Heptagonal Triangles and Their Companions

Forum Geometricorum Volume 9 (009) 15 148. FRUM GEM ISSN 1534-1178 Heptagonal Triangles and Their ompanions Paul Yiu bstract. heptagonal triangle is a non-isosceles triangle formed by three vertices of a regular heptagon. Its angles are π, π and 4π. s such, there is a 7 7 7 unique choice of a companion heptagonal triangle formed by three of the remaining four vertices. Given a heptagonal triangle, we display a number of interesting companion pairs of heptagonal triangles on its nine-point circle and rocard circle. mong other results on the geometry of the heptagonal triangle, we prove that the circumcenter and the Fermat points of a heptagonal triangle form an equilateral triangle. The proof is an interesting application of Lester s theorem that the Fermat points, the circumcenter and the nine-point center of a triangle are concyclic. 1. The heptagonal triangle T and its companion heptagonal triangle T is one with angles π 7, π 7 and 4π 7. Its vertices are three vertices of a regular heptagon inscribed in its circumcircle. mong the remaining four vertices of the heptagon, there is a unique choice of three which form another (congruent) heptagonal triangle T. We call this the companion of T, and the seventh vertex of the regular heptagon the residual vertex of T and T (see Figure 1). In this paper we work with complex number coordinates, and take the unit circle Figure 1. heptagonal triangle and its companion in the complex plane for the circumcircle of T. y putting the residual vertex at 1, we label the vertices of T by = ζ 4, = ζ, = ζ, Publication ate: June, 009. ommunicating Editor: ntreas P. Hatzipolakis. The author thanks the Editor for suggesting the configuration studied in 8, leading to, among other results, Theorem 0 on six circles concurring at the Feuerbach point of the heptagonal triangle.

16 P. Yiu and those of T by = ζ 3, = ζ 6, = ζ 5, where ζ := cos π 7 + isin π 7 is a primitive 7-th root of unity. We study the triangle geometry of T, some common triangle centers, lines, circles and conics associated with it. We show that the Simson lines of,, with respect to T are concurrent (Theorem 4). We find a number of interesting companion pairs of heptagonal triangles associated with T. For example, the medial triangle and the orthic triangle of T form such a pair on the nine-point circle (Theorem 5), and the residual vertex is a point on the circumcircle of T. It is indeed the Euler reflection point of T. In the final section we prove that the circumcenter and the Fermat points form an equilateral triangle (Theorem ). The present paper can be regarded as a continuation of ankoff-garfunkel [1].. Preliminaries.1. Some simple coordinates. learly, the circumcenter of T has coordinate 0, and the centroid is the point G = 1 3 (ζ + ζ + ζ 4 ). Since the orthocenter H and the nine-point center N are points (on the Euler line) satisfying we have G : GN : NH = : 1 : 3, H = ζ + ζ + ζ 4, N = 1 (ζ + ζ + ζ 4 ). (1) This reasoning applies to any triangle with vertices on the unit circle. The bisectors of angles,, of T intersect the circumcircle at,, respectively. These form a triangle whose orthocenter is the incenter I of T (see Figure ). This latter is therefore the point I = ζ 3 ζ 5 + ζ 6. () Similarly, the external bisectors of angles,, intersect the circumcircle at,, respectively. Identifying the excenters of T as orthocenters of triangles with vertices on the unit circle, we have I a = (ζ 3 + ζ 5 + ζ 6 ), I b = ζ 3 + ζ 5 ζ 6, (3) I c = ζ 3 + ζ 5 + ζ 6. Figure shows the tritangent circles of the heptagonal triangle T... Representation of a companion pair. Making use of the simple fact that the complex number coordinates of vertices of a regular heptagon can be obtained from any one of them by multiplications by ζ,..., ζ 6, we shall display a companion pair of heptagonal triangle by listing coordinates of the center, the residual vertex and the vertices of the two heptagonal triangles, as follows.

Heptagonal triangles and their companions 17 I a I I b I c Figure. The tritangent centers enter: Residual vertex: P Q Rotation Vertices Rotation Vertices ζ 4 P + ζ 4 (Q P) ζ 3 P + ζ 3 (Q P) ζ P + ζ(q P) ζ 6 P + ζ 6 (Q P) ζ P + ζ (Q P) ζ 5 P + ζ 5 (Q P).3. While we shall mostly work in the cyclotomic field Q(ζ), 1 the complex number coordinates of points we consider in this paper are real linear combinations of ζ k for 0 k 6, (the vertices of the regular heptagon on the circumcircle of 1 See orollary 3 for an exception.

18 P. Yiu T). The real coefficients involved are rational combinations of c 1 = ζ + ζ6 = cos π 7, c = ζ + ζ 5 = cos 4π 7, c 3 = ζ3 + ζ 4 = cos 6π 7. Note that c 1 > 0 and c, c 3 < 0. n expression of a complex number z as a real linear combination of ζ 4, ζ, ζ (with sum of coefficients equal to 1) actually gives the absolute barycentric coordinate of the point z with reference to the heptagonal triangle T. For example, ζ 3 = c ζ 4 + c ζ + 1 ζ, ζ 5 = c 1 ζ 4 + 1 ζ c 1 ζ, ζ 6 = 1 ζ 4 c 3 ζ + c 3 ζ, 1 = c ζ 4 c 3 ζ c 1 ζ. We shall make frequent uses of the important result. Lemma 1 (Gauss). 1 + (ζ + ζ + ζ 4 ) = 7i. Proof. lthough this can be directly verified, it is actually a special case of Gauss famous theorem that if ζ = cos π n + isin π n for an odd integer n, then n 1 ζ k = k=0 For a proof, see [, pp.75 76]. { n if n 1 (mod 4), ni if n 3 (mod 4)..4. Reflections and pedals. Lemma. If α, β, γ are unit complex numbers, the reflection of γ in the line joining α and β is γ = α + β αβγ. Proof. s points in the complex plane, γ has equal distances from α and β as γ does. This is clear from γ α = β(1 αγ) = βγ(γ α), γ β = α(1 βγ) = αγ(γ β). orollary 3. (1) The reflection of ζ k in the line joining ζ i and ζ j is ζ i +ζ j ζ i+j k. () The pedal (orthogonal projection) of ζ k on the line joining ζ i and ζ j is 1 (ζi + ζ j + ζ k ζ i+j k ). (3) The reflections of in, in, and in are the points = ζ + ζ ζ 6, = ζ + ζ 4 ζ 5, (4) = ζ ζ 3 + ζ 4.

Heptagonal triangles and their companions 19 3. oncurrent Simson lines The Simson line of a point on the circumcircle of a triangle is the line containing the pedals of the point on the sidelines of the triangle. Theorem 4. The Simson lines of,, with respect to the heptagonal triangle T are concurrent. Figure 3. Simson lines Proof. The pedals of on is the midpoint of ; similarly for those of on and on. We tabulate the coordinates of the pedals of,, on the sidelines,, respectively. These are easily calculated using orollary 3. 1 ( 1 + ζ + ζ + ζ 3 1 ) (ζ + ζ 4 1 ) (ζ ζ + ζ 3 + ζ 4 ) 1 (ζ + ζ ζ 4 + ζ 6 1 ) ( 1 + ζ + ζ 4 + ζ 6 1 ) (ζ + ζ4 ) 1 (ζ + ζ 1 ) ( ζ + ζ + ζ 4 + ζ 5 1 ) ( 1 + ζ + ζ4 + ζ 5 ) We check that the Simson lines of,, all contain the point 1. For these, it is enough to show that the complex numbers (ζ+ζ +ζ 3 )(1 + ζ + ζ 4 ), (ζ +ζ 4 +ζ 6 )(1 + ζ + ζ 4 ), (ζ+ζ 4 +ζ 5 )(1 + ζ + ζ ) are real. These are indeed ζ + ζ 6, ζ + ζ 5, ζ 3 + ζ 4 respectively.

130 P. Yiu Remark. The Simson line of, on the other hand, is parallel to (see Figure 3). This is because the complex number coordinates of the pedals of, namely, 1 + ζ + ζ ζ 3, 1 + ζ + ζ 4 ζ 6, 1 + ζ + ζ 4 ζ 5, all have the same imaginary part 1 4 (ζ ζ6 + ζ ζ 5 ζ 3 + ζ 4 ). 4. The nine-point circle 4.1. companion pair of heptagonal triangles on the nine-point circle. s is well known, the nine-point circle is the circle through the vertices of the medial triangle and of the orthic triangle. The medial triangle of T clearly is heptagonal. It is known that T is the only obtuse triangle with orthic triangle similar to itself. The medial and orthic triangles of T are therefore congruent. It turns out that they are companions. Theorem 5. The medial triangle and the orthic triangle of T are companion heptagonal triangles on the nine-point circle of T. The residual vertex is the Euler reflection point E (on the circumcircle of T). H 1 1 E N 0 1 0 0 Figure 4. companion pair on the nine-point circle If the angles of an obtuse angled triangle are α β < γ, those of its orthic triangle are α, β, and γ π. The two triangles are similar if and only if α = γ π, β = α and γ = β. From these, α = π, β = π and γ = 4π. This shows that the triangle is heptagonal. The equilateral 7 7 7 triangle is the only acute angled triangle similar to its own orthic triangle.

Heptagonal triangles and their companions 131 Proof. (1) The companionship of the medial and orthic triangles on the nine-point circle is clear from the table below. enter: N = 1 (ζ + ζ + ζ 4 ) Residual vertex: E = 1 ( 1 + ζ + ζ + ζ 4 ) Rotation Medial triangle Rotation rthic triangle ζ 4 0 = 1 (ζ + ζ ) ζ 3 1 = 1 (ζ + ζ ζ 3 + ζ 4 ) ζ 0 = 1 (ζ + ζ 4 ) ζ 6 1 = 1 (ζ + ζ + ζ 4 ζ 6 ) ζ 0 = 1 (ζ + ζ4 ) ζ 5 1 = 1 (ζ + ζ + ζ 4 ζ 5 ) a H 1 1 H b E H a 0 b N c 1 0 0 H c Figure 5. The Euler reflection point of T () We show that E is a point on the reflection of the Euler line in each of the sidelines of T. In the table below, the reflections of are computed from the simple fact that a, b, c are rhombi. n the other hand, the reflections of H in the sidelines can be determined from the fact that HH a and have the same midpoint, so do HH b and, HH c and. The various expressions for E given in the rightmost column can be routinely verified.

13 P. Yiu Line Reflection of Reflection of H E = a = ζ + ζ H a = ζ 6 ( c 1 c c 3 ) a + ( c c 3 )H a b = ζ + ζ 4 H b = ζ5 ( c 1 c c 3 ) b + ( c 1 c 3 )H b c = ζ + ζ 4 H c = ζ 3 ( c 1 c c 3 ) c + ( c 1 c )H c Thus, E, being the common point of the reflections of the Euler line of T in its sidelines, is the Euler reflection point of T, and lies on the circumcircle of T. 4.. The second intersection of the nine-point circle and the circumcircle. Lemma 6. The distance between the nine-point center N and the -excenter I a is equal to the circumradius of the heptagonal triangle T. Proof. Note that I a N = +ζ+ζ +ζ 4 complex number. = 3+1+(ζ+ζ +ζ 4 ) 4 = 3+ 7i This simple result has a number of interesting consequences. 4 is a unit Proposition 7. (1) The midpoint F a of NI a is the point of tangency of the ninepoint circle and the -excircle. () The -excircle is congruent to the nine-point circle. (3) F a lies on the circumcircle. H 1 I a 1 F a E N 0 1 0 0 Figure 6. The -Feuerbach point of T

Heptagonal triangles and their companions 133 Proof. (1) y the Feuerbach theorem, the nine-point circle is tangent externally to each of the excircles. Since NI a = R, the circumradius, and the nine-point circle has radius 1 R, the point of tangency with the -excircle is the midpoint of NI a, i.e., F a = I a + N = + 3(ζ + ζ + ζ 4 ). (5) 4 This proves (1). () It also follows that the radius of the -excircle is 1 R, and the -excircle is congruent to the nine-point circle. (3) Note that F a = 1+3+6(ζ+ζ +ζ 4 ) 8 = 1+3 7i 8 is a unit complex number. Remark. The reflection of the orthic triangle in F a is the -extouch triangle, since the points of tangency are (ζ 3 + ζ 5 + ζ 6 ) + ζ3, (ζ3 + ζ 5 + ζ 6 ) + ζ5, (ζ3 + ζ 5 + ζ 6 ) + ζ6 (see Figure 6). 4.3. nother companion pair on the nine-point circle. enter: N = 1 (ζ + ζ + ζ 4 ) Residual vertex: F a = 1 4 ( + 3(ζ + ζ + ζ 4 )) Rot. Feuerbach triangle Rot. ompanion ζ 3 F b = 1 4 (ζ + ζ + ζ 3 + ζ 4 ζ 6 ) ζ 4 F a = 1 4 (3ζ + ζ + 4ζ 4 + ζ 5 + ζ 6 ) ζ 6 F e = 1 4 (ζ + ζ + ζ 4 ζ 5 + ζ 6 ) ζ F b = 1 4 (4ζ + 3ζ + ζ 3 + ζ 4 + ζ 5 ) ζ 5 F c = 1 4 (ζ + ζ ζ 3 + ζ 4 + ζ 5 ) ζ F c = 1 4 (ζ + 4ζ + ζ 3 + 3ζ 4 + ζ 6 ) Proposition 8. F e, F a, F b, F c are the points of tangency of the nine-point circle with the incircle and the -, -, -excircles respectively (see Figure 7). Proof. We have already seen that F a = 1 N + 1 I a. It is enough to show that the points F e, F b, F c lie on the lines NI, NI b, NI c respectively: F e = (c 1 c 3 ) N + ( c 1 c 3c 3 ) I, F b = (c c 3 ) N + ( c 1 3c c 3 ) I b, F c = (c 1 c ) N + ( 3c 1 c c 3 ) I c. Proposition 9. The vertices F a, F b, F c of the companion of F b F e F c are the second intersections of the nine-point circle with the lines joining F a to,, respectively. Proof. F a = c F a (c 1 + c 3 ), F b = c 3 F a (c 1 + c ), F c = c 1 F a (c + c 3 ).

134 P. Yiu F b I a F c F a F b N I F e I b F a F c I c Figure 7. nother companion pair on the nine-point circle 5. The residual vertex as a Kiepert perspector Theorem 10. is a Kiepert perspector of the heptagonal triangle. Proof. What this means is that there are similar isosceles triangles,, with the same orientation such that the lines,, all pass through the point. Let be the intersection of the lines and, that of and, and that of and (see Figure 8). Note that,, and are all parallelograms. From these, = ζ 4 ζ 5 + ζ 6, = ζ ζ 3 + ζ 5, = ζ + ζ 3 ζ 6.

Heptagonal triangles and their companions 135 Figure 8. as a Kiepert perspector of T It is clear that the lines, and all contain the point. The coordinates of,, can be rewritten as = ζ + ζ = ζ4 + ζ = ζ + ζ4 + ζ ζ + ζ4 ζ + ζ ζ4 (1 + (ζ + ζ + ζ 4 )), (1 + (ζ + ζ + ζ 4 )), (1 + (ζ + ζ + ζ 4 )). Since 1 + (ζ + ζ + ζ 4 ) = 7i (Gauss sum), these expressions show that the three isosceles triangles all have base angles arctan 7. Thus, the triangles,, are similar isosceles triangles of the same orientation. From these we conclude that is a point on the Kiepert hyperbola. orollary 11. The center of the Kiepert hyperbola is the point K i = 1 (ζ3 + ζ 5 + ζ 6 ). (6) Proof. Since is the intersection of the Kiepert hyperbola and the circumcircle, the center of the Kiepert hyperbola is the midpoint of H, where H is the orthocenter of triangle (see Figure 9). This has coordinate as given in (6) above. Remark. K i is also the midpoint of I a.

136 P. Yiu H K i Figure 9. The Kiepert hyperbola of T Since X = 1 is antipodal to the Kiepert perspector = 1 on the circumcircle, it is the Steiner point of T, which is the fourth intersection of the Steiner ellipse with the circumcircle. The Steiner ellipse also passes through the circumcenter, the -excenter, and the midpoint of HG. The tangents at I a and X pass through H, and that at passes through Y = 1 (1 (ζ3 + ζ 5 + ζ 6 )) on the circumcircle such that XNY is a parallelogram (see Lemma 1). H I a N Y G X = 1 Figure 10. The Steiner ellipse of T

Heptagonal triangles and their companions 137 6. The rocard circle 6.1. The rocard points. Proposition 1 (ankoff and Garfunkel). The nine-point center N is the first rocard point. K N Figure 11. The rocard points of the heptagonal triangle T Proof. The relations 1 (ζ + ζ + ζ 4 ) ζ 4 = ( c 1 3c c 3 )(4 + ζ + ζ + ζ 4 ) (ζ ζ 4 ), 7 1 (ζ + ζ + ζ 4 ) ζ = ( c 1 c 3c 3 )(4 + ζ + ζ + ζ 4 ) (ζ ζ), 7 1 (ζ + ζ + ζ 4 ) ζ = ( 3c 1 c c 3 )(4 + ζ + ζ + ζ 4 ) (ζ 4 ζ ) 7 show that the lines N, N, N are obtained by rotations of,, through the same angle (which is necessarily the rocard angle ω). This shows that the nine-point center N is the first rocard point of the heptagonal triangle T. Remark. It follows that 4 + ζ + ζ + ζ 4 = 14(cos ω + isinω). Proposition 13. The symmedian point K has coordinate (1+(ζ+ζ +ζ 4 )) 7 = i 7.

138 P. Yiu Proof. It is known that on the rocard circle with diameter K, NK = ω. From this, K = 1 (cos ω isinω) N cos ( ω = 1 i ) N 7 by Lemma 1. = (4 + ζ3 + ζ 5 + ζ 6 ) 7 = 7 (1 + (ζ + ζ + ζ 4 )) = i 7 ζ + ζ + ζ 4 orollary 14. The second rocard point is the Kiepert center K i. Proof. y Proposition 13, the rocard axis K is along the imaginary axis. Now, the second rocard point, being the reflection of N in K, is simply 1 (ζ3 + ζ 5 + ζ 6 ). This, according to orollary 11, is the Kiepert center K i. Since is along the real axis, it is tangent to the rocard circle. 6.. companion pair on the rocard circle. enter: 1 7 (1 + (ζ + ζ + ζ 4 )) Residual vertex: = 0 Rot. First rocard triangle Rot. ompanion ζ 3 ω = 1 7 ( 4c 1 c 8c 3 ) ( ζ 5 ) ζ 4 1 7 ( 4c 1 c 8c 3 ) ζ ζ 6 ω = 1 7 ( 8c 1 4c c 3 ) ( ζ 3 1 ) ζ 7 ( 8c 1 4c c 3 ) ζ 4 ζ 5 ω = 1 7 ( c 1 8c 4c 3 ) ( ζ 6 ) ζ 1 7 ( c 1 8c 4c 3 ) ζ Since ζ 5 is the midpoint of the minor arc joining ζ and ζ, the coordinate of the point labeled ω shows that this point lies on the perpendicular bisector of. Similarly, ω and ω lie on the perpendicular bisectors of and respectively. Since these points on the rocard circle, they are the vertices of the first rocard triangle. The vertices of the companion are the second intersections of the rocard circle with and the lines joining to,, respectively. Proposition 15. The first rocard triangle is perspective with at the point 1 (see Figure 1). Proof. 1 = ( 3c 1 c c 3 ) ω + c 1 ζ 4, = ( c 1 3c c 3 ) ω + c ζ, = ( c 1 c 3c 3 ) ω + c 3 ζ.

Heptagonal triangles and their companions 139 K ω ω 1 ω Figure 1. regular heptagon on the rocard circle 7. companion of the triangle of reflections We have computed the coordinates of the vertices of the triangle of reflections in (4). It is interesting to note that this is also a heptagonal triangle, and its circumcenter coincides with I a. The residual vertex is the reflection of in I a. enter: I a = (ζ 3 + ζ 5 + ζ 6 ) Residual vertex: = (ζ 3 + ζ 5 + ζ 6 ) Rotation Triangle of reflections Rotation ompanion ζ 4 = ζ + ζ ζ 6 ζ 3 = 1 + ζ 4 ζ 6 ζ = ζ + ζ 4 ζ 5 ζ 6 = 1 + ζ ζ 5 ζ = ζ ζ 3 + ζ 4 ζ 5 = 1 + ζ ζ 3 The companion has vertices on the sides of triangle, = (1 + c 1 )ζ c 1 ζ ; = (1 + c )ζ c ζ 4 ; = (1 + c 3 )ζ 4 c 3 ζ. It is also perspective with T. Indeed, the lines,, are all perpendicular to the Euler line, since the complex numbers 1 + ζ ζ 3 ζ 4 ζ + ζ + ζ 4, 1 + ζ 4 ζ 6 ζ ζ + ζ + ζ 4, 1 + ζ ζ 5 ζ ζ + ζ + ζ 4 are all imaginary, being respectively (ζ ζ 5 ), (ζ 3 ζ 4 ), (ζ ζ 6 ).

140 P. Yiu I a E Figure 13. The triangle of reflections of T Proposition 16. The triangle of reflections is triply perspective with T. Proof. The triangle of reflection is clearly perspective with at the orthocenter H. Since,, are all parallel (to the imaginary axis), the two triangles are triply perspective ([3, Theorem 381]). In other words, is also perspective with. In fact, the perspector is the residual vertex : = (1 + c 1 ) 1 + ( + c 1 )ζ, = (1 + c ) 1 + ( + c )ζ, = (1 + c 3 ) 1 + ( + c 3 )ζ 4. Remark. The circumcircle of the triangle of reflections also contains the circumcenter, the Euler reflection point E, and the residual vertex.

Heptagonal triangles and their companions 141 8. partition of T by the bisectors Let I I I be the cevian triangle of the incenter I of the heptagonal triangle T =. It is easy to see that triangles I, I and I are also heptagonal. Each of these is the image of the heptagonal triangle under an affine mapping of the form w = αz + β or w = αz + β, according as the triangles have the same or different orientations. Note that the image triangle has circumcenter β and circumradius α. I I I I Figure 14. Partition of T by angle bisectors Each of these mappings is determined by the images of two vertices. For example, since and I have the same orientation, the mapping f 1 (z) = αz +β is determined by the images f 1 () = and f 1 () = ; similarly for the mappings f and f 3. ffine mapping f 1 (z) = (ζ + ζ 4 )z ζ 5 I f (z) = (1 + ζ + ζ 3 + ζ 4 )z (1 + ζ 3 + ζ 6 ) I f 3 (z) = (1 + ζ + ζ 4 + ζ 5 )z (1 + ζ 3 + ζ 5 ) I Thus, we have I = f 1 () = ζ 3 ζ 5 + ζ 6, I = f () = 1 + ζ + ζ ζ 3 + ζ 5, I = f 3 () = 1 + ζ + ζ 4 ζ 5 + ζ 6. Note also that from f ( I ) = I, it follows that I = 1 + ζ ζ 3 + ζ 4 ζ 6. Remark. The affine mapping that associates a heptagonal triangle with circumcenter c and residual vertex d to its companion is given by w = d c dc cd z + d c d c.

14 P. Yiu 8.1. Four concurrent lines. simple application of the mapping f 1 yields the following result on the concurrency of four lines. Proposition 17. The orthocenter of the heptagonal triangle I lies on the line and the perpendicular from I to. I I I H I Figure 15. Four concurrent altitudes Proof. Since has orthocenter H = ζ + ζ + ζ 4, the orthocenter of triangle I is the point H = f 1 (H) = (1 + ζ 4 ) = (ζ + ζ 5 )ζ. This expression shows that H lies on the radius. Now, the vector H I is given by I H = ( 1 + ζ + ζ ζ 3 + ζ 5 ) + (1 + ζ 4 ) = ζ + ζ ζ 3 + ζ 4 + ζ 5. n the other hand, the vector is given by ζ ζ 4. To check that H I is perpendicular to, we need only note that (ζ + ζ ζ 3 + ζ 4 + ζ 5 )(ζ ζ 4 ) = (ζ ζ 6 ) + (ζ ζ 5 ) + (ζ 3 ζ 4 ) is purely imaginary. Remark. Similarly, the orthocenter of I lies on the -altiude of, and that of I on the -altitude.

Heptagonal triangles and their companions 143 8.. Systems of concurrent circles. Proposition 18. The nine-point circles of I and (the isosceles triangle) are tangent internally at the midpoint of. N f (N) I M Figure 16. Two tangent nine-point circles Proof. The nine-point circle of the isosceles triangle clearly contains the midpoint M of. Since triangle is also isosceles, the perpendicular from to passes through M. This means that M lies on the nine-point circle of triangle I. We show that the two circles are indeed tangent at M. The nine-point center of I is the point f (N) = 1 (ζ + ζ + ζ 3 + ζ 4 + ζ 6 ). n the other hand, the nine-point center of the isosceles triangle is the point Since N = 1 (ζ + ζ 3 + ζ 6 ). M = ζ + ζ 6 = (1 c 4c 3 )f (N) + (c + 4c 3 )N as can be verified directly, we conclude that the two circles are tangent internally.

144 P. Yiu Theorem 19. The following circles have a common point. (i) the circumcircle of I, (ii) the nine-point circle of I, (iii) the -excircle of I, (iv) the nine-point circle of I. I a I I I Figure 17. Four concurrent circles Proof. y Proposition 7(3), the first three circles concur at the -Feuerbach point of triangle I, which is the point f (F a ) = 1 4 (ζ + ζ + ζ 4 ζ 5 + ζ 6 ). It is enough to verify that this point lies on the nine-point circle of I, which has center ( ζ + ζ + ζ 4 ) f 3 = ζ + ζ + ζ 3 + ζ 4 + ζ 6, and square radius 1 4 1 + ζ + ζ 4 + ζ 5 = 1 ( 3(ζ + ζ 6 ) + (ζ + ζ 5 ) + (ζ 3 + ζ 4 ) ). 4 This is exactly the square distance between f (F a ) and the center, as is directly verified. This shows that f (F a ) indeed lies on the nine-point circle of I.

Heptagonal triangles and their companions 145 Theorem 0. Each of the following circles contains the Feuerbach point F e of T : (i) the nine-point circle of T, (ii) the incircle of T, (iii) the nine-point circle of the heptagonal triangle I, (iv) the -excircle of I, (v) the -excircle of the heptagonal triangle I, (vi) the incircle of the isosceles triangle I I. I a I N N I I I F e I I c Figure 18. Six circles concurrent at the Feuerbach point of T Proof. It is well known that the nine-point circle and the incircle of T are tangent to each other internally at the Feuerbach point F e. It is enough to verify that this point lies on each of the remaining four circles. (iii) and (iv) The -excircle of I is the image of the -excircle of under the affine mapping f 1. It is therefore enough to check that f 1 (F b ) = F e : f 1 (F b ) = 1 4 (ζ + ζ4 )(ζ + ζ + ζ 3 + ζ 4 ζ 6 ) ζ 5 = 1 4 (ζ + ζ + ζ 4 ζ 5 + ζ 6 ) = F e.

146 P. Yiu (v) The heptagonal triangle I is the image of under the mapping f. It can be verified directly that W = 1 4 (ζ ζ + 3ζ 3 + 3ζ 5 ) ζ 6 is the point for which f (W) = F e. The square distance of W from the -excenter I a = (ζ 3 + ζ 5 + ζ 6 ) is the square norm of W I a = 1 4 ( ζ + ζ + ζ 3 + ζ 5 ). n easy calculation shows that this is 1 16 ( ζ + ζ + ζ 3 + ζ 5 )(ζ + ζ 4 + ζ 5 ζ 6 ) = 1 4 = r a. It follows that, under the mapping f, F e lies on the -excircle of I. Q I I I M I P F e I Figure 19. The incircle of an isosceles triangle (vi) Since I and I I are isosceles triangles, the perpendicular bisectors of and I are the bisectors of angles I I and I I respectively. It follows that the incenter of the isosceles triangle I I coincides with the circumcenter of triangle I, which is the point I = (1 + ζ 3 + ζ 5 ) from the affine mapping f 3. This incircle touches the side I I at its midpoint M, the side I at the midpoint Q of I, and the side I at the orthogonal projection P of on (see Figure 19). simple calculation shows that PMQ = 3π 7. To show that F e lies on the same circle, we need only verify that PF e Q = 4π 7. To this end, we first determine some complex number coordinates: P = 1 (ζ + ζ ζ 3 + ζ 4 ), Q = 1 ( 1 + ζ + ζ4 ζ 5 + ζ 6 ). Now, with F e = 1 4 (ζ + ζ + ζ 4 ζ 5 + ζ 6 ), we have Q F e = (ζ 4 + ζ 6 )(P F e ). From the expression ζ 4 + ζ 6 = ζ (ζ + ζ 6 ), we conclude that indeed PF e Q = 4π 7.

Heptagonal triangles and their companions 147 9. theorem on the Fermat points Lemma 1. The perpendicular bisector of the segment N is the line containing X = 1 and Y = 1 (1 (ζ3 + ζ 5 + ζ 6 )). Proof. (1) omplete the parallelogram I a HX, then X = + H I a = (ζ + ζ + ζ 4 ) + (ζ 3 + ζ 5 + ζ 6 ) = 1 is a point on the circumcircle. Note that N is the midpoint of I a X. Thus, NX = NI a = R = X. This shows that X is on the bisector of N. () omplete the parallelogram NI a Y, with Y = + I a N. Explicitly, Y = 1 (1 (ζ3 + ζ 5 + ζ 6 )). ut we also have X +Y = (+H I a )+(+I a N) = ( N I a )+(+I a N) = +N. This means that XNY is a rhombus, and NY = Y. From (1) and (), XY is the perpendicular bisector of N. H I a F a N K K i Y L X Figure 0. The circumcenter and the Fermat points form an equilateral triangle Theorem. The circumcenter and the Fermat points of the heptagonal triangle T form an equilateral triangle.

148 P. Yiu Proof. (1) onsider the circle through, with center at the point L := 1 3 (ζ3 + ζ 5 + ζ 6 ). This is the center of the equilateral triangle with as a vertex and K i = 1 (ζ3 + ζ 5 + ζ 6 ) the midpoint of the opposite side. See Figure 0. () With X and Y in Lemma 1, it is easy to check that L = 1 3 (X + Y ). This means that L lies on the perpendicular bisector of N. (3) Since K i is on the rocard circle (with diameter K), K i is perpendicular to the line KK i. It is well known that the line KK i contains the Fermat points. 3 Indeed, K i is the midpoint of the Fermat points. This means that L is lies on the perpendicular bisector of the Fermat points. (4) y a well known theorem of Lester (see, for example, [5]), the Fermat points, the circumcenter, and the nine-point center are concyclic. The center of the circle containing them is necessarily L, and this circle coincides with the circle constructed in (1). The side of the equilateral triangle opposite to is the segment joining the Fermat points. orollary 3. The Fermat points of the heptagonal triangle T are the points F + = 1 3 (λ + λ )(ζ 3 + ζ 5 + ζ 6 ), F = 1 3 (λ + λ)(ζ 3 + ζ 5 + ζ 6 ), where λ = 1 ( 1 + 3i) and λ = 1 ( 1 3i) are the imaginary cube roots of unity. Remarks. (1) The triangle with vertices I a and the Fermat points is also equilateral. () Since I a = R, each side of the equilateral triangle has length 3 R. (3) The Lester circle is congruent to the orthocentroidal circle, which has HG as a diameter. (4) The rocard axis K is tangent to the -excircle at the midpoint of I a H. References [1] L. ankoff and J. Garfunkel, The heptagonal triangle, Math. Mag., 46 (1973) 7 19. [] K. Ireland and M. Rosen, lassical Introduction to Modern Number Theory, nd edition, 1990, Springer-Verlag. [3] R.. Johnson, dvanced Euclidean Geometry, 199, over reprint 007. [4]. Kimberling, Encyclopedia of Triangle enters, available at http://faculty.evansville.edu/ck6/encyclopedia/et.html. [5] P. Yiu, The circles of Lester, Evans, Parry, and their generalizations, preprint. Paul Yiu: epartment of Mathematical Sciences, Florida tlantic University, 777 Glades Road, oca Raton, Florida 33431-0991, US E-mail address: yiu@fau.edu 3 The line joining the Fermat points contains K and Ki.