{} 2 a , b , c a = a , b = ar , c = ar

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Geometric Sequence and Triangles Motivation: M. N. D., N.

1 About

Let a, b, c the sides of a triangle ABC such that

{

a,b,c

}

(in this order) is a geometric sequence. This means that we have a ratio r such that

a=a, b=ar, c=ar². Figure 1 gives an example with the ratio r = 1.2.

A

B a C

ar² ar

Fig. 1: Ratio r = 1.2

We will see that the Golden Section plays an important role in special cases of these triangles.

2 Special triangles 2.1 Equilateral triangle

In the trivial case of r = 1 we have the equilateral triangle (Fig. 2).

A

B a C

a a

Fig. 2: Equilateral triangle

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2.2 Right triangles

For r > 1 we have a < b < c. The theorem of Pythagoras yields:

ar²

( )

² ⁼

^{( )}

^ar ² ⁺^a²

r⁴ =r²+1 r² =¹⁺₂⁵ =Φ

r= Φ

The Golden Section Φ=¹⁺₂⁵ comes in (Walser 2001 and Walser 2013). We get the large Right Golden Triangle of Figure 3.

A

B a C

a a

Fig. 3: Large Right Golden Triangle

For r < 1 we have a > b > c. The theorem of Pythagoras yields in this case:

a² =

( )

ar ²⁺

( )

^ar² ²

1=r²+r⁴ r² =_Φ¹

r= _Φ¹ We get the small Right Golden Triangle (Fig. 4).

A

B a C

a ¹ a ¹

Fig. 4: Small Right Golden Triangle

The small Golden Triangle fits into the large Golden Triangle (Fig. 5). It has the same shape like the large Right Golden Triangle, but a different orientation.

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A

B a C

A

Fig. 5: Both Right Golden triangles

3 General

3.1 Range of the ratio r

The triangle inequality gives the condition a + b > c. Hence a+ar>ar²

1+r>r² r<Φ On the other side we have a < b + c. That means a<ar+ar² 1<r+r² r>_Φ¹ Therefore we have the range _Φ¹ <r<Φ.

Note that the ratios of the small Right Golden Triangle, the equilateral triangle, and the large Right Golden Triangle are within this range:

Φ1 < _Φ¹ <1< Φ<Φ

Together with the range bounds they are a geometric sequence.

3.2 Two fix points

For the general triangle we let two of the three vertices of the triangle fix and study the locus of the third point. There are three cases:

(i) B and C fix, A variable (ii) C and A fix, B variable (iii) A and B fix, C variable

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3.2.1 B and C fix

In a Cartesian coordinate system we choose B

( )

−¹₂,0 ^,^C

( )

¹2,0 ^{, and}^{A x,}

^{( )}

^y (Fig. 6).

For the sides of the triangle we get:

a=1

b=

( )

x−¹₂ ²⁺^y²

c=

( )

x+¹₂ ²⁺^y² The geometric sequence condition is:

r= ^b_a = ^c_b ⇒ b² =ac This yields the implicit equation for the locus of A:

x−¹₂

( )

²⁺^y² ⁼

( )

^x⁺¹2 ²⁺^y²

x−¹₂

( )

²⁺^y²

⎛⎝⎜ ⎞

⎠⎟

2−

( )

x+¹₂ ²⁺^y² ⁼⁰ or

f x,

( )

y ⁼^x⁴ ⁻^2x³⁺¹₂^x²⁺^2x²^y²⁻³₂^x⁻²^xy²⁻₁₆³ ⁻¹₂^y²⁺^y⁴ ⁼⁰ We have a curve of degree 4.

A

B C 1

1

x y

Fig. 6: B and C fix

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For x = 0 we find the equilateral triangle (Fig. 7).

A

B C 1

1

x y

Note that the side AB of the equilateral triangle is tangential to the curve. For the proof we use the gradient:

∇f x,

( )

y ⁼ ⁴^x³⁻⁶^x²⁺^x⁺⁴^xy² ⁻²³⁻^2y² 4x²y−4xy−y+4y³

⎡

⎣

⎢⎢

⎢

⎤

⎦

⎥⎥

⎥

∇f

( )

0, ₂³ ⁼ ₋ ⁻3³²⁻²⁴³ 2 +4

( )

₂³ ³

⎡

⎣

⎢⎢

⎤

⎦

⎥⎥

= −3 3

⎡

⎣⎢

⎢

⎤

⎦⎥

⎥ The vector −3

3

⎡

⎣⎢

⎢

⎤

⎦⎥

⎥ is orthogonal to the curve, but also orthogonal to the side AB of the equilateral triangle.

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For x= ¹₂ we get the large Right Golden triangle (Fig. 8).

A

B C 1

1

x y

Fig. 8: Right Golden Triangle

We find also the small Right Golden Triangle (Fig. 9). Here we have x=−¹₂+ ¹

Φ² ≈ −0.118

A

B C 1

1

x y

1 1

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Finally we find the Golden Section in different cases (Fig. 10 and 11). The major is in- dicated in blue and the minor in red.

B C 1

1

x y

Fig. 10: Golden Section

B C 1

1

x y

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3.2.2 C and A fix

Now we choose C

( )

−¹₂,0 ^and^A

( )

¹2,0 (Fig. 12).

a=

( )

x+¹₂ ²⁺^y² b=1

c=

( )

x−¹₂ ²⁺^y² The geometric sequence condition is again:

r= ^b_a = ^c_b ⇒ b² =ac This yields the implicit equation for the locus of B:

1=^⎛_⎝⎜

( )

x+¹₂ ²⁺^y²^⎞_⎠⎟^⎛_⎝⎜

( )

^x⁻¹2 ²⁺^y²^⎞_⎠⎟

1 1

A

B

x y

C

Fig. 12: C and A fix

The curve looks like an ellipse, but this cannot be, since it is a curve of degree 4. Never- theless the difference to the ellipse with the same semi axes is very small (Fig. 13).

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1 1

x y

Fig. 13: Curve compared with ellipse

Again we find an equilateral triangle (Fig. 14), Right Golden triangles (Fig. 15) and the golden Section (Fig. 16).

1 1

A B

x y

C

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1 1

A B

x y

C

1

1 1

A x

y

C

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3.2.3 A and B fix

Now we choose A

( )

−¹₂,0 ^and^B

( )

¹2,0 (Fig. 17).

a=

( )

x−¹₂ ²⁺^y² b=

( )

x+¹₂ ²⁺^y² c=1

The geometric sequence condition is again:

r= ^b_a = ^c_b ⇒ b² =ac This yields the implicit equation for the locus of C:

x+¹₂

( )

²⁺^y² ⁼

( )

^x⁻¹2 ²⁺^y²

x+¹₂

( )

²⁺^y²

⎛⎝⎜ ⎞

⎠⎟

2 =

( )

x−¹₂ ²⁺^y²

A B

–1

1

x y

Fig. 17: A and B fix The curve is symmetric to the curve of Figure 6.

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3.3 The three curves

Figure 18 shows the three curves in a triangular lattice. We see a lot of Golden Sections.

Fig. 18: The tree curves References

Walser, Hans (2001): The Golden Section. Translated by Peter Hilton and Jean Peder- sen. The Mathematical Association of America 2001. ISBN 0-88385-534-8.

Walser, Hans (6. Auflage). (2013). Der Goldene Schnitt. Mit einem Beitrag von Hans Wußing über populärwissenschaftliche Mathematikliteratur aus Leipzig.

Leipzig: Edition am Gutenbergplatz. ISBN 978-3-937219-85-1.