🧠:[<think>Okay, let's try to tackle this problem. We need to find all real numbers ( k ) such that the inequality[a^3 + b^3 + c^3 + d^3 + 1 geqslant k(a + b + c + d)]holds for all ( a, b, c, d in [-1, +infty) ).Hmm, so we have four variables here, each ranging from -1 to infinity. The inequality has to be true for all possible values of these variables in that range. The left side is the sum of their cubes plus 1, and the right side is ( k ) times the sum of the variables. We need to find the maximum possible ( k ) such that this inequality is always satisfied.First, maybe I should think about the case when all variables are equal. If we set ( a = b = c = d = t ), then the inequality becomes:[4t^3 + 1 geqslant 4kt]Dividing both sides by 4:[t^3 + frac{1}{4} geqslant kt]So, we need this inequality to hold for all ( t geq -1 ). Maybe analyzing this single-variable case can give some insight into the original problem? Not sure if that's the right approach, but let's see.Alternatively, maybe considering symmetry. Since the variables are symmetric, perhaps the minimal case occurs when all variables are equal? Or maybe when some variables take the minimal value of -1? Let's test some specific cases.First, let's check when all variables are -1. Then the left-hand side (LHS) becomes:[(-1)^3 + (-1)^3 + (-1)^3 + (-1)^3 + 1 = (-4) + 1 = -3]The right-hand side (RHS) is ( k(-1 -1 -1 -1) = -4k ). So the inequality becomes:[-3 geqslant -4k implies 4k geq 3 implies k leq frac{3}{4}]Wait, so if all variables are -1, then ( k ) must be at most 3/4. But we need the inequality to hold for all variables in the given range. So this gives an upper bound of ( k leq 3/4 ).But we need to check if ( k = 3/4 ) works for all possible ( a, b, c, d ). Let's test another case. Suppose all variables are 0. Then LHS is ( 0 + 0 + 0 + 0 + 1 = 1 ), and RHS is ( k(0 + 0 + 0 + 0) = 0 ). So 1 ≥ 0, which is always true. Doesn't give us any information about ( k ).Another case: suppose one variable is -1, and the others are 0. Let's say ( a = -1 ), ( b = c = d = 0 ). Then LHS is ( (-1)^3 + 0 + 0 + 0 + 1 = -1 + 1 = 0 ), and RHS is ( k(-1 + 0 + 0 + 0) = -k ). The inequality becomes ( 0 geq -k implies k geq 0 ). So ( k ) must be non-negative. But from the previous case, ( k leq 3/4 ). So possible ( k ) is between 0 and 3/4? But we need the maximum ( k ), so probably 3/4. Wait, but need to verify more cases.Wait, the problem is asking for all real numbers ( k ), but I think the answer is the maximal ( k ), but the problem says "all real numbers k" such that the inequality holds. But maybe there's a range? Wait, but if the inequality holds for some ( k ), then it will hold for any smaller ( k ), because the RHS becomes smaller. So the set of all real numbers ( k ) for which the inequality holds is ( (-infty, k_0] ), where ( k_0 ) is the maximum such value. Therefore, the problem is essentially asking for ( k_0 ).So the answer is likely ( boxed{dfrac{3}{4}} ). But we need to verify this.Wait, let's check another case where some variables are greater than -1. For example, suppose two variables are -1 and two variables are 0. Then LHS is ( 2(-1)^3 + 2(0)^3 + 1 = -2 + 0 + 1 = -1 ), and RHS is ( k(-1 -1 + 0 + 0) = -2k ). The inequality becomes ( -1 geq -2k implies 2k geq 1 implies k geq 1/2 ). But we already have ( k leq 3/4 ), so this gives a lower bound.But since we need the inequality to hold for all cases, the maximum ( k ) is the minimal upper bound given by all these cases, which so far is 3/4.But let's check another case where variables are not just -1 or 0. For example, take one variable as -1 and others as something else.Let’s take ( a = -1 ), and ( b = c = d = t geq -1 ). Then the inequality becomes:Left side: ( (-1)^3 + 3t^3 + 1 = -1 + 3t^3 + 1 = 3t^3 )Right side: ( k(-1 + 3t) )So the inequality is ( 3t^3 geq k(-1 + 3t) ).We need this to hold for all ( t geq -1 ). Let's set ( k = 3/4 ) and see if it holds.So substitute ( k = 3/4 ):Left side: ( 3t^3 )Right side: ( (3/4)(-1 + 3t) = -3/4 + (9/4)t )So inequality becomes:( 3t^3 + 3/4 - (9/4)t geq 0 )Multiply both sides by 4/3 to simplify:( 4t^3 + 1 - 3t geq 0 )So ( 4t^3 - 3t + 1 geq 0 )Let’s analyze this function ( f(t) = 4t^3 - 3t + 1 ).First, check at t = -1: ( 4(-1)^3 - 3(-1) + 1 = -4 + 3 + 1 = 0 ). So equality holds here.At t = 0: ( 0 - 0 + 1 = 1 geq 0 ).At t = 1: ( 4 - 3 + 1 = 2 geq 0 ).Derivative: ( f’(t) = 12t^2 - 3 ). Setting derivative to zero: ( 12t^2 - 3 = 0 implies t^2 = 1/4 implies t = 1/2 ) or ( t = -1/2 ).Check t = 1/2: ( 4*(1/8) - 3*(1/2) +1 = 0.5 - 1.5 +1 = 0 ).Check t = -1/2: ( 4*(-1/2)^3 - 3*(-1/2) +1 = 4*(-1/8) + 3/2 +1 = -0.5 + 1.5 +1 = 2 geq 0.So the function ( f(t) ) has roots at t = -1 and t = 1/2, and a local minimum at t = 1/2. Wait, but at t =1/2, f(t) = 0. So between t = -1 and t = 1/2, how does the function behave?From t = -1 to t =1/2, the function starts at 0, goes up to 2 at t = -1/2, then decreases to 0 at t =1/2. Then after t=1/2, it increases again. So between t = -1 and t =1/2, the function is non-negative? Let's check t =0, which is 1. Yes. So overall, the function is non-negative for all t ≥ -1. Therefore, with k=3/4, the inequality holds when a=-1 and others equal to t ≥ -1.That's a good sign.Another test case: set three variables at -1 and one variable at something else. Let’s say a = b = c = -1, d = t.Then LHS: ( 3(-1)^3 + t^3 + 1 = -3 + t^3 + 1 = t^3 -2 )RHS: ( k(-1 -1 -1 + t) = k(t - 3) )So inequality: ( t^3 - 2 geq k(t - 3) )Set k=3/4:Left side: t^3 -2Right side: (3/4)(t - 3) = (3/4)t - 9/4Inequality: t^3 -2 - (3/4)t + 9/4 ≥ 0Simplify: t^3 - (3/4)t + (9/4 - 2) = t^3 - (3/4)t + 1/4 ≥ 0Multiply by 4: 4t^3 - 3t +1 ≥0Wait, that's the same polynomial as before! So same analysis applies. We saw that this is non-negative for t ≥ -1. So when t ≥ -1, 4t^3 -3t +1 ≥0, so the inequality holds. Therefore, this case also works for k=3/4.Hmm, interesting. So in the previous two cases, even when setting variables to -1 and others to t, the inequality reduces to the same polynomial, which we saw is non-negative.Another test case: set two variables to some t, and others to -1. Let's say a = b = -1, c = d = t.Then LHS: 2*(-1)^3 + 2*t^3 +1 = -2 + 2t^3 +1 = 2t^3 -1RHS: k(-1 -1 + t + t) = k(2t -2)Inequality: 2t^3 -1 ≥ k(2t -2)Set k=3/4:Left: 2t^3 -1Right: (3/4)(2t -2) = (3/2)t - 3/2Inequality: 2t^3 -1 - (3/2)t + 3/2 ≥0Simplify: 2t^3 - (3/2)t + 1/2 ≥0Multiply by 2: 4t^3 -3t +1 ≥0Again, same polynomial. So same conclusion. Therefore, in these cases, the inequality holds for k=3/4.So far, all test cases where variables are set to -1 and others to t result in the same polynomial 4t^3 -3t +1 ≥0, which is non-negative for t ≥ -1. Therefore, k=3/4 seems to hold.But let's check another case where variables are not -1. For example, take all variables equal to t where t is something else. Let's say a = b = c = d = t. Then the inequality becomes:4t^3 +1 ≥ k(4t)We need this to hold for all t ≥ -1. Let's test t=1:Left: 4 +1 =5Right: 4kSo 5 ≥4k ⇒ k ≤5/4. But we already have k=3/4, which is less than 5/4, so this is okay.But maybe t is negative? Let's take t approaching -1. Let’s set t=-1:Left: 4*(-1)^3 +1 = -4 +1 = -3Right:4k*(-1)= -4kSo the inequality is -3 ≥ -4k ⇒4k ≥3 ⇒k ≥3/4. Which is exactly our upper bound. So here, equality is required.So, if k=3/4, then equality holds when t=-1.Another test case: take t=-0.5. Then:Left:4*(-0.5)^3 +1=4*(-1/8)+1= -0.5 +1=0.5Right:4*(3/4)*(-0.5)=3*(-0.5)=-1.5So 0.5 ≥-1.5, which is true.Wait, but this is a case where t is between -1 and 0. So the inequality holds here.What about t approaching infinity? Let’s take t approaching +∞. Then the left side is 4t^3, which grows much faster than the right side 4kt. So for large t, the inequality will hold as long as the leading coefficient on the left is greater than the right. Since 4t^3 dominates 4kt, the inequality holds for any k when t is large. So no problem there.Another test case: take one variable as something greater than -1, others as -1. Wait, we already did similar cases.Alternatively, take two variables as x and two as y. Maybe set a = b = x, c = d = y. Then the inequality becomes:2x^3 + 2y^3 +1 ≥ k(2x + 2y)But this might complicate things. Maybe not necessary.Alternatively, consider the case where variables are different. For example, a mix of -1, 0, and positive numbers.But maybe the minimal case is when variables are at their minimal value, i.e., -1. But we saw that when all variables are -1, the inequality gives k ≤3/4. And when variables are increased from -1, the left side grows (since cubes increase as variables increase beyond -1) and the right side is linear. So maybe the critical case is when all variables are -1.But we have to check all possibilities. Suppose three variables are -1, and the fourth is something else. Wait, we already did that. It resulted in the same polynomial.Alternatively, set a variable to a positive number. Let's take a = 2, and others = -1. Then:Left: 2^3 + 3*(-1)^3 +1 =8 -3 +1=6Right:k(2 + (-1)*3)=k(2 -3)= -kSo inequality:6 ≥ -k ⇒k ≥ -6. But since k is supposed to be at most 3/4 and non-negative (from previous cases), this doesn't affect our upper bound.Alternatively, take a =1, others=-1. Then:Left:1 +3*(-1)^3 +1=1 -3 +1=-1Right:k(1 -3)=k(-2)So inequality:-1 ≥ -2k ⇒2k ≥1 ⇒k ≥1/2. Which is okay since k=3/4 is larger than 1/2.Another case: set two variables to 1, two to -1:Left:2*(1) + 2*(-1)^3 +1=2 -2 +1=1Right:k(2*1 +2*(-1))=k(0)=0So 1 ≥0, which holds.So far, in all test cases, k=3/4 seems to satisfy the inequality.Alternatively, perhaps we can use Lagrange multipliers or take derivatives with respect to each variable. But since there are four variables, that might be complicated.Alternatively, consider that for each variable, the inequality can be considered individually. Let's consider each term. Suppose we can show that for each variable ( x in [-1, infty) ), ( x^3 + frac{1}{4} geq frac{3}{4}x ). Then, summing over all variables, we get:[sum_{i=1}^4 left( x_i^3 + frac{1}{4} right) geq frac{3}{4} sum_{i=1}^4 x_i][implies sum x_i^3 + 1 geq frac{3}{4} sum x_i]Which is exactly our inequality with ( k = 3/4 ). Therefore, if we can prove that for each ( x in [-1, infty) ), ( x^3 + 1/4 geq (3/4)x ), then the original inequality holds.So let's check this single-variable inequality: ( x^3 + frac{1}{4} geq frac{3}{4}x )Rearranging: ( x^3 - frac{3}{4}x + frac{1}{4} geq 0 )Multiply by 4 to eliminate denominators: ( 4x^3 - 3x + 1 geq 0 )Wait, this is the same polynomial as before! So we need to check if ( 4x^3 -3x +1 geq0 ) for all ( x geq -1 ).Let’s analyze this function ( f(x) = 4x^3 -3x +1 ).First, check at x = -1: ( 4*(-1)^3 -3*(-1) +1 = -4 +3 +1 =0 )At x =0: 0 -0 +1=1 ≥0At x=1:4 -3 +1=2≥0Find critical points. Take derivative:( f’(x) =12x^2 -3 )Set to zero:12x² -3=0 ⇒x²=3/12=1/4⇒x=1/2 or x=-1/2Evaluate f at x=1/2:4*(1/8) -3*(1/2) +1= 0.5 -1.5 +1=0At x=-1/2:4*(-1/2)^3 -3*(-1/2) +1=4*(-1/8)+3/2 +1= -0.5 +1.5 +1=2 ≥0So the function has roots at x=-1 and x=1/2, with a local maximum at x=-1/2 (f(-1/2)=2) and a local minimum at x=1/2 (f(1/2)=0). For x ≥ -1, the function is non-negative since:- For x ∈ [-1,1/2], the function starts at 0, goes up to 2 at x=-1/2, then decreases back to 0 at x=1/2.- For x ≥1/2, the function increases to infinity.Hence, ( 4x^3 -3x +1 geq0 ) for all x ≥ -1. Therefore, the single-variable inequality holds, and so summing over all variables gives the original inequality with k=3/4.Therefore, k=3/4 is indeed the maximum value such that the inequality holds for all a,b,c,d in [-1, ∞). Hence, the answer is boxed{dfrac{3}{4}}.Final Answerboxed{dfrac{3}{4}}</think>To find all real numbers ( k ) such that the inequality[a^3 + b^3 + c^3 + d^3 + 1 geqslant k(a + b + c + d)]holds for all ( a, b, c, d in [-1, +infty) ), we start by considering specific cases and analyzing the polynomial involved.1. Case when all variables are -1: - The left-hand side (LHS) becomes (-3) and the right-hand side (RHS) becomes (-4k). - The inequality (-3 geq -4k) gives ( k leq frac{3}{4} ).2. Case when one variable is -1 and others are 0: - The LHS is 0 and the RHS is (-k). - The inequality (0 geq -k) gives ( k geq 0 ).3. Case when one variable is -1 and others are ( t ): - The inequality reduces to ( 3t^3 geq k(-1 + 3t) ). - Setting ( k = frac{3}{4} ) and simplifying gives ( 4t^3 - 3t + 1 geq 0 ), which holds for all ( t geq -1 ).4. Case when three variables are -1 and one is ( t ): - The inequality reduces to ( t^3 - 2 geq k(t - 3) ). - Setting ( k = frac{3}{4} ) and simplifying gives ( 4t^3 - 3t + 1 geq 0 ), which holds for all ( t geq -1 ).5. General single-variable analysis: - For each variable ( x in [-1, infty) ), the inequality ( x^3 + frac{1}{4} geq frac{3}{4}x ) holds. - This simplifies to ( 4x^3 - 3x + 1 geq 0 ), which is non-negative for all ( x geq -1 ).By verifying these cases and analyzing the polynomial ( 4x^3 - 3x + 1 ), we conclude that the maximum value of ( k ) such that the inequality holds for all ( a, b, c, d in [-1, +infty) ) is ( frac{3}{4} ).Thus, the final answer is (boxed{dfrac{3}{4}}).,<think>Okay, so I need to find all real numbers k such that the inequality a³ + b³ + c³ + d³ + 1 is greater than or equal to k times (a + b + c + d) for all a, b, c, d in the interval [-1, +∞). Hmm, let's start by understanding what the problem is asking.First, the variables a, b, c, d can be any real numbers from -1 up to positive infinity. The inequality has to hold for all possible combinations of these variables in that range. So, my goal is to determine the maximum possible value of k such that this inequality is always true. If I can find such a k, that would be the answer.Since the inequality is symmetric in a, b, c, d, maybe I can consider cases where all variables are equal, or maybe set some variables to certain values to find constraints on k. But before jumping into that, let me think about possible strategies.One common approach for inequalities involving multiple variables is to use the method of Lagrange multipliers, but since this is an Olympiad problem, there might be a more straightforward method, perhaps using calculus or considering critical points where the inequality becomes tight (i.e., equality holds). Alternatively, maybe using convexity or other properties of the functions involved.Let me first test some boundary cases. Since the variables can be as low as -1, maybe the minimum of the left-hand side (LHS) minus the right-hand side (RHS) occurs at some of these boundary points. For example, when a, b, c, d are at their minimum value, -1. Let me compute the LHS and RHS in that case.If a = b = c = d = -1:LHS = (-1)^3 + (-1)^3 + (-1)^3 + (-1)^3 + 1 = (-1) + (-1) + (-1) + (-1) + 1 = -4 + 1 = -3RHS = k*(-1 + -1 + -1 + -1) = k*(-4)So the inequality becomes -3 ≥ -4k. Solving for k: -3 ≥ -4k ⇒ 3 ≤ 4k ⇒ k ≤ 3/4. Wait, so when all variables are -1, we get that k must be less than or equal to 3/4. Hmm, that gives an upper bound for k. But is this the tightest constraint?But maybe there are other cases where k needs to be even smaller. Let's test other cases.Another boundary case is when some variables are -1 and others are larger. Let me try setting three variables to -1 and one variable to something else. Let's say a = b = c = -1, and d is a variable that can vary in [-1, ∞). Then compute LHS - RHS as a function of d and find the minimal k such that this expression is non-negative for all d ≥ -1.So, let's set a = b = c = -1, d = t (where t ≥ -1). Then:LHS = (-1)^3 + (-1)^3 + (-1)^3 + t³ + 1 = (-3) + t³ + 1 = t³ - 2RHS = k*(-1 + -1 + -1 + t) = k*(t - 3)So the inequality becomes t³ - 2 ≥ k(t - 3) for all t ≥ -1.We need t³ - 2 - k(t - 3) ≥ 0 for all t ≥ -1.Let me denote f(t) = t³ - kt + (3k - 2). So f(t) ≥ 0 for all t ≥ -1.To find the minimal k such that f(t) ≥ 0 for all t ≥ -1, we need to ensure that f(t) is non-negative in that interval. Let's analyze this function.First, check at t = -1:f(-1) = (-1)^3 - k*(-1) + 3k - 2 = -1 + k + 3k - 2 = 4k - 3To have f(-1) ≥ 0: 4k - 3 ≥ 0 ⇒ k ≥ 3/4.Wait, that's interesting. So when three variables are -1 and the fourth is -1, we need k ≤ 3/4 (from the previous case), but when three variables are -1 and the fourth is t, at t = -1, we get that k must be ≥ 3/4. So this suggests that the only possible k is 3/4? Because k has to be both ≤ 3/4 and ≥ 3/4. So k = 3/4.But wait, is that sufficient? Let me verify. If k = 3/4, then f(t) = t³ - (3/4)t + (3*(3/4) - 2) = t³ - (3/4)t + (9/4 - 8/4) = t³ - (3/4)t + 1/4.We need to check if t³ - (3/4)t + 1/4 ≥ 0 for all t ≥ -1.Let's check t = -1:(-1)^3 - (3/4)*(-1) + 1/4 = -1 + 3/4 + 1/4 = -1 + 1 = 0. So equality holds here.What about t approaching infinity? As t → ∞, t³ dominates, so f(t) tends to infinity, which is positive.Now, check if there are any minima in between t ≥ -1 where f(t) might be negative. Take derivative of f(t):f'(t) = 3t² - 3/4. Set to zero:3t² - 3/4 = 0 ⇒ t² = 1/4 ⇒ t = ±1/2.But since t ≥ -1, critical points at t = 1/2 and t = -1/2.Compute f(-1/2):(-1/2)^3 - (3/4)*(-1/2) + 1/4 = (-1/8) + 3/8 + 1/4 = ( (-1 + 3)/8 ) + 2/8 = (2/8) + 2/8 = 4/8 = 1/2 ≥ 0.Compute f(1/2):(1/2)^3 - (3/4)*(1/2) + 1/4 = 1/8 - 3/8 + 2/8 = (1 - 3 + 2)/8 = 0/8 = 0.So at t = 1/2, f(t) = 0. So at t = 1/2, the function touches zero, and at t = -1/2, it's 1/2. Thus, between t = -1 and infinity, the minimal value of f(t) is zero. Therefore, when k = 3/4, the inequality holds for this case.But we need to check other cases as well. For example, if some variables are set to different values.Wait, earlier when all variables are -1, we had k ≤ 3/4. When three variables are -1 and the fourth is t, setting t = -1 forced k ≥ 3/4. So combining these two, k must be exactly 3/4. But we need to confirm that k = 3/4 works for all possible a, b, c, d in [-1, ∞).Alternatively, maybe there's another case where even with k = 3/4, the inequality might not hold. Let's try some other configurations.Suppose two variables are -1, and two variables are something else. Let's say a = b = -1, c = d = t. Then:LHS = (-1)^3 + (-1)^3 + t³ + t³ +1 = -1 -1 + 2t³ +1 = 2t³ -1RHS = k*(-1 + -1 + t + t) = k*(2t - 2)So the inequality becomes 2t³ -1 ≥ k*(2t - 2)Substituting k = 3/4:Left-hand side: 2t³ -1Right-hand side: (3/4)(2t - 2) = (3/4)(2(t - 1)) = (3/2)(t - 1) = (3/2)t - 3/2So inequality: 2t³ -1 ≥ (3/2)t - 3/2Bring all terms to left:2t³ - (3/2)t -1 + 3/2 ≥ 0 ⇒ 2t³ - (3/2)t + 1/2 ≥ 0Multiply both sides by 2 to eliminate fractions:4t³ - 3t + 1 ≥ 0Check if this is true for t ≥ -1.Let me check t = -1:4*(-1)^3 -3*(-1) +1 = -4 +3 +1 = 0t = 0:4*0 -0 +1 = 1 ≥0t = 1:4 -3 +1 = 2 ≥0What about critical points? Take derivative:12t² -3. Setting to zero:12t² -3 =0 ⇒ t²= 3/12=1/4 ⇒ t=±1/2Compute at t=1/2:4*(1/2)^3 -3*(1/2) +1 = 4*(1/8) - 3/2 +1 = 0.5 -1.5 +1=0At t=-1/2:4*(-1/2)^3 -3*(-1/2)+1=4*(-1/8)+ 3/2 +1= -0.5 +1.5 +1=2 ≥0So the function 4t³ -3t +1 is non-negative for all t ≥ -1. So equality holds at t= -1 and t=1/2, otherwise it's positive. Therefore, the inequality holds for k=3/4 in this case.Another test case: set one variable to -1 and the others to 0. Let's see.Let a = -1, b = c = d =0.LHS = (-1)^3 +0 +0 +0 +1 = -1 +1 =0RHS = k*(-1 +0 +0 +0) = -kSo inequality: 0 ≥ -k ⇒ k ≥0Well, since we already have k=3/4 from previous constraints, which is greater than 0, this is satisfied.Another case: set all variables to 0. Then LHS =0 +0 +0 +0 +1=1, RHS =k*(0+0+0+0)=0. So 1 ≥0, which is always true. Doesn't impose any constraint on k.What about variables greater than -1. Let's try a case where some variables are positive. For example, set a=b=c=d=1.LHS =1 +1 +1 +1 +1=5RHS =k*(1+1+1+1)=4kSo 5 ≥4k ⇒k ≤5/4. But since we have k=3/4 from earlier, which is less than 5/4, that's okay. So this doesn't conflict.But wait, perhaps if variables are large, the inequality could force k to be smaller? Wait, but as variables become large, the left-hand side is cubic in variables, which grows much faster than the linear term on the right. So for large positive a, b, c, d, the LHS will dominate, so the inequality will hold for any k, as long as k isn't super large. But in our earlier test with a=b=c=d=1, we get k ≤5/4. But since we already have k=3/4 from other cases, that's okay. But maybe if variables are set to some other values?Wait, maybe if some variables are large positive and some are -1. Let's try mixing. Let a = t (large), and b=c=d=-1.Then:LHS = t³ + (-1)^3 + (-1)^3 + (-1)^3 +1 = t³ -3 +1 = t³ -2RHS = k*(t + (-1) + (-1) + (-1)) =k*(t -3)So the inequality is t³ -2 ≥k(t -3)We need this to hold for all t ≥ -1.But as t approaches infinity, t³ dominates, so LHS grows like t³, RHS grows like kt. Since t³ grows faster, the inequality will hold for large t regardless of k. However, we need to check for all t ≥ -1.If k=3/4, then the inequality becomes t³ -2 ≥ (3/4)(t -3)Multiply both sides by 4: 4t³ -8 ≥3t -9Bring all terms to left:4t³ -3t +1 ≥0Wait, same as before. Wait, but in this case, t is varying. Wait, this is similar to the previous case where three variables are -1 and one is t. Wait, actually, this is the same function. So, as before, the inequality holds for k=3/4 here.So maybe 3/4 is indeed the correct value.Another thought: perhaps using the method of Lagrange multipliers to find the minimum of (a³ + b³ + c³ + d³ +1) -k(a + b + c + d). If we can show that the minimum is non-negative for all a,b,c,d in [-1, ∞), then k is valid.To find the minimum, take partial derivatives with respect to each variable and set them to zero.But since the function is separable, each variable contributes a term of the form x³ -k x. So, for each variable x (a,b,c,d), the function is x³ -k x + constant (the +1 from the original expression). Wait, actually, the total expression is sum_{x}(x³ -k x) +1. So the total minimum is the sum of the minima of each x³ -k x over x in [-1, ∞), plus 1.Therefore, if we can find the minimum of f(x) =x³ -k x over x ≥ -1 for each variable, then sum them and add 1, and set this total minimum to be greater or equal to zero.So, since all variables are symmetric, the minimum of the sum is four times the minimum of f(x) for a single variable x in [-1, ∞), plus 1.Therefore, if we can find the minimum of f(x) =x³ -k x over x ≥ -1, then multiply by 4 and add 1, and set that ≥0.So let's compute min_{x ≥ -1} (x³ -k x).To find the minimum, take derivative: f’(x)=3x² -k.Set derivative to zero: 3x² -k=0 ⇒x=±√(k/3). But since x ≥ -1, we need to check if these critical points are in the domain.If x=√(k/3) is in the domain, which it is as long as √(k/3) ≥ -1, which is always true since square roots are non-negative. Similarly, x=-√(k/3) is a critical point only if -√(k/3) ≥ -1 ⇒√(k/3) ≤1 ⇒k ≤3.So, if k ≤3, then both critical points x=√(k/3) and x=-√(k/3) are within the domain? Wait, x ≥ -1. So x=-√(k/3) is in the domain only if -√(k/3) ≥ -1 ⇒√(k/3) ≤1 ⇒k ≤3.So for k ≤3, the critical points are at x=±√(k/3). For k >3, x=-√(k/3) would be less than -1, so the critical point at x=-√(k/3) is not in the domain. Therefore, for k >3, the critical points within the domain are only x=√(k/3) and the boundary x=-1.Therefore, to compute the minimum of f(x)=x³ -k x over x ≥-1, we need to consider:Case 1: k ≤3. Then critical points at x=±√(k/3). But x must be ≥-1. If k ≤3, then -√(k/3) ≥ -1. So both critical points are in the domain. Then the minimum can occur at x=-√(k/3) or at x=-1.Case 2: k >3. Then critical points x=√(k/3) and x=-√(k/3) where x=-√(k/3) < -1, so only critical point in domain is x=√(k/3), and we also check boundary x=-1.But since we are looking for the minimum, we need to evaluate f(x) at all critical points and boundaries.But this seems complicated. Maybe instead, let's consider that for each x in [-1, ∞), the function x³ -k x has a minimum. To find the minimum, compare the value at critical points and endpoints.But since we need the minimum over four variables, each contributing such a term, it's quite involved. However, since all variables are symmetric, perhaps the minimal occurs when all variables are at their minimal points. But this might not necessarily be the case.Alternatively, perhaps the minimal value of the entire expression occurs when as many variables as possible are set to their minimal value, which is -1. Since setting variables to lower values might decrease the LHS more than the RHS. Let me check.Suppose we set some variables to -1. Let's say m variables are set to -1, and 4 - m variables are set to some other value. Then the LHS would be m*(-1)^3 + sum_{remaining variables} x³ +1, and the RHS would be k*(m*(-1) + sum x).But depending on how the variables are set, the balance between LHS and RHS could change. For example, when we set variables to -1, the LHS decreases by 1 per variable, and the RHS decreases by k per variable. But if k is positive, then decreasing the RHS (since subtracting k per variable) would make the inequality easier to hold. Wait, but actually, when variables are set to -1, the LHS becomes smaller (because x³ at -1 is -1), but the RHS becomes smaller as well (since each term is -k). So the net effect is that the LHS - RHS is (-1 + k) per variable set to -1. Therefore, if we set a variable to -1, LHS - RHS becomes (original LHS - original RHS) + (-1 + k). So depending on whether k is greater than 1 or not, setting variables to -1 may help or hurt.Wait, if k >1, then -1 +k is positive, so setting variables to -1 increases LHS - RHS, which is good because we need LHS - RHS ≥0. If k <1, then -1 +k is negative, so setting variables to -1 decreases LHS - RHS, making the inequality harder to hold. So if k <1, we need to be careful when variables are set to -1.But in our earlier case, when all variables are -1, LHS - RHS = -3 - (-4k) = -3 +4k. For this to be ≥0, we need 4k -3 ≥0 ⇒k ≥3/4. But earlier we saw that when three variables are -1 and one is t, setting t=-1 forced k ≤3/4. Wait, maybe there's some inconsistency here.Wait, no, actually, when all variables are -1, LHS - RHS = (-4 +1) - k*(-4) = (-3) +4k. So -3 +4k ≥0 ⇒k ≥3/4. However, when three variables are -1 and one is t, at t=-1, LHS - RHS = (-3 + (-1) +1) -k*(-4) = (-3 -1 +1) +4k = (-3) +4k. Wait, that's the same as the all variables -1 case. So in that case, same condition. But when we set three variables to -1 and let the fourth vary, we found that k must be at least 3/4 and at most 3/4, hence exactly 3/4. Then, in other cases, such as two variables set to -1, etc., with k=3/4, the inequality still holds. So maybe 3/4 is indeed the answer.But to confirm, perhaps consider when variables are set to some value between -1 and higher. Let's try another test case: set a = b = c = d = t, where t ∈ [-1, ∞). Then the inequality becomes 4t³ +1 ≥4kt.Let me set k=3/4, so the inequality is 4t³ +1 ≥3t.Check for t ∈ [-1, ∞).At t=-1: LHS=4*(-1)^3 +1 =-4 +1=-3; RHS=3*(-1)=-3. So -3 ≥-3, equality holds.At t=0: LHS=0 +1=1; RHS=0. So 1 ≥0, holds.At t=1: LHS=4 +1=5; RHS=3. So 5 ≥3, holds.What about critical points? Take t such that 4t³ +1 -3t=0.Find t where 4t³ -3t +1=0. Let's check possible roots.Try t= -1: 4*(-1)^3 -3*(-1)+1= -4 +3 +1=0, so t=-1 is a root. Factor out (t +1):Divide 4t³ -3t +1 by (t +1):Using polynomial division or synthetic division:Coefficients: 4, 0, -3, 1Divide by t +1:-1 | 4   0   -3   1   |    -4   4   -1   -----------------     4  -4   1   0So the quotient is 4t² -4t +1. Therefore, 4t³ -3t +1=(t +1)(4t² -4t +1).Set 4t² -4t +1=0. Using quadratic formula:t=(4 ±√(16 -16))/8=(4 ±0)/8=0.5. So double root at t=0.5.Thus, 4t³ -3t +1=(t +1)(2t -1)^2.Therefore, the equation 4t³ -3t +1=0 is satisfied at t=-1 and t=1/2 (double root). Therefore, for t ≥-1:4t³ -3t +1=(t +1)(2t -1)^2 ≥0, since (t +1) ≥0 for t ≥-1, and (2t -1)^2 is always non-negative. Therefore, 4t³ -3t +1 ≥0 for all t ≥-1. Hence, equality holds at t=-1 and t=1/2. So when all variables are equal, the inequality holds with equality at t=-1 and t=1/2.Therefore, this suggests that k=3/4 is indeed the minimal k such that the inequality holds for all a,b,c,d in [-1, ∞).But to be thorough, perhaps check another mixed case. Let's set two variables to 1/2 and two variables to -1. Let's compute LHS and RHS.Set a = b =1/2, c = d =-1.LHS = (1/2)^3 + (1/2)^3 + (-1)^3 + (-1)^3 +1 = (1/8 +1/8) + (-1 -1) +1 = (1/4) -2 +1 = 1/4 -1 = -3/4RHS = k*(1/2 +1/2 + (-1) + (-1)) =k*(1 -2)=k*(-1)So the inequality becomes -3/4 ≥ -k ⇒k ≥3/4. Since we have k=3/4, equality holds here. So this is another case where equality is achieved. Therefore, k=3/4 is indeed the critical value where multiple cases reach equality.Therefore, combining all these cases, the maximal possible k for which the inequality holds for all a, b, c, d in [-1, ∞) is 3/4. Hence, the answer is k=3/4.Final AnswerThe maximum real number ( k ) is (boxed{dfrac{3}{4}}).</think>To find all real numbers ( k ) such that the inequality ( a^3 + b^3 + c^3 + d^3 + 1 geq k(a + b + c + d) ) holds for all ( a, b, c, d in [-1, +infty) ), we analyze various cases and boundary conditions.1. Case where all variables are -1: [ a = b = c = d = -1 ] [ text{LHS} = (-1)^3 + (-1)^3 + (-1)^3 + (-1)^3 + 1 = -4 + 1 = -3 ] [ text{RHS} = k(-1 -1 -1 -1) = -4k ] The inequality becomes ( -3 geq -4k ), leading to ( k leq frac{3}{4} ).2. Case where three variables are -1 and one variable ( t ): [ a = b = c = -1, quad d = t geq -1 ] [ text{LHS} = (-1)^3 + (-1)^3 + (-1)^3 + t^3 + 1 = t^3 - 2 ] [ text{RHS} = k(-1 -1 -1 + t) = k(t - 3) ] The inequality becomes ( t^3 - 2 geq k(t - 3) ). Setting ( t = -1 ): [ (-1)^3 - 2 geq k(-1 - 3) implies -3 geq -4k implies k geq frac{3}{4} ] For ( k = frac{3}{4} ), checking the function ( t^3 - frac{3}{4}t + frac{1}{4} ) shows it is non-negative for all ( t geq -1 ).3. Case where two variables are -1 and two variables are 0: [ a = b = -1, quad c = d = 0 ] [ text{LHS} = (-1)^3 + (-1)^3 + 0 + 0 + 1 = -2 + 1 = -1 ] [ text{RHS} = k(-1 -1 + 0 + 0) = -2k ] The inequality becomes ( -1 geq -2k implies k geq frac{1}{2} ).4. Case where variables are large: As variables grow large, the cubic term dominates, ensuring the inequality holds for any ( k ).5. Verification for ( k = frac{3}{4} ): - For ( a = b = c = d = t ), the inequality becomes ( 4t^3 + 1 geq 3t ), which holds for all ( t geq -1 ). - For mixed values, such as two variables at ( frac{1}{2} ) and two at ( -1 ), the inequality holds with equality.After verifying all cases, the maximum value of ( k ) ensuring the inequality holds for all ( a, b, c, d in [-1, +infty) ) is (boxed{dfrac{3}{4}}).]

🔑:When a=b=c=d=-1, we have -3 geqslant k(-4), so k geqslant frac{3}{4}. When a=b=c=d=frac{1}{2}, we have 4 times frac{1}{8}+1 geqslant kleft(4 times frac{1}{2}right), so k leqslant frac{3}{4}. Therefore, k=frac{3}{4}. Now we prove thata^{3}+b^{3}+c^{3}+d^{3}+1 geqslant frac{3}{4}(a+b+c+d)holds for any a, b, c, d in[-1,+infty). First, we prove that 4 x^{3}+1 geqslant 3 x, x in[-1,+infty). In fact, by (x+1)(2 x-1)^{2} geqslant 0, we get 4 x^{3}+1 geqslant 3 x, x in[-1,+infty). Therefore,4 a^{3}+1 geqslant 3 a, 4 b^{3}+1 geqslant 3 b, 4 c^{3}+1 geqslant 3 c, 4 d^{3}+1 geqslant 3 d,Adding the above four inequalities, we obtain the desired inequality (1).Thus, the real number k=frac{3}{4}.

🧠:[<think>Okay, let's see. The problem is asking which of the numbers 29998, 29999, 30000, or 30001 can be expressed as the sum of the products of three consecutive natural numbers taken two at a time. Hmm. Alright, let's break this down.First, let's understand what "the sum of the products of three consecutive natural numbers taken two at a time" means. So, if we have three consecutive natural numbers, say n, n+1, and n+2, we need to take their products two at a time and then add those products together. So, the products two at a time would be:- n*(n+1)- n*(n+2)- (n+1)*(n+2)Then, the sum would be n(n+1) + n(n+2) + (n+1)(n+2). Let me compute that expression first.Expanding each term:1. n(n+1) = n² + n2. n(n+2) = n² + 2n3. (n+1)(n+2) = n² + 3n + 2Adding them up:(n² + n) + (n² + 2n) + (n² + 3n + 2)= n² + n + n² + 2n + n² + 3n + 2= 3n² + (1n + 2n + 3n) + 2= 3n² + 6n + 2So, the sum is 3n² + 6n + 2. Therefore, the problem is asking which of the given numbers can be written in the form 3n² + 6n + 2 for some natural number n.Our target is to find n such that 3n² + 6n + 2 equals one of the options: 29998, 29999, 30000, 30001.Let me write the equation:3n² + 6n + 2 = K, where K is one of the options.So, we can rearrange this equation:3n² + 6n + (2 - K) = 0But since we're looking for natural numbers n, perhaps it's easier to solve for n in terms of K and see if n is a natural number.Alternatively, let's express this quadratic equation as:3n² + 6n = K - 2Divide both sides by 3:n² + 2n = (K - 2)/3Then, complete the square on the left side:n² + 2n + 1 = (K - 2)/3 + 1So,(n + 1)² = (K - 2 + 3)/3 = (K + 1)/3Therefore,(n + 1)² = (K + 1)/3So, (K + 1)/3 must be a perfect square. Since n is a natural number, n + 1 must be a positive integer, so (K + 1)/3 must be a perfect square.Therefore, to check each option, we can compute (K + 1)/3 and see if it's a perfect square. Let's do that for each option:(A) 30001:(30001 + 1)/3 = 30002/3 ≈ 10000.666... Not an integer, so no.(B) 30000:(30000 + 1)/3 = 30001/3 ≈ 10000.333... Not an integer.(C) 29999:(29999 + 1)/3 = 30000/3 = 10000. That's an integer. Now check if 10000 is a perfect square. Yes, 100² = 10000. Therefore, n + 1 = 100, so n = 99. Which is a natural number. So this works.(D) 29998:(29998 + 1)/3 = 29999/3 ≈ 9999.666... Not an integer.Therefore, the answer is (C) 29999.Wait, but let me double-check. If n = 99, then let's compute 3n² + 6n + 2.Compute n=99:3*(99)^2 + 6*99 + 2First, 99^2: 99*99. Let's compute that. 100*100 = 10000, subtract 100 + 99 = 199, so 10000 - 199 = 9801. So 99² = 9801.Then:3*9801 = 294036*99 = 59429403 + 594 = 2999729997 + 2 = 29999Yes, that's correct. So indeed, when n = 99, the sum is 29999. Therefore, the correct answer is (C) 29999.But just to be thorough, let's check the other options in case I made a mistake.For example, let's check (B) 30000.Suppose (K +1)/3 = 30000 +1 /3 = 30001 /3 ≈ 10000.333... Not a perfect square. So no.For (A) 30001: (30001 +1)/3=30002/3≈10000.666... Not integer.For (D) 29998: (29998 +1)/3=29999/3≈9999.666... Not integer.So yes, only (C) gives a perfect square. So (C) is the answer.Final Answerboxed{C}</think>To determine which of the given numbers can be expressed as the sum of the products of three consecutive natural numbers taken two at a time, we start by considering three consecutive natural numbers ( n ), ( n+1 ), and ( n+2 ). We need to compute the sum of their products taken two at a time:[n(n+1) + n(n+2) + (n+1)(n+2)]Expanding each term and combining them, we get:[n(n+1) + n(n+2) + (n+1)(n+2) = n^2 + n + n^2 + 2n + n^2 + 3n + 2 = 3n^2 + 6n + 2]We need to check if this expression equals any of the given numbers: 29998, 29999, 30000, or 30001. We solve for ( n ) by setting up the equation:[3n^2 + 6n + 2 = K]Rearranging and completing the square, we get:[(n + 1)^2 = frac{K + 1}{3}]We check each option to see if (frac{K + 1}{3}) is a perfect square:- Option A (30001): (frac{30001 + 1}{3} = frac{30002}{3} approx 10000.666) (not an integer)- Option B (30000): (frac{30000 + 1}{3} = frac{30001}{3} approx 10000.333) (not an integer)- Option C (29999): (frac{29999 + 1}{3} = frac{30000}{3} = 10000) (a perfect square, (100^2))- Option D (29998): (frac{29998 + 1}{3} = frac{29999}{3} approx 9999.666) (not an integer)For option C, (10000) is a perfect square, so (n + 1 = 100) implies (n = 99). Substituting (n = 99) into the expression confirms the result:[3(99^2) + 6(99) + 2 = 29999]Thus, the correct answer is (boxed{C}).,<think>Okay, so I have this problem here that says: Among the numbers 29998, 29999, 30000, 30001, the one that can be expressed as the sum of the products of three consecutive natural numbers taken two at a time is ( ). The options are A) 30001, B) 30000, C) 29999, D) 29998.Hmm, let me try to understand what this means. The problem is asking which of these numbers can be written as the sum of the products of three consecutive natural numbers taken two at a time. So, first, I need to figure out what exactly "the sum of the products of three consecutive natural numbers taken two at a time" means.Let me break it down. If we have three consecutive natural numbers, say n, n+1, n+2. Then, taking their products two at a time would mean multiplying each pair and then adding those products together. So, for three numbers, there are three possible pairs: n*(n+1), n*(n+2), and (n+1)*(n+2). Then, summing those three products. Let me check:Products two at a time:1. n(n+1)2. n(n+2)3. (n+1)(n+2)Sum = n(n+1) + n(n+2) + (n+1)(n+2)Is that right? Yes, that seems to be the case. So, the problem is asking which of the given numbers can be expressed in that form for some natural number n. So, I need to compute the sum n(n+1) + n(n+2) + (n+1)(n+2) and see if it equals one of those four numbers. Then, the correct answer is the option that matches.First, maybe I can simplify that expression algebraically to find a formula for the sum, which might make it easier to compute for any n, and then solve for n such that the sum is one of the given numbers.Let's compute the sum step by step:Sum = n(n+1) + n(n+2) + (n+1)(n+2)First, expand each term:1. n(n+1) = n² + n2. n(n+2) = n² + 2n3. (n+1)(n+2) = n² + 3n + 2Now, sum them up:n² + n + n² + 2n + n² + 3n + 2Combine like terms:n² + n² + n² = 3n²n + 2n + 3n = 6nAnd the constant term is 2.So, total sum = 3n² + 6n + 2Therefore, the sum of the products of three consecutive natural numbers taken two at a time is 3n² + 6n + 2. So, the problem is to find n such that 3n² + 6n + 2 equals one of the given numbers: 29998, 29999, 30000, 30001.So, we can set up the equation:3n² + 6n + 2 = Kwhere K is one of the given options. Then, we need to see if there exists a natural number n for which this equation holds for each K. The correct answer is the K for which such an n exists.Alternatively, perhaps solving for n in terms of K and seeing if n is a natural number.Let me rearrange the equation:3n² + 6n + 2 = KSubtract K from both sides:3n² + 6n + (2 - K) = 0This is a quadratic equation in terms of n. For n to be a natural number, the discriminant must be a perfect square, and the roots must be positive integers.The quadratic equation is: 3n² + 6n + (2 - K) = 0The discriminant D is:D = 6² - 4*3*(2 - K) = 36 - 12*(2 - K) = 36 - 24 + 12K = 12 + 12K = 12(K + 1)So, discriminant D = 12(K + 1)For the equation to have integer solutions, D must be a perfect square. Let's denote D = m², where m is a positive integer. Therefore,12(K + 1) = m²So, 12 divides m². Since 12 = 4*3, and 4 is a square, 3 must divide m². Therefore, 3 divides m, so m = 3k for some integer k. Then,12(K + 1) = (3k)² = 9k²Divide both sides by 3:4(K + 1) = 3k²So, 4(K + 1) must be divisible by 3. Therefore, 3 divides 4(K + 1). Since 4 and 3 are coprime, 3 divides (K + 1). Therefore, K + 1 must be divisible by 3. Therefore, K ≡ 2 mod 3.Looking at the options:A) 30001: 30001 mod 3: 3+0+0+0+1=4, 4 mod 3=1. So 30001≡1 mod 3.B) 30000: 3+0+0+0+0=3, 3 mod 3=0. So 30000≡0 mod 3.C) 29999: 2+9+9+9+9=38, 38 mod 3=2. So 29999≡2 mod 3.D) 29998: 2+9+9+9+8=37, 37 mod 3=1. So 29998≡1 mod 3.So, from the above, only K=29999 (option C) is ≡2 mod 3. Wait, but the equation said K +1 must be divisible by 3. Therefore, K +1 ≡0 mod 3 ⇒ K≡2 mod 3. So yes, only 29999 ≡2 mod3. Therefore, 29999 +1=30000, which is divisible by 3. So, only K=29999 satisfies the condition that 4(K +1) must be divisible by 3. So, K=29999 is the only candidate.But let's check if that's the case. So, if K=29999, then K+1=30000. Then, 4*(30000)=120000. So, 120000=3k² ⇒ k²=40000 ⇒ k=200. Therefore, k=200, which is an integer. Therefore, D=12(K +1)=12*30000=360000, which is 600². Therefore, the discriminant is a perfect square, so the quadratic equation has integer roots.Now, let's solve the quadratic equation for K=29999:3n² + 6n + (2 - 29999) = 0 ⇒ 3n² + 6n - 29997 = 0Divide all terms by 3:n² + 2n - 9999 = 0Using quadratic formula:n = [-2 ± sqrt(4 + 4*9999)] / 2Compute discriminant:sqrt(4 + 39996) = sqrt(40000) = 200Therefore, n = [-2 ± 200]/2We discard the negative root because n is a natural number:n = (-2 + 200)/2 = 198/2 = 99Therefore, n=99 is a natural number. Therefore, K=29999 is achievable with n=99. Therefore, the answer is C) 29999.Wait, but let me check for the other options as well, just to be thorough.Take K=30000:Then, 3n² +6n +2=30000 ⇒ 3n² +6n=29998 ⇒n² +2n=29998/3 ≈ 9999.333...But 29998 divided by 3 is not an integer, so n² + 2n would not be an integer, but n is a natural number, so n² +2n must be an integer. Therefore, no solution here.Similarly, K=30001:3n² +6n +2=30001 ⇒3n² +6n=30001 -2=29999 ⇒n² +2n=29999/3≈9999.666..., which is not integer. So, no solution.K=29998:3n² +6n +2=29998 ⇒3n² +6n=29996 ⇒n² +2n=29996/3≈9998.666..., which is not integer. So, again, no solution.Therefore, only K=29999 yields integer n=99. Therefore, the correct answer is C) 29999.But just to double-check, let's compute the sum when n=99.Three consecutive numbers: 99, 100, 101.Compute the products two at a time:99*100 = 990099*101 = 9999100*101 = 10100Sum: 9900 + 9999 + 10100 = Let's compute step by step.First, 9900 + 9999 = 19899Then, 19899 + 10100 = 29999Yes, exactly. So, the sum is 29999, which is option C. Therefore, the answer is C) 29999.Final Answerboxed{C}</think>Given the numbers 29998, 29999, 30000, and 30001, we need to determine which one can be expressed as the sum of the products of three consecutive natural numbers taken two at a time.First, we consider three consecutive natural numbers ( n ), ( n+1 ), and ( n+2 ). The sum of their products taken two at a time is:[n(n+1) + n(n+2) + (n+1)(n+2)]Expanding and simplifying this expression:[n(n+1) + n(n+2) + (n+1)(n+2) = n^2 + n + n^2 + 2n + n^2 + 3n + 2 = 3n^2 + 6n + 2]We need to find ( n ) such that ( 3n^2 + 6n + 2 ) equals one of the given numbers. Setting up the quadratic equation for each number:[3n^2 + 6n + 2 = K]Rearranging gives:[3n^2 + 6n + (2 - K) = 0]The discriminant ( D ) of this quadratic equation must be a perfect square for ( n ) to be a natural number:[D = 12(K + 1)]For ( D ) to be a perfect square, ( 12(K + 1) ) must be a perfect square. This implies ( K + 1 ) must be divisible by 3. Checking the given numbers:- ( 30001 equiv 1 mod 3 )- ( 30000 equiv 0 mod 3 )- ( 29999 equiv 2 mod 3 )- ( 29998 equiv 1 mod 3 )Only ( 29999 equiv 2 mod 3 ), so ( K + 1 = 30000 ) is divisible by 3. Solving for ( n ) when ( K = 29999 ):[3n^2 + 6n - 29997 = 0 implies n^2 + 2n - 9999 = 0]The discriminant is:[sqrt{4 + 4 cdot 9999} = sqrt{40000} = 200]Thus, the roots are:[n = frac{-2 pm 200}{2}]Taking the positive root:[n = frac{198}{2} = 99]Verifying with ( n = 99 ):- The numbers are 99, 100, 101- Products: ( 99 times 100 = 9900 ), ( 99 times 101 = 9999 ), ( 100 times 101 = 10100 )- Sum: ( 9900 + 9999 + 10100 = 29999 )Thus, the correct answer is (boxed{C}).]

🔑:3. C.Notice that, for three consecutive natural numbers n-1, n, n+1, the sum of their pairwise products has the formn(n-1)+n(n+1)+(n-1)(n+1)=3 n^{2}-1 text {. }The above expression leaves a remainder of 2 when divided by 3, while 30001 and 29998 leave a remainder of 1 when divided by 3, and 30000 is divisible by 3. Thus, we can exclude options mathrm{A}, mathrm{B}, and mathrm{D}.In fact, 29999 leaves a remainder of 2 when divided by 3, and99 times 100+100 times 101+101 times 99=29999 .

🧠:[<think>Okay, let's tackle this problem step by step. The question is about whether the condition "sin α = √2/2" is a sufficient, necessary, both, or neither condition for "α = 45°" in a triangle ABC where α is an interior angle. The options are A, B, C, D as listed.First, I need to recall some trigonometry basics. The sine of 45° is indeed √2/2. So if an angle α in a triangle is 45°, then sin α must equal √2/2. That seems straightforward. So, does that mean that sin α = √2/2 implies α is 45°? Well, wait a second. The sine function has the same value for multiple angles. Specifically, sin θ = sin (180° - θ). So, if sin α = √2/2, then α could be 45° or 135°, right?But hold on, in a triangle, all interior angles are between 0° and 180°, and the sum of the three angles is 180°. So, if α is an interior angle of triangle ABC, it must satisfy 0° < α < 180°. However, 135° is an obtuse angle. So, could α be 135° in a triangle? Let's check.If α were 135°, then the sum of the other two angles would have to be 45°, since 180° - 135° = 45°. That's possible. For example, a triangle with angles 135°, 30°, and 15° would work. So, in such a triangle, sin 135° is equal to sin 45°, which is √2/2. Therefore, even if sin α = √2/2, α could be 45° or 135°, depending on the triangle.Therefore, "sin α = √2/2" is not enough to conclude that α is exactly 45°, because there's another possible angle (135°) within the range of interior angles of a triangle that also has the same sine value. However, if α is 45°, then sin α must be √2/2. So, the equation sin α = √2/2 is a necessary condition for α being 45°, but it's not sufficient because it doesn't uniquely determine α as 45°.Wait, but let's clarify the terms here. A necessary condition is something that must be true for the statement to hold. A sufficient condition is something that, if true, guarantees the statement. So, if we're looking at the condition "sin α = √2/2" for "α = 45°", then:- If α = 45°, then sin α = √2/2. So, sin α = √2/2 is a necessary condition for α = 45°. Because in order for α to be 45°, sin α has to be √2/2. So, necessity is established.- However, sin α = √2/2 is not a sufficient condition for α = 45°, because as we saw, α could also be 135°, which is another angle in the possible range of angles for a triangle, and would still satisfy sin α = √2/2. So, sufficiency is not met.Therefore, the condition is necessary but not sufficient. So the answer should be B.But wait, let me check if there are any other angles besides 45° and 135° in the range (0°, 180°) where sin α = √2/2. The sine function is positive in the first and second quadrants, so between 0° and 180°, the solutions to sin α = √2/2 are indeed 45° and 135°. There are no others. So, in the context of a triangle, where angles are between 0° and 180°, these are the only two possibilities.Hence, if we know that sin α = √2/2, then α could be either 45° or 135°, so we can't conclude it's exactly 45°, hence the condition isn't sufficient. But if α is 45°, then sin α must be √2/2, so the condition is necessary.Therefore, the answer is B: A necessary but not sufficient condition.But let me verify once more. Let's say we have two scenarios:1. If α is 45°, then sin α is √2/2. So, the condition is necessary.2. If sin α is √2/2, then α can be 45° or 135°, so the condition is not sufficient to conclude that α is 45°.Therefore, "sin α = √2/2" is a necessary condition for α being 45°, but not a sufficient one. So, B is correct.Alternatively, if the question was about "α is 45°" being a condition for "sin α = √2/2", then "α is 45°" would be a sufficient condition but not necessary, because sin α can be √2/2 even if α is 135°. But the question is phrased the other way around: "sin α = √2/2" is the condition for "α = 45°" to be... So, we need to see what role the equation plays in determining α's measure.In logical terms, if we have statement P: "α = 45°", and statement Q: "sin α = √2/2". The question is, is Q a sufficient, necessary, etc., condition for P?So, P implies Q, because if α is 45°, then sin α is √2/2. So, Q is a necessary condition for P. But Q does not imply P, because Q can be true (sin α = √2/2) even if P is false (α is 135°). Therefore, Q is necessary but not sufficient for P. Hence, answer B.Yes, that makes sense. So, the answer is B.Final Answerboxed{B}</think>Given that ( alpha ) is an interior angle of ( triangle ABC ), we need to determine whether the condition ( sin alpha = frac{sqrt{2}}{2} ) is a sufficient, necessary, both, or neither condition for ( alpha = 45^circ ).1. Necessary Condition: If ( alpha = 45^circ ), then ( sin alpha = frac{sqrt{2}}{2} ) must hold. This establishes that ( sin alpha = frac{sqrt{2}}{2} ) is a necessary condition for ( alpha = 45^circ ).2. Sufficient Condition: However, ( sin alpha = frac{sqrt{2}}{2} ) does not uniquely determine ( alpha ) as ( 45^circ ). The sine function has the same value for ( 45^circ ) and ( 135^circ ) (i.e., ( sin 45^circ = sin 135^circ = frac{sqrt{2}}{2} )). Since ( alpha ) could be ( 135^circ ) (an obtuse angle) and still satisfy ( sin alpha = frac{sqrt{2}}{2} ), the condition is not sufficient.Thus, ( sin alpha = frac{sqrt{2}}{2} ) is a necessary but not sufficient condition for ( alpha = 45^circ ).[boxed{B}],<think>Okay, so I have this problem here. Let me read it again. It says: Given that α is an interior angle of triangle ABC, then "sin α = √2/2" is a condition for "α = 45°" to be which of the options A, B, C, or D. The options are about whether the condition is sufficient, necessary, both, or neither. Hmm.Alright, let me start by recalling some trigonometry basics. The sine of 45 degrees is indeed √2/2. So if α is 45°, then sin α must be √2/2. That makes sense. But is the converse true? If sin α is √2/2, does that necessarily mean α is 45°? Wait, no. Because sine of 135° is also √2/2. But in a triangle, all angles are between 0° and 180°, right? Since it's an interior angle of a triangle, α has to be greater than 0° and less than 180°. However, in a triangle, the sum of all angles is 180°, so each angle must be less than 180°, and greater than 0°, obviously.So, if α is an interior angle of a triangle, and sin α = √2/2, then α could be 45° or 135°, right? Because 135° is also in the range of possible angles for a triangle, as long as the other two angles adjust accordingly. For example, if one angle is 135°, the other two angles would have to add up to 45°, so they'd be acute angles. That's possible. So, if we have sin α = √2/2, α could be 45° or 135°, but since α is part of a triangle, we need to check if 135° is possible.So, does that mean that "sin α = √2/2" allows for two possibilities: 45° or 135°? If that's the case, then saying "sin α = √2/2" is not enough to conclude that α is exactly 45°, because there's another possibility. Therefore, the condition "sin α = √2/2" is necessary for α to be 45°, because if α is 45°, then sin α must be √2/2. However, it's not sufficient, because there's another angle, 135°, which also has the same sine value.Therefore, the condition is necessary but not sufficient. So, the answer should be B: A necessary but not sufficient condition.Wait, but let me double-check. Let's consider that in a triangle, if α is 135°, then the other two angles must add up to 45°, which is possible. For example, a triangle with angles 135°, 30°, 15° would satisfy that. So, in such a triangle, the angle α = 135°, which has sin α = √2/2. Therefore, sin α = √2/2 does not uniquely determine that α is 45°, it could also be 135°, but in a valid triangle.Therefore, the condition "sin α = √2/2" is necessary for α to be 45°, because if α is 45°, then sin α must equal that. But it's not sufficient because sin α could equal that even if α is 135°. So, necessary but not sufficient. Therefore, answer B.Alternatively, let's think about the definitions:A sufficient condition means that if the condition is met, then the statement must be true. But here, sin α = √2/2 doesn't guarantee α = 45°, since it could be 135°, so it's not sufficient.A necessary condition means that if the statement is true (α = 45°), then the condition must hold (sin α = √2/2). Which is true. If α is 45°, then sin α is indeed √2/2. So the condition is necessary.Therefore, the answer is B.But wait, just to be thorough, is there any case where in a triangle, α could be 135°, but sin α is something else? No, because sin 135° is √2/2. So, if in a triangle, an angle is 135°, its sine is √2/2. Therefore, the condition sin α = √2/2 is both possible for 45° and 135°, but since 135° is allowed in a triangle (as long as the other angles compensate), then yes, there are two possibilities. Therefore, the condition is necessary but not sufficient. So answer B.I think that's solid. Let me check if there's another angle in the range 0° to 180° with sin α = √2/2. 45° and 135°, yes. Those are the two in the unit circle that have that sine value. So no others. So in the context of a triangle, which can have angles up to 180°, but actually less than 180°, and greater than 0°, so 135° is acceptable if the other angles are small enough.Therefore, the answer is B.Final Answerboxed{B}</think>Given that ( alpha ) is an interior angle of ( triangle ABC ), we need to determine whether the condition ( sin alpha = frac{sqrt{2}}{2} ) is sufficient, necessary, both, or neither for ( alpha = 45^circ ).1. Necessary Condition: If ( alpha = 45^circ ), then ( sin alpha = frac{sqrt{2}}{2} ) must hold. This makes ( sin alpha = frac{sqrt{2}}{2} ) a necessary condition for ( alpha = 45^circ ).2. Sufficient Condition: However, ( sin alpha = frac{sqrt{2}}{2} ) does not uniquely determine ( alpha ) as ( 45^circ ). Since ( sin 135^circ = frac{sqrt{2}}{2} ), ( alpha ) could also be ( 135^circ ) in a valid triangle (as long as the other angles sum to ( 45^circ )). Therefore, ( sin alpha = frac{sqrt{2}}{2} ) is not a sufficient condition for ( alpha = 45^circ ).Thus, the condition ( sin alpha = frac{sqrt{2}}{2} ) is necessary but not sufficient for ( alpha = 45^circ ).[boxed{B}]]

🔑:Since α is an interior angle of triangle ABC, it follows that "sin α= frac{sqrt{2}}{2}" Rightarrow "α=45^{circ} or α=135^{circ}", and "α=45^{circ}" Rightarrow "sin α= frac{sqrt{2}}{2}", therefore, "sin α= frac{sqrt{2}}{2}" is a necessary but not sufficient condition for "α=45^{circ}". Hence, the correct option is boxed{B}. "sin α= frac{sqrt{2}}{2}" Rightarrow "α=45^{circ} or α=135^{circ}", "α=45^{circ}" Rightarrow "sin α= frac{sqrt{2}}{2}", from which the result can be derived. This question examines the judgment of sufficient conditions, necessary and sufficient conditions, necessary conditions, the basic knowledge of trigonometric functions, reasoning and proof ability, computational solving ability, and the idea of reduction and transformation, making it a basic question.

🧠:[<think>Okay, so I have this triangle ABC. The sides opposite angles A, B, and C are a, b, c respectively. They gave me that 2 sin A = a cos B, and that side b is √5. The first part is to find sin C when c = 2. The second part is to find the maximum area of triangle ABC.Let me start with part (1). Hmm, given 2 sin A = a cos B. I need to relate this to the sides and angles. Maybe I can use the Law of Sines or the Law of Cosines here? Let's recall:Law of Sines: (a/sin A) = (b/sin B) = (c/sin C) = 2R (where R is the circumradius).Law of Cosines: c² = a² + b² - 2ab cos C, similarly for other angles.Given that 2 sin A = a cos B. Let me see. Maybe I can express sin A and cos B in terms of sides?From Law of Sines, sin A = (a / (2R)), but maybe that's not helpful yet. Alternatively, from Law of Cosines, cos B = (a² + c² - b²)/(2ac). Wait, but they also give b = √5 and c = 2 for part (1). Let me plug in those values.So first, for part (1), c = 2, b = √5. Let's write down the given equation: 2 sin A = a cos B.Let me express cos B using the Law of Cosines. For angle B, which is opposite side b (which is √5), we have:cos B = (a² + c² - b²)/(2ac)But c is 2, b is √5. So substituting:cos B = (a² + (2)² - (√5)²)/(2a*2) = (a² + 4 - 5)/(4a) = (a² - 1)/(4a)So cos B = (a² - 1)/(4a)Now, sin A can be expressed via Law of Sines. From Law of Sines, (a/sin A) = (b/sin B) = (c/sin C)We can relate sin A to side a and side c (since c is 2) and sin C. But maybe it's better to express sin A in terms of side lengths. Alternatively, maybe express sin A in terms of another angle.Alternatively, since we have a relationship involving sin A and cos B, maybe use some trigonometric identities or relations between angles. Remember that in a triangle, angles add up to π. So A + B + C = π. So perhaps express A in terms of B and C? Hmm, not sure.Wait, maybe use the Law of Sines to express sin A as (a / (2R)), but we don't know R. Alternatively, from Law of Sines, (a/sin A) = (b/sin B). So sin A = (a sin B)/b. Since b is √5, sin A = (a sin B)/√5.So plugging into the given equation: 2*(a sin B)/√5 = a cos BSimplify: 2 sin B / √5 = cos BSo divide both sides by cos B (assuming cos B ≠ 0):2 sin B / (√5 cos B) = 1 => 2 tan B / √5 = 1 => tan B = √5 / 2Therefore, tan B = √5 / 2. So angle B is arctangent of √5 / 2.From this, we can find sin B and cos B. Since tan B = opposite/adjacent = √5 / 2, so we can imagine a right triangle where the opposite side is √5 and adjacent is 2, so hypotenuse is √( (√5)^2 + 2^2 ) = √(5 + 4) = √9 = 3.Therefore, sin B = opposite/hypotenuse = √5 / 3, and cos B = adjacent/hypotenuse = 2/3.So, sin B = √5 / 3, cos B = 2/3.Wait, earlier we had 2 sin A = a cos B, and we found cos B = 2/3, so 2 sin A = a*(2/3) => sin A = a/3.But from Law of Sines: a/sin A = b/sin B. Since b = √5 and sin B = √5/3, then:a/sin A = √5 / (√5/3) = 3. So a/sin A = 3 => a = 3 sin A.But earlier we have sin A = a/3. So substituting into a = 3 sin A, we get a = 3*(a/3) => a = a. Which is a tautology. Hmm, so that doesn't give us new information. Maybe need another approach.Wait, but if we have sin A = a/3, and from the Law of Sines, (a/sin A) = 3, so (a/sin A) = 3 => sin A = a/3. So that's consistent.But how do we find a? Maybe we can use the Law of Cosines for angle A or angle C.Wait, given that we need to find sin C. Let me recall that we have sides a, b=√5, c=2. So if we can find angle C, we can find sin C.Alternatively, since we have sides b and c, maybe use Law of Cosines to find angle A or something else.Wait, perhaps we can find angle B first, since we found tan B = √5/2, so angle B is known. Then since angles sum to π, we can relate angle A and C. But we need more information.Alternatively, use the Law of Cosines on angle B. Wait, we already found cos B = 2/3. Let me verify that.Earlier, using the Law of Cosines on angle B:cos B = (a² + c² - b²)/(2ac) = (a² + 4 - 5)/(4a) = (a² -1)/(4a). Then we also found from 2 sin A = a cos B that cos B = 2 sin A / a. Then through another route, using sin A = (a sin B)/b, which gave us 2 sin A = a cos B => 2*(a sin B)/√5 = a cos B => 2 sin B / √5 = cos B. Then from tan B = √5 / 2, which led us to sin B = √5 /3 and cos B = 2/3. Therefore, cos B is 2/3. Therefore, equating the two expressions for cos B:From Law of Cosines: (a² -1)/(4a) = 2/3So solve for a:Multiply both sides by 4a: (a² -1) = (8a)/3Multiply both sides by 3: 3a² -3 = 8aBring all terms to one side: 3a² -8a -3 =0Now solve the quadratic equation: 3a² -8a -3 =0Using quadratic formula: a = [8 ±√(64 + 36)]/6 = [8 ±√100]/6 = [8 ±10]/6So two solutions:a = (8 +10)/6 =18/6=3or a=(8 -10)/6=(-2)/6= -1/3But a is a length, so must be positive. Thus, a=3.So side a is 3. Now that we know a=3, b=√5, c=2.Now, to find sin C. Using Law of Sines: (c/sin C)= (a/sin A). We know c=2, a=3, and sin A = a/3 =3/3=1. Wait, sin A=1? That would mean angle A is 90 degrees. Wait, but if angle A is 90 degrees, then the triangle is right-angled at A. Let's check if this makes sense.If angle A is 90 degrees, then side a (opposite angle A) is the hypotenuse. So in a right-angled triangle, the hypotenuse is the longest side. Here, a=3, b=√5≈2.236, c=2. So yes, a=3 is the longest side. That makes sense.So angle A is 90 degrees, so sin A =1. Then from the earlier equation sin A =a/3=3/3=1, which checks out.Now, angle C. Since the triangle is right-angled at A, angles B and C must add up to 90 degrees. So angle C =90 - angle B.We already found angle B's sine and cosine: sin B=√5/3, cos B=2/3. Therefore, angle B is arcsin(√5/3). Therefore, angle C=90 - angle B.So sin C = sin(90 - B)=cos B=2/3. Wait, that's straightforward. Because sin(90 - x)=cos x, so sin C=cos B=2/3. Therefore, sin C=2/3.But let me verify using Law of Sines. According to Law of Sines, (c/sin C)=(a/sin A). So c=2, a=3, sin A=1. Therefore:2/sin C=3/1 => sin C=2/3. Yes, that's correct. So sin C=2/3.Therefore, the answer to part (1) is 2/3.Now, moving to part (2): Find the maximum area of triangle ABC.So they don't specify c here, so c is variable? Wait, in part (1), they fixed c=2, but in part (2), it's a general triangle with the given conditions: 2 sin A = a cos B, b=√5, and we need to find the maximum area.So we need to consider varying sides a and c, but keeping b=√5 and satisfying 2 sin A = a cos B. The area of the triangle is (1/2)*b*c*sin A, or alternatively (1/2)*a*b*sin C, or (1/2)*a*c*sin B. Maybe express area in terms of variables we can control.But first, let's recall from the previous part that we had a=3, c=2, but in part (2), c can vary. Wait, but in part (1), c was fixed, but in part (2), it's the maximum area over all possible such triangles. So we need to find the maximum area possible for triangle ABC where b=√5 and 2 sin A = a cos B.To approach this, let's try to express the area in terms of a single variable and then maximize it.First, let's recall that in any triangle, the area can be expressed as (1/2)*a*b*sin C, or (1/2)*b*c*sin A, or (1/2)*a*c*sin B. Since we have relations involving angles A and B, maybe express area in terms of angle A or B.Alternatively, express variables in terms of a single parameter.From the given condition: 2 sin A = a cos B.From Law of Sines: a/sin A = b/sin B => a = (b sin A)/sin B. Since b=√5, so a = (√5 sin A)/sin B.Substitute into the given equation:2 sin A = a cos B = (√5 sin A / sin B) * cos BSimplify:2 sin A = (√5 sin A cos B)/sin BDivide both sides by sin A (assuming sin A ≠ 0, which it can't be in a triangle):2 = (√5 cos B)/sin B => 2 = √5 cot B => cot B = 2/√5 => tan B = √5/2Wait, this is the same as in part (1). So tan B = √5/2, so angle B is fixed? Wait, but in part (1), we had c=2, but here in part (2), we are to find the maximum area. If tan B is fixed, then angle B is fixed, so angle B is arctan(√5/2), which would mean that angle B is fixed, but that conflicts with the idea that the triangle can vary to maximize area.Wait, perhaps I made a mistake here. Let's see:Starting again: given 2 sin A = a cos B.From Law of Sines: a = (b / sin B) * sin A. Since b=√5, so a = √5 sin A / sin B.Plug into 2 sin A = a cos B:2 sin A = (√5 sin A / sin B) * cos BCancel sin A (assuming sin A ≠0):2 = (√5 / sin B) * cos B => 2 = √5 cot B => cot B = 2/√5 => tan B = √5/2So tan B is fixed, so angle B is fixed. Therefore, angle B is a constant value, arctan(√5/2). So regardless of other sides, angle B remains the same? That's interesting.Wait, but then if angle B is fixed, and side b is fixed (√5), then the triangle is determined up to the other sides, but angle B is fixed.Wait, but if angle B is fixed, then the triangle's shape is fixed, so the area would depend on some scaling factor. But wait, side b is fixed as √5. So perhaps the triangle is determined by angle B fixed and side b fixed, so the other sides can vary but angle B is fixed. Wait, this seems contradictory.Wait, in the problem statement, for part (2), it's still the same triangle ABC with the given conditions 2 sin A = a cos B and b=√5. So angle B is fixed as arctan(√5/2), so all such triangles have angle B fixed, side b fixed. So the other sides a and c can vary, but subject to the Law of Sines and Law of Cosines, and the given condition.But if angle B is fixed, then sides a and c are related. From Law of Sines, a/sin A = c/sin C = b/sin B. Since b and angle B are fixed, sin B is fixed. So a = (b / sin B) sin A, c = (b / sin B) sin C.But since angles A + B + C = π, angle C = π - A - B. So angle C is determined once angle A is known.Therefore, the triangle is determined by angle A, with angle B fixed. Therefore, the sides a and c can be expressed in terms of angle A.Therefore, the area can be expressed in terms of angle A. So let's try that.Area = (1/2)*a*c*sin B. Since angle B is fixed, sin B is a constant. From earlier, sin B = √5 / 3. Because in part (1), we found sin B = √5 /3. Let me verify that.Earlier, we found tan B = √5/2, so in a right triangle with opposite √5 and adjacent 2, hypotenuse is 3, so sin B = √5/3, cos B = 2/3.Yes, so sin B = √5/3. Therefore, area = (1/2)*a*c*(√5/3) = (√5 /6)*a*c.Alternatively, express a and c in terms of angle A.From Law of Sines:a = (b / sin B) sin A = (√5 / (√5/3)) sin A = 3 sin ASimilarly, c = (b / sin B) sin C = 3 sin CBut angle C = π - A - B, so sin C = sin(π - A - B) = sin(A + B) [since sin(π - x) = sin x]Wait, sin(π - A - B) = sin(C) = sin(A + B) because A + B + C = π, so C = π - A - B. But sin(C) = sin(A + B). However, sin(A + B) = sin π - C = sin C. Wait, maybe this is a bit circular.But angle C = π - A - B, so sin C = sin(A + B). Let's use angle relations. Since B is fixed, angle C = π - A - B, so sin C = sin(A + B). Let me express sin(A + B):sin(A + B) = sin A cos B + cos A sin BWe can substitute this into the expression for c:c = 3 sin C = 3 [sin A cos B + cos A sin B]But we know from the given condition that 2 sin A = a cos B. Since a = 3 sin A (from Law of Sines above), cos B = 2 sin A / a = 2 sin A / (3 sin A) = 2/3. So cos B = 2/3, as found earlier.Also, sin B = √5 /3. Therefore:sin C = sin(A + B) = sin A*(2/3) + cos A*(√5/3)Therefore, c = 3 [ (2/3 sin A) + (√5 /3 cos A) ] = 2 sin A + √5 cos ATherefore, c = 2 sin A + √5 cos ASo now, a = 3 sin A, c = 2 sin A + √5 cos AThus, the area is (√5 /6)*a*c = (√5 /6)*(3 sin A)*(2 sin A + √5 cos A)Let me compute this:(√5 /6)*3 sin A*(2 sin A + √5 cos A) = (√5 /6)*3*(2 sin² A + √5 sin A cos A) = (√5 /2)*(2 sin² A + √5 sin A cos A)Simplify:= (√5 /2)*(2 sin² A + √5 sin A cos A) = √5 sin² A + (5/2) sin A cos ATherefore, area = √5 sin² A + (5/2) sin A cos ANow, to find the maximum of this expression with respect to angle A. Since angle A must satisfy A > 0, and angle C = π - A - B > 0, so A + B < π. Since B is fixed, angle A < π - B.Given that angle B is arctan(√5/2), let's compute angle B numerically to understand the constraints. tan B = √5 /2 ≈ 2.236 /2 ≈1.118, so angle B ≈ 48.19 degrees. Therefore, angle A must be less than 180 - 48.19 ≈ 131.81 degrees. So A ∈ (0, ~131.81°).But since the problem is in radians, but we can work in radians if needed, but perhaps it's easier to keep it symbolic.Anyway, to maximize the area expression: √5 sin² A + (5/2) sin A cos A.Let me denote θ = A for simplicity. So area = √5 sin² θ + (5/2) sin θ cos θ.We need to find the maximum of this function. Let's compute the derivative with respect to θ and set it to zero.First, let me rewrite the area as:Area(θ) = √5 sin² θ + (5/2) sin θ cos θCompute derivative dA/dθ:dA/dθ = 2√5 sin θ cos θ + (5/2)(cos² θ - sin² θ)Set derivative equal to zero:2√5 sin θ cos θ + (5/2)(cos² θ - sin² θ) = 0Multiply both sides by 2 to eliminate denominator:4√5 sin θ cos θ + 5(cos² θ - sin² θ) = 0Let me note that sin(2θ) = 2 sin θ cos θ, and cos(2θ) = cos² θ - sin² θ.So rewrite the equation:4√5*(1/2) sin 2θ + 5 cos 2θ = 0 => 2√5 sin 2θ +5 cos 2θ=0Let me write this as:2√5 sin 2θ +5 cos 2θ=0Let me divide both sides by cos 2θ (assuming cos 2θ ≠0):2√5 tan 2θ +5=0 => tan 2θ = -5/(2√5) = -√5/2Therefore, 2θ = arctan(-√5/2) + kπ. But since θ is an angle in a triangle, θ ∈ (0, π - B). Therefore, 2θ ∈ (0, 2π - 2B). Let's find the principal value.arctan(-√5/2) is in the fourth quadrant, but since tangent is periodic with π, we can write 2θ = π - arctan(√5/2) or 2θ = 2π - arctan(√5/2), but considering θ must be positive and less than π - B.But angle B is arctan(√5/2), so arctan(√5/2) ≈ 48.19°, so π - B ≈ 131.81°, so 2θ < 263.62°, which is less than 2π.But let's find 2θ such that tan 2θ = -√5/2. The principal solution is 2θ = -arctan(√5/2), but since tangent is periodic with π, we can write 2θ = π - arctan(√5/2). Because tan(π - x) = -tan x. So tan(π - arctan(√5/2)) = -tan(arctan(√5/2)) = -√5/2. Therefore, 2θ = π - arctan(√5/2)Therefore, θ = (π - arctan(√5/2))/2But arctan(√5/2) is angle B, so θ = (π - B)/2Therefore, angle A = (π - B)/2Therefore, the maximum area occurs when angle A = (π - B)/2.Now, compute the area at this angle.First, compute sin A and cos A.Angle A = (π - B)/2, so let me denote φ = (π - B)/2.Therefore, φ = (π/2 - B/2)So sin φ = sin(π/2 - B/2) = cos(B/2)cos φ = cos(π/2 - B/2) = sin(B/2)Therefore, sin A = cos(B/2), cos A = sin(B/2)Now, compute the area:Area = √5 sin² A + (5/2) sin A cos A = √5 cos²(B/2) + (5/2) cos(B/2) sin(B/2)Simplify:First term: √5 cos²(B/2)Second term: (5/2)*(1/2) sin B = (5/4) sin B (since sin B = 2 sin(B/2) cos(B/2))Wait, but (5/2) cos(B/2) sin(B/2) = (5/2)*(1/2) sin B = (5/4) sin BTherefore, Area = √5 cos²(B/2) + (5/4) sin BBut sin B = √5/3, so substitute:Area = √5 cos²(B/2) + (5/4)*(√5/3) = √5 cos²(B/2) + (5√5)/12Now, need to express cos²(B/2) in terms of known values. Recall that cos²(B/2) = (1 + cos B)/2We know cos B = 2/3, so:cos²(B/2) = (1 + 2/3)/2 = (5/3)/2 = 5/6Therefore, Area = √5*(5/6) + (5√5)/12 = (5√5)/6 + (5√5)/12 = (10√5 +5√5)/12 = (15√5)/12 = (5√5)/4Therefore, the maximum area is (5√5)/4.Alternatively, we can verify this by another method. Let's check.Alternatively, express the area in terms of a single variable.Earlier, we had a =3 sin A, c=2 sin A +√5 cos AThus, Area = (√5 /6)*a*c = (√5 /6)*(3 sin A)*(2 sin A +√5 cos A)Which simplifies to:(√5 /6)*(6 sin² A + 3√5 sin A cos A) = (√5 /6)*(6 sin² A + 3√5 sin A cos A)= (√5)* sin² A + (3*5)/6 sin A cos A = √5 sin² A + (5/2) sin A cos ASame as before.Then, we found that the maximum occurs when angle A=(π - B)/2. Therefore, substituting back, the maximum area is (5√5)/4.Alternatively, perhaps use another method. For example, since angle B is fixed, and side b is fixed, the triangle is determined by sides a and c. But perhaps using calculus to maximize the area expression.Alternatively, use Lagrange multipliers with constraints.But since we already did the calculus approach and found the maximum area as (5√5)/4, I think that's the answer.But let me check with another approach. Suppose we fix side b=√5 and angle B=arctan(√5/2). Then, the area of the triangle is (1/2)*a*c*sin B. Since sin B=√5/3, Area=(1/2)*a*c*(√5/3)= (√5/6)*a*c. So to maximize area, we need to maximize a*c.But a and c are related by the Law of Cosines and the given condition.But from the given condition, 2 sin A = a cos B. Also, from Law of Sines, a=3 sin A, so substituting back, we have a=3 sin A, so 2 sin A = 3 sin A cos B => 2 = 3 cos B => cos B=2/3, which is consistent with earlier findings.But in this case, since angle B is fixed, perhaps a and c can be expressed in terms of angle A, and then a*c can be maximized.We had earlier:a=3 sin Ac=2 sin A +√5 cos ATherefore, a*c=3 sin A*(2 sin A +√5 cos A)=6 sin² A +3√5 sin A cos ATherefore, to maximize a*c, we need to maximize 6 sin² A +3√5 sin A cos A.Which is equivalent to maximizing 2 sin² A + √5 sin A cos A, which is the same as before. So if we let f(A)=2 sin² A +√5 sin A cos A, then the maximum of f(A) will give maximum area.To find the maximum of f(A), we can write this as a single sinusoidal function.Express f(A) =2 sin² A +√5 sin A cos A.We can use double angle identities:sin² A = (1 - cos 2A)/2sin A cos A = (sin 2A)/2Thus:f(A) =2*(1 - cos 2A)/2 +√5*(sin 2A)/2 = (1 - cos 2A) + (√5/2) sin 2ATherefore, f(A)=1 - cos 2A + (√5/2) sin 2AThis can be expressed as 1 + [ -cos 2A + (√5/2) sin 2A ]Let me write the expression in brackets as R sin(2A - φ), where R is the amplitude and φ is the phase shift.Let me compute R:The amplitude of -cos 2A + (√5/2) sin 2A is √[(-1)^2 + (√5/2)^2] = √[1 + 5/4] = √(9/4) = 3/2.Therefore, -cos 2A + (√5/2) sin 2A = (3/2) sin(2A - φ), where φ is such that:sin φ = (-1)/ (3/2) = -2/3, and cos φ = (√5/2)/(3/2) = √5/3Therefore, φ = arcsin(-2/3). But since cos φ is positive, φ is in the fourth quadrant. So φ = -arcsin(2/3) or φ = 2π - arcsin(2/3). But since we can write:sin(2A - φ) = sin 2A cos φ - cos 2A sin φSo matching coefficients:- cos 2A + (√5/2) sin 2A = sin 2A cos φ - cos 2A sin φTherefore,cos φ = √5/2*(2/3) = √5/3 (?)Wait, no. Let me check:Wait, in our expression:- cos 2A + (√5/2) sin 2A = (3/2) sin(2A - φ)Expanding RHS:(3/2)[sin 2A cos φ - cos 2A sin φ] = (3/2) sin 2A cos φ - (3/2) cos 2A sin φComparing with LHS:-1 cos 2A + (√5/2) sin 2ATherefore,Coefficient of sin 2A: (3/2) cos φ = √5/2 => cos φ = √5/3Coefficient of cos 2A: - (3/2) sin φ = -1 => sin φ = 2/3Therefore, φ is the angle with sin φ=2/3 and cos φ=√5/3. Therefore, φ = arcsin(2/3). Therefore, the expression becomes:- cos 2A + (√5/2) sin 2A = (3/2) sin(2A - φ)Therefore, f(A)=1 + (3/2) sin(2A - φ)The maximum value of sin(2A - φ) is 1, so the maximum of f(A) is 1 + 3/2 =5/2.Therefore, maximum of f(A)=5/2.Therefore, maximum a*c=6*(5/2)=15? Wait, wait. Wait, f(A)=2 sin² A +√5 sin A cos A. Which we expressed as 1 + (3/2) sin(2A - φ), which has maximum 1 + 3/2=5/2.Therefore, maximum value of f(A)=5/2.Thus, maximum a*c=6*(5/2)=15?Wait, no, because a*c=3 sin A*(2 sin A +√5 cos A)=6 sin² A +3√5 sin A cos A=3*(2 sin² A +√5 sin A cos A)=3*f(A)=3*(5/2)=15/2.Ah, right. So maximum a*c=15/2. Therefore, maximum area=(√5 /6)*a*c=(√5 /6)*(15/2)= (15√5)/12= (5√5)/4, which matches the previous result.Therefore, the maximum area is indeed (5√5)/4.So, part (1): sin C=2/3, part (2): maximum area=5√5/4.Final Answer(1) boxed{dfrac{2}{3}} (2) boxed{dfrac{5sqrt{5}}{4}}</think>Given a triangle ( triangle ABC ) with sides opposite angles ( A ), ( B ), and ( C ) being ( a ), ( b ), and ( c ) respectively, and the conditions ( 2 sin A = a cos B ) and ( b = sqrt{5} ):(1) Finding ( sin C ) when ( c = 2 ):1. Use the Law of Cosines for angle ( B ): [ cos B = frac{a^2 + c^2 - b^2}{2ac} = frac{a^2 + 4 - 5}{4a} = frac{a^2 - 1}{4a} ]2. From the given condition ( 2 sin A = a cos B ), substitute ( cos B ): [ 2 sin A = a cdot frac{a^2 - 1}{4a} implies 2 sin A = frac{a^2 - 1}{4} ]3. Use the Law of Sines ( frac{a}{sin A} = frac{b}{sin B} ): [ sin A = frac{a sin B}{sqrt{5}} ]4. Substitute ( sin A ) into the equation: [ 2 cdot frac{a sin B}{sqrt{5}} = frac{a^2 - 1}{4} ]5. Solve for ( tan B ): [ 2 sin B / sqrt{5} = cos B implies tan B = frac{sqrt{5}}{2} ]6. Find ( sin B ) and ( cos B ): [ sin B = frac{sqrt{5}}{3}, quad cos B = frac{2}{3} ]7. Use the Law of Sines again to find ( a ): [ a = 3 sin A implies sin A = frac{a}{3} ]8. Solve the quadratic equation for ( a ): [ 3a^2 - 8a - 3 = 0 implies a = 3 ]9. Use the Law of Sines to find ( sin C ): [ frac{c}{sin C} = frac{a}{sin A} implies frac{2}{sin C} = 3 implies sin C = frac{2}{3} ](2) Finding the maximum area of ( triangle ABC ):1. Express the area in terms of ( a ) and ( c ): [ text{Area} = frac{sqrt{5}}{6} cdot a cdot c ]2. Express ( a ) and ( c ) in terms of angle ( A ): [ a = 3 sin A, quad c = 2 sin A + sqrt{5} cos A ]3. Substitute into the area formula: [ text{Area} = frac{sqrt{5}}{6} cdot 3 sin A cdot (2 sin A + sqrt{5} cos A) ]4. Simplify and maximize the expression: [ text{Area} = sqrt{5} sin^2 A + frac{5}{2} sin A cos A ]5. Use trigonometric identities and calculus to find the maximum value: [ text{Maximum Area} = frac{5sqrt{5}}{4} ] Final Answers(1) (boxed{dfrac{2}{3}}) (2) (boxed{dfrac{5sqrt{5}}{4}}),<think>Okay, let's try to solve this problem step by step. So we have triangle ABC with sides opposite angles A, B, C being a, b, c respectively. The given conditions are 2 sin A = a cos B and b = √5. Then part (1) asks for sin C when c = 2, and part (2) asks for the maximum area of triangle ABC.Starting with part (1). Let me recall some trigonometric laws that might help here. The Law of Sines relates the sides and angles: a/sin A = b/sin B = c/sin C. The Law of Cosines relates sides with angles: c² = a² + b² - 2ab cos C, and similarly for the other angles. Maybe these can be useful here.Given that 2 sin A = a cos B. Let me write that down:2 sin A = a cos B. Hmm. Let's see. From the Law of Sines, a = 2R sin A, where R is the circumradius. But I don't know if that's helpful here. Alternatively, since we have a relationship between sin A and cos B, maybe we can express them in terms of sides using the laws.From the Law of Sines, sin A = (a / (2R)), but maybe that's not directly helpful. Let's think about substituting sin A and cos B using the Law of Cosines.Wait, cos B can be expressed via the Law of Cosines: cos B = (a² + c² - b²)/(2ac). And sin A can be expressed from the Law of Sines: sin A = (a / (2R)), but again, maybe not helpful. Alternatively, since from Law of Sines, sin A / a = sin B / b = sin C / c. So sin A = (a / b) sin B? Wait, no: from Law of Sines, sin A / a = sin B / b, so sin A = (a / b) sin B.But given that 2 sin A = a cos B, substituting sin A from Law of Sines gives:2*(a / b) sin B = a cos BSimplify both sides. Divide both sides by a (assuming a ≠ 0, which it is a side length so it's positive):2*(1 / b) sin B = cos BMultiply both sides by b:2 sin B = b cos BBut we know that b = √5. So:2 sin B = √5 cos BDivide both sides by cos B:2 tan B = √5Therefore, tan B = √5 / 2So angle B has tan inverse of √5/2. Let's compute sin B and cos B from this.If tan B = √5 / 2, then we can imagine a right triangle where the opposite side is √5 and adjacent side is 2, so hypotenuse is √( (√5)^2 + 2^2 ) = √(5 + 4) = √9 = 3.Therefore, sin B = opposite / hypotenuse = √5 / 3, and cos B = adjacent / hypotenuse = 2 / 3.Okay, so now we have sin B = √5/3 and cos B = 2/3.Let me note that down.Now, going back to the given equation 2 sin A = a cos B. We can substitute cos B here.So 2 sin A = a*(2/3)Simplify:2 sin A = (2a)/3Divide both sides by 2:sin A = a/3Hmm. From the Law of Sines, sin A = a / (2R), where R is the circumradius, but again, maybe not helpful yet. Alternatively, from Law of Sines, sin A / a = sin B / b. Since we know sin B = √5/3 and b = √5, then sin A / a = (√5/3)/√5 = 1/3. Therefore, sin A = a*(1/3). Which matches the equation we just found: sin A = a/3. So that's consistent.So from here, sin A = a/3. But we can use Law of Sines to relate a and sin A. Wait, Law of Sines says a/sin A = b/sin B. We know b = √5 and sin B = √5/3. Therefore:a / sin A = √5 / (√5/3) = √5 * 3 / √5 = 3. So a / sin A = 3. But we also found that sin A = a/3. Let's check that.If a / sin A = 3, then sin A = a / 3, which is exactly what we had. So that's consistent. Therefore, this gives us a relationship, but maybe we need more information to find a or angle C.But we are given that c = 2. So maybe we can use Law of Cosines or Law of Sines involving c and angle C.Given that c = 2, let's try Law of Sines. From Law of Sines:a / sin A = c / sin CWe have c = 2, so:a / sin A = 2 / sin CBut we already know that a / sin A = 3 (from earlier), so:3 = 2 / sin C => sin C = 2/3.Wait, that seems straightforward. Wait, is that correct? Let me verify.Law of Sines: a/sin A = b/sin B = c/sin C = 2R. So since a/sin A = 3, then all those ratios equal 3. Therefore, c/sin C = 3 => sin C = c / 3 = 2/3. So sin C = 2/3. That's part (1). Hmm, that seems too easy. Let me check again.Wait, given that a/sin A = 3, and c/sin C = 3, so sin C = c/3. Since c = 2, sin C = 2/3. So yes, that's correct.Wait, but is there any possibility that angle C could be in the second quadrant, making sin C = 2/3 still valid? Since in a triangle, all angles are between 0 and π, so sin C is positive regardless. So 2/3 is the correct answer. So part (1) answer is 2/3. That seems done.But let me check using another approach to confirm. Let's use the Law of Cosines.We know sides b = √5, c = 2, and need to find sin C. If we can find angle C, then sin C is straightforward. Alternatively, we can find side a first.From the Law of Cosines, for angle B:cos B = (a² + c² - b²)/(2ac)We already found cos B = 2/3, c = 2, b = √5. So:2/3 = (a² + 4 - 5)/(2*a*2)Simplify numerator:a² + 4 -5 = a² -1Denominator: 4aSo:2/3 = (a² -1)/(4a)Multiply both sides by 4a:(8a)/3 = a² -1Multiply both sides by 3:8a = 3a² -3Bring all terms to one side:3a² -8a -3 =0Quadratic equation: 3a² -8a -3 =0. Let's solve for a.Using quadratic formula:a = [8 ±√(64 + 36)] /6 = [8 ±√100]/6 = [8 ±10]/6We have two solutions:a = (8 +10)/6 = 18/6 = 3a = (8 -10)/6 = (-2)/6 = -1/3Since a is a side length, it must be positive. So a =3.Therefore, a = 3. Now, with a =3, we can use the Law of Sines to find sin C.Law of Sines: a/sin A = c/sin C. But we already found that a/sin A =3, so 3 = c/sin C => sin C = c/3 =2/3. So same result. Thus, part (1) answer is 2/3. That seems solid.Now, moving to part (2): Find the maximum area of triangle ABC.The area of a triangle can be given by (1/2)ab sin C, or using other combinations of sides and angles. Since we need the maximum area, we can express the area in terms of variables and then find its maximum.Given that in the problem, the conditions are 2 sin A = a cos B and b=√5. So we have to work under these constraints. The previous part had c=2, but part (2) is general.We need to maximize the area of triangle ABC given b=√5 and 2 sin A = a cos B.First, let's recall that the area can be expressed in several ways:Area = (1/2)ab sin C = (1/2)bc sin A = (1/2)ac sin B.Perhaps we can express the area in terms of variables we can control, then use calculus or another method to find the maximum.Alternatively, maybe we can express the area in terms of one variable and then maximize it.Let me think. Let's try to relate variables using the given condition.From the given equation: 2 sin A = a cos B.Earlier, in part (1), we found that when c=2, we had a=3, sin C=2/3. But now, c is variable, so perhaps we need to express a in terms of c, or find relations between the sides and angles.First, let's see if we can express a in terms of other variables. From part (1), when we had c=2, we found a=3. But now, we need to consider variable c.But let's recall that in part (1), we used both the given equation and the Law of Cosines/Sines. Maybe similar steps can be applied here.Let's start again. Given 2 sin A = a cos B. From Law of Sines, sin A = a / (2R), but maybe that's not helpful yet. Wait, from Law of Sines, a/sin A = b/sin B = c/sin C = 2R.But given that 2 sin A = a cos B, let's manipulate this equation. Let's express sin A and cos B in terms of sides.From Law of Sines: sin A = (a / (2R)), but I don't know R. Alternatively, sin A can be expressed in terms of angle B or C since angles sum to π.Wait, angles in a triangle sum to π, so A + B + C = π. Therefore, A = π - B - C. Maybe using trigonometric identities here.But maybe it's better to use the relationship between angles and sides.From the given equation: 2 sin A = a cos B.From Law of Sines, sin A = (a / (2R)), so 2*(a / (2R)) = a cos B => (a/R) = a cos B => 1/R = cos B. Hmm, but unless R is known, this might not help. Alternatively, perhaps express cos B via Law of Cosines.Law of Cosines for angle B: cos B = (a² + c² - b²)/(2ac). And we have b = √5, so cos B = (a² + c² -5)/(2ac).Given that 2 sin A = a cos B, substitute cos B:2 sin A = a*( (a² + c² -5)/(2ac) )Simplify:2 sin A = (a² + c² -5)/(2c )Multiply both sides by 2c:4c sin A = a² + c² -5Now, from Law of Sines, sin A = (a / (2R)), but maybe another relation is better. Alternatively, from Law of Sines: sin A / a = sin B / b = sin C / c. So sin A = (a / b) sin B. Let me use this.Given that sin A = (a / b) sin B, and b = √5, so sin A = (a / √5) sin B. Substitute into the equation:4c*(a / √5) sin B = a² + c² -5Hmm. So 4c*(a / √5) sin B = a² + c² -5. Not sure if this helps. Alternatively, perhaps express sin B in terms of sides.From Law of Cosines, maybe we can relate angles. Alternatively, use the Law of Sines to express sin B in terms of side b. From Law of Sines, sin B = (b / (2R)) = (√5 / (2R)), but again, R is unknown.Alternatively, express sin B via another angle. Since A + B + C = π, maybe express angle B in terms of A and C.Alternatively, perhaps we can express the area in terms of variables and then use calculus to maximize it.The area of the triangle is (1/2)ab sin C, but since we have b fixed at √5, maybe express in terms of a and sin C. Alternatively, (1/2)*b*c*sin A. Let's see.Area = (1/2)*b*c*sin A. Since b = √5, that's (1/2)*√5*c*sin A.But we have the equation 2 sin A = a cos B. So sin A = (a/2) cos B. Therefore, Area = (1/2)*√5*c*(a/2) cos B = (√5 a c cos B)/4.Hmm, but this seems more complex. Maybe another approach.Alternatively, express the area as (1/2)*a*b*sin C. Wait, but we need to relate variables here. Let me think.Alternatively, using sides and angles. Since we have two conditions: 2 sin A = a cos B and b = √5. Let's try to express all variables in terms of a single variable.Let me consider the Law of Cosines and Law of Sines together. Let's try to express angle B in terms of angle A, or vice versa.From A + B + C = π, so C = π - A - B. Maybe we can express C in terms of A and B.But perhaps first, let's find relations from the given equation 2 sin A = a cos B.From Law of Sines, a / sin A = b / sin B => a = (b / sin B) sin A. Since b = √5, so a = (√5 / sin B) sin A.Given that 2 sin A = a cos B, substitute a from above:2 sin A = (√5 / sin B) sin A * cos BAssuming sin A ≠ 0 (which it can't be in a triangle), we can divide both sides by sin A:2 = (√5 / sin B) * cos BWhich simplifies to:2 = √5 * (cos B / sin B) = √5 cot BTherefore, cot B = 2 / √5 => tan B = √5 / 2.Wait a second! This is exactly what we found in part (1). So tan B = √5 / 2, regardless of the value of c? Wait, but in part (1), c was given as 2, but here in part (2), c is variable. However, according to this derivation, tan B is always √5 / 2, regardless of c? That seems contradictory.Wait, hold on. Let me check this again.Given 2 sin A = a cos B. From Law of Sines, a = (√5 / sin B) sin A. Substitute into 2 sin A = a cos B:2 sin A = (√5 / sin B) sin A cos BCancel sin A (assuming sin A ≠ 0):2 = (√5 / sin B) cos BWhich gives:2 = √5 * (cos B / sin B) = √5 cot B => cot B = 2 / √5 => tan B = √5 / 2Therefore, angle B is fixed, with tan B = √5 / 2. Therefore, angle B is fixed, regardless of the value of c. That's interesting. So in any triangle ABC with the given conditions, angle B is fixed, with tan B = √5 / 2. Therefore, angle B is constant, and sides a and c vary such that this condition holds.So angle B is fixed, so sin B and cos B are fixed. From tan B = √5 / 2, as in part (1), we can compute sin B = √5 / 3 and cos B = 2 / 3.Therefore, angle B is fixed, so sides a and c must adjust such that the given condition holds, but angle B remains fixed. Therefore, in part (2), when we need to maximize the area, it's under the condition that angle B is fixed, and sides a and c vary accordingly.Wait, if angle B is fixed, then perhaps the triangle can be considered with fixed side b = √5 and fixed angle B, and the other sides a and c varying. Then, the area would depend on sides a and c and angle A or C. Let me think.Alternatively, since angle B is fixed, and side b is fixed, maybe using the formula for area in terms of two sides and the included angle. The area can be written as (1/2)*a*c*sin B. Since angle B is fixed, sin B is fixed (sin B = √5 / 3). Therefore, the area is proportional to a*c. So to maximize the area, we need to maximize a*c.Therefore, if we can express a*c in terms of a single variable and maximize it, that would give the maximum area.Alternatively, since angle B is fixed, maybe we can use the Law of Sines to relate a and c. From Law of Sines:a / sin A = c / sin C = b / sin B = √5 / (√5 /3 ) = 3. So a = 3 sin A and c = 3 sin C.But angle A + angle B + angle C = π. Since angle B is fixed, angle A + angle C = π - B. Let’s denote angle A = α, then angle C = π - B - α.Therefore, a = 3 sin α, c = 3 sin (π - B - α ) = 3 sin (B + α ). Wait, no, angle C = π - B - α, so sin C = sin (π - B - α ) = sin (B + α ). Wait, sin (π - x ) = sin x, so sin (π - B - α ) = sin (B + α ). Hmm, not sure. Wait, π - B - α = π - (B + α ). So sin (π - (B + α )) = sin (B + α ). Because sin (π - θ ) = sin θ. So yes, sin C = sin (B + α ).Therefore, c = 3 sin (B + α ). Then, the product a*c = 3 sin α * 3 sin (B + α ) = 9 sin α sin (B + α ). We need to maximize this expression with respect to α.But angle B is fixed. Let’s recall that angle B has tan B = √5 /2, so B = arctan(√5 /2 ). Let’s compute sin B and cos B again for clarity:From tan B = √5 /2, we have opposite = √5, adjacent = 2, hypotenuse = 3, so sin B = √5 /3, cos B = 2 /3.Therefore, angle B is fixed, sin B = √5 /3, cos B = 2/3.So, a*c = 9 sin α sin (B + α ). Let's denote θ = α for simplicity. Then, the product a*c = 9 sin θ sin (B + θ ). We need to maximize this over θ, where θ >0 and θ + B < π (since angle C = π - B - θ must be positive).Let’s expand sin θ sin (B + θ ). Using the identity sin A sin B = [cos (A - B ) - cos (A + B ) ] / 2.Therefore, sin θ sin (B + θ ) = [ cos (θ - (B + θ )) - cos (θ + (B + θ )) ] / 2 = [ cos (-B ) - cos (2θ + B ) ] / 2 = [ cos B - cos (2θ + B ) ] / 2.Therefore, a*c = 9 * [ cos B - cos (2θ + B ) ] / 2 = (9/2) [ cos B - cos (2θ + B ) ].To maximize a*c, we need to minimize cos (2θ + B ), since cos B is a constant. The minimum value of cos (2θ + B ) is -1, but we need to check if 2θ + B can reach π (where cos π = -1). However, θ must satisfy θ >0 and θ < π - B (since angle C = π - B - θ must be positive). Therefore, 0 < θ < π - B.Therefore, 0 < θ < π - B => B < 2θ + B < 2(π - B ) + B = 2π - B.But cos (2θ + B ) has a minimum at 2θ + B = π, which is achievable if π - B is in the range of possible 2θ + B. Let’s check if 2θ + B can reach π.If 2θ + B = π, then θ = (π - B ) / 2. But θ must be less than π - B, so (π - B ) / 2 < π - B => π - B < 2π - 2B => 0 < π - B, which is true since B < π.Therefore, θ = (π - B ) / 2 is allowed. Therefore, the minimum of cos (2θ + B ) is -1, achieved when 2θ + B = π, θ = (π - B ) / 2.Therefore, the maximum value of a*c is (9/2)(cos B - (-1)) = (9/2)(cos B +1 ).Since cos B = 2/3, so:Maximum a*c = (9/2)(2/3 +1 ) = (9/2)(5/3 ) = (9*5)/(2*3 ) = 45/6 = 15/2.Therefore, maximum area is (1/2)*a*c*sin B = (1/2)*(15/2)*(√5 /3 ) = (15/4)*(√5 /3 ) = (15√5 ) / 12 = 5√5 / 4.Wait, let me verify that step-by-step.We found that maximum a*c =15/2.Then, area = (1/2)*a*c*sin B.So:Area = (1/2)*(15/2)*(√5 /3 ) = (15/4)*(√5 /3 ) = (15√5 ) / 12 = simplifies to (5√5)/4.Yes, because 15 divided by 3 is 5, so 5√5 /4.Therefore, the maximum area is 5√5 /4.Alternatively, let's check with another approach. If angle B is fixed, then the triangle is determined by sides a and c, with angle B between them. Wait, no. Wait, side b is opposite angle B, so actually, sides a and c are adjacent to angle B. Wait, actually, in triangle ABC, side a is opposite angle A, side b opposite angle B, etc. So if angle B is fixed, and side b is fixed at √5, then the triangle is determined by sides a and c, with angle B between sides a and c.Wait, no. Wait, in triangle ABC, angle B is between sides a and c. Because angle B is at vertex B, between sides a (opposite angle A) and c (opposite angle C). Wait, actually, in standard notation, side a is opposite angle A, side b opposite angle B, side c opposite angle C. Therefore, angle B is between sides a and c.Therefore, sides a and c form angle B between them, which is fixed. Therefore, the triangle with fixed side b = √5 and fixed angle B can be considered as a triangle with sides a and c enclosing angle B. Therefore, the area is (1/2)*a*c*sin B. Since sin B is fixed (√5 /3 ), then to maximize the area, we need to maximize a*c. Therefore, maximum area is (1/2)*(maximum a*c)*sin B. Which is exactly the approach we took earlier, leading to 5√5 /4.Alternatively, in such a triangle with two sides a and c enclosing a fixed angle B, the product a*c is maximized when the triangle is isoceles, i.e., when a = c. Wait, is that the case?Wait, in general, for a triangle with fixed angle B between sides a and c, the product a*c is maximized when a = c? Let me verify.Suppose we have sides a and c with angle B between them. The product a*c can be expressed in terms of the sides. From the Law of Cosines, b² = a² + c² - 2ac cos B. But here, b is fixed at √5. So √5² = 5 = a² + c² - 2ac*(2/3 ). So 5 = a² + c² - (4/3 )ac.We need to maximize a*c under the constraint 5 = a² + c² - (4/3 )ac.This is a constrained optimization problem. Let me set x = a, y = c. Then, the constraint is x² + y² - (4/3 )xy =5. We need to maximize xy.Use method of Lagrange multipliers. Let’s define the function f(x,y)=xy, and the constraint g(x,y)=x² + y² - (4/3 )xy -5=0.The Lagrangian is L = xy - λ (x² + y² - (4/3 )xy -5 ).Taking partial derivatives:∂L/∂x = y - λ (2x - (4/3 ) y ) =0∂L/∂y = x - λ (2y - (4/3 ) x ) =0∂L/∂λ = -(x² + y² - (4/3 )xy -5 )=0From first equation: y = λ (2x - (4/3 ) y )From second equation: x = λ (2y - (4/3 ) x )Let me write these as:1. y = 2λ x - (4/3 ) λ y2. x = 2λ y - (4/3 ) λ xLet me rearrange equation 1:y + (4/3 ) λ y = 2λ xy (1 + (4/3 ) λ ) = 2λ xSimilarly, equation 2:x + (4/3 ) λ x = 2λ yx (1 + (4/3 ) λ ) = 2λ ySo from equation 1 and 2, we have:From equation1: x = [ y (1 + (4/3 ) λ ) ] / (2λ )From equation2: y = [ x (1 + (4/3 ) λ ) ] / (2λ )Substitute equation1 into equation2:y = [ [ y (1 + (4/3 ) λ ) / (2λ ) ] * (1 + (4/3 ) λ ) ] / (2λ )Simplify:y = y * (1 + (4/3 ) λ )² / (4λ² )Assuming y ≠0 (which it is a side length, so positive), divide both sides by y:1 = (1 + (4/3 ) λ )² / (4λ² )Multiply both sides by 4λ²:4λ² = (1 + (4/3 ) λ )²Take square roots (considering positive roots since λ is a multiplier related to positive variables):2λ = 1 + (4/3 ) λSolve for λ:2λ - (4/3 )λ =1(6λ -4λ)/3 =1(2λ)/3=1 => λ=3/2So λ=3/2. Substitute back into equation1:y (1 + (4/3 )(3/2 )) = 2*(3/2 )xCompute inside the parentheses:1 + (4/3 )(3/2 ) =1 + 2 =3Therefore:3y =3x => y =xTherefore, when a =c, the product a*c is maximized. Therefore, maximum occurs when a =c.Therefore, when a =c, the triangle is isoceles with a =c. Let's verify this.So when a =c, from the constraint equation:5 =a² + a² - (4/3 )a² =2a² - (4/3 )a² = (6/3 -4/3 )a²=(2/3 )a² => a²= (5)*(3/2 )=15/2 =>a=√(15/2 )Therefore, a =c=√(15/2 ). Then, the product a*c=15/2, which matches our earlier result. Therefore, maximum a*c=15/2.Thus, maximum area is (1/2)*a*c*sin B=(1/2)*(15/2 )*(√5 /3 )=(15/4)*(√5 /3 )= (15√5 )/12=5√5 /4.Therefore, the maximum area is 5√5 /4.Alternatively, confirm using another approach.Alternatively, when a =c, angle A = angle C. Since angles sum to π, angle A + angle B + angle C = π. Since angle A = angle C, so 2 angle A + angle B = π. Therefore, angle A = (π - angle B )/2. Then, from Law of Sines, a/sin A =b/sin B. Since a =c, and angle A=angle C.So a=3 sin A (from Law of Sines: a=3 sin A ), and angle A=(π - B )/2. Therefore, sin A= sin( (π -B )/2 )=sin( π/2 - B/2 )=cos(B/2 ). Therefore, a=3 cos(B/2 ). Then, area=(1/2)*a*c*sin B=(1/2)*(9 cos²(B/2 ))*(√5 /3 ). Let's compute this.First, cos(B/2 ). From angle B where tan B=√5 /2, so we can compute cos B=2/3. Using the identity cos B=2 cos²(B/2 ) -1. Therefore,2 cos²(B/2 ) =1 +cos B=1 +2/3=5/3 => cos²(B/2 )=5/6 => cos(B/2 )=√(5/6 ). Therefore, a=3*√(5/6 )=3*(√30)/6=√30 /2. Then, a*c= (√30 /2 )²=30/4=15/2. Same as before. Then area=(1/2)*15/2*√5 /3= (15/4)*(√5 /3 )=5√5 /4. Same result.Thus, regardless of the method, the maximum area is 5√5 /4. Therefore, part (2) answer is 5√5 /4.Final Answer(1) boxed{dfrac{2}{3}} (2) boxed{dfrac{5sqrt{5}}{4}}</think>Given the problem in triangle ABC with sides opposite to angles A, B, and C being a, b, and c respectively, and the conditions 2sin A = acos B and b = sqrt{5}:(1) Finding sin C when c = 2:1. From the given condition 2sin A = acos B, we use the Law of Sines and Law of Cosines.2. Using the Law of Sines, sin A = frac{a}{2R} and cos B = frac{a^2 + c^2 - b^2}{2ac}.3. Substituting cos B into the given equation and simplifying, we find tan B = frac{sqrt{5}}{2}.4. Calculating sin B and cos B from tan B = frac{sqrt{5}}{2}, we get sin B = frac{sqrt{5}}{3} and cos B = frac{2}{3}.5. Using the Law of Sines again, a = 3sin A and sin A = frac{a}{3}.6. Applying the Law of Sines for c = 2, we find sin C = frac{2}{3}.(2) Finding the maximum area of triangle ABC:1. The area of triangle ABC can be expressed as frac{1}{2}absin C or frac{1}{2}acsin B.2. Given 2sin A = acos B and b = sqrt{5}, we use the fixed angle B with tan B = frac{sqrt{5}}{2}.3. The maximum area is achieved when a = c (isosceles triangle), maximizing the product ac.4. Using the Law of Cosines and solving the quadratic equation, we find a = c = sqrt{frac{15}{2}}.5. Calculating the maximum area using frac{1}{2}acsin B, we get the maximum area as frac{5sqrt{5}}{4}. Final Answer(1) boxed{dfrac{2}{3}} (2) boxed{dfrac{5sqrt{5}}{4}}]

🔑:(1) Given 2sin A=acos B, we have frac{sin A}{a} = frac{sin B}{b}. Since b= sqrt {5}, we have 2sin B= sqrt {5}cos B.This implies that tan B= frac{sqrt {5}}{2}.Thus, sin B= frac{sqrt {5}}{3}.Given that c=2, we have sin C= frac{csin B}{b}= frac{2}{3}.(2) From part (1), we have cos B= frac{2}{3}.Therefore, 5=a^{2}+c^{2}- frac{4}{3}acgeqslant 2ac- frac{4}{3}ac= frac{2}{3}ac.This implies that acleqslant frac{15}{2}.Thus, the maximum area of triangle ABC is: frac{1}{2}times frac{15}{2}times frac{sqrt {5}}{3}=boxed{frac{5 sqrt {5}}{4}}.

🧠:[<think>Okay, so I need to write the vector x as a combination of vectors p, q, and r. That means I need to find scalars a, b, and c such that:x = a*p + b*q + c*rGiven the vectors:x = {-1; 7; -4}p = {-1; 2; 1}q = {2; 0; 3}r = {1; 1; -1}So, substituting each component, this gives me a system of equations. Let me write them out.First, for the first component (the top one):-1 = a*(-1) + b*2 + c*1Second component (middle):7 = a*2 + b*0 + c*1Third component (bottom):-4 = a*1 + b*3 + c*(-1)So, putting these into equations:1. -1 = -a + 2b + c2. 7 = 2a + 0b + c3. -4 = a + 3b - cNow, I need to solve this system of equations for a, b, c.Let me write the equations again for clarity:Equation 1: -a + 2b + c = -1Equation 2: 2a + c = 7Equation 3: a + 3b - c = -4Hmm, maybe I can use substitution or elimination. Let's see.From Equation 2, we can express c in terms of a:c = 7 - 2aThat's helpful. Then I can substitute c = 7 - 2a into Equations 1 and 3.Starting with Equation 1:- a + 2b + c = -1Substitute c:- a + 2b + (7 - 2a) = -1Combine like terms:- a - 2a + 2b + 7 = -1Which is:-3a + 2b + 7 = -1Subtract 7 from both sides:-3a + 2b = -8Let me note that as Equation 1a:Equation 1a: -3a + 2b = -8Now, Equation 3:a + 3b - c = -4Again, substitute c = 7 - 2a:a + 3b - (7 - 2a) = -4Distribute the negative sign:a + 3b -7 + 2a = -4Combine like terms:3a + 3b -7 = -4Add 7 to both sides:3a + 3b = 3Divide both sides by 3:a + b = 1Let me call this Equation 3a:Equation 3a: a + b = 1Now, I have two equations:1a. -3a + 2b = -83a. a + b = 1Now, let's solve these two equations with two variables.From Equation 3a: a = 1 - bSubstitute a = 1 - b into Equation 1a:-3*(1 - b) + 2b = -8Multiply out:-3 + 3b + 2b = -8Combine terms:-3 + 5b = -8Add 3 to both sides:5b = -5Divide by 5:b = -1Now, substitute b = -1 into Equation 3a:a + (-1) = 1So, a = 1 + 1 = 2Now, we have a = 2 and b = -1. Let's find c from earlier:c = 7 - 2a = 7 - 2*2 = 7 - 4 = 3Wait, c = 3? Let me check if these values satisfy all original equations.Check Equation 1:- a + 2b + c = -2 + 2*(-1) + 3 = -2 -2 +3 = (-4) +3 = -1, which matches the RHS.Equation 2: 2a + c = 4 + 3 = 7, which is correct.Equation 3: a + 3b - c = 2 + 3*(-1) -3 = 2 -3 -3 = -4, which is correct.So, the solution is a=2, b=-1, c=3.Therefore, the decomposition is:x = 2p - q + 3rWait, let me confirm:Compute 2p: 2*{-1;2;1} = {-2;4;2}Compute -q: -1*{2;0;3} = {-2;0;-3}Compute 3r: 3*{1;1;-1} = {3;3;-3}Now, add them together:2p + (-q) + 3r = (-2 + (-2) + 3; 4 + 0 + 3; 2 + (-3) + (-3))Wait, hold on. Let's compute each component:First component: (-2) + (-2) +3 = (-4) +3 = -1 ✔️Second component: 4 +0 +3 =7 ✔️Third component:2 + (-3) + (-3)= 2 -6= -4 ✔️Yes, that gives x = {-1;7;-4}, which is correct. So the coefficients are a=2, b=-1, c=3.But wait, in the problem statement, they might expect the answer in terms of decomposition. So x = 2p - q +3r.Let me double-check if there is another way to approach this, maybe using matrix methods. Let's see.Expressed as a matrix equation: A*[a;b;c] = x, where A is the matrix with columns p, q, r.So matrix A:Columns are p, q, r:Row 1: -1, 2, 1Row 2: 2, 0, 1Row 3: 1, 3, -1So the system is:[-1 2 1][a] [-1][ 2 0 1][b] = [7 ][1 3 -1][c] [-4]Which is the same as the equations we had earlier. So solving it via substitution gave us a=2, b=-1, c=3.Alternatively, using matrix inversion or row operations. Let me verify using matrix inversion.But since I already have the solution, maybe just check.But given that substitution worked and the check confirms the solution, I think this is correct.So the final decomposition is x = 2p - q + 3r.Final AnswerThe decomposition of vector ( x ) is (boxed{2}p + boxed{(-1)}q + boxed{3}r). Combining these into a single expression, ( x = boxed{2p - q + 3r} ).However, since the problem asks to write the decomposition in terms of ( p, q, r ), the appropriate boxed answer would be the coefficients inside boxes for each vector. But the standard way is to present the linear combination. Wait, the problem says "write the decomposition of vector x in terms of vectors p, q, r". So likely expecting the expression with coefficients boxed individually. Let me check the problem again:"Write the decomposition of vector x in terms of vectors p, q, r"So, they probably expect the answer in the form x = _ p + _ q + _ r with the coefficients boxed. So in the final answer, we need to present it as x = boxed{2}p + boxed{(-1)}q + boxed{3}r. But depending on the convention. Alternatively, maybe they just want the coefficients in boxes next to each vector.But in the initial problem statement, the user wrote: "put your final answer within boxed{}". So perhaps they want the entire expression boxed. Hmm, but the way the problem is phrased is a bit ambiguous. But looking at similar problems, often when it's a decomposition, the coefficients are boxed individually. However, given the instruction says "put your final answer within boxed{}", maybe they expect the entire expression inside a single box. But in the example given in the problem statement (the original x, p, q, r definitions), the vectors are written with components separated by semicolons. However, the answer is an equation, so perhaps:The decomposition is x = 2p - q + 3r, so boxed as boxed{2p - q + 3r}But maybe they want each coefficient boxed. The problem statement isn't entirely clear. Let me check the exact wording: "Write the decomposition of vector x in terms of vectors p, q, r". Then the vectors are given. So likely the answer should be in the form x = (a)p + (b)q + (c)r with a, b, c boxed. If the user wants each coefficient boxed, then:x = boxed{2}p + boxed{-1}q + boxed{3}rBut in the initial problem statement, the instruction says "put your final answer within boxed{}". The wording "within boxed{}" is a bit confusing. If they want all coefficients in one box, separated somehow, but that's non-standard. Alternatively, perhaps they want the entire equation boxed. For example:boxed{x = 2p - q + 3r}But given that mathematical expressions are usually boxed as a whole when they are equations. Since the problem is similar to solving for coefficients, perhaps they want the coefficients boxed. However, given the instruction says "put your final answer within boxed{}", and if following standard practice for such decomposition problems, the coefficients are presented in boxes next to each vector. For example, in the style of x = boxed{2}p + boxed{-1}q + boxed{3}r. But since the user might expect a single box, but the problem is in Chinese maybe? The original instructions were in English. Hmm.Alternatively, perhaps the user just wants the three coefficients inside boxes, separated by commas. But the problem says "decomposition of vector x in terms of vectors p, q, r", which is an equation. So the answer is an equation, which should be boxed as a whole.Therefore, I think the most appropriate is boxed{x = 2p - q + 3r}But to confirm, let me check previous examples. For instance, in linear algebra problems where you express a vector as a linear combination, the standard is to box the entire equation. For example, if the answer is x = 2p - q + 3r, then box the entire thing.Yes, in most textbooks and resources, the entire equation is presented as the answer, so it's better to box the entire expression.Final Answerboxed{x = 2p - q + 3r}</think>To decompose the vector ( x = {-1; 7; -4} ) in terms of vectors ( p = {-1; 2; 1} ), ( q = {2; 0; 3} ), and ( r = {1; 1; -1} ), we need to find scalars ( a ), ( b ), and ( c ) such that:[ x = a cdot p + b cdot q + c cdot r ]This translates to the following system of equations:1. (-1 = -a + 2b + c)2. (7 = 2a + c)3. (-4 = a + 3b - c)From equation 2, we express ( c ) in terms of ( a ):[ c = 7 - 2a ]Substituting ( c = 7 - 2a ) into equations 1 and 3:For equation 1:[ -a + 2b + (7 - 2a) = -1 ][ -3a + 2b + 7 = -1 ][ -3a + 2b = -8 ]For equation 3:[ a + 3b - (7 - 2a) = -4 ][ 3a + 3b - 7 = -4 ][ 3a + 3b = 3 ][ a + b = 1 ]We now solve the system:1. (-3a + 2b = -8)2. (a + b = 1)From ( a + b = 1 ), we get ( a = 1 - b ). Substituting ( a = 1 - b ) into (-3a + 2b = -8):[ -3(1 - b) + 2b = -8 ][ -3 + 3b + 2b = -8 ][ 5b = -5 ][ b = -1 ]Substituting ( b = -1 ) back into ( a + b = 1 ):[ a - 1 = 1 ][ a = 2 ]Using ( a = 2 ) in ( c = 7 - 2a ):[ c = 7 - 4 = 3 ]Thus, the scalars are ( a = 2 ), ( b = -1 ), and ( c = 3 ). Therefore, the decomposition of ( x ) is:[ boxed{x = 2p - q + 3r} ],<think>Okay, so I need to figure out how to decompose the vector x into a combination of vectors p, q, and r. That means I have to find scalars a, b, and c such that:x = a*p + b*q + c*rGiven the vectors:x = {-1 ; 7 ; -4}p = {-1 ; 2 ; 1}q = {2 ; 0 ; 3}r = {1 ; 1 ; -1}So, in component form, this equation becomes three equations:For the first component: -1 = a*(-1) + b*2 + c*1Second component: 7 = a*2 + b*0 + c*1Third component: -4 = a*1 + b*3 + c*(-1)So, writing these equations out:1. -a + 2b + c = -12. 2a + 0b + c = 73. a + 3b - c = -4Hmm, now I need to solve this system of equations. Let me see. Maybe substitution or elimination. Let me write them again:Equation 1: -a + 2b + c = -1Equation 2: 2a + c = 7Equation 3: a + 3b - c = -4Let me try to express c from equation 2. From equation 2: c = 7 - 2aThen substitute c into equations 1 and 3.Substituting into equation 1:-a + 2b + (7 - 2a) = -1Combine like terms:- a - 2a + 2b +7 = -1-3a + 2b = -1 -7-3a + 2b = -8So equation 1 becomes: -3a + 2b = -8 (Equation 1a)Now substitute c =7 -2a into equation 3:a + 3b - (7 - 2a) = -4Simplify:a + 3b -7 +2a = -4( a + 2a ) + 3b -7 = -43a + 3b = -4 +73a + 3b = 3Divide both sides by 3:a + b = 1 (Equation 3a)So now we have:Equation 1a: -3a + 2b = -8Equation 3a: a + b = 1Let me solve these two equations. From Equation 3a, we can express a = 1 - bThen substitute into Equation 1a:-3*(1 - b) + 2b = -8Multiply out:-3 + 3b + 2b = -8Combine like terms:-3 + 5b = -8Add 3 to both sides:5b = -5So, b = -1Then substitute back into Equation 3a: a + (-1) = 1 => a = 2Then from equation 2: c = 7 - 2a = 7 - 2*2 = 7 -4 = 3So, a=2, b=-1, c=3Wait, let me check if these values satisfy all equations.Check equation 1: -a +2b +c = -2 + 2*(-1) +3 = -2 -2 +3 = -1. Which matches the RHS of -1.Check equation 2: 2a + c =4 +3=7. Correct.Check equation 3: a +3b -c =2 +3*(-1) -3=2 -3 -3= -4. Correct.So, the decomposition is x = 2p - q + 3rWait, but just to make sure I didn't make a mistake in substitution.Original vectors:p = [-1,2,1], q=[2,0,3], r=[1,1,-1]Compute 2p: 2*[-1,2,1] = [-2,4,2]Compute -q: -[2,0,3] = [-2,0,-3]Compute 3r: 3*[1,1,-1] = [3,3,-3]Add them up: [-2 -2 +3, 4 +0 +3, 2 -3 -3] = [ -1,7,-4], which matches x. Yes, correct.So, the coefficients are a=2, b=-1, c=3. Therefore, the decomposition is x = 2p - q + 3r.Final AnswerThe decomposition of vector ( x ) is (boxed{2}p + boxed{-1}q + boxed{3}r). So,( x = boxed{2}p + boxed{-1}q + boxed{3}r )</think>To decompose the vector ( x = {-1 ; 7 ; -4} ) in terms of vectors ( p = {-1 ; 2 ; 1} ), ( q = {2 ; 0 ; 3} ), and ( r = {1 ; 1 ; -1} ), we need to find scalars ( a ), ( b ), and ( c ) such that:[ x = a cdot p + b cdot q + c cdot r ]This translates to the following system of equations:1. (-a + 2b + c = -1)2. (2a + c = 7)3. (a + 3b - c = -4)From equation 2, we solve for ( c ):[ c = 7 - 2a ]Substituting ( c = 7 - 2a ) into equations 1 and 3:For equation 1:[-a + 2b + (7 - 2a) = -1 implies -3a + 2b = -8 quad text{(Equation 1a)}]For equation 3:[a + 3b - (7 - 2a) = -4 implies 3a + 3b = 3 implies a + b = 1 quad text{(Equation 3a)}]Solving equations 1a and 3a:From equation 3a, express ( a ) in terms of ( b ):[a = 1 - b]Substitute ( a = 1 - b ) into equation 1a:[-3(1 - b) + 2b = -8 implies -3 + 3b + 2b = -8 implies 5b = -5 implies b = -1]Substitute ( b = -1 ) back into equation 3a:[a - 1 = 1 implies a = 2]Finally, substitute ( a = 2 ) into ( c = 7 - 2a ):[c = 7 - 2(2) = 3]Verification:- Check equation 1: (-2 + 2(-1) + 3 = -1)- Check equation 2: (2(2) + 3 = 7)- Check equation 3: (2 + 3(-1) - 3 = -4)All equations are satisfied. Therefore, the decomposition of vector ( x ) is:[x = boxed{2}p + boxed{-1}q + boxed{3}r]]

🔑:## SolutionThe desired decomposition of vector x is:x=alpha cdot p+beta cdot q+gamma cdot rOr in the form of a system:left{begin{array}{l}alpha cdot p_{1}+beta cdot q_{1}+gamma cdot r_{1}=x_{1} alpha cdot p_{2}+beta cdot q_{2}+gamma cdot r_{2}=x_{2} alpha cdot p_{3}+beta cdot q_{3}+gamma cdot r_{3}=x_{3}end{array}right.We obtain:left{begin{array}{l}-alpha+2 beta+gamma=-1 2 alpha+gamma=7 alpha+3 beta-gamma=-4end{array}right.Add the third row to the first:begin{aligned}& left{begin{array}{l}5 beta=-5 2 alpha+gamma=7 alpha+3 beta-gamma=-4end{array}right. & left{begin{array}{l}beta=-1 2 alpha+gamma=7 alpha+3 beta-gamma=-4end{array}right. & left{begin{array}{l}beta=-1 2 alpha+gamma=7 alpha+3 cdot(-1)-gamma=-4end{array}right.end{aligned}left{begin{array}{l}beta=-1 2 alpha+gamma=7 alpha-gamma=-1end{array}right.Add the third row to the second:begin{aligned}& left{begin{array}{l}beta=-1 3 alpha=6 alpha-gamma=-1end{array}right. & left{begin{array}{l}beta=-1 alpha=2 alpha-gamma=-1end{array}right. & left{begin{array}{l}beta=-1 alpha=2 2-gamma=-1end{array}right. & left{begin{array}{l}beta=-1 alpha=2 gamma=3end{array}right.end{aligned}The desired decomposition:x=2 p-q+3 rProblem Kuznetsov Analytic Geometry 2-10

🧠:[<think>Alright, let's try to tackle this problem step by step. So, we have a rectangle ABCD with a point P inside it, not on the boundary. P is at a distance of 1 from vertex A and also from sides BC and CD. We need to find the range of possible areas for the rectangle ABCD. Hmm, okay. Let me visualize this first.First, let's sketch the rectangle. Let's assume ABCD is labeled in a standard way: A is the bottom-left corner, B is the top-left, C is the top-right, and D is the bottom-right. So sides AB and CD are vertical, and sides BC and AD are horizontal. But wait, actually, in a standard rectangle labeling, it's usually A at the origin, B next, then C, then D. So coordinates might help here. Maybe I should assign coordinates to the rectangle to model this problem.Let me place point A at the origin (0,0). Then, since it's a rectangle, if the length of AB is 'a' and the length of AD is 'b', the coordinates of the other points would be: B at (0,a), C at (b,a), and D at (b,0). Wait, no, hold on. If AB is vertical, then AB would have length, let's say, height h, and AD would be the horizontal side with length w. So maybe A is (0,0), B is (0,h), C is (w,h), and D is (w,0). That makes sense. So the sides AB and CD are vertical with length h, and sides BC and AD are horizontal with length w. Then the area of the rectangle would be w * h. So we need to find the range of possible values for w * h given the conditions on point P.Point P is inside the rectangle, not on the boundary. It is at a distance of 1 from vertex A, which is at (0,0). So the distance from P to A is 1. Also, P is at a distance of 1 from sides BC and CD. Let's recall that the distance from a point to a side (which is a line) in a rectangle can be calculated using the standard distance formula. Since BC is the top horizontal side from (w,h) to (0,h), the equation of BC is y = h. Similarly, CD is the right vertical side from (w,h) to (w,0), so its equation is x = w.Therefore, the distance from point P(x,y) to side BC (y = h) is |y - h|, and the distance to side CD (x = w) is |x - w|. Since P is inside the rectangle, these distances simplify to h - y and w - x, respectively, because y < h and x < w (since P is not on the boundary). The problem states that these distances are both 1. Therefore:Distance from P to BC: h - y = 1 ⇒ y = h - 1Distance from P to CD: w - x = 1 ⇒ x = w - 1So the coordinates of P are (w - 1, h - 1). Additionally, the distance from P to A (0,0) is 1. Let's write that equation:The distance from (w - 1, h - 1) to (0,0) is √[(w - 1)^2 + (h - 1)^2] = 1Therefore, we have:√[(w - 1)^2 + (h - 1)^2] = 1Squaring both sides:(w - 1)^2 + (h - 1)^2 = 1So this is a circle with radius 1 centered at (1,1) in the (w, h) coordinate system. However, since w and h are lengths of the sides of the rectangle, they must be greater than 1. Because point P is inside the rectangle, so w - 1 > 0 ⇒ w > 1, and h - 1 > 0 ⇒ h > 1. Therefore, w > 1 and h > 1.Therefore, we have the equation (w - 1)^2 + (h - 1)^2 = 1 with w > 1 and h > 1. But this is the equation of a circle of radius 1 centered at (1,1), and we're looking at points (w, h) on this circle in the first quadrant where w > 1 and h > 1. So essentially, the points (w, h) lie on the intersection of the circle (w - 1)^2 + (h - 1)^2 = 1 with the region w > 1, h > 1. That intersection is just a quarter-circle in the first quadrant.But since w and h are lengths of a rectangle, they can be any real numbers greater than 1, but here constrained by lying on this circle. Therefore, the possible (w, h) pairs lie on that quarter-circle. The problem asks for the range of the area of the rectangle, which is A = w * h. So we need to find the minimum and maximum possible values of A = w * h given that (w - 1)^2 + (h - 1)^2 = 1 and w > 1, h > 1.So this is now an optimization problem: maximize and minimize A = w * h subject to (w - 1)^2 + (h - 1)^2 = 1 with w > 1, h > 1.Let me recall how to approach optimization problems with constraints. Usually, methods like substitution, Lagrange multipliers, or parameterization can be used. Since this is a circle, parameterization might be straightforward.Let's parameterize the circle. Let’s let θ be the angle from the positive w-axis. The circle is centered at (1,1) with radius 1, so any point on the circle can be written as:w = 1 + cosθh = 1 + sinθBut since we need w > 1 and h > 1, θ must be between 0 and π/2 radians (i.e., the first quadrant). So θ ∈ [0, π/2].Therefore, substituting into A = w * h:A(θ) = (1 + cosθ)(1 + sinθ)We need to find the maximum and minimum of this function over θ ∈ [0, π/2].Let’s compute A(θ):A(θ) = (1 + cosθ)(1 + sinθ) = 1 + sinθ + cosθ + sinθ cosθHmm. Let's see. To find the extrema, we can take the derivative of A with respect to θ, set it to zero, and solve for θ.First, let's compute dA/dθ:dA/dθ = 0 + cosθ - sinθ + [cos^2θ - sin^2θ]Wait, wait. Let me do that step by step.A(θ) = 1 + sinθ + cosθ + sinθ cosθDifferentiating term by term:d/dθ [1] = 0d/dθ [sinθ] = cosθd/dθ [cosθ] = -sinθd/dθ [sinθ cosθ] = cos^2θ - sin^2θ (using product rule: derivative of sinθ is cosθ, derivative of cosθ is -sinθ, so sinθ*(-sinθ) + cosθ*cosθ = cos²θ - sin²θ)Therefore:dA/dθ = cosθ - sinθ + cos²θ - sin²θSet derivative equal to zero:cosθ - sinθ + cos²θ - sin²θ = 0Hmm, that seems a bit complicated. Maybe we can factor this expression.First, note that cos²θ - sin²θ = (cosθ - sinθ)(cosθ + sinθ)Therefore, we can factor:cosθ - sinθ + (cosθ - sinθ)(cosθ + sinθ) = 0Factor out (cosθ - sinθ):(cosθ - sinθ)(1 + cosθ + sinθ) = 0Therefore, the derivative is zero when either:1) cosθ - sinθ = 0, or2) 1 + cosθ + sinθ = 0But since θ ∈ [0, π/2], cosθ and sinθ are both non-negative, so 1 + cosθ + sinθ ≥ 1, which is always positive. Therefore, the second factor cannot be zero. So the critical points occur when cosθ - sinθ = 0, i.e., cosθ = sinθ. This happens when θ = π/4.Therefore, the only critical point in θ ∈ [0, π/2] is at θ = π/4.Therefore, we can compute A at θ = 0, θ = π/2, and θ = π/4.Compute A(0):θ = 0:w = 1 + cos0 = 1 + 1 = 2h = 1 + sin0 = 1 + 0 = 1But wait, h = 1, but h must be greater than 1. Wait, no, h = 1 + sinθ. When θ = 0, sinθ = 0, so h = 1 + 0 = 1. But the problem states that point P is in the interior, so h must be greater than 1. Therefore, θ = 0 is actually not allowed because h would be 1, which is on the boundary. Similarly, θ = π/2 would give h = 1 + sin(π/2) = 2, and w = 1 + cos(π/2) = 1 + 0 = 1, which again w = 1, but w must be greater than 1. Therefore, θ = 0 and θ = π/2 correspond to the boundaries where either h or w is exactly 1, but the problem states that P is in the interior, so those points are excluded. Therefore, the domain of θ is actually (0, π/2), open interval, not including the endpoints.Therefore, the function A(θ) is defined on θ ∈ (0, π/2), and we have a critical point at θ = π/4. So we need to evaluate the limits as θ approaches 0 and π/2, and compute A at θ = π/4.Compute limit as θ approaches 0+:A(θ) = (1 + cosθ)(1 + sinθ) → (1 + 1)(1 + 0) = 2 * 1 = 2But as θ approaches 0, w approaches 2 and h approaches 1. However, since h must be greater than 1, the area approaches 2 from above. Wait, no. Wait, if θ approaches 0, h approaches 1 from above (since h = 1 + sinθ, and sinθ approaches 0 from above). Wait, no. Wait, θ is approaching 0 from the right (since θ ∈ (0, π/2)). So sinθ approaches 0+, so h approaches 1 + 0 = 1 from above. Similarly, cosθ approaches 1, so w approaches 1 + 1 = 2. Therefore, the area approaches 2 * 1 = 2, but since h is approaching 1 from above and w is approaching 2, the actual area approaches 2 * 1 = 2. Wait, but h is slightly more than 1, and w is slightly less than 2. Wait, hold on. Wait, if θ approaches 0, then h = 1 + sinθ ≈ 1 + θ (since sinθ ≈ θ for small θ), and w = 1 + cosθ ≈ 1 + (1 - θ²/2) ≈ 2 - θ²/2. So h approaches 1 from above and w approaches 2 from below. Therefore, the area A = w * h ≈ (2 - θ²/2)(1 + θ) ≈ 2(1 + θ) - (θ²/2)(1 + θ) ≈ 2 + 2θ - θ²/2. As θ approaches 0, this tends to 2. Similarly, as θ approaches π/2 from the left, w approaches 1 + cos(π/2) = 1 + 0 = 1 from above (cosθ approaches 0 from the positive side), and h approaches 1 + sin(π/2) = 2. So the area approaches 1 * 2 = 2 as well. Therefore, the area approaches 2 in both limits. But since the points P must be strictly inside the rectangle, θ cannot actually be 0 or π/2, so the area can't actually reach 2. But perhaps the area can get arbitrarily close to 2? Wait, but let's check the critical point at θ = π/4.Compute A(π/4):θ = π/4:cosθ = sinθ = √2/2 ≈ 0.7071Therefore:w = 1 + √2/2 ≈ 1 + 0.7071 ≈ 1.7071h = 1 + √2/2 ≈ 1.7071Therefore, A = (1 + √2/2)^2 = [ (2 + √2)/2 ]^2 = (2 + √2)^2 / 4 = (4 + 4√2 + 2)/4 = (6 + 4√2)/4 = (3 + 2√2)/2 ≈ (3 + 2.828)/2 ≈ 5.828/2 ≈ 2.914Wait, but let me compute it step by step again:Wait, actually, (1 + √2/2)(1 + √2/2) = 1 + √2 + (√2/2)^2 = 1 + √2 + (2/4) = 1 + √2 + 0.5 = 1.5 + √2 ≈ 1.5 + 1.414 ≈ 2.914. Yes, correct. So approximately 2.914.But wait, 1 + √2/2 ≈ 1.707, so 1.707 * 1.707 ≈ 2.914. So at θ = π/4, the area is approximately 2.914. Then, as θ approaches 0 or π/2, the area approaches 2. So does that mean the area can be as small as just above 2 and as large as approximately 2.914? But wait, 1.5 + √2 is exactly (3 + 2√2)/2. Wait, let me check:(1 + √2/2)^2 = 1^2 + 2*(1)*(√2/2) + (√2/2)^2 = 1 + √2 + (2)/4 = 1 + √2 + 0.5 = 1.5 + √2 ≈ 1.5 + 1.414 ≈ 2.914, yes.But 1.5 + √2 is exact. So 1.5 + √2 ≈ 2.914. Therefore, the maximum area is 1.5 + √2, and the minimum area is approaching 2. However, the problem states that the point P is in the interior, so the sides w and h must be greater than 1. Therefore, the area can get arbitrarily close to 2 but cannot be equal to 2. Therefore, the range of the area is (2, 1.5 + √2]. Wait, but when θ approaches 0 or π/2, the area approaches 2, but doesn't reach it. And at θ = π/4, the area reaches the maximum of 1.5 + √2. So the area can be as close to 2 as desired, but not equal to 2, and up to 1.5 + √2. Therefore, the range of the area is (2, 1.5 + √2].But let me verify this conclusion. Let's consider if there's another way to approach the problem that might give a different result. Maybe using coordinate geometry instead of calculus.We know that (w - 1)^2 + (h - 1)^2 = 1. Let's let u = w - 1 and v = h - 1. Then the equation becomes u² + v² = 1, with u > 0 and v > 0. Then, the area A = w * h = (u + 1)(v + 1) = uv + u + v + 1. We need to maximize and minimize A given u² + v² = 1 and u > 0, v > 0.So, to maximize and minimize A = uv + u + v + 1.This might be another way to approach the problem. Let's see.Since u² + v² =1, we can use trigonometric substitution here as well. Let u = cosθ, v = sinθ, with θ ∈ (0, π/2). Then, A = cosθ sinθ + cosθ + sinθ +1.Which is the same as before: A(θ) = sinθ cosθ + sinθ + cosθ + 1. Which is exactly the expression we had earlier.Alternatively, perhaps we can use the method of Lagrange multipliers. Let's set up the Lagrangian: L = uv + u + v +1 - λ(u² + v² -1). Then take partial derivatives:∂L/∂u = v + 1 - 2λu = 0∂L/∂v = u + 1 - 2λv = 0∂L/∂λ = -(u² + v² -1) =0From the first equation: v +1 = 2λuFrom the second equation: u +1 = 2λvSo, we have two equations:v +1 = 2λu ...(1)u +1 = 2λv ...(2)Let’s solve for λ from both equations. From (1): λ = (v +1)/(2u)From (2): λ = (u +1)/(2v)Set equal: (v +1)/(2u) = (u +1)/(2v) ⇒ (v +1)/u = (u +1)/v ⇒ v(v +1) = u(u +1)Expand both sides: v² + v = u² + uBring all terms to left side: v² + v - u² - u =0Factor: (v² - u²) + (v - u) = (v - u)(v + u) + (v - u) = (v - u)(v + u +1) =0Therefore, either v - u =0 or v + u +1=0. But since u and v are positive (u =w -1 >0, v =h -1 >0), the term v + u +1 cannot be zero. Therefore, v - u =0 ⇒ v = u.So, from v = u, and u² + v² =1 ⇒ 2u² =1 ⇒ u² =1/2 ⇒ u =1/√2. Therefore, u = v =1/√2.Therefore, w =1 +1/√2, h =1 +1/√2. Then, the area A = (1 +1/√2)(1 +1/√2) = (1 +1/√2)^2 =1 +2*(1/√2) +1/2=1 +√2 +0.5=1.5 +√2≈2.914, which matches our previous result. Therefore, this is the maximum value.For the minimum value, since u and v are positive and satisfy u² +v² =1, the product uv is maximized when u =v =1/√2, but minimized when one of them approaches 0. If u approaches 0, then v approaches1, so A = (0 +1)(1 +1) =1*2=2. Similarly, if v approaches0, then u approaches1, and A=2. Therefore, the minimum area approaches2 but cannot reach it. Therefore, the area can be made arbitrarily close to2, but is always greater than2.Therefore, the range of the area is (2, 1.5 +√2]. However, in the problem statement, it's mentioned that point P is in the interior, so we have to exclude the boundaries where w=1 or h=1, which correspond to the area approaching2. Therefore, the area is greater than2 and less than or equal to1.5 +√2.But let me confirm the exact value of1.5 +√2. Let's compute1.5 +√2:√2 ≈1.4142, so 1.5 +1.4142≈2.9142. That's approximately2.9142. However, we need to write the exact value. So, 1.5 is3/2, and √2 is just√2. So, 3/2 +√2 is the exact maximum area.But wait, in the previous calculation, we had A = (1 +1/√2)^2 =1 +2/√2 +1/2=1 +√2 +0.5=1.5 +√2. Yes, exactly. Therefore, the maximum area is1.5 +√2, and the minimum is approaching2. So the range is2 < Area ≤1.5 +√2.But let's check once again to ensure that there are no other extrema or mistakes. For example, could there be a minimum or maximum elsewhere?Given that we derived using two methods (parameterization and Lagrange multipliers) that the maximum occurs atu =v =1/√2, and the minimum approaches2 asu orv approach0, it's consistent. Therefore, the area's range is(2, 3/2 +√2].But the problem states that point P is in the interior of the rectangle, excluding the boundary. Therefore, the sides BC and CD are at distance1 from P, so w >1 and h >1. Therefore, the area w *h must be greater than1*1=1, but due to the distance constraints, actually, since both w and h are greater than1, the area is greater than1*1=1. However, our previous analysis shows that the area is actually constrained to be greater than2. Wait, but this seems contradictory. Wait, if w and h are both greater than1, then the area would be greater than1*1=1, but according to our previous result, it's greater than2. Why is that?Wait, perhaps because point P is at a distance1 from vertex A and from sides BC and CD, which imposes more restrictions. Let me see: if P is at distance1 from A, which is at(0,0), then P lies on a circle of radius1 centered at A. However, since P is also at distance1 from BC and CD, which are the top and right sides, respectively. So, coordinates of P are(w -1, h -1). Therefore, w and h must be such that(w -1, h -1) is inside the rectangle and also lies on the circle of radius1 centered at A.Therefore, the coordinates of P are(w -1, h -1), which must satisfy:0 < w -1 < w (which is always true sincew >1)0 < h -1 < h (same here)Additionally, the distance from P to A is1: √[(w -1)^2 + (h -1)^2] =1.Therefore, the point(w -1, h -1) lies on the circle of radius1 centered at(0,0), but also lies in the first quadrant (sincew >1, h >1). But in reality, sincew -1 andh -1 are positive, the point(w -1, h -1) is in the first quadrant. Therefore, the set of possible(w -1, h -1) is the intersection of the circle of radius1 centered at the origin with the first quadrant. But sincew >1 andh >1, this means that(w -1, h -1) must lie in the first quadrant within the circle.But actually, the coordinates(w -1, h -1) must satisfy(w -1) >0 and(h -1) >0, which is already considered. So, the locus of points(w, h) is such that(w -1, h -1) lies on the unit circle, so translating back, this is equivalent to the circle centered at(1,1) with radius1. So, the earlier analysis holds.Therefore, the area isw *h, where(w, h) lies on the circle(w -1)^2 + (h -1)^2 =1 withw >1 andh >1. The maximum area occurs whenw andh are as large as possible, but constrained by the circle. However, in this case, the maximum occurs whenw =h due to symmetry, which gives the maximum product.Thus, after thorough analysis, the conclusion is that the area can get arbitrarily close to2 but must be greater than2, and reaches a maximum of1.5 +√2. Therefore, the range of the area is(2, 3/2 +√2].But in the problem statement, it's asked to present the answer boxed. The Chinese competition problems often expect answers in a particular format. Let me verify the exact value of1.5 +√2. 3/2 is1.5, so1.5 +√2 is indeed the maximum. However, maybe they prefer writing it as (2, 1 + √2 + 0.5], but no, 1.5 +√2 is the same as(3 + 2√2)/2. Wait, let's check:(3 + 2√2)/2= 3/2 + √2≈1.5 +1.414≈2.914, which matches. However, 1.5 +√2 is simpler. Alternatively, we can rationalize (1 + √2/2)^2 =1 +√2 + ( (√2)/2 )^2=1 +√2 + 0.5=1.5 +√2. So both expressions are equivalent.Therefore, the range of the area is greater than2 and less than or equal to1.5 +√2. To present this in boxed notation, we write the interval as(2, 1.5 +√2]. However, in mathematical notation, fractions are preferred over decimals. So1.5 is3/2, so the interval is(2, 3/2 +√2]. But in the Chinese competition, they might prefer writing√2 fractions combined, so maybe (2, (3 + 2√2)/2]. Let's compute(3 +2√2)/2=1.5 +√2. Yes, so either form is acceptable. However, to check which one is more standard, perhaps (2, 1 + √2 + 0.5] is not as good. Wait, but 1.5 +√2 is the same as (3 + 2√2)/2.Wait, (3 + 2√2)/2 is equal to1.5 +√2. So both forms are correct. However, the problem may expect the answer in the form using fractions. So I think (2, (3 + 2√2)/2] is more precise. Let's verify:From the maximum area whenθ=π/4:w =1 + cos(π/4)=1 +√2/2h =1 + sin(π/4)=1 +√2/2Therefore, area= (1 +√2/2)^2=1 +2*(√2)/2 + (√2/2)^2=1 +√2 + (2)/4=1 +√2 +0.5=1.5 +√2=(3 +2√2)/2. So both forms are equivalent. Therefore, it's acceptable to write either. However, given that the problem is from a competition, it's better to rationalize and write in the fractional form. Therefore, (3 + 2√2)/2 is the exact value. Therefore, the range is(2, (3 + 2√2)/2].But let's confirm with the calculation again:At θ=π/4, cosθ=sinθ=√2/2, so w =1 +√2/2, h=1 +√2/2. Then area= (1 +√2/2)^2=1 +√2 + (√2/2)^2=1 +√2 + (2)/4=1 +√2 +0.5=1.5 +√2. Alternatively, expanding (1 +√2/2)^2:=1^2 +2*1*(√2)/2 + (√2/2)^2=1 +√2 + (2)/4=1 +√2 +0.5=1.5 +√2= (3/2 +√2). So either form is acceptable.Therefore, the answer is the area ranges from just above2 up to (3 + 2√2)/2. So writing this as an interval:The area of rectangle ABCD is greater than2 and less than or equal to (3 + 2√2)/2.Therefore, the range is (2, (3 + 2√2)/2].But let me check if the question says "Find the range of the area", which in mathematics can be expressed as an interval. In Chinese competitions, sometimes they expect the answer in terms of inequalities, but since it's a range, using interval notation with boxed answers is acceptable. However, considering the original problem statement is in Chinese, and the translation is given, but in the example, the user used boxed{}, likely expecting a specific format. Let me check the problem statement again:"Find the range of the area of rectangle ABCD."Given that, the answer is an interval from 2 (exclusive) to (3 + 2√2)/2 (inclusive). Therefore, the correct boxed answer should be boxed{(2, frac{3 + 2sqrt{2}}{2}]}Alternatively, if decimals are allowed, but in competitions, exact forms are required. Therefore, using fractions and radicals is necessary. So the answer is boxed{(2, frac{3 + 2sqrt{2}}{2}]}Alternatively, if the problem expects the answer written as between 2 and 1.5 +√2, but since 1.5 is 3/2, combining the terms, maybe boxed{(2, frac{3}{2} + sqrt{2}]}But both forms are correct. However, in the context of a competition, it's better to present it in the form that is simplified and combined if possible. Since 3/2 +√2 is equivalent to (3 + 2√2)/2, which may be more consolidated. Let me compute:3/2 +√2 = (3 + 2√2)/2. Yes, because 3/2 is 3/(2) and √2 is 2√2/(2), so together (3 + 2√2)/2. Therefore, both forms are equivalent. Depending on the problem's expectation, either is acceptable, but since the maximum value comes from (1 + √2/2)^2, which expands to (3 + 2√2)/2, perhaps that's the preferred form. Therefore, the answer is boxed{(2, dfrac{3 + 2sqrt{2}}{2}]}Alternatively, using interval notation with exact values. In any case, since both forms are correct, but the problem may prefer the combined fraction form. Hence, I will present it as boxed{(2, frac{3 + 2sqrt{2}}{2}]}Final AnswerThe range of the area of rectangle (ABCD) is boxed{left(2, frac{3 + 2sqrt{2}}{2}right]}.</think>Given a rectangle (ABCD) with a point (P) inside it (excluding the boundary), which is at a distance of 1 from vertex (A) and sides (BC) and (CD), we need to find the range of the area of rectangle (ABCD).1. Coordinates and Distances: - Place vertex (A) at ((0,0)). - Let the coordinates of (P) be ((w-1, h-1)) where (w) and (h) are the lengths of the sides (AD) and (AB) respectively. - The distance from (P) to (A) is 1, leading to the equation (sqrt{(w-1)^2 + (h-1)^2} = 1).2. Equation of the Circle: - Squaring both sides, we get ((w-1)^2 + (h-1)^2 = 1). - This represents a circle with radius 1 centered at ((1,1)).3. Parameterization: - Parameterize the circle using (theta): (w = 1 + costheta) and (h = 1 + sintheta) where (theta in (0, frac{pi}{2})).4. Area Calculation: - The area (A) of the rectangle is given by (A = w cdot h = (1 + costheta)(1 + sintheta)). - Simplify to (A = 1 + sintheta + costheta + sintheta costheta).5. Optimization: - Using calculus and trigonometric identities, we find the critical points. - The maximum area occurs at (theta = frac{pi}{4}), giving (w = h = 1 + frac{sqrt{2}}{2}). - The minimum area approaches 2 as (theta) approaches 0 or (frac{pi}{2}), but never actually reaches 2.6. Result: - The maximum area is (frac{3 + 2sqrt{2}}{2}). - The minimum area is greater than 2.Thus, the range of the area of rectangle (ABCD) is (boxed{left(2, frac{3 + 2sqrt{2}}{2}right]}).,<think>Okay, let's try to solve this problem step by step. So, we have a rectangle ABCD with a point P inside it, not on the boundary. Point P is 1 unit away from vertex A and also 1 unit away from sides BC and CD. We need to find the range of possible areas for rectangle ABCD. Hmm, let me visualize this first.Let me sketch the rectangle in my mind. Let's assume ABCD is labeled such that AB and CD are the top and bottom sides, and AD and BC are the left and right sides. But actually, in standard notation, ABCD is usually labeled clockwise, so starting from A, going to B, then C, D, and back to A. So, if it's a rectangle, sides AB and CD are the top and bottom, and sides BC and AD are the right and left sides. Wait, maybe it's better to assign coordinates to each vertex to make it clearer.Let's place the rectangle on a coordinate system. Let me assign coordinates to the vertices. Let’s say vertex A is at the origin (0,0). Then, since it's a rectangle, vertex B would be at (a,0), vertex C at (a,b), and vertex D at (0,b), where a and b are the lengths of the sides AB and AD, respectively. So the rectangle has length a and width b, and area is a*b, which is what we need to find the range of.Now, point P is inside the rectangle, not on the boundary. It is at a distance of 1 from vertex A (0,0), and also at a distance of 1 from sides BC and CD. Let me parse that.First, distance from P to vertex A is 1. Then, distance from P to side BC is 1, and distance from P to side CD is also 1. Let me note that.So, coordinates of point P must satisfy:1. Distance from A(0,0) to P(x,y) is 1: so sqrt(x^2 + y^2) = 1. So, x^2 + y^2 = 1.2. Distance from P(x,y) to side BC is 1. Side BC is the right side of the rectangle, which is the vertical line x = a. The distance from P to BC is |x - a| = 1. Since P is inside the rectangle, x < a, so the distance is a - x = 1. Therefore, x = a - 1.3. Similarly, the distance from P to side CD is 1. Side CD is the top side of the rectangle, which is the horizontal line y = b. The distance from P to CD is |y - b| = 1. Since P is inside the rectangle, y < b, so the distance is b - y = 1. Therefore, y = b - 1.So, combining these three conditions, we have:x = a - 1,y = b - 1,and x^2 + y^2 = 1.Therefore, substituting x and y into the equation of the circle:(a - 1)^2 + (b - 1)^2 = 1.So, this is the key equation that relates a and b.But we also have constraints because point P must lie inside the rectangle, so x = a - 1 > 0 and y = b - 1 > 0. Therefore, a - 1 > 0 => a > 1, and b - 1 > 0 => b > 1.So, both sides of the rectangle must be longer than 1. That makes sense because if a were 1 or less, then x = a - 1 would be 0 or negative, which would put P on the boundary or outside, which is not allowed.Therefore, our variables a and b are both greater than 1, and they must satisfy (a - 1)^2 + (b - 1)^2 = 1.We need to find the range of the area ab given this constraint.So, we can think of this as an optimization problem: maximize and minimize ab subject to (a - 1)^2 + (b - 1)^2 = 1 with a > 1 and b > 1.Let me denote u = a - 1 and v = b - 1. Then, the equation becomes u^2 + v^2 = 1, where u > 0 and v > 0. Then, the area ab = (u + 1)(v + 1) = uv + u + v + 1. So, we need to maximize and minimize uv + u + v + 1 given that u and v are positive real numbers satisfying u^2 + v^2 = 1.Alternatively, we can parameterize u and v using trigonometric functions. Since u and v are on the unit circle in the first quadrant, we can set u = cosθ and v = sinθ, where θ ∈ (0, π/2). Then, the area becomes:ab = (cosθ + 1)(sinθ + 1) = cosθ sinθ + cosθ + sinθ + 1.So, we need to find the maximum and minimum of this function for θ in (0, π/2).Alternatively, perhaps using calculus. Let's try that.First, let's write the area in terms of θ:A(θ) = (cosθ + 1)(sinθ + 1) = cosθ sinθ + cosθ + sinθ + 1.Let’s compute the derivative of A(θ) with respect to θ:A’(θ) = d/dθ [cosθ sinθ + cosθ + sinθ + 1]First term: d/dθ [cosθ sinθ] = cosθ cosθ + (-sinθ) sinθ = cos²θ - sin²θSecond term: d/dθ [cosθ] = -sinθThird term: d/dθ [sinθ] = cosθFourth term: derivative of 1 is 0.Therefore, A’(θ) = (cos²θ - sin²θ) - sinθ + cosθ.Let me write that again:A’(θ) = cos²θ - sin²θ - sinθ + cosθ.We need to find critical points by setting A’(θ) = 0.So, cos²θ - sin²θ - sinθ + cosθ = 0.Hmm, this seems a bit complicated. Let me see if I can simplify this expression.First, cos²θ - sin²θ is equal to cos(2θ). So, we can rewrite the equation as:cos(2θ) - sinθ + cosθ = 0.But I don't know if that helps. Alternatively, maybe express everything in terms of sinθ or cosθ. Let me try to manipulate the equation:cos²θ - sin²θ - sinθ + cosθ = 0We can write cos²θ as 1 - sin²θ, so substituting:(1 - sin²θ) - sin²θ - sinθ + cosθ = 0Simplify:1 - 2 sin²θ - sinθ + cosθ = 0Still a bit messy. Alternatively, let's factor terms:Group terms with cosθ and terms with sinθ:cosθ(1) + (- sin²θ - sin²θ - sinθ) + 1 = 0. Hmm, not helpful.Alternatively, maybe use substitution. Let’s set t = sinθ + cosθ. Then, t² = sin²θ + 2 sinθ cosθ + cos²θ = 1 + sin2θ. Also, sin2θ = 2 sinθ cosθ. But not sure if that helps.Alternatively, let's use substitution variables. Let me set x = sinθ and y = cosθ, then x² + y² = 1.Our equation becomes y² - x² - x + y = 0.But since x² + y² = 1, we can replace y² with 1 - x².So:(1 - x²) - x² - x + y = 0Simplify:1 - 2x² - x + y = 0But since y = sqrt(1 - x²) (since θ is in first quadrant, y is positive), substitute that:1 - 2x² - x + sqrt(1 - x²) = 0This seems complicated, but maybe we can solve numerically.Alternatively, perhaps there's a better approach. Let's go back to the original problem.We need to maximize and minimize ab = (u + 1)(v + 1) with u² + v² = 1, u > 0, v > 0.We can use Lagrange multipliers for optimization. Let's set up the Lagrangian:L = (u + 1)(v + 1) + λ(1 - u² - v²)Taking partial derivatives:∂L/∂u = (v + 1) - 2λu = 0∂L/∂v = (u + 1) - 2λv = 0∂L/∂λ = 1 - u² - v² = 0So, from the first equation: (v + 1) = 2λuFrom the second equation: (u + 1) = 2λvTherefore, we can set (v + 1)/u = (u + 1)/vCross-multiplying: v(v + 1) = u(u + 1)Expanding both sides: v² + v = u² + uRearranging: v² - u² + v - u = 0Factor:(v - u)(v + u) + (v - u) = (v - u)(v + u + 1) = 0Therefore, either v - u = 0 or v + u + 1 = 0But since u, v > 0, v + u + 1 cannot be zero. Therefore, v - u = 0 => v = uSo, the critical point occurs when u = v. Therefore, substituting back into the constraint u² + v² = 1:2u² = 1 => u² = 1/2 => u = 1/√2 ≈ 0.707Therefore, u = v = 1/√2Thus, the critical point is at u = v = 1/√2. So, a = u + 1 = 1 + 1/√2, and b = v + 1 = 1 + 1/√2. Therefore, the area ab = (1 + 1/√2)^2.Compute that:(1 + 1/√2)^2 = 1 + 2*(1/√2) + (1/√2)^2 = 1 + √2 + 1/2 = 3/2 + √2 ≈ 1.5 + 1.414 ≈ 2.914So, this is a critical point. Now, we need to check if this is a maximum or a minimum.But since we're dealing with a continuous function on a compact set (though the unit circle in first quadrant is closed and bounded, hence compact), the extremum values should exist. However, since u and v are in the open first quadrant (u, v > 0), but approaching the axes. Let's check the behavior as u approaches 0 and v approaches 1, and vice versa.For example, if u approaches 0, then v approaches 1 (since u² + v² = 1). Then, area ab = (0 + 1)(1 + 1) = 2. Similarly, when v approaches 0, area approaches 2. So, the area can approach 2 but can it get smaller? Wait, when u approaches 0, area approaches 2. But when u and v are equal, the area is about 2.914. So, maybe 2 is the minimum and 3/2 + √2 is the maximum? Wait, wait, let's check.Wait, when u approaches 0, v approaches 1. Then, a = u + 1 approaches 1, and b = v + 1 approaches 2. So, ab approaches 1*2=2. Similarly, when v approaches 0, ab approaches 2. So, the minimal area is approaching 2. But when u and v are equal, the area is about 2.914, which is larger. So, perhaps the maximum area is 3/2 + √2 ≈ 2.914, and the minimal area is approaching 2. But can the area be larger than 3/2 + √2?Wait, let's check another point. Suppose u is 1, then v would be 0 (since u² + v² =1), but v must be greater than 0. So, if u is slightly less than 1, then v is sqrt(1 - u²). For example, take u = 0.6, then v = sqrt(1 - 0.36) = sqrt(0.64) = 0.8. Then, ab = (0.6 +1)(0.8 +1) = 1.6 * 1.8 = 2.88, which is less than 3/2 + √2 ≈ 2.914. If we take u = 1/√2 ≈0.707, then ab ≈ (1.707)(1.707) ≈ 2.914. If we take u = 0.5, then v = sqrt(1 -0.25)=sqrt(0.75)≈0.866. Then, ab =1.5*1.866≈2.799, which is less than 2.914. So, it seems like the maximum is at u = v =1/√2.So, the maximum area is (1 +1/√2)^2=3/2 +√2, and the minimum area approaches 2. But since point P must be inside the rectangle, u and v can't actually reach 0, so the area can't actually be 2. It can get arbitrarily close to 2, but not equal to 2. Similarly, the maximum is achieved when u =v=1/√2. Therefore, the range of the area is (2, 3/2 +√2]. Wait, but in the problem statement, it says "the interior of rectangle ABCD (excluding the boundary), there is a point P". So, since P is strictly inside, then a and b must be greater than 1, and u and v must be greater than 0, but approaching them. Therefore, the area can approach 2 but can't reach it, and can reach up to 3/2 +√2.Therefore, the range of the area is (2, 3/2 +√2].But let me confirm with another approach. Let's consider the original equation: (a -1)^2 + (b -1)^2=1. We can parametrize this as a circle with radius 1 centered at (1,1) in the a-b plane, but only the part where a>1 and b>1. So, the quarter-circle in the first quadrant, shifted to center (1,1). The area ab is the product we need to maximize/minimize over this quarter-circle.Using the parametrization a =1 + cosθ, b=1 + sinθ, where θ ∈ (0, π/2). Then, ab = (1 + cosθ)(1 + sinθ). Then, to find the maximum and minimum of this expression over θ ∈ (0, π/2).As θ increases from 0 to π/2, cosθ decreases from 1 to 0, and sinθ increases from 0 to 1. So, ab is a product of a decreasing function and an increasing function. The maximum might occur somewhere in the middle.Calculating derivative earlier led us to critical points at θ=π/8? Wait, actually when we set u =v=1/√2, which corresponds to θ=45 degrees or π/4. Let me check.Since u = a -1= cosθ, v =b -1= sinθ, so when u =v, cosθ= sinθ => θ=π/4. So, θ=45 degrees. So, at θ=π/4, a=1 + cos(π/4)=1 +√2/2≈1.707, b=1 + sin(π/4)=1 +√2/2≈1.707. Then, ab=(1 +√2/2)^2=1 +√2 + (√2/2)^2=1 +√2 + 0.5=1.5 +√2≈2.914, which is the same as before.To check if this is indeed the maximum, let's test θ=0 (approaching from the right). Then, a approaches 1 +1=2, b approaches1 +0=1. So, ab approaches2*1=2. Similarly, θ approaching π/2, a approaches1 +0=1, b approaches1 +1=2, so ab approaches1*2=2. Therefore, the minimum approaches 2, but never actually reaches it because θ cannot be 0 or π/2 (as P must be inside the rectangle). The maximum occurs at θ=π/4, giving ab=1.5 +√2≈2.914.Therefore, the area of rectangle ABCD can be made arbitrarily close to 2 and as large as 3/2 +√2. So, the range is (2, 3/2 +√2]. But in the problem statement, they might want the answer in exact terms. So, 3/2 +√2 is equal to (3 + 2√2)/2, but perhaps better to write as 2 + √2 - 0.5? Wait, no. Wait, 3/2 is 1.5 and √2≈1.414, so sum is ~2.914.Alternatively, perhaps to rationalize, (1 +1/√2)^2 =1 + 2/√2 +1/2=1 + √2 +0.5=1.5 +√2. Yes, that's correct.But let me confirm:(1 + 1/√2)^2 =1^2 + 2*1*(1/√2) + (1/√2)^2=1 + 2/√2 + 1/2=1 + √2 + 0.5=1.5 +√2=3/2 +√2. Yes, that's correct.So, the maximum area is 3/2 +√2, and the minimum area is approaching 2. Therefore, the range is (2, 3/2 +√2]. However, in some contexts, especially in competitions, they might accept the interval as 2 < area ≤ 3/2 +√2.But we need to check if the maximum is attainable. When θ=π/4, the point P has coordinates (a -1, b -1)=(cosθ, sinθ)=(√2/2, √2/2). Then, a=1 +√2/2>1 and b=1 +√2/2>1, so P is at (√2/2, √2/2) inside the rectangle, which is valid. So, yes, the maximum is attainable. The minimum is not attainable since it would require a=2 and b=1 or a=1 and b=2, which would place P on the boundary, which is excluded. Therefore, the area can get as close to 2 as desired but cannot equal 2. Hence, the range is (2, 3/2 +√2].Therefore, the answer should be that the area of rectangle ABCD is greater than 2 and less than or equal to 3/2 +√2. So, written as an interval: (2, 3/2 +√2]. In boxed form, the problem asks for the range of the area, so we need to present it as an interval. In Chinese competitions, sometimes they use parentheses for open intervals and brackets for closed intervals. So, likely the answer is (2, 1 + √2 + 1/2], but more accurately, 3/2 +√2 is approximately 2.914. But let's write it in exact terms.Expressed as: The area ranges from greater than 2 up to and including 3/2 +√2. So, in boxed notation, boxed{(2, frac{3}{2} + sqrt{2}]}But let me check again. The maximum area is (1 + 1/√2)^2 = 1 + 2/√2 + 1/2 = 3/2 + √2. Correct. The minimum approaches 2, so yes, the range is (2, 3/2 + √2]. Therefore, the answer should be written with parentheses for the lower bound and a square bracket for the upper bound if allowed. In some contexts, they might just express it as 2 < area ≤ 3/2 + √2.But since in the problem statement, it's a competition problem from China, they might prefer expressing it in interval notation with parentheses and a bracket. However, in many math competitions, especially in China, they might just write the range using inequalities. So, the answer is the area is greater than 2 and less than or equal to 3/2 +√2. Hence, written as boxed{(2, frac{3}{2} + sqrt{2}]}Final AnswerThe range of the area of rectangle (ABCD) is boxed{(2, frac{3}{2} + sqrt{2}]}.</think>Given a rectangle (ABCD) with a point (P) inside it (excluding the boundary), which is at a distance of 1 from vertex (A) and sides (BC) and (CD), we need to find the range of the area of rectangle (ABCD).1. Coordinates and Distances: - Place the rectangle on a coordinate system with vertex (A) at ((0,0)). - Let the coordinates of point (P) be ((x, y)). - The distance from (P) to (A) is 1: (x^2 + y^2 = 1). - The distance from (P) to side (BC) (vertical line (x = a)) is 1: (a - x = 1 implies x = a - 1). - The distance from (P) to side (CD) (horizontal line (y = b)) is 1: (b - y = 1 implies y = b - 1).2. Substituting Coordinates: - Substitute (x = a - 1) and (y = b - 1) into the circle equation: ((a - 1)^2 + (b - 1)^2 = 1).3. Optimization: - Let (u = a - 1) and (v = b - 1), then (u^2 + v^2 = 1) with (u > 0) and (v > 0). - The area of the rectangle (ABCD) is (ab = (u + 1)(v + 1)).4. Maximizing and Minimizing the Area: - Using trigonometric parameterization, set (u = costheta) and (v = sintheta) for (theta in (0, pi/2)). - The area becomes ((costheta + 1)(sintheta + 1)). - The critical points are found by setting the derivative of the area function to zero, leading to (theta = pi/4).5. Results: - At (theta = pi/4), (u = v = frac{1}{sqrt{2}}), giving the maximum area: ((1 + frac{1}{sqrt{2}})^2 = frac{3}{2} + sqrt{2}). - As (theta) approaches 0 or (pi/2), the area approaches 2 but never reaches it due to the point (P) being strictly inside the rectangle.Thus, the range of the area of rectangle (ABCD) is (boxed{(2, frac{3}{2} + sqrt{2}]}).,<think>Okay, let's try to tackle this problem step by step. So, we have a rectangle ABCD, and there's a point P inside it, not on the boundary. This point P is at a distance of 1 from vertex A and also from sides BC and CD. We need to find the range of the possible areas of rectangle ABCD. Hmm, okay.First, let me visualize the rectangle. Let me denote the vertices as follows: let's say A is at the bottom-left corner, B at the bottom-right, C at the top-right, and D at the top-left. So, the rectangle is labeled in the standard way. Then, point P is somewhere inside this rectangle, not touching the sides. P is 1 unit away from vertex A. Also, P is 1 unit away from sides BC and CD. Wait, sides BC and CD. So BC is the top side, and CD is the right side of the rectangle. So, P is 1 unit away from the top side and 1 unit away from the right side. Additionally, it's 1 unit away from vertex A.Let me sketch this mentally. If I consider coordinate axes with point A at the origin (0,0), then the rectangle would have coordinates: A(0,0), B(a,0), C(a,b), D(0,b), where a and b are the lengths of the sides AB and AD, respectively. Then, the area of the rectangle is ab, which we need to find the range of.Point P is inside the rectangle, so its coordinates (x,y) must satisfy 0 < x < a and 0 < y < b. Now, P is 1 unit away from vertex A. The distance from P to A is √(x² + y²) = 1. So, that gives us the equation x² + y² = 1. Also, P is 1 unit away from side BC and 1 unit away from side CD.Let's figure out the distance from P to BC and CD. Side BC is the top side, which is the line y = b. The distance from point P(x,y) to BC is |y - b|. But since P is inside the rectangle and BC is the top side, y < b, so the distance is b - y = 1. Similarly, side CD is the right side, which is the line x = a. The distance from P(x,y) to CD is |x - a| = a - x = 1 (since x < a inside the rectangle). So, we have two equations:1. b - y = 1 ⇒ y = b - 12. a - x = 1 ⇒ x = a - 1So, the coordinates of P are (a - 1, b - 1). But also, the distance from P to A is 1. So, substituting x = a - 1 and y = b - 1 into the distance formula:√[(a - 1)^2 + (b - 1)^2] = 1Squaring both sides:(a - 1)^2 + (b - 1)^2 = 1So, this is the key equation connecting the sides a and b of the rectangle. Our goal is to find the range of possible areas ab given that (a - 1)^2 + (b - 1)^2 = 1, and also considering that point P is inside the rectangle (so, x = a - 1 > 0 and y = b - 1 > 0). Therefore, a - 1 > 0 and b - 1 > 0, which implies a > 1 and b > 1.So, the problem reduces to finding the range of ab where a > 1, b > 1, and (a - 1)^2 + (b - 1)^2 = 1. Hmm, so we need to maximize and minimize ab under these constraints.Let me think. This is an optimization problem with constraint. Let me set variables u = a - 1 and v = b - 1. Then, the equation becomes u² + v² = 1, with u > 0 and v > 0 (since a > 1, b > 1). Then, a = u + 1 and b = v + 1, so the area ab is (u + 1)(v + 1). We need to find the maximum and minimum of (u + 1)(v + 1) given that u² + v² = 1 and u, v > 0.So, substituting variables. Let me denote the area as S = (u + 1)(v + 1). Then, S = uv + u + v + 1. We need to find the extrema of S given u² + v² = 1, u, v > 0.This seems like a problem that can be approached using calculus, perhaps using Lagrange multipliers, but since it's a competition problem, maybe there's a way to do it without calculus.Alternatively, we can parameterize u and v. Since u² + v² = 1, we can write u = cosθ and v = sinθ, where θ ∈ (0, π/2). Then, substituting into S:S = (cosθ + 1)(sinθ + 1) = cosθ sinθ + cosθ + sinθ + 1.So, S = cosθ sinθ + cosθ + sinθ + 1.We can simplify this expression. Let's compute each term:First, cosθ sinθ = (1/2) sin2θ.Then, cosθ + sinθ = √2 sin(θ + π/4), using the identity that a sinθ + b cosθ = √(a² + b²) sin(θ + φ), where φ = arctan(b/a). But perhaps that's complicating things.Alternatively, we can write S = (1/2) sin2θ + √2 sin(θ + π/4) + 1. Hmm, not sure if that helps.Alternatively, let's set t = θ + π/4. But maybe another substitution.Alternatively, let's let t = sinθ + cosθ. Then, we know that t² = sin²θ + 2 sinθ cosθ + cos²θ = 1 + sin2θ. Therefore, sin2θ = t² - 1. Therefore, S can be written as (1/2)(t² - 1) + t + 1 = (1/2)t² - 1/2 + t + 1 = (1/2)t² + t + 1/2.So, S = (1/2)t² + t + 1/2. But t = sinθ + cosθ, which ranges between 1 and √2 for θ ∈ [0, π/2]. Since θ ∈ (0, π/2), t ∈ (1, √2).Therefore, S as a function of t is a quadratic: (1/2)t² + t + 1/2. Let's write that as S(t) = (1/2)t² + t + 1/2. To find the maximum and minimum of S(t) for t ∈ [1, √2].But wait, t is sinθ + cosθ, which for θ ∈ [0, π/2], t reaches its maximum at θ = π/4, where t = √2, and minimum at θ approaching 0 or π/2, where t approaches 1. So, t is in [1, √2].So, S(t) is a quadratic in t. Let's analyze it. The quadratic is S(t) = (1/2)t² + t + 1/2. The derivative of S with respect to t is t + 1, which is always positive for t ≥ 0. Therefore, S(t) is an increasing function of t on the interval [1, √2]. Therefore, the minimum of S occurs at t = 1, and the maximum at t = √2.Calculating S(1) = (1/2)(1) + 1 + 1/2 = 0.5 + 1 + 0.5 = 2.Calculating S(√2) = (1/2)(2) + √2 + 1/2 = 1 + √2 + 0.5 = 1.5 + √2 ≈ 1.5 + 1.414 ≈ 2.914.But wait, let's check the exact values. Since t ranges from 1 to √2, then S(t) ranges from 2 to 1.5 + √2? Wait, but S(t) = (1/2)t² + t + 1/2.At t = 1: (1/2)(1) + 1 + 1/2 = 0.5 + 1 + 0.5 = 2.At t = √2: (1/2)(2) + √2 + 1/2 = 1 + √2 + 0.5 = 1.5 + √2 ≈ 1.5 + 1.414 ≈ 2.914.But actually, S(t) is strictly increasing, so the area S(t) ranges from 2 to 1.5 + √2. But wait, 1.5 + √2 is approximately 2.914, but let's keep it exact.But wait, is that correct? Let me check with another approach. Let's compute S when θ = 0. If θ approaches 0, then u = cosθ approaches 1, and v = sinθ approaches 0. Then, a = u + 1 approaches 2, b = v + 1 approaches 1. Then, area ab approaches 2*1 = 2. Similarly, when θ approaches π/2, u approaches 0, v approaches 1, a approaches 1, b approaches 2, so area approaches 2. So the minimal area is 2, and the maximum occurs somewhere in between.Wait, but when θ = π/4, then u = v = √2/2 ≈ 0.707. Then, a = √2/2 + 1 ≈ 1.707, b = √2/2 + 1 ≈ 1.707. Then, area ab ≈ (1.707)^2 ≈ 2.914, which is approximately (1.5 + √2) ≈ 1.5 + 1.414 ≈ 2.914. So that matches. Therefore, the area ranges from 2 to 1.5 + √2. But 1.5 + √2 is (3/2 + √2). But let's compute exactly:When θ = π/4, t = sinθ + cosθ = √2. Then, S(t) = (1/2)(2) + √2 + 1/2 = 1 + √2 + 0.5 = 3/2 + √2. So the maximum is 3/2 + √2. Therefore, the area S ranges from 2 to 3/2 + √2. Therefore, the range of the area is [2, 3/2 + √2].But let me verify this. Let me check when θ = π/4, so u = v = √2/2. Then, a = √2/2 + 1, b = √2/2 + 1. Then, ab = (√2/2 + 1)^2. Let's compute that:(√2/2 + 1)^2 = (1 + √2/2)^2 = 1^2 + 2*1*(√2)/2 + (√2/2)^2 = 1 + √2 + (2/4) = 1 + √2 + 1/2 = 3/2 + √2. So that's correct.Similarly, when θ approaches 0 or π/2, ab approaches 2. Therefore, the minimal area is 2, and the maximum area is 3/2 + √2.But wait, the problem states that point P is in the interior of the rectangle, excluding the boundary. So, does that affect the endpoints? Because when θ approaches 0 or π/2, the coordinates of P approach (a - 1, b - 1), which would be (1,0) or (0,1) if a approaches 2 or b approaches 2. But since P must be in the interior, coordinates must satisfy 0 < x < a and 0 < y < b, so x = a -1 > 0 ⇒ a >1, and y = b -1 >0 ⇒ b>1. Therefore, even when θ approaches 0, a approaches 2 (so x approaches 1, which is still less than a approaching 2), and y approaches 0, but since b approaches 1, y = b -1 approaches 0, but since b>1, actually, as θ approaches 0, b approaches 1 from above? Wait, no. Wait, when θ approaches 0, u = cosθ approaches 1, so a = u +1 approaches 2, and v = sinθ approaches 0, so b = v +1 approaches 1. But since θ is in (0, π/2), b = v +1 is always greater than 1. So even as θ approaches 0, b approaches 1 from above. Therefore, y = b -1 approaches 0 from above. So P approaches (1,0), but since x approaches 1 and a approaches 2, so x is always less than a. Similarly, y approaches 0, but since b approaches 1, y approaches 0 from above, so P remains in the interior. However, the problem states "excluding the boundary", but when θ is exactly 0 or π/2, P would be on the boundary. But since θ is in (0, π/2), P is strictly inside. Therefore, the minimal area is approached as θ approaches 0 or π/2, giving ab approaching 2. So the range of area is (2, 3/2 + √2]. Wait, but do we include the endpoints?Wait, if θ approaches 0, then a approaches 2, b approaches 1, so ab approaches 2. Similarly, θ approaches π/2, a approaches 1, b approaches 2, ab approaches 2. But since a and b are greater than 1, the minimal area is approached but not achieved. Wait, but hold on. If θ is allowed to be 0, then a would be 2, b would be 1, but then point P would be at (1,0), which is on the boundary. But the problem states that P is in the interior, so such cases are excluded. Therefore, the minimal area is greater than 2, and the maximum area is achieved when θ = π/4, which is 3/2 + √2. Therefore, the area can approach 2 but never actually reach it. So the range should be (2, 3/2 + √2].But let's check this. Suppose we parameterize θ in (0, π/2). Then, ab can take any value in (2, 3/2 + √2]. Because when θ approaches 0 or π/2, ab approaches 2, but since θ cannot be 0 or π/2 (as that would place P on the boundary), ab cannot be exactly 2. However, when θ = π/4, ab reaches 3/2 + √2, which is attainable because θ = π/4 is allowed (as P is at (√2/2 + 1 -1, √2/2 +1 -1) = (√2/2, √2/2), which is inside the rectangle as long as a and b are greater than 1, which they are when θ = π/4: a = √2/2 +1 ≈1.707, b = same). So in that case, P is inside the rectangle.Therefore, the range of the area is (2, 3/2 + √2]. However, the problem might consider the closure, meaning including 2 as the infimum, even though it's not attained. But in competitions, sometimes they accept intervals including the limit points even if they're not achieved. Let me check the problem statement again.Problem says: "In the interior of rectangle ABCD (excluding the boundary), there is a point P, which is at a distance of 1 from vertex A and sides BC, CD. Find the range of the area of rectangle ABCD."So, since P must be in the interior, the minimal area cannot actually be 2, but can get arbitrarily close. So the range is all real numbers greater than 2 and up to 3/2 + √2. Therefore, the answer should be written as (2, 3/2 + √2], but depending on the competition's expectations, they might write it as [2 + something, but no. Since 2 is not attainable, but is the infimum. However, in Chinese competitions, sometimes they might express the range including the limit. But let me check again.Wait, when we did the parametrization, we set u = a -1 and v = b -1, so u and v are positive real numbers with u² + v² = 1. Therefore, a and b are greater than 1, and the area ab is (u +1)(v +1). Since u and v are on the unit circle in the first quadrant, the minimal value of ab is approached as either u approaches 1 and v approaches 0 or vice versa. So in the limit, ab approaches (1 +1)(0 +1) = 2, but since u and v cannot be 1 and 0 (as that would put P on the boundary), ab cannot be 2. Hence, the area is greater than 2 and up to 3/2 + √2. Therefore, the range is (2, 3/2 + √2]. However, in Chinese math competitions, they often use square brackets for intervals even if it's an open interval, but we need to be precise. Alternatively, maybe they consider the closure. Wait, but strictly speaking, since point P is in the interior, the area cannot be exactly 2. So the answer should be that the area is greater than 2 and less than or equal to 3/2 + √2. So in box notation, we write (2, 3/2 + √2].But let me verify again. Suppose we consider a rectangle with area approaching 2. For example, take a = 2 - ε, b = 1 + ε, for very small ε. Wait, no. Wait, if a approaches 2, then u = a -1 approaches 1, so u = 1 - δ, v = sqrt(1 - u²) = sqrt(1 - (1 - δ)^2) ≈ sqrt(2δ - δ²) ≈ sqrt(2δ). So as δ approaches 0, v approaches 0. Then, b = v +1 approaches 1. So ab approaches (2 - δ)(1 + sqrt(2δ)) ≈ 2*1 = 2. So ab approaches 2 from above. Therefore, the area can get as close to 2 as desired but cannot be equal to 2. So the lower bound is open. The upper bound is achieved when θ = π/4. So the maximum area is 3/2 + √2. Therefore, the range is (2, 3/2 + √2].But let's compute 3/2 + √2 in decimal to check. √2 ≈1.414, so 1.5 +1.414 ≈2.914. So the area can be as close to 2 as we like, but not equal to 2, and as large as approximately 2.914. So the answer is the open interval from 2 to 3/2 + √2, including 3/2 + √2. Therefore, written as (2, 3/2 + √2].But in the problem statement, it's asked to "find the range of the area", and in Chinese competitions, sometimes they expect the answer in terms of exact expressions, not decimal approximations. So 3/2 + √2 is exact. Therefore, the range is from greater than 2 up to and including 3/2 + √2. Hence, the answer is (2, 3/2 + √2].But the problem is from a competition, so maybe they expect the answer in a particular format. Let me check the original problem statement again: "Find the range of the area of rectangle ABCD." Since the problem is translated, but in the original, it might expect the answer as an interval between two values. In the solution above, through substitution and calculus, we found that the area ranges from just above 2 up to 3/2 + √2. So the answer is 2 < area ≤ 3/2 + √2. Therefore, in boxed form, we can write the interval as (2, frac{3}{2} + sqrt{2}].But in Chinese math competitions, they often use square brackets for closed intervals. However, since 2 is not attainable, it should be an open interval on the left. So the correct range is (2, 3/2 + √2], which in LaTeX is written as boxed{(2, frac{3}{2} + sqrt{2}]}.But let me check once more with another method. Let's use Lagrange multipliers. Let S = ab, with a >1, b>1, and (a -1)^2 + (b -1)^2 =1. Let me set f(a,b) = ab, and the constraint g(a,b) = (a -1)^2 + (b -1)^2 -1 =0.Using Lagrange multipliers, we set the gradient of f equal to λ times gradient of g.Compute gradients:∇f = (b, a)∇g = (2(a -1), 2(b -1))So, we have:b = λ * 2(a -1)a = λ * 2(b -1)And the constraint (a -1)^2 + (b -1)^2 =1.From first equation, λ = b/(2(a -1))From second equation, λ = a/(2(b -1))Therefore, b/(2(a -1)) = a/(2(b -1)) ⇒ b/(a -1) = a/(b -1)Cross-multiplying: b(b -1) = a(a -1)So, b² - b = a² - a ⇒ b² - a² - b + a =0 ⇒ (b -a)(b +a) - (b -a) =0 ⇒ (b -a)(b +a -1)=0Therefore, either b -a =0 or b +a -1=0.Case 1: b -a =0 ⇒ b =a.Substitute into the constraint:(a -1)^2 + (a -1)^2 =1 ⇒ 2(a -1)^2 =1 ⇒ (a -1)^2 =1/2 ⇒ a -1= ±1/√2. Since a >1, we take the positive root: a =1 +1/√2. Then, since b =a, then b=1 +1/√2. Then, the area is ab= (1 +1/√2)^2=1 +2*(1/√2)+1/2=1 + √2 +0.5= 3/2 +√2. This is the maximum area.Case 2: b +a -1=0 ⇒a +b =1. But since a >1 and b >1, a +b >2, which contradicts a +b=1. Therefore, this case is impossible. Therefore, the only critical point is when a =b=1 +1/√2, leading to the maximum area.Therefore, the minimal area is approached as a approaches 2 and b approaches1, or vice versa, but since a and b are constrained by (a -1)^2 + (b -1)^2=1, the minimal area is approached when either (a -1) approaches1 and (b -1) approaches0, leading to a approaches2 and b approaches1, hence area approaches2. Similarly, if (a -1) approaches0 and (b -1) approaches1, area approaches1*2=2. But in these cases, point P approaches the boundary, but since P must be in the interior, the area cannot actually be 2. Therefore, the range is (2, 3/2 +√2].Therefore, the answer is boxed{(2, frac{3}{2} + sqrt{2}]}Final AnswerThe range of the area of rectangle (ABCD) is boxed{(2, frac{3}{2} + sqrt{2}]}.</think>Given a rectangle (ABCD) with a point (P) inside the interior (excluding the boundary), which is at a distance of 1 from vertex (A) and sides (BC) and (CD), we need to find the range of the area of rectangle (ABCD).1. Coordinates and Distances: - Place vertex (A) at the origin ((0,0)). - Let the coordinates of point (P) be ((x, y)). - The distance from (P) to (A) is (sqrt{x^2 + y^2} = 1), giving the equation (x^2 + y^2 = 1). - The distance from (P) to side (BC) (top side) is (b - y = 1), so (y = b - 1). - The distance from (P) to side (CD) (right side) is (a - x = 1), so (x = a - 1).2. Substitution and Constraint: - Substituting (x = a - 1) and (y = b - 1) into the distance equation gives ((a - 1)^2 + (b - 1)^2 = 1). - Let (u = a - 1) and (v = b - 1). Then (u^2 + v^2 = 1) with (u > 0) and (v > 0).3. Area Calculation: - The area of the rectangle (ABCD) is (ab = (u + 1)(v + 1)). - Expressing the area in terms of (u) and (v), we get (S = uv + u + v + 1).4. Optimization: - Parameterize (u) and (v) using trigonometric identities: (u = costheta) and (v = sintheta) for (theta in (0, frac{pi}{2})). - Substitute into the area formula: (S = (costheta + 1)(sintheta + 1)). - Simplify to find the range: (S = costheta sintheta + costheta + sintheta + 1). - Using trigonometric identities and calculus, determine that the area (S) is maximized at (theta = frac{pi}{4}) and approaches the minimum value as (theta) approaches 0 or (frac{pi}{2}).5. Results: - The maximum area occurs at (theta = frac{pi}{4}), giving (S = frac{3}{2} + sqrt{2}). - The minimum area approaches 2 as (theta) approaches 0 or (frac{pi}{2}), but never actually reaches 2 due to the interior constraint of point (P).Thus, the range of the area of rectangle (ABCD) is (boxed{(2, frac{3}{2} + sqrt{2}]}).,<think>Okay, let's see. I need to find the range of the area of rectangle ABCD given that there's a point P inside the rectangle (not on the boundary). Point P is 1 unit away from vertex A and also 1 unit away from sides BC and CD. Hmm, that sounds a bit tricky. Let me try to visualize this.First, let me draw rectangle ABCD. Let's name the vertices in order: A, B, C, D. So, sides AB, BC, CD, DA. Now, the point P is inside the rectangle, so it's not touching any of the sides. P is at a distance of 1 from vertex A. Also, P is 1 unit away from sides BC and CD. Wait, sides BC and CD are adjacent. BC is one of the vertical sides, and CD is the bottom side if we consider the rectangle in standard position. So, point P is 1 unit away from BC (which is a vertical side) and 1 unit away from CD (the horizontal side). Also, it's 1 unit away from vertex A.Let me assign coordinates to the rectangle to make it easier. Let's place vertex A at the origin (0,0). Then, if the rectangle has length AB = a and height AD = b, the coordinates of the other vertices would be: B(a,0), C(a,b), D(0,b). Now, point P is inside the rectangle, so its coordinates (x,y) must satisfy 0 < x < a and 0 < y < b.Given that P is 1 unit away from vertex A (0,0). The distance from P to A is sqrt(x^2 + y^2) = 1. So, we have x^2 + y^2 = 1. Then, P is also 1 unit away from sides BC and CD. Side BC is the vertical side at x = a, so the distance from P to BC is a - x. Similarly, side CD is the horizontal side at y = 0 (wait, no. CD is from (a,b) to (0,b). Wait, hold on. If A is (0,0), B is (a,0), C is (a,b), D is (0,b). Then side CD is from (a,b) to (0,b), which is the top horizontal side. Wait, no. If the rectangle is labeled in order ABCD, then moving from A to B to C to D, then yes, CD is the top horizontal side. Wait, but then the distance from P to CD would be the vertical distance from P to y = b. Similarly, distance from P to BC (the side at x = a) is a - x. But the problem states that P is 1 unit away from sides BC and CD, so:Distance from P to BC: a - x = 1Distance from P to CD: b - y = 1So, from these two equations, we can express x and y in terms of a and b:x = a - 1y = b - 1But we also have the distance from P to A is 1. Since A is (0,0), the distance is sqrt(x^2 + y^2) = 1. So substituting x and y:sqrt((a - 1)^2 + (b - 1)^2) = 1Therefore, squaring both sides:(a - 1)^2 + (b - 1)^2 = 1So that's a relationship between a and b. But we need to find the range of the area of the rectangle, which is a * b. So, given that (a - 1)^2 + (b - 1)^2 = 1, find the possible values of a * b.But wait, also, since P is inside the rectangle, not on the boundary, then x and y must satisfy 0 < x < a and 0 < y < b. But x = a - 1 and y = b - 1. Therefore:0 < a - 1 < a => 0 < a - 1 => a > 1 and a - 1 < a => which is always true. Similarly, 0 < b - 1 < b => b > 1. So both a and b must be greater than 1.So, we have a > 1, b > 1, and (a - 1)^2 + (b - 1)^2 = 1. We need to find the range of the product a * b.Let me set variables to simplify. Let u = a - 1 and v = b - 1. Then u > 0, v > 0, and u^2 + v^2 = 1. The area of the rectangle is (u + 1)(v + 1) = uv + u + v + 1.So, we need to maximize and minimize uv + u + v + 1 given that u and v are positive real numbers such that u^2 + v^2 = 1.Hmm. Let me think about how to approach this. Let's consider uv + u + v + 1. Maybe we can express this in terms of u and v and use the constraint u^2 + v^2 = 1.Alternatively, parametrize u and v. Since u^2 + v^2 = 1, we can write u = cosθ and v = sinθ, where θ is in the first quadrant (since u, v > 0). Then, the area becomes:(cosθ + 1)(sinθ + 1) = cosθ sinθ + cosθ + sinθ + 1.Simplify this expression. Let's compute each term:cosθ sinθ = (1/2) sin2θcosθ + sinθ = sqrt(2) sin(θ + 45°)But perhaps we can write the entire expression as:cosθ sinθ + cosθ + sinθ + 1 = (1/2) sin2θ + sqrt(2) sin(θ + 45°) + 1But maybe that's complicating things. Alternatively, let's consider the expression:Area = (cosθ + 1)(sinθ + 1) = cosθ sinθ + cosθ + sinθ + 1We can denote S = sinθ + cosθ and P = sinθ cosθ. Then, since S^2 = sin^2θ + cos^2θ + 2 sinθ cosθ = 1 + 2P, so P = (S^2 - 1)/2.Therefore, Area = P + S + 1 = (S^2 - 1)/2 + S + 1 = (S^2 - 1 + 2S + 2)/2 = (S^2 + 2S + 1)/2 = (S + 1)^2 / 2So Area = (sinθ + cosθ + 1)^2 / 2Interesting. So since S = sinθ + cosθ, which has a maximum of sqrt(2) when θ = 45°, and minimum of 1 when θ approaches 0° or 90°, but wait, θ is between 0 and 90°, but since u and v are positive, θ is in (0, 90°). Wait, when θ approaches 0°, sinθ approaches 0 and cosθ approaches 1, so S approaches 1. Similarly, when θ approaches 90°, sinθ approaches 1 and cosθ approaches 0, so S approaches 1. So the minimum of S is 1 (attained at the limits), and the maximum is sqrt(2) (attained at θ = 45°).Therefore, S ranges from 1 to sqrt(2). Then, Area = (S + 1)^2 / 2. So when S = 1, Area = (1 + 1)^2 / 2 = 4/2 = 2. When S = sqrt(2), Area = (sqrt(2) + 1)^2 / 2. Let's compute that:(sqrt(2) + 1)^2 = 2 + 2 sqrt(2) + 1 = 3 + 2 sqrt(2). Then, divided by 2, it's (3 + 2 sqrt(2))/2.But wait, so the Area ranges from 2 to (3 + 2 sqrt(2))/2? Wait, but is that correct?Wait, hold on. If θ can go from 0 to π/2, but when θ approaches 0, u approaches 1 and v approaches 0, but u and v are defined as u = cosθ and v = sinθ, so when θ approaches 0, u approaches 1, v approaches 0. However, in our original problem, u and v must be positive, but as θ approaches 0, v approaches 0. However, in the problem statement, point P is inside the rectangle, so v = b - 1 must be greater than 0, so b - 1 > 0 => b > 1. Similarly, u = a - 1 > 0 => a > 1. But if θ approaches 0, then v approaches 0, which would mean b approaches 1. But since b must be greater than 1, θ cannot actually reach 0. Similarly, θ cannot reach π/2. So θ is in (0, π/2), open interval. Therefore, S = sinθ + cosθ is in (1, sqrt(2)), so Area is in (2, (3 + 2 sqrt(2))/2). But wait, when θ approaches 0 or π/2, S approaches 1, so Area approaches (1 + 1)^2 / 2 = 2. But since θ can't actually reach 0 or π/2, the area can't actually be 2. However, in the problem statement, point P is strictly inside the rectangle, so a and b must be greater than 1. Therefore, the area can get arbitrarily close to 2 but never actually reach 2. Similarly, the maximum area is when θ = 45°, so S = sqrt(2), then Area = (sqrt(2) + 1)^2 / 2. Let me compute that:(sqrt(2) + 1)^2 = 2 + 2 sqrt(2) + 1 = 3 + 2 sqrt(2)Therefore, Area = (3 + 2 sqrt(2)) / 2 ≈ (3 + 2.828)/2 ≈ 5.828 / 2 ≈ 2.914.But wait, so the area ranges from greater than 2 up to (3 + 2 sqrt(2))/2. However, the problem says "Find the range of the area of rectangle ABCD." So in the competition problem, do they expect an open interval (2, (3 + 2√2)/2) or including the endpoints? But in the problem statement, it says "in the interior of rectangle ABCD (excluding the boundary)", so point P is strictly inside, so a and b must be greater than 1, so the area can't actually be 2 or (3 + 2√2)/2. Wait, but when θ = 45°, then u = v = √2/2, so a = u + 1 = 1 + √2/2, b = v + 1 = 1 + √2/2. Then, the area is (1 + √2/2)^2 = 1 + √2 + ( (√2)/2 )^2 = 1 + √2 + (2/4) = 1 + √2 + 0.5 = 1.5 + √2 ≈ 1.5 + 1.414 ≈ 2.914. Wait, but earlier we had that Area = (sqrt(2) +1)^2 / 2 ≈ (2.414)^2 / 2 ≈ 5.828 / 2 ≈ 2.914, which is the same. Wait, but hold on, when θ = 45°, u = v = √2/2 ≈ 0.707, so a = 1 + √2/2 ≈ 1.707, b = same. Then, area is a * b ≈ 1.707 * 1.707 ≈ 2.914, which matches. So in that case, the maximum area is (1 + √2/2)^2? Wait, let's compute a * b:a = 1 + u = 1 + cosθb = 1 + v = 1 + sinθSo the area is (1 + cosθ)(1 + sinθ). Then, when θ = 45°, cosθ = sinθ = √2/2, so:(1 + √2/2)(1 + √2/2) = 1 + √2 + ( (√2)/2 )^2 = 1 + √2 + (2)/4 = 1 + √2 + 0.5 = 1.5 + √2 ≈ 1.5 + 1.414 ≈ 2.914.Alternatively, we can write this as ( (sqrt(2) + 2)/2 ) * ( (sqrt(2) + 2)/2 ) ?Wait, let's see:1 + √2/2 = (2 + √2)/2So ( (2 + √2)/2 )^2 = (4 + 4√2 + 2)/4 = (6 + 4√2)/4 = (3 + 2√2)/2 ≈ (3 + 2.828)/2 ≈ 5.828/2 ≈ 2.914, which matches the earlier calculation. So the maximum area is (3 + 2√2)/2.But then, when θ approaches 0°, the area approaches (1 + 1)(1 + 0) = 2, but since θ can't be 0°, the area can't be 2. Similarly, as θ approaches 90°, the area approaches (1 + 0)(1 + 1) = 2. So the area can get arbitrarily close to 2 but can't reach it. However, in competition problems, sometimes they consider the closure, i.e., including the endpoints, even if the point is in the interior, maybe by taking limits. But according to the problem statement, the point is in the interior, so we need to make sure that P is strictly inside. Therefore, a and b must be greater than 1, so the minimal area is greater than 2 and the maximum area is (3 + 2√2)/2. But the problem says "Find the range of the area of rectangle ABCD." So in the answer, should it be written as (2, (3 + 2√2)/2 ) or [2, (3 + 2√2)/2 ]? But strictly speaking, since a and b have to be greater than 1, the area must be greater than 2. However, in some math competitions, they might accept the closed interval if the area can approach those limits. Let's check.Suppose we consider the case when θ approaches 0°, then a approaches 1 + 1 = 2 (wait, no. Wait, u = cosθ, which approaches 1 as θ approaches 0. So a = u + 1 = cosθ + 1 approaches 2? Wait, hold on, no. Wait, original substitution was u = a - 1, v = b - 1. Then, u^2 + v^2 = 1, so u = cosθ, v = sinθ. Then, a = u + 1 = cosθ + 1, b = v + 1 = sinθ + 1. So as θ approaches 0°, cosθ approaches 1, so a approaches 2, and sinθ approaches 0, so b approaches 1. Wait, but earlier I thought a = 1 + u, but u = cosθ. Wait, no. Wait, substitution:We had u = a - 1 and v = b - 1. So a = u + 1 and b = v + 1. Then, u^2 + v^2 = 1. So if we parametrize u = cosθ and v = sinθ, then a = 1 + cosθ, b = 1 + sinθ. Therefore, when θ approaches 0°, cosθ approaches 1, so a approaches 2, and sinθ approaches 0, so b approaches 1. But b must be greater than 1, so θ can't be 0°, so b approaches 1 from above. Similarly, when θ approaches 90°, a approaches 1 and b approaches 2.Therefore, in both cases, as θ approaches 0° or 90°, the area a * b approaches 2 * 1 = 2. But since b approaches 1 from above and a approaches 2 from below, or vice versa, actually:Wait, when θ approaches 0°, a approaches 1 + 1 = 2, and b approaches 1 + 0 = 1. But since θ is approaching 0°, and u = cosθ approaches 1, and v = sinθ approaches 0, so a approaches 2, but in reality, since θ is approaching 0°, but θ can't be 0°, so a approaches 2 from below? Wait, no. Wait, if θ is approaching 0°, cosθ approaches 1 from below? Wait, cosθ is decreasing from 1 to 0 as θ goes from 0° to 90°, so cosθ is 1 at θ = 0°, and decreases to 0. But θ is approaching 0°, so cosθ approaches 1 from below? Wait, cosθ is continuous, so as θ approaches 0° from the right, cosθ approaches 1 from below? Wait, actually, no. For θ in (0°, 90°), cosθ is positive and decreasing from 1 to 0. So as θ approaches 0°, cosθ approaches 1 from above? Wait, no. Wait, when θ is 0°, cosθ is 1. If θ is approaching 0° from the positive side (i.e., θ slightly larger than 0°), then cosθ is slightly less than 1. So cosθ approaches 1 from below. Therefore, a = 1 + cosθ approaches 2 from below. Similarly, sinθ approaches 0 from above, so b = 1 + sinθ approaches 1 from above. Therefore, a approaches 2 from below, and b approaches 1 from above. Therefore, the product a * b approaches 2 * 1 = 2 from below. Wait, no. If a approaches 2 from below (approaching 2 but less than 2) and b approaches 1 from above (approaching 1 but greater than 1), then a * b approaches 2 * 1 = 2, but since a is slightly less than 2 and b is slightly more than 1, their product is slightly less than 2? Wait, 1.99 * 1.01 = approx 2.0099. Wait, that's over 2. Wait, no, if a approaches 2 from below, like 2 - ε, and b approaches 1 from above, 1 + ε. Then, (2 - ε)(1 + ε) = 2 + 2ε - ε - ε^2 = 2 + ε - ε^2 ≈ 2 + ε. So as ε approaches 0, the product approaches 2 from above. Wait, this is confusing.Wait, let me compute:Let’s take θ approaching 0°. Let’s set θ = ε, where ε is a small positive angle. Then, cosε ≈ 1 - ε²/2, and sinε ≈ ε. Therefore:a = 1 + cosε ≈ 1 + 1 - ε²/2 = 2 - ε²/2b = 1 + sinε ≈ 1 + εSo the product a * b ≈ (2 - ε²/2)(1 + ε) ≈ 2(1 + ε) - (ε²/2)(1 + ε) ≈ 2 + 2ε - ε²/2 - ε³/2 ≈ 2 + 2ε - (ε²)/2As ε approaches 0, this approaches 2 + 0 - 0 = 2. But since ε is positive, the term 2ε is positive, so the product is slightly larger than 2. Wait, so when θ approaches 0°, the area approaches 2 from above. Similarly, when θ approaches 90°, the area also approaches 2 from above.Wait, so in both cases, as θ approaches 0° or 90°, the area approaches 2 from above. Therefore, the minimal area is 2, but it's not attainable. However, in the problem statement, since the point P is inside the rectangle, a and b must be greater than 1, so the area can get as close to 2 as desired but never actually reaches 2. Therefore, the range of the area should be (2, (3 + 2√2)/2]. Wait, but when θ = 45°, the area is (3 + 2√2)/2, which is the maximum. Is this attainable? Let's check.If θ = 45°, then u = cos45° = √2/2 ≈ 0.707, v = sin45° = √2/2. Then, a = 1 + √2/2 ≈ 1.707, b = 1 + √2/2 ≈ 1.707. The coordinates of point P would be x = a - 1 = √2/2 ≈ 0.707, y = b - 1 = √2/2 ≈ 0.707. So P is at (√2/2, √2/2). The distance from P to A is sqrt( (√2/2)^2 + (√2/2)^2 ) = sqrt( 0.5 + 0.5 ) = sqrt(1) = 1, which satisfies the condition. Also, distance from P to BC (x = a) is a - x = (1 + √2/2) - √2/2 = 1, and distance to CD (y = b) is b - y = (1 + √2/2) - √2/2 = 1. So P is exactly 1 unit away from A, BC, and CD. However, in the problem statement, it says "in the interior of rectangle ABCD (excluding the boundary)", but P is at (√2/2, √2/2), which is inside the rectangle since 0 < √2/2 < a and 0 < √2/2 < b (because a and b are 1 + √2/2 ≈ 1.707). Therefore, in this case, the maximum area is attainable when θ = 45°, so the maximum area is (3 + 2√2)/2, which is included in the range. However, the minimal area is not attainable since it requires θ approaching 0° or 90°, but even in those limits, the area approaches 2 but doesn't reach it. Therefore, the range should be (2, (3 + 2√2)/2].But in some contexts, especially in competitions, they might write the range as [2, (3 + 2√2)/2], considering the infimum and supremum. However, given that the problem specifies the point is in the interior, not on the boundary, the minimal area is not achieved, so the interval should be open at the lower bound and closed at the upper bound. But I need to check if when θ = 45°, point P is strictly inside the rectangle. Since a = 1 + √2/2 ≈ 1.707, and P is at x = √2/2 ≈ 0.707, which is less than a, and y = √2/2 ≈ 0.707, which is less than b ≈ 1.707, so yes, P is strictly inside. Therefore, the maximum area is attainable, so the upper bound is included. The lower bound is not attainable, so it's excluded. Therefore, the range is (2, (3 + 2√2)/2].But let's confirm with another approach. Maybe using Lagrange multipliers.We need to maximize and minimize the function f(a, b) = a * b subject to the constraint (a - 1)^2 + (b - 1)^2 = 1 and a > 1, b > 1.Set up Lagrangian: L = ab - λ[(a - 1)^2 + (b - 1)^2 - 1]Take partial derivatives:∂L/∂a = b - 2λ(a - 1) = 0∂L/∂b = a - 2λ(b - 1) = 0∂L/∂λ = -( (a - 1)^2 + (b - 1)^2 - 1 ) = 0From the first equation: b = 2λ(a - 1)From the second equation: a = 2λ(b - 1)So, substituting b from the first equation into the second:a = 2λ(2λ(a - 1) - 1 )Let me compute:a = 2λ(2λ(a - 1) - 1 ) = 2λ*(2λ a - 2λ - 1 )= 4λ² a - 4λ² - 2λBring all terms to the left:a - 4λ² a + 4λ² + 2λ = 0a(1 - 4λ²) + 4λ² + 2λ = 0Similarly, let's try to express a in terms of λ. Alternatively, we can divide the first equation by the second equation to eliminate λ.From first equation: b = 2λ(a - 1)From second equation: a = 2λ(b - 1)Divide the first equation by the second:b / a = [2λ(a - 1)] / [2λ(b - 1)] => b / a = (a - 1)/(b - 1)Cross-multiplying: b(b - 1) = a(a - 1)So, b² - b = a² - aRearranged: b² - a² - b + a = 0Factor: (b - a)(b + a) - (b - a) = 0 => (b - a)(b + a - 1) = 0Therefore, either b - a = 0 or b + a - 1 = 0.Case 1: b = aThen, substitute into the constraint (a - 1)^2 + (a - 1)^2 = 1 => 2(a - 1)^2 = 1 => (a - 1)^2 = 1/2 => a - 1 = ±1/√2. Since a > 1, we take the positive: a = 1 + 1/√2. Then, b = 1 + 1/√2. Therefore, the area is (1 + 1/√2)^2 = 1 + 2/√2 + 1/2 = 1 + √2 + 0.5 = 1.5 + √2 ≈ 2.914, which matches our previous result for the maximum area.Case 2: b + a - 1 = 0 => a + b = 1. But since a > 1 and b > 1, a + b > 2, so this case is impossible. Therefore, the only critical point is when a = b = 1 + 1/√2, which gives the maximum area.To confirm if this is a maximum, we can check the second derivative or test points around it, but intuitively, since the minimal area is approached as the rectangle becomes very long and thin, tending towards 2, and the maximum occurs at the symmetric case, this critical point is indeed the maximum.Therefore, the area can be as small as just above 2 and as large as (3 + 2√2)/2. However, in exact terms, (3 + 2√2)/2 can be simplified as (3/2) + √2. Alternatively, we can rationalize it as (sqrt(2) + 1)^2 / 2.But let me compute (3 + 2√2)/2:3 + 2√2 divided by 2 is 3/2 + √2. Yes, since 3/2 + √2 ≈ 1.5 + 1.414 ≈ 2.914, which matches.Therefore, the range of the area is all real numbers greater than 2 and up to (3 + 2√2)/2. Since in competition problems, they often expect the answer in a simplified radical form and specifying the interval. Since the maximum is attainable, it's included, and the minimum is not, so the range is (2, (3 + 2√2)/2]. However, in some contexts, intervals for areas are written using inequalities. So the area S satisfies 2 < S ≤ (3 + 2√2)/2.But let me check if (3 + 2√2)/2 is correct. Wait, let's calculate:We had Area = (1 + cosθ)(1 + sinθ). When θ = 45°, cosθ = sinθ = √2/2, so:(1 + √2/2)(1 + √2/2) = (1 + √2/2)^2 = 1 + √2 + ( (√2)/2 )^2 = 1 + √2 + (2)/4 = 1 + √2 + 0.5 = 1.5 + √2. Which is indeed equal to (3 + 2√2)/2. Since 1.5 is 3/2 and √2 is 2√2/2, so 3/2 + 2√2/2 = (3 + 2√2)/2. Correct.Therefore, the range of the area is greater than 2 and less than or equal to (3 + 2√2)/2. Therefore, the answer is that the area ranges from 2 (exclusive) to (3 + 2√2)/2 (inclusive).However, in Chinese competitions, sometimes they prefer to write the range using interval notation with square brackets even if the endpoint is not achieved, but considering the problem's constraints (point inside the rectangle), strictly speaking, it should be an open interval on the lower bound. But to confirm, let's check the problem statement again:"In the interior of rectangle ABCD (excluding the boundary), there is a point P, which is at a distance of 1 from vertex A and sides BC, CD. Find the range of the area of rectangle ABCD."Since the point is in the interior, the conditions are strict inequalities: distances to BC, CD, and A are exactly 1, but since P is in the interior, the coordinates must satisfy 0 < x < a and 0 < y < b. However, the distances to BC and CD are 1, so x = a - 1 and y = b - 1. Hence, a - 1 > 0 and b - 1 > 0, so a > 1 and b > 1. Therefore, when θ approaches 0°, a approaches 2 and b approaches 1, but b must remain greater than 1. Therefore, the minimal area is approached as a approaches 2 and b approaches 1 from above, making the product approach 2, but not reach it. Similarly, the maximum is achieved when θ = 45°, so the area is (3 + 2√2)/2.Therefore, the correct range is 2 < Area ≤ (3 + 2√2)/2. In boxed form, we can write this as boxed{(2, frac{3 + 2sqrt{2}}{2}]}. However, competition problems might present it using inequalities, such as boxed{2 < text{Area} leq frac{3 + 2sqrt{2}}{2}}.But sometimes, in Chinese competitions, they might express the range using interval notation with square brackets even if the endpoint is not attainable, but I think in this case, since the problem specifies the point is in the interior, they would expect the open interval for the lower bound. Let me check previous problems from the same competition to see the style.But given that I don't have access to previous problems, I'll go with the strict interpretation: since the minimal area is not attainable, use an open interval for the lower bound and closed for the upper. Therefore, the range is (2, (3 + 2√2)/2], which in boxed form is boxed{(2, frac{3 + 2sqrt{2}}{2}]}. Alternatively, if they require inequalities, boxed{2 < text{Area} leq frac{3 + 2sqrt{2}}{2}}.However, in the original problem statement, when they say "Find the range of the area", sometimes in Chinese math competitions, they express the answer as an interval using square brackets even if it's not attainable, but given that the point is in the interior, it's safer to use parentheses for the lower bound. So I think the correct answer is boxed{left(2, frac{3 + 2sqrt{2}}{2}right]}.But let me check the maximum area value once again to ensure there's no miscalculation. The maximum occurs when a = b = 1 + √2/2. Then, the area is (1 + √2/2)^2. Let's compute:(1 + √2/2)^2 = 1^2 + 2*1*(√2/2) + (√2/2)^2 = 1 + √2 + (2)/4 = 1 + √2 + 0.5 = 1.5 + √2 ≈ 1.5 + 1.4142 ≈ 2.9142. On the other hand, (3 + 2√2)/2 ≈ (3 + 2.8284)/2 ≈ 5.8284/2 ≈ 2.9142, which matches. So yes, (3 + 2√2)/2 is correct.Therefore, the final answer should be written as boxed{left(2, dfrac{3 + 2sqrt{2}}{2}right]} or using inequality notation. Since the problem is from a Chinese competition, and considering standard answer formats, they might present it as boxed{(2, frac{3 + 2sqrt{2}}{2}]}.Final AnswerThe range of the area of rectangle (ABCD) is boxed{left(2, dfrac{3 + 2sqrt{2}}{2}right]}.</think>Given a rectangle (ABCD) with a point (P) inside its interior (excluding the boundary), which is at a distance of 1 from vertex (A) and sides (BC) and (CD), we need to find the range of the area of rectangle (ABCD).1. Assigning Coordinates: - Place vertex (A) at ((0,0)), (B) at ((a,0)), (C) at ((a,b)), and (D) at ((0,b)). - Point (P) has coordinates ((x,y)) such that (0 < x < a) and (0 < y < b).2. Distance Conditions: - Distance from (P) to (A) is (sqrt{x^2 + y^2} = 1). - Distance from (P) to (BC) (x = a) is (a - x = 1), so (x = a - 1). - Distance from (P) to (CD) (y = b) is (b - y = 1), so (y = b - 1).3. Substituting Coordinates: - Substituting (x = a - 1) and (y = b - 1) into the distance equation: ((a - 1)^2 + (b - 1)^2 = 1).4. Parametrization: - Let (u = a - 1) and (v = b - 1). Then (u^2 + v^2 = 1) with (u > 0) and (v > 0). - The area of the rectangle is ((u + 1)(v + 1)).5. Maximizing and Minimizing the Area: - Using trigonometric parametrization (u = costheta) and (v = sintheta) for (theta in (0, frac{pi}{2})). - The area becomes ((1 + costheta)(1 + sintheta)).6. Simplifying the Area Expression: - The area expression simplifies to (frac{(sintheta + costheta + 1)^2}{2}). - The maximum value occurs at (theta = 45^circ), giving (frac{(3 + 2sqrt{2})}{2}). - The minimum value approaches 2 as (theta) approaches (0^circ) or (90^circ), but is not attainable due to the interior condition.7. Conclusion: - The range of the area of rectangle (ABCD) is all real numbers greater than 2 and up to (frac{3 + 2sqrt{2}}{2}).The final answer is (boxed{left(2, dfrac{3 + 2sqrt{2}}{2}right]}).]

🔑:Solve as shown in Figure 10, introduce variables x, y.Then x^{2}+y^{2}=P A^{2}=1, and x+y>P A=1.Thus, S_{text {rectangle } A B C D}=(1+x)(1+y)=1+x+y+x y>2 text {, }and when x (or y) rightarrow 0, S rightarrow 2.Also, (x+y)^{2} leqslant 2left(x^{2}+y^{2}right)=2, then x+y leqslant sqrt{2},and x y leqslant frac{x^{2}+y^{2}}{2}=frac{1}{2}.Therefore, S_{text {rectangle } A B C D} leqslant frac{3}{2}+sqrt{2}.The equality holds if and only if x=y=frac{sqrt{2}}{2}.Thus, 2<S_{text {rectangle } A B C D} leqslant frac{3}{2}+sqrt{2}.