Math problem, last people I can think of

[Kedirekin]

Here's the result I got for spheres in a pizza box.

cols: int( (2*b/d) - 1 )/2
rows: int( 2*b/(sqrt(3)*d) )

So, for example, if the (square) box has dimension 8*d:

cols: int( (2*8*d/d) -1 )/2 = 15/2 = 7.5
rows: int( 2*8*d/(sqrt(3)*d) ) = int(16/1.732) = 9

For box with dimension 8.5*d:

cols: int( (2*8.5*d/d) -1 )/2 = 16/2 = 8
rows: int( 2*8.5*d/(sqrt(3)*d) ) = int(17/1.732) = 9

And finally for box with dimension 8.7*d:

cols: int( (2*8.7*d/d) -1 )/2 = 16/2 = 8
rows: int( 2*8.7*d/(sqrt(3)*d) ) = int(17.4/1.732) = 10


PS. I had to sit down with a small box lid and a pile of nickles before I could figure this out.


For anyone who has difficulty visualizing half a column:

O O O O O O
O O O O O
O O O O O O
O O O O O
O O O O O O
O O O O O

O O O O O O
O O O O O O
O O O O O O
O O O O O O
O O O O O O
O O O O O O

[Christos]

Originally posted by r6d2
I created a spreadsheet to tally all possible combinations of numbers from 2 to 100 (only one instance each pair), a total of 4950 rows.

Then added the product and the sum as columns. Then I counted how many times all possible products appear. I got 173 different products. I filtered out all pairs for which the products appear only once in the table.

I'm stuck there...

Ok I made a prog that calculated all the possible sums (what "Plus" sees on his paper) that can't be expressed as a sum of 2 primes (see my post above)...
Here they are...
It should help you there are only 83:
11 93 137 177
17 95 139 179
23 97 141 181
27 101 143 182
29 103 145 183
35 105 147 184
37 107 149 185
41 109 151 187
47 111 153 188
51 113 155 189
53 115 157 190
57 117 159 191
59 119 161 192
65 121 163 193
67 123 165 195
71 125 167 196
77 127 169 197
79 129 171 198
83 131 173 199
87 133 174 200
89 135 175

[r6d2]

I think I narrowed it a bit more. These are the "plus" values whose corresponding products can be expressed in more than one way, thus suposedly mult looked at them, used his product (which he knows) and solved the problem...

11 17 23 27 29 35 37 41 47 53

[Christos]

Now lets narrow it down to pairs. All you have to do is to check for every corresponding product its corresponding sum (look at things from "Mult's" point of view). If you find one which has only one corresponding sum in the range you specified then it's possible...

For example:

11=6+5...
1st Corresponding product=6*5=30
30=15*2=10*3=6*5
Corresponding sums: 15+2=17, 10+3=13, 6+5=11
11 is in the list you specified and 17 is in so 6 and 5 are not the answer.
We want only one of the corresponding sums of the product to be in the list so "Mult" can decide which one...

11=7+4
2nd Corresponding product=7*4=28
28=14*2=7*4
Corresponding sums: 14+2=16, 7+4=11
11 is in, 16 is not...
So 7,4 is a possible answer.

11=8+3
3rd Corresponding product=3*8=24
24=2*12=3*8=6*4
Corresponding sums: 12+2=14, 8+3=11, 6+4=10
11 is in, 14 is not, 10 is not...
So 8,3 is a possible answer.

In the end the correct answer can be found if we remember Plus' last words:
"In that case I know too."

Remember that he can't see the product but just by seeing the sum and knowing that Mult reached an answer with certainty then he finds the answer also.
So the sum he sees has only one possible corresponding product that has only one corresponding sum in our list.

For example:
As we so before there are 2 possible combinations with a sum of 11 (7,4 and 8,3) so the sum can't be 11 because then Plus' last words would not be true.
So Neither 7,4 or 8,3 is the answer

[r6d2]

Either you are getting some marbles and a pizza box to do the real thing, or you are writing lots of numbers on a sheet of paper. Come on, did everybody give up? :)

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[fccHandler]

My favorites were the books of Martin Gardner. I can't remember the titles, but he wrote lots of books on "recreational mathematics" and logic puzzles. He also did a wonderful annotated Alice in Wonderland. ;)

[r6d2]

Originally posted by Kedirekin
Here's the result I got for spheres in a pizza box.The problem with this approach is that sometimes it is better to use the dice arrangement for the marbles. For instance when b = 2*d you can see that clearly.

For d << b your formula is more appropriate, but the general case is not so simple. There are some thresholds to determine.

@Manao,

Please don't post the answer to the two integers problem yet, but can you tell me if I am close with those candidates I have so far for the sums?

[Kedirekin]

I can't speak for Manao, but I did get the same set of 10 sums by (I assume) a different method.

Basically I found all the combinations of 3 prime numbers that give one and only one factoring that isn't in Christos's list of sums that can't be expressed as the sum of two primes (and is that a mouthful).

For example, the primes 2, 2 and 7 give the product 28, which has factors 2*14 and 4*7. 11 (4+7) is in Christos's list, but 16 (2+14) is not. Therefore 4 and 7 are two numbers that allow Mult to know the answer. Unfortunately they are not numbers that allow Plus to know the answer, because 2 and 9 (which add to 11) would also allow Mult to know the answer.

The possible factors I got were:
2, 9
2, 25
2, 49
4, 7
4, 13
4, 19
4, 23
4, 31
4, 37
4, 43
4, 47

There is of course the remote possibility that the two numbers factor into 4 or more primes. I didn't go that far.

And unfortunately I cannot find an easy way to factor all the products of all possible addends that add up to the 10 sums; it's just too much work.


Fixed some numbers in my list. Guess my list of sums isn't exactly the same - I don't have 29, 37 or 53, but I do have 51.

[Christos]

Originally posted by r6d2
Either you are getting some marbles and a pizza box to do the real thing, or you are writing lots of numbers on a sheet of paper. Come on, did everybody give up? :)


I have exams so my free time is close to zero... :) But I check this thread every day... :p
I Finally found some time to write a program that does exactly what I discribed in my previous post...
Here are the results...

results.txt (http://users.ntua.gr/el99605/results.txt)

So the numbers are 4 and 13 (and the sum is 17 by the way:))

It took less than a second for my P4 to find the answer :p

P.S.: As I said this was done in a hurry and I had no time to check so please verify...
I could be wrong... :)

[r6d2]

Originally posted by Christos
So the numbers are 4 and 13Congrats, Christos! You know, I still don't get it. My brain only has a 6502 CPU it seems.

[Christos]

Originally posted by r6d2
You know, I still don't get it. My brain only has a 6502 CPU it seems.

Did you read carefully my 3 previous posts? If you could tell me what isn't clear, I could explain...

[r6d2]

Originally posted by Christos
Did you read carefully my 3 previous posts? If you could tell me what isn't clear, I could explain...My 6502 is parsing one word at a time. :)

But, no, I did not understand s**t, but let me try harder. It has been a tough week.

[Christos]

Originally posted by r6d2
My 6502 is parsing one word at a time. :)

But, no, I did not understand s**t, but let me try harder. It has been a tough week.

Well here is why 4 and 13 is the right answer. Maybe it can help you understand how I got it if you reverse the process?
Plus sees 17 (4+13) on his paper and Mult sees 52 (4*13) on his.

So Mult says: "I don't know the answer" because:
52=26*2
52=13*4
(Nothing else possible)

Plus says: "I Knew it" because:
17=15+2
17=14+3
17=13+4
17=12+5
17=11+6
17=10+7
17=9+8

And so Plus knows for sure that Mult sees either:
15*2=30
14*3=42
13*4=52
12*5=60
11*6=66
10*7=70
9*8=72
For all of those possible products there are more than one factorings, so plus knows beforehand that Mult could not guess correctly. Hence he says "I knew it".

Now as we so before Mult knows that the numbers are either 4,13 or 2,28 and he says "Then I know" as soon as Plus says "I knew it". Why?
Because if it were 2 and 28 then Plus would have 30 (2+28) and he could not say "I knew it" (***meaning "I was sure that you could not find the numbers based on the sum I had", meaning "The sum I have can not be expressed as a sum of 2 primes"***) because 30 could come from 23+7 so Mult would have 161 which would cause him to say "I know the answer".

Now the hard part...
As soon as Mult says "Then I know the answer" Plus says "Then I know it too". Why?
Because for for all the possible combinations that give 17 (15+2, 14+3, 13+4, 12+5, 11+6, 10+7, 9+8) only 13,4 could lead Mult to say "Then I now the answer" as soon as Plus said "I knew it".
Why?
Well for example if we consider 15,2 Mult would have 30 (15*2) which could come from:
15*2=30
10*3=30
6*5=30

But
15+2=17
10+3=13
6+5=11
So when Plus would still say "I knew it" but Mult would not be able to decide which one was it. (Because neither can be expressed as the sum of 2 primes). Therefore he could not say "Then I know"

If the same happens with 14,3 and 12,5 and 11,6 and 10,7 and 9,8 then my answer is correct ?

That's it...
I hope I didn't make things worse... :)

P.S: Any suggestions for the hats and prisoners problem? :p

[Kedirekin]

For what it is worth, I checked 4 and 13 by hand, and it does satisfy all the criteria.


Numbers Sum Possibles Possible Factors Sum of Is Sum of Possible
From Sum Product Factors Primes? Solution

4 13 17 2 15 30 2 15 17 no no
6 5 11 no
10 3 13 yes

3 14 42 3 14 17 no no
6 7 13 yes
21 2 23 no

4 13 52 4 13 17 no yes
2 26 28 yes

5 12 60 5 12 17 no no
10 6 16 yes
15 4 19 yes
20 3 23 no

6 11 66 6 11 17 no no
3 22 25 yes
2 33 35 no

7 10 70 7 10 17 no no
14 5 19 yes
35 2 37 no

8 9 72 8 9 17 no no
4 18 22 yes
2 36 38 yes
3 24 27 no

52 is the only product that has one and only one factoring that cannot be expressed as the sum of two primes.

[Christos]

Originally posted by Kedirekin
For what it is worth, I checked 4 and 13 by hand, and it does satisfy all the criteria.

52 is the only product that has one and only one factoring that cannot be expressed as the sum of two primes.

Thanks Kedirekin...
I was right then... 4 and 13 are the right numbers :)
My P4 got it right :)
I thing your post does a much better job explaining the answer than mine...
Cheers...

@r6d2
Look at the table Kedirekin posted and look at the column titled:"Is sum of primes?"
The "hard part" I was talking about in my post is that the correct combination must have only one "no" in that column...

[r6d2]

I'll have to digest that with ease. :)

Now, ready for the marbles in a cube? Well, noboody has solved the marbles in a pizza box yet. :)

After those two, we can go for marbles in a cilinder!

[r6d2]

OK, let's leave the marbles thing aside for a while. Let's get back to this intuition thing.
Suppose you have two jars of the same volume. One contains coke and the other one contains pepsi. You also have a small glass. You fill the glass with coke and pour it into the pepsi jar. You shake the pepsi jar so the two fluids mix perfectly.

Then, you fill the glass with the contents of the pepsi jar, and pour it into the coke jar. Thus the two jars end up having the same volume again.

Which jar contains a bigger portion of the other one? The coke one has more pepsi than the pepsi one has coke or is it the other way around?

[jel]

my guess is that the pepsi jar will contain more coke than the coke jar will contain pepsi, as the liquid mixed with the coke *could* contain pepsi and coke.

[r6d2]

Originally posted by jel
as the liquid mixed with the coke *could* contain pepsi and coke. I don't quite get your reasoning, but remember you shaked the pepsi jar to get a perfect mix.

[Christos]

Since the jars had the same volume of coke (the first one)and pepsi(the second one) then both will end up having the same amound of the other in the end...

Example:
*Start
Jar1:600ml Coke
Jar2:600ml Pepsi
Glass:300ml

*Move 1 (Fill glass with coke and poor it in the pepsi jar)
Jar1:300ml Coke
Jar2:600ml Pepsi + 300ml Coke (1/3Coke-Pepsi mix)

*Move 2 (Fill glass with the mix)
Glass:300ml of mix so... 100ml Coke + 200ml Pepsi
Jar2:400ml Pepsi + 200ml Coke (1/3Coke-Pepsi mix)

*Move 3 (Pour into the Coke jar)
Jar1:400ml Coke + 200ml Pepesi (1/3Pepsi-Coke mix)

[jel]

hmmm...yup christos is right ... mathematically ... but instinctively it has the same effect as the 'rope' question a little while back ... :confused:
@r6d2 ... did you edit your previous post or have i completely lost the plot...

[r6d2]

I didn't edit the post but to fix a typo, just 2 minutes after posting it, which plot did you miss?

BTW, Christos, you're as good with beverages as you are with mysterious numbers... Solution not easily according to intuition, right?

A derivative problem...
What is the size of the glass needed to end up with 50% mix on each jar?

[jel]

which plot did you miss? all of the above :rolleyes:

...back to the marble question. are the 'boundaries' you are referring to:
if the distance from the tangental edge of the last marble in any given length, is less than the radius, then this will occur -
0 0 0 0 0 0
0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0

and if the distance is greater than the radius, this will occur -
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0

[r6d2]

Originally posted by jel
...back to the marble question. are the 'boundaries' you are referring to:It's more like this:

When an integer number of marbles fit on the first row, the second row may be better arranged diagonally. This depends on the relationship between b and d. For instance, if b = 2*d, it is better to arrange them like if they were dice, vertically. Here you have one of the thresholds.

The second one is when b = n*d + d/2, where it will always be better diagonally. The general solution takes into account any values of b and d.

[sysKin]

Originally posted by r6d2
A derivative problem...
What is the size of the glass needed to end up with 50% mix on each jar?Hah, let me give you the answer at leat once. As big as the jar :P or bigger if you like the "half-empty/half-full" problems :D

No really, you need to mix them all, not leaving anything.

Radek

[r6d2]

Originally posted by sysKin
Hah, let me give you the answer at leat once.Well, that one was easy. :)

What about this one:
How many times do you have to repeat the process with the small glass to get a 50% mix?

[sysKin]

Originally posted by r6d2
What about this one:
How many times do you have to repeat the process with the small glass to get a 50% mix?How about: infinite number of times? Any time you do it, mixture gets closer and closer to 50% but never actually gets there.

:)

[samlar]

What has all this got to do with the price of a can of beans

[r6d2]

Originally posted by sysKin
How about: infinite number of times?Well, what you're saying is that if the glass has the volume of the jar, it takes 1 time, and if not, it takes an infinite number of times? Doesn't intuition tell you that there should be a relationship between the volume of the jar and the glass, which determines the number of times?

BTW, on the can of beans... We haven't got there yet. But it would be an interesting problem to determine how many beans fit into a can. :D

[fccHandler]

All this reminds me of an article I read on CNN:

http://www.cnn.com/2004/TECH/science/02/16/science.candy.reut/index.html

Try M&M's in your pizza box, instead of marbles. :D

[r6d2]

Originally posted by fccHandler
Try M&M's in your pizza box, instead of marbles. :D LOL!

That's astonishing. For centuries mankind has thought that spheres are the optimum shape for a given volume, and thus intuitively it was considered that the arrangement of them as bulk would be optimum too!

In fact, planets are sphere shaped because it minimises the momentum of inertia and thus the energy of the physical body. So, what's true for planets isn't true for associated subatomic particles? Great discovery indeed!

[Manao]

Sorry for not answering sooner, I was in holidays.

Christos : you got it right, that's the answer.

[r6d2]

Well, seems that there is no more interest in marbles in boxes, or the solution is beyond the scope of this forum. :D:D:D

Let's go for a simpler ride?
Coke cans have a standard volume of 350 ml and are made of aluminium. Suppose the can is a perfect cylinder and ignore the stacking top and bottom irregularities.

Can you design a container which uses less aluminium and still contain the same volume?

[Manao]

A ball which have the volume of the cylinder, hence a radius of 4.37 cm ( the formula of the volume being 4*pi*r^3 / 3 )

[wmansir]

Originally posted by r6d2
Coke cans have a standard volume of 350 ml and are made of aluminium. Suppose the can is a perfect cylinder and ignore the stacking top and bottom irregularities.

Can you design a container which uses less aluminium and still contain the same volume?

Ha! I can design a container that holds almost 10 times that volume and still uses less aluminium!















http://216.58.174.219/internetclassifiedsws/ecook/images/Kyle%20and%20the%20Amazing%203%20Liter.jpg

:p

And no, that is not me.

[r6d2]

Originally posted by Manao
A ball which have the volume of the cylinder, hence a radius of 4.37 cm ( the formula of the volume being 4*pi*r^3 / 3 ) OK, but your answer is merely speculative... :) How much aluminium is saved?

Or in other words, how many more containers could be built?

And also, as a derivative problem, if you have to make it a cylinder anyway (balls wouldn't be functional really, you could not put one on the table once open) :), which would be its dimensions and how much aluminium is saved?

[rudeboymcc]

you could make it flat bottomed so it stays upright! but it will be a bitch to drink, you'll have to put it completely upside down to get to the last part!

[Manao]

I wasn't speculative, I was affirmative ;)

You should have specified we should keep a cylinder, because the sphere is known to be the optimal volume in consideration to the ratio area / volume.

Anyway, with a cylinder, let's r be its radius, h its height.

Its volume is pi*r*r*h, its area is 2*pi*r*h + 2*pi*r*r. We want 2*pi*r*h minimal, with pi*r*r*h = V constant.

So we have h = V / (pi * r * r), and we want to minimize 2*pi*r*(h + r) = 2*pi*r*(V / (pi*r*r) + r)

The derivate is 4*pi*r - 2 * V / (r*r), we want it to be zero, hence, r^3 = V / (2 * pi) and r = 3.82 cm, and h = 7.64. We remark that 2 * r = h, hence, it fits perfectly a cube ( which is not quite surprising ) and it differs from usual cans, which are thinner.

I don't know the exact height and radius of a can, but I guess it's 3.3 / 10, hence a total area of 279 cm². With 3.82 and 7.64, it's 275 cm². Not much of a gain ( 1.5 % )

[vio]

Originally posted by r6d2
Well, seems that there is no more interest in marbles in boxes, or the solution is beyond the scope of this forum. :D:D:D



I think it is beyond the scope of this forum. I think there is a prize if you can get sphere packing conjecture above a certain value.

I'm still wrapping my mind around the colored hat question.

[Manao]

To end ( I hope ) the marbles in the pizza box :

We put the marbles like that :0 0 0 0 0 3
0 0 0 0 2
0 0 0 0 0 1So we have to know :
- the number of odd and even rows ( n_o, and n_e )
- the number of marbles in an odd row ( n_mo )
- the number of marbles in an even row ( n_me )

And the total number of marbles will be n_o * n_mo + n_e * n_me

----

First, the number of rows ( n_r ) :
The distance between the middle of two consecutive rows is d * sqrt(3) / 2. But we also have to count the radius of the marbles in the first and last rows. So, if we have n_r rows, we have a total height of d * ( 1 + (n_r - 1) * sqrt(3) / 2 ). Hence, n_r = int(( b / d - 1 ) * 2 / sqrt(3) + 1). Let's check :
d = 1, b = 1 -> n_r = 1 ( height : 1 )
d = 1, b = 2 -> n_r = 2 ( height : 1.866 )
d = 1, b = 9 -> n_r = 10 ( height : 8.794 )

----

Now, from n_r, we can deduce n_o and n_e :
n_o = int( (n_r + 1) / 2 )
n_e = int( n_r / 2 )

Check :
n_r = 1 + 2 * k -> n_o = k + 1, n_e = k
n_r = 2 * k -> n_o = k, n_e = k

----

Finally, we have to compute n_me and n_mo. The length of an odd row is l_o = n_mo * d, so n_mo simply equals to int( b / d )
The length of and even row is l_e = n_me * d + d / 2, hence n_me = int(b / d - 0.5)

Check :
b = 2, d = 1 : n_mo = 2, n_me = 1
b = 2.5, d = 1 : n_mo = 2, n_me = 2

----

So the final formula ( if no mistakes were made ) :
int((int(( b / d - 1 ) * 2 / sqrt(3) + 1) + 1)/ 2) * int(b / d) + int(int(( b / d - 1 ) * 2 / sqrt(3) + 1) / 2) * int(b / d - 0.5)

But hey, I forgot another bundary effect : it may happened that for the last row, putting another odd row instead of an even row is better, like that : 0 0 0 0 0
0 0 0 0 0
0 0 0 0
0 0 0 0 0
0 0 0 0
0 0 0 0 0When does this happen ? When b is between d * (k * sqrt(3) + 2) and d * ((k+1) * sqrt(3) + 1) ( height wondition ), and when b is between k * d and (k + 0.5) * d ( in order to have even rows shorter than odd ones )

Such a number, for example, is 2 ( thanks r6d2 for pointing me to that ). That makes the formula even uglier, so I won't modify it.

[r6d2]

@manao,

Good work on the ideal can thing!
Originally posted by Manao
But hey, I forgot another bundary effect : it may happened that for the last row, putting another odd row instead of an even row is better, like that : 0 0 0 0 0
0 0 0 0 0
0 0 0 0
0 0 0 0 0
0 0 0 0
0 0 0 0 0I think this may happen with several of the top rows, not just the last one. Now that would make the formula even uglier.

[r6d2]

Consider movies like Armageddon and Sudden Impact, which are categorised by some people practically as comedies from a physics standpoint. :)

Now let's speculate about the following scenario.
The Earth (mass M) revolves around the Sun at a speed v (~30 km/s). An asteroid (mass m) is in course of collision with Earth along the tangent of its orbit, on the same plane, at the same speed and in the same direction.

Suppose the timing of the trajectories are set up in such a way that both Earth and the asteroid will collide at the precise point where the radius of the Earth orbit forms a 90º angle with the asteroid's path.

After the collision, the asteroid and Earth form one single body of mass M + m.

What happens with Earth and its movement, from a physical perspective?

(i.e., ignore the casualties, the zone of impact, and such. Just think about the orbit. Does it stay the same?)

[Manao]

We don't add any quantity of movement ( I'm not sure how it is called in english, it's the product of the mass by the motion vector ), so we won't modify the orbit, nor the period of revolution.

And for the marbles, yes you're right, you may have several such rows ( but always less than 7, because sqrt(3) * 8 / 2 < 7, so you could put 8 rows with offsetinstead of the 7 ).

[easybeat]

The proposed collision wouldn't occur. there's no way Arnie, Sly and Bruce would let that happen. (Actually I here Arnie's a bit busy at the moment banning same sex marriages so we'll have to make do with Bruce and Sly)

[duartix]

I don't understand why it shouldn't change orbit Manao.
I guess from my intuition that the new body Earth+Asteroid would suffer a route deviation of about:
(mAsteroid*vAsteroid) / (mEarth*vEarth)

And I believe quantity of movement is caled momentum. (Can someone please confirm?)

[Manao]

Maybe I misunderstood the way the asteroid is meeting the Earth, but r6d2 says " An asteroid (mass m) is in course of collision with Earth along the tangent of its orbit, on the same plane, at the same speed and in the same direction " so for me the asteroid doesn't crash on the earth, it just happens to be at the same place at the same moment, with the same speed and direction than the Earth.

[r6d2]

Well, the asteroid does crash, but it does is "softly", sort of begins to ride along with Earth. Actually the point of collision cannot be the exact point of the right angle, because both Earth and the asteroid have a radius, but the simplification does not affect the result.

Manao is right, there is no change in momentum (and yes, that's how is it called). In fact, we have that:
M*v[M] + m*v[m] = (M+m)*v[M+m]

Notation: x[y] reads property x of body y, i.e., x sub yThe left side of the equation is before the impact, and the right side is after the impact. Since the original velocity v is the same for both masses, this implies that v[M+m] = v, i.e, the orbit is maintained.

However, intuition suggests that this is only valid during the instant immediately after the impact. After that, since the Earth now weights more, the attraction force from the Sun will be higher, and since it is no moving any faster, it should begin to "fall" in a spiral into the Sun.

It this reasoning correct?

[Manao]

No. The Earth will be heavier, so indeed the attraction of the sun will be higher. But, the Earth will be heavier, so the centrifugal force will be higher, and will compensate exactly the attraction force.

The falling in spiral would occur only if the energy of the Earth wasn't constant, hence if there was something to slow it down. No planet / asteroid falls in spiral, it's always revolving in ellipse ( if it doesn't touch the sun itself )

[mpucoder]

That's what I always heard, the orbit around a body of a certain mass depends on velocity alone. This is why all geosynchronous satellites, regardless of mass, are at the same altitude.
But what about the Moon? Earth now has more mass, so the gravitational attraction between Earth and Moon increases. Causing a slight decay in the Lunar orbit, and an increase in the tidal friction, slowing Earth's rotation.

[Kedirekin]

Originally posted by r6d2
An asteroid ... is in course of collision with Earth along the tangent of its orbit, on the same plane, at the same speed and in the same direction.


You do realize that, from an orbital mechanics standpoint, you've described an impossibility. If the asteroid is in a tangent to earth's orbit and moving at the same speed and direction, it is in the same orbit and can never collide.

In effect, what you've described can only happen if the asteroid is already resting on earth's surface.