386 MATHEMATICSMAGAZINE In the revolutionof a graphy = f(xl) aboutthe y-axis, we let F(l X2, X3) = f (x + X2 + X) . Using sphericalcoordinates the integralbecomes I jlv/1 + (f/(p))2 p2 sin 0 dpdqds= 47rp2 + ( (p))2p dO O a a Formula(4) follows. For revolution aboutthe x-axis, the surface y2 + y2 + y2 = f2(X) can be parame- terizedusing sphericalcoordinates as x = t, Y = f(t) sin cos , Y2 = f(t) sin sin0, 3 = f(t) cos for a < t < b, 0 < 0 < r, 0 < 0 < 2r. Formula(4) follows aftera straightforward calculationof the determinantin (5). Epilogue Now thatwe have extendedthe resultsto fourdimensions, we mightcon- sider extendingthem furtherto n dimensions.In fact, in higherdimensions there are otherways to revolvegraphs about axes. For example,in four dimensionswe could considerthe doublerotation, where the graphis firstrevolved about the y-axis in three dimensionsand thenthe resultingsurface is revolvedabout the x-axis in four dimen- sions. REFERENCES 1. E. A. Abbott,Flatland, Harperand Row Publishers,NY, 1983. 2. D. Davis, TheNature and Power of Mathematics,Princeton University Press, Princeton,NJ, 1993. 3. J. H. Hubbardand B. B. Hubbard,Vector Calculus, Linear Algebra, and Differential Forms, Prentice Hall, Upper Saddle River,NJ, 1999. 4. G. J. Porter,k-volume in ln and the GeneralizedPythagorean Theorem, Amer. Math. Monthly 103 (1996), 252-256. A VectorApproach to Ptolemy'sTheorem ERWIN JUST NORMAN SCHAUMBERGER Bronx Community College of the City University of New York Bronx, NY 10453 [email protected] Ptolemy'sTheorem [1] statesthat the productof the diagonalsof a cyclic quadrilateral (a quadrilateralthat can be inscribedin a circle) is equalto the sum of the productsof its oppositesides. Ourpurpose is to provePtolemy's Theorem by incorporatingthe use of vectors,an approachwhich we have neverbefore seen. The notionthat we might succeed in this effortoccurred to us afterobserving that Ptolemy's Theorem may be used to provethe followingresult [2]: THEOREM 1. Suppose a circle contains point A of parallelogram ABCD and in- tersects side AB, side AD, and diagonal AC in points E, G, and F, respectively. Then IAFIIACI = AEIIABI+ IAGIIADI. Mathematical Association of America is collaborating with JSTOR to digitize, preserve, and extend access to Mathematics Magazine ® www.jstor.org VOL.VOL77, 77 NO.IO 5,,DCEBR208 DECEMBER 2004 387 Proof We readily deduce that AABC ~ AGFE, from which it follows that IACI _ ABI IADI IGEI IFGI IEFI We apply Ptolemy's Theorem to quadrilateralAEFG to obtain IAFIIGEI= IAEIIFGI+ IAGIIEFI. Multiplying through gives IACI IABI IADI IAFIIGEI =AC IAEIIFGI + IAGIIEFI IGEI IFGI IADIJEFI' which yields the required conclusion. U It was our hope that if we could avoid using Ptolemy's Theorem in the proof of Theorem 1, then perhaps we could use Theorem 1 to deduce Ptolemy's Theorem. By incorporating a vector approach, Theorem 1 can indeed be proved independently of Ptolemy's Theorem. This is described in the body of the proof of Theorem 2. (Sub- sequently, we found another proof of Theorem 1 that does not use Ptolemy's Theo- rem [3]). It turns out that, unlike in Theorem 1, none of the points of the parallelogram used in the proof of Theorem 2 need be exterior to the circle. THEOREM2. (PTOLEMY'STHEOREM) Let ABCD be a cyclic quadrilateral. Then IACIIBDI= IABIICDI+ IADIIBCI. D 388 MATHEMATICSMAGAZINE Let F be a pointon chordAC. Let E and G be pointson chordsAB andAD (extended if necessary)such thatquadrilateral AEFG is a parallelogram.Let P be the point on the circle for whichAP is a diameter.Then LABP,LACP, and LADPare rightangles or, in the case thatdiameter AP coincideswith one of the chordsAB, AC, OrAD, two of these anglesare right angles. In eithercircumstance, it follows that IAFI ACI =AP-A =A . (a + A) = A- A + At- A- = AEIIABI + IAGIADI. Since IAGI= IEFI,this can be rearrangedas IAFI AEI IEFIEF IACIIBDII =IABIICD EI + ADIIBCI (1) IBDI ICDI IBCI' Since LFEA _ LBCD and LAFE =I LDAC = LDBC, we have AAEF ~ ABCD, from which we obtain IAFI_AEI IEFI ' IBDI ICDO IBC We can now concludefrom (1) and(2) that IACIIBDI= IABIICDI+ JADIIBCI, which completesthe proof. REFERENCES 1, NathanAltshiller Court,College Geometry,Barnes & Noble, 1952, p. 128. 2. H. S. M. Coxeter and S. L. Greitzer,Geometry Revisited, Random House, 1967, p. 43. 3. M. N. Aref and William Werick, Problemsand solutions in Euclidean Geometry,Dover Publications,1968, p. 134, Problem 13. ProjectedRotating Polygons DIMITRIOS KODOKOSTAS Department of Computer Science Technological Education Institute of Larissa Larissa,41110 Greece [email protected] [email protected] Take any number of lines 11,12, ..,., In through the center of a circle; pick a point M of the circle and projectit perpendicularlyonto each line, creatingpoints M1,M2, ..., Mn.Connect the pointsby segmentsto form a polygon M M2... Mn as in FIG- URE1). Its shapedepends on M, right?The surprisinganswer is thatit does not! The readercan have fun verifyingthis fact with the use of dynamicmathematical softwarelike the Geometer'sSketchpad. Seeing the polygon dance aroundthe circle can be a realjoy. .