Three Natural Homoteties of The Nine-Point Circle
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1 Forum Geometricorum Volume 13 (2013) FRUM GEM ISS Three atural omoteties of The ine-point ircle Mehmet Efe kengin, Zeyd Yusuf Köroğlu, and Yiğit Yargiç bstract. 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 ninepoint circle under specific homoteties, and that their centers form the anticevian triangle of the nine-point center. We also construct two concentric circles associated 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 with angles α, β, γ, circumcenter, orthocenter, and nine-point center, we let M a, M b, M c be the midpoints of the sides,,, a, b, c the pedals of on, on, on respectively. onsider also the reflections of in, of in, and of in. ur first result (Theorem 3 below) is about the pedal circles of,, with respect to triangle. onstruct the circle through,,, and let a be the second intersection of this circle with the line. Proposition 1. a and are the isogonal conjugates in triangle. a a a a Figure 1 Proof. learly the lines a and are isogonal with respect to angle, since and are isogonal conjugates. lso, a = 2 a = 2 = 2 = a = a. Publication Date: ovember 12, ommunicating Editor: Paul Yiu.
2 210 M. E. kengin, Z. K. Köroğlu and Y. Yargiç Therefore, the lines and a are symmetric in the external bisector of angle, and so are isogonal with respect to angle. Similarly, and a are isogonal with respect to angle. This shows that and a are isogonal conjugates. The points and a have a common pedal circle, with center at the midpoint a of a. Proposition 2. a is parallel to. M c c b M b a X a a a a a a a Figure 2 Proof. LetX a, a, a be the pedals of a on,, respectively. From M b a = = M c a a, we have a a//m b M c //. Therefore the cyclic quadrilateral a a a X a, having a pair of parallel sides, must be a symmetric trapezoid. ow, ax a a = a a = = β = a a X a. The second equality is valid because a and are isogonal with respect to, and the last one because, X a, a, a are concyclic. It follows that a a = ax a = a a. Similarly, a a = a a. Therefore, a a a a is a parallelogram, and a a is parallel to a a, and also to, being all perpendicular to. SinceM b and M c are the midpoints of and, we have = M b a a = 2 a Therefore, a is parallel to. = 2 a =.
3 Three natural homoteties of the nine-point circle 211 Theorem 3. The pedal circle of (and a ) is the image of the nine-point circle of under the homothety h(,t a ), wheret a = 2sinβsinγ cosα. Proof. The circle a a a is homothetic to the nine-point circle b M b M c at since a = b = a = a = a. M b M c The ratio of homothety is t a = = 2 a = 2Rsinβsinγ 2 M a 2Rcosα = sinβsinγ cosα.. Since the center a of the pedal circle of and a is the midpoint of a, the line a intersects at its midpoint, the nine-point center of. nalogously let b, c be the second intersections of the circles, with the lines, respectively. The common pedal circle of and b has center b the midpoint of b and that of and c has center c the midpoint of c. These pedal circles are images of the nine-point circle under the homotheties h(,t b ) andh(,t c ) with t b = sinγsinα cosβ and t c = sinαsinβ cos γ Theorem 4. a b c is the anticevian triangle of the nine-point. respectively. b c c b a a Figure 3 Proof. Since a is the midpoint of a and the nine-point center is the midpoint of, by Proposition 2,,, a are collinear. Similarly, b and c are on the cevians and respectively. We show that the line b c, c a,
4 212 M. E. kengin, Z. K. Köroğlu and Y. Yargiç a b contain,, respectively. From this the result follows. It is enough to show that b c contains. For this, note that,, c are collinear because + c = 2 + c = 2α+(180 ) = 2α+(180 2α) = 180. Similarly, b,, and are collinear. Therefore, the midpoints of b and c, namely, b and c, are collinear with. 2. Two concentric circles associated with six pedals Let be the triangle bounded by the lines a a, b b, and c c. Theorem 5. The incenter of triangle is the orthocenter of the orthic triangle a b c, and the incircle touches the sides at the midpointsp a,p b,p c of the segments a a, b b, c c respectively. c b P c U a P b c o b c a b a P a a a a Figure 4.
5 Three natural homoteties of the nine-point circle 213 Proof. We first claim that the segments a a, b b, c c have equal lengths. ote that the homothety h(, t a ) maps b, c to a, a respectively. ence, a a = t a b c = sinβsinγ 2Rsinαcosα = 4Rsinαsinβsinγ. cosα Since this expression is symmetric inα,β,γ, it also gives the lengths of b b and c c. ote that the corresponding sidelines of triangles and the orthic triangle a b c are parallel. The two triangles are homothetic. y parallelism, b c = c a = a b = b c. Therefore, c b is an isosceles triangle with b = c. Since b b = c c, we deduce that P b = P c. Similarly, P c = P a and P a = P b. ence, P a, P b, P c are the points of tangency of the incircle of triangle with its sides. ext we claim that a, a and P a all lie on a line perpendicular to a a. Let U a be the midpoint of. Since U a is parallel to, it is perpendicular to b c. s b = c, the line U a is the perpendicular bisector of b c. The homothety h(, t a ) maps U a b c into a a a a, and a a is the perpendicular bisector of a a. Therefore, it passes through the midpoint P a of a a. Since a a is perpendicular to b c, it passes through the orthocenter of the orthic triangle a b c. The same is true for the other two lines b b and c c, which are the perpendiculars to the sides and at the points P b and P c respectively. Therefore, the incenter of is the orthocenter of the orthic triangle. Remarks. (1) The common length of a a, b b, c c is also the perimeter of the orthic triangle, being 4Rsinαsinβsinγ = R(sin2α+sin2β +sin2γ). (2) The orthocenter of the orthic triangle is the triangle center X(52) in [3]. orollary 6. The lines a a, b b, c c are concurrent at o. Theorem 7. The six pedals b, c, c, a, a, b lie on a circle with center o. Proof. From Theorem 5, we have o P a = o P b = o P c. lso recall from the proof of the same theorem, the segments a a, b b, c c have equal lengths. Therefore, o a a, o b b, and o c c are congruent isosceles triangles, and o is the center of a circle containing these six pedals (see Figure 4). Theorem 8. The triangles,, and P a P b P c are perspective at the symmedian point of triangle Proof. (1) Since c b and c b are isosceles triangles, c b and c b are parallel, and the triangles b c and are homothetic (see Figure 5). ow, c b = c b = b a = α = c b. Similarly, b c = c b. Therefore, c and b are tangents from to the circumcircle of triangle b c. The line is a symmedian of triangle b c. Since and b c are homothetic at, the same line is a
6 214 M. E. kengin, Z. K. Köroğlu and Y. Yargiç c b P c b P b c K c a b a P a a Figure 5. symmedian of triangle, and it contains the symmedian point K of triangle. The same reasoning shows that and also contain K. Therefore, triangles and are perspective at K. (2) In triangle, a a is antiparallel to since a a = a b = a b The the reflection of triangles a a in the bisector of angle is homothetic to. Therefore, the median P a of triangle a a is the same as the symmedian K; similarly for P b and P c. The three lines are concurrent at the symmedian pointk. 3. triangle bounded by three radical axes Let L a, L b, L c be the radical axes of the nine-point circle with the pedal circles of,, respectively. These lines bound a triangle Q a Q b Q c. The vertex Q a is the radical center of the nine-point circle and the pedal circles of and ; similarly for the verticesq b andq c.
7 Three natural homoteties of the nine-point circle 215 b Q a c b c Q c Q b a a Figure 6 Lemma 9. LetJ a be the midpoint of. The linej a M a is perpendicular toq b Q c and contains the midpoint of. J a P M a Figure 7. Proof. Since is the midpoint of, the segment J a is parallel to and therefore to M a. Furthermore, J a = 1 2 = M a. It follows that J a M a intersects at its midpoint.
8 216 M. E. kengin, Z. K. Köroğlu and Y. Yargiç Proposition 10. Given triangle with incentral triangle DEF, extend and to P and Q such that P = = Q. Let T be the midpoint PQ, and M the midpoint of the arc of the circumcircle. (a) The linetm is perpendicular EF. (b)t and T are parallel to DF andde respectively. P M F T D I E Q E Y Figure 8. Proof. (a) y the angle bisector theorem, E = bc bc a+b, F = a+c. Therefore, E F = a+c a+b = Q P, showing that PQ is parallel to EF (see Figure 8). n the other hand, the circumcircles of and PQ intersect at and M, which is the center of the rotation taking the oriented segmentsp andq into each other (see [4, p.5]). Since M = M, M is the center of this rotation. ence, MT is the perpendicular bisector ofpq. We conclude thatmt andef are perpendicular to each other. (b) We show that T is parallel to DF. LetY be the intersection of the linest and. pplying Menelaus theorem to trianglepq with transversalty, we have Y YQ QT TP P Y = 1 = YQ = P = c a = Y Q = c c a. Therefore, Y = c(a+b) c a. ow, DF intersects at E such that E is the E external bisector of angle E. E = c a = E = c c a. It follows that E = c E c a b. From these, Y = b a+b = F. Therefore,T is parallel to DF. The same reasoning shows thatt is parallel to DE. Remark. Proposition 10 remains valid if P and Q are chosen on the rays and instead, ande,f are external bisectors.
9 Three natural homoteties of the nine-point circle 217 Theorem 11. The orthocenter of triangleq a Q b Q c is the midpoint of. Q a b M c D a M b c D c Q c D b a M a Q b Figure 9 Proof. It is enough to prove that Q a M a is parallel to. Let D a = b c, D b = c a, D c = a b. We claim that D b D c is perpendicular to. The points D b and D c have equal powers with respect to the nine-point circle of and the circumcircle of. Therefore, the line D b D c is the radical axis of the these two circles. The circumcenter of is the reflection of in, and form a parallelogram with,,, with as the common midpoint of the diagonals. Therefore is the line joining the centers of the nine-point circle and the center of the circle, and is perpendicular to the radical axis D b D c. The line also contains the center a of the circle Γ a. Therefore the radical axesq b Q c andd b D c are parallel, and is perpendicular to Q b Q c. ow we show that Q a M a is parallel to.
10 218 M. E. kengin, Z. K. Köroğlu and Y. Yargiç It is easy to see that D a D b D c is the incentral triangle of a b c. (If triangle is obtuse, then the two bisectors not corresponding to obtuse angle have to be replaced by external bisectors; see Remark following Proposition 10). pplying Proposition 10 to the orthic triangle a b c, the lines Q a Q b and Q a Q c are parallel to D a D b and D a D c respectively, and the midpoint of the arc b a c is M a, the midpoint of. Therefore, Q a M a is perpendicular to D b D c, which is parallel to Q b Q c. The lines Q a M a, Q b M b, Q c M c are the altitudes of the triangle Q a Q b Q c. ut these lines are parallel to,, respectively. They are concurrent at the midpoint of. Remark. The midpoint of is the triangle center X(140) in [3]. References [1] [2] R. onsberger, Episodes in ineteenth and Twentieth entury Euclidean Geometry, Math. ssoc. merica, [3]. Kimberling, Encyclopedia of Triangle enters, available at [4] Y. Zhao, Three Lemmas in Geometry, available at: web.mit.edu/yufeiz/www/olympiad/three geometry lemmas.pdf. Mehmet Efe kengin: Istanbul Lisesi (Istanbul igh School), Türkocaği addesi, o:4, Fatih Istanbul, Turkey address: mehmetefeakengin@hotmail.com Zeyd Yusuf Köroğlu: Istanbul Lisesi (Istanbul igh School), Türkocaği addesi, o:4, Fatih Istanbul, Turkey address: zeyd.yusuf@gmail.com Yiğit Yargiç: Ludwig Maximillian University of Munich, Geschwister-Scholl-Platz 1, München, Germany address: yigityargic@hotmail.com
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