A Construction of the Golden Ratio in an Arbitrary Triangle

Transcription

1 Forum Geometricorum Volume 18 (2018) FORUM GEOM ISSN onstruction of the Golden Ratio in an rbitrary Triangle Tran Quang Hung bstract. We have known quite a lot about the construction of the golden ratio in the special triangles. In this article, we shall establish a construction of the golden ratio in arbitrary triangle using two symmedians and give a synthetic proof for this. Given triangle inscribed in circle (ω) center O. (i) the symmedians and F. (ii) the ray F meets (ω) at the point. (iii) the perpendicular line from to O meets and at Q and R respectively (See Figure 1). R Q (ω) F O Figure 1 roposition 1. Q divides Rin the golden ratio. We give three lemmas to prove this proposition. Lemma 2. Given convex cyclic quadrilateral. Two diagonals and intersect at. Then ublication ate: June 20, ommunicating Editor: aul Yiu. The author is grateful to rofessor aul Yiu for his help in the preparation of this article.

2 240 Q. H. Tran. roof. y inscribed angles are equal in the cyclic quadrilateral, we have the similar triangles and.from this, we get the ratios (1) and From (1) and (2), we obtain This finishes the proof.. (2). Figure 2 Lemma 3 (tolemy s Theorem [1]). For a cyclic quadrilateral, the sum of the products of the two pairs of opposite sides equals the product of the diagonals. Using concept of homogeneous barycentric coordinates [9], we give and prove the following lemma Lemma 4. Let be a triangle inscribed in circle (ω). is a point inside triangle. has homogeneous barycentric coordinates (x : y : z). EF is cevian triangle of. Ray EF meets (ω) at Q. Then yq xq + zq.

3 construction of the golden ratio in an arbitrary triangle 241 (ω) E R Q F Figure 3 roof. ecause has homogeneous barycentric coordinates (x : y : z) so E(x : 0:z) and F (x : y :0). Thus we have the ratio E E z x. (3) Extend ray QE to meet (ω) again at R. From Lemma 2, we have From (3) and (4), we deduce E E Q R. (4) Q R z x. Thus, zq x R. (5) Similarly, we have the identity Q R y x. Thus, yq x R. (6) Using (5) and (6) and Lemma 3, we consider the expression

4 242 Q. H. Tran Therefore yq zq x Q R R x Q R R R R x R x xq. yq xq + zq. This completes the proof of Lemma 4 (See Figure 4). G R Q (ω) F S O Figure 4 roof of roposition 1. The line R meets (ω) again at G. From the similar triangles Q QG and RG R. We have the ratios Therefore, Q Q G, R R G. R Q Q R. (7)

5 construction of the golden ratio in an arbitrary triangle 243 ecause QR is perpendicular to O so QR is antiparallel line, this follows from QR, thus QR. We get Q R. (8) From (7) and (8), we deduce R Q. (9) Let symmedians and F meet at S then S( 2 : 2 : 2 ) see [9]. pply Lemma 4 for triangle with S and ray F meet (ω) at, we have 2 This is equivalent to (10) Note that, by Lemma 3,. (11) From (10) and (11) we have This means From (9) and (12), we obtain ( ) 1 1. (12) R Q Q R 1. This enough to show that the ratio R 5+1 Q 2 which is such the golden ratio. This completes the proof of roposition 1. References [1]. ogomolny, tolemy s Theorem, Interactive Mathematics Miscellany and uzzles, [2] T. O. ao, Q. H. Ngo, and. Yiu, Golden sections in an isosceles triangle and its circumcircle, Global Journal of dvanced Research on lassical and Modern Geometries, 5 (2016) [3]. aunić and. Yiu, Regular polygons and the golden section, Forum Geom., 16 (2016) [4] K. Hofstetter, simple construction of the golden section, Forum Geom., 2 (2002)

6 244 Q. H. Tran [5] G. Odom and J. van de raats, Elementary roblem 3007, mer. Math. Monthly, 90 (1983) 482; solution, 93 (1986) 572. [6] M. ietsch, The golden ratio and regular polygons, Forum Geom., 17 (2017) [7] Q. H. Tran, The golden section in the inscribed square of an isosceles right triangle, Forum Geom., 15 (2015) [8] Q. H. Tran, nother simple construction of the golden ratio in an isosceles triangle, Forum Geom., 17 (2017) [9]. Yiu, Introduction to the Geometry of the Triangle, Florida tlantic University Lecture Notes, 2001; with corrections, 2013, available at Tran Quang Hung: High school for Gifted students, Hanoi University of Science, Hanoi National University, Hanoi, Vietnam address: