Advanced book on Mathematics Olympiad

Geometry and Trigonometry 645

ξ=

dx

ds

,η=

dy

ds

,ζ=

dz

ds

.

The fact that the curve is closed simply implies that

∫L

0

ξds=

∫L

0

ηds=

∫L

0

ζds= 0.

Pick an arbitrary great circle of the unit sphere, lying in some planeαx+βy+γz=0.
To show that the spherical image of the curve intersects the circle, it suffices to show that
it intersects the plane. We compute

∫L

0

(αξ+βη+γ ζ )ds= 0 ,

which implies that the continuous functionαξ+βη+γζvanishes at least once (in fact,
at least twice since it takes the same value at the endpoints of the interval). The equality

αξ(s)+βη(x)+γ ζ (s)= 0

is precisely the condition that(ξ(s), η(x), ζ (s))is in the plane. The problem is solved.

Remark.The spherical image of a curve was introduced by Gauss.
(K. Löwner)

644.We use Löwner’s theorem, which was the subject of the previous problem. The total
curvature is the length of the spherical image of the curve. In view of Löwner’s theorem,
it suffices to show that a curveγ(t)that intersects every great circle of the unit sphere
has length at least 2π.
For eacht, letHtbe the hemisphere centered atγ(t). The fact that the curve intersects
every great circle implies that the union of all theHt’s is the entire sphere. We prove the
conclusion under this hypothesis. Let us analyze how the covered area adds up as we
travel along the curve. Looking at Figure 87, we see that as we add to a hemisphereHt 0
the hemisphereHt 1 , the covered surface increases by the portion of the sphere contained
within the dihedral angle formed by two planes. The area of such a “wedge’’ is directly
proportional to the length of the arc of the great circle passing throughγ(t 0 )andγ(t 1 ).
When the arc is the whole great circle the area is 4π, so in general, the area is numerically
equal to twice the length of the arc. This means that as we move along the curve fromtto
t+ t, the covered area increases by at most 2‖γ′(t)‖. So after we have traveled along
the entire curve, the covered area has increased by at most 2

∫

C‖γ

′(t)‖dt(Cdenotes the

curve). For the whole sphere, we should have 2

∫

C‖γ

′(t)‖dt≥ 4 π. This implies that the

length of the spherical image, which is equal to

∫

C‖γ

′(t)‖dt, is at least 2π, as desired.