Download Girard`s Theorem: Triangles and

Rewley House June 2013
Tau vs. Pi
Girard’s Theorem: Triangles and
τ
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The angles in a triangle sum to 180◦ unless the triangle is drawn on the surface of a sphere. In this
case the angle sum depends on the spherical radius. We can examine this difference in the context
of the theorem of Albert Girard, published in a treatise in 1626 (which also saw the first use of the
abbreviations sin and cos):
Girard’s Theorem A spherical triangle, drawn
! on the surface of a sphere of radius r and having
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angles A, B and C, has area r2 A + B + C − τ .
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Of course, this is usually stated with a π in place of τ/2; and this is an instance where some will
declare the π version of the formula to be much more elegant and ‘natural’. But we might instead say
“How suggestive that τ/2 is! What result of circle or spherical geometry does that come from?” And
this is a natural and elegant question to ask!
The proof of Girard’s theorem is rather easy once we establish what we mean by ‘spherical triangle’
and how it relates to ‘lunes’.
A spherical triangle is created by the pairwise
intersections of three great circles. A great
circle is the circle around the surface of the
sphere created by slicing through it with a
plane through the centre of the sphere. We
can imagine the sphere sinking into the sea at
an angle: the great circle is the water mark at
the point where the centre touches the surface
of the sea (as illustrated on the right).
Two great circles, provided they are not the same one, intersect at two points on the surface of the
sphere. These points are antipodal and they describe a crescent on the surface which is called a lune.
It stretches exactly half-way around the sphere, from pole to pole (not necessarily the North and South
poles).
A third great circle will cut this crescent into two spherical triangles. In the figure below, at (a), the
triangle ABC has been created by the intersection of a great circle, call it QAB , through A and B,
another, QCB , through C and B, and a third, QAC , through A and C. The circles QAB and QCB form
the lune shown in the figure at (b), and QAC cuts this into the red triangle ABC and the blue ‘bottom’
triangle.
(a)
(b)
(c)
If we follow QAB and QCB round the ‘back’ of the sphere their intersection creates a second lune
identical in shape and area to the first. QAC cuts this lune into two triangles as well and these triangles
are congruent to the first pair but located antipodally (see (c) in the figure).
Now we have the ingredients for Girard’s theorem.
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If the sphere has radius r then its surface area is 2τr2 . If two great circles meet in a lunar angle θ,
0 < θ ≤ τ, then the proportion of surface area which is occupied by the lune they create is θ/τ. So we
have
θ
area of lune with lunar angle θ = × 2τr2 = 2r2 θ.
τ
(We have a situation similar to area of circle sector subtended by angle θ = 21 r2 θ. The special case
θ = τ gives the area of a circle ‘naturally’ as 12 τr2 . In the same way, the special case of a lunar angle
of τ gives a complete spherical surface: 2τr2 .)
Now we extend our previous lunar dissection: our three intersecting great circles give us three antipodal pairs of lunes, all passing through, and duplicating, triangle ABC. In the figure below our original
pair of lunes is shown at (a), with the lunar angle being the angle B. At (b), the lunar angle is taken
to be C and the original triangle ABC is now paired with one (green) stretching over the top of the
sphere. And at (c), the lunar angle is A and the second (yellow) triangle extends to the right.
(a)
(b)
(c)
Denote by LA , LB and LC the areas of the three lunes with angles A, B and C, respectively. Recall that
these areas each have an antipodal duplicate. Denote by T the area of triangle ABC; this also has its
antipodal duplicate.
Add up all the duplicate pairs of lunar areas: you get the whole sphere but with T counted three times
and its antipodal duplicate likewise counted three times:
2LA + 2LB + 2LC = 2r2 τ + 4T
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so T =
LA + LB + LC − r2 τ
2
1 2
2r A + 2r2 B + 2r2C − r2 τ
=
2
!
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2
= r A + B + C − τ , which is Girard’s Theorem.
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It seems fitting that the factor of 1/2 remains in the formula: acknowledging its heritage in the duplication which has been the key to the whole proof.
And if this is an elegant version of Girard’s Theorem then so is the following corollary: rewrite the
theorem as
T
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A + B + C = τ + 2.
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r
Let the radius of our sphere go to infinity while A, B and C remain fixed. Then we get the result: sum
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of angles of triangle in the plane = τ, the 1/2 acknowledging a heritage in spherical geometry.
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Acknowledgement:
My account of Girard’s Theorem follows closely that of John C. Polking which you can find, with some very
nice animations of lunar decompositions (but with not a τ in sight!) at his web site math.rice.edu/ polking/.
Robin Whitty, May 2013
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