Lecture 3
Regular polygons in S 2
A polygon in S 2 is said to be ‘equilateral ’ (resp. ‘equiangular ’) if all its sides
have same length (resp. if all its angles are equal).
Definition : A polygon is said to be ‘regular ’ if it is convex, equilateral and
equiangular. A proper regular polygon of n sides is called a ‘regular n-gon’.
Construction of regular polygons in S 2 : Fix 0 < r < π2 and n ≥ 3. Let C
denote the circle which is the boundary of the disc B((0, 0, 1) , r) contained in S 2 .
Then C - as a circle in a plane - has c := (0, 0, cos r) as its center, and radius sin r.
Let P1 , . . . , Pn be n points in C which occur clockwise such that ]{Pi − c , Pi+1 − c} =
2π
∀ i = 1, . . . , n. Put Pn+1 := P1 . Let ℘n,r denote the convex polygon with
n
P1 , . . . , Pn as its vertices. By construction, the rotation, ρ 2π about the z-axis is a
n
symmetry of ℘n,r . So, ℘n,r is an equilateral, equiangular n-gon. Any two convex
polygons constructed as above are congruent to each other for a fixed n ≥ 3 and fixed
0 < r < π2 .
Let a be the length of a side of ℘n,r . Then
a = a(n, r) = arccos < P1 , P2 >
= arccos{< P10 , P20 > + cos2 r}
where P10 , P20 are orthogonal projections of P1 , P2 respectively on the xy-plane.
∴ a = arccos{sin2 r cos
2π
+ cos2 r}
n
i.e.,
n
πo
a = a(n, r) = arccos 1 − 2 sin2 r sin2
n
Now we compute the angle θ = θ(n, r) of ℘n,r . Recall that n ≥ 3.
Put u :=
P3 − < P3 , P2 > P2
P1 − < P1 , P2 > P2
and v :=
,
k P1 − < P1 , P2 > P2 k
k P3 − < P3 , P2 > P2 k
Then
P1 = cos a P2 + sin a u ,
P3 = cos a P2 + sin a v &
1
(1)
θ = θ(n, r) = ]{u, v}.
∴ < P1 , P3 >= cos2 a + sin2 a cos θ.
But < P1 , P3 >= 1 − 2 sin2 r sin2
2π
n
.
Thus
− cos2 a
1 − 2 sin2 r sin2 2π
n
cos θ =
sin2 a
2 sin2 r sin2 2π
n
= 1−
sin2 a
2 sin2 r sin2 ( 2π
n
= 1−
4 sin2 r sin2 πn 1 − sin2 r sin2
2 cos2 πn
.
= 1−
1 − sin2 r sin2 πn
π
n
Therefore,
)
(
2 cos2 πn
θ = θ(n, r) = arccos 1 −
1 − sin2 r sin2
π
n
(2)
Remarks :
1. When n = 2 and r = π2 , the digons De1 ,α (α ∈ (0, π)) give a family of regular
polygons which are not congruent to each other.
2. When 0 < r <
π
2
and n = 2, one gets degenerate polygons.
Theorem 4.1 Any regular n-gon in S 2 is congruent to ℘n,r for unique r ∈ (0, π2 ).
Proof : Let ℘0 be any regular n-gon in S 2 . As ℘0 is a proper polygon, n ≥ 3 holds.
Let a0 be the length of a side of ℘0 . By Lemma 3.3, na0 < 2π. i.e., a0 ∈ (0, 2π
).
n
Consider the function f : (0, π2 ) → [0, π] defined by f (r) := arccos 1 − 2 sin2 r sin2
Then f (r) > 0 ∀ 0 < r <
π
2
and f (r) −→
2π
n
as r % π2 .
Also,
2 sin(2r) sin2
f (r) =
sin(f (r))
0
2
π
n
> 0.
π
n
.
) is a homeomorphism.
∴ f : (0, π2 ) → (0, 2π
n
), ∃ ! r ∈ (0, π2 ) such that a0 = f (r). Thus a0 = a(n, r) for unique
Since a0 ∈ (0, 2π
n
r ∈ (0, π2 ).
Now we prove that ℘0 is congruent to ℘n,r . Let P10 , . . . , Pn0 (resp. P1 , . . . , Pn ) be
the vertices of ℘0 (resp. of ℘) which occur in a cyclic order. Let θ 0 (resp. θ) be the
angle of ℘0 (resp. of ℘) at its vertices. If θ 0 < θ (resp. > θ), then by the Lemma
of Cauchy (Lemma 3.4), d(P10 , Pn0 ) < d(P1 , Pn ) ( resp. > d(P1 , Pn ) ) which contradicts
that a0 = a(n, r). So, θ 0 = θ. Applying Cauchy’s lemma again to the convex polygons [P1 , P2 . . . . , Pj ] & [P10 , P20 . . . . , Pj0 ], we get d(P10 , Pj0 ) = d(P1 , Pj ) ∀ j = 2, . . . , n.
Similarly, d(Pi0 , Pj0 ) = d(Pi , Pj ) ∀ i, j ∈ {1, . . . , n}. By Theorem 1.3, there exists an
isometry ϕ of S 2 such that ϕ(℘0 ) = ℘n,r .
Proposition 4.2 Let ℘ be a regular n-gon in S 2 having side a, angle θ and area A.
Then ∃ ! r ∈ (0, π2 ) such that ℘ is inscribed in a circle of radius r. Further, equations
sin a2
(i) r = r(n, a) = arcsin
,
sin πn

q
sin2 2θ − cos2 πn
,
(ii) r = r(n, θ) = arcsin 
sin 2θ sin πn
q
cos2
(iii) r = r(n, A) = arcsin 
cos
2π−A
2n
2π−A
2n
(3)
(4)
− cos2 πn
sin πn


(5)
hold, and any regular n-gon is determined (uniquely upto congruence) by any one of
three : a ∈ (0, π), θ ∈ (0, π), A ∈ (0, 2π).
Proof : By the Theorem 4.1, there exists unique r ∈ (0, π2 ) such that ℘ is congruent
to ℘n,r .
(i)
π
by equation (1).
cos a = 1 − 2 sin2 r sin2 ,
n
Therefore,
sin2 a2
1 − cos a
2
sin r =
=
.
(6)
2 sin2 πn
sin2 πn
Thus,
sin a2
r = arcsin
sin πn
3
.
(ii) From equation (2) we have,
2 cos2 πn
cos θ = 1 −
1 − sin2 r sin2
π
n
.
cos2 πn
1 − cos θ
=⇒
=
2
1 − sin2 r sin2
=⇒ 1 − sin2 r sin2
(7)
π
n
.
π
n
θ
2
cos2
=
n
sin2
π .
Then,
2 θ
2 π
sin
−
cos
2 n
sin2 r =
sin2 2θ sin2 πn
and (ii) follows.
(iii) From equations (6) and (7) we get,
2 cos2 πn
cos θ = 1 −
.
cos2 a2
Therefore,
2 cos2 πn
− (n − 2)π.
A = n θ − (n − 2)π = n arccos 1 −
cos2 a2
=⇒
cos
2 cos2 πn
∴ 1−
= − cos
cos2 a2
A − 2π
+π
n
2π − A
n
2 cos2 πn
=1−
.
cos2 a2
2
= − 2 cos
2π − A
2n
−1 .
This implies that
cos2 πn
a
cos
=
2
cos2 2π−A
2n
2
Note above that A < 2π & n ≥ 3 imply
4
2π−A
2n
∈ (0, π2 ).
(8)
Thus,
cos a =
2 cos2 πn
− 1.
cos2 2π−A
2n
(9)
Then,
sin a2
sin r =
sin πn
p
1 − cos2 a2
=
sin πn
r
π
cos2 n
1 − cos2 2π−A
( 2n )
=
sin πn
q
cos2
=
cos
2π−A
2n
2π−A
2n
[by (8)]
− cos2
sin πn
π
n
.
and (iii) is proved.
Finally, it can be verified that the functions in (i), (ii), (iii) are strictly monotone functions and hence ℘ is determined upto congruence by any one of the entities
a, θ, A.
Corollary 4.3 Let (℘˜k )k∈N be a sequence of proper regular polygons in S 2 such that
℘˜k has k vertices ∀ k, (area(℘˜k )) −→ A0 as k −→ ∞. For each k ∈ N, let rk be the
radius of the circle in which ℘˜k is inscribed. Then
"p
#
A0 (4π − A0 )
(i)
lim rk = arcsin
k−→∞
2π
(ii)
lim ( perimeter(℘˜k )) =
k−→∞
p
A0 (4π − A0 ).
Proof :
(i) By (5),
2
sin rk =
2π−Ak
− cos2 πk
2k
k
cos2 2π−A
sin2 πk
2k
cos2
5
=
4π−Ak
sin A2kk
2k
k
cos2 2π−A
sin2 πk
2k
sin
∴
2
lim sin rk
k−→∞
4π − Ak Ak k 2
= lim
·
·
k−→∞
2k
2k π 2
=
[∵
lim
θ−→0
sin θ
= 1]
θ
(4π − A0 )A0
.
4π 2
This implies that
#
"p
A0 (4π − A0 )
.
lim rk = arcsin
k−→∞
2π
"p
#
A0 (4π − A0 )
(ii) Put r0 = arcsin
.
2π
Now, each ℘˜k is a regular k-gon inscribed in a circle of radius rk in S 2 , and (rk ) −→ r0
as k −→ ∞ by (i). Hence, (perimeter(℘˜k )) converges to the perimeter of the circle
of radius r0 in S 2 .
p
∴ lim ( perimeter(℘˜k )) = 2π sin r0 = A0 (4π − A0 ).
k−→∞
6