A Note on Reflections

Transcription

1 Forum Geometricorum Volume 14 (2014) FORUM GEOM SSN Note on Reflections Emmanuel ntonio José García bstract. We prove some simple results associated with the triangle formed by the reflections of a point in the midpoints of the sides of a given triangle. Let be a given triangle with midpoints, M b, of the sides,, respectively. onsider a point P and its reflections,, in, M b, respectively. Proposition 1. Triangle is oppositely congruent to at the complement (inferior) of P. M b P G Q Figure 1 Proof. Let G be the centroid of triangle. onsider triangle P. t has the segment as a median, and so has centroid G. f Q is the midpoint of, then PQis another median of triangle P. Therefore, G divides PQin the ratio PG : GQ =2:1, and Q is the complement of P. Similarly, the same point Q is the midpoint of and. t follows that is oppositely congruent to at Q. Publication Date: ugust 26, ommunicating Editor: Paul iu.

2 156 E.. J. García Let,, be the reflections of,, in the sidelines,, respectively. Proposition 2. The circle through,, has center O and contains the P and its reflection in O. Mb P O Figure 2 Proof.,, are the reflections of P in the perpendicular bisectors of,, respectively. Each of these points is equidistant with P from the circumcenter O of triangle. Therefore the circle through P, center O, also contains the reflection of P in O. Theorem 3. f the Euler lines of P, P, P are concurrent at S, then the Euler lines of,, are concurrent at the superior (anticomplement) of S. Proof. Triangle is a translation of triangle P. =( + P ) = P =( + P ) =. learly the two Euler lines of the two triangles are parallel. Since the centroid of is the superior of the centroid of P, every point on the Euler line of is the superior of a point on the Euler line of P. The same is true for the pairs, P and, P. t follows that if the Euler lines of P, P, P are concurrent at S, then those of,, are concurrent at the superior (anticomplement) of S. Remark. t is well known that for P =, the incenter, the Euler lines of,, are concurrent at the Schiffler point (21) on the Euler line of (see [4]). t follows that the Euler lines of,, are concurrent at the superior (anticomplement) of (21) (see Figure 3). This is the triangle center (2475) of [5], also on the Euler line of.

3 note on reflections 157 S GS O Figure 3 On the other hand, iu [8] has noted that, for P =, the Euler lines of,, are concurrent at the cevian quotient Q/, where Q is the Spieker center, the inferior (complement) of. Lemma 4. For P =, the incenter, the line intersects the Euler line of triangle on the side. T c T b G a T a K G a T a a Figure 4 Proof. Let a be the center of the excircle tangent to at T a. Denote by r and r a the inradius and radius of the -excircle. We shall make use of the formulas r = Δ s and r a = Δ s a, where Δ and s are the area and semiperimeter of triangle. Since and have a common midpoint, is a parallelogram. Therefore, and are perpendicular to a and a respectively. From

4 158 E.. J. García this we note that is the orthocenter of triangle a. onsequently, a is the orthocenter of triangle. Let G a be the point dividing in the ratio G a : G a =2:1. This is the centroid of triangle, and a G a is the Euler line. Extend a G a to intersect T a at G a. Since the line T a contains the antipode of T a on the incircle, it is parallel to. Therefore, G a : G at a = TG a : G a = 2:1. Let a intersect at K. onsider triangle T a a with on T a a, K on a, and G a on T a.wehave G a G at a T a ak a K = 2 1 r r a r ra = arr a 2Δ (r a a r)δ = a 1 s 1 s a 1 s a 1 =1. s y eva s theorem,, a G a and T ak are concurrent. This means that and the Euler line a G a of triangle intersect on. Theorem 5. For P =, the incenter, the Euler lines of triangles,, are concurrent at the cevian quotient Q/, where Q is the Spieker center. c b Q a Figure 5

5 note on reflections 159 Proof. The line contains the Spieker center Q as its midpoint. Denote by the cevian triangle of Q. y Lemma 4 above, the Euler line of triangle is a. Similarly, the Euler lines of and are the lines b and c.now, a b c is the anticevian triangle of, and is the cevian triangle of S. These lines are concurrent at the cevian quotient Q/ (see [7, 8.3]). Remark. For Q = the Spieker center, the cevian quotient Q/ is the triangle center (191) of [5]. t is the reflection of in the Schiffler point (21) (see Remark after Theorem 3 above). Theorem 6 (ollings). The circles ( ), (), ( ) are concurrent at a point on the circumcircle of, which is the superior of the center of the rectangular hyperbola through,,, and P. M H H P G P M Figure 6 Proof. ollings [1] actually shows that if is the superior (anticomplementary) triangle of, and P is the superior of P, then the circles ( ), (), ) and () intersect at the center M of the rectangular hyperbola through,,, P (see Figure 3). Since,,, P are the superiors of,,, P respectively, M is the superior of the center M of the rectangular hyperbola through,,, P. This is a point on the nine-point circle of. t follows that M is a point on the circumcircle.

6 160 E.. J. García For example, if P =, the incenter, the rectangular hyperbola through,,, (and H) has center the Feuerbach point F. The common point of the circles ( ), (), ( ) is the inferior of F. This is the point (100). Peter Moses has informed us [2], among other things, that triangle is perspective to the mid-arc triangle at (100) (see Figure 7). The vertices of the mid-arc triangle are the intersections of the angle bisectors with the circumcircle. Since the inferiors (complements) of,, are the midpoints,, of,, respectively, it is enough to prove that is perspective with the mid-arc triangle of the medial triangle M b. F M b N G (100) Figure 7 Proposition 7. Let,, be the midpoints of,, respectively, and M a, M b, M c the midpoints of the the arcs M b,, M b of the nine-point circle of triangle not containing, M b,. The triangles and M am b M c are perspective at the Feuerbach point of triangle. Proof. The nine-point circle of triangle is tangent to the incircle at the Feuerbach point F. Let M a be the midpoint of the arc M b of the nine-point circle (not containing ). FM c = 1 2 M bf = 1 2 M b = 1. (1) 2 On the other hand, if is the midpoint of, then the circle through, and T c is the nine-point circle of triangle, and it passes through the Feuerbach point F as well (see Remark below), and FM b = T b = T b = 1. (2) 2

7 note on reflections 161 M a T b T c N M b F T a Figure 8 t follows from (1) and (2) that, and F are collinear. The same reasoning shows that M b,, and F are collinear, so are, and F. Therefore, the triangles and M b are perspective at the Feuerbach point. Remark. The nine-point circles of,,, and are concurrent at the center of the rectangular hyperbola through,,,. This is the Feuerbach point F. See Theorem 6 above. synthetic proof of this fact can be found in [6]. References [1] S. N. ollings, Reflections on reflections 2, Math. Gazette, 58 (1974) 264. [2] E.. J. García, DGEOM, messages 1208, pril 2, [3] E.. J. García, DGEOM, messages 1226, pril 3, [4]. P. Hatzipolakis, F. M. van Lamoen,. Wolk, and P. iu, oncurrency of four Euler lines, Forum Geom., 1 (2001) [5]. Kimberling, Encyclopedia of Triangle enters, available at [6] J. Vonk, The Feuerbach point and reflections of the Euler line, Forum Geom., 9 (2009) [7] P. iu, ntroduction to the Geometry of the Triangle, Florida tlantic University Lecture Notes, 2001; with corrections, 2013, available at [8] P. iu, DGEOM, message 1227, pril 3, Emmanuel ntonio José García: Libertad, 26, Los Hoyitos, El Seibo, Dominican Republic address: emmanuelgeogarcia@gmail.com