Heptagonal Triangles and Their Companions

Heptagonal Triangles and TheirCompanions

Paul YiuDepartment of Mathematical Sciences,

Florida Atlantic University,

[email protected]

MAA Florida Sectional MeetingFebruary 13–14, 2009

Florida Gulf Coast University1

2 P. Yiu

Abstract. A heptagonal triangle is a non-isoscelestriangle formed by three vertices of a regular hep-tagon. Its angles areπ7 , 2π

7 and 4π7 . 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 anumber of interesting companion pairs of heptag-onal triangles on its nine-point circle and Brocardcircles. Among other results on the geometry ofthe heptagonal triangle, we prove that the circum-center and the Fermat points of a heptagonal tri-angle form an equilateral triangle. The proof is aninteresting application of Lester’s theorem that theFermat points, the circumcenter and the nine-pointcenter of a triangle are concyclic.

The heptagonal triangle 3

The heptagonal triangleT and its circumcircle

c

a

b

B

C

A

O

4 P. Yiu

The companion ofT

D = residual vertex

B

C

A′

A

C′

B′

O


Relation between the sides of the heptagonal triangle

a

a

a

b

b

c

b2 = a(c + a)

b

a

b

b

c

c

c2 = a(a + b)

6 P. Yiu

Archimedes’ construction (Arabic tradition)

b a cAB

HK

AABK

C

E

F

H


Division of segment: a neusis constructionLet BHPQ be a square, with one sideBH sufficiently extended.

Draw the diagonalBP . Place a ruler throughQ, intersecting the diagonalBP atT , and

the sideHP atE, and the lineBH atA such that the trianglesAHE andTPQ have equal

areas. Then,

b a c

B HK A

Q PL

T

E

GSP

b2 = a(c + a), c2 = (a + b)b.

8 P. Yiu

The orthic triangle

The heptagonal triangle, apart from the equilateral triangle, isthe only triangle similar its own orthic triangle.

H

B1

C1

A1

C

A B


Analysis:

H

A′

B′

C′

A

B C

(a)

H

A′

B′

C′

A

B C

A′ =

Case (a):ABC acute:

A′ = π − 2A, B′ = π − 2B, C ′ = π − 2C.

{A′, B′, C ′} = {A, B, C} =⇒ A = B = C =π

3.

Case (b): AngleC obtuse:

A′ = 2A, B′ = 2B, C ′ = 2C − π.

{A′, B′, C ′} = {A, B, C} =⇒ {A, B, C} = {π

7,

2π

7,

4π

7}.

10 P. Yiu

H

B1

C1

A1

C

A B

The orthic triangle of the heptagonal triangle hassimilarity factor 1

2 , because . . .


H

B1

C1

A1

C

A B

B0

C0

A0

N

because its vertices are on thenine-point circle of T,which also contains the vertices of themedial triangle,i.e., the midpoints of the sides ofT.

12 P. Yiu

H

B1

C1

A1

C

A B

B0

C0

A0

N

?

The orthic triangle and the medial triangle arecompanion heptagonal triangles.

What can we say about theresidual vertex?


This residual vertex also lies on the circumcircle ofT.

H

B1

C1

A1

C

A B

B0

C0

A0

N

To justify this and probe more into the geometry of the hep-tagonal triangle, we make use of . . .

14 P. Yiu

Complex Coordinates

1 = D

ζ = B

C = ζ2

A′= ζ3

ζ4= A

ζ5= C′

ζ6= B′

O

ζ = primitive 7-th root of unity

ζ6 + ζ5 + ζ4 + ζ3 + ζ2 + ζ + 1 = 0.


Points on the Euler line ofT

B

C

A

O

H

N

G

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

Center Notation Coordinatescircumcenter O 0centroid G 1

3(ζ + ζ2 + ζ4)nine-point center N 1

2(ζ + ζ2 + ζ4)orthocenter H ζ + ζ2 + ζ4

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Companionship of medial and orthic triangles

D

B

C

A′

A

C′

B′

O

H

B0

A0

C0

B1

A1

C1

N

E

Center: N = 1

2(ζ + ζ2 + ζ4)

Residual vertex: E = 1

2(−1 + ζ + ζ2 + ζ4)

Rotation Medial triangle Rotation Orthic triangleζ4 A0 = 1

2(ζ + ζ2) ζ3 C1 = 1

2(ζ + ζ2 − ζ3 + ζ4)

ζ B0 = 1

2(ζ2 + ζ4) ζ6 A1 = 1

2(ζ + ζ2 + ζ4 − ζ6)

ζ2 C0 = 1

2(ζ + ζ4) ζ5 B1 = 1

2(ζ + ζ2 + ζ4 − ζ5)

Why isE a point on the circumcircle ofT?


Residual vertexE= Euler reflection point ofT= Intersection of the reflections of the Euler line

in the three sidelines ofT.

B

C

A

O

H

B0

A0

C0

B1

A1

C1

N

E

18 P. Yiu

D

B

C

A′

A

B′

O

H

B0

A0

C0

B1

A1

C1

N

EHa

Hb

Hc

Remark. The reflections ofH in the sides are the antipodes ofthe vertices of the companion ofH

′.


The second intersection of the nine-point circle and the circum-circle

D

B

C

A′

A

C′

B′

O

H

B0

A0

C0

B1

A1

C1

N

E

?

This is a point related to . . .

20 P. Yiu

. . . the tritangent circles ofT, i.e., theincircle andexcircles.

B

C

A′

C′

B′

O

Ia

Ib

Ic

I

−A′

−C′

−B′


Feuerbach theorem: The nine-point circle is tangent tothe incircle internally atFe and tothe excircles atFa, Fb, Fc.

A

B

C

O

N

Fa

Ia

Ib

Fb

Ic

Fc

I

Fe

22 P. Yiu

The second intersection of the circumcircle and the nine-pointcircle isFa, the point of tangency of the nine-point circle withtheA-excircle.

D

B

C

A′

A

C′

B′

O

H

B0

A0

C0

B1

A1

C1

N

E

Fa

Ia


Anothercompanion pair of heptagonal triangleson the nine-point circle: triangleFbFeFc and triangleF ′

aF′bF

′c,

whereF ′a, F ′

b, F ′c are the intersections ofAFa, BFa, CFa with

the nine-point circle.

A

B

C

O

N

Fa

Ia

Ib

Fb

Ic

Fc

I

Fe

F ′

a

F ′

b

F ′

c

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 (Brocardangleω).

B

C

A

O

Brocard

B

C

A

O

Brocard


B

C

A

O

K

Brocard

Brocard

The two Brocard points are equidistant fromO.If K is thesymmedian point of the triangle, then

the circle with diameterOK contains the two Brocard points.Each of the two chords makes an angleω with OK.

26 P. Yiu

B

C

A

O

N

B

C

A

O

?

For the heptagonal triangleT,(i) the Brocard angleω satisfies

cotω =√

7;

(ii) one of the Brocard points is the nine-point centerN

(Bankoff-Garfunkel,Mathematics Magazine, 46 (1973) 7–19).


B

C

A

O

N

K

?

N =1

2(ζ + ζ2 + ζ4) =⇒ K =

2√7· i

? = −1

2(ζ3 + ζ5 + ζ6).

Later, we shall identify ? as an interesting triangle centerofT.

Meanwhile,

28 P. Yiu

a companion pair of heptagonal triangleson the Brocard circle:

D

B

C

A′

A

C′

B′

O

K

A−ω

C−ω

B−ω−

12

The vertices are the second intersections ofOA, OB, OC,OA′, OB′, OC ′ with the nine-point circle.

A−ωB−ωC−ω is called the first Brocard triangle ofT. TheBrocard circle is also called theseven-point circle, containingO, K, two Brocard points, three vertices of Brocard triangle.


Theresidual vertex of T

D

B

C

A

O

A′′

B′′

C′′

If similar isosceles trianglesA′′BC, B′′CA, C ′′AB

of vertical angles2ω are constructed on the sides,thenAA′′, BB′′, CC ′′ intersect atD on the circumcircle.

What this means is that . . .

30 P. Yiu

the midpoint ofDH

= 12(1 + ζ + ζ2 + ζ4) = −1

2(ζ3 + ζ5 + ζ6).

Note that this is thesecond Brocard point of T we met before:

B

C

A

O

N

K

?

N =1

2(ζ + ζ2 + ζ4) =⇒ K =

2√7· i

? = −1

2(ζ3 + ζ5 + ζ6).

But more importantly, it is


the centerKi of theKiepert hyperbola.

D

B

C

A

O

A′′

B′′

C′′

Ki

H

32 P. Yiu

Ki is also the midpoint between the twoFermat points of T:

D

B

C

A

O

A′′

B′′

C′′

KiFermat

Fermat

H


Recall the Fermat points

A

B C

F+

C′

A′

B′

Fermat point

A

B C

F−

C′′

A′′

B′′ Negative Fermat point

34 P. Yiu

There are many wonderful properties of the Fermat points ofa triangle. For example,

F

Y

Z

X

Y

A

B C


Lester Theorem: the circumcenter, the nine-point center, andthe Fermat points are concyclic.

O

N

A

B C

F+

F−

Lester’s circle

36 P. Yiu

Theorem In the heptagonal triangleT, the circumcenter andthe Fermat points form an equilateral triangle.

D

B

C

A

O

KiFermat

Fermat

H


Proof. (1) The line joining the Fermat pointsalso contains the symmedian pointK,and is perpendicular toOKi.

D

B

C

A

O

KKi

Fermat

Fermat

H

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(2) The perpendicular bisector ofON intersectsthe circumcircle atX = −1 andY = 1

2(1 − (ζ3 + ζ5 + ζ6)) on the circumcircle.(3) This perpendicular bisector intersectsOKi at

L = −13(ζ

3 + ζ5 + ζ6).

B

C

A

O

N

K

Ki

X

Y

H

Fa

L


(4) Consider the Lester circle. SinceKi is the midpoint be-tween the Fermat points, the center of the Lester circle is the in-tersection ofOKi with the perpendicular bisector ofON . Thisis the pointL.

B

C

A

O

N

K

Ki

X

Y

H

Fa

L

Fermat

Fermat

40 P. Yiu

(5) SinceOL = 23 · OKi, the centroid and the circumcenter

of the triangleOF+F− coincide.

This shows that triangleOF+F− is equilateral.

B

C

A

O

N

K

Ki

H

Fa

L

F+

F−

Heptagonal Triangles and Their Companions

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