ēf) 2ca(e f)(ē f) = 0 (ē f) (e f) (ē f) + ((ē f)+(e f))(e f ēf) 2(e f)(ē f) = 0 (b2 c 2 )(e f) (e f) a(b c)(ē f) (ē f) + (b c)(e f)(e f ēf) (e f)(ē f) = 0 z = 1 c(ē f) (a+b+c) e f ēf 2 2(e f) 2(ē f) Note that this is symmetric in a,b,c. This means that the three perpendiculars froma 1 tobc,b 1 toca, andc 1 toab are concurrent. The point of concurrency is the orthopoleqof the lineef with respect to triangleabc.

7 Orthopoles, flanks, and Vecten points 407 Corollary 8. Let the circumcircle of triangle ABC be the unit circle. The orthopole Q of the line EF with respect to triangle ABC has complex coordinate and conjugate q = 1 c(ē f) (a+b+c) e f ēf 2 2(e f) 2(ē f) q = +bc+ca 2. Orthopoles and flank lines e f (ē f) + e f ēf 2(e f) Theorem 9. Erect a square outwardly from each side of triangleabc: AC a C b B, BA b A c C and CB c B a A with centers C o, A o, B o respectively. For flank lines B a C a,c b A b,a c B c construct the orthopoles Q a, Q b, Q c respectively. The lines AQ a,bq b,cq c concur in Vecten point X 485. The proof of the theorem relies on the following lemma. Lemma 10. The orthopoleq a of the flank lineb a C a with respect to triangleabc and points A,A o are collinear. (15) Proof. (See Figure 1) With no loss of generality we shall say that the circumcircle of triangle ABC is the union circle. b a = a + (c a)i, c a = a + (a b)i, ba = c+(c a)i ac, c a = b+(a b)i By Corollary 8 find the orthopoleq a as substituteb a,c a for e,f in (15). q a = 1 2 (a+b+c) c( b a c a ) 2(b a c a ) b a c a b a c a 2( b a c a ) q a = 1 2 (a+b+c) a(b+c) 2bc +c(a b) 2 i(a b)(b c)(c a) 2(2a b c) +b(c a)2 2(a(b+c) 2bc) (16) q a = +bc+ca 2a b c +c(a b) 2 i(a b)(b c)(c a) + 2(a(b+c) 2bc) b(c a)2 (2a b c) (17) The pointsa,a o,q a are collinear if and only if ([5, p. 65], [6, p. 37]) Use (16), (17), (1) andā = 1/a. q a a a o a = η R i.e. q a a a o a = q a ā ā o ā (q a a)(ā o ā) ( q a ā)(a o a) = (a b)(c a)(bc a2 )(b c) 2 (i 2 +1) (b+c 2a)(2bc ca) = 0

8 408 T. Zaharinov Proof of Theorem 9. (See Figure 1) By Lemma 10 applied to all three vertices of triangle ABC: Q a AA o, Q b BB o and Q c CC o. But from Theorem 1, the linesaa o,bb o,cc o concur in the Vecten point X 485. Theorem 11. Erect a square inwardly from each side of triangleabc : BC b C aa, CA ca b B and AB ab cc with centers C o, A o, B o respectively. For flank lines B ac a, C b A b, A cb c construct the orthopoles Q a, Q b, Q c respectively. The lines AQ a,bq b,cq c concur in Inner Vecten point X 486. Lemma 12. The orthopoleq a of the flank lineb ac a with respect to triangleabc and points A,A o are collinear. (See Figure 2) Proof. The proof is the same as the proof of Lemma 10. With no loss of generality we shall say that the circumcircle of triangleabc is the union circle. b a = a (c a)i,c a = a (a b)i, b a = c (c a)i ca, c a = b (a b)i q a = 1 2 (a+b+c) a(b+c) 2bc +c(a b) 2 +i(a b)(b c)(c a) 2(2a b c) +b(c a)2 2(a(b+c) 2bc) (18) Proof of Theorem 11. (See Figure 2) By Lemma 12 applied to all three vertices of triangle ABC: Q a AA o, Q b BB o and Q c CC o. But from Theorem 3, the linesaa o,bb o,cc o concur in the Inner Vecten point X 486. Lemma 13. [4, p. 105] Erect squares outwardly from sides of triangle ABC: AC a C b B, BA b A c C andcb c B a A with centersc o,a o,b o. Erect a squareb a B a1 C a1 C a with centera o1 outwardly fromb a C a. The points A o1,a o and A are collinear. Proof. With no loss of generality, we can say that the circumcircle C of triangle ABC is the union circle. b a = a + (c a)i, c a = a + (a b)i, b a = c+(c a)i ac, c a = b+(a b)i b a1 = b a +(b a c a )i = 3a b c+(c a)i ba1 = 3bc ca +(c a)i c a o1 = 1 2 (c a +b a1 ) = 1 2 (4a b c (b c)i) ā o1 = 4bc ca a(b c)i. The points A,A o,a o1 are collinear if and only if ([5, p. 65], [6, p. 37]) a o1 a a o a = η R i.e. a o1 a a o a = āo1 ā ā o ā

9 Orthopoles, flanks, and Vecten points 409 Figure 4. (a o1 a)(ā o ā) (a o a)(ā o1 ā) = 2a b c (b c)i 2 2bc ca a(b c)i a(b+c)+a(b c)i 2bc b+c 2a+(b c)i 2 = 0 Theorem 14. Erect a square outwardly from each side of triangleabc: AC a C b B, BA b A c C and CB c B a A. Construct the orthopoles Q a1 for BC with respect to flank triangle AB a C a ; Q b1 for CA with respect to flank triangle BC b A b ; Q c1 for AB with respect to flank triangle CA c B c. The lines AQ a1, BQ b1, CQ c1 concur in Vecten point X 485. Proof. (See Figure 4) LetC o,a o,b o be centers of the squaresac a C b B,BA b A c C, andcb c B a A respectively. Erect squaresb a B a1 C a1 C a,c b C b1 A b1 A b,a c A c1 B c1 B c outwardly fromb a C a,c b A b,a c B c with centers A o1,b o1,c o1 respectively. Triangle ABC is a flank triangle for triangle AB a C a. From Lemma 10 the orthopole Q a1 of the flank line BC with respect to triangle AB a C a and points A,A o1 are collinear. From Lemma 14 points A o1,a o and A are collinear. Follow Q a1,a o1,a,a o are collinear. By analogy Q b1,b o1,b,b o and Q c1,c o1,c,c o are collinear. From Theorem 1 the last three lines concur in the Vecten point X 485.

10 410 T. Zaharinov References [1] C. Kimberling, Encyclopedia of Triangle Centers, [2] R. A. Johnson, Advanced Euclidean Geometry, New York, 1960 [3] F. M. van Lamoen, Friendship Among Triangle Centers, Forum Geom., 1 (2001) 1 6. [4] Jean-Louis Ayme, La Figure De Vecten [5] T. Andreescu and D. Andrica, Complex Numbers from A to... Z, Birkhäuser, Boston, [6] R. Malcheski, S. Grozdev, and K. Anevska, Geometry Of Complex Numbers, Sofia, Bulgaria, [7] T. Zaharinov, The Simson triangle and its properties, Forum Geom., 17 (2017) Todor Zaharinov: Darvenitsa j.k., 1756 Sofia, Bulgaria address: zatrat@v.bg

The Simson Triangle and Its Properties

Forum Geometricorum Volume 7 07 373 38 FORUM GEOM ISSN 534-78 The Simson Triangle and Its Properties Todor Zaharinov bstract Let be a triangle and P P P 3 points on its circumscribed circle The Simson