Triangles with Given Incircle and Centroid

Transcription

1 Forum eometriorum Volume 11 (2011) FORUM EOM SSN Triangles with iven nirle and entroid aris amfilos bstrat. n this artile we study the set of triangles sharing a ommon inirle and a ommon entroid. t is shown that these triangles have their verties on a oni easily onstrutible from the given data. t is also shown that the irumirles of all these triangles are simultaneously tangent to two fixed irles, hene their enters lie also on a oni. 1. ntrodution The objet of study here is on triangles sharing a ommon inirle and entroid. t belongs to the wider subjet of (one-parameter) families of triangles, initiated by the notion of poristi triangles, whih are triangles sharing a ommon inirle and also a ommon irumirle [7, p.22]. There is a onsiderable literature on poristi triangles and their variations, whih inlude triangles sharing the same irumirle and Euler irle [16], triangles sharing the same inirle and Euler irle [6], or inirle and orthoenter [11, vol., p.15], or irumirle and entroid [8, p.6]. The starting point of the disussion here, and the ontent of 2, is the exploration of a key onfiguration onsisting of a irle and a homothety f. This onfiguration generates in a natural way a homography H and the oni = H(). 3 looks a bit loser at homography H and explores its properties and the properties of the oni. 4 uses the results obtained previously to the ase of the inirle and the homothety with ratio k = 2 entered at the entroid of the triangle of referene to explore the kind of the oni = f( ) on whih lie the verties of all triangles of the poristi system. 5 disusses the lous of irumenters of the triangles of the poristi system loating, besides the assumed fixed inirle, also another fixed irle 0, alled seondary, to whih are tangent all Euler irles of the system. Finally, 6 ontains misellaneous fats and remarks onerning the problem. 2. The basi onfiguration The basi onfiguration of this study is a irle and a homothety f entered at a point and having a ratio k 1. n the following the symbol p X throughout denotes the polar line of point X with respet to, this line beoming tangent to when X. Lemma 1 handles a simple property of this onfiguration. Lemma 1. Let be a irle and a point. Let further be a point on the irle and L a line parallel to the tangent p at and not interseting the irle. Then there is exatly one point on L suh that the tangents {t,t } to from and line interset on p equal segments M = M. ubliation ate: February 25, ommuniating Editor: aul Yiu.

2 28. amfilos p L E J K p p M t p E t' Figure 1. basi lemma Figure 1 suggests a proof. Let {E,} be the intersetions of lines E = L p and = L p E. raw from the tangents {t =,t = J} to interseting line p respetively at and. Let also M be the intersetion M = p. Then M = M. This follows immediately from the onstrution, sine by the definitions made the penil of lines (,J,K,E) at is harmoni. Thus, this penil defines on p a harmoni division and, sine the intersetion of lines {p,l} is at infinity, M is the middle of. onversely, the equality M = M implies that the penil of four lines,, M, L is harmoni. f E is the intersetion E = L p then p E oinides with M and p E E p. Thus is uniquely determined as = L p E, where E = L p. Remarks. (1) oint K, defined as the intersetion K = p, is on the diameter of through. This follows from the fat that K is on the polar of and also on the polar of E. Hene its polar is p K = E. (2) n other words the lemma says that there is exatly one triangle with vertex L suh that the irumsribed to triangle has as median for side. onsequently if this happens also for another vertex and is the median also of then oinides with the entroid of and the third line is also a median of this triangle. The next proposition explores the lous of point M = M() uniquely determined from in the previous lemma by letting vary on the irle and taking the parallel L to p to be idential with the homotheti to p line L = f(p ) (see Figure 2). roposition 2. onsider a irle and a homothety f with ratio k 1 and enter at a point. For eah point let L = f(p ) be the homotheti image of p and E = L p, M = p E p. Then M desribes a oni as moves on the irle. The proof of the proposition is given using homogeneous artesian oordinates with the origin set at and the x-axis identified with line, where is the enter

3 Triangles with given inirle and entroid 29 ' E L ' p E * p N p M Figure 2. Lous of M=p p E of the given irle. n suh a system irle is represented by the equation x 2 +y 2 2uxz+(u 2 r 2 )z 2 = 0 ( x y z ) 1 0 u x y = 0. u 0 u 2 r 2 z Here (u, 0) are the orresponding artesian oordinates of the enter and r is the radius of the irle. The polar p of point (0,0,1) is omputed from the matrix and is represented by equation ( ) u u 0 u 2 r 2 x y = 0 ux + (u 2 r 2 )z = 0. z The homothety f is represented by the matrix x y k k 0 x y. z z Hene the tangent p at (x,y,z) and its image L under f have orrespondingly oeffiients ( ) x y z 1 0 u = ( x uz y ux + (u 2 r 2 )z ), u 0 u 2 r 2 and ( 1 k (x uz) 1 k y ux + (u2 r 2 )z ). The homogeneous oordinates of E = p L are then omputed through the vetor produt of the oeffiients of these lines whih is: ( 1 k y(u2 r 2 ), u( ux + (u 2 r 2 )z) + 1 k (x uz)(u2 r 2 ), u k y).

4 30. amfilos The oeffiients of the polar p E are then seen to be ( 1 k y(u2 r 2 ) u( ux + (u 2 r 2 )z) + 1 k (x uz)(u2 r 2 ) u k y) 1 0 u u 0 u 2 r 2 = ( 1 k y(u2 r 2 ) + u2 k y u( ux + (u2 r 2 )z) + 1 k (x uz)(u2 r 2 ) 0 = ( 1 k yr2 u( ux + (u 2 r 2 )z) + 1 k (x uz)(u2 r 2 ) 0 ) = ( yr 2 (u 2 (1 k) r 2 )x + u(u 2 r 2 )(k 1)z 0 ). The homogeneous oordinates of M = p E p are again omputed through the vetor produt of the oeffiients of the orresponding lines: x uz yr 2 y (u 2 (1 k) r 2 )x + u(u 2 r 2 )(k 1)z ux + (u 2 r 2 )z 0 ((u 2 (1 k) r 2 )x + u(u 2 r 2 )(k 1)z)(ux (u 2 r 2 )z) = ( yr 2 )(ux (u 2 r 2 )z) ( ux + (u 2 r 2 )z)(r 2 z + u(1 k)(uz x)) = (u2 (1 k) r 2 )x + u(u 2 r 2 )(k 1)z yr 2 u(1 k)x (r 2 + u 2 (1 k))z = u2 (1 k) r 2 0 u(u 2 r 2 )(k 1) 0 r 2 0 x y. u(1 k) 0 (r 2 + u 2 (1 k)) z The matrix produt in the last equation defines a homography H and shows that the lous of points M is the image = H() of the irle under this homography. This ompletes the proof of roposition 2 Remarks. (3) The properties of the oni are of ourse tightly onneted to the properties of the homography H appearing at the end of the proposition and denoted by the same letter H = u2 (1 k) r 2 0 u(u 2 r 2 )(k 1) 0 r 2 0. u(1 k) 0 (r 2 + u 2 (1 k)) These properties will be the subjet of study in the next setion. (4) omposing with the homothety f we obtain the lous of = f(m) whih is the homotheti oni = f( ) = (f H)(). t is this oni rather, than, that relates to our original problem. Sine though the two onis are homotheti, work on either leads to properties for both of them. (5) Figure 2 underlines the symmetry between these two onis. n it denotes the homotheti image = f() of. Using this irle instead of and the inverse homothety g = f 1 we obtain a basi onfiguration in whih the roles of and are interhanged. )

5 Triangles with given inirle and entroid 31 (6) f and point = f(m) is outside the irle then triangle onstruted by interseting p with the tangents from has, aording to the lemma, line as median. nversely, if a triangle is irumsribed in and has its median M passing through and divided by it in ratio M = k, then, again aording to the lemma, it has M on the oni. n partiular if a triangle is irumsribed in and has two medians passing through then it has all three of them passing through, the ratio k = 2 and the middles of its sides are points of the oni. ' * M Figure 3. Lous of entroids (7) The previous remark implies that if ratio k 2 then every triangle irumsribed in and having its median M passing through has its other medians interseting at the same point. Figure 3 illustrates this remark by displaying the lous of the entroid of triangles whih are irumsribed in, their median M passes through and is divided by it in ratio k but later does not oinide with the entroid ( i.e., k 2). t is easily seen that the lous of the entroid in suh a ase is part of a oni whih is homotheti to with respet to and in ratio k = 2+k 3. Obviously for k = 2 this oni ollapses to the point and triangles like irumsribed in have all their verties on. (8) Figure 4 shows a ase in whih the points whih are outside fall into four onneted ars of a hyperbola. n this example k = 2 and, by the symmetry of the ondition, it is easily seen that if a point is on one of these ars then the other verties of are also on respetive ars of the same hyperbola. The fat to notie here is that onis and are defined diretly from the basi onfiguration onsisting of the irle and the homothety f. t is though not possible for every point of to be vertex of a triangle irumsribed in and with entroid at. summarize the results obtained so far in the following proposition. roposition 3. ll triangles sharing the same inirle and entroid have their side-middles on a oni and their verties on a oni = f( ) homotheti to by the homothety f entered at the entroid with ratio k = 2.

6 32. amfilos E M Figure 4. rs: = f(m) and outside The proof follows from the previous remarks and the fat that in this setting the (anti-)homothety f has enter at and ratio k = 2. y Lemma 5 below, the median M of side oinides with the polar p E of point E = L p, where L is the parallel to from. The oni generated by point M, as in Lemma 5, passes now through the middle of but also through the middles of the other sides, sine the onfiguration, as already remarked, is independent of the preferred vertex and depends only on the irle and the homothety f (see Figure 5). E p p E * M Figure 5. oni The oni = f( ) whih is homotheti to will pass through all three verties of triangle. Sine the same argument an be applied to any triangle sharing with the same inirle and entroid it follows that all these triangles have their verties on this oni. Figure 6 displays two triangles sharing the same inirle and entroid and having their verties on oni in a ase in whih this happens to be ellipti. Note that in this ase the entire oni onsists of verties of triangles irumsribing and having their entroid at. This follows

7 Triangles with given inirle and entroid 33 from the fat that in this ase does not interset (see 4) and by applying then the onelet s porism [3, p.68]. ' ' * J M ' ** Figure 6. The two onis and efore going further into the study of these onis devote the next setion to a ouple of remarks onerning the homography H. 3. The homography H roposition 4. The homography H defined in the previous setion is uniquely haraterized by its properties: (i) H fixes points {,} whih are the diameter points of on the x-axis, (ii) H maps points {S,T } to {S,T }. Here {S,T } are the diameter points of on a parallel to the y-axis and {S,T } are their orthogonal projetions on the diameter of = f() whih is parallel to the y-axis. The proof of the proposition (see Figure 7) follows by applying the matrix to the oordinate vetors of these points whih are (u+r,0,1), (u r,0,1), S(u,r,1), T(u, r,1), S (ku,r,1) and T (ku, r,1), and using the well-known fat that a homography is uniquely determined by presribing its values at four points in general position [2, Vol., p.97]. Remarks. (1) y its proper definition, line XX for X and X = H(X) = H() is tangent to irle at X. (2) The form of the matrix H implies that the oni is symmetri with respet to the x-axis and passes through points and having there tangents oiniding respetively with the tangents of irle. Thus and are verties of and oinides with the auxiliary irle of if this is a hyperbola. n the ase is an ellipse, by the previous remark, follows that it lies entirely outside hene later is the maximal irle insribed in the ellipse.

8 34. amfilos ' E p E S S' p ' J M E' x T T' p Figure 7. Homography H : {,,S, T } {,,S, T } (3) The kind of depends on the loation of the line L 0 mapped to the line at infinity by H. L 0 p U t X' U S S' p ' J p X X T T' X' Figure 8. XX is tangent to This line is easily determined by applying H to a point (x,y,z) and requiring that the resulting point (x,y,z ) has z = 0, thus leading to the equation of a line parallel to y-axis: u(1 k)x (r 2 + u 2 (1 k))z = 0. The oni, depending on the number n of intersetion points of L 0 with irle, is a hyperbola (n = 2), ellipse (n = 0) or beomes a degenerate parabola onsisting of the pair of parallel tangents to at {,}.

9 Triangles with given inirle and entroid 35 E p ' J x M E' p * ** Figure 9. Homotheti onis and (4) The tangent p X of the irle at X maps via H to the tangent t X at the image point X = H(X) of. n the ase is a hyperbola this implies that eah one of its asymptotes is parallel to the tangent p U of at an intersetion point U L 0 (see Figure 8). (5) Figure 9 displays both onis and = f( ) in a ase in whih these are hyperbolas and suggests that irle is tangent to the asymptotes of. This is indeed so and is easily seen by first observing that H maps the enter of to the enter J of = f(). n fat, line ST maps by H to S T (see Figure 10) and line is invariant by H. Thus H() = J. Thus all lines through map under H to lines through J. n partiular the antipode U of U on line U maps to a point H(U ) on the tangent to U and U maps to the point at infinity on p U. Thus line UU maps to U J whih oinides with p U and is parallel to the asymptote of. Sine and are homotheti by f line U J is an asymptote of thereby proving the laim. (6) The arguments of the last remark show that line L 0 is the symmetri with respet to the enter of (see Figure 10) of line U V whih is the polar of point J with respet to irle. They show also that the intersetion point K of U V with the x-axis is the image via H of the axis point at infinity. n easy alulation using the matries of the previous setion shows that these remarks about the symmetry of {UV,p J } and the loation of K is true also in the ases in whih J is inside and there are no real tangents from it to. (7) The homography H demonstrates a remarkable behavior on lines parallel to the oordinate axes (see Figure 11). s is seen from its matrix it preserves the point at infinity of the y-axis hene permutes the lines parallel to this axis. n partiular, the arguments in (5) show that the parallel to the y-axis from is simply orthogonally projeted onto the parallel through point J. More generally points X moving on a parallel L to the y-axis map via H to points X L = H(L),

10 36. amfilos L 0 p U U S V' S' p ' V T U' T' J Figure 10. Loation of asymptotes suh that the line XX passes through a point X L depending only on L and being harmoni onjugate with respet to {,} to the intersetion X 0 of L with the x-axis. N U X S U' S' X' ' X L X 0 K J x V V' T T' L 0 L L' Figure 11. Mapping parallels to the axes (8) Regarding the parallels to the x-axis and different from it their images via H are lines passing through K and also through their intersetion N with the parallel to y-axis through J (see Figure 11). The x-axis itself is invariant under H and the ation of this map on it is ompletely determined by the triple of points {,,} and their images {,,J}. Next lemma and its orollary give an insight into the differene of a general ratio k 2 from the entroid ase, in whih k =

11 Triangles with given inirle and entroid 37 2, by fousing on the behavior of the tangents to from = f(m) and their intersetions with the variable tangent p of irle. Lemma 5. onsider a hyperbola and its auxiliary irle. raw two tangents {t,t } parallel to the asymptotes interseting at a point J. Then every tangent p to the auxiliary irle intersets lines {t,t } orrespondingly at points {,R} and the oni at two points {S,T } of whih one, S say, is the middle of R. F H J K t' t p R E S p * Figure 12. Hyperbola property The proof starts by defining the polar FE of J with respet to the oni. This is simultaneously the polar of J with respet to irle sine (,H,,J) = 1 are harmoni with respet to either of the urves (see Figure 12). Take then a tangent of at as required and onsider the intersetion point of with the polar EF. The polar p of with respet to the irle passes through J (by the reiproity of relation pole-polar) and is parallel to tangent p. esides (E,F,K,) = 1 build a harmoni division, thus the penil of lines at J : J(E,F,K,) defines a harmoni division on every line it meets. pply this to the tangent p. Sine JK is parallel to this tangent, S is the harmoni onjugate with respet to {R,} of the point at infinity of line p. Hene it is the middle of R. orollary 6. Under the assumptions of Lemma 5, in the ase oni is a hyperbola, point M = p is the ommon middle of segments {NO,KL,} on the tangent p of irle. Of these segments the first NO is interepted by the asymptotes of, the seond KL is interepted by the oni and the third is interepted by the tangents to from.

12 38. amfilos The proof for the first segment N O follows diretly from Lemma 5. The proof for the seond segment KL results from the well known fat [4, p.267], aording to whih KL and NO have ommon middle for every seant of the hyperbola. The proof for the third segment follows from the definition of and its properties as these are desribed by Lemma 1. ' N M J O L p K * ** Figure 13. ommon middle M Remark. (9) When the homothety ratio k = 2 point oinides with K (see Figure 13) and point L oinides with. nversely if K and L are points of then the orresponding points f 1 () and f 1 () are middles of the sides, k = 2 and is the entroid of. t even suffies one triangle satisfying this identifiation of points to make this onlusion. 4. The kind of the oni n this setion examine the kind of the oni by speializing the remarks made in the previous setions for the ase of ratio k = 2, shown equivalent to the fat that there is a triangle irumsribed in having its entroid at, its sidemiddles on oni and its verties on = f( ). First notie that irle = f() is the Nagel irle ([13]), its enter J = f() is the Nagel point ([10, p. 8]), its radius is twie the radius r of the inirle and it is tangent to the irumirle. Line is the Nagel line of the triangle ([15]). The homography H defined in the seond setion obtains in this ase (k = 2, u = ) the form. H = 3u2 r 2 0 3u(u 2 r 2 ) 0 r u 0 (r 2 + 3u 2 )

13 Triangles with given inirle and entroid 39 s notied in 2 the line L 0 sent to the line at infinity by H is orthogonal to line and its x-oordinate is determined by x 0 = r2 + 3u 2, 3u where u is the distane of from the inenter. The kind of oni is determined by the loation of x 0 relative to the inirle. n the ase x 0 u < r u > r 3 we obtain hyperbolas. n the ase u < r 3 we obtain ellipses and in the ase u = r 3 we obtain two parallel lines orthogonal to line. Sine the hyperboli ase was disussed in some extend in the previous setions here examine the two other ases. ' E ' J ' Figure 14. The ellipti ase First, the ellipti ase haraterized by the ondition u < r 3 and illustrated by Figure 14. n this ase it is easily seen that irle = f() enloses entirely irle and sine is the maximal insribed in irle the points of this oni are all on the outside of irle. Thus, from all points of this oni there exist tangents to the irle defining triangles with inirle and entroid. esides these ommon elements triangles share also the same Nagel point whih is the enter J of the ellipse. Note that this ellipse an be easily onstruted as a oni passing through five points {,,,, }, where {, } are the diametral points of its Nagel irle on. Figure 15 illustrates the ase of the singular oni, whih an be onsidered as a degenerate parabola. From the ondition u = r 3 follows that and = f() are tangent at one of the diametral points {,} of and the tangent there arries two of the verties of the triangle.

14 40. amfilos E p ' Figure 15. The singular ase Figure 16 displays two partiular triangles of suh a degenerate ase. The isoseles triangle haraterized by the ratio of its sides = 3 2 and the rightangled haraterized by the ratio of its orthogonal sides = 4 3, whih is similar to the right-angled triangle with sides {3,4,5}. ' ' ' ' Figure 16. Two speial triangles t an be shown [3, p. 82] that triangles belonging to this degenerate ase have the property to posses sides suh that the sum of two of them equals three times the length of the third. This lass of triangles answers also the problem [9] of finding all triangles suh that their Nagel point is a point of the inirle. 5. The lous of the irumenter n this setion study the lous 2 desribed by the irumenter of all triangles sharing the same inirle and entroid. Sine the homothety f, entered at with ratio k = 2, maps the irumenter O to the enter E u of the Euler irle

15 Triangles with given inirle and entroid 41 E, the problem redues to that of finding the lous 1 of points E u. The lue here is the tangeny of the Euler irle E to the inirle at the Feuerbah point F e of the triangle ([12]). Next proposition indiates that the Euler irle is also tangent to another fixed irle 0 (S,r 0 ), whose enter S (see Figure 17 and Figure 19) is on the Nagel line. all this irle the seondary irle of the onfiguration (or of the triangle). s will be seen below this irle is homotheti to the inirle with respet to the Nagel point N a and at a ertain ratio κ determined from the given data. 0 E E u S N a F e Figure 17. nvariant distane S roposition 7. Let {E u,f e, N a } be orrespondingly the enter of the Euler irle, the Feuerbah and the Nagel point of the triangle. Let be the seond intersetion point of line F e N a and the inirle. Then the parallel E u from E u to line intersets the Nagel line at a point S suh that segments SN a, S have onstant length for all triangles sharing the same inirle and entroid. s a result all these triangles have their Euler irles E simultaneously tangent to the inirle and a seond fixed irle 0 entered at S. The proof proeeds by showing that the ratio of oriented segments κ = N as N a = N a N a = N a N a F a N a N a F a is onstant. Sine the inirle and the Euler irle are homotheti with respet to F e, point, being the intersetion of line F e N a with the parallel to from E u, is on the Euler irle. Hene the last quotient is the ratio of powers of N a with respet

16 42. amfilos to the two irles: the variable Euler irle and the fixed inirle. enoting by r and R respetively the inradius and the irumradius of triangle ratio κ an be expressed as κ = N ae u 2 (R/2) 2 N a 2 r 2. The omputation of this ratio an be arried out using standard methods ( [1, p.103], [17, p.87]). slight simplifiation results from the fat ([17, p.30]) that homothety f maps the inenter to the Nagel point N a, produing the onstellation displayed in Figure 18, in whih H is the orthoenter and N is the point on line O suh that N : N O = 1 : 3 and N = EuNa 2. O E u N' H N a Figure 18. onfiguration s symmetry Thus, the two lengths needed for the determination of κ are N a E u = 2 N and N a = 3. n the next four equations O is given by Euler s relation ([17, p.10]),, O result by a standard alulation ([1, p.111], [14, p.185]), and N results by applying Stewart s theorem to triangle O [5, p.6]. O 2 = R(R 2r), O 2 = 1 9 (9R2 + 2r 2 + 8Rr 2s 2 ), 2 = 1 9 (5r2 + (s 2 16Rr)), N 2 = 1 16 (6r2 32Rr + 2s 2 + R 2 ), where s is the semiperimeter of the triangle. Using these relations ratio κ is found to be κ = 3r2 + (s 2 16Rr) 2(4r 2 + (s 2 16Rr)), whih using the above expression for u 2 = 2 beomes κ = (3u)2 2r 2 2((3u) 2 r 2 ). y our assumptions this is a onstant quantity, thereby proving the proposition.

17 Triangles with given inirle and entroid 43 The denominator of κ beomes zero preisely when 3 u = r. This is the ase when point S (see Figure 17) goes to infinity and also the ase in whih, aording to the previous setion, the lous of the verties of is a degenerate parabola of two parallel lines. Exluding this exeptional ase of infinite κ, to be handled below, last proposition implies that segments S,SN a (see Figure 17) have (onstant) orresponding lengths SN a = κ N a, S = κ r. Hene the Euler irle of is tangent simultaneously to the fixed inirle as well as to the seondary irle 0 with enter at S and radius equal to r 0 = κ r. n the ase κ = 0 the seondary irle ollapses to a point oiniding with the Nagel point of the triangle and the Euler irle of passes through that point. roposition 8. The enters E u of the Euler irles of triangles whih share the same inirle (,r) and entroid with r 3 are on a entral oni 1 with one fous at the inenter of the triangle and the other fous at the enter S of the seondary irle. The kind of this oni depends on the value of κ as follows. (1) For κ > 1 whih orresponds to 3 u < r oni 1 is an ellipse similar to and has great axis equal to r 0 r 2, where r 0 the radius of the seondary irle. (2) The other ases orrespond to values of κ < 0, 0 < κ < 1 2 and κ = 0. n the first two ases the oni 1 is a hyperbola similar to the onjugate one of the hyperbola. (3) n the ase κ = 0 the Euler irles pass through the Nagel point and the oni 1 is a retangular hyperbola similar to. '' ' 2 ' S N a 1 ' * Figure 19. xes ratio of The fat that the lous 1 of points E u is part of the entral oni with foi at {,S} and great axis equal to r 0 r 2 is a onsequene of the simultaneous tangeny of the Euler irles with the two fixed irles and 0 ([11, vol., p.42]). n the ase

18 44. amfilos ' F e ' S W N a E u ' Figure 20. xes ratio of 1 (ellipti ase) κ > 1 it is readily verified that the Euler irles ontain irle and are ontained in 0 and the sum of the distanes of E u from {,S} is r 0 r. n the ase κ < 0 the Nagel point is between S and and the Euler irle ontains and is outside 0. Thus the differene of distanes of E u from {,S} is equal to r 0 + r. n the ase 0 < κ < 1 2 the Euler irle ontains both irles and 0 and the differene of distanes of E u from {,S} is r r 0. Finally in the ase κ = 0 the differene of the distanes from {,S} is equal to r. This and the ondition 0 = κ = (3u) 2 2r 2 imply easily that 1 is a retangular hyperbola. To prove the similarity to in the ase κ > 1 it suffies to ompare the ratios of the axes of the two onis. y the remarks of 3, in this ase ( = u < r 3 ), the inirle is the maximal irle ontained in and the ratio of its axes is (see Figure 19) q = 1 2. Here is the hord parallel to the great axis and equal to the diameter of the inirle and 1 2 is the intersetion of this hord with the inirle. t is readily seen that this ratio is equal to 1 N2 a = 1 (3u)2. The orresponding ratio for 2 r 2 1 an be realized by onsidering the speial position of for whih the orresponding point E u beomes a vertex of 1 i.e., the position for whih the projetion W of E u on line is the middle of the segment S (see Figure 20). y this position of the ratio of axes of oni 1 is realized as EuW E u. y the resulting similarity

19 Triangles with given inirle and entroid 45 at this is equal to FeNa r F e = 2 Na 2 r = 1 (3u)2, thereby proving the laim in r 2 the ase 3 u < r. n the ase 3 u > r, i.e., when is a hyperbola, it was seen in Remark (5) of 3 that the asymptotes of the hyperbola are the tangents to from the Nagel point N a (see Figure 21). The proof results in this ase by taking the position of the variable triangle in suh a way that the enter of the Euler irle goes to infinity. n this ase points {,} defining the parallels {,E u } tend respetively to points { 0, 0 }, whih are the projetions of {,S} on the asymptote and the parallels tend respetively to { 0,S 0 }, whih are parallel to an asymptote of the oni 1. nalogously is verified also that the other asymptote of 1 is orthogonal to the orresponding other asymptote of. This proves the laim on the onjugay of 1 to. E E u K 0 N a S 0 0 F e Figure 21. xes ratio of 1 (hyperboli ase) Remarks. (1) The fat that the interval ( 1 2,1) represents a gap for the values of κ is due to the formula representing κ as a Moebius transformation of (3u) 2, whose the asymptote parallel to the x-axis is at y = 1 2. (2) n easy exploration of the same formula shows that {, 0 } are always disjoint, exept in the ase u = 2r 3, in whih κ = 1 3 and the two irles beome externally tangent at the Spieker point S p of the triangle, whih is the middle of N a. turn now to the singular ase orresponding to = u = r 3 for whih κ beommes infinite. Following lemma should be known, inlude though its proof for the sake of ompleteness of the exposition.

20 46. amfilos Lemma 9. Let irle, be tangent to line L 0 at its point and line L 1 be parallel to L 0 and not interseting. From a point H on L 1 draw the tangents to whih interset L 0 at points {K,L}. Then L K is onstant. Figure 22, illustrating the lemma, represents an interesting onfiguration and the problem at hand gives the opportunity to list some of its properties. (1) irle, passing through the enter of the given irle and points {K,L}, has its enter on the line H. (2) The other intersetion points {,R} of the tangents {HL,HK} with irle define line R, whih is symmetri to L 0 with respet to line H. (3) uadrangle LRK, whih is insribed in, is an isoseles trapezium. (4) oints (,M,O,H) = 1 define a harmoni division. Here M is the diametral of and O = L 0 R. (5) oint M, defined above, is an exenter of triangle HLK. (6) oints (,N,,F) = 1 define also a harmoni division, where N is the other intersetion point with of and F = L 1. L 1 H F R L 0 L O K ' N M Figure 22. L K is onstant (1) is seen by drawing first the medial lines of segments {L,K} whih meet at and define there the irumenter of LK. Their parallels from {L,K} respetively meet at the diametral M of on the irumirle of LK. Sine they are orthogonal to the bisetors of triangle HLK they are external bisetors of its angles and define an exenter of triangle HLK. (2), (3), (5) are immediate onsequenes. (4) follows from the standard way ([5, p.145]) to onstrut the polar of a point H with respet to a oni. (6) follows from (4) and the parallelism of lines {L 0,L 1,O,NM}. The initial laim is a onsequene of (6). This laim is also equivalent to the orthogonality of irle to the irle with diameter F.

21 Triangles with given inirle and entroid 47 L 1 ' L 0 L2 L E u ' R J N a S T M K F e Figure 23. Lous of E u for = r 3 roposition 10. Let (,r) be a irle with enter at point and radius r. Let also be situated at distane = r 3 from. Then the following statements are valid: (1) The homothety f entered at and with ratio k = 2 maps irle to irle (N a,2r) of radius 2r and tangent to at its intersetion point with line suh that : = 4 : 3. (2) Let T be the diametral point of on and be an arbitrary point on the line L 0, whih is orthogonal to T at T. Let be the triangle formed by drawing the tangents to from and interseting them with the parallel L 1 to L 0 from. Then is the entroid of. (3) The enter E u of the Euler irle of, as varies on line L 0, desribes a parabola with fous at and diretrix L 2 orthogonal to at a point J suh that J : = 3 : 4. Statement (1) is obvious. Statement (2) follows easily from the definitions, sine the middles {L,K} of {,} respetively define line KL whih is tangent to, orthogonal to and passes through the enter N a of (see Figure 23). enoting by the intersetion of K with and omparing the similar triangles and KN a one identifies easily with. To prove (3) use first Feuerbah s theorem ([12]), aording to whih the Euler irle is tangent to the inirle at a point F e. Let then E u R be the radius of the Euler irle parallel to. Line RF e passes through N a, whih is the diametral of on irle and whih oinides with the Nagel point of the triangle. This follows from Thales theorem for the similar isoseles triangles F e N a and

22 48. amfilos E u F e R. oint R projets on a fixed point S on. This follows by first observing that RSF e is a yli quadrangle, points {F e,s} viewing R under a right angle. t follows that N a S N a = N a R N a F u whih is equal to N a K N a L = 1 4 later, aording to previous lemma, being onstant and independent of the position of on line L 0. Using the previous fats we see that E u = E u F e F e = E u F e r = E u R r. Let point J on be suh that SJ = r and line L 2 be orthogonal to at J. Then the projetion of E u on L 2 satisfies E u = E u, implying that E u is on the parabola with fous at and diretrix L 2. The laim on the ratio J : = 3 : 4 follows trivially. 6. Misellanea n this setion disuss several aspets of the strutures involved in the problem at hand. start with the determination of the perspetor of the oni in baryentri oordinates. The lue here is the inidene of the oni at the diametral points {,} of the Nagel irle on the Nagel line (see Figure 24). These points an be expressed as linear ombinations of {, } : = p + p, = q + q. N a ' ** Figure 24. Triangle oni Here the equality is meant as a relation between the baryentri oordinates of the points and on the right are meant the normalized baryentri oordinates = 1 3 (1,1,1), = 1 2s (a,b,), where {a,b,} denote the side-lengths of triangle and s denotes its half-perimeter. From the assumptions follows that p p = 2(r u) = 2r + 3u, q q = + u) = 2(r 2r + 3u. The equation of the oni is determined by assuming its general form αyz + βzx + γxy = 0,

23 Triangles with given inirle and entroid 49 and omputing {α,β,γ} using the inidene ondition at and. This leads to the system of equations implying αp y p z + βp z p x + γp x p y = 0, αq y q z + βq z q x + γq x q y = 0, (α,β,γ) = λ((q x q y p z p x q z q x p x p y ),(q y q z p x p y q x q y p y p z ),(q z q x p y p z q y q z p z p x )), where the baryentri oordinates of and are given by p x p y = p + p 2r + 3u = r u a b, p 3 z 1 s q x q y = q + q 2r + 3u = 1 1 r + u a b. 3 1 s q z Making the neessary alulations and elliminating ommon fators we obtain (α,β,γ) = (( b)((3u(s a)) 2 r 2 (2s 3a) 2 ),, ), the dots meaning the orresponding formulas for β,γ resulting by yli permutation of the letters {a,b,}. The next proposition depits another aspet of the onfiguration, related to a ertain penil of irles passing through the (fixed) Spieker point of the triangle. roposition 11. Let (,r) be a irle at with radius r and a point. Let also be a triangle having as inirle and as entroid. Let further { E (E u,r E ), 0 (S,r 0 )} be respetively the Euler irle and the seondary irle of and {E u,f e,n a,s p } be respetively the enter of the Euler irle, the Feuerbah point, the Nagel point and the Spieker point of. Then the following statements are valid. (1) The penil of irles generated by and 0 has limit points {S p,t }, where T is the inverse of S p with respet to. (2) f is the other intersetion point of the Euler irle E with the line F e N a, then irle t tangent to the radii {F e,e u } respetively at points {F e, } belongs to the penil of irles J whih is orthogonal to and passes through {S p,t }. (3) The polar p T of T with respet to irles {, 0 } as well as the oni is the same line UV whih passes through S p. (4) n the ase κ > 1 the oni is an ellipse tangent to 0 at its intersetion points with line p T = UV. The orthogonality of irle t to the three irles {, 0, E } follows from its definition. This implies also the main part of the seond laim. To omplete the proof of the two first laims it suffies to identify the limit points of the penil of irles generated by and 0. For this use an be made of the fat that these two points are simultaneously harmoni onjugate with respet to the diameter points of the irles and 0 on line. enoting provisorily the intersetion points of t with

24 50. amfilos V E E u T S S p N a M F e * 0 U t Figure 25. irle penil assoiated to (, r) and line by {S,T } and their distane by d last property translates to the system of equations r 2 = S ( S + d), r 2 0 = SS ( SS + d). Eliminating d from these equations, setting x = S, S = SN a N a = (κ 1)N a = (1 κ)(3u), and r 0 = κ r, we obtain after some alulation, equation 6ux 2 + (9u 2 + 4r 2 )x + 6ur 2 = 0. One of the roots of this equation is x = 3u 2, identifying S with the Spieker point S p of the triangle. This ompletes the proof of the first two laims of the proposition. The third laim is an immediate onsequene of the first two and the fat that is bitangent to 0 at its diametral points with line. The fourth laim results from a trivial verifiation of {U,V } using the oordinates of the points. Remark. Sine points {, } remain fixed for all triangles onsidered, the same happens for every other point X of the Nagel line whih an be written as a linear ombination X = λ + µ of the normalized baryentri oordinates of {, } and with onstants {λ, µ} whih are independent of the partiular triangle. lso fixing suh a point X and expressing as a linear ombination of {,X} would imply the onstany of. This, in partiular, onsidering triangles with the

25 Triangles with given inirle and entroid 51 same inirle and Nagel point or the same inirle and Spieker point would onvey the disussion bak to the present one and fore all these triangles to have their verties on the oni. Referenes [1] T. ndreesu and. ndria, omplex Numbers From to Z, irkhaeuser, oston, [2] M. erger, eometry, 2 volumes, Springer, erlin, [3] O. ottema, Topis in Elementary eometry, Springer Verlag, Heidelberg, [4] J. arnoy, éométrie nalytique, authier-villars, [5] H. S. M. oxeter and S. L. reitzer, S. eometry Revisited, M, [6] R. rane, nother oristi System of Triangles, mer. Math. Monthly, 33 (1926) [7] W. allatly, The modern geometry of the triangle, Franis Hodgson, London, [8]. ibert, seudo-ivotal ubis and oristi Triangles, 2010, available at [9] J. T. roenmann, roblem 1423, rux Math., 25 (1989) 110. [10] R. Honsberger, Episodes in Nineteenth and Twentieth entury Eulidean eometry, M, [11] J. Koehler, Exeries de eoemetrie nalytique, 2 volumes, authier-villars, aris, [12] J.. Sott, n areal view of Feuerbah s theorem, Math. azette, 86 (2002) [13] S. Sigur, Sigur/resoures/pitures.pdf [14].. Smith, Statis and the moduli spae of triangles, Forum eom., 5 (2005) [15] J. Vonk, On the Nagel line and a prolifi polar triangle, Forum eom., 8 (2008) [16] J. H. Weaver, system of triangles related to a poristi system, mer. Math. Monthly, 31 (1928) [17]. Yiu, ntrodution to the eometry of the Triangle, Florida tlanti University Leture Notes, aris amfilos: epartment of Mathematis, University of rete, rete, reee address: pamfilos@math.uo.gr