SOME PROPERTIES OF THE BROCARD POINTS OF A CYCLIC QUADRILATERAL

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1 SOME PROPERTIES OF THE ROR POINTS OF YLI QURILTERL IMITR ELEV bstract. In this article we have constructed the rocard points of a cyclic quadrilateral, we have found some of their properties and using these properties we have proved the problem of.. Zaslavsky. 1. The Problem lexey Zaslavsky, rocard s points in quadrilateral [4]. Given a convex quadrilateral. It is easy to prove that there exists a unique point P such that P = P = P. We will call this point rocard point (r()) and the respective angle rocard angle (φ()) of broken line. Note some properties of rocard s points and angles: φ() = φ() if is cyclic; if is harmonic then φ() = φ(). Thus there exist two points P, Q such that P = P = P = P = Q = Q = Q = Q. These points lie on the circle with diameter OL where O is the circumcenter of, L is the common point of its diagonals and P OL = QOL = φ(). Problem. Let be a cyclic quadrilateral, P 1 = r(), P 2 = r(), P 3 = r(), P 4 = r(), Q 1 = r(), Q 2 = r(), Q 3 = r(), Q 4 = r(). Then S P1 P 2 P 3 P 4 = S Q1 Q 2 Q 3 Q 4. I first saw this problem 2-3 years ago in an article by lexei Myakishev (see [3]). I did not consider solving it then, but my pupils built a structure very similar (seemingly) to the described. Thus Zaslavsky s problem served as a stimulus (accelerator) to develop a good article (related to isogonal conjugate points), which they presented on international events. I sincerely hope that this article will appear on the pages of Kvant. Now they are university students, but again I saw the problem in the Journal of lassical Geometry. t first I thought that using the developed article I would figure out the solution quickly, but I did not. new construction came out. 2. Some Properties Of rocard Points Of a Triangle. Let P and Q be the first and the second rocard points of a. We will prove that the rocard points and the three vertices, and define three pairs of similar triangles, and three pairs of triangles with equal areas (Fig. 1). 1

2 2 IMITR ELEV P Fig. 1. Proposition 1. If P and Q are the first and the second rocard points of a, then: (i) P Q, P Q and P Q; (ii) S P = S Q, S P = S Q and S P = S Q. ϕ Q P P a Q Q a P c Fig. 2. Proof. If we denote with ϕ the rocard angle (Fig. 2), then P = Q = ϕ, and from the construction ([1]) of the rocard points of we have P = Q = 180. Hence P Q. We construct the altitudes P P c, P P a, QQ c and QQ a. Then, having P Q, P c P Q a Q and Q c P P a Q, we have the following equations: S P = P P a = QQ a P P a = Q S Q QQ c P P c QQ c P P Q = 1. For the other pairs of triangles, the proof is analogous. We are going to find, similar to these properties, for the rocard points of a cyclic quadrilateral. Q c 3. onstruction Let the quadrilateral be a cyclic, = E and let for exactitude is between and E, = F and is between and F. We denote = α and = β.

3 SOME PROPERTIES OF THE ROR POINTS OF YLI QURILTERL 3 In order to construct the rocard points of the quadrilateral, we firstly construct the points M 1, M 2, M 3, M 4 and N 1, N 2, N 3, N 4, as shown in Table 1: M 1 M 1 and M 1 N 1 N 1 and N 1 M 2 M 2 and M 2 N 2 N 2 and N 2 M 3 M 3 and M 3 N 3 N 3 and N 3 M 4 M 4 and M 4 N 4 N 4 and N 4 Table 1 Now we construct the intersection points of the pairs of circumcircles of the triangles, as shown in Table 2: M 1 M 3 M 2 P 1 P 2 M 4 P 4 P 3 Table 2 N 1 N 3 N 4 Q 1 Q 4 N 2 Q 2 Q 3 Proposition 2. The points P 1, P 2, P 3, P 4 and Q 1, Q 2, Q 3, Q 4 are the rocard points of the quadrangle. F β α N 4 Q 4 ϕ 4 E Fig. 3. N 3 Proof. Let Q 4 be the intersection point of the circumcircles of triangles N 3 and N 4 (Table 2). We denote Q 4 = ϕ 4 (Fig. 3). Since N 4 and N 3 (Table 1), then N 4 = β and N 3 = α. So: Q 4 = β Q 4 = β N 4 Q 4 = β (β ϕ 4 ) = ϕ 4, Q 4 = α Q 4 = α (180 Q 4 ϕ 4 ) = α (180 (180 α) ϕ 4 ) = ϕ 4 Hence Q 4 = r() is a rocard point and ϕ 4 = φ() is a rocard angle in the quadrilateral. s is cyclic quadrilateral, hence ϕ 4 =

4 4 IMITR ELEV φ(). For the other points, the proof is analogous. If we denote the rocard angles in with ϕ 1, ϕ 2, ϕ 3 and ϕ 4 then we have: (1) ϕ 1 = φ() = φ() = P 1 = P 1 = = P 1 = Q 1 = Q 1 = Q 1, ϕ 2 = φ() = φ() = P 2 = P 2 = = P 2 = Q 2 = Q 2 = Q 2, ϕ 3 = φ() = φ() = P 3 = P 3 = = P 3 = Q 3 = Q 3 = Q 3, ϕ 4 = φ() = φ() = P 4 = P 4 = = P 4 = Q 4 = Q 4 = Q Some properties of the rocard points of a cyclic quadrilateral Proposition 3. The triads of points, P 1, M 1 ;, P 2, M 2 ;, P 3, M 3 ;, P 4, M 4 ;, Q 1, N 1 ;, Q 2, N 2 ;, Q 3, N 3 and, Q 4, N 4 are collinear. Proof. Q 4 + Q 4 N 4 = (α 180 ) + α = 180 (Fig. 3). For the other triads of points, the proof is analogous. Proposition 4. The fours of lines M 1, M 2, M 3, M 4 and N 1, N 2, N 3, N 4 are concurrent. β N 1 α Q 3 Q 4 N 4 = Q 0 Q 1 E Fig. 4. N 3 Proof. Let N 4 N 1 = (Fig. 4). Then using N 1 N 4, E and N 1 (see Table 1) we have the following equations: N 4 = N 1 N 4 = N 4 N 1 N 4 = E = E

5 SOME PROPERTIES OF THE ROR POINTS OF YLI QURILTERL 5 Now, let N 3 = Q 0. From the similarity of N 3Q 0 N 4Q 0, N 3 E and also because N 4 we have consecutively the following equations: Q 0 Q 0 N 4 = N 3 N 4 = E. E.N 4 = E E E = E Then = Q 0 N 4 Q 0 N and Q 0. In the same way N 2. So the 4 lines N 1, N 2, N 3, N 4 intersect in a point, and similarly the lines M 1, M 2, M 3, M 4 have a common point P 0. ecause E and the Law of Sines used for E, we attain: N 4 = E = E2 E 2 = sin2 β sin 2 α. In the same way we have the following equations: (2) = = P 0 = P 0 = sin2 β N 4 N 2 P 0 M 3 P 0 M 1 sin 2 α and = = P 0 = P 0 = sin2 α N 1 N 3 P 0 M 4 P 0 M 2 sin 2 β Now we will show that the points P 0, and the vertices,, and define four pairs of triangles with equal areas (Fig. 5). P 0 Fig. 5. Proposition 5. If P 0 and are the intersection points of the fours of lines M 1, M 2, M 3, M 4 and N 1, N 2, N 3, N 4 (see Proposition 4), then: S P0 = S Q0, S P0 = S Q0, S P0 = S Q0 and S P0 = S Q0.

6 6 IMITR ELEV M 4 β P 1 P4 P 0 Q 4 α N 4 Q 1 ϕ 4 Fig. 6. Taking into account that the fours of points, P 0, P 4, M 4 and,, Q 4, N 4 (Fig. 6) are collinear (Proposition 3 and Proposition 4) and equations (2) we attain: (3) S P0 = P 0 M 4 S M4 = sin 2 α sin 2 α + sin 2 β S M 4, S Q0 = N 4 S N4 = sin 2 β sin 2 α + sin 2 β S N 4, Since M 4 = N 4 = 180 α (see Table 1) and M 4 = N 4 = ϕ 4 (see (1)), then M 4 N 4 and (4) S M4 S N4 = 2 N 2 4 = sin2 β sin 2 α (Law of Sines for N 4) Using equations (3) and (4) we calculate S P0 S Q0 = sin2 α sin 2 β SM 4 S N4 = 1. The proof, of the areas equality, of the other triangle pairs is similar. Now we can consider the quadrangles P 1 P 2 P 3 P 4 and Q 1 Q 2 Q 3 Q 4. We will prove that the points P 0, and the vertices of quadrangles P 1 P 2 P 3 P 4 and Q 1 Q 2 Q 3 Q 4 define four pairs of triangles with equal areas (Fig. 7).

7 SOME PROPERTIES OF THE ROR POINTS OF YLI QURILTERL 7 P 4 P 1 P 3 P P0 0 Q 3 Q 4 Q 1 P 2 Q 2 Fig. 7. Proposition 6. If P 0 and are the intersection points of the fours of lines M 1, M 2, M 3, M 4 and N 1, N 2, N 3, N 4 (see Proposition 4), then: S P1 P 2 P 0 = S Q1 Q 2, S P2 P 3 P 0 = S Q2 Q 3, S P3 P 4 P 0 = S Q3 Q 4, S P4 P 1 P 0 = S Q4 Q 1. Proof. Firstly we will show that the areas of the pairs of triangles P 1, Q 1 and P 2, Q 2 are equal (Fig. 8). The proof is similar to that of Proposition 1. Since P 1 = Q 1 = ϕ 1 and P 1 = Q 1 (see equations (1)), hence P 1 Q 1. We construct the altitudes P 1 P b, P 1 P c, Q 1 Q b and Q 1 Q c. P c Q c P Pb b P 1 P 0 Q b P 2 Q 2 Q 1 ϕ 1 Fig. 8. Then using P 1 Q 1, P 1 P b Q 1 Q c and P 1 P c Q 1 Q b, we have the following equations: S P1 S Q1 =.P 1P c.q 1 Q b = Q 1Q c P 1 P b P1P c Q 1 Q b = Q 1 P 1 P1 Q 1 = 1. So we attain: S P1 = S Q1 and S P2 = S Q2 (analogous).

8 8 IMITR ELEV Since the triads of points, P 1, P 0 ;, P 0, P 2 ;, Q 1, and,, Q 2 are collinear (Propositions 3 and 4) and S P0 = S Q0 (Proposition 5), hence S P0 P 2 = S Q0 Q 2 and S P0 P 1 = S Q0 Q 1. We calculate S P1 P 2 P 0 = P 0P 1 P 0 S P 0 P 2 = S P 0 P 1 S P0 S P0 P 2 and S Q1 Q 2 = Q 1 S Q 2 = S Q 1 S Q0 Q S 2, therefore S P1 P 2 P 0 = S Q1 Q 2. Q0 The proof, of the areas equality, of the other pairs of triangles is similar. From Proposition 6 immediately follows onsequence S P1 P 2 P 3 P 4 = S Q1 Q 2 Q 3 Q 4. We should note that another solution is given by handan anerjee in his blog [2]. 5. Two additional properties bove we have proved that P 1 Q 1 and S P1 = S Q1. In fact, we have also P 1 Q 1 and S P1 = S Q1. In general, each pair of rocard points P i and Q i, in a cyclic quadrilateral and its corresponding vertices, define two pairs of triangles with equal areas (to compare see Proposition 1 and Fig. 1). P 4 P 1 P 3 P 0 P 2 P 0 Q 3 Q 3 Q 4 Q 4 Q 1 Q 2 Fig. 9.

9 SOME PROPERTIES OF THE ROR POINTS OF YLI QURILTERL 9 On the other hand, the points P 0 and not only stole the pairs of triangles with equal areas related to vertices of the quadrilateral (see Proposition 5 and Fig. 5), but they have and those in quadrilaterals P 1 P 2 P 3 P 4 and Q 1 Q 2 Q 3 Q 4 (see Proposition 6 and Fig. 7). Naturally arise the question, where are the pairs of similar triangles associated with P 0, and vertices of quadrilaterals, P 1 P 2 P 3 P 4 and Q 1 Q 2 Q 3 Q 4. It turns out that they appear in a very curious way. In Fig. 9 we show that the triangles defined by the point P 0 and quadrilateral P 1 P 2 P 3 P 4 are similar to the triangles determined by a point and quadrilateral. onversely, the triangles defined by the point and quadrilateral Q 1 Q 2 Q 3 Q 4 are similar to the triangles determined by a point P 0 and quadrilateral (Fig. 9). Proposition 7. If P 0 and are the intersection points of the fours of lines M 1, M 2, M 3, M 4 and N 1, N 2, N 3, N 4 (see Proposition 4), then: P 1 P 2 P 0, P 2 P 3 P 0, P 3 P 4 P 0, P 4 P 1 P 0 and Q 1 Q 2 P 0, Q 2 Q 3 P 0, Q 3 Q 4 P 0, Q 4 Q 1 P 0. Proof. We will prove that Q 1 Q 2 P 0 (Fig. 8). The points Q 1 and Q 2 lie on the circumcircle of the triangle N 1 (Table 2) therefore Q 1 Q 2 = Q 1 = ϕ 4. ut Q 2 Q 1 = + Q 1 = β ϕ 1 + ϕ 2 (see equations (1)) and P 0 = 180 ( P 0 + P 1 ) = 180 (180 β ϕ 2 + ϕ 1 ) = β ϕ 1 + ϕ 2, namely Q 2 Q 1 = P 0. Hence Q 1 Q 2 P 0. The similarity of the other pairs of triangles we prove similarly. From Proposition 7 follows, that P 1 + P 3 = Q 2 + Q 4 = + = 180 (see Fig. 9), which means that quadrilaterals P 1 P 2 P 3 P 4 and Q 1 Q 2 Q 3 Q 4 are cyclic. I apply the problems of my students. nton elev, Kaloyan ucovsky, 6. ppendix Problem 1. If in the quadrangle is inscribed ellipse with foci F 1 and F 2, and the isogonal conjugated points of F 1 and F 2 with respect to,,, are 1, 1, 1, 1 and 2, 2, 2, 2, then the quadrangles and are congruent. Problem 2. In the quadrangle is inscribed a circle with center O. If the isogonal conjugated points of O with respect to,,, are 1, 1, 1 and 1, then is a parallelogram. If, then is a rhombus. If around the quadrangle is described a circle, then is a rectangle.

10 10 IMITR ELEV The construction, which reveals how quadrangles and merged into a parallelogram (when F 1 get near to F 2 ) is really fascinating. We did it using GeoGebra. References [1]. V. kopyan and.. Zaslavsky. Geometry of conics. Vol. 26 of Mathematical World. merican Mathematical Society, Providence, RI, [2]. anerjee. rocard points in a cyclic quadrilateral. May 25, url: [3]. G. Myakishev. Elementary geometry and comuter url: [4].. Zaslavsky. rocard s points in quadrilateral. Journal of classical geometry:72, 1, Sofia High School in onstruction, rchitecture and Geodesy Hristo otev. address: dbelev@gbg.bg