SNOWFLAKES AND THE GEOMETRIC SERIES
1. The Geometric series
The finite geometric series formula says that if a is any integer different than 1
or 0 then we have
an+1 − 1
1 + a + a2 + · · · + an =
.
a−1
The infinite geometric series formula says that if a is a number with |a| < 1 then
1
.
1−a
Problem 0 Show that the finite geometric series formula is correct. Use the
finite geometric series formula to convince yourself that the infinite geometric series
formula is correct.
The two formulas above have lots of uses and are very important. We hope that
the problems below will convince you of this.
1 + a + a2 + · · · + an + · · · =
2. The Area of a Snowflake
We will constuct two complicated geometric sets and try to find their areas and
perimeters.
We start with the inner snowflake. To constuct the inner snowflake, we first
constuct a family of polygons, I1 , I2 , I3 , . . . as follows:
I1 is an equilateral triangle.
I2 is obtained from I1 by adding an equilateral triangle to the middle third of
each side of I1 .
I3 is obtained from I2 by adding an equilateral triangle to the middle third of
each side of I2 .
In general, In+1 is obtained from In by adding an equilateral triangle to the
middle third of each side if In .
The inner snowflake SI consists of all points that belong to some In .
The second set that we construct is the outer snowflake. To construct it, we
start by making a family of polygons, O1 , O2 , O3 , . . . as follows:
O1 is a regular hexagon.
O2 is obtained from O1 by removing an equilateral triangle from the middle third
of each side of O1 .
O3 is obtained from O2 by removing an equilateral triangle from the middle third
of each side of O2 .
In general, On+1 is obtained from On by removing an equilateral triangle from
the middle third of each side of On .
The outer snowflake SO consists of all points that belong to all On .
Before we continue we will need some results from geometry.
1
2
SNOWFLAKES AND THE GEOMETRIC SERIES
Problem 1 Let 4BAC be an isosceles triangle with ∠BAC = 120◦ . Let E, F
be the points that trisect BC. Then 4AEF is an equilateral triangle, and the two
triangles 4AEB and 4AF C are congruent isosceles triangles with 120◦ angles.
Problem 2 Show the following two things:
• If T is an equilateral
triangle with side
of length a, then the altitude of T
√
√
has length a 2 3 , and the area of T is 43 a2 .
• If R is an isosceles triangle with√120◦ angle with two sides of length a then
the third side of R has length a 3.
We now construct I1 , I2 , I3 , . . . and O1 , O2 , O3 , . . . such that
I1 ⊂ I2 ⊂ I3 ⊂ · · · ⊂ O3 ⊂ O2 ⊂ O1 .
Let O1 be a regular hexagon with side 1 and let I1 be an equilateral triangle
inscribed in O1 . Then O1 \I1 consists of three isosceles √
120◦ triangles with short
side 1. Then by Problem 1 the sides of I1 have lenght 3.
Our general procedure for constructing polygons was already described above.
For each n, On \In will consists of a family of congruent isosceles 120◦ triangles and
On+1 \In+1 is obtained from On \In by removing an equilateral triangle from the
middle third of each side of each isosceles 120◦ triangles. Notice that Problem 2
guarantees that thus process obtains from a set of isosceles 120◦ triangles to a new
set of 120◦ triangles. See the next page for a picture.
Let
sn = length of a side of In .
tn = area of equilateral triangle with side sn .
mn = number of sides of In .
an = area of In .
Sn = length of a side of On .
Tn = area of equilateral triangle with side Sn .
Mn = number of sides of On .
An = area of On .
Notice that we have the following formulas:
1
sn+1 = sn ,
3
mn+1 = 4mn ,
an+1 = an + mn tn+1 ,
1
Sn+1 = Sn ,
3
Mn+1 = 4Mn ,
An+1 = An − Mn Tn+1 .
Since an equilateral triangle of side s can be decomposed into nine equilateral
triangles of side 3s we have
tn+1 =
tn
,
9
and Tn+1 =
Also
a1 = area(I1 ) = t1 .
Tn
.
9
SNOWFLAKES AND THE GEOMETRIC SERIES
3
4
SNOWFLAKES AND THE GEOMETRIC SERIES
Problem 3 Why do we have A1 = 6T1 ?
Problem 4 Show that area(I1 ) = 21 area(O1 ).
Problem 5 Make a table of values of mn , tn , mn−1 tn , Mn , Tn , Mn−1 Tn .
Problem 6 Using the table write down formulas for area(In+1 ) and area(On+1 )
and show that both the inner snowflake and the outer snowflake have area 45 .
Problem 7 Calculate the length of the perimeter of In . Calculate the length of
the perimeter of On .
Problem 8 Show that the inner snowflake is not equal to the outer snowflake.
3. Other Problems
Problem 9 Use the geometric series formula to show that 0.99999999 . . . = 1.
Similarly find rational numbers that are equal to 0.123123123123 . . ., and 4.5151515151 . . ..
Problem 10 Find the following sum
1 1 1 1 1 1 1 1 1
1 + + + + + + + + + + ···
2 2 4 4 4 8 8 8 8
Problem 11 The people in the ancient city of Babylon decided to build a tower
by stacking infinitely many cubes on top of each other. The first cube is to have
1
1
side 1 foot, the second cube is to have side √
feet, the third cube √
feet and so
3
3
2
3
on. Do you think that they will be able to build this tower?
Problem 12 Let us consider arithmetic modulo 5. That is let us think of the five
numbers 0, 1, 2, 3, 4 and let us define addition and multiplication on these numbers
by using the usual addition and multiplication and then taking the remainder of
dividing by 5. Let us use symbols ⊕ and ⊗ for these operations. For example,
3 + 4 = 7, but the remainder of 7 when I divide by 5 is 2 therefore 3 ⊕ 4 = 2.
4 · 4 = 16, but the remainder of 16 when I divide by 5 is 1 therefore 4 ⊗ 4 = 1.
Compute 2 ⊕ 2, 4 ⊕ 1, 3 ⊕ 3, 0 ⊕ 4. In fact, complete the following multiplicative
table for system for arithmetic modulo 5:
⊗ 0
0
1
2
3
4
1
2
3
4
We will use power notation as follows a⊗0 = 1, a⊗1 = a, a⊗2 = a ⊗ a, a⊗3 =
a ⊗ a ⊗ a and so on. Show that 1 = 2⊗4 , 2 = 2⊗1 , 3 = 2⊗3 , 4 = 2⊗2 . Use this and
the finite geometric series formula to show that
1 ⊕ 2 ⊕ 3 ⊕ 4 = 0.
Similarly, show that
1⊗2 ⊕ 2⊗2 ⊕ 3⊗2 ⊕ 4⊗2 = 0.
Do this for exponents ⊗3, ⊗4, ⊗5, ⊗6, ⊗7, ⊗8 and guess a general formula.
Problem 13 Use arithmetic module 5 to show that
11111 + 12111 + 13111 + 14111
is a multiple of 5.