PROPERTIES OF SQUARES

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June Merry Richardson
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1 PROPERTIES OF SQUARES The square is oe of the simplest two-dimesioal geometric figures. It is recogized by most pre-idergarters through programs such as Sesame Street, the comic strip Spoge Bob Square Pats, or simply playig with blocs. We wat here to loo at the geeral properties of the square ad develop some of the relevat formulas describig areas ivolvig this figure. The simplest defiitio of a square is that it is a figure cosistig of four straight lie edges of legth each coected to each other by four right agle vertees. Epressed i terms of side-legth ad coectig agle, the four vertees are foud at- [,]=[, 0 +(-)/], =,,3,4 Here 0 represets the agle the = side maes with respect to the -ais. Settig 0 = -/4 we get the followig figure-

2 The area of this square is simply A S = with the distace to ay of the vertees from the square ceter beig /sqrt(). If we place a smaller square of sides c iside a larger square of sides =a+b such that the vertees of the smaller triagle just touch the sides of the larger triagle, we get the followig patter- From the figure we see that the area of the four orage right triagles just equals the differece i area of the larger to the smaller square. Hece we have- ab 4( ) ( a b) c O multiplyig thigs out we arrive at the famous Pythagorea Theorem a b c Pythagoras of Samos ( BC) was a famous Gree mathematicia who fouded a school which attempted to coect music, the movemet of stars ad plaets, ad the fate of ma to umbers. It is ot sure whether Pythagoras arrived at his theorem from earlier Babyloia wor or whether it was a idepedet discovery. Nevertheless, his

3 aoucemet called for a major celebratio at his school icludig the sacrificig of a oe. Oe ca use the Pythagorea Theorem to determie the ratio of the two parts of a square created by cuttig the square by a straight lie y=+ which itersects two eighborig edges. Oe has the followig picture- Here the area A =ab/ ad the area A = -A. Hece the ratio- R A A [ ] ab This ratio becomes R= whe a=b= as epected. Uder that coditio the two areas are equal so that the resultat diagoal has legth sqrt() accordig to the Pythagorea theorem. Should the straight lie cross opposite sides of the square the sub-areas will form two trapezoids. The area ratio becomes-

4 R ( c d ) Here c is the distace o the left side of the square to that at the right side, both measure from the lie y=-/ up. Whe c=d=/, oe gets two equal area rectagles of area / each. The dividig lie will have legth. We et loo at some patters costructed form a collectio of may squares. Oe of the first which comes to mid is to start with a uit side-legth square ad superimpose a first geeratio of four smaller squares of side-legth ½ as show. Follow this with a secod, third, ad fourth geeratio each time halvig the square side-legth. This produces the patter- This patter ehibits a four-fold symmetry ad is of fiite size. The total area of the patter will be- 49 Area= 4{ } whe tae through the fourth geeratio. Oe ca eted the calculatio to the ifiite geeratio i which case we recogize the area has the slightly larger value of-

5 7 Area total= The etire patter just fits ito a large square of side-legth 3 ad total area 9. From this we also see that the blac squares tae up a area of 7/3 compared to the remaiig area iside the large outer boudig square of 0/3. Aother iterestig patter is foud by addig oto a uit side-legth square a smaller square of side-legth ½ at its upper right. Follow this by addig a third square of side legth ¼ as show. Cotiuig the process each time halvig the side-legth for the et square, oe produces the followig patter- It is ow possible to place quarter circles of decreasig radius ito these squares to get the spiral structure show. This spiral is cotiuous ad has the same derivative i goig betwee squares but has a chage i curvature at these trasitios This same shortcomig is ecoutered with the Fiboacci Spiral which loos very much as the preset spiral but starts by breaig a rectagle ito squares of side-legth matchig the golde ratio =[+sqrt(5)]/. I eamiig the above picture, oe could as to what poit does the spiral coverge. Fidig the precise value first stumped us. However, a hour of thought made us realize that the solutio is straight forward. To get the value you simply start addig up all the movemets startig with =0. This yields-

/ 4 /6 / 64 / 56 (/ 4) The y value is gotte by the followig additio- 0 ( ) 4 5 y / (/8){ / 4 /6..} 8 ( ) 0 4 5 Note that the covergece poit [,y]=[0.,0.

6 / 4 /6 / 64 / 56 (/ 4) The y value is gotte by the followig additio- 0 ( ) 4 5 y / (/8){ / 4 /6..} 8 ( ) Note that the covergece poit [,y]=[0.,0.4] of the spiral lies alog a radial lie passig through the origi [0,0] ad iclied at agle arcta() relative to the ais. The square also maes its appearace i architecture ad art. The moder artist who was especially cocered with squares was Joseph Albers. Here are two eamples of his wor- We coclude our discussios o squares by looig at a old problem studets ecouter i their elemetary statics course. The questio is what is the miimum overlap required betwee staced square blocs to eep a tower of such blocs from topplig. Ay child playig with blocs will have figured out the solutio ituitively but it is ot util studets become familiar with ceters of mass, momets, ad reactio forces that they will be able to give a mathematical proof for what it taes to eep the tower stable. We will assume that we are dealig with a set of cubical woode blocs of equal size ad uiform desity, Also we will cosider oly stacig i a fied vertical plae. Clearly for two blocs of edge legth the impedig topplig occurs whe the ceter of mass of the upper bloc lies just above the edge of the lower bloc. That is, a / overhag is the maimum allowed. O addig more square blocs, the requiremet is that the effective

7 ceter of mass of the upper blocs lie eactly above the edge of the + square. We have a picture which loos lie this- The origi of the coordiate coicides with the ceter of mass of the first bloc so that the mass ceter of the secod bloc lies at =- = /. At impedig topplig =/. Now to determie the effective mass ceter i the directio of the first blocs we use the formula- m m all with m m So for the first two blocs we have 4 ] [0 ad for the first three blocs we fid- / 6 / 5 3 ] 4 3 [ so

8 From the tred, we epect 4 to go as /8. To test this out we write- 3 [0 ] so We ca geeralize these results by otig the maimum overhag of the group of top blocs at the th bloc relative to the + bloc will be /. This result shows that the preset tower ca have a total overhag greater tha sice the harmoic series diverges. Ideed ay tower of five or more blocs will a shift i betwee the top ad bottom bloc of more tha. /8 Jauary 0, 07 Iauguratio Day

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