mathematics hl ib type 1
TRANSCRIPT
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In this portfolio it will be studied the patterns of cutting dierent obect with
dierent dimensions in the ma!imum amount of pieces with a "n# number of
cuts$ %efore starting the study two important things must be e!plained$ The
&rst one is that n will always be a positi'e integer so n()* the reason is that in
this study we will not analy+e the possibility of ha'ing negati'e cuts$ Thesecond clari&cation there is to do e!press what a cut is* a cut is an obect in the
dimension of the obect we are "cutting# ,1* which will di'ide the obect into
dierent parts$ For e!ample if we cut a cube-./0 we use planes-/0* not
lines-1/0 or points-)/0* but when we cut a circle-/0 we will use a line -1/0$ 2e
will not use lines or planes which are bent* using them would change all the
results gotten$
1 dimension
To start the in'estigation we should analy+e what is the ma!imum number of
pieces by cutting a 1 dimensional obect$ The reason we don3t start with ) is
because an obect with this amount of dimension doesn3t e!ist in practical
world as this is a point$
4 1 dimension obect is a straight line* therefore we will s5etch a line being cut
by 1* and &nally . point gi'ing the ma!imum amount of parts* but we will see
the possibility of how the line could not achie'e this$
In this picture we see the line being cut 1 time gi'ing parts$
In this picture the line is being cut time gi'ing . parts$
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I
n this picture the line is being cut . times gi'ing 6 parts$
4s notice when doing a number of cuts* the number of pieces is e7ual the
number of cuts 8 1* when tryin% to ha&e the ma$imum number of
pieces'
These tabulated results will help us to gi'e a general rule which relates number
of cuts -n0 and ma!imum number of segments* which we will call 9: 'alues 9 'alues$1 .. 6
;ust by loo5ing at it is clear that the general rule formula is gi'en by<
( ) n*1
If the line would ha'e cut without reaching the ma!imum number of segments
the result would ha'e been dierent$ For e!ample let us assume the second
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point would been put o'er the &rst one would be done li5e this<
In this picture we used lines rather than dots because it would be complicated
to show two dots cutting in the same place$
Here* although two cuts ha'e been drawn* there are ust segments$ For this
reason it is important that in this study the ma!imum amount of segments or
parts is used to create rules$
dimensions
:ow to continue the study we ha'e to see what happens when a dimension
obect* li5e a circle* is cut dierent number of times$ I will s5etch the circle
being cut from = chords* gi'ing the ma$imum number of parts$ To create the
ma!imum parts in a circle e'ery chord must cut the biggest number of e!isting
parts$ Here we will use a circle* but we could ha'e used any other dimensional obects such as a s7uare$
In this picture we see a circle cut in by 1 chord$
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In this picture we see a circle cut in 6 by chords$
In this picture we see a circle cut in > by . chords$
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In this picture we see a circle cut in 11 by 6 chords$
Finally* in this picture we see a circle cut in 1 parts by = chords$
4fter ha'ing made sure that this drawing show the ma!imum amount of parts
we can get by cutting them with dierent number of chords we can tabulate
the results to &nd using technology a general formula which relates number of
cuts -n0 and ma!imum number of parts -which we will call @0
:umber of cuts -n0 Ma!imum number of parts -@01 6. >6 11
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= 1 This can be done using technology* meaning that we will use a graphical
calculator to &nd out the formula* this is done by typing the number of cuts and
ma!imum number of parts in the table labeled A1 and A -this table is found by
clic5ing 9T4T9 and after pressing B/IT on TI. or TI60
4fter ha'ing imported these numbers in the table we ha'e to decide whichdegree function it is to get a general rule$ In one dimension the general rule
was a polynomial with a degree 1* therefore we could hypothesi+e to use for
the general rule in dimensions a polynomial of degree * so we press on the
calculator 9tats* after we go on C4AC and we go on Duadreg were we set to use
the data in table in A1 and A * the calculator will gi'e us this answer<
This means that the general rule is gi'en by< @E1
2n ²+
1
2n+1 * where @ is the
ma!imum number of parts we can get by cutting a circle with n cuts$
To &nd out the recursi'e rule we don3t need to use a calculator or technology*
there is a longer and manual way which will be good to use because as it will
show us other features of the pattern between dierent dimensions$ This method wor5s by writing the 'alues of @ we get with n cuts after we
calculate the change between two consecuti'e 'alues and we repeat this
procedure till the changes are all e7ual
This table will e!plain it better* and it will gi'e us the 'alues of the ma!imum
possible parts we can get with n cuts* @$
n 'alues
-number of
cuts0
1 . 6 =
@ 'alues 6 > 11 1
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1 -change
10
-6,0 . ->,60 6 -11,>0 = -1,110
-change
0
1 -.,0 1 -6,.0 1 -=,60
4fter ha'ing &nished the table to &nd the recursi'e rule we need to do a second
table* but in this one @ is replaced by a polynomial of degree * the choice of
the degree is gi'en by the number of changes done in the &rst tableG these two
numbers must be e7ual$
n 'alues 1 . 6 =an8bn8c a8b8c 6a8b8c a8.b8c 1a86b8c =a8=b8c
1 -change
10
.a8b =a8b >a8b a8b
-change
0
a a a
:ow we arri'e to the &nal step* we need to wor5 out a* b and c* to do so we
associate e'ery place from the &rst table to their corresponding places in the
second table* so we get e'ery number in terms of a* b and c -for e!ample
aE10* after we start sol'ing e'ery e7uation from the last change to ha'e ust
one un5nown<
aE1
aE 1
2
4t this point we 5now the 'alue of a* so we can pass to the ne!t step* wor5ing
out the 'alue with two un5nown* one of them being a<
.a8bE
.!- 1
208bE
bE
4−3
2
bE 1
2
Final step* wor5ing out the e7uation with . un5nowns* two of them being a and
b* so now 5nows 'alues
a8b8cE
1
28
1
28cE
cE1
4s a &nal answer we get1
2 n8 1
2 n81 ust li5e the pre'ious case$
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4nother thing* which I belie'e is more important than this results* which we
already got with the calculator* is that in this way we clearly see that to &nd the
ne!t 'alue of @ we ha'e to do the addition of the 'alue we recei'e with 1 cut
less for @ and 9* to ma5e it more clear<
F-@*n0 is a the ma!imum number of parts we can get in dimension producedwith n cuts then* F-@*n0E F-@*n,108F-9*n,10 * to ma5e an e!ample from the
table* let us ta5e nE6 and dimension so the 6th 'alue of @ will be gi'en by
the .rd 'alue of @ and the .rd 'alue of 9
F-@*60E F-@*.08F-9*.0
E>86
E11
11 is e!actly the answer we were loo5ing for$ %ut does this mean this
conecture is true for e'ery 'alue of n?
F -@* n0E1
2 n8
1
2n81
F-9* n,10E-n,1081
F-@* n,10E 1
2-n,108
1
2-n,1081
F-@* n,108 F-9* n,10E
1
2-n,108
1
2-n,10818-n,1081E
12
n,n 8 12
8 12
n, 12
818n,181E
1
2n8
1
2n81
This conecture has been pro'en right* meaning it will wor5 for e'ery 'alue of n$
In the conecture we use F-@*n0E F-@*n,108F-9*n,10* but what would the
formula be if we don3t want to use n,1?
In this case we would write @-n0EJ89-n0* where J is an algebraic e!pression in
terms of nG to &nd n this procedure must be used<
1
2 n8
1
2n81 EJ8n81
JE 1
2 n8
1
2n81,n,1E
1
2 n,
1
2n8$
9o @-n0E 1
2 n,
1
2n89-n0
! dimension
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The continue the in'estigation we will pass to a . dimensional obect* we will
use a cubeG in some of the s5etches where I will show how the cube will be cut
to get the ma!imum number of parts the cuts aren3t clearG I will try e!plain how
I drawn it and how I thought about it to facilitate the understanding of thes5etches$ 4fter ha'ing seen the s5etches I will tabulate the result to &nd a
general formulae which will gi'e us the ma!imum number of parts* which we
will nominate P* gi'en a number of cuts$
In this picture we see the cube cut by 1 plane into parts$
In this picture we see the cube cut by planes into 6 parts
$
In this picture we see the cube cut by . planes into parts$
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In these pictures we see the cube cut by 6 planes into 1= parts$
9een from top seen from
bottom
In this picture we see the cube cut by = planes into parts$
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9een from top 9een from bottom
4s said before the last two cuts are 'ery confusing but to ma5e it easier to
understand what I did* I will e!plain the logic behind these cuts$
For the 6th cut we need to thin5 about the circle we cut before using . chords
and therefore recei'ing > pieces* though we "transform# the circle in a s7uare*
as you remember I wrote before that we didn3t need to use a circle we could
ha'e used any dimensional plane and the rules and parts would be the same$
To get the ma!imum number of parts using 6 planes in the cube I imitate the
.rd chord I drawn on the / s5etch$
In this picture we see the cube seen from the top and from the side* the red*
the blue and green line represent the 1st . planes the 6th is drawn in both
drawings e!actly li5e the .rd chord in the dimensional obects* in this way we
can get 1= pieces from the cube$
9een from top 9een from side
The =th plane does as well follow this idea of imitating the 6th chord in the /
drawing$
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:ow we will see a table showing the number of cuts* planes* and the
corresponding P* number of ma!imum parts$ 4fter this we will create a general
rule using technology and the method table we used before$
:umber of cuts -n0 Ma!imum number of parts -P01 6. 6 1==
Following the procedure e!plained in the pre'ious chapter we will use the
calculator to &nd a general formula$ These numbers must be imported on the
calculator in the columns* I will use A1 and A but you can chose any$ 2e will
chose that the calculator must &nd the result with a polynomial of degree .* wechose this number because of the precedents studies in 1d and d the general
formula was polynomial in degree respecti'ely of 1 and so we hypothesis the
degree of the formula is e7ual to the dimension we are wor5ing on$
Here we see two screenshots of the calculator -TI60 showing answer$
2e might want to change the decimal into fractions for simplicity<
aE)$1>E1K bE) cE)$......E=K
4ccording to the calculator the answer is PE 1
6n.8
5
6n 81$
This formula gi'es us the ma!imum amount of parts created for a gi'en n* but
it doesn3t show us any patterns with the cutting of obects in 1/ and /$
Therefore I rather use the method I showed you before* creating a table of
'alues and &nding the change between two consecuti'e numbers$
n 'alues 1 . 6 =P 'alues 6 1=
1 -change
10
-6,0 6 -,60 > 1=,0 11 -,1=0
-change
0
-6,0 . ->,.0 6 -11,>0
. -change
.0
1 -.,0 1 -6,.0
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To &nd out the general rule formula we should do a second table and replacing
P 'alues by the polynomial an.8bn8cn8d* but because we already 5now the
'alues of a* b* c and d doing it would not help us$
From this table though there is a clear pattern which will help us with all the
in'estigation* we see that the change between two consecuti'e 'alues of
ma!imum parts which can obtained with n cuts* gi'es us the 'alue of the
dimension of the obect we are cutting minus 1* for e!ample 1 in this table
corresponds to @ 'alues and corresponds to 9 'alues$ 2e could ha'e
noticed it also on the &rst table that we did* but it may ha'e been ust a
coincidence* while now although not pro'en it seems li5e a real pattern$
4nother way to &nd a P 'alue is to apply a conecture* which is 'ery similar to
the conecture we found in /* F-P*n0EF-P*n,108F-@*n,10 where F-P*n0 is thema!imum number of parts we can get by cutting a .d obect with n cuts$
This formula is gi'en by the table on top* an e!ample lets use nE* so for the
ne!t term of n -.0 we will be e!pecting P to be according to the table and the
general rule and the drawings* let us see this conecture to see if it gi'es us the
same answer$
F-P*.0EF-P*08F-@*0E
686E
Therefore the answer is the right one* but this isn3t pro'ing the conecture is
right* it might ust be for coincidence that it gi'es us in this case the same
results as the general rule* let us try to pro'e it for any positi'e 'alues of n$
F-P*.0E 1
6n.8
5
6n 81$
F-P*n,10E1
6-n,10.8
5
6-n,10 81
F-@*n,10E 1
2-n,108
1
2-n,1081
F-P*n,108F-@*n,10E
1
6-n,10.8
5
6-n,10 818
1
2-n,108
1
2-n,1081E
1
6n. ,
3
6n8
3
6n L
1
68
5
6n ,
5
6 8 1 8
1
2n,n 8
1
28
1
2n ,
1
28
1E
1
6n.8
5
6n8 1
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This calculations show us that we could use this conecture is true* we could try
tough to not use any :,1 and ust ha'e a formula with F-P*n0 and f-@*n0* where
P e7uals @ 8 $
PE 1
6n.8
5
6n 81$
@E 1
2-n08
1
2-n081
EP,@E
1
6n.8
5
6n 81,
1
2-n08
1
2-n081
1
6n.,
1
2n8
1
3n$
4s said in the pre'ious chapter @EJ89* so we could write down the e7uation
where PE8J89 which would gi'e this<
PE 1
6n.,
1
2n8
1
3n 8
1
2 n,
1
2n 8n 8 1E
1
6n.8
5
6n 81$
" dimensions
In this section we will continue the study in a more theoretical way* here we will
analy+e a 6 dimensions obect$ 4t this time we do not 5now how to show in a
picture 6 dimensions* but there has been some hypothetical hyperspaces which
are in 6 / such as Binstein spacetime$ Nsing the data we ha'e the pre,
collected* we can wor5 out when 6/ obect will be separetaed with n cuts* how
many part we can get* the solutions D$
The &rst step is &nding the 'alues of D* to do this we will use the table showing
the dierences we used in . / and we can hypothesi+e that the 'alues of P are
e7ual to the change of two consecuti'e 'alues of D* the hypothesis is gi'en
that for any other dimension this is true$
n 'alues 1 . 6 =P 'alues 6 1=
1 -change
10
-6,0 6 -,60 > 1=,0 11 -,1=0
-change
0
-6,0 . ->,.0 6 -11,>0
. -change
.0
1 -.,0 1 -6,.0
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Our new table will start with for n e7ual to 1 the 'alue of D will be * as for an
obect in any dimension* with 1 cut* the obect will be di'ided into * this part
will be e!plained in the last chapter more accurately$
The second 'alue of D* so for nE* will be the &rst 'alue 8 the &rst 'alue of P*
therefore for nE D E 8E6$ 2e continue till we ha'e the 'alues for the &rst= cuts* which are shown in the following table<
n 'alues 1 . 6 =D 'alues 6 1 .1P 'alues
1 -change
10
6 1=
-change
0
-6,0 6 -,60 > 1=,0 11 -,1=0
. -change.0
-6,0 . ->,.0 6 -11,>0
6 -change
60
1 -.,0 1 -6,.0
2ith this table we can &nd a general rule for the 6/ obects* using the
calculator* or by hand$ I will enter the n 'alues and the D 'alues on the A1 A
table in the calculator and after e!pecting a result of a function of degree 6* the
reason of why we are e!pecting is that in the pre'ious dimension their
respecti'e general function3s degree was corresponding to the number of
dimensions$
2e can change the numbers into decimals for simplicity<
)$)61>E1K6 ,)$)....E,1K1 )$6=...E11K6 )$=....E>K1
The calculator gi'es this result<1
24n6,
1
12n.8
11
24n 8
7
12n 81$
The recursi'e formula to &nd the 'alues of D is the same as the one we used
to create the table<
F-D*n0EF-D*n,108F-P*n,10 where F-D*n0 is the ma!imum number of parts we
can get by cutting a 6d obect with n cuts$
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In the pre'ious chapters* after the conecture I always added an e!ample
showing that for a gi'en 'alue of n the result was the same using this
conecture or the drawings$ For this dimension it is pointless doing this as the
'alue we ha'e* are calculated than5s to the formula* but we can still chec5 the
formula mathematicallyF-D*n0E
1
24n6,
1
12n.8
11
24n 8
7
12n 81$
F-P*n,10E 1
6-n,10.8
5
6-n,108 1
F-D*n,10E 1
24-n,106,
1
12-n,10.8
11
24-n,10 8
71
12-n,10 81$
F-D*n,108F-P*n,10E
1
24-n,106,
1
12-n,10.8
11
24-n,10 8
7
12-n,10 81 8
1
6-n,10.8
5
6-n,108
1E1
24n6 ,
4
24n. 8
6
24n ,
4
24n8
1
24 ,
1
12n. 8
3
12n,
3
12 n 8
1
128
11
24n ,
22
24n 8
11
24 8
7
12n ,
7
12 8 1 8
1
6n. ,
3
6n8
3
6n L
1
68
5
6n ,
5
6 8 1 E
1
24n6,
1
12n.8
11
24n 8
7
12n 81
4s we did in the pre'ious chapter we will try to create a formula which gi'es us
the 'alue of D* using the e7uation found the pre'ious chapter and a secondary
e7uation which we will wor5 on* so in this case we ha'e to &nd DE8P* we will
wor5 out by using simple calculations$
F-D*n0E 1
24n6,
1
12n.8
11
24n 8
7
12n 81
F-P*n0 E1
6-n0.8
5
6-n08 1
ED,PE
1
24n6,
1
12n.8
11
24n 8
7
12n 81 ,
1
6-n0.8
5
6-n08 1E
1
24n6,
13
12n.8
11
24n ,
1
4 n$
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4s stated before PE8J89* so we could rewrite D as<
DE 1
24n6,
13
12n.8
11
24n ,
1
4 n 8
1
6n.,
1
2n8
1
3n 8
1
2 n ,
1
2
n8n81
2here the green is G red is G blue is J and blac5 is 9$
Further studies
2e analy+ed the pattern of cutting an obect of / dimensions* / being a
positi'e integer* n times* though our study &nished with / being e7ual to 6G my
7uestion is what would the conecture be or the general rule be when we will
use a / dimension?
I will start by wor5ing out a conecture< I will &rst show what e'ery conecture
we used for the pre'ious dimension was* after I will wor5 out a general rule$
/imension Conecture -where n is the number of
chords 0 F-@*n0E F-@*n,108F-9*n,10 where F-@*n0 is a the
ma!imum number of parts we can get in
dimension produced with n cuts then. F-P*n0EF-P*n,108F-@*n,10 where F-P*n0 is the
ma!imum number of parts we can get by
cutting a .d obect with n cuts$
6 F-D*n0EF-D*n,108F-P*n,10 where F-D*n0 is thema!imum number of parts we can get by
cutting a 6d obect with n cuts
From this table we can immediately notice that to &nd the ma!imum number of
parts done with -n0 cuts in a gi'en dimension we ust need to ta5e the sum of
the ma!imum number of parts we can obtain in the same dimension with -n,10
cuts and the ma!imum number of parts we can obtain in the gi'en dimension
minus one with -n,10 cuts$
This ma5es us understand that the conecture for any dimension is simply< F-/*
n0EF-/*n,108F-/,1* n,10* where F-D*n0 is the ma!imum number of parts we
can get by cutting a / d obect with n cuts and / stands for the number of
dimension* which has to be a positi'e integer$
The problem of this formula is that the only ways to 5now the ma!imum
number of parts in a dimension you ha'e to 5now the number of cuts minus 1
in the same dimension and the number of cuts minus 1 in the dimension,1$
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Imagine you want to calculate the ma!imum number of parts you can obtain by
cutting a 1)) dimensional obect using 1)) cuts$ This is not impossible* but
'ery long$
To ma5e it easier therefore I want to create a formula where it is possible todirectly calculate the ma!imum number of parts ust by 5nowing how many
cuts and in what /imension
It is clear that the formulas to &nd the ma!imum number of parts for
hyperspaces in dierent dimensions ha'e something in common* for e!ample
their formula all ha'e a degree which is e7ual to the dimension* and the
coeQcient of the highest degree is always1
D !G we can notice that they all
end up with a constant of 1$ I &rst researched all their similitude to &nd a
general formula* but after a while* I changed path$
@ather than loo5ing at the formula for each dimension I started to loo5 at the
ma!imum number of parts I could obtain from each dimension and there I
found a formula$
To e!plain it* I will need a lot of graphical support* I will mainly use e!cel tables$
First I will show e'ery ma!imum result for the &rst cuts in dimensions$ The
data it ta5en by adding* ust li5e the recursi'e rule said< P -/* n0EP-/*n,108P-/,1* n,10
I will start now by tal5ing about patters which can be easily noticed$ The &rst
one is that when the / ≥ n the results are ust n -green cells0$ The second
pattern which can be seen is that when /En,1 the result will gi'e n
,1 -yellowcells0$
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Table 4
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From here we can deduce that for a cut* the dierence from the green part to
the yellow part is 1$
Aet us analy+e further the changed between dimensions for a gi'en 'alue of
cuts$Aet us watch at the column for the .rd cut and its dierences* we start from the
dierence from the green +one to the yellow +one* so it always starts with 1$
/ierences< -,>E01 G ->,6E0 . G -6,1E0.$
I will now show a table of dierences* this will show the dierence between the
ma!imum number of parts as you increase of dimension<
4 'alue* for e!ample 1 cut in ) dimension* is gi'en by the dierence of the
'alues of 1 Cut 1 / and 1 Cut ) / found in table 1$
Aoo5ing at the dierences from dimension to dimension in the same column we
can &nd a pattern* this brings us to the similarity between the dierences in a
column to the Pascal triangle and the binomial theorem$Pascal triangle represents a series of number that follow a rule which says that
e'ery 'alue of a line is e7ual to the sum of the two 'alues abo'e it<
4 G %
C G / G B
2here /E48%
if we start from 1<
1
1 G 1
1 G G 1
1 G . G . G 1
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Table %
:E)
:E1
:E
:E.
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2e can e!press any 'alue of the Pascal triangle by using the binomial theorem*
using the formula N !
T !
( N
−T
)!
where T is the position of the 'alue you want to
&nd -) to :0$
2e can now notice that the dierence between a 'alue which is a power of two
- in the green +one0 and another 'alue is the sum of the 'alues in the Pascal
triangles line :En$
Imagine now that we are loo5ing for the ma!imum number of parts we can get
from cutting a 6 dimensional obect times* from the table we can see that the
result is =>* but how could we do it without the table?
2e said that when / ≥ n the result is n* so it is part of the green +one$* so to
calculate a 'alue which is not a power of * we can simply subtract the sum of
all the dierences using Pascal triangle* to be more clear* for this e!ample we
ha'e n being si!* so we can loo5 the 'alues of the line :E of the Pascal
triangle< 1 G G 1=G ) G 1= G G 1$
:ow as we now that our /En, * we 5now that the cell we are loo5ing for itsfounded two abo'e the green +one$
To &nd the total dierence therefore we will use the &rst numbers of the
serie< 1 and * and subtract the addition to the number in the green +one*
which correspond to $
,-180E6,>E=>
This is what we would get by loo5ing at the table$
To create a real general formula we should write something where from a
power of is subtracted the sum from 1 to a 'alue of a line of the Pascal
triangle$ 4ny 'alue of the Pascal triangle can be e!pressed by the binomial
theorem$ 2e seen that for cutting the 6 / obect with chords we needed the
sum of the &rst two 'alues of the Pascal triangle of the line :E* 1 and which
How many pieces?Francesco Chiocchio Page 1
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written in binomial way are6!
0!(6 !) and
6 !
1 !(5 !) $ 9o the formula would loo5
something li5e this< , ∑i=0
1
6 !
i ! (6−i )!
%ut now let us generali+e it for any dimension or cuts< n, ∑i=0
n− D−1n !
i ! (n−i ) !
#eometrical e$planation
/uring all the study we loo5ed at the ma!imum number of parts of ahyperspace in dierent dimensions we could get by cutting it* we seen somepatterns but ne'er e!plained it* so in this chapter I will focus on thegeometrical e!planation of the results and the patters$
First we must 5now that points in an hyperspace are determined as n E Y !i * !* !. *Z[ * for istance in dimension e'ery point can be determined using twonumbers - ! G y 0* in ./ a plane is identi&ed as - ! G y G + 0 or in 6 /* for instanceBinstein space time* the points are obtained - ! G y G + Gt0$
2hen we cut the hyperspace for n ≤ /* e'erything we do* is we &rst cut the\!3 'alues in two* after the \y3 and so on* depending how many dimensions weare wor5ing on$
For e!ample let us ta5e a s7uare in dimension* it is de&ned as - J * 0 whenwe do the &rst cut we create two parts which are de&ned as - J1 * 0 and - J *
0 when we cut the s7uare with a second chord we will ha'e 6 parts which canbe de&ned li5e so < - J1 * 1 0*- J1 * 0*- J * 10*- J * 0:ow we do not ha'e any other dimensions to split into two so our .rd chord will*when trying to obtain the ma!imum number of parts* di'ide by two as manyparts as it possibly can* which we 5now is .$ This will gi'e us a > parts which
can be de&ned li5e this <
- J1 * 1 0* - J1 * 1 03* - J1 * 0* - J1 * 03* - J * 10*- J * 103 * - J * 0
The ne!t chord will cut 6 of these parts by two so we would ha'e somethig li5ethis <- J1 * 1 0* - J1 * 1 03* - J1 * 1 033* - J1 * 0* - J1 * 033* - J1 * 0333 *- J * 10* - J *
103 *- J * 1033* - J * 10333 * - J * 0$ This procedure can be applied to any 'alue of n$2e can apply the same rule for a . dimensional obect such as a cube* which is
de&ned as - ! * y * +0 to cut all the dimension by we need . cuts* and weobtain . parts<
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