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Some Properties of Rectangle and a Random Point SOME PROPERTIES OF RECTANGLE AND A RANDOM POINT QUANG HUNG TRAN Abstract. We establish a relationship between the two important central lines of the triangle, the Euler line and the Brocard axis, in a configuration with an arbitrary rectangle and a random point. The classical Cartesian coordinate system method shows its strength in these theorems. Along with that, some related problems on rectangles and a random point are proposed with similar solutions using Cartesian coordinate system. 1. Introduction In the geometry of triangles, the Euler line [8] plays an important role and is almost the most classic concept in this field. The Euler line passes through the centroid, orthocenter, and circumcenter of the triangle. The symmedian point [9] of a triangle is the concurrent point of its symmedian lines, the Brocard axis [10] is the line connecting the circumcircle and the symmedian point of that triangle. The Brocard axis is also a central line that plays an important role in the geometry of triangle. The coordinate method was invented by Ren´eDescartes from the 17th century [1,2], up to now, the classical coordinate method of Descartes is still one of the most important method of mathematics and geometry. In this paper, we apply Cartesian coordinate system to solve an interesting the- orem for an arbitrary rectangle and a random point in which two important central lines in the triangle are mentioned as follows. The idea of using the Cartesian coor- dinate system is also used in a similar way to solve the new problems we introduced in Section 3. 2. Main theorem and proof Theorem 1. Let ABCD be a rectangle with center I. Let P be a random point in its plane. arXiv:2106.10810v1 [math.HO] 21 Jun 2021 1) Euler line of triangles P AB and PCD meet at Q. Euler line of triangles P BC and P AD meet at R. Then, line QR goes through center I. 2) Brocard axis of triangles P AB and PCD meet at M. Brocard axis of triangles P BC and P AD meet at N. Then, line MN goes through P . Proof. We will use Cartesian coordinate for the solutions. Since ABCD is an arbitrary rectangle and P is a random point, we can take P (0; 0), A(a; b), B(c; b), C(c; d), and D(a; d) for all real numbers a, b, c, and d. Date: June 22, 2021. 2010 Mathematics Subject Classification. 51M04, 51-03. 1 2 QUANG HUNG TRAN We get perpendicular bisectors da, db, dc, and dd of PA, PB, PC, and PD, respectively, are a2 + b2 a (2.1) d : y = − · x; a 2b b b2 + c2 c (2.2) d : y = − · x; b 2b b c2 + d2 c (2.3) d : y = − · x; c 2d d a2 + d2 a (2.4) d : y = − · x: d 2d d Thus circumcenters Oa, Ob, Oc, and Od of triangles P AB, P BC, PCD, and P DA, respectively, are a + c −ac + b2 (2.5) O = d \ d = ; ; a a b 2 2b −bd + c2 b + d (2.6) O = d \ d = ; ; b b c 2c 2 a + c −ac + d2 (2.7) O = d \ d = ; ; c c d 2 2d −bd + a2 b + d (2.8) O = d \ d = ; : d d a 2a 2 1) (See Figure1). We have that centroid Ga, Gb, Gc, and Gd of triangles P AB, P BC, PCD, and P DA, respectively, are a + c 2b (2.9) G = ; ; a 3 3 2c b + d (2.10) G = ; ; b 3 3 a + c 2d (2.11) G = ; ; c 3 3 2a b + d (2.12) G = ; : d 3 3 Therfore, Euler lines a, b, c, and d of triangles P AB, P BC, PCD, and P DA, respectively, are b2 + ac b2 + 3ac (2.13) : y = − · x; a b ab + bc bc2 + bd2 + b2d + c2d bc + cd (2.14) : y = − · x; b c2 + 3bd c2 + 3bd d2 + ac d2 + 3ac (2.15) : y = − · x; c d ad + cd SOME PROPERTIES OF RECTANGLE AND A RANDOM POINT 3 y Q D(a, d) C(c, d) b a+c b+d I 2 , 2 d A(a, b) B(c, b) c x P (0, 0) a R Figure 1. Rectangle with Euler lines bd2 + a2b + a2d + b2d ab + ad (2.16) : y = − · x: d a2 + 3bd a2 + 3bd We get the intersections a2c − abd + ac2 − bcd 2abc + 2acd (2.17) Q = \ = ; ; a c 3ac − bd 3ac − bd −2abd − 2bcd abc + acd − b2d − bd2 (2.18) R = \ = ; ; b d ac − 3bd ac − 3bd and then the line connecting points Q and R is b + d (2.19) QR : y = − x + b + d: a + c a+c b+d Now it not hard to see that line PQ goes through center I 2 ; 2 . 2) (See Figure2). Using the barycentric coordinates of symmedian point in [9], we have that the coordinates of symmedian points Sa, Sb, Sc, and Sd of triangles P AB, P BC, PCD, and P DA, respectively, are (2.20) PA2 · B + PB2 · A + AB2 · P a2c + ab2 + ac2 + b2c a2b + 2b3 + bc2 S = = ; ; a PA2 + PB2 + AB2 2a2 − 2ac + 2b2 + 2c2 2a2 − 2ac + 2b2 + 2c2 15 4 QUANG HUNG TRAN y N D(a, d) C(c, d) βb βd A(a, b) B(c, b) P (0, 0) x βc βa M Figure 2. Rectangle with Brocard axis (2.21) PB2 · C + PC2 · B + BC2 · P b2c + 2c3 + cd2 b2d + bc2 + bd2 + c2d S = = ; ; b PB2 + PC2 + BC2 2b2 − 2bd + 2c2 + 2d2 2b2 − 2bd + 2c2 + 2d2 (2.22) PC2 · D + PD2 · C + CD2 · P a2c + ac2 + ad2 + cd2 a2d + c2d + 2d3 S = = ; ; c PC2 + PD2 + CD2 2a2 − 2ac + 2c2 + 2d2 2a2 − 2ac + 2c2 + 2d2 (2.23) PD2 · A + PA2 · D + DA2 · P 2a3 + ab2 + ad2 a2b + a2d + b2d + bd2 S = = ; : d PD2 + PA2 + DA2 2a2 + 2b2 − 2bd + 2d2 2a2 + 2b2 − 2bd + 2d2 Therfore, Brocard axis βa, βb, βc, and βd of triangles P AB, P BC, PCD, and P DA, respectively, are b4 + a2b2 + a2c2 + b2c2 −b4 − ac3 + a2c2 − a3c − 2ab2c (2.24) β : y = + · x; a 2bc2 + 2a2b − 4abc bc3 + a3b − abc2 − a2bc (2.25) bc4 + b2d3 + b3c2 + b3d2+ +c2d3 + c4d + bc2d2 + b2c2d −cd3 − b3c + bcd2 + b2cd β : y = + · x; b 2c4 + 2bd3 − 2b2d2 + 2b3d + 4bc2d c4 + bd3 − b2d2 + b3d + 2bc2d SOME PROPERTIES OF RECTANGLE AND A RANDOM POINT 5 d4 + a2c2 + a2d2 + c2d2 −d4 − ac3 + a2c2 − a3c − 2acd2 (2.26) β : y = + · x; c 2a2d + 2c2d − 4acd a3d + c3d − ac2d − a2cd (2.27) a2b3 + a2d3 + a4b + a4d+ +b2d3 + b3d2 + a2bd2 + a2b2d −ab3 − ad3 + abd2 + ab2d β : y = + · x: d 2a4 + 2bd3 − 2b2d2 + 2b3d + 4a2bd a4 + bd3 − b2d2 + b3d + 2a2bd We get the intersections (2.28) −a3bd + a3c2 − a2bcd + a2c3− a3bc + a3cd + a2bc2 + a2c2d+ 3 2 2 2 3 3 2 3 2 0−ab d − ab d − abc d − abd − +ab c + ab cd + abc + abcd +1 3 2 2 3 3 3 3 3 2 2 3 B −b cd − b cd − bc d − bcd +ac d + acd − b d − b d C M = βa\βc = B ; C ; B 2a3c − 2a2c2 − 4abcd+ 2a3c − 2a2c2 − 4abcd+ C B C B C B +2ac3 − 2b3d − 2b2d2 − 2bd3 +2ac3 − 2b3d − 2b2d2 − 2bd3 C B C B C B C B C (2.29) @ A −a3bd + a3c2 − a2bcd + a2c3− a3bc + a3cd + a2bc2 + a2c2d+ 3 2 2 2 3 3 2 3 2 0−ab d − ab d − abc d − abd − +ab c + ab cd + abc + abcd +1 3 2 2 3 3 3 3 3 2 2 3 B −b cd − b cd − bc d − bcd +ac d + acd − b d − b d C N = βb\βd = B ; C ; B 2a3c + 2a2c2 + 4abcd+ 2a3c + 2a2c2 + 4abcd+ C B C B C B +2ac3 − 2b3d + 2b2d2 − 2bd3 +2ac3 − 2b3d + 2b2d2 − 2bd3 C B C B C B C B C and then the line@ connecting points M and N is A b2d3 + b3d2 − abc3 − acd3 − ab3c − ac3d− −a2bc2 − a2c2d − a3bc − a3cd − abcd2 − ab2cd (2.30) MN : y = · x: −a2c3 − a3c2 + abd3 + ab2d2 + ab3d + bcd3+ +bc3d + a3bd + b2cd2 + b3cd + abc2d + a2bcd It is easy to see that MN goes through P (0; 0). This completes the proof. 6 QUANG HUNG TRAN 3. Some others theorems on rectangle and a random point We introduce some more theorems on rectangle and a random point, all of them can be solved by Cartesian coordinate as we work above. Theorem 2 (Generalization of Theorem1) . Let A1B1C1D1 and A2B2C2D2 be two rectangles with the same center I. Let P be a random point in its plane. Euler lines of triangles PA1B1 and PC1D1 meet at Q. Euler lines of triangles PA2D2 and PB2C2 meet at R. Then, three points Q, R, and I are collinear (See Figure 3). B1 A1 A2 Q B2 P I R D2 C2 D1 C1 Figure 3. SOME PROPERTIES OF RECTANGLE AND A RANDOM POINT 7 Theorem 3. Let ABCD be a rectangle with center I. Let P be a random point in its plane. Let Ha, Hb, Hc, and Hd be the orthocenters of triangles P AB, P BC, PCD, and P DA, respectively. Let Q and R be the midpoints of HaHc and HbHd, respectively. Then, three points Q, R, and I are collinear (See Figure4). Hd B C R I H Hc a P Q Hb A D Figure 4. 8 QUANG HUNG TRAN Theorem 4. Let ABCD be a rectangle with center I. Let P be a random point in its plane. Let Oa, Ob, Oc, and Od be the circumcenters of triangles P AB, P BC, PCD, and P DA, respectively.
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