Solving Polynomial Equations 12: Polygonal subdivision with triangles AND quadrilaterals
Wild Egg Maths
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Since the Catalan numbers count subdivisions of a polygon into triangles, and the Fuss or Fuss-Catalan numbers count subdivisions into quadratilerals, we have logically decided to try to see if the elements in our "solving general cubic equations" array correspond to subdivision of a polygon into triangles AND quadrilaterals.
What could be more natural and reasonable? But is mathematics itself natural and reasonable? Let's see, by rolling up our sleaves and using our geometric skills, along with our sense of symmetry, to do some elementary, but interesting counting!
This is part of our investigation of general polynomial equations and how to solve them looking at this classical problem with some fresh eyes.
Also a small correction, in that 0th column the Catalan number after 14 is 42 (not 15!)
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00:01
hello everyone i'm norman wahlberger
00:04
and thank you for joining me on this
00:06
adventure
00:07
where we're exploring solving polynomial
00:10
equations in a novel way
00:12
we have created a interesting array
00:17
uh which are essentially coefficients of
00:20
some
00:21
double power series or rational function
00:24
series
00:25
that are associated with the cubic
00:27
equation of a general kind
00:29
and here are the um
00:33
are the coefficients that um we have at
00:36
least in the top
00:37
corner and what we'd like to do is
00:39
explain this and in our last
00:41
video we suggested that because we know
00:45
that this
00:46
first this
00:49
first column here is associated
00:53
to the catalan story and that this first
00:57
row
00:58
is associated to the fuse catalan
01:01
story so let's just write that down so
01:04
that's the catalan
01:06
series and this one here are the fuse
01:11
catalan numbers
01:14
and the catalan story
01:17
is basically involving the number of
01:22
divisions of a polygon
01:28
into triangles into triangles
01:33
while this one is number of
01:36
divisions into quadrilaterals
01:43
okay now that has made us at least have
01:46
the initial guess
01:48
and uh change color here
01:51
our guess is that
01:54
maybe the general
01:57
entry
02:01
counts
02:04
number of divisions
02:08
of a polygon
02:12
into triangles
02:16
and quadrilaterals
02:21
maybe okay and that's what we would like
02:24
to explore today so we would like to
02:26
do some counting and see if we can get
02:29
some of these
02:29
numbers okay so maybe a first
02:33
place to start with would be to try to
02:36
clarify what the sizes of the polygons
02:39
that
02:40
are going to be appearing here so if we
02:42
go back up here
02:43
then if we go down the first column
02:46
the polygons involved are
02:51
well this is a this one here is a
02:53
triangle
02:55
and then this one here is a
02:59
quadrilateral and this one here is a
03:01
pentagon
03:03
one two three four five etcetera so
03:05
they're going down by one as we go in
03:07
that direction
03:08
while in the fuse catalan story
03:12
what we've got here um this one
03:14
represents uh
03:15
subdivision of a quadrilateral
03:18
and this 3 represents subdivision of a
03:21
hexagon
03:22
and the 12 subdivision into an octagon
03:27
eight sides so kind of going up by two
03:30
as we proceed in this direction okay
03:33
and why is that well like let's let's
03:36
say we're interested in
03:37
trying to explain this five here right
03:39
this five is going to be um
03:42
counting the number of ways of
03:44
subdividing
03:46
a polygon where we start maybe with the
03:48
triangle and go up by
03:51
two or we start with the quadrilateral
03:54
and go
03:55
up by one okay so that five corresponds
03:59
to the five let's say the five
04:02
should correspond to subdivisions
04:06
of a pentagon
04:11
and more generally if we kind of make a
04:13
little table here
04:15
um corresponding to the table that i
04:18
have up here
04:20
um right here we should have a two gun
04:23
and then there was a
04:24
four gun and the sixth gun and eight
04:26
gone and ten gone so on
04:27
while the catalan story went down by one
04:29
two three four
04:31
fives and so on so correspondingly um
04:34
the entries should be five six seven
04:37
going down by one as we add a triangle
04:39
because if you have some polygon
04:42
and you add a triangle like that one
04:44
there's a new triangle
04:46
then what you've done is you've
04:48
increased the number of vertices by one
04:51
okay but on the other hand if you have a
04:53
polygon and you
04:55
add a quadrilateral like that one there
04:59
then you increased the number of
05:01
vertices by two
05:02
so if we go along the rows we should get
05:04
three five seven nine
05:06
eleven go down the column six seven
05:09
eight nine
05:10
go along the rows four six eight ten
05:12
twelve etcetera
05:14
okay so our aim today is to try to see
05:18
if
05:19
um we can count the number of ways of
05:21
subdividing
05:22
these polygons
05:26
subdivide we're going to subdivide them
05:28
into certain number of triangles
05:31
and a certain number of quadrilaterals
05:34
okay
05:35
and hopefully we will end up getting
05:38
these numbers
05:38
up here all right so the great place to
05:41
start is with this five so let's say
05:44
five that corresponds to a pentagon so
05:46
there's a pentagon
05:50
and i might draw it that way i'm going
05:53
to
05:54
put my polygons in a certain fashion
05:57
here
05:57
um so that the the top
06:01
is kind of flat that's maybe like a
06:03
distinguished uh
06:04
top okay how many ways can we subdivide
06:07
this into
06:07
a triangle and a quadrilateral well
06:10
we've already talked about that
06:11
so there's
06:14
five ways one corresponding that line
06:17
another one say corresponding to that
06:19
line
06:20
another one corresponding to
06:24
that line and another one corresponding
06:27
to
06:29
that line and one more
06:34
which one do we have we've got anime
06:36
from there maybe
06:37
here okay so there are five ways
06:43
and that
06:47
explains our five and we've already
06:48
talked about that
06:51
okay so what's the next simplest number
06:53
to work on
06:54
let's try to go for this 21
06:57
which would be
07:01
corresponding to this six down here
07:05
we're going to subdivide a hexagon
07:08
into one quadrilateral and
07:12
two triangles okay back to blue
07:17
okay so we've got a hexagon
07:20
and we want to think about
07:24
subdividing it into two triangles and
07:27
one quadrilateral
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okay um well let's count how many ways
07:33
we can do this
07:35
so maybe we can choose a vertex and from
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that vertex we could
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peel off a triangle on sort of either
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side
07:43
right so there's a triangle right there
07:46
there's another triangle there and then
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what remains is is uh
07:50
is our quadrilateral okay and
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how many ways are there doing this well
07:57
there's
07:58
six possible ways of doing that because
08:00
it all depends on the choice of this
08:02
initial vertex once we've chosen that
08:04
initial vertex
08:06
well then the the other two lines are
08:09
determined so for example uh
08:12
another one would be
08:15
let's say we chose that vertex and then
08:17
we'd have lines emanating like this
08:19
there would be our two triangles and
08:21
there would be our
08:22
quadrilateral and uh that would be
08:25
another example
08:26
okay so there's six like that but
08:29
there's probably more
08:31
okay so let's see how many more can we
08:33
figure out
08:34
so we're gonna have to stick the two
08:35
triangles in another way of
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choosing a triangle would be uh or maybe
08:41
two triangles
08:42
is to um say
08:45
uh choose two adjacent triangles like
08:49
this so there's a triangle and there's a
08:50
triangle
08:51
and then there's a quadrilateral sort of
08:53
opposite it
08:55
okay so
08:58
we have rotational invariance that's the
09:00
basic um
09:01
advantage that we have here with this
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geometrical approach is that
09:05
once we have one configuration we can
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rotate it just like we did in the
09:08
previous case right to get other
09:09
configurations
09:11
so if we rotate this one we have to ask
09:13
well how many different
09:15
configurations can we get with the
09:17
rotations and you see that this picture
09:19
here is determined essentially by
09:20
say this side here this this is a unique
09:23
side it's sort of the free
09:24
side of this quadrilateral centered
09:27
there and if we determine such a
09:30
aside then there's going to be six
09:33
possible ways so there's six possible
09:35
ways
09:35
of arranging things like that
09:39
how many other ways are there well
09:42
let's see uh
09:46
we could make the triangles like this
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on either side of a kind of a central
09:53
rectangle
09:57
now such
10:00
a configuration is
10:04
we can still rotate it okay but once we
10:06
rotate it by 180 degrees we'll
10:08
get back the same rotation in the same
10:10
configuration sort
10:11
so that means that there's actually not
10:13
going to be six possible
10:15
rotations of this there's only going to
10:17
be three of them
10:19
okay one way of thinking about that is
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it's determined by this axis
10:22
like that kind of an axis that
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determines it and there are these other
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opposite
10:27
axes like this three different axes so
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three different possible configurations
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that look like this
10:33
how many do we have so far we have uh
10:35
six plus six plus three that's
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- we need six more um if we're going
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to get 21 which is what we're sort of
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aiming for but let's let's
10:43
um not uh think about that too much
10:45
let's just think about trying to get all
10:47
of them
10:47
so what other kinds of configurations
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how else can we put in these um
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these triangles um
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put them on those sides there we put two
10:56
together like this
11:06
oh actually you see you see this this
11:07
configuration here
11:09
it has actually a kind of a mirror kind
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of thing right
11:13
so suppose i look at this hexagon
11:17
again and still look at this as our
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quadrilateral
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um but instead of subdividing the other
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quadrilateral with a
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diagonal line the way we just did we
11:26
could just do it this way
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okay so we have two triangles like this
11:32
so there's going to be also six ways of
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doing that because we can rotate
11:37
uh this one then again it's also
11:39
determined by this sort of
11:40
special edge here and we can rotate it
11:42
in all of six ways and
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these six are going to be different from
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the previous six so these are these two
11:47
are gonna be different kinds because the
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the nature of the
11:51
uh dividing the line is gonna be
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different
11:57
so how many do we have now we have 6
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plus 6 plus 6 that's 18
12:01
plus 3 is 21. hmm
12:05
that's good are there any others
12:11
well i don't think there are i think
12:13
this is it the total
12:15
equals 21.
12:20
so we can go back up to our number and
12:23
we can take that one off
12:24
i think we convinced ourselves that
12:27
there are 21 ways
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of subdividing a hexagon into
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one quadrilateral and two triangles
12:37
okay the next one that's the easiest
12:39
probably this one here
12:40
- what's that that's corresponding to
12:44
two quadrilaterals and one triangle
12:48
okay just one triangle so that's the
12:51
next one and that corresponds to
12:53
this one here that's a 7-gon
12:57
all right so let's look at a 7-gon
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okay so we'll put the top there and we
13:07
have
13:08
one two three
13:11
four five six okay there's seven gone
13:16
and what we want to do we want to
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uh basically subdivide this
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into one triangle and
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two quadrilaterals i might make a note
13:28
of that over here one triangle and two
13:30
quadrilaterals that's our aim
13:33
okay so um well
13:36
maybe maybe we could start with a
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triangle which is sort of like a
13:39
boundary triangle there's there's a
13:41
triangle
13:42
okay and then we have
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what's left is a hexagon that we need to
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subdivide
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into two quadrilaterals
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okay so that's a bit familiar because
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this number
14:00
here
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is three corresponds to the number of
14:05
ways of dividing a
14:06
hexagon into uh two
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quadrilaterals okay so let me just
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convince myself that so if we here's a
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hexagon here
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remaining down here and we need to
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subdivide it into two quadrilaterals so
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you know we could do that
14:22
in um a number of ways like basically
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sort of like using it there's a diagonal
14:28
okay that subdivises into two pieces
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and there's going to be three of those
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and then this defining vertex here which
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sort of positions the external triangle
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can be rotated
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around in six ways so it looks to me
14:45
like there's three times six
14:47
which is eighteen possible ways of um
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doing that where the triangle that we've
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got is one of these external triangles
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that has two
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sides the dark sternum
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okay what about let's maybe go over here
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one two three four five six seven
15:07
there's another seven got
15:08
now we have to put our triangle in a
15:10
different spot so it's not
15:12
going to have two sides
15:15
external maybe just one side external so
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let's put the external side right here
15:19
okay so there's going to be an external
15:20
slide there
15:21
so the other the other pieces have to be
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quadrilaterals so i have to have four
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vertices one two three four so that
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would be one possibility and
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one two three that would be another
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possibility
15:39
like that would be really the only
15:40
possibility right if we were going to
15:43
also no that didn't have um
15:46
no that's not that wasn't right pardon
15:48
me let me reverse that
15:50
um yeah
15:53
i should have gone like this because i
15:54
said we're going to use this as an
15:56
external um
15:57
side okay so that's the subdivision if
15:59
we're going to use this as an external
16:00
side
16:01
then the other vertex of the triangle
16:03
has to be the opposite one
16:04
in order that we're going to have two
16:06
quadrilaterals on either side
16:09
so how many ways are there doing that
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well it all depends on the initial
16:13
uh side right this we can rotate that
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around in
16:17
seven different ways so i guess seven
16:24
okay um that's when there's
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one external
16:35
side of the triangle okay what if
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there's no
16:38
external sides of the triangle
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so maybe it starts here
16:45
so then there's got to be a
16:47
quadrilateral on one side one two
16:50
three four maybe like that
16:53
um one two three
16:57
well i don't know if we can do that
16:58
right
17:03
ah i know what my mistake is i made a
17:04
mistake didn't i
17:06
which you probably noticed you're
17:07
probably wondering how long it'll take
17:09
him to figure this out
17:10
okay so um when i was counting this one
17:12
here this was not quite right was it
17:14
because i said there was there were
17:15
three ways of subdividing this external
17:18
hexagon that was correct but there's not
17:20
six ways of rotating this round because
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we're in a seven
17:23
gone now okay there's seven ways okay
17:26
that should be three times seven that
17:28
should be twenty-one
17:30
now it makes sense now our total is 21
17:33
plus
17:33
7 is 28. okay
17:39
yeah okay someone makes mistakes
17:41
counting wants to check it has to go
17:42
back let's do it again no problem
17:44
okay so now we're happy we have
17:48
verified this one right here this is 28.
17:53
we're kind of on the right track okay
17:55
but uh we're still not entirely
17:57
convinced we don't want to bet a lot of
17:59
money on this
18:00
at this stage okay but let's carry on
18:02
this next one would probably be this 84.
18:05
that 84 represents let's see
18:08
one quadrilateral and
18:12
three triangles and it corresponds to
18:16
subdivision of a
18:18
another septagon okay so now we're
18:22
looking at the seventh gone again but
18:25
now we have
18:25
one quadrilateral and three triangles
18:33
hopefully we're going to get what was it
18:34
80
18:37
84 okay
18:40
so i'm gonna get getting some practice
18:42
with geometry sort of symmetries
18:45
and also some basic accounting and lots
18:48
of opportunities for making mistakes and
18:50
then correcting them
18:52
okay so let's draw our seven gone again
18:57
and now we need one quadrilateral and
19:01
three triangles so maybe we can
19:04
concentrate on our
19:06
quadrilateral okay so
19:09
how can we stick that quadrilaterally in
19:10
various places well
19:13
let's say we're going to use these four
19:16
points
19:17
okay there's a quadrilateral
19:20
and then we have
19:23
a one two three four five a five gone
19:26
left over that we need to
19:28
subdivide into three triangles
19:33
okay um that we already know
19:37
because that's that five there
19:40
that five is the number of ways of
19:43
subdividing
19:44
a pentagon into three triangles okay
19:47
so the number of ways of doing this well
19:50
there's gonna be
19:51
five corresponding the number of ways of
19:54
subdividing this pentagon for example
19:56
like this one here and then that one
19:57
there so there will be
19:59
three triangles and there's our
20:01
quadrilateral
20:03
and uh so five ways of subdividing that
20:05
exterior pentagon and then
20:07
how many ways of rotating this
20:10
first quadrilateral round well the
20:12
quadrilaterals determine say by that
20:13
side there
20:14
which centers it and so there's going to
20:16
be seven of those
20:17
so 35 possible ways of
20:21
sticking the quadrilateral in that way
20:27
um okay that's good what about another
20:30
way
20:31
three four five six seven how about if
20:35
we
20:36
put in the quadrilateral like this
20:40
okay so there's the quadrilateral now
20:42
there's a triangle on the
20:44
one side and then we have two possible
20:46
ways of of
20:48
subdividing the external quadrilateral
20:51
into into two more triangles
20:53
corresponding to two possible diagonals
20:55
okay so there's
20:57
two ways of doing that and
21:01
how many ways of sticking in that first
21:03
quadrilateral
21:04
well it was determined by um
21:08
by you know these two sides here
21:13
um well maybe we could say uh
21:16
it's that configuration is determined by
21:19
this this vertex here because if we know
21:20
this vertex then we know that external
21:22
triangle right there
21:23
which determines the quadrilateral and
21:25
then there's two ways of filling the
21:26
rest out
21:27
so i would say there's 2 times 7 which
21:29
is 14
21:31
configurations that look like that
21:39
how about if we if we put in the the
21:42
quadrilateral
21:43
is slightly in a slightly sort of
21:45
different position so we still have this
21:47
as a side
21:48
but we decide to go down like this okay
21:52
then that'll be the external triangle
21:54
then we'll have to subdivide one of
21:55
these things
21:56
well i think you can see that that's
21:58
going to be the same as what we have
22:00
just rotate one of these around we're
22:02
going to get one of these
22:03
so this is just the same okay so
22:06
that's not actually going to contribute
22:11
okay what other kinds of ways are there
22:13
of
22:14
putting in a quadrilateral into our
22:18
seven garden well another way
22:21
would be let's see um suppose we
22:25
went like this and
22:28
uh one two
22:33
three four how about this kind of shape
22:36
okay
22:37
it's a little bit different from what we
22:39
had before because now
22:41
there's the three triangles on either
22:43
side
22:45
or on three sides of this quadrilateral
22:50
which is in there so this such a
22:52
configuration is determined by
22:54
that side which determines it and
22:56
there's seven possible ways of rotating
22:57
that around
22:58
so there'll be seven of those
23:03
how many uh do we have so far we've got
23:04
uh numbers of sevens five plus two plus
23:07
one that's eight times
23:08
seven which is 56 we want to get to 84
23:11
which is uh 12. so we need four more
23:16
four more sevens i think okay
23:20
what other possible ways are there one
23:23
two
23:23
three four five six seven
23:28
um well we could have
23:32
maybe a configuration where we take two
23:35
adjacent sides
23:36
forming part of the quadrilateral so
23:38
something like this
23:40
so the quadrilaterals are embedded like
23:42
this where it has
23:44
two external sides which are kind of
23:46
adjacent
23:47
okay there's the quadrilateral
23:50
and then there's going to be a triangle
23:53
on the boundary there and then
23:55
this other thing can be divided in one
23:59
of two ways into
24:01
triangles so
24:04
how many ways altogether of doing that
24:07
kind of thing
24:08
well if we think of starting with this
24:10
vertex this vertex here
24:12
generates those two edges
24:16
and so there's uh we'd rotate that all
24:19
around
24:20
and um rotate that all around
24:25
and for each one of those though there's
24:29
still
24:29
sort of two possible ways of creating
24:31
this this extra diagonal
24:33
because we've got to also have gone this
24:34
way right
24:37
so i think there's times seven times two
24:39
so there's 14 ways there
24:44
but you see there's a kind of a symmetry
24:45
so
24:47
we have to be thinking geometrically
24:49
here so uh
24:50
there's an another variant of that so if
24:52
i choose the same
24:54
two sides i could stick this
24:57
quadrilateral in on this in this fashion
25:00
here sort of leaning to the left instead
25:01
of leaning to the right in some sense
25:03
so now the external triangle is there
25:06
and then we have to subdivide this
25:08
this one here further into two triangles
25:10
so this is a lot like this sort of the
25:12
mirror kind of symmetry
25:14
so there's going to be also 7 times 2
25:16
equals 14 ways
25:20
are there any more are there any more
25:23
well first of all this
25:24
the total so far is 35 plus
25:28
14 plus 7 plus 14
25:31
plus 14 which is 42
25:35
plus um plus 42 that is 84.
25:39
uh before we circle it um are there any
25:41
other ways of
25:44
sticking a quadrilateral
25:48
so if the quadrilateral has one external
25:52
side it was
25:55
uh this is sort of one external side for
25:58
the quadrilateral
25:58
if it has three external sides it's got
26:02
to
26:02
look like this one that's three external
26:04
sides
26:05
um if it has two external sides they can
26:08
be sort of
26:09
separate like like like this one or they
26:12
can be together
26:13
like these two so the only possibility
26:16
is
26:17
could there be a quadrilateral with no
26:19
external sides
26:22
sort of um so like let's say we we we
26:25
just went boundary boundary
26:29
but there's not enough vertices if there
26:30
was an eight gun we could do that but
26:32
there's there's no there's no um way of
26:35
putting in a quadrilateral that has
26:36
sort of four
26:39
sides because there's only going to be
26:41
three triangles that are that are used
26:43
so there's none like this so i think
26:45
this total really is legit
26:48
there's 84. okay that is good so now we
26:51
can tick off
26:54
one more
26:57
okay great so i think we've we've gotten
27:01
our confidence level
27:02
much higher right now i think we're at
27:05
around the 90
27:06
to confidence level maybe even a little
27:07
bit more for optimistic kinds
27:10
but there's still a lot of work to be
27:11
done here right and you can see that
27:13
this job gets more onerous as we get to
27:16
these bigger configurations
27:17
and i think it's a great exercise for
27:19
you okay to start thinking about these
27:21
other ones like how about this 165
27:23
this 180 or this 330 can you actually
27:27
count
27:28
the number of subdivisions of the
27:30
corresponding
27:32
polygons and see if you can get those
27:34
numbers
27:35
uh for the results
27:38
okay obviously if you can't then we need
27:41
to uh
27:42
do something different here you know we
27:44
need to modify this but if you can get
27:46
those
27:47
then that's great and maybe some of you
27:49
will be able to you know go further and
27:51
and do some more accounting so this is
27:54
at a very sort of hands-on level
27:56
we're just experimenting but it's giving
27:58
us a lot of confidence
28:01
we do not have ticks besides these ones
28:03
area but we have we have a fair amount
28:05
of confidence
28:07
so obviously there's still a lot more to
28:09
do but i think we're
28:10
happening in a very good direction here
28:12
because we're we're combining the
28:14
the catalan story with the foose catalan
28:16
story providing a
28:17
larger framework and the reason why
28:20
we're getting this
28:21
is not because we've dreamt it up but
28:23
this is a consequence of our
28:25
investigation into essentially the
28:26
general cubic equation
28:28
so this rich combinatorial situation is
28:31
emerging from this purely algebraic
28:37
investigation