Heptagonal Triangles and Their Companions Paul Yiu Department of Mathematical Sciences, Florida Atlantic University,

[email protected]

MAA Florida Sectional Meeting February 13–14, 2009 Florida Gulf Coast University 1 2 P. Yiu Abstract. A heptagonal triangle is a non-isosceles triangle formed by three vertices of a regular hep- π 2π 4π tagon. Its angles are 7 , 7 and 7 . As such, there is a unique choice of a companion heptagonal tri- angle formed by three of the remaining four ver- tices. Given a heptagonal triangle, we display a number of interesting companion pairs of heptag- onal triangles on its nine-point circle and Brocard circles. Among other results on the geometry of the heptagonal triangle, we prove that the circum- center and the Fermat points of a heptagonal tri- angle 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. The heptagonal triangle 3 The heptagonal triangle T and its circumcircle

C

B a

b c

O

A 4 P. Yiu The companion of T

C

B

A′

O D = residual vertex

A

B′

C′ The heptagonal triangle 5 Relation between the sides of the heptagonal triangle

b a

b c

a c a b

c b

a b

b2 = a(c + a) c2 = a(a + b) 6 P. Yiu Archimedes’ construction (Arabic tradition)

K H B A b a c

C

B AA K H

F

E The heptagonal triangle 7 Division of segment: a neusis construction Let BHPQ be a square, with one side BH sufficiently extended. Draw the diagonal BP . Place a ruler through Q, intersecting the diagonal BP at T , and the side HP at E, and the line BH at A such that the triangles AHE and TPQ have equal areas. Then,

B K H A

b a c

E

T

Q L P GSP

b2 = a(c + a), c2 =(a + b)b. 8 P. Yiu The orthic triangle The heptagonal triangle, apart from the equilateral triangle, is the only triangle similar its own orthic triangle.

H

A1

B1 C

A C1 B The heptagonal triangle 9 Analysis:

A A

C′ B′ A′ B C

′ C H B′

H B ′ C A ′ (a) A =

Case (a): ABC acute:

A′ = π 2A, B′ = π 2B, C′ = π 2C. − − − π A′,B′,C′ = A, B, C = A = B = C = . { } { } ⇒ 3 Case (b): Angle C obtuse:

A′ =2A, B′ =2B, C′ =2C π. − π 2π 4π A′,B′,C′ = A, B, C = A, B, C = , , . { } { } ⇒ { } { 7 7 7 } 10 P. Yiu

H

A1

B1 C

A C1 B

The orthic triangle of the heptagonal triangle has 1 similarity factor 2, because ... The heptagonal triangle 11

H

A1

B1 C

N

B0 A0

A C0 C1 B

because its vertices are on the nine-point circle of T, which also contains the vertices of the medial triangle, i.e., the midpoints of the sides of T. 12 P. Yiu

H

A1

? B1 C

N

B0 A0

A C0 C1 B

The orthic triangle and the medial triangle are companion heptagonal triangles.

What can we say about the residual vertex? The heptagonal triangle 13 This residual vertex also lies on the circumcircle of T.

H

A1

B1 C

N

B0 A0

A C0 C1 B

To justify this and probe more into the geometry of the hep- tagonal triangle, we make use of . . . 14 P. Yiu Complex Coordinates

C = ζ2

ζ = B

′ A = ζ3

O 1 = D

ζ4 = A

ζ6 = B′

ζ5 = C′ ζ = primitive 7-th root of unity

ζ6 + ζ5 + ζ4 + ζ3 + ζ2 + ζ +1=0. The heptagonal triangle 15 Points on the Euler line of T H

C B

N G

O

A

OG : GN : NH =2:1:3.

Center Notation Coordinates circumcenter O 0 1 2 4 centroid G 3(ζ + ζ + ζ ) 1 2 4 nine-point center N 2(ζ + ζ + ζ ) orthocenter H ζ + ζ2 + ζ4 16 P. Yiu Companionship of medial and orthic triangles H

B1

A1 C

B A0 E N ′ A C1

B0

C0 O D

A

B′

C′

1 2 4 Center: N = 2 (ζ + ζ + ζ ) Residual vertex: E = 1 ( 1+ ζ + ζ2 + ζ4) 2 − Rotation Medial triangle Rotation Orthic triangle 4 1 2 3 1 2 3 4 ζ A0 = 2 (ζ + ζ ) ζ C1 = 2 (ζ + ζ ζ + ζ ) 1 2 4 6 1 2 − 4 6 ζ B0 = 2 (ζ + ζ ) ζ A1 = 2 (ζ + ζ + ζ ζ ) 2 1 4 5 1 2 4 − 5 ζ C0 = (ζ + ζ ) ζ B1 = (ζ + ζ + ζ ζ ) 2 2 − Why is E a point on the circumcircle of T? The heptagonal triangle 17 Residual vertex E = Euler reflection point of T = Intersection of the reflections of the Euler line in the three sidelines of T.

H

B1

A1 C

A0 B E N

C1

B0

C0 O

A 18 P. Yiu

H

B1

A1 C Hb

A0 B Ha E N

′ A C1

B0

C0 O D

Hc A

B′

Remark. The reflections of H in the sides are the antipodes of the vertices of the companion of H′. The heptagonal triangle 19 The second intersection of the nine-point circle and the circum- circle H

B1

? A1 C

A0 B E N

′ A C1

B0

C0 O D

A

′ B

C′ This is a point related to . . . 20 P. Yiu . . . the tritangent circles of T, i.e., the incircle and excircles.

Ia

− ′ C C −B′ B

A′ I

Ib

O

−A′

B′

C′

Ic The heptagonal triangle 21 Feuerbach theorem: The nine-point circle is tangent to the incircle internally at Fe and to the excircles at Fa, Fb, Fc.

Ia

Fa C

N B Fe I Fb

Ib Fc

O

A

Ic 22 P. Yiu The second intersection of the circumcircle and the nine-point circle is Fa, the point of tangency of the nine-point circle with the A-excircle.

H Ia B1

Fa A1 C

A0 B E N ′ A C1

B0

C0 O D

A

B′

C′ The heptagonal triangle 23 Another companion pair of heptagonal triangles on the nine-point circle: triangle FbFeFc and triangle Fa′Fb′Fc′, where Fa′, Fb′, Fc′ are the intersections of AFa, BFa, CFa with the nine-point circle.

′ Fb Ia

Fa ′ C Fc B Fe N I Fb

Ib Fc ′ Fa O

A

Ic 24 P. Yiu The Brocard points Given a triangle there are two unique points (Brocard points) for each of which the three marked angles are equal (Brocard angle ω).

C C

B B Brocard Brocard

O O

A A The heptagonal triangle 25

C

K B

Brocard Brocard

O

A

The two Brocard points are equidistant from O. If K is the symmedian point of the triangle, then the circle with diameter OK contains the two Brocard points. Each of the two chords makes an angle ω with OK. 26 P. Yiu

C C

B B ?

N

O O

A A

For the heptagonal triangle T, (i) the Brocard angle ω satisfies cot ω = √7; (ii) one of the Brocard points is the nine-point center N (Bankoff-Garfunkel, Mathematics Magazine, 46 (1973) 7–19). The heptagonal triangle 27 C

B K

N ?

O

A

1 2 N = (ζ + ζ2 + ζ4) = K = i 2 ⇒ √7 · 1 ? = (ζ3 + ζ5 + ζ6). −2 Later, we shall identify ? as an interesting triangle center of T. Meanwhile, 28 P. Yiu a companion pair of heptagonal triangles on the Brocard circle:

C

K B A−ω

′ A C−ω

B−ω − 1 2 D O

A

′ B

C′

The vertices are the second intersections of OA, OB, OC, OA′, OB′, OC′ with the nine-point circle. A ωB ωC ω is called the first Brocard triangle of T. The Brocard− − circle− is also called the seven-point circle, containing O, K, two Brocard points, three vertices of Brocard triangle. The heptagonal triangle 29 The residual vertex of T C′′

C B

O D

A′′ A B′′

If similar isosceles triangles A′′BC, B′′CA, C′′AB of vertical angles 2ω are constructed on the sides, then AA′′, BB′′, CC′′ intersect at D on the circumcircle. What this means is that . . . 30 P. Yiu the midpoint of DH 1 2 4 1 3 5 6 = 2(1 + ζ + ζ + ζ ) = 2(ζ + ζ + ζ ). Note that this is the second− Brocard point of T we met before: C

B K

N ?

O

A

1 2 N = (ζ + ζ2 + ζ4) = K = i 2 ⇒ √7 · 1 ? = (ζ3 + ζ5 + ζ6). −2 But more importantly, it is The heptagonal triangle 31

the center Ki of the Kiepert hyperbola.

C′′

H

C

B Ki

O D

A′′ A B′′ 32 P. Yiu

Ki is also the midpoint between the two Fermat points of T:

C′′

H

C

B Fermat Ki

Fermat

O D

A′′ A B′′ The heptagonal triangle 33 Recall the Fermat points

B′ ′′ F− A A A

C′

F+ ′′ C B C

B C

A′ Fermat point B′′ Negative Fermat point 34 P. Yiu There are many wonderful properties of the Fermat points of a triangle. For example,

Y

A

Z Y

F

B C

X The heptagonal triangle 35 Lester Theorem: the circumcenter, the nine-point center, and the Fermat points are concyclic.

A F−

Lester’s circle

O

F+ N

B C 36 P. Yiu Theorem In the heptagonal triangle T, the circumcenter and the Fermat points form an equilateral triangle.

H

C

B Fermat Ki

Fermat

D O

A The heptagonal triangle 37 Proof. (1) The line joining the Fermat points also contains the symmedian point K, and is perpendicular to OKi.

H

C

B Fermat K Ki

Fermat

D O

A 38 P. Yiu (2) The perpendicular bisector of ON intersects the circumcircle at X = 1 and Y = 1(1 (ζ3 + ζ5 + ζ6)) on the circumcircle. − 2 − (3) This perpendicular bisector intersects OKi at L = 1(ζ3 + ζ5 + ζ6). −3 H

C Fa

K B N Ki Y

L

X O

A The heptagonal triangle 39

(4) Consider the Lester circle. Since Ki is the midpoint be- tween the Fermat points, the center of the Lester circle is the in- tersection of OKi with the perpendicular bisector of ON. This is the point L.

H

C Fa

Fermat K B Ki Y N Fermat L

X O

A 40 P. Yiu (5) Since OL = 2 OK , the centroid and the circumcenter 3 · i of the triangle OF+F coincide. −

This shows that triangle OF+F is equilateral. − H

C Fa

F+ B K

Ki N F− L

O

A