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The Simson Triangle and Its Properties

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The Simson Triangle and Its Properties Forum Geometricorum b Volume 17 (2017) 373–381. b b FORUM GEOM ISSN 1534-1178 The Simson Triangle and Its Properties Todor Zaharinov Abstract. Let ABC be a triangle and P1, P2, P3 points on its circumscribed circle. The Simson triangle for P1, P2, P3 is the triangle bounded by their Sim- son lines with respect to triangle ABC. We study some interesting properties of the Simson triangle. 1. Introduction The following theorem is often called Simson’s theorem. (see [2, p. 137, Theo- rem 192]) Theorem 1 (Wallace-Simson line). The feet of the perpendiculars to the sides of triangle from a point are collinear, if and only if the point is on the circumscribed circle of the triangle. This is Simson line (or Wallace-Simson line). Definition (Simson triangle). The Simson triangle is the triangle bounded by the Simson lines of three points on the circumscribed circle of a fixed triangle. It is degenerate if the three Simson lines are concurrent. ℓ3 A N1 P2 K P3 N3 ℓ1 B C N2 ℓ2 P1 Figure 1 Theorem 2 (Orthopole, (see [2, p. 247, Theorem 406])). If perpendiculars are dropped on a 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 line. Publication Date: June 28, 2017. Communicating Editor: Paul Yiu. 374 T. Zaharinov 2. Notations ABC is a triangle of reference. The circumcircle K of △ABC has center O and radius R. P1 ∈K, P2 ∈K, P3 ∈K. l1, l2, l3 are the Simson lines of P1, P2, P3 with respect to ABC. N1N2N3 is the Simson triangle bounded by these Simson lines: N1 = l2 ∩ l3, N2 = l3 ∩ l1, N3 = l1 ∩ l2. The circumcircle of △N1N2N3 is K2 with center O2 and radius R2. The orthocenter of △ABC is H. The orthocenter of △P1P2P3 is H1. The orthocenter of △N1N2N3 is H2. The center of nine-point circle for ABC is E. We will use complex numbers in the proofs. By u we shall denote the complex number, corresponding to point U. Without loss of generality, we take the circumcircle K to be the unit circle. Then R = 1 and O = 0. a · a¯ = b · ¯b = c · c¯ = p1 · p¯1 = p2 · p¯2 = p3 · p¯3 = 1. h = a + b + c; e = 1/2(a + b + c); h1 = p1 + p2 + p3. Lemma 3. Let V and W be points on the unit circle. The orthogonal projection of a point P onto the line ℓ = VW is given by 1 p = (v + w + p − vwp¯). ℓ 2 In particular, if P is also on the unit circle, then 1 vw pℓ = v + w + p − . 2 p Proof. Write the orthogonal projection as pℓ = (1−t)v+tw for some real number t. The vector p − (1 − t)v − tw is perpendicular to BC. This means that (p − (1 − t)v − tw)(¯v − w¯)+(¯p − (1 − t)¯v − tw¯)(v − w) = 0. 1 1 Since v and w are on the unit circle, v¯ = v , w¯ = w . We have 1 1 1 − t t (p − (1 − t)v − tw) − + p¯ − − (v − w) = 0. v w v w From this, v − w − p + vwp¯ t = , 2(v − w) and the orthogonal projection is 1 p = (1 − t)v + tw = (v + w + p − vwp¯). ℓ 2 1 1 bc If P is also on the unit circle, then p¯ = p , and pℓ = 2 v + w + p − p . The Simson triangle and its properties 375 Proposition 4. Let P be a point on the unit circumcircle of triangle ABC. The equation of its Simson line is abc 2abcz¯ − 2pz + p2 +(a + b + c)p − (bc + ca + ab) − = 0. (1) p Proof. Let P be a point on the unit circumcircle. Its projections on the side lines of triangle ABC are, by Lemma 3, 1 bc pa = b + c + p − , 2 p 1 ca pb = c + a + p − , 2 p 1 ab pc = a + b + p − . 2 p The line joining pb and pc has equation 1 ca 1 ab z pb pc z 2 c + a + p − p 2 a + b + p − p 0= z¯ p¯b p¯c = 1 1 1 1 p 1 1 1 1 p z¯ 2 c + a + p − ca 2 a + b + p − ab 1 1 1 1 1 1 (b − c)(p − a)(2 abcpz¯ − 2p2z + p3 +(a + b + c)p2 − (bc + ca + ab)p − abc) = 4abcp2 2 abc (b − c)(p − a)(2abcz¯ − 2pz + p +(a + b + c)p − (bc + ca + ab) − p ) = . 4abcp Therefore, the equation of the Simson line of P is given by (1) above. Proposition 5. The intersection of the Simson lines of two points P , Q ∈K is the point with coordinates 1 abc p + q + a + b + c + . (2) 2 pq Proof. Let ℓp, ℓq be the Simson lines of two points P , Q on the unit circumcircle of ABC. Their intersection is given by the solution of abc 2abcz¯ − 2pz + p2 +(a + b + c)p − (bc + ca + ab) − = 0, (3) p abc 2abcz¯ − 2qz + q2 +(a + b + c)q − (bc + ca + ab) − = 0. (4) q Subtracting (4) from (3), we obtain 1 1 −2(p − q)z +(p2 − q2)+(a + b + c)(p − q) − abc − = 0. p q Dividing by 2(p − q), we obtain z as given in (2) above. 376 T. Zaharinov 3. Simson triangle Theorem 6. The Simson triangle N1N2N3 is directly similar to triangle P1P2P3 (see Figure 1). Proof. For three points P1, P2, P3 on K, by Proposition 5, the pairwise intersec- tions of their Simson lines are 1 abc n1 = p2 + p3 + a + b + c + , 2 p2p3 1 abc n2 = p3 + p1 + a + b + c + , 2 p3p1 1 abc n3 = p1 + p2 + a + b + c + , 2 p1p2 the vertices of the Simson triangle. Note that 1 abc 1 abc n2 − n3 = p3 + p1 + a + b + c + − p1 + p2 + a + b + c + 2 p3p1 2 p1p2 1 abc(p2 − p3) = p3 − p2 + 2 p1p2p3 abc − p1p2p3 = (p2 − p3). 2p1p2p3 abc−p1p2p3 Since the factor k := 2p1p2p3 is symmetric in p1, p2, p3, we conclude that N2N3 = k · P2P3, N3N1 = k · P3P1, N1N2 = k · P1P2, and the triangles N1N2N3 and P1P2P3 are directly similar. |abc−p1p2p3| Corollary 7. The Simson triangle N1N2N3 has circumradius 2 , and circumcenter at the midpoint of the segment joining the orthocenters H of ABC and H1 of P1P2P3 (see Figure 2). Proof. Since the Simson triangle is similar to P1P2P3 with similarity factor k = |abc−p1p2p3| 2 , it clearly has circumradius k. The orthocenters of triangles ABC and P1P2P3 are the points h = a + b + c and h1 = p1 + p2 + p3. With these, we rewrite h1 + h 1 abc h1 + h 1 abc − p1p2p3 n1 = + − p1 = + . 2 2 p2p3 2 2 p2p3 Therefore, h1 + h 1 abc − p1p2p3 n1 − = = k, 2 2 p2p3 since |p2| = |p3| = 1 . The same relation holds if n1 is replaced by n2 and n3. This h1+h shows that the Simson triangle has circumcenter 2 , which is the midpoint of H1 and H. It also confirms independently that the circumradius is k. The Simson triangle and its properties 377 ℓ3 A N1 P2 K P3 H1 H O2 N3 ℓ1 B C N2 ℓ2 P1 Figure 2 4. The Simson triangle and orthopoles Lemma 8. Let V and W be points on the unit circumcircle of triangle ABC, A1 the orthogonal projection of A onto VW . The perpendicular from A1 to BC has equation z a − (v + w) (a2 − bc)+ a(v + w) − vw z¯ − + + = 0. (5) bc 2vw 2abc Proof. By Lemma 3, the coordinates a1 of A1 and its complex conjugate are 1 vw a1 = v + w + a − , 2 a −a2 + a(v + w)+ vw a¯1 = . 2avw By Lemma 3 again, the coordinates a2 of the orthogonal projection A2 of A1 onto BC, together with its complex conjugate, are 1 a2 = (b + c + a1 − bca¯1) 2 a2bc − abc(v + w)+(a2 − bc + 2ca + 2ab)vw + avw(v + w) − v2w2 = , 4avw −a2bc + abc(v + w) − (a2 − bc − 2ca − 2ab)vw − avw(v + w)+ v2w2 a¯2 = . 4abcvw 378 T. Zaharinov The line A1A2 contains a point with coordinates z if and only if z a1 a2 0= z¯ a¯1 a¯2 1 1 1 f(a, b, c, v, w)g(a,b,c,v,w,z) = , 8a2bcv2w2 where f(a, b, c, v, w) := a2bc − abc(v + w) − (a2 + bc − 2ca − 2ab)vw − avw(v + w)+ v2w2, g(a,b,c,v,w,z):= 2abcvwz¯ − 2avwz + a2bc − abc(v + w) +(a2 − bc)vw + avw(v + w) − v2w2. Therefore, the equation of the perpendicular is g(a,b,c,v,w,z) = 0. Dividing by 2abcvw, we obtain the equation (5). Now we consider the construction in Lemma 8 beginning with all three vertices of △ABC. This results in the three lines z a − (v + w) (a2 − bc)+ a(v + w) − vw z¯ − + + = 0, bc 2vw 2abc z b − (v + w) (b2 − ca)+ b(v + w) − vw z¯ − + + = 0, ca 2vw 2abc z c − (v + w) (c2 − ab)+ c(v + w) − vw z¯ − + + = 0. ab 2vw 2abc The intersection of the last two lines is given by z z b − (v + w) c − (v + w) − + + − ca ab 2vw 2vw (b2 − ca)+ b(v + w) − vw (c2 − ab)+ c(v + w) − vw + − = 0, 2abc 2abc (b − c)z b − c (b − c)(a + b + c + v + w) − + + = 0, abc 2vw 2abc abc Multiplying by b−c , we obtain 1 abc z = a + b + c + v + w + .
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