Prime

So the program does exhaustive search of all possible divisions with division into 2 method. Then it sorts the results and weeds out duplicates. That is not exactly same as finding mathematical formula for the number of possible arrangements. Still, an efficient algorithm that would traverse all possible combinations seems like a non-trivial task in itself. It could also evolve into a formula. There are two ways to get duplicate solutions. One is where the algorithm arrives to the identical division pattern through different sequences of divisions. Another – where same sets of rectangles can be arranged in different ways inside the square. An algorithm avoiding just the first kind of duplicate solutions (arriving to the same division pattern more than once) seems like an interesting programming task in itself. Do you have such an algorithm? I’ll make an illustration: Let’s say we had to divide square into n parts. A recursive algorithm could be something like: RecAlg(area) 1: Go through all possible ways to break area into 2. __1.1: For each break down: ____1.1.1: If area1 > 1/n then call RecAlg(area1) ____1.1.2: If area2 > 1/n then call RecAlg(area2) ____1.1.3: Store division results __1.2: Store division result 2: Add up all results. … Such algorithm would go through all possible divisions. However, it is not very efficient. It would stumble into exact same division many times over. And the larger is n, more inefficient the algorithm becomes. On the bright side, the order (number of division variations) for such algorithm is easy to estimate. For a division into n, there are [n/2] valid ways to divide a square into two parts and 2*[n/2] ways to divide rectangle into two parts. (Those ways are: [1/n, (n-1)/n], [2/n, (n-2)/n], and so on.) Thus the Order for division of square into n parts would be: O(n) = O(n-1)*O(1) + O(n-2)*O(2) + … + O(n/2)*O(n/2). For example, let’s divide square into 5. There are two ways [5/2] to make the initial split of a square into 2. Those are: (1/5, 4/5) and (2/5, 3/5). Say, program has processed the (1/5, 4/5) split finding the number of variations O(4)*O(1). It moves on to process (2/5, 3/5) split. When algorithm divides the 2/5 area into two rectangles as shown in FIG. 3, it has stumbled into variations already processed, since the same arrangement could have resulted when processing (1/5, 4/5) split. Question: can you devise an algorithm that would “visit” each different division pattern exactly once? Then the Order of such algorithm would be a half way to the formula. Second half, possibly more difficult, to figure out the formula for: in how many different ways a given collection of rectangles could be fit into a square. After that, to solve the OP with stipulation no irrational dimensions, all we’d have left is to prove that division into 2 method covers all rational number solutions. If that’s not the case, the problem stands unshaken. We’d have to look for other methods for solution.

PRECISE ANSWER! I can only add an illustration:

Where did isosceles triangles come from? a, b, and c could be three entirely different numbers.

I am still unclear, how does the algorithm choose the coordinates of each cut? Does it decide up front how many smaller rectangles each larger rectangle is going to have? That seems to lead to many hidden duplicate solutions. Does your program actually calculate the dimensions of the rectangles in each set?

OK, so if I understand it right, the algorithm cuts 1/n area from the remaining area on each division. Then, of course, you are guaranteed rational dimensions. With every cut, the algorithm reuses one of the existing rectangle sides, which is of a rational length. And since 1/n is a rational number it can be obtianed by multiplication of two rational numbers, or two irrational, but not the mix. So if one side is rational, the other one has to be also. What I don't understand, how that algorithm would produce all possible divisions? For example, a square divided into 4 squares as shown below? If algoritm cuts off 1/4 of the total area on each division, and then leaves that correctly sized cut untouched by consecutive divisions, it would never achieve the arrangement as above. On the other hand, if the algorithm cuts the area into equal (or whatever other relation) parts on each consecutive step, then we are back to what I am pointing out -- the ambiguity (not uniquely defined operation.) Say, algorithm first drew the AB line through the middle. On the next step it drew CO line down the middle. Then it would have to change the coordinates of the AB line -- slide it down. Then after drawing DO line, it would have to change the position of the AB again. My illustration in post 87 refers to your algorithm on its n-th step. Does your algorithm actually verify that each rectangle has an area of 1/n and calculate the exact dimensions of all rectangles, or does it simply subdivide the square?

A trilateral pyramid has all three angles at the apex equal to 90 degrees each: AHB=AHC=BHC=90; The side edges are equal a, b, and c: AH = a; BH = b; CH = c. Find the volume of the pyramid. If you recall, volume of a pyramid is equal to one third of its height (dropped from the apex to the base) times the area of the base. I remember this one from 9th or 10th grade, when math teacher gave it to the class and 10 minutes to solve it

That line of reasoning does not take into account the OP requirement that all subdividing rectangles must be of equal area. For example: Suppose we have an instance of divison of a unit square into 5 equal area rectangles. The area of each rectangle is 1/5. Then we designate one of the small rectangles for further subdivision into two. Since the square is going to be divided into 6 rectangles, the area of each must be 1/6. Thus the rectangle designated for subdivison must be expanded from 1/5 to 2*(1/6)=1/3, while the remaining 4 rectangles must shrink from area of 1/5 to 1/6. In general terms, operation of changing subdivision of n into n+1 must expand the area of one rectangle from 1/n to 2/(n+1) and shrink others from 1/n to 1/(n+1). I did not see anywhere in this topic, HOW the subdivison operation changes dimensions of individual rectangles. I gave an example of ambiguity of subdivision in my post 87. (I mistakingly refered to it as post 89 in few posts above.) Once we decide exactly how the dimensions of rectangles are changed, we could go ahead with a proof or rationality. It could be based on the fact that a rational number can only be a sum of other rational numbers.

Sorry, I missed that point in my previous reply. The example that you reffer to, actually sports irrational dimensions. I calculate the sides for the rectangles at the edges of the square are: (1 +- 51/2/3)/2. Square root of 5 is an irrational number. All other rectangles inside that square have irrational dimensions too. I have posted a general proof that any arrangement of that type, where the four rectangles at the edges are identical, will have only irrational dimensions. (See my post 89 part II.)

Here is an example of unambiguous construction. The square was subdivided in the order shown with the numbers in blue. (In a spiral pattern.) If we were to deconstruct it, we'd have to proceed in the reverse order. I.e. the only side that could be removed first is the one between 6 and 7. Removing of any other side would leave hanging segments and the square would no longer be divided into rectangles. In that spiral division arrangement, the dimensions of rectangles are determined uniquely as shown by formulas, where n is the number of rectangles inside the square. (In this example n = 7). The formulas also show exactly how the dimensions would be adjusted with further subdivision, or reverse process of deconstruction. This pattern yields only one instance of subdivision. However, other patterns of subdividing square are not so restrictive and lead to ambiguity as shown in post 89. To make those useful, we'd have to describe exactly how the remaining rectangles must be adjusted when dividing one of them into two. (The illustration was drawn approximately, without actual measurments. So some rectangles appear bigger.)

I have accepted the modification to the OP, and I'm discarding irrational solutions. My point is, how can you be sure that "subdivision method" adopted here covers all rational solutions? Here is a direct quote from the OP. Whereas, "subdivision method" adapted here is geared toward finding different arrangements of rectangles inside the square -- not finding sets of rectangles with different dimensions that can fill the square. I did not see your program, but I get an impression that the division operation is ambiguous. See illustration in my post 89, where I show how the same division can result in two different arrangements. The reverse is also possible, different division would result in the same set of rectangles. Can you prove that subdivision always yields a set of rational-dimensioned rectangles? (Not that I'm arguing, it's not true. But I think, a proof is in order.) If I understand right, here you support my argument. Which is subdivision method may not be sufficient to find all solutions. However, I am not completely sure myself. It may, or may not be. For certain, it needs more work, since it does not specify exactly how it re-arranges the dimensions of the rectangles with each new division. And it requires all those proofs I mentioned above. The proofs will help to understand the problem better and help find working algorithm.

It seems to me, the only way to disambiguate this problem, is to use Chuck's suggestion that the circles lie in perpendicular planes. Otherwise, draw a circle with radius 20. Draw any perpendicular line segment of any length intersecting a radius of that circle, use it as a radii for another circle. Or draw a tangent line to the first circle, use that as a radius for another circle. In such ways two circles may intersect whichever way, or one could be enclosed entirely inside another, or they could be completely outside each other and you can extend their radii to intersect and show that they are at the right angle to each other.

That actually happened in my town. I saw it on the local news.

I haven't seen this one yet:

A side note. The same rule holds for any number N, not just 9. That is the remainder from the division of any square by N is one of the remainders from division of squares of 0 through [N/2] (integer part of N/2) by N. E.g. remainder of division of any square by 5 would be 0, 1, or 4. But never 2, or 3.

First, couple of corrections to my previous post: 1. The illustration of two different arrangements resulting from the same subdivision operation may not be a "proof" that "subdivision into two" method does not cover all variations. Still, it identifies an algorithmic problem. And the method remains unproved. 2. The equation in part II should be x*y = 1/n, not x*y = 1/n2. Correspondingly, n=4 yields the only rational solution. Let's examine possible dimensions for the subdividing rectangles. Suppose, we subdivide the square of side 1 into n rectangles of equal area. The area of each rectangle is 1/n. Then the smallest any side can be is 1/n. (A side smaller than that, would necessitate the other side to be of length greater than 1, the side of the original square.) The shorter side of rectangle can be up to (1/n)1/2, making it a square. The next smallest size for a rectangle side is 1/(n-1). No dimensions between 1/n and 1/(n-1) are possible. Such rectangle would leave less than 1/n distance between its end and one of the square sides, making it impossible to fill that area. If there is a rectangle of the side 1/(n-1) there must be one and only one rectangle of side 1/n. Next size up is (n-1)/n(n-2) for n > 2 as shown on the illustration below. Although, it would not be the smaller side for n = 3. There may not be any sizes between that and 1/(n-1), since the latter cannot be subdivided. The (n-1)/n(n-2) must necessarily come in a package with 1/(n-1) and 1/n. (FIG. 1) Incidentally, in Flowstoneknight arrangement, the side of a size (1 – (1 – 1/n)1/2)/2 could be the next size after 1/(n-1), but it is an irrational number, which was disqualified in the earlier posts. Next rational size that I see is 1/(n-2) as shown in FIG. 2. That one can come with two (1/n)x1 rectangles, or with one (2/n)x(1/2). That makes things more complicated… The above analysis shows that rectangles come in certain increments and have some forced relations. An algorithm could find all possible dimensions for an n-subdivision, and traverse (perhaps recursively) all possible sets of those dimensions, that would pack the area. The problem with that approach is dealing with remaining free space, which is not a single rectangle.

What does that mean though? Every mile per hour with respect to the ground? Treadmill? If OP stipulates the plane accelerates with respect to ground, what's the riddle?

This math problem runs deep. I'm afraid, dividing one of the n rectangles into two to obtain an n+1 combination is not the solution. 1. It does not cover all possible arrangements even among rational-number solutions. 2. Where is the proof that after expanding one of the rectangles for division and adjusting dimensions of all other rectangles, we don't get irrational dimension solutions? (Although, that proof seems to be within reach.) 3. The arrangement offered by Flowstoneknight (Post49) cannot be obtained by dividing rectangles into two method. It can be proven algebraically to yield irrational number dimensions. However, there are other, more complex arrangements of the sort, where dimensions are not necessarily irrational numbers. 4. Since the OP requires finding a number of different sets of dimensions for each division into n equal areas, regardless of their order inside the subdivided square, the division into two method does not seem to render any orderly way of weeding out duplicate solutions. I. Let's start with illustration (and proof) for the point (1) above: Suppose we divided the square into 9 equal squares with the dimensions (1/3)x(1/3) each (FIG 1). Now let's obtain a 10-division arrangement by dividing the square in the middle into two as shown. All rectangle dimensions must be adjusted. We can do it by shrinking the top and bottom row of 3 squares each, into two rows of 3 (1/3)x(3/10) rectangles and adjusting squares in the middle into 4 (1/4)x(4/10) rectangles. (FIG 2.) Or we can adopt an alternative scheme of shrinking/expanding: Shrink the row at the top into 3 (1/3)x(3/10) rectangles; The lower two on the left side – into 2 (7/20)x(2/7); The rightmost 2 at the bottom – into 2 (5/14)x(14/50); And the remaining 3 in the middle – into 3 (5/21)x(21/50). (FIG. 3) Thus we have two very distinguishable arrangements resulting from the very same division of a rectangle into two. With more complex arrangements of rectangles the number of variations resulting from the same operation increases. II. Let's prove algebraically that the subdivision found by Flowstoneknight (Post49) yields irrational dimensions. Suppose, the length of each rectangle at the edge of the square is x and the width is y. Let's further assume that the square was subdivided into n parts (at least 5, but could be more by dividing the middle square, or even the rectangles at the edges.) Then we have: x+y=1; x*y = 1/n2; Solving we find: x = (1 +- (1 – 4/n2)1/2)/2 The expression under the square root can only yield rational number when n = 2. However, the same type of arrangement does not have to follow such a neat symmetrical pattern. For example, it could sport unequal rectangles at the edges: It seems plausible that such an arrangement is possible. Whereas, to calculate the relation of the dimensions to the side of the original square leads to large equations and higher degree polynomials. And we cannot say (without further research) that those would not yield rational solutions. And again, that arrangement cannot be obtained by "division into two" method. If we search for solution by using "subdivision into two" method, we must prove that all other arrangements would lead to irrational dimensions (if such is the case.) Then each rectangle must define its relation to all others and identify all different methods of re-arranging rectangles when one is divided. Alternatively, we could try identify some minimal allowed increment in dimensions of sub-rectangles, if such exist. I say, if we can even devise an algorithm capable of finding the solution for a given n-division – that could be worthy of publishing in a respectable mathematical journal.

Vimil has the answer again! Isn't it interesting, how program's need to propagate defines its tail size?

The OP was somewhat ambiguous, while creating all that diversion. And so it may be a fair game. Especially so taking into the account the consecutive posts: Prof. T rejected Yoruichi-san's guess that photon was unlikely to reach the Earth. And yet based the "correct answer" on photon's bouncing around. Of course, for my answer, the OP would have to worded slightly differently. On the other hand, the "time standing still" for objects moving at the speed of light is more common-knowledge than photon's behavior.

The speed of light is about 300000 km/sec to an observer. However, when object travels at the speed of light, time does not elapse for that object. (As far as that object is concerned, it got from here to there in no time at all.) So I stand by my answer: zero time elapses for the traveling photon.

I was talking about the speed of the rotation of the wheels, not airplane ground speed. Sorry, if that was unclear.

I must note one more thing that I omitted in previous posts. In the model, where the conveyor matches the wheels rotation, and the plane does not take off -- it is not just friction that counteracts the force of the jets. As the plane edges forward, the belt has to move faster to match the wheels' rotation. The wheels rotate faster. There is an acceleration in RPMs. Acceleration requires force. As speed becomes comparable to the speed of light the acceleration becomes more and more costly in terms of energy required. People build humangous reactors requiring massive amounts of energy to accelerate just a small atomic particle to speeds nearing the speed of light. Against acceleration of airplaine's wheels to some considerable speed, jet engine has not a prayer. And at some point the friction in the axels would become enough to set the airplaine on fire. Pulling conveyor belt from under the plane is not the most effective way to stop it. But theoretically, or, at least, logically, it can be done.

I'll disagree that the type of the engine is relevant in any case. If it's internal combustion engine that rotates wheels, then the problem comes when the plane takes off an loses its friction. Otherwise that engine would work as well as a jet. I don't see any paradox, nor the explanation. (But I didn't go through the entire topic.) However, we may have a concensus here. If the belt matches speed of the plane (whatever that means) -- then the plane takes off, and I still don't see what's the riddle there. Is it, whether the plane's engines can overcome a little bit of additional friction? If the belt matches the rotation of the wheels -- the plane does not take off. Although, in practice, you'd have to make the belt a chain, the wheels sprockets, and rotate the belt fast enough so that the friction in the axels matches the jets' thrust. (Or when the system starts reaching relativistic effects -- approaching speed of light.) Either way the riddle seems simple. The real puzzle is, why this riddle raises so much passion and heated argument? Not necessarily on this forum, I mean, in general. It is all over the internet. All people around the world divided into two camps. And the war is about to break out with airplanes taking off moving conveyors and dropping logical bombs on the enemy.