On the Complement of the Schiffler Point

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1 Forum Geometricorum Volume 5 (005) FORUM GEOM ISSN On the omplement of the Schiffler Point Khoa Lu Nguyen bstract. onsider a triangle with excircles (), ( ), (), tangent to the nine-point circle respectively at F a, F b, F c. onsider also the polars of,, with respect to the corresponding excircles, bounding a triangle XY Z. We present, among other results, synthetic proofs of (i) the perspectivity of XY Z and F af b F c at the complement of the Schiffler point of, (ii) the concurrency at the same point of the radical axes of the nine-point circles of triangles,, and. 1. Introduction onsider a triangle with excircles ( ), ( ), ( ). It is well known that the nine-point circle (W ) is tangent externally to the each of the excircles. Denote by F a, F b, and F c the points of tangency. onsider also the polars of the vertices with respect to ( ), with respect to ( ), and with respect to ( ). These are the lines a a, b b, and c c joining the points of tangency of the excircles with the sidelines of triangle. Let these polars bound a triangle XY Z. See Figure 1. Juan arlos Salazar [1] has given the following interesting theorem. Theorem 1 (Salazar). The triangles XY Z and F a F b F c are perspective at a point on the Euler line. Darij Grinberg [3] has identified the perspector as the triangle center X 44 of [6], the complement of the Schiffler point. Recall that the Schiffler point S is the common point of the Euler lines of the four triangles I, I, I, and, where I is the incenter of. Denote by,, the midpoints of the sides,, respectively, so that is the medial triangle of, with incenter I which is the complement of I. Grinberg suggested that the lines XF a, YF b and ZF c are the Euler lines of triangles I, I and I respectively. The present author, in [10], conjectured the following result. Theorem. The radical center of the nine-point circles of triangles, and is a point on the Euler line of triangle. Subsequently, Jean-Pierre Ehrmann [1] and Paul Yiu [13] pointed out that this radical center is the same point S, the complement of the Schiffler point S. In this paper, we present synthetic proofs of these results, along with a few more interesting results. Publication Date: October 18, 005. ommunicating Editor: Paul Yiu. The author is extremely grateful to Professor Paul Yiu for his helps in the preparation of this paper.

2 150 K. L. Nguyen X b c c F b F c S W O b c b F a a Z a a Y Figure 1.. Notations a, b, c Lengths of sides,, R, r, s ircumradius, inradius, semiperimeter r a, r b, r c Exradii O, G, W, H, ircumcenter, centroid, nine-point center, orthocenter I, F, S, M Incenter, Feuerbach point, Schiffler point, Mittenpunkt P omplement of P in triangle,, Midpoints of,, 1, 1, 1 Points of tangency of incircle with,,,, Excenters F a, F b, F c Points of tangency of the nine-point circle with the excircles a, a, a Points of tangency of the -excircle with the lines,, ; similarly for b, b, b and c, c, c W a, W b, W c Nine-point centers of,, M a, M b, M c Midpoints of,, X b b c c ; similarly for Y, Z X b, X c Orthogonal projections of on and on ; similarly for Y c, Y a, Z a, Z b J a Midpoint of arc of circumcircle not containing ; similarly for J b, J c K a b F b c F c ; similarly for K b, K c

3 On the complement of the Schiffler point 151 b c c H S O b c a b a a Figure. 3. Some preliminary results We shall make use of the notion of directed angle between two lines. Given two lines a and b, the directed angle (a, b) is the angle of counterclockwise rotation from a to b. It is defined modulo 180. We shall make use of the following basic properties of directed angles. For further properties of directed angles, see [7]. Lemma 3. (i) For arbitrary lines a, b, c, (a, b)+(b, c) (a, c) mod 180. (ii) Four points,,, D are concyclic if and only if (, ) =(D, D). Lemma 4. Let (O) be a circle tangent externally to two circles (O a ) and (O b ) respectively at and. IfPQ is a common external tangent of (O a ) and (O b ), then the quadrilateral P Q is cyclic, and the lines P, Q intersect on the circle (O). Proof. Let Pintersect (O) at K. Since (O) and (O a ) touch each other externally at, OK is parallel to O a P. On the other hand, O a P is also parallel to O b Q as they are both perpendicular to the common tangent PQ. Therefore KO is parallel

4 15 K. L. Nguyen K O O b O a P Figure 3 Q to O b Q in the same direction. This implies that K,, Q are collinear since (O b ) and (O) touch each other at externally at. Therefore (PQ,Q)= 1 (QO b,o b )= 1 (KO,O)=(K,)=(P,), and P Q is cyclic. We shall make use of the following results. Lemma 5. Let be a triangle inscribed in a circle (O), and points M and N lying on and respectively. The quadrilateral NM is cyclic if and only if MN is perpendicular to O. Theorem 6. The nine-point circles of,,, and intersect at the point F a. F a a Figure 4.

5 On the complement of the Schiffler point 153 Proposition 7. The circle with diameter a M a contains the point F a. F a a M a M b M c Figure 5. Proof. Denote by M b and M c the midpoints of and respectively. The point F a is common to the nine-point circles of, and. See Figure 5. We show that ( a F a,f a M a )=90. ( a F a,f a M a )=( a F a,f a M b )+(M b F a,f a M a ) =( a M c,m c M b)+(m b, M a ) = ( M c,m c M b ) (, ) = ((, )+(, )) = Some properties of triangle XY Z In this section we present some important properties of the triangle XY Z Homothety with the excentral triangle. Since YZ and are both perpendicular to the bisector of angle, they are parallel. Similarly, ZX and XY are parallel to and respectively. The triangle XY Z is therefore homothetic to the excentral triangle. See Figure 7. We shall determine the homothetic center in Theorem 11 below.

6 154 K. L. Nguyen 4.. Perspectivity with. onsider the orthogonal projections P and P of and X on the line. Wehave c P : P b =(s c)+c cos :(s b)+b cos = s b : s c by a straightforward calculation. X P P Figure 6. On the other hand, c P : P b =cotx c b :cotx b c ( =cot 90 ) ( :cot 90 ) =tan :tan = 1 s c : 1 s b =s b : s c. It follows that P and P are the same point. This shows that the line X is perpendicular to and contains the orthocenter H of triangle. The same is true for the lines Y and ZX. The triangles XY Z and are perspective at H The circumcircle of XY Z. pplying the law of sines to triangle X c,we have X =(s b) sin ( 90 ) sin =(s b)cot = r a. It follows that HX =Rcos + r a =R + r. See Figure 4. Similarly, HY = HZ = R + r. Therefore, triangle XY Z has circumcenter H and circumradius R + r.

7 On the complement of the Schiffler point 155 X b c c b H b c a Z a a Y Figure The Taylor circle of the excentral triangle onsider the excentral triangle with its orthic triangle. The orthogonal projections Y a and Z a of on and, Z b and X b of on and, together with X c and Y c of on and are on a circle called the Taylor circle of the excentral triangle. See Figure 8. Proposition 8. The points X b, X c lie on the line YZ. Proof. The collinearity of a, X b, X c follows from ( a X b,x b )=( a, ) =( a,)+(, ) =90 +(,) =(X c, )+(,) =(X c, ) =(X c X b,x b ). Similarly, X b is also on the line YZ, and Z a, Z b are on the line XY, Y c, Y a are on the line XZ. Proposition 9. The line Y a Z a contains the midpoints, of,, and is parallel to.

8 156 K. L. Nguyen X b c Y c Z b c Z a b Y a c a b Z a X b a X c Y Figure 8. Proof. Since, Y a,, Z a are concyclic, (Y a,y a Z a )=(, Z a )= =(,Y a). Therefore, the intersection of and Y a Z a is the circumcenter of the right triangle Y a, and is the midpoint of. Similarly, the intersection of and Y a Z a is the midpoint of. Proposition 10. The line X contains the midpoint of. Proof. Since the diagonals of the parallelogram Y a XZ a bisect each other, the line X passes through the midpoint of the segment Y a Z a. Since Y a Z a and are parallel, with on Z a and on Y a, the same line X also passes through the midpoint of the segment. Theorem 11. The triangles XY Z and are homothetic at the Mittenpunkt M of triangle, the ratio of homothety being R + r : R. Proof. The lines X, Y, Z contain respectively the midpoints of,, of,,. They intersect at the common point of,,, the Mittenpunkt M of triangle. This is the homothetic center of the triangles XY Z and. The ratio of homothety of the two triangle is the same as the ratio of their circumradii.

9 On the complement of the Schiffler point 157 Theorem 1. The Taylor circle of the excentral triangle is the radical circle of the excircles. Proof. The perpendicular bisector of Y c Z b is a line parallel to the bisector of angle and passing through the midpoint of. This is the -bisector of the medial triangle. Similarly, the perpendicular bisectors of Z a X c and X b Y a are the other two angle bisectors of the medial triangle. These three intersect at the incenter of the medial triangle, the Spieker center of. It is well known that S p is also the center of the radical circle of the excircles. To show that the Taylor circle coincides with the radical circle, we show that they have equal radii. This follows easily from X c Z a = r a sin cos 6. Proofs of Theorems 1 and cos = r a sin = r a. We give a combined proof of the two theorems, by showing that the line XF a is the radical axis of the nine-point circles (W b ) and (W c ) of triangles and. In fact, we shall identify some interesting points on this line to show that it is also the Euler line of triangle I XF a as the radical axis of (W b ) and (W c ). Proposition 13. X lies on the radical axis of the circles (W b ) and (W c ). Proof. y Theorem 1, XZ a XZ b = XY a XY c. Since Y c, Y a are on the ninepoint circle (W b ) and Z a, Z b on the the circle (W c ), X lies on the radical axis of these two nine-point circles. Since Z a and Y a are perpendicular to and, and and XY Z are homothetic, is the orthocenter of triangle XY a Z a. It follows that X is the orthocenter of Y a Z a. Since (Y a,y a )=(Z a,z a )=90, the triangle Y a Z a has circumcenter the midpoint M a of. It follows that XM a is the Euler line of triangle Y a Z a. Proposition 14. M a lies on the radical axis of the circles (W b ) and (W c ). Proof. Let M b and M c be the midpoints of and respectively. See Figure 9. Note that these lie on the nine-point circles (W b ) and (W c ) respectively. Since,,, are concyclic, we have =. pplying the homthety h(, 1 ), we have the collinearity of M a,, M c, and of M a,, M b, Furthermore, M a M a M c = M a M a M b. This shows that M a lies on the radical axis of (W b ) and (W c ).

10 158 K. L. Nguyen b c M b M c W b W c c H O b b c a Z W a a a Figure 9. Proposition 15. X, F a, and M a are collinear. Proof. We prove that the Euler line of triangle Y a Z a contains the point F a. The points X and M a are respectively the orthocenter and circumcenter of the triangle. Let a be the antipode of a on the -excircle. Since X has length r a and is perpendicular to, X a is a parallelogram. Therefore, X a contains the midpoint M a of. y Proposition 7, ( a F a,f a M a )=90. learly, ( a F a,f a a )=90. This means that F a, M a, and a are collinear. The line containing them also contains X. Proposition 16. XF a is also the Euler line of triangle Y a Z a. Proof. The circumcenter of Y a Z a is clearly M a. On the other hand, since is the orthocenter of triangle XY a Z a, X is the orthocenter of triangle Y a Z a. Therefore the line XM a, which also contains F a, is the Euler line of triangle Y a Z a. 6.. XF a as the Euler line of triangle I.

11 On the complement of the Schiffler point 159 Proposition 17. M a is the orthocenter of triangle I. H a G 1 a M a Figure 10. Proof. Let H a be the orthocenter of I. Since H a is perpendicular to I,it is parallel to. Similarly, H a is parallel to. Thus, H a is a parallelogram, and is the midpoint of H a. onsider triangle H a which has M a and for the midpoints of two sides. The intersection of M a H a and is the centroid of the triangle, which coincides with G. Furthermore, GH a : GM a = G : G =: 1. Hence, M a is the orthocenter of I. Proposition 18. K a is the circumcenter of I. Proof. y Lemma 4, the points F b, F c, b and c are concyclic, and the lines b F b and c F c intersect at a point K a on the nine-point circle, which is the midpoint of the arc not containing. See Figure 11. The image of K a under h(g, ) is J a, the circumcenter of I. It follows that K a is the circumcenter of I. Proposition 19. K a lies on the radical axis of (W b ) and (W c ). Proof. Let D and E be the second intersections of K a F b with (W b ) and K a F c with (W c ) respectively. We shall show that K a F b K a D = K a F c K a E. Since c, F c, F b, b are concyclic, we have K a F c K a c = K a F b K a b = k, say. Note that c E c F c = c Z a c Z b = (s a) sin ( + ) tan cos.

12 160 K. L. Nguyen b c c K a F b F c W b b c F a a a a Figure 11. Since ( ) and (W ) extouch at F c,wehave KaFc cf c = R r a. Therefore, c E = K af c c E c F c K a c c F c K a F c K a c = R (s ( ) a) sin + r a k tan cos = R(s a) sin ( + ) k s tan tan cos. Similarly, b D = R(s ( ) a) sin + K a b k s tan tan cos. Since sin ( + ) ( ) =sin +, it follows that b D K a b = ce K a c. Hence, DE is parallel to b c. From K a F b K a b = K a F c K a c,wehavek a F b K a D = K a F c K a E. This shows that K a lies on the radical axis of (W b ) and (W c ). orollary 0. K a lies on the line XF a Proof of Theorems 1 and. We have shown that the line XF a is the radical axis of (W b ) and (W c ). Likewise, YF b is that of (W c ), (W a ), and ZF c that of (W a ), (W b ). It follows that the three lines are concurrent at the radical center of the three circles. This proves Theorem 1.

13 On the complement of the Schiffler point 161 We have also shown that the line XF a is the image of the Euler line of I under the homothety h(g, 1 ); similarly for the lines YF b and ZF c. Since the Euler lines of I, I, and I intersect at the Schiffler point S on the Euler line of, the lines XF a, YF b, ZF c intersect at the complement of the Schiffler point S, also on the same Euler line. This proves Theorem. 7. Some further results Theorem 1. The six points Y, Z, b, c, F b, F c are concyclic. X b c c K a F b F c W b b c F a a Z J a a a X a Y V Figure 1. Proof. (i) The points b, c, F b, F c are concyclic and the lines b F b, c F c meet at K a. Let X a be the circumcenter of K a b c. Since F b and F c are points on K a b and K a c, and F b b c F c is cyclic, it follows from Lemma 5 that K a X a is perpendicular to F b F c. Hence X a is the intersection of the perpendicular from K a to F b F c and the perpendicular bisector of. Since triangle K a b c is similar to K a F c F b, and b c = b + c, its circumradius is b + c R F b F c = 1 (R +rb )(R +r c ).

14 16 K. L. Nguyen Here, we have made use of the formula F b F c = b + c (R +rb )(R +r c ) R from []. (ii) simple angle calculation shows that the points Y, Z, b, c are also concyclic. Its center is the intersection of the perpendicular bisectors of b c and YZ. The perpendicular bisector of b c is clearly the same as that of. Since YZis parallel to, its perpendicular is the parallel through H (the circumcenter of XY Z) to the bisector of angle. (iii) Therefore, if this circumcenter is V, then J a V = H =R cos. (iv) To show that the two circle F b b c F c is the same as the circle in (ii), it is enough to show that V lies on the perpendicular bisector of F b F c. This is equivalent to showing that VW is perpendicular to F b F c. To prove this, we show that K a WVX a is a parallelogram. pplying the Pythagorean theorem to triangle b X a,wehave 4 X a =(R +r b)(r +r c ) (b + c) =R +4R(r b + r c )+4r b r c (b + c) =R +4R R(1 + cos )+4s(s a) (b + c) =R (1 + 4(1 + cos )) a =R (1 + 4(1 + cos ) 4sin ) =R (1 + cos ). This means that X a = R (1 + cos ), and it follows that X a V = V X a = J + JV X a =R(1 cos )+Rcos R (1 + cos ) = R = K aw. Therefore, VW, being parallel to K a X a, is perpendicular to F b F c. Denote by a the circle through these 6 points. Similarly define b and c. orollary. The radical center of the circles a, b, c is S. Proof. The points X and F a are common to the circles b and c. The line XF a is the radical axis of the two circles. Similarly the radical axes of the two other two pairs of circles are YF b and ZF c. The radical center is therefore S. Proposition 3. The line X a is perpendicular to YZ.

15 On the complement of the Schiffler point 163 Proof. With reference to Figure 8, note that b Y a : b X = b sin ( + ) sin : b c sin = b :(b + c) = b :(b + c) sin + sin + sin + sin ( + ) sin + sin sin( + )+sin = b : c = b : b a. This means that X a is parallel to Y c, which is perpendicularto and YZ. orollary 4. XY Z is perspective with the extouch triangle a b c, and the perspector is the orthocenter of XY Z. Remark. This is the triangle center X 7 of [6]. Proposition 5. The complement of the Schiffler point is the point S which divides HW in the ratio HS : S W =(R + r) : R. X b c c K a c F c H K b S W F a F b b K c a b Z a a Y Figure 13. Proof. We define K b and K c similarly as K a. Since K b and K c are the midpoints of the arcs and, K b K c is perpendicular to the -bisector of,

16 164 K. L. Nguyen and hence parallel to YZ. The triangle K a K b K c is homothetic to XY Z. The homothetic center is the common point of the lines XK a, YK b, and ZK c, which are XF a, YF b, ZF c. This is the complement of the Schiffler point. Since triangles K a K b K c and XY Z have circumcenters W, H, and circumradii R and R + r, this homothetic center S divides the segment HW in the ratio given above. References [1] J.-P. Ehrmann, Hyacinthos message 10564, October 1, 004. [] L. Emelyanov and T. Emelyanova, note on the Feuerbach point, Forum Geom., 1 (001) [3] D. Grinberg, Hyacinthos message 1034, ugust 31, 004. [4] D. Grinberg, Hyacinthos message 1056, October 1, 004. [5] D. Grinberg, Hyacinthos message 10587, October 3, 004 [6]. Kimberling, Encyclopedia of Triangle enters, available at [7] R.. Johnson, dvanced Euclidean Geometry, 195, Dover reprint. [8] K. L. Nguyen, Hyacinthos message 10384, September, 5, 004. [9] K. L. Nguyen, Hyacinthos message 1050, September, 3, 004. [10] K. L. Nguyen, Hyacinthos message 10563, October 1, 004. [11] K. L. Nguyen, Hyacinthos message 10913, November 8, 004. [1] J.. Salazar, Hyacinthos message 1033, ugust 9, 004. [13] P. Yiu, Hyacinthos message 10565, October 1, 004. Khoa Lu Nguyen: 806 andler Dr, Houston, Texas, , US address: treegoner@yahoo.com

Forum Geometricorum Volume 3 (2003) FORUM GEOM ISSN Harcourt s Theorem. Nikolaos Dergiades and Juan Carlos Salazar

Forum Geometricorum Volume 3 (2003) 117 124. FORUM GEOM ISSN 1534-1178 Harcourt s Theorem Nikolaos Dergiades and Juan arlos Salazar bstract. We give a proof of Harcourt s theorem that if the signed distances