Who can attempt this?

?

Where are our mathematicians?

Cc. Durentt the mathematician, mathmagician, mathschic, fynestboi, karmanaut, sagamite, agentofchange1, agentofallah, darkhorizon, teempakguy, geniusdavid, madmatician, mathemagician, ladokuntlad. Etc.

Cc. Proffemi

.

Frontpage

My heuristic guess is 14. And if I'm correct, I don't think it is fair to characterise this as the hardest challenge ever.

I can prepare a mathematical proof later, if you wish to know how I worked it out.
-----------------------------------------------------------------------------------------------
Self correction: My previous answer is totally wrong. Having been fully awakened, I can now appreciate the complexity of the problem better. I was previously under the impression that the solution was the two smallest possible pyramids, but it is now clear to me that this sum will in no way form a larger pyramid. Let's thinks about it again and give it another try

timonski:
Cc. Durentt the mathematician, mathmagician, mathschic, fynestboi, karmanaut, sagamite, agentofchange1, agentofallah, darkhorizon, teempakguy, geniusdavid, madmatician, mathemagician, ladokuntlad. Etc.

680

http://www.showmyiq.com/forum/viewtopic.php?f=20&t=20

ladokuntlad:


680

http://www.showmyiq.com/forum/viewtopic.php?f=20&t=20

Can you formally prove it though?

timonski:
Who can attempt this?

timonski:
Cc. Durentt the mathematician, mathmagician, mathschic, fynestboi, karmanaut, sagamite, agentofchange1, agentofallah, darkhorizon, teempakguy, geniusdavid, madmatician, mathemagician, ladokuntlad. Etc.

First of all, by just looking at your diagram, I think the only logical way one can use these balls to form pyramids is by decreasing the lowest layer of balls by an increasing, number starting from 2, everytime one builds an additional layer above it.

So working from the topmost (not the lowest layer), I can deduce that the formula for pyramids are n₀ + {n₁+d} + {n₂+(d+1)} + {n₃+(d+2)} ........................+ {n_infinity +(d+(infintity+1))}

Where n₀ = 1

All subsequent n_x = {n_{x - 1} + (d+y)} where x and y are >/= 1.

And d =2.



After creating the formula, now to solve the problem:

The miminal number of cannonballs to form a tetrahedral pyramid is n₀ + {n₁+d}.

So that is basically 1 + {1+2}. Which is equal to 4.

Since the question states the other pyramid must be of a different size, then we have to move up to the next minimal number of cannonballs to form a tetrahedral pyramid, which is n₀ + {n₁+d} + {n₂+(d+1)}.

So that is basically 1 + {1+2} + {3+3}. Which is equal to 10.

So the minimum number of cannonballs to form a tetrahedral pyramid using 2 smaller pyramids of different sizes is 14 cannonballs.



Fcking hell! I am smart!

Wait o. Let me recheck. I need Excel spreadsheet for the calculation.


Revised:

Now although that formula is interesting, the picture misled me. It appears the bigger pyramid cannot have 10 cannonballs in its base as the picture misleadingly suggests, it must have 9 (hence the pyramid will be unstable in real life, it can only be that stable in a picture).

I assumed wrongly but logically that there must be 10 balls at the base of the bigger/combined pyramid.

It must be 9 cannonbal base because both the smaller pyramids are 9 each, hence bigger one has to have 18 cannonballs as my Excel calculations states, not 19.

Using my Excel calculations, it appears the increasing size (starting from the lowest) of tetrahedral pyramids are: 4, 9, 18, 31, 48 etc.

So to find the minimum number of cannonballs to form a tetrahedral pyramid using 2 smaller pyramids of different sizes, we need to add all these smaller increasing size of tetrahedral pyramids (e.g. [4+9], [9+18], [4+48] etc) until we find an addition that meets a far larger size of tetrahedral pyramids.

I am yet to see a fit, so I propose that it is impossible.

AgentOfAllah:


Can you formally prove it though?

My calculations gave the proof.

ladokuntlad:

680
http://www.showmyiq.com/forum/viewtopic.php?f=20&t=20

Based on the picture, it seems my calculation and this one is actually wrong.

Sagamite:




First of all, by just looking at your diagram, I think the only logical way one can use these balls to form pyramids is by decreasing the lowest layer of balls by 2 everytime one builds an additional layer above it.

So working from the topmost (not the lowest layer), I can deduce that the formula for pyramids are n₀ + (n₁+d) + (n₂+(d+1)) + (n₃+(d+2) ........................+ (n_infinity +(d+(infintity-1)))

Where n₀ = 1

All subsequent n_x = {n_{x - 1} + (d+y)} where x and y are >/= 1.

And d =2.



After creating the formula, now to solve the problem:

The miminal number of cannonballs to form a tetrahedral pyramid is n₀ + (n₁+d).

So that is basically 1 + (1+2). Which is equal to 4.

Since the question states the other pyramid must be of a different size, then we have to move up to the next minimal number of cannonballs to form a tetrahedral pyramid, which is n₀ + (n₁+d) + (n₂+(d+1)).

So that is basically 1 + (1+2) + (3+3). Which is equal to 10.

So the minimum number of cannonballs to form a tetrahedral pyramid using 2 smaller pyramids of different sizes is 12 cannonballs.



Fcking hell! I am smart!

Wait o. Let me recheck. I need Excel spreadsheet for the calculation.


Revised:

Now although that formula is interesting, the picture misled me. It appears the bigger pyramid cannot have 10 cannonballs in its base as the picture misleadingly suggests, it must have 9 (hence the pyramid will be unstable in real life, it can only be that stable in a picture).

I assumed wrongly but logically that there must be 10 balls at the base of the bigger/combined pyramid.

It must be 9 cannonbal base because both the smaller pyramids are 9 each, hence bigger one has to have 18 cannonballs as my Excel calculations states, not 19.

Using my Excel calculations, it appears the increasing size of tetrahedral pyramids are: 4, 9, 18, 31, 48 etc.

So to find the minimum number of cannonballs to form a tetrahedral pyramid using 2 smaller pyramids of different sizes, we need to add all these smaller increasing size of tetrahedral pyramids (e.g. [4+9], [9+18], [4+48] etc) until we find an addition that meets a far larger size of tetrahedral pyramids.

I am yet to see a fit, so I propose that it is impossible.

Lol dude, you've just confused yourself! The last answer, 680 (i.e. 120 + 560), which was provided by ladokuntlad was correct. He just didn't show a mathematical proof for it. I don't think there is yet a mathematical proof, I spent time thinking about it, but I haven't successfully developed one yet.

Let me clarify the question for you:
A tetrahedral pyramid, as the name implies, is a tetra (four) faced pyramid.

If we must assume anything, it should necessarily be that the cannon balls are of the same material, shape and volume, thus, such a tetrahedral pyramid of cannon balls, if 7 lines were to be drawn from their edges to intersect at their vertices, can only form four faces (including the base) of the same equilateral triangle.

Now, we are told that two tetrahedral pyramids, each having a total of 10 balls were combined to form one larger pyramid in which all 20 balls were used, and you suggest the larger pyramid cannot have up to 20 balls. This is incorrect, the larger pyramid has exactly 20 balls and you will see why in a moment.

First, let's examine how a tetrahedral pyramid of cannon balls is formed. We proceed from the base. We know the base is an equilateral triangle with cannon balls of equal shape, material and volume. Such a triangle is formed if we stack the balls horizontally in an increasing order of 1 ball per line (the same way billiard balls are arranged as the example below)

O
OO
OOO
OOOO
What does the sum of this arrangement look like to you? It looks to me like:

Sum_base = 1+2+3+4+5.....N. Here, 'N' is the number of rows. This is simply the sum up to the Nth term of an arithmetic series

Sum = N/2 ((2a + (N-1)d), where d, sequential difference = 1, and a, the first term = 1. This term can be simplified thus:

Sum = (N² + N)/2 = Sum_base. Now we know the sum of the base of the pyramid, so we are getting somewhere.

Now, to get a pyramid, you will have to vertically stack a preceding equilateral triangle value over the base until the last ball as shown below
O L1
+
O
OO L2
+
O
OO
OOO L3

+
O
OO
OOO L4
OOOO

Notice the relationship between the number of balls on one side and the number of vertical stacks you will need..that's right, they are equal! So if you had four balls on one side of the triangle, you'll have four rows of vertical stacks. 5 side balls = 5 vertical stacks etc. This is helpful because if we started arranging from the top (single) ball on the tip of the pyramid, then the position of the subsequent layer (L2) is also equal to the number of balls on the side of the triangle the makes up that layer, which is 2. Triangle in L3 has 3 side balls...and so on!

Now we can formulate our tetrahedral pyramid formula from two useful pieces of information.
1) Number of balls in a triangle is:
Sum_base = (N² + N)/2, where N is number of side balls

2) Number of side balls, N, is equal to row number of triangle in a tetrahedral pyramid, given we start counting from the tip.

So the total number of cannon balls in any tetrahedral pyramid can be given by:

N_tet = S{1,N} (N² + N)/2. Where S{1,N} is the summation symbol sigma from N=1 to N.

This can be simplified thus:
1/2 ( S{1,N} N² + S{1,N} N)

But S{1,N} N² = Sum of squares identity = N(N+1)(2N+1)/6, and S{1,N} N = (N² + N)/2 again

Simplifying, you get:

N_tet = N(N+1)(N+2)/6 or N+2 [size=16pt]C[/size] 3

This formula gives you the expected number of cannon balls in the pyramid for any given value of N.

Supposing we plug N = 3, you will find that the number of cannon balls needed is 10. This means you can form a complete tetrahedral pyramid with 10 cannon balls. Hurrah!

If you had an identical pyramid with another ten cannon balls, can you form a valid complete pyramid by combining them, provided you have to use all the balls?

By combining, we have 20 balls. So is there any N that will give us 20? If such a number exists, then it is a complete tetrahedral pyramid of cannon balls. Since N = 3 gave us 10 cannon balls, we can only go higher, so lets try N = 4

(4)(4+1)(4+2)/6 = 4*5*6/6 = 20! So you see, these identical stack of cannon balls will definitely form another stack of tetrahedral pyramid.

The next question wants us to find two lowest values for N_a and N_b such that N_a =/= N_b =/= 0, whereby if you added

(N_a+2 [size=16pt]C[/size] 3) + (N_b+2 [size=16pt]C[/size] 3), it will produce a third value, (N_c+2 [size=16pt]C[/size] 3), such that N_c is an integer not equal to zero.

120 + 560 = 680 actually satisfies this condition, where N_a = 8, N_b = 14 and N_c = 15

You can easily use the tetrahedral sum formula to generate an infinitely long vector from N = 1 to N = infinity, and then manually search for three tetrahedral numbers a, b, c such that:
a+b=c. While this process is based on trial and error, and can be tedious, it will eventually produce results, which is how ladokuntlad was able to determine that 680 (i.e. 120 + 560) was right. With the help of a computer algorithm and a good programmer, I was able to determine another set of numbers that satisfy this condition:

2,699,004 + 12, 794, 200 = 15, 493, 204,

Where N_a = 252, N_b = 424 and N_c = 452.

My challenge is this though: Can you come up with a formal proof/algorithm with which we can easily derive any such set of numbers if we wanted to?

The answer is simply 30

timonski:
Who can attempt this?

The answer is simply 30.

AgentOfAllah:


Lol dude, you've just confused yourself! The last answer, 680 (i.e. 120 + 560), which was provided by ladokuntlad was correct. He just didn't show a mathematical proof for it. I don't think there is yet a mathematical proof, I spent time thinking about it, but I haven't successfully developed one yet.

Let me clarify the question for you:
A tetrahedral pyramid, as the name implies, is a tetra (four) faced pyramid.

If we must assume anything, it should necessarily be that the cannon balls are of the same material, shape and volume, thus, such a tetrahedral pyramid of cannon balls, if 7 lines were to be drawn from their edges to intersect at their vertices, can only form four faces (including the base) of the same equilateral triangle.

Now, we are told that two tetrahedral pyramids, each having a total of 10 balls were combined to form one larger pyramid in which all 20 balls were used, and you suggest the larger pyramid cannot have up to 20 balls. This is incorrect, the larger pyramid has exactly 20 balls and you will see why in a moment.

First, let's examine how a tetrahedral pyramid of cannon balls is formed. We proceed from the base. We know the base is an equilateral triangle with cannon balls of equal shape, material and volume. Such a triangle is formed if we stack the balls horizontally in an increasing order of 1 ball per line (the same way billiard balls are arranged as the example below)

O
OO
OOO
OOOO
What does the sum of this arrangement look like to you? It looks to me like:

Sum_base = 1+2+3+4+5.....N. Here, 'N' is the number of rows. This is simply the sum up to the Nth term of an arithmetic series

Sum = N/2 ((2a + (N-1)d), where d, sequential difference = 1, and a, the first term = 1. This term can be simplified thus:

Sum = (N² + N)/2 = Sum_base. Now we know the sum of the base of the pyramid, so we are getting somewhere.

Now, to get a pyramid, you will have to vertically stack a preceding equilateral triangle value over the base until the last ball as shown below
O L1
+
O
OO L2
+
O
OO
OOO L3

+
O
OO
OOO L4
OOOO

Notice the relationship between the number of balls on one side and the number of vertical stacks you will need..that's right, they are equal! So if you had four balls on one side of the triangle, you'll have four rows of vertical stacks. 5 side balls = 5 vertical stacks etc. This is helpful because if we started arranging from the top (single) ball on the tip of the pyramid, then the position of the subsequent layer (L2) is also equal to the number of balls on the side of the triangle the makes up that layer, which is 2. Triangle in L3 has 3 side balls...and so on!

Now we can formulate our tetrahedral pyramid formula from two useful pieces of information.
1) Number of balls in a triangle is:
Sum_base = (N² + N)/2, where N is number of side balls

2) Number of side balls, N, is equal to row number of triangle in a tetrahedral pyramid, given we start counting from the tip.

So the total number of cannon balls in any tetrahedral pyramid can be given by:

N_tet = S{1,N} (N² + N)/2. Where S{1,N} is the summation symbol sigma from N=1 to N.

This can be simplified thus:
1/2 ( S{1,N} N² + S{1,N} N)

But S{1,N} N² = Sum of squares identity = N(N+1)(2N+1)/6, and S{1,N} N = (N² + N)/2 again

Simplifying, you get:

N_tet = N(N+1)(N+2)/6 or N+2 [size=16pt]C[/size] 3

This formula gives you the expected number of cannon balls in the pyramid for any given value of N.

Supposing we plug N = 3, you will find that the number of cannon balls needed is 10. This means you can form a complete tetrahedral pyramid with 10 cannon balls. Hurrah!

If you had an identical pyramid with another ten cannon balls, can you form a valid complete pyramid by combining them, provided you have to use all the balls?

By combining, we have 20 balls. So is there any N that will give us 20? If such a number exists, then it is a complete tetrahedral pyramid of cannon balls. Since N = 3 gave us 10 cannon balls, we can only go higher, so lets try N = 4

(4)(4+1)(4+2)/6 = 4*5*6/6 = 20! So you see, these identical stack of cannon balls will definitely form another stack of tetrahedral pyramid.

The next question wants us to find two lowest values for N_a and N_b such that N_a =/= N_b =/= 0, whereby if you added

(N_a+2 [size=16pt]C[/size] 3) + (N_b+2 [size=16pt]C[/size] 3), it will produce a third value, (N_c+2 [size=16pt]C[/size] 3), such that N_c is an integer not equal to zero.

120 + 560 = 680 actually satisfies this condition, where N_a = 8, N_b = 14 and N_c = 15

You can easily use the tetrahedral sum formula to generate an infinitely long vector from N = 1 to N = infinity, and then manually search for three tetrahedral numbers a, b, c such that:
a+b=c. While this process is based on trial and error, and can be tedious, it will eventually produce results, which is how ladokuntlad was able to determine that 680 (i.e. 120 + 560) was right. With the help of a computer algorithm and a good programmer, I was able to determine another set of numbers that satisfy this condition:

2,699,004 + 12, 794, 200 = 15, 493, 204,

Where N_a = 252, N_b = 424 and N_c = 452.

My challenge is this though: Can you come up with a formal proof/algorithm with which we can easily derive any such set of numbers if we wanted to?

Nope, mate.

I have highlighted in red where the argument fails.

You cannot place a cyclical cannonball on two other cyclical cannonball except you are David Blaine.

Not even if you supported the base cannonballs with a giant, specially-manufactured snooker rack.

Secondly that layer will not form a tetrahedral pyramid, so you have broken the given rule.

I even doubt one can place 3 cannonballs on 4 cannonballs, so I suspect the structure of the next layer too is unrealistic.

Again, a 4 cannonballs layer is not tetrahedral.

Whatever pyramid you are able to achieve with this, even if we ignore its structural infeasibility, will not be a tetrahedral pyramid.

Sagamite:


Nope, mate.

I have highlighted in red where the argument fails.

You cannot place a cyclical cannonball on two other cyclical cannonball except you are David Blaine.

Not even if you supported the base cannonballs with a giant, specially-manufactured snooker rack.

I don't think you read my comment attentively, I gave the equation for the total number of spheres in each layer of cannon ball, and it is:

T_base = (N² + N)/2, so if the top layer, the tip of the pyramid is 1 sphere, the second layer will have N = 2, which means T_base = (2² + 2)/2= 3. Provided N is an integer, this equation does not allow for a 2-sphere layer, so I don't know where you got the impression that I implied you can balance a spherical cannon ball on two others.

Secondly that layer will not form a tetrahedral pyramid, so you have broken the given rule.

I think you've grossly misunderstood the nature of the problem. Of course you cannot form a tetrahedral pyramid with a single triangular layer. Such an absurd construct was never suggested anywhere. What has been said is that given a tetrahedral pyramid with N number of layers, counted latitudinally from the base to the tip, the total number of balls needed to form such a pyramid is: N(N+1)(N+2)/6, of which, each layer, proceeding from the base to the tip, will have total number of balls equal to (N_i²+N)/2 for {i = N, N-1, N-2, N-3, ... 3, 2, 1}
See the images below for more

[img]http://3.bp..com/_oFB9sXQiV5Q/S9muKpbd6oI/AAAAAAAAGbg/TqIrMAaaVlo/s800/Pyramid.JPG[/img]

I even doubt one can place 3 cannonballs on 4 cannonballs, so I suspect the structure of the next layer too is unrealistic.

Refer to the foregoing explanation

Again, a 4 cannonballs layer is not tetrahedral.

You seem confused. Of course, you cannot have a 4 cannon balls layer, it wouldn't satisfy the T_base equation, but you can have 4 layers of triangulated cannon balls to form a tetrahedral pyramid. If you refer to the second image above, we have 5 layers there. But if you knock off the bottom one, we'll be having 4.

Whatever pyramid you are able to achieve with this, even if we ignore its structural infeasibility, will not be a tetrahedral pyramid.

I hope with the images, you can now appreciate better, the feasibility of the tetrahedral pyramid of cannon balls. The base is a triangle and simultaneously a face. Add that face to the three others that must necessarily form from the three sides of the basal triangular plane, and you have a pyramidal tetrahedral!

Sagamite:




First of all, by just looking at your diagram, I think the only logical way one can use these balls to form pyramids is by decreasing the lowest layer of balls by an increasing, number starting from 2, everytime one builds an additional layer above it.

So working from the topmost (not the lowest layer), I can deduce that the formula for pyramids are n₀ + {n₁+d} + {n₂+(d+1)} + {n₃+(d+2)} ........................+ {n_infinity +(d+(infintity+1))}

Where n₀ = 1

All subsequent n_x = {n_{x - 1} + (d+y)} where x and y are >/= 1.

And d =2.



After creating the formula, now to solve the problem:

The miminal number of cannonballs to form a tetrahedral pyramid is n₀ + {n₁+d}.

So that is basically 1 + {1+2}. Which is equal to 4.

Since the question states the other pyramid must be of a different size, then we have to move up to the next minimal number of cannonballs to form a tetrahedral pyramid, which is n₀ + {n₁+d} + {n₂+(d+1)}.

So that is basically 1 + {1+2} + {3+3}. Which is equal to 10.

So the minimum number of cannonballs to form a tetrahedral pyramid using 2 smaller pyramids of different sizes is 14 cannonballs.



Fcking hell! I am smart!

Wait o. Let me recheck. I need Excel spreadsheet for the calculation.


Revised:

Now although that formula is interesting, the picture misled me. It appears the bigger pyramid cannot have 10 cannonballs in its base as the picture misleadingly suggests, it must have 9 (hence the pyramid will be unstable in real life, it can only be that stable in a picture).

I assumed wrongly but logically that there must be 10 balls at the base of the bigger/combined pyramid.

It must be 9 cannonbal base because both the smaller pyramids are 9 each, hence bigger one has to have 18 cannonballs as my Excel calculations states, not 19.

Using my Excel calculations, it appears the increasing size (starting from the lowest) of tetrahedral pyramids are: 4, 9, 18, 31, 48 etc.

So to find the minimum number of cannonballs to form a tetrahedral pyramid using 2 smaller pyramids of different sizes, we need to add all these smaller increasing size of tetrahedral pyramids (e.g. [4+9], [9+18], [4+48] etc) until we find an addition that meets a far larger size of tetrahedral pyramids.

I am yet to see a fit, so I propose that it is impossible.

lol. Mehn, It is possible to solve. Try harder.