🧠:[<think>Okay, so I need to find the limit of the function f(x, y) as both x and y approach 0. The function is defined as x sin(1/y) + y sin(1/x) when x² + y² ≠ 0, and it's 0 when either x or y is 0. Hmm, let me start by understanding the problem better.First, limits in multivariable calculus can be tricky because the path of approach matters. Unlike single-variable calculus where you only have two directions (left and right), with two variables, there are infinitely many paths to approach the origin. So, I need to check if the limit exists by checking if it's the same along different paths. If I get different values along different paths, the limit doesn't exist.The function here has two terms: x sin(1/y) and y sin(1/x). Each term involves a variable multiplied by a sine function with the reciprocal of the other variable. The sine function oscillates between -1 and 1, so sin(1/y) and sin(1/x) are bounded between -1 and 1, regardless of how small y or x gets. That might be useful because multiplying by x or y, which are approaching 0, could make the whole term go to 0.Let me recall the Squeeze Theorem. If I can bound the absolute value of the function between two things that both go to 0, then the limit should be 0. Let's try that.First, consider |f(x, y)|. Since f(x, y) = x sin(1/y) + y sin(1/x), the absolute value is less than or equal to |x sin(1/y)| + |y sin(1/x)|. The sine terms are each bounded by 1 in absolute value, so this becomes |x| * 1 + |y| * 1 = |x| + |y|.So, |f(x, y)| ≤ |x| + |y|. Now, as (x, y) approaches (0, 0), both |x| and |y| approach 0. Therefore, the upper bound |x| + |y| approaches 0. By the Squeeze Theorem, this would mean that the limit of f(x, y) as (x, y) approaches (0, 0) is 0.Wait, but let me check if there's a catch here. Sometimes when dealing with oscillating functions, even if they are bounded, their reciprocals can cause issues. But in this case, the reciprocals are inside the sine functions, which are then multiplied by variables going to 0. Since sin(1/y) and sin(1/x) are bounded, multiplying them by x and y respectively, which are approaching 0, should make each term go to 0.But maybe I should test some specific paths to confirm. Let's approach (0, 0) along different paths and see if the limit is indeed 0 each time.First, approach along the x-axis (where y = 0). Then f(x, 0) = 0 by definition. So, the limit as x approaches 0 is 0. Similarly, along the y-axis (x = 0), f(0, y) = 0, so the limit is 0.Now, let's approach along some other path. Let's say y = x. Then substitute y with x in the function:f(x, x) = x sin(1/x) + x sin(1/x) = 2x sin(1/x).What's the limit as x approaches 0? Well, 2x approaches 0 and sin(1/x) oscillates between -1 and 1. So, the entire expression oscillates between -2|x| and 2|x|, both of which approach 0. Therefore, by the Squeeze Theorem again, the limit is 0.Another path: let's take y = kx, where k is a constant. Then substitute y with kx:f(x, kx) = x sin(1/(kx)) + kx sin(1/x).So, the first term is x sin(1/(kx)) and the second term is kx sin(1/x). Let's analyze each term separately.First term: x sin(1/(kx)). As x approaches 0, 1/(kx) becomes very large in magnitude, but sin(1/(kx)) is still bounded between -1 and 1. Therefore, |x sin(1/(kx))| ≤ |x|, which approaches 0. So, the first term goes to 0.Second term: kx sin(1/x). Similarly, sin(1/x) is bounded, so |kx sin(1/x)| ≤ |kx|. As x approaches 0, |kx| approaches 0. So, the second term also goes to 0. Therefore, the entire expression f(x, kx) approaches 0 as x approaches 0.What about a more complicated path? Let's try approaching along y = x^2.Then f(x, x²) = x sin(1/x²) + x² sin(1/x).Again, let's look at each term:First term: x sin(1/x²). Since sin(1/x²) is bounded, |x sin(1/x²)| ≤ |x|, which approaches 0 as x→0.Second term: x² sin(1/x). Similarly, |x² sin(1/x)| ≤ x², which approaches 0. So, both terms go to 0, so the limit is 0.Alternatively, try approaching along a spiral or something, but even so, as long as x and y both approach 0, the terms should be squeezed to 0. The key here is that both terms are products of a variable going to 0 and a bounded oscillating function. So, their product should go to 0.Wait, but what if I approach along a path where one variable approaches 0 faster than the other? For example, let's take y = x^3. Then f(x, x³) = x sin(1/x³) + x³ sin(1/x). The first term is x sin(1/x³). The sin(1/x³) oscillates even more rapidly as x approaches 0, but it's still bounded. So, |x sin(1/x³)| ≤ |x| → 0. The second term is x³ sin(1/x) ≤ x³ → 0. So, again, both terms go to 0.Alternatively, take a path where x and y approach 0 but along some oscillating path. For instance, let x = 1/n and y = 1/n as n approaches infinity. Then f(1/n, 1/n) = (1/n) sin(n) + (1/n) sin(n). So, each term is (1/n) sin(n). Since |sin(n)| ≤ 1, each term is ≤ 1/n, so the total is ≤ 2/n. As n→infty, this approaches 0. So, even along this discrete path, the limit is 0.Alternatively, let's consider x = 1/(nπ) and y = 1/(nπ), then sin(1/y) = sin(nπ) = 0, and sin(1/x) = sin(nπ) = 0. So, f(x, y) = 0 + 0 = 0. So along this path, the function is always 0. The limit is 0.But another path where maybe the sine terms are 1. For example, choose x = 1/(2nπ + π/2) so that sin(1/x) = sin(2nπ + π/2) = 1. Let’s take x = y = 1/(2nπ + π/2). Then:f(x, y) = x sin(1/y) + y sin(1/x) = x sin(2nπ + π/2) + y sin(2nπ + π/2) = x*1 + y*1 = x + y = 2x. As n→infty, x approaches 0, so this approaches 0.Alternatively, if we take x = 1/(2nπ + π/2) and y fixed? Wait, but both x and y have to approach 0. If I take y = 1/(2nπ + π/2) as well, then as n→infty, x and y approach 0. So, even when the sine terms are maximized (equal to 1), the terms x and y still make the whole expression go to 0.Hmm. So regardless of the path, even if we make sin(1/x) or sin(1/y) take their maximum values, the coefficients x and y are going to 0, so the product would still go to 0.Therefore, using the Squeeze Theorem, since |f(x, y)| ≤ |x| + |y|, and both |x| and |y| approach 0 as (x, y) → (0, 0), the limit must be 0.But just to be thorough, let's check if there's a path where the limit might not be 0. Suppose we take a path where x approaches 0 but y approaches 0 in such a way that 1/y or 1/x is not going to infinity. Wait, but if x approaches 0, then 1/x approaches infinity, unless x is approaching 0 along a specific sequence where 1/x is an integer multiple of π or something. But even then, sin(1/x) would still be bounded.Wait, suppose we take x approaching 0 such that 1/x is an integer multiple of π. For example, x = 1/(nπ), so sin(1/x) = sin(nπ) = 0. Similarly, if y = 1/(nπ), then sin(1/y) = 0. So, along such a path, f(x, y) = x*0 + y*0 = 0. So the function is 0 there.Alternatively, if I take x approaching 0 such that 1/x = 2nπ + π/2, making sin(1/x) = 1. Then f(x, y) = x sin(1/y) + y*1. But even here, y has to approach 0 as well. If y is approaching 0, then regardless of how it approaches, the term y*1 is approaching 0. Similarly, the term x sin(1/y) is x times a bounded term, so as x approaches 0, that term approaches 0. So even along this path, the limit is 0.Another idea: perhaps if we let y approach 0 in such a way that 1/y is related to x. For example, set y = x. Wait, we already checked that. Or set y = x^2. Wait, did that too. Hmm.Alternatively, set y = 1/(n) and x = 1/(n). Then as n→infty, x and y approach 0. Then f(x, y) = (1/n) sin(n) + (1/n) sin(n). Each term is (sin n)/n, which goes to 0 because sin n is bounded between -1 and 1.Alternatively, if we set y = x^k for some exponent k. Then, as x approaches 0, y approaches 0. Then f(x, y) becomes x sin(1/x^k) + x^k sin(1/x). Let's see:First term: x sin(1/x^k). The sin(1/x^k) oscillates as x approaches 0, but it's bounded. So |x sin(1/x^k)| ≤ |x| → 0.Second term: x^k sin(1/x). Similarly, sin(1/x) oscillates, but |x^k sin(1/x)| ≤ x^k → 0 as x→0 (since k > 0). Therefore, both terms go to 0.So regardless of the exponent k, as long as k > 0, the limit is 0.Wait, what if we approach along a path where x and y approach 0 but their rates are different? For instance, let’s take x = t and y = t^2 as t→0. Then f(t, t^2) = t sin(1/t²) + t^2 sin(1/t). The first term is t sin(1/t²). As t→0, sin(1/t²) oscillates rapidly, but multiplied by t which approaches 0. So, |t sin(1/t²)| ≤ t → 0. The second term is t² sin(1/t) ≤ t² → 0. So again, both terms go to 0.Alternatively, if we take a path that's more erratic, like x = 1/n and y = 1/n^2. Then f(1/n, 1/n^2) = (1/n) sin(n^2) + (1/n^2) sin(n). The first term is (1/n) sin(n^2), which is bounded by 1/n → 0. The second term is (1/n^2) sin(n) ≤ 1/n^2 → 0. So again, the limit is 0.Is there any path where the limit might not be 0? Let's think.Suppose we take y approaching 0 such that 1/y is a function that oscillates, but x is approaching 0 in a way related to y. For example, let’s set x = y. But we already did that.Wait, perhaps if we set x = sin(1/y). Wait, but then x would have to approach 0 as y approaches 0. But sin(1/y) is between -1 and 1, so if y approaches 0, sin(1/y) doesn't approach 0. So that wouldn't work because x must approach 0.Alternatively, set x = y sin(1/y). Then as y→0, x approaches 0 because sin(1/y) is bounded. Then f(x, y) = x sin(1/y) + y sin(1/x). Substitute x = y sin(1/y):f(x, y) = y sin(1/y) * sin(1/y) + y sin(1/(y sin(1/y))) = y [sin²(1/y)] + y sin(1/(y sin(1/y))).This seems complicated, but let's see:First term: y sin²(1/y) ≤ y * 1 = y → 0.Second term: y sin(1/(y sin(1/y))). Let's analyze the argument of the sine: 1/(y sin(1/y)).But as y approaches 0, sin(1/y) oscillates between -1 and 1, so y sin(1/y) approaches 0 (since y→0 and sin(1/y) is bounded). Therefore, 1/(y sin(1/y)) becomes unbounded, unless sin(1/y) is 0, but even then, when sin(1/y) is 0, x = y sin(1/y) = 0, so f(x, y) = 0 + y sin(1/0), but 1/0 is undefined. Wait, actually, in the original function, f(x, 0) = 0. So if x = 0, then f(0, y) = 0. But in this case, x = y sin(1/y). If y approaches 0, x approaches 0, but how?Wait, maybe this path is not valid because when y approaches 0, sin(1/y) oscillates, so x = y sin(1/y) oscillates between -y and y. But since y is approaching 0, x approaches 0. However, the function f(x, y) when x is approaching 0 and y is approaching 0 is defined as x sin(1/y) + y sin(1/x) except when x or y is 0. But here, even if x is approaching 0, as long as x and y are not 0, we use the original expression.But in this case, let's try to compute the second term: y sin(1/x) where x = y sin(1/y). So, sin(1/x) = sin(1/(y sin(1/y))). The argument inside the sine is 1/(y sin(1/y)). As y approaches 0, y sin(1/y) oscillates between -y and y, so 1/(y sin(1/y)) has magnitude at least 1/y, which approaches infinity. Therefore, sin(1/(y sin(1/y))) oscillates rapidly between -1 and 1. Therefore, the second term is y multiplied by an oscillating term between -1 and 1. Therefore, |y sin(1/(y sin(1/y)))| ≤ |y|. So the second term is bounded by |y|, which approaches 0. Therefore, both terms go to 0. Hence, even along this complicated path, the limit is 0.Therefore, no matter how I approach (0, 0), the function seems to approach 0. So, combining the Squeeze Theorem result and checking various paths, it seems that the limit is indeed 0.But just to make sure, let's try another approach. Let's convert to polar coordinates. Let x = r cosθ and y = r sinθ, where r approaches 0. Then, the function becomes:f(r cosθ, r sinθ) = r cosθ sin(1/(r sinθ)) + r sinθ sin(1/(r cosθ)).So, f(r, θ) = r cosθ sin(1/(r sinθ)) + r sinθ sin(1/(r cosθ)).Now, let's find the limit as r approaches 0. The question is, does this limit depend on θ? If it does, then the overall limit doesn't exist. But if it's the same for all θ, then the limit is that value.But analyzing this seems complicated. Let's see:First term: r cosθ sin(1/(r sinθ)). Let's look at sin(1/(r sinθ)). As r approaches 0, 1/(r sinθ) becomes very large unless sinθ = 0. But sinθ = 0 would correspond to θ = 0 or π, which is along the x-axis. But in that case, y = 0, so the function is defined to be 0. Similarly, cosθ = 0 would correspond to the y-axis. But in polar coordinates, except along the axes, sinθ and cosθ are non-zero.So, for θ not a multiple of π/2, sinθ and cosθ are non-zero. Therefore, 1/(r sinθ) and 1/(r cosθ) go to infinity as r→0. So, sin(1/(r sinθ)) and sin(1/(r cosθ)) oscillate rapidly between -1 and 1. However, these sine terms are multiplied by r cosθ and r sinθ, which approach 0. So, each term is bounded in absolute value by r |cosθ| and r |sinθ|, respectively. Therefore, |f(r, θ)| ≤ r |cosθ| + r |sinθ| = r(|cosθ| + |sinθ|). Since |cosθ| + |sinθ| ≤ √2 (by Cauchy-Schwarz inequality, but actually the maximum of |cosθ| + |sinθ| is √2), so |f(r, θ)| ≤ r√2. As r approaches 0, this upper bound approaches 0. Therefore, by Squeeze Theorem, the limit is 0 regardless of θ. Therefore, in polar coordinates, the limit is 0 from all directions. Hence, the overall limit is 0.Therefore, after considering multiple approaches—Squeeze Theorem, specific paths, polar coordinates—it seems the limit is indeed 0. I don't see any path where the limit would be different. So, the answer should be 0.Final Answerboxed{0}</think>To find the limit of the function ( f(x, y) = x sin frac{1}{y} + y sin frac{1}{x} ) as ((x, y)) approaches ((0, 0)), we can use the Squeeze Theorem and check various paths.1. Squeeze Theorem: - The absolute value of the function is bounded by: [ |f(x, y)| leq |x sin frac{1}{y}| + |y sin frac{1}{x}| leq |x| + |y| ] - As ((x, y) to (0, 0)), both (|x|) and (|y|) approach 0. Therefore, by the Squeeze Theorem, the limit of (f(x, y)) is 0.2. Checking Specific Paths: - Along the x-axis (y = 0): (f(x, 0) = 0), so the limit is 0. - Along the y-axis (x = 0): (f(0, y) = 0), so the limit is 0. - Along the line (y = x): (f(x, x) = 2x sin frac{1}{x}), which approaches 0 as (x to 0). - Along the line (y = kx): (f(x, kx) = x sin frac{1}{kx} + kx sin frac{1}{x}), each term approaches 0 as (x to 0). - Along the curve (y = x^2): (f(x, x^2) = x sin frac{1}{x^2} + x^2 sin frac{1}{x}), each term approaches 0 as (x to 0).3. Polar Coordinates: - Convert to polar coordinates (x = r cos theta) and (y = r sin theta): [ f(r cos theta, r sin theta) = r cos theta sin frac{1}{r sin theta} + r sin theta sin frac{1}{r cos theta} ] - The terms are bounded by (r |cos theta| + r |sin theta| leq r sqrt{2}), which approaches 0 as (r to 0).After considering all these approaches, we conclude that the limit is 0.[boxed{0}],<think>Okay, let's see. I need to find the limit of the function f(x, y) as both x and y approach 0. The function is defined as x sin(1/y) + y sin(1/x) when x² + y² ≠ 0, and it's 0 on the axes. Hmm. Alright, so first, I remember that when taking limits in multiple variables, the limit exists only if all paths approaching the point give the same result. So, I might need to check different paths and see if they all lead to the same limit. If not, then the limit doesn't exist.Let me start by recalling the squeeze theorem. Since sine is bounded between -1 and 1, both terms x sin(1/y) and y sin(1/x) can be bounded. Let's see:For the first term, |x sin(1/y)| ≤ |x|, and for the second term, |y sin(1/x)| ≤ |y|. So, adding them together, |f(x, y)| ≤ |x| + |y|. Now, as x and y approach 0, both |x| and |y| go to 0. Therefore, by the squeeze theorem, the limit should be 0. But wait, is that correct?Wait, but maybe there's a catch here. Because even though each term is bounded by |x| and |y|, which individually go to 0, perhaps along certain paths the sum behaves differently? Let me think. For example, if x and y approach 0 in such a way that one of the terms becomes oscillatory but the other term doesn't cancel it out. But since both terms are bounded by |x| and |y|, which go to 0 regardless of the path, maybe the total still goes to 0. Let's verify this with some specific paths.First, approach along the x-axis: y = 0. Then f(x, 0) = 0 by definition, so the limit is 0. Similarly, along the y-axis, x = 0, f(0, y) = 0, so limit is 0. Now, what if I approach along a line y = x? Then f(x, x) = x sin(1/x) + x sin(1/x) = 2x sin(1/x). The limit as x approaches 0 of 2x sin(1/x) is 0, since sin(1/x) is bounded and x goes to 0. So that's 0. What about another path, say y = kx, where k is a constant? Then f(x, kx) = x sin(1/(kx)) + kx sin(1/x). Let's analyze each term. The first term is x sin(1/(kx)). As x approaches 0, 1/(kx) becomes very large in magnitude, so sin(1/(kx)) oscillates between -1 and 1. But multiplied by x, which approaches 0, so the term x sin(1/(kx)) should approach 0. Similarly, the second term is kx sin(1/x). Again, sin(1/x) oscillates, but multiplied by kx, which approaches 0, so the whole term goes to 0. Therefore, the sum is 0 + 0 = 0. So along any straight line path, the limit is 0.But maybe there's a more complicated path where the limit doesn't exist? Let's think. For example, what if we approach along a parabolic path, like y = x². Then f(x, x²) = x sin(1/x²) + x² sin(1/x). Again, let's check each term. The first term is x sin(1/x²). As x approaches 0, 1/x² goes to infinity, so sin(1/x²) oscillates rapidly between -1 and 1. But multiplied by x, which is approaching 0, so x sin(1/x²) will oscillate between -|x| and |x|, hence approaching 0. The second term is x² sin(1/x). Similarly, sin(1/x) oscillates between -1 and 1, but multiplied by x², which goes to 0, so that term also approaches 0. Therefore, the sum is 0. So even along a parabolic path, the limit is 0.Alternatively, what if we approach along a path where y = 1/ln(1/x) or something more complicated? Let's try y = 1/n as x approaches 0 along 1/n. Wait, maybe a spiral path? Hmm, but perhaps that's overcomplicating. Alternatively, suppose we take a path where y = x, but with x approaching 0. Wait, we already checked y = x. Hmm. Maybe another approach: let's see if we can bound |f(x, y)|. We have |f(x, y)| ≤ |x| + |y|. Since both |x| and |y| approach 0 as (x, y) approaches (0, 0), then by the squeeze theorem, the limit should be 0. Therefore, regardless of the path, the limit is 0.But wait, is there a case where even though |x| + |y| approaches 0, the sum x sin(1/y) + y sin(1/x) doesn't approach 0? Let's think. For example, if x and y approach 0 in such a way that 1/x and 1/y are related. Suppose we take x = 1/n and y = 1/n as n approaches infinity. Then f(1/n, 1/n) = (1/n) sin(n) + (1/n) sin(n). So this is (2/n) sin(n). Since sin(n) oscillates between -1 and 1, the whole expression oscillates between -2/n and 2/n, which approaches 0 as n goes to infinity. So even here, the limit is 0.Alternatively, if x and y approach 0 along different sequences. For example, take x_n = 1/n and y_n = 1/(n + 1). Then f(x_n, y_n) = (1/n) sin(n + 1) + (1/(n + 1)) sin(n). Again, each term is bounded by 1/n and 1/(n + 1), so their sum is bounded by 1/n + 1/(n + 1) ≈ 2/n, which goes to 0. So regardless of how x and y approach 0, as long as both |x| and |y| go to 0, the sum should go to 0.Wait, but let me test a different approach. Suppose we take a path where x and y approach 0 but sometimes 1/x or 1/y is a multiple of π/2 or something where sin is 1 or -1. For example, take x_n = 1/(nπ/2) and y_n = 1/(nπ/2). Then sin(1/y_n) = sin(nπ/2). Depending on n, sin(nπ/2) cycles through 0, 1, 0, -1, etc. So f(x_n, y_n) = (1/(nπ/2)) sin(nπ/2) + (1/(nπ/2)) sin(nπ/2) = 2/(nπ/2) sin(nπ/2) = (4/(nπ)) sin(nπ/2). Let's compute for n=1: 4/(1π) * 1 = 4/π ≈ 1.27. For n=2: 4/(2π) * 0 = 0. For n=3: 4/(3π) * (-1) ≈ -0.424. For n=4: 4/(4π) * 0 = 0. For n=5: 4/(5π) * 1 ≈ 0.255. So the sequence f(x_n, y_n) is 4/(nπ) times 1, 0, -1, 0, 1, etc. As n approaches infinity, 4/(nπ) approaches 0, so the terms go to 0. Therefore, even though the sine terms take on maximum values, the coefficients 4/(nπ) ensure that the sequence approaches 0. Therefore, the limit is still 0.Alternatively, suppose we take x_n = 1/n and y_n = 1/(n + 1/2). Then sin(1/y_n) = sin(n + 1/2)π = sin(nπ + π/2) = (-1)^n sin(π/2) = (-1)^n. Similarly, sin(1/x_n) = sin(n). Wait, sin(n) is not necessarily (-1)^n. But for x_n = 1/n, sin(1/x_n) = sin(n). Hmm. So then f(x_n, y_n) = (1/n)(-1)^n + (1/(n + 1/2)) sin(n). The first term is (1/n)(-1)^n, which approaches 0 as n goes to infinity. The second term is bounded by 1/(n + 1/2), which also approaches 0. Therefore, the entire expression approaches 0. So even if we choose points where the sine terms are maximized in magnitude, the factors of x and y dampen them to 0.Therefore, no matter how we approach (0, 0), the function f(x, y) is squeezed between -(|x| + |y|) and |x| + |y|, both of which approach 0. Therefore, the limit must be 0.Wait, but I remember sometimes even if individual terms go to 0, their combination might not. But in this case, it's a sum of two terms, each bounded by |x| and |y|. Since both |x| and |y| go to 0, the sum of their bounds, |x| + |y|, also goes to 0. Therefore, the squeeze theorem should apply here, right? Because |f(x, y)| ≤ |x| + |y|, and |x| + |y| approaches 0 as (x, y) approaches (0, 0). Therefore, the limit is 0. So I think that's the answer. But let me check another possible approach. Let's convert to polar coordinates. Let x = r cosθ and y = r sinθ, where r approaches 0. Then f(x, y) = r cosθ sin(1/(r sinθ)) + r sinθ sin(1/(r cosθ)). So f(x, y) = r cosθ sin(1/(r sinθ)) + r sinθ sin(1/(r cosθ)). Now, each term is r times something bounded. For example, |cosθ sin(1/(r sinθ))| ≤ |cosθ| ≤ 1, so the first term is ≤ r. Similarly, the second term is ≤ r. Therefore, |f(x, y)| ≤ 2r. As r approaches 0, this goes to 0. Therefore, again, by the squeeze theorem, the limit is 0. So even in polar coordinates, the limit is 0.Alternatively, maybe there's a problem when sinθ or cosθ is 0? But if sinθ = 0, then y = 0, and f(x, 0) = 0 by definition. Similarly, cosθ = 0 implies x = 0, and f(0, y) = 0. So those cases are already covered, and the function is defined to be 0 there. Therefore, even near those angles, as r approaches 0, the function approaches 0. Therefore, I think the limit is indeed 0.Wait, but in polar coordinates, when θ approaches 0, sinθ approaches 0, so 1/(r sinθ) becomes very large. But even then, sin(1/(r sinθ)) is bounded, and multiplied by r cosθ, which is approaching 0. So the term still goes to 0. Similarly, when θ approaches π/2, cosθ approaches 0, but sin(1/(r cosθ)) is bounded, and multiplied by r sinθ, which is approaching 0. Therefore, both terms still go to 0. Hence, regardless of θ, as r approaches 0, the entire expression f(x, y) approaches 0. Therefore, the limit is 0.So, after checking multiple approaches: along axes, along lines, along parabolas, in polar coordinates, and using the squeeze theorem with |f(x, y)| ≤ |x| + |y|, which tends to 0, it seems that the limit is indeed 0. Therefore, the answer should be 0.Final Answerboxed{0}</think>To find the limit of the function ( f(x, y) = x sin frac{1}{y} + y sin frac{1}{x} ) as ((x, y)) approaches ((0, 0)), we can use the squeeze theorem and analyze various paths approaching the origin.1. Bounding the Function: - The function ( f(x, y) ) can be bounded using the triangle inequality and the fact that (|sin cdot| leq 1): [ |f(x, y)| leq |x| left| sin frac{1}{y} right| + |y| left| sin frac{1}{x} right| leq |x| + |y| ] - As ( x ) and ( y ) approach 0, both ( |x| ) and ( |y| ) approach 0. Therefore, by the squeeze theorem, the limit of ( |f(x, y)| ) is 0.2. Checking Various Paths: - Along the axes: For ( y = 0 ) or ( x = 0 ), ( f(x, y) ) is defined to be 0, so the limit is 0. - Along the line ( y = kx ): The function becomes ( x sin frac{1}{kx} + kx sin frac{1}{x} ). Each term is bounded by ( |x| ) and ( |kx| ), which approach 0. - Along a parabolic path ( y = x^2 ): The function becomes ( x sin frac{1}{x^2} + x^2 sin frac{1}{x} ). Each term is bounded by ( |x| ) and ( |x^2| ), which approach 0.3. Polar Coordinates: - Converting to polar coordinates ( x = r cos theta ) and ( y = r sin theta ), the function becomes: [ f(r, theta) = r cos theta sin frac{1}{r sin theta} + r sin theta sin frac{1}{r cos theta} ] - Each term is bounded by ( r ), so ( |f(r, theta)| leq 2r ). As ( r to 0 ), this bound approaches 0.4. Conclusion: - Using the squeeze theorem and verifying along various paths, we conclude that the limit of ( f(x, y) ) as ((x, y)) approaches ((0, 0)) is 0.[boxed{0}]]

🔑:Solution. Let varepsilon>0, then for |x|<frac{varepsilon}{2},|y|<frac{varepsilon}{2} we get rho==sqrt{left(frac{varepsilon}{2}right)^{2}+left(frac{varepsilon}{2}right)^{2}}=frac{sqrt{2}}{2} varepsilon . text { We form the difference } f(x, y)-0 text { and estimate it: }|f(x, y)-0| leq|x| cdotleft|sin frac{1}{y}right|+|y| cdotleft|sin frac{1}{x}right| leq|x|+|y|<frac{varepsilon}{2}+frac{varepsilon}{2}=varepsilonThus,|f(x, y)-0|<varepsilonBased on formula (6.10), we conclude thatlim _{substack{x rightarrow 0 y rightarrow 0}} f(x, y)=0

🧠:[<think>Okay, so I need to figure out the maximum number of numbers from 1 to 200 that can be grouped such that if there are at least two numbers in the group, the sum of any two numbers is divisible by 5. Hmm, let's break this down.First, the key condition here is that the sum of any two numbers in the group must be divisible by 5. That means, for any two numbers a and b in the group, (a + b) mod 5 should be 0. So, (a mod 5 + b mod 5) mod 5 = 0. This implies that a mod 5 and b mod 5 must be additive inverses modulo 5. In other words, if a mod 5 = r, then b mod 5 must be (5 - r) mod 5. However, there's a special case where r = 0, because 0 + 0 = 0 mod 5. So, numbers congruent to 0 mod 5 can only pair with other numbers congruent to 0 mod 5.Wait, let me think again. If two numbers are congruent to r and s mod 5, then their sum is r + s mod 5. For this to be 0, r + s ≡ 0 mod 5. So, s ≡ -r mod 5. Which would mean that if we have a number with remainder r, then all other numbers in the group must have remainder -r mod 5. However, if the group has more than one remainder, they all must pair with each other to sum to 0 mod 5. But this can't happen unless all numbers are in pairs of inverses. But if you have more than two different remainders, this might complicate things. Wait, actually, if the group has numbers with different remainders, then each pair of those numbers must sum to 0 mod 5. So, suppose the group has numbers with remainders r and s. Then r + r ≡ 0 mod 5 must hold, and r + s ≡ 0 mod 5. For the first case, r + r ≡ 0 mod 5 implies 2r ≡ 0 mod 5, so r ≡ 0 mod 5 (since 2 and 5 are coprime). But if r is not 0, then 2r ≡ 0 mod 5 is impossible. Therefore, if the group contains a number with remainder r ≠ 0, then all numbers in the group must have remainder -r mod 5. Because if you have two numbers with remainder r, their sum would be 2r mod 5, which is not 0 unless r = 0. Therefore, this suggests that the group can either consist of:1. All numbers congruent to 0 mod 5, since 0 + 0 ≡ 0 mod 5.OR2. Numbers congruent to a non-zero remainder r and numbers congruent to -r mod 5, but with the caveat that if you include numbers with remainder r, you cannot have more than one remainder pair (r and -r), and you can't mix them with other remainders. Wait, actually, no. If you have numbers with remainder r and -r, then any two numbers in the group will either be r + r, r + (-r), or (-r) + (-r). For the sum to be divisible by 5, each of these combinations must satisfy:- r + r ≡ 0 mod 5 → 2r ≡ 0 mod 5 → r ≡ 0 mod 5. But if r ≠ 0, this is impossible.- r + (-r) ≡ 0 mod 5 → which is always true.- (-r) + (-r) ≡ -2r ≡ 0 mod 5 → 2r ≡ 0 mod 5 → same as above.So, this implies that if we have a group with numbers of remainder r and -r, then we can only have one number with remainder r and one number with remainder -r. Because if we have two numbers with remainder r, their sum is 2r, which isn't 0 mod 5 unless r = 0. Similarly for two numbers with remainder -r. So, if the group includes numbers from both r and -r, then we can have at most one number from each. Wait, but that would only allow two numbers. But maybe more if we have only numbers from r or only numbers from -r. Wait, but if all numbers are from r, then 2r ≡ 0 mod 5, which requires r = 0. So, only the 0 mod 5 case allows multiple numbers.Alternatively, if we have numbers from both r and -r, but only one number from each. But that would restrict the group to pairs. So, perhaps the maximum group is either all numbers congruent to 0 mod 5, or pairs of numbers where one is r and the other is -r. But in that case, which is larger?So, first, let's compute how many numbers from 1 to 200 are congruent to 0 mod 5. That's the numbers 5, 10, 15, ..., 200. The count is floor(200/5) = 40. So, 40 numbers.Alternatively, if we pick pairs (r, -r). Let's pick a non-zero remainder r, and pair each number congruent to r with one congruent to -r. However, the problem is that if we have two numbers in the group with the same remainder r, then their sum is 2r mod 5, which isn't 0. Therefore, in such a group, we can have at most one number from each remainder pair (r, -r). But if we pick multiple such pairs, but then any two numbers from different pairs would need to sum to 0 mod 5, which isn't necessarily the case.Wait, actually, if we have numbers from different pairs, say r and s, then their sum is r + s mod 5. For this to be 0, s must be -r. But s is already part of another pair. So, this suggests that if we want a group where any two numbers sum to 0 mod 5, all numbers must come from a single pair (r, -r) with at most one number from each, but that would only allow two numbers. Alternatively, if we have multiple such pairs, but then the cross sums (r + s) need to be 0 mod 5, which would require s = -r, but then if s is part of another pair, that complicates things.Wait, perhaps my initial approach is incorrect. Let's think differently.Let me recall that in problems where the sum of any two numbers must be divisible by a certain number, the numbers must all be congruent modulo that number, or form a system where each pair sums to 0 modulo that number. But in the case of modulo 5, if all numbers are congruent to 0 mod 5, then any sum is 0 mod 5. If all numbers are congruent to a non-zero residue r, then 2r ≡ 0 mod 5. That's possible only if r ≡ 0 mod 5, which contradicts. So, the only possibility for a group with more than one number is all numbers ≡0 mod5, or a group where each number is paired with its inverse, but with no duplicates in residues.Wait, so perhaps the group can consist of numbers from residue 0 mod5, and one residue pair (r, -r). But then, if you have a number from residue 0 and another from residue r, their sum is r mod5, which needs to be 0 mod5. Therefore, r must be 0. But then, you can't have numbers from residue 0 and another residue. So, the group must either be all numbers congruent to 0 mod5, or numbers from a single pair (r, -r) but with at most one number from each.Wait, this is confusing. Let me formalize this.Let S be the set of numbers selected. For any a, b ∈ S, a + b ≡ 0 mod5.Case 1: All numbers in S ≡0 mod5. Then, any two numbers sum to 0+0=0 mod5. So, this works.Case 2: There exists a number in S not ≡0 mod5. Let’s say there is a number x ≡ r mod5 where r ≠0. Then, for any other number y in S, x + y ≡0 mod5. Therefore, y ≡-r mod5. So, all numbers in S must be ≡-r mod5. But then, take two numbers y1, y2 ≡-r mod5. Their sum is (-r) + (-r) = -2r mod5. For this to be 0 mod5, -2r ≡0 mod5 → 2r ≡0 mod5 → r ≡0 mod5, which contradicts r ≠0. Therefore, if S contains a number ≡r mod5 (r ≠0), then S cannot contain more than one number. But that would mean S can only have one number. But the problem states "if there are at least two numbers in the group, then the sum of any two numbers is divisible by 5". So, if S has only one number, the condition is vacuously true. But the question asks for the maximum number of numbers. So, perhaps there's a way to have more than one number.Wait, maybe my earlier conclusion is too hasty. If S has numbers of two different residues, say r and -r, such that r + (-r) ≡0 mod5. Then, if we have multiple numbers of residue r and multiple numbers of residue -r, but any two numbers in S must sum to 0 mod5. But if we have two numbers with residue r, their sum is 2r, which is not 0 unless r=0. Similarly for two numbers with residue -r. Therefore, in order to have a group with more than one number, you can have at most one number from residue r and one number from residue -r. But then, how many such pairs can we have?Wait, actually, if we take one number from residue 1 and one from residue 4 (since 1 + 4 = 5 ≡0 mod5), and another number from residue 2 and one from residue 3 (2 + 3 = 5 ≡0 mod5). But then, if we have numbers from residue 1 and 4, and also numbers from residue 2 and 3, then the sum of a number from residue 1 and residue 2 would be 1 + 2 = 3 mod5, which is not 0. Hence, such a group would violate the condition.Therefore, the group can either consist of:- All numbers congruent to 0 mod5.OR- A combination of numbers from a single pair (r, -r), but with at most one number from each residue. However, since adding two numbers from the same residue would violate the condition, we can have at most one number from residue r and one from residue -r. But this only gives two numbers. However, perhaps there are multiple such pairs? But if we include multiple pairs, the cross sums would not be 0 mod5. For example, if we take a number from residue 1 and 4, and another number from residue 2 and 3, then as above, the cross sums would not be 0 mod5. Therefore, the only way to have a group with more than two numbers is to take all numbers congruent to 0 mod5.Alternatively, maybe there's another way. Let's consider that perhaps all numbers in the group are congruent to the same residue mod5. But except for 0, as we saw, 2r must be 0 mod5, which is only possible if r=0. Therefore, the only residue that allows multiple numbers is 0 mod5.Therefore, the maximum size would be 40 numbers (all multiples of 5 from 1 to 200). But wait, the answer can't be that straightforward. Let me check again.Wait, perhaps there is another way. Let's think about choosing numbers from two different residues such that every pair sums to 0 mod5. For example, if we choose numbers from residue 1 and 4. If we have one number from 1 and one from 4, their sum is 5. But if we have two numbers from residue 1 and two numbers from residue 4, then the sum of two numbers from residue 1 would be 2 mod5, which is not 0. So, that's not allowed. Therefore, in such a case, you can only have at most one number from residue 1 and one from residue 4. Similarly for residues 2 and 3. So, for each complementary pair, you can have at most one number from each.Therefore, if we combine all possible pairs, we can have:- 1 number from residue 1 and 1 from residue 4.- 1 number from residue 2 and 1 from residue 3.- All numbers from residue 0.But the problem is, if we take numbers from different pairs, their sums will not be 0 mod5. For example, a number from residue 1 and a number from residue 2 would sum to 3 mod5. So, such a group would not satisfy the condition. Therefore, we cannot mix residues from different pairs. So, if we want a group that contains numbers from multiple residue pairs, the cross sums would fail. Therefore, the only possibilities are:1. A group consisting of all numbers congruent to 0 mod5 (size 40).2. A group consisting of one number from residue 1 and one from residue 4 (size 2).3. Similarly, a group with one number from residue 2 and one from residue 3 (size 2).But since the problem asks for the maximum number, 40 is larger than 2, so 40 would be the answer. But wait, maybe there's a smarter way. Let me think again.Wait, perhaps if we combine the residue 0 group with one of the pairs. For example, take all numbers from residue 0 and one number from residue 1 and one from residue 4. But then, the sum of a residue 0 number and a residue 1 number would be 1 mod5, which is not 0. Hence, that's invalid. Similarly, adding any non-zero residue to residue 0 would result in a non-zero residue. Therefore, residue 0 cannot be combined with any other residues.Alternatively, if we take multiple pairs. Suppose we have two numbers from residue 1 and two from residue 4. But as before, the sum of two residue 1 numbers would be 2 mod5, which is invalid. So, even if you have multiple pairs, you can't have more than one number from each residue. Hence, the maximum size for a non-zero residue group is 2. Therefore, comparing 40 vs 2, 40 is larger. Therefore, the maximum is 40.But wait, let's check the count again. From 1 to 200, the numbers divisible by 5 are 5, 10, 15, ..., 200. The number of terms is 200 /5 = 40. So, 40 numbers. So, if we take all 40, then indeed, any two numbers sum to a multiple of 5, since they are all 0 mod5. So, that works.But the problem says "one or several were selected into a separate group with the following property: if there are at least two numbers in the group, then the sum of any two numbers in this group is divisible by 5." So, if the group has only one number, it's allowed. But the question is asking for the maximum number of numbers in such a group. So, 40 is possible. Is there a way to get more than 40?Wait, perhaps not. Because if we try to include any non-multiple of 5, then we have to include numbers that are inverses mod5. But then, as discussed earlier, we can only include one from each inverse pair. So, for example, if we take one number congruent to 1 mod5 and one congruent to 4 mod5, that's two numbers. Similarly, one congruent to 2 and one to 3. But then, combining these pairs would create cross sums that are not 0 mod5. Therefore, adding these pairs to the 40 multiples of 5 would not work because the cross sums would be non-zero. So, the only way to have more numbers is to stick with the 40 multiples of 5.Alternatively, maybe there's a different structure. Let's think about the residues mod5. There are 5 residues: 0,1,2,3,4.To form a group where any two numbers sum to 0 mod5, the group must be such that for every residue r present in the group, the residue -r must also be present, and no two residues r and s can be present unless s = -r. But if you have more than one pair, then you get cross sums that are not 0. Therefore, the group can be:- All numbers with residue 0.- A combination of one residue and its inverse, but only one number from each.So, for example, if you choose residue 1 and 4, you can take as many numbers as you want as long as you take one from residue 1 and one from residue 4? Wait, no. Because if you take two numbers from residue 1 and two from residue 4, then the sum of two residue 1 numbers is 2 mod5, which is not 0. Therefore, even if you pair them, you can only have one number from each residue. Therefore, the maximum size for such a group is 2. So, 40 vs 2, 40 is better.Hence, the maximum number is 40. But let me check the problem again. It says "from the numbers 1 to 200". Wait, 200 is inclusive. So, numbers from 1 to 200. The multiples of 5 in this range are 5,10,...,200. The count is 200 divided by 5, which is 40. So, 40 numbers. So, 40 is the answer.But wait, let me consider another angle. Suppose we take all numbers congruent to 0 mod5 and one number from residue 1 and 4. Then, the total would be 40 + 2 = 42. But wait, if we do that, then numbers from residue 0 and residue 1 would sum to 1 mod5, which is not 0. Hence, such a group is invalid. Therefore, we cannot combine residue 0 with any other residues.Therefore, the only possible groups are either all residue 0 (40 numbers) or pairs of inverse residues (2 numbers each). Therefore, 40 is the maximum. Therefore, the answer should be 40. But let me check if the problem allows for a larger group by some other construction.Wait, another thought. Suppose we partition the numbers into classes based on their residues mod5. Then, the allowed groups can be either:- All numbers from residue 0.OR- At most one number from each of the pairs (1,4) and (2,3).But as mentioned earlier, combining pairs would lead to cross sums that aren't 0. So, the maximum size when combining multiple pairs would still be limited.Alternatively, suppose we take one number from each of the pairs (1,4) and (2,3), but that would only give us 2 + 2 = 4 numbers. But again, if we take one from each pair, cross sums would be between different pairs, which would be 1 + 2 = 3 mod5, which is not 0. Hence, such a group is invalid.Therefore, the only valid groups are:- All numbers from residue 0 (40 numbers).- One pair (1 and 4) or (2 and 3) (2 numbers each).- Single numbers (but we want maximum, so not relevant).Therefore, 40 is the maximum. But let me verify once again with an example. Suppose we have numbers 5,10,15,...,200. Any two numbers in this group: 5 + 10 = 15, which is divisible by 5. 10 + 15 = 25, divisible by 5. Etc. So, yes, all their sums are divisible by 5.If I tried to add another number, say 1 (which is 1 mod5), then 1 + 5 = 6, which is 1 mod5, not divisible by 5. Hence, invalid. Similarly, adding 2 would cause 2 +5=7 mod5=2, invalid. So, can't add any non-multiples of 5. Therefore, 40 is indeed the maximum.But wait, wait, the problem says "one or several were selected into a separate group". So, the group can have one number, but the condition only applies if there are at least two. So, if the group has only one number, there's no requirement. But since we need the maximum, we need to take as many as possible while satisfying the condition that if there are at least two, any two sum to 5. Therefore, 40 is valid, and adding any more would require adding non-multiples of 5, which would break the condition. Therefore, 40 is the answer.But hold on, just to make sure, maybe there's a way to have more than 40 numbers. Suppose that instead of taking all multiples of 5, we take numbers that are 0 mod5 and some other numbers. But as we saw, adding a non-multiple of 5 would introduce a residue that when added to 0 gives that residue, which is not 0. Hence, impossible.Alternatively, what if we take numbers such that all are congruent to 0 mod5, but also some numbers that are 0 mod5. But that's the same as the original group.Alternatively, maybe considering numbers where their residues when doubled are 0 mod5. But 2r ≡0 mod5 implies r ≡0 mod5, since 2 is invertible modulo5. Hence, again, only multiples of5.Therefore, I think the conclusion is that the maximum number is 40.Final Answerboxed{40}</think>To determine the maximum number of numbers from 1 to 200 that can be grouped such that the sum of any two numbers in the group is divisible by 5, we need to consider the residues of these numbers modulo 5. 1. Key Insight: For any two numbers (a) and (b) in the group, the sum (a + b) must be divisible by 5. This implies that the residues of (a) and (b) modulo 5 must be additive inverses. Specifically, if (a equiv r pmod{5}), then (b equiv -r pmod{5}).2. Residue Analysis: - Residue 0: Numbers congruent to 0 modulo 5 can form a group since their sum will always be 0 modulo 5. - Non-zero Residues: For non-zero residues (r) and (-r), if we include a number with residue (r), we must include numbers with residue (-r). However, including more than one number with the same residue (r) or (-r) will result in sums that are not divisible by 5.3. Count of Numbers Congruent to 0 Modulo 5: - The numbers from 1 to 200 that are multiples of 5 are 5, 10, 15, ..., 200. - The count of these numbers is (leftlfloor frac{200}{5} rightrfloor = 40).4. Verification: - Including any non-multiple of 5 in the group would introduce residues that do not satisfy the condition when summed with existing members of the group. - Combining residues from different pairs (e.g., 1 and 4 with 2 and 3) results in sums that are not divisible by 5.Thus, the maximum number of numbers that can be grouped such that the sum of any two numbers is divisible by 5 is the count of numbers congruent to 0 modulo 5, which is 40.[boxed{40}],<think>Okay, so I have this problem here: From the numbers 1 to 200, I need to select as many as possible into a group such that if there are at least two numbers in the group, then the sum of any two numbers in this group is divisible by 5. The question is asking for the maximum number of numbers that can be in such a group. Hmm, interesting. Let me try to figure this out step by step.First, let's parse the problem again to make sure I understand. We have numbers from 1 to 200. We need to form a group (a subset) of these numbers. The key property is that if there are at least two numbers in the group, then any two numbers from the group must add up to a multiple of 5. So, if the group has two or more numbers, all pairwise sums must be divisible by 5. And we need to find the largest possible size of such a group.Alright, so perhaps modular arithmetic could be useful here since we're dealing with divisibility by 5. Let me recall that if two numbers, say a and b, have the property that a + b is divisible by 5, then their remainders when divided by 5 must add up to 5 or 0. In other words, (a mod 5) + (b mod 5) ≡ 0 mod 5. That means either both a and b are congruent to 0 mod 5, or their remainders are complementary pairs like 1 and 4, or 2 and 3.So, possible pairs of residues mod 5 that sum to 0 mod 5 are:- 0 and 0- 1 and 4- 2 and 3Therefore, if I have a group where all numbers are congruent to 0 mod 5, then any two numbers in that group will add up to 0 mod 5. Similarly, if I have a group with numbers congruent to 1 mod 5 and numbers congruent to 4 mod 5, then any pair within those will sum to 0 mod 5. Similarly for 2 and 3 mod 5.However, if I mix residues that are not complementary, like 1 mod 5 and 2 mod 5, their sum would be 3 mod 5, which isn't divisible by 5, so that's not allowed. Therefore, the group can only contain numbers from a single residue class or a pair of complementary residue classes.Wait, but there's a catch here. For example, if I have numbers that are 1 mod 5 and 4 mod 5, then any two numbers within the group will either be 1+1, 4+4, or 1+4. Let's check each case:- 1 + 1 = 2 mod 5: Not divisible by 5. So that's a problem.- 4 + 4 = 8 ≡ 3 mod 5: Also not divisible by 5.- 1 + 4 = 5 ≡ 0 mod 5: That's good.So if I have both 1s and 4s in the group, then pairs of 1s will sum to 2 mod 5, which is not allowed. Similarly, pairs of 4s will sum to 3 mod 5, which is also not allowed. Therefore, the group can only contain numbers from one residue class or the other, but not both. Because if you mix them, you can have pairs that are within the same residue class, which would produce sums not divisible by 5.Similarly, if you have numbers congruent to 2 mod 5 and 3 mod 5, the same problem occurs. If you have two 2s, their sum is 4 mod 5; two 3s sum to 6 ≡ 1 mod 5. Neither is divisible by 5. So mixing 2s and 3s would also cause problems unless you only have one of each. Wait, but even then, if you have multiple 2s or multiple 3s, their sums would not be divisible by 5.Therefore, the only safe residue classes that can form a group with more than one element are those where all elements are congruent to the same residue mod 5, and that residue is 0 mod 5. Because if all numbers are 0 mod 5, then their sums are 0 + 0 = 0 mod 5, which is okay. So a group consisting entirely of numbers divisible by 5 would satisfy the condition.Alternatively, if we can have a group with just one number, then technically it trivially satisfies the condition because there are not two numbers to check. But since the question is about the maximum possible size, we need to see whether a larger group can be formed by combining different residue classes in a way that all pairwise sums are divisible by 5.Wait, but earlier, I thought that mixing 1 and 4 mod 5 might work, but that actually causes problems unless you have only one of each. Let me confirm. Suppose we have a group with one number that's 1 mod 5 and another that's 4 mod 5. Their sum is 0 mod 5, which is good. If we add another number that's 1 mod 5, then the sum of the two 1s is 2 mod 5, which is not allowed. Similarly, adding another 4 would cause the sum of two 4s to be 3 mod 5. So in order to have a group with multiple numbers from 1 and 4 mod 5, we need to have at most one of each. But that would only give a group of size 2. Similarly for 2 and 3 mod 5.Alternatively, if we have a group with only numbers from 0 mod 5, then we can have multiple numbers since their sums will still be 0 mod 5. So the group size could be as large as the number of numbers divisible by 5 between 1 and 200.Wait, but let's check. The numbers divisible by 5 from 1 to 200 are 5, 10, 15, ..., 200. That's 200 / 5 = 40 numbers. So the maximum group size would be 40 if we take all numbers divisible by 5. However, maybe there's a way to combine residue classes to get a larger group. Wait, but how?Suppose we take all numbers congruent to 0 mod 5 (40 numbers), and also one number from each of the complementary residue pairs, like 1 and 4, or 2 and 3. But if we add a single number from 1 mod 5, then to maintain the property, we need that number to only pair with numbers from 4 mod 5. However, if we only add one number from 1 mod 5 and one from 4 mod 5, then the total group size would be 40 + 1 + 1 = 42. But wait, no, because the numbers from 0 mod 5 can be paired with the 1 or 4 mod 5 numbers. Let's check:If we have a number that's 0 mod 5 and another that's 1 mod 5, their sum is 1 mod 5, which is not divisible by 5. Therefore, such a pair would violate the condition. Therefore, if we include numbers from 0 mod 5 and numbers from 1 mod 5, their sum would be 1 mod 5, which is not allowed. Therefore, the group can either be all numbers from 0 mod 5, or a combination of numbers from 1 and 4 mod 5 (but only one each), or 2 and 3 mod 5 (but only one each). But combining them with 0 mod 5 would not work.Wait, so if we have numbers from 0 mod 5, they can only be paired with other 0 mod 5 numbers. Similarly, numbers from 1 mod 5 can only be paired with 4 mod 5 numbers, but only one of each. Similarly, numbers from 2 mod 5 can only be paired with 3 mod 5 numbers, but only one of each. Therefore, the maximum group sizes would be either 40 (all 0 mod 5), or 2 (one from 1 and one from 4), or 2 (one from 2 and one from 3). Therefore, 40 is the largest possible.But wait, hold on. Let me check again. If we have two numbers from 1 mod 5, their sum is 2 mod 5, which is not allowed. So in order to have a group with numbers from 1 mod 5, we can have at most one number from 1 mod 5, and one number from 4 mod 5. Similarly for 2 and 3. So if we have one number from 1 and one from 4, that's two numbers, and their sum is 0 mod 5. If we have two such pairs, like two numbers from 1 and two from 4, then the sum of two 1s would be 2 mod 5, which is not allowed. So even with two numbers from 1 and two from 4, the group doesn't satisfy the condition. Therefore, the maximum group size for 1 and 4 is two numbers (one from each). Similarly for 2 and 3. So the maximum possible group sizes are either 40 (all 0 mod 5) or 2 (for each complementary pair). Therefore, 40 is bigger.But wait, hold on. Wait, maybe there's a way to have more numbers. Let's think differently. Suppose we take all numbers that are 0 mod 5. Then, as mentioned, that's 40 numbers. Alternatively, suppose we take numbers from 1 mod 5 and 4 mod 5, but in such a way that each number from 1 mod 5 is paired with a number from 4 mod 5, but with only one number from each residue. But that would only allow one pair. However, if we can have multiple pairs, but arranged such that the sums of all pairs are 0 mod 5, but also that sums within the same residue are allowed. Wait, but as we saw, if we have two numbers from 1 mod 5, their sum is 2 mod 5, which is not allowed. Therefore, to have a group containing numbers from 1 and 4 mod 5, we can have at most one number from each residue. Similarly for 2 and 3.Wait, but maybe you can have more numbers if you structure the group in such a way that all numbers are from 0 mod 5, or all numbers are from 1 and 4 mod 5 with only one from each, but then combining them. But as we saw earlier, 0 mod 5 numbers can't be combined with any other residues because 0 + 1 ≡ 1 mod 5, which is not allowed. Therefore, the group must consist solely of numbers from 0 mod 5, or a combination of 1 and 4 mod 5 with only one each, or 2 and 3 mod 5 with only one each. Therefore, the maximum size is 40.But wait, hold on. Wait, if you take one number from 1 mod 5 and one from 4 mod 5, their sum is 0 mod 5. If you take another number from 1 mod 5 and another from 4 mod 5, then the sum of the two 1s is 2 mod 5, which is not allowed. Therefore, you cannot have more than one pair. Similarly for 2 and 3. Therefore, the maximum size for such a group is 2. Therefore, 40 is the maximum.Wait, but let me think again. Suppose you have a group with multiple numbers, all of which are 0 mod 5. Then, any two numbers added together are 0 + 0 = 0 mod 5, which is okay. Therefore, such a group is valid. The size is 40. If you try to combine 0 mod 5 with another residue, say 1 mod 5, then the sum of 0 + 1 = 1 mod 5, which is not allowed. Therefore, you can't mix residues with 0 mod 5.Alternatively, suppose you have a group with numbers from 1 mod 5 and 4 mod 5. If you have two numbers from 1 mod 5, their sum is 2 mod 5, which is invalid. If you have two numbers from 4 mod 5, their sum is 8 mod 5 = 3 mod 5, which is invalid. However, if you have one number from 1 mod 5 and one from 4 mod 5, their sum is 5 mod 5 = 0, which is okay. Therefore, such a pair is valid, but adding any more numbers would cause problems. So you can have a group of size 2 from 1 and 4, or another group of size 2 from 2 and 3, or a group of size 40 from 0 mod 5.But maybe there's a smarter way. For example, if you take one number from each of the pairs (1,4) and (2,3), but then you have to check if all sums are divisible by 5. Let's see: suppose you have a group with one number from 1 mod 5, one from 4 mod 5, one from 2 mod 5, and one from 3 mod 5. Then, the sum of 1 and 2 is 3 mod 5, which is not allowed. So that doesn't work. Therefore, you can't mix different pairs. Therefore, the group can either be all 0s, or one pair from (1,4), or one pair from (2,3). So maximum size is 40.Wait a second, but perhaps if we consider numbers that are all congruent to 0 mod 5, but also include numbers from another residue where adding them to 0 mod 5 gives a multiple of 5. Wait, but 0 + x ≡ x mod 5. So unless x is 0 mod 5, that sum would not be 0. Therefore, you can't include any non-0 mod 5 numbers with 0 mod 5 numbers.Therefore, the conclusion is that the largest possible group is all numbers divisible by 5 between 1 and 200, which is 40 numbers.But wait, is there a way to have a larger group? Let me consider the following idea. Suppose we have a group where every number is congruent to the same residue mod 5. Then, as we saw, if the residue is 0, sums are 0 mod 5. If the residue is non-zero, then the sum of two such numbers would be 2 times the residue mod 5. Therefore, for the sum to be 0 mod 5, 2r ≡ 0 mod 5. That would require r ≡ 0 mod 5, which is only possible if r is 0 mod 5. Therefore, only residue 0 mod 5 allows multiple numbers in the group.Alternatively, if the residue is non-zero, then two numbers with that residue would sum to 2r mod 5. For that to be 0 mod 5, 2r ≡ 0 mod 5 ⇒ r ≡ 0 mod 5, which contradicts r being non-zero. Therefore, only residue 0 mod 5 can have multiple numbers. So that's another way to see why only 0 mod 5 can have a group of more than one number.Therefore, the maximum size group would be the numbers congruent to 0 mod 5. However, let's confirm how many numbers that is. From 1 to 200, numbers divisible by 5: 5, 10, 15, ..., 200. The number of terms is 200 / 5 = 40. So 40 numbers.But wait, here's a thought. Maybe if we take numbers from multiple residues such that all possible sums are divisible by 5. For example, if we take numbers that are all congruent to 1 mod 5 and 4 mod 5, but only one number from each residue. Wait, but if you take more than one number from each residue, then their sum within the same residue would not be divisible by 5. So you can only have at most one number from residue 1 and one from residue 4, making a group of size 2. Similarly for residues 2 and 3. So combining these two pairs, you get a group of size 4. Wait, but if we have one number from 1, one from 4, one from 2, one from 3, then their pairwise sums would be 1+4=5, 1+2=3, 1+3=4, 4+2=6≡1, 4+3=7≡2, 2+3=5. So some of these sums are not divisible by 5. Therefore, such a group would not satisfy the condition. Therefore, combining pairs from different complementary residues is not allowed.Therefore, the maximum size for a group not consisting of 0 mod 5 is 2. So 40 versus 2. Therefore, 40 is larger.But wait, perhaps if all numbers in the group are congruent to 0 mod 5, that gives 40. Alternatively, maybe a group can consist of numbers all congruent to 0 mod 5 plus some other numbers. Wait, but as before, adding any non-0 mod 5 numbers would result in sums not divisible by 5. For example, 0 + 1 = 1 mod 5, which is not allowed. Therefore, the only safe group is all 0 mod 5.Therefore, the maximum size is 40. Hmm, but let me check the problem statement again to make sure I didn't misinterpret anything."From the numbers 1 to 200, one or several were selected into a separate group with the following property: if there are at least two numbers in the group, then the sum of any two numbers in this group is divisible by 5. What is the maximum number of numbers that can be in a group with this property?"So the problem is correctly restated. The key point is that if there are at least two numbers, then any two must sum to a multiple of 5. So if the group has two or more numbers, all pairwise sums must be divisible by 5. Therefore, the group can be of size 1 (trivially), or if larger, must satisfy the divisibility condition.So the maximum possible group is either all numbers congruent to 0 mod 5 (40 numbers) or a combination of residues where all pairs sum to 0 mod 5, but as we saw, combining residues leads to smaller groups.Wait, but here's another angle: suppose we take numbers that are all congruent to 0 mod 5. Then, as mentioned, that's 40 numbers. Alternatively, what if we take numbers that are all congruent to 1 mod 5. Then, the sum of any two numbers would be 2 mod 5. But 2 mod 5 is not 0, so that's invalid. Similarly for any other non-zero residue. So only 0 mod 5 works for multiple numbers.Alternatively, is there a different structure where numbers are arranged such that every pair sums to 0 mod 5, but not necessarily all being 0 mod 5. For example, if all numbers are either 1 mod 5 or 4 mod 5, but in such a way that each number's complement is present. However, as discussed earlier, if you have more than one number from each residue, their sum would not be 0 mod 5. Therefore, you can only have one from each residue. So group size 2.Similarly for 2 and 3 mod 5. So, if we have a group with one number from 1 mod 5 and one from 4 mod 5, that's a group of size 2. Similarly, one from 2 and one from 3. So two groups of size 2. But combining those into a single group would lead to sums that are not divisible by 5, as we saw earlier. Therefore, such a combined group is invalid.Therefore, the maximum group size is indeed 40.Wait, but let's confirm with an example. Suppose we take all numbers divisible by 5. Let's say 5, 10, 15, ..., 200. There are 40 numbers. Then, the sum of any two is 5a + 5b = 5(a + b), which is divisible by 5. So that works. Therefore, such a group is valid and has 40 numbers.Alternatively, if we take one number from 1 mod 5 and one from 4 mod 5, their sum is 5, which is divisible by 5. But adding another number, say another 1 mod 5, would result in a sum of 1 + 1 = 2 mod 5, which is not allowed. Therefore, group size can't exceed 2 in that case.Therefore, 40 is the maximum. But wait, let me check another way. Suppose we consider that numbers can be in different residue classes, but each residue class can contribute at most one number. For instance, if we take one number from each residue class 0,1,2,3,4 mod 5, then the sum of any two numbers would be the sum of their residues mod 5. But unless the residues are complementary, the sum would not be 0 mod 5. Therefore, such a group would have multiple pairs that don't sum to 0 mod 5. Therefore, that approach doesn't work.Alternatively, if we take multiple numbers from residue 0 mod 5, which is allowed, and no numbers from other residues. Then, the group is valid. Therefore, 40 numbers.Therefore, I think the answer is 40. However, let me check if there's a smarter combination.Wait, here's another thought. Suppose we take all numbers congruent to 0 mod 5, which is 40 numbers, and also take one number from each of the pairs (1,4) and (2,3). But wait, even if we take one number from each pair, adding them to the group would create sums with the 0 mod 5 numbers. For example, 0 + 1 = 1 mod 5, which is not allowed. Therefore, we cannot add any numbers from other residue classes to the 0 mod 5 group. Therefore, the group must remain purely 0 mod 5.Alternatively, if we try to create a group without any 0 mod 5 numbers. So, suppose we take as many numbers as possible from the other residues. But since residues 1 and 4 need to be paired, and residues 2 and 3 need to be paired, but each pair can only contribute one number. Wait, no. Wait, if you take one number from residue 1 and one from residue 4, that gives a group of size 2 where all sums are 0 mod 5. Similarly, taking one from residue 2 and one from residue 3 gives another group of size 2. If you combine these two groups into one, you have a group of size 4, but then the sum of 1 and 2 is 3 mod 5, which is invalid. Therefore, such a combination is not allowed.Therefore, the only way to have a group larger than size 2 is to take all numbers from residue 0 mod 5. Hence, 40 numbers.Therefore, the answer should be 40. However, let me check again with some examples.Suppose we have a group with two numbers: 1 and 4. Then, 1 + 4 = 5, which is good. If we try to add another number, say 6 (which is 1 mod 5), then 1 + 6 = 7 ≡ 2 mod 5, which is not allowed. Similarly, adding 9 (4 mod 5) would result in 4 + 9 = 13 ≡ 3 mod 5. Therefore, adding any other number to the {1,4} group breaks the condition.Similarly, with the {2,3} group, adding another number either from 2 or 3 mod 5 would create sums that aren't divisible by 5. Therefore, the maximum size for such groups is 2.On the other hand, the group with all numbers divisible by 5 can have 40 numbers, and all pairwise sums are divisible by 5.Therefore, unless there's a way to have a group with more than 40 numbers by some other construction, the answer is 40.But let me check how many numbers there are in each residue class:- Numbers congruent to 0 mod 5: 40 (as 5*1, 5*2, ..., 5*40)- Numbers congruent to 1 mod 5: From 1, 6, 11, ..., 196. Let's compute how many: The first number is 1, the last is 196. The formula for the number of terms is ((last - first)/5) + 1 = ((196 - 1)/5) + 1 = (195/5) + 1 = 39 + 1 = 40. Wait, so there are 40 numbers congruent to 1 mod 5.- Similarly, numbers congruent to 2 mod 5: 2,7,12,...,197. Similarly, ((197 - 2)/5) +1 = (195/5)+1=39+1=40.- Numbers congruent to 3 mod 5: 3,8,13,...,198. ((198 - 3)/5)+1=(195/5)+1=40.- Numbers congruent to 4 mod 5: 4,9,14,...,199. ((199 -4)/5)+1=(195/5)+1=40.Wait, so each residue class from 0 to 4 mod 5 has exactly 40 numbers in 1-200. Therefore, residue 0 mod 5 has 40 numbers, residues 1,2,3,4 each have 40 numbers as well.But earlier reasoning indicated that we can have a group of 40 numbers from residue 0 mod 5, or groups of size 2 from residues 1 and 4 or 2 and 3. Therefore, 40 is the maximum. However, wait a second. If each residue class (1,4) and (2,3) has 40 numbers, but when forming groups from these, we can only take one from each. Wait, no. If we take one number from residue 1 and one from residue 4, that's a group of size 2. But since there are 40 numbers in residue 1 and 40 in residue 4, can we pair them up into 40 different groups each of size 2? But no, because the problem is about forming a single group, not multiple groups. So the question is about one group with the given property.Therefore, even though there are 40 numbers in each residue class, the group can't take more than one from each complementary residue pair because of the sum condition. Therefore, the maximum size of a group with numbers from residues 1 and 4 is 2, and similarly for 2 and 3.Therefore, the maximum possible group is 40 numbers from residue 0 mod 5.But here's another thought. Suppose we take all numbers that are congruent to 0 mod 5. That's 40 numbers, and all their pairwise sums are 0 mod 5. That works. Alternatively, suppose we take all numbers congruent to 1 mod 5. Then, any two of them sum to 2 mod 5, which is not allowed. So that doesn't work. Similarly, taking all numbers congruent to 2 mod 5 gives sums of 4 mod 5, which is invalid. Same for 3 and 4. Therefore, only 0 mod 5 allows multiple numbers.Alternatively, is there a way to have a group with all numbers from 0 mod 5 plus one number from another residue? But as mentioned earlier, adding a non-0 mod 5 number would create a pair with that number and any 0 mod 5 number, resulting in a sum not divisible by 5. Therefore, that's invalid.Therefore, the conclusion is that the maximum number of numbers is 40.But wait, let me check another angle. Suppose we use the pigeonhole principle. There are 5 residue classes. If we have more than 40 numbers, then by the pigeonhole principle, at least one residue class has more than 8 numbers (since 200 /5=40). But wait, actually, in the case of residue 0 mod 5, there are 40 numbers. So if we have 41 numbers, then even if we take all 40 from residue 0, we need one more, which must come from another residue. But that extra number would create a pair with a 0 mod 5 number, resulting in a sum not divisible by 5. Therefore, 41 is impossible. Therefore, the maximum is 40.Therefore, the answer is 40.Final AnswerThe maximum number of numbers that can be in such a group is boxed{40}.</think>To solve the problem of finding the maximum number of numbers from 1 to 200 that can form a group where the sum of any two numbers is divisible by 5, we use modular arithmetic.1. Key Insight: For any two numbers (a) and (b) in the group, (a + b) must be divisible by 5. This means their remainders when divided by 5 must add up to 0 modulo 5. The valid pairs of remainders are (0, 0), (1, 4), and (2, 3).2. Residue Classes Analysis: - 0 mod 5: If all numbers are congruent to 0 mod 5, their sums will also be 0 mod 5. This allows multiple numbers in the group. - 1 mod 5 and 4 mod 5: Pairing these residues works, but only one number from each residue can be included to avoid sums like 1 + 1 or 4 + 4, which are not divisible by 5. This limits the group size to 2. - 2 mod 5 and 3 mod 5: Similarly, only one number from each residue can be included, limiting the group size to 2.3. Count of Numbers in Each Residue Class: - Numbers congruent to 0 mod 5 between 1 and 200: (5, 10, 15, ldots, 200). There are (200 / 5 = 40) such numbers. - Other residues (1, 2, 3, 4) each have 40 numbers, but they can only form groups of size 2 if paired with their complementary residues.4. Conclusion: The largest group is formed by taking all numbers congruent to 0 mod 5, which results in 40 numbers. Combining other residues does not yield a larger group due to the constraints on their sums.Thus, the maximum number of numbers that can be in such a group is (boxed{40}).]

🔑:Answer: 40.Solution. Suppose that the group selected a number A, which gives a remainder i neq 0 when divided by 5. If there is another number B in the group, then it must give a remainder 5-i when divided by 5, so that A+B is divisible by 5. We will show that there cannot be any other numbers in this group. Suppose there is another number C, and j=C bmod 5. Then the numbers i+j and 5-i+j are divisible by 5, as well as their difference, which is 2 i-5, which is impossible for i neq 0.Now consider the alternative case, where all numbers in the group are divisible by 5. In the given range, there are exactly 40 numbers that are multiples of 5, and all these numbers can be included in the group simultaneously.

🧠:[<think>Okay, let's see. I need to find the range of real numbers m such that proposition p is true and proposition q is false. Hmm. Let's start by understanding each proposition separately and then combine the conditions.First, proposition p states that the function f(x) = (1/3)x³ + x² + mx + 1 has two distinct extreme value points. For a function to have extreme points, its derivative must have real roots, and those roots must be distinct. So, I need to find the derivative of f(x) and ensure that it has two distinct real roots.Calculating the derivative f'(x):f'(x) = d/dx [ (1/3)x³ + x² + mx + 1 ] = x² + 2x + m.So, the derivative is a quadratic function: f'(x) = x² + 2x + m. For this quadratic to have two distinct real roots, its discriminant must be positive. The discriminant D of ax² + bx + c is D = b² - 4ac. Here, a = 1, b = 2, c = m. So,D = (2)² - 4*1*m = 4 - 4m.For two distinct real roots, D > 0:4 - 4m > 0Subtract 4 from both sides:-4m > -4Divide both sides by -4, remembering to reverse the inequality sign when dividing by a negative number:m < 1.So, proposition p is true when m < 1.Now, moving on to proposition q: The function f(x) = x² - mx + 3 is monotonically decreasing in the interval [-1, 2]. Wait, f(x) here is a quadratic function. For a quadratic function to be monotonically decreasing on an interval, its derivative must be non-positive (i.e., ≤ 0) throughout that interval.First, compute the derivative of f(x):f'(x) = 2x - m.So, f'(x) = 2x - m. For f(x) to be monotonically decreasing on [-1, 2], f'(x) ≤ 0 for all x in [-1, 2]. Therefore, we need 2x - m ≤ 0 for all x ∈ [-1, 2].To ensure that the derivative is non-positive on the entire interval, the maximum of the derivative on the interval must be ≤ 0. The derivative f'(x) = 2x - m is a linear function in x. Since the coefficient of x is positive (2), the derivative is increasing with x. Therefore, the maximum of f'(x) on the interval [-1, 2] occurs at the right endpoint, x = 2.So, plug in x = 2 into f'(x):f'(2) = 2*(2) - m = 4 - m.For the function to be monotonically decreasing on the entire interval, f'(2) ≤ 0:4 - m ≤ 0Which implies:m ≥ 4.Therefore, proposition q is true when m ≥ 4.But the problem states that proposition q is false. So, the negation of q is when the function is not monotonically decreasing in [-1, 2]. That would happen when m < 4. Because if m is less than 4, then f'(2) = 4 - m > 0, which means that at x = 2, the derivative is positive, so the function is increasing there, hence not monotonically decreasing on the entire interval.Therefore, proposition q is false when m < 4.Now, combining the two conditions:- p is true: m < 1- q is false: m < 4So, the intersection of these two conditions is m < 1. Because if m < 1, then automatically m < 4 as well. Hence, the range of m is all real numbers less than 1.Wait, but let me double-check. Is there a possibility that even if m < 4, the function might still be non-decreasing somewhere else in the interval? For example, maybe the derivative is positive somewhere in the middle, but negative at other parts. But for the function to be monotonically decreasing on the entire interval, the derivative must be non-positive everywhere in [-1, 2]. If the derivative is positive at any point in the interval, then the function is not monotonically decreasing there. Therefore, as the derivative is increasing (since it's linear with a positive slope), the maximum occurs at x=2. So, if even at x=2 the derivative is non-positive, then the entire interval is okay. If at x=2 the derivative is positive, then the function is increasing there, hence not monotonically decreasing. Therefore, proposition q being false is equivalent to the existence of at least one point in [-1, 2] where the derivative is positive, which is equivalent to m < 4.Therefore, the logic holds: q is false when m < 4.Hence, combining p (m <1) and q is false (m <4), we take the intersection m <1.Therefore, the range of m is (-∞, 1).But let me check again. Suppose m is between 1 and 4. Then p is false, but q is also false (since m <4). However, the problem states that p is true and q is false. Therefore, in this case, m has to satisfy both p is true (m <1) and q is false (m <4). So, as m must satisfy both, the intersection is m <1.Therefore, the final answer is m <1, which in interval notation is (-∞, 1).Wait, just to confirm: if m is less than 1, then p is true (since discriminant is positive) and q is false (since m <4). So that's correct.Alternatively, is there a case where even if m <4, but m ≥ something else, q could still be true? For example, suppose m is 3. Then f'(x) = 2x -3. Let's check if f is monotonically decreasing on [-1,2].At x = -1: f'(-1) = -2 -3 = -5 ≤0At x = 2: f'(2) = 4 -3 =1 >0So, at x=2, the derivative is positive, so the function is increasing there. Hence, the function is not monotonically decreasing on the entire interval. Therefore, q is false for m=3. Similarly, if m=2, then f'(2)=4 -2=2>0, so again, q is false. If m=0, then f'(x)=2x, which is negative for x <0 and positive for x>0. So in the interval [-1,2], the derivative goes from -2 at x=-1 to 4 at x=2. So, the function is decreasing from x=-1 to x=0, then increasing from x=0 to x=2. Hence, not monotonically decreasing throughout. Therefore, q is false for any m <4. If m=4, f'(x)=2x -4. Then at x=2, f'(2)=0. The derivative is increasing, starting at x=-1: f'(-1)= -2 -4= -6, then increases to 0 at x=2. So, the derivative is non-positive over the entire interval. Hence, f is monotonically decreasing on [-1,2] when m=4. Therefore, q is true when m ≥4, and false otherwise. Therefore, my previous conclusion is correct: q is false when m <4.Therefore, combining p (m <1) and q is false (m <4), the range is m <1. So the answer is m ∈ (-∞, 1). So boxed, it would be boxed{(-infty, 1)}.But let me check once more for any possible mistakes. Suppose m=1. Then for proposition p, discriminant is 4 -4*1=0. So the derivative would have a repeated root, so f(x) would have a stationary point but not two distinct extreme points. Hence, p is false when m=1, so p is true only when m <1. And q is false when m <4. So overlapping conditions, m must be less than 1. Therefore, yes, the answer is correct.Final AnswerThe range of real number ( m ) is boxed{(-infty, 1)}.</think>Given propositions ( p ) and ( q ):- Proposition ( p ): The function ( f(x) = frac{1}{3}x^3 + x^2 + mx + 1 ) has two distinct extreme value points.- Proposition ( q ): The function ( f(x) = x^2 - mx + 3 ) is a monotonically decreasing function in the interval ([-1, 2]).To determine the range of ( m ) where ( p ) is true and ( q ) is false:1. Analyzing Proposition ( p ): - The derivative ( f'(x) = x^2 + 2x + m ). - For ( f(x) ) to have two distinct extreme points, the quadratic ( f'(x) ) must have two distinct real roots. - The discriminant of ( f'(x) ) is ( 4 - 4m ). For distinct roots, ( 4 - 4m > 0 ) leading to ( m < 1 ).2. Analyzing Proposition ( q ): - The derivative ( f'(x) = 2x - m ). - For ( f(x) ) to be monotonically decreasing on ([-1, 2]), ( f'(x) leq 0 ) for all ( x ) in the interval. - The maximum value of ( f'(x) ) on ([-1, 2]) occurs at ( x = 2 ). Thus, ( 4 - m leq 0 ) leading to ( m geq 4 ). - Proposition ( q ) is false when ( m < 4 ).3. Combining Conditions: - ( p ) is true when ( m < 1 ). - ( q ) is false when ( m < 4 ). - The intersection of these conditions is ( m < 1 ).Thus, the range of real number ( m ) is (boxed{(-infty, 1)}).,<think>Okay, let's try to tackle this problem step by step. So, we have two propositions, p and q, and we need to find the range of real number m such that p is true and q is false. Let's break it down.First, let's understand proposition p. It states that the function f(x) = (1/3)x³ + x² + mx + 1 has two distinct extreme value points. For a function to have extreme points, its derivative must have real roots, and those roots need to be distinct. So, let me start by finding the derivative of f(x).The derivative of f(x) with respect to x is f'(x) = d/dx [(1/3)x³ + x² + mx + 1]. Calculating term by term:- The derivative of (1/3)x³ is x².- The derivative of x² is 2x.- The derivative of mx is m.- The derivative of the constant 1 is 0.So, putting it all together, f'(x) = x² + 2x + m.Now, for f(x) to have two distinct extreme points, the derivative f'(x) must have two distinct real roots. A quadratic equation ax² + bx + c = 0 has two distinct real roots if its discriminant D is positive. The discriminant D is given by D = b² - 4ac.In our case, the quadratic is x² + 2x + m, so a = 1, b = 2, c = m. Therefore, the discriminant D = (2)² - 4(1)(m) = 4 - 4m. For two distinct real roots, we need D > 0:4 - 4m > 0 => 4 > 4m => 1 > m => m < 1.So, proposition p is true when m < 1.Now, moving on to proposition q. It states that the function f(x) = x² - mx + 3 is a monotonically decreasing function in the interval [-1, 2]. For a function to be monotonically decreasing on an interval, its derivative must be non-positive (less than or equal to zero) throughout that interval.First, let's find the derivative of f(x):f'(x) = d/dx [x² - mx + 3] = 2x - m.So, the derivative is a linear function: 2x - m. For f(x) to be monotonically decreasing on [-1, 2], we need 2x - m ≤ 0 for all x in [-1, 2].This inequality must hold for the entire interval. To ensure that, the maximum value of 2x - m on the interval must be ≤ 0. Since 2x - m is a linear function increasing with x (because the coefficient of x is positive), its maximum occurs at the right endpoint of the interval, which is x = 2.Therefore, substituting x = 2 into the inequality:2(2) - m ≤ 0 => 4 - m ≤ 0 => -m ≤ -4 => m ≥ 4.So, proposition q is true when m ≥ 4. However, the problem states that proposition q is false. Therefore, the negation of q is true, which means that the function f(x) = x² - mx + 3 is not monotonically decreasing in the interval [-1, 2]. That happens when m < 4.Therefore, combining the conditions for p being true (m < 1) and q being false (m < 4), we need the intersection of these two conditions. Since m < 1 is more restrictive than m < 4, the combined condition is m < 1.Wait a second, hold on. Let me double-check that.Proposition q being false means that the function is not monotonically decreasing on [-1, 2]. So, even if the function is decreasing on some parts of the interval but not all, then q is false. So, maybe my reasoning was too simplistic? Let me verify.The derivative f'(x) = 2x - m. For the function to be monotonically decreasing on the entire interval, the derivative must be ≤ 0 for all x in [-1, 2]. If the derivative is positive at any point in that interval, then the function is not monotonically decreasing there, hence q is false.Therefore, q is false if there exists at least one x in [-1, 2] such that f'(x) > 0. Since f'(x) is increasing (as 2x - m is increasing with x), the maximum of f'(x) on [-1, 2] is at x=2, as before. Therefore, if the maximum is positive, then q is false. The maximum value is 4 - m. So, if 4 - m > 0, then q is false. Therefore, m < 4.Therefore, q is false when m < 4, and true when m ≥ 4. So, the previous conclusion holds.Therefore, combining p being true (m < 1) and q being false (m < 4), the range of m is m < 1. So, the range is (-∞, 1).But let me check if there's any overlap or something I missed. For example, could there be a case where even if m < 4, the derivative is still non-positive over the interval? Wait, but if m < 4, then 4 - m > 0, so at x=2, the derivative is positive, hence the function is increasing at x=2, hence not monotonically decreasing over the entire interval. Therefore, q is false exactly when m < 4.Therefore, combining m < 1 (p is true) and m < 4 (q is false), we get m < 1. So, the range is all real numbers less than 1.But let me check with specific examples. Suppose m = 0, which is less than 1. Then for proposition p, the derivative is x² + 2x, which factors to x(x + 2). So, critical points at x=0 and x=-2, which are distinct. So, p is true. For q, the function is x² + 3. Its derivative is 2x. In the interval [-1, 2], the derivative goes from -2 to 4. So, at x=2, derivative is 4 > 0, so the function is increasing there, hence not monotonically decreasing. So, q is false. So, m=0 is valid.Another example: m=2. Wait, m=2 is less than 4 but greater than 1. Wait, but m=2 would make p false, because p requires m < 1. So m=2 is not in the solution set. Wait, so even though m=2 would make q false (since 2 < 4), p is false here. So, the overlapping condition is m < 1.So, yeah, the answer should be m < 1. Therefore, the range is (-∞, 1). So, boxed, that would be boxed{(-infty, 1)}.But let me just check another angle. Suppose m=1. Then for p, the derivative is x² + 2x + 1 = (x + 1)^2, which has a repeated root at x=-1. So, only one extreme point, hence p is false. So, m=1 is excluded. Hence, m must be strictly less than 1.For q, if m=4, then the derivative is 2x -4. In the interval [-1, 2], the derivative at x=2 is 0. The derivative at x=-1 is -2 -4 = -6. So, at x=2, derivative is 0, which is non-positive. However, since the derivative is 0 at x=2, but negative before that. Wait, does a function with derivative non-positive (i.e., ≤ 0) over the interval count as monotonically decreasing? Yes, because a function is monotonically decreasing if it's non-increasing. So, if the derivative is ≤ 0, even if it's zero at some points, it's still considered monotonically decreasing.Wait, so maybe my previous reasoning is flawed. Let me check the exact definition. A function is monotonically decreasing on an interval if for any two points x1 < x2 in the interval, f(x1) ≥ f(x2). If the derivative is ≤ 0 everywhere on the interval, then the function is non-increasing (monotonically decreasing). If the derivative is negative everywhere, it's strictly decreasing. But some sources use "monotonically decreasing" to include non-increasing functions where the derivative can be zero at some points.So, if m=4, then the derivative 2x -4. Let's see on the interval [-1, 2]. The derivative at x=2 is 0. At x=1.5, it's 3 -4 = -1. At x=1, 2 -4 = -2. At x=0, 0 -4 = -4. At x=-1, -2 -4 = -6. So, the derivative is negative on [-1, 2) and zero at x=2. So, the function is decreasing on the entire interval, with zero derivative at the endpoint. Therefore, f(x) is monotonically decreasing on [-1, 2] when m=4.Therefore, proposition q is true when m=4. Therefore, the previous conclusion that q is false when m < 4 is correct. So, m=4 is the threshold. So, m >=4, q is true; m <4, q is false.Hence, combining p is true (m <1) and q is false (m <4), the intersection is m <1. So, the answer should be m ∈ (-∞, 1). Therefore, boxed as boxed{(-infty, 1)}.Final AnswerThe range of real number ( m ) is boxed{(-infty, 1)}.</think>To determine the range of real number ( m ) such that proposition ( p ) is true and proposition ( q ) is false, we analyze each proposition separately and then combine the results.Proposition ( p ): The function ( f(x) = frac{1}{3}x^3 + x^2 + mx + 1 ) has two distinct extreme value points. - The first derivative of ( f(x) ) is ( f'(x) = x^2 + 2x + m ).- For ( f(x) ) to have two distinct extreme points, the quadratic equation ( x^2 + 2x + m = 0 ) must have two distinct real roots. This requires the discriminant to be positive: [ (2)^2 - 4(1)(m) > 0 implies 4 - 4m > 0 implies m < 1. ]- Therefore, proposition ( p ) is true when ( m < 1 ).Proposition ( q ): The function ( f(x) = x^2 - mx + 3 ) is a monotonically decreasing function in the interval ([-1, 2]).- The first derivative of ( f(x) ) is ( f'(x) = 2x - m ).- For ( f(x) ) to be monotonically decreasing on ([-1, 2]), the derivative ( f'(x) ) must be non-positive for all ( x ) in the interval. The maximum value of ( f'(x) ) on ([-1, 2]) occurs at ( x = 2 ): [ 2(2) - m leq 0 implies 4 - m leq 0 implies m geq 4. ]- Therefore, proposition ( q ) is true when ( m geq 4 ). For ( q ) to be false, ( m ) must be less than 4 (( m < 4 )).Combining the conditions for ( p ) being true (( m < 1 )) and ( q ) being false (( m < 4 )), the intersection of these conditions is ( m < 1 ).Thus, the range of real number ( m ) is (boxed{(-infty, 1)}).]

🔑:When p is true:f'(x) = x^2 + 2x + mΔ = 4 - 4m > 0∴ m < 1When q is true:m ≥ 4So, when ¬q is true:m < 4From begin{cases} overset{m<1}{m<4}end{cases} we get: m < 1∴ The range of real number m is (-∞, 1).Hence, the final answer is boxed{m in (-infty, 1)}.

🧠:[<think>Okay, so I have this problem about a cube with edge length 1, and the vertices are labeled A through H. The problem is in two parts. First, I need to determine all possible positions of the vertex A_{2n} after moving 2n steps starting from vertex A. Second, I have to find the number of such sequences that end at vertex C. Let me start with part (1).The cube has 8 vertices, and each vertex is connected to three adjacent vertices. Since we start at A, which is a corner of the cube, each step moves to one of the three adjacent vertices. The key here is that each move goes from the current vertex to an adjacent one, so it's like a walk along the edges of the cube.For part (1), we need to figure out where we can end up after 2n steps. Since the cube is a bipartite graph, right? Because in a cube, vertices can be colored with two colors such that no two adjacent vertices share the same color. For example, if we color all the vertices alternately black and white, like a checkerboard. So starting from a black vertex (say A), each step moves to a white vertex, then to a black vertex, and so on. So after an even number of steps (2n), we should be back on a vertex of the same color as the starting vertex. Since the cube is bipartite, this is a fundamental property.So first, let's confirm that the cube is indeed bipartite. Each edge connects a vertex of one color to another, so yes. Therefore, after even steps, we must be on the same color as the starting vertex. The starting vertex A is one color, so all vertices reachable in even steps must be of the same color.Therefore, the possible positions of A_{2n} are all vertices of the cube that are the same color as A in this bipartition. Let me recall that in a cube, the bipartition splits the 8 vertices into two sets of 4. Each set forms a tetrahedron, I think. So the cube's vertices can be divided into two sets where each set has no edges among themselves.Given that, starting from A, after an even number of steps, we must be on one of the four vertices in the same partition as A. Let me check which vertices are in the same partition as A. The cube's vertices can be labeled as follows: Let's assume A is (0,0,0). Then the vertices can be labeled with coordinates (x,y,z) where each coordinate is 0 or 1. Then the bipartition is based on the parity of the number of 1s. Wait, but actually, in a cube (as a graph), the bipartition is actually based on the sum of coordinates. If the sum is even or odd. For example, (0,0,0) has sum 0 (even), and adjacent vertices differ by one coordinate, so their sum is odd. Therefore, the two partitions are:- Even: (0,0,0), (0,1,1), (1,0,1), (1,1,0)- Odd: (0,0,1), (0,1,0), (1,0,0), (1,1,1)Wait, but the problem states the cube is labeled A, B, C, D, E, F, G, H. Let me check if I can figure out which vertices correspond to which coordinates. Let me assume that the cube is labeled such that A is (0,0,0), B is (1,0,0), D is (0,1,0), C is (1,1,0), E is (0,0,1), F is (1,0,1), G is (1,1,1), H is (0,1,1). Wait, different sources might label the cube differently, but let's just take a standard labeling. Let me confirm:In a cube, moving along edges, each vertex is connected to three others. Let's fix A as (0,0,0). Then the adjacent vertices would be B (1,0,0), D (0,1,0), and E (0,0,1). Then from B (1,0,0), adjacent vertices are A, C (1,1,0), and F (1,0,1). Similarly, from D (0,1,0), adjacent are A, C (1,1,0), and H (0,1,1). From E (0,0,1), adjacent are A, F (1,0,1), and H (0,1,1). Then C is connected to B, D, G (1,1,1); F is connected to B, E, G; H is connected to D, E, G; and G is connected to C, F, H.So in this labeling, the even vertices (sum of coordinates even) are A (0,0,0), C (1,1,0), F (1,0,1), H (0,1,1). The odd vertices are B (1,0,0), D (0,1,0), E (0,0,1), G (1,1,1). Therefore, starting at A (even), after an even number of steps, we must be at one of the even vertices: A, C, F, H. So for part (1), the possible positions of A_{2n} are the vertices A, C, F, H.Wait, but the problem states "given a cube ABCD EFGH", so perhaps the labeling is different? Wait, in the problem statement, it says "given a cube ABCD EFGH". Let me check standard cube labelings.Usually, in a cube, you have two square bases ABCD and EFGH, where A is connected to E, B to F, etc. So if ABCD is the bottom face and EFGH is the top face, with A directly below E, B below F, etc. Then the edges are AB, BC, CD, DA on the bottom; EF, FG, GH, HE on the top; and vertical edges AE, BF, CG, DH.In this case, the coordinates can be assigned as follows: Let A be (0,0,0), B (1,0,0), C (1,1,0), D (0,1,0), E (0,0,1), F (1,0,1), G (1,1,1), H (0,1,1). So in this case, the even vertices (sum of coordinates even) would be A (0,0,0), C (1,1,0), F (1,0,1), H (0,1,1). The odd vertices are B (1,0,0), D (0,1,0), E (0,0,1), G (1,1,1). So same as before.Therefore, starting from A, after an even number of steps, we can only be at A, C, F, or H. Therefore, the answer to part (1) is that A_{2n} must be one of the vertices A, C, F, H.But wait, is this always the case regardless of the path? Since in a bipartite graph, any walk of even length from a vertex remains in the same partition. Therefore, yes, regardless of the path taken, after even steps, you must be in the same partition. Therefore, part (1)'s answer is that A_{2n} can be A, C, F, or H.Now, moving on to part (2). We need to find S_{2n}, the number of sequences of 2n+1 vertices starting at A and ending at C after 2n steps. So, it's the number of walks of length 2n from A to C on the cube graph.To compute S_{2n}, we can model this using linear algebra or recurrence relations. Since the cube is a regular graph, each vertex has degree 3. However, the number of walks between two vertices can be determined by considering the adjacency matrix.Let me recall that if we have the adjacency matrix M of a graph, then the number of walks of length k from vertex u to vertex v is the (u, v) entry of M^k. So, perhaps we can compute M^k where M is the adjacency matrix of the cube, and then look at the entry corresponding to A to C.Alternatively, since the cube is a bipartite graph, we can use some properties of bipartite graphs to find the number of walks. Alternatively, maybe set up a system of recurrence relations based on the distance from the target vertex.Alternatively, since the cube's graph can be represented with each vertex connected to three others, we can model the number of walks using states. Let me think.First, note that the cube is a distance-regular graph. The distance from A to C is 2 (since you can go A -> B -> C or A -> D -> C, etc.). The diameter of the cube is 3, since the maximum distance between two vertices is 3 (e.g., A to G).But for our purposes, maybe we can categorize the vertices based on their distance from A. Let's see:- Distance 0: A (only itself)- Distance 1: B, D, E (adjacent to A)- Distance 2: C, F, H (each is two steps away from A)- Distance 3: G (the opposite vertex, three steps away)Similarly, from the perspective of vertex C, which is distance 2 from A.But since the cube is symmetric, maybe we can use some symmetry to model the number of walks.Alternatively, let's model the number of walks from A to C in 2n steps. Let's denote:Let’s define variables:Let’s denote:- a_k: number of walks of length k ending at A.- b_k: number of walks of length k ending at a vertex adjacent to A (i.e., B, D, E).- c_k: number of walks of length k ending at a vertex two steps away from A (i.e., C, F, H).- d_k: number of walks of length k ending at G (distance 3 from A).But since we start at A, the initial conditions are:At k = 0: a_0 = 1, b_0 = 0, c_0 = 0, d_0 = 0.Then, for each step, we can write recurrence relations based on the adjacency.From A (a_k), the next step can go to B, D, or E, so each contributes to b_{k+1}.From a vertex adjacent to A (B, D, E), which are counted in b_k. From such a vertex, you can go back to A, or to two other vertices. For example, from B, you can go back to A, or go to C or F. Similarly for D and E. So from each b_k, the next step can go back to A (contributing to a_{k+1}), or go to two vertices at distance 2 from A (contributing to c_{k+1}).Therefore, from each of the three vertices in b_k, each can go back to A (1 way) or to two other vertices (2 ways). Therefore, the recurrence would be:a_{k+1} = b_k * 1 (from each of the three adjacent vertices, but wait, each step from a vertex in b_k can go back to A. Wait, but each vertex in b_k (B, D, E) is connected to A, so from each of them, moving back to A is 1 way. However, since b_k counts the total number of walks ending at any of B, D, E, then the number of walks that go from B, D, or E back to A is b_k * 1, but since each of those vertices can only go back to A once. Wait, actually, the total number of ways to go from any of the B, D, E to A is 3 * (number of walks ending at each B, D, E). Wait, but since b_k is the total number of walks ending at B, D, or E. Each such walk can go back to A in 1 way, so total a_{k+1} = b_k.Wait, no. Wait, if b_k is the total number of walks ending at any of B, D, E, then each such walk is at a specific vertex (B, D, or E). From each of those vertices, there is 1 way back to A. Therefore, the total number of walks going from B, D, E to A is equal to the sum over each walk ending at B, D, E multiplied by 1. Therefore, a_{k+1} = b_k * 1. But since each walk ending at B, D, or E can go back to A, so if there are, say, x walks ending at B, y at D, z at E, then the total ways to go back to A would be x + y + z = b_k. Therefore, a_{k+1} = b_k.Similarly, from each vertex in b_k (B, D, E), you can go to two other vertices that are two steps away from A. For example, from B, you can go to C or F; from D, you can go to C or H; from E, you can go to F or H. Each of these contributes to c_{k+1}. Therefore, from each vertex in b_k, there are two ways to go to c_{k+1} vertices, so total contribution is b_k * 2. Hence, c_{k+1} = 2 * b_k.Wait, but hold on. If you are at vertex B, you can go to C or F. Similarly, from D to C or H, and from E to F or H. But these are all different vertices. However, c_k counts the total number of walks ending at C, F, or H. So, for each step from B, D, E, you can go to two different vertices in C, F, H. Therefore, the total number of walks to C, F, H is 2 * b_k. So yes, c_{k+1} = 2 * b_k.Now, what about walks ending at G (distance 3). From which vertices can you reach G? G is connected to C, F, H. So, to get to G, you must come from C, F, or H. Each of those is in c_k. From each vertex in c_k (C, F, H), you can go to G, as well as two other vertices. For example, from C, you can go to B, D, or G. But B and D are in b_k (adjacent to A), and G is in d_k. Similarly, from F, you can go to B, E, or G; and from H, you can go to D, E, or G. Therefore, from each vertex in c_k, there is 1 way to go to G, so total contribution to d_{k+1} is c_k * 1. Therefore, d_{k+1} = c_k.Additionally, from c_k (C, F, H), you can also go back to two vertices adjacent to A. For example, from C, you can go back to B or D. So, each vertex in c_k can go to two vertices in b_{k+1}. Therefore, the contribution to b_{k+1} is c_k * 2. So:b_{k+1} = (number of ways from a_k to b_{k+1}) + (number of ways from c_k to b_{k+1}) + (number of ways from d_k to b_{k+1})Wait, wait. Let's re-examine.Wait, actually, from a_k (walks ending at A), you can go to b_{k+1} (B, D, E). So, from A, there are 3 neighbors: B, D, E. Therefore, the number of walks going from A to its neighbors is 3 * a_k. Therefore, b_{k+1} = 3 * a_k + ... ?Wait, hold on. I think I messed up the previous reasoning. Let's start over.Let me formalize this.We have four states:- a_k: number of walks of length k ending at A.- b_k: number of walks of length k ending at B, D, or E.- c_k: number of walks of length k ending at C, F, or H.- d_k: number of walks of length k ending at G.We need to find recurrence relations between these.From state a_k:From A, you can go to B, D, E. So, each walk ending at A can proceed to B, D, or E. Therefore, the number of walks going from A to B, D, E is 3 * a_k. Therefore, this contributes 3 * a_k to b_{k+1}.From state b_k:Each walk ending at B, D, or E can go back to A, or go to two other vertices. For example, from B, you can go back to A or go to C or F. Similarly for D and E. Therefore, each walk ending at B, D, E has 1 way back to A and 2 ways to C, F, H. Also, from B, D, E, is there a way to go to G? No, because G is three steps away from A. Wait, from B, you can go to C or F, which are two steps away, and from C or F you can go to G. But from B directly, you can't go to G. Similarly for D and E.Therefore, from each walk ending at B, D, E (total b_k walks), each can go back to A (1 way) or go to two vertices in C, F, H (2 ways). Therefore, the total contributions:From b_k to a_{k+1}: b_k * 1.From b_k to c_{k+1}: b_k * 2.From state c_k:Each walk ending at C, F, or H can go to two vertices adjacent to A (B, D, E) or go to G. For example, from C, you can go to B, D, or G. Similarly, from F, you can go to B, E, or G; from H, you can go to D, E, or G. Therefore, from each vertex in c_k, there are two ways to go back to b_{k+1} and one way to go to d_{k+1}.Therefore, the contributions:From c_k to b_{k+1}: c_k * 2.From c_k to d_{k+1}: c_k * 1.From state d_k:Each walk ending at G can go to C, F, or H. Each of those is in c_{k}. Therefore, from G, you can go back to C, F, or H. So, three ways. Therefore, contributions:From d_k to c_{k+1}: d_k * 3.Putting all together, the recurrence relations are:a_{k+1} = b_k * 1b_{k+1} = 3 * a_k + 2 * c_kc_{k+1} = 2 * b_k + 3 * d_kd_{k+1} = c_k * 1Wait, let me check:- a_{k+1} comes from b_k moving back to A. So yes, a_{k+1} = b_k.- b_{k+1} comes from a_k moving to B, D, E (3*a_k) and from c_k moving back to B, D, E (2*c_k). Wait, no. Wait, from c_k, each walk ending at C, F, H can go to two vertices in B, D, E, so total contributions 2*c_k. So yes, b_{k+1} = 3*a_k + 2*c_k.- c_{k+1} comes from b_k moving to C, F, H (2*b_k) and from d_k moving to C, F, H (3*d_k). Wait, from d_k, each walk ending at G can go to three vertices: C, F, H. So yes, from d_k, each contributes 3*d_k to c_{k+1}. Therefore, c_{k+1} = 2*b_k + 3*d_k.- d_{k+1} comes from c_k moving to G (1*c_k). So yes, d_{k+1} = c_k.But wait, in the previous step, from c_k, each walk can go to G in 1 way per vertex, but since c_k is the total number of walks ending at C, F, or H, each of those can go to G once. So total contribution is c_k * 1. Wait, but if c_k is the number of walks ending at C, F, or H, then each such walk is at a specific vertex. Each of C, F, H has one edge to G, so from each of the three vertices, you can go to G. Therefore, if there are, say, x walks ending at C, y at F, z at H, then the total number of ways to go to G is x + y + z = c_k. Therefore, yes, d_{k+1} = c_k.So summarizing:a_{k+1} = b_kb_{k+1} = 3a_k + 2c_kc_{k+1} = 2b_k + 3d_kd_{k+1} = c_kBut since d_{k+1} = c_k, we can substitute d_{k} = c_{k-1}. Therefore, we can eliminate d from the equations.So replacing d_k with c_{k-1}:c_{k+1} = 2b_k + 3c_{k-1}Similarly, let's see if we can express everything in terms of a, b, c.But let's see:From a_{k+1} = b_k, so b_k = a_{k+1}Similarly, from b_{k+1} = 3a_k + 2c_k, and since a_{k} = b_{k-1}, then:b_{k+1} = 3b_{k-1} + 2c_kFrom c_{k+1} = 2b_k + 3c_{k-1}So now, we have two equations:1. b_{k+1} = 3b_{k-1} + 2c_k2. c_{k+1} = 2b_k + 3c_{k-1}But perhaps we can find a recurrence relation solely in terms of c_k or b_k.Alternatively, let's express c_{k} in terms of previous terms.But since we also have a relation involving d, but we've substituted d.Alternatively, let's try writing the equations for b and c.From equation 1: b_{k+1} - 3b_{k-1} = 2c_kFrom equation 2: c_{k+1} - 3c_{k-1} = 2b_kSo now, we can write:From equation 1: c_k = (b_{k+1} - 3b_{k-1}) / 2From equation 2: 2b_k = c_{k+1} - 3c_{k-1}Substituting c_k from equation 1 into equation 2:2b_k = c_{k+1} - 3c_{k-1}But c_{k+1} = (b_{k+2} - 3b_k)/2 (from equation 1, shifted by 1)And c_{k-1} = (b_k - 3b_{k-2})/2Therefore,2b_k = [ (b_{k+2} - 3b_k)/2 ] - 3*(b_k - 3b_{k-2})/2Multiply both sides by 2:4b_k = (b_{k+2} - 3b_k) - 3*(b_k - 3b_{k-2})Expand the right-hand side:4b_k = b_{k+2} - 3b_k - 3b_k + 9b_{k-2}Simplify:4b_k = b_{k+2} - 6b_k + 9b_{k-2}Bring all terms to left-hand side:4b_k + 6b_k - 9b_{k-2} - b_{k+2} = 0Which simplifies to:10b_k - 9b_{k-2} - b_{k+2} = 0Rearranged:b_{k+2} = 10b_k - 9b_{k-2}That's a linear recurrence relation of order 2 (but with indices shifted). Let me write it as:b_{k} = 10b_{k-2} - 9b_{k-4}Wait, no. Wait, if we set m = k + 2, then:b_{m} = 10b_{m-2} - 9b_{m-4}So the recurrence relation for b is a fourth-order recurrence? Wait, actually, the equation is:b_{k+2} - 10b_k + 9b_{k-2} = 0Which is a linear recurrence with characteristic equation r^4 - 10r^2 + 9 = 0. Wait, but since the indices are k+2, k, k-2, so stepping by 2. Let me set m = k, then the equation becomes:b_{m+2} - 10b_{m} + 9b_{m-2} = 0Which is a linear recurrence with a step of 2. To solve this, we can substitute t = m/2, but that might complicate. Alternatively, let's consider the recurrence as a second-order linear recurrence with terms separated by two steps.Let me denote for even and odd indices separately. However, since our initial steps start at k=0, and we need to compute up to k=2n, maybe the parity isn't an issue here. Alternatively, find the characteristic equation.Let’s let’s assume a solution of the form b_k = r^k. Then substituting into the recurrence:r^{k+2} - 10r^k + 9r^{k-2} = 0Divide both sides by r^{k-2}:r^4 - 10r^2 + 9 = 0Let’s set x = r^2, then the equation becomes:x^2 - 10x + 9 = 0Solving:x = [10 ± sqrt(100 - 36)] / 2 = [10 ± sqrt(64)] / 2 = [10 ± 8]/2Thus, x = (10 + 8)/2 = 18/2 = 9, and x = (10 - 8)/2 = 2/2 = 1.Therefore, the roots are x = 9 and x = 1, so r^2 = 9 and r^2 = 1. Therefore, r = ±3 and r = ±1.Therefore, the general solution for b_k is:b_k = A*(3)^k + B*(-3)^k + C*(1)^k + D*(-1)^kBut because the recurrence is of order 2 in terms of steps of 2, we can write the solution as:b_k = α*(3)^{k/2} + β*(-3)^{k/2} + γ*(1)^{k/2} + δ*(-1)^{k/2}But since k must be even or odd, but in our case, the steps start from k=0, and since we are dealing with walks of even length 2n, perhaps k here is even. Wait, actually, in the original problem, we are dealing with walks of length 2n, so k would be even. Wait, let me clarify.Wait, in our initial setup, a_k, b_k, c_k, d_k are defined for any k, but in the problem, we need S_{2n} which is the number of walks of length 2n from A to C, which is c_{2n}.But in our state variables, c_k counts the number of walks of length k ending at C, F, or H. Therefore, S_{2n} is c_{2n} divided by 3? Wait, no. Wait, if c_{2n} is the total number of walks ending at C, F, or H, then to get the number of walks ending specifically at C, we need to divide by 3, since by symmetry, the number of walks to each of C, F, H should be the same.But wait, is that the case? Since the cube is symmetric, yes. Starting from A, which is symmetric with respect to C, F, and H. Therefore, the number of walks ending at each of C, F, H should be equal. Therefore, S_{2n} = c_{2n} / 3.Therefore, if we can find c_{2n}, then divide by 3 to get S_{2n}.Alternatively, maybe we can adjust the state variables to count only walks ending at C, but that complicates the recurrence relations since the symmetries would be broken. So perhaps it's better to compute c_{2n} and then divide by 3.But let's see. Let's proceed with solving the recurrence relation for b_k, which is:b_{k+2} - 10b_k + 9b_{k-2} = 0But this seems a bit complicated. Alternatively, let's go back to the original state equations and try to find a recurrence for c_k.Alternatively, let's see if we can find a recurrence for c_{k} directly.From earlier, we have:c_{k+1} = 2b_k + 3d_kBut d_k = c_{k-1}So, c_{k+1} = 2b_k + 3c_{k-1}Also, from b_{k} = 3a_{k-1} + 2c_{k-1}But a_{k-1} = b_{k-2}Therefore, b_k = 3b_{k-2} + 2c_{k-1}So substituting into c_{k+1}:c_{k+1} = 2*(3b_{k-2} + 2c_{k-1}) + 3c_{k-1}= 6b_{k-2} + 4c_{k-1} + 3c_{k-1}= 6b_{k-2} + 7c_{k-1}But b_{k-2} can be expressed in terms of a_{k-3} and c_{k-3}:From b_{k-2} = 3a_{k-3} + 2c_{k-3}But a_{k-3} = b_{k-4}Therefore, b_{k-2} = 3b_{k-4} + 2c_{k-3}This seems to be getting more complex. Maybe another approach is needed.Alternatively, note that the cube's adjacency matrix eigenvalues can be used to compute the number of walks. The cube graph is a well-known graph with eigenvalues that can be calculated.The cube graph has eigenvalues 3, 1, 1, 1, -1, -1, -1, -3. Because the cube is a bipartite regular graph, its eigenvalues are symmetric around 0. The number of walks of length k from vertex u to v can be expressed as a linear combination of the eigenvalues raised to the power k.But perhaps this is a more advanced method, but maybe manageable.The number of walks of length k from A to C is equal to the (A, C) entry of M^k, where M is the adjacency matrix.To compute this, we can diagonalize M and then compute the entry.However, since I might not remember all the eigenvectors, maybe another approach.Alternatively, use generating functions or exploit the symmetry.Alternatively, consider that the cube is a Cayley graph for the group Z_2^3, with generators corresponding to the standard basis vectors. The number of walks can be computed using the Fourier transform on the group, but this might be overkill.Alternatively, note that the problem is similar to a 3-dimensional simple symmetric random walk on the hypercube, but restricted to the cube's structure.Alternatively, use recursion.Let me try to compute the initial terms and see if I can find a pattern.We have:At k = 0:a_0 = 1 (at A)b_0 = 0c_0 = 0d_0 = 0k = 1:a_1 = b_0 = 0b_1 = 3a_0 + 2c_0 = 3*1 + 2*0 = 3c_1 = 2b_0 + 3d_0 = 0 + 0 = 0d_1 = c_0 = 0k = 2:a_2 = b_1 = 3b_2 = 3a_1 + 2c_1 = 0 + 0 = 0c_2 = 2b_1 + 3d_1 = 2*3 + 0 = 6d_2 = c_1 = 0k = 3:a_3 = b_2 = 0b_3 = 3a_2 + 2c_2 = 3*3 + 2*6 = 9 + 12 = 21c_3 = 2b_2 + 3d_2 = 0 + 0 = 0d_3 = c_2 = 6k = 4:a_4 = b_3 = 21b_4 = 3a_3 + 2c_3 = 0 + 0 = 0c_4 = 2b_3 + 3d_3 = 2*21 + 3*6 = 42 + 18 = 60d_4 = c_3 = 0k = 5:a_5 = b_4 = 0b_5 = 3a_4 + 2c_4 = 3*21 + 2*60 = 63 + 120 = 183c_5 = 2b_4 + 3d_4 = 0 + 0 = 0d_5 = c_4 = 60k = 6:a_6 = b_5 = 183b_6 = 3a_5 + 2c_5 = 0 + 0 = 0c_6 = 2b_5 + 3d_5 = 2*183 + 3*60 = 366 + 180 = 546d_6 = c_5 = 0Hmm, interesting. So c_{2} = 6, c_{4} = 60, c_{6} = 546,...Similarly, S_{2n} = c_{2n} / 3. Let's check:For n=1 (2n=2), c_2 = 6, so S_2 = 6 / 3 = 2. Is that correct?From A, a walk of 2 steps to C. The possible walks are A -> B -> C, A -> D -> C, A -> E -> ... wait, no. Wait, starting at A, first step to B, D, or E. From B, next step can be A, C, or F. To get to C in two steps, from A -> B -> C or A -> D -> C or A -> E -> ... but E is connected to F and H, not to C. Wait, hold on. Wait, from A, first step is to B, D, or E. From B, next step can be A, C, or F. From D, next step can be A, C, or H. From E, next step can be A, F, or H. Therefore, the walks of length 2 from A to C are:A -> B -> CA -> D -> CTherefore, there are 2 walks. So S_2 = 2. Which matches c_2 / 3 = 6 / 3 = 2.Similarly, for n=2 (2n=4), c_4 = 60, so S_4 = 60 / 3 = 20. Let's verify with a smaller example.Walks of length 4 from A to C. This is more complex, but let's see.Alternatively, note that the numbers seem to follow a pattern: c_2 = 6, c_4 = 60, c_6 = 546. Let's see the ratios:60 / 6 = 10546 / 60 = 9.1Hmm, not obvious. Let's compute the next term:k=7:a_7 = b_6 = 0b_7 = 3a_6 + 2c_6 = 3*183 + 2*546 = 549 + 1092 = 1641c_7 = 2b_6 + 3d_6 = 0 + 0 = 0d_7 = c_6 = 546k=8:a_8 = b_7 = 1641b_8 = 3a_7 + 2c_7 = 0 + 0 = 0c_8 = 2b_7 + 3d_7 = 2*1641 + 3*546 = 3282 + 1638 = 4920So c_8 = 4920, so S_8 = 4920 / 3 = 1640.Looking at the sequence for S_{2n}: 2, 20, 182, 1640,...Check if this sequence is known. Let me compute the terms:n=1: 2n=2: 20n=3: 182n=4: 1640Looking up the sequence 2, 20, 182, 1640 in the OEIS (Online Encyclopedia of Integer Sequences)... Hmm, since I can't actually browse, let me see if there's a pattern.Compute the ratios between terms:20 / 2 = 10182 / 20 = 9.11640 / 182 ≈ 9.010...Hmm, approximately multiplying by 9 each time, but not exactly. Let's see the differences:From 2 to 20: +18From 20 to 182: +162From 182 to 1640: +1458These differences are 18, 162, 1458, which are 18, 18*9, 162*9, so each term is 9 times the previous difference. Therefore, the sequence might follow a recurrence relation S_{n} = 9*S_{n-1} + something?Wait, let's check:Assume S_{n} = x*S_{n-1} + y*S_{n-2}For n=2: 20 = x*2 + y*0 (assuming S_0=1?)Wait, maybe S_0 would be the number of walks of length 0 from A to C, which is 0 except S_0=1 if n=0? Wait, confusion.Wait, in the problem statement, S_{2n} is the number of walks of length 2n from A to C. So for n=0, S_0=0 (since A≠C). But in our earlier calculation, c_0=0, which corresponds to S_0=0. However, in the state variables, a_0=1 (walks of length 0 ending at A), and others 0.But let's see:Given S_1=2 (for n=1, 2n=2), S_2=20 (for n=2, 2n=4), S_3=182 (n=3, 2n=6), S_4=1640 (n=4, 2n=8).Let’s check the ratios between S_{n} and S_{n-1}:20 / 2 = 10182 / 20 = 9.11640 / 182 ≈ 9.010...Approaching 9. So perhaps the recurrence relation is S_{n} = 10*S_{n-1} - 9*S_{n-2}Check:For S_2=20: 10*S_1 - 9*S_0 = 10*2 -9*0=20. Correct.S_3=10*20 -9*2=200 -18=182. Correct.S_4=10*182 -9*20=1820 -180=1640. Correct.Yes! So the recurrence relation is S_{n} = 10*S_{n-1} -9*S_{n-2}, with initial conditions S_0=0 (since for n=0, 2n=0 steps, we're at A, not C), S_1=2.But wait, our original S_{2n} corresponds to n >=1, but in terms of the recurrence, S_n corresponds to 2n steps. Wait, actually, in our terms, the index in S_{2n} is the number of steps divided by 2. Let me clarify:Let me redefine m = n, so S_{2m} is the number of walks of length 2m from A to C. Then, we observed that S_{2m} = 10*S_{2(m-1)} -9*S_{2(m-2)}.So this is a linear recurrence relation with characteristic equation r^2 -10r +9=0. Solving:r = [10 ± sqrt(100 - 36)] / 2 = [10 ± 8]/2 => r=9 or r=1.Therefore, the general solution is S_{2m} = α*9^m + β*1^m.Using the initial conditions:For m=1: S_2=2 = α*9 + β*1For m=2: S_4=20 = α*81 + β*1Solve the system:9α + β = 281α + β = 20Subtract the first equation from the second:72α = 18 => α = 18 /72 = 1/4Then substitute back into first equation:9*(1/4) + β = 2 => 9/4 + β = 8/4 => β = -1/4Therefore, the general solution is:S_{2m} = (1/4)*9^m - (1/4)*1^m = (9^m -1)/4But wait, for m=1: (9 -1)/4 = 8/4=2. Correct.For m=2: (81 -1)/4=80/4=20. Correct.For m=3: (729 -1)/4=728/4=182. Correct.For m=4: (6561 -1)/4=6560/4=1640. Correct.Therefore, the formula is S_{2m} = (9^m -1)/4But since m corresponds to n in the problem statement, where S_{2n} is the number of walks of length 2n from A to C. Therefore, replacing m with n:S_{2n} = (9^n -1)/4Wait, but let's check the indices. The problem states "Find S_{2n}". So in terms of the answer, they want it expressed in terms of n. So yes, since m is the same as n here.Wait, but let me confirm:In the problem statement, part (2) says "Find S_{2n}". So if we denote the number of steps as 2n, then S_{2n} = (9^n -1)/4.But let me check the calculation:For n=1, S_2=(9^1 -1)/4=8/4=2. Correct.For n=2, S_4=(9^2 -1)/4=(81 -1)/4=80/4=20. Correct.Yes, this formula works.Alternatively, we can write 9^n as (3^2)^n =3^{2n}, but that might not help. So the answer is (9^n -1)/4.Alternatively, factor it as (3^{2n} -1)/4.But in the problem statement, it's 2n steps, so it's more natural to write it in terms of 3^{2n}:(3^{2n} -1)/4 = (9^n -1)/4.Either form is acceptable. But since 9^n is simpler, I think (9^n -1)/4 is the answer.Therefore, part (2) answer is S_{2n} = (9^n -1)/4.But let me verify once more with another term. For n=3, S_6=(9^3 -1)/4=729 -1=728; 728/4=182. Which matches our earlier calculation. Correct.Therefore, the final answer is S_{2n} = frac{9^n -1}{4}Final Answer(1) The possible positions of vertex (A_{2n}) are (boxed{A}), (boxed{C}), (boxed{F}), and (boxed{H}).(2) The number (S_{2n}) is (boxed{dfrac{9^n - 1}{4}}).</think>Given a cube (ABCDEFGH) with edge length 1, we need to determine the possible positions of vertex (A_{2n}) after (2n) steps starting from vertex (A), and find the number of sequences ending at vertex (C). Part (1)The cube is a bipartite graph, meaning its vertices can be divided into two sets such that no two vertices within the same set are adjacent. Starting from vertex (A), which is in one set, after an even number of steps, we must remain in the same set. The vertices in the same set as (A) (even parity) are (A), (C), (F), and (H). Therefore, the possible positions of (A_{2n}) are:[boxed{A}, boxed{C}, boxed{F}, boxed{H}] Part (2)To find (S_{2n}), the number of sequences ending at (C) after (2n) steps, we use recurrence relations and properties of the cube graph. The key steps are:1. Recurrence Relations: We define states for the number of walks ending at different vertices and derive recurrence relations.2. Symmetry and Solving Recurrence: Using symmetry and solving the recurrence relations, we find that the number of walks ending at (C) follows the recurrence (S_{2n} = 10S_{2(n-1)} - 9S_{2(n-2)}) with initial conditions (S_0 = 0) and (S_2 = 2).3. General Solution: Solving the recurrence, we find the general solution (S_{2n} = frac{9^n - 1}{4}).Thus, the number (S_{2n}) is:[boxed{dfrac{9^n - 1}{4}}],<think>Okay, so I need to solve this problem about a cube with edge length 1, and vertices labeled A through H. Let me first make sure I understand the problem correctly.We have a cube, and the set P consists of its eight vertices. The problem defines a sequence of ordered vertices starting at A (which is A0), and each subsequent vertex Ai+1 must be adjacent to Ai. The sequence has 2n+1 vertices, so after 2n moves, we end up at A_{2n}. The first part asks for all possible positions of A_{2n}, and the second part wants the number of such sequences that end at vertex C.Starting with part (1): Determine all possible positions of vertex A_{2n}.Hmm, since the cube is a 3-dimensional structure, each vertex is connected to three edges. So from any vertex, there are three adjacent vertices. But since we can't stay in the same place, each move goes to one of three neighbors. However, since we're dealing with sequences of even length (2n moves), starting at A, where can we end up?I remember that in graph theory, the cube can be represented as a graph where each vertex is connected to three others. The cube's graph is bipartite. That is, the vertices can be divided into two sets such that every edge connects a vertex from one set to the other. In a cube, this corresponds to the two sets being the "even" and "odd" vertices based on their coordinates. For example, if we assign coordinates to the cube's vertices, starting with A at (0,0,0), then all vertices with an even number of 1's in their coordinates (like A, which is (0,0,0)) are in one set, and those with an odd number of 1's (like C, which is (1,1,0)) are in the other set.Since the cube is bipartite, any walk (a sequence of moves along edges) starting at a vertex in one set will alternate between the two sets. Therefore, after an even number of steps (2n), we must be back in the starting set. Since A is in the even set, after 2n steps, we must also be in the even set. So all possible positions of A_{2n} must be the vertices in the same bipartition as A. Let me verify.Let's label the cube's vertices with coordinates. Let me recall the standard labeling:- A: (0,0,0)- B: (1,0,0)- C: (1,1,0)- D: (0,1,0)- E: (0,0,1)- F: (1,0,1)- G: (1,1,1)- H: (0,1,1)So, the "even" vertices (with even number of 1s) are A, C, F, H. Wait: A is (0,0,0) has 0 ones (even). C is (1,1,0) has two ones (even). F is (1,0,1) has two ones (even). H is (0,1,1) has two ones (even). The others, B, D, E, G, have one or three ones (odd). So the bipartition is correct.Therefore, after an even number of steps, we must be at an even vertex. So the possible positions of A_{2n} are the even vertices: A, C, F, H.But wait, the problem is in 3D space. Maybe there's more to it? For example, in a cube, even if you start at A, can you reach all the even vertices in 2n steps? Let me see. For example, starting at A, can we reach C, F, or H in 2n steps?But wait, in a cube, the graph is connected, so given enough steps, you can reach any vertex in the same bipartition. Since the cube is bipartite, the distance between two vertices in the same partition must be even. So the minimum number of steps to go from A to C is 2 (since they are diagonally opposite on a face). Similarly, A to F is also two steps (A to B to F, or A to E to F). A to H is two steps (A to D to H). So all these even vertices are at even distances from A, so for any even number of steps (2n), you can reach them, as long as n is at least 1. But since n can be any positive integer (assuming n is a positive integer; the problem doesn't specify, but since the sequence is 2n+1, n must be at least 0. If n=0, the sequence is just A0, so A0 is A. But the problem probably considers n as a positive integer here, but maybe not. Wait, 2n+1 vertices, so n can be 0, leading to 1 vertex, which is trivial. But for part (1), I think they just want the positions after 2n steps, regardless of n's value. So for any n, starting at A, after 2n steps, you can be at any even vertex. Therefore, the answer to part (1) is that A_{2n} can be at any of the four even vertices: A, C, F, H.But let me double-check. Suppose n=0, then the sequence is just A0 = A, which is allowed. For n=1, we have 3 vertices: A0, A1, A2. So starting at A, after 2 steps, where can we be? Each step moves to an adjacent vertex. So from A (0,0,0), first step to B, D, or E. Then from there, the second step. Let's enumerate:If first step is to B (1,0,0), second step can be to A, C, or F. Similarly, starting from A, go to D (0,1,0), second step can be to A, C, or H. Starting from A, go to E (0,0,1), second step can be to A, F, or H. Therefore, after two steps, possible positions are A, C, F, H. So indeed, even for n=1, A2 can be at any of the four even vertices.Therefore, for part (1), the answer is that A_{2n} can be at any of the four vertices A, C, F, H.Now part (2): Let S_{2n} be the number of sequences such that A_{2n}=C. Find S_{2n}.So we need to count the number of walks of length 2n starting at A and ending at C on the cube graph. Since the cube is a well-known graph, maybe we can use linear algebra or recurrence relations for the number of walks.In general, the number of walks of length k between two vertices in a graph can be found by taking the k-th power of the adjacency matrix and looking at the corresponding entry.But let's think about the cube's structure. Each vertex has three neighbors. Since the cube is bipartite, as we saw, walks from A to C must have even length. Let me consider how to model the number of walks.Alternatively, we can model the number of walks using recurrence relations. Let me define:Let’s denote by a_k the number of walks of length k ending at A,c_k the number ending at C,f_k the number ending at F,h_k the number ending at H.Similarly, since the other vertices (B, D, E, G) are in the other partition, walks of even length cannot end there, so we can ignore them for now.But since the cube is symmetric, the number of walks ending at C, F, or H should be the same due to the cube's symmetry. Let me confirm.A is a corner, and C, F, H are each the opposite corner on a face or the space diagonal? Wait, in the cube:- The distance from A to C is 2 (they are diagonally opposite on a face).- The distance from A to F is 2 (they are diagonally opposite on a different face).- The distance from A to H is 2 (diagonally opposite on another face).Wait, but the space diagonal from A to G is distance 3, which is odd. So C, F, H are all at distance 2 from A, which is even. So they are in the same partition. So due to the cube's symmetry, the number of walks from A to C, A to F, or A to H should be the same. So c_k = f_k = h_k for all k.But the number of walks ending at A is different. For example, starting at A, after 0 steps, you're at A. After 1 step, you can't be at A. After 2 steps, you can be back at A.So perhaps, we can define variables:Let’s denote a_k as the number of walks of length k starting at A and ending at A,and b_k as the number of walks of length k starting at A and ending at any of C, F, H.Due to symmetry, each of C, F, H will have the same number of walks, so c_k = f_k = h_k = b_k / 3. Wait, but if we define b_k as the total number of walks ending at any of the three vertices, then each individual one is b_k / 3. But maybe it's easier to consider a_k and b_k where b_k is the number of walks ending at each of the other even vertices.Wait, perhaps not. Let me think. Let me suppose that at each step, from a vertex, you can move to three neighbors. If we are at A, the number of ways to stay at A in the next step is zero, but the number of ways to go to each neighbor is 1. Wait, no. For a walk, each step is moving along an edge. So the recurrence relations need to consider the number of ways to reach a vertex.Let me think in terms of recurrence relations.Suppose that a_k is the number of walks of length k ending at A,and c_k is the number ending at C (similarly for F and H, but due to symmetry, they are equal, so let's just use c_k for one of them, but maybe it's better to group them).Wait, actually, since from A's perspective, C, F, H are symmetric. Similarly, from A's perspective, all three neighbors (B, D, E) are symmetric. So perhaps the number of walks ending at C, F, H after k steps are all equal.Alternatively, we can model this using two variables: a_k and d_k, where a_k is the number of walks ending at A, and d_k is the number of walks ending at any of the "distance 2" vertices (C, F, H). Then, since each of C, F, H has the same number of walks, we can write d_k = 3 * c_k, where c_k is the number of walks ending at each of them.But perhaps it's better to have two variables: a_k (walks ending at A) and b_k (walks ending at C, F, H). Then, since the total number of walks of length k is a_k + 3*b_k (since there are three vertices each contributing b_k walks). But how do we set up the recurrence?From A, to reach A in k steps, you must have been at one of its neighbors (B, D, E) in step k-1. Each neighbor has a certain number of walks ending there at step k-1. Similarly, to reach C in k steps, you must have been at one of C's neighbors in step k-1. Let's see.Wait, each step, you can move to a neighbor. So to get to a vertex X at step k, you sum the number of walks ending at each neighbor of X at step k-1.But since we need to track walks ending at A and walks ending at C, F, H, perhaps we can write recurrence relations for a_k and c_k (assuming c_k is the number for each of C, F, H). Wait, but let's think.Let me think of the cube's adjacency. The vertex A (0,0,0) is connected to B (1,0,0), D (0,1,0), and E (0,0,1).The vertex C (1,1,0) is connected to B (1,0,0), D (0,1,0), and G (1,1,1).Similarly, F (1,0,1) is connected to B (1,0,0), E (0,0,1), and G (1,1,1).H (0,1,1) is connected to D (0,1,0), E (0,0,1), and G (1,1,1).And G (1,1,1) is connected to C, F, H.Wait, so maybe to model the recurrence relations:To compute a_k (walks ending at A), we need to consider walks that ended at B, D, or E at step k-1, since those are the neighbors of A. Similarly, to compute c_k (walks ending at C), we need to consider walks that ended at B, D, or G at step k-1.But since we have symmetry, perhaps we can define variables based on the distance from A or something else.Alternatively, let's model the cube graph as two sets: Set 0 (even) contains A, C, F, H, and Set 1 (odd) contains B, D, E, G.Since the cube is bipartite, all walks from Set 0 must go to Set 1 on odd steps and back to Set 0 on even steps.Therefore, for even k, the walk is in Set 0, and for odd k, in Set 1.Let me define variables:Let’s denote:- a_k: number of walks of length k ending at A.- c_k: number of walks of length k ending at C.Similarly, due to symmetry, f_k = h_k = c_k.But maybe better to group them. Let me denote:For even k:- a_k: walks ending at A.- b_k: walks ending at each of C, F, H. So total walks ending at C, F, H is 3*b_k.For odd k:- d_k: walks ending at each of B, D, E. Since from Set 0, on odd steps, you are in Set 1, which has B, D, E, G. Wait, but G is also in Set 1. Wait, G is (1,1,1), which has three 1s, so odd. So Set 1 is B, D, E, G.But walks starting at A, after odd steps, are in Set 1. So for odd k, the walks end at one of B, D, E, G. However, G is different from B, D, E in terms of distance from A. G is the vertex opposite to A, the space diagonal. Whereas B, D, E are adjacent to A.Wait, perhaps we need more variables. Maybe define two variables for Set 1: one for the vertices adjacent to A (B, D, E) and one for the vertex opposite to A (G). Let's say:For odd k:- m_k: number of walks ending at each of B, D, E.- g_k: number of walks ending at G.So total walks for odd k is 3*m_k + g_k.Now, let's try to build recurrence relations.Starting with k=0:- a_0 = 1 (starting at A).- c_0 = 0, since we haven't moved.For k=1 (odd):- From A, you can go to B, D, E. So m_1 = 1 (each of B, D, E can be reached in 1 way), and g_1 = 0 (can't reach G in 1 step).Thus, total walks of length 1: 3*1 + 0 = 3.For k=2 (even):- To end at A, you must have come from B, D, or E. Each of B, D, E has m_{1} =1 walks ending there. From each of these, you can go back to A or to two other vertices. Wait, but the number of ways to get to A at step 2 is the sum over all neighbors of A of the number of ways to get to that neighbor at step 1.Since each neighbor (B, D, E) has m_1 =1, and from each neighbor, you can go back to A or to two other vertices. But for the count, the number of walks ending at A at step 2 is equal to the sum over neighbors of A of the number of walks ending at those neighbors at step 1.So a_2 = m_1(B) + m_1(D) + m_1(E) = 1 + 1 + 1 = 3.Similarly, to compute b_2 (number of walks ending at each of C, F, H), we need to consider how to get to C, F, H from step 1.For example, to get to C at step 2, you must have been at B, D, or G at step 1. But at step 1, we were only at B, D, E, or G. Wait, at step 1 (k=1), the walk is at Set 1: B, D, E, G.From Set 1, to get to C (Set 0), you can come from B, D, or G.Similarly, for F, you can come from B, E, or G.For H, you can come from D, E, or G.But since at step 1, the walks are at B, D, E, G with counts m_1 =1 for B, D, E and g_1 =0 for G.Thus, to compute b_2 (walks ending at C, F, H each):For C: can come from B, D, or G. But at step 1, G has 0 walks. So walks ending at C at step 2 is m_1(B) + m_1(D) + g_1 =1 +1 +0=2.Similarly, for F: comes from B, E, G: m_1(B) + m_1(E) + g_1=1 +1 +0=2.For H: comes from D, E, G: m_1(D) + m_1(E) + g_1=1 +1 +0=2.Therefore, each of C, F, H has 2 walks, so b_2 =2. Therefore, total walks ending at C, F, H is 3*2=6.Thus total walks at step 2: a_2 +3*b_2=3+6=9, which is equal to 3^2=9, since each step has 3 choices, but actually, it's the number of walks of length 2, which should be 3*3=9. Correct.Similarly, for k=3 (odd):To compute m_3 and g_3.First, walks ending at B, D, E (m_3) and G (g_3).To end at B at step 3, you must have come from A or C or F at step 2.But at step 2, the walks are in Set 0: A, C, F, H.From Set 0, moving to Set 1.From A, you can go to B, D, E.From C, you can go to B, D, G.From F, you can go to B, E, G.From H, you can go to D, E, G.But walks ending at B at step 3 can come from A, C, or F.Similarly, the number of walks ending at B is a_2 (walks ending at A) *1 (A to B) + b_2 (walks ending at C) *1 (C to B) + b_2 (walks ending at F) *1 (F to B). But since a_2=3, and b_2=2 for each of C, F, H. Wait, but H doesn't connect to B. So specifically:Number of walks ending at B:From A: a_2 ways to be at A, then go to B: 3 *1=3? Wait, no.Wait, no: to compute the number of walks ending at B at step 3, we need to look at all Set 0 vertices that are adjacent to B and sum the number of walks ending at those vertices at step 2 multiplied by the number of ways to go from there to B.But B is adjacent to A, C, F.Therefore, walks ending at B at step 3 = (walks ending at A at step 2) *1 + (walks ending at C at step 2)*1 + (walks ending at F at step 2)*1.Since a_2=3, and b_2=2 (for C and F each), so 3 +2 +2=7.Wait, but hold on. Wait, in step 2, walks ending at A: a_2=3, walks ending at C: b_2=2, walks ending at F: b_2=2, walks ending at H: b_2=2.Therefore, walks ending at B: from A (3), from C (2), from F (2). Total 3 +2 +2=7.But since we have symmetry, walks ending at D and E will be similar. For example, walks ending at D would come from A, C, H: 3 +2 +2=7, and walks ending at E would come from A, F, H: 3 +2 +2=7. Therefore, m_3=7.Similarly, walks ending at G at step 3 come from C, F, H. Each of these has b_2=2, so walks ending at G: 2 +2 +2=6. Therefore, g_3=6.Therefore, total walks at step 3: 3*m_3 + g_3=3*7 +6=27=3^3, which is correct.So we can see a pattern here. Let's try to generalize the recurrence relations.For even k=2n:- a_{2n} = number of walks ending at A.- b_{2n} = number of walks ending at each of C, F, H.For odd k=2n+1:- m_{2n+1} = number of walks ending at each of B, D, E.- g_{2n+1} = number of walks ending at G.We need to find the relations between these.From even to odd:To compute m_{2n+1} (walks ending at each of B, D, E):Each of B, D, E can be reached from A and two other Set 0 vertices.For example, B can be reached from A, C, F.Similarly, D can be reached from A, C, H.E can be reached from A, F, H.But due to symmetry, each of B, D, E is symmetric in the cube with respect to A.Therefore, the number of walks ending at each of B, D, E is equal. So:m_{2n+1} = a_{2n} + 2*b_{2n}Similarly, walks ending at G (g_{2n+1}) can be reached from C, F, H. Each of these has b_{2n} walks, so:g_{2n+1} = 3*b_{2n}From odd to even:To compute a_{2n+2} and b_{2n+2} from m_{2n+1} and g_{2n+1}.a_{2n+2} is the number of walks ending at A, which can be reached from B, D, E. Each of these has m_{2n+1} walks. So:a_{2n+2} = 3*m_{2n+1}Similarly, walks ending at each of C, F, H (b_{2n+2}) can be calculated as follows.For example, C can be reached from B, D, G.Each of B, D has m_{2n+1} walks, and G has g_{2n+1} walks.Therefore, walks ending at C: m_{2n+1} + m_{2n+1} + g_{2n+1} = 2*m_{2n+1} + g_{2n+1}But due to symmetry, this is the same for F and H. Therefore:b_{2n+2} = 2*m_{2n+1} + g_{2n+1}So summarizing the recurrence relations:For even steps:a_{2n+2} = 3*m_{2n+1}b_{2n+2} = 2*m_{2n+1} + g_{2n+1}For odd steps:m_{2n+1} = a_{2n} + 2*b_{2n}g_{2n+1} = 3*b_{2n}Now, let's substitute the expressions for m and g into the equations for a and b.From the odd step equations:m_{2n+1} = a_{2n} + 2*b_{2n}g_{2n+1} = 3*b_{2n}Substitute these into the even step equations:a_{2n+2} = 3*(a_{2n} + 2*b_{2n})b_{2n+2} = 2*(a_{2n} + 2*b_{2n}) + 3*b_{2n} = 2*a_{2n} +4*b_{2n} +3*b_{2n} = 2*a_{2n} +7*b_{2n}So now we have:a_{2n+2} = 3*a_{2n} +6*b_{2n}b_{2n+2} = 2*a_{2n} +7*b_{2n}This is a system of linear recurrences. We need to solve this system to find a closed-form expression for b_{2n}, since S_{2n} is the number of walks ending at C, which is b_{2n}.Alternatively, maybe express everything in terms of a and b.Alternatively, notice that a_{2n} and b_{2n} satisfy a recurrence relation. Let's see if we can find a second-order recurrence for b.First, let's compute the initial terms.We have:For n=0 (k=0):a_0 =1b_0 =0For n=1 (k=2):a_2 =3b_2 =2For n=2 (k=4):a_4 =3*a_2 +6*b_2 =3*3 +6*2=9 +12=21b_4 =2*a_2 +7*b_2 =2*3 +7*2=6 +14=20For n=3 (k=6):a_6=3*a4 +6*b4=3*21 +6*20=63 +120=183b6=2*a4 +7*b4=2*21 +7*20=42 +140=182Similarly, for n=4 (k=8):a8=3*183 +6*182=549 +1092=1641b8=2*183 +7*182=366 +1274=1640Hmm, interesting. The numbers for a and b are very close, differing by 1. For example:At n=1: a2=3, b2=2 (diff 1)n=2: a4=21, b4=20 (diff 1)n=3: a6=183, b6=182n=4: a8=1641, b8=1640So it seems that a_{2n} = b_{2n} +1.Wait, let's check:At n=1: a2=3, b2=2: 3=2+1, yes.n=2: a4=21=20 +1, yes.n=3:183=182 +1, yes.n=4:1641=1640 +1, yes.So perhaps in general, a_{2n} = b_{2n} +1.If this is true, then we can substitute this into the recurrence relations.Assuming a_{2n} = b_{2n} +1.Let me check if this holds.Assume a_{2n} = b_{2n} +1.Then, substitute into the recurrence:From a_{2n+2}=3*a_{2n} +6*b_{2n}=3*(b_{2n} +1) +6*b_{2n}=3*b_{2n} +3 +6*b_{2n}=9*b_{2n} +3From b_{2n+2}=2*a_{2n} +7*b_{2n}=2*(b_{2n} +1) +7*b_{2n}=2*b_{2n} +2 +7*b_{2n}=9*b_{2n} +2But from our assumption, a_{2n+2}=b_{2n+2}+1.So from above, a_{2n+2}=9*b_{2n} +3, and b_{2n+2}=9*b_{2n} +2.Then, a_{2n+2}=b_{2n+2} +1=(9*b_{2n} +2)+1=9*b_{2n} +3, which matches. Therefore, the relation a_{2n}=b_{2n}+1 holds for all n≥0.Therefore, we can use this to reduce the system to a single recurrence.Since a_{2n} = b_{2n} +1, we can substitute into the expression for b_{2n+2}:b_{2n+2}=2*a_{2n} +7*b_{2n}=2*(b_{2n} +1) +7*b_{2n}=2*b_{2n} +2 +7*b_{2n}=9*b_{2n} +2Therefore, the recurrence for b is:b_{2n+2} =9*b_{2n} +2With initial condition b0=0, b2=2.Wait, but let's check for n=1:b_{2*1+2}=b4=9*b2 +2=9*2 +2=20, which matches.Similarly, b6=9*b4 +2=9*20 +2=182, which is correct.So the recurrence is linear and nonhomogeneous. The homogeneous part is b_{2n+2} -9*b_{2n}=2.To solve this recurrence, first solve the homogeneous equation:b_{2n+2} -9*b_{2n}=0Characteristic equation: r^2 -9=0 => r=3, r=-3So the general solution to the homogeneous equation is:b_{2n}^{(h)}=A*3^{2n} +B*(-3)^{2n}=A*9^{n} +B*9^{n}= (A +B)*9^{n}Wait, (-3)^{2n}=9^{n}, so actually:b_{2n}^{(h)}=A*9^{n} +B*9^{n}= (A+B)*9^{n}But this suggests that the homogeneous solution is a multiple of 9^n. However, since the homogeneous equation is of second order, but our step is every 2n, maybe I need to adjust.Wait, actually, the recurrence is in terms of b_{2n+2} and b_{2n}, which is a second-order recurrence with step 2. However, if we let m=n+1, then the recurrence is:b_{2m} =9*b_{2(m-1)} +2Which is a first-order linear recurrence in terms of m. Wait, no. Because it's defined every 2 steps. Wait, actually, the index is 2n, so perhaps treating it as a first-order recurrence with indices at even numbers.Yes, actually, since the equation relates b_{2n+2} to b_{2n}, it's a first-order linear recurrence in terms of n. So we can treat it as:b_{k+2} =9*b_k +2, where k=2n.But this is a linear nonhomogeneous recurrence relation. The standard method is to find the homogeneous solution and a particular solution.We have the homogeneous solution as b^{(h)}_k = C*9^{k/2}For the particular solution, since the nonhomogeneous term is constant (2), we can try a constant solution b^{(p)}_k = D.Substituting into the recurrence:D=9*D +2=> D -9D=2 => -8D=2 => D= -2/8= -1/4.Therefore, the general solution is:b_k = C*9^{k/2} -1/4Applying the initial conditions. Let's recall that k=2n.Given b0=0 when k=0 (n=0):0 = C*9^{0} -1/4 => C -1/4=0 => C=1/4Next, check with b2=2 when k=2 (n=1):b2= (1/4)*9^{1} -1/4= (9/4 -1/4)=8/4=2, which matches.Therefore, the general solution is:b_{2n} = (1/4)*9^{n} -1/4= (9^{n} -1)/4Therefore, the number of walks ending at C after 2n steps is b_{2n}= (9^{n} -1)/4.But wait, let's verify with previous values.For n=1 (k=2):b2=(9^1 -1)/4=(9-1)/4=8/4=2. Correct.For n=2 (k=4):b4=(9^2 -1)/4=(81 -1)/4=80/4=20. Correct.For n=3 (k=6):b6=(9^3 -1)/4=(729 -1)/4=728/4=182. Correct.Therefore, this formula holds.Thus, S_{2n}= (9^{n} -1)/4.But let me check for n=0, which would give S0=(1 -1)/4=0, which is correct since there are no walks of length 0 ending at C (it starts at A).Hence, the answer to part (2) is S_{2n}= (9^n -1)/4.But let me write it in LaTeX notation: frac{9^n - 1}{4}Alternatively, frac{9^n -1}{4} = frac{3^{2n} -1}{4}Alternatively, the problem might expect an expression in terms of powers of 3 or something else, but since 9 is 3 squared, either form is acceptable.But since 9^n is straightforward, probably that's fine.Therefore, part (2) answer is (9^n -1)/4.Final Answer(1) The possible positions of vertex ( A_{2n} ) are ( boxed{A} ), ( boxed{C} ), ( boxed{F} ), and ( boxed{H} ).(2) The number ( S_{2n} ) is ( boxed{dfrac{9^n - 1}{4}} ).</think>Given a cube (ABCDEFGH) with edge length 1, we need to determine the possible positions of vertex (A_{2n}) and the number of sequences ending at vertex (C) after (2n) steps. Part (1)The cube's graph is bipartite, dividing vertices into two sets based on the parity of their coordinates. Starting from vertex (A) (an even vertex), any walk of even length (2n) must end in the same bipartition set. The even vertices are (A, C, F, H). Therefore, the possible positions of (A_{2n}) are:[boxed{A}, boxed{C}, boxed{F}, boxed{H}] Part (2)To find the number (S_{2n}) of sequences ending at (C), we use recurrence relations. Define (a_{2n}) as the number of walks ending at (A) and (b_{2n}) as the number of walks ending at each of (C, F, H). The recurrence relations are:[a_{2n+2} = 3a_{2n} + 6b_{2n}][b_{2n+2} = 2a_{2n} + 7b_{2n}]With initial conditions (a_0 = 1) and (b_0 = 0), we find (a_{2n} = b_{2n} + 1). Solving the recurrence for (b_{2n}), we get:[b_{2n} = frac{9^n - 1}{4}]Thus, the number (S_{2n}) is:[boxed{dfrac{9^n - 1}{4}}]]

🔑:12. (1) Establish a spatial rectangular coordinate system with A B, A D, and A E as the x, y, and z axes, respectively. ThenA(0,0,0), B(1,0,0), cdots, H(0,1,1) text {. }If A_{i+1} and A_{i} are adjacent vertices, then A_{i+1} only changes one of the coordinates of A_{i} from 0 to 1, or from 1 to 0.Thus, the sum of the three coordinates of A_{i+1} and A_{i} have different parities.Noting that A_{0}=A(0,0,0).Therefore, the sum of the three coordinates of A_{2 n} must be even.Hence, it can only be A(0,0,0) or C(1,1,0) or F(1,0,1) or H(0,1,1).(2) Consider the relationship between S_{2 n+2} and S_{2 n}.Suppose left(A_{0} A_{1} cdots A_{2 n} A_{2 n+1} A_{2 n+2}right) is a sequence of 2 n+3 ordered vertices satisfying the condition, i.e.,A_{0}=A, A_{2 n+2}=C text {. }When A_{2 n}=C, there are S_{2 n} sequences left(A_{0} A_{1} cdots A_{2 n}right), and A_{2 n+1} has three possibilities, so there are 3 S_{2 n} such sequences of 2 n+3 ordered vertices.When A_{2 n} is one of the vertices A, F, or H, A_{2 n+1} has only two possibilities. Since there are 3^{2 n} sequences of 2 n+1 ordered vertices left(A_{0} A_{1} cdots A_{2 n}right), there are 2left(3^{2 n}-S_{2 n}right) such sequences of 2 n+3 ordered vertices.Thus, S_{2 n+2}=3 S_{2 n}+2left(3^{2 n}-S_{2 n}right)=S_{2 n}+2 times 3^{2 n}.Given S_{2}=2, we havebegin{array}{l}S_{2 n}=left(S_{2 n}-S_{2 n-2}right)+cdots+left(S_{4}-S_{2}right)+S_{2} =2left(3^{2 n-2}+3^{2 n-4}+cdots+3^{2}right)+2 =2left(9^{n-1}+9^{n-2}+cdots+1right)=frac{9^{n}-1}{4} .end{array}（Proposed by Li Dayuan, Liu Hongkun, Ye Shengyang, and Lian Shifan）

🧠:[<think>Okay, let's see. I need to solve this system of inequalities involving sine and cosine functions. The inequalities are:1. sin x + cos y ≥ √22. sin y + cos z ≥ √23. sin z + cos x ≥ √2Hmm, so three inequalities with three variables: x, y, z. All are real numbers. The problem is in the field of real numbers, so we need to find all real numbers x, y, z that satisfy all three inequalities simultaneously.First thought: sine and cosine functions have maximum values of 1. So, sin x and cos y each can be at most 1. So their sum can be at most 2. But here the inequality is that their sum is at least √2, which is approximately 1.414. That seems possible, but the question is when can their sum be at least √2? But maybe there's a more restrictive condition here because of the cyclic nature of the inequalities. Each inequality involves the next variable in a cycle: x related to y, y to z, z to x.Wait, but the right-hand side of each inequality is √2, which is actually the maximum value of sin θ + cos θ for some angle θ. Let me check that. For example, sin θ + cos θ can be rewritten as √2 sin(θ + π/4) using the identity a sin θ + b cos θ = √(a² + b²) sin(θ + φ), where φ = arctan(b/a) or something like that. Let me verify:sin θ + cos θ = √2 sin(θ + π/4). Let's compute the right-hand side:√2 sin(θ + π/4) = √2 [sin θ cos(π/4) + cos θ sin(π/4)] = √2 [sin θ (√2/2) + cos θ (√2/2)] = √2*(√2/2)(sin θ + cos θ) = (2/2)(sin θ + cos θ) = sin θ + cos θ. So yes, that identity holds. Therefore, sin θ + cos θ = √2 sin(θ + π/4). The maximum value of this expression is √2, achieved when sin(θ + π/4) = 1, i.e., θ + π/4 = π/2 + 2πk, so θ = π/4 + 2πk. Similarly, the minimum value is -√2.But in our problem, each inequality is of the form sin(variable) + cos(next variable) ≥ √2. Since the maximum of sin α + cos β is √2 (if both sin α and cos β can reach 1 at the same time), but here α and β are different variables, so maybe it's possible that each inequality must achieve exactly √2, otherwise if one is greater than √2, but then the others might not? Wait, but the maximum possible is √2, so actually each inequality must be equal to √2? Because if you have sin x + cos y ≥ √2, but the maximum possible is √2, so equality must hold for all three inequalities. Therefore, all three inequalities must be equalities. That seems like a key insight.So the system reduces to:1. sin x + cos y = √22. sin y + cos z = √23. sin z + cos x = √2Therefore, each of these must be equal to √2. So now, I need to solve these equations.From the first equation: sin x + cos y = √2. Since both sin x and cos y are at most 1, the only way their sum is √2 is if both sin x and cos y are equal to 1. Because 1 + 1 = 2, which is greater than √2. Wait, wait. Wait, √2 is approximately 1.414, which is less than 2. Wait, hold on. Wait, sin x can be 1 and cos y can be 1, then sin x + cos y = 2, which is greater than √2. So that's possible. But if the maximum of sin x + cos y is 2, but the inequality is ≥ √2, so there are solutions where sin x and cos y can be less than 1, as long as their sum is at least √2. But earlier, I thought that since each inequality is part of a cyclic system, maybe the only solution is when each term is maximized. But perhaps that's not the case.Wait, but maybe the system forces each inequality to achieve equality. Let me think. Suppose that in the first inequality, sin x + cos y > √2. Then, for the second inequality, sin y + cos z ≥ √2, and the third, sin z + cos x ≥ √2. If each is greater than √2, is that possible?But since each function is periodic, maybe there are multiple solutions. But maybe due to the cyclic dependencies, the only solution is when all variables are set such that each sine and cosine term is 1. Let's explore that.Suppose sin x = 1 and cos y = 1. Then, sin x = 1 implies x = π/2 + 2πk, and cos y = 1 implies y = 2πn. Then, moving to the second equation: sin y + cos z = √2. If y = 2πn, then sin y = 0. Therefore, 0 + cos z = √2. But cos z cannot exceed 1, so this is impossible. Therefore, that approach doesn't work.Wait, so that's a problem. If we try to set sin x = 1 and cos y = 1, then the second equation fails. Therefore, perhaps we need a different approach.Alternatively, maybe each pair sin x and cos y must be such that their sum is exactly √2. Because if the sum is greater than √2, perhaps the next equation can't hold. Let me check.Suppose sin x + cos y = √2. Then, from the first equation. Let's recall that the maximum of sin θ + cos θ is √2, but here it's sin x + cos y. So since x and y are different variables, maybe there's a way to set x and y such that sin x + cos y = √2. But how?Alternatively, maybe all variables are set such that sin a = cos b for each pair. For example, in the first equation, sin x = cos y. If sin x = cos y, then sin x + cos y = 2 sin x. If we set 2 sin x = √2, then sin x = √2 / 2, which is sin x = 1/√2, so x = π/4 + 2πk or 3π/4 + 2πk. But then cos y = sin x = √2/2, so y = ±π/4 + 2πn. But then moving to the second equation: sin y + cos z = √2. If y = π/4 + 2πn, then sin y = √2/2, so √2/2 + cos z = √2 => cos z = √2 - √2/2 = √2/2. Therefore, z = ±π/4 + 2πm. Then the third equation: sin z + cos x = √2. If z = π/4 + 2πm, then sin z = √2/2. And cos x: if x = π/4 + 2πk, then cos x = √2/2. So √2/2 + √2/2 = √2, which holds. Similarly, if z = -π/4 + 2πm, sin z = -√2/2, but then cos x would need to be √2 - (-√2/2) = (3/2)√2, which is impossible because cos x cannot exceed 1. So z must be π/4 + 2πm. Similarly, if x is 3π/4 + 2πk, then cos x = -√2/2. Then the third equation: sin z + cos x = √2. sin z + (-√2/2) = √2 => sin z = (3/2)√2, which is impossible. Therefore, x must be π/4 + 2πk, y = π/4 + 2πn, z = π/4 + 2πm. Let me verify this solution.If x = π/4 + 2πk, then sin x = √2/2. Then cos y = √2/2, so y = ±π/4 + 2πn. But if we take y = π/4 + 2πn, then sin y = √2/2. Then the second equation: sin y + cos z = √2/2 + cos z = √2, so cos z = √2 - √2/2 = √2/2, so z = ±π/4 + 2πm. But as above, if z = -π/4 + 2πm, then sin z = -√2/2, which would make the third equation: sin z + cos x = -√2/2 + √2/2 = 0 < √2, which doesn't satisfy the inequality. Therefore, z must be π/4 + 2πm. Then cos x = √2/2, as x is π/4 + 2πk, so sin z + cos x = √2/2 + √2/2 = √2, which works.Similarly, if we take y = -π/4 + 2πn, then cos y = cos(-π/4) = √2/2, which satisfies the first equation. Then sin y = sin(-π/4) = -√2/2. Then the second equation: sin y + cos z = -√2/2 + cos z ≥ √2. So cos z must be ≥ √2 + √2/2 = (3/2)√2 ≈ 2.121, which is impossible because cos z ≤ 1. Therefore, y cannot be -π/4 + 2πn. Therefore, y must be π/4 + 2πn.Similarly, if we consider other quadrants, but since we need sin x and cos y to sum to √2, which is a positive value, so both sin x and cos y need to be positive. Therefore, x must be in the first or second quadrant (where sin is positive), and y must be in the first or fourth quadrant (where cos is positive). But given the periodicity, we can focus on the principal values.Therefore, the solution seems to be x = π/4 + 2πk, y = π/4 + 2πn, z = π/4 + 2πm for integers k, n, m. Let me check if this satisfies all three equations.First equation: sin x + cos y = sin(π/4) + cos(π/4) = √2/2 + √2/2 = √2. Check.Second equation: sin y + cos z = √2/2 + √2/2 = √2. Check.Third equation: sin z + cos x = √2/2 + √2/2 = √2. Check.Therefore, this works. Now, could there be other solutions where sin x + cos y = √2 without both sin x and cos y being √2/2? Let's see. Suppose sin x = a, cos y = b, with a + b = √2. What's the maximum possible value of a + b? Wait, actually, since sin x can be at most 1, and cos y can be at most 1, the maximum sum is 2. But in our case, we need a + b = √2. So, for example, sin x could be 1, cos y = √2 - 1 ≈ 0.414. Then sin x + cos y = 1 + (√2 -1) = √2. Similarly, sin x could be 0.9, cos y = √2 - 0.9 ≈ 0.514, which is possible because cos y can be 0.514. So there might be other solutions where sin x and cos y are not both √2/2. However, then we need to check if such solutions can satisfy the other equations. Let's explore this possibility.Suppose in the first equation, sin x = 1 and cos y = √2 - 1. Then x = π/2 + 2πk, and y = arccos(√2 -1) + 2πn or y = -arccos(√2 -1) + 2πn. Then, moving to the second equation: sin y + cos z ≥ √2. Let's compute sin y. If y = arccos(√2 -1), then sin y = sqrt(1 - (√2 -1)^2). Let's compute that:(√2 -1)^2 = (2 - 2√2 +1) = 3 - 2√2. So sin y = sqrt(1 - (3 - 2√2)) = sqrt(1 -3 + 2√2) = sqrt(-2 + 2√2). Let's compute -2 + 2√2 ≈ -2 + 2.828 ≈ 0.828, so sqrt(0.828) ≈ 0.91. So sin y ≈ 0.91. Then sin y + cos z ≈ 0.91 + cos z. To make this sum ≥ √2 ≈ 1.414, we need cos z ≥ 1.414 - 0.91 ≈ 0.504. So z must be in the range where cos z ≥ 0.504, which is z ∈ [-arccos(0.504), arccos(0.504)] + 2πm. Then moving to the third equation: sin z + cos x. Since x is π/2 + 2πk, cos x = 0. So sin z + 0 ≥ √2. But sin z ≤1, so sin z ≥ √2 ≈1.414 is impossible. Therefore, this path leads to a contradiction. Therefore, such a solution where sin x =1 and cos y =√2 -1 doesn't work because the third equation can't be satisfied.Similarly, if we tried another combination where sin x and cos y are not equal to √2/2, we might run into similar issues. Let's test another possibility. Suppose sin x = √2/2 + ε and cos y = √2/2 - ε for some small ε. Then their sum is still √2. But then sin y would be sqrt(1 - (cos y)^2). Let's compute cos y = √2/2 - ε. Then (cos y)^2 = ( (√2/2)^2 - 2*(√2/2)*ε + ε² ) = (0.5 - √2 ε + ε²). Therefore, sin y = sqrt(1 - 0.5 + √2 ε - ε²) = sqrt(0.5 + √2 ε - ε²). For small ε, this is approximately sqrt(0.5) + (√2 ε)/(2 sqrt(0.5)) - ... ≈ √(0.5) + (√2 ε)/(√2) ) = √(0.5) + ε. So sin y ≈ √(0.5) + ε. Then in the second equation: sin y + cos z ≈ √(0.5) + ε + cos z ≥ √2. So cos z ≈ √2 - √(0.5) - ε ≈ √2 - (√2)/2 - ε = (√2)/2 - ε. Then, moving to the third equation: sin z + cos x. cos x = sqrt(1 - (sin x)^2) = sqrt(1 - ( (√2/2 + ε)^2 )) ≈ sqrt(1 - (0.5 + √2 ε + ε²)) ≈ sqrt(0.5 - √2 ε - ε²) ≈ √(0.5) - (√2 ε)/(2√0.5) ≈ √0.5 - ε. Therefore, sin z + cos x ≈ sqrt(1 - (cos z)^2) + √0.5 - ε. Compute cos z ≈ (√2)/2 - ε, so (cos z)^2 ≈ 0.5 - √2 ε + ε². Therefore, sin z ≈ sqrt(1 - 0.5 + √2 ε - ε²) ≈ sqrt(0.5 + √2 ε - ε²) ≈ √0.5 + (√2 ε)/(2√0.5) ≈ √0.5 + ε. Therefore, sin z + cos x ≈ (√0.5 + ε) + (√0.5 - ε) = √0.5 + √0.5 = √2. So this would satisfy the third equation. But wait, this seems to hold for small ε? But does this actually work?Wait, but we need to check all equations. If we perturb ε a little, but in reality, the variables are interdependent. Let me formalize this. Suppose:sin x = √2/2 + εcos y = √2/2 - εThen, from the second equation:sin y + cos z ≥ √2But sin y = sqrt(1 - (cos y)^2) = sqrt(1 - (√2/2 - ε)^2 ) ≈ sqrt(1 - (0.5 - √2 ε + ε²)) ≈ sqrt(0.5 + √2 ε - ε²) ≈ sqrt(0.5) + (√2 ε)/(2 sqrt(0.5)) = √(0.5) + εSo sin y ≈ √0.5 + εThen, sin y + cos z ≈ √0.5 + ε + cos z ≥ √2Therefore, cos z ≈ √2 - √0.5 - ε ≈ √2 - (√2)/2 - ε = (√2)/2 - εThen, cos z ≈ √2/2 - εTherefore, sin z = sqrt(1 - (cos z)^2) ≈ sqrt(1 - (0.5 - √2 ε + ε²)) ≈ sqrt(0.5 + √2 ε - ε²) ≈ √0.5 + εThen, the third equation:sin z + cos x ≈ (√0.5 + ε) + cos x ≥ √2But cos x = sqrt(1 - (sin x)^2) ≈ sqrt(1 - (0.5 + √2 ε + ε²)) ≈ sqrt(0.5 - √2 ε - ε²) ≈ √0.5 - (√2 ε)/(2√0.5) - ... ≈ √0.5 - εTherefore, sin z + cos x ≈ (√0.5 + ε) + (√0.5 - ε) = √0.5 + √0.5 = √2. So equality holds.Therefore, even if we perturb ε a bit, the third equation still holds. Wait, but in reality, this is only a first-order approximation. If ε is not small, the higher-order terms may affect the result. Let me test with a specific ε.Let’s take ε = 0.1. Then:sin x = √2/2 + 0.1 ≈ 0.7071 + 0.1 = 0.8071cos y = √2/2 - 0.1 ≈ 0.7071 - 0.1 = 0.6071Compute sin y: sqrt(1 - (0.6071)^2) ≈ sqrt(1 - 0.3685) ≈ sqrt(0.6315) ≈ 0.7946Then, sin y + cos z ≈ 0.7946 + cos z ≥ √2 ≈ 1.4142Thus, cos z ≈ 1.4142 - 0.7946 ≈ 0.6196Then, sin z = sqrt(1 - (0.6196)^2) ≈ sqrt(1 - 0.384) ≈ sqrt(0.616) ≈ 0.785Then, cos x = sqrt(1 - (0.8071)^2) ≈ sqrt(1 - 0.6514) ≈ sqrt(0.3486) ≈ 0.5904Then, sin z + cos x ≈ 0.785 + 0.5904 ≈ 1.3754 < √2 ≈ 1.4142. So this does not satisfy the third inequality. Therefore, the approximation breaks down when ε is not very small. Therefore, such a solution doesn't hold. Therefore, maybe the only solution is when ε = 0, i.e., sin x = cos y = √2/2, leading to x = π/4 + 2πk, y = π/4 + 2πn, etc. Alternatively, perhaps there's a symmetrical solution where all variables are π/4 + 2πk, but also other symmetrical solutions? Let's check another angle. Suppose x = y = z = π/4. Then:First equation: sin(π/4) + cos(π/4) = √2/2 + √2/2 = √2. Check.Second equation: same. Check.Third equation: same. Check.So that works. Similarly, if all variables are 5π/4, let's see:sin(5π/4) = -√2/2, cos(5π/4) = -√2/2. Then sin x + cos y = -√2/2 + (-√2/2) = -√2 < √2. Doesn't satisfy. So no.What if variables are 3π/4?sin(3π/4) = √2/2, cos(3π/4) = -√2/2. Then sin x + cos y = √2/2 + (-√2/2) = 0 < √2. Doesn't work.What if x = π/4, y = -π/4, z = π/4.First equation: sin(π/4) + cos(-π/4) = √2/2 + √2/2 = √2. Check.Second equation: sin(-π/4) + cos(π/4) = -√2/2 + √2/2 = 0 < √2. Doesn't work.Therefore, only when all variables are π/4 + 2πk does the system hold. Similarly, what if variables are in other quadrants but shifted by 2π? For example, x = π/4 + 2π, y = π/4 + 2π, etc. Same result. So the general solution is x = π/4 + 2πk, y = π/4 + 2πn, z = π/4 + 2πm for integers k, n, m.But wait, let me check another possibility. Suppose x = π/4, y = π/4, z = π/4. That works. What if x = 9π/4 (which is equivalent to π/4 + 2π), y = π/4 - 2π, z = π/4 + 4π. All these are just co-terminal angles with π/4, so effectively the same solution. Therefore, the solutions are all triples where each variable is congruent to π/4 modulo 2π. So x ≡ π/4 (mod 2π), y ≡ π/4 (mod 2π), z ≡ π/4 (mod 2π).Therefore, the solution set is:x = π/4 + 2πk,y = π/4 + 2πn,z = π/4 + 2πm,where k, n, m are integers.I think that's the only solution. Let me verify once more with different values. Suppose x = π/4, y = π/4 + 2π, z = π/4 - 2π. Then sin x = sin(π/4) = √2/2, cos y = cos(π/4 + 2π) = cos(π/4) = √2/2. So first equation holds. Similarly, second equation: sin y = sin(π/4 + 2π) = √2/2, cos z = cos(π/4 - 2π) = cos(π/4) = √2/2. So second equation holds. Third equation: sin z = sin(π/4 - 2π) = sin(π/4) = √2/2, cos x = cos(π/4) = √2/2. So all equations hold. Therefore, shifting any variable by multiples of 2π still works. Therefore, the general solution is as above.Are there any other solutions? Let's suppose that one variable is π/4 + 2πk, and another variable is π/4 + 2πn, but the third variable is something else. Let's say x = π/4, y = π/4, z = something else.From the second equation: sin y + cos z = √2/2 + cos z = √2 ⇒ cos z = √2 - √2/2 = √2/2 ⇒ z = ±π/4 + 2πm. If z = π/4 + 2πm, then third equation holds. If z = -π/4 + 2πm, then sin z = -√2/2, cos x = √2/2 ⇒ sum is 0 < √2. Doesn't work. Therefore, z must be π/4 + 2πm. Therefore, even if two variables are set to π/4, the third must be as well. Therefore, all variables must be π/4 modulo 2π.Therefore, the only solutions are where x, y, z are all equal to π/4 plus integer multiples of 2π.Thus, the solution set is:x = π/4 + 2πk,y = π/4 + 2πn,z = π/4 + 2πm,for integers k, n, m.Hence, that's the answer.Final AnswerThe solutions are all triples where each variable is an odd multiple of π/4. Thus, the system is satisfied when ( x = frac{pi}{4} + 2pi k ), ( y = frac{pi}{4} + 2pi n ), and ( z = frac{pi}{4} + 2pi m ) for integers ( k, n, m ). Therefore, the solution is:boxed{x = frac{pi}{4} + 2pi k, y = frac{pi}{4} + 2pi n, z = frac{pi}{4} + 2pi m} quad text{for integers } k, n, m.</think>To solve the system of inequalities:begin{aligned}& sin x + cos y geq sqrt{2} & sin y + cos z geq sqrt{2} & sin z + cos x geq sqrt{2}end{aligned}we start by noting that the maximum value of (sin theta + cos theta) is (sqrt{2}), which occurs when (theta = frac{pi}{4} + 2pi k) for integer (k). Given the cyclic nature of the inequalities, we consider the possibility that each inequality must achieve equality for the system to hold.1. For (sin x + cos y = sqrt{2}), the maximum value is achieved when both (sin x) and (cos y) are (frac{sqrt{2}}{2}). This happens when (x = frac{pi}{4} + 2pi k) and (y = frac{pi}{4} + 2pi n).2. Similarly, for (sin y + cos z = sqrt{2}), we need (sin y = frac{sqrt{2}}{2}) and (cos z = frac{sqrt{2}}{2}), leading to (y = frac{pi}{4} + 2pi n) and (z = frac{pi}{4} + 2pi m).3. For (sin z + cos x = sqrt{2}), we need (sin z = frac{sqrt{2}}{2}) and (cos x = frac{sqrt{2}}{2}), leading to (z = frac{pi}{4} + 2pi m) and (x = frac{pi}{4} + 2pi k).By substituting these values back into the original inequalities, we verify that each equation holds true. Any deviation from these values leads to contradictions or fails to satisfy the inequalities. Thus, the solutions must be such that each variable is (frac{pi}{4}) plus an integer multiple of (2pi).The final solution is:[boxed{x = frac{pi}{4} + 2pi k, y = frac{pi}{4} + 2pi n, z = frac{pi}{4} + 2pi m} quad text{for integers } k, n, m.],<think>Alright, so I need to solve this system of inequalities involving sine and cosine functions. Let me write them down again to make sure I have them right:1. sin x + cos y ≥ √22. sin y + cos z ≥ √23. sin z + cos x ≥ √2Hmm, okay. Each inequality has a sine of one variable and a cosine of another, all connected in a cyclic way: x with y, y with z, z with x. Interesting. The right side for each is √2, which is approximately 1.414. But I know that both sine and cosine functions have maximum values of 1. So, sin x and cos y each can be at most 1. Therefore, the left-hand side of each inequality, which is the sum of a sine and a cosine, can be at most 2. But here, they need to be at least √2. However, the maximum possible for each is 2, but the question is about when they are at least √2. But maybe the key here is that √2 is actually the maximum of sin θ + cos θ for a single variable θ. Wait, let me check that.For a single variable θ, the maximum value of sin θ + cos θ is indeed √2. Because if you consider sin θ + cos θ = √2 sin(θ + π/4), so the maximum is √2 and the minimum is -√2. So, each of these inequalities is of the form sin a + cos b ≥ √2. Since the maximum possible value of sin a + cos b would be 1 + 1 = 2, but maybe there's a combination where sin a and cos b can be maximized such that their sum is at least √2. But if sin a + cos b ≥ √2, and since the maximum of sin a + cos b (if a and b are independent variables) is 2, but perhaps there are specific relationships between a and b that can make their sum reach exactly √2? Wait, but sin a + cos b can be greater than √2 even if a and b are different variables. For example, if sin a is 1 and cos b is 1, then their sum is 2, which is greater than √2. So, but maybe the system here requires each of these inequalities to hold simultaneously. So, perhaps there are certain constraints on x, y, z such that all three inequalities are satisfied.But since each inequality requires a sum of sin and cos to be at least √2, which is a pretty high threshold. Let me think about when sin a + cos b can be equal to or greater than √2.If sin a + cos b ≥ √2, then since sin a ≤ 1 and cos b ≤ 1, the only way their sum can be √2 or more is if both sin a and cos b are as large as possible. For example, if sin a = 1 and cos b = 1, then their sum is 2, which is greater than √2. Alternatively, if sin a is slightly less than 1, say sin a = √2/2, then cos b would have to be at least √2 - √2/2 = √2/2. But maybe the minimal case where equality holds is when sin a = cos b = √2/2, but that sum would be √2. However, since sin a and cos b can be larger, the sum can be larger. But perhaps the key is that for each inequality, sin a and cos b must each be at least √2/2? Wait, not necessarily. For example, if sin a is 1, then cos b can be √2 - 1 ≈ 0.414, which is still positive. So, there are different possibilities.But maybe the system requires that all three inequalities hold. So, perhaps we need to find x, y, z such that:sin x + cos y ≥ √2,sin y + cos z ≥ √2,sin z + cos x ≥ √2.Given that each of these has a sum involving sin of one variable and cos of another, perhaps there is some symmetry here. Let me check if there's a solution where all inequalities become equalities. That is:sin x + cos y = √2,sin y + cos z = √2,sin z + cos x = √2.If such a solution exists, maybe that's the only solution? Or maybe there are multiple solutions. Let me think.Suppose we set each sin term and cos term to be equal to √2/2. Because if sin x = √2/2 and cos y = √2/2, then their sum is √2. Similarly for the others. But sin x = √2/2 implies that x = π/4 + 2πk or 3π/4 + 2πk for some integer k. Similarly, cos y = √2/2 implies that y = ±π/4 + 2πn for some integer n. Similarly for the other variables.But if we take x = π/4, y = -π/4 (since cos(-π/4) = √2/2), then sin x = √2/2, cos y = √2/2, so their sum is √2. Then for the second equation, sin y + cos z = √2. But sin y would be sin(-π/4) = -√2/2. Then we have -√2/2 + cos z = √2, which would require cos z = (3√2)/2, which is impossible because cosine cannot exceed 1. Therefore, that approach doesn't work.So, maybe setting sin x and cos y both to √2/2 doesn't work because then the next equation fails. Hmm. So, perhaps we need a different approach.Alternatively, if we set sin x = 1 and cos y = √2 - 1 ≈ 0.414. Then sin x + cos y = 1 + (√2 -1) = √2. Similarly, if sin y = 1, then cos z must be √2 -1, and sin z =1, then cos x must be √2 -1. But this would require x, y, z to be such that sin x =1, sin y=1, sin z=1, and cos y=√2 -1, cos z=√2 -1, cos x=√2 -1.But sin x =1 implies that x = π/2 + 2πk. Then cos x would be cos(π/2 + 2πk) = 0. But we need cos x = √2 -1 ≈ 0.414. But cos x can't be both 0 and 0.414. Contradiction. Therefore, this approach also fails.So, perhaps we need all the sin terms and cos terms to be 1. For example, sin x =1, cos y=1. Then sin x + cos y =2 ≥ √2. Similarly, sin y=1, cos z=1, and sin z=1, cos x=1. But then, if sin x =1, then x= π/2 + 2πk. Then cos x=0, but the third inequality requires sin z + cos x ≥ √2. Since cos x=0, sin z must be ≥ √2, which is impossible because sin z ≤1. So, this approach also fails.Hmm. So, trying to set sin or cos to 1 doesn't work because it causes contradictions in the other equations. So maybe another approach is needed.Let me recall that for a single variable θ, sin θ + cos θ reaches its maximum at √2 when θ= π/4 + 2πk. Similarly, its minimum is -√2. But here, we have different variables in each sine and cosine. So, maybe if we set x, y, z such that each variable is π/4 + 2πk, but then let's check.If x = y = z = π/4, then sin x = √2/2, cos y = √2/2, so sin x + cos y = √2. Similarly for the other equations. So, all three inequalities would hold as equalities. So, x = y = z = π/4 + 2πk is a solution. Similarly, adding multiples of 2π to each variable would also work.But are there other solutions?Suppose x, y, z are all equal to π/4 + 2πk. Then yes, each inequality becomes √2/2 + √2/2 = √2. So equality holds. So that's a solution.But maybe there are other solutions where variables are different?Suppose x = π/4, y = -π/4, z = 7π/4. Let me check.sin x = sin(π/4) = √2/2cos y = cos(-π/4) = √2/2. So first inequality holds.sin y = sin(-π/4) = -√2/2cos z = cos(7π/4) = √2/2. So sin y + cos z = -√2/2 + √2/2 = 0 < √2. So the second inequality fails. So that's no good.Alternatively, if x = π/2, then sin x =1. Then cos y needs to be at least √2 -1 ≈0.414. So y could be any number where cos y ≥ √2 -1. Similarly, sin y must then be ≥ √2 - cos z, and so on. But this seems complicated.Alternatively, maybe all variables must be at angles where their sine and cosine values are maximized. But since each equation has a sine and cosine of different variables, maybe there's a cyclic dependency. Let me think.Suppose that sin x = cos y, sin y = cos z, sin z = cos x. Then each inequality would be sin x + sin x = 2 sin x ≥ √2, which implies sin x ≥ √2/2. Similarly for the others. So, sin x ≥ √2/2, sin y ≥ √2/2, sin z ≥ √2/2, and also cos y = sin x, cos z = sin y, cos x = sin z.So, if cos y = sin x, then sin x = cos y = sin(π/2 - y). Therefore, x = π/2 - y + 2πk or x = π/2 + y + 2πk.Similarly, cos z = sin y implies z = π/2 - y + 2πm or z = π/2 + y + 2πm.But this seems to lead to a system of equations. Let's try to set x = π/2 - y, z = π/2 - y, etc., but this might get too convoluted.Alternatively, suppose that x = y = z. Then the system becomes:sin x + cos x ≥ √2,three times. But sin x + cos x = √2 sin(x + π/4), which has a maximum of √2. Therefore, equality holds when x + π/4 = π/2 + 2πk, i.e., x = π/4 + 2πk. Therefore, if x = y = z = π/4 + 2πk, then all three inequalities hold as equalities. So that's a solution.But are there solutions where x, y, z are not all equal?Suppose x, y, z differ by multiples of π/2. Let me try x = π/4, y = π/4, z = π/4. Then all equations hold as equalities. If x = 5π/4, which is π/4 + π, then sin x = -√2/2, cos y would need to be √2 - (-√2/2) = (3√2)/2, which is impossible. So that doesn't work.Alternatively, suppose x = π/4, y = π/4 + 2π, z = π/4 + 2π. Then they are still solutions.Alternatively, consider shifting all variables by 2π, which doesn't change their sine and cosine values, so those are also solutions.But are there other solutions?Suppose that one variable is π/4 + 2πk, and another is -π/4 + 2πm, but then as I tried earlier, this might not satisfy all three inequalities.Alternatively, consider that for each inequality, the maximum possible value is 2, but we need only that each is at least √2. So, maybe there are regions where each inequality is satisfied, and we need to find the intersection of those regions.But since the inequalities are cyclic, perhaps the only solutions are those where all variables are π/4 + 2πk, but let me check.Wait, let's consider the first inequality: sin x + cos y ≥ √2.The maximum of sin x is 1, so cos y must be at least √2 -1 ≈0.414. Similarly, since cos y ≥ √2 - sin x.But if sin x is large, then cos y can be smaller. However, we have to satisfy all three inequalities. So, for example, if sin x is 1, then cos y must be at least √2 -1. Then, in the second inequality, sin y + cos z ≥ √2. Since cos y ≥ √2 -1, then y is within the region where cos y is at least √2 -1, so y ∈ [-arccos(√2 -1), arccos(√2 -1)] + 2πn. Then, sin y in that interval. Let's compute arccos(√2 -1).But √2 ≈1.414, so √2 -1 ≈0.414. arccos(0.414) is approximately 65.5 degrees or π/2.8 radians. So y is within approx ±1.14 radians. Then sin y in that interval would be between sin(-1.14) ≈ -0.912 and sin(1.14) ≈0.912. Therefore, sin y can be up to ~0.912, so cos z must be at least √2 -0.912 ≈1.414 -0.912 ≈0.502. Then z must be within arccos(0.502) ≈59.9 degrees or ~1.045 radians. Then in the third inequality, sin z + cos x ≥√2. sin z is up to ~0.861 (sin(1.045) ≈0.861), so cos x must be at least √2 -0.861 ≈0.553. Then x must be within arccos(0.553) ≈56.4 degrees or ~0.984 radians. Then in the first inequality, sin x + cos y ≥√2. sin x is up to ~0.836 (sin(0.984)≈0.836), so cos y must be at least √2 -0.836≈0.578. But earlier we had cos y ≥0.414. So this would require cos y ≥0.578, which is a stricter condition. Then y must be within arccos(0.578)≈54.7 degrees or ~0.954 radians. Then sin y is up to ~0.816. So cos z must be at least √2 -0.816≈0.598. Then z must be within arccos(0.598)≈53.1 degrees. Then sin z ≈0.799. So cos x must be at least √2 -0.799≈0.615. Then x within arccos(0.615)≈51.3 degrees. Then sin x≈0.780. Then cos y ≥√2 -0.780≈0.634. So y within arccos(0.634)≈50.7 degrees. sin y≈0.773. cos z≥√2 -0.773≈0.641. z within arccos(0.641)≈49.9 degrees. sin z≈0.764. cos x≥√2 -0.764≈0.650. x within arccos(0.650)≈48.5 degrees. sin x≈0.750. cos y≥√2 -0.750≈0.664. y within arccos(0.664)≈47.5 degrees. sin y≈0.737. cos z≥√2 -0.737≈0.677. z within arccos(0.677)≈46.2 degrees. sin z≈0.722. cos x≥√2 -0.722≈0.692. x within arccos(0.692)≈45.0 degrees. sin x≈0.707. cos y≥√2 -0.707≈0.707. So y within arccos(0.707)=45 degrees. Then sin y=√2/2≈0.707. cos z≥√2 -0.707≈0.707. z within 45 degrees. sin z≈0.707. cos x≥√2 -0.707≈0.707. x within 45 degrees. So now, we've narrowed it down to 45 degrees. Then sin x=√2/2, cos y=√2/2, so x=π/4, y=π/4. Then similarly, z=π/4. Therefore, the only solution is when all variables are π/4 + 2πk.Therefore, through this iterative narrowing down, it seems that the only solution is when x = y = z = π/4 + 2πk, where k is an integer. So the system essentially forces all variables to be at π/4 modulo 2π.Let me verify this.If x = y = z = π/4, then:sin(π/4) + cos(π/4) = √2/2 + √2/2 = √2. So equality holds for all three inequalities. Similarly, if we add 2πk to each variable, since sine and cosine are 2π-periodic, the values remain the same.Now, suppose that x = π/4 + 2πk, y = π/4 + 2πm, z = π/4 + 2πn. Then, since the inequalities are separate, even if k, m, n are different integers, the sines and cosines would still evaluate to √2/2, so the sum would still be √2. Therefore, the solutions are all triples where each variable is π/4 modulo 2π. So x ≡ π/4 mod 2π, y ≡ π/4 mod 2π, z ≡ π/4 mod 2π.But wait, does this mean that x, y, z must each be congruent to π/4 modulo 2π, but independently? For example, x could be π/4 + 2πk, y could be π/4 + 2πm, z could be π/4 + 2πn, where k, m, n are integers. So, each variable is independently allowed to be π/4 plus any multiple of 2π. So, as long as each variable is in that form, the inequalities hold as equalities.But are there solutions where the variables are not exactly π/4 + 2πk, but still satisfy all three inequalities?Suppose x = π/4 + ε, y = π/4 + δ, z = π/4 + γ, where ε, δ, γ are small perturbations. Let's see if such perturbations can exist while still satisfying all inequalities.First inequality: sin(π/4 + ε) + cos(π/4 + δ) ≥ √2.Using Taylor expansion:sin(π/4 + ε) ≈ sin(π/4) + ε cos(π/4) - (ε²/2) sin(π/4) = √2/2 + ε√2/2 - (ε²/2)(√2/2).cos(π/4 + δ) ≈ cos(π/4) - δ sin(π/4) - (δ²/2) cos(π/4) = √2/2 - δ√2/2 - (δ²/2)(√2/2).Adding these together:√2/2 + ε√2/2 - (ε²/2)(√2/2) + √2/2 - δ√2/2 - (δ²/2)(√2/2) ≈ √2 + (ε - δ)√2/2 - (ε² + δ²)(√2)/4.So, to first order, the perturbation (ε - δ)√2/2 must be non-negative. Similarly, for the inequality to hold, the linear term (ε - δ) must be ≥0. Similarly, the other inequalities would give:Second inequality: sin(π/4 + δ) + cos(π/4 + γ) ≥ √2.Expanding similarly:≈ √2/2 + δ√2/2 - (δ²/2)(√2/2) + √2/2 - γ√2/2 - (γ²/2)(√2/2) ≈ √2 + (δ - γ)√2/2 - (δ² + γ²)(√2)/4.So, the linear term (δ - γ) must be ≥0.Third inequality: sin(π/4 + γ) + cos(π/4 + ε) ≥ √2.Expanding:≈ √2/2 + γ√2/2 - (γ²/2)(√2/2) + √2/2 - ε√2/2 - (ε²/2)(√2/2) ≈ √2 + (γ - ε)√2/2 - (γ² + ε²)(√2)/4.So, the linear term (γ - ε) must be ≥0.So, putting the three inequalities together:1. ε - δ ≥0,2. δ - γ ≥0,3. γ - ε ≥0.From the first, ε ≥ δ.From the second, δ ≥ γ.From the third, γ ≥ ε.Combining these: ε ≥ δ ≥ γ ≥ ε. Therefore, ε = δ = γ.So, the only solution to these three inequalities is ε = δ = γ. So, all perturbations must be equal. Then, substituting back into the first inequality:The quadratic term would be - (ε² + δ²)(√2)/4, but since ε = δ = γ, this becomes - (2ε²)(√2)/4 = - (ε²√2)/2. Therefore, the total expression is √2 + 0 - (ε²√2)/2. For the inequality to hold, we need:√2 - (ε²√2)/2 ≥ √2 ⇒ - (ε²√2)/2 ≥0 ⇒ ε² ≤0 ⇒ ε=0.Therefore, the only solution in the neighborhood of π/4 is ε=δ=γ=0. Therefore, there are no other solutions near π/4 except the exact point where all variables are π/4 +2πk.Therefore, this suggests that the only solutions are those where x, y, z are all equal to π/4 modulo 2π.But wait, let's check another case where two variables are π/4 and the third is different. Suppose x=π/4, y=π/4, z= something else.First inequality: sin(π/4) + cos(π/4) = √2. So that's okay.Second inequality: sin(π/4) + cos z ≥ √2. So √2/2 + cos z ≥ √2 ⇒ cos z ≥ √2/2 ≈0.707. So z ∈ [-π/4, π/4] modulo 2π.Third inequality: sin z + cos(π/4) ≥ √2. sin z + √2/2 ≥ √2 ⇒ sin z ≥ √2/2 ≈0.707. Therefore, z ∈ [π/4, 3π/4] modulo 2π.So, combining these two requirements for z: z must be in both [-π/4, π/4] and [π/4, 3π/4] modulo 2π. The only intersection point is z=π/4. Therefore, z must be π/4. So, even if we fix x and y as π/4, z must also be π/4. So, this again leads to z=π/4 +2πk.Therefore, even if two variables are set to π/4, the third must be π/4 as well.Alternatively, suppose x=5π/4, which is π/4 + π. Then sin x = sin(5π/4) = -√2/2. Then cos y must be ≥√2 - (-√2/2)= (3√2)/2 ≈2.121, which is impossible. So, no solution here.Alternatively, suppose x=3π/4. Then sin x=√2/2. Then cos y ≥√2 - √2/2= √2/2. So y ∈ [-π/4, π/4] +2πn. Then, moving to the second inequality: sin y + cos z ≥√2. sin y is sin of y which is in [-π/4, π/4], so sin y ∈ [-√2/2, √2/2]. Therefore, cos z must be ≥√2 - sin y ≥ √2 - √2/2= √2/2. So, cos z ≥√2/2, meaning z ∈ [-π/4, π/4] +2πm. Then the third inequality: sin z + cos x ≥√2. sin z is in [-√2/2, √2/2], and cos x = cos(3π/4)= -√2/2. Therefore, sin z - √2/2 ≥√2 ⇒ sin z ≥ (3√2)/2, which is impossible. Therefore, x=3π/4 does not lead to a solution.Therefore, if x deviates from π/4, even if we adjust y and z accordingly, we can't satisfy all three inequalities. This further suggests that the only solutions are when all variables are π/4 +2πk.Another approach: consider that each inequality sin a + cos b ≥√2. For each inequality to hold, we must have both sin a and cos b as large as possible. The maximum sum is 2, but we need at least √2. However, given the cyclic nature of the system, perturbing one variable affects the next equation.Alternatively, think of this as a system where each equation depends on two variables cyclically. If we assume that all variables are equal, we get the solution x=y=z=π/4 +2πk. If variables are not equal, does a solution exist?Suppose x=π/4, y=π/4 +2π, z=π/4 +4π. Then all equations still hold. So, adding multiples of 2π to individual variables is allowed. But they still are essentially π/4.Alternatively, suppose x=π/4, y=π/4, z=9π/4 (which is π/4 +2π). Then sin z=sin(π/4)=√2/2, cos x=cos(π/4)=√2/2. So the third inequality holds. So, variables can individually be shifted by 2π and still work.But if we try to shift one variable by π/2, say x=π/4 +π/2=3π/4. Then sin x=√2/2, cos y needs to be ≥√2 -√2/2=√2/2. So y ∈ [-π/4, π/4] +2πn. Then, sin y + cos z ≥√2. sin y is at most √2/2, so cos z must be ≥√2 -√2/2=√2/2. So z ∈ [-π/4, π/4] +2πm. Then, third inequality: sin z + cos x. sin z is at most √2/2, cos x=cos(3π/4)= -√2/2. So sin z -√2/2 ≥√2 ⇒ sin z ≥ (3√2)/2. Impossible. So again, no solution.Therefore, any deviation from x=π/4 +2πk seems to break the system. Hence, the only solution is x=y=z=π/4 +2πk, where k is any integer. But since the variables are independent, each can have their own integer multiple of 2π. So the general solution is x=π/4 +2πk, y=π/4 +2πm, z=π/4 +2πn, for integers k, m, n.Alternatively, writing in terms of congruences:x ≡ π/4 mod 2π,y ≡ π/4 mod 2π,z ≡ π/4 mod 2π.Therefore, the system has solutions only when all three variables are congruent to π/4 modulo 2π. Therefore, the solution set is:x = π/4 + 2πk,y = π/4 + 2πm,z = π/4 + 2πn,where k, m, n ∈ ℤ.To confirm, let's substitute back into the original inequalities.First inequality: sin(π/4 +2πk) + cos(π/4 +2πm) = sin(π/4) + cos(π/4) = √2/2 + √2/2 = √2. So equality holds.Similarly, the other two inequalities also hold as equalities. Therefore, these are indeed solutions.Now, we need to check if there are any other solutions. Suppose there exists a solution where at least one variable is not π/4 modulo 2π. For example, suppose x=π/4, y=π/4, but z≠π/4 +2πn. Then, from the third inequality: sin z + cos x ≥√2. Since cos x=cos(π/4)=√2/2, we need sin z ≥√2 -√2/2=√2/2. Therefore, z must be in [π/4, 3π/4] modulo 2π. Similarly, from the second inequality: sin y + cos z ≥√2. Since sin y=sin(π/4)=√2/2, we need cos z ≥√2 -√2/2=√2/2. Therefore, z must be in [-π/4, π/4] modulo 2π. Combining these two, z must be in both [π/4, 3π/4] and [-π/4, π/4]. The only intersection is at z=π/4. Therefore, z must be π/4 +2πn. Hence, even if we fix two variables as π/4, the third must be π/4. Hence, no other solutions exist.Therefore, the only solutions are when all three variables are equal to π/4 modulo 2π. Hence, the solution set is as mentioned earlier.Final AnswerThe system of inequalities is satisfied when all variables are equal to (frac{pi}{4}) plus an integer multiple of (2pi). Thus, the solutions are:[boxed{x = frac{pi}{4} + 2kpi, quad y = frac{pi}{4} + 2mpi, quad z = frac{pi}{4} + 2npi}]where (k), (m), and (n) are integers.</think>To solve the system of inequalities:[begin{aligned}& sin x + cos y geq sqrt{2} & sin y + cos z geq sqrt{2} & sin z + cos x geq sqrt{2}end{aligned}]we start by noting that the maximum value of (sin theta + cos theta) for a single variable (theta) is (sqrt{2}), which occurs when (theta = frac{pi}{4} + 2kpi) for any integer (k). Given the cyclic nature of the system, we consider the possibility that all variables (x), (y), and (z) are such that each inequality becomes an equality. This leads us to check if (x = y = z = frac{pi}{4} + 2kpi) satisfies all three inequalities. Substituting (x = y = z = frac{pi}{4}) into each inequality, we get:- (sin frac{pi}{4} + cos frac{pi}{4} = frac{sqrt{2}}{2} + frac{sqrt{2}}{2} = sqrt{2})- (sin frac{pi}{4} + cos frac{pi}{4} = sqrt{2})- (sin frac{pi}{4} + cos frac{pi}{4} = sqrt{2})Thus, all inequalities hold as equalities. To check if there are other solutions, we consider perturbations around (frac{pi}{4}) and find that any deviation leads to contradictions in the system. Additionally, analyzing the intervals for each variable shows that the only intersection point where all conditions are satisfied is when each variable is (frac{pi}{4}) modulo (2pi).Therefore, the solutions to the system of inequalities are:[boxed{x = frac{pi}{4} + 2kpi, quad y = frac{pi}{4} + 2mpi, quad z = frac{pi}{4} + 2npi}]where (k), (m), and (n) are integers.]

🔑:SOLUTION. By adding all three given inequalities, we obtain the inequality(sin x+cos x)+(sin y+cos y)+(sin z+cos z) geqq 3 sqrt{2}On the other hand, for any real number t we have { }^{3}begin{aligned}sin t+cos t & =sqrt{2}left(frac{sqrt{2}}{2} sin t+frac{sqrt{2}}{2} cos tright)=sqrt{2}left(sin t cos frac{pi}{4}+cos t sin frac{pi}{4}right)= & =sqrt{2} sin left(t+frac{pi}{4}right) leqq sqrt{2}end{aligned}since sin left(t+frac{pi}{4}right) leqq 1; equality in the second line (2) occurs if and only if t+frac{pi}{4}= =frac{pi}{2}+2 k pi, or t=frac{pi}{4}+2 k pi for some integer k. By adding three estimates (2) for t=x, t=y and t=z, we obtain the inequality(sin x+cos x)+(sin y+cos y)+(sin z+cos z) leqq 3 sqrt{2}Note that inequality (3) holds for any triple of real numbers (x, y, z), while the opposite inequality (1) holds for every solution of the original system. Thus, we find that inequality (1) can only be satisfied as an equality, which, according to the previous, occurs if and only if the numbers x, y, z are of the formx=frac{pi}{4}+2 k_{1} pi, quad y=frac{pi}{4}+2 k_{2} pi, quad z=frac{pi}{4}+2 k_{3} pi quadleft(k_{1}, k_{2}, k_{3} in mathbb{Z}right) .By substituting into the original system, we can easily verify that each such triple of numbers (x, y, z) is indeed a solution, as for it we havesin x=cos x=sin y=cos y=sin z=cos z=frac{sqrt{2}}{2} .(The mention of the check was necessary because inequality (1) is only a consequence of the given system: we could not rule out that for some triple of numbers (x, y, z) (1) holds, but one of the three original inequalities does not.)Note. Inequality (2) can be easily obtained from the inequality between the arithmetic and quadratic mean:sin t+cos t leqq|sin t|+|cos t| leqq 2 sqrt{frac{sin ^{2} t+cos ^{2} t}{2}}=sqrt{2}To solve the problem, we can also use other consequences of the given system instead of (1). For example, rewrite the first of the given inequalities in the form sin x geqq sqrt{2}-cos y. If this inequality holds, [^1] then the squared inequality sin ^{2} x geqq(sqrt{2}-cos y)^{2} also holds, since sqrt{2}-cos y>0 for any y in mathbb{R}. By adding three inequalitiessin ^{2} x geqq(sqrt{2}-cos y)^{2}, sin ^{2} y geqq(sqrt{2}-cos z)^{2}, sin ^{2} z geqq(sqrt{2}-cos x)^{2}we obtain the inequality after some rearrangements0 geqq(1-sqrt{2} cos x)^{2}+(1-sqrt{2} cos y)^{2}+(1-sqrt{2} cos z)^{2}from which everything needed follows.We can start solving the problem in yet another way. First, square each of the three given inequalities (this is a consequential modification, as the smaller (right) side of the inequality is positive) and then add them. After a simple rearrangement, the inequality2 sin x cos y+2 sin y cos z+2 sin z cos x geqq 3is obtained. To verify the general validity of the opposite inequality, we estimate each of the three terms on the left side from above using the "classical" inequality 2 u v leqq u^{2}+v^{2} (which is strict if u neq v):begin{aligned}& 2 sin x cos y+2 sin y cos z+2 sin z cos x leqq & leqqleft(sin ^{2} x+cos ^{2} yright)+left(sin ^{2} y+cos ^{2} zright)+left(sin ^{2} z+cos ^{2} xright)=3 .end{aligned}(Thus, we derive the necessary conditions sin x=cos y, sin y=cos z, sin z=cos x, under which the solution of the original system is already a trivial task.)

🧠:[<think>Okay, so I need to find the volume of a cone when given the slant height and the lateral area. Hmm, let's see. First, let me recall the formulas related to cones. The lateral area (which is also called the curved surface area) of a cone is given by πrl, where r is the radius and l is the slant height. The volume of a cone is (1/3)πr²h, where h is the height. The problem states that the slant height (l) is 5, and the lateral area is 20π. So I can start by using the lateral area formula to find the radius (r). Let's write down the formula for lateral area:Lateral Area = πrlPlugging in the known values:20π = π * r * 5Hmm, okay, so I can simplify this equation. Let me divide both sides by π first to cancel out π:20 = 5rThen divide both sides by 5:r = 20 / 5 = 4So the radius is 4 units. Now that I have the radius and the slant height, I can find the height (h) of the cone using the Pythagorean theorem. In a cone, the slant height (l), radius (r), and height (h) form a right-angled triangle, where l is the hypotenuse. So:l² = r² + h²We can rearrange this to solve for h:h² = l² - r²Plugging in the known values:h² = 5² - 4² = 25 - 16 = 9Then taking the square root of both sides:h = √9 = 3So the height is 3 units. Now that I have the radius and the height, I can plug these into the volume formula:Volume = (1/3)πr²hSubstituting the values:Volume = (1/3)π*(4)²*3Calculating that step by step:4² = 16Then 16 * 3 = 48Then 48 * (1/3) = 16So the volume is 16π. Let me just check my steps again to make sure I didn't make any mistakes.First, lateral area formula: πrl = 20π. Solved for r and got 4. Then used Pythagorean theorem to find h, ended up with h=3. Then volume formula. All steps check out. The arithmetic seems correct. 4 squared is 16, multiplied by 3 gives 48, divided by 3 gives 16. So yeah, the volume should be 16π. Wait, just thinking if there's another way to approach this. Let me verify again. If the slant height is 5, radius 4, then h must be 3 because 3-4-5 is a Pythagorean triplet. That's a classic right triangle, so that makes sense. Then volume is 1/3 π r squared h. So 1/3 * π * 16 * 3. The 3s cancel, leaving 16π. Yep, that seems right. I don't think I missed anything here. The problem was straightforward once I remembered the formulas. So I think the answer is 16π.Final AnswerThe volume of the cone is boxed{16pi}.</think>Given the slant height ( l = 5 ) and the lateral area ( 20pi ) of a cone, we need to find its volume.First, we use the formula for the lateral area of a cone:[text{Lateral Area} = pi r l]Substituting the given values:[20pi = pi r cdot 5]Dividing both sides by (pi):[20 = 5r]Solving for ( r ):[r = frac{20}{5} = 4]Next, we use the Pythagorean theorem to find the height ( h ) of the cone. The slant height ( l ), radius ( r ), and height ( h ) form a right-angled triangle:[l^2 = r^2 + h^2]Substituting the known values:[5^2 = 4^2 + h^2]Simplifying:[25 = 16 + h^2]Solving for ( h^2 ):[h^2 = 25 - 16 = 9]Taking the square root:[h = sqrt{9} = 3]Finally, we calculate the volume of the cone using the formula:[text{Volume} = frac{1}{3} pi r^2 h]Substituting the values ( r = 4 ) and ( h = 3 ):[text{Volume} = frac{1}{3} pi (4)^2 cdot 3]Simplifying:[text{Volume} = frac{1}{3} pi cdot 16 cdot 3 = 16pi]The volume of the cone is boxed{16pi}.,<think>Okay, let's see. I need to find the volume of a cone. They gave me the slant height and the lateral area. Hmm, volume of a cone is (1/3)πr²h, right? So I need the radius and the height. But how do I get those from the slant height and lateral area?Wait, the lateral area is given as 20π. The formula for lateral (or curved) surface area of a cone is πrl, where r is the radius and l is the slant height. Oh, so they gave me l = 5 and lateral area = 20π. Let me write that down.Lateral area = πrl = 20π. So if I plug in l = 5, then πr*5 = 20π. Divide both sides by π: 5r = 20. Then divide by 5: r = 4. Okay, so the radius is 4. Got that.Now, slant height l is related to the radius r and height h by the Pythagorean theorem. Since the cone is a right circular cone, the slant height is the hypotenuse of the right triangle formed by r, h, and l. So, l² = r² + h².We know l = 5 and r = 4. So plug in those values: 5² = 4² + h². That's 25 = 16 + h². Subtract 16 from both sides: h² = 9. Take the square root: h = 3. Okay, so the height is 3.Now, plug r and h into the volume formula. Volume = (1/3)πr²h. So that's (1/3)π*(4)²*3. Let me compute that. 4 squared is 16. 16 times 3 is 48. Then (1/3) of 48 is 16. So the volume is 16π.Wait, let me double-check. Slant height 5, radius 4, so height 3. Yes, because 3-4-5 triangle. Lateral area: π*4*5=20π, which matches. Volume: (1/3)π*16*3=16π. Yep, seems right. So I think 16π is the answer.Final AnswerThe volume of the cone is boxed{16pi}.</think>Given the slant height ( l = 5 ) and the lateral area ( 20pi ) of a cone, we need to find the volume.First, we use the formula for the lateral (curved) surface area of a cone:[pi r l = 20pi]Substituting ( l = 5 ):[pi r cdot 5 = 20pi]Dividing both sides by ( pi ):[5r = 20]Solving for ( r ):[r = 4]Next, we use the Pythagorean theorem to find the height ( h ). The slant height ( l ), radius ( r ), and height ( h ) form a right triangle:[l^2 = r^2 + h^2]Substituting ( l = 5 ) and ( r = 4 ):[5^2 = 4^2 + h^2][25 = 16 + h^2]Solving for ( h ):[h^2 = 9][h = 3]Finally, we calculate the volume of the cone using the formula:[text{Volume} = frac{1}{3} pi r^2 h]Substituting ( r = 4 ) and ( h = 3 ):[text{Volume} = frac{1}{3} pi (4)^2 (3)][text{Volume} = frac{1}{3} pi cdot 16 cdot 3][text{Volume} = 16pi]The volume of the cone is (boxed{16pi}).]

🔑:Let's denote the radius of the base of the cone as r. Then, the lateral area S_{text{lateral}} = pi r times 5 = 20pi, therefore r = 4, therefore the height of the cone h= sqrt{5^2 - r^2} = 3, therefore the volume of the cone V= frac{1}{3}pi r^2 h = frac{1}{3} times 16pi times 3 = 16pi. Hence, the answer is: boxed{16pi}. This problem involves calculating the base radius using the lateral area formula, then finding the height of the cone, and finally substituting into the volume formula to find the volume. This question tests knowledge of the formulas for the lateral area and volume of a cone, and is considered a basic problem.