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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
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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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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)
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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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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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