The Geometry Aspect of M. C. Escher’s Circle Limit III by Thuan Huynh.

The Geometry Aspect of M. C. Escher’s Circle Limit III

by Thuan Huynh

About M. C. Escher Maurits Cornelius Escher was born on

17th June, 1898 in Leeuwarden, Netherlands.

He studied at the School of Architecture and Decorative Arts in Haarlem. However, he gave up architecture in favor of graphic arts .

At first, his works depicted landscapes using impossible perspectives.

After 1937, he discovered the “open gate” of mathematics and made a few attempts in his early works to satisfy his urge for filling the plane.

His fame slowly spread, and during the 1950s, articles on his work appeared. His works began to be displayed in science museums rather than art galleries.

He died in 27th March, 1972 at the age of 73.

Fig. 2. Circle Limit III.

December 1959.

According to Escher’s measurement, the angle between the bounding circle and the white circle is precisely

My goal is to calculate mathematically.

080

Fig. 2. Escher’s “framework”

Fig. 3. Fig. 4.

2

31

Xat -

,X and Xat angles with Triangles

AB O and ACO ianglessimilar tr The 2 1AB O and ACO Triangles 2 1

ofn calculatio The

XO=CO

XO=BO

XO=CO

BO=OX :have weso

C.O of circle aon liedX and CLet

B.O radius of circle aon lied X and BLet

C.O radius of circle aon lied X and CLet

C.O radius of circle aon lied X and CB,Let

2. figure From

22 2

333

111

22 2

2 2

33

11

22

1=AO=AX=AX=AX

:have weThus

1. radius andA center with circle bounding aon lied X Since

) )(AX X)(O2cos (-) (AX + ) X (O=)(AO

:have weCosine of Law

by the then AO opposite is Xat angle whose,AOX From

45=AOO<

:have then weA,center at angle 45at rotateddisk Let the

. Ocenter at AO distances and

.X O

:radius theshows also 2 figure and 1, radius hasA center with circle aLet

. - and anglesat A)(center circle

unit a ofboundary meet the all they and ,60 of angles equalat other

each cross 2 figure from circles theseof arcsrelevant theAssumeAlso,

3231

1

11 12

12

112

1

1111

02 1

0

vv

vv

0

(3)

2

)13(

BO-AO=

2

CO=

3

AO =>

)CA =BO-AO(for )

12sin(

BO-AO=

)4

sin(

CO=

)3

2 sin(

AO

:have weSine of

Law by the 4.Thus figurein asly respectiveCA and CO,AO sides toopposite

12

and 4

,3

2 of angles thehas ACO thesopoint,common a toradiustheir

between angle the toequals circles gintesectin wobetween t angle theBecause

(2) ) )(BO(2cos +) (BO+1=

) )(BO2(-cos -) (BO+1=

) O)(X)(AX )-cos( (2-) O (X +) (AX=) (AO

:have wecosine of Law

By the .AO toopposite is Xat - angle whose,AO XSimilarly,

(1) ))(CO(2cos-)(CO+ 1=) (AO=>

2211

222211

11

1

22

2

22

2

2 2 22

222

22

2

2222

12

12

1

(7) 2)+ (4cos +2cos=BO

) 2+ (4cos +2cos- =CO

:have weBO and CO positivefor equations theseSolving

(6) 0=2-))(BO(4cos-) (BO

0=2-) )(CO(4cos+) (CO

:BO and COfor equations quadratic yield (5) and (2) (1), From

) (BO2

3=) (AO

(5) ) (CO2

3=) (AO

:AO and AO findcan we(4) and (3) From

(4)

2

)13(

AB=

2

BO=

3

AO

:yields

ly,respective BO and AO sides toopposite 4

and

3

2 angle with AB OSimilarly,

22

21

21

22

2

12

1

21

22

22

21

21

21

22

222

4

1)+32)(-6(-) 2)+61)(-3( =

4

)CO-(BO=cos =>

2-) )(BO(4cos-) (BO =2-) )(CO(4cos+) (CO

:let we(6) From

(8) 1)+32)(-6( =) (CO

2)+61)(-3(=) (BO

: thatfollowsIt

2=) )(CO(BO

:have we(7) From

2)BO-6(=) BO-2(AO=1)CO-3(

:have we(3) From

12

22

212

1

21

22

1 2

2221

woodcut. theof

tsmeasuremen actual get the to20.5cmby result hesemultiply t

weso41cm, ofdiameter has circle bounding sEscher' Since

.66059750 AB

.33940250BO

.69662140BO

2.2104024AO

1.3572189AO

1.8047860 BO

1.1081646CO

:find alsocan wehere From

79.97=(0.17416)cos= =>

3

1

2

1

2

1

0(-1)

ReferencesDunham, Douglas. “Some Math Behind M. C. Escher’s

Circle Limit

Patters”. d.umn.edu.14 April. 2013.

http://www.d.umn.edu/~ddunham/umdmath09.pdf

Schattschneider, Doris. M. C. Escher: Visions of Symmetry. New

York: Harry N. Abram, Inc., 2004. Print.

Schattschneider, Doris, and Michele Emmer. M. C. Escher’s

Legacy. New York: Springer, 2003. Print.