# Observations on Cooperative-Learning Group Assignments

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Observations on Cooperative-Learning Group Assignments Russell Marcus Chauncey Truax Post-Doctoral Fellow Department of Philosophy, Hamilton College 198 College Hill Road Clinton NY 13323 (315) 859-4056 (office) (315) 381-3125 (home) (315) 335-4711 (mobile) October 14, 2009 Abstract: Good cooperative-learning lessons are naturally appropriate in philosophy classes, and can be productive and fun.
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The uses of homogeneous barycentric coordinates in plane euclidean geometry
Paul Yiu
Abstract.The notion of homogeneous barycentric coordinates provides a powerful tool of analysing problems in plane geometry. In this paper, we explain the advantages over the traditional use of trilinear coordi-nates, and illustrate its powerfulness in leading to discoveries of new and interesting collinearity relations of points associated with a triangle.
1. Introduction In studying geometric properties of the triangle by algebraic methods, it has been customary to make use of trilinear coordinates. See, for examples, [1], [2], [3], [4], [5]. With respect to a Þxed triangleABC(of side lengthsa,b,c, and opposite anglesα,β,γ), the trilinear coordinates of a point is a triple of numbers proportional to the signed distances of the point to the sides of the triangle. The late Jesuit mathematician Maurice Wong has given [5] a synthetic construction of the point with trilinear coordinatescotα: cotβ: cotγ, and more generally, in [4] 2n2n2n points with trilinear coordinatesa x:b y:c zfrom one with trilinear coordi-natesx:y:zwith respect to a triangle with sidesa,b,c. On a much grandiose scale, Kimberling [2], [3] has given extensive lists of centres associated with a triangle, in terms of trilinear coordinates, along with some collinearity relations. The present paper advocates the use of homogeneous barycentric coordinates instead. The notion of barycentric coordinates goes back to Mo®bius. See, for example, [6]. With respect to a Þxed triangleABC, we write, for every pointPon the plane containing the triangle, 1 P= ((P BC)A+ (P CA)B+ (P AB)C), ABC and refer to this as the barycentric coordinate ofP. Here, we stipulate that the area of a triangleXY Zbe zero ifX,Y,Zare collinear, positive if the orientation of the vertices is counter - clockwise, and negative otherwise. In terms of barycentric coordinates, there is the basic area formula.
Int. J. Math. Educ. Sci. Technol., 31 (2000) 569 Ð 578. 569
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Proposition 1(Bottema [7]).If the vertices of a triangleP1P2P3have homoge neous coordinatesPi=xiA+yiB+ziC, then the area of the triangle is x1y1z1 x2y2z2.   x3y3z3 Often, it is convenient to consider the homogeneous coordinates ofP, namely, P BC:P CA:P AB. Here is a useful mechanical interpretation: the homogeneous barycentric coordi-nates ofPare ÒweightsÓ atA,B,Csuch that the Òcenter of massÓ is precisely at P. For example, every pointPon the lineBChas homogeneous coordinates of the form 0 :CP:P B, in terms of signed lengths of directed segments. It is clear that a point with homogeneous barycentric coordinatesx:y:z has trilinear coordinates(x/a) : (y/b) : (z/c). Conversely, a point with trilinear coordinatesu:v:whas homogeneous barycentric coordiantesau:bv:cw.
2. Traces of a point on the sides of a triangle 2.1. Coordinates of traces.The Þrst advantage of homogeneous barycentric coor-dinates is that the coordinates of the traces of a point on the sides of the reference triangle can be read off easily. LetP=x:y:zin homogeneous barycentric coordinates. The linesAP,BP,CPintersect the linesBC,CA,ABrespectively at the pointsX,Y,Zwith homogeneous coordinates X= 0 :y:z, Y=x: 0 :z,(1) Z=x:y: 0. See Þgure 1. Conversely, ifX,Y,Zhave homogeneous coordinates given by (1) above, then the ceviansAX,BY, andCZare concurrent at a point with homoge-neous barycentric coordinatesx:y:z. A
B
Z
P
Y
X Figure 1. The traces of a point
C
Here, ÔnormalizedÕ means dividing the expression by the sum of the coefÞcients. From this we conclude that the Nagel pointNadivides the segmentIGexternally in the ratioINa:NaG=2 : 3. This is, of course, a well known result. See, for example [8].
2.3notable centres.. Some In Table 1, we list the homogeneous barycentric coor-dinates of some notable centres associated with a triangleABC, with sidesa,b,c, 1 and semiperimeters= (a+b+c). 2
(sa)A+ (sb)B+ (sc)Cnormalized s(A+B+C)(aA+bB+cC) normalized s(3G)(2s)Inormalized 3G2I.
= = = =
: : :
2.2example: the Nagel point.. An The Nagel pointNais the point of concurrency of the cevians joining each vertex to the point of contact of the excircle on its opposite side. The existence of this point is clear from noticing that the points of contactX,Y,Zhave homogeneous coordinates
X Y Z
The uses of homogeneous barycentric coordinates
= = =
C
sc, sc, 0.
(2)
Na
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Figure 2. The Nagel point
B
: : :
0 sa sa
A
sb 0 sb
In barycentric coordinates, this is
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centre Symbol Homogeneous barycentric coordinates CentroidG1:1:1 IncentreI a:b:c IAa:b:c ExcentresIBa:b:c ICa:b:c Gergonne pointGe(sb)(sc) : (sc)(sa) : (sa)(sb) Nagel pointN sa:sb:sc 2 2 2 Symmedian pointK a:b:c 2 2 2 KAa:b:c 2 2 2 Exsymmedian pointsKBa:b:c 2 2 2 KCa:b:c 2 2 2 2 2 2 2 2 2 2 2 2 CircumcentreO a(b+ca) :b(c+ab) :c(a+bc) 2 2 2 2 2 2 OrthocentreH(a+bc)(c+ab) 2 2 2 2 2 2 :(b+ca)(a+bc) 2 2 2 2 2 2 :(c+ab)(b+ca) Table 1. Homogeneous barycentric coordinates of some notable points.
3. Multiplication 3.1. Multiplication of points on a line.The following proposition provides a sim-ple construction for the product of two points on a segment, and leads to the fruitful notion of multiplication of points of the plane not on any of the lines deÞning the reference triangle.
Proposition 2.LetX1,X2be two points on the lineBC, distinct from the vertices B,C, with homogeneous coordinates0 :y1:z1and0 :y2:z2. Fori= 1,2, complete parallelogramsAKiXiHiwithKionABandHionAC. The line joining the intersections ofBH1andCK2, and ofBH2andCK1, passes through the vertexA, and intersectsBCat a pointXwith homogeneous coordinates0 : y1y2:z1z2.
K2
K1
B
A
H1
H2
X1XX2 Figure 3. Multiplication of points
C
The uses of homogeneous barycentric coordinates
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Proof.Consider the ceviansBH1andCK2. Clearly, CH1CX1y1 = =, H1A X1B z1 and AK2CX2y2 = =. K2B X2B z2 By CevaÕs theorem, the unique pointXonBCfor which the ceviansAX,BH1, andCK2are concurrent is given by BX CH1AK2 ∙ ∙= 1. XC H1A K2B From this,BX:XC=z1z2:y1y2, andXis the point0 :y1y2:z1z2. A similar calculation shows that this is the same point for which the cevians AX,BH2, andCK1are concurrent.3.2. Multiplication of points in a plane.Consider two pointsPi,i= 1,2, with nonzero homogeneous barycentric coordinatesxi:yi:zi. By applying Proposi-tion 2 to the traces on each of the three sides of the reference triangle, we obtain three points
X= 0 :y1y2:z1z2, Y=x1x2: 0 :z1z2,(3) Z=x1x2:y1y2: 0. The ceviansAX,BY,CZintersect at a point with homogeneous barycentric co-ordinates x1x2:y1y2:z1z2. We shall denote this point byP1P2, and call it the product ofP1andP2(with respect to triangleABC).
3.3. An abelian group structure.The multiplication of points considered above clearly deÞnes an abelian group structure on the setGof all points with nonzero homogeneous barycentric coordinates, i.e., points not on any of the lines deÞning the reference triangle. The centroidGis the multiplicative identity, since it has homogeneous barycentric coordinates1 : 1 : 1. The inverse of a pointP= x:y:zis precisely its isotomic conjugate ([4]), with homogeneous barycentric 1 coordinates1/x: 1/y: 1/z. For this reason, we shall denote byPthe isotomic conjugate ofP.
3.4. Isogonal conjugates.The isogonal conjugate ([4]) of a pointP=x:y:z (with nonzero homogeneous barycentric coordinates) is the point 2 2 2 a b c P:= : . x y z 2 2 2 Note thatPP=a:b:cis the symmedian point of the triangle,. This 2 and can be constructed from the incentreIasI=II. It is also the isogonal conjugate of the centroidG= 1 : 1 : 1.
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3.5. Examples.A second advantage of the use of homogeneous barycentric coor-dinates is that a factorization of the coordinates entails a construction procedures of the point in question in terms of simpler ones. Consider, for example, the prob-lem of locating in the interior of triangleABCa point whose distances from the sides are in the proportion of the respective exradii. This is the point with trilinear coordinates 1 1 1 rA:rB:rC= : :, sa sb sc and appears asX57in [2]. In homogeneous barycentric coordinates, this is the point 1 1 1 a:b:c. sa sb sc As such, it can be constructed as the productIGeof the incentreIand the Gergonne pointGe(see table 1).
4. The square root construction 4.1square root of a point.. The LetPbe a point in the interior of triangleABC, with tracesX,Y,Zon the sides. The square root ofPis an interior pointQ such thatQQ=P. To locate such a pointQ, construct circles with respective diametersBC,CA, andAB, intersecting the respective perpendiculars to the sides    throughX,Y,Z(respectively) atX,Y,Z. Construct the bisectors of the right       anglesBX C,CY A,AZ Bintersecting the sidesBC,CA,ABatX,Y,Z.   The ceviansAX,BY, andCZare concurrent at the requisite square rootQ. This follows easily from the lemma below.
Lemma 3.LetXbe a point on a segmentBC. Suppose the perpendicular through  Xintersects the circle with diameterBCatX. Construct the bisector of the right   angleBX Cto intersectBCatX. Then   2 BX BX =. X C XC
B
 X
X X Figure 4. Square root of a point
C
The uses of homogeneous barycentric coordinates
   Proof.SinceX Xbisects the right angleBX C, we have   BX X B =.   X C X C It follows that   2  2 BX X B BCBX BX = = =.  2 X C X C BCXC XC This completes the proof of the lemma.
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4.2. Examples.For example, given a triangle, to construct the pointQwhose dis-tances from the sides are proportional to the square roots of the lengths of these sides. The trilinear coordinates ofQbeinga:b:c, the homogeneous 3/2 3/2 3/2 barycentric coordinates are given bya:b:cpoint is the square. This 3 3 3 root of the pointa:b:c, which isIK. As another example, consider the point with homogeneous coordinates α β γ sin : sin : sin. 2 2 2 Since α(sb)(sc) (sa)(sb)(sc)a 2 sin = =, 2sbc abc a we have α β γ a b c 2 2 2 sin : sin : sin = : :, 2 2 2sa sb sc and the point in question can be constructed as the geometric mean of the incentre I=a:b:cand the Gergonne pointGe= 1/(sa) : 1/(sb) : 1/(sc).
5. More examples of collinearity relations A third advantage of homogeneous barycentric coordinates is the ease of obtain-ing interesting collinearity relations of points associated with a triangle. We have already seen one example in§is another interesting example of the use2.2. Here of homogeneous coordinates.
5.1. Equalintercept point.Given a triangle, to construct a pointPthe three lines through which parallel to the sides of a given point cut out equal intercepts. In general, the line throughP=xA+yB+zCparallel toBCcuts out an intercept of length(1x)a. It follows that the three intercepts parallel to the sides are equal if and only if 1 1 1 1x: 1y: 1z= : :. a b c The right hand side clearly gives the homogeneous barycentric coordinates of the isotomic conjugate of the incentreI. It follows that 1 1 1 I= [(1x)A+ (1y)B+ (1z)C] = (3GP). 2 2 1 From this,P= 3G2I, and can be easily constructed as the point dividing the 11 segmentI Gexternally in the ratioI P:P G= 3 :2NoteÞgure 5. . See
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1 that an easy application of Proposition 1 shows that the line joiningP,GandI does not containIunless the triangle is isosceles. A
B
P
I G
1 I
Figure 5. Equal-intercept point
C
In fact, many of the collinearity relations in [2], [3] can be explained by manip-ulating homogeneous barycentric coordinates. We present a few more examples. 5.2point with trilinear coordinates. The b+c:c+a:a+b.Consider, for example, the Ôsimplest unnamed centreÕX37in [2], with trilinear coordinatesb+c:c+a: a+b. In homogeneous barycentric coordinates, this isa(b+c) :b(c+a) :c(a+b). Thus,X37=IP, wherePis the point with homogeneous barycentric coordinates b+c:c+a:a+b. This happens to be the pointX10of [2], the Spieker centre of the triangle, the incentre of the triangle formed by the midpoint of the sides of the given triangle. Without relying on this piece of knowledge, a direct, simple calculation with the barycentric coordinates leads to an easy construction of this point. It is indeed the point which divides the segmentIGexternally in the ratio 3 :1.
P= (b+c)A+ (c+a)B+ (a+b)Cnormalized = (a+b+c)(A+B+C)(aA+bB+cC) normalized = (2s)(3G)(2s)Inormalized 1 = (3GI). 2 5.3. The Mittenpunkt.Consider the MittenpunktX9with trilinear coordinatessa:sb:sc, or homogeneous barycentric coordinatesa(sa) :b(sb) : c(sc). While this can certainly be interpreted as the productINaof the incentre Iand the Nagel pointNa, we consider the barycentric coordinates:
=
a(sa)A+b(sb)B+c(sc)Cnormalized 2 2 2 s(aA+bB+cC)(a A+b B+c C) normalized.
The uses of homogeneous barycentric coordinates
577
This shows thatX9is on the line joining the incentreIto the symmedian point 2 2 2 K=a:b:c. Noting that 2 2 2 2 a+b+c= 2[s(4R+r)r],(4) whereRandrare respectively the circumradius and the inradius of the triangle, (see [8]), we have the symmedian pointKdividing the segmentIX9externally in 2 the ratio(4R+r)r:s. While this certainly leads to a construction ofX9, it is much easier to locateX9by Þnding another collinearity relation. Since a(sa) + (sb)(sc) = (sa)(sb) + (sb)(sc) + (sc)(sa) is a symmetric function ina,b,c, say,f(a, b, c), we also have
=
a(sa)A+b(sb)B+c(sc)Cnormalized f(a, b, c)G[(sb)(sc)A+ (sc)(sa)B+ (sa)(sb)C] normalized
From this it follows thatX9is collinear with the centroidGand the Gergonne pointGecan therefore be located as the intersection of the two lines. It IKand GGe. The fact thatX9lies on both of these lines is stated in [2], along with other lines containing the same point.
5.4the Nagel and Gergonne points.. Isogonal conjugates of The isogonal conju-gateNof the Nagel point has homogeneous barycentric coordinates a 2 2 2 a b c 2 2 2 : : =a(sb)(sc) :b(sc)(sa) :c(sa)(sb). sa sb sc Making use of the formula α(sb)(sc) 2 sin =, 2bc we write this in barycentric coordinates:       α β γ 2 2 2 N=asinA+bsinB+csinCnormalized a 2 2 2 =a(1cosα)A+b(1cosβ)B+c(1cosγ)Cnormalized = 2sI((acosα)A+ (bcosβ)B+ (ccosγ)C) normalized =RIrOnormalized. Here, we have made use of   2rs (acosα)A+ (bcosβ)B+ (ccosγ)C=O. R A similar calcuation shows that the isogonal conjugateGof the Gergonne point, e namely, 2 2 2 G=a(sa) :b(sb) :c(sc) e 1 has barycentric coordinate(RI+rO). This means that the isogonal conjugates R+r of the Gergonne point and the Nagel point divide the segmentOIharmonically in
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