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Title: Lines in the plane, slopes, and Euler

1
Lines in the plane,slopes,and Eulers
formulaby Tal Harel
2
Part 1Lines in the plane and decompositions of
graphs

Sylvester, 1893 Prove that it is not possible
to arrange any finite number of real points so
that a right line through every two of them shall
pass through a third, unless they all lie in the
same right line
Proved by Gallai.

Theorem 1 In any configuration of n points in
the plane,
not all on a line, there is a line which contains
exactly two
of the points.
Proof Let P be the given set of points and
consider the
set L of all lines which pass through at least
two points of
P. Among all pairs (P, l) with P not on l, choose
a pair
(P0,l0) such that P0 has the smallest distance to
l0, with Q
being the point on l0 closest to P0 (that is, on
the line
through P0 vertical to l0).

Claim This line l0 does it!
If not, then l0 contains at least 3 point of P,
and thus 2 of
them, say P1 and P2, lie on the same side of Q.
It follows that the distance of P1 to the line l1
determined
by P0 and P2 is smaller than the distance of P0
to l0, and
this contradicts our choice for l0 and P0.
?

Proof used metric axioms (short distance) and
order axioms (point lies between two points on a
line) of the real plane.
Do we really need these properties beyond the
usual incidence axioms of points and lines?

- No 2-point line
- Therefore (Sylverster-Gallai), cannot be
embedded in a real plane
6

Shown by Coxeter the order axioms will suffice
for a proof of the Sylverster-Gallai theorem.
One can devise a proof that does not use any
metric properties.
We will see a proof of the Sylverster-Gallai
theorem using Eulers formula.
Later

Theorem 2 Let P be a set of n 3 points in the
plane, not
all on a line. Then the set of L lines passing
through at
least two points contains at least n lines.
Proof
n 3 Nothing to show.
Induction on n - We assume correctness for n and
need to prove correctness for n 1
Let P n 1. There exists a line
containing exactly two points p and q of P.
Consider the set P P q and the set L of
lines determined by P
If the points of P do not all lie on a single
line, then by induction L n and hence L
n 1 because of the additional line in L.

If the points in P are all on a single line,
then we have the pencil
Which results in precisely n 1 lines.
?

Theorem 3 Let X be a set of n 3 elements, and
let
A1, , Am be proper subsets of X, such that
every pair of
elements of X is contained in precisely one set
Ai.
Then m n holds.
Proof (Motzkin \ Conway)
For let rx be the number of sets Ai
containing x.
(By the assumptions 2 rx m)
If , then rx Ai because the Ai
sets containing x
and an element of Ai must be distinct. Suppose m
lt n,
then mAi lt n rx and thus m(n - Ai) gt n(m
rx) for
, and we come to a contradiction
?

How does all of this relate to graph theory?
Think of the following statement
If we decompose a complete graph Kn into m
cliques different from Kn, such that every edge
is in a unique clique, then m n.
Let X correspond to the vertex set of Kn and the
sets Ai to the vertex sets of the cliques, then
the statements are identical
Next, we want to decompose Kn into complete
bipartite graphs such that again every edge is in
exactly one of these graphs.
There is an easy way to do this

Number the vertices from 1 to n.
Take the complete bipartite graph joining 1 to
all other vertices. We obtain the graph K1,n-1
which is called a star
Join 2 to vertices 3 to n, resulting in a star
K1,n-2.
We go on, and decompose Kn into stars
K1,n-1, K1,n-2, , K1,1. The decomposition uses
n 1 complete bipartite graphs.
Can we do better, i.e less graphs?
No, according to Graham and Pollak, which leads
us to

Theorem 4 If Kn is decomposed into complete
bipartite subgraphs H1, , Hm, then m n 1.
Proof Let the vertex set of Kn be 1, , n,
and let Lj, Rj
be the defining vertex sets of the complete
bipartite graph
Hj, j 1, , m.
To every vertex i we associate a variable xi.
Since
H1, , Hm decompose Kn, we find (1)
Now, we suppose the theory is false m lt n 1.

Then the system of linear equations
Has fewer equations than variables, hence there
exists a
non-trivial solution c1, , cn. From (1) we
infer
But this implies
A contradiction, and the proof is complete
?

14
Part 2The slope problem

Theorem (Scott, 1970)
If n 3 points in the plane do not lie on one
single line, then
they determine at least n 1 different slopes,
where
equality is possible only if n is odd and n 5.

Proof (a big one)
Has 6 steps
(1) Show for n 2m
(2) Define permutation model
(3) Crossing moves
(4) Touching and ordinary moves
(5) Facts about the moves
(6) Putting it all together

16
(1) The case n 3 is trivial. For any set of
n odd points n 2m 1 5 (m 2) (Not all on
a line) we can find a subset of n 1 2m
points, not all on a line, which already
determines n 1 slopes Thus, it suffices to show
that every even set of n 2m points in the plane
(m 2) determines at least n slopes. So for the
following stages, we only consider configurations
of n 2m points in the plane that determines t
2 different slopes.

17
(2) We construct a periodic sequence of
permutations - Start with some direction on the
plane that is not one of the configurations
slopes - Number the points 1, , n in the order
in which they appear in the 1-dimensional
projection in this direction - The first
permutation p0 123n represents the order of
the points for our starting direction. - The
direction moves counterclockwise. This changes
the projection and the permutation. Changes in
the order of the projected points appear exactly
when the direction passes one of the
configuration slopes

4
1
3
5
6
2
18

- Changes in the permutations are far from
random or arbitrary By performing a 180 degrees
rotation of the direction, we obtain a sequence
of permutations
p0 -gt p1 -gt p2 -gt -gt pt - 1 -gt pt
Which has the following special properties
The sequence starts with p0 123n and ends with
pt n321.
The length t of the sequence is the number of
slopes of the point configuration.
In the course of the sequence, every pair i lt j
is switched exactly once. This means that on the
way from p0 to pt, only increasing substrings are
reversed.
Every move consists in the reversal of one or
more disjoint increasing substrings (one or more
lines in the direction we pass)

19
By continuing the circular motion around
the configuration, one can view the sequence as a
part of a two-way infinite, periodic sequence of
permutations -gt p-1 -gt p0 -gt -gt pt -gt pt1
-gt -gt p2t -gt - pit is the reverse of pi
for all i. - pi2t pi for all . We
will see that every sequence with the above
properties (and t 2) must have length t n.

20
(3) We divide each permutation into a left half
and a right half of equal size m n / 2, we
count the letters (number labels) that cross the
imaginary barrier in the middle. We call pi -gt pi
1 a crossing move if one of the substrings it
reverses does involve letters from both sides of
the barrier. A crossing move has order d if it
moves 2d letters across the barrier the
crossing string has exactly d letters on one side
and at least d letters on the other side. p2
213564 -gt 265314 p3 is a crossing move of
order 2. 652341 -gt 654321 is a crossing
move of order 1.

21
In the course of the sequence p0 -gt p1 -gt -gt
pt, each of the letters 1,2,,n has to cross the
barrier at least once. This implies that, if the
orders of the c crossing moves are d1, d2,,dc,
then we have This also implies that we have at
least 2 crossing moves, since a crossing move
with 2di n occurs only if all the points are on
one line, i.e for t 1. Geometrically, a
crossing move corresponds to the direction of a
line of the configuration that has less than m
points on each side.

(4) A touching move is a move that reverses
some string that is adjacent to the central
barrier, but does not cross it. For example p4
625314 -gt 652341 p5 is a touching
move. Geometrically, a touching move corresponds
to the slope of a line of the configuration that
has exactly m Points on one side, and hence at
most m 2 points on the other side. Moves that
are neither touching nor crossing will be called
ordinary moves. For example p1 213546 -gt
213564 p2. So every move is either crossing,
touching or ordinary, and we use the letters
T,C,O to denote the types of moves.
23

C(d) will denote a crossing move of order d.
Thus for our small example we get p0 T gt p1
O gt p2 C(2) gt p3 O gt p4 T gt p5 C(1) gt
p6 Or even shorter we can record this sequence
as T, O, C(2), O, T, C(1)
24
(5) To complete the proof, we need these 2
facts Between any two crossing moves, there is
at least one touching move. Between any crossing
move of order d and the next touching move, there
are at least d 1 ordinary moves. In fact,
after a crossing move of order d the barrier
is contained in a symmetric decreasing substring
of length 2d, with d letters on each side of the
barrier. For the next crossing move the central
barrier must be brought into an increasing
substring of length at least 2. But only touching
moves affect whether the barrier in in
an increasing substring. This yields the first
fact.

For the second fact, note that with each ordinary
move (reversing some increasing substring) the
decreasing 2d-string can get shortened by only
one letter on each side. And, as long as the
decreasing string has at least 4 letters, a
touching move is impossible. This yields
the second fact.
26

(6) The T-O-C pattern of the infinite sequence
of permutations, as derived in (2), is obtained
by repeating over and over again the T-O-C
pattern of length t of the sequence p0 -gt -gt
pt. Thus with the facts of (5) we see that in the
infinite sequence of moves, each crossing move of
order d is embedded into a T-O-C pattern of
the type () of length 1 (d 1) 1 (d
1) 2d. In the infinite sequence, we may
consider a finite segment of length t that starts
with a touching move. This segment consists of
substrings of the type (), plus possibly extra
inserted Ts.
27

This implies that its length t satisfies which
completes the proof ?
28
Part 3Three applications of Eulers formula

Eulers formula (Scott, 1750)
If G is a connected plane graph with n vertices,
e edges
and f faces, then
n e f 2

Proof Let be the edge set of a
spanning tree for G,
that is, of a minimal sub graph that connects all
the vertices
of G.
- Does not contain a cycle because of the
minimality assumptions.
We construct the dual graph G of G we put a
vertex into
the interior of each face of G, and connect two
such
vertices of G by edges that correspond to common
boundary edges between the corresponding faces.
If there
are several common boundary edges, then we draw
several connecting edges in the dual graph.
- G may have multiple edges, even if G is
simple.

Consider the collection of edges in
the dual graph
that corresponds to edges in E T. The edges in
T
connect all the faces, since T does not have a
cycle but
also T does not contain a cycle, since otherwise
it would
separate some vertices of G inside the cycle from
vertices
outside.
- Cannot be, because T is a spanning sub graph,
and
the edges of T and of T do not intersect.
For every tree the number of vertices is one
larger than the
number of edges. To see this, choose one vertex
as the
root, and direct all edges away from the root
this yields a
bijection between the non-root vertices and the
edges, by
matching each edge with the vertex it points at.

Applied to the tree T this yields n eT 1,
while for the tree
T it yields f eT 1. Adding both equations
we get
n f (eT 1) (eT 1) e 2
?

We will show some consequences of Eulers
formula, and
also three beautiful proofs that have Eulers
formula at
their core
A proof of the Sylvester-Gallai theorem
A theorem on two-colored point configurations
- Both of these use Eulers formula alongside
other arithmetic relationships between basic
graph parameters.
3. Picks theorem about areas of elementary
triangles.

First, we show some important local
consequences of
Eulers formula
- Let G be any simple planar graph with n gt 2
vertices.
G has a vertex of degree at most 5.
G has at most 3n 6 edges.
If the edges of G are two-colored, then there is
a vertex of G with at most two color-changes in
the cyclic order of the edges around the vertex.
Note For each of the three statements, we
assume that G is connected.

Proof
Every face has at least 3 sides (since G is
simple), so
(fk is the number of k-faces)
f f1 f2 f3 and 2e f1 2f2 3f3
yield
f f3 f4 f5 and 2e 3f3 4f4 5f5
and thus 2e 3f 0.
Now if each vertex has degree at least 6, then
(ni is the number of vertices of degree i in G)
n n0 n1 n2 and 2e n1 2n2 3n3
imply
n n6 n7 n8 and 2e 6n6 7n7 8n8
and thus 2e 6n 0.

Taking both inequalities together, we get
6(e n f) (2e 6n) 2(2e 3f) 0
and thus e n f, contradicting Eulers
formula.
(B) As in the first step of part (A), we obtain
2e 3f 0 and
thus
3n 6 3e 3f e
from Eulers formula.
Let c be the number of corners where color
changes occur. Suppose the statement is false,
then we have
c 4n corners with color changes, since at
every vertex there is an even number of changes.
Now every face with 2k or 2k 1 sides has at
most 2k such corners.

So we conclude that
using the inequalities from before again.
So we have e n f, again contradicting
Eulers formula. ?

1. The Sylvester-Gallai theorem, revisited
The Sylvester-Gallai theorem Given any set of n
3
points in the plane, not all on one line, there
is always a line
that contains exactly two of the points.
Proof (Sylvester-Gallai via Euler)

2. Monochromatic lines
Theorem Given any finite configuration of
black and white points in the plane, not all
on one line, there is always a monochromatic
line a line that contains at least two points of
one color and none of the other
Proof
skipped

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