Scott - Some Recent Discoveries in Elementary Geometry

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SOMERECENT ISCOVERIESN ELEMENTARYEOMETRY

Some recent discoveries in elementary geometry

PAUL SCOTT

Introduction

I have alwaysdisagreedwith governmentministerswho assert thatit is a

good idea to separateteaching from research at the universities. For I

believe thatwith mathematics,researchandgood teaching go handin hand.

And if we interpret research' n its widest terms of questioning,discovery,involvement,etc, thenI believe that the same holds true at the school level.

Mathematicsneeds to be taughtas a vibrant, iving subjectwhichchallengesthe intellectand the imagination,not as a dustycollection of historicalfacts.

Further,asking questionsis just as much fun as answering hem!

I have recently been challenged in my readingwith a numberof new

insights and discoveries in the area of elementary geometry. We shall

surveysome of these,with the emphasison ideas rather hanproofs.

Pythagoras' heorem

Everyone knows the classical theorem of Pythagoras:given a triangle

ABC with a right angle atC, and side lengthsa, b andc as in Figure1, then2

+ b22

a + b = c

Problem1. Is it possible to generalise Pythagoras'theorem to

3-space? Whatwould such a generalisationstate? Try t now!

The answer is, given any right tetrahedron,with right angles at D, and

face areasa, r, rand 8 as in Figure2, then

a 2+ ?2

2= .

This 'obvious' analogue appearedin a paper by Parthasarathy1] in1978 and was recently rediscoveredby Ward [2]. It has an analogue in

A Ic

c b

YBA

B a C

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FIGURE FIGURE2



THEMATHEMATICALAZETTE

general dimension. It is fairly easy to verify the 3-dimensional case bycalculatingthe areas. In fact, I have since discovered that the formula for

the area of the sloping face appeared n 1910 in an old geometrybook [3],and that the Pythagorean generalization is not unknown to (some!)statisticians,expressed n termsof projections.

Theregulartetrahedronand triangle inequalities

In 1985, Murray Klamkin A

looked at the following geometricconfiguration. Suppose we have a

regular tetrahedron ABCD (with

unit edges say) and a point P withPA = a, PB = b, PC = c, and /PD = d. (SeeFigure3.) /

Klamkin felt that there shouldbe some relationshipbetween thenumbersa, b, c andd, andmanaged B I

to show that the geometricsituation FIGURE

occursif andonly if

(a2 + b2 + c2 + d2)2 > 3(a4 + b4 + c4 + d4). (1)

Surprisingly, his resultis relatedto the triangle nequality. Supposethat

a, b, c areside-lengthsof a triangle. Then

b + c > a, c + a > b, a+ b > c.

Assumingthata, b and c arepositive, this is equivalent o

(b + c - a)(c + a - b)(a + b - c) > 0

and

(a+ b +

c) (b+ c -

a)(c+ a -

b)(a+ b -

c)> 0.

Multiplyingout, this gives

(a2 + +b2 42) > 2 a + b4 + c4), (2)

and the similaritybetween(1) and(2) is clear*.

What is not so clearis just why these inequalitiesoccurin such differentcontexts. However,we caneasily show that the triangle nequalityholds foreach threeof the a, b, c and d in Figure3. Forexample,considerb, c and d.For fixed values of b and c, rotating APBC about BC shows that d is

maximal when P lies in the plane DBC, in the position where PBDC is aquadrilateralwith diagonals PD and BC. Now applying to quadrilateralPBDC the classical theoremof Ptolemythat 'theproductof the diagonalsisless than the sum of the products of the opposite edges' givesd. I < b.1 + c.1. This is therequired riangle nequality.

It is interesting hatusing Heron'sformula or the areaA of the triangle,(2) is equivalentto16A2> 0.

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SOMERECENT ISCOVERIESN ELEMENTARYEOMETRY

This led Klamkin o consider

Problem2. Suppose we are given n segments having lengths

al, a2, ... , an (n > 3), is there a simple condition whichimpliesthatany threesegments orm the sides of a triangle?

Using mathematical induction and some calculation, Klamkin [4]showedin 1987 that:

If a,, a2, ... , a arepositive andn > 3, and

(a,2 + a22 + ... + an2)2 > (n - 1)(al4 + a24 + ... an4),

theneach tripletai, aj,ak(i ? j, j ? k, k ? i) gives lengthsfor the sides of

a triangle.

Question: Is itpossible tofind a condition that six segmentscanbe assembled to give a tetrahedron?

Points and lines in a triangle

The simple triangle has a surprisingly large number of interestingproperties. For example,it is known that the mediansAX, BY,CZof AABC

meet in acommonpoint,thecentroid,M say (Figure4).

Noting that Area(AABX) Area(AACX), and Area(IAMBX)

Area(AMCX),it is easy to deduce that the three smallertrianglesAMBC,AMCAand AMABall have the same area. Now supposethatM is not the

centroid, but a point for which the three above triangles have the same

perimeter. Call such a pointM an isoperimetricpoint of AABC.

A A

z y z y

B X C B X C

FIGURE FIGURE

Problem 3. Does every trianglehave an isoperimetricpoint? If

not,which

trianglesdo? What

propertiesdoes this

isoperimetricpoint have? How are the perimetersof the originaltriangleandthe smallertrianglerelated?

Notice the simple idea. Actually, the solution is quite complicated,as

Veldkamp [5] found in 1985. (Perimeter problems are traditionallydifficult!) For example, a trianglehas an isoperimetricpoint if and only if

none of its anglesexceed 2 arcsin4/5 - 106? 15' 37".

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THEMATHEMATICALAZE'I"

Theconcurrenceof the medians is a specialcase of a moregeneralresultabout the cevians of a triangle. A cevian is defined to be a line segment

from a vertex of a triangleto the side opposite it. With the notation ofFigure 5, the classical theorem of Ceva (1678) states thatAX, BY, CZ areconcurrentn pointP if andonly if

BX.CY.AZ

XC. YA.ZB

or

al .bl.cl * =. (3)a2.b2.c2

All of the nice concurrenceresults aboutmedians, altitudes,angle bisectorsarespecialcases of this result.

|Problem 4. Can we generalize Ceva's theorem to three dimensions? |

Another 'obvious' generalisation! The answerwas providedby Landy[6] in 1988. The secret lies in finding the correctinterpretation f Ceva'stheorem. Supposethat masses ml, m2, m3 areplacedat the verticesA, B, C

respectively. Now let points X, Y,Z be placed at the centres of masses of

m2,m3; m3, ml; ml, m2respectively.

TakingmomentsaboutX, Y,Z respectively,we requireaim2 = a2m3, bIm3 = b2ml, clml = c2m2.

Thus

a, m3 b, ml cl m)

a2 m2 b2 m3 c2 ml

and multiplicationof these equations gives (3). In other words, Ceva'stheorem holds when and only when the pointsX, Y,Z occur as centres ofmass in the aboveway.

With the naturaldefinitionof a cevian for a tetrahedron,he likely (and

correct)generalisationbecomes:

The four cevians of a tetrahedron re concurrent f andonly ifwe can assign masses to the vertices in such a way that eachcevian base point lies at the centre of mass of the surroundingface.

Napoleon revisited

Sometimes a startlingnew proof is given for an old theorem. Take forexamplethe prettyresultattributedprobably alsely) to Napoleon:

Napoleon's Theorem. If equilateral rianglesare erected externally(orinternally)on the three sides of any triangle, then their centres form an

equilateral riangle. (See Figure6.)

The traditional proof is not trivial, and begins by examining the

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SOMERECENTDISCOVERIESN ELEMENTARYEOMETRY

circumcircles of the externaltriangles. It is of interestin that it establishes

along the way that the lines AA',BB', CC' are concurrent n a point F, the

Fermat or Steiner point of AABC. B'This is the point at which P lieswhen the sum of the lengths C'PA + PB + PC is minimal. A

However, in 1988, Rigby [7]demonstrated a nice alternative

proof involving a tessellation of the

plane in which equilateral triangles / / Csurround congruent (shaded)triangles (Figure7(a)).

In outline (see Figure 7(b),which can be thought of as beingsuperimposedupon Figure7(a)):

(1) The centres of the small

equilateral triangles of the Atessellation form an equilateraltriangle lattice (unbroken FIGURE

lines).(2) The centres of the otherequilateral rianglesof the tessellationlie at the

centres of the latticetriangles.

(3) Hence all the centres of the equilateral triangles form a smaller

equilateral riangularattice,andNapoleon'stheorem follows.

Analogues of Napoleon's theorem can be derived when, for example,similartrianglesare erected on the sides of the given triangle.

Question. What happens in Napoleon's theorem when the

original triangle degenerates to a line segment?

FIGURE (b)

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FIGURE(a)



THEMATHEMATICALAZETTE

FIGURE 8

Dissections

It is well known that a squarecan be dissected into a finite number ofsmallersquares,with no two of the smallersquaresbeing equal in size (see

Figure8, where the numbersgive the side-lengths). The 1940 paper[8] is

quite difficult, but has a nice bibliographyof results up to that time. You

may like to think about the lowest possible numberof component squares(in fact 21), and whether thereis a 3-dimensionalanalogueof the problem.In a differentdirection:

|Problem5. Perhaps here aregeneralisationsn theplane? lThe obvious candidate is the equilateral triangle: can an equilateral

trianglebe tiled by a finite numberof smallerunequalequilateral riangles?In 1948 Tutte [9] showed that the answer to this questionis 'No', but more

recently n 1981 a strongerresultappeared:

No convex region can be tiled by a finite number of smaller

unequalequilateral riangles Buchman[10]).

In each of the above cases the tiling can take place if two equalequilateraltrianglesare allowed.

Finally,let us think of tilinga rectanglewithrectangles. (Thatshouldbe

simple!) If all the component rectangles have integer side lengths, then

clearlyso does the large rectangle.

Problem 6. If each of the component rectangleshas at least one

integerside length,what can be saidabout the largerectangle?

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SOMERECENT ISCOVERIESNELEMENTARYEOMETRY

Put in this form it is clear what the 'expected' (and correct)answer is.But it is the question which seems to me to be creative and imaginative.

Perhaps n our mathematics eachingwe shouldplace less emphasison theanswers,and more on the questions! If you areinterested,StanWagon [11(1987) establishes the answer to this question in fourteen differentways!And the large rectangledoes have at least one integerside length.

References

1. K. R. Parthasrathy,A generalization of Ceva's theorem to higherdimension,AmericanMathematicalMonthly95, pp. 137-140(1978).

2. I. Ward,The tritetrule, Math.Gaz.79 (July 1995)pp. 380-382.

3. R. J. T. Bell, Coordinategeometry of three dimensions, Macmillan

(1910).

4. M. S. Klamkin, Simultaneous triangle inequalities, Mathematics

Magazine60 (1987) pp. 236-237.

5. G. R. Veldkamp, The isoperimetric point and the point(s) of equaldetour in a triangle,American MathematicalMonthly,92 (1985) pp.546-558.

6. S. Landy, A generalizationof Ceva's theorem to higher dimension,

AmericanMathematicalMonthly95 (1988) pp. 137-140.7. J. F. Rigby, Napoleonrevisited,Journalof Geometry33 (1988) pp. 129-

146.

8. R. L. Brooks,C. A. B. Smith,A. H. Stone, W. T. Tutte,The dissectionof rectangles into squares,Duke Mathematical Journal 7 (1940) pp.312-340.

9. W. T. Tutte, The dissection of equilateraltriangles into equilateral

triangles, Proceedings of the Cambridge Philosophical Society 44

(1948) pp.463-482.

10. E. Buchman,The impossibilityof tiling a convex region with unequalequilateraltriangles,American MathematicalMonthly 88 (1981) pp.748-753.

11. S. Wagon,Fourteenproofsof a result abouttiling a rectangle,American

MathematicalMonthly94 (1987) pp. 601-617.

PAUL SCOTT

Department f PureMathematics,University fAdelaide,SouthAustralia5005

Failureofthepigeonhole rinciple?Bradley ones, he worldNo. 19fromCroydon,willbe the owestranked layer

in the 32-man field contestingthe televised phase of the EmbassyWorld

ChampionshipnSheffield.

HughandJoycePorteousawthisconundrumn theGuardian,5 March 997.

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Scott - Some Recent Discoveries in Elementary Geometry