Geometry. Elements of Projective Geometry. - School...

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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.

Transcript of Geometry. Elements of Projective Geometry. - School...

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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.

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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.

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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.

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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

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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.

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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

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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:

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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.

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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

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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

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Geometry 10: cumulative recap.

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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

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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,

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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.

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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.

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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

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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.

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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.

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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

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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

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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

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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

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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.

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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

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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

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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

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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

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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

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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

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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

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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 ).

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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

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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.

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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+ +

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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,

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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

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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

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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

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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

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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, ,

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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’

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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.

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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

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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

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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

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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

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.

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, .

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Solution. Express via sides of the pentagon,

,

,

.

, or,

Or,

, since .

A

B

CO

N

M

K

D

E

LF

G

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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

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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’

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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.

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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

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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)

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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

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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

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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

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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

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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.

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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

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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’

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the reflection of F2 in l. Then for any point X on l, |XF1| - |XF2| ≤

|F1F2’|.

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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

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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.

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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

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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.

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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.

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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

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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.

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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

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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)

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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.

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, 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

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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

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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*

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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.