The Heptagonal Triangle

Paul Yiu

Abstract.

1. The heptagonal triangle T Consider the triangles formed by choosing three among the vertices of a given regular heptagon. Since the diagonals of the regular heptagon have only two possi- ble lengths, 1 the scalene triangles so formed are all congruent, each having angles π 2π 4π 7 , 7 and 7 . We call these heptagonal triangles. L. Bankoff and J. Garfunkel [1] have given a number of interesting properties of the heptagonal triangle, so has Z. Cerin [2]. In this paper we systematically make use of complex number coordinates to study the triangle geometry of the heptagonal triangle, some com- mon triangle centers, lines, and circle associated with it. Among other things, we shall prove that the circumcenter and the Fermat points form an equilateral triangle (Theorem 16). C B

A′

O D

A

B′ C′

Figure 1. The heptagonal triangle T

2π 2π Consider the primitive 7-th root of unity ζ := cos 7 + i sin 7 , and denote by T the heptagonal triangle with vertices A = ζ4, B = ζ, C = ζ2 of a regular heptagon inscribed the unit circle of the complex plane. Among the remaining vertices of the regular heptagon, only ′ ′ ′ A = ζ3, B = ζ6, C = ζ5

1If each side of the regular heptagon has length a, the diagonal lengths b < c satisfy b2 −a2 = ac, c2 − b2 = ab and c2 − a2 = bc. See [1]. 1 2 P. Yiu form a heptagonal triangle. We label the seventh vertex D. Thus, D = 1. Clearly, the circumcenter O of T has coordinate 0, and the centroid is the point 1 2 4 G = 3 (ζ + ζ + ζ ). Since the orthocenter H and the nine-point center N are points (on the Euler line) satisfying

OG : GN : NH =2:1:3, we have

H = ζ + ζ2 + ζ4, 1 N = (ζ + ζ2 + ζ4). (1) 2 Lemma 1 (Gauss). 1 + 2(ζ + ζ2 + ζ4)= √7i.

2. Rotations A rotation is effected by the multiplication of a unit complex number. For ex- ample, the reflection of A in the side BC is the point A∗ obtained by rotating the vector BA about B by an angle 4π . Since cos 4π + i sin 4π = ζ−2, we − 7 − 7 − 7 have

∗ − − A = B + ζ 2(A B)= ζ + ζ 2(ζ4 ζ)= ζ + ζ2 ζ6. − − − For later use, we record the coordinates of the vertices of the triangle of reflections as ∗ A = ζ + ζ2 ζ6, ∗ − B = ζ2 + ζ4 ζ5, (2) ∗ − C = ζ ζ3 + ζ4. − Similarly, the coordinate of the incenter I satisfies the equation − ζ + ζ 1(I ζ)= ζ2 + ζ2(I ζ2). − − It is easy to verify that I = ζ3 ζ5 + ζ6. (3) − The same method gives the coordinates of the excenters.

Lemma 2. The excenters of the heptagonal triangle T are the points 3 5 6 Ia = (ζ + ζ + ζ ), − 3 5 6 Ib = ζ + ζ ζ , (4) − 3 5 6 Ic = ζ + ζ + ζ . − ∗ ∗ ∗ Proposition 3. The triangle of reflections A B C has circumcenter Ia and cir- cumradius 2R. Heptagonal triangles 3

Proof. ∗ 4 6 A Ia = (1 + ζ + ζ ), − − ∗ 5 B Ia = (1 + ζ + ζ ), − − ∗ 2 3 C Ia = (1 + ζ + ζ ) − − all have square norm equal to 2. 

The companion heptagonal triangle of A∗B∗C∗ has vertices on the sides of triangle ABC. The seventh vertex is the reflection of O in Ia.

3. The nine-point circle and its intersections with the circumcircle 3.1. A regular heptagon on the nine-point circle. As is well known, the nine-point circle is the circle through the vertices of the medial triangle and of the orthic triangle. The vertices of the medial triangle are ζ + ζ2 ζ2 + ζ4 ζ + ζ4 A0 = , B0 = , C0 = . 2 2 2 The vertices of the orthic triangle are the midpoints of the segments AA∗, BB∗, CC∗. These are the points

1 2 4 6 A1 = (ζ + ζ + ζ ζ ), 2 − 1 2 4 5 B1 = (ζ + ζ + ζ ζ ), 2 − 1 2 3 4 C1 = (ζ + ζ ζ + ζ ). 2 − Proposition 4. The vertices of the medial and orthic triangles are six vertices of a regular heptagon B0C0C1A0B1A1E on the nine-point circle.

Proof. With the coordinate of the nine-point N given by (1), we have B0 N = 1 ζ2 − ζ; C0 N = = ζ(B0 N). Similarly, C1 N = ζ(C0 N), A0 N = − 2 − − 2 − − − − ζ(C1 N), B1 N = ζ(A0 N), and A1 N = ζ(B1 N). If we put − − 1 − 2 4 − − E := N + ζ(A1 N)= ( 1+ ζ + ζ + ζ ), then ζ(E N)= B0 N. This − 2 − − − shows that B0C0C1A0B1A1E is a regular heptagon.  Proposition 5. E is the Euler reflection point of the heptagonal triangle T. Proof. The reflections of the circumcenter O in the sidelines are the points ∗ O = B + C O = ζ + ζ2, a − ∗ 2 4 Ob = C + A O = ζ + ζ , ∗ − O = A + B O = ζ + ζ4. c − 4 P. Yiu H

B1

A1 C

B A0 E N ′ A C1 B0

C0 O D

A

B′

C′ Figure 2.

Those of the orthocenter H in the sidelines are ∗ ∗ 6 Ha = A + A H = ζ , ∗ ∗ − − H = B + B H = ζ5, b − − ∗ ∗ 3 Hc = C + C B = ζ . ∗ ∗ ∗ ∗ ∗ ∗ − − The lines OaHa , Ob Hb , Oc Hc are the reflections of the Euler line OH in the sidelines. The following expressions show that E is the common point of these lines.

1 1+ ζ + ζ6 1 ζ ζ6 ( 1+ ζ + ζ2 + ζ4)= ( ζ6)+ − − (ζ + ζ2) 2 − 2 · − 2 · 1+ ζ2 + ζ5 1 ζ2 ζ5 = ( ζ5)+ − − (ζ2 + ζ4) 2 · − 2 · 1+ ζ3 + ζ4 1 ζ3 ζ4 = ( ζ3)+ − − (ζ + ζ4). 2 · − 2 · This is a point on the circumcircle since 1+ ζ + ζ2 + ζ4 1+ ζ6 + ζ5 + ζ3 − − = 1. 2 · 2 

3.2. The second intersection of the nine-point circle and the circumcircle.

Lemma 6. The distance between the nine-point center N and the A-excenter Ia is equal to the circumradius of the heptagonal triangle T. Heptagonal triangles 5

2+ζ+ζ2+ζ4 Proof. Note that Ia N = . It follows that − 2 1 NI2 = (2 + ζ + ζ2 + ζ4)(2 + ζ6 + ζ5 + ζ3) = 1. a 4 

This simple result has a number of interesting consequences.

Proposition 7. (1) The midpoint Fa of NIa is the point of tangency of the nine- point circle and the A-excircle; (2) The A-excircle is congruent to the nine-point circle. (3) Fa lies on the circumcircle.

H Ia B1

Fa A1 C

A0 B E N ′ A C1 B0

C0 O D

A

B′

C′ Figure 3.

Proof. (1) By the Feuerbach theorem, the nine-point circle is tangent externally to each of the excircles. Since NIa = R, the circumradius, and the nine-point circle 1 has radius 2 R, the point of tangency of with the A-excircle is the midpoint of NIa, i.e., I + N 2 + 3(ζ + ζ2 + ζ4) F = a = . (5) a 2 4 This proves (1). 1 (2) It also follows that the radius of the A-excircle is 2 R, and the A-excircle is congruent to the nine-point circle. 6 P. Yiu

(3) A simple calculation shows that 2 4 6 5 3 ∗ 2 + 3(ζ + ζ + ζ ) 2 + 3(ζ + ζ + ζ ) FaF = = 1, a  4  4  and Fa is a point on the circumcircle. 

Remark. The reflection of the orthic triangle in Fa is the A-extouch triangle, since the points of tangency are ζ3 ζ5 ζ6 (ζ3 + ζ5 + ζ6)+ , (ζ3 + ζ5 + ζ6)+ , (ζ3 + ζ5 + ζ6)+ . − 2 − 2 − 2 See Figure 3. 3.3. Another regular heptagon on the nine-point circle. On the nine-point circle, there is a regular heptagon with Fa as a vertex. These vertices, in an orientation opposite to T, have coordinates 1 F = (2 + 3(ζ + ζ2 + ζ4)), a 4 −1 1 2 4 5 6 Fe = N + ζ (Fa N)= (2ζ + ζ + ζ ζ + ζ ), − 4 − −2 1 2 3 4 5 Fc = N + ζ (Fa N)= (ζ + 2ζ ζ + ζ + ζ ), − 4 − ′ −3 1 2 4 5 6 F = N + ζ (Fa N)= (3ζ + 2ζ + 4ζ + ζ + ζ ), a − 4 −4 1 2 3 4 6 Fb = N + ζ (Fa N)= (ζ + ζ + ζ + 2ζ ζ ), − 4 − ′ −5 1 2 3 4 6 F = N + ζ (Fa N)= (2ζ + 4ζ + ζ + 3ζ + ζ ), c − 4 ′ −6 1 2 3 4 5 F = N + ζ (Fa N)= (4ζ + 3ζ + ζ + 2ζ + ζ ). b − 4 See Figure 4.

Proposition 8. Fe, Fa, Fb, Fc are the points of tangency of the nine-point circle with the incircle and the A-, B-, C-excircles respectively.

Proof. It is enough to show that the points Fe, Fb, Fc lie on the lines NI, NIb, NIc respectively. These are clear from 6 3 4 2(6 + 5ζ + 5ζ + 7ζ + 7ζ )Fe = ( 3 + 8ζ2 + 8ζ5 3ζ3 3ζ4)N + (1 + 2ζ + 2ζ6 + 3ζ3 + 3ζ4)I, − − − − 6 3 4 2(2 + ζ + ζ ζ ζ )Fb − − 6 3 4 6 =(2+ ζ + ζ 2ζ 2ζ )N +(2+ ζ + ζ )Ib, − − 6 3 4 2(2 + ζ + ζ ζ ζ )Fc − − 2 5 3 4 2 5 = (3ζ + 3ζ + 4ζ + 4ζ )N +(2+ ζ + ζ )Ic. −  Heptagonal triangles 7

′ Ia Fb

Fa ′ C Fc B Fe N I Fb

Ib Fc ′ Fa O

A

Ic

Figure 4.

′ ′ ′ Proposition 9. The points Fa, Fb, Fc are the second intersections of the nine-point circle with the lines joining Fa to A, B, C respectively. Proof.

6 2 5 ′ (7 + 6ζ + 6ζ + ζ + ζ )Fa 6 6 2 5 =(5+ ζ + ζ )Fa + (2 + 5ζ + 5ζ + ζ + ζ )A, 2 5 ′ (3 + 2ζ + 2ζ )Fb 6 3 4 6 3 4 = (2 5ζ 5ζ 3ζ 3ζ )Fa + ( 1 + 3ζ + 3ζ + ζ + ζ )B, − − − − − 6 3 4 ′ (6 + ζ + ζ + 3ζ + 3ζ )Fc 6 3 4 6 2 5 = (2 2ζ 2ζ + ζ + ζ )Fa + (4 + 3ζ + 3ζ + 2ζ + 2ζ )C. − −  8 P. Yiu

′ ′ ′ Remark. The triangles FbFeFc and FaFbFc are similar to the heptagonal triangle T.

4. The Kiepert perspector on the circumcircle We establish an interesting property of the vertex D = 1, which has important consequences on further properties of the heptagonal triangle T. Theorem 10. D is a Kiepert perspector of the heptagonal triangle ABC.

C′′

C B

A′

O D

A′′ A B′′

B′ C′

Figure 5. D as a Kiepert perspector of T

Proof. What this means is that there are similar isosceles triangles A′′BC, B′′CA, C′′AB with the same orientation such that the lines AA′′, BB′′, CC′′ all pass through the point D. Let A′′ be the intersection of the lines AD and A′B′, B′′ that of BD and B′C′, and C′′ that of CD and C′A′ (see Figure 5). Note that AC′B′A′′, BAC′B′′, and A′B′CC′′ are all parallelograms. From these, ′′ A = ζ4 ζ5 + ζ6, ′′ − B = ζ ζ3 + ζ5, ′′ − C = ζ2 + ζ3 ζ6. − It is clear that the lines AA′′, BB′′ and CC′′ all contain the point D. Heptagonal triangles 9

The coordinates of A′′, B′′, C′′ can be rewritten as 2 2 ′′ ζ + ζ ζ ζ A = + − (1 + 2(ζ + ζ2 + ζ4)), 2 2 · 4 2 4 2 ′′ ζ + ζ ζ ζ B = + − (1 + 2(ζ + ζ2 + ζ4)), 2 2 · 4 4 ′′ ζ + ζ ζ ζ C = + − (1 + 2(ζ + ζ2 + ζ4)). 2 2 · Since 1 + 2(ζ + ζ2 + ζ4) = √7i (Gauss sum), these expressions show that the three isosceles triangles all have base angles arctan √7. Thus, the triangles A′′BC, B′′CA, C′′AB are similar isosceles triangles of the same orientation. From these we conclude that D is a point on the Kiepert hyperbola.  Corollary 11. The center of the Kiepert hyperbola is the point

1 3 5 6 Ki = (ζ + ζ + ζ ). (6) −2 Proof. Since D is the intersection of the Kiepert hyperbola and the circumcircle, the center of the Kiepert hyperbola is the midpoint of DH, where H is the ortho- center of triangle ABC. This has coordinate as given in (6) above. 

Remark. Ki is also the midpoint of OIa. 5. The Brocard circle 5.1. The Brocard points. Proposition 12 (Bankoff and Garfunkel). The nine-point center N is the first Bro- card point.

C

K B

N A′

D O

A

B′

C′

Figure 6. The Brocard points of the heptagonal triangle T 10 P. Yiu

Proof. The relations

1 4+ ζ + ζ2 + ζ4 (ζ + ζ2 + ζ4) ζ4 = (ζ ζ4), 2 − 12 + 6(ζ2 + ζ5) + 2(ζ3 + ζ4) · − 1 4+ ζ + ζ2 + ζ4 (ζ + ζ2 + ζ4) ζ = (ζ2 ζ), 2 − 12 + 6(ζ3 + ζ4) + 2(ζ + ζ6) · − 1 4+ ζ + ζ2 + ζ4 (ζ + ζ2 + ζ4) ζ2 = (ζ4 ζ2) 2 − 12 + 6(ζ + ζ6) + 2(ζ2 + ζ5) · − can be easily verified. In each case, the denominator of the fraction on the right hand side is a real number. This means that the lines NA, NB, NC are obtained by rotation of BA, CB, AC through the same angle (which is necessarily the Brocard angle ω). This shows that the nine-point center N is the first Brocard point of the heptagonal triangle T. 

Remark. It follows that 4+ ζ + ζ2 + ζ4 = √14(cos ω + i sin ω).

It is known that on the Brocard circle with diameter OK, ∠NOK = ω. −

i K = 1 N  − √7 · 2(4 + ζ3 + ζ5 + ζ6) ζ + ζ2 + ζ4 = 7 · 2 2 = (1 + 2(ζ + ζ2 + ζ4)). 7

Note that by Lemma 1, the Brocard axis OK is along the imaginary axis. Now, 1 3 the second Brocard point, being the reflection of N in OK, is simply 2 (ζ + 5 6 − ζ + ζ ). This, according to Corollary 11, is the Kiepert center Ki. Since OD is along the real axis, it is tangent to the Brocard circle.

5.2. A regular heptagon on the Brocard circle.

Proposition 13. The first Brocard triangle (whose vertices are the intersections of the Brocard circle with the perpendicular bisectors of the sides) is also a heptago- nal triangle. It is perspective with ABC at the point 1 . − 2

Proposition 14. The circumcenter O of the heptagonal triangle T, together with the second intersections of the Brocard circle and the lines joining O to A, B, C, A′, B′, C′, form a regular heptagon. Heptagonal triangles 11

C

B K A0

A′ B0 C0

D O

A

B′

C′

Figure 7. A regular heptagon on the Brocard circle

Proof. 1 1 (1 + 2(ζ + ζ2 + ζ4))(1 ζ)= (1 3ζ 3ζ6 ζ2 ζ5) ζ4, 7 − 7 − − − − · 1 1 (1 + 2(ζ + ζ2 + ζ4))(1 ζ2)= (1 3ζ2 3ζ5 ζ3 ζ4) ζ, 7 − 7 − − − − · 1 1 (1 + 2(ζ + ζ2 + ζ4))(1 ζ3)= (1 ζ ζ6 3ζ3 3ζ4) ( ζ5), 7 − 7 − − − − · − 1 1 (1 + 2(ζ + ζ2 + ζ4))(1 ζ4)= (1 ζ ζ6 3ζ3 3ζ4) ζ2, 7 − 7 − − − − · 1 1 (1 + 2(ζ + ζ2 + ζ4))(1 ζ5)= (1 3ζ2 3ζ5 ζ3 ζ4) ( ζ6), 7 − 7 − − − − · − 1 1 (1 + 2(ζ + ζ2 + ζ4))(1 ζ6)= (1 3ζ 3ζ6 ζ2 ζ5) ( ζ3). 7 − 7 − − − − · − 

6. Proposition 15. The perpendicular bisector of the segment ON is the line con- taining X = 1 and Y = 1 (1 (ζ3 + ζ5 + ζ6)). − 2 − Proof. (1) Complete the parallelogram OIaHX, then 2 4 3 5 6 X = O + H Ia = (ζ + ζ + ζ ) + (ζ + ζ + ζ )= 1 − − is a point on the circumcircle. Also, NX = NIa = R = OX. This shows that X is on the bisector of ON. 12 P. Yiu

(2) Complete the parallelogram ONIaY , with Y = O + Ia N. Explicitly, Y = 1 (1 (ζ3 + ζ5 + ζ6)). But we also have − 2 − X + Y = (O + H Ia)+ (O + Ia N) = (2 N Ia)+ (O + Ia N)= O + N. − − · − − This means that OXNY is a rhombus, and XY is the perpendicular bisector of ON. 

H Ia

C Fa

B N K Ki Y

A′ L

X D O

A

B′

C′

Figure 8. The circumcenter and the Fermat points form an equilateral triangle

We conclude by proving a surprising theorem involving the Fermat points. Theorem 16. The circumcenter and the Fermat points of the heptagonal triangle T form an equilateral triangle. Proof. (1) Consider the circle through O, with center at the point 1 L := (ζ3 + ζ5 + ζ6). −3 This is the center of the equilateral triangle with O as a vertex and Ki the midpoint of the opposite side. Heptagonal triangles 13

1 (2) With X and Y in Proposition 15, it is easy to check that L = 3 (X + 2Y ). This means that L lies on the perpendicular bisector of ON. (3) Since Ki is on the Brocard circle (with diameter OK), OKi is perpendicular to the line KKi. It is well known that the line KKi contains the Fermat points. Indeed, Ki is the midpoint of the Fermat points. This means that L is lies on the perpendicular bisector of the Fermat points. (4) By a well known theorem of Lester, 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 O is the segment joining the Fermat points. 

Corollary 17. The Fermat points of the heptagonal triangle T are the points

1 2 3 5 6 F+ = (λ + 2λ )(ζ + ζ + ζ ), 3 1 2 3 5 6 F− = (λ + 2λ)(ζ + ζ + ζ ), 3

1 2 1 where λ = 2 ( 1+ √3i) and λ = 2 ( 1 √3i) are the imaginary cube roots of unity. − − −

Remarks. (1) The triangle with vertices Ia and the Fermat points is also equilateral. 2 (2) Since OIa = √2R, each side of the equilateral triangle has length 3 R. (3) The Lester circle is congruent to the orthocentroidal circle, whichq has HG as a diameter. (4) The Brocard axis OK is tangent to the A-excircle at the midpoint of IaH.

7. Some further properties (1) Since X = 1 is antipodal to the Kiepert perspector D = 1 on the circum- circle, 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 circum- center, the A-excenter, and the midpoint of HG. The tangents at Ia and X pass through H, and that at O passes through Y . (2) The triangle of reflections has its circumcenter at Ia. Its circumcircle con- tains D = 1 and E the Euler reflection point, and O. 14 P. Yiu

D A∗

C

B∗

Ia

C

E B A A′

B

D O C∗

A

B′

C′

Figure 9. The heptagonal triangle T

8. The reflection triangle

∗ A = ζ + ζ2 ζ6, − = (ζ2 + ζ5 + ζ3 + ζ4) 1+(2+ ζ + ζ6)ζ; ∗ · B = ζ2 + ζ4 ζ5, − = (ζ + ζ6 + ζ3 + ζ4) 1+(2+ ζ2 + ζ5)ζ2; · B =1+ ζ4 ζ6, − =(1+ ζ2 + ζ5)ζ2 (ζ2 + ζ5)ζ4; ∗ − C = ζ ζ3 + ζ4 − = (ζ + ζ6 + ζ2 + ζ5) 1+(2+ ζ3 + ζ4)ζ4; · A =1+ ζ2 ζ3, − =(1+ ζ + ζ6)ζ (ζ + ζ6)ζ2; − C =1+ ζ ζ5, − =(1 + ζ3 + ζ4)ζ4 (ζ3 + ζ4)ζ; − D = 2(ζ3 + ζ5 + ζ6), − Heptagonal triangles 15

The triangle of reflections is triply perspective with T.

Triple perspectivity.

A′B′C′ perspector ABC H BCA infinite point of 1z CAB D

A′

C′

H

B O′ ζ

a b C

c 0 A = 1

′ D = ζ4 B

ζ6

ζ5

9. Proposition 18. The Simson lines of the vertices of the companion heptagonal triangle are concurrent.

10. Given a real cyclotomic number a(ζ + ζ6)+ b(ζ2 + ζ5)+ c(ζ3 + ζ4), we seek another one so that the product is rational 16 P. Yiu

Proposition 19. If d := (a+b+c)3 7(ab2 +bc2 +ca2 +abc) = 0, the reciprocal of the real cyclotomic number a(ζ +−ζ6)+ b(ζ2 + ζ5)+ c(ζ3 +6 ζ4) is 1 ′ ′ ′ a (ζ + ζ6)+ b (ζ2 + ζ5)+ c (ζ3 + ζ4) , d
where ′ a = (c a)2 bc, ′ − − b = (a b)2 ca, ′ − − c = (b c)2 ab. − − References [1] L. Bankoff and J. Garfunkel, The heptagonal triangle, Math. Mag., 46 (1973) 7–19. [2] Z. Cerin, 2005. [3] C. Kimberling, Encyclopedia of Triangle Centers, available at http://faculty.evansville.edu/ck6/encyclopedia/ETC.html.

Paul Yiu: Department of Mathematical Sciences, Florida Atlantic University, Boca Raton, Florida 33431-0991, USA E-mail address: [email protected]