Models of Incidence Planes – Straight-Fan Planes

Video Series: Incidence Geometry (Tutorials with Step-by-Step Proofs)

(D4M) — Here is the video transcript for this video.

hi everyone welcome back in this episode
models of incident planes uh we’re going to talk about straight fan planes
and before we get started i wanted to mention that this episode is part of the
series incidents geometries tutorials with step-by-step proofs so let’s do
some math so first we’re going to uh quickly review the incident axioms
for any two points there exists a unique line passing through them
for every line are just two distinct points on it
um given uh there exists three distinct points that are non-collinear
um and so any uh thing with action one two and three i’m gonna call an instance
geometry now that’s very brief uh summary right
there but uh if you want more check out the first episode on the series where i
go through these and i explain uh what the series is all about

once you assume axiom one two and three we can then go and prove these 11
theorems here and we did this in the previous episodes check it out
uh link is below in the description uh something that we also talked about
in a previous episode were these three parallel properties
and we’ve done lots of models so far and so um in this episode we’re going to
talk about straight fans you know if you’re not familiar with
what the elliptic parallel property is in the euclidean and the hyperbolic
parallel properties definitely want to check out those those episodes all right
so a straight fan a straight fan plane is an incident plane in other words
axioms a1 82 and a3 hold and the additional condition um
all but one point is incident with exactly one line um

now i think that shabo should be a with um all uh is in is an incident plane
with all but one point incident with exactly one straight line
uh so that should say with there all right so let me just sketch an incident
plane here um so we can think of all the points in the plane around this line
here l except for uh one point and so this will be all the points in the
geometry and i usually like to name this line l and line p but
all the points in the geometry are on this line except for one point
and we’re going to use this definition here to talk about a quad quadrangle
so four distinct points no three of which are collinear is said to be
um are said to be a quad quadrangle as compared to a triangle

okay so um what we’re going to come up here with is our first theorem here an
incident plane that satisfies the elliptic parallel property
is either going to be a straight fan plane or
a quadrangle has to exist in the plane so uh this is the um um
probably the most challenging thing to talk about in this episode here but it’s
not that bad actually um but let me go back here for just a
moment and see if we can sketch a scan a straight straight fan plane so as i was
mentioning here all the points in the geometry are on this
line here l so let’s call this say here a a b c d at e
and let’s say this is point f here and it’s off the line l so this is an
example of a straight um straight fan plane
and this would be a line in the plane and there would be some more lines so we

would have a line through a and f b and f c and f d and f and e and f
this is the point f now i’m drawing the line like there’s
lots of points on it but it’s really just there for intuitive purposes
uh the only line on this uh the only points on this line right here af
are a and f so it’s just a and f and line l is made up of the points a b c
d and e and nothing else so i’m just drawing it here for
illustrative purposes but really the only points on this line are a b c d and e
and the only points on this line here a f are a and f
and same thing with b f c f d f e f so this would be a straight fan plane
um all the axioms actions a1 a2 and a3 hold

and all but one point is incident with one line and that’s the line l
so we can see that every line has at least two points on it
so we could talk about the line bf cf df and ef
and these are all lines that just have d and f e and f
so as you can see here every line has at least two points on it
and there exists uh three collinear points three non-collinear points for
example i can take a b and f and that’s three non-linear points right there all
right so this is a geometry right here uh incidence geometry and it’s actually
a straight fan geometry also so [Music] um notice that any line i take through f

right here is actually going to not be parallel to l
no matter which line i take through f if i take this one this one this one
this one and this one they all have at least one point in common with l so
there are no parallel lines through here um so
let’s go here with our theorem here now so if we satisfy the elliptic parallel
property then we’re either going to have a straight fan plane
or quadrangle exists in the plane so let’s see how this works
so we’re going to assume the elliptic parallel property holds and there’s no
quad quadrangle exists at all in other words any time i take four points
at least three of those have to be collinear
and because this right here says if you take four points
um any three of which are going to be non-collinear so um

we’re gonna assume that doesn’t exist so and notice that three-point geometry
that we talked about before is a straight fan so if you’re not sure
what three-point geometry is uh go back and check out that episode
where we talked about models and three-point geometry was one of the
models that we talked about there but it’s a straight fan um
so the next one here is to consider is what if we have four points
so uh take our four points um in the instant plane here all right so
since no quadrangle exists we can assume that
the three points here are incident with some line
otherwise if if you if you take the four points
and no matter which of the three you take and they’re non-colinear then that
would be a quadrangle but we’re saying none of none of them exist right so that
means we have to be able to find three of the four

that are collinear i’ll keep a drawing over here as we’re going along here
so we have line l and we have a b and c online on line l
so because uh no quadrangles exist and we’re we’re having at least four
points in our geometry now um because we already talked about the
three point the three point you know this already works for it’s a straight fan
there should be a dash in there anyways um we got these three points here
they’re all collinear with line l now we’re assuming that um the euclidean the
elliptic parallel property holds and there’s no quadrangles
so our goal now is to show that we must have a straight fan then
so to show that we have a straight fan um we need to know that all the points
are on this line and uh there’s only one point not off the line right
so since uh all right so we got a b and c all right so now by theorem four

if you remember back to theorem four and theorem form holds as soon as you
have a one a two and x and e three right but a four said any line there exist at
least one point not on it so i’ll put the point up here and call it p
so theorem four says any line there’s some point not on the
line okay so we got our point p now now what i’m going to do is i’m going to
assume for a contradiction that we don’t have a straight fan in other words
there’s another point not on the line so remember a straight fan says all the
points are on one line uh except for one all right well what if we have two so
let’s take a second point and i’m going to assume it’s true that’s not on l also
so let’s assume that we don’t have a straight fan here
so there’s our q there okay now uh the line that goes through p and
q remember a2 holds right so there has to be a line through a2

as there has to be a line through being through p and q
so the lines p and q and l are incident um and the re
and either uh in one of the three points abc or some other point
so because um we have the elliptic parallel property holding
so if i pick a line through p it cannot be parallel to l
so the the line through p and q here uh is not parallel to l in other words
they have to be incident at some point and
by theorem one actually they’re incident at a unique point
so if i draw a line through here it has to hit line l and we don’t know where
the reason why it has to hit line l is because uh the elliptic parallel
property is holding so there’s these two lines cannot be
parallel all right so they so this line has to hit l somewhere

let’s call that point um a prime um and you know a prime could be a or it
could be b or c but at some point on line l
so i’m going to assume the intersection point and when i say v
it’s because uh by theorem one these two lines these two non-parallel lines have
to intersect at at a unique point which we’re calling a prime
now a prime might be a in which case a prime is not b and c
or a prime might be b in which case a and c would not be equal to a prime
or this line might intersect a prime might be c in which case a and b
wouldn’t be a prime so no matter where this line intersects l
two of these three points right here is not equal to the intersection point i’m
going to assume that those two points are b and c
so i’m assuming that a prime is somewhere on this line but it’s not b and c

now if it is b then i will choose a and c here
or if a prime the intersection point is c then i would choose a and b here
so some combination of these three two of these three
is not going to be the intersection point and i’m just going to say b and c
are on the line l and it’s not the intersection point this line and this
line don’t intersect at b c okay so uh that’s what i mean right
there by that all right so now because no quadrangle exists and when i look at
the four points b c p and q remember so we have four points here no
none of them are equal to each other b and c are not equal to each other p is
not c p is not b and the reason why is because these two
points are on line l and p is not and b and c are not q because
uh that would put q on line now which it’s not and p and q are not equal to
each other because we’re assuming there’s another point q that’s not on

the line l right that’s our assumption that we’re
still trying to get a contradiction for all right so anyways we got four
distinct points now because no quadrangle exists of the four points
at least three are incident with a line so when i look at these four points here
there has to be a line going through all three of them but
line l goes through b and c there’s only one line that goes through b and c
so that would put q on l and p on l but p p and q are not on line l right so
if i take these four points because of the fact that no quadrangle exists
three are on a line this could be points p q and b or p q and c
um both cases are impossible since these lines only have one intersection point
in common which is not b or c so therefore there exact exist exactly

one point if we make up this point q we run into these problems here and we just
get a contradiction so p cannot exist i’m sorry so q cannot exist so p is the
only point not on the line l and that means that we have a straight fan
all right so there’s how that works uh if you know you have the elliptic
parallel property you either have a straight fan or you have a quadrangle
all right so next um every straight fan satisfies the
elliptic parallel property so let’s see why that’s true
so in a straight fa straight fan plane all points are instant with one line
call it line l so all points are on this line except for one
which i usually like to call p so clearly every line
because this is off the line and these points are on the line we got lines
going through them so we can have axiom a2 satisfied

and because every line i draw through p is going to intersect line l
right so every line must be incident with this line
and so there are no parallel lines and as soon as you say there’s no
parallel lines you have the elliptic parallel property holding so
it then follows that the elliptic parallel property must hold all right good
so um one more thing um what if you have finitely many number of points
right so let’s say i take 20 points well in a straight fan plane what that
means is we actually have 20 lines so we have the same number of points and the
same number of lines and this makes intuitive sense because
if we take all the points on a line and one point not on the line
then what this point doing what this point is doing is it’s really counting

the lines there’ll be one line here two line three line four line five line
and then the line l itself so in a way p is sort of counting the
number of lines here notice there is one line with all the
points on it except for one so n minus one points
and only one point p not on the line right so that gives us the number of lines
sorry so um yeah that’s all there is pretty much two of that to that right there
um actually though i was thinking here is i asked this question in the last
episode and i wanted to revisit it i asked is there an incident geometry with
infinitely many points and the reason why i asked this question is because all
the models that we’ve given so far had finite number of points
and so i answered that question with a handshake plane
let’s see if we can answer it again with a straight fan plane
is there an incident geometry with infinitely many points so let’s take our

points to be the script p that’s the points
and let’s call it one two three and so on
so these are the points in the geometry and script l will be the lines
um and so what will be the lines so we’ll make up a line that has 2 3 4
and so on that’ll be one line all the natural numbers except one
and then now another line will be 1 2 and then 1 3 and 1 4 and so on
so this is a set of lines this line has all these points on it infinitely many
and each of these lines just have two points on them

so this is the lines and the points and by incident i just mean a point is
incident with a line if it’s in the uh set uh making up the line
all right and so as we can see every um every line has at um at least two points
on it and there exists three non-collinear points so we can take for example um
you know one two and three are three non-collinear points
um there’s no line where all three of them are on it because it would have to
have two and three and one on it uh this one doesn’t have a one on it and all of
these lines only have two points on it so there’s no line going through all
three of these so axiom a3 holds and then axiom a1 holds is there’s only
one line through any two points so if i pick any two of these points
from these lines right here that’ll be a unique line and then i have this line

right here all right um yeah so axioms a1 a2 and a3
hold and this is a straight fan plane and it has infinitely many points on it
and so there’s another example of an incident geometry with infinitely many
points and so we’ve answered that question twice now
all right um hey let’s do some more math click right here and we’ll do some more
geometry i’ll see you there

About The Author
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David A. Smith (Dave)

Mathematics Educator

David A. Smith is the CEO and founder of Dave4Math. His background is in mathematics (B.S. & M.S. in Mathematics), computer science, and undergraduate teaching (15+ years). With extensive experience in higher education and a passion for learning, his professional and academic careers revolve around advancing knowledge for himself and others. His work helps others learn about subjects that can help them in their personal and professional lives.

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