March 2015 Challenge Problem

Brandon Doylebdoyle13@student.monroecc.edu

03.06.15

Let us start with a basic square as I have drawn below with a side length of 1 unit.

Now in each of the following two interations I have drawn below we add a square 1/3 theformers size to the central 1/3 of each side formed, thus yielding a shape like that of thesnowflake curve. Note that filled in black regions are presumed to be empty space.

Thus we see that the overall limit to the area confined within the perimeter formed as weadd more and more squares is at most 2 square units since

12 = a2 + b2 = 1 = 2a2

= 12

= a2

=

2

2= a, (1)

but we must of course double this since a represents half a side length and therefore AT =2

2 = 2. For an illustration please refer to the following figure.

2a =

2

Figure 1: The maximum area AT possible within this shape is (2a)2 =

(2)2

= 2 units2.

Now let us examine further why this is the case. If we draw a line from the very center ofthe containing square in Figure 1 in the direction of its vertices to the farthest reaches ofthe squares we have yet added it has a length Lv given by

Lv =1

2+

kn=1

(1

3

)n, (2)

which is of course a geometric sequence with a convergence given byr

1 r , such that r =1

3.

Thus we will examine what happens to this series as k . We have

Lv =1

2+ lim

k

[k

n=1

(1

3

)n]

=1

2+

(13

)1 (1

3

)=

1

2+

1

2= 1. (3)

If we apply what we have come across in (3) to the results of (1) we indeed see that

limk

2

[1

2+

kn=1

(1

3

)n]2=(

2)2

= 2. (4)

Now we will examine what happens to the squares that become trapped inside (somethingthat cannot happen with triangles) in order to confirm our results. Inside each square, asmore and more squares are added, the space slowly becomes completely filled. Please referto Figure 2 for an illustration.

March 2015 Challenge Problem Math Deparament, MCC

1/9 units

Figure 2: The branching of interior squares that became trappedin Figure 1 after 3 more iterations.

We see that the area as the iterations continue fills almost completely in, so if we wereto write a summation representing this squares area, we should see that it does not exceed1/91/9 = 1/81units2 after an infinite number of iterations. Since all of the squares are identi-cal, we must worry more about the number of each as the iterations continue. For instance, itis obvious that the sequence will continue and we will have squares of 1/27, 1/81, 1/243, 1/729the originals width and so on.

Every time, however, notice that we add five sets of four squares, with the exception ofthe first. So we have a sum that looks like

4

(1

33

)2+ 5 4

(1

34

)2+ 5 5 4

(1

35

)2+ 5 5 5 4

(1

36

)2+ + 4 5n

(1

3n+3

)2, (5)

which we may rewrite as

limk

[4

kn=0

5n(

1

3n+3

)2]=

1

81. (6)

But how much area is confined within the 8 largest branches of squares that look like whatI have drawn in Figure 2? We can add up this area as yet another geometric seriesto arriveat

8j=0

1

3j

(4n=0

5n

32n+6

)= 32

j=0

n=0

5n

32n+6+j

=8

81

j=0

(1

3

)j=

8

81

1

1 13

=4

27. (7)

Now lets try something different. The series representing the total area contained in allsquares after each iteration which I will denote A(n) is below.

A(n) = 1 + 4

(1

31

)2

1

+ 4 5(

1

32

)2

2

+ 4 5 5(

1

33

)2

3

+ + 4 5n1(

1

3n

)2

n

.

March 2015 Challenge Problem Math Deparament, MCC

Thus, rewritten as a summation we have

A(n) = 1 + 4n

k=1

5k1(

1

3k

)2, (8)

which looks a lot like what we found in (6). I have created a table showing the area containedin the first 10 iterations below.

1 2 3 4 5 6 7 8 9 10

1.444 1.691 1.829 1.905 1.947 1.971 1.984 1.991 1.995 1.997

Figure 3: The area A(n) contained in all squares for n = {1, , 10}.

So we see that the overall area numerically approaches 2, as we expected. Since the answersare required as fractions in lowest terms I have included another table below with the exactfractional values.

1

139

2

13781

3

1333729

4

124976561

5

11497359049

6

1047257531441

7

94878134782969

8

8570281743046721

9

772887853387420489

10

69638031173486784401

Figure 4: The exact fractional values.

Now how about the perimeter of this shape? We should probably expect this series to divergesince we are adding squares on each iteration, like Kochs Snowflake. We arrive at a seriesthat looks like

P(n) = 4 1 + 4 51(

1

31

)

1

+ 4 52(

1

32

)

2

+ 4 53(

1

33

)

3

+ + 4 5n(

1

3n

)

n

= 4 + 4n

k=1

(5

3

)k. (9)

Thus we have yet another geometric series. But this one is different since r 1, and sincegeometric series only converge for r < 1, we indeed will arrive at an infinite amount ofperimeter after an infinite number of iterations (as expected). Thus

limn

[4 + 4

nk=1

(5

3

)k]=. (10)

I have drawn up another table with the perimeter values associated with the first 10 iterationsbelow.

March 2015 Challenge Problem Math Deparament, MCC

1323

2

1969

3

108827

4

576481

5

29792243

6

151876729

7

7681282187

8

38668846561

9

1941315219683

10

9730195659049

Figure 5: The exact fractional values of P(n) for n = {1, , 10}.

Thus we see in figure 5 that the perimeter summation does indeed diverge to as n

March 2015 Challenge Problem Math Deparament, MCC

References

[1] James Gleick. Chaos: Making A New Science. New York: Penguin, 1987. 98-100. Print.

March 2015 Challenge Problem Math Deparament, MCC