[unreality]

A collection of random problems I've come up with over the past few days... [please use spoilers when answering!]

(1) A large sphere of radius 'r' is rolling around the interior of a massive cube-shaped room, chasing around a smaller sphere. The room's volume is thousands of times bigger than the larger sphere and completely empty (other than the spheres). The smaller sphere's radius is some constant multiplied by 'r'. The smaller sphere is rolling away from the larger sphere, trying to escape... it quickly realizes that the only possible escape is to wait it out against the edge or in the corner.

(1a) What is this constant's upper limit if the smaller sphere can fit against the edge of the floor and a wall and not touch the larger sphere?

(1b) What is this constant's upper limit if the smaller sphere can fit snugly in the corner of the room (ie, the floor and two walls) and not touch the larger sphere?

(2) You have a 10 by 10 by 10 cube, made of 1000 unit cubes. You also have a "punching device", which can be aimed at a specific surface unit square and pointed perpindicular to that surface square. This "punches out" and gets rid of all of the cubes in that column. If this is hard to understand, I can elaborate further. The question is, what is the maximum number of "punches" you can use to completely eliminate all of the cube?

(3) You are playing hide-and-go-seek-tag with a friend (problems of this type also translate into scenarios where two separated parties are looking to meet up and are simultaneously moving around). You and your friend are in an area that can be described as a circle with a circumference of 100. You are at one point on the circle, your friend is somewhere else. Assume that if someone is going in one direction at a constant speed, they won't change directions. Also, your speed is 'v', your friend's speed is 'w', and v>w>0. All of this applies for both sub-problems.

(3a) It's a tossup whether your friend is going to stay put, move counterclockwise or move clockwise. Because you don't know which of those three options your friend will choose, and the game has to end sometime, you HAVE to move, either clockwise or counterclockwise (you intend to flip a coin to choose which way to go). What is the average time it will take for you to catch your friend?

(3b) In this scenario, you know that your friend is NOT going to stay put... so you have to choose, between moving (either counterclockwise or clockwise) of course, or staying put and waiting. What is the average time it will take for you and your friend to meet up in both cases (ie, with you moving too or with you staying put)? So, what should you do - stay put or move? And what is your answer to that if v=5 and w=3? For what set of speeds (v,w) does it not matter whether you go or stay (ie, both options have the same avg time) and what does this have to do with the golden ratio?

~~~

thanks and good luck

[Guest]

Let the larger sphere's radius be 1 for simplicity. Then the desired constant is the smaller sphere's radius(x). The right triangle formed by the combined radii of the 2 spheres when they are tangent and sides parallel to the walls and floor has 2 sides of 1 - x and one side of 1 + x. Therefore, 2(1 - x) ^2 = (1 + x) ^ 2. expanding and combining like terms yields x ^ 2 - 6x + 1 = 0. Quadratic formula > x = 3 +/- 2(sqrt(2)). Since x < 1, x = 3 - 2(sqrt(2)).

[Guest]

Use R and r. The larger right isosceles triangle has a hypotenuse of R + r + sqrt ® and each leg is R. Pythagorus says then that r + sqrt ® + R = R * sqrt (2). Combining terms, r/R = (sqrt (2) - 1)/1 + sqrt (2)

[Guest]

I think I understand no.2, but I'm not sure about the solution...

Using my "common sense", it seems as though the worst case scenario would be removing as many columns of 10 squares as possible (in other words, just sticking to one face and punching every cube. Therefore, the best case would be removing as many one squared columns as possible. The 10 x 10 x 10 is made of 10 1 x 10 x 10 prisms in the xy plane stacked together through the z axis. Then (this is the part I'm not sure about) it would be best to clear off as many cubes as possible in the z direction, without lowering the maximum of each 1 x 10 x 10 prism. That would be equivalent to shooting 9 x 9 = 81 times in the z direction, leaving a stack of "L" shapes, 10 units thick. Then, procede to remove each L's rod that points upwards individually, requiring 10 x 9 = 90 punches. Then use 100 puches to destroy the remaining 10 x 10 x 1 prism, for a total of 81+90+100 = 271 punches...

[unreality]

Let the larger sphere's radius be 1 for simplicity. Then the desired constant is the smaller sphere's radius(x). The right triangle formed by the combined radii of the 2 spheres when they are tangent and sides parallel to the walls and floor has 2 sides of 1 - x and one side of 1 + x. Therefore, 2(1 - x) ^2 = (1 + x) ^ 2. expanding and combining like terms yields x ^ 2 - 6x + 1 = 0. Quadratic formula > x = 3 +/- 2(sqrt(2)). Since x < 1, x = 3 - 2(sqrt(2)).

I will tell you how I solve 1a:

Imagine a "great circle" slice-cut cross-section of the large sphere when it's snug against the edge (but not corner, I had to use a whole 3d cube for 1b). The radius of this circle, r, equals that of the larger sphere. Now imagine it enclosed in a perfect square such that the four tips of the circle touch the square... that is, inscribe a circle within a square.

Now, the diameter of the circle is 2r, and this is also the side length of the square.

Thus the diagonal of the square is 2r*root2

Think of the diameter of the large circle again, rotated such that it's aligned with the diagonal of the square. The remaining portions of the diagonal of the square are found by subtraction:

2r*root2 - 2r

and divide by two to get just one of the lengths:

r*root2 - r

which equals:

r*(root2 - 1)

Now this is the length of the line segment from the corner of the square to the edge of the circle. If you imagine the smaller circle filling this area so the edges touch, this length is twice that of the smaller circle's radius...

so (root2 - 1)/2 is the constant

HOWEVER, at this point I realized my error. It's quite simple. When I drew my picture, I drew the corner of the smaller circle kind of squattish by accident and it fit snugly into the corner. For this reason I forgot about the extra bit of space between THAT circle and the square corner. Thus we can say for certain that the constant is SMALLER than

(root2 - 1)/2

but the above is NOT the upper limit since it's a bit bigger than the upper limit.

Your method gives the correct answer.

Generalizing it with r just to make it official:

distance between the centers of the circles: r+x (x is our smaller radius)

this forms the hypothesis of a right triangle. The sides of the triangle are both r-x... thus it's a 45.45.90 triangle

(r-x)*root2 = (r+x)

r*root2 - x*root2 = r + x

r*root2 - r = x + x*root2

r*(root2 - 1) = x*(1 + root2)

x = r * (root2 - 1)/(root2 + 1)

Thus the constant is (root2 - 1)/(root2 + 1)

you'll find that this is the same number you got... ie, 3 - 2*root2 ... just in different form.

x = approx = .1715728753*r

edit: Simon Legree got 1a too ;D

Edited July 5, 2009 by unreality

[Guest]

I hope you can understand my explanations.

1a. Are you wondering what the answer is? Or are you sharing the love with these puzzles while you know the answer? Because I want to check my answers.

Because the large sphere will fit snug against the wall, a ray can be drawn inside a square, with the dimensions of the square being r (because the sphere contacts it at r height) and the radius of the ray being r.

A diagonal line can be drawn through the ray, from corner to corner, with the length being r+x where x is the distance from the end of the ray to said corner. A^2 + B^2 = C^2, r^2 + r^2 = 2r^2, C = r+x = squareroot-of-2*r.

Squareroot-of-2 = 1.414.

x = .414r

Against one wall and the floor, the smaller sphere would have to be less than .414r. (to be more exact, x would be squareroot-of-2 minus 1)

1b. As far as I can tell, when a sphere is up against a corner, the three touch points will all be r distance from the respective seams of the walls and floor. I'm not too good with 3-dimensional angles, but a plane needs to be drawn at a 45 degree angle (from sides, not above) through the sphere into the smaller sphere, and then the corner. Then a 2 dimensional figure can be drawn...(am drawing it now). It looks a lot like 1a.

AH HA! In stead of having r as the length of the square sides, it needs to be squareroot-of-2*r because it is at a 45 degree angle. But what is the radius I should use? It seems that I should use r for the radius. A^2 + B^2 = C^2, (squareroot-of-2*r)^2 + (squareroot-of-2*r)^2 = 4r^2, C = 2r.

The diagonal is 2r, with radius r, so the small sphere needs a radius of less than r/2. x = 1/2.

The results are confirmed by the fact that there is more room in a corner than a side.

2. Maximum eh? Minimum would be 100(knocking out 10 each time). Hmmm.

I got it! Working in a crisscross pattern on a single level of 100 cubes at a time.

10 across, 9 down, 9 across, 8 down, 8 across...

One set of 100 could take 19 punches. Times 9 rows is 171 punches. For the final plane of 100, you punch 1 out at a time. Totaling 271. That is the most inefficient method I can come up with.

3.(a and b) I think this has no proper solution. You did not provide a range of sight, and I am presuming that once you reach the edge of the circle, you rebound at the appropriate respective angle (based on the size of the area).

In this way, with unlimited angles, there is a possibility of an infinite game. And from childhood experience, a game can last for so long that you need to call it off (although the randomized speed, directions, and obstructions provide insurmountable variables).

As such, there is such a wide (reaching and including infinity) time that the game can be played, that an average cannot be found.

Even with robots, with set ranged sensors and speed, the multitude of angles would provide too many differences.

I hope these solve them to your satisfaction. I spent a good hour on it.

EDIT: Looking up, it seems to me that staying up all night for fireworks has hampered my concentration. I still keep my answers for 2 and 3.

Edited July 5, 2009 by Out4Blood4

[unreality]

I will tell you how I solve 1a:

Imagine a "great circle" slice-cut cross-section of the large sphere when it's snug against the edge (but not corner, I had to use a whole 3d cube for 1b). The radius of this circle, r, equals that of the larger sphere. Now imagine it enclosed in a perfect square such that the four tips of the circle touch the square... that is, inscribe a circle within a square.

Now, the diameter of the circle is 2r, and this is also the side length of the square.

Thus the diagonal of the square is 2r*root2

Think of the diameter of the large circle again, rotated such that it's aligned with the diagonal of the square. The remaining portions of the diagonal of the square are found by subtraction:

2r*root2 - 2r

and divide by two to get just one of the lengths:

r*root2 - r

which equals:

r*(root2 - 1)

Now this is the length of the line segment from the corner of the square to the edge of the circle. If you imagine the smaller circle filling this area so the edges touch, this length is twice that of the smaller circle's radius...

so (root2 - 1)/2 is the constant

HOWEVER, at this point I realized my error. It's quite simple. When I drew my picture, I drew the corner of the smaller circle kind of squattish by accident and it fit snugly into the corner. For this reason I forgot about the extra bit of space between THAT circle and the square corner. Thus we can say for certain that the constant is SMALLER than

(root2 - 1)/2

but the above is NOT the upper limit since it's a bit bigger than the upper limit.

Your method gives the correct answer.

Generalizing it with r just to make it official:

distance between the centers of the circles: r+x (x is our smaller radius)

this forms the hypothesis of a right triangle. The sides of the triangle are both r-x... thus it's a 45.45.90 triangle

(r-x)*root2 = (r+x)

r*root2 - x*root2 = r + x

r*root2 - r = x + x*root2

r*(root2 - 1) = x*(1 + root2)

x = r * (root2 - 1)/(root2 + 1)

Thus the constant is (root2 - 1)/(root2 + 1)

you'll find that this is the same number you got... ie, 3 - 2*root2 ... just in different form.

x = approx = .1715728753*r

edit: Simon Legree got 1a too ;D

I have fleshed out my original solution to 1a and corrected it. So there are two methods of solution so far for 1a: making a 45-45-90 triangle with the centers of the two circles inside a snug square and using the relationship of side-to-hypotenuse (ie, pythagoras) to calculate x.

Or my original solution, corrected:

square snug around circle with radius r (this is cross-section of the larger sphere when it's flush against the wall). Smaller circle in bottom right corner (also flush against wall).

2r = diameter of large circle = side of square

2r*root2 = diagonal of square

(2r*root2 - 2r)/2 = r(root2 - 1) = distance from corner of square to the "corner" of the large circle

call this length 'd', that is... r(root2 - 1) = d

now enclose the SMALLER circle in a snug square. The radius of this is x, which is what we want to find.

With the same method used before, 2x*root2 = diagonal of this smaller corner square

this diagonal equals the diameter of the small circle (2x) plus twice the distance from one corner of the square to one corner of the circle (call this z, so 2z)

then: 2x*root2 = 2x + 2z

2x*root2 - 2x = 2z

x*root2 - x = z

x*(root2 - 1) = z

Now, d is MOST of the diagonal of this smaller square... it is one edge of the circle to the opposite edge and ONE corner length... so d = 2x + z and d-2x = z

so we can replace 'z' with 'd-2x'

x*(root2 - 1) = d - 2x

x*(root2 - 1) + 2x = d

x*(root2 - 1 + 2) = d

x*(root2 + 1) = d

so x = d/(root2 + 1)

but we've already calculated d to be r*(root2 - 1)

so x = r*(root2 - 1)/(root2 + 1)

so the constant is:

(root2 - 1) / (root2 + 1)

which also equals 3 - 2*root2 if you use one variant of the right-triangle method

I hope you can understand my explanations.

1b. As far as I can tell, when a sphere is up against a corner, the three touch points will all be r distance from the respective seams of the walls and floor. I'm not too good with 3-dimensional angles, but a plane needs to be drawn at a 45 degree angle (from sides, not above) through the sphere into the smaller sphere, and then the corner. Then a 2 dimensional figure can be drawn...(am drawing it now). It looks a lot like 1a.

AH HA! In stead of having r as the length of the square sides, it needs to be squareroot-of-2*r because it is at a 45 degree angle. But what is the radius I should use? It seems that I should use r for the radius. A^2 + B^2 = C^2, (squareroot-of-2*r)^2 + (squareroot-of-2*r)^2 = 4r^2, C = 2r.

The diagonal is 2r, with radius r, so the small sphere needs a radius of less than r/2. x = 1/2.

The results are confirmed by the fact that there is more room in a corner than a side.

You can use the pythagoras method here to get a similar result as the correct answer to 1a.

sqrt(3 * (r-x)^2) = (r+x)

root3 * (r-x) = (r+x)

solving for x gives: x = r* (root3 - 1) / (root3 + 1)

While you can use my solution method with the corner lengths of inscribed squares and cubes, 'next's Pythagoras method is the quickest, and gives the answers for number one:

1a: (root2 - 1) / (root2 + 1)

1b: (root3 - 1) / (root3 + 1)

2. Maximum eh? Minimum would be 100(knocking out 10 each time). Hmmm.

I got it! Working in a crisscross pattern on a single level of 100 cubes at a time.

10 across, 9 down, 9 across, 8 down, 8 across...

One set of 100 could take 19 punches. Times 9 rows is 171 punches. For the final plane of 100, you punch 1 out at a time. Totaling 271. That is the most inefficient method I can come up with.

I believe that this is the maximum but I have yet to prove it - gimme a bit and I'll get back to you and 'next' on this one

3.(a and b) I think this has no proper solution. You did not provide a range of sight, and I am presuming that once you reach the edge of the circle, you rebound at the appropriate respective angle (based on the size of the area).

In this way, with unlimited angles, there is a possibility of an infinite game. And from childhood experience, a game can last for so long that you need to call it off (although the randomized speed, directions, and obstructions provide insurmountable variables).

As such, there is such a wide (reaching and including infinity) time that the game can be played, that an average cannot be found.

Even with robots, with set ranged sensors and speed, the multitude of angles would provide too many differences.

incorrect... by circle I meant the mathematical definition of a circle, which is locus of points equidistant from a center. That is, I meant the people can only travel around the EDGE of the circle. I guess I didn't make that clear enough - I'll see if the edit time is up yet... if I can't edit anymore, then clarification: for #3, the people can only travel around the edge of the circle!!!!!!!! That's what I meant by "clockwise" and "counterclockwise"... ie, you walk a circular arc path.

That should make it much easier

[Guest]

incorrect... by circle I meant the mathematical definition of a circle, which is locus of points equidistant from a center. That is, I meant the people can only travel around the EDGE of the circle. I guess I didn't make that clear enough - I'll see if the edit time is up yet... if I can't edit anymore, then clarification: for #3, the people can only travel around the edge of the circle!!!!!!!! That's what I meant by "clockwise" and "counterclockwise"... ie, you walk a circular arc path.

That should make it much easier

Yes that would make things easier! Well I thought that the "area" was where they were playing, and they rotated clockwise and counterclockwise through 360 degrees.

Well it looks like my math was off. I'l try and figure out 3 in a bit.

[Guest]

Possible solution for 1a:

This case reduces to a 2-dimensional problem with circles instead of spheres.

The larger circle, assuming it is in the corner (0,0), has equation (x-Y)^2 + (y-Y)^2 = Y^2.

The smaller circle, also "in the corner", has equation (x-cY)^2 + (y-cY)^2 = (cY)^2

Also, the radii of both circles lie on the line y = x. Thus, we can determine c by setting y = x in both equations. The smaller solution from the larger circle should equal to the larger solution from the larger circle (tangential circles). Equate the two solutions and solve for c:

2(x-Y)^2 = Y^2

x= Y(1 - 1/sqrt(2))

2(x-cY)^2 = (cY)^2

x = cY(1 + 1/sqrt(2))

Y(1-1/sqrt(2)) = cY(1+1/sqrt(2))

c = (sqrt(2) - 1)/(sqrt(2) + 1) = 0.1715728...

ETA solution for 1b:

Moving to 3 dimensions, the same logic still applies since the formula for the surface of a sphere "in the corner" is (x-Y)^2 + (y-Y)^2 + (z-Y)^2 = Y^2. Similar calculations show c = (sqrt(3) - 1)/(sqrt(3) + 1) in this case.

Edited July 6, 2009 by Bamafan

[unreality]

bamafan: correct Good job {and congrats to everyone that has solved #1 so far - so far we know of three different correct methods of solution, with bamafan giving us the third}

as of yet, I haven't had enough time to try to make a solid "proof" for the 10x10x10 cube punching problem... after I get off the computer right now I'll think on it... we'll see if anyone else can come up with equivalent/better/proven solutions

As for #3, nobody has cracked this one yet

[Guest]

I'll give 3a a try:

(1) 1/3 of the time v and w will approach each other => closing speed = v+w.

(2) 1/3 of the time v will move toward w => closing speed = v.

(3) 1/3 of the time v and w will move in the same direction => closing speed = v-w.

For each scenario, the time taken for the two to meet will be equal to the starting distance between v and w divided by the speed with which the distance between them closes.

The starting distance between v and w is a random variable following the uniform distribution on the interval [0, 100]. The expected value of this distribution is 50.

Thus, the average time for v to "catch" w is equal to 1/3 * 50/(v+w) + 1/3 * 50/v + 1/3 * 50/(v-w) = 50/3 * (3v^2 - w^2)/(v^3 - v*w^2), where v and w represent the speeds of the two participants.

[unreality]

I'll give 3a a try:

(1) 1/3 of the time v and w will approach each other => closing speed = v+w.

(2) 1/3 of the time v will move toward w => closing speed = v.

(3) 1/3 of the time v and w will move in the same direction => closing speed = v-w.

For each scenario, the time taken for the two to meet will be equal to the starting distance between v and w divided by the speed with which the distance between them closes.

The starting distance between v and w is a random variable following the uniform distribution on the interval [0, 100]. The expected value of this distribution is 50.

Thus, the average time for v to "catch" w is equal to 1/3 * 50/(v+w) + 1/3 * 50/v + 1/3 * 50/(v-w) = 50/3 * (3v^2 - w^2)/(v^3 - v*w^2), where v and w represent the speeds of the two participants.

correct

My own method of solution is essentially the same... I split it into sixths, each one being d and 100-d, when it all adds up however it becomes 100-d + d + 100-d + d + 100-d + d, which averages to 100/6 which reduces to 50/3, which is what you got for the distance constant (unitwise, the rest becomes speed^2 / speed^3 which is time per distance, multiplying by the distance constant gives average time - for "physics" problems I always check everything with units)

Anyway, yes, congrats - 3a has been solved ;D

also, sorry I keep getting sidetracked from proving #2, I've been teaching myself Common Lisp

[Guest]

If you stay put, your friend reaches you in 50/w time

If you move, there are 4 cases possible;

2 cases (clockwise and counter clockwise when both move in same direction

2 cases (clockwise and counter clockwise when both move in opposite direction

Average time then in case of moving is

1/4[100/(v+w) + 100/(v-w)] = 50v/(v²-w²)

So, the decision depends on whether 50/w or 50v/v²-w² is greater

equating both sides, we have

v²-w² = vw

Let v = kw

then;

k²w²-w² = kw²

or

k²-1 = k

solving for k, we get k = (1+root5)/2

or k = 1.618

So, for all set of speeds where v = 1.618w, it doesnt matter whether you move or not

For k < 1.618, it is better to stay put

For k > 1.618 it is better to move

If v = 5 and w = 3, since v/w = k = 1.666, it is better to move

Didnt know what golden ratio was, but I checked on the internet and turns out that that the golden ratio is such that for a line segment from a to b, if a+b is to a as a is to b and we get the golden ratio of a/b = k = 1.618 (which is what I found here)

Edited July 6, 2009 by DeeGee

[Guest]

Re. #2:

I believe that this is the maximum but I have yet to prove it - gimme a bit and I'll get back to you and 'next' on this one

it's pretty easy to get to 550: knock out the 2nd row of the top layer. Pick off the cubes in the exposed row one by one (11 "moves", 20 cubes). Repeat 50 times. And there's an even more inefficient way too...

[unreality]

DeeGee: yep you got 3b

d3k3: when you "punch" out a column, you do an entire column of unit cubes, to a total of 10 unit cubes destroyed if all of the column is intact, 9 if there's a missing unit cube from another punch, etc. You can't only do one cube in and stop. In other words, you point it perpindicular to the face of a unit cube and "punch it" and it creates a line to infinty, destroying all cubes on that line. So even if there's gaps in there, it punches them all out.

So if I understand you correctly, you first knock off a surface row that's one away from an edge row. Then you say you hit off each of those unit cubes that are in the isolated row... You can't do that since the puncher would punch out all 9 in that row. I guess you could say it's not "small" enough to fit in order to one cube at a time.

I know I didn't explain it perfectly, it's kind of hard to do it without a picture and I don't have any 3d-picture-making tools and I'm not very artistically inclined

edit: it may help to think of cubes not being destroyed but rather being "off" or "on".

They all start out "on". A punch to a row will turn all "on" cubes in that row to the "off" position, and the ones that are currently "off" will stay "off"

Edited July 6, 2009 by unreality

[Guest]

DeeGee: yep you got 3b

d3k3: when you "punch" out a column, you do an entire column of unit cubes, to a total of 10 unit cubes destroyed if all of the column is intact, 9 if there's a missing unit cube from another punch, etc. You can't only do one cube in and stop. In other words, you point it perpindicular to the face of a unit cube and "punch it" and it creates a line to infinty, destroying all cubes on that line. So even if there's gaps in there, it punches them all out.

So if I understand you correctly, you first knock off a surface row that's one away from an edge row. Then you say you hit off each of those unit cubes that are in the isolated row... You can't do that since the puncher would punch out all 9 in that row. I guess you could say it's not "small" enough to fit in order to one cube at a time.

I know I didn't explain it perfectly, it's kind of hard to do it without a picture and I don't have any 3d-picture-making tools and I'm not very artistically inclined

edit: it may help to think of cubes not being destroyed but rather being "off" or "on".

They all start out "on". A punch to a row will turn all "on" cubes in that row to the "off" position, and the ones that are currently "off" will stay "off"

what you're saying is that the device "punches" in both directions? In that case, I can't do better than your 271. Interestingly, if you always attack the smallest face, it takes 270 punches, or one fewer. In fact it will always be one fewer for any nxnxn cube...

[unreality]

Didnt know what golden ratio was, but I checked on the internet and turns out that that the golden ratio is such that for a line segment from a to b, if a+b is to a as a is to b and we get the golden ratio of a/b = k = 1.618 (which is what I found here)

oh it's so much more than that You seem well-versed in mathematics so to be honest I'm really surprised that you can't have heard of it before - I can't begin to list all of its amazing properties and the unusual places it pops up. IMO it's up there with 'pi' and 'e'