VOL.76, NO. 1, FEBRUARY
2003
15
A Taleof ThreeCircles
CHARLES
GREGORY
1. DELMAN
GALPERIN
EasternIllinois University
Charleston, IL61920
cfcid@eiu.edu, cfgg@eiu.edu
Everyone knows that the sum of the angles of a triangle formed by three lines in the
plane is 180?, but is this still true for curvilinear triangles formed by the arcs of three
circles in the plane? We invite the reader to experiment enough to see that the angle
sum indeed depends on the triangle, and that no general pattern is obvious. We give
a complete analysis of the situation, showing along the way, we hope, what insights
can be gained by approaching the problem from several points of view and at several
levels of abstraction.
We begin with an elementary solution using only the most basic concepts of Euclidean geometry. While it is direct and very short, this solution is not complete, since it
works only in a special case. The key to another special case turns out to be a model of
hyperbolic geometry, leading us to suspect that the various manifestations of the problem lie on a continuum of models of geometries with varying curvature. This larger
geometric framework reveals many beautiful unifying themes and provides a single
method of proof that completely solves the original problem. Finally, we describe a
very simple formulation of the solution, whose proof relies on transformationsof the
plane, a fitting ending we think, since a transformationmay be regarded as a change
in one's point of view. The background developed earlier informs our understanding
of this new perspective, and allows us to give a purely geometric description of the
transformationsneeded.
For the reader who is unfamiliar with the classical noneuclidean geometries, in
which the notions of line and distance are given new interpretations,we provide an
overview that is almost entirely self-contained. Such a reader will be introduced to
such things as angle excess, stereographic projection, and even a sphere of imaginary
radius. For the reader who is familiar with the three classical geometries, we offer
some new ways of looking at them, which we are confident will reveal some surprises.
Three problems
Consider three circles in the plane intersecting transversely (that is, with no circles
tangent to each other) at a common point, P, as in FIGURE1. What is the sum of the
measures of the angles of triangle of circular arcs ABC? Answer: 180?! (The picture
gives a hint, but we will spell out the solution shortly.)
Next consider FIGURE2, showing three circles whose centers are collinear. Now
two (obviously congruent) curvilinear triangles are formed. What can be said about
the sum of the angle measures in this case? Answer: This time, the sum is less than
180?!
Finally, consider three circles that intersect in the patternof a generic Venn diagram,
as in FIGURE3. The boundary of the common intersection of their interiors is a convex
curvilinear triangle; the sum of its angles is greater than 180?, because the straightsided triangle with the same vertices lies inside it. What about the other six curvilinear
16
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Figure1 Threecircles througha common point
Figure2 Threecircles with collinear centers
triangles formed by these circles? Answer: As the readermight guess, although it is far
from obvious from the diagram, the sum in this case is also always greater than 180?!
Figure3 Threecircles in Venn diagramposition
The solution to the first problem admits an elementary proof. Consider triangle
ABC in FIGURE1. Note that by symmetry, the angles labeled with the letter a have
the same measure, as do those labeled by B and by y. We then see that 2(a + B + y) =
360?, hence a + B + y = 180?.
The solution to the second problem also admits a simple proof, although a more advanced geometric idea is needed. Namely, the half-plane lying above the line through
the three centers may be considered as the upper half-plane model of hyperbolic geometry. In this different sort of geometry, semicircles with centers on the boundary line
take the place of lines in Euclidean geometry, and angles between these (hyperbolic)
lines are computed using the Euclidean angles between the semicircles. A well-known
VOL.76, NO. 1, FEBRUARY
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17
resultin hyperbolicgeometrystatesthatthe sum of the angle measuresof any hyperbolic triangleis alwaysless than180?.
For the thirdproblem,we will also show that the circles are lines in a model of
geometry,this time sphericalgeometry,in whichthe anglesumof anytriangleis more
than180?.Moregenerally,we consideranyconfigurationof threecirclesthatintersect
in pairsandgive a simplecriterionfor decidinginto whichgeometrythey fall.
The generaltheorem
Three circles, provided they intersect in the way we have described,determine
threelines by theirpoints of pairwiseintersection.It may surpriseyou to learnthat
these threelines are eitherconcurrentor parallel.The positionof the point common
to these threelines, as describedin the following lemma,is the key to determining
whichgeometrywill answerthe questionaboutthe angle sum.
LEMMA.Let cl, c2, and c3 be three circles in the plane, with any two of them
intersectingin two distinctpoints.Let 112,113,and 23be the lines determinedby these
pairwiseintersections.Thenthe lines lij} are eitherconcurrentor parallel(in which
case we considerthem to be concurrentat infinity).Furthermore,
there are exactly
threepossibilitiesfor the locationof the pointof concurrency,P:
1. P lies on all three circles (FIGURE4a);
2. P lies outsideall threecircles(FIGURE
4b), possiblyat infinity;or,
3. P lies insideall threecircles(FIGURE
4c).
Finally,if P lies outsideall threecircles,butnot at infinity,the six tangentsfromP
to the threecirclesall havethe samelength.Thecirclecenteredat P withthiscommon
to all threeof the originalcircles.If P lies at infinity,
lengthas radiusis perpendicular
the line throughthe centersof all threecirclesplaysthis role.
P
a
b
c
Figure4 The three possibilitiesfor the location of point P
Proof Supposea line throughP intersectsa circlec in two pointsA and B (which
neednotbe distinct).Thepowerof P withrespectto c is definedas the signedproduct
(PA)(PB), wherePA denotesthe directeddistancefromP to A. Weinvitethe reader
to prove,usingsimilartriangles,thatthe powerdoes not dependon the line chosen.(A
proofmay be foundin CoxeterandGreitzer[2, Theorem2.11].)
The powerof P withrespectto c is positiveif P is outsidec, negativeif P is inside
and
0 if P lies on c. Viewing PA and PB as vectors,we can equivalentlydefine
c,
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the poweras the scalarproductPA ?PB; since the vectorsareparallel,the cosine of
the anglebetweenthemis ?1 accordingto whethertheypointin the sameor opposite
directions.
If two circlesintersectat pointsA andB, we deduce(by consideringline PA, say)
that a point P has the same powerwith respectto both circles if and only if it lies
<-)
on line AB. Supposefirstthat112 and 123are not paralleland let P be theirpoint of
intersection.ThenP has the samepowerwithrespectto all threecircles;hence, P lies
on 113andthe lines areconcurrent.If, on the otherhand,two of the lines areparallel,
an argumentby contradictionshowsthatall threemustbe.
If the lines intersect,then,as we havejust observed,the powerof P with respectto
all threecirclesis the same.Denotingthis commonpowerby P, we have:
Case 1. If P = 0, then P lies on all threecircles.
Case 2. If P > 0, then P lies outsideall threecircles.
Case 3. If P < 0, then P lies insideall threecircles.
Finally,if P lies outside,thenthe lengthof a tangentfrom P to anyof the circlesis
/-P. The circlewith this radiusandcenterP is orthogonalto all threecircles.1
As an interestingdigression,we note thatfor a generalpairof circles,which need
not intersect,a point has the same power with respectto both if and only if it lies
on a particularline, called theirradicalaxis, orthogonalto the segmentjoining their
centers.In the courseof computingthe equationof this line in Cartesiancoordinates,
CoxeterandGreitzer[2, Section2.2] also provethe beautifulfact that,if the equation
of a circle is put in the standardform F(x, y) = (x - a)2 + (y - b)2 - r2 = 0, then
the powerof any point (x, y) with respectto thatcircle is just F(x, y)! This resultis
not unexpected,since the poweris constanton circlesconcentricwith the given one.
We are now preparedto state the theorem.Trianglemeans a curvilineartriangle
thatis not subdividedby any arcs of the threecircles. In case one, only the triangle
thatdoes not includeP as a vertexis considered.
THEOREM.
Let cl, c2, andc3 be threecirclesin the plane,with eachpairintersecting in two distinctpoints.Let 12,113,and123be the lines determinedby thesepairwise
intersections,withcommonpointP. Thenthe sumof the anglesof anytriangleformed
by the threecirclesis determinedaccordingto the threecases of the lemma:
1. if P lies on all threecircles,the sumis equalto 180?;
2. if P lies outsideall threecircles,the sumis less than180?;and,
3. if P lies insideall threecircles,the sumis greaterthan180?.
The proof of the theoremrequiressome knowledgeof the classicalnoneuclidean
geometries,to which we now turnour attention.The readerwho is alreadyfamiliar
with these is invitedto skip aheadto the section in which the theoremis provedas
follows: For any of the possible configurations,we give a conformalmap thattakes
our threecircles to the lines of one of the threeclassical geometries;since the map
is conformal,the angle sum is preserved,andthus foundto be equalto, less than,or
greaterthan180?accordingly.
A quick tour of three geometries
A geometryis an abstractmathematicalsystem in which the undefinednotions of
point and line are assumedto behavein accordancewith certainaxioms.(Euclidean
VOL.76, NO. 1, FEBRUARY
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19
andnon-EuclideanGeometries,by MarvinGreenberg[4], presentsa lively accountof
thehistoricaldevelopmentof the classicalgeometries.EdwinMoise'stext,Elementary
Geometryfrom an Advanced Standpoint [7], is anothervery comprehensive reference.)
In practice,we visualize a geometryby workingwith a model, which is described
moreconcretely.Objectsin the modelrepresentpointsandlines in sucha way thatthe
axiomsof the systemarefulfilled.
Althoughthereis muchto say aboutthe classicalgeometries,we continueourtale
of threecirclesandfocus on the sum of anglesin a triangle.
The Cartesianplane,consistingof pairsof real numbers,is a model for Euclidean
geometry,providedwe adopt the usual concept of distance.Each pair of numbers
representsa point,while lines are the solutionsets of linearequations.Using algebra
to studylines,points,andcirclesis calledtheanalyticmethod.Thatthe anglemeasures
in a trianglesumto two rightanglescan be derivedeitherfromEuclid'saxiomsor an
analyticapproach.
The Cartesianplane and its physicalapproximations,such as tabletops,in which
parallellines remainequidistantandmeeta commontransversalline at the sameangle
on the same side, have historicallybeen describedas flat (as opposedto a sphereor
othercurvedsurface);thus, Euclideangeometryis said to be flat, or have curvature
zero.
Any surfaceon whicha distancefunction,or metric,has been definedis a modelof
some geometry.(Thesenotionsmay be extendedto higherdimensionsas well.) The
lines on this surface,called geodesics, are the curvesthatlocally realizethe shortest
distancesbetweentheirpoints.
Spherical geometry A spheresittingin R3, wherewe know how to measuredistances,inheritsa metricthatmeasuresdistancesalongthe surface.Here,the lines are
the great circles-those circles cut by planes throughthe sphere'scenter.The word
local in the definitionof geodesicis important;you mustchoosethe shortway around.
The geodesic natureof greatcircles can be understoodintuitivelyby notingthatthe
distancealong an arc is proportionalto the centralangle it subtends,andthatcentral
anglesobey a sortof triangleinequality:wherethreeplanesmeet,the sum of anytwo
of the face anglesis greaterthanthe third.Thus,a pathmadeup of very small (think
infinitesimal)arcswill be shortestif all the arcslie in the sameplanethroughthe center.A rigorousproof, as suggestedby the precedingdiscussion,requiresintegration
and other concepts from calculus. (Geometryfrom a Differential Viewpoint by John
McCleary[6] is an accessibleintroductionto the applicationof calculusto geometry.)
Of particularimportanceto us is the sum of the angles of a spherical(geodesic)
triangle.Ourcalculationswill be nicerif we measureanglesin radians,which we do
fromnow on. Someexperimentation,
whichwe invitethereaderto do, suggeststhatthe
angle sum of a sphericaltrianglealwaysexceeds 7rand decreaseswith the triangle's
area,approaching,but not reaching,7r as this area approacheszero (and approaching, but not reaching,57ras the trianglefills up the whole sphere).This observation
suggeststhatwe focus on the amountby which the angle sum of a sphericaltriangle
exceeds 7, which we call its angle excess. We now provethat a triangle'sexcess is
alwayspositiveby preciselyexaminingits relationshipwith the triangle'sarea.
Severalfacts will lead us to the correctrelationship.First, notice that the angle
excess is additive:if a triangleis subdividedinto two smallertriangles,the excesses of
the componenttrianglesaddup to the excess of the whole, as the readercan calculate.
Second,congruenttrianglesclearlyhavethe sameangleexcess. Thesearethe essential
propertiesof area:the areasof the partsof a subdividedregionaddup to the areaof
the whole, andcongruentregionshaveequalarea!Moreover,a sphere,like the plane,
is homogeneous:anytrianglecan be rigidlymoved,by rotationsandreflectionsof the
MAGAZINE
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20
sphere, to any other position without changing distances or angles (or area). Together,
these facts suggest that, for a spherical triangle, angle excess is a constant multiple of
area.
/\
b
a
Figure5 The sector swept out by a spherical angle
To prove that angle excess is proportional to area, observe that the angle between
two great circles is proportionalto the area of the sector they bound. (Note the essential
role of homogeneity here!) On a sphere of radius R, the area of the sector swept out by
an angle a is (a/7r)(47rR2) = 4R2a. (See FIGURE5a. For simplicity, we have used
the same symbol to representboth the angle and its measure.) Let A representthe area
of a triangle with angles a, ,B,and y. The sectors swept out by a, ,B, and y cover the
sphere redundantly;the triangle and its antipodal image are each covered three times,
while the remainder of the sphere is covered exactly once (FIGURE5b). Thus, with a
little algebra, we obtain the formula
A
=
a +b+y-7
R2'
In particular,on a sphere of unit radius, the angle excess is exactly equal to the area.
Hyperbolic geometry A sphere is said to have constant positie curvature, and it is
easy to imagine that a small sphere has large curvature,while a large sphere has small
curvature.What would it mean for a surface to have constant negative curvature?This
question will lead us to a model where the sum of the angles in a triangle is always
less than
tanwo right angles.
To see how we might describe a surface of negative curvature, e start with some
formal manipulations on the equation x2 + y2 + z2 = R2. As the constant R2 increases
toward +oo, the curvatureof the sphere described by the equation diminishes toward
zero. We may view the Euclidean plane as a sphere of infinite radius. What happens
when the constant term passes infinity and reappearson the negative end of the number
line?
We will call a surface satisfying an equation of the form x2 + y2 + z2 = R2 =
(i R)2, which has been aptly described as a "sphere of imaginary radius," a pseudosphere. Of course, the equation will have no solutions unless we expand the domain of
our variables from the real to the complex numbers, and its "radius"makes no sense
unless we expand our notion of distance to include imaginary numbers.
Distance and angles in Cartesian space are measured via the dot product, familiar
from multivariablecalculus. This product is an example of a symmetric, bilinear form,
a type of operationthat plays a large role in many branches of mathematics. In addition,
2003
VOL.76, NO. 1, FEBRUARY
21
the dot productis positivedefinite:for any vectorv, v v > 0, and v v = 0 if and
only if v = 0. The lengthof a vectorv, denotedIvl,is definedas v/. v andthe angle
0 betweentwo vectorsis determinedby the formulacos(0) = v w/(Ivl wi).
If we allowthe coordinatesof ourvectorsto be anycomplexnumbers,the dotproduct remainsa symmetric,bilinearform, althoughit is no longerpositive definite.A
formwith thesepropertiesis called a pseudo-metric.It is this formthatwe choose to
measuredistancesandanglesin complexCartesianspace,C', allowingthese quantities to havecomplexvalues.
Returningto the equationx2 + y2 + z2 = -R2, we consider,to obtain a twodimensionalsurface,only those solutionsfor whichx and y are real. The remaining
variable,z, is then forced to be purelyimaginary,so we let z = it. To picturethe
pseudosphere,we map the solutionset of our equationto a surfacein R3, using the
map (x, y, it) -+ (x, y, t). This image is a two-sheeted hyperboloid, the solution set
of the equationx2 + y2 - t2 = _R2. The sheetsareanalogousto the two hemispheres
of an ordinarysphere,the origin (0, 0, 0) may be thoughtof as the center,and the
points(0,,0, R) as the NorthandSouthpoles.
In orderto measuredistancesand angles,we mustrememberthatour space is an
image of a subspaceof C3, wherethe actualdistancesandanglesaremeasuredusing
the dot product.Since a vector (x, y, t) actuallyrepresentsthe vector (x, y, it), its
actuallengthis vX2 + y2 + (it)2 = /x2 + y2 - t2. Similarly,the anglebetweentwo
vectors (xl, yi, tl) and (x2, Y2,t2) must be based on the form x1x2 + yly2 - t1t2.Some
versionof the
readersmay recognizethis as a Lorentzmetric,the three-dimensional
pseudo-metricof relativisticspace-time.(Fora readableandphysicallymotivated,but
advanced, introduction to pseudo-metrics, see Semi-Riemannian Geometry, by Barret
O'Neill [8].) The apparentdistancesandanglesin ourpicturearedistortedfromtheir
actualvalues;for example,the truelengthof everyradialvectorfromthe originto the
surfaceis the imaginarynumberi R!
To measuredistancesandangleson the pseudosphereitself, we applythe pseudometric to its tangent vectors. For any point P in IR3,the tangent space at P is the copy
of R3 consistingof all vectorsemanatingfrom P. The tangentline at P to a curve
throughP on the surfaceis a one-dimensionalsubspaceof this tangentspace. The
tangentplane to a surfaceat P is the two-dimensionalspace composedof all these
lines.
It is useful(andintuitive)to writecoordinatesandotherquantitiesrelatedto tangent
real-valuedfunction
vectorsin termsof differentialexpressions.If f is a differentiable,
on R3, studentsknowhow to computeVf, anduse it to finddirectionalderivativesby
takingdot products.Wepreferanothervocabulary:the differentialof f is the function
that takes a tangent vector Vp, at a point P, to the real number Vf(P)
?Vp. This gives
the best linearapproximation
to the changein the valueof f thatresultsfromstarting
at P andtravellingwith a displacementvp.
In particular,the Cartesiancoordinates,x, y, and t in our model may be viewed
as functionsof P E IR3,and dx, dy, and dt, are the differentialsof these functions.
If we suppressthe vectorargumentof these differentials,we can use dx, dy, anddt
as the first,second,and thirdcoordinatesof a generaltangentvectorwith respectto
a parallelcoordinatesystem based at P. (In FIGURE6a, the linearapproximations
dx, dy, and dt happento be exact, since orthogonalprojectiononto an axis is a
linearfunction.)Thus we may convenientlywrite the (Lorentz)length of a tangent
vectoras the differentialexpressionvdx2 + dy2 - dt2. This gives us the metricwe
need in orderto talk aboutthe lines of this geometry,which are the geodesicsof this
metric.
A beautifulfactis thatthis differentialexpressionis realandpositive,whenapplied
to vectors tangentto the pseudosphere.Since the length of a path is computedby
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MATHEMATICS
integratingthe lengths of its tangent vectors, paths between points on the pseudosphere
have real, positive length, and it makes sense to talk about geodesics as paths of locally
minimal length.
To see that dx2 + dy2 - dt2 > 0, it helps to use cylindrical coordinates. As an orthonormal basis for the tangent space to R3 at the point with cylindrical coordinates
(r, 0, t), take a radial unit vector, a circumferentialunit vector in the counter-clockwise
direction, and a vertical unit vector; with respect to this basis, the coordinates of a tangent vector are (dr, rdO, dt). (The factor of r in the circumferential coordinate results
from the fact that changing the central angle by a small amount changes the distance
travelled by r times that amount. See FIGURE6b.) The expression dx2 + dy2, the
squared length of the vector's horizontal component, is equal to dr2 + r2 d02. For
a vector tangent to the hyperboloid, dt2 < dr2, since the radial slope of the hyperboloid is less than that of the cone to which it is asymptotic. (Algebraically, it follows
from the equation r2 - t2 = -R2 that r2 < t2, and by differentiating we calculate that
dt2 = (r2/t2) dr2.) Thus dx2 + dy2 - dt2 = r2 d02 + (dr2 - dt2) > 0.
i," rdQO
x
,
y
Y
,t
..r:
(0,0, t)
t
(0,0,0)
*(0,0,0)
a
b
Figure6 Comparisonof two bases for the tangent space at P
The geodesics turn out to be the intersections of the pseudosphere with planes
through the origin, analogous to those on an ordinary sphere, although it is harder to
see why this is so, and we won't prove it here. The pseudosphere is also homogeneous,
and the angle sum of a triangle satisfies a completely analogous formula:
a +
+ y - r= _
A
R2'
It follows that this sum is always less than ir. Of course, there is a different pseudosphere for each value of R. By analogy with the sphere, one might guess that a small
value of R yields a pseudosphere of large negative curvature, while a large value of
R gives a pseudosphere that is so little curved as to be nearly flat. However great or
small the curvature,the sum of the angles in a triangle is still less than two right angles.
(A book by B. A. Dubrovin, A. T. Fomenko, and S. P. Novikov [3] gives a thorough
discussion of both the sphere and pseudosphere.)
Each sheet of the pseudosphere is a model of a geometric object called a hyperbolic
plane. It would be nice to be able to see this plane looking more like a plane, without
having to work with an object as complicated as the Lorentz metric. There is a planar
map of the pseudosphere that shows angles accurately, obtained by a method called
stereographic projection. A similar map is also available for the ordinary sphere. Any
VOL. 76, NO. 1, FEBRUARY
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23
mapthatpreservesangle measureis called conformal.Since our tale of threecircles
involvesanglemeasure,conformalmapswill be a powerfultool.
Stereographicprojection
The sphere To obtaina conformalmappingof the ordinarysphereonto the plane,
projectfromanypointof the sphereonto the planetangentto its antipodalpoint.Any
such map, or its inversemap from the plane to the sphere,is called a stereographic
projection.Everycircleon the sphere(notjust the greatcircles)projectsstereographically to a circle or line (whichmay be thoughtof as a circle throughoo) in the plane,
and it maps to a line if and only if it passes throughthe point of projection(which
mapsto oo). Conversely,every line or circle in the plane is the image of a circle on
the sphere.(See FIGURE7, and note therethatthe centerof a circle on the plane is
not, in general,the projectionof the centerof the correspondingcircle on the sphere,
butratherthe apexof the cone tangentto it.) Thesevaluablepropertiesmaybe proven
by elementaryarguments.(Theargumentsareoutlinedandnicely illustratedin Hilbert
and Cohn-Vossen's classic book, Geometry and the Imagination [5, pp. 248-251].)
Figure7 Stereographicprojectionpreservesangles and takes circles to circles or lines
To see that stereographicprojectionfrom the sphereto the plane is conformal,
considerFIGURE8a, which shows the cross-sectionof the spherecut off by a plane
throughthe pointof projection,N, its antipodalpoint, S, and anotherpoint P on the
sphere;Jr(P) is the projectedimage of P, and A is the pointwherethe cross-section
of the planetangentto the sphereat P intersectsthe tangentplaneat S.
An angle with vertexat P in the plane tangentto the sphereis cut by two planes
intersectingalongline NP. Since L7r(P)PA _ LP7r(P)A,theseplanescut the same
angleat 7r(P) in the horizontalplanethroughS, by symmetry.To see this, imaginean
angleformedby two stiff planesof paper;if you snip at the sameangleto the spinein
eitherdirection,the angleyou makeis the same.(Theprojectedimage of the angleis
its reflectionin the planethroughA thatis perpendicular
to NPn(P).)
FIGURE8b illustratesthe local behaviorof stereographicprojection.If Q is a
similarto
nearbypointon the tangentplanethroughP, then AN QP is approximately
ANnr(P)r(Q). Therefore,stereographicprojectionis linearlyapproximatedat each
point P by a dilation(uniformscaling).The dilationfactorvarieswith the latitudeof
P, increasingwithoutboundas P approachesN, with a minimumvalueof 1 at S.
24
MATHEMATICS
MAGAZINE
N
N
-(Q)
S
A
c(P)
7n(P)
b
a
Figure8 Stereographicprojection is conformaland locally approximatedby a dilation
The pseudosphere To obtain a conformal map of the pseudosphere onto the plane,
project each point of the hyperboloid model onto the horizontal plane through the
South pole, (0, 0, -R) (that is, the plane t = -R), via the line connecting it to the
North pole, (0, 0, R), as in FIGURE9. This projection maps the southern hemisphere
onto the interior of the disk of radius 2R centered at the origin, while the northern
hemisphere, except for the North pole, goes onto the disk's exterior. The disk's interior
is known as a Poincare Disk model of a hyperbolic plane, and its boundaryis called the
circle at infinity.It can be shown that each geodesic maps to a circle or line orthogonal
to the circle at infinity (with the points of intersection removed); the geodesics through
the poles go to lines. (See B. A. Dubrovin, A. T. Fomenko, and S. P. Novikov [3] for a
proof.)
.
%
, :,
/1
\
* '
/
*
. -'\
Circle at infinity /I
:
</
/ I -- - - - "~Poi
DiskModel
Poincare DiskModel
Figure9
Projectionof the hyperboloidmodel of the pseudosphere onto a plane
The conformality of the projection of the hyperboloid model cannot be demonstratedby elementary geometric argumentsbecause the angles on the hyperboloid are
measured using a different bilinear form than the one used to measure angles in the
plane. So to prove that angle measure is preserved, we must resort to calculation to
compare the angles between tangent vectors to curves, before and after projection. The
readerwho wishes to just believe us and avoid the technicalities involved may skip the
VOL
76 NO.
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76,
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25
calculation below without any loss of continuity. For those who wish to venture in, the
proof provides a nice application of calculus techniques to geometry.
Proof that projection to the Poincare disk model is conformal We restrict our
attention to the southern hemisphere; the proof for the northernhemisphere is similar.
Let Jr denote the projection map, extended to the region t < R, and viewed as a map
onto R12by ignoring the last coordinate (-R) in the image. Suppose curves a and B
intersect at P = (x, y, t). The derivative of 7r at P, Dr (P), carries vectors tangent to
a and P at P to vectors tangent to their projected images at r (P) (by the chain rule).
Let (vp, Wp)L denote the Lorentz product of tangent vectors vp and wp at P, that is,
((xl, Yl, tl), (x2, Y2, t2))L = X1X2+ Y1Y2 tt2.
We sketch a proof that, if vp and wp are tangent to the hyperboloid, then Dr (P)(vp) D7(P)(wp) = [4R2/(R - t)2](vp, Wp)L. In other words, the dot product of the images is just scaled by a constant factor from the Lorentz product of the preimages. It
follows that the angle, 0, between vp and Wp, which is determined by
(Vp, Wp)L
=
cOos
1
(Vp,
Vp)L
1
(Wp,
is equal to the angle, if, between Dr(P)(vp)
mined by
cos 41 =
Dn(P)(vp)
[D7r(P)(vp) . D7r(P)(vp)]
Wp)
and DZr(P)(wp), which is deter-
. D7r(P)(wp))
1
[DJr(P)(Wp) . D7T(P)(Wp)]2
since the scaling factor cancels out. In contrast to the sphere, the scaling factor decreases as P moves away from S (t decreases), with a maximumvalue of 1 at S.
N=(O,O,R)
2R
g(P). :---
/
t=-R
R-t
P=(x,y,t)/
r
Figure10
Effectof the projectionn on the radialcoordinate
Referring to similar triangles, as in FIGURE10, we see that the image under 7T of
the point P with cylindrical coordinates (r, 0, t) is the point whose polar coordinates
are (2R, 0). The derivative at P of the conversion from Cartesian to cylindrical coordinates takes (dr, rdO, dt) to (dr, dO, dt). If 0 : R3 -> IR2is the map (r, 0, t) (2R 0), its derivative (or Jacobian, if you prefer) at (r, 0, t) is
2R
Do (r, 0, t) =(Rt
'0
o
1
2Rr\
(R-t)2
0'
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MAGAZINE
MATHEMATICS
Applying D (r, 0,t) to (dr, dO, dt), followed by the derivative at 0(r,0, t) =
(2R, 0) of the conversion from polar back to Cartesian coordinates, we calculate (thanks to the ever-valuable chain rule) that D7r(P)(dr, rdO, dt) = 2R (dr +
t, rd0).
For any point P on the hyperboloid, r2 - t2 = -R2, and for any vector tangent to the hyperboloid at P, dt = (r/t)dr. After some simplification using these
substitutions, we find that D7r(P)(dr, rd0, (rt)dr)=
2R(-(R/t)dr, rd). The
reader can now check that for two tangent vectors at P, D7r(P)(vp) ?Dr(P)(wp) =
[4R2/(R - t)2](Vp, Wp)L, as stated.
Sequences of stereographic projections Stereographic projection is an indispensable tool for transforming geometric models without changing the angles between
geodesics. In particular,it provides us with anothercelebrated model of the hyperbolic
plane called the Poincare half-plane. To obtain it, first project the Poincare disk of radius two from a horizontal plane onto the southern hemisphere of the unit sphere (and
its complement, including the point at oo, onto the other hemisphere). Then project
onto any vertical plane tangent to the equator.The image of the circle at infinity under
this sequence of projections is called the line at infinity. Each half-plane is an image
of a hyperbolic plane, as in FIGURE11.
Figure 11 Conformal transformationsbetween the Poincare disk, hemisphere, and
Poincar6half-plane models of hyperbolicgeometry
By a sequence of two stereographicprojections from different points, we also obtain
a conformal map of the Euclidean plane in which the images of the geodesics are either
lines or circles. See FIGURE12.
In summary, we now have maps of the plane, sphere, and pseudosphere in which
the geodesics are represented by lines and circles. Just as a map of the earth must be
distorted in order to print it on the flat pages of an atlas, our maps distort the true
distances between points in the geometric objects they depict. The art of map-making
revolves around choosing a projection whose particulartype of distortion allows the
map to be useful for its intended purpose. For example, the famous Mercatorprojection
of the earth's surface is useful to navigators because it accurately depicts the compass
VOL.76,
76, NO. 1, FEBRUARY
FEBRUARY2003
2003
27
A
I
I
I aI
I
, I:I I..".
II,
$I,I
t.r
Figure12
Euclideangeometry on the sphere
bearing between any two points. (McCleary [6, chapter 8biS] gives a nice discussion of
mapprojections.)
Since we are interested in the properties of angle measure, we have chosen to view
our circles through conformal maps, which render the true angles between smooth
curves. The question remains: Given a trio of circles that intersect in pairs, how can
we interpret the configuration as a geodesics on a map of one of the geometries we
have studied? Is such an interpretationeven possible?
Choosinga geometry:a proofof the theorem
Recall that P is the intersection of the three lines determined by the pairwise intersections of the three circles. We will show that, depending on the location of P relative
to the circles, there is a conformal map that takes the circles to lines of a standard
geometric model-a model for which we know a great deal about angle sums.
Case 1: P lies on all three circles. In this case consider a sphere tangent to the plane
with South pole at P, and a second plane tangent to this sphere at the North pole, as
in FIGURE12. The images of our three circles on the sphere are circles through the
South pole. If we project again, this time from the South pole of the sphere onto the
second plane, these circles are taken to straightlines. Thus, cl, c2, and c3 are geodesics
in a conformal planar model of Euclidean geometry, and the sum of the angles of the
triangle they form is 180?.
Case 2: P lies outside all three circles. We already studied the special case where
the three lines determined by pairwise intersections of the circles are parallel; this
happens when the centers of all three circles are collinear, and that line was seen to be
the boundary line of a hyperbolic plane. Having dispensed with that case, we assume
that the lines are concurrent at a point P lying outside of all three circles. In this case,
the Lemma presents us with a circle, d, that is orthogonal to all three circles. Thus,
each arc of the circles lying inside or outside d is a geodesic in a model of a hyperbolic
plane, and thus the sum of the angles of any triangle they form is less than 180?.
Case 3: P lies inside all three circles. Consider the family of spheres (of varying
size) tangent to the plane with South pole at P. Consider how the area of the stereo-
28
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MATHEMATICS
graphicprojectionof the disk boundedby cl onto each of these spherescomparesto
the surfaceareaof the sphere:if the sphereis very small, the projectedregion will
haveareamorethanhalf thatof the sphere(in fact, it will includethe entireSouthern
hemisphere);if the sphereis very large,the projectedareawill be less thanhalf the
surface.This is just a matterof having cl lie inside or outside the preimageof the
equatoron the plane.(See FIGURE13.) Allowingthe sphereto varycontinuously,we
see thatthereis a uniquesphere,S, suchthatthe projectedareaon S is exactlyhalfthe
surfacearea.(Foran alternativeargument,see the thirdremarkbelow.)Consequently,
the imageon S of cl is a greatcircle.
Figure13 The family of spheres tangent to the plane at P
Weclaimthatthe imageson S of c2 andC3arealso greatcircles.It sufficesto prove
this for c2, as the same argumentworksfor C3.To do so, observethat the image of
112understereographicprojectionis a meridian,thatis, a greatcirclepassingthrough
the Northand Southpoles of the sphere.Since the image of cl is also a greatcircle,
the images of the pointsof intersectionof cl with 112are antipodal.Since the points
of cl n 112 also lie on c2, it follows thatthe image of c2 is a greatcircle. FIGURE14
illustratesthis.
Figure14 The images of the circles on the sphere S are great circles
VOL.76, NO. 1, FEBRUARY
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29
We havethusshownthatthe imageson S of the threecirclesaregreatcircles;that
is, theyaresphericalgeodesics.Thusc1, c2, andc3 aregeodesicsin a conformalplanar
modelof sphericalgeometry,andthe sumof the anglesof anytriangleformedby them
is greater than 180?.1
REMARK.ThatEuclideangeometryoccursonly in the instancethatpoint P lies
exactly on the circles is an illustrationof the fact that our familiarflat geometryis
just a single point in the spectrumof geometries,from the sphereof largeradiusR,
to the flatplane,where R is effectivelyinfinite,to the pseudosphere,whose radiusis
imaginaryin the modelwe haveshown.
REMARK.Using the sequenceof stereographicprojectionsdescribedearlier,we
may transformthe generalpicturefor Case 2 into the half-planepicture.
REMARK.Althoughwe like the continuityargumentin Case 3 of the Theorem,it
may be avoidedas follows. Let s = /-'P(P). The sphereS in the aboveproofis the
one thathasradiuss/2. To showthis,consider,for eachof the circlesc1, c2, andC3,the
chordthroughP thatis perpendicular
to theradiusthroughP. Eachof thesechordshas
length2s andmidpointP. Let e be the circle centeredat P with radiuss. This circle
to the equatorof S, and cl, c2, and C3,whichintersecte at
projectsstereographically
diametricallyoppositepoints,also projectto greatcircles.This alternativeargument
emphasizesthe parallelsbetweenCases2 and3.
REMARK. The theorem remains true if circle is understood in its more general
sense to includestraightlines as well.
We invite the readerto extendthe theoremto the limitingcases in which two or
moreof the circlesare tangent.The possibilitiesareillustratedin FIGURE15. Case I
I
Figure15 The possibilitiesfor two circles to be tangent
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of the figureleadsto Euclideangeometry,whenviewedwiththe rightmap:Onevertex
of thetriangleis thrownoutto infinityby the mapthattakesthe circlesto straightlines;
hence,two sides of the trianglebecomeparallellines. The possibilitiesin Case II are
hyperbolic,with one or more verticesof the trianglelying on the circle (or line) at
infinity.Whatis the measureof an angle whose vertexlies on the circle at infinity?
Note thatnone of the limitingcases is spherical.
Changingyour point of view: a transformational
approach
Weareindebtedto KeithBurs for pointingoutthefollowingveryelegantformulation
andproofof the Theorem.His proofuses the groupof Mobiustransformations
of the
extendedcomplexplane(thatis, the plane,regardedas the field of complexnumbers,
areinvertible,bicontinuous,and
togetherwith a pointat oo). Mobiustransformations
and
take
conformal,
generalizedcircles (with lines regardedas circlesthroughoo) to
circles. Moreover,given any two orderedsets consistingof threepoints each, there
is a (unique)Mobiustransformation
takingeach pointof one set to the corresponding
of
the
other.
(Theexcellent,veryreadabletexton geometryfroma Kleinianpoint
point
of view by Brannan,Esplen,and Graycontainsa thoroughdiscussionof the Mobius
groupandits geometricproperties[1, Chapter5].)
In particular,and this is all we will need, thereis a Mobiustransformation
taking
with the requiredproperties
any given pointto oo. As we have seen, a transformation
can be constructedby composinga pairof stereographicprojections.
Considera curvilineartriangleABC formedby the three circles cl, c2, and c3.
Withoutloss of generality,assumethat A lies on cl and c2, and let A' be the other
pointof intersectionof cl andc2. We thenhavethreepossibilitiesfor the positionsof
A and A' with respectto the thirdcircle, C3,which correspondto the threecases of
the LemmaandTheorem:A' lies on C3;A' lies on the oppositeside of c3 from A; or
A' lies on the same side of C3as A. By simplychangingourpointof view, placingA'
at oc, we can discernby inspectionhow the angle sum of triangleABC comparesto
180?.Althoughthis approachdoes not demonstratethe underlyingglobalgeometryin
whichthe circlesaregeodesics,it does havethe advantageof beingdelightfullysimple
anddirect.
THEOREM.(ALTERNATIVE
FORMULATION)Let cl, c2, and c3 be three circles in
the plane, with each pairintersectingin two distinctpoints.Then exactly one of the
followingthreeconditionsholds anddeterminesthe sum of the anglesof anytriangle
formedby the threecircles:
1. the intersectionof each pairof circlescontainsa pointof the thirdcircle,in which
case the sumof the anglesis 180?;
2. the intersectionof eachpairof circleslies entirelyinsideor outsidethe thirdcircle,
in whichcase the sum of the anglesis less than180?;or,
3. the intersectionof each pairof circlescontainsone pointinside andone pointoutside the thirdcircle,in whichcase the sum of the anglesis greaterthan180?.
Proof. Let A, B, and C be the verticesof a triangleformedby cl, C2,andC3,with
{A, A'} = cl n c2, as in the paragraphprecedingthe statementof the theorem.The
angle sum of the triangledoes not dependon which pairof circles we consider,so it
sufficesto showthatthis sumis determinedby the positionsof A andA' relativeto c3.
gu,thattakes A' to oo. Underthis transformation,
Apply a M6biustransformation,
the images of cl and c2, which pass throughA', are lines. If A' lies on C3,then the
imageof C3is also a line, hence the angle sum of the image of triangleABC is 180?.
31
31~~~~~~~~~~~
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VOL.76, NO. 1, FEBRUARY
Case1
A
BL(B)
a(C)
Case 2
'(AC
Case 3
Figure16 The effect of a Mobius transformation,IL,takingA' to oo
If A and A' lie on the same side of C3,then the image of c3 is a circle with the image of
A outside it, since oo is outside; hence the angle sum of the image of triangle ABC is
less than 180?. If A and A' lie on opposite sides of C3,then the image of c3 is a circle
with the image of A inside it; hence the angle sum of the image of triangle ABC is
O
greater than 180?. The three possibilities are illustrated in FIGURE16.
Conclusion
Each of the classical geometries, Euclidean, spherical, and hyperbolic, has a variety
of conformal representationson the plane, obtained by stereographic projection. Rec-
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MATHEMATICS
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enabledus to classify the generalizedtriangleswhose
ognizingthese representations
sides areeithersegmentsor circulararcsas belongingto one of threetypes of geometries,allowingus to makethe correctconclusionaboutthe sum of the angles.Finally,
of the extendedplane,we were ableto
by using a groupof conformaltransformations
refineoursolutionto be verysimple,althoughperhapsless informative.Oursuccessis
just one exampleof the valueof studyingnoneuclideangeometriesandtransformation
groups.
Acknowledgments. In addition to Keith Bums, we would like to thank Duane Broline, Leo Comerford,Bob
Foote, Yuri Ionin, and Rosemary Schmalz for their helpful comments. We would also like to thank an early
refereefor pointingout the argumentin the thirdremark.
REFERENCES
1. D. Brannan,M. Esplen, and J. Gray,Geometry,CambridgeUniversityPress, Cambridge,1999.
2. H. S. M. Coxeterand S. L. Greitzer,GeometryRevisited,RandomHouse, New York, 1967.
3. B. A. Dubrovin,A. T. Fomenko,and S. P. Novikov, Modem Geometry-Methods and Applications.PartI. The
geometryof surfaces, transformationgroups,andfields, GraduateTextsin Mathematics# 93, Springer-Verlag,
New York, 1992.
4. M. J. Greenberg,Euclideanand non-EuclideanGeometries,Freeman,New York, 1993.
5. D. Hilbertand S. Cohn-Vossen,Geometryand the Imagination,Chelsea, New York, 1952.
6. J. McCleary,Geometryfrom a DifferentiableViewpoint,CambridgeUniversityPress, Cambridge,1994.
7. E. Moise, ElementaryGeometryfrom an AdvancedStandpoint,Addison-Wesley,Reading, 1990.
8. B. O'Neill, Semi-RiemannianGeometry,withApplicationsto Relativity,Academic Press, New York, 1983.
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GLENN APPLEBY
SANTA CLARA UNIVERSITY
SANTACLARA,CA 95053