Isogonal Conjugacy Through a Fixed Point Theorem

Transcription

1 Forum Geometricorum Volume 16 ( FORUM GEOM ISSN Isogonal onjugacy Through a Fixed Point Theorem Sándor Nagydobai Kiss and Zoltán Kovács bstract. For an arbitrary point in the plane of a fixed triangle, we study the affine combination of orthogonal projections onto the sidelines of the triangle, with coefficients the absolute barycentric coordinates of with respect to the triangle. We prove that this map is affine if and only if lies not on the circumcircle, and the only fixed point of this map is the isogonal conjugate of. 1. Introduction In this paper we investigate affine combinations of orthogonal projections onto lines in the plane. In general, this map has the form φ: R R,P φ(p =P + b, where R is a matrix (generally not regular and b is a vector. This map is a collineation, not necessarily injective. Nevertheless, the image set of φ is an affine subspace. The image space is R if and only if is regular, and we call φ an affine map in this case, as usual. (In this case φ is a bijective collineation. Specifically, we fix a triangle, and for R, consider the affine combination of orthogonal projections with respect to the sides of, with coefficients the absolute barycentric coordinates of with respect to. We call this map the weighted pedal map φ (see Section. We prove that the weighted pedal map φ is affine if and only if does not lie on the circumcircle. If is on the circumcircle, then the image space is the Simson line of the antipodal point of. In the affine case we prove that the only fixed point of φ is the isogonal conjugate of the point. Note, the isogonal conjugate of a point in the plane of the triangle is constructed geometrically by reflecting the lines,, and about the angle bisectors at,, and. The three reflected lines then concur at the isogonal conjugate ([4]. Finally, we investigate the image of the circumcircle in special cases. Throughout this paper we work with barycentric coordinates. We follow conventions in [3]. Let be the fundamental triangle with angles,, and side lengths = a, = b, = c. Every finite point P in the plane has a unique absolute barycentric coordinates (u, v, w for which P = u + v + w and u + v + w =1. For any λ R \{0}, (u,v,w =λ(u, v, w are the homogeneous (or relative barycentric coordinates of P and we write P =(u : v : w. Publication Date: pril 6, 016. ommunicating Editor: Paul Yiu.

2 17 S. N. Kiss and Z. Kovács Let S = b +c a = bc cos, S = a +c b = ac cos, S = a +b c = ab cos. S denotes twice the area of triangle. We shall freely use the following identities: S = b c S = c a S = a b S, S = a S + S S = b S + S S = c S + S S, S = S S + S S + S S, S = a S + b S + c S, S S = a S + b S c S.. The weighted pedal map Definition. Let be a point in the plane of the triangle. Denote by (α, β, γ the absolute barycentric coordinates of with respect to the triangle, i.e. = α + β + γ, α + β + γ =1. Define the weighted pedal map φ : R R by φ (P =απ a (P +βπ b (P +γπ c (P, (1 where π a, π b, π c denote the orthogonal projections onto the sidelines a, b, c respectively. From the point of view of linear algebra, φ is an affine combination of orthogonal projections onto the sidelines of the triangle. In particular, if is the centroid of the triangle, then φ is the average of orthogonal projections onto the sidelines. From a geometrical point of view φ (P has the same barycentric coordinates with respect to the pedal triangle P a P b P c of P as has with respect to, provided P is not on the circumcircle. In the latter case the pedal triangle degenerates, but affine combinations of points P a, P b and P c are still meaningful. Pc Pb P P Pa Figure 1. onstruction of φ (P (part I

3 Isogonal conjugacy through a fixed point theorem 173 Pc Pb D P Figure. onstruction of φ (P (part II Figures 1 show the construction of φ (P. Let the eva triangle of the point (Figure 1. The only problem is to construct points,, such that = P a P, c = P c P, a = P a P. b because the eva point of the triangle is P = φ (P. Figure shows the obvious construction of the point, where D P c and D P b. Let the point P in the plane of the triangle have absolute barycentric coordinates (u, v, w, i.e. P = u + v + w, where u + v + w =1. It is well-known (see e.g. [5, p. 5] that π a (P = (a v + us +(a w + us a, ( π b (P = (b w + vs +(b u + vs b, (3 π c (P = (c u + ws +(c v + ws c. (4 Substituting from (-(4 we get ( φ (P = (β + γu + β S b v + γ S ( + α S a u + β S b c w v +(α + βw ( + α S a u +(γ + αv + γ S c w. (5 The sum of coefficients of,, in (5 is 1. Thus, (5 gives the absolute barycentric coordinates of φ (P. Theorem 1. φ is an affine map if and only if is not on the circumcircle.

4 174 S. N. Kiss and Z. Kovács Proof. φ is an affine map if and only if the image of a triangle is a triangle. We give a necessary and sufficient condition that {φ (,φ (,φ (} to be an affine independent point set. From (5 we get φ ( =(1 α + αs a + αs a =(a (β + γ :αs : αs φ ( = βs b +(1 β + βs b =(βs :(γ + αb : βs φ ( = γs c + γs c +(1 γ =(γs : γs :(α + βc. It follows that φ is affine if and only if (β + γa αs αs Δ= βs (γ + αb βs 0. (6 γs γs (α + βc direct expansion of the determinant gives Δ=S (α + β + γ(βγa + γαb + αβc = S (βγa + γαb + αβc. We know that βγa + γαb + αβc =0 is the equation of the circumcircle [5, p. 63]. (6 is equivalent to not lying on the circumcircle. Theorem. Suppose that is on the circumcircle. The image space of the map φ is the Simson line of the antipodal point of. Proof. First we prove that if is on the circumcircle then the matrix defined in (6 is of rank, i.e. the image space is a line. For an indirect proof, suppose that all the minors of the matrix in (6 have zero determinant. For example, (1 αa αs βs (1 βb =(1 α(1 βa b αβa b cos = a b (γ + αβ sin =0. nalogously, α + βγ sin =0and β + γαsin =0. y addition we get (α + β + γ +(αβ sin + βγ sin + αγ sin =0, }{{}}{{} 1 0 which is a contradiction. The center of the circumcircle is The antipodal point of is O = =O = ( a S S, b S S, c S S ( a S S α, b S S. (7 β, c S S γ.

5 Isogonal conjugacy through a fixed point theorem 175 s( φ ( c b O a Figure 3. Image space of φ in the degenerate case The orthogonal projection of onto the lines,, are =(0:S α + a γ : S α + a β, =(S β + b γ :0:S β + b α, =(S γ + c β : S γ + c α :0, respectively (see Figure 3. If, then the Simson line of passes through and. (If =, then = = and we use points and, but everything goes through analogously. We check that φ ( =(a (β + γ : αs : αs lies on the line. Indeed, a (β + γ αs αs 0 a γ + αs a β + αs b γ + βs 0 b α + βs = a S (α + β + γ(βγa + γαb + αβc }{{} 0 =0. It is easy to give the inverse transformation of φ, where is not on the circumcircle. Let φ (P =u + v + w, where u + v + w =1. From (5 we get u = 1 u β S b γ S c Δ v γ + α γ S c w α + β, β S b

6 176 S. N. Kiss and Z. Kovács where Δ=S (βγa + γαb + αβc. fter expansion: u = 1 Δb c (αb c + βγs u +(S γ c S βv +(S β b S } {{ } λ c γw }{{} λ b = 1 ( (αb Δb c c + γβs u + λ c βv + λ b γw, (8 nalogously, with λ a = S α a S we get v = 1 ( λc Δc a αu +(c a β + S γαv + λ a γw (9 w = 1 ( λb Δa b αu + λ a βv +(a b γ + S αβγw. (10 In the case where is the circumcenter, φ is central similarity with scale factor 1/. This fact follows from the next theorem. Theorem 3. Let O denote the circumcenter of the triangle. The image of the circumcircle under the map φ O is the nine-point circle of. It is possible to determine the equation of the image of the circumcircle using the inverse transformation principle, by substituting (8, (9, (10 into the equation of the circumcircle. fter a long calculation we get the equation of the nine-point circle. Here we outline a geometric proof. Proof. Let k be the circumcircle. Since φ O is an affine map, φ O (k is an ellipse. We display below 6 points of the nine-point circle on φ O (k. From these it follows that φ O (k is the nine-point circle. First of all, with (u, v, w =(1, 0, 0 for, and (α, β, γ given by (7 for O, we have, by (5, ( b S + c S φ O ( = S + S S S + S S S ( S + S S = S + S S S + S S S = ( S S S + S S S + S S S = + H, where H is the orthocenter of. Similarly, φ O ( = +H and φ O ( = +H. Note that the points +H, +H, + are on the nine-point circle. Let =O, =O and =O the antipodal points of, and on the circumcircle. fter substitution into (5, making use of (7, we get φ O ( = +, φ O ( = +, φ O ( = +. These images are the midpoints of the sides of triangle ; they lie on the ninepoint circle.

7 Isogonal conjugacy through a fixed point theorem fixed point theorem for the weighted pedal map Theorem 4. Suppose is not on the circumcircle. The only fixed point of the weighted pedal map φ is the isogonal conjugate of : φ ( =. Proof. We prove that the fixed point has barycentric coordinates (a βγ : b γα : c αβ with respect to the triangle. P is a fixed point of the map (1 if and only if (γ + βu + β S b v + γ S c w = u α S a u +(α + γv + γ S c w = v α S a u + β S v +(α + βw = w b i.e. αu + β S b v + γ S c w =0 α S a u βv + γ S c w =0 (11 γ S a u + β S v γw =0. b The matrix of this homogeneous system of linear equations has rank, because it has zero determinant, but α βs b α S a β = αβs a b 0. It means that solutions are generated with one nonzero vector. With direct substitution it is easy to see that (a βγ,b αγ, c αβ solves (11. Remark. list of isogonal conjugate pairs is given in [4]; see also [1]. If is the centroid of the triangle, than φ is the average of the orthogonal projections π a, π b and π c. In this case is the symmedian point, and geometric analysis of Theorem 4 gives a well-known property of the symmedian point, namely, the symmedian point is the centroid of its pedal triangle, and it is the only point with this property (see, for example, []. We conclude this note with an application of Theorem 4. Theorem 5. If the orthocenter H does not lie on the circumcircle, then the image of the circumcircle k by the weighted pedal map φ H is an inscribed ellipse. Proof. Note, the isogonal conjugate of the orthocenter is the circumcenter O, which is the only fixed point of φ H. Thus φ (H is an ellipse with center O. Let F denote the intersection point of tangent lines to the circumcircle ( at and (see Figure 4. F has the absolute barycentric coordinates a b S, S, c S.

8 178 S. N. Kiss and Z. Kovács F E O D Figure 4. n inscribed ellipse fter calculation we get φ (F = = ( a S S, b S S, S S S. This is the antipode of on the circumcircle. Repeating this construction for and, we get points,. Since k is the inscribed circle of the tangential triangle DEF with center O, φ (k is an inscribed ellipse of the triangle with center φ (O =O. fter point reflection in point O, φ (k remains fixed and this fact gives the statement. References [1]. Kimberling, Encyclopedia of Triangle enters, []. Pohoata, short proof of Lemoine s theorem, Forum Geom., 8 ( [3].. Ungar, arycentric alculus in Euclidean and Hyperbolic Geometry, comparative introduction, World Scientific Publishing o. Pte. Ltd., Hackensack, NJ, 010. [4] E. W. Weisstein, Isogonal conjugate, [5] P. Yiu, Introduction to the geometry of the triangle, 00; S. Nagydobai Kiss: onstatin râncuşi Technology Lyceum, Satu Mare, Romania address: d.sandor.kiss@gmail.com Zoltán Kovács: Institute of Mathematics and omputer Science, University of Nyíregyháza, Nyíregyháza, Sóstói ú. 31/b, Hungary address: kovacs.zoltan@nye.hu