Heptagonal Triangles and TheirCompanions
Paul YiuDepartment of Mathematical Sciences,
Florida Atlantic University,
MAA Florida Sectional MeetingFebruary 13–14, 2009
Florida Gulf Coast University1
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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 5
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)
The heptagonal triangle 7
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.
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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
The heptagonal triangle 9
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}.
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H
B1
C1
A1
C
A B
The orthic triangle of the heptagonal triangle hassimilarity factor 1
2 , because . . .
The heptagonal triangle 11
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.
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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?
The heptagonal triangle 13
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 . . .
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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.
The heptagonal triangle 15
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?
The heptagonal triangle 17
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
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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 heptagonal triangle 19
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 . . .
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. . . the tritangent circles ofT, i.e., theincircle andexcircles.
B
C
A′
C′
B′
O
Ia
Ib
Ic
I
−A′
−C′
−B′
The heptagonal triangle 21
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
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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
The heptagonal triangle 23
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
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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
The heptagonal triangle 25
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.
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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).
The heptagonal triangle 27
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,
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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.
The heptagonal triangle 29
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 . . .
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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
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Ki is also the midpoint between the twoFermat points of T:
D
B
C
A
O
A′′
B′′
C′′
KiFermat
Fermat
H
The heptagonal triangle 33
Recall the Fermat points
A
B C
F+
C′
A′
B′
Fermat point
A
B C
F−
C′′
A′′
B′′ Negative Fermat point
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There are many wonderful properties of the Fermat points ofa triangle. For example,
F
Y
Z
X
Y
A
B C
The heptagonal triangle 35
Lester Theorem: the circumcenter, the nine-point center, andthe Fermat points are concyclic.
O
N
A
B C
F+
F−
Lester’s circle
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Theorem In the heptagonal triangleT, the circumcenter andthe Fermat points form an equilateral triangle.
D
B
C
A
O
KiFermat
Fermat
H
The heptagonal triangle 37
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
The heptagonal triangle 39
(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