VOL. 78, NO. 4, OCTOBER 2005
311
andthen
(P3,
The finalresultwill be the same.This shows thatthe ray emergingafterthreereflections is parallelto the originalray.
Finally,we considerthe specialoccurrenceof a rayenteringthe cubecomerparallel
to one of its faces. This case canbe viewedas if the reflectionstakeplaceon a billiard
table,with a ball hittingsuccessivelytwo concurrentsides of the table;the answeris
the same.
Conclusion We have shown that (almost)every ray emergingfrom a cube corner
systemafterreflectingoff its threeplanewalls has the samedirectionas the incoming
ray thatproducedit. Thanksto cleverapplicationsof this magic mirrorproperty,our
travelis saferand we can measurethe distanceall the way to the Moon. If you ever
encounterSmartDust,rememberthe reflectivepropertybehindits design.
REFERENCES
last revised
1. J. S. Allen, About bicycle reflectors contents, <http://www.bikexprt.com/bicycle/reflectors/>,
March 14, 2003, and all links, last revised April 27, 2002.
2. J. E. Fallerand E. J. Wampler,The lunarlaser reflector,ScientificAmerican,38, March 1970, 38-50.
3. J. M. Kahn, R. H. Katz, and K. S. J. Pister, Emerging challenges: mobile networking for "SmartDust,"
J. Comm.and Networks2:3 (2000), 188-196.
4. M. S. Scholl, Ray trace througha corner-cuberetroflectorwith complex reflectioncoefficients, J. Opt. Soc.
Amer A 12 (1995), 1589-1592.
5. Chun-ChingShih, Depolarizationeffect in a resonatorwith corner-cubereflectors,J. Opt. Soc. Amer A 13
(1996), 1378-1384.
6. L. Zhou, J. M. Kahn,and K. S. J. Pister,Corner-cuberetro-reflectorsbased on structure-assistedassembly for
free-spaceoptical communication,IEEEJ. Microelectromech.Syst. 12:3 (2003), 233-242.
SphericalTrianglesof Area7r
and Isosceles Tetrahedra
JEFF BROOKS
JOHN STRANTZEN
La Trobe University
Victoria3086, Australia
[email protected]
[email protected]
Thereis a beautifultheoremin sphericalgeometrythatis not well known,and that
doesn't appear in most texts on the subject.
The theorem says that for any spherical triangle of area nr on the unit sphere, four
congruent copies of it tile the sphere; that is, the copies can be positioned so that
any two triangles meet at an edge, with vertices meeting vertices, and the union of
these four triangles is the sphere [1; 3, p. 44; 6, p. 94]. The configuration is shown in
FIGURE1, which may aid readers in guessing the proof.
What appears to be less known (in fact, the authors can find no reference to it) is
that the four vertices on the sphere determined by this tiling form the vertices of an
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MATHEMATICSMAGAZINE
312
isosceles tetrahedron(oppositeedges of the tetrahedronare equal) [2]. Hence, every
such tiling of the sphereis obtainedas the projectionfromthe centerof an inscribed
isosceles tetrahedron.
Background We introducesome preliminarydefinitionsandresultsthatthe reader
may alreadyknow,in whichcase it is possibleto skip ahead.Givena point P on the
unit sphere,the intersectionof the line joining the centerof the sphereto P meets the
sphereat a diametricallyoppositepoint P'. The pair P and P' are called antipodal
points. We define a line on the sphereto be a greatcircle, thatis, the intersectionof
the spherewith a planecontainingthe centerof the sphere.The complementof a line
is two disjointopenhemispheres.
Giventwo distinctnonantipodalpoints, P and Q, thereis a uniqueline containing
them. This line is the intersectionof the spherewith the plane containingP and Q
andthe centerof the sphere.We denoteit by lPQ. We let PQ denotethe smallerof the
arcson the line thatconnectsP to Q. It follows thatlengthof PQ is less than7r.We
denotethe fact thattwo arcsPQ andRS have equallengthby PQ = RS, blurringthe
distinctionbetweena segmentandits measurement.
GiventhreedistinctnoncollinearpointsA, B, C, no two of whichareantipodal,we
define the spherical triangle AABC to be the union of the arcs AB, BC, and CA. We
callAB the edge of the trianglejoiningvertexA to vertexB, similarlyfor BC andCA.
Such a trianglenecessarilylies in an open hemisphere.Clearlyeach edge of AABC
determinesan open hemispherecontainingthe remainingvertex.The intersectionof
thesethreeopenhemispheresis the interiorof AABC.The areaof AABCis definedto
be the areaof its interior.It shouldbe intuitiveas to whatis meantby the threeangles
of AABC,andwe denoteby ZCAB,ZABC,and LBCAthe measurein radiansof these
theorem
threeangles.Themainresultthatwe use to proveourtheoremis a remarkable
attributedto Girard.Girard'stheoremsays that
Area(AABC)
Weeks[8] gives an elegantproofof this result.
Finally,we need to say whatwe meanby two trianglesbeing congruent.We adopt
theEuclideannotion,whichsaysthattwo trianglesarecongruentif thereis anisometry
betweenthe triangles.The isometriesof the sphereareof threetypes, a reflectionin a
line, a productof two reflections(this is equivalentto a rotationaboutan axis through
the antipodalpoints where the lines meet), and a reflectionfollowed by a rotation
to the line of reflection(a glide reflection)[5].
perpendicular
Reflectionsandglidereflectionsdo notpreserveorientation,whichwe wish to do in
thisdiscussion.Hence,for us two trianglesarecongruentif one canbe rotatedontothe
other,see [7]. The usualtheoremsfor congruenceof sphericaltriangles-SSS, SAS,
ASA, andAAA-will determineuniquenessup to isometry,so we mustshow thatthe
copies of ourtriangleareall rotationsof it.
Proving the theorem To demonstrateourresult,let AABCbe a sphericaltriangleof
arear. Choose an edge (say AB), let M be the midpoint of that edge, as in FIGURE1.
From our definition of a triangle it follows that MC\{M, C} lies in the interior of
/ABC. Rotate AABC by r about the axis through M and its antipodalpoint M'. Under
this rotation A and B interchange and C maps to a fourth distinct point D. It is clear
that C and D are distinct, since the length of MC is less than 7. Also C, M, and D lie
on the line determined by M, M', and C.
It is not difficult to see that MC has a length greater than 7/2 otherwise AABC
would fit inside a quadrant (as the length of AB is less than r) and hence have area
313
VOL. 78, NO. 4, OCTOBER 2005
C
M
A
M
B
D
Figure1
strictly less than 7r. Hence, M' is the midpoint of CD. The points C, M, and D lie on
the line determined by M, M', and C and M'D = M'C. Also M, M', and A lie on
line IAB, with A and B lying in distinct open hemispheres determined by line 1CD.
Consider the four triangles AABC, ABAD, ACDA, and ADCB. Clearly any two
meet at a common edge with vertices meeting vertices. If we partition ACDA into
ACM'A and AAM'D, and ADCB into ADM'B and ABM'C, we see that the three
triangles ACM'A, AABC, and ABM'C tile the closed hemisphere determined by line
AB and C. Similarly AAM'D, ABAD, and ADM'B tile the closed hemisphere determined by line lAB and D. It follows that AABC, ABAD, ACDA, and ADCB tile the
sphere.
We have AABC congruent to ABAD, as they are rotations of each other by n about
the axis through M, M'. Also ACDA is congruent to ADCB for the same reason. By
Girard's theorem, we have ZCAB+ ZABC+ ZBCA = 2r (since Area AABC = 7r).
Also, ZCAB + ZDAC+ IBAD = 27r (sum of angles around vertex A). But ZBAD =
ZABC(AABC is congruent to ABAD) hence ZDAC = ZBCA.
It now follows that if we rotate AABC by 7 about the midpoint N of AC, then A and
C change places and B maps to D (DA = CB as ACDA is congruent to ADCB). Thus
AABC is congruent to ACDA and so the four triangles are congruent. In particular
AB
which means that the tetrahedronformed by the
vertices ABCD is isosceles.
Now the great circle through M, M', and the midpoint N of AC has MN = M'N
it follows that MN = r/2.
(AABC is congruent to ACDA), and as MN
We therefore deduce that the medial triangle of AABC is an octant [1, 4].
Constructions This proof gives a method for drawing such triangles on a sphere:
Mark two antipodal points on the sphere. Center congruent arcs of length less than r
on these points in such a way that the arcs do not lie on the same great circle. Connect
the end points of the arcs in the appropriateway to obtain the tiling. Varying the length
of the arcs and the angle they make with each other gives all possible tilings.
To construct an isosceles tetrahedron, note that its faces must be four congruent
acute angled triangles. Take any acute angled triangle ABC and rotate it by r about an
axis through the midpoint of AB perpendicular to the plane of the triangle. Let D be
MATHEMATICSMAGAZINE
314
the image of C. ThenAB < CD. Now rotateAABD aboutAB so thatthe image of D
ABCD'is isosceles.
is D' andAB = CD'. It follows thattetrahedron
Acknowledgment. The authorswould like to thank Paul Pontikis for the typing, JeanetteVarrentiand Marcel
Jacksonfor the electronicpicture.
REFERENCES
1. GrantCairns, MargaretMcIntyre,and John Strantzen,Geometricproofs of some recent results of Yang Lu,
66:4 (October 1993), 263-265.
this MAGAZINE
2. N. A. Court,ModernPure Solid Geometry,Chelsea PublishingCo., New York, 1964.
3. H. L. Davies, Packings of sphericaltriangles, Proceedings of The Colloquiumon Convexity,Copenhagen,
August 1965, pp. 42-51.
4. W. J. M'Clelland and T. Preston,A Treatiseon Spherical TrigonometrywithApplicationsto Spherical Geometry Part I, MacmillanPublishingCo., New York, 1907, and Part II, 1909.
5. PatrickJ. Ryan, Euclideanand Non-EuclideanGeometry,An AnalyticApproach,CambridgeUniversityPress,
Cambridge,London,New York,New Rochelle, Melbourne,Sydney, 1986.
6. D. M. Y. Sommerville, Division of space by congruenttriangles and tetrahedra,Proceedings of The Royal
Society of Edinburgh,43 (1923) pp. 85-116.
7. I. Todhunterand J. G Leatham,Spherical Trigonometry,Macmillan and Co. Limited, St. Martin's Street,
London, 1943.
8. JeffreyR. Weeks, TheShape of Space, 2nd ed., MarcelDecker, Inc., New York,Basel, 2002.
A ButterflyTheoremfor Quadrilaterals
SIDNEY KUNG
University of North Florida
Jacksonville, FL32224
[email protected]
One of the moreappealingtheoremsin planegeometryis the butterflytheorem:
BUTTERFLYTHEOREM.Through the midpoint I of a chord AC of a circle, two
other chords EF and HG are drawn. If EG and HF intersect AC at M and N, respectively, then IM = IN.
Since 1815, whenthis theoremappearedas a proposedproblemin the Gentleman's
Diary (1815, pp. 39-40; see also [1, p. 195]) it has attractedmanylovers of mathematics,some of whomhaveproducedsimpleandelegantproofs,while othersdevised
variousgeneralizations.In a delightfulandwell-documentedarticle,Bankoff[1] discusses the butterflytheoremfor circlesandsome variants.In particular,on p. 207 one
findsan "areamethod"appliedto provethatif I is a pointanywhereon the chordAC
(as in FIGURE1), IA = a, IC = c, IM = m, IN = n, then:
m
Our
In this note, we give a similarproof of a butterflytheoremfor quadrilaterals.
proofdependsprimarilyuponthe followingpropertiesfor areasof triangles:
P1 If K is the intersection of the lines XY and UV, V
A(UXY)
(FIGURE2a), then
where A(UXY) denotes the area of triangle UXY.