Forum Geometricorum Volume 13 (2013) FORUM GEOM ISSN Pedal Polygons. Daniela Ferrarello, Maria Flavia Mammana, and Mario Pennisi

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1 Forum Geometricorum Volume 13 (2013) FORUM GEOM ISSN edal olygons Daniela Ferrarello, Maria Flavia Mammana, and Mario ennisi Abstract. We study the pedal polygon H n of a point with respect to a polygon, where the points H i are the feet of the perpendiculars drawn from to the sides of. In particular we prove that if is a quadrilateral which is not a parallelogram, there exists one and only one point for which the points H i are collinear. 1. Introduction Consider a polygon A n and call it. Let be a point and leth i be the foot of the perpendicular from to the line A i A i+1, i = 1,2,...,n (with indices i taken modulon). The pointsh i usually form a polygon H n, which we call the pedal polygon of with respect to, and denote byh(see Figure 1). We call the pedal point. See ([2, p.22]) for the notion of pedal triangle. A 6 H 5 H 6 A 4 H 4 Figure 1 In this article we find some properties of the pedal polygonhof a point with respect to. In particular, when is a triangle we find the points such that the pedal triangle H is a right, obtuse or acute triangle. When is a quadrilateral which is not a parallelogram, we prove that there exists one and only one point for which the points H i are collinear. Moreover, we find the points for which ublication Date: September October 1, Communicating Editor: aul Yiu.

2 154 D. Ferrarello, M. F. Mammana and M. ennisi the pedal quadrilateralhof has at least one pair of parallel sides. We also prove that, in general, there exists one and only one pedal point with respect to which H is a parallelogram. In the last part of the paper, we find some properties of the pedal polygon H in the general case of a polygonwithnsides. 2. roperties of the pedal triangle Let be a triangle. The pedal triangle of the circumcenter of is the medial triangle of ; the one of the orthocenter is the orthic triangle of ; the one of the incenter is the Gergonne triangle of (i.e., the triangle whose vertices are the points in which the incircle of touches the sides of ). Theorem 1. [2, p.41] If is a triangle, the points H i are collinear if and only if lies on the circumcircle of. The line containing the points H i is called Simson line of the point with respect to(see Figure 2). O Figure 2. Theorem 2. [1, p.108] The pointsfor which the pedal trianglehis isosceles are all and only the points that lie on at least one of the Apollonius circles associated to the vertices of. The Apollonius circle associated to the vertex A i is the locus of points such that A i+1 : A i+2 = A i A i+2 : A i A i+1. The three Apollonius circles are coaxial and they intersect in the two isodynamic points of the triangle,i 1 andi 2. Therefore, the isodynamic points of are the only points whose pedal triangles with respect to are equilateral (see Figure 3).

3 edal polygons 155 I 1 I 2 Figure 3 We will find now the points whose pedal triangle is a right, acute, obtuse triangle. Let be a point (see Figure 4) and let A i A i+1 = a i+2, A i = x i, H i H i+1 = h i+2. Since the quadrilateral A i H i H i+2 is cyclic, h i = x i sina i ([2, p.2]). O Figure 4

4 156 D. Ferrarello, M. F. Mammana and M. ennisi By the ythagorean theorem and its converse, the pedal triangle of is right in H i if and only if x 2 i sin 2 A i = x 2 i+1sin 2 A i+1 + x 2 i+2sin 2 A i+2. By the law of sines, this is equivalent to a 2 ix 2 i = a 2 i+1x 2 i+1 + a 2 i+2x 2 i+2. (1) This relation represents the locus γ i of points for which the triangle H is right in H i. Therefore, the locus of points whose pedal triangle is a right triangle is γ 1 γ 2 γ 3. Observe thatγ i contains the pointsa i+1 anda i+2 ; moreover,γ i and γ i+1 intersect only in the point A i+2. We verify now that γ 1 is a circle. Set up an orthogonal coordinate system such that (1,0) and ( 1,0); let (a,b) and (x,y). The relation (1) becomes: 4((x a) 2 +(y b) 2 ) = ((a+1) 2 +b 2 )((x 1) 2 +y 2 )+((a 1) 2 +b 2 )((x+1) 2 +y 2 ). Simplifying, we obtain the equation of a circle: (a 2 +b 2 1)(x 2 +y 2 )+4by (a 2 +b 2 1) = 0. Moreover, it is not hard to verify that the tangents to the circumcircle of in the points and pass through the center of γ 1. Analogously the same holds for γ 2 and γ 3. We can then state that γ i is a circle passing through the pointsa i+1 anda i+2 ; the tangents to the circumcircle ofin the points A i+1 and A i+2 pass through the center C i of γ i ; moreover, γ i and γ i+1 are tangent in A i+2. Then, if C 1 C 2 C 3 is the tangential triangle of, γ i is the circle with center C i passing througha i 1 anda i+1 (see Figure 5). γ 2 C 3 C 2 O γ 3 γ 1 C 1 Figure 5 Observe that, by the law of cosines, the angle inh i of the pedal triangle of is obtuse if and only if: a 2 ix 2 i > a 2 i+1x 2 i+1 +a 2 i+2x 2 i+2,

5 edal polygons 157 i.e., the point lies inside the circle γ i. Thus, we have established the following theorem. Theorem 3. The pedal triangle of a point is (a) a right triangle if and only if lies on one of the circles γ i, (b) an obtuse triangle if and only if is inside one of the circlesγ i, (c) an acute triangle if and only if is external to all the circles γ i. 3. roperties of the pedal quadrilateral Let be a cyclic quadrilateral. The pedal quadrilateral of the circumcenter of is the Varignon parallelogram of, and the one of the anticenter ([6, p.152]) is the principal orthic quadrilateral of ([5, p.80]). Let be a tangential quadrilateral. The pedal quadrilateral of the incenter of is the contact quadrilateral of, i.e., the quadrilateral whose vertices are the points in which the incircle oftouches the sides of. For a generic quadrilateral, we consider the problem of finding the pedal points for which the pointsh i are collinear. It is easy to verify that if has only one pair of parallel sides, there is only one pedal point for which the points H i are collinear. is the common point to the lines containing opposite and non parallel sides of, and the points H i lie on the perpendicular from to the parallel sides of. On the other hand, if is a parallelogram, there is no point with respect to which the points H i are collinear. A 4 H4 A 6 S Figure 6 Suppose now that is a quadrilateral without parallel sides (see Figure 6). Let be the common point to the lines and A 4, and A 6 the common point

6 158 D. Ferrarello, M. F. Mammana and M. ennisi to and A 4. Consider the four triangles A 6,, A 4 A 6, A 4, and let C 1,C 2,C 3,C 4 be their circumcircles, respectively. If the pedal point lies on one of the circles C i, then, by Theorem 1, at least three of the points H i are collinear. It follows that the four points H i are collinear if and only if lies in every C i. The four circles are concurrent in the Miquel point of the quartet of lines containing the sides of ([3, p.82]). Thus, we have established the following theorem. Theorem 4. If is a quadrilateral, that is not a parallelogram, there exists one and only one pedal point with respect to which the points H i are collinear. We call this point the Simson point of the quadrilateral, and denote it by S. We call the Simson line of the line containing the points H i. Observe that the pointsh i determine a quadrilateral if and only if C 1 C 2 C 3 C 4. Theorem 5. Ifis a quadrilateral which is not a parallelogram, the reflections of the Simson point with respect to the lines containing the sides of are collinear and the linelcontaining them is parallel to the Simson line. K 2 K 4 K 3 K 1 A 4 H4 A 6 S Figure 7 roof. The theorem is trivially true ifhas one pair of parallel sides. Suppose that is without parallel sides (see Figure 7). Let K i, i = 1,2,3,4, be the reflection of S with respect to the line A i A i+1. The points S, H i and K i are collinear and SH i = H i K i, thenk i is the image ofh i under the homothetyh(s,2). Then, since the pointsh i are collinear (Theorem 4), the pointsk i are also collinear. Moreover, the linelcontaining the points K i is parallel to the Simson line of.

7 edal polygons 159 Conjecture. If is a cyclic quadrilateral without parallel sides, the line l passes through the anticenter H of, and the Simson line bisects the segment SH. K 1 K 3 H M K 2 S H 4 A 4 A 6 K 4 Figure 8 The conjecture was suggested by using a dynamic geometry software (see Figure 8). However, we have been unable to prove it. If is a cyclic quadrilateral with a pair of parallel sides, then is an isosceles trapezoid. The line l coincides with the Simson line, i.e., the line joining the midpoints of the bases of, and passes through the anticenter of. In this case the Simson line contains the segmentsh. We now find the points whose pedal quadrilaterals have at least on pair of parallel sides. If is a parallelogram, then the points whose pedal quadrilaterals have at least one pair of parallel sides are all and only the points of the diagonals of. Suppose now thatis not a parallelogram. We prove that the locus of the point whose pedal quadrilateral has the sides H 4 and parallel is the circle S (see Figure 9). First observe that S is a point with respect to whom H 4 and are parallel because the points H i are collinear. Set up now an orthogonal coordinate system such that ( 1,0) and (1,0); let (a,b), A 4 (c,d) and

8 160 D. Ferrarello, M. F. Mammana and M. ennisi (x,y). If H 4 and are parallel, then lies on the circleγ of equation: (hd+kb)x 2 +(hd+kb)y 2 (hk 4bd)y = hd+kb, whereh = a 2 +b 2 1 andk = c 2 +d 2 1. Note that the points and are onγ, andγ is the circle S. H4 A 4 A 6 S Figure 9 Analogously we can prove that the points whose pedal quadrilateral has the sides and H 4 parallel is the circle A 4 S. Therefore we have established the following theorem. Theorem 6. The points whose pedal quadrilaterals have at least one pair of parallel sides are precisely those on the circles S and A 4 S. H 1 H4 A 4 S Figure 10 In general, the circles S and A 4 S intersect at two points, the Simson point S and one other point (see Figure 10). The pedal quadrilateral of is

9 edal polygons 161 a parallelogram. We call the parallelogram point of. Observe that if is a parallelogram the parallelogram point is the intersection of the diagonals of. Ifis cyclic, the pedal quadrilateral of the circumcentero ofis the Varignon parallelogram of. Therefore, the parallelogram point of is O. It follows that if is cyclic, the Simson point is the intersection point of the circles O and A 4 O, other thano. 4. Some properties of the pedal polygon Let be a polygon with n sides. Consider the pedal polygon H of a point with respect to. We denote by Q i the quadrilateral H i A i+1 H i+1, for i = 1,2,...,n. Since the angles in H i and in H i+1 are right, Q i cannot be concave. Lemma 7. If ABCD is a convex or a crossed quadrilateral such that ABC and CDA are right angles, then it is cyclic. Moreover, its circumcenter is the midpoint of AC and its anticenter is the midpoint of BD. B D H B A O H C A O C D Figure 11 roof. Let ABCD be a convex or a crossed quadrilateral with ABC and CDA right angles (see Figure 11). Then, it is cyclic with the diagonal AC as diameter. (When it is a crossed quadrilateral it is inscribed in the semicircle with diameter AC). Then its circumcenter O is the midpoint of AC. Consider the maltitudes with respect to the diagonals AC and BD. The maltitude througho is perpendicular to the chordbd of the circumcircle, then it passes through the midpoint H of BD. But also the maltitude relative to AC passes through H. Then, the anticenter of the quadrilateral is H. Note that the maltitudes of a crossed quadrilateral ABCD are concurrent because they are also the maltitudes of the cyclic convex quadrilateral ACBD.

10 162 D. Ferrarello, M. F. Mammana and M. ennisi By Lemma 7, the quadrilaterals Q i are cyclic. Denote by O i and A i the circumcenter and the anticenter ofq i respectively. We callo 1 O 2...O n the polygon of the circumcenters of with respect to and denote it by c (). We call A 1 A 2...A n the polygon of the anticenters of with respect to and we denote it with a (). Theorem 8. The polygon c () is the image of under the homothety h (, 1 2). roof. By Lemma 7, the circumcentero i ofq i is the midpoint ofa i (see Figure 12 for a pentagon ). O 1 H 5 O 2 O 5 O 3 O 4 H 4 A 4 Figure 12 Note that by varying the polygons c () are all congruent to each other (by translation). Theorem 9. The polygon a () is the medial polygon of H, with vertices the midpoints of the segmentsh i H i+1 fori = 1,2,...,n. A 1 H 5 A 2 A 5 H 4 A A 3 4 A 4 Figure 13

11 edal polygons 163 roof. By Lemma 7, the anticenter A i of Q i is the midpoint of H i H i+3 (see Figure 13 for a pentagon). Corollary 10. (a) Ifis a triangle, a () is the medial triangle of H. (b) If is a quadrilateral, a () is the Varignon parallelogram of H. Theorem 11. IfHis cyclic, the Euler lines of the quadrilateralsq i are concurrent at the circumcenter of H. roof. The Euler line of the quadrilateral Q i passes through the circumcenter O i ofq i and through the anticentera i ofq i, that is the midpoint ofh i H i+3, then it is the perpendicular bisector of a side ofh(see Figure 14 for a quadrilateral). H 4 O 1 A 1 A 4 A 4 O 4 O 2 A 2 A 3 O 3 Figure 14 Corollary 12. If is a triangle, the Euler lines of the quadrilaterals Q i are concurrent at the circumcenter of H (see Figure 15). Remark. If is a quadrilateral and H is not cyclic, the Euler lines of the quadrilateralsq i bound a quadrilateral affine to H ([4, p.471]). References [1] O. Bottega, Topics in Elementary Geometry, second edition, Springer, [2] H.S.M. Coxeter and S.L. Greitzer, Geometry revisited, Math. Assoc. America, [3] R. Honsberger, Episodes in nineteenth and twentieth century Euclidean geometry, Math. Assoc. America, [4] M. F. Mammana and B. Micale, Quadrilaterals of Triangle Centres, The Mathematical Gazette, 92 (2008)

12 164 D. Ferrarello, M. F. Mammana and M. ennisi O 1 A 1 A 3 A 2 O 3 O 2 Figure 15 [5] M. F. Mammana, B. Micale and M. ennisi, Orthic Quadrilaterals of a Convex Quadrilateral, Forum Geom., 10 (2010) [6]. Yiu, Notes on Euclidean Geometry, Florida Atlantic University Lecture Notes, 1998; available at math.fau.edu/yiu/geometry.html. Daniela Ferrarello: Department of Mathematics and Computer Science, University of Catania, Viale A. Doria 6, 95125, Catania, Italy address: ferrarello@dmi.unict.it Maria Flavia Mammana: Department of Mathematics and Computer Science, University of Catania, Viale A. Doria 6, 95125, Catania, Italy address: fmammana@dmi.unict.it Mario ennisi: Department of Mathematics and Computer Science, University of Catania, Viale A. Doria 6, 95125, Catania, Italy address: pennisi@dmi.unict.it

Example G1: Triangles with circumcenter on a median. Prove that if the circumcenter of a triangle lies on a median, the triangle either is isosceles

1 Example G1: Triangles with circumcenter on a median. Prove that if the circumcenter of a triangle lies on a median, the triangle either is isosceles or contains a right angle. D D 2 Solution to Example