Some Constructions of the Golden Ratio in an Arbitrary Triangle

TL;DR

This paper presents novel geometric constructions of the golden ratio within any triangle utilizing symmedians and the nine-point circle, expanding the methods for ratio-based geometric properties.

Contribution

It introduces new constructions of the golden ratio in arbitrary triangles using symmedians and the nine-point circle, which were not previously documented.

Findings

New geometric constructions of the golden ratio in triangles

Use of symmedians and nine-point circle in ratio constructions

Generalization to any arbitrary triangle

Abstract

We establish some new constructions of the golden ratio in an arbitrary triangle using symmedians and nine-point circle.

Figures

Equations

$$\frac{PA}{PC}=\frac{AB\cdot AD}{CB\cdot CD}.$$

$$\frac{PA}{PB}=\frac{AD}{BC}$$

$$\frac{PB}{PC}=\frac{AB}{CD}.$$

$$\frac{PA}{PC}=\frac{AB\cdot AD}{CB\cdot CD}.$$

$$\frac{CA}{yQB}=\frac{BC}{xQA}+\frac{AB}{zQC}.$$

$$\frac{EA}{EC}=\frac{z}{x}.$$

$$\frac{EA}{EC}=\frac{AQ\cdot AR}{CQ\cdot CR}.$$

$$\frac{AQ\cdot AR}{CQ\cdot CR}=\frac{z}{x}.$$

$$zQC=x\frac{AQ\cdot AR}{CR}.$$

$$\frac{AQ\cdot AR}{BQ\cdot BR}=\frac{y}{x}.$$

$$yQB=x\frac{AQ\cdot AR}{BR}.$$

$$\frac{CA}{yQB}-\frac{AB}{zQC}$$

$$=\frac{CA\cdot RB-AB\cdot RC}{xAQ\cdot AR}$$

$$=\frac{BC\cdot RA}{xAQ\cdot AR}$$

$$=\frac{BC}{xQA}.$$

$$\frac{CA}{yQB}=\frac{BC}{xQA}+\frac{AB}{zQC}.$$

$$\frac{RA}{RS}=\frac{AP}{SC}$$

$$\frac{QS}{QA}=\frac{SB}{AP}.$$

$$\frac{QS}{RS}\cdot \frac{RA}{QA}=\frac{SB}{SC}.$$

$$\frac{RA}{QA}=\frac{AC}{AB}.$$

$$SB=PC,SC=PB.$$

$$\frac{QS}{RS}=\frac{AB\cdot PC}{CA\cdot PB}.$$

$$BC\cdot PA=AB\cdot PC-CA\cdot PB.$$

$$\frac{RQ}{RS}=\frac{QS}{RS}-1=\frac{AB\cdot PC}{CA\cdot PB}-1=\frac{AB\cdot PC-AC\cdot PB}{AC\cdot PB}=\frac{BC\cdot PA}{CA\cdot PB}.$$

$$\frac{CA}{C{A}^2\cdot PB}=\frac{BC}{B{C}^2\cdot PA}+\frac{A{B}^2}{AB\cdot PC}.$$

$$\frac{1}{CA\cdot PB}=\frac{1}{BC\cdot PA}+\frac{1}{AB\cdot PC}.$$

$$\frac{1}{CA\cdot PB}=\frac{1}{BC\cdot PA}+\frac{1}{CA\cdot PB+BC\cdot PA}.$$

$$\frac{BC\cdot PA}{CA\cdot PB}-\frac{CA\cdot PB}{BC\cdot PA}=1.$$

$$\frac{RQ}{RS}-\frac{RS}{RQ}=1.$$

$$\frac{RQ}{RS}=\frac{\sqrt{5}+1}{2}$$

$$\frac{RQ}{RS}=\frac{BC\cdot PA}{CA\cdot PB}$$

$$\frac{RS}{RQ}=\frac{CA\cdot PB}{BC\cdot PA}.$$

Full text

1. Introduction

The golden ratio often appear in regular polygons [3, 4, 7, 9, 10] and less in the isosceles triangle [2, 11]. The author introduced a construction of the golden ratio in an arbitrary triangle with two symmedians in [12]. Continuing with this idea, we shall introduce some new other constructions of the golden ratio in an arbitrary triangle with symmedians and nine-point circle. Some constructions can be considered as the generalization of classical construction of George Odom in [6].

2. The lemmas

We recall the lemmas with proofs from [12].

Lemma 1**.**

Given a convex cyclic quadrilateral $ABCD$. Two diagonals $AC$ and $BD$ interesect at $P$. Then

[TABLE]

Proof.

From the equality of the corresponding angles in the cyclic quadrilateral, we have the similar triangles $\triangle PAB\sim\triangle PDC$ and $\triangle PAD\sim\triangle PBC$. From this, we get the ratios

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and

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From (1) and (2), we obtain

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The proof is complete. ∎

Lemma 2** (Ptolemy’s theorem [1]).**

For a cyclic quadrilateral, the sum of the products of the two pairs of opposite sides equals the product of the diagonals.

This is the famous theorem of plane geometry that can be found in [1].

Using the concept of homogeneous barycentric coordinates [8], we give and prove the following lemma.

Lemma 3**.**

Let $ABC$ be a triangle inscribed in a circle $(\omega)$. $P$ is a point inside triangle $ABC$. $P$ has homogeneous barycentric coordinates $(x:y:z)$. $DEF$ is a cevian triangle of $P$. Ray $EF$ meets $(\omega)$ at $Q$. Then

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Proof.

If $P$ has homogeneous barycentric coordinates $(x:y:z)$, we get that the barycentric coordinates of $E$, $F$ are $(x:0:z)$ and $(x:y:0)$ respectively. Thus we have the ratio

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Let the ray $QE$ meet $(\omega)$ second time at $R$. From Lemma 1, we have

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From (3) and (4), we deduce

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Thus

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Similarly, we have the identity

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Thus

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Using (5), (6), and Lemma 2, we consider the expression

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Therefore,

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This completes the proof of Lemma 3. ∎

3. Constructions and proofs

Construction 1**.**

Given a triangle $ABC$ with the circumcircle $(\omega)$.

• i)

Constructing the symmedians $BE$ and $CF$.

• ii)

Ray $EF$ meets $(\omega)$ at $P$.

• iii)

The parallel line from $P$ to $BC$ meets $AB$ and $AC$ at $Q$ and $R$ respectively and meets $(\omega)$ again at $S$.

Proposition 1**.**

$R$* divides $QS$ into the golden ratio.*

Proof.

It follows from $\triangle ARP\sim\triangle SRC$ and $\triangle AQP\sim\triangle SQB$ that

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and

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Therefore,

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Because $QR\parallel BC$,

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Also by $PS\parallel BC$, we have $PSCB$ is an isosceles trapezoid. Hence,

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From (7), (8) and (9), we deduce

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Note that, by Lemma 2,

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From (10) and (11), we infer that

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Let $BE$ meet $CF$ at $L$, then $L$ has barycentric coordinates $L(BC^{2}:CA^{2}:AB^{2})$, see [8]. Apply Lemma 3 for triangle $ABC$ with $L$ and ray $EF$ meet $(\omega)$ at $P$, we obtain

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This is equivalent to

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From (11) and (13), we have

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This means

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From (12) and (14), we secure

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This is enough to show that the ratio

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which is such the golden ratio. This completes our proof. ∎

Remark**.**

This construction can be considered as the generalization of the construction [2] and [6]. If $ABC$ is isosceles at $A$, we obtain the construction in [2]. If $ABC$ is equilateral, we obtain the construction of George Odom in [6].

Construction 2**.**

Given a triangle $ABC$ with the circumcircle $(\omega)$.

• i)

Constructing the symmedians $AD$ and $CF$.

• ii)

Ray $DF$ meets $(\omega)$ at $P$.

• iii)

The parallel line from $P$ to $BC$ meets $AB$ and $AC$ at $Q$ and $R$ respectively and meets $(\omega)$ again at $S$.

Proposition 2**.**

$R$* divides $SQ$ into the golden ratio.*

Proof.

The proof is similar to the proof of Proposition 1

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or

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Let the symmedians $AD$ and $CF$ meet at $L$, then $L(BC^{2}:CA^{2}:AB^{2})$, see [1]. Apply Lemma 2 for triangle $ABC$ with $L$ and ray $DF$ meet $(\omega)$ at $P$, we have

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This is equivalent to

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Note that, Lemma 2 then

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From (16) and (17) we have

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This means

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From (15) and (19), we deduce that

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this is enough for deriving equality

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which is such the golden ratio. This completes our proof. ∎

Construction 3**.**

Given a triangle $ABC$ with the nine-point center $N$. $D$, $E$, and $F$ are the midpoints of $BC$, $CA$, and $AB$, respectively. Consider the circle $(\omega)$ with diameter $BC$.

• i)

The perpendicular line from $D$ to $ND$ meets circle $(\omega)$ at $S$ and $T$.

• ii)

Consider the circle $(\Omega)$ with center $N$ passing through $S$ and $T$.

• iii)

Ray $EF$ meets circle $(\Omega)$ at $G$.

Proposition 3**.**

$E$* divides $FG$ into the golden ratio.*

Proof.

We shall show that

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Because $N$ is the nine-point center of triangle $ABC,$ thus $N$ must be the circumcenter of triangle $DEF$. Also, $DEFB$ is a parallelogram, so $EF=BD=DT$.

Let $EF$ meet $(\Omega)$ again at $L$. By symmetry and the intersecting chords theorem,

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This proves that $E$ divides $FG$ in the golden ratio. Similarly $F$ divides $EL$ into the golden ratio. We complete the proof. ∎

Remark**.**

This construction also can be considered as another generalization of the classical construction of George Odom in [6]. If $ABC$ is an equilateral triangle, we obtain the construction in [6].

Construction 4**.**

Given a triangle $ABC$.

• i)

Consider the symmedian $BE$.

• ii)

Let $F$ be a point on segment $AE$ such that $\frac{FE}{EC}=\frac{1}{5}$.

• iii)

The parallel line from $F$ to $BE$ meets $AB$ at $G$.

• iv)

The perpendicular bisectors of $AG$ and $BC$ meet at $K$.

• v)

The circle with center $K$ passing though $A$ meets $BC$ at $MN$.

Proposition 4**.**

$M$* divides $NB$ into the golden ratio.*

Proof.

Because $BE$ is symmedian of $ABC$, we conclude that $\frac{EA}{EC}=\frac{BA^{2}}{BC^{2}}$. $EC=5\cdot EF$ or in other words

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$K$ lies on perpendicular bisector of $AG$, circle $(K)$ passes through $G$, we have

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From (20) and (21), we obtain

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Because $K$ lies on perpendicular bisector of $BC$, so $BM=CN$ and we deduce that

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From (22) and (23), we get

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this is equivalent to

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or

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or

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which is such the golden ratio. This completes our proof. ∎

Remark**.**

This construction can be considered as generalization of the construction of Dao Thanh Oai in [3]. If $ABC$ is equilateral triangle or right isosceles, we obtain the construction in [3].

Acknowledgement**.**

The author is grateful Alexander Skutin for his proofreading and his suggestion of the nice projective viewpoint for the proofs of Proposition 1 and Proposition 2.