May 5, 2013 Math 10
Geometry. Elements of Projective Geometry.
Projective geometry is a fascinating subject, and the one widely used
now, mainly in rendering the computer graphics, but also in making
maps, in art – painting, etc. The systematic development of this subject
requires a full college-level course, but its most basic theorems can be
proved by the methods of Euclid geometry; in fact, three of the four
are so old that no other methods were available at the time of their
discovery. All these theorems
deal either with collinearity
(certain sets of points lying on a
line) or concurrence (certain sets
of lines passing through a point).
The spirit of projective
geometry begins to emerge as
soon as we notice that, for many
purposes, parallel lines behave
like concurrent lines.
Axioms and basic definitions of projective plane geometry.
Undefined Terms: point, line, incident
Axiom 1. Any two distinct points are incident with exactly one line.
Axiom 2. Any two distinct lines are incident with at least one point.
Axiom 3. There exist at least four points, no three of which are
collinear.
Axiom 4. The three diagonal points of a complete quadrangle are never
collinear.
Axiom 5. (Desargues' Theorem) If two triangles are perspective from
a point, then they are perspective from a line.
Axiom 6. If a projectivity on a pencil of points leaves three distinct
points of the pencil invariant, it leaves every point of the pencil
invariant.
Definition. A pencil of points, in projective geometry, is a set of all
the lines in a plane passing through a point, or in three dimensions, a
pencil is all the planes passing through a given line.
Definition. A set of points is collinear if every point in the set is
incident with the same line. Points incident with the same line are said
to be collinear.
Definition. Lines incident with the same point are said to be
concurrent.
Definition. A complete quadrangle is a set of four points, no three of
which are collinear, and the six lines incident with each pair of these
points. The four points are called vertices and the six lines are called
sides of the quadrangle.
Example: Complete quadrangle ABCD has vertices
A, B, C, and D. The sides are AB, AC, AD, BC, BD, and CD.
Two sides of a complete quadrangle are opposite if the point incident to both lines is not a vertex.
Example: Complete quadrangle ABCD has three
pairs of opposite sides AB and CD, AC and BD, and AD and BC.
We have considered in the past the theorem of Pappus of Alexandria,
which is the oldest of the basic projective theorems.
Pappus theorem. If A, C, E are three points
on one line, B, D and F on another, and if
three lines, AB, CD, EF, meet DE, FA, BC,
respectively, then the three points of
intersection, L, M, N, are collinear.
There are many ways to prove this theorem;
the one we used was based on using the
center of mass. This shows close connection between the barycentric
coordinates, which are introduced using the method of center of mass,
and those in projective geometry.
Another remarkable theorem on concurrence was discovered by Blaise
Pascal. Pascal published this theorem in 1640, when he was sixteen
years old. Pappus theorem is a special case of Pascal’s.
Polygons and Hexagons
A polygon may be defined as a figure consisting of a number of points
(called vertices) and an equal number of line segments (called sides),
namely a cyclically ordered set of points in a plane, with no three
successive points collinear, together with the line segments joining
consecutive pairs of the points. In other words, a polygon is a closed
broken line lying in a plane. A polygon having n vertices and n sides is
called an n-gon (meaning literally "n-angle"). Thus we have a pentagon (
n = 5 ), a hexagon ( n = 6 ), and so on. In fact, the Greek name for the
number n is used except when n = 3 or 4. In these two simple cases it is
customary to use the Latin forms triangle and quadrangle rather than
"trigon" and "tetragon" (although "trigon" survives in the word
"trigonometry"). Note that there is a difference between a quadrangle
and a "quadrilateral". In projective geometry, where the sides are
whole lines instead of mere segments, both terms are used with
distinct meanings. Figure below shows different types of quadrangles.
Now consider hexagon ABCDEF. Two vertices of a hexagon are said to
be adjacent, alternate, or opposite, when they are separated by one
side, two sides, or three sides, respectively. Thus, in a hexagon
ABCDEF, F and B are adjacent to A, E and C are alternate to A, and D
is opposite to A. The line joining two opposite vertices is called a
diagonal. Thus ABCDEF has three diagonals: AD, BE, CF. Similarly the
hexagon ABCDEF has three pairs of opposite sides: AB and DE, BC and
EF, CD and FA.
A given hexagon can be named ABCDEF in twelve ways: Any one of its
six vertices can be named A, either of the two adjacent vertices can
be named B, and the rest are then determined by the alphabetical
order. Six given points, no three collinear, can be named A, B, C, D, E, F
in 6! = 720 ways. Each way determines a hexagon ABCDEF having the
six given points for its vertices. Hence the number of distinct
hexagons determined by the six points is 720:12 = 60.
Figure above shows three of the sixty hexagons determined by six
points on a circle. Although we are accustomed to the first ("convex")
kind, we must not forget or neglect the other fifty-nine possible
hexagons that can be derived from the same six points.
Pappus hexagon theorem (Pappus rephrased). If each set of three
alternate vertices of a hexagon is a set of three collinear points, and
the three pairs of opposite sides intersect, then the three points of
intersection are collinear.
Pascal theorem. If a hexagon is inscribed in a conic (ellipse, parabola,
hyperbola), then the three points at which the pairs of opposite sides
meet, lie on a straight line.
One possible situation illustrating
Pascal’s theorem is shown in the
Figure on the right. It bears
direct similarity to the Pappus
hexagon construction, except that
all vertices of the hexagon now lie
on a circle rather than on two
straight lines.
The theorem is clearly projective. If it holds for one kind of conics,
it holds for any other, as one can
be obtained from the other by a
coordinate transformation. In particular, if it holds for a circle, it also
holds for a pair of intersecting straight lines. As a projective property,
it then also holds for a pair of parallel lines. This is why it is said to be
a generalization of the theorem of Pappus. Below we consider the case
when the six points lie on a circle.
Pascal theorem for circle. If the vertices of a hexagon lie on a circle,
and the three pairs of opposite sides intersect, then the three points
of intersection are collinear.
Proof. Pascal’s original proof has been lost. Some even claim that Pascal
actually did not publish any proof. A number of various Euclidean
proofs have appeared since then, and one wonders whether any of them
is an original Pascal’s proof. The most standard proof (Coxeter,
Greitzer) is based on the multiple use of the Menelaus theorem.
Perhaps the simplest proof was published by Jan van Yzeren in 1993 in
The American Mathematical Monthly (Mathematical Association of
America) 100 (10): 930–931.
There are 60 possible different realizations of the Pascal’s hexagon.
We will only consider the instance shown in the figure above, the
others can be proven similarly by using the same method.
A6
O
A1
A2
A3
A4A5
C3
C1
C2
B1
B2
Consider the hexagon A1A2A3A4A5A6 inscribed in a circle as shown in
the figure. We need to prove that intersection points of the non-
adjacent sides, C1 = A1A2A4A5, C2 = A2A3A5A6, and C3 = A3A4 A1A6,
are collinear. Construct a circle circumscribed around the triangle
A2A5C2. Denote points where this circle intersects A1A2 and A2A4 by B1
and B2, respectively. Using the arcs of the circle, or the supplementary
angles inscribed in the opposite arcs, one can show that triangles
A1A4C3 and B1B2C2 have respectively parallel sides, that is, they are
perspective from the point C1. Therefore, C1, C2 and C3 are collinear.
Desargues theorem.
If two specimens of a figure, composed of points and lines, can be put
into correspondence in such a way that pairs of corresponding points
are joined by concurrent lines, we say that the two specimens are
perspective from a point. If the correspondence is such that pairs of
corresponding lines meet at collinear points, we say that the two
specimens are perspective from a line. In the spirit of projective
geometry, Desargues's two-triangle theorem states that if two
triangles are perspective from a point, they are perspective from a
line.
The geometrical theory of perspective was inaugurated by the
architect Filippo Brunelleschi (1377-1446), who designed the octagonal
dome of the cathedral in Florence, and also the Pitti Palace. A deeper
study of the same theory was undertaken by another architect, Girard
Desargues (1591-1661), whose "two-triangle" theorem was later found
to be just as important as Pappus's. It can actually be deduced from
Pappus's; but the details are complicated, and it can be far more easily
deduced from the Menelaus's Theorem.
To avoid complications arising from the possible occurrence of parallel
lines, let us be content to rephrase it as follows:
Theorem (Desargues). If two triangles are perspective from a point,
and if their pairs of corresponding sides meet, then the three points
of intersection are collinear.
We have a theorem of pure incidence. Figures show two of the many
ways in which the diagram can be drawn. Here APQR and APQ'R' are
perspective from 0 and their pairs of corresponding sides meet at D,
E, F. For a proof, we can apply Menelaus Theorem to the three triads
of points DR'Q', EPR', FQ'P on the sides of the three triangles
obtaining
,
,
.
After multiplying these three
expressions together and doing a
modest amount of cancellation, we are
left with
, whence D, E,
F are collinear, as desired. Desargues's
theorem is easily seen to imply its
converse: that if two triangles are
perspective from a line, they are
perspective from a point.
Stereometric proof.
AB
C
D
M
NP A
B
C
D
M
NP
AB
C
D
M
NP A
B
C
D
M
NP
1 2
3 4
Exercise 1. Plane passing through midpoints (
M, N, P, Q) of four edges of a regular
tetrahedron forms a square cross section. It is
easy to draw, just connect these points. Let us
prove that MNPQ is really a square. MN and PQ
are equal and parallel to each other as midlines
of triangle with the same base. The same is
true for MQ and NP. And also MN=NP=PQ=QM
because DC=AB.
, but the same can be calculated for MP,
.
Therefore, quadrilateral MNPQ has equal sides. Also, MN is parallel to
QP and NP is parallel to MQ so diagonals of this quadrilateral are also
equal. Therefore, MNPQ is a square.
Exercise 2. Draw the cross section of a pyramid by a plane passing
through three points on a faces of the pyramid.
AB
C
D
M
NP
Q
Excercise. Draw a plane passing through midpoints M, N and Q of
edges A’D’, D’C’ and AB respectively of a
cube ABCDA’B’C’D’. Find the area of the
section created by the plane you drew (the
section is a part of the plane contained
inside the cube) if the side of the cube is 2
inches long.
A
A’
B
B’
C
C’
D
D’
N
M
P
Geometry 10: cumulative recap.
Dihedral and polyhedral angels.
A plane is divided by a line lying in it into
two parts, called half-planes, and the line is
called the edge of each of these half-planes.
A figure in space formed by two half- planes
(P and Q ) which have the same edge (AB) is
called dihedral angle. AB is an edge and P
and Q are faces of the dihedral angle.
If, from an arbitrary point D on the edge AB
of a dihedral angle a ray perpendicular to the
edge is drawn in each of the faces, then the
angle CDE, formed by these two rays is called
a linear angle of the dihedral angle. The
measure of the linear angle of a dihedral angle
does not depend on the position of its vertex
on the edge. The plane containing a linear angle
of a dihedral angle is perpendicular to the edge since it contains two
lines perpendicular to it. Similarly to angles in plane geometrically
dihedral angles can be vertical, supplementary, etc.
Polyhedral angles are formed by the intersection
of three or more planes at a common vertex
point. A polyhedral angle is called convex if it
lies on one side of the plane of each of its faces.
B
AQ
P
B
A
CD
E
Law of cosines for a trihedral angle.
Where and are
planar angles of the faces of the trihedral
angle, and is the linear angle of the dihedral
angle of the edge .
Proof. From an arbitrary point C on the edge
draw two rays perpendicular to this edge. A
and B are points of intersection of these rays
with edges and . Using the law of cosines for the triangle
ABC we have,
, while from the triangle AOB,
Subtracting the first equation from the second,
(1)
Triangles OCB and OCA are the right triangles, therefore
and (2)
From (1) and (2)
where we used
,
,
,
. Hence,
Euler's Formula.
There is another "Euler's Formula" about complex numbers, this one is
about the one used in Geometry and Graphs. For any polyhedron that
doesn't intersect itself, the
Number of Faces
plus Number of Vertices (corner points)
minus Number of Edges
always equals 2. This can be written F + V - E = 2.
Example. A cube has 6 faces, 8 vertices and 12 edges:
6+8-12=2.
To see why this works, imagine taking the cube and adding
an edge (say from corner to corner of one face).
You will have an extra edge, plus an extra face:
7 + 8 - 13 = 2.
Likewise if you included another vertex (say halfway along a
line) you would get an extra edge, too:
6 + 9 - 13 = 2.
It turns out, rather beautifully, that Euler's formula is true for pretty
much every polyhedron. The only polyhedra for which it doesn't work
are those that have holes running through them like those considered
in the examples later on. These polyhedra are called non-simple, in
contrast to the ones that don't have holes, which are called simple.
Non-simple polyhedra might not be the first to spring to mind, but
there are many of them out there.
Example With Platonic Solids
Consider 5 Platonic Solids (Note: Euler's Formula can be used to prove
that there are only 5 Platonic Solids):
Name Shape Faces Vertices Edges F+V-E
Tetrahedron
4 4 6 2
Cube
6 8 12 2
Octahedron
8 6 12 2
Dodecahedron
12 20 30 2
Icosahedron
20 12 30 2
The Sphere.
All Platonic Solids (and many other solids) are like a
Sphere – they can be reshaped so that they become a
Sphere (move their corner points, then curve their faces
a bit). They can be inscribed in a sphere, or an ellipsoid.
For this reason we know that F+V-E = 2 for a sphere (one can not
simply say a sphere has 1 face, and 0 vertices and edges, for F+V-E=1).
What if we joined up two opposite corners of the
icosahedron? It is still an icosahedron (but no longer
convex). In fact it looks a bit like a drum where
someone has stitched the top and bottom together.
Now, there would be the same number of edges and
faces ... but one less vertex:
F + V - E = 1
It doesn't always add to 2!
The reason it didn't work was that this new shape is basically
different: that joined bit in the middle means that two vertices get
reduced to 1.
Euler Characteristic
So, F+V-E can equal 2, or 1, and maybe other values, so the more
general formula is
F + V - E = χ
Where χ is called the "Euler Characteristic".
Here are a few examples:
Name Shape χ
Sphere
2
Torus
0
Mobius strip
0
Finally, the Euler Characteristic can also be less
than zero.
This is the "Cubohemioctahedron": It has 10
Faces (it may look like more, but some of the
"inside" faces are really just one face), 24 Edges
and 12 Vertices, so:
F + V - E = -2.
In fact the Euler Characteristic is a basic idea in Topology (the study
of the Nature of Space).
The proof
Despite the formula's name, it wasn't in fact Euler who came up with the first complete
proof. Its history is complex, spanning 200 years and involving some of the greatest
names in maths, including René Descartes (1596 - 1650), Euler himself, Adrien-Marie
Legendre (1752 - 1833) and Augustin-Louis Cauchy (1789 - 1857).
It's interesting to note that all these mathematicians used very different approaches to
prove the formula, each striking in its ingenuity and insight. It's Cauchy's proof,
though, that I'd like to give you a flavour of here. His method consists of several stages
and steps. The first stage involves constructing what is called a network.
Forming a network
Imagine "removing" just one face of the polyhedron pointing upward, leaving the edges
and vertices around it behind, so that you have an open "box". Next imagine that you
can hold onto the box and pull the edges of the missing face away from one another. If
you pull them far enough the box will flatten out, and become a network of points and
lines in the flat plane. The series of diagrams below illustrates this process as applied to
a cube.
As you can see from the diagram above, each face of the
polyhedron becomes an area of the network surrounded by
edges, and this is what we'll call a face of the network.
These are the interior faces of the network. There is also
an exterior face consisting of the area outside the network;
this corresponds to the face we removed from the
polyhedron. So the network has vertices, straight edges
and polygonal faces.
Figure
1:
Turning the cube into a network.
Figure 2: The network has faces, edges and
vertices.
When forming the network we neither added nor removed any vertices, so the network
has the same number of vertices as the polyhedron — V. The network also has the
same number of edges — E — as the polyhedron. Now for the faces; all the faces of the
polyhedron, except the "missing" one, appear "inside" the network. The missing face
has become the exterior face which stretches away all round the network. So, including
the exterior face, the network has F faces. Thus, you can use the network, rather than
the polyhedron itself, to find the value of V - E + F. We'll now go on to transform our
network to make this value easier to calculate.
Transforming the Network
There are three types of operation which we can perform upon our network. We'll
introduce three steps involving these.
Step 1 We start by looking at the polygonal faces of the network and ask: is there a
face with more than three sides? If there is, we draw a diagonal as shown in the
diagram below, splitting the face into two smaller faces.
Figure 3: Dividing faces. In the end we are left with triangular faces.
We repeat this with our chosen face until the face has been broken up into triangles.
If there is a further face with more than three sides, we use Step 1 on that face until it
too has been broken up into triangular faces. In this way, we can break every face up
into triangular faces, and we get a new network, all of whose faces are triangular. We
illustrate this process by showing how we would transform the network we made from a
cube.
Figure 4: This is what happens to the cube's network as we repeatedly perform Step 1.
We go back to Step 1, and look at the network we get after performing Step 1 just
once. Now, by drawing a diagonal we added one edge. Our original face has become
two faces, so we have added one to the number of faces. We haven't changed the
number of vertices. The network now has V vertices, E + 1 edges and F + 1 faces. So
how has V - E + F changed after we performed Step 1 once? Using what we know about
the changes in V, E and F we can see that V - E + F has become V - (E + 1) + (F + 1).
Now we have
V - (E + 1) + (F + 1) = V - E - 1 + F + 1 = V - E + F.
So V - E + F has not changed after Step 1! Because each use of Step 1 leaves V - E +
F unchanged, it is still unchanged when we reach our new network made up entirely of
triangles! The effect on V - E + F as we transform the network made from the cube is
shown in the table below.
Round V E F V - E + F
(a) 8 12 6 2
(b) 8 13 7 2
(c) 8 14 8 2
(d) 8 15 9 2
(e) 8 16 10 2
(f) 8 16 11 2
We now introduce Steps 2 and 3. They will remove faces from around the outside of the
network, reducing the number of faces step by step. Once we begin to do this the
network probably won't represent a polyhedron anymore, but the important property of
the network is retained.
Step 2 We check whether the network has a face which shares only one edge with the
exterior face. If it does, we remove this face by removing the one shared edge. The
area which had been covered by our chosen face becomes part of the exterior face, and
the network has a new boundary. This is illustrated by the diagram below for the
network made from the cube.
Figure 5: Removing faces with one external edge.
Now, we will take V, E and F to be the numbers of vertices, edges and faces the
network made up of triangular faces had before we performed Step 2. We now look at
how the number V - E + F has changed after we perform Step 2 once. We have
removed one edge, so our new network has E - 1 edges. We have not touched the
vertices at all, so we still have V vertices. The face we used for Step 2 was merged with
the exterior face, so we now have F - 1 faces. So V - E + F has become V - (E -
1) + (F - 1) and
V - (E - 1) + (F - 1) = V - E + 1 + F - 1 = V - E + F.
So once again V - E + F has not changed.
Step 3 We check whether our network has a face which shares two edges with the
exterior face. If it does, we remove this face by removing both these shared edges and
their shared vertex, so that again the area belonging to our chosen face becomes part
of the exterior face. This is illustrated below in the case of the network made from the
cube, as it is after performing Step 2 twice.
Figure 6: Removing faces with two external edges.
As we did before we now take V, E and F to be the numbers of vertices, edges and
faces of the network we're starting with. Now how has the number V - E + F been
affected by Step 3? We have removed one vertex — the one between the two edges —
so there are now V - 1 vertices. We have removed two edges, so there are now E - 2
edges. Finally, our chosen face has merged with the exterior face, so we now have F - 1
faces. So V - E + F has become (V - 1) - (E - 2) - (F - 1) and
(V - 1) - (E -2) + (F - 1) = V - 1 - E + 2 + F - 1 = V - E + F.
So once more V - E + F has not changed.
The secret of the proof lies in performing a sequence of Steps 2 and 3 to obtain a very
simple network. Recall that we had repeatedly used Step 1 to produce a network with
only triangular faces. This network will definitely have a face which shares exactly one
edge with the exterior face, so we take this face and perform Step 2. We can perform
Step 2 on several faces, one at a time, until a face sharing two edges with the exterior
face appears. We can then perform Step 3 using this face. We carry on performing
Steps 2 and 3, and keep removing faces in this way.
There are two important rules to follow when doing this. Firstly, we must always
perform Step 3 when it's possible to do so; if there's a choice between Step 2 and Step
3 we must always choose Step 3. If we do not, the network may break up into separate
pieces. Secondly, we must only remove faces one at a time. If we don't we may end up
with edges sticking out on their own into the exterior face, and we'll no longer have a
proper network. To illustrate the process, we'll perform several steps on the cube
network, continuing from where we left it in the last diagram.
Figure 7: Applying our algorithm to the network of the cube.
Now we can ask ourselves one or two questions. Does this process of removing faces
ever stop, and, if it does, what are we left with? A little consideration will show you that
it must stop — there are only finitely many faces and edges we can remove — and that
when it does, we are left with a single triangle. You can see some diagrams describing
the whole process for the network formed from a dodecahedron (recall that this was
one of the Platonic solids introduced earlier).
Now look at the numbers of vertices, edges and faces present in our final network —
the single triangle. We have V=3, E=3, and F=2 — we must still include the exterior
face. Now
V - E + F = 3 - 3 + 2 = 2.
Throughout the whole process, starting with the complete polyhedron and ending with a
triangle, the value of V - E + F has not changed. So if V - E + F = 2 for the final
network, we must also have V - E + F = 2 for the polyhedron itself! The proof is
complete!
Beyond polyhedra
Consequences of Euler's formula extend far beyond the world of polyhedra. One
example are something very small: computer chips. Computer chips are integrated
circuits, made up of millions of minute components linked by millions of conducting
tracks. These are reminiscent of our networks above, except that usually it is not
possible to lay them out in a plane without some of the conducting tracks — the edges
— crossing. Crossings are a bad thing in circuit design, so their number should be kept
down, but figuring out a suitable arrangement is no easy task. Euler's polyhedron
formula, with its information on networks, is an essential ingredient in finding solutions.
Now let's move to the very large: our universe. To this day cosmologists have not
agreed on its exact shape. Pivotal to their consideration is topology, the mathematical
study of shape and space. In the 19th century mathematicians discovered that all
surfaces in three-dimensional space are essentially characterized by the number of
holes they have: our simple polyhedra have no holes, a doughnut has one hole, etc.
Euler's formula does not work for polyhedra with holes, but mathematicians discovered
an exciting generalisation. For any polyhedron, V - E + F is exactly 2 minus 2 times the
number of holes! It turns out that this number, called the Euler characteristic, is crucial
to the study of all three-dimensional surfaces, not just polyhedra. Euler's formula can
be viewed as the catalyst for a whole new way of thinking about shape and space.
Lines and planes. Axioms and basic properties.
For any plane in space there exist
points belonging to that plane and not
belonging to it.
If two points of a line lie in a
given plane, then every point of
the line lies in this plane (the
line belongs to the plane).
If two different planes have a
common point, they intersect in
a line passing through that point.
Through every three points not lying on the same line one can
draw a plane and such plane is unique.
For any plane the axioms of plane geometry hold true.
Corollaries.
Through a line and a point outside it (or through two intersecting
lines, or through two parallel lines) there can be drawn one and
only one plane.
Through each line in space, infinitely many planes can be drawn: a
plane can be rotated about every line lying in this plane.
Skew lines.
Two lines can be positioned in space in
such a way that no plane can be drawn
through them. In the figure, no plane
a
C
A
B
can be drawn through these two lines, since otherwise there would
exist two planes passing through the line AB and the point C. Of
course, two lines not lying in the same plane do not intersect each
other no matter how far they are extended. However they are not
called parallel, the term being reserved for those lines which, being in
the same plane, do not intersect each other no matter how far they
are extended.
Two lines not lying in the same plane are called skew lines.
Exercises.
1. Explain why three-legged chairs standing on flat floor are always
stable, while many four-legged ones totter.
2. Prove that through any two points in space infinitely many planes
can be drawn.
A line and a plane parallel to each other.
A plane and a line not lying in that plane are called parallel if they do
not intersect each other no matter how far they are extended.
Theorem. If a given line (AB) does not lie in a given plane (P), but is
parallel to a line (CD) that lie in it, then the given line is parallel to the
plane.
Proof. Through AB and CD, draw the plane R and assume that the line AB intersect
the plane P. Then the intersection point, being a point of line AB, lies in the plane R
that contains AB, and at the same time it lies in the plane P. Then, the intersection
point, being in both planes R and P, must lie on the line CD of intersection of these
two planes. But this is impossible, since AB||CD by hypothesis. Thus the assumption
that line AB intersects plane P is false.
Theorem. If a given line (AB) is parallel to a
given plane (P), then it is parallel to the
A B
C D
P
R
intersection line (CD) of the given plane with every plane (R) containing
the given line.
Proof. Lines AB and CD lie in the same plane (R) and do not intersect each other,
since (CD) belongs to plane (P), which is parallel to (AB) and therefore has no
common points with it.
Corollary.
If a line (AB) is parallel to each of the two
intersecting planes (P and Q), then it is parallel
to their intersection line (CD).
If two lines (A and CD) are parallel to a third
one (EF), then they are parallel to each other.
Parallel planes.
Two planes are called parallel if they do not have
common points (or, equivalently, if they do not
intersect each other no matter how far they are
extended).
Theorem. If two intersecting lines (AB and
AC) of one plane (P) are respectively parallel
to two lines (A’B’ and A’C’) of another plane
(P’), then these planes are parallel.
Proof. Suppose that the planes P and P’ intersect
along a certain line (DE). Then both lines AB and CD,
which are parallel to the plane P’, are also parallel to the intersection line DE. Since
all three lines AB, AC and DE belong to the same
plane P, this is impossible.
Theorem. If two parallel planes (P and Q)
are intersected by another plane (R), then
A
B
C
DQ
P
AB
C
A’ B’
C’E
D
P
P’
P
Q
R
A
BC
C
D
the intersection lines (AB and CD) are parallel.
Proof. Both lines AB and CD belong to the same plane R and have no common points,
since they belong to parallel planes (P and Q).
Theorem. The segments (AC and
BD) cut off by parallel planes (P and Q)
on parallel lines are congruent.
Proof. Quadrilateral ABCD is a parallelogram
belonging to the plane which contains the
lines AB and CD.
Theorem. Two
angles (BAC and
B’A’C’) whose respective sides are parallel and
have the same direction are congruent and lie
either in parallel planes (P and Q) or in the
same plane.
Proof. Consider the case when angles are not in the same plane Mark arbitrary
congruent segments |AB|=|A’B’|, |AC|=|A’C’| on the sides of the given angles BAC
and B’A’C’. Segments |AA’|, |BB’| and |CC’| are parallel and congruent because
AA’BB’ and AA’CC’ are parallelograms. Hence, BB’CC’ is also a parallelogram, and
triangles ABC and A’B’C’ are congruent.
Problems.
1. Through a given point M, not lying on either of
two given skew lines (a and b) find a line
intersecting each of the given lines.
2. Through a given point M not lying in a given plane
(p), find a plane parallel to the given one.
Corollary.
P
P’
A
B
C
A’
B’
C’
P
Q
A B
C D
M
a
b
Through each point not lying in a given plane (P) there exists a
unique plane parallel to the given one.
A line perpendicular to a plane.
Theorem. If a given line (AO) intersecting a given plane (p) is
perpendicular to two lines (OB and OC) drawn in the plane through the
intersection point (O), then the given line is perpendicular to any other
line (OD) drawn in the plane p through
the same intersection point O.
Proof. On the extension of the line AO, mark
the segment OA’ congruent to AO. On the
plane M draw any line intersecting the three
lines, drawn from the point O, at some points B,
D and C. Connect these points with A and A’ by
straight segments. We thus obtain several
triangles, which we examine in the following
order. First, consider the triangles ABC and
A’BC. They are congruent, since BC is their
common side, BA=BA’ as two slants to the line AA’ whose feet are the same
distance away from the foot O of the perpendicular BO and CA=CA’ for the same
reason. It follows from the congruence of these triangles, that ABC=A’BC.
Next, consider the triangles ADB and A’DB. They are congruent, since BD their
common side, BA=BA’ and ABD=A’BD. It follows from the congruence of these
triangles, that DA=DA’. Finally, consider the triangle ADA’. It is isosceles, and
therefore its median DO is perpendicular to the base AA’.
Definition: A line is called perpendicular to the plane if it intersects
the plane and forms a right angle with every line lying in the plane and
passing through the intersection point. In this case one would also say
that the plane is perpendicular to the line. A line intersecting the
plane, but not perpendicular to it, is called oblique to this plane, or a
slant.
A
A’
O
BD
C
M
Theorem. Through every point (A) lying on a given plane (AB), a plane
perpendicular to the line can be drawn and such a plane is unique.
Proof. Draw any two plane M and N through
the line AB and at the point A, erect
perpendicular to AB inside these planes: AC
in the plane M, and AD in the plane N. The
plane P, passing through the line AC and AD
is perpendicular to AB. Conversely, every
plane perpendicular to AB at the point A
must intersect M and N along lines
perpendicular to AB, i. e. AC and AD
respectively. Therefore every plane
perpendicular to AB at the point A coincide
with P.
Corollary.
All lines perpendicular to a given line (AB) at a given point (A) lie
in the same plane, namely the plane (P) perpendicular to the given
line at the given point.
Indeed, the plane passing through any two lines perpendicular to AB at A is
perpendicular to AB at the point A and therefore coincides with the plane P.
Through every point (C) not lying on a given line (AB) one can draw
a plane perpendicular to the given line and such plane is unique.
Draw an auxiliary plane M through C and the line AB, and drop from C the
perpendicular CA to AB inside the plane M. Every plane perpendicular to AB
and passing through C must intersect the plane M along a line perpendicular
to AB, i. e. along the line CA. Therefore such a plane must coincide with the
plane P passing through the point A and perpendicular to the line AB.
At every point (A) of a given plane (P) , a perpendicular can be
erected and such a line is unique.
P
M N
B
A
D Ca b
In a plane P, draw two arbitrary lines a and b passing through A. Every line
perpendicular to P at the point A will be perpendicular to each a and b, and
therefor will lie in the plane M perpendicular to a at A, and in the plane N
perpendicular to b at A. Thus it will coincide with the intersection line AB of
the planes M and N.
Comparing the perpendicular and slants.
When from the same point A, a perpendicular AB and a slant AC to the
same plane P not passing through A are drawn, the segment BC,
connecting the feet of the perpendicular and the slant is called the
projection of the slant to the plane P.
Theorem.
If from the same point (A)not lying in the given plane (P), a
perpendicular AB and any slants (AC, AD, AE…) to this plane drawn,
then:
1. Slants with congruent
projection are congruent,
2. The slant with the grater
projection is grater.
Indeed, rotating the right
triangles ABC and ABD around
the leg BD, we can superimpose their planes with the plane of the
triangle ABE. Then all the perpendicular AB and all the slants will
fall into the same plane, and all their projection will lie in the same
line. Then the conclusions of the theorem follow from corresponding
results in plane geometry.
A
B
C D
E
P
Pappus theorem. If A, C, E are three points on one line, B, D and F on
another, and if three lines, AB, CD, EF, meet DE, FA, BC, respectively,
then the three points of intersection, L, M, N, are collinear.
This is one of the most important theorems
in planimetry, and plays important role in
the foundations of projective geometry.
There are a number of ways to prove it. For
example, one can consider five triads of
points, LDE, AMF, BCN, ACE and BDF, and
apply Menelaus theorem to each triad. Then,
appropriately dividing all 5 thus obtained
equations, we can obtain the equation
proving that LMN are collinear, too, also by
the Menelaus theorem. However, one can
prove the Pappus theorem directly, using
the method of point masses.
The Pappus hexagon formulation. If the six
vertices of a hexagon lie alternately on two
lines, then the three points of intersection of pairs of opposite sides
are collinear.
The dual of Pappus theorem states that given one set of concurrent
lines A, B, C, and another set of concurrent lines a, b, c, then the lines
x, y, z defined by pairs of points resulting from pairs of intersections
A∩b and a∩B, A∩c and a∩C, B∩c and b∩C are concurrent. (Concurrent
means that the lines pass through one point.)
The Pappus configuration is the configuration of 9 lines and 9 points
that occurs in Pappus's theorem, with each line meeting 3 of the points
and each point meeting 3 lines. This configuration is self dual.
(http://en.wikipedia.org/wiki/Pappus's_hexagon_theorem;
http://www.cut-the-knot.org/pythagoras/Pappus.shtml ).
Instead of simply proving the theorem, consider the following problem.
Problem. Using only pencil and straightedge, continue the line to the
right of the drop of ink on the
paper without touching the drop.
Solution by the Method of the Center of Mass.
Construct a triangle OAB, which encloses the drop, and with the vertex
O on the given line (OD). Let O1 be the crossing point of (OD) and the
side AB. Let us now load vertices A and B of the triangle with point
masses mA and mB, such that their center of mass (COM) is at the point
O1. Then, each point of the (Cevian) segment OO1 is the center of mass
of the triangle OAB for some point mass mO loaded on the vertex O.
The (Cevian) segments from vertices A and B, which pass through the
center of mass of the triangle C, connect each of these vertices with
the center of mass of the other two vertices on the opposite side of
the triangle, OB and OA, respectively.
For the mass mO1 loaded on the vertex O, the center of mass of the
triangle is C1, and the centers of mass of the sides OA and OB are A1
and B1, respectively. Similarly, C2, A2 and B2 are those for the mass mO2
on the vertex O. The center of mass of the side AB is always at the
point O1, independent of mass mO.
If we can show that segments A1B2 and A2B1 cross the given line (OD)
at the same point, D, then our problem is solved, as we can draw
Cevians BA2 and AB2, whose crossing points are on the segment OO1 on
the other side of the drop, by sequentially drawing Cevians BA1 and AB1
and segments A1B2, B1A2, Figure 1(a).
Let us load vertices O, A and B with masses mO1+ mO2, 2mA and 2mB,
respectively, Figure 1(b). The center of mass of OAB is now at some
point C, in-between C1 and C2 (actually, it is not important where it is on
O D
the line OO1). Let us now move point masses mO1 and mA to their center
of mass A1 on the side OA, mO2 and mB to their center of mass B2 on
the side OB, and mA and mB to their center of mass O1 on the side AB.
Now masses are at the vertices of the triangle A1B2O1 with the same
center of mass, C, Figure 1(c). Consequently, the crossing point D of
segments A1B2 and OO1 is the center of mass for masses mO1+mA and
mO2+mB placed at points A1 and B2, respectively. Point C then is the
center of mass for mO1+mO2+mA +mB at point D and mA +mB at point O1,
Figure 1(e). Repeating similar arguments for the triangle A2B1O1, Figure
1(d,f), we see that point D is also the crossing point of segments A1B2
and OO1. Therefore, D is the crossing point of all three segments,
A1B2, A2B1 and OO1, which completes the proof.
A1
B1
O
mA
mB
A2
B2
A
B
m mO1 O2,
C1 C2
O1D
(a)
(c)
(b)
C
A1
B1
O
2mA
2mB
A2
B2
A
B
m mO1 O2+O1D
C
A1
B1
O
mA+mB
A2
B2
A
B
O1D
m mO2 B+
m mO1 A+(d)
C
A1
B1
O
mA+mB
A2
B2
A
B
O1D
m mO1 B+
m mO2 A+
(e)
C
A1
B1
O
mA+mB
A2
B2
A
B
O1D
m mO1 B+m mO2 A+ +
(f)
C
A1
B1
O
mA+mB
A2
B2
A
B
O1D
m mO1 B+m mO2 A+ +
Solving problems using the Archimedes’ method of center of mass.
Recap. Let us assume that a system of geometric points,
has masses associated with each point.
The total mass of the system is . By
definition, the center of mass of such system is point M, such that
For the case of just two massive points, { } and { } this
reduces to
, the Archimedes famous lever rule.
Given the coordinate system with the origin O, we can specify position
of any geometric point A by the vector, connecting the origin O
with this point. For the system of point masses, ,
located at geometric points , position of a point mass
is specified by the vector connecting the origin with point where
the mass is located.
It can be easily proven using the COM definition given above that the
position of the COM of the system, M, is given by
, or,
An important property of the COM immediately follows from the
above. If we add a point to the system, the resultant COM
is the COM of the system of two points: the new point and the point with mass placed at the COM of the first n points,
Menelaus theorem. Points A’, B’ and C’ on the sides, or on the lines that
suitably extend the sides BC, AC, and AB, of triangle ABC, are collinear
if and only if,
Menelaus's theorem provides a criterion for
collinearity, just as Ceva's theorem provides a
criterion for concurrence.
Proof. The statement could be proven with, or
without using the method of point masses.
First, assume the points are collinear and
consider rectangular triangles obtained by
drawing perpendiculars onto the line A’B’. Using
their similarity, one has
Wherefrom the statement of the theorem is obtained by multiplication
(Coxeter & Greitzer).
Alternatively, let us load points A, A’ and C in the upper Figure with the
point masses m1, m2 and m3, respectively. We select m1, m2 and m3, such
that B’ is the COM of m1(A) and m3 (C), and B is the COM of m2(A’) and
m3 (C). The COM of all 3 masses belongs to both segments AB and A’B’,
which means that it is at point C’. Then,
A
B
CB’
A’
C’
A
B
CB’
A’
C’
hA
hB
hC
hA
hB
hC
Wherefrom the Menelaus theorem is obtained by multiplication. The
case shown in the lower figure is considered in a similar way.
Homework review.
Problem. Given two non-collinear vectors, and , show that
a.
is perpendicular to and
is
perpendicular to . Find and .
Solution.
Consider the figure. Vector can be
decomposed into a sum of two vectors, one of
which, , is parallel to , and the other, ,
is perpendicular to it, . The length of
is
. Therefore,
, where we have used that
is
vector of length 1 directed along . Therefore, vector
is the component of vector , perpendicular to . Using
the Pythagorean Theorem for the triangle CAA1, its length is
.
These results could also have been obtained by applying the dot
product,
, and squaring ,
.
Problem. Given two non-collinear vectors, and , show that
b. For any vector ,
.
A
BC
b
aA1
c
Solution.
Consider the figure. Given two
perpendicular vectors, and
, we can decompose
vector
,
where we again used that
is a vector of length 1 along and
is
a vector of length 1 along . Using the definition of the dot
product, , ,
we can rewrite,
, where we have to substitute
and
, which was found in the
previous problem. So we have,
Wherefrom
follows.
Problem. Write equation of the line making an angle with the
vector and passing through the origin, (0,0).
Solution.
We know how to write equation of a line perpendicular to the given
vector. Let us find vector of unit length, which makes angle
A
BC
b
a
A1
cbperp
D
D1
D2
90-with the given vector . The
equation of the line perpendicular to and
passing through the origin is obtained by writing
the dot product of the vector connecting
the origin with an arbitrary point on the line
with the vector : . This line will
make angle with the given vector . How
do we find vector ?
Similarly to what we’ve done in the previous problem, we decompose it
into two components: one along and another along vector
, which is perpendicular to and has the same length .
Wherefrom
, or,
, , and the equation of the
line, , is
Y
XO
l
a=(a ,a )x y
A1
e
A
apepr=(a ,-a )y x
E
Archimedes method of center of mass.
Let us assume that a system of geometric points, has
masses associated with each point. The total mass of
the system is . By definition, the center of
mass of such system is point M, such that
For the case of just two massive points, { } and { } this
reduces to
, the Archimedes famous lever rule.
Heuristic properties of the Center of Mass.
1. Every system of finite number of point
masses has unique center of mass (COM).
2. For two point masses, and , the COM
belongs to the segment connecting these
points; its position is determined by the
Archimedes lever rule: the point’s mass
times the distance from it to the COM is
the same for both points, .
3. The position of the system’s center of
mass does not change if we move any
subset of point masses in the system to the
center of mass of this subset. In other
words, we can replace any number of point
masses with a single point mass, whose
mass equals the sum of all these masses and which is positioned
at their COM.
Given the coordinate system with the origin O, we can specify position
of any geometric point A by the vector, connecting the origin O
with this point. For the system of point masses, ,
located at geometric points , position of a point mass
is specified by the vector connecting the origin with point where
the mass is located.
It can be easily proven using the COM definition given above that the
position of the COM of the system, M, is given by
, or,
An important property of the COM immediately follows from the
above. If we add a point to the system, the resultant COM
is the COM of the system of two points: the new point and the point with mass placed at the COM of the first n points,
Problem. Prove that the medians of an arbitrary triangle ABC are
concurrent (cross at the same point M).
Problem. Prove that the altitudes of an
arbitrary triangle ABC are concurrent
(cross at the same point H).
Problem. Prove that the bisectors of an
arbitrary triangle ABC are concurrent
(cross at the same point O).
Problem. Prove Ceva’s theorem.
A
B
CM
A’C’
B’
Solution. Let us begin with Ceva’s theorem. Given triangle ABC, prove
that three Cevians, AA’, BB; and CC’ are concurrent if and only if,
Let us first prove that if Cevians are concurrent, then this equation
holds. Load the vertices A, B, C of the triangle with masses ,
respectively, such that A’ is the center of mass (COM) of masses
and at points B and C,
and B’ is the center of mass (COM)
of masses and at points A and C,
. The crossing point, M,
of the two Cevians, AA’ and BB’ is the COM of the triangle ABC. If
Cevian CC’ passes through this point, then C’ is the center of mass
(COM) of masses and at points A and B, so
. Therefore,
Conversely, if the above equation holds, then
, so C’
is the center of mass (COM) of masses and . Therefore, Cevian
CC’ passes through the center of mass, M, which is the crossing point
of AA’ and BB’.
The first three statements – that medians, altitudes and bisectors are
concurrent – follow straightforwardly from Ceva’s theorem. They can
also be proven directly, by appropriately choosing the masses
so that the crossing point becomes the COM of the triangle.
For medians, one needs equal loading, . For bisectors,
one can use mass equal to the length of the opposite side, ,
, . For the altitudes,
,
,
.
Problem. Prove that the orthocenter, centroid and circumcenter of any
triangle ABC are collinear (lie on the Euler’s line), and centroid divides
the distance from the orthocenter to the circumcenter in the ratio 2:1.
Homework review.
Problem. Three equilateral triangles are
erected externally on the sides of an arbitrary
triangle ABC. Show that triangle O1O2O3
obtained by connecting the centers of these
equilateral triangles is also an equilateral
triangle (Napoleon’s triangle, see Figure).
Solution. Denote |AB| = c, |BC| = a, |AC| = b.
Let us find the side . Express
, or,
.
Note, that
, and
. Also, ,
, and
, where . Then,
, or,
,
.
Now, using the Law of cosines, , and the Law of
sines,
, where R is the radius of the circumcircle, we obtain
. Obviously, the same expression holds
for the sides and . Hence, triangle is equilateral.
A
B
C
O
A’C’
B’
O1
O2
O3
Problem. Let O be the circumcenter (a center of the circle
circumscribed around) and H be the orthocenter (the intersection
point of the three altitudes) of a triangle
ABC. Prove, that .
Solution.
Let , and , be the diameters of
the circumcircle of the triangle ABC.
Then, quadrilaterals ABA1B1 and BCB1C1
are rectangles (they are made of pairs of
inscribed right triangles whose hypotenuse
are the corresponding diameters), and AHCB1 is parallelogram.
Therefore,
and
. Now,
.
Homework review.
Problem. Given triangle ABC, find the locus of points M such that
. Using this finding, prove that
three altitudes of the triangle ABC are concurrent (i.e. all three
intersect at a common crossing point, the orthocenter of the triangle
ABC).
Solution. Let M be an arbitrary point
on the plane. Express (see Figure)
, ,
.
Then, obviously,
A
B
C
M
b
c
a
A
B
CH
b
c
a
O
A1
C1
B1
Hence, all points M on the plane satisfy the given vector condition.
Now, let H be the crossing point of the two altitudes of the triangle,
and
. Then, by the definition of an
altitude. However, we have just proved that for any point, H included,
. Therefore, , and
is also an altitude.
Problem. Let A, B and C be angles of a triangle ABC.
a. Prove that
.
b. *Prove that for any three numbers, m,n,p,
Solution. Let vectors m, n, p be parallel to
, and , respectively, as in the
Figure. Then,
, wherefrom
immediately follows
.
Problem. Prove that if for vectors a and b, |a+b| = |a-b|, then a b.
Solution 1. Consider the vector addition
parallelogram ABCD shown in the Figure.
Since its diagonals have equal length,
|AC| = |a+b| = |a-b| = |DB|, the
parallelogram is a rectangle (this is
because the diagonals divide it into pairs
of congruent isosceles triangles).
Solution 2. (a+b)2 - (a-b)2 = 4(a∙b) = 4 ab => =>
.
O
A
D
a
b
C
a+b a-b
B
A
B
C
-Cm
p
b
c
an-B
-A
Problem. Show that for any two non-collinear
vectors and in the plane and any third
vector in the plane, there exist one and only
one pair of real numbers (x,y) such that can
be represented as .
Solution. Let us draw parallelogram OACB, whose diagonal is the
segment OC, , and the sides OA and OB are parallel to the
vectors and , respectively. Since , there exists number x, such
that . Similarly, there exists number y, such that .
Then, .
Problem. Derive the formula for the scalar
(dot) product of the two vectors, and
, , using their
representation via two perpendicular vectors of
unit length, and , directed along the X and
the Y axis, respectively.
Solution. It is clear from the Figure that
.
Problem. Consider the following problem. Point A’
divides the side BC of the triangle ABC into two
segments, BA’ and A’C, whose lengths have the ratio
|BA’|:|A’C| = m:n. Express vector via vectors
and . Find the length of the Cevian AA’ if the sides
of the triangle are |AB| = c, |BC| = a, and |AC| = b.
Solution. It is clear from the Figure, that
, and
. Therefore,
O A
B
a
b
C
c
A(a ,a )x y
O X
Y
ex
ey ab
B(b ,b )x y
A
B
C
A’m
n
b
ca
.
Or, we can obtain the same result as
.
For the length of the segment AA’ we have,
. Using the
Law of cosines, we write , and obtain the final
result,
.
Or, equivalently,
.
Substituting m + n = a, we obtain the Stewart’s theorem (Coxeter,
Greitzer, exercise 4 on p. 6).
If AA’ is a median, then |BA’|:|A’C| = 1:1, i.e. m = n = 1, and we have,
,
(AA’ is a median).
If AA’ is a bisector, |BA’|:|A’C| = c:b, i.e. m = c, n = b, and we obtain
, needed for the solution of the homework problem
for the Sep. 30 class, as well as
(AA’ is a bisector).
Problem. In the pentagon ABCDE, M, K, N and L are the midpoints of
the sides AE, ED, DC, and CB, respectively. F and G are the midpoints
of thus obtained segments MN and KL (see Figure). Show that the
segment FG is parallel to AB and its length is ¼ of that of AB, .
Solution. Express via sides of the pentagon,
,
,
.
, or,
Or,
, since .
A
B
CO
N
M
K
D
E
LF
G
A B
C
A B
C D
Method of coordinates: vectors, translations.
A vector is a quantity that has magnitude (length) and
direction. Thus, a vector represents a translation. It
is represented by a directed line segment, which is a
segment with an arrow indicating the direction of
movement, pointing from the tail (the initial point) to
the head (the terminal point). Unlike a ray, a directed
line segment has a specific length.
If the tail is at point A and the head is at point B,
the vector from A to B is written as . Vectors may
also be labeled as a single letter with an arrow,
, or a single bold face letter, such as vector v.
The length (magnitude) of a vector v is written |v|.
Length is always a non-negative real number.
Addition (subtraction) of vectors.
One can formally define an operation of
addition on the set of all vectors (in the
space of vectors). For any two vectors,
and , such an operation
results in a third vector, , such
that three following rules hold,
(1)
(2)
(3)
Vectors (translations) also have the following property (which is not
necessarily true for the so-called “pseudo-vectors”, such as
representing rotations instead of translations; rotations do not satisfy
commutativity (1) either),
, (4)
which allows to define the subtraction
of vectors. The vector is the
vector with the same magnitude as
but pointing in the opposite direction.
We define the subtraction as an
addition of this opposite pointing vector:
Can you see how the vector in the figure below is equal to ?
Notice how this is the same as stating that like with
subtraction of scalar numbers.
Addition and subtraction of vectors can
be simply illustrated by considering the
consecutive translations. They allow decomposing any vector into a
sum, or a difference, of a number (two, or more) of other vectors.
Problems:
1. AA’, BB’ and CC’ are the medians of the triangle
ABC. Prove that
2. If for vectors and
then .
A
B
C
A’C’
B’
Multiplication with a scalar
Given a vector and a real number (scalar) , we can define vector
as follows. If λ is positive, then is the vector whose direction is the
same as the direction of and whose length is times the length of .
In this case, multiplication by simply stretches (if ) or
compresses (if ) the vector .
If, on the other hand, is negative, then we have to take the opposite
of before stretching or compressing it. In other words, the vector
points in the opposite direction of , and the length of is
times the length of . No matter the sign of , we observe that the
magnitude of is times the magnitude of
Multiplication with a scalar satisfies many of the same properties as
the usual multiplication.
1. (distributive law, form 1)
2. (distributive law, form 2)
3.
4.
5.
In the last formula, the zero on the left is the number 0, while the
zero on the right is the vector , which is the unique vector whose
length is zero.
If for some scalar , then we say that the vectors and are
parallel. If is negative, some people say that and are anti-parallel.
Math-9 recap. The Law of Cosines.
For any triangle ABC,
.
To prove it, we consider right triangles
formed by the height AM,
, ,
,
Dot (scalar) product of vectors.
One can formally define an operation of scalar multiplication on
vectors, consistent with the following definition of length, or
magnitude of a vector,
. (8)
and the following properties, which hold if , , and are real vectors
and r is a scalar.
The dot product is commutative:
The dot product is distributive over
vector addition:
The dot product is bilinear:
A B
C
A
B
C
a
b
c
h
M
When multiplied by a scalar value, dot product satisfies:
These last two properties follow from the first two. Now, we recall the
law of cosines, and it is clear from the drawing at the right, that
. (9)
Two non-zero vectors a and b are orthogonal if and only if a • b = 0.
Unlike multiplication of ordinary numbers, where if ab = ac, then b
always equals c unless a is zero, the dot product does not obey the
cancellation law: if a • b = a • c and a ≠ 0, then we can write: a • (b − c)
= 0 by the distributive law; the result above says this just means that
a is perpendicular to (b − c), which still allows (b − c) ≠ 0, and therefore
b ≠ c.
The coordinate representation of vectors.
Let vector a define translation A->B under
which an arbitrary point A with coordinates
(xA,yA) on the (X,Y) coordinate plane is
displaced to a point B with coordinates (xB,yB)
= (xA+ax,yA+ay) on this plane. Then, vector a is
fully determined by the pair of numbers, (ax,
ay), which specify displacements along X and
Y axis, respectively. If O is the point of
origin in the coordinate plane, (xO,yO) = (0, 0),
it is clear, that
, (5)
from where follows the coordinate notation of vectors, = (xA,yA),
= (xB,yB), = (ax, ay) = a. Since numbers ax and ay denote magnitudes
of translation along the X and Y axes, respectively, corresponding to
the displacement by a vector a, it can be represented as a sum of
these two translations, a = ax ex + ay ey , or, (ax, ay) =(ax, 0) + (0, ay), or,
, (6)
where ex = and ey = are vectors of length 1, called unit vectors.
The length (magnitude) of a vector, in coordinate representation, is
, a = |a| =
. (7)
As you can see from the diagram in the Figure above, the length of a
vector can be found by forming a right triangle and utilizing the
Pythagorean Theorem or by using the Distance Formula.
The vector in the figure translates 6 units to the right and 4 units
upward. The magnitude of the vector is from the Pythagorean
Theorem, or from the Distance Formula:
The direction of a vector is determined by the angle it makes with a
horizontal line. In the diagram above, to find the direction of the
vector (in degrees) we will utilize trigonometry. The tangent of the
angle formed by the vector and the horizontal line (the one drawn
parallel to the x-axis) is 4/6 (opposite/adjacent).
The vector operations are easy to express in terms of these
coordinates. If and , their sum is simply
as illustrated in the figure below. It is also easy to see that
And, for any scalar ,
Coordinate representation of vectors. The dot product.
The coordinate representation for the dot
product is (a∙b) = axbx + ayby.
According to the definition
given above, this must be
equivalent to (a∙b) = ab cos
( ), which could be
straightforwardly verified
from the drawing on the
right.
This follows immediately from the formula for
the cosine of a difference of two angles,
,
.
O
A
B
CA(x ,y )
A A
B(x ,y )B B
O X
Y
A(a ,a )x y
B(b ,b )x y
O X
Y
b
Math-9 recap. Derive expressions for the sine and the cosine of the
sum of two angles (see Figure),
Solution. Consider the circle of a
unit radius, |OB| = |OC| = 1, in the
Figure. Then, , ,
, , etc.
Consequently,
.
Similarly,
.
A(x ,y )A A
B(x ,y )B B
O X
Y
B(x ,0)
1
B
A(x ,0)
1
A
C
C1
C2
Geometry 9 recap: curves of the second degree.
Definition of ellipse, hyperbola and parabola: Directrix and focus
Parabola is the locus of points such that the ratio of the distance to a
given point (focus) and a given line (directrix) equals 1.
Ellipse can be defined as the
locus of points P for which the
distance to a given point (focus
F2) is a constant fraction of the
perpendicular distance to a given
line, called the directrix,
|PF2|/|PD| = e < 1.
Hyperbola can also be defined as
the locus of points for which the
ratio of the distances to one
focus and to a line (called the directrix) is a constant e. However, for a
hyperbola it is larger than 1, |PF2|/|PD| = e > 1. This constant is the
eccentricity of the hyperbola. By symmetry a hyperbola has two
directrices, which are parallel to the conjugate axis and are between it
and the tangent to the hyperbola at a vertex.
In order to show that the above definitions indeed those of an ellipse
and a hyperbola, let us obtain relation between the x and y coordinates
of a point P (x,y) satisfying the definition. Using axes shown in the
Figure, with focus F2 on the X axis at a distance l from the origin and
choosing the Y-axis for the directrix, we have
Finally, we thus obtain,
Which is the equation of an ellipse for and of a hyperbola
for . In each case the center is at
and
, and the semi-axes are
and
, which brings the
equation to a canonical form,
We also obtain the following relations between the eccentricity e and
the ratio of the semi-axes, a/b:
, or,
, where
plus and minus sign correspond to the case of a hyperbola and an
ellipse, respectively.
Optical property of ellipse, parabola and hyperbola
If a ray of light is reflected in a mirror, then the reflection angle
equals the incidence angle. This is related to the Fermat principle,
which states that the light always travels
along the shortest path. It is clear from the
Figure that of all reflection points P on the
line l (mirror) the shortest path between
points F1 and F2 on the same side of it is such
that points F1, P, and the reflection of F2 in l,
F’2, lie on a straight line.
The optical property of the ellipse. Suppose
a line l is tangent to an ellipse at a point P.
Then l is the bisector of the exterior angle
F1PF2 (and its perpendicular at point P is the
bisector of F1PF2). A light ray passing through
one focus of an elliptical mirror will pass
through another focus upon reflection.
Proof. Let X be an arbitrary point of l
different from P. Since X is outside the ellipse, we have XF1 +XF2 > PF1
+PF2, i.e., of all the points of l the point P has the smallest sum of the
distances to F1 and F2. This means that the angles formed by the lines
PF1 and PF2 with l are equal.
The optical property of the parabola.
Suppose a line l is tangent to a parabola at a
point P. Let P’ be the projection of P to the
directrix. Then l is the bisector of the angle
FPP’ (see figure).
If a point light source, such as a small light
bulb, is placed in the focus of a parabolic
mirror, the reflected light forms plane-parallel beam (spotlights).
Proof. Let point P belong to a parabola and l’ be a bisector of the angle
FPP’, where |PP’| is the distance to the directrix l. Then, for any point
Q on l’, |FQ| = |QP’| ≥ |QQ’|. Hence, all points Q on l, except for Q = P,
are outside the parabola, so l’ is tangent to the parabola at point P.
Consider the following problem. Given two lines, l
and l’, and a point F not on any of those lines, find
a point P on l such that the (signed) difference of
distances from it to l’ and F is maximal (#2 in the
homework). As seen in the figure, for any P’ on l
the distance to l’, |P’L’| ≤ |P’L| ≤ |P’F| + |FL|, where |FL| is the distance
from F to l’. Hence, |P’L’| - |P’F| ≤ |FL| and the difference is largest
(=|FL|) when point P belongs to the perpendicular FL from point F to l’.
The optical property of the hyperbola. Suppose a line l is tangent to
a hyperbola at a point P; then l is the bisector of
the angle F1PF2, where F1 and F2 are the foci of
the hyperbola (see figure).
Proof. Let point P belong to a hyperbola and l’ be
a bisector of the angle F1PF2. Let F1’ be the
reflection of F1 in l’. Then, for any point Q on l’,
|QF1| = |QF1’|, and |QF2| - |QF1| = |QF2| - |QF1’|
≤ |F2F1’|, again by the triangle inequality. Hence, all points Q on l’,
except for Q = P, are in-between the branches of the hyperbola, so l’ is
tangent to the hyperbola at point P.
Consider the following problem. Given line l and points
F1 and F2 lying on different sides of it, find point P on
the line l such that the absolute value of the
difference in distances from P to points F1 and F2 is
maximal (#1 in the homework). As above, let F2’ be
P
L
Fl
l’
P’
L’
the reflection of F2 in l. Then for any point X on l, |XF1| - |XF2| ≤
|F1F2’|.
Additional properties of ellipse, parabola and hyperbola
The optical property yields elementary proofs of some amazing results.
The isogonal property of conics. From any
point P outside an ellipse draw two tangents
to the ellipse, with tangency points X and Y.
Then the angles F1PX and F2PY are equal (F1
and F2 are the foci of the ellipse).
Proof. Let F1’, F2’ be the reflections of F1
and F2 in PX and PY, respectively (see
Figure).
Then PF1’ = PF1 and PF2’ = PF2. Moreover, the points F1, Y and F2 lie on a
line (because of the optical property). The same is true for the points F2, X and F1’. Thus F2F1’ = F2X + XF1 = F2Y + YF1 = F2’F1. Thus, the
triangles PF2F1’ and PF1F2’ are equal (having three equal sides).
Therefore, ∠F2PF1 + 2∠F1PX =∠F2PF1’ = ∠F1P F2’= ∠F1PF2 + 2∠F2PY.
Hence ∠ F1PX =∠ F2PY, which is the desired result.
Consequence 1. The line F1P is the bisector
of the angle XF1Y.
A similar property holds for the hyperbola
Proof follows the same reasoning as for the
ellipse. Two cases should be considered
separately: when the tangency points X and
Y are either on different branches of the
hyperbola, or on the same branch, as in the
Figure.
Consequence 2. The locus of points from which a given ellipse is seen
at a right angle (i.e., the tangents to the ellipse drawn from such a
point are perpendicular) is a circle centered at the center of the
ellipse (see Figure).
Proof. Let F1, F2 be the foci of the ellipse
and suppose that the tangents to the
ellipse at X and Y intersect in P. Let F1’ be
the reflection of F1 in PX (see Figure). Then
∠XPY =∠F1’PF2 and F1’F2 = F1X+F2X, i.e., the
length of the segment F1’F2 equals the
major axis of the ellipse (the length of the
rope tying the goat). The angle F1’PF2 is
right if and only if |F1’P|2 +|F2P|2 = | F1’F2|2
(by the Pythagorean theorem). Therefore
XPY is a right angle if and only if |F1P|2 + |F2P|2 equals the square of
the major axis of the ellipse, (2a)2. But it is not difficult to see that
this condition defines a circle. Indeed, suppose F1 has Cartesian
coordinates (x1, y1), and F2 has coordinates (x2, y2). Then the
coordinates of the desired points P (x, y) satisfy the condition
But since the coefficients of x2 and y2 are equal (to 2) and the
coefficient of xy is zero, the set of points satisfying this condition is a
circle. By virtue of symmetry, its center is the midpoint of the
segment F1F2.
For the hyperbola such a circle does not always exist. When the angle
between the asymptotes of the hyperbola is acute, the radius of the
circle is imaginary. If the asymptotes are perpendicular, then the
circle degenerates into the point which is the center of the hyperbola.
Definition of ellipse, hyperbola and parabola: Tangent circles.
Ellipse is the locus of centers of all circles
tangent to two given nested circles (F1,R)
and (F2,r). Its foci are the centers of these
given circles, F1 and F2, and the major axis
equals the sum of the radii of the two
circles, 2a = R+r (if circles are externally
tangential to both given circles, as shown in
the figure), or the difference of their radii
(if circles contain smaller circle (F2,r)).
Consider circles (F1,R) and (F2,r) that are
not nested. Then the loci of the centers O
of circles externally tangent to these two
satisfy |OF1| - |O F2| = R - r.
Hyperbola is the locus of the centers of
circles tangent to two given non-nested
circles. Its foci are the centers of these
given circles, and the vertex distance 2a equals the difference in radii
of the two circles.
As a special case, one given circle may be a point located at one focus;
since a point may be considered as a circle of zero radius, the other
given circle—which is centered on the other focus—must have radius
2a. This provides a simple technique for constructing a hyperbola. It
follows from this definition that a tangent line to the hyperbola at a
point P bisects the angle formed with the two foci, i.e., the angle F1PF2.
Consequently, the feet of perpendiculars drawn from each focus to
such a tangent line lies on a circle of radius a that is centered on the
hyperbola's own center.
X
O
F1
Y
0
R
r’
r
r’
F2
X
O
F1
Y
0
R
r’
r
r’
F2
If the radius of one of the given circles is zero, then it shrinks to a
point, and if the radius of the other given circle becomes infinitely
large, then the “circle” becomes just a straight line.
Parabola is the locus of the centers of circles passing through a given
point and tangent to a given line. The point is the focus of the parabola,
and the line is the directrix.
Optical property of ellipse, parabla and hyperbola (continued).
If a light source is placed at one focus of an
elliptic mirror, all light rays on the plane of the
ellipse are reflected to the second focus. Since
no other smooth curve has such a property, it
can be used as an alternative definition of an
ellipse. (In the special case of a circle with a
source at its center all light would be reflected
back to the center.) If the ellipse is rotated
along its major axis to produce an ellipsoidal mirror (specifically, a
prolate spheroid), this property will hold for all rays out of the source.
Alternatively, a cylindrical mirror with elliptical cross-section can be
used to focus light from a linear fluorescent
lamp along a line of the paper; such mirrors are
used in some document scanners. 3D elliptical
mirrors are used in the floating zone furnaces
to obtain locally high temperature needed for
melting of the material for the crystal growth.
Sound waves are reflected in a similar way, so in
a large elliptical room a person standing at one
focus can hear a person standing at the other focus remarkably well.
In the 17th century, Johannes Kepler discovered that the orbits along
which the planets travel around the Sun are ellipses with the Sun at
one focus, in his first law of planetary motion.
The method of coordinates. Ellipse.
An ellipse is a smooth closed curve which is
symmetric about its horizontal and vertical axes.
The distance between antipodal points on the
ellipse, or pairs of points whose midpoint is at the
center of the ellipse, is maximum along the major
axis or transverse diameter, and a minimum along
the perpendicular minor axis or conjugate
diameter.The semi-major axis (denoted by a in
the figure) and the semi-minor axis (denoted by b in the figure) are
one half of the major and minor axes, respectively. These are
sometimes called (especially in technical fields) the major and minor
semi-axes, the major and minor semiaxes, or major radius and minor
radius.
The foci of the ellipse are two
special points F1 and F2 on the
ellipse's major axis and are
equidistant from the center point.
The sum of the distances from
any point P on the ellipse to those
two foci is constant and equal to
the major axis ( PF1 + PF2 = 2a ).
Each of these two points is called
a focus of the ellipse.
Excersise. Prove that the sum of the distances from any point inside
the ellipse to the foci is less — and from any point outside the ellipse is
greater — than the length of the major axis.
Consider the locus of points such that the sum of the distances to two
given points A(-f,0) and B(f,0) is the same for all points P(x,y). It is an
ellipse with the foci A and B. This distance is equal to the length of the
major axis of the ellipse, 2a. Then, for every point on the ellipse,
,
Where from, by squaring this equation twice, we obtain for an ellipse,
.
The equation of an ellipse whose major and minor axes coincide with
the Cartesian axes is
. The area enclosed by an ellipse is πab,
where (as before) a and b are one-half of the ellipse's major and minor
axes respectively.
The eccentricity of an ellipse, usually denoted by ε or e, is the ratio of
the distance between the two foci, to the length of the major axis or e
= 2f/2a = f/a. For an ellipse the eccentricity is between 0 and 1 (0 < e < 1). When it is 0 the foci coincide with the center point and the figure
is a circle. As the eccentricity tends toward 1, the ellipse gets a more
elongated shape. It tends towards a line segment if the two foci
remain a finite distance apart and a parabola if one focus is kept fixed
as the other is allowed to move arbitrarily far away. The distance ae
from a focal point to the centre is called the linear eccentricity of the
ellipse (f = ae).
Other definitions of the ellipse.
Each focus F of the ellipse is associated with a line parallel to the
minor axis called a directrix. Refer to the figure. The distance from
any point P on the ellipse to the focus F is a constant fraction of that
point's perpendicular distance to the directrix resulting in the equality,
e =PF/PD < 1.
In stereometry, an ellipse is defined as a plane curve
that results from the intersection of a cone by a
plane in a way that produces a closed curve. Circles
are special cases of ellipses, obtained when the
cutting plane is orthogonal to the cone's axis.
Consider a circle with the center O, which is
internally tangent to two given circles with
centers F1 and F2 and radii R and r,
respectively, such that one is inside the other
(see figure). The sum of the distances from
O to the centers F1 and F2, equals |OF1| + |O
F2| = R + r, and is independent of the position
of the circle (O,r’). Therefore, the locus of
centers of all such circles [internally tangent
to two given nested circles (F1,R) and (F2,r)] is
an ellipse with foci F1 and F2.
Consider circles (F1,R) and (F2,r) that are not
nested. Then the loci of the centers O of
circles externally tangent to these two
satisfy |OF1| - |O F2| = R - r.
The method of coordinates. Parabola.
Consider locus of points equidistant from a
given point F and a given line l, which does not
contain this point. It is a parabola with focus F
and the directrix l. The line perpendicular to
the directrix and passing through the focus
(that is, the line that splits the parabola
through the middle) is called the "axis of
Y
O X
y = kx2
(0,a)
(0,-a) y = -a
X
O
F1
Y
0
R
r’
r
r’
F2
X
O
F1
Y
0
R
r’
r
r’
F2
symmetry". The point on the axis of symmetry that intersects the
parabola is called the "vertex", and it is the point where the curvature
is greatest.
The easiest way to show that the above definition indeed corresponds
to a parabola, is by using the method of coordinates. Indeed, let line l
be parallel to the X-axis and intersect the Y-axis at y = -a, and focus F
(0, a) lie on the Y-axis at the same distance a from the origin. Then for
any point on the parabola according to the
definition above,
, wherefrom
.
If the directrix is a line y = a, and the focus has
some general coordinates, F(xF, yF), then the
points on the parabola satisfy the equation
, where
and
. If the directrix is parallel to the Y-
axis, then parabola’s equation becomes x = k (y -
yF)2+b.
Parabolas can open up, down, left, right, or in
some other arbitrary direction. Any parabola
can be repositioned and rescaled to fit exactly
on any other parabola — that is, all parabolas
are similar. How do you think the equation of a
parabola with the directrix y = ax + b, at an
arbitrary angle atan(a) to the (X,Y) axes, looks
like? In order to solve this problem, we need to
learn how to find a distance from the point
P(x,y) to a line y = ax + b.
Y
O X
y = k(x- )2x +bF
(0,a)
( ,y )xF F
y = a
Y
O X
(0,b)
( ,y )xF F
y = ax+b
(-B/a,0)
Alternatively (in stereometry), parabola is
defined as a conic section, similar to the
ellipse and hyperbola. Parabola is a unique
conic section, created from the intersection
of a right circular conical surface and a plane
parallel to a generating straight line of that
surface. The parabola has many important
applications, from automobile headlight
reflectors to the design of ballistic missiles.
They are frequently used in physics,
engineering, and many other areas.
The method of coordinates. Hyperbola.
Hyperbola can be defined as the locus of
points where the absolute value of the
difference of the distances to the two foci is
a constant equal to 2a, the distance between
its two vertices. This definition accounts for
many of the hyperbola's applications, such as
trilateration; this is the problem of
determining position from the difference in
arrival times of synchronized signals, as in
GPS.
Similarly to the case of an ellipse, we write,
,
where (f,0) and (-f,0) are the positions of the
two foci, which we chose to lie on the X-axis.
Squaring the above equation twice, we arrive
at the canonical equation for the parabola,
The asymptotes of the hyperbola
(red curves) are shown as blue
dashed lines and intersect at the
center of the hyperbola, C. The two
focal points are labeled F1 and F2,
and the thin black line joining them
is the transverse axis. The
perpendicular thin black line
through the center is the
conjugate axis. The two thick black
lines parallel to the conjugate axis
(thus, perpendicular to the
transverse axis) are the two
directrices, D1 and D2. The
eccentricity e equals the ratio of
the distances from a point P on the
hyperbola to one focus and its
corresponding directrix line (shown
in green). The two vertices are
located on the transverse axis at
±a relative to the center. So the
parameters are: a — distance from
center C to either vertex;
b — length of a perpendicular
segment from each vertex to the
asymptotes; c — distance from
center C to either Focus point, F1
and F2, and θ — angle formed by
each asymptote with the
transverse axis.
, or,
, where .
A hyperbola consists of two disconnected curves called its arms or
branches. At large distances from the center, the hyperbola
approaches two lines, its asymptotes, which intersect at the
hyperbola's center. A hyperbola approaches its asymptotes arbitrarily
closely as the distance from its center increases, but it never
intersects them; however, a degenerate hyperbola consists only of its
asymptotes. Consistent with the symmetry of the hyperbola, if the
transverse axis is aligned with the x-axis of
a Cartesian coordinate system, the slopes of
the asymptotes are equal in magnitude but
opposite in sign, ±b⁄a, where b=a×tan(θ) and
where θ is the angle between the transverse
axis and either asymptote. The distance b
(not shown) is the length of the
perpendicular segment from either vertex to
the asymptotes.
A conjugate axis of length 2b, corresponding
to the minor axis of an ellipse, is sometimes
drawn on the non-transverse principal axis;
its endpoints ±b lie on the minor axis at the height of the asymptotes
over/under the hyperbola's vertices. Because of the minus sign in some
of the formulas below, it is also called the imaginary axis of the
hyperbola.
If b = a, the angle 2θ between the asymptotes equals 90° and the
hyperbola is said to be rectangular or equilateral. In this special case,
the rectangle joining the four points on the asymptotes directly above
and below the vertices is a square, since the lengths of its sides 2a =
2b.
If b = a, the angle 2θ between the asymptotes equals 90° and the
hyperbola is said to be rectangular or equilateral. In this special case,
the rectangle joining the four points on the asymptotes directly above
and below the vertices is a square, since the lengths of its sides 2a =
2b.
If the transverse axis of any hyperbola is aligned with the x-axis of a
Cartesian coordinate system and is centered on the origin, the equation
of the hyperbola can be written as
A hyperbola aligned in this way is called an "East-West opening
hyperbola". Likewise, a hyperbola with its transverse axis aligned with
the y-axis is called a "North-South opening hyperbola" and has equation
Every hyperbola is congruent to the origin-centered East-West opening
hyperbola sharing its same eccentricity ε (its shape, or degree of
"spread"), and is also congruent to the origin-centered North-South
opening hyperbola with identical eccentricity ε — that is, it can be
rotated so that it opens in the desired direction and can be translated
(rigidly moved in the plane) so that it is centered at the origin. For
convenience, hyperbolas are usually analyzed in terms of their centered
East-West opening form.
Other definitions of the hyperbola.
The above definition may also be expressed in terms of tangent circles.
The center of any circles externally tangent to two given circles lies on
a hyperbola, whose foci are the centers of the given circles and where
the vertex distance 2a equals the difference in radii of the two
circles. As a special case, one given circle may be a point located at one
focus; since a point may be considered as a circle of zero radius, the
other given circle—which is centered on the other focus—must have
radius 2a. This provides a simple technique for constructing a
hyperbola. It follows from this definition that a tangent line to the
hyperbola at a point P bisects the angle formed with the two foci, i.e.,
the angle F1P F2. Consequently, the feet of perpendiculars drawn from
each focus to such a tangent line lies on a circle of radius a that is
centered on the hyperbola's own center.
In stereometry, hyperbola can be defined
as a conic cross-section, similar to parabola
and ellipse. Namely, hyperbola is the curve
of intersection between a right circular
conical surface and a plane that cuts
through both halves of the cone. For the
other major types of conic sections, the
ellipse and the parabola, the plane cuts through only one half of the
double cone. If the plane is parallel to the axis of the double cone and
passes through its central apex, a degenerate hyperbola results that is
simply two straight lines that cross at the apex point.
Directrix and focus
Similarly to an ellipse, a hyperbola can be defined as the locus of points
for which the ratio of the distances to one focus and to a line (called
the directrix) is a constant ε. However, for a hyperbola it is larger
than 1. This constant is the eccentricity of the hyperbola. By symmetry
a hyperbola has two directrices, which are parallel to the conjugate
axis and are between it and the tangent to the hyperbola at a vertex.
Reciprocation of a circle*
The reciprocation of a circle B in a circle C always yields a conic section such as a
hyperbola. The process of "reciprocation in a circle C" consists of replacing every
line and point in a geometrical figure with their corresponding pole and polar,
respectively. The pole of a line is the inversion of its closest point to the circle C,
whereas the polar of a point is the converse, namely, a line whose closest point to C
is the inversion of the point.
The eccentricity of the conic section obtained by reciprocation is the ratio of the
distances between the two circles' centers to the radius r of reciprocation circle
C. If B and C represent the points at the centers of the corresponding circles,
then
Since the eccentricity of a hyperbola is always greater than one, the center B must
lie outside of the reciprocating circle C.
This definition implies that the hyperbola is both the locus of the poles of the
tangent lines to the circle B, as well as the envelope of the polar lines of the points
on B. Conversely, the circle B is the envelope of polars of points on the hyperbola,
and the locus of poles of tangent lines to the hyperbola. Two tangent lines to B
have no (finite) poles because they pass through the center C of the reciprocation
circle C; the polars of the corresponding tangent points on B are the asymptotes of
the hyperbola. The two branches of the hyperbola correspond to the two parts of
the circle B that are separated by these tangent points.
Review the following problems from recent homework.
1. A and B are given points, k is a given constant, SMAB is the area of
triangle MAB. Prove that the locus of points M, such that
, is a pair of straight lines.
2. Find the equation of the locus of points equidistant from two
lines, and , where a, b, m, n are real
numbers.