4 10 7 3 = 4 (one above correct move)
5 15 (divisible by 3) 10 5 = 5
6 21 (divisible by 3) 14 7 = 7
7 28 19 9 = 10 (one above correct move)
8 36 (divisible by 3) 24 12 = 12
9 45 (divisible by 3) 30 15 = 15
10 55 37 18 = 19 (one above correct move)
11 66 (divisible by 3) 44 22 = 22
12 78 (divisible by 3) 52 26 = 26
13 91 61 30 = 31 (one above correct move)
14 105 (divisible by 3) 70 35 = 35
15 120 (divisible by 3) 80 40 = 40
My deduction from the above table, is that the number of counters which are a multiple of 3, give the correct minimum number of moves. Other number of counters which are not multiple of 3, give 1 above the correct minimum number of moves.
From this information, I am going to try derive a formula to give the minimum number of moves.
Deriving an equation for minimum number of moves
Base,b=1,2,3,4,…..
Number of counters, nb= b+(b-1)+(b-2)+(b-3)+…….+1
Number of moves=m
If nb is divisible by 3,then (If nb mod 3 = 0)
m = nb/3
else
m = (nb-1)/3
Example Example
b =8 b =13
nb =8+7+6+5+4=3+2+1 =36 nb =13+12+11+10+9+8+7+6+5+4+3+2+1 = 91
nb is divisible by 3, therefore n is not divisible by 3,therefore
m =nb/3 m = (nb-1)/3
m = 36/3 =12 m = (91-1)3 = 30
note:= nb = 1+2+3+4+…+(b-1)+ b
Therefore, we use ‘Arithmetic Progression’. Where first term, a =1, common difference, d =1
nb = (b/2) *(2a + (b-1)d) = [b *(b+1)] / 2
Therefore, number of counters in the triangle, nb = [b*(b+1)] / 2
TESTING EQN FOR NUMBER OF COUNTERS
Number of counters in triangle which has a base of 4 counters is denoted as n4
n4 = 4*(4+1) = 10 counters
2
If number of counters is a multiple of 3, then
(i.e nb mod 3 = 0 i.e remainder = 0 when dividing by 3 )
moves = (number of counters)/3
[ b*( b+1)] / 2
m = _____________
3
If number of counters is NOT a multiple of 3, then
moves = [(number of counters)- 1] / 3
{[ b*( b+1)] / 2} - 1
m = __________________
3
You use the minus 1 in this formula because it is seen from the investigating table 2, the counters which are not a multiple of 3, give 1 above the minimum move.
Multiplying out
m= [b2+b] - 2 x 1
-
3
m= b2+b-2
6
m= (b-1)(b+2)
6
TESTING THE EQUATIONS
Using values of base, b between 1 and 7
Let b=1 {[1(1+1)]/2} - 1
m = ____________ = 0/3 = 0
3
Let b=2 m = [2(2+1)/2] / 3 = 3/3 = 1 move
Let b=3 m = [3(3+1)/2] / 3 = 6/3 = 2 moves
Let b=4 {[4(4+1)]/2} - 1
m = ____________ = (10 – 1)/3 = 3 moves
3
Let b=5 m = [5(5+1)/2] / 3 = 30/3 = 5 moves
Let b=6 m = [6(6+1)/2] / 3 = 21/3 =7 moves
Let b=7 {[7(7+1)]/2} - 1
m = ____________ = (28 – 1)/3 = 9 moves
3
We notice by testing the equations, that the number of counters which are multiples of 3, give the exact minimum number of moves. Where as the number of counters, which are not multiples of 3, we have to take away 1 from the number of counters before dividing it by 3.
INVESTIGATING A PATTERN BETWEEN TRIANGLE CORNERS
AND THE HEXAGON
Have I convinced myself that the equations I have produced, do give me the least number of moves. NO!!!
I am now going to investigate the structure and relationship between the 3 corner triangles and the hexagon produced within the moving arrow.
TABLE 3
From table 3 and diagram 3 (structure of arrow head), it is seen that in each cycle of 3 moves, the hexagon base, bH increases by 1 each time. Also hexagon is always one more counter than the top triangle base,b1.(ie bH + 1).
Let’s look at the cycle with base of 6,7,8 to give us a clearer indication of what is happening in the arrow head.
INVESTIGATING ARROW HEAD STRUCTURE
I noticed that at least two corners of the triangle remained unchanged, every 3 moves (cycle). Redrawing the arrow head such that 2 corner triangles which remain the same in each cycle, ie facing the 2 corner triangles at the bottom 2 corners of the base of the arrow head.
Diagram 3
Base=3
Number of counters = 6
Number of moves = 2
Base=4
Number of counters = 10
Number of moves = 3
Base=5
Number of counters = 15
Number of moves = 5
Base=6
Number of counters = 21
Number of moves = 7
Base=7
Number of counters = 28
Number of moves = 9
Base=8
Number of counters = 36
Number of moves = 12
Base=9
Number of counters = 45
Number of moves = 15
Base=10
Number of counters = 55
Number of moves = 18
Base=11
Number of counters = 66
Number of moves = 22
Base=12
Number of counters = 78
Number of moves = 26
OBSERVING TRIANGLE OF BASE 6
Base=6
Number of counters = 21
Number of moves = 7
We notice that the base of arrowhead, b is divisible by 3 exactly and the two base triangles (b1 and b2) have 2 counters each and the hexagon has a base of 2 counters, bH. The top triangle has a base b3 of 1 counter. The pattern of base of arrow head is P= b1 + bH + b2 = b/3 = c is a whole number.
Therefore c=b/3 ie b = 3c
We know the number of moves is the total number of counters in the 3 corner triangles.
So deriving an equation
Number of moves equals sum of counters in 2 bottom triangles added to number of counters in the top triangle.
m = 2[2 + 1] +1
Since the counters in the corner triangles produce a natural number sequence, therefore
m = 2[c + (c-1) + ……..+ 1] + [(c-1) + (c-2) +……..+ 1]
starts at (c-1) because the top triangle base(b3) starts at 1 less than the bottom two triangles base (b1 and b2)
Using arithmetic progression sequence,
m = 2 c(c+1) + (c-1)[(c-1) + 1]
2 2
= 2 (c^2 + c) + c^2 -c
2 2
= 2c^2 +2c + c^2 - c
2
= c(2c + 2 + c – 1)
2
m = c(3c + 1)
2
Sustituting c = b/3
m = (b/3)[3(b/3) + 1]
2
m = b(b + 1) ………………………..(1)
6
OBSERVING TRIANGLE OF BASE 7
Base=8
Number of counters = 36
Number of moves = 12
Base of arrow head has increased by 1 (b + 1) and is not divisible by 3 exactly. Therefore
b = 3c +1, where c = (b – 1)/3.
Triangle base b1 and b2 are still the same. Top triangle base, b3 has 2 counters, 1 less than the hexagon base, bH, which has increased to 3 counters (bH = c + 1)
m = 2 [c +(c-1) +(c-1) …….+ 1] + [c +(c-1) +(c-2) +…..+1]
Since all 3 corner triangles have the same number of counters,
so both equations start with c.
Using arithmetic progression
m = 2 c(c+1) + c(c+1)
2 2
= 2c^2 + 2c + c^2 + c
2
m = 3c(c + 1)
2
Sustituting c = (b – 1)/3
m = (b – 1)(b + 2) ……………….(2)
6
OBSERVING TRIANGLE OF BASE 8
Base=8
Number of counters = 36
Number of moves = 12
If b = 3c + 2, where c =(b – 1)/3 the number of counters in the 2 bottom triangles are still the same.
m = 2 [c+(c-1)+(c-2)+……+1] + [(c+1)+c+(c-1)+…..+1]
Top triangle base,b3 has increased by 1,whilst
Bottom 2 triangle bases are the same
Using arithmetic progression,
m = 2 c(c+1) + (c+1)[(c+1)+1]
- 2
Substituting c = b –2
2
m = b(b + 1) …………………(3)
6
DERIVING THE GENERAL METHOD
-
Top triangle base is always 1 less than the base of hexagon (bH – 1).
- Bottom triangle bases are always equal during 3 sequences at a time.
-
Since sum of triangle is b(b +1), we know the sequence of base of the triangle from b,
2
b + 1, b + 2, ie 3 cases for the base of the triangle.
- We want to consider for 2 smaller triangles in the base (b), which are the same. To do this
we divide b into 3 parts, 2 parts for the 2 corner triangles and 1 part for the hexagon. We have discussed that the base of the triangle (b) will not always be exactly divisible by 3. So we take the factor modulus into account, ie considering the whole part of the divisibility only (ie disregard the decimal points).
- To find the sum of counters of 2 bottom triangles of the base of the arrowhead (b), we
must say b/3 as [b/3] we consider only the whole part.
Therefore total number of counters of 2 base triangles are equal to
n1+2 = 2 [b/3]([b/3] + 1) Sum of 2 triangles in the base b
2
We know top triangle is always 1 less counter than the hexagon base (bH) gives bH – 1.
where bH = b – 2[b/3]
Therefore bH-1 =bH-2[b/3]-1
b3 = bT – 2[b/3] - 1 base of top corner triangle
Expressing this as sum an arithmetic sequence.
n3 = ∑ 1= b3/2 (2a + (b3-1)d)
n3 = b3(b3 + 1) /2
n3= (b –2[b/2] –1)*[(b –2 [b/2]-1)+1]
2
We know number moves must be equal to n1+2 +n3, bottom Triangle plus Top triangle.
n1+2 n3
m= 2([b/3] ([b/3]+1) + ( b – 2 [b/3]-1) [(b –2 [b/3] –1)+1]
2 2
Multiplying out
m = 2 [b/3]([b/3]+1) + ( b – 2 [b/3]-1) [(b –2 [b/3] –1)+1]
2 2
m = ½{ 2 [b/3] ^2 +2 [b/3]+ (b –2 [b/3] –1) (b –2[b/3])}
m = ½{ 2 [b/3] ^2 + 2 [b/3]+ b^2-2b[b/3] – b+2[b/3] + 4[b/3]^2 –2b[b/3]}
m = ½{ 6 [b/3]^2 +4[b/3]-4b[b/3]+b^2-b}
Proving 3 formulas deriving 3 moves in each cycle calculated
The least number of moves in each case.
Prove By Induction
b = 3c derived m = [b (b+1)]/6 where c = b/3
b = 3c+1 derived m = [b (b+1) (b+2)]/6 where c = (b –1)/3
b = 3c+2 derived m = [b(b+1)] /3 where c = (b-2)/3
PROVING BY INDUCTION
(1) We know it is true for m =[ b(b+1)]/6 where b=3c
The next base is b’=3c+1,we have to prove equation (2)
m’= (b’-1)(b+2) is also true.
6
At this stage one more base is added to the top triangle from the case where all 3 corner triangle bases were equal, but at this stage the top corner triangle has increased by one base.
Therefore adding c to both sides
m+c=b(b+1) + c
6
substituting 3c into b since b=3c
m+c=b(b+1) + c
6
m+c=3c(3c+1) +c = 3c(3c+1)+6c
6 6
m+c=3c(3c+3)
6
where b ’=3c+1
But we know b’=3c+1 , since 3c=(3c+1)-1 and 3c+3=(3c+1)+2
Since 3c= b’–1
b’ –1
For the equation m+c=(bT-1)(bT+2)= m’
2
we get the multiples of the numerator
So this proves that if equation (1) works for b=3c, then equation (2) works for the next base
b’=3c+1
Assuming b’=3c+1 and equation m’=[(b-1)(b+2)]/6 are true.
Need to prove for the next base, b=b+1 is true for m=[b(b+1)]/6
Again the moves, m is formed from m’ by adding c+1 counters to the top triangle.
m’+c+1=(b-1)(b+2) + c+1
6
replacing b with 3c+1 m’+c+1=(3c+2)(c+1) = m
2
Using assumption b=3c+1 and b=b’+1
We get
b= b(b+1)
6
We have proved that if equation(2) works for any b=3c+1, then equation(3) will work for the b’.
Prove that b=3c+2
If we look at the pattern of the cycle,3c, 3c+1, 3c+2, whilst in our induction we add c+1 to our equations ,according to this principle we assume the next cyclic move would be 3c+3.
