Squaring the Circle Revisited

TL;DR

This paper presents new geometric constructions for squaring the circle, achieving accuracy to three and nine decimal places, improving upon previous methods by Ramanujan and Dixon.

Contribution

The authors introduce two novel constructions for squaring the circle, one accurate to three decimal places and another to nine decimal places, enhancing geometric approximation techniques.

Findings

Construction correct to three decimal places

Construction correct to nine decimal places

Improved approximation accuracy over previous methods

Abstract

Squaring the circle is impossible, but it can be squared approximately. Ramanujan gave a construction correct to eight decimal places. In his book Mathographics, Dixon gave constructions correct to three decimal places. In this article, we provide a new construction correct to three decimal places and another correct to nine decimal places.

Equations

$$MH=\sqrt{NM\cdot MB}=\sqrt{3/\sqrt{5}\cdot (3/\sqrt{5}+1)}=\sqrt{6/5\cdot (1+\varphi )},$$

$$\frac{6/5\cdot (1+\varphi )}{\pi }\approx \mathrm{1.000015.}$$

$$\displaystyle\frac{63}{25}\bigg{(}1+\frac{5}{2}\cdot\frac{15\sqrt{5}-7}{269}\bigg{)}$$

$$\pi$$

$$E{F}^2=E{B}^2-B{F}^2=4D{B}^2-A{D}^2=4\cdot (4/5{)}^2-(1/5{)}^2=63/25.$$

$$\frac{\frac{63}{25}\big{(}1+\frac{5}{2}\cdot\frac{15\sqrt{5}-7}{269}\big{)}}{\pi}\ \approx\ 1.000000000068.$$

Full text

1. History

Squaring the circle is a problem proposed by ancient geometers who asked whether it was possible to construct a square which has the same area with a given circle using only rulers and compasses. The method is called ruler-and-compass construction or classical construction. Nowadays, we know that this task is impossible, but the problem of squaring the circle had such a long history that many circle-squarers attempted and failed. Because it is possible to approximate the area of a circle with inscribing regular polygons to any degree of precision, and polygons can also be squared, it is reasonable to expect that a circle can be squared. While many Greek mathematicians and philosophers including Anaxagoras, Hippocrates of Chios, and Oenopides attempted a solution, mathematicians after the $16^{th}$ century tried to prove its impossibility. However, it was not until the $19^{th}$ century that considerable progress was made. In 1837, Pierre Wantzel [5] proved that lengths that can be constructed must be the zero of a polynomial. This result is a breakthrough towards proving the impossibility of a solution; that is, it suffices to prove that $\pi$ is not the solution of all polynomials or transcendental, in other words. Almost 50 years later, Lindemann [2] proved the transcendence of $\pi$ and thus, closed the problem.

However, mathematicians now have a new hobby to square the circle ’approximately’. The most famous work is due to Ramanujan [3] who constructed $\frac{355}{113}$, an approximation of $\pi$ correct to $6$ decimal places. Later, Ramanujan [4] gave another construction correct to $8$ decimal places, where he used $\sqrt[4]{9^{2}+19^{2}/22}$ to approximate $\pi$ (Figure 1). Because there are rational numbers arbitrarily close to $\pi$, we can construct lengths approximating $\pi$ up to any degree of precision. Unfortunately, there is a trade-off between the level of precision and the complexity of the construction. Higher precision often requires many more steps involved; hence, the best construction should balance these two factors. In 1991, Dixon gave constructions for $6/5\cdot(1+\phi)\approx 3.1416$, where $\phi=\frac{1+\sqrt{5}}{2}$, the golden ratio [1]. In his book Mathographics, Dixon stated that his method was quite complicated. In this paper, we first show Dixon’s method then show a quicker method to construct $6/5\cdot(1+\phi)$. Finally, we construct $\sqrt{\frac{63}{25}(1+\frac{5}{2}\frac{15\sqrt{5}-7}{269})}$, whose square approximates $\pi$ correct to $9$ decimal places, a new record!

2. Basic Constructions

We list several basic constructions for the convenience of later discussions.

1. (1)

Given a line $\ell$ and a point, it is possible to draw a line that goes through the point and perpendicular to $\ell$. 2. (2)

Let $n\in\mathbb{N}$. Given a line segment of length $a$, it is possible to divide the line into $n$ equal segments. So, it is possible to construct a new segment with length equal $ra$ for all $r\in\mathbb{Q}$. 3. (3)

Given a line segment, it is possible to draw a circle taking the line segment as one of its diameter. 4. (4)

Given two line segments with lengths $a$ and $b$, we are able to construct segments of lengths $a\cdot b$ and $a/b$. 5. (5)

Given a line segment of length $a$, we are able to construct a segment of length $\sqrt{a}$.

All of these basic constructions can be found in Mathographics by Dixon [1]; for conciseness, we do not illustrate these constructions. We would like to define an efficient construction.

Definition 2.1**.**

Given a constructible length $a$, an efficient construction of $a$ must satisfy three requirements:

• •

be easy to follow and ’not too time-consuming’.

• •

not involve sub-construction of huge lengths; else, we need a BIG piece of paper!

• •

not involve sub-construction of too small lengths; else, we cannot see them!

In this paper, we construct squares of almost the same area with the unit circle (radius 1).

3. Square the Circle Using $6/5\cdot(1+\phi)$

A naive way to construct $\sqrt{6/5(1+\phi)}$ would be to construct $\phi$, then add $1$ to it, then multiply the whole length by $6/5$, and finally take the square root of the resulting length. However, such a construction is neither efficient nor elegant.

3.1. Dixon’s method

ABCDEFGHIK

The yellow square and the yellow circle almost have the same area.

Here are the steps:

1. (1)

Let $AB=1$. Draw $AD\perp AB$ with $AD=1$. Let $C$ be the midpoint of $AD$. 2. (2)

Draw the circle centered at $C$, radius $CB$, which cuts the extended $AD$ at $E$. 3. (3)

Draw the circle centered at $D$, radius $DE$, which cuts the extended $DE$ at $F$. 4. (4)

Pick $H$ on the extended $EF$ such that $HF=1$. 5. (5)

Pick $G$ on $EF$ such that $GF=3AF/10$. 6. (6)

Draw the circle, which takes $HG$ to be one of its diameter. 7. (7)

Pick $I$ on the circle so that $IF\perp GH$. Double $FI$ to have $FK=\sqrt{6/5\cdot(1+\phi)}$, our desired quantity.

We now prove that $FK=\sqrt{6/5\cdot(1+\phi)}$.

Proof.

We have $CE=CB=\sqrt{AB^{2}+AC^{2}}=\sqrt{1+(1/2)^{2}}=\frac{\sqrt{5}}{2}$. Hence, $DF=DE=CE+CD=(\sqrt{5}+1)/2=\phi$. Because $GF=3AF/10$ and $AF=DF+AD=1+\phi$, $GF=\frac{3(1+\phi)}{10}$. Because $\triangle GHI$ is right at $I$, we have $FK=2IF=2\sqrt{GF\cdot FH}=2\sqrt{\frac{3(1+\phi)}{10}}=\sqrt{6/5\cdot(1+\phi)}$. ∎

3.2. Our method

ABCDMNH

The yellow square and the yellow circle almost have the same area.

Here are the steps:

1. (1)

Let $AB=1$. Construct $BC\perp AB$ and $BC=2$. 2. (2)

Draw the circle centered at $A$, radius $AC$, which cuts the extended $AB$ at $D$. 3. (3)

Let $M$ on $AD$ such that $DM=2AD/5$. 4. (4)

Let $N$ be such that $ND=DM/2$. 5. (5)

Draw the circle taking $NB$ to be one of its diameter. 6. (6)

Let $H$ be on the circle such that $MH\perp NB$. Then $MH=\sqrt{6/5\cdot(1+\phi)}$, our desired quantity.

We now prove that $MH=\sqrt{6/5\cdot(1+\phi)}$.

Proof.

Because $D$ and $C$ are on the same circle centered at $A$, $AD=AC=\sqrt{1^{2}+2^{2}}=\sqrt{5}$. Because $DM=2AD/5,DM=2/\sqrt{5}$ and $MA=3/\sqrt{5}$. So, $MN=3/\sqrt{5}$ and $MB=3/\sqrt{5}+1$. Since $NB$ is the diameter of a circle and $H$ is a point on the circle, $\triangle NHB$ is right at $H$. Therefore,

[TABLE]

as desired. ∎

Remark 3.1**.**

We compute the error

[TABLE]

So, for every one million parts, we are off by about fifteen parts. This approximation is very good!

Our construction is easier and thus, quicker as can be seen from the number of steps and the figures themselves. In particular, step (5) in Dixon’s method is what makes it inefficient.

4. Square the Circle Using \frac{63}{25}\big{(}1+\frac{5}{2}\cdot\frac{15\sqrt{5}-7}{269}\big{)}

Note that

[TABLE]

Therefore, our approximation is correct to $9$ decimal places. Our goal is to construct a length of \big{(}\frac{63}{25}\big{(}1+\frac{5}{2}\cdot\frac{15\sqrt{5}-7}{269}\big{)}\big{)}^{1/2} efficiently. Of course, the naive way would be to construct a length of $\sqrt{5}$ and then lengthen it by $15$ times to then divide it by $269$. You must be very patient and must have a BIG piece of paper to do so! The two below figures show how we construct $\frac{\sqrt{63}}{5}$, $\frac{\sqrt{15\sqrt{5}-7}}{5}$, and $\frac{\sqrt{269}}{8}$ much more efficiently. Once we have these lengths, we can construct \big{(}\frac{63}{25}\big{(}1+\frac{5}{2}\cdot\frac{15\sqrt{5}-7}{269}\big{)}\big{)}^{1/2} with basic constructions.

ABCDEFGHIKLMNOPQRSTU

Here are the steps:

Left figure:

1. (1)

Let $AB=AC=1$. Let $D$ be on $AB$ so that $AD=AB/5$. 2. (2)

Draw a semicircle with radius $DB$, which intersects $BC$ at $E$. Pick $F$ on the semicircle such that $BF=AD$. 3. (3)

Draw a semicircle with diameter $DB$. Pick $G$ on the semicircle so that $BG=EA$. 4. (4)

Let $H$ be such that $HB\perp BC$ and $HB=1$. 5. (5)

Let $I$ be on $CH$ such that $CI=3CH/5$. On extended $CH$, pick $K$ so that $IK=1$. 6. (6)

Draw a semicircle with diameter $CK$. Let $L$ be on the semicircle so that $IL\perp CK$. Pick $M$ so that $I$ is the midpoint of $LM$. 7. (7)

Pick $N$ on $IL$ so that $NI=DG$. 8. (8)

Draw a semicircle with diameter $LM$. Pick O on the semicircle so that $NO\perp LM$.

Right figure:

1. (1)

Let $RQ=1$ and $PR=1/4$. 2. (2)

Let $SQ=3/2$ and $SQ\perp RQ$. 3. (3)

Let $T$ be the midpoint of $SP$. Draw $UT=SR$ and $UT\perp PS$. Connect $PU$.

We will prove that $EF=\frac{\sqrt{63}}{5}$, $ON=\frac{\sqrt{15\sqrt{5}-7}}{5}$, and $PU=\frac{\sqrt{269}}{8}$. But for now, assume that they are true. By item (2) and item (4) in Section 2, we can construct a length of $\frac{5}{8}\cdot\frac{NO}{PU}=\sqrt{\frac{15\sqrt{5}-7}{269}}$. By the same items, we can construct $\frac{5}{2}\cdot\frac{15\sqrt{5}-7}{269}+1$. Then, by item (5), we construct $\sqrt{\frac{5}{2}\cdot\frac{15\sqrt{5}-7}{269}+1}$. Finally, by item (4), we construct $\frac{\sqrt{63}}{5}\sqrt{\frac{5}{2}\cdot\frac{15\sqrt{5}-7}{269}+1}$.

Proof.

We have

[TABLE]

So, $EF=\frac{\sqrt{63}}{5}$.

Since $GB=EA=ED-AD=4/5-1/5=3/5$, we know that $DG=\sqrt{DB^{2}-GB^{2}}=\sqrt{(4/5)^{2}-(3/5)^{2}}=\sqrt{7}/5$. Also, $CH=\sqrt{BC^{2}+BH^{2}}=\sqrt{2^{2}+1^{2}}=\sqrt{5}$ and so, $CI=3CH/5=3/\sqrt{5}$. Because $\triangle LCK$ is right at $L$, $LI=\sqrt{CI\cdot IK}=\sqrt{3}/\sqrt[4]{5}$. So, $LN=LI-NI=LI-DG=\sqrt{3}/\sqrt[4]{5}-\sqrt{7}/5$ and $NM=NI+IM=LI+DG=\sqrt{3}/\sqrt[4]{5}+\sqrt{7}/5$. Because $\triangle LOM$ is right at $O$, NO=\sqrt{LN\cdot NM}=\sqrt{\big{(}\sqrt{3}/\sqrt[4]{5}-\sqrt{7}/5\big{)}(\sqrt{3}/\sqrt[4]{5}+\sqrt{7}/5)}=\frac{\sqrt{15\sqrt{5}-7}}{5}.

Lastly, we prove that $PU=\frac{\sqrt{269}}{8}$. We have $RS^{2}=RQ^{2}+QS^{2}=1^{2}+(3/2)^{2}=13/4$, and $PS^{2}=PQ^{2}+SQ^{2}=61/16$. Hence, $TP^{2}=PS^{2}/4=61/64$. Therefore, $PU^{2}=TP^{2}+TU^{2}=TP^{2}+RS^{2}=13/4+61/64=269/64$, and so $PU=\frac{\sqrt{269}}{8}$. ∎

Our construction of \big{(}\frac{63}{25}\big{(}1+\frac{5}{2}\cdot\frac{15\sqrt{5}-7}{269}\big{)}\big{)}^{1/2} is at the same level of complexity of Ramanujan’s construction of $\sqrt[8]{9^{2}+19^{2}/22}$ and is correct to $9$ decimal places (one more than Ramanujan’s).

Remark 4.1**.**

We compute the error

[TABLE]

So, for every one tera $(10^{12})$ parts, we are off by about 68 parts. This approximation is extremely good!