2. WHY IS THIS IMPORTANT …
One application of
Spherical Geometry is
that pilots and ship
captains need to
navigate the earth to
reach their
destination.
For example, did you know that the shortest flying distance
from Florida to the Philippine Islands is a path across
Alaska?
The Philippines are South of Florida – so why is flying
North to Alaska a short-cut?
The answer is that Florida, Alaska, and the Philippines are
3. THE BASICS!
• Sphere
• A set of points in three dimensional space equidistant
from the center.
• Radius
• Distance from the center to the points on the sphere
• Diameter
• Twice the radius
4. POINTS ON SPHERE
• Points are only on the surface of a sphere
• Points are the same concept we are used to
studying in absolute and Euclidean
geometry
• Special pairs of points:
• Called antipodal or polar
• those points that correspond to opposite ends
of a diameter of the sphere. Example: N & S
poles
5. WHAT ARE LINES?
• Shortest path on the sphere between two points on the
sphere’s surface
• Known as great circles or arcs
A great circle is a circle on a
sphere which divides the sphere in 2
equal hemispheres; it contains a
diameter of the sphere and hence the
center as well
6. PLANES INTERSECTING
WITH SPHERE
• 3 cases involving planes and spheres
• The plane misses the sphere
• The plane hits the sphere at 1 point
• Known as a tangent plane
• The plane and the sphere meet in a circle
7. 7
PLANES INTERSECTING
SPHERE
• The circle of intersection will be largest
when the plane passes through the
center of the sphere
• Paradoxically, larger radius means less
distance traveled on the surface (closest to
Euclidean distance)
8. 9
PLANE OF THE SPHERE
• There is only ONE plane that exists in spherical
geometry
• The plane is the surface of the sphere
9. DISTANCE
• Given two points A and B on the sphere, the distance AB*
between them is length of the shortest arc-segment which
connects them.
• Because of the way we defined spherical ‘lines’, this arc-
segment must be an arc of a great circle.
• Joining our points A and B to the center of the sphere, to
create a 'wedge' shape. We can measure the angle at the
center - call it θ.
• The length of our arc is then rθ.
10
12. ANGLES
• Imagine two great circles on the sphere. Choose one
of the two points where they intersect. The angle
between the two arcs is then defined to be the angle
between the two planes containing the great circles.
13
13. ADDITIONAL ANGLE DEFINITION
• Imagine a flat, transparent
plane which just touches the
sphere at the point we wish to
measure our angle. Looking at
the plane from directly above,
arcs on the sphere coincide
with lines on the plane. The
angle between our spherical
lines is then the angle between
the corresponding lines on the
plane
14
14. TRIANGLES
• A spherical triangle consists of three vertices, and three arc-segments which
join them.
• Note that the lines do not have to be the shorter arcs.
• Also, if any two points are antipodal, we must specify which great circle to be the arc
joining them (as there are an infinite number of great circles containing both points).
15
15. QUESTION …
16
Does the sum of the
angles of a triangle
equal 180 in spherical
geometry??
16. ANSWER ….
• The angle sum will never be 180°
• Always 180° < A + B + C < 540°
17
A
B
C
17. GIRARD’S THEOREM FOR TRIANGULAR AREA
• Consider a (small) spherical triangle with vertices A, B, and
C noncollinear
• Note that A, B, and C are the intersections of arcs a, b, and
c so that we have polar points
A’,B’, and C’ also
• A and A’ are the endpoints of two lunes, call them Lα and
Lα’. Similarly for points B, B’ and C, C’.
18
18. GIRARD’S THEOREM CON’T
• A(Lα U Lα’) + A(Lβ U Lβ’) + A(Lγ U Lγ’)
• = A(sphere) + 2A(ΔABC) +2A(ΔA’B’C’)
• since Lβ and Lγ each overcount the area of ΔABC.
(Similarly for ΔA’B’C’ with lunes Lβ’ and Lγ’)
• 2(2α+2β+2γ) = 4π + 4A(ΔABC)
• α+β+γ = π + A(ΔABC) and thus
A(ΔABC) = α+β+γ - π (Girard’s Theorem)
19
19. SPHERICAL EXCESS
• Given a spherical ΔABC with angles α, β, and γ, the
quantity α+β+γ-π is called the spherical excess of
ΔABC.
• Girard’s theorem tells us that area of a spherical polygon
is intimately linked with the spherical excess.
20
21. THEOREM
• AAA Congruence Criterion
• If two triangles have three angles of one congruent, respectively, to the three angles
of the other, the triangles are congruent
22
23. DEFINITION
The cross product is defined as the vector perpendicular to two given
vectors A and B separated by an angle and is shown by:
||A x B|| = ||A|| ||B|| sin
The magnitude of the cross product equals the area of the parallelogram
that the initial two vectors span
24. PROPERTIES
Anti-commutative: a x b = -b
x a
Distributive over addition: a x (b + c) = a
x b + a x c
Compatible with scalar multiplication: (ra) x b =
a x (rb) = r(a x b)
Not associative, but satisfies the Jacobi identity: a x (b x c)
+ b x (c x a) + c x (a x b) = 0
Cross product is only valid in R3 and R7
26. MATRIX SOLVING METHOD
The first row is the standard basis vectors and must appear in the
order given here. The second row is the components of u [u1
u2 u3] and the third row is the components of v [v1 v2 v3].
Given two vectors u and v, u x v equals the determinant of the
matrix.
28. VECTOR CROSS PRODUCT
Cross Product Applet
||a x b|| = ||a|| ||b|| sin()
Two non-zero vectors a and b are parallel iff a x b = 0
29. GEOMETRIC APPLICATIONS
3-D
Volume of a parallelepiped with sides of length a, b, c equals the
magnitude of the scalar triple product:
V = |a · (b x c)|
30. TORQUE & ANGULAR
MOMENTUM
Torque is the measure of how much a force acting
on an object will cause that object to rotate. Ø is
T= r x F = r F sin(ø)
If a particle with linear momentum p is at a position r with respect to
some point, then its angular momentum L is the cross product of r and
p
L = r x p
31. TORQUE EXAMPLE
F
R
T= R x F = RFsin()
is the angle between the location of
the applied force and the point where
the radius meets this force.
In this case, = 90, which means that
the torque is just the product of the
radius and the applied force because
sin(90) = 1.
32. LORENTZ FORCE
The Lorentz Force F exerted on a charged particle in an
electromagnetic field equals the electric charge q of the particle times
the electric field E plus the cross product of the velocity v of the
particle and the magnetic field B
F = q (E + v x B)
33. LORENTZ FORCE RHR
Align the thumb of the right hand
in the direction of the velocity v.
Then, point the index finger of the
right hand in the direction of the
magnetic field B. Now, the palm of
the right hand points in the
direction of the Lorentz force F.
Lorentz Force Applet
46. AN ORTHOGONAL PROJECTION
Note: If the calculations above are correct, then
will be an orthogonal set.
As a check, compute
Since the line segment in the figure on the previous
slide between y and is perpendicular to L, by
construction of , the point identified with is the
closest point of L to y. Sli
de
6.2
-
46
ˆ ˆ
{y,y y}
ŷi(y - ŷ) = 8
4
é
ë
ê
ù
û
úi -1
2
é
ë
ê
ù
û
ú = -8+8 = 0
ŷ
ŷ ŷ
47. ORTHONORMAL SETS
• A set {u1,…,up} is an orthonormal set if it is an
orthogonal set of unit vectors.
• If W is the subspace spanned by such a set, then
{u1,…,up} is an orthonormal basis for W, since the
set is automatically linearly independent, by
Theorem 4.
Sli
de
6.2
-
47
49. ORTHONORMAL SETS
Thus {v1, v2, v3} is an orthogonal set.
Also,
which shows that v1, v2, and v3 are unit vectors.
Thus {v1, v2, v3} is an orthonormal set.
Since the set is linearly independent, its three
vectors form a basis for . See the figure on the
next slide.
Sli
de
6.2
-
49
v2
iv3
=1/ 396 -8 / 396 + 7 / 396 = 0
v1
iv1
= 9 /11+1/11+1/11=1
v2
iv2
=1/ 6 + 4 / 6 +1/ 6 =1
v3
iv3
=1/ 66 +16 / 66 + 49 / 66 =1
3
52. ORTHONORMAL SETS
Theorem 7: Let U be an matrix with
orthonormal columns, and let x and y be in
.
Then
a.
b.
c. if and only if
Properties (a) and (c) say that the linear
mapping preserves lengths and
orthogonality.
Sli
de
6.2
-
52
m n
n
Ux = x
(Ux)i(Uy) = xiy
(Ux)i(Uy) = 0 xiy = 0
x x
U
57. Great circle is also known
as orthodrome or Riemannian circle.
•Definition Of Great circle: A great circle of a sphere is
the intersection of the sphere and a plane which passes
through the center point of the sphere.
58. SMALL CIRCLE OF A SPHERE IS DEFINED
AS THE INTERSECTION OF A SPHERE AND A
PLANE, IF THE PLANE DOES NOT PASS
THROUGH THE CENTER OF THE SPHERE
61. SLIDESMANIA
M
WHAT IS INCIDENCE
GEOMETRY?
❖ Incidence geometry is geometry involving
only points, lines, and the incidence relation.
It ignores the relations of betweenness and
congruence.
62. SLIDESMANIA
M
MODELS OF INCIDENCE
GEOMETRY
Definition: Let 𝒕𝟏 be any line and let 𝑨 and
𝑩 be any points that do not lie on 𝒕𝟏. If 𝑨=𝑩
or if the segment 𝑨𝑩 contains no point lying
on 𝒕𝟏, we say that 𝐴 and 𝐵 are on the same
side of 𝒍. If 𝑨𝑩 contains a point of 𝒕𝟏, we say
that 𝑨 and 𝑩 are on opposite side of 𝒍.
63. SLIDESMANIA
M
4
● Incidence Axiom 1 (IA1). For every pair of points
𝑃 and 𝑄 there exists a unique line 𝑙 incident with 𝑃
and 𝑄.
● Incidence Axiom 2 (IA2). For every line 𝑙 there
exist at least two distinct points incident with 𝑙.
● Incidence Axiom 3 (IA3). There exist three
distinct points 𝐴, 𝐵, 𝐶 not simultaneously incident with
a common line 𝑙.
INCIDENCE
AXIOMS
64. SLIDESMANIA
M
PROPOSITION 8.1. IF 𝑙, 𝑚 ARE DISTINCT
LINES AND ARE NOT PARALLEL, THEN 𝑙
AND 𝑚 MEET AT A
UNIQUE POINT.
Proof. The lines 𝑙, 𝑚 must intersect since they are not
parallel. Suppose they meet at two distinct points
𝑃, 𝑄. Then there are two lines 𝑙, 𝑚 through points 𝑄.
This is contradictory to Incidence Axiom 1. So
𝑙 and 𝑚 meet at a unique point.
5
65. SLIDESMANIA
M
PROPOSITION 8.2. THERE EXIST THREE
DISTINCT LINES NOT THROUGH ANY
COMMON POINT.
Proof. Let 𝐴, 𝐵, 𝐶 be three distinct points not collinear. Since
any two distinct points determine a unique line, the three
lines 𝐴𝐵, 𝐴𝐶, BC must be distinct. We claim that the three
lines have no common point. Suppose they have a common
point 𝑃. We must have 𝑃 ≠ 𝐴 (Otherwise, 𝑃 = 𝐴 implies that
𝐴 lies on the line 𝐵𝐶 since 𝑃 is a common point of
𝐴𝐵, 𝐴𝐶, 𝐵𝐶. Thus 𝐴, 𝐵, 𝐶 are collinear, a contradiction.) Then
two lines 𝐴𝐵, 𝐴𝐶 pass through points 𝑃, 𝐴. This is
contradictory to Incidence Axiom 1.
6
66. SLIDESMANIA
M
PROPOSITION 8.3. FOR EVERY LINE
THERE IS AT LEAST ONE POINT
NOT LYING ON IT.
Proof. Let 𝐴, 𝐵, 𝐶 be three distinct points not
collinear by Incidence Axiom 3. Suppose
there is a line 𝑙 which has no point outside
𝑙, i.e., 𝑙 contains every point. Then 𝑙 contains
all 𝐴, 𝐵, 𝐶. This means that 𝐴, 𝐵, 𝐶 are
collinear, a contradiction.
7
67. SLIDESMANIA
M
PROPOSITION 8.4. FOR EVERY POINT THERE IS
AT LEAST ONE LINE
NOT PASSING THROUGH IT.
8
Proof. Let 𝑙, 𝑚, 𝑛 be three distinct lines not
concurrent by Proposition 8.2. Given an
arbitrary point 𝑃; then one of 𝑙, 𝑚, 𝑛 does not
pass-through 𝑃.
68. SLIDESMANIA
M PROPOSITION 8.5. FOR EVERY
POINT 𝑃 THERE EXIST AT LEAST
TWO LINES THROUGH 𝑃.
9
• Proof. Let 𝐴, 𝐵, 𝐶 be three points not
collinear. If 𝑃 is outside the line 𝐴𝐵, then the
lines 𝐴𝑃, 𝐵𝑃 pass through point 𝑃 and
• must be distinct. Otherwise, 𝐴𝑃= 𝐵𝑃 implies
that 𝐴, 𝐵, 𝑃 are collinear; so 𝑃 is on the line 𝐴𝐵
, contradicting to that 𝑃 is outside 𝐴𝐵 If 𝑃 is
on the line 𝐴𝐵, then 𝐴𝐵, 𝐶𝑃 are two distinct
lines passing through 𝑃.
69. SLIDESMA
NIA
M
Model #1:
Points: A, B, C
•Lines: {A, B}, {A, C}, {B, C}
•Point lies on 𝑙 if the letter belongs to the set 𝑙.
10
MODELS OF INCIDENCE GEOMETRY
Model #2:
Points: A, B, C, D
•Lines: {A, B}, {A, C}, {A, D}, {B, C}, {B, D}, {C,
D}
•Point lies on 𝑙 if the letter belongs to the set 𝑙.
70. SLIDESMANIA
M
Points: Ordered pairs of real numbers (𝑥, 𝑦)
• Lines: Triples of real numbers (𝑎, 𝑏, 𝑐) so that either 𝑎 ≠ 0 or 𝑏 ≠ 0. It
is the set of all points (𝑥, 𝑦) that satisfy the equation 𝑎𝑥 + 𝑏𝑦 + 𝑐 = 0.
• Point lies on 𝑙 if it is solution of the 𝑙′𝑠 equation.
11
MODEL #3 CARTESIAN
PLANE:
• Points: unordered pairs {(𝑥, 𝑦, z), (−𝑥, −𝑦−, z)}, where (𝑥, 𝑦, z) lies on
the unit sphere
• Lines are sets of points {(𝑥, 𝑦, z), −𝑥, −𝑦−, z)} that are parts of great
circles on the unit sphere.
• Point lies on the line if both (𝑥, 𝑦, z), (−𝑥, −𝑦−, z) lie on the
corresponding great circle.
Model #4 (Real projective plane:
71. SLIDESMANIA
M
Points: ordered pairs of real numbers (x, y), where y > 0.
• Lines:
o Subsets of vertical lines that consist of points (x, y), with y >
0
o Semicircles whose centers are points (x, 0), where x is a real
number
MODEL #5 (HYPERBOLIC
PLANE):
Model #6 (Projective completion of model#2):
Points: A, B, C, D, E, F, G
• {A, B, E}, {A, C, F}, {A, D, G}, {B, C, G}, {B, D, F}, {C, D, E}, {E, F, G}
• Point lies on 𝑙 if the letter belongs to the set 𝑙.
72. SLIDESMANIA
M
● Models of affine geometry (3 incidence geometry axioms +
Euclidean PP) are called affine planes and examples are
Model #2
Model #3 (Cartesian plane).
● Model of (3 incidence axioms + hyperbolic PP) is Model #5
(Hyperbolic plane).
● Models of projective geometry are called projective planes.
73. SLIDESMANIA
M
ISOMORPHISM OF
MODELS
Two models of incidence geometry are said to be
isomorphic if there is a one-to-one correspondence
P ↔ P' between the points of the models and a
one-to-one correspondence l ↔ l’ between the
lines of the models such that P lies on l if and only
if P' lies on l’. Such a correspondence is called an
isomorphism from one model onto the other.
74. SLIDESMANIA
M
EXAMPLE
Consider a set {a, b, c} of three letters, which we will call lines now. The points will
be those subsets that contain exactly two letters {a, b}, {a, c} and {b, c}. Let
incidence be the set membership. For instance, point {a, b} is on the lines a and b,
but not on the line c. This model seems to be structurally the same as the three-
point model in Example 1, except the change of notations. An explicit isomorphism
is given as follows:
● A ↔ {a, b},
● {A, B} ↔ b,
B ↔ {b, c},
{B, C} ↔ c,
C ↔ {a, c},
{A, C} ↔ a.
75. SLIDESMANIA
M
EXAMPLE
20
● Consider the axiomatic system:
Axioms 1: There exist exactly three distinct Fe’s in this system
Axioms 2: Any two distinct Fe’s belong to exactly one Fo
Axioms 3: Not all Fe’s belong to the same Fo.
Axioms 4: Any two distinct Fo’s contain at least one Fe that belongs to both
We will interpret the Fe’s as elemet, Fo’s as sets, and “belongs to” means set
containment.
Let elements be
𝑥, 𝑦, 𝑎𝑛𝑑 𝑧
Lets the sets be
𝐹=(𝑥, 𝑦), 𝐺=(𝑥, 𝑧), 𝐻=(𝑦, 𝑧)
This model is isomorphic to the model from the example, where one-to-one
correspondence is given as
Bob ↔ 𝑥 Ted ↔ 𝑦 Carol ↔ 𝑧
Entertainment ↔ 𝐹 = (𝑥, 𝑦) Finance ↔ 𝐻 = (𝑦,𝑧) Refreshments ↔ 𝐺 = (𝑥, 𝑧)
77. LESSON 5-5
INEQUALITIES IN TRIANGLES
OBJECTIVE:
To use inequalities involving angles
and sides of triangles
78. Theorem 5-10
If a triangle is scalene, then the largest angle
lies opposite the longest side
and the smallest angle
lies opposite the shortest side.
17”
X
Y Z
29”
32”
Example 1: List the angles
from smallest to largest
Z Y X
79. Theorem 5-11(Converse of Theorem 5-10)
If a triangle is scalene, then the longest side
lies opposite the largest angle,
and the shortest side
lies opposite the smallest angle.
R
Q
S
30°
Example 2: In QRS, list
the sides from smallest
to largest
SR QS QR
80. Example 3:
In TUV, which side is the shortest?
58°
U
V
62°
Use sum to find mT.
mT = 60°, so U is smallest
Therefore VT is shortest
T
81. Theorem 5-12 The Triangle Inequality Theorem
The sum of the lengths of any two sides of a
triangle is
greater than the length of the third side.
82. Example 4:
Can a triangle have sides with the given lengths?
Explain.
a) 3ft., 7ft., 8ft. b) 3cm., 6cm., 10cm.
Yes, 3 + 7 > 8 NO, 3 + 6 < 10
83. Example 5:
A triangle has sides of lengths 8cm and 10cm.
Describe the lengths possible for the third side.
Let x = the length of the 3rd side.
The sum of any 2 sides must be
greater than the 3rd.
84. x + 8 > 10
x > 2
x + 10 > 8
x > -2
8 + 10 > x
18 > x
x < 18
So, x must be longer than 2cm
& shorter than 18cm.
So, there are 3 possibilities.
2 < x < 18
85. Lesson 3-3: Triangle Inequalities 85
Triangle Inequality
The smallest side is across from the smallest angle.
The largest angle is across from the largest side.
is thesmallest angle, is thesmallest side.
A BC
is thelargest angle, is thelargest side.
B AC
54
37
89
B
C
A
86. Lesson 3-3: Triangle Inequalities 86
TRIANGLE INEQUALITY –
EXAMPLES…
For the triangle, list the angles in order from least to greatest measure.
C
A
B
5 cm
, ,
.
arg arg .
AB isthesmallest side C smallest angle
BC isthel est side Ais
Angles in order from least to grea
the
tes
l est angle
t C B A
87. Lesson 3-3: Triangle Inequalities 87
TRIANGLE INEQUALITY –
EXAMPLES…
For the triangle, list the sides in order from shortest to longest measure.
8x-10
7x+6
7x+8
C
A
B
(7x + 8) ° + (7x + 6 ) ° + (8x – 10 ) ° = 180°
22 x + 4 = 180 °
22x = 176
X = 8
m<C = 7x + 8 = 64 °
m<A = 7x + 6 = 62 °
m<B = 8x – 10 = 54 °
64 °
62 °
54 °
, ,
.
arg .
B isthesmallest angle AC shortest side
C isthel est angle ABi
Sides in order from smallest to
sthe
long
longest s
est AC BC AB
ide
88. Lesson 3-3: Triangle Inequalities 88
The perpendicular segment from a point to a line is
the shortest segment from the point to the line.
Corollary 1:
The perpendicular segment from a point to a plane
is the shortest segment from the point to the plane.
Corollary 2:
If one angle of a triangle is larger than a second angle,
then the side opposite the first angle is larger than the
side opposite the second angle.
Converse:
Converse Theorem & Corollaries
89. Lesson 3-3: Triangle Inequalities 89
TRIANGLE INEQUALITY
THEOREM:
The sum of the lengths of any two sides of a triangle is
greater than the length of the third side.
c
b
a
B
C
A
a + b > c
a + c > b
b + c > a
Example: Determine if it is possible to draw a triangle with side
measures 12, 11, and 17. 12 + 11 > 17 Yes
11 + 17 > 12 Yes
12 + 17 > 11 Yes
Therefore a triangle can be drawn.
90. Lesson 3-3: Triangle Inequalities 90
FINDING THE RANGE OF THE
THIRD SIDE:
Since the third side cannot be larger than the other two added
together, we find the maximum value by adding the two sides.
Since the third side and the smallest side cannot be larger than the
other side, we find the minimum value by subtracting the two sides.
Example: Given a triangle with sides of length 3 and 8, find the range of
possible values for the third side.
The maximum value (if x is the largest
side of the triangle) 3 + 8 > x
11 > x
The minimum value (if x is not that largest
side of the ∆) 8 – 3 > x
5> x
Range of the third side is 5 < x < 11.
