Download Golden Sections of Triangle Centers in the Golden Triangles

Forum Geometricorum
Volume 16 (2016) 119–124.
FORUM GEOM
ISSN 1534-1178
Golden Sections of Triangle Centers
in the Golden Triangles
Emmanuel Antonio José Garcı́a and Paul Yiu
Abstract. A golden triangle is one whose vertices are among the vertices of a
regular pentagon. There are two kinds of golden triangles, short and tall, which
are isosceles triangles with vertical angles 108◦ and 36◦ respectively. We consider some basic triangle centers of a short and tall golden triangles sharing one
vertex and with the same circumcircle, and exhibit pairs of basic triangle centers
divided in the golden ratio by another triangle center.
As is well known, the golden ratio naturally occurs in the regular√pentagon, as
the ratio of the length of a diagonal d and a side a: ϕ := ad = 12 ( 5 + 1). The
intersection of two diagonals divides each in the golden ratio. If ABCDE is a
regular pentagon, and the diagonals AD and BE intersect at P (see Figure 1),
then
BP
DA
DP
BE
=
= ϕ,
=
= ϕ.
BP
PE
DP
PA
For later use, we note the following simple trigonometric ratios from Figure 1:
ϕ
1
d/2
a/2
= ,
sin 18◦ =
=
.
cos 36◦ =
a
2
d
2ϕ
A
A
a
a
a
a
d−a
P
B
a
d−a
E
Bs
Cs
d
d
O
d
O
a
C
D
Figure 1. The golden section
Bt
a
Ct
Figure 2. Short and tall
golden triangles
Given a regular pentagon, the subtriangles with vertices among those of the
pentagon are all isosceles. They fall into two types:
(i) those with three adjacent vertices of the pentagon have angles 108◦ , 36◦ , 36◦ ,
which we call short golden triangles,
Publication Date: April 1, 2016. Communicating Editor: Nikolaos Dergiades.
120
E. A. J. Garcı́a and P. Yiu
(ii) those with only two adjacent vertices of the pentagon have angles 36◦ , 72◦ ,
72◦ , which we call tall golden triangles.
In this note we consider golden sections in the two kinds of golden triangles.
For purpose of comparison, we consider a pair of short and tall golden triangles
inscribed in the same regular pentagon ABs Bt Ct Cs (see Figure 2). The short
golden triangle Ts := ABs Cs has sides d, a, a; the tall golden triangle Tt :=
ABt Ct has sides a, d, d. They share the same circumcenter O. Denote by R their
common circumradius. Note that the areas Δi , i = s, t, of the golden triangles are
in the golden ratio:
1
ad sin 72◦
d
Δt
= 12 2
= = ϕ.
◦
Δs
a
2 a sin 108
For i = s, t, since the golden triangle Ti is isosceles, its triangle centers are all
on the (common) perpendicular bisector of the side Bi Ci . We shall call this the
center line of the golden triangles; It contains the midpoints Fi of Bi Ci (see Figure
3).
Hs
A
Is
Bs
Fs
Cs
P
O
It
Ht
Bt
Ft
Ct
Figure 3
Here are some simple constructions of the basic triangle centers of Ts and Tt .
(1) The incenter Is of Ts is the intersection of the center line with the perpendicular of Bs Bt at Bs ; it is also the reflection of O in the side Bs Cs . From this, the
inradius of Ts is
rs = Fs Is = OFs = R cos 72◦ =
R
.
2ϕ
Golden sections of triangle centers in the golden triangles
121
(2) The incenter It of Tt is the intersection of the diagonals Bt Cs and Ct Bs . It
is also the reflection of A in the side Bs Cs . From this, the inradius of Tt is
a
sin2 36◦
tan 36◦ = R sin 36◦ tan 36◦ = R ·
2
cos 36◦
ϕ 2
1− 2
R
4 − ϕ2
= (3ϕ − 4).
= R·
=
R
·
ϕ
2ϕ
2
2
rt =
(3) Let Hs be the orthocenter of Ts . Clearly ∠Hs OBs = ∠Is OBs = 72◦ . Since
Hs is the isogonal conjugate of O in Ts , ∠Hs Bs O = 2∠Is Bs O = 2 · 36◦ = 72◦ .
Therefore, triangle Hs Bs O is a (tall) golden triangle, and
d
OHs
= = ϕ =⇒ OHs = ϕR.
OBs
a
Also, by the angle bisector theorem,
Bs H s
Hs Is
=
= ϕ.
Is O s
Bs O s
This shows that Is divides Hs O in the golden ratio.
Since Bs A bisects angle Hs Bs Is , the same reasoning shows that A divides Hs Is
in the golden ratio.
(4) In the tall golden triangle Tt , the orthocenter Ht is the intersection of the
center line with the perpendicular to Bs Bt at Bt . Note that Bt Ht = 2R cos 72◦ =
R
2R · a/2
d = ϕ.
Since Ht is the isogonal conjugate of O in Tt , by the angle bisector theorem,
OIt
Bt O
=
= ϕ.
It Ht
Bt H t
Therefore, It divides OHt in the golden ratio
(5) Since O and It are the reflections of Is and A in Bs Ct , OIt = AIs , and
Is O
Is O
=
= ϕ.
OIt
AIs
Therefore, O divides Is It in the golden ratio.
(6) In the tall golden triangle, O divides AHt in the golden ratio.
2 · R cos 36◦
AHt
=
= 2 cos 36◦ = ϕ.
AO
R
We summarize these results in the following proposition.
Proposition 1. Let Ti , i = s, t be golden triangles sharing a common vertex A
and the same circumcircle with center O. Let Hi and Ii denote the orthocenter and
incenter of Ti .
(a) The incenter Ii divides Hi O or OHi in the golden ratio, according as i =
s, t.
(b) The circumcenter O divides each of Is It and AHt in the golden ratio.
(c) A divides Hs Is in the golden ratio.
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E. A. J. Garcı́a and P. Yiu
Some observations by Nikolaos Dergiades:
(i) O is the midpoint of Is Ht .
(ii) If A is the antipode of A on the circumcircle, the triangles Is Hs Bs and OBs A
are similar to Ts , and since Is Bs = OBs = R, we have Bs Hs = Bs A = Rϕ.
(iii) The triangle It Hs Bs has a right angle at Bs , and since Is Hs = Is Bs , Is is the
midpoint of Hs It .
(iv) The segments OIs = R
ϕ , Is Bs = R, Bs Hs = Rϕ are in geometric progression
(with common ratio ϕ). Since ϕ = 1 + ϕ1 , we have Bs Hs = Bs Is + OIs . This
means that the circles Bs (Hs ), Is (Hs ) and O(Is ) are concurrent at a point D which
lies on the line OBs (see Figure 4).
Hs
A
Is
Bs
Cs
Fs P
O
It
D
Ht
Bt
Ft
Ct
A
Figure 4
For i = s, t, the incircle of Ti is tangent to the side Bi Ci at its midpoint Fi .
Since this midpoint also lies on the nine-point circle of Ti , it is the Feuerbach
point of Ti . The nine-point circle of Ti , i = s, t, also contains the midpoints Mi,b .
Mi,c of the sides ACi and ABi .
Proposition 2. (a) Fs divides Ft A in the golden ratio.
(b) The incenter It divides Fs Ft in the golden ratio.
Proof. (a) Let P be the intersection of the diagonals ACt and Bs Cs (see Figure 3).
Since Bs Cs and Bt Ct are parallel,
Ft A
Ct A
= ϕ.
=
Ft Fs
Ct P
Therefore, Fs divides Ft A in the golden ratio.
(b) Since It is the intersection of the diagonals Bs Ct and Bt Cs ,
ϕ.
Fs It
It Ft
=
Bs Cs
Bs Ct
=
Golden sections of triangle centers in the golden triangles
123
Fs
A
A
Ns
Ns
Ms,c
Is
Fs
Fs
Bs
Ft
Cs
Bs
Cs
Mt,b
Mt,c
O
O
Nt
Bt
Ms,b
Is
Ft
Figure 5
Nt
Ct
Bt
Ft
Ct
Figure 6
Proposition 3. (a) For the short golden triangle Ts with nine-point center Ns , the
incenter Is divides Fs Ns in the golden ratio.
(b) For the tall golden triangle Tt , the nine-point center Nt divides Ft O in the
golden ratio (See Figure 5).
R
R
. Therefore, FFssNIss = r2s = ϕ, and Is
Proof. (a) The inradius of Ts is rs = 2ϕ
divides Fs Ns in the golden ratio.
(b) Ft O = R cos 36◦ = R2 · ϕ. Therefore, FFttNOt = ϕ, and Nt divides Ft O in the
golden ratio.
Proposition 4. For {i, j} = {s, t}, the nine-point center Ni of Ti is the reflection
of Fj in the center O.
Proof. (a) Since Is is the reflection of O in Bs Cs ,
R
,
OFs = Fs Is = rs =
2ϕ
R
R 1
ϕ
R
+ =
+ 1 = R · = R cos 36◦ = Ft O.
ONs = OFs + Fs Ns =
2ϕ
2
2 ϕ
2
Therefore, Ns is the reflection of Ft in O.
(b) Since Nt − Ft = Ns − Fs ,
Nt = Ns − Fs + Ft = 2 · O − Ft − Fs + Ft = 2 · O − Fs
is the reflection of Fs in O.
Therefore, the nine-point center Ns is the antipode of Ft on the circle, center
O, passing through Ft , which is the inscribed circle of the regular pentagon. It
follows that ∠Ns Ms,b Ft and ∠Ns Ms,b Ft are right angles. This means that Ft Ms,b
and Ft Ms,c are tangents to the nine-point circle of Ts at Ms,b and Ms,c respectively
(see Figure 6). The line Ft Ms,b passes through Mt,b , which divides Ft Ms,b in the
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E. A. J. Garcı́a and P. Yiu
golden ratio. Similarly, Ft Ms,c is the tangent at Ms,c and is divided in the golden
ratio by Mt,c .
The same reasoning also leads to the following.
(i) The points Ms,b , Fs , Mt,c are collinear, and Fs divides Ms,b Mt,c in the golden
ratio. Furthermore, the line containing them is tangent to the nine-point circle of
Tt at Mt,c .
(ii) The points Ms,c , Fs , Mt,b are collinear, and Fs divides Ms,c Mt,b in the
golden ratio. Furthermore, the line containing them is tangent to the nine-point
circle of Tt at Mt,b .
We conclude this note with a few more division in the golden ratio with points
in Figure 5. The simple proofs are omitted.
For i = s, t, let Fi be the antipode of Fi on the nine-point circle of Ti . Then
(a) A divides Fs Ns in the golden ratio,
(b) Ft divides each of the segments ANt and Ft Ns in the golden ratio,
(c) O divides Fs Ft in the golden ratio,
(d) Is divides Fs Ft in the golden ratio.
Statement (d) follows from Proposition 2(b) and a translation by R along the
center line.
Figure 7 summarizes the golden sections in this note, each indicated by a longer
solid segment followed by a shorter dotted segment. The endpoints and the division
points are indicated on the “center line”.
Hs
Fs
A
Ns
Is
Fs
Ft
O
Nt
It
Ht
Ft
Figure 7
References
[1] M. Bataille, Another simple construction of the golden section, Forum Geom., 11 (2011) 55.
[2] K. Hofstetter, A simple construction of the golden section, Forum Geom., 2 (2002) 65–66.
[3] J. Niemeyer, A simple construction of the golden section, Forum Geom., 11 (2011) 53.
Emmanuel Antonio José Garcı́a: CEDI Bilingual School, Camila Henrı́quez Ureña, 20, Santo
Domingo, Dominican Republic.
E-mail address: [email protected]
Paul Yiu: Department of Mathematical Sciences, Florida Atlantic University, 777 Glades Road,
Boca Raton, Florida 33431-0991, USA
E-mail address: [email protected]