On the Schiffler center

Transcription

1 Forum Geometricorum Volume 4 (2004) FORUM GEOM ISSN On the Schiffler center Chrles Ths Abstrct. Suppose tht ABC is tringle in the Eucliden plne nd I its incenter. Then the Euler lines of ABC, IBC, ICA, nd IAB concur t point S, the Schiffler center of ABC. In the min theorem of this pper we give projective generliztion of this result nd in the finl prt, we construct Schiffler-like points nd lot of other relted centers. Other results in connection with the Schiffler center cn be found in the rticles [] nd [3].. Introduction We recll some formuls nd tools of projective geometry, which will be used in 2. Although we focus our ttention on the rel projective plne, it will be convenient to work in the complex projective plne P... Suppose tht (x,x 2 ) re projective coordintes on complex projective line nd tht two pirs of points re given s follows: P nd P 2 by the qudrtic eqution x 2 +2bx x 2 + cx 2 2 =0 () nd Q nd Q 2 by x 2 +2b x x 2 + c x 2 2 =0. (2) Then the cross-rtio (P P 2 Q Q 2 ) equls iff c 2bb + c =0. (3) Proof. Put t = x x 2 nd ssume tht t,t 2 (t, t 2 respectively) re the solutions of () ((2) respectively), divided by x 2 2. Then (t t 2 t t 2 )= is equivlent to 2(t t 2 + t t 2 )=(t + t 2 )(t + t 2 ) or 2( c + c )=( 2b 2b )( ), which gives (3)..2.. Consider tringle ABC in the complex projective plne P nd ssume tht l is line in P, not through A, B, orc. Put AB l = M C,BC l = M A, nd CA l = M B nd determine the points M C, M A, nd M B by (ABM CM C )=(BCM AM A )=(CAM BM B )=, then AM A, BM B, nd CM C concur t point Z, the so-clled triliner pole of l with regrd to ABC. Publiction Dte: June 28, Communicting Editor: J. Chris Fisher.

2 86 C. Ths Proof. If A =(, 0, 0), B =(0,, 0), C =(0, 0, ), nd l is the unit line x + x 2 + x 3 =0, then M C =(,, 0), M A =(0,, ), M B =(, 0, ), nd M C = (,, 0), M A = (0,, ),M B = (, 0, ), nd Z is the unit point (,, ) The triliner pole Z C of the unit-line l with regrd to ABQ, where A = (, 0, 0),B =(0,, 0), nd Q =(A, B, C), hs coordintes (2A + B + C, A + 2B + C, C). Proof. The point Z C is the intersection of the line QM C nd BM QA, with M C = (,, 0), nd M QA the point of QA, such tht (Q AM QA M QA) =, with M QA = QA l. We find for M QA the coordintes (2A + B + C, B, C), nd strightforwrd clcultion completes the proof..3. Consider non-degenerte conic C in the complex projective plne P, nd two points A, Q, not on C, whose polr lines with respect to C, intersect C t T,T 2, nd I,I 2 respectively. Then Q lies on one of the lines l,l 2 through A which re determined by (AT AT 2 l l 2 )=(AI AI 2 l l 2 )=. Proof. This follows immeditely from the fct tht the pole of the line AQ with respect to C is the point T T 2 I I For ny tringle ABC of P nd line l not through vertex, the Desrgues- Sturm involution theorem ([7, p.34], [8, p.63]) provides one-to-one correspondence between the involutions on l nd the points P in P tht lie neither on l nor on side of the tringle. Specificlly, the conics of the pencil B(A, B, C, P ) intersect l in pirs of points tht re interchnged by n involution with fixed points I nd J. Conversely, P is the fourth intersection point of the conics through A, B, nd C tht re tngent to l t I nd J. The point P cn esily be constructed from A, B, C, I, nd J s the point of intersection of the lines AA, nd BB, where A is the hrmonic conjugte of BC l with respect to I nd J, nd B is the hrmonic conjugte of AC l with respect to I nd J..5. Denote the pencil of conics through the four points A,A 2,A 3, nd A 4 by B(A,A 2,A 3,A 4 ), nd ssume tht l is line not through A i, i =,...,4. Put M 2 = A A 2 l, nd let M 2 be the hrmonic conjugte of M 2 with respect to A nd A 2, nd define the points M 23, M 34, M 3, M 4, nd M 24 likewise. Let X, Y, nd Z be the points A A 2 A 3 A 4, A 2 A 3 A A 4, nd A A 3 A 2 A 4 respectively. Finlly, let I nd J be the tngent points with l of the two conics of the pencil which re tngent t l. Then the eleven points M 2, M 3, M 4, M 23, M 24, M 34, X, Y, Z, I, nd J belong to conic ([8, p.09]). Proof. We prove tht this conic is the locus C of the poles of the line l with regrd to the conics of the pencil B(A,A 2,A 3,A 4 ). But first, let us prove tht this locus is indeed conic: if we represent the pencil by F + tf 2 =0, where F =0 nd F 2 =0re two conics of the pencil, the eqution of the locus is obtined by eliminting t from two liner equtions which represent the polr lines of two points of l, which gives qudrtic eqution. Then, cll A 3 the point which is the

3 On the Schiffler center 87 hrmonic conjugte of A 3 with respect to M 2 A 3 l nd M 2, nd consider the conic of the pencil through A 3: the pole of l with respect to this conic clerly is M 2, which mens tht M 2, nd thus lso M ij, is point of the locus. Next, X, Y nd Z re points of the locus, since they re singulr points of the three degenerte conics of the pencil. And finlly, I nd J belong to the locus, becuse they re the poles of l with regrd to the two conics of the pencil which re tngent to l..6. Consider gin tringle ABC in P, nd point P not on side of ABC. The Cev tringle of P is the tringle with vertices AP BC, BP CA, nd CP AB. Exmple: with the nottion of.2. the Cev tringle of Z is M A M B M C. Next, ssume tht I nd J re ny two (different) points, not on side of ABC, on line l, not through vertex, nd tht P is the point which corresponds (ccording to.4) to the involution on l with fixed points I nd J. Let H A H B H C be the Cev tringle of P, let A (B, nd C respectively) be the hrmonic conjugte of PA l (PB l, nd PC l respectively) with respect to A nd P (B nd P, nd C nd P, respectively), nd let M A M B M C be the Cev tringle of the triliner pole Z of l with regrd to ABC. Then there is conic through I, J, nd the triples H A H B H C,A B C, nd M A M B M C. This conic is known s the eleven-point conic of ABC with regrd to I nd J ([7, pp ]). Proof. Apply.5 to the pencil B(A, B, C, P ). 2. The min theorem Theorem. Let ABC be tringle in the complex projective plne P, l be line not through vertex, nd I nd J be ny two (different) points of l not on side of the tringle. Choose C to be one of the four conics through I nd J tht re tngent to the sides of tringle ABC, nd define Q to be the pole of l with respect to C. If Z, Z A, Z B, nd Z C re the triliner poles of l with respect to the tringles ABC, QBC, QCA, nd QAB respectively, while P, P A, P B, nd P C respectively, re the points determined by these tringles nd the involution on l whose fixed points re I nd J (see.4), then the lines PZ, P A Z A, P B Z B, nd P C Z C concur t point S P. Proof. We choose our projective coordinte system in P s follows : A(, 0, 0), B(0,, 0), C(0, 0, ), nd l is the unit line with eqution x + x 2 + x 3 =0. The point P hs coordintes (α, β, γ). Two degenerte conics of the pencil B(A, B, C, P ) re (CP,AB) nd (BP,CA), which intersect l t the points ( α, β,α + β), (,, 0) nd ( α, α + γ, γ), (, 0, )) respectively. Joining these points to A, we find the lines (α + β)x 2 + βx 3 =0, x 3 =0nd γx 2 +(α + γ)x 3 =0, x 2 =0, or s qudrtic equtions (α + β)x 2 x 3 + βx 2 3 =0nd γx2 2 +(α + γ)x 2x 3 =0respectively. Therefore, the lines AI nd AJ re given by kx lx 2x 3 + mx 2 3 =0whereby k, l, nd m re solution of (see.): { βk (α + β)l =0 (α + γ)l + γm =0,

4 88 C. Ths nd thus (k, l, m) =(γ(α + β),βγ,β(α + γ)). Next, the lines through A which form together with AI, AJ nd with AB, AC n hrmonic qudruple, re determined by px qx 2x 3 + rx 2 3 =0with p, q, r solutions of (see gin.) { β(α + γ)p 2βγq + γ(α + β)r =0 q =0, nd thus these lines re given by γ(α + β)x 2 2 β(α + γ)x2 3 =0. In the sme wy, we find the qudrtic eqution of the two lines through B (C, respectively) which form together with BI, BJ nd with BC, BA (with CI, CJ nd with CA, CB respectively) n hrmonic qudruple : α(β + γ)x 2 3 γ(β + α)x2 =0 (β(γ +α)x 2 α(γ +β)x2 2 =0respectively). The intersection points of these three pirs of lines through A, B, nd C re the poles Q, Q 2, Q 3, Q 4 of l with respect to the four conics through I nd J tht re tngent to the sides of tringle ABC (see.3) nd their coordintes re Q (A, B, C),Q 2 ( A, B, C),Q 3 (A, B, C), nd Q 4 (A, B, C), where A = α(β + γ), B = β(γ + α), C = γ(α + β). For now, let us choose for Q the point Q (A, B, C). The coordintes of the points Z, Z A, Z B, nd Z C re (,, ), (A, A +2B + C, A + B +2C), (2A + B + C, B, A + B +2C), nd (2A + B + C, A +2B + C, C) (see.2.2). Now, in connection with the point P C, remrk tht (AP C l (QB l) IJ)=. But (Q 2 Q 4 l (Q Q 3 l) IJ)= nd Q 2 Q 4 = Q 2 B, Q Q 3 = Q B, so tht AP C l = Q 2 B l, nd since Q 2 B hs eqution Cx + Ax 3 =0, the point AP C l hs coordintes (A, C A, C) nd the line AP C hs eqution Cx 2 +(C A)x 3 =0. In the sme wy, we find the eqution of the line BP C : Cx +(C B)x 3 =0, nd the common point of these two lines is the point P C with coordintes (B C, A C, C). Finlly, the line P C Z C hs eqution : C(B + C)x C(A + C)x 2 +(A 2 B 2 )x 3 =0, nd cyclic permuttion gives us the equtions of P A Z A nd P B Z B. Now, P A Z A, P B Z B, nd P C Z C re concurrent if the determinnt B 2 C 2 A(C + A) A(B + A) B(C + B) C 2 A 2 B(A + B) C(B + C) C(A + C) A 2 B 2 is zero, which is obviously the cse, since the sum of the rows gives us three times zero. Then, the line PZ hs eqution (β γ)x +(γ α)x 2 +(α β)x 3 =0. But A 2 = α(β + γ), B 2 = β(γ + α), nd C 2 = γ(α + β), so tht (B 2 C 2 )( A 2 + B 2 + C 2 )=2αβγ(β γ), nd PZ hs lso the following eqution (B 2 C 2 )( A 2 + B 2 + C 2 )x +(C 2 A 2 )(A 2 B 2 + C 2 )x 2 +(A 2 B 2 )(A 2 + B 2 C 2 )x 3 =0. For PZ, P A Z A, nd P B Z B to be concurrent, the following determinnt must vnish :

5 On the Schiffler center 89 (B 2 C 2 )( A 2 + B 2 + C 2 ) (C 2 A 2 )(A 2 B 2 + C 2 ) (A 2 B 2 )(A 2 + B 2 C 2 ) B 2 C 2 A(C + A) A(B + A) B(C + B) C 2 A 2 B(A + B) =(B + C)(C + A)(A + B)(A(B C)( A 2 + B 2 + C 2 )( A + B + C) + B(C A)(A 2 B 2 + C 2 )(A B+ C)+C(A B)(A 2 + B 2 C 2 )(A + B C)) =0. We my conclude tht PZ,P A Z A, P B Z B, nd P C Z C re concurrent. completes the proof. This Remrks. () If Q is chosen s the point Q 2 (Q 3,orQ 4, respectively), then A (B, orc respectively) must be replced by A ( B, or C respectively) in the foregoing proof. (2) The coordintes of the common point S P of the lines PZ,P A Z A,P B Z B, nd P C Z C re (A A+B+C B+C, B A B+C C+A, C A+B C A+B ). (3) Of course, when we work in the rel (complexified) projective plne P with rel tringle ABC, rel line l nd rel point P, the points Q nd S P, re not lwys rel. Tht depends on the vlues of α, β, nd γ nd thus on the position of the point P in the plne. For instnce, in exmple 5.5 of 5, the points Q nd S P will be imginry. (4) The conic through A, B, C, nd through the points I, J on l hs eqution α(β + γ)x 2 x 3 + β(γ + α)x 3 x + γ(α + β)x x 2 =0 or A 2 x 2 x 3 + B 2 x 3 x + C 2 x x 2 =0. Indeed, eliminting x from this eqution nd from x + x 2 + x 3 =0, gives us γ(α + β)x γβx 2x 3 + β(γ + α)x 2 3 =0, which determines the lines AI nd AJ (see the proof of the theorem). The pole of the line l with respect to this conic is the point Y (β + γ, γ + α, α + β), which clerly is point of the line PZ. We denote this conic by (Y ). (5) The locus of the poles of the line l with respect to the conics of the pencil B(A, B, C, P ) is the conic with eqution βγx 2 + γαx αβx 2 3 α(γ + β)x 2 x 3 β(α + γ)x 3 x γ(β + α)x x 2 =0. It is the eleven-point conic of tringle ABC with regrd to I nd J (see.6): it is the conic through the points M A (0,, ), M B (, 0, ), M C (,, 0), AP BC = H A (0,β,γ), BP CA = H B (α, 0,γ), CP AB = H C (α, β, 0), A (2α + β + γ,β,γ),b (α, α +2β + γ,γ),c (α, β, α + β +2γ),I, nd J. The pole of the line l with regrd to this conic is the point Y (2α + β + γ, α +2β + γ, α + β +2γ), which is lso point of the line PZ. We denote this conic by (Y ). Here is n lterntive formultion of the min theorem. Theorem. Let ABC be tringle in the complex projective plne P, l be line not through vertex, nd I nd J be ny two (different) points of l not on side

6 90 C. Ths of the tringle. Denote by Q the pole of l with respect to one of the four conics through I nd J tht re tngent to the sides of the tringle. If Y, Y A, Y B, nd Y C re the poles of l with respect to the conics determined by I, J, nd the triples ABC, QBC, QCA, nd QAB respectively, while Y, Y A, Y B, nd Y C re the respective poles with respect to their eleven-point conics with regrd to I nd J, then YY, Y A Y A, Y BY B, nd Y CY C concur t point S. 3. The Eucliden cse In this section we give pplictions of the min theorem in the Eucliden plne Π. Throughout the following sections, we only consider generl rel tringle ABC in Π, i.e., the side-lengths, b, nd c re distinct nd the tringle hs no right ngle. Corollry. Let ABC be tringle in Π nd ssume tht l is the line t infinity of Π. Suppose tht P coincides with the orthocenter H of ABC; then the conics of the pencil B(A, B, C, H) re rectngulr hyperbols nd the involution on l, determined by H (see.4), becomes the bsolute (or orthogonl) involution with fixed points the cyclic points (or circle points) J nd J of Π. The four conics through J, J nd tngent to the sidelines of ABC re now the incircle nd the excircles of ABC, nd the points Q = Q, Q 2, Q 3, Q 4 become the incenter I, nd the excenters I A (the line II A contins A), I B, nd I C, respectively. Next, the points Z, Z A, Z B, nd Z C, re the centroids of ABC, IBC, ICA, nd of IAB respectively. Finlly, P A, P B, P C re the orthocenters H A, H B, H C of IBC, ICA, nd IAB respectively. Then the lines HZ, H A Z A, H B Z B, nd H C Z C concur t point S H. Remrk tht HZ, H A Z A, H B Z B, nd H C Z C re the Euler lines of the tringles ABC, IBC, ICA, nd IAB, respectively. The point of concurrence of these Euler lines is known s the Schiffler point S ([9]), but we prefer in this pper the nottion S H, since it results from setting P = H. In connection with Remrks 4 nd 5 of the foregoing section, nd gin working with l s the line t infinity nd J, J the cyclic points, the conic (Y ) becomes the circumcircle (O) of ABC, (Y ) becomes its nine-point circle (O ), nd OO is the Euler line. In connection with Remrk 5, we recll tht the locus of the centers of the rectngulr hyperbols through A, B, C (nd H) is the nine-point circle (O ) of ABC nd tht, for ech point U of the circumcircle (O), the midpoint of HU is point of (O ) (nd O is the midpoint of HO on the Euler line). The min theorem llows us to generlize the foregoing corollry s follows: Corollry 2. Let ABC be tringle nd let l be the line t infinity in Π. Choose generl point P (i.e., not on sideline of ABC, not on l nd different from the centroid of ABC) nd cll J, J the tngent points on l of the two conics of the pencil B(A, B, C, P ) which re tngent to l (these re the centers of the prbols through A, B, C nd P ). Denote by Q the center of one of the four conics through J nd J, which re tngent t the sidelines of ABC. Next, Z is the centroid

7 On the Schiffler center 9 of ABC nd Z A,Z B,Z C re the centroids of the tringles QBC, QCA, QAB respectively. Finlly, P A (P B, nd P C respectively) is the fourth common point of the two prbols through Q, B, C (through Q, C, A, nd through Q, A, B respectively) nd tngent to l t J nd J. Then the lines PZ, P A Z A, P B Z B, nd P C Z C concur t point S P. 4. The use of triliner coordintes From now on, we work with triliner coordintes (x,x 2,x 3 ) with respect to the rel tringle ABC in the Eucliden plne Π ([2, 5]): A, B, C, nd the incenter I of ABC, hve coordintes (, 0, 0), (0,, 0), (0, 0, ), nd (,, ) respectively. The line t infinity l hs eqution x + bx 2 + cx 3 =0, where, b, c re the sidelengths of ABC. The orthocenter H, the centroid Z, the circumcenter O, nd the center of the nine-point circle O, hve triliner coordintes ( cos A, cos B, ) ( cos C,, b, ) c, (cos A, cos B,cos C), nd (bc( 2 b c 2 (b 2 c 2 ) 2 ),c(b 2 c 2 + b 2 2 (c 2 2 ) 2 ),b(c 2 2 +c 2 b 2 ( 2 b 2 ) 2 )) respectively. The equtions of the circumcircle (O) nd the nine-point circle (O ) re x 2 x 3 +bx 3 x +cx x 2 =0nd x 2 sin 2A+x2 2 sin 2B +x2 3 sin 2C 2x 2x 3 sin A 2x 3 x sin B 2x x 2 sin C =0. The Schiffler point S = S H (the common point ( of the Euler lines ) of ABC, IBC, ICA, nd IAB) hs triliner coordintes +b+c b+c, b+c c+, +b c +b. If T is point of Π, not on sideline of ABC, reflect the line AT bout the line AI, nd reflect BT nd CT bout the corresponding bisectors BI nd CI. The three reflections concur in the isogonl conjugte T of T, nd T hs ( ) triliner coordintes (t 2 t 3,t 3 t,t t 2 ) or t, t 2, t 3 if T hs triliner coordintes (t,t 2,t 3 ). Exmples: the circumcenter O is the isogonl conjugte of the orthocenter H, nd the centroid Z is the isogonl conjugte of the Lemoine point (or symmedin point) K(, b, c). Let us now interpret the min theorem (or Corollry 2) in the Eucliden cse using triliner coordintes, with l : x +bx 2 +cx 3 =0s line t infinity nd with P (α, β, γ) generl point of Π. In fct, the only thing tht we hve to do, is to replce in the proof of the min theorem the eqution x +x 2 +x 3 =0of l,byx + bx 2 +cx 3 =0, nd strightforwrd clcultion gives us the following triliner coordintes for the point Q: ( bcα(bβ + cγ), cβ(cγ + α), bγ(α + bβ)) = (A, B, C). Next, the points Z, Z A, Z B, nd Z C re the centroids of ABC, QBC, QCA nd QAB with triliner coordintes (, b, ) c, (bca,c(a+2bb+cc),b(a+ bb+2cc)), (c(2a+bb+cc),cb,(a+bb+2cc)), (b(2a+bb+cc),(a+ 2bB + cc),bc), respectively. Now, for the points P A,P B,P C, gin fter strightforwrd clcultion, we find the coordintes: P A (bca,c(cc A),b(bB A)),P B (c(cc bb),cb,(a bb)) nd P C (b(bb cc),(a cc),bc). And finlly, we find the triliner coordintes of the point S P, corresponding to Q: ( A( A + bb + cc), bb + cc B(A bb + cc), cc + A ) C(A + bb cc). A + bb

8 92 C. Ths Remrk tht we find for the cse P (α, β, γ) =H( cos A, cos B, cos C ): A = bc bcα(bβ + cγ) = cos A ( b cos B + c cos C )= bc(b cos C+c cos B) cos A cos B cos C bc = cos A cos B cos C = B = C nd Q(A, ( B, C) =I(,, ), ) while since A = B = C, we get for S H the coordintes +b+c b+c, b+c c+, +b c +b, which gives us the Schiffler point S. Let us lso clculte the triliner coordintes of the points Y nd Y, defined bove s the centers of the conic (Y ) through A, B, C, J nd J, nd of the conic (Y ) through the midpoints of the sides of ABC nd through J, J (or the elevenpoint conic of ABC with regrd to J nd J ; remrk tht J nd J re the cyclic points only when P = H): (Y ) hs eqution α(bβ + cγ)x 2 x 3 + β(cγ + α)x 3 x + γ(α + bβ)x x 2 =0 nd center Y (bc(bβ + cγ),c(cγ + α),b(α + bβ)), (Y ) hs eqution βγx 2 + bγαx2 2 + cαβx2 3 α(γc + bβ)x 2x 3 β(α + cγ)x 3 x γ(bβ + α)x x 2 =0nd center Y (bc(2α + bβ + cγ),c(α +2bβ + cγ),b(α + bβ +2cγ)). Remrk tht Q = P Y, with the nottion (x,x 2,x 3 ) (y,y 2,y 3 ) = ( x y, x 2 y 2, x 3 y 3 ). Recll tht the coordinte trnsformtion between triliner coordintes (x,x 2,x 3 ) with regrd to ABC nd triliner coordintes (x,x 2,x 3 ) with regrd to the medil tringle M A M B M C, is given by ([5, p.207]): x bx 2 = 0 b c 0 c x x 2. cx 3 b 0 x 3 Now, this gives for (x,x 2,x 3 ) the coordintes of the point Y,if(x,x 2,x 3 ) re the coordintes (α, β, γ) of P nd it gives for (x,x 2,x 3 ) the coordintes of Y if (x,x 2,x 3 ) re the coordintes of Y. Moreover, ABC nd its medil tringle re homothetic. As corollry, we hve tht if P (Y, respectively) is tringle center X(k) for ABC (for the definition of tringle center, see [5, p.46]), then Y (Y respectively) is center X(k) for M A M B M C. 5. Applictions In this section we choose P (α, β, γ) s tringle center of the tringle ABC nd clculte the coordintes of the corresponding points Y, Y, Q nd S P (sometimes Y nd S P re not given). Remrk tht P must be different from the centroid Z of ABC. The tringle centers re tken from Kimberling s list : X(), X(2),..., X(2445) (list until 29 Mrch 2004, see [6]). When we found the points Y, Y, Q or S P in this list, we give the number X( ) nd if possible, the nme of the center. But, without doubt, we overlooked some centers nd more points Y, Y, Q, S P thn indicted will occur in Kimberling s list. Severl times, only the first triliner coordinte is given: the second nd the third re obtined by cyclic permuttions.

9 On the Schiffler center The first exmple is of course: P (α, β, γ) =H( cos A, cos B, cos C )=X(4) (orthocenter), Y = O(cos A, cos B,cos C) =X(3) (circumcenter), Y = O (bc( 2 b c 2 (b 2 c 2 ) 2 ),, )=X(5) (nine-point center), Q = I(, (, ) = X() (incenter), ) nd S H = S +b+c b+c,, = X(2) (Schiffler point) P (α, β, γ) =I(,, ) = X(), Y =( b+c, c+ b, +b c )=X(0) (Spieker point = incenter of the medil tringle M A M B M C ), Y =( 2+b+c,, ) =X(25) (Spieker point of the medil tringle), Q =( bc(b + c),, ), nd S I =( bc(b + c) bc(b+c)+b c(c+)+c b(+b),, ). b c(c+)+c b(+b) 5.3. P (α, β, γ) =K(, b, c) =X(6) (Lemoine point), Y =( b2 +c 2,, ) =X(4) = Lemoine point of medil tringle, Y =( 22 +b 2 +c 2,, ), Q =( b 2 + c 2,, ), nd S K =( b 2 + c 2 b 2 +c 2 +b c c 2 +b 2 b c c,, ). 2 +b P (α, β, γ) =( ( +b+c),, )=X(7) (Gergonne point), Y =( + b + c, b + c, + b c) =X(9) (Mittenpunkt = Lemoine point of the excentrl tringle I A I B I C = Gergonne point of medil tringle), Y =(bc((b+c) (b c) 2 ),, )=X(42) (Mittenpunkt of medil tringle), Q =(, b, c )=X(366), nd S X(7) =( + b+ c,, ). b+ c 5.5. P (α, β, γ) =( b c, c, b )=X(00), Y =(bc(b c) 2 ( + b + c),, )=X() (Feuerbch point = X(00) of medil tringle), Y =(bc(( b) 2 ( + b c)+(c ) 2 ( b + c)),, ) (Feuerbch point of medil tringle), nd Q =( bc(b c)( + b + c),, ). In the foregoing exmples, the coordintes of the point S P re mostly rther complicted. Another method is to strt with the coordintes of the point Q: if (k, l, m) re the triliner coordintes of Q, then short clcultion shows tht it corresponds with the point P ( ( 2 k 2 +b 2 l 2 +c 2 m 2 ),, ) nd S P becomes the point (k k+bl+cm bl+cm,, ). Finlly, the coordintes of Y nd Y re (k 2 ( 2 k 2 + b 2 l 2 + c 2 m 2 ),, ), nd (bc( 2 k 2 (b 2 l 2 + c 2 m 2 ) (b 2 l 2 c 2 m 2 ) 2 ),, ), respectively. Here re some exmples.

10 94 C. Ths 5.6. Q(k, l, m) =K(, b, c) =X(6)(Lemoine point), P = ( ( 4 +b 4 +c 4 ),, ) = X(66) = X(22) (X(22) is the Exeter point), Y =( 3 ( 4 + b 4 + c 4 ),, )=X(206) (X(66) of medil tringle), Y =(bc( 4 (b 4 + c 4 ) (b 4 c 4 ) 2 ),, )(X(206) of medil tringle), nd S X(66) =( ( 2 +b 2 +c 2 ) b 2 +c,, )=( cos A 2 b 2 +c,, )=X(76) Q(k, l, m) =H( P =( Y =( S P =( )=X(4) (orthocenter),, ), ), cos A, cos B, cos C ( 2 cos 2 + b2 A cos 2 + c2 B cos 2 C cos 2 A ( 2 cos 2 A + b2 cos 2 B + cos A cos B cos C cos 2 A,, ). c2 cos 2 C ),, ), nd 5.8. Q(k, l, m) =( b+c,, )=X(0) (Spieker point), P =(,, )=X(596), ( ( (b+c) 2 +(c+) 2 +(+b) 2 ) ) Y = (b+c) 2 ( (b + c) 2 +(c + ) 2 +( + b) 2 ),, (X(596) of medil tringle), nd b+c S P =( 2+b+c, c+ +2b+c, +b +b+2c ). We lso cn strt with the coordintes of the point Y (p, q, r), then P =( p+bq+cr,, ), Y (bc(bq + cr),, ), nd Q = p( p+bq+cr) P Y =(,, ). Here re some exmples Y (p, q, r) = I(,, ) = X(), P = ( +b+c, b+c b, +b c c ) = X(8) (Ngel point), Y = ( b+c,, ) = X(0) (Spieker point = incenter of medil tringle), ( ) +b+c Q =,, = X(88), nd S P =(A A+bB+cC bb+cc,, ) with Q(A, B, C) Y = K(, b, c) = X(6) (Lemoine point), P =( 2 +b 2 +c 2,, )=( cosa,, )=X(69), 2 Y =( b2 +c 2,, )=X(4) (Lemoine point of medil tringle), nd Q =( 2 + b 2 + c 2,, ). 5.. Y =( 2+b+c,, )=X(25) (Spieker point of medil tringle), P =( b+c,, )=X(0) (Spieker point), Y =( 2+3b+3c,, ) (X(25) of medil tringle), nd Q =(bc (b + c)(2 + b + c),, ).

11 On the Schiffler center 95 References [] L. Emelynov nd T. Emelyenov, A note on the Schiffler point, Forum Geom., 3 (2003) 3 6. [2] W. Glltly, The Modern Geometry of the Tringle, 2nd ed. 93, Frncis Hodgson, London. [3] A. P. Htzipolkis, F. M. vn Lmoen, B. Wolk, nd P. Yiu, Concurrency of four Euler lines, Forum Geom., (200) [4] R. Honsberger, Episodes in Nineteenth nd Twentieth Century Eucliden Geometry, Mth. Assoc. of Americ, 995. [5] C. Kimberling, Tringle centers nd centrl tringles, Congressus Numerntium, 29 (998) 285. [6] C. Kimberling, Encyclopedi of Tringle Centers, vilble t [7] D. Pedoe, A Course of Geometry for Colleges nd Universities, Cmbridge Univ. Press, 970. [8] P. Smuel, Projective Geometry, Springer Verlg, 988. [9] K. Schiffler, G. R. Veldkmp, nd W. A. vn der Spek, Problem 08, Crux Mth., (985) 5; solution 2 (986) Chrles Ths: Deprtment of Pure Mthemtics nd Computer Algebr, Krijgsln 28-S22, B Gent, Belgium E-mil ddress: chrles.ths@ugent.be