MATH 3581 — College Geometry — Spring 2010 — Solutions to Homework Assignment # 3 B E A C F D.

MATH 3581 — College Geometry — Spring 2010 — Solutions to Homework Assignment # 3

4 2

Page 20, # 6-7. The four-line geometry contains four lines, with any two lines

uniquely determining a point. Thus, there are 4-choose-2 = 6 points in

this space. Also,

C

§1.4. Finite Geometries.

this space is uniform, with three points on each line.

Letting the four lines be labeled 1, 2, 3, 4, the three points

on each line can be listed in a table. The following table

is similar to the one on

page 20 of the text.

1

2

3

4

ABC

ADE

BDF

CEF

B

E

A

C

F

D

From this table, we can see the dual of the statement "every two lines determine a point"

is not true, since not every two points determine a line. For example, the points and

are noncollinear, si

C D

nce these two points do not occur together on any of the four lines

listed in the table. We can also see from this table that every point lies on exactly two

lines, and we can show this by checking tha

t every two lines determine a point:

1 2 , , , , , 1 3 , , , , , and similarly

we compute 1 4 , 2 3 , 2 4 , and 3 4 .

Since every two lines intersect, there are no parallel lines in this space.

A B C A D E A A B C B D F B

C D E F

Page 21, # 16-17. The four-point geometry is the dual of the four-line geometry, with

4 points and 6 lines in space, each two points uniquely determining a line, and three

lines meeting at each point. Denoting the points as , , , , we have the six lines

labeled as 1 , 2 , 3 , 4 , 5 , and 6 .

We can also put this information in a table by

constructing the dual of the table given above:

1 2

A B C D

AB AC AD BC BD CD

A

3

1 4 5

2 4 6

3 5 6

B

C

D

Just as the table for the four-line geometry shows there are pairs of points

which do not lie on a common line, so does the table for the four-point

geometry show there are lines with no point in common. In fact, there

are three such pairs of parallel lines: 1 and 6, 2 and 5, 3 and 4.

If we were to follow the postulates of projective geometry, where every

pair of lines has a point of intersection, then letting 1 6, 2 5,

and 3 4 would define the three diagonal points for the complete

quadrangle .

X Y

Z

ABCD

Gino Fano (1871-1952) studied under Felix Klein as part of the Erlanger program,

translating the work done by Klein's group into Italian. He held a position at the

University

§1.5. Fano's Geometry.

of Turin during the first third of the 20th century, having to move to

Switzerland after Mussolini came into power. He was one of the first to imagine

geometry as an abstract structure in which "points" and "lines" did not necessarily

have their usual geometric interpretations:

"As a basis of our study we assume an arbitrary collection of entities

of an arbitrary nature, entities which for brevity we shall call points,

and this quite independently of their nature."

David Hilbert and others followed Fano's example when they created their

axiomatic systems for geometry during the first half of the 20th century.

Fano is also generally created for being the first person to consider axiomatic systems

for structures containing only finitely many objects, i.e. finite geometries. Initially

he was interested in the three-dimenisonal case, eventually determining the smallest

self-dual 3D projective geometry is one containing 15 points, 15 planes, and 35 lines.

This geometry, denoted as PG(3,2), is still being studied, with new results discovered

each year.

If we restrict to a single plane we get PG(2,2) in which there are 7 points and 7 lines,

and this is the geometry used in the next set of exercises. (Actually, in the plane the

simplest case is PG(2,1), consisting of the 3 vertices and 3 sides of a triangle.)

This homework set ends with a study of PG(2,3).

Postulates for PG(2,2) (Fano's Geometry) :

FG1) There exists at least one line in space.

FG2) Each line in space contains exactly 3 points.

FG3) Not all points lie on the same line. (Hence there must be at least 4 points.)

FG4) Any two points uniquely determine a line. (Thus we need at least 6 lines.)

FG5) Every two lines intersect in exactly one point.

Note FG4 and FG5 are duals of each other. The reason why they are both listed is

because FG5 can be weakened slightly to say "any two lines have at least one point

in common," and then combine this statement with FG4 to prove FG5.

As discussed in class, it is possible to prove the dual statement is true for each of

Fano's postulates, so that every point lies on exactly 3 lines. We can then also prove

there are exactly 7 lines and 7 points in PG(2,2).

, # 7. If we consider the first three Euclidean postulates as being

E1) points exist, E2) lines exist, and E3) planes exist, then automatically

FG1), FG3), and FG4) are all satisfied, with FG4) being the same as E2).

However, FG2) will not be satisfied, since in Euclidean geometry the

Postulate of Measure guarantees infinitely many points on each line,

and FG5) will not be satisfied due to the Parallel Postulate.

# 8. At right is the depiction of PG(2,2) as given in

Figure 1.8 (c), at the bottom of page 22 of the text,

in which each point has been labeled 1 through 7.

To find a triangle in this geometry which uses 4 as a vertex,

what is needed are two other points, not both on the same line as 4.

Since there are three lines through 4, we can pick any two lines (three ways to do this),

and then pick one of the other two points on each of the two chosen lines. Thus, there

should be a total of 3 2 2 12 different triangles using point 4 as one of its vertices:

456, 452, 416, 412, 457, 453, 417

, 413, 467, 463, 427, and 423. Note three of these

triangles use the "circular" line containing 2, 5, 7.

# 9. The one-line geometry containing exactly three points

satisfies FG1) and FG2), but if that's all there is in this space

then FG3) is not satsified.

# 10. The four-point geometry satisfies FG1) and FG3) but not FG2)

since there are only two points on each line.

# 11. The four-line geometry satisfies FG1), FG2), and FG3), but not FG4)

since there are points which do not determine a line.

1 2 3

4

56

7

Girard Desargues (1591-1661) was an architect and an engineer who used geometry

as an engineering tool. For example, he designed and installed a system near Pa

§1.6. Desargues's Perspective Geometry.

ris for

raising water, using an ingenious mechanism called an epicycloidal wheel, based on the

epicycloid from analytic geometry. The story goes that he was helping an artist friend

with problems of perspection, and he inscribed what is now known as Desargues's

Theorem on the back of one of his friend's woodcuts. This was not discovered until

over 200 years later, and it was after this point that he began to be included in lists of

important French mathematicians. Sometime late in his life he did publish a paper on

geometry, cryptically entitled . Brocard later suggested this stood for DALG Des

Arg .ues, Lyons, Geometrie

The finite geometry based on Desargues's Theorem,

illustrated at right, contains 10 lines and 10 points.

This is not a projective geometry because not every

two points determine a line (in the illustration

is not a line) and not every two lines determine a

point (for example lines 7 and 10 do not intersect).

However, this geometry is self-dual, using the

concepts of poles and polars.

In t

AE��

he given chart, point is the pole for line 1,

and line 1 is the polar for . This is because

every line through (and there are three of them,

5, 6, and 7) does not intersect line 1, and line 1 is

A

A

A

the only line with this property.

(Note no polar contains its pole.)

Moreover, every line that does not contain

does intersect 1 at some point.

We also see in the illustration that is

perspe

A

ABC

ctive to from point .

These triangles are also perspective from line 10

, , , which

illustrates Desargues's Theorem.

The pole

DEF J AD BE CF

GHK BC EF AC DF AB DE

J

��

��

is called the center of perspectivity,

while the polar 10 is called the axis of perspectivity.

Pole

A

B

C

D

E

F

G

H

K

J

Polar

1 = [EFG]

2 = [DFH]

3 = [DEK]

4 = [BCG]

5 = [ACH]

6 = [ABK]

7 = [ADJ]

8 = [BEJ]

9 = [CFJ]

10 = [GHK]

(a) 1 3 8 , and so the two triangles perspective from uses these

six points as vertices, with and not belonging to the same triangle. Thus, we have four

choices for the tri

E FG DK BJ E

F G

��

angles perspective from : and , and ,

and , or and . However, we also see that the lines , , , and

do not exist in this geometry, whereas

E FDB GKJ FDJ GKB FKB

GDJ FKJ GDB FB FK GD

GB FD

��

�� 2, 9, 4, and 10. Thus,

the correct choice is and (the green triangles in the illustration) are perspective

from , because these two triangles are also perspective from

FJ GB GK

FDJ GKB

E

��

the line 5, which is the polar

for . To check this, note that , , , and 5 .E FD GK H FJ GB C DJ KB A ACH ��

D

F

K

A H

G

B

C

E

J

10

7

5

(b) 1 4 10 , line 7 is the polar for , and ,

, . Therefore and (the red triangles in the illustration)

are perspective

G EF BC HK ADJ G BK CH A

EK FH D EB FC J BEK CFH

��

��

both from the pole and the polar 7. Whenever we use Desargues's Theorem

to create one of these finite geometries, there will always be ten pairs of triangles, each pair

perspective from one of the t

G

en poles as well as its corresponding polar. We have found

three such pairs so far in this exercise, and you can have fun finding the other 7.

(c) 9 , 4 5 9, 1 2 9, 7 8 9.

Missing from this list

CFJ C F J

are lines 3, 6, and 10, and so we have 3 parallel to 9, 6 parallel to 9, and

10 parallel to 9, where in finite geometries "parallel" means the two lines do not intersect at any

of the given points listed in the geometry.

(d) From part (c) we have lines 3 and 6 both parallel to 9, but 3 and 6 are not parallel to

each other because 3 6 . Thus, in Desargues's finite geometry parallelism is not

a tra

K nsitive relation. In fact, this gives an example of an equivalence relation defined on the

set of lines in Desargues's finite geometry (two lines being equivalent if and only if they are

parallel), and this equivalence relation is symmetric but neither transitive nor reflexive (no

line is parallel to itself). This illustrates why Desargues's geometry is non-Euclidean,

because parallelism is transitive in Euclidean geometry for the set of lines in a given plane.

Consider the geometry illustrated at right, for which the

following statements are true:

# 9. Each line contains exactly 4 points.

# 10. There are 10 points and 5 lines.

#

§1.7. More Finite Geometries.

11. Every two lines uniquely determine a point, and so

there are no parallel (non-concurrent) lines.

# 12. Not every two points uniquely determine a line, as

illustrated by the dotted line.

Here are the duals of these statements:

9 . Each point is on exactly 4 lines.

10 . There are 5 points and 10 lines.

11 . Every two points uniquely determine a line, and so

there are n

o noncollinear points.

12 . Not every two lines uniquely determine a point.

The figure at right satisfies these dual statements.

However, neither of these geometries is a projective geometry because nei

ther

geometry is self-dual (the number of points is not the same as the number of lines).

Consider the geometry illustrated at right,

for which the following statements are true:



# 11. Each line has exactly one line which is parallel.

# 12. Not every two lines uniquely determine a point.




11 . Fo

r each point there is exactly one point such that is not a line.

12 . Not every two points uniquely determine a line.

P Q PQ

��

This geometry is self-dual,but not a projective geometry because of statements

15 and 16. If we try to make this figure projective by dropping the word "Not"

from 16 then we get 4 points and 6 lines, and if we drop "Not" from 16 then

we get 6 points and 4 lines, so in either case we lose the property of being

self-dual. Thus, the 4-point geometry and the 4-line geometry, while they are

duals of

each other, do not give projective geometries because each fails to be

self-dual.

Consider the geometry illustrated at right,

for which the following statements are true:



# 11. Each line has exactly one line which is parallel.

# 12. Not every two lines uniquely determine a point.

Note also there are exactly three pairs of parallel lines,

and for each point there are exactly three other points

which

P

are noncollinear with .




11 . For each point there is exactly one point such that

P

P Q

is not a line.

12 . Not every two points uniquely determine a line.

We should also have exactly three pairs of points, , , , , and , ,

which do not form lines, but the other twelve pairs of thes

PQ

P Q R S T U

��

e six points do form

lines. Also, for each of the twelve lines there should be exactly three other

lines which do not intersect the given line. (Continued on next page)

Starting with the first line, let 1 , i.e. the four points on line 1 are , , , .

Not all points are on 1, so there is a fifth point , and there must be 4 lines joining to

each of , , , and .

ABCD A B C D

E E

A B C D

, or on any li

Rather than listing lines 2 , 3 , etc., we know

there must be two new points each on these four lines (since there are four points on

each line, and we can't use for example , D

EAxx EBxx

B C

13 2

ne containing other than 1,

because 1 is the line uniquely determined by any of the point pairs , , , , , ).

13 12Labeling the 13 points through , there are 78 pairs of points.

212

A

AB AC AD BC BD CD

A M C

of these pairs contain , and we know is on exactly 4 lines. Thus I will label

lines 1 through 4 as the lines containing , then let 5 through 7 be the lines which

(along with 1) contain , then us

A A

A

B e 8 through 10 to take care of , and finally 11

through 13 to take care of . (continued on next page)

C

D

This finite geometry is the next simplest projective geometry after Fano's

geometry PG(2,2). The postulates of PG(2,3) are the same as for PG(2,2) except for

the number of points on each lin

PG(2,3).

e:

There exists at least one line. Every line has exactly 4 points.

Not all points lie on the same line (hence there must be at least 5 points).

Any two points uniquely determine a line (thus we need at least 4 more lines,

and at least 8 more points). Every two lines intersect in exactly one point.

From Fano's Theorem (1.13 on page 31 of the text) we know there must be

1 3 23 13 points and 13 lines in this geometry, so we could also describe

PG(2,3), not in terms of its postulates, but in terms of its properties:

There are exactly 13 points and 13 lines in space.

Each line contains exactly 4 points, and each point lies on exactly 4 lines.

Every two points uniquely determine a line, and any two lines intersect in a point.

Here is the dual of the Solomon’s Seal geometry depicted on the preceding page.

Note there are only six points in this geometry. As in graph theory, the other places where it looks like two lines intersect are not actually points in the space.

Listed below are the 13 lines with initial point assignments as suggested by the

discussion on the preceding page.

Here are the 78 possible point pairs, and I have crossed out the ones which have

already been assigned.

ABACADAEAFAGAHAIAJAKALAM

BCBDBEBFBGBHBIBJBKBLBM

CDCECFCGCHCICJCKCLCM

DEDFDGDHDIDJDKDLDM

EFEGEHEIEJEKELEM

FGFHFIFJFKFLFM

GHGIGJGKGLGM

HIHJHKHLHM

IJIKILIM

JKJLJM

KLKMLM

1 = [ABCD]2 = [AEFG]3 = [AH I J]4 = [AKLM]5 = [BE x x]6 = [BF x x]7 = [BG x x]8 = [CE x x]9 = [CF x x]

10 = [CG x x]11 = [DE x x]12 = [DF x x]13 = [DG x x]

Notice in the “Sudoku” puzzle above I used the row EFG from line 2 and ran it

down three times in the column for lines 5 through 13. I could try that again using HIJ and KLM from lines 3 and 4, but once I but those in lines 5 through 7 I won’t be able to put them in the same

order in the remaining six lines. However, I could permute them, say IJH

for lines 8, 9, 10 and JHI for lines 11, 12, 13. Then all that remains is to

determine the permutations of KLM in the last six lines.

Here is my final solution:

1 = [ABCD]2 = [AEFG]3 = [AH I J]4 = [AKLM]5 = [BEHK]6 = [BF I L]7 = [BGJM]8 = [CE I M]9 = [CFJK]

10 = [CGHL]11 = [DEJL]

12 = [DFHM]13 = [DG I K]

Once I obtain my final solution I can double-check, verifying that each point

is on exactly 4 lines:

1 2 3 4

1 5 6 7

1 8 9 10

1 11 12 13

2 5 8 11

2 6 9 12

2 7 10 13

3 5 10 12

3 6 8 13

3 7 9 11

4 5 9 13

4 6 10 11

4 7 8 12

A

B

C

D

E

F

G

H

I

J

K

L

M

You can also see in the above list that each of the numbers 1 through 13 is used exactly 4 times, and the pattern of numbers given above is the

same as the pattern of letters listed at left.Of course, this makes sense because this

geometry is self-dual.

Now what does this geometry look like?Proceed to the next page and get prepared

to be blown away . . .

A

A

B

CD

B

C

E

D

F

G

H

I

J

K

L

M

PG(2,3)

It took me about an hour of searching the internet, but I finally found a graph of PG(2,3) which I have reproduced here. The graph I found looks nicer than mine, but it is in black and white, while I have color-coded my graph to make it easier to see all the lines. If I can find that website again I will post the link to it on our website.

It this graph, the red circles are supposed to both represent the same point A, the green circles both represent the point B, and so on. The black line around the outside is the line 1 = [ABCD]. The other three lines through A (2, 3, 4) have been colored in red, the lines through B have been colored in green, and the other lines are similarly color coded. I used different shades of blue and yellow/orange for the lines through C and D so you can tell which line is which. Note some of these are actually loops in the graph, for example the line 10 = [GCHL].

MATH 3581 — College Geometry — Spring 2010 — Solutions to Homework Assignment # 3 B E A C F D.

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Transcript of MATH 3581 — College Geometry — Spring 2010 — Solutions to Homework Assignment # 3 B E A C F D.