Forum Geometricorum b Volume 13 (2013) 209–218. b b FORUM GEOM ISSN 1534-1178 Three Natural Homoteties of The Nine-Point Circle Mehmet Efe Akengin, Zeyd Yusuf Koro¨ glu,˘ and Yigit˘ Yargic¸ Abstract. Given a triangle with the reflections of its vertices in the opposite sides, we prove that the pedal circles of these reflections are the images of nine- point circle under specific homoteties, and that their centers form the anticevian triangle of the nine-point center. We also construct two concentric circles asso- ciated with the pedals of these reflections on the sidelines, and study the triangle bounded by the radical axes of these pedal circles with the nine-point circle. 1. Three pedal circles Given a triangle ABC with angles α, β, γ, circumcenter O, orthocenter H, and nine-point center N, we let Ma, Mb, Mc be the midpoints of the sides BC, CA, AB, Ha, Hb, Hc the pedals of A on BC, B on CA, C on AB respectively. Consider also the reflections A′ of A in BC, B′ of B in CA, and C′ of C in AB. Our first result (Theorem 3 below) is about the pedal circles of A′, B′, C′ with respect to triangle ABC. Construct the circle through O, B, C, and let Oa be the second intersection of this circle with the line AO. ′ Proposition 1. Oa and A are the isogonal conjugates in triangle ABC. A H O B Ha C ′ Ba Ca ′ A Oa Figure 1 Proof. Clearly the lines AOa and AH are isogonal with respect to angle A, since O and H are isogonal conjugates. Also, ′ ′ a a ∠A BCa = 2∠A ACa = 2∠HAB = 2∠OAC = ∠O OC = ∠O BC. Publication Date: November 12, 2013. Communicating Editor: Paul Yiu. 210 M. E. Akengin, Z. K. Koro¨ glu˘ and Y. Yargic¸ Therefore, the lines A′B and OaB are symmetric in the external bisector of an- gle B, and so are isogonal with respect to angle B. Similarly, A′C and OaB are isogonal with respect to angle C. This shows that A′ and Oa are isogonal conju- gates. ′ The points A and Oa have a common pedal circle, with center at the midpoint a ′ N of A Oa. ′ Proposition 2. OaA is parallel to OH. A Hb Mc Mb Hc N O H Ha Xa B C Ba ′ ′ Ca Ba Ca ′ A a N Oa Figure 2 ′ ′ Proof. Let Xa, Ba, Ca be the pedals of Oa on BC, CA, AB respectively. From AMb AO AMc ′ = = ′ , ABa AOa ACa ′ ′ ′ ′ we have BaCa//MbMc//BC. Therefore the cyclic quadrilateral BaCaHaXa, having a pair of parallel sides, must be a symmetric trapezoid. Now, ∠ ′ ∠ ′ ∠ ′ ∠ ′ CaXaOa = CaBOa = CBA = β = CaOaXa. ′ The second equality is valid because OaB and A B are isogonal with respect ′ to B, and the last one because B, Xa, Oa, Ca are concyclic. It follows that ′ ′ ′ ′ ′ ′ ′ CaOa = CaXa = BaHa. Similarly, BaOa = CaHa. Therefore, CaOaBaHa ′ ′ is a parallelogram, and BaHa is parallel to OaCa, and also to CH, being all per- pendicular to AB. Since Mb and Mc are the midpoints of AC and AB, we have AO AMb AC AH AH = ′ = · ′ = · = ′ . AOa ABa 2 ABa 2 AHa AA ′ Therefore, OaA is parallel to OH. Three natural homoteties of the nine-point circle 211 ′ Theorem 3. The pedal circle of A (and Oa) is the image of the nine-point circle h 2 sin β sin γ of ABC under the homothety (A, ta), where ta = cos α . ′ ′ Proof. The circle BaBaCa is homothetic to the nine-point circle HbMbMc at A since AB AA′ AO AB′ AC′ a = = a = a = a . AHb AH AO AMb AMc The ratio of homothety is ′ AA 2 · AHa 2R sin β sin γ sin β sin γ ta = = = = . AH 2 · OMa 2R cos α cos α . a ′ ′ Since the center N of the pedal circle of A and Oa is the midpoint of OaA , the line AN a intersects OH at its midpoint N, the nine-point center of ABC. Analogously let Ob, Oc be the second intersections of the circles OCA, OAB ′ with the lines BO, CO respectively. The common pedal circle of B and Ob has b ′ ′ c center N the midpoint of ObB and that of C and Oc has center N the midpoint ′ of OcC . These pedal circles are images of the nine-point circle under the homoth- h h sin γ sin α sin α sin β eties (B,tb) and (C,tc) with tb = cos β and tc = cos γ respectively. Theorem 4. N aN bN c is the anticevian triangle of the nine-point N. B′ ′ C Nb A Nc Ob Oc H N O B C A′ Na Oa Figure 3 a ′ Proof. Since N is the midpoint of OaA and the nine-point center N is the mid- b c point of OH, by Proposition 2, A, N, Na are collinear. Similarly, N and N are on the cevians BN and CN respectively. We show that the line N bN c, N cN a, 212 M. E. Akengin, Z. K. Koro¨ glu˘ and Y. Yargic¸ N aN b contain A, B, C respectively. From this the result follows. It is enough to b c ′ show that N N contains A. For this, note that B , A, Oc are collinear because ′ ′ ∠B AB + ∠BAOc = 2∠B AC + ∠BOOc ◦ = 2α + (180 − ∠BOC) ◦ = 2α + (180 − 2α) ◦ = 180 . ′ ′ ′ Similarly, Ob, A, and C are collinear. Therefore, the midpoints of ObB and OcC , namely, N b and N c, are collinear with A. 2. Two concentric circles associated with six pedals ′′ ′′ ′′ Let A B C be the triangle bounded by the lines BaCa, CbAb, and AcBc. Theorem 5. The incenter of triangle A′′B′′C′′ is the orthocenter of the orthic triangle HaHbHc, and the incircle touches the sides at the midpoints Pa, Pb, Pc of the segments BaCa, CbAb, AcBc respectively. A′′ Bc Cb C′ A B′ Pc Pb Ua Hc O Hb N H Ho Ac B Ab Ha C B′′ Ca Pa Oa Ba N a ′ A C′′ Figure 4. Three natural homoteties of the nine-point circle 213 Proof. We first claim that the segments BaCa, CbAb, AcBc have equal lengths. Note that the homothety h(A, ta) maps Hb, Hc to Ba, Ca respectively. Hence, sin β sin γ B C = t · H H = · 2R sin α cos α = 4R sin α sin β sin γ. a a a b c cos α Since this expression is symmetric in α, β, γ, it also gives the lengths of CbAb and AcBc. Note that the corresponding sidelines of triangles A′′B′′C′′ and the orthic trian- gle HaHbHc are parallel. The two triangles are homothetic. By parallelism, ′′ ′′ ∠A AbAc = ∠HcHaA = ∠CHaHb = ∠AbAcC . ′′ ′′ ′′ Therefore, A AcAb is an isosceles triangle with A Ab = A Ac. Since AbCb = ′′ ′′ ′′ ′′ ′′ AcBc, we deduce that A Pb = A Pc. Similarly, B Pc = B Pa and C Pa = ′′ C Pb. Hence, Pa, Pb, Pc are the points of tangency of the incircle of triangle A′′B′′C′′ with its sides. Next we claim that Ha, Na and Pa all lie on a line perpendicular to BaCa. Let Ua be the midpoint of AH. Since NUa is parallel to OA, it is perpendicular to BbHc. As NHb = NHc, the line NUa is the perpendicular bisector of HbHc. a a The homothety h(A, ta) maps UaHbNHc into HaBaN Ca, and HaN is the perpendicular bisector of BaCa. Therefore, it passes through the midpoint Pa of a BaCa. Since HaN is perpendicular to HbHc, it passes through the orthocenter b of the orthic triangle HaHbHc. The same is true for the other two lines HbN and c ′′ ′′ ′′ ′′ HcN , which are the perpendiculars to the sides C A and A B at the points Pb ′′ ′′ ′′ and Pc respectively. Therefore, the incenter of A B C is the orthocenter of the orthic triangle. Remarks. (1) The common length of BaCa, CbAb, AcBc is also the perimeter of the orthic triangle, being 4R sin α sin β sin γ = R(sin2α + sin2β + sin2γ). (2) The orthocenter of the orthic triangle is the triangle center X(52) in [3]. a b c Corollary 6. The lines N Ha, N Hb, N Hc are concurrent at Ho. Theorem 7. The six pedals Ab, Ac, Bc, Ba, Ca, Cb lie on a circle with center Ho. Proof. From Theorem 5, we have HoPa = HoPb = HoPc. Also recall from the proof of the same theorem, the segments BaCa, CbAb, AcBc have equal lengths. Therefore, HoBaCa, HoCbAb, and HoAcBc are congruent isosceles triangles, and Ho is the center of a circle containing these six pedals (see Figure 4). ′′ ′′ ′′ Theorem 8. The triangles ABC, A B C , and PaPbPc are perspective at the symmedian point of triangle ABC Proof. (1) Since AAcAb and ABcCb are isosceles triangles, BcCb and AcAb are parallel, and the triangles ACbBc and ABC are homothetic (see Figure 5). Now, ′′ ′′ ∠A BcCb = ∠A AcAb = ∠HbHaC = α = ∠BcACb. ′′ ′′ ′′ Similarly, ∠A CbBc = ∠BcACb. Therefore, A Bc and A Cb are tangents from ′′ ′′ A to the circumcircle of triangle ACbBc. The line A A is a symmedian of triangle ′′ ACbBc. Since ABC and ACbBc are homothetic at A, the same line A A is a 214 M. E. Akengin, Z. K. Koro¨ glu˘ and Y. Yargic¸ A′′ Bc Cb B′ C′ A Pc Pb Hb Hc H K O Ac B Ha Ab C C′′ Ba Pa Ca B′′ A′ Figure 5. symmedian of triangle ABC, and it contains the symmedian point K of triangle ABC.