💡First, I need to determine the initial distance between the two trains. Both trains are moving towards each other, so their relative speed is the sum of their individual speeds. I'll start by converting the speeds from kilometers per hour to meters per second to maintain consistent units.Next, I'll calculate the relative speed by adding the speeds of both trains. Once I have the relative speed, I can use the given time until they meet to find the total distance they cover together. This total distance includes the initial distance between the trains plus the lengths of both trains.Finally, I'll subtract the combined lengths of the trains from the total distance covered to find the initial distance between them.

💡Alright, so I have this problem about an arithmetic sequence. Let me try to figure it out step by step. First, the problem says that the sum of the first n terms is Sₙ. It also gives me that 2a₁ + a₁₃ = -9. I need to find S₉. The options are A: -27, B: 27, C: -54, D: 54.Okay, let's recall what an arithmetic sequence is. It's a sequence where each term after the first is obtained by adding a constant difference, which we usually denote as d. So, the nth term of an arithmetic sequence can be written as aₙ = a₁ + (n-1)d.Given that, let's write down what a₁₃ is. Since a₁₃ is the 13th term, it should be a₁ + 12d. So, a₁₃ = a₁ + 12d.Now, the problem says 2a₁ + a₁₃ = -9. Let's substitute a₁₃ with a₁ + 12d. That gives us:2a₁ + (a₁ + 12d) = -9Let me simplify that:2a₁ + a₁ + 12d = -9 3a₁ + 12d = -9Hmm, I can factor out a 3 from the left side:3(a₁ + 4d) = -9Divide both sides by 3:a₁ + 4d = -3Okay, so that's a nice equation: a₁ + 4d = -3. I'll keep that in mind.Now, I need to find S₉, which is the sum of the first 9 terms. The formula for the sum of the first n terms of an arithmetic sequence is Sₙ = n/2 * (2a₁ + (n-1)d). Alternatively, it can also be written as Sₙ = n*(a₁ + aₙ)/2.Let me use the second formula because I think it might be easier here. So, S₉ = 9*(a₁ + a₉)/2.Wait, but I don't know a₉ yet. Let me find a₉. Since a₉ is the 9th term, it's a₁ + 8d.So, a₉ = a₁ + 8d.Therefore, S₉ = 9*(a₁ + a₁ + 8d)/2 S₉ = 9*(2a₁ + 8d)/2 S₉ = 9*(a₁ + 4d)Oh, interesting! So, S₉ simplifies to 9*(a₁ + 4d). But from earlier, I found that a₁ + 4d = -3. So, substituting that in:S₉ = 9*(-3) S₉ = -27So, the sum of the first 9 terms is -27. Looking back at the options, that's option A.Wait, let me double-check to make sure I didn't make any mistakes. Starting from 2a₁ + a₁₃ = -9, which becomes 2a₁ + (a₁ + 12d) = -9, leading to 3a₁ + 12d = -9. Dividing by 3 gives a₁ + 4d = -3. Then, for S₉, I used the formula and found it's 9*(a₁ + 4d), which is 9*(-3) = -27. That seems correct.Alternatively, I can use the other sum formula: Sₙ = n/2*(2a₁ + (n-1)d). For n=9, that would be:S₉ = 9/2*(2a₁ + 8d) S₉ = (9/2)*(2(a₁ + 4d)) S₉ = 9*(a₁ + 4d) Again, substituting a₁ + 4d = -3, S₉ = 9*(-3) = -27.Yep, same result. So, I'm confident that the answer is -27, which is option A.**Final Answer**boxed{A}

💡Alright, so I've got this problem with four conclusions, and I need to figure out how many of them are correct. Let me take them one by one.**Conclusion 1:** The contrapositive of the proposition "If ( x = y ), then ( sin x = sin y )" is true.Okay, first, what's a contrapositive? If I remember correctly, the contrapositive of "If P, then Q" is "If not Q, then not P." So, applying that here, the contrapositive would be "If ( sin x neq sin y ), then ( x neq y )." Is this statement true? Well, sine is a periodic function, right? So, ( sin x = sin y ) doesn't necessarily mean ( x = y ). For example, ( sin(0) = sin(pi) ), but ( 0 neq pi ). So, if ( sin x neq sin y ), does that mean ( x neq y )? Hmm, actually, yes. Because if ( x = y ), then ( sin x ) must equal ( sin y ). So, if ( sin x ) is not equal to ( sin y ), then ( x ) can't be equal to ( y ). So, the contrapositive is true. So, conclusion 1 is correct.**Conclusion 2:** The number 2015 expressed in octal is ( 1037_8 ).Alright, I need to convert 2015 from decimal to octal. Let me recall how to do that. You divide the number by 8 and keep track of the remainders.2015 divided by 8 is 251 with a remainder of 7.251 divided by 8 is 31 with a remainder of 3.31 divided by 8 is 3 with a remainder of 7.3 divided by 8 is 0 with a remainder of 3.So, writing the remainders from last to first, it's 3 7 3 7, which is ( 3737_8 ). But the conclusion says ( 1037_8 ). That doesn't match. So, conclusion 2 is incorrect.**Conclusion 3:** The negation of the proposition "If ( x^2 = 1 ), then ( x = 1 )" is: "If ( x^2 = 1 ), then ( x neq 1 )."Hmm, negation of an implication. The original statement is "If P, then Q." The negation of that is "P and not Q." So, the negation should be " ( x^2 = 1 ) and ( x neq 1 )." But the conclusion says "If ( x^2 = 1 ), then ( x neq 1 )." That's not the correct negation. The correct negation is not another implication but a conjunction. So, conclusion 3 is incorrect.**Conclusion 4:** A necessary but not sufficient condition for "Plane ( alpha ) parallel to plane ( beta )" is "There exist three non-collinear points in plane ( alpha ) that are equidistant from plane ( beta )."Okay, so a necessary condition means that if ( alpha ) is parallel to ( beta ), then those three points must exist. Is that true? If two planes are parallel, then every point in one plane is equidistant from the other plane. So, certainly, there exist three non-collinear points in ( alpha ) that are equidistant from ( beta ). So, it's a necessary condition.Is it sufficient? That is, if there exist three non-collinear points in ( alpha ) equidistant from ( beta ), does that mean ( alpha ) is parallel to ( beta )? Well, three non-collinear points define a plane. If all three are equidistant from ( beta ), then the plane ( alpha ) must be parallel to ( beta ). Wait, is that true? Because if three non-collinear points are equidistant from another plane, doesn't that force the entire plane to be parallel?Wait, actually, no. If the three points are equidistant, but not necessarily in the same direction. Wait, no, in 3D space, if three non-collinear points are all at the same distance from another plane, then the plane containing these three points must be parallel to the other plane. Because otherwise, the distances would vary. So, actually, this condition is both necessary and sufficient. Therefore, the conclusion says it's necessary but not sufficient, which is incorrect. So, conclusion 4 is incorrect.So, out of the four conclusions, only conclusion 1 is correct. Therefore, the answer is 1.

💡Okay, so I'm trying to understand this Wythoff game that was invented in the year 31907 by mathematician Wythoff. It's a two-player game with two piles of objects, and players take turns removing objects according to specific rules. The goal is to take the last object and win. Let me try to break this down step by step.First, the rules of the game are:1. A player can take any number of objects from one pile, but not none. So, you have to take at least one object from one pile.2. A player can take the same number of objects from both piles, but again, at least one object from each.The player who takes the last object wins. Okay, so it's similar to some other impartial games I've heard about, like Nim or Kayles, but with a specific rule about taking from both piles.I remember that in many such games, there are winning and losing positions. A winning position is one where the current player can force a win no matter what the opponent does, and a losing position is one where no matter what the current player does, the opponent can force a win.So, maybe in this Wythoff game, there are specific pairs of pile sizes that are losing positions. If a player can move to one of these losing positions, they can force the opponent into a losing spot.I think the key here is to find these losing positions. Maybe there's a pattern or a formula that defines them. I recall something about the golden ratio being involved in Wythoff's game. The golden ratio is approximately 1.618, and it's often denoted by the Greek letter phi (φ).Let me try to recall the exact sequences involved. I think the losing positions are given by pairs (a_n, b_n), where a_n is the floor of n times phi, and b_n is the floor of n times phi squared. Since phi squared is equal to phi plus one, approximately 2.618.So, for each n, starting from 0, we have:- a_n = floor(n * phi)- b_n = floor(n * phi^2)And these pairs (a_n, b_n) are the losing positions. If a player is faced with one of these pairs, they are in a losing position if both players play optimally.Let me test this with some small values to see if it makes sense.For n = 0:- a_0 = floor(0 * phi) = 0- b_0 = floor(0 * phi^2) = 0So, the pair (0, 0) is a losing position. That makes sense because if there are no objects left, the game is over.For n = 1:- a_1 = floor(1 * phi) = floor(1.618) = 1- b_1 = floor(1 * phi^2) = floor(2.618) = 2So, the pair (1, 2) is a losing position.For n = 2:- a_2 = floor(2 * phi) = floor(3.236) = 3- b_2 = floor(2 * phi^2) = floor(5.236) = 5So, the pair (3, 5) is a losing position.For n = 3:- a_3 = floor(3 * phi) = floor(4.854) = 4- b_3 = floor(3 * phi^2) = floor(7.854) = 7So, the pair (4, 7) is a losing position.Wait a minute, I see a pattern here. The difference between b_n and a_n seems to be n. Let's check:For n = 1: 2 - 1 = 1For n = 2: 5 - 3 = 2For n = 3: 7 - 4 = 3Yes, that's consistent. So, b_n = a_n + n.This seems to be a key property. So, for each n, the losing position is (a_n, a_n + n), where a_n is the floor of n times phi.Now, let's see if this holds for higher n.For n = 4:- a_4 = floor(4 * phi) = floor(6.472) = 6- b_4 = floor(4 * phi^2) = floor(10.472) = 10So, the pair (6, 10) is a losing position.Difference: 10 - 6 = 4, which is equal to n. Good.For n = 5:- a_5 = floor(5 * phi) = floor(8.090) = 8- b_5 = floor(5 * phi^2) = floor(13.090) = 13So, the pair (8, 13) is a losing position.Difference: 13 - 8 = 5, which is equal to n. Perfect.Okay, so this seems to hold. Now, how does this help in determining the winning strategy?If a player can move the game to one of these losing positions, they can force the opponent into a losing spot. So, the strategy is to always move to a losing position.But how does one do that? Let's say the current position is (k, l). The player needs to determine if (k, l) is a losing position or not. If it's not, they can make a move to convert it into a losing position.To do this, the player can calculate the values of a_n and b_n and see if the current position matches any of these pairs. If it doesn't, they can adjust the piles to reach one of these pairs.For example, suppose the current position is (5, 8). Is this a losing position? Let's check the pairs we've calculated:n=1: (1,2)n=2: (3,5)n=3: (4,7)n=4: (6,10)n=5: (8,13)So, (5,8) is not one of these pairs. Therefore, it's a winning position. The player can make a move to convert it into a losing position.How? They can either remove objects from one pile or both piles. Let's see:If they remove 2 objects from the first pile: (5-2, 8) = (3,8). Is (3,8) a losing position? Looking at our list, n=2 is (3,5), n=3 is (4,7), n=4 is (6,10), so no.If they remove 3 objects from the first pile: (5-3,8) = (2,8). Not a losing position.If they remove 4 objects from the first pile: (5-4,8) = (1,8). Not a losing position.If they remove 5 objects from the first pile: (0,8). Not a losing position.Similarly, removing from the second pile:Remove 1 object: (5,8-1)=(5,7). Not a losing position.Remove 2 objects: (5,6). Not a losing position.Remove 3 objects: (5,5). Not a losing position.Remove 4 objects: (5,4). Not a losing position.Remove 5 objects: (5,3). Not a losing position.Remove 6 objects: (5,2). Not a losing position.Remove 7 objects: (5,1). Not a losing position.Remove 8 objects: (5,0). Not a losing position.Alternatively, removing the same number from both piles:Remove 1 object: (4,7). Ah, (4,7) is a losing position for n=3. So, by removing 1 object from both piles, the player can move from (5,8) to (4,7), which is a losing position for the opponent.Therefore, the strategy is to identify if the current position is a losing position. If not, make a move to reach the nearest losing position.This seems to be a solid strategy. But how does one calculate these losing positions for larger numbers? It would be impractical to list them all out manually.I think the key is to use the properties of the golden ratio and the sequences a_n and b_n. Since a_n = floor(n * phi) and b_n = a_n + n, we can calculate these for any n.But wait, phi is an irrational number, so the sequences a_n and b_n are also irrational sequences. However, they have interesting properties, such as being complementary and covering all natural numbers without overlap.This means that every natural number appears exactly once in either a_n or b_n, but not both. So, for any given number, it will be either in the a sequence or the b sequence, but not both.This is useful because it ensures that for any position (k, l), one of the numbers will be in the a sequence and the other in the b sequence, or vice versa, depending on the value of n.But how does this help in determining the move? Let's think.If the current position is (k, l), and assuming k < l without loss of generality, we can check if k is in the a sequence. If it is, then the corresponding b_n would be k + n. If l equals b_n, then it's a losing position. If not, we can adjust the piles to reach the losing position.Alternatively, if k is not in the a sequence, we can find the nearest a_n less than k and see if l equals b_n. If not, we can make a move to reach (a_n, b_n).This seems a bit abstract. Maybe there's a more straightforward way to calculate the move.I recall that in Wythoff's game, the losing positions are also known as P-positions, and the winning positions are N-positions. The P-positions are those where the previous player can force a win, and N-positions are those where the next player can force a win.Given that, the strategy is to move to a P-position if you're in an N-position, and if you're in a P-position, any move you make will take you to an N-position, allowing the opponent to win.So, the key is to identify whether the current position is a P-position or an N-position.To do this, we can use the formula involving the golden ratio. For a given position (k, l), we can calculate the difference between the two piles, say d = |k - l|. Then, we can calculate the floor of d * phi and see if it equals the smaller pile.Wait, let me think again. If we have two piles, k and l, with k < l, then the difference is d = l - k. If k equals floor(d * phi), then the position is a P-position.Let me test this with our earlier examples.Take n=1: (1,2). d=1. floor(1 * phi)=1. So, k=1 equals floor(d * phi). Therefore, it's a P-position.n=2: (3,5). d=2. floor(2 * phi)=3. So, k=3 equals floor(d * phi). P-position.n=3: (4,7). d=3. floor(3 * phi)=4. k=4 equals floor(d * phi). P-position.n=4: (6,10). d=4. floor(4 * phi)=6. k=6 equals floor(d * phi). P-position.n=5: (8,13). d=5. floor(5 * phi)=8. k=8 equals floor(d * phi). P-position.Yes, this seems to hold. So, the formula is:For a position (k, l), with k < l, calculate d = l - k. If k = floor(d * phi), then it's a P-position; otherwise, it's an N-position.This is a useful criterion. So, to determine if a position is losing, we can use this formula.Now, how does one use this to make a move? Suppose we're in an N-position, and we want to move to a P-position.Given (k, l), with k < l, and k ≠ floor(d * phi), we need to find a move that will take us to a P-position.One way is to calculate the desired k' such that k' = floor(d' * phi), where d' is the new difference after the move.But this might be a bit involved. Alternatively, we can use the fact that from any N-position, there is exactly one move to a P-position.Wait, is that true? Let me think.In Wythoff's game, from any N-position, there is exactly one move to a P-position. This is because the P-positions are arranged in such a way that they are spaced out by the golden ratio, ensuring that only one move can lead to a P-position.So, if we can find that one move, we can force the opponent into a losing position.But how?Let me try to formalize this.Given a position (k, l), with k < l, and it's an N-position, we need to find a move that will take us to a P-position.Let d = l - k.If k ≠ floor(d * phi), then we need to adjust either k or l or both to reach a P-position.One approach is to calculate the target P-position that is reachable from (k, l).Since from any N-position, there is exactly one P-position reachable, we can find it by:1. If k is not in the a sequence, find the next a_n less than k, and adjust l accordingly.2. If l is not in the b sequence, find the corresponding a_n and adjust k accordingly.Alternatively, we can use the formula involving the golden ratio to calculate the move.Let me try to outline the steps:1. Given (k, l), with k < l.2. Calculate d = l - k.3. Calculate m = floor(d * phi).4. If k = m, then it's a P-position.5. If k < m, then the move is to take (m - k) objects from the larger pile, resulting in (k, l - (m - k)).6. If k > m, then the move is to take (k - m) objects from the smaller pile, resulting in (k - (k - m), l) = (m, l).Wait, let me test this with an example.Take the position (5,8). We know it's an N-position because (4,7) is a P-position, and we can move to it by removing 1 from both piles.Let's apply the steps:1. k = 5, l = 8.2. d = 8 - 5 = 3.3. m = floor(3 * phi) = floor(4.854) = 4.4. Since k = 5 > m = 4, we need to take (k - m) = 1 object from the smaller pile.5. So, the new position is (5 - 1, 8) = (4,8). Wait, but (4,8) is not a P-position.Wait, that's not correct. Because (4,7) is a P-position, not (4,8).Hmm, maybe my approach is flawed.Let me try another way.Given (5,8), we know that (4,7) is a P-position. So, the move is to remove 1 from both piles.But according to the steps I outlined, I ended up at (4,8), which is not a P-position. So, perhaps the method needs adjustment.Maybe the correct approach is to calculate the difference d = l - k, find m = floor(d * phi), and then adjust the larger pile to make the smaller pile equal to m.Wait, let's try that.Given (5,8):1. d = 8 - 5 = 3.2. m = floor(3 * phi) = 4.3. Since k = 5 > m = 4, we need to adjust the larger pile to make the smaller pile equal to m.4. So, we need to make the smaller pile 4, which means removing 1 from the smaller pile: (5 - 1, 8) = (4,8).But (4,8) is not a P-position. The P-position would be (4,7).Hmm, so this approach doesn't directly lead to a P-position. Maybe I need to consider both piles.Alternatively, perhaps the correct move is to adjust the larger pile to make the difference equal to m.Wait, let's think differently.Given (k, l), with k < l, and it's an N-position, we can calculate the desired P-position as (m, m + d'), where d' is the new difference.But I'm getting confused here.Let me look for another approach.I remember that in Wythoff's game, the losing positions are pairs where the two piles are (a_n, b_n) = (floor(n * phi), floor(n * phi^2)).Given that, for any position (k, l), if k is in the a sequence and l is the corresponding b_n, then it's a P-position.So, to move to a P-position, we need to find n such that either k = a_n and l = b_n, or l = a_n and k = b_n.But since we're assuming k < l, we only need to consider k = a_n and l = b_n.So, given (k, l), we can check if k is in the a sequence. If it is, then l should be b_n = a_n + n.If l is not equal to b_n, then we can adjust the piles to reach (a_n, b_n).But how?Let me try with (5,8).We need to find n such that a_n = 5.From our earlier list:n=1: a_1=1n=2: a_2=3n=3: a_3=4n=4: a_4=6n=5: a_5=8Wait, a_5=8, which is greater than 5. So, there is no n where a_n=5. Therefore, 5 is not in the a sequence.Similarly, 8 is in the a sequence for n=5.So, if k=5 is not in the a sequence, then we need to find the n such that a_n < k < a_{n+1}.Looking at our list:a_3=4, a_4=6. So, 5 is between a_3 and a_4.Therefore, n=3.Then, b_n = a_n + n = 4 + 3 = 7.So, the P-position corresponding to n=3 is (4,7).Therefore, from (5,8), we can move to (4,7) by removing 1 from both piles.That works.So, the general strategy is:1. Given (k, l), with k < l.2. Find n such that a_n <= k < a_{n+1}.3. The corresponding P-position is (a_n, b_n) = (a_n, a_n + n).4. If l = b_n, then it's already a P-position.5. If l > b_n, then remove (l - b_n) from the larger pile.6. If l < b_n, then remove (k - a_n) from the smaller pile.Wait, let's test this.Take (5,8):1. k=5, l=8.2. Find n such that a_n <=5 <a_{n+1}.From our list:a_3=4, a_4=6. So, n=3.3. P-position is (4,7).4. l=8 >7.5. So, remove (8 -7)=1 from the larger pile.6. New position: (5,7). Wait, but (5,7) is not a P-position.Wait, that's not correct. Because (4,7) is the P-position, not (5,7).Hmm, maybe I need to adjust both piles.Alternatively, since we need to reach (4,7), we can remove 1 from both piles: (5-1,8-1)=(4,7).Yes, that works.So, perhaps the correct approach is:If l > b_n, remove (l - b_n) from the larger pile and also remove the same amount from the smaller pile to reach (a_n, b_n).Wait, but that might not always be possible.Alternatively, perhaps the correct move is to remove (k - a_n) from the smaller pile and (l - b_n) from the larger pile, but ensuring that the amounts removed are the same if we're taking from both piles.Wait, this is getting complicated.Let me try to formalize the strategy.Given (k, l), with k < l.1. Find n such that a_n <= k <a_{n+1}.2. The target P-position is (a_n, b_n) = (a_n, a_n + n).3. If l = b_n, then it's already a P-position.4. If l > b_n, then remove (l - b_n) from the larger pile and also remove (k - a_n) from the smaller pile, but ensuring that the amounts removed are the same if we're taking from both piles.Wait, but in our example, (5,8):n=3, a_n=4, b_n=7.l=8 >7.So, we need to remove 1 from the larger pile to make it 7.But we also need to adjust the smaller pile to make it 4.Since the smaller pile is 5, we need to remove 1 from it as well.Therefore, the move is to remove 1 from both piles, resulting in (4,7).Yes, that works.So, the general strategy is:If l > b_n, remove (l - b_n) from the larger pile and (k - a_n) from the smaller pile, but since we can only remove the same amount from both piles, we need to ensure that (l - b_n) = (k - a_n).In our example, (5,8):l - b_n =8 -7=1k - a_n=5 -4=1So, we can remove 1 from both piles.If l - b_n ≠ k - a_n, then we cannot make such a move, and we need to adjust differently.Wait, but in Wythoff's game, from any N-position, there is exactly one move to a P-position, which can be either removing from one pile or both piles.So, perhaps the correct approach is:Given (k, l), with k < l.1. Find n such that a_n <= k <a_{n+1}.2. The target P-position is (a_n, b_n).3. If l = b_n, it's already a P-position.4. If l > b_n, then remove (l - b_n) from the larger pile.5. If l < b_n, then remove (k - a_n) from the smaller pile.But wait, in our example, l > b_n, so we remove (l - b_n)=1 from the larger pile, resulting in (5,7). But (5,7) is not a P-position.So, that approach doesn't work.Alternatively, perhaps we need to remove (k - a_n) from the smaller pile and (l - b_n) from the larger pile, but only if (k - a_n) = (l - b_n).In our case, (k - a_n)=1 and (l - b_n)=1, so we can remove 1 from both piles.If they are not equal, then we cannot remove the same amount from both piles, so we need to remove from one pile.Wait, but in that case, how do we ensure that we reach a P-position?I think the key is that from any N-position, there is exactly one move to a P-position, which can be either removing from one pile or both piles.So, perhaps the strategy is:Given (k, l), with k < l.1. Find n such that a_n <= k <a_{n+1}.2. The target P-position is (a_n, b_n).3. If l = b_n, it's already a P-position.4. If l > b_n, then remove (l - b_n) from the larger pile.5. If l < b_n, then remove (k - a_n) from the smaller pile.But in our example, l > b_n, so we remove (l - b_n)=1 from the larger pile, resulting in (5,7). But (5,7) is not a P-position.Wait, that's a problem.Alternatively, perhaps the correct move is to remove (k - a_n) from the smaller pile, which would make it a_n, and then adjust the larger pile accordingly.But in our case, (k - a_n)=1, so removing 1 from the smaller pile gives us (4,8). But (4,8) is not a P-position.Hmm.Wait, maybe I need to consider that the P-position is (a_n, b_n), and to reach it, we might need to remove objects from both piles.In our example, (5,8) to (4,7), we remove 1 from both piles.So, the move is to remove 1 from both piles.Therefore, the strategy is:Given (k, l), with k < l.1. Find n such that a_n <= k <a_{n+1}.2. The target P-position is (a_n, b_n).3. Calculate the difference between k and a_n: d1 = k - a_n.4. Calculate the difference between l and b_n: d2 = l - b_n.5. If d1 == d2, then remove d1 from both piles.6. If d1 != d2, then remove the larger difference from the corresponding pile.Wait, let's test this.For (5,8):n=3, a_n=4, b_n=7.d1=5-4=1d2=8-7=1Since d1=d2=1, remove 1 from both piles to get (4,7).Perfect.Another example: (6,10).n=4, a_n=6, b_n=10.It's already a P-position.Another example: (7,10).n=4, a_n=6, b_n=10.k=7 > a_n=6.d1=1, d2=0.Since d1 != d2, we need to remove the larger difference, which is d1=1, from the smaller pile.So, remove 1 from the smaller pile: (6,10), which is a P-position.Yes, that works.Another example: (3,5).n=2, a_n=3, b_n=5.It's a P-position.Another example: (4,7).n=3, a_n=4, b_n=7.It's a P-position.Another example: (2,5).n=1, a_n=1, b_n=2.k=2 > a_n=1.d1=1, d2=5-2=3.Since d1 != d2, remove the larger difference, which is d2=3, from the larger pile.So, remove 3 from the larger pile: (2,2).But (2,2) is not a P-position. Wait, that's a problem.Wait, (2,2) is actually a losing position because if you have two piles of equal size, the player can take all from one pile and win. Wait, no, in Wythoff's game, taking all from one pile is allowed, but the player who takes the last object wins. So, if you have (2,2), the player can take 2 from one pile and win.Wait, but (2,2) is actually a winning position because the player can take both objects from one pile and win.So, moving to (2,2) is not a P-position.Wait, so my strategy is flawed.Let me think again.Given (2,5):n=1, a_n=1, b_n=2.k=2 > a_n=1.d1=1, d2=5-2=3.Since d1 != d2, we need to remove the larger difference, which is d2=3, from the larger pile.So, remove 3 from the larger pile: (2,2).But (2,2) is a winning position, not a P-position.So, this approach doesn't work.Wait, perhaps the correct move is to remove from the smaller pile.If we remove d1=1 from the smaller pile: (1,5).Is (1,5) a P-position?n=1: (1,2)n=2: (3,5)So, (1,5) is not a P-position.Alternatively, remove d1=1 from the smaller pile and d2=3 from the larger pile, but since d1 != d2, we can't remove the same amount from both piles.So, perhaps the correct move is to remove d1=1 from the smaller pile, resulting in (1,5), which is an N-position, but not a P-position.Wait, that's not helpful.Alternatively, perhaps the correct move is to remove d2=3 from the larger pile, resulting in (2,2), which is a winning position.But that's not a P-position.Hmm, this is confusing.Wait, maybe I need to consider that from (2,5), the correct move is to remove 2 from the larger pile, resulting in (2,3).Is (2,3) a P-position?n=1: (1,2)n=2: (3,5)No, (2,3) is not a P-position.Alternatively, remove 1 from both piles: (1,4).Is (1,4) a P-position?n=1: (1,2)n=2: (3,5)No, (1,4) is not a P-position.Alternatively, remove 3 from the larger pile: (2,2). Not a P-position.Alternatively, remove 4 from the larger pile: (2,1). Not a P-position.Wait, maybe I'm missing something.Let me check the P-positions again.n=1: (1,2)n=2: (3,5)n=3: (4,7)n=4: (6,10)n=5: (8,13)So, for (2,5), the P-positions around it are (1,2), (3,5), (4,7).So, from (2,5), the possible moves are:- Remove from the smaller pile: (1,5), (0,5)- Remove from the larger pile: (2,4), (2,3), (2,2), (2,1), (2,0)- Remove from both piles: (1,4), (0,4), (1,3), (0,3), etc.Looking for P-positions among these:(1,2) is a P-position, but (1,5) is not.(3,5) is a P-position, but (2,5) is not.(4,7) is a P-position, but (2,5) is not.So, from (2,5), the only way to reach a P-position is to move to (3,5), but that would require adding objects, which is not allowed.Wait, that can't be.Alternatively, perhaps the correct move is to remove 2 from the larger pile to make it (2,3), but (2,3) is not a P-position.Alternatively, remove 3 from the larger pile to make it (2,2), which is a winning position.Wait, I'm stuck.Maybe the issue is that (2,5) is an N-position, and the only move to a P-position is to (3,5), but that would require increasing the smaller pile, which is not allowed.Wait, that can't be.Wait, perhaps I made a mistake in identifying the P-positions.Let me recalculate the P-positions up to n=5.n=0: (0,0)n=1: (1,2)n=2: (3,5)n=3: (4,7)n=4: (6,10)n=5: (8,13)Yes, that's correct.So, from (2,5), the possible moves are:- Remove from the smaller pile: (1,5), (0,5)- Remove from the larger pile: (2,4), (2,3), (2,2), (2,1), (2,0)- Remove from both piles: (1,4), (0,4), (1,3), (0,3), etc.Looking for P-positions among these:(1,2) is a P-position, but (1,5) is not.(3,5) is a P-position, but (2,5) is not.(4,7) is a P-position, but (2,5) is not.So, none of the moves from (2,5) lead directly to a P-position. That can't be right because in Wythoff's game, from any N-position, there should be exactly one move to a P-position.Wait, maybe I'm missing something.Wait, perhaps the P-positions are not just the ones I listed, but also their mirrors. So, for example, (2,1) is also a P-position because it's the mirror of (1,2).But in our earlier list, we only listed (k, l) with k < l. So, perhaps (2,1) is also a P-position.But in that case, from (2,5), we can move to (2,1) by removing 4 from the larger pile, but that's not a valid move because you can't remove 4 from the larger pile without also removing from the smaller pile if you're taking from both.Wait, no, you can remove 4 from the larger pile alone, resulting in (2,1).But (2,1) is a P-position because it's the mirror of (1,2).So, perhaps the correct move is to remove 4 from the larger pile, resulting in (2,1), which is a P-position.Yes, that makes sense.So, the strategy is:Given (k, l), with k < l.1. Find n such that a_n <= k <a_{n+1}.2. The target P-position is (a_n, b_n).3. If l = b_n, it's already a P-position.4. If l > b_n, then remove (l - b_n) from the larger pile.5. If l < b_n, then remove (k - a_n) from the smaller pile.But in the case of (2,5):n=1, a_n=1, b_n=2.k=2 > a_n=1.d1=1, d2=5-2=3.Since d1 != d2, we need to remove the larger difference, which is d2=3, from the larger pile.So, remove 3 from the larger pile: (2,2). But (2,2) is not a P-position.Wait, but earlier I thought that (2,1) is a P-position.So, perhaps the correct move is to remove 4 from the larger pile to reach (2,1), which is a P-position.But how does that fit into the strategy?Maybe the strategy needs to consider both the difference and the possibility of mirroring.Alternatively, perhaps the correct approach is to calculate the difference d = l - k, then calculate m = floor(d * phi), and if k != m, then remove (k - m) from the smaller pile or (l - (m + d)) from the larger pile.Wait, let's try that.Given (2,5):d =5 -2=3m = floor(3 * phi)=4Since k=2 < m=4, we need to remove (m - k)=2 from the larger pile.So, remove 2 from the larger pile: (2,3).But (2,3) is not a P-position.Wait, that's not helpful.Alternatively, if k < m, remove (m - k) from the larger pile.But in this case, removing 2 from the larger pile gives (2,3), which is not a P-position.Alternatively, perhaps the correct move is to remove (m - k) from the smaller pile.But m=4, k=2, so remove 2 from the smaller pile: (0,5). That's not a P-position.Wait, this is getting too confusing.Maybe I need to refer back to the initial strategy.Given (k, l), with k < l.1. Find n such that a_n <= k <a_{n+1}.2. The target P-position is (a_n, b_n).3. If l = b_n, it's already a P-position.4. If l > b_n, remove (l - b_n) from the larger pile.5. If l < b_n, remove (k - a_n) from the smaller pile.But in our case, (2,5):n=1, a_n=1, b_n=2.l=5 > b_n=2.So, remove (5 -2)=3 from the larger pile: (2,2). Not a P-position.But (2,1) is a P-position, which is the mirror of (1,2).So, perhaps the correct move is to remove 4 from the larger pile to reach (2,1).But how does that fit into the strategy?Maybe the strategy needs to consider that sometimes the P-position is the mirror of the calculated one.Alternatively, perhaps the correct approach is to calculate both (a_n, b_n) and (b_n, a_n) and see which one is reachable.But that complicates things.Wait, perhaps the issue is that in my earlier list, I only considered (a_n, b_n) with a_n < b_n, but in reality, the P-positions include both (a_n, b_n) and (b_n, a_n).So, for n=1, the P-positions are (1,2) and (2,1).Similarly, for n=2, (3,5) and (5,3).So, in that case, from (2,5), we can move to (2,1) by removing 4 from the larger pile.Yes, that makes sense.So, the strategy should consider both (a_n, b_n) and (b_n, a_n) as P-positions.Therefore, given (k, l), with k < l.1. Find n such that a_n <= k <a_{n+1}.2. The target P-positions are (a_n, b_n) and (b_n, a_n).3. If l = b_n, then it's already a P-position.4. If l > b_n, then remove (l - b_n) from the larger pile to reach (k, b_n).5. If l < b_n, then remove (k - a_n) from the smaller pile to reach (a_n, l).But in our case, (2,5):n=1, a_n=1, b_n=2.l=5 > b_n=2.So, remove (5 -2)=3 from the larger pile: (2,2). Not a P-position.But (2,1) is a P-position, which is the mirror.So, perhaps the correct move is to remove (l - a_n)=5 -1=4 from the larger pile to reach (2,1).Yes, that works.So, the general strategy is:Given (k, l), with k < l.1. Find n such that a_n <= k <a_{n+1}.2. The target P-positions are (a_n, b_n) and (b_n, a_n).3. If l = b_n, it's already a P-position.4. If l > b_n, then remove (l - a_n) from the larger pile to reach (k, a_n).5. If l < b_n, then remove (k - a_n) from the smaller pile to reach (a_n, l).Wait, let's test this.For (2,5):n=1, a_n=1, b_n=2.l=5 > b_n=2.So, remove (5 -1)=4 from the larger pile: (2,1), which is a P-position.Yes, that works.Another example: (5,8).n=3, a_n=4, b_n=7.l=8 > b_n=7.So, remove (8 -4)=4 from the larger pile: (5,4). Wait, but (5,4) is not a P-position.Wait, that's not correct.Wait, (5,8) to (4,7) is the correct move.But according to this strategy, we remove (8 -4)=4 from the larger pile, resulting in (5,4), which is not a P-position.Hmm, that's a problem.Wait, perhaps the correct move is to remove (l - b_n)=1 from the larger pile, resulting in (5,7). But (5,7) is not a P-position.Alternatively, remove (k - a_n)=1 from the smaller pile, resulting in (4,8). Not a P-position.Wait, but we know that (4,7) is a P-position, so the correct move is to remove 1 from both piles.So, perhaps the strategy needs to consider removing from both piles when possible.Therefore, the correct approach is:Given (k, l), with k < l.1. Find n such that a_n <= k <a_{n+1}.2. The target P-positions are (a_n, b_n) and (b_n, a_n).3. If l = b_n, it's already a P-position.4. If l > b_n, then remove (l - b_n) from the larger pile.5. If l < b_n, then remove (k - a_n) from the smaller pile.6. If neither of these leads to a P-position, then remove (k - a_n) from both piles.Wait, in the case of (5,8):n=3, a_n=4, b_n=7.l=8 > b_n=7.So, remove (8 -7)=1 from the larger pile: (5,7). Not a P-position.Then, check if removing (k - a_n)=1 from both piles leads to a P-position: (4,7). Yes, it does.So, the strategy should be:Given (k, l), with k < l.1. Find n such that a_n <= k <a_{n+1}.2. The target P-positions are (a_n, b_n) and (b_n, a_n).3. If l = b_n, it's already a P-position.4. If l > b_n, then remove (l - b_n) from the larger pile.5. If l < b_n, then remove (k - a_n) from the smaller pile.6. If neither of these leads to a P-position, then remove (k - a_n) from both piles.But this seems a bit convoluted.Alternatively, perhaps the correct approach is to calculate the difference d = l - k, then calculate m = floor(d * phi), and if k != m, then remove (k - m) from the smaller pile or (l - (m + d)) from the larger pile.Wait, let's try that.Given (2,5):d=3, m=floor(3 * phi)=4.k=2 < m=4.So, remove (4 -2)=2 from the larger pile: (2,3). Not a P-position.Alternatively, remove (4 -2)=2 from the smaller pile: (0,5). Not a P-position.Wait, that's not helpful.Alternatively, perhaps the correct move is to remove (m - k)=2 from the larger pile: (2,3). Not a P-position.Hmm.Wait, perhaps the correct formula is to remove (m - k) from the larger pile if k < m, or remove (k - m) from the smaller pile if k > m.But in our case, k=2 < m=4, so remove (4 -2)=2 from the larger pile: (2,3). Not a P-position.But we know that (2,1) is a P-position, so perhaps the correct move is to remove 4 from the larger pile.But how does that fit into the formula?I think I'm overcomplicating this.Let me try to summarize the strategy based on what I've learned:1. Identify if the current position is a P-position or an N-position.2. If it's a P-position, any move will lead to an N-position, so the opponent can win.3. If it's an N-position, there is exactly one move to a P-position.4. To find that move, calculate the difference d = l - k.5. Calculate m = floor(d * phi).6. If k = m, it's a P-position.7. If k < m, remove (m - k) from the larger pile.8. If k > m, remove (k - m) from the smaller pile.But in our example, (2,5):d=3, m=4.k=2 < m=4.So, remove (4 -2)=2 from the larger pile: (2,3). Not a P-position.But we know that (2,1) is a P-position, so perhaps the correct move is to remove 4 from the larger pile.Wait, maybe the formula needs to be adjusted.Alternatively, perhaps the correct move is to remove (m - k) from the larger pile if k < m, but in this case, that leads to (2,3), which is not a P-position.Alternatively, perhaps the correct move is to remove (m - k) from the smaller pile.But that would be removing 2 from the smaller pile: (0,5). Not a P-position.Hmm.Wait, maybe the formula is not directly applicable when k is not in the a sequence.Perhaps the correct approach is to find the nearest P-position by either removing from one pile or both piles.Given that, from (2,5), the possible P-positions are (1,2), (3,5), (4,7), etc.So, the closest P-position is (3,5), but that requires increasing the smaller pile, which is not allowed.Alternatively, the P-position (2,1) is reachable by removing 4 from the larger pile.So, the move is to remove 4 from the larger pile, resulting in (2,1).Yes, that works.Therefore, the strategy is:Given (k, l), with k < l.1. Find n such that a_n <= k <a_{n+1}.2. The target P-positions are (a_n, b_n) and (b_n, a_n).3. If l = b_n, it's already a P-position.4. If l > b_n, then remove (l - a_n) from the larger pile to reach (k, a_n).5. If l < b_n, then remove (k - a_n) from the smaller pile to reach (a_n, l).Wait, let's test this.For (2,5):n=1, a_n=1, b_n=2.l=5 > b_n=2.So, remove (5 -1)=4 from the larger pile: (2,1), which is a P-position.Yes, that works.For (5,8):n=3, a_n=4, b_n=7.l=8 > b_n=7.So, remove (8 -4)=4 from the larger pile: (5,4). Wait, but (5,4) is not a P-position.Wait, that's not correct.Wait, (5,4) is the same as (4,5), which is not a P-position.But we know that (4,7) is a P-position, so the correct move is to remove 1 from both piles.So, perhaps the strategy needs to consider removing from both piles when possible.Therefore, the correct approach is:Given (k, l), with k < l.1. Find n such that a_n <= k <a_{n+1}.2. The target P-positions are (a_n, b_n) and (b_n, a_n).3. If l = b_n, it's already a P-position.4. If l > b_n, then remove (l - b_n) from the larger pile.5. If l < b_n, then remove (k - a_n) from the smaller pile.6. If neither of these leads to a P-position, then remove (k - a_n) from both piles.But in the case of (5,8):n=3, a_n=4, b_n=7.l=8 > b_n=7.So, remove (8 -7)=1 from the larger pile: (5,7). Not a P-position.Then, check if removing (k - a_n)=1 from both piles leads to a P-position: (4,7). Yes, it does.Therefore, the strategy should be:Given (k, l), with k < l.1. Find n such that a_n <= k <a_{n+1}.2. The target P-positions are (a_n, b_n) and (b_n, a_n).3. If l = b_n, it's already a P-position.4. If l > b_n, then remove (l - b_n) from the larger pile.5. If l < b_n, then remove (k - a_n) from the smaller pile.6. If neither of these leads to a P-position, then remove (k - a_n) from both piles.This seems to cover all cases.So, in summary, the strategy to win Wythoff's game is:1. Identify if the current position (k, l) is a P-position or an N-position.2. If it's a P-position, any move will lead to an N-position, so the opponent can win.3. If it's an N-position, calculate the move to reach the nearest P-position.4. To calculate the move: a. Find n such that a_n <= k <a_{n+1}. b. The target P-positions are (a_n, b_n) and (b_n, a_n). c. If l = b_n, it's already a P-position. d. If l > b_n, remove (l - b_n) from the larger pile. e. If l < b_n, remove (k - a_n) from the smaller pile. f. If neither of these leads to a P-position, remove (k - a_n) from both piles.By following this strategy, a player can always force the opponent into a losing position, ensuring victory.

💡Okay, so I have this problem where points A(2,2) and C(8,4) are two vertices of an isosceles right-angled triangle ABC. The right angle is at point B, which lies on the x-axis. I need to find the x-coordinate of point B. Hmm, let me think about how to approach this.First, since ABC is an isosceles right-angled triangle with the right angle at B, that means the legs AB and BC are equal in length. So, AB = BC. Also, because B is on the x-axis, its coordinates must be (x, 0) for some x.I think I can use the distance formula to express AB and BC in terms of x and then set them equal to each other. The distance formula between two points (x1, y1) and (x2, y2) is sqrt[(x2 - x1)^2 + (y2 - y1)^2]. Let me write down the expressions for AB and BC.AB is the distance between A(2,2) and B(x,0):AB = sqrt[(x - 2)^2 + (0 - 2)^2] = sqrt[(x - 2)^2 + 4]BC is the distance between B(x,0) and C(8,4):BC = sqrt[(8 - x)^2 + (4 - 0)^2] = sqrt[(8 - x)^2 + 16]Since AB = BC, I can set these two expressions equal to each other:sqrt[(x - 2)^2 + 4] = sqrt[(8 - x)^2 + 16]To eliminate the square roots, I'll square both sides:(x - 2)^2 + 4 = (8 - x)^2 + 16Let me expand both sides:Left side: (x^2 - 4x + 4) + 4 = x^2 - 4x + 8Right side: (64 - 16x + x^2) + 16 = x^2 - 16x + 80Now, set them equal:x^2 - 4x + 8 = x^2 - 16x + 80Subtract x^2 from both sides:-4x + 8 = -16x + 80Now, let's solve for x. I'll add 16x to both sides:12x + 8 = 80Subtract 8 from both sides:12x = 72Divide both sides by 12:x = 6Wait, so the x-coordinate of B is 6? Let me check if this makes sense.If B is at (6,0), then AB is the distance from (2,2) to (6,0):AB = sqrt[(6 - 2)^2 + (0 - 2)^2] = sqrt[16 + 4] = sqrt[20] = 2*sqrt(5)BC is the distance from (6,0) to (8,4):BC = sqrt[(8 - 6)^2 + (4 - 0)^2] = sqrt[4 + 16] = sqrt[20] = 2*sqrt(5)Yes, AB equals BC, so it is an isosceles right-angled triangle. Also, since the legs are equal and the right angle is at B, this should satisfy all the conditions.I think I did everything correctly. I set up the distance formulas, squared both sides to eliminate the square roots, expanded, simplified, and solved for x. The result seems to check out when I plug it back in.So, the x-coordinate of B is 6.**Final Answer**boxed{6}

💡Okay, so I have this problem where I need to evaluate the quantity ( x_1^3 + x_2^3 + x_3^3 ) where ( x_1, x_2, x_3 ) are the roots of the polynomial ( P(x) = x^3 - 3x - 1 ). Hmm, let me think about how to approach this.First, I remember that for polynomials, especially cubic ones, Vieta's formulas relate the roots to the coefficients. Maybe I can use that here. Vieta's formulas say that for a cubic equation ( x^3 + ax^2 + bx + c = 0 ), the sum of the roots ( x_1 + x_2 + x_3 = -a ), the sum of the products ( x_1x_2 + x_1x_3 + x_2x_3 = b ), and the product of the roots ( x_1x_2x_3 = -c ).Looking at the given polynomial ( P(x) = x^3 - 3x - 1 ), I can rewrite it as ( x^3 + 0x^2 - 3x - 1 ). So, comparing to the general form, ( a = 0 ), ( b = -3 ), and ( c = -1 ).So, applying Vieta's formulas:- ( x_1 + x_2 + x_3 = -a = 0 )- ( x_1x_2 + x_1x_3 + x_2x_3 = b = -3 )- ( x_1x_2x_3 = -c = 1 )Alright, so I have these three pieces of information. Now, I need to find ( x_1^3 + x_2^3 + x_3^3 ). I recall there's a formula for the sum of cubes of roots. Let me try to remember it.I think it's something like ( x_1^3 + x_2^3 + x_3^3 = (x_1 + x_2 + x_3)^3 - 3(x_1 + x_2 + x_3)(x_1x_2 + x_1x_3 + x_2x_3) + 3x_1x_2x_3 ). Let me check if that makes sense.Wait, actually, I think that's the expansion of ( (x_1 + x_2 + x_3)^3 ). Let me write it out:( (x_1 + x_2 + x_3)^3 = x_1^3 + x_2^3 + x_3^3 + 3(x_1^2x_2 + x_1^2x_3 + x_2^2x_1 + x_2^2x_3 + x_3^2x_1 + x_3^2x_2) + 6x_1x_2x_3 ).Hmm, that's more complicated. Maybe there's a better way. I think there's an identity that relates ( x_1^3 + x_2^3 + x_3^3 ) directly to the symmetric sums.Oh, right! There's an identity called the sum of cubes, which can be expressed in terms of the elementary symmetric sums. The formula is:( x_1^3 + x_2^3 + x_3^3 = (x_1 + x_2 + x_3)^3 - 3(x_1 + x_2 + x_3)(x_1x_2 + x_1x_3 + x_2x_3) + 3x_1x_2x_3 ).Wait, that seems similar to what I wrote earlier. Let me plug in the values I have.Given that ( x_1 + x_2 + x_3 = 0 ), ( x_1x_2 + x_1x_3 + x_2x_3 = -3 ), and ( x_1x_2x_3 = 1 ), let's substitute these into the formula.First, ( (x_1 + x_2 + x_3)^3 = 0^3 = 0 ).Next, ( 3(x_1 + x_2 + x_3)(x_1x_2 + x_1x_3 + x_2x_3) = 3 times 0 times (-3) = 0 ).Finally, ( 3x_1x_2x_3 = 3 times 1 = 3 ).Putting it all together:( x_1^3 + x_2^3 + x_3^3 = 0 - 0 + 3 = 3 ).Wait, that seems straightforward. But let me double-check if there's another way to approach this, just to be sure.Another approach could be using the original polynomial equation. Since each root ( x_i ) satisfies ( x_i^3 - 3x_i - 1 = 0 ), we can express ( x_i^3 ) as ( 3x_i + 1 ).So, for each root:( x_1^3 = 3x_1 + 1 )( x_2^3 = 3x_2 + 1 )( x_3^3 = 3x_3 + 1 )Now, if I sum these up:( x_1^3 + x_2^3 + x_3^3 = 3x_1 + 1 + 3x_2 + 1 + 3x_3 + 1 )Combine like terms:( x_1^3 + x_2^3 + x_3^3 = 3(x_1 + x_2 + x_3) + 3 )From Vieta's formula, we know ( x_1 + x_2 + x_3 = 0 ), so:( x_1^3 + x_2^3 + x_3^3 = 3(0) + 3 = 3 )Okay, so both methods give me the same result, which is 3. That makes me more confident that the answer is correct.Just to recap, the two methods I used were:1. Using the identity for the sum of cubes in terms of symmetric sums.2. Directly using the fact that each root satisfies the original polynomial equation to express ( x_i^3 ) in terms of ( x_i ) and then summing them up.Both approaches led me to the same conclusion, so I think it's safe to say that ( x_1^3 + x_2^3 + x_3^3 = 3 ).I don't see any mistakes in my reasoning, but let me just think if there's another perspective or if I missed something.Wait, another thought: since the polynomial is cubic, and all roots are real, maybe there's a way to find the roots explicitly and then compute their cubes. But that seems complicated because solving a cubic can be messy, and I don't think it's necessary here since the methods I used already gave me the answer.Also, considering that the polynomial is ( x^3 - 3x - 1 ), which is similar to the equation for the roots of the cosine function in some trigonometric identities, but I don't think that's relevant here.No, I think my initial methods are solid. I used two different approaches, both leading to the same result, so I'm pretty confident.**Final Answer**The value of ( x_1^3 + x_2^3 + x_3^3 ) is boxed{3}.