🧠:[<think>Alright, so I have this inequality to prove: given that a, b, c are all positive numbers, show that a² + b² + c² + (1/a + 1/b + 1/c)² is greater than or equal to 6√3. Then I also need to find the values of a, b, c where equality holds. Hmm, okay. Let me start by thinking about inequalities I know and how they might apply here.First, the problem involves both a² + b² + c² and (1/a + 1/b + 1/c)². So maybe I can apply some sort of Cauchy-Schwarz inequality here, or AM-GM inequality. Let me recall those. The Cauchy-Schwarz inequality states that for any real numbers, (sum a_i²)(sum b_i²) ≥ (sum a_i b_i)². The AM-GM inequality says that the arithmetic mean is greater than or equal to the geometric mean, with equality when all numbers are equal.Looking at the terms, there's a² + b² + c², which is a sum of squares, and then the square of a sum of reciprocals. Maybe if I consider the Cauchy-Schwarz inequality on these terms? Let me try that. If I set up two vectors, one being (a, b, c) and the other being (1/a, 1/b, 1/c), then by Cauchy-Schwarz:(a² + b² + c²)(1/a² + 1/b² + 1/c²) ≥ (1 + 1 + 1)² = 9.But in the problem, we have (1/a + 1/b + 1/c)² instead of (1/a² + 1/b² + 1/c²). So that's a different term. Hmm, perhaps I need a different approach.Alternatively, maybe I can use the AM-GM inequality on the terms a², b², c² and then on 1/a, 1/b, 1/c. Let me try that. The AM-GM for three terms says that (x + y + z)/3 ≥ (xyz)^(1/3). So for a² + b² + c², the AM would be (a² + b² + c²)/3 ≥ (a² b² c²)^(1/3). Similarly, for (1/a + 1/b + 1/c), maybe if I apply AM-GM to those terms, (1/a + 1/b + 1/c)/3 ≥ (1/(a b c))^(1/3). Then squaring both sides would give [(1/a + 1/b + 1/c)/3]^2 ≥ (1/(a b c))^(2/3). Hmm, but how does this help me?Wait, the original expression is a² + b² + c² + (1/a + 1/b + 1/c)². Maybe if I can bound each part separately using AM-GM and then combine them? Let's see. If I denote S = a² + b² + c² and T = (1/a + 1/b + 1/c)², then I need to show that S + T ≥ 6√3. But how to relate S and T?Alternatively, perhaps using the Cauchy-Schwarz inequality in a different way. For example, the Cauchy-Schwarz inequality for the vectors (a, b, c) and (1/a, 1/b, 1/c) gives:(a*(1/a) + b*(1/b) + c*(1/c))² ≤ (a² + b² + c²)(1/a² + 1/b² + 1/c²)But the left side is (1 + 1 + 1)² = 9, so that gives 9 ≤ (a² + b² + c²)(1/a² + 1/b² + 1/c²). Hmm, but again, this gives a relation between a² + b² + c² and the sum of reciprocals squared, not the square of the sum of reciprocals. So maybe that's not directly helpful.Wait, but maybe if I can relate (1/a + 1/b + 1/c)² to (1/a² + 1/b² + 1/c²). Let's see, expanding (1/a + 1/b + 1/c)² gives 1/a² + 1/b² + 1/c² + 2(1/(ab) + 1/(ac) + 1/(bc)). So (1/a + 1/b + 1/c)² is equal to the sum of reciprocals squared plus twice the sum of reciprocal products. So perhaps I can bound this in terms of the sum of reciprocals squared? But not sure.Alternatively, maybe use Hölder's inequality? Hölder's inequality generalizes Cauchy-Schwarz and might relate different exponents. For example, Hölder's inequality with exponents p, q such that 1/p + 1/q = 1. Maybe if I set up terms to apply Hölder. Let me recall Hölder's inequality: (sum |x_i y_i|) ≤ (sum |x_i|^p)^(1/p) (sum |y_i|^q)^(1/q). Hmm, not sure how to apply that here directly.Alternatively, maybe consider substitution. Let me suppose that a = b = c. Since the inequality is symmetric in a, b, c, maybe the minimum occurs when all variables are equal. Let's check that. If a = b = c = k, then the expression becomes 3k² + (3/k)^2 = 3k² + 9/k². Then we need to find the minimum of 3k² + 9/k². Let's set f(k) = 3k² + 9/k². Take derivative: f’(k) = 6k - 18/k³. Set to zero: 6k - 18/k³ = 0 ⇒ 6k = 18/k³ ⇒ k^4 = 3 ⇒ k = 3^(1/4). Then the minimum is f(3^(1/4)) = 3*(3^(1/2)) + 9/(3^(1/2)) = 3*sqrt(3) + 9/sqrt(3) = 3 sqrt(3) + 3 sqrt(3) = 6 sqrt(3). So when a = b = c = 3^(1/4), the expression equals 6 sqrt(3). So equality holds at least when a = b = c = 3^(1/4). So maybe this is the case when equality holds. Now, I need to check whether the inequality holds in general, and that this is indeed the minimum.So if I can show that for any a, b, c > 0, a² + b² + c² + (1/a + 1/b + 1/c)² ≥ 6 sqrt(3), with equality when a = b = c = 3^(1/4).Alternatively, perhaps use the AM-GM inequality on the entire expression. Let's see, the expression is S + T where S = a² + b² + c² and T = (1/a + 1/b + 1/c)^2. If I can express S + T in terms that can apply AM-GM, maybe.Alternatively, use the inequality between arithmetic mean and geometric mean on S and T. Let me consider S + T ≥ 2 sqrt(S*T). But then 2 sqrt(S*T) would be a lower bound. But we need S + T ≥ 6 sqrt(3). So if we can show that 2 sqrt(S*T) ≥ 6 sqrt(3), then that would imply sqrt(S*T) ≥ 3 sqrt(3), so S*T ≥ 27. But is this true?From the Cauchy-Schwarz inequality earlier, we have S*(sum 1/a²) ≥ 9. So if T = (sum 1/a)^2, and sum 1/a² is part of T. Hmm, perhaps not straightforward. Let's compute S*T. Wait, S*T = (a² + b² + c²)*(1/a + 1/b + 1/c)^2. Not sure if that helps.Alternatively, maybe use Cauchy-Schwarz in the following way: Let me write (a² + b² + c² + (1/a + 1/b + 1/c)^2) as the sum of two terms, and then try to bound them. But perhaps I need to break it down further.Wait, another idea: Let me use the AM-GM inequality on each variable separately. For each variable a, maybe combine a² and 1/a², but in the problem, the reciprocal terms are summed first and then squared. Hmm, that complicates things.Alternatively, fix two variables and apply inequalities on them. Wait, this might get too complicated. Let me think again about the case when a = b = c. If that gives the minimum, then maybe by symmetry, the minimum occurs at a = b = c, and thus the inequality holds. But I need to confirm that for any deviation from a = b = c, the expression increases. How can I formally show that?Perhaps use the method of Lagrange multipliers. Let me set up the function f(a, b, c) = a² + b² + c² + (1/a + 1/b + 1/c)² and find its minimum under the constraint a, b, c > 0. Taking partial derivatives with respect to a, b, c, setting them to zero.Compute ∂f/∂a = 2a - 2(1/a + 1/b + 1/c)(1/a²). Similarly for ∂f/∂b and ∂f/∂c. Setting each partial derivative to zero:For a: 2a - 2(1/a + 1/b + 1/c)(1/a²) = 0 ⇒ 2a = 2(1/a + 1/b + 1/c)(1/a²) ⇒ a = (1/a + 1/b + 1/c)(1/a²) ⇒ a^3 = (1/a + 1/b + 1/c). Similarly for b and c:For b: b^3 = (1/a + 1/b + 1/c)For c: c^3 = (1/a + 1/b + 1/c)Therefore, from the three equations, we get a^3 = b^3 = c^3 = (1/a + 1/b + 1/c). Thus, a = b = c. Let me denote a = b = c = k. Then, substituting back into the equation, we get k^3 = 3/k. Multiply both sides by k: k^4 = 3 ⇒ k = 3^(1/4). So this critical point occurs at a = b = c = 3^(1/4), which matches our earlier trial. Now, to confirm that this is indeed a minimum, we can check the second derivative or consider that the function tends to infinity as any variable approaches 0 or infinity, so the critical point must be a minimum. Therefore, the minimal value is 6√3, achieved when a = b = c = 3^(1/4). Therefore, the inequality holds.But wait, I need to make sure that there are no other critical points where a ≠ b ≠ c. But given the symmetry, if we assume a = b = c gives the only critical point, then that's the minimum. Alternatively, suppose that two variables are equal and the third is different. Let's say a = b ≠ c. Then, compute the partial derivatives and see if such a critical point can exist. But this might be complicated. However, since the problem is symmetric in a, b, c, it's reasonable to assume that the minimum occurs at a = b = c.Alternatively, another approach: use the inequality of arithmetic and geometric means on the two terms a² + b² + c² and (1/a + 1/b + 1/c)^2. Let me denote S = a² + b² + c² and T = (1/a + 1/b + 1/c)^2. Then, by AM-GM, (S + T)/2 ≥ sqrt(S*T). So S + T ≥ 2 sqrt(S*T). To show that 2 sqrt(S*T) ≥ 6√3, which would require sqrt(S*T) ≥ 3√3, so S*T ≥ 27. If I can show that S*T ≥ 27, then the result would follow. Let's check if S*T ≥ 27.But from Cauchy-Schwarz, we have S*(sum 1/a²) ≥ 9. So S*(sum 1/a²) ≥ 9. But T = (sum 1/a)^2. Using Cauchy-Schwarz again on sum 1/a: (sum 1/a)^2 ≤ 3*(sum 1/a²). Therefore, sum 1/a² ≥ T/3. Then, combining with the previous inequality S*(sum 1/a²) ≥ 9 ⇒ S*(T/3) ≥ 9 ⇒ S*T ≥ 27. So yes, S*T ≥ 27. Therefore, S + T ≥ 2*sqrt(S*T) ≥ 2*sqrt(27) = 2*3√3 = 6√3. Hence, the inequality holds. Equality occurs when S = T and also when Cauchy-Schwarz inequalities used become equalities. For the first Cauchy-Schwarz, S*(sum 1/a²) ≥ 9, equality when a² : b² : c² = 1/a² : 1/b² : 1/c² ⇒ a^4 = b^4 = c^4 ⇒ a = b = c. For the second Cauchy-Schwarz, (sum 1/a)^2 ≤ 3*(sum 1/a²), equality holds when 1/a = 1/b = 1/c ⇒ a = b = c. Therefore, equality occurs only when a = b = c. Then, as before, a = b = c = 3^(1/4). Therefore, the minimum is achieved at a = b = c = 3^(1/4), and the inequality is proven.So, putting it all together, by applying Cauchy-Schwarz and AM-GM inequalities, we can show that the given expression is minimized when a, b, c are equal to the fourth root of 3, and the minimal value is 6√3.Final AnswerThe inequality holds with equality when ( a = b = c = sqrt[4]{3} ), so the values are boxed{a = b = c = sqrt[4]{3}}.</think>Given that (a), (b), and (c) are all positive numbers, we need to prove that (a^2 + b^2 + c^2 + left( frac{1}{a} + frac{1}{b} + frac{1}{c} right)^2 geqslant 6 sqrt{3}) and determine the values of (a), (b), and (c) for which equality holds.First, we consider the case where (a = b = c). If (a = b = c = k), the expression becomes:[3k^2 + left( frac{3}{k} right)^2 = 3k^2 + frac{9}{k^2}]We need to find the minimum of the function (f(k) = 3k^2 + frac{9}{k^2}). Taking the derivative and setting it to zero:[f'(k) = 6k - frac{18}{k^3} = 0 implies 6k = frac{18}{k^3} implies k^4 = 3 implies k = sqrt[4]{3}]Substituting (k = sqrt[4]{3}) back into the expression:[3(sqrt[4]{3})^2 + frac{9}{(sqrt[4]{3})^2} = 3sqrt{3} + 3sqrt{3} = 6sqrt{3}]This shows that the minimum value is (6sqrt{3}) when (a = b = c = sqrt[4]{3}).To confirm this is the minimum for all positive (a), (b), and (c), we use the Cauchy-Schwarz inequality. For vectors ((a, b, c)) and ((frac{1}{a}, frac{1}{b}, frac{1}{c})):[(a^2 + b^2 + c^2)left( frac{1}{a^2} + frac{1}{b^2} + frac{1}{c^2} right) geq (1 + 1 + 1)^2 = 9]Additionally, using the Cauchy-Schwarz inequality on ((frac{1}{a} + frac{1}{b} + frac{1}{c})^2 leq 3 left( frac{1}{a^2} + frac{1}{b^2} + frac{1}{c^2} right)), we get:[left( frac{1}{a} + frac{1}{b} + frac{1}{c} right)^2 leq 3 left( frac{1}{a^2} + frac{1}{b^2} + frac{1}{c^2} right)]Combining these results:[(a^2 + b^2 + c^2) left( frac{1}{a^2} + frac{1}{b^2} + frac{1}{c^2} right) geq 9][left( frac{1}{a} + frac{1}{b} + frac{1}{c} right)^2 leq 3 left( frac{1}{a^2} + frac{1}{b^2} + frac{1}{c^2} right)]Thus,[(a^2 + b^2 + c^2) cdot frac{1}{3} left( frac{1}{a} + frac{1}{b} + frac{1}{c} right)^2 geq 9 implies (a^2 + b^2 + c^2) left( frac{1}{a} + frac{1}{b} + frac{1}{c} right)^2 geq 27]By AM-GM inequality:[a^2 + b^2 + c^2 + left( frac{1}{a} + frac{1}{b} + frac{1}{c} right)^2 geq 2 sqrt{(a^2 + b^2 + c^2) left( frac{1}{a} + frac{1}{b} + frac{1}{c} right)^2} geq 2 sqrt{27} = 6sqrt{3}]Equality holds when (a = b = c = sqrt[4]{3}). Final AnswerThe inequality holds with equality when ( a = b = c = sqrt[4]{3} ), so the values are boxed{a = b = c = sqrt[4]{3}}.,<think>Okay, so I need to prove that for positive numbers a, b, c, the expression a² + b² + c² + (1/a + 1/b + 1/c)² is greater than or equal to 6√3. And then find when equality holds. Hmm, let's start by understanding what techniques might be useful here. The problem involves squares and reciprocals, so maybe inequalities like Cauchy-Schwarz or AM-GM could help. Let me recall those.First, the Cauchy-Schwarz inequality states that for any real numbers, (Σx_i y_i)² ≤ (Σx_i²)(Σy_i²). But here, we have a combination of squares of variables and squares of their reciprocals. Maybe I can apply Cauchy-Schwarz in some way here. Alternatively, the AM-GM inequality says that the arithmetic mean is greater than or equal to the geometric mean. Since we have terms like a² and (1/a)², maybe I can pair them or use AM-GM on some combination.Let me first look at the expression: a² + b² + c² + (1/a + 1/b + 1/c)². This is the sum of squares of a, b, c plus the square of the sum of reciprocals. Hmm, expanding the second part might be useful. Let me try expanding (1/a + 1/b + 1/c)². That would give 1/a² + 1/b² + 1/c² + 2/(ab) + 2/(ac) + 2/(bc). So the entire expression becomes:a² + b² + c² + 1/a² + 1/b² + 1/c² + 2/(ab) + 2/(ac) + 2/(bc).Now, the expression is a combination of squares and reciprocal cross terms. Maybe if I group the squares together and the cross terms together. So, we have (a² + 1/a²) + (b² + 1/b²) + (c² + 1/c²) + 2/(ab) + 2/(ac) + 2/(bc).For each variable, a² + 1/a² is a term. I know that by AM-GM, a² + 1/a² ≥ 2, since the arithmetic mean is (a² + 1/a²)/2 ≥ sqrt(a² * 1/a²) = 1. So each of those three terms is at least 2, so their sum is at least 6. Then, the cross terms are 2/(ab) + 2/(ac) + 2/(bc). If I factor out the 2, that's 2(1/(ab) + 1/(ac) + 1/(bc)).Hmm, but 1/(ab) + 1/(ac) + 1/(bc) can be written as (1/a + 1/b + 1/c)(1/(ab) + 1/(ac) + 1/(bc))... Wait, maybe that's complicating things. Alternatively, maybe apply AM-GM on the cross terms. Let me think.Alternatively, if I consider variables x = a, y = b, z = c, then the cross terms are 2/(xy) + 2/(xz) + 2/(yz). Maybe I can use AM-GM on these terms. For example, the arithmetic mean of 1/(xy), 1/(xz), 1/(yz) is [1/(xy) + 1/(xz) + 1/(yz)] / 3, and the geometric mean is (1/(xy) * 1/(xz) * 1/(yz))^(1/3). Let's compute the geometric mean: (1/(x^2 y^2 z^2))^(1/3) = 1/(x^(2/3) y^(2/3) z^(2/3)) ) = 1/( (xyz)^(2/3) ). Hmm, not sure if that helps directly.Alternatively, maybe use Cauchy-Schwarz on the terms. Let me recall that (sum u_i v_i)^2 ≤ (sum u_i^2)(sum v_i^2). Maybe if I set u_i as a, b, c and v_i as 1/a, 1/b, 1/c. Then (sum u_i v_i)^2 = (1 + 1 + 1)^2 = 9. And (sum u_i^2)(sum v_i^2) = (a² + b² + c²)(1/a² + 1/b² + 1/c²). So this gives 9 ≤ (a² + b² + c²)(1/a² + 1/b² + 1/c²). So that's an inequality we can get from Cauchy-Schwarz, but how does that relate to our original expression?Our original expression is a² + b² + c² + (1/a + 1/b + 1/c)^2. Wait, but we just expanded (1/a + 1/b + 1/c)^2 as the sum of reciprocals squared plus cross terms. So maybe there's a way to combine the Cauchy-Schwarz result with the cross terms.Alternatively, maybe use Holder's inequality? Holder's inequality generalizes Cauchy-Schwarz and might be useful when dealing with products. Holder's inequality states that (Σa_i b_i c_i) ≤ (Σa_i^p)^{1/p} (Σb_i^q)^{1/q} (Σc_i^r)^{1/r}} where 1/p + 1/q + 1/r = 1. Not sure if that's directly applicable here.Alternatively, maybe consider symmetry. If the inequality is symmetric in a, b, c, then perhaps the minimum occurs when a = b = c. Let's check that. Suppose a = b = c = k. Then the expression becomes 3k² + (3/k)^2 = 3k² + 9/k². Let's compute this as a function of k. Let f(k) = 3k² + 9/k². To find the minimum, take derivative: f’(k) = 6k - 18/k³. Set to zero: 6k - 18/k³ = 0 => 6k = 18/k³ => k^4 = 3 => k = 3^(1/4). Then, compute f(k): 3*(3^(1/4))² + 9/(3^(1/4))² = 3*3^(1/2) + 9/3^(1/2) = 3√3 + 9/√3 = 3√3 + 3√3 = 6√3. So when a = b = c = 3^(1/4), the expression equals 6√3. Therefore, if the minimum occurs at a = b = c, then the inequality holds. But we need to confirm that this is indeed the minimum, and that for all positive a, b, c, the expression is at least 6√3.But how do we prove that the expression is minimized at a = b = c? Since the problem is symmetric in a, b, c, it's often the case that the extremum occurs when all variables are equal. So maybe we can use the method of Lagrange multipliers to confirm this, but that might be complicated. Alternatively, use the inequality of arithmetic and geometric means on the entire expression.Alternatively, consider applying the Cauchy-Schwarz inequality in some way. Let's see, the original expression is a² + b² + c² + (1/a + 1/b + 1/c)². Let me denote S = a² + b² + c² and T = (1/a + 1/b + 1/c). Then the expression is S + T². So maybe we can find a lower bound for S + T².Alternatively, note that S and T are related via Cauchy-Schwarz. From Cauchy-Schwarz, (a² + b² + c²)(1/a² + 1/b² + 1/c²) ≥ (1 + 1 + 1)^2 = 9. So S * (1/a² + 1/b² + 1/c²) ≥ 9. But T = (1/a + 1/b + 1/c), and from Cauchy-Schwarz again, (1/a + 1/b + 1/c)^2 ≤ (1 + 1 + 1)(1/a² + 1/b² + 1/c²) = 3(1/a² + 1/b² + 1/c²). Therefore, T² ≤ 3(1/a² + 1/b² + 1/c²). So 1/a² + 1/b² + 1/c² ≥ T² / 3. Plugging this into the previous inequality: S * (T² / 3) ≥ 9 ⇒ S * T² ≥ 27 ⇒ T² ≥ 27 / S. Therefore, our expression is S + T² ≥ S + 27/S. So now, we can consider S + 27/S. The function f(S) = S + 27/S has its minimum when derivative is zero: f’(S) = 1 - 27/S² = 0 ⇒ S² = 27 ⇒ S = 3√3. Therefore, the minimum of S + 27/S is 3√3 + 27/(3√3) = 3√3 + 9/√3 = 3√3 + 3√3 = 6√3. Therefore, S + T² ≥ 6√3. So this shows that the original expression is at least 6√3. Therefore, the inequality holds.Wait, but is this approach valid? Let me check the steps again. We started with S = a² + b² + c² and T = 1/a + 1/b + 1/c. Then, using Cauchy-Schwarz, S * (1/a² + 1/b² + 1/c²) ≥ 9. Then, using Cauchy-Schwarz again on T, T² ≤ 3(1/a² + 1/b² + 1/c²), hence 1/a² + 1/b² + 1/c² ≥ T² / 3. Then substituting into the first inequality, S * (T² / 3) ≥ 9 ⇒ S * T² ≥ 27 ⇒ T² ≥ 27 / S. Therefore, S + T² ≥ S + 27/S. Then, since S is positive, the function S + 27/S has a minimum at S = sqrt(27) = 3√3, and the minimum value is 6√3. Therefore, the entire expression is ≥ 6√3. Therefore, this chain of inequalities leads us to the desired result. So that's a valid approach.But now, we need to check when equality holds. For equality to hold in the final step, we need S + 27/S = 6√3, which requires S = 3√3. Then, equality in T² ≥ 27/S requires T² = 27/S = 27/(3√3) = 9/√3 = 3√3. So T = sqrt(3√3) = (3√3)^{1/2} = 3^{3/4} * 3^{1/4} = Hmm, maybe better to express T in terms of S. Wait, T² = 3√3, so T = sqrt(3√3) = (3)^{3/4}.But we need to track back through the equalities. For equality in Cauchy-Schwarz, when do we have equality? In the first Cauchy-Schwarz inequality, (a² + b² + c²)(1/a² + 1/b² + 1/c²) ≥ 9, equality holds when a² : b² : c² = 1/a² : 1/b² : 1/c², i.e., when a² = b² = c². Therefore, when a = b = c. Similarly, in the second application of Cauchy-Schwarz on T² ≤ 3(1/a² + 1/b² + 1/c²), equality holds when 1/a = 1/b = 1/c, which again implies a = b = c. Therefore, all the equalities hold when a = b = c. Therefore, when a = b = c, we have equality.But earlier, when we considered a = b = c = k, we found that the expression equals 6√3 when k = 3^{1/4}. So, the equality holds when a = b = c = 3^{1/4}. Let me confirm this. Let k = 3^{1/4}. Then a = b = c = k, so a² = (3^{1/4})² = 3^{1/2} = √3. So each of a², b², c² is √3, so their sum is 3√3. Then, 1/a + 1/b + 1/c = 3/k = 3 / 3^{1/4} = 3^{1 - 1/4} = 3^{3/4}. Then, (1/a + 1/b + 1/c)^2 = (3^{3/4})² = 3^{3/2} = 3√3. Therefore, the entire expression is 3√3 + 3√3 = 6√3, which matches. So equality holds when a = b = c = 3^{1/4}.Therefore, the proof is complete. The inequality holds with equality when a, b, c are all equal to the fourth root of 3, i.e., 3^{1/4}.Final AnswerThe inequality holds with equality when ( a = b = c = sqrt[4]{3} ), so the values are boxed{sqrt[4]{3}}.</think>Given that (a), (b), (c) are all positive numbers, we need to prove that (a^{2} + b^{2} + c^{2} + left( frac{1}{a} + frac{1}{b} + frac{1}{c} right)^{2} geqslant 6 sqrt{3}) and determine the values of (a), (b), (c) for which the equality holds.First, we expand the expression (left( frac{1}{a} + frac{1}{b} + frac{1}{c} right)^{2}) to get:[left( frac{1}{a} + frac{1}{b} + frac{1}{c} right)^{2} = frac{1}{a^{2}} + frac{1}{b^{2}} + frac{1}{c^{2}} + 2left( frac{1}{ab} + frac{1}{ac} + frac{1}{bc} right)]Thus, the original expression becomes:[a^{2} + b^{2} + c^{2} + frac{1}{a^{2}} + frac{1}{b^{2}} + frac{1}{c^{2}} + 2left( frac{1}{ab} + frac{1}{ac} + frac{1}{bc} right)]Using the Cauchy-Schwarz inequality, we know:[(a^{2} + b^{2} + c^{2})left( frac{1}{a^{2}} + frac{1}{b^{2}} + frac{1}{c^{2}} right) geq (1 + 1 + 1)^2 = 9]Let (S = a^{2} + b^{2} + c^{2}) and (T = frac{1}{a} + frac{1}{b} + frac{1}{c}). By another application of Cauchy-Schwarz on (T):[T^2 leq 3left( frac{1}{a^{2}} + frac{1}{b^{2}} + frac{1}{c^{2}} right)]Thus,[frac{1}{a^{2}} + frac{1}{b^{2}} + frac{1}{c^{2}} geq frac{T^2}{3}]Substituting this into the earlier inequality:[S cdot frac{T^2}{3} geq 9 implies S cdot T^2 geq 27 implies T^2 geq frac{27}{S}]Therefore, the expression becomes:[S + T^2 geq S + frac{27}{S}]The function (f(S) = S + frac{27}{S}) is minimized when (S = 3sqrt{3}), leading to:[f(S) = 3sqrt{3} + frac{27}{3sqrt{3}} = 3sqrt{3} + 3sqrt{3} = 6sqrt{3}]To check when equality holds, we need (a = b = c). Let (a = b = c = k). Then:[3k^2 + left( frac{3}{k} right)^2 = 3k^2 + frac{9}{k^2}]Minimizing this, we find (k = sqrt[4]{3}). Substituting back, we verify the equality holds when (a = b = c = sqrt[4]{3}).Thus, the values of (a), (b), (c) for which equality holds are (boxed{sqrt[4]{3}}).]

🔑:Proof: Since a, b, c are all positive numbers,By the basic inequality, we havebegin{cases} a^{2}+b^{2}geqslant 2ab b^{2}+c^{2}geqslant 2bc c^{2}+a^{2}geqslant 2acend{cases}Therefore, a^{2}+b^{2}+c^{2}geqslant ab+bc+ac, (1),Similarly, frac {1}{a^{2}}+ frac {1}{b^{2}}+ frac {1}{c^{2}}geqslant frac {1}{ab}+ frac {1}{bc}+ frac {1}{ac}, (2),Thus, a^{2}+b^{2}+c^{2}+(frac {1}{a}+ frac {1}{b}+ frac {1}{c})^{2}, (3),geqslant ab+bc+ac+3 frac {1}{ab}+3 frac {1}{bc}+3 frac {1}{ac},geqslant 6 sqrt {3},Therefore, the original inequality holds.The equality holds if and only if a=b=c,And if and only if a=b=c, (ab)^{2}=(bc)^{2}=(ac)^{2}=3, the equality in (3) holds.That is, the equality holds if and only if a=b=c=sqrt[4]{3}.Therefore, the final answer is boxed{a=b=c=sqrt[4]{3}}.

🧠:[<think>Okay, so I need to solve this equation: 1 + log(2^x + 1) = log2 + log(4^x + 9). Hmm, let me start by recalling some logarithm properties. First, the number 1 on the left side. Maybe I can express that as a logarithm to make the equation easier. Since log base 10 of 10 is 1, right? So log(10) = 1. Alternatively, if the logs here are natural logarithms, then 1 would be ln(e). But the problem doesn't specify the base. Wait, in many math problems, if the base isn't specified, log could be base 10. But sometimes, especially in higher mathematics, it might be natural logarithm. Hmm. However, since there's a log2 term on the right side, maybe it's base 10. Because log2 in base 10 is just a number, whereas in natural logs, it would be ln2. Let me check the equation again. The right side has log2 + log(4^x +9). If it's base 10, then log2 is about 0.3010. If it's base e, then ln2 is about 0.693. But maybe the base doesn't matter here; perhaps we can proceed without specifying the base. Wait, but let's see. If the logs are all the same base, then maybe we can manipulate them.Alternatively, perhaps I should use properties of logarithms to combine terms. Let's start by moving all the log terms to one side. So subtract log2 from both sides and subtract log(4^x +9) from both sides? Wait, maybe not. Let's see.Original equation: 1 + log(2^x +1) = log2 + log(4^x +9). Let me rewrite 1 as log(10) if it's base 10, or ln(e) if natural. Since the right side has log2, I think it's safer to assume base 10. So 1 = log(10). Therefore, the left side becomes log(10) + log(2^x +1). Then, using the property that log(a) + log(b) = log(ab), we can combine these two terms. So left side becomes log[10*(2^x +1)]. Similarly, the right side is log2 + log(4^x +9), which can be combined into log[2*(4^x +9)]. So the equation simplifies to:log[10*(2^x +1)] = log[2*(4^x +9)]Since if log(a) = log(b), then a = b, provided that a and b are positive. So, we can set the arguments equal:10*(2^x +1) = 2*(4^x +9)Now, let's simplify this equation. First, divide both sides by 2:5*(2^x +1) = 4^x +9Expand the left side:5*2^x +5 = 4^x +9Bring all terms to one side:4^x -5*2^x + (9 -5) = 0Wait, let me check. Let me rearrange the equation:5*2^x +5 = 4^x +9Subtract 5*2^x and 5 from both sides:4^x -5*2^x + (9 -5) = 0Wait, 9 -5 is 4, so:4^x -5*2^x +4 =0Hmm, 4^x is (2^2)^x = 2^(2x) = (2^x)^2. So let me let y = 2^x. Then, 4^x = y^2. So substituting into the equation:y^2 -5y +4 =0This is a quadratic equation in terms of y. Let's solve for y:y^2 -5y +4 =0Factoring:(y -1)(y -4) =0So y =1 or y=4But y =2^x, so:If y=1, then 2^x=1 => x=0If y=4, then 2^x=4 => x=2Now, we need to check these solutions in the original equation because sometimes when we manipulate logarithmic equations, we might introduce extraneous solutions.First, check x=0:Original equation: 1 + log(2^0 +1) = log2 + log(4^0 +9)Compute each term:2^0 =1, so left side inside log:1 +1=2. So left side:1 + log(2). Right side: log2 + log(1 +9)= log2 + log(10). So left side is 1 + log2, right side is log2 +1. Since log10=1. Therefore, both sides are 1 + log2. So x=0 is a valid solution.Now check x=2:Left side:1 + log(2^2 +1)=1 + log(4 +1)=1 + log5Right side:log2 + log(4^2 +9)=log2 + log(16 +9)=log2 + log25So left side:1 + log5, right side: log2 + log25. Let's compute these values.Note that log25 is log(5^2)=2log5. So right side: log2 + 2log5. Left side:1 + log5.Wait, let's see if these are equal. Let's express 1 as log10. So left side: log10 + log5 = log(10*5)=log50. Right side: log2 + log25 = log(2*25)=log50. Therefore, both sides equal log50, so x=2 is also a valid solution.Therefore, both solutions are valid.So the solutions are x=0 and x=2.But let me verify once again, step by step, in case I made a mistake.Original equation:1 + log(2^x +1) = log2 + log(4^x +9)For x=0:Left side:1 + log(1 +1)=1 + log2Right side:log2 + log(1 +9)=log2 + log10= log2 +1. So same as left side. Correct.For x=2:Left side:1 + log(4 +1)=1 + log5Right side:log2 + log(16 +9)=log2 + log25= log2 + log(5^2)=log2 + 2log5But 1 + log5 vs log2 + 2log5. Wait, does 1 + log5 equal log2 + 2log5?Wait, 1 is log10, so left side is log10 + log5 = log(10*5)=log50Right side is log2 + log25=log(2*25)=log50. So yes, they are equal. Therefore, both solutions are valid.So the answer is x=0 and x=2.But wait, let me check if there's any restriction on the domain of the original equation. The arguments inside the logarithms must be positive.For log(2^x +1): since 2^x is always positive, 2^x +1 is always greater than 1, so positive. Similarly, 4^x +9 is always positive. Also, log2 and log(4^x +9) are defined. Therefore, all terms are defined for all real x. Hence, both solutions x=0 and x=2 are valid.Therefore, the solutions are x=0 and x=2.Final AnswerThe solutions are boxed{0} and boxed{2}.</think>To solve the equation (1 + log left(2^{x} + 1right) = log 2 + log left(4^{x} + 9right)), we start by expressing the constant term 1 as a logarithm. Assuming the logarithm is base 10, we rewrite 1 as (log 10). This allows us to combine the logarithmic terms on both sides using the property (log a + log b = log(ab)).Rewriting the equation:[log 10 + log left(2^{x} + 1right) = log 2 + log left(4^{x} + 9right)]Combining the logarithms on both sides:[log left[10 left(2^{x} + 1right)right] = log left[2 left(4^{x} + 9right)right]]Since the logarithms are equal, their arguments must be equal:[10 left(2^{x} + 1right) = 2 left(4^{x} + 9right)]Dividing both sides by 2:[5 left(2^{x} + 1right) = 4^{x} + 9]Expanding and rearranging terms:[5 cdot 2^{x} + 5 = 4^{x} + 9][4^{x} - 5 cdot 2^{x} + 4 = 0]Letting (y = 2^{x}), we substitute (4^{x} = y^2):[y^2 - 5y + 4 = 0]Factoring the quadratic equation:[(y - 1)(y - 4) = 0]Thus, (y = 1) or (y = 4). Converting back to (x):- If (y = 1), then (2^{x} = 1) which gives (x = 0).- If (y = 4), then (2^{x} = 4) which gives (x = 2).Checking both solutions in the original equation confirms they are valid. Therefore, the solutions are (boxed{0}) and (boxed{2}).,<think>Okay, let me try to solve this equation step by step. The equation given is:1 + log(2^x + 1) = log 2 + log(4^x + 9)Hmm, so we have logarithms here with different bases? Wait, the problem doesn't specify the base of the logarithm. In most math problems, if the base isn't mentioned, it's usually base 10. But sometimes in calculus or other contexts, it might be natural logarithm. But given the presence of log 2 on the right side, maybe it's base 10? Wait, log 2 in base 10 is just a number, approximately 0.3010. But actually, the base might not matter here because if it's consistent across all terms. Wait, but 4^x is the same as (2^2)^x = 2^(2x). Maybe we can write everything in terms of 2^x. Let me note that 4^x = (2^x)^2. So maybe substitution would help here.Let me start by recalling logarithmic properties. The left side is 1 + log(2^x + 1). The 1 in front could be written as log(10) if it's base 10, but maybe the problem is using base 10. Wait, but if the logarithms are base 10, then 1 is log(10). Alternatively, maybe the logs are base 2? Hmm, but the right side has log 2. If log is base 10, then log 2 is log base 10 of 2. If log is base 2, then log 2 would be 1. But the left side has a 1 added to log(2^x +1). If logs are base 2, then 1 would be log base 2 of 2. Let me check both possibilities.Wait, perhaps we can figure out the base by seeing if the equation is dimensionally consistent. Let me suppose that the logs are base 10. Then 1 is a constant, and log(2^x +1) is log base 10. On the right side, log 2 (base 10) plus log(4^x +9) (base 10). Alternatively, if the logs are natural logs, then 1 is just 1, and log 2 would be ln 2. But in any case, the properties of logarithms should hold regardless of the base. Let me try to manipulate the equation using logarithm rules.First, let's recall that log a + log b = log(ab). Also, 1 can be written as log 10 if it's base 10, or log e if it's natural, but since the right side has log 2, which would be log base 10 of 2, maybe we need to assume base 10. Alternatively, maybe all logs are base 2. Let's see.Wait, if all logs are base 2, then log 2 would be 1. Let me check that. Let's assume logs are base 2. Then the equation becomes:1 + log₂(2^x +1) = log₂ 2 + log₂(4^x +9)But log₂ 2 is 1, so that simplifies to:1 + log₂(2^x +1) = 1 + log₂(4^x +9)Subtracting 1 from both sides:log₂(2^x +1) = log₂(4^x +9)Then, since the logs are equal and the logarithm function is injective (one-to-one), their arguments must be equal:2^x +1 = 4^x +9But 4^x is (2^x)^2, so let's set y = 2^x. Then the equation becomes:y + 1 = y² + 9Rearranged: y² - y + 8 = 0Wait, that quadratic equation would have discriminant D = (-1)^2 - 4*1*8 = 1 -32 = -31. Negative discriminant, so no real solutions. Hmm, but this is a problem from a competition, so maybe there's a real solution. Therefore, maybe my assumption about the base is incorrect.Alternatively, if the logs are base 10. Let's try that again.Original equation:1 + log(2^x +1) = log 2 + log(4^x +9)If logs are base 10, then 1 is log 10. So:log 10 + log(2^x +1) = log 2 + log(4^x +9)Using log a + log b = log(ab):Left side: log[10*(2^x +1)]Right side: log[2*(4^x +9)]Therefore:log[10*(2^x +1)] = log[2*(4^x +9)]Since log is injective, the arguments must be equal:10*(2^x +1) = 2*(4^x +9)Divide both sides by 2:5*(2^x +1) = 4^x +9Again, let's substitute y = 2^x. Then 4^x = (2^x)^2 = y². So:5(y + 1) = y² +9Expand left side:5y +5 = y² +9Bring all terms to one side:y² -5y +4 =0Factor:Looking for two numbers that multiply to 4 and add to -5. Hmm, factors of 4 are 1 &4, 2&2. -1 and -4 multiply to 4 and add to -5.Therefore:(y -1)(y -4) =0 → y=1 or y=4But y=2^x, so:If y=1: 2^x=1 → x=0If y=4: 2^x=4 → x=2Now, check for validity in original equation.First, check x=0:Left side: 1 + log(2^0 +1) =1 + log(1 +1)=1 + log 2Right side: log 2 + log(4^0 +9)= log 2 + log(1 +9)= log2 + log10In base 10, log10=1, so right side is log2 +1. Which is same as left side. So x=0 is a solution.Check x=2:Left side:1 + log(2^2 +1)=1 + log(4 +1)=1 + log5Right side: log2 + log(4^2 +9)= log2 + log(16 +9)= log2 + log25Now, log25=log(5^2)=2 log5. So right side is log2 +2 log5Left side is 1 + log5. Let's see if these are equal.Assuming logs are base 10, 1= log10. So left side is log10 + log5=log(10*5)=log50Right side: log2 + log25=log(2*25)=log50. So they are equal. Therefore, both x=0 and x=2 are solutions.Wait, but when I assumed base 2, there was no solution, but with base 10, there are two solutions. So the problem must be assuming base 10. Since in competition problems, unless specified, log is often base 10. But let's check the problem statement again.The original problem is written as:1 + log(2^x +1) = log2 + log(4^x +9)But in Chinese or other contexts, sometimes log is base 10 by default. Also, in the original competition problem, since it's from a secondary technical school, it's likely base 10. So x=0 and x=2 are solutions. Wait, but the problem says "Solve the equation", so both solutions are valid.But let's verify again with base e. Wait, if logs were natural logarithms, then 1 would be ln e. Let's check:Left side:1 + ln(2^x +1) = ln e + ln(2^x +1) = ln[e*(2^x +1)]Right side: ln2 + ln(4^x +9) = ln[2*(4^x +9)]Set equal:e*(2^x +1) = 2*(4^x +9)But this equation would be more complicated. Let's try substituting y=2^x:e*(y +1) = 2*(y² +9)Which is a quadratic in y:2y² - e y + (18 -e)=0The discriminant would be e² -4*2*(18 -e)= e² -8*(18 -e)= e² -144 +8e ≈ (2.718)^2 +8*2.718 -144 ≈ 7.389 +21.744 -144≈ -114.867. Negative, so no real solutions. Therefore, if logs were natural, there would be no solution. So the problem must be assuming base 10. Therefore, the solutions are x=0 and x=2.But let me double-check if x=0 is valid. Plugging into original equation:Left side:1 + log(1 +1)=1 + log2Right side:log2 + log(1 +9)=log2 + log10= log2 +1Which is same as left side. So x=0 is valid.x=2:Left side:1 + log(4 +1)=1 + log5≈1 +0.69897≈1.69897Right side:log2 + log25≈0.3010 +1.39794≈1.69894Which is approximately equal, considering rounding errors. So both solutions are valid.Therefore, the solutions are x=0 and x=2.But wait, the problem is from a competition, and sometimes there might be restrictions. Let me check if there are any restrictions on the domain.The arguments inside the logarithms must be positive. For log(2^x +1): 2^x +1 is always positive, since 2^x >0, so 2^x +1 >1>0. For log(4^x +9): 4^x +9 is also always positive. So all arguments are positive for any real x, so both solutions are valid.Therefore, the solutions are x=0 and x=2.But let me check again the steps to make sure I didn't make a mistake.Starting with the equation:1 + log(2^x +1) = log2 + log(4^x +9)Assuming log is base10, converted 1 to log10, combined the logs:log[10*(2^x +1)] = log[2*(4^x +9)]Then equate the arguments:10*(2^x +1) = 2*(4^x +9)Divide by 2:5*(2^x +1) = 4^x +9Substitute y=2^x:5(y +1) = y² +9 → y² -5y +4=0 → y=1, y=4 → x=0, x=2Yes, all steps correct. So both solutions are valid.Therefore, the answer is x=0 and x=2.Final AnswerThe solutions are boxed{0} and boxed{2}.</think>Given the equation:[ 1 + log left(2^{x} + 1right) = log 2 + log left(4^{x} + 9right) ]We assume the logarithms are base 10. Rewriting 1 as (log 10):[ log 10 + log left(2^{x} + 1right) = log 2 + log left(4^{x} + 9right) ]Using the property of logarithms (log a + log b = log(ab)):[ log left[10 left(2^{x} + 1right)right] = log left[2 left(4^{x} + 9right)right] ]Since the logarithms are equal, their arguments must be equal:[ 10 left(2^{x} + 1right) = 2 left(4^{x} + 9right) ]Dividing both sides by 2:[ 5 left(2^{x} + 1right) = 4^{x} + 9 ]Let ( y = 2^x ). Then ( 4^x = (2^x)^2 = y^2 ):[ 5(y + 1) = y^2 + 9 ]Rearranging into a quadratic equation:[ y^2 - 5y + 4 = 0 ]Factoring:[ (y - 1)(y - 4) = 0 ]Thus, ( y = 1 ) or ( y = 4 ). Converting back to ( x ):- If ( y = 1 ), then ( 2^x = 1 ) which gives ( x = 0 ).- If ( y = 4 ), then ( 2^x = 4 ) which gives ( x = 2 ).Checking both solutions in the original equation confirms they are valid. Therefore, the solutions are:[ boxed{0} text{ and } boxed{2} ]]

🔑:B3. By applying the rule for the sum of logarithms, we transform the given equation into log 10left(2^{x}+1right)= log 2left(4^{x}+9right). We antilogarithmize the equation to the form 10left(2^{x}+1right)=2left(4^{x}+9right) and, by introducing a new unknown 2^{x}=t, transform it into the equation t^{2}-5 t+4=0. The solutions to the equation are t_{1}=1 and t_{2}=4. We then calculate the values of the original unknowns of the equation: x_{1}=0 and x_{2}=2.# dUFA 21st competition in mathematics for students of secondary technical and vocational schools National competition, May 15, 2021 Solutions to problems for 4th yearbegin{array}{|c|c|c|}hline mathrm{A} 1 & mathrm{~A} 2 & mathrm{~A} 3 hline mathrm{E} & mathrm{C} & mathrm{C} hlineend{array}

🧠:[<think>Okay, so I need to find the limit as x approaches 0 of [(1 - x)^6 * sin(3x) - sin(3x)] divided by 3x². Hmm, let's break this down step by step. First, maybe I can factor out sin(3x) from the numerator. Let me try that.So the numerator is (1 - x)^6 * sin(3x) - sin(3x). If I factor sin(3x), it becomes sin(3x) * [(1 - x)^6 - 1]. Then the denominator is 3x². So the expression simplifies to [sin(3x) * ((1 - x)^6 - 1)] / (3x²). That seems a bit simpler. Now, maybe I can split this into two parts: sin(3x)/3x² multiplied by [(1 - x)^6 - 1]. Wait, but splitting limits only works if both limits exist. Hmm, maybe I need to use some approximation or Taylor series here.Let me recall that when x approaches 0, sin(3x) is approximately 3x, right? Because sin(z) ≈ z when z is near 0. So sin(3x) ≈ 3x. So maybe I can replace sin(3x) with 3x plus higher order terms. Then the expression becomes [3x * ((1 - x)^6 - 1)] / (3x²) = [ (3x) * ((1 - x)^6 - 1) ] / (3x²). The 3x cancels with the denominator's 3x², leaving [( (1 - x)^6 - 1 ) / x]. Then the limit as x approaches 0 of [(1 - x)^6 - 1]/x.Wait, so maybe that simplifies the problem. So the original limit is equivalent to the limit of [(1 - x)^6 - 1]/x as x approaches 0. Let me check that again. Because sin(3x) ≈ 3x, so replacing sin(3x) with 3x gives us [3x * ((1 - x)^6 - 1)] / 3x², which simplifies to [ (1 - x)^6 - 1 ] / x. So now I need to compute that limit.Alternatively, maybe using L’Hospital’s Rule? Let's see. If I take the limit as x approaches 0 of [(1 - x)^6 - 1]/x. Let's plug in x = 0. Then (1 - 0)^6 - 1 = 1 - 1 = 0, and denominator is 0, so it's 0/0 indeterminate form. So yes, L’Hospital’s Rule applies here. Taking derivative of numerator and denominator.The derivative of the numerator: d/dx [ (1 - x)^6 - 1 ] = 6(1 - x)^5*(-1) - 0 = -6(1 - x)^5. The derivative of the denominator is 1. So applying L’Hospital’s Rule gives the limit as x approaches 0 of -6(1 - x)^5 / 1 = -6(1 - 0)^5 = -6*1 = -6. Therefore, the original limit would be -6.Wait, but let me confirm if replacing sin(3x) with 3x was valid. Because in the original expression, we have sin(3x) in the numerator and x² in the denominator. When we approximate sin(3x) as 3x, we are effectively using the first term of the Taylor series. The next term would be - (3x)^3 / 6, which is -27x³ / 6 = -9x³/2. So the error term is O(x³). But when we multiply by ((1 - x)^6 - 1), which is O(x), as we saw, since (1 - x)^6 - 1 ≈ -6x + higher terms. So sin(3x) * ((1 - x)^6 -1 ) would be (3x - 9x³/2 + ...) * (-6x + ...) = 3x*(-6x) + higher terms. So leading term is -18x², and then the higher terms are O(x³). So when we divide by 3x², we get -18x² / 3x² = -6, and the higher terms go to zero. Therefore, replacing sin(3x) with 3x is valid here because the higher-order terms vanish in the limit. So the answer should indeed be -6.Alternatively, another approach: expand (1 - x)^6 using the binomial theorem. Let's compute (1 - x)^6. That is 1 - 6x + 15x² - 20x³ + 15x⁴ - 6x⁵ + x⁶. Then subtract 1: (1 - x)^6 - 1 = -6x + 15x² - 20x³ + 15x⁴ - 6x⁵ + x⁶. Then multiply by sin(3x). So the numerator is [ -6x + 15x² - 20x³ + ... ] * [3x - (27x³)/6 + ...]. Let's multiply these out term by term.First, multiply -6x by 3x: -18x². Then -6x multiplied by the next term in sin(3x), which is - (27x³)/6: that gives +6x * 27x³ /6 = +27x⁴. Wait, but maybe we can just keep terms up to x² or x³ since higher terms will vanish when divided by x².Wait, let me see. The numerator is (1 - x)^6 * sin(3x) - sin(3x) = sin(3x)[(1 - x)^6 - 1]. We can write the expansion as sin(3x) multiplied by the expansion of (1 - x)^6 -1.As we found, (1 - x)^6 -1 = -6x + 15x² - 20x³ + ... So then sin(3x) is 3x - (27x³)/6 + ... = 3x - (9x³)/2 + ... So multiplying these two:(-6x + 15x² - 20x³) * (3x - (9x³)/2) = let's do term by term:First, multiply -6x by 3x: -18x².Then -6x * (-9x³)/2: +27x⁴ / 2.Then 15x² * 3x: 45x³.15x² * (-9x³)/2: -135x^5 / 2.Then -20x³ * 3x: -60x^4.-20x³ * (-9x³)/2: +90x^6.So combining terms:-18x² + (45x³) + (27x⁴/2 -60x⁴) + (-135x^5/2 +90x^6).So simplifying:-18x² +45x³ + (-60x⁴ +13.5x⁴) + ...= -18x² +45x³ -46.5x⁴ +...But since we are going to divide by 3x², let's see:Numerator is -18x² +45x³ -46.5x⁴ +... So divided by 3x² gives:(-18x²)/(3x²) + (45x³)/(3x²) + (-46.5x⁴)/(3x²) + ... = -6 +15x -15.5x² + ... Then take the limit as x approaches 0, which is -6. So this also confirms that the limit is -6.Alternatively, using the original approach where we split sin(3x)/3x² as sin(3x)/(3x²). Wait, if we write the original expression as [sin(3x) * ((1 - x)^6 -1)] / (3x²) = [sin(3x)/3x] * [((1 - x)^6 -1)/x]. Because 3x² is 3x * x, so [sin(3x)/3x] * [((1 - x)^6 -1)/x]. Then we have the limit as x approaches 0 of [sin(3x)/3x] * [((1 - x)^6 -1)/x].We know that as x approaches 0, sin(3x)/3x approaches 1, since sin(z)/z approaches 1 as z approaches 0. Here z =3x. So the first part is 1. Then the second part is [((1 - x)^6 -1)/x] as x approaches 0. Which we already calculated using L’Hospital’s Rule as -6. Therefore, the limit is 1 * (-6) = -6.So all approaches lead to the same answer, -6. Therefore, the final answer should be boxed{-6}.Final Answerboxed{-6}</think>To determine the limit of the function:[lim _{x rightarrow 0} frac{(1-x)^{6} sin 3x - sin 3x}{3x^{2}}]we start by factoring out (sin 3x) from the numerator:[sin 3x left[(1 - x)^6 - 1right]]This simplifies the expression to:[frac{sin 3x left[(1 - x)^6 - 1right]}{3x^2}]We can split this into two parts:[left(frac{sin 3x}{3x}right) left(frac{(1 - x)^6 - 1}{x}right)]As (x rightarrow 0), (frac{sin 3x}{3x} rightarrow 1). Therefore, we need to evaluate the limit of the second part:[lim_{x rightarrow 0} frac{(1 - x)^6 - 1}{x}]Using the binomial expansion of ((1 - x)^6):[(1 - x)^6 = 1 - 6x + 15x^2 - 20x^3 + 15x^4 - 6x^5 + x^6]Subtracting 1 and dividing by (x):[frac{(1 - x)^6 - 1}{x} = frac{-6x + 15x^2 - 20x^3 + 15x^4 - 6x^5 + x^6}{x} = -6 + 15x - 20x^2 + 15x^3 - 6x^4 + x^5]Taking the limit as (x rightarrow 0):[lim_{x rightarrow 0} left(-6 + 15x - 20x^2 + 15x^3 - 6x^4 + x^5right) = -6]Thus, the original limit is:[left(lim_{x rightarrow 0} frac{sin 3x}{3x}right) left(lim_{x rightarrow 0} frac{(1 - x)^6 - 1}{x}right) = 1 times (-6) = -6]Therefore, the final answer is:[boxed{-6}],<think>Okay, let's see. I need to find the limit as x approaches 0 of [(1 - x)^6 * sin(3x) - sin(3x)] divided by (3x^2). Hmm. Alright, first, maybe I can factor out sin(3x) from the numerator? Let's try that.So the numerator is (1 - x)^6 * sin(3x) - sin(3x). If I factor sin(3x) out, that would leave me with sin(3x) * [(1 - x)^6 - 1]. So the expression becomes [sin(3x) * ((1 - x)^6 - 1)] / (3x^2). Now, I need to simplify this. Let's see. As x approaches 0, sin(3x) behaves like 3x, right? Because sin(theta) ≈ theta when theta is near 0. So maybe I can use the approximation sin(3x) ≈ 3x for small x. That might help. Also, the term (1 - x)^6. Maybe I can expand that using the binomial theorem. Let's recall that (1 + a)^n = 1 + na + n(n -1)a^2/2! + ... So here, a is -x and n is 6. So expanding (1 - x)^6:(1 - x)^6 = 1 - 6x + (6*5)/2 x^2 - (6*5*4)/6 x^3 + ... But maybe I don't need all the terms. Since the denominator is x^2, I probably need to expand up to the x^2 term to see if the x terms cancel out. Let's do that.So (1 - x)^6 = 1 + (-6x) + [6*5/2]x^2 + higher order terms. That is 1 - 6x + 15x^2 - ... So subtracting 1 from that gives (1 - x)^6 -1 = -6x + 15x^2 - ... Therefore, the numerator becomes sin(3x) * (-6x + 15x^2 - ...). Let's substitute sin(3x) with its approximation 3x - (3x)^3/6 + ... So sin(3x) ≈ 3x - (27x^3)/6 = 3x - (9/2)x^3. Hmm. But maybe for the purposes of multiplying by (-6x + 15x^2), since we are dividing by x^2, we need to keep terms up to x^3?Wait, let's think step by step. The numerator is [sin(3x)] * [(1 - x)^6 -1]. Let's denote A = sin(3x) and B = (1 - x)^6 -1. Then, A ≈ 3x - (27/6)x^3 = 3x - (9/2)x^3. And B ≈ -6x + 15x^2 - 20x^3 + ... (Wait, actually, let me check the expansion again for (1 - x)^6. The coefficients can be calculated from the binomial coefficients.)Wait, (1 - x)^6 expansion:The binomial coefficients for n = 6 are:1, 6, 15, 20, 15, 6, 1.But with alternating signs because it's (1 - x)^6:1 - 6x + 15x^2 - 20x^3 + 15x^4 - 6x^5 + x^6.Therefore, (1 - x)^6 -1 = -6x + 15x^2 -20x^3 +15x^4 -6x^5 +x^6. So up to the x^2 term, it's -6x +15x^2.So if we take B ≈ -6x +15x^2, then A*B would be [3x - (9/2)x^3] * [-6x +15x^2].Let's multiply these two:First term: 3x * (-6x) = -18x^2Second term: 3x * 15x^2 = 45x^3Third term: (-9/2)x^3 * (-6x) = 27x^4Fourth term: (-9/2)x^3 *15x^2 = -135/2 x^5But since we are dividing by 3x^2, and we need to take the limit as x approaches 0, terms higher than x^2 in the numerator would vanish when divided by x^2. Wait, but actually, the numerator is A*B, which is a product of two expressions, each expanded to certain terms. So when multiplied, the leading terms would be up to x^3 or higher.Wait, maybe I need to keep terms up to x^3 in the numerator to see if, when divided by x^2, there's a term that remains.Wait, let's do the multiplication properly.A = sin(3x) ≈ 3x - (27x^3)/6 = 3x - (9/2)x^3B = (1 - x)^6 -1 ≈ -6x +15x^2 -20x^3So A*B = [3x - (9/2)x^3] * [-6x +15x^2 -20x^3]Multiply term by term:First, 3x * (-6x) = -18x^23x *15x^2 = 45x^33x * (-20x^3) = -60x^4Then, -(9/2)x^3 * (-6x) = 27x^4-(9/2)x^3 *15x^2 = - (135/2)x^5-(9/2)x^3 * (-20x^3) = 90x^6So combining all terms:-18x^2 +45x^3 -60x^4 +27x^4 - (135/2)x^5 +90x^6Simplify like terms:-18x^2 +45x^3 + (-60x^4 +27x^4) + ... which is -18x^2 +45x^3 -33x^4 + ... Therefore, the numerator is approximately -18x^2 +45x^3 -33x^4 + higher order terms.Then, the entire expression is [ -18x^2 +45x^3 -33x^4 + ... ] / (3x^2) = [ -18x^2 / 3x^2 ] + [45x^3 /3x^2 ] + [ -33x^4 /3x^2 ] + ... Simplify each term:-18/3 = -645x^3 /3x^2 = 15x-33x^4 /3x^2 = -11x^2So the expression becomes -6 +15x -11x^2 + ... Therefore, as x approaches 0, the higher order terms (15x, -11x^2, etc.) go to 0, so the limit is -6.Wait, but let me check that again. The numerator after expansion is -18x^2 +45x^3 -33x^4 + ..., divided by 3x^2. So dividing term by term:First term: -18x^2 /3x^2 = -6Second term:45x^3 /3x^2 =15xThird term: -33x^4 /3x^2 = -11x^2So when x approaches 0, 15x approaches 0 and -11x^2 approaches 0, so the limit is indeed -6. Therefore, the answer should be -6. But let's verify this another way to make sure.Alternatively, perhaps using L’Hospital’s Rule? Let's see. The original limit is [(1 - x)^6 sin3x - sin3x]/(3x^2) as x approaches 0.First, plug x =0 into the numerator and denominator. The numerator becomes (1 -0)^6 sin0 - sin0 =0 -0 =0. The denominator is 0. So it's 0/0, so L’Hospital can be applied.Take derivative of numerator and denominator.First, derivative of denominator: d/dx [3x^2] =6x.Derivative of numerator: d/dx [ (1 -x)^6 sin3x - sin3x ]First term: derivative of (1 -x)^6 sin3x. Use product rule.Let’s denote u = (1 -x)^6, v = sin3x.u’ =6(1 -x)^5*(-1) = -6(1 -x)^5v’= 3cos3xSo derivative is u’v + uv’ = -6(1 -x)^5 sin3x + (1 -x)^6 *3cos3xThen subtract derivative of sin3x, which is 3cos3x. So total derivative:-6(1 -x)^5 sin3x +3(1 -x)^6 cos3x -3cos3xThus the derivative of the numerator is:-6(1 -x)^5 sin3x +3(1 -x)^6 cos3x -3cos3xEvaluate this at x approaching 0. Let's plug x=0:First term: -6(1 -0)^5 sin0 = -6*1*0=0Second term:3(1 -0)^6 cos0 =3*1*1=3Third term: -3cos0= -3*1= -3So total derivative at x=0 is 0 +3 -3=0. Denominator derivative at x=0 is 6*0=0. So still 0/0. Apply L’Hospital again.Now take second derivatives.Derivative of denominator is 6x, derivative is 6.Derivative of numerator: derivative of [ -6(1 -x)^5 sin3x +3(1 -x)^6 cos3x -3cos3x ]Let’s differentiate term by term.First term: -6(1 -x)^5 sin3xLet u = -6(1 -x)^5, v = sin3xu’ = -6*5(1 -x)^4*(-1) =30(1 -x)^4v’= 3cos3xDerivative is u’v + uv’ =30(1 -x)^4 sin3x + (-6(1 -x)^5)(3cos3x)=30(1 -x)^4 sin3x -18(1 -x)^5 cos3xSecond term: 3(1 -x)^6 cos3xLet u=3(1 -x)^6, v=cos3xu’=3*6(1 -x)^5*(-1)= -18(1 -x)^5v’= -3sin3xDerivative: u’v +uv’ = -18(1 -x)^5 cos3x +3(1 -x)^6*(-3sin3x)= -18(1 -x)^5 cos3x -9(1 -x)^6 sin3xThird term: -3cos3xDerivative: -3*(-3sin3x)=9 sin3xPutting it all together, the second derivative of the numerator is:[30(1 -x)^4 sin3x -18(1 -x)^5 cos3x] + [ -18(1 -x)^5 cos3x -9(1 -x)^6 sin3x ] +9 sin3xCombine like terms:Terms with sin3x:30(1 -x)^4 sin3x -9(1 -x)^6 sin3x +9 sin3xTerms with cos3x:-18(1 -x)^5 cos3x -18(1 -x)^5 cos3x = -36(1 -x)^5 cos3xSo total second derivative:[30(1 -x)^4 sin3x -9(1 -x)^6 sin3x +9 sin3x] -36(1 -x)^5 cos3xNow evaluate at x=0:First, sin0=0, cos0=1.Sin3x at x=0 is sin0=0, cos3x at x=0 is cos0=1.So substituting x=0:Terms with sin3x: all terms with sin3x become 0 because sin0=0.So the entire first bracket becomes 0.Terms with cos3x: -36(1 -0)^5 *1 = -36*1*1= -36Therefore, the second derivative of the numerator at x=0 is -36.The second derivative of the denominator is 6.Therefore, applying L’Hospital’s Rule twice, the limit is -36 /6 = -6.So that confirms the earlier result. The limit is -6. Therefore, the answer is boxed{-6}.Final Answerboxed{-6}</think>To determine the limit of the function:[lim _{x rightarrow 0} frac{(1-x)^{6} sin 3x - sin 3x}{3 x^{2}}]we start by factoring out (sin(3x)) from the numerator:[frac{sin(3x) left[(1 - x)^6 - 1right]}{3x^2}]Next, we use the binomial expansion for ((1 - x)^6):[(1 - x)^6 = 1 - 6x + 15x^2 - 20x^3 + cdots]Subtracting 1 from this expansion gives:[(1 - x)^6 - 1 = -6x + 15x^2 - 20x^3 + cdots]We also use the approximation (sin(3x) approx 3x - frac{9}{2}x^3) for small (x). Multiplying this approximation by the expansion of ((1 - x)^6 - 1):[sin(3x) left[(1 - x)^6 - 1right] approx left(3x - frac{9}{2}x^3right) left(-6x + 15x^2 - 20x^3right)]Expanding the product and keeping terms up to (x^3):[(3x)(-6x) + (3x)(15x^2) + left(-frac{9}{2}x^3right)(-6x) = -18x^2 + 45x^3 + 27x^4 - cdots]Dividing by (3x^2):[frac{-18x^2 + 45x^3 + 27x^4 - cdots}{3x^2} = -6 + 15x + 9x^2 - cdots]As (x rightarrow 0), the higher-order terms vanish, leaving the limit as (-6).To confirm, we apply L'Hospital's Rule twice. After differentiating the numerator and denominator twice, we find the second derivative of the numerator at (x = 0) is (-36) and the second derivative of the denominator is (6). Thus, the limit is:[frac{-36}{6} = -6]The final answer is:[boxed{-6}]]

🔑:}For (x=0) an indeterminate form (frac{0}{0}) arises. According to l'Hôpital's rule, the limit in this case is given by[lim _{x rightarrow 0} frac{f(x)}{g(x)}=lim _{x rightarrow 0} frac{f^{prime}(x)}{g^{prime}(x)}]Twice differentiating the numerator and the denominator (not using the quotient rule!) yields:[begin{aligned}f^{prime prime}(x) & =36left(x^{5}-5 x^{4}+10 x^{3}-10 x^{2}+5 x-1right) cos 3 x- & -3left(3 x^{6}-18 x^{5}+35 x^{4}-20 x^{3}-15 x^{2}+22 x-10right) sin 3 x g^{prime prime}(x) & =6end{aligned}]Thus, the value[lim _{x rightarrow 0} frac{f^{prime prime}(x)}{g^{prime prime}(x)}=-6]is the limit of the initial expression.

🧠:[<think>Okay, so I have this problem here about complex numbers. Let me try to understand it step by step. The problem says:We have a set of distinct complex numbers ( z_i ), where ( i = 1, 2, ldots, n ). These numbers satisfy the inequality:[min_{i neq j} |z_i - z_j| geq max_i |z_i|]And we need to find the maximum possible ( n ) and, for this ( n ), describe all sets that satisfy the condition.First, let me parse what this condition means. The minimum distance between any two distinct complex numbers in the set is at least as large as the maximum modulus (distance from the origin) of any number in the set. So, if the farthest any number is from the origin is some value ( R ), then every pair of numbers must be at least ( R ) apart from each other.The question is asking for the largest possible ( n ) such that this condition can be satisfied, and also to characterize all such sets when ( n ) is maximum.Let me start by considering some simple cases. Maybe if all the numbers are on the real line? Or arranged in some symmetric configuration in the complex plane? Since complex numbers can be plotted in the plane, maybe there's a geometric interpretation here.First, let's denote ( R = max_i |z_i| ). Then the condition becomes that any two distinct ( z_i ) and ( z_j ) must satisfy ( |z_i - z_j| geq R ).So, all the points lie within or on the circle of radius ( R ) centered at the origin, and every pair of points is at least ( R ) apart from each other.To maximize ( n ), we need to fit as many points as possible inside (or on) a circle of radius ( R ) such that the distance between any two points is at least ( R ).This seems similar to circle packing problems, where we want to place circles of radius ( r ) inside a larger circle of radius ( R ) without overlapping. Here, the points are like centers of circles with radius ( R/2 ), since if two points are at least ( R ) apart, then circles of radius ( R/2 ) around them don't overlap. So, it's equivalent to packing non-overlapping circles of radius ( R/2 ) within a circle of radius ( R + R/2 = 3R/2 ). Wait, but the original points are within radius ( R ), so maybe the centers of the circles (points ( z_i )) are within radius ( R ), and each circle has radius ( R/2 ), so the entire packing must fit within a circle of radius ( R + R/2 = 3R/2 ). Hmm, but I need to think if this is the right analogy.Alternatively, perhaps it's simpler. Let's suppose that all points are on the boundary of the circle of radius ( R ), i.e., all have modulus exactly ( R ). Then, if we arrange them equally spaced on the circumference, the distance between adjacent points is ( 2R sin(pi/n) ). For this distance to be at least ( R ), we require ( 2R sin(pi/n) geq R ), so ( sin(pi/n) geq 1/2 ). Thus, ( pi/n geq pi/6 ), so ( n leq 6 ). Therefore, if all points are equally spaced on the circle, then the maximum ( n ) would be 6, since for ( n = 6 ), the distance between adjacent points is exactly ( R ), and for ( n = 7 ), the distance would be less than ( R ).But the problem doesn't specify that the points have to lie on the boundary. They just need to have modulus at most ( R ). So perhaps some points can be inside the circle, allowing more points to be placed? But if points are inside, then their distance to other points might be smaller. Wait, but the minimum distance between any two points must be at least ( R ), so even if some points are inside, they still need to be at least ( R ) away from all other points. However, the maximum modulus is ( R ), so if a point is inside, its distance from the origin is less than ( R ), but its distance from other points still has to be at least ( R ).Hmm. Let me visualize this. Suppose we have a point at the origin. Then, any other point must be at least ( R ) away from the origin. But if the origin is one of the points, then the maximum modulus ( R ) is the distance from the origin to the farthest point, which would be at least ( R ). Wait, but if the origin is included, then the other points must lie on or outside the circle of radius ( R ). But the problem states that the maximum modulus is ( R ). So, if we have a point at the origin, then the other points must be exactly at distance ( R ), but then the distance from the origin to those points is ( R ), which is okay. However, the distance between the origin and any other point is ( R ), which satisfies the minimum distance condition (since ( R geq R )), but the minimum distance between any two points is the minimum of ( R ) (distance from origin to others) and the distances between the other points. If the other points are on the circle of radius ( R ), then the distance between them must also be at least ( R ). So, similar to the previous case where all points are on the circle. However, if we include the origin, we might have to check if we can fit more points.Suppose we have one point at the origin and other points on the circle of radius ( R ). The distance from the origin to any other point is ( R ), which is okay. The distance between any two points on the circle must be at least ( R ). So, how many points can we place on the circle such that the distance between any two is at least ( R )? Let's think.If we have points on the circle of radius ( R ), the chord length between two points is ( 2R sin(theta/2) ), where ( theta ) is the angle between them. For the chord length to be at least ( R ), we need ( 2R sin(theta/2) geq R ), so ( sin(theta/2) geq 1/2 ). Therefore, ( theta/2 geq pi/6 ), so ( theta geq pi/3 ). Therefore, the angle between any two adjacent points must be at least ( pi/3 ), which allows at most ( 2pi / (pi/3) ) = 6 ) points on the circle. So, with the origin included, we can have 1 + 6 = 7 points. Wait, but is that correct?Wait, if we have the origin and 6 points on the circle of radius ( R ), each adjacent pair on the circle is separated by angle ( pi/3 ). Then, the distance between two adjacent points on the circle is ( 2R sin(pi/6) = 2R * 1/2 = R ). So, that's good. The distance from the origin to each point is ( R ), which meets the minimum distance requirement. However, we need to check that the minimum distance between any two points is at least ( R ). The distances between points on the circle are either ( R ) (adjacent) or larger (non-adjacent). For example, the distance between points separated by two steps would be ( 2R sin(pi/3) = sqrt{3} R ), which is more than ( R ). Similarly, the distance from the origin to any other point is ( R ). Therefore, the minimum distance is ( R ), which satisfies the condition. Therefore, in this case, with 1 + 6 = 7 points, the condition is satisfied.But wait, hold on. If we have the origin and six points on the circle, then the minimum distance between any two points is ( R ), which is equal to the maximum modulus (since the points on the circle have modulus ( R )). So, this would satisfy the inequality ( min |z_i - z_j| geq max |z_i| ), since ( R geq R ).But the problem states "distinct complex numbers", so maybe 7 is possible? But is this correct?Wait, let me check again. If we have one point at the origin and six points on the circle of radius ( R ), spaced 60 degrees apart. Then, the distance from the origin to each circle point is ( R ), which is equal to the max modulus. The distance between adjacent circle points is ( R ), as we saw. So, all the distances are either ( R ) or more. Therefore, the minimum distance is ( R ), and the maximum modulus is ( R ). So, ( min |z_i - z_j| = R geq max |z_i| = R ). Therefore, equality holds. So, this configuration works.But can we add another point? Let's see. If we try to add a seventh point on the circle, but we already have six points equally spaced. Wait, but in the previous case, with the origin, we already have seven points. Wait, no: 1 at the origin and 6 on the circle, total 7. If we try to add an eighth point, where could it go?If we try to put another point on the circle, the chord length between it and its neighbors would be less than ( R ), since six points on a circle already are spaced 60 degrees apart. Adding a seventh point on the circle would require angles less than 60 degrees between some points, resulting in chord lengths less than ( R ). Therefore, we can't add another point on the circle.Alternatively, could we place another point inside the circle? Let's say we try to place a point somewhere inside. However, the distance from this new point to all existing points (the origin and the six circle points) must be at least ( R ). Let's see. Suppose we try to place a point inside the circle. The distance from this point to the origin must be at least ( R ), but if the point is inside the circle, its modulus is less than ( R ). So, the distance from the origin to this point is less than ( R ), which violates the minimum distance requirement. Therefore, any point inside the circle would be too close to the origin. Therefore, we can't place any points inside the circle.Alternatively, could we place another point on the circle? As I thought earlier, adding a seventh point on the circle would make the chord lengths between adjacent points less than ( R ). So, that's not possible.Alternatively, maybe arranging points differently, not equally spaced? Maybe using a different configuration where some points are further apart. But since the circle is the most efficient way to pack points around a central point, I think equally spaced is optimal. So, perhaps 7 is the maximum.But wait, before I get ahead of myself, let me check this. Let's see: suppose we don't put a point at the origin. Then, we can have six points on the circle, each separated by 60 degrees, with distance ( R ) between adjacent points, which gives a minimum distance of ( R ). The maximum modulus is ( R ). So, in this case, n=6. If we add the origin, we get n=7, as above.But is 7 possible? Let me check again.Suppose one point at the origin (modulus 0) and six points on the circle of radius R (modulus R). The distances from the origin to the circle points are R, which is equal to the maximum modulus. The distances between the circle points are R (for adjacent ones) and larger for non-adjacent ones. So, the minimum distance in the entire set is R. So, the condition is satisfied because min distance is R, and max modulus is R. So, R >= R holds.Therefore, n=7 is possible. But can we have n=8?To have n=8, we would need eight points where each pair is at least R apart, and all are within modulus R. Let's see. If we tried to place eight points, some would have to be inside the circle, but as before, points inside the circle would be too close to the origin if the origin is already occupied, or if the origin is not occupied, then maybe?Wait, suppose we don't place a point at the origin. Then, can we have eight points on or inside the circle of radius R, each pair at least R apart?If all eight points are on the circle, then as previously, the chord length between adjacent points would have to be at least R, which would require the angular separation to be at least 60 degrees. But eight points on the circle would require each adjacent pair to be separated by 45 degrees, leading to chord lengths of 2R sin(22.5°) ≈ 2R * 0.383 ≈ 0.766R, which is less than R. So, that's too small. Therefore, eight points on the circle won't work.Alternatively, if some points are inside the circle. Let's say we have seven points on the circle and one inside. But the inside point must be at least R away from all seven circle points. However, the distance from any interior point to a point on the circle is at most 2R (if diametrically opposite), but to be at least R away from all circle points, the interior point would have to lie outside the circle of radius R/2 centered at each circle point. Wait, that's getting complicated. Alternatively, consider that for an interior point to be at least R away from all seven circle points, it must lie in the intersection of the exteriors of seven circles of radius R centered at each of the seven circle points. But the original circle has radius R, so the intersection of these exteriors within the original circle would be empty. Because any point inside the original circle is within R of at least one of the circle points (since the circle points are spread out). Wait, maybe not necessarily. Let me think.Suppose we have seven points on the circle. Let me imagine arranging seven points on the circle, each separated by about 51.4 degrees (since 360/7 ≈ 51.4). Then, if we try to place an eighth point somewhere inside the circle, we need it to be at least R away from all seven points on the circle. Is this possible?The distance from the center to each circle point is R. The distance from the interior point to a circle point would be sqrt(R^2 + d^2 - 2Rd cosθ), where d is the distance from the center to the interior point, and θ is the angle between them. Wait, actually, using the law of cosines: if the interior point is at distance r from the center, and the angle between the interior point and a circle point is θ, then the distance between them is sqrt(R^2 + r^2 - 2Rr cosθ). To have this distance ≥ R, we need:sqrt(R^2 + r^2 - 2Rr cosθ) ≥ RSquaring both sides:R^2 + r^2 - 2Rr cosθ ≥ R^2Simplify:r^2 - 2Rr cosθ ≥ 0r(r - 2R cosθ) ≥ 0Since r ≥ 0, this implies that r - 2R cosθ ≥ 0 ⇒ r ≥ 2R cosθBut since the interior point is within the original circle (r ≤ R), then 2R cosθ ≤ R ⇒ cosθ ≤ 1/2 ⇒ θ ≥ 60° or θ ≤ -60°.But if we have seven points equally spaced on the circle, the angle between any two adjacent points is 360/7 ≈ 51.4°, which is less than 60°. Therefore, for any interior point, the angle to the nearest circle point is less than 51.4°, which is less than 60°, so cosθ > 1/2. Therefore, to satisfy r ≥ 2R cosθ, we would need r ≥ 2R cosθ > 2R*(1/2) = R. But since the interior point must have r ≤ R, this is impossible. Therefore, there is no such interior point that is at least R away from all seven circle points. Therefore, n=8 is impossible.Therefore, if we have seven points on the circle and one at the center, but as we saw earlier, the center point would be at distance R from the circle points, but adding another point inside is impossible. Wait, but in the previous case with the origin, we had seven points: one at origin and six on the circle. To get eight points, we would need another point either on the circle or inside. But adding another point on the circle would require reducing the angular separation, leading to distances less than R. Adding a point inside is impossible as shown. Therefore, n=7 seems to be the maximum.But wait, in the case with one point at the origin and six on the circle, we have seven points. What if we don't place a point at the origin? Then, can we have seven points on the circle? Wait, if we have seven points on the circle, each separated by about 51.4 degrees, the chord length between adjacent points would be 2R sin(25.7°) ≈ 2R * 0.438 ≈ 0.876R, which is less than R. Therefore, the minimum distance would be less than R, violating the condition. Therefore, seven points on the circle without a central point would not satisfy the condition.Therefore, the only way to have seven points is to have one at the origin and six on the circle. Let me verify again. The distances between the origin and the circle points are R, the distances between adjacent circle points are R, and the distances between non-adjacent circle points are larger than R. Therefore, the minimum distance is R, which equals the maximum modulus. So, the condition is satisfied.Now, can we have more than seven? As discussed earlier, adding an eighth point seems impossible. Therefore, n=7 is the maximum.But let me check another configuration. Suppose we don't use the origin but arrange points in some other symmetric configuration. For example, two concentric circles. Suppose we have some points on an inner circle and some on an outer circle. Let me see.Suppose we have a point on the outer circle (radius R) and some points on an inner circle (radius r < R). The distance between a point on the inner circle and a point on the outer circle must be at least R. Let's calculate the minimum distance between a point on the inner circle and a point on the outer circle. The minimum distance would be R - r if they are colinear with the origin. Wait, if they are in the same direction, the distance is R - r. For this to be at least R, we need R - r ≥ R ⇒ -r ≥ 0 ⇒ r ≤ 0, which is impossible. Therefore, points on an inner circle cannot coexist with points on the outer circle, since the distance between them would be less than R. Therefore, all points must be on the outer circle or include the origin.Alternatively, if all points are on two concentric circles, but then the distance between points on different circles would be problematic. So, this approach seems invalid.Alternatively, arranging points in a regular polygon but not on a circle. For example, a regular hexagon with one point at the center. Wait, that's the configuration we already considered: one at the center and six on the hexagon vertices. So, that's seven points.Alternatively, a different polygon? For example, a pentagon with two points inside? Let's see. If we have a pentagon on the circle, each point is 72 degrees apart. The chord length is 2R sin(36°) ≈ 1.175R, which is greater than R. So, the minimum distance is 1.175R, which is more than R. Then, the maximum modulus is R. Then, min distance is 1.175R ≥ R, which satisfies the condition. So, in this case, n=5 on the circle. Then, if we add two points inside, but again, points inside would have to be at least R away from all five circle points and from each other. But similar to before, the distance from an interior point to a circle point is at least R. However, using similar reasoning as before, the interior points would need to be at a distance ≥ 2R cosθ from the center, where θ is half the angular separation. For a pentagon, angular separation is 72°, so θ=36°. Then, 2R cos36° ≈ 2R * 0.809 ≈ 1.618R. But the interior points can only be at most R away from the center, so this is impossible. Therefore, no interior points can be added.Therefore, perhaps the pentagon with five points on the circle has minimum distance ~1.175R, which is greater than R. But since the problem requires min distance ≥ max modulus (R), this configuration works. But then n=5. But we already have a configuration with n=7. So, the pentagon is not the maximum.Wait, but the problem asks for the maximum possible n. So, 7 is better. So, returning, the configuration with one point at the origin and six on the circle gives n=7, which satisfies the condition.But perhaps there's another configuration with more points? Let's think differently. Suppose we have multiple points arranged in a three-dimensional structure, but since we're dealing with complex numbers (which are in 2D), that's not applicable.Alternatively, place points at the vertices of a regular hexagon with one at the center. That gives seven points. Alternatively, place points at the vertices of a regular hexagon without the center. That gives six points. So, seven is better.Alternatively, maybe use a star-shaped configuration or something else. Let me consider a star polygon, like a five-pointed star (decagram). But the points would still need to lie on the circle, and the distances might vary. For example, in a five-pointed star (density 2), the chord lengths between adjacent points are shorter. So, probably not helpful.Alternatively, place three points on the circle forming an equilateral triangle (each 120 degrees apart), and then place other points inside. But again, the interior points would be too close to the circle points.Wait, suppose we have three points on the circle forming an equilateral triangle (each 120 degrees apart). The distance between any two circle points is ( sqrt{3} R approx 1.732 R ). Then, if we place another point somewhere inside. Let's say at the center. The distance from the center to each circle point is R. So, the minimum distance becomes R, and the maximum modulus is R. So, that gives four points. But we can do better with the hexagon plus center.Alternatively, place four points on the circle (square), each 90 degrees apart. The distance between adjacent points is ( sqrt{2} R approx 1.414 R ). If we add the center, then the minimum distance is R (from center to circle points), and the distances between circle points are 1.414 R. So, the minimum distance is R, so the condition is satisfied. So, n=5. But again, the hexagon plus center gives n=7, which is better.So, the key seems to be that the regular hexagon (6 points) plus the center gives seven points with minimum distance R. Any other configuration either has fewer points or fails the distance requirement.Therefore, tentatively, the maximum n is 7, achieved by one point at the origin and six points forming a regular hexagon on the circle of radius R.But wait, the problem says "all sets satisfying the condition". So, perhaps there are other configurations besides the regular hexagon plus center? For example, maybe slightly perturbing the points on the circle as long as the minimum distance remains R. But I think the regular hexagon is the only way to have six points on the circle with adjacent distances exactly R. Any other configuration would either have points closer together or further apart. But to maximize the number of points, we need them as tightly packed as possible given the minimum distance requirement. So, the regular hexagon is optimal.Alternatively, could we have a different configuration where some points are not on the same circle but still satisfy the distance and modulus conditions? For example, suppose we have two points on a circle of radius R, separated by 180 degrees, so distance 2R. Then, place other points in such a way that they are at least R away from all points. But this seems difficult. For example, two points opposite each other on the circle. Then, adding a third point anywhere on the circle would be at least R away from one of them but only 2R cos(theta) away from the other, where theta is the angle from the third point to one of the existing points. To have 2R cos(theta) ≥ R ⇒ cos(theta) ≥ 0.5 ⇒ theta ≤ 60°. So, the third point would need to be within 60 degrees of one of the existing points. But then, the distance between the third point and the nearby existing point would be 2R sin(theta/2). If theta is 60°, then distance is R. If theta is less, the distance is less. So, placing the third point 60 degrees away from one of the existing points gives distance R. Then, we have three points: two opposite each other and one at 60 degrees from one of them. Then, the minimum distance is R. The maximum modulus is R. Then, adding a fourth point. Where? Maybe opposite the third point, 60 degrees from the other existing point. Then, the fourth point is 60 degrees from the second original point, forming a square? Wait, no. If the original two points are at 0° and 180°, the third at 60°, then the fourth could be at 180° + 60° = 240°, which is 60° from 180°. Then, the distance between 60° and 240° is... Let me calculate. The angle between them is 180°, so the distance is 2R. Between 0° and 240° is 240°, distance is 2R sin(120°) = 2R*(√3/2) = √3 R ≈ 1.732 R. Between 60° and 240° is 180° - 60° = 120°, so distance is 2R sin(60°) = √3 R. So, minimum distance is R (between adjacent points). So, this gives four points with min distance R. Then, could we add more points? If we add a fifth point at 120°, which is 60° away from 60°, the distance between 60° and 120° is 2R sin(30°) = R. Then, the fifth point at 120°, sixth at 180° + 120° = 300°, and so on. Essentially, building up a hexagon by adding points 60° apart. So, yes, this leads back to the regular hexagon configuration. Therefore, regardless of how we start, the optimal configuration ends up being the regular hexagon plus center.Therefore, it seems that the maximum n is 7, achieved by the regular hexagon with a center point. Moreover, any such set must be similar to this configuration, possibly rotated or scaled, but with one point at the center and six points forming a regular hexagon on the circle of radius R.But let me check if there are other configurations. Suppose we have a different polygon, like a regular pentagon with two points inside. But as we saw earlier, points inside can't be at least R away from all the circle points. Similarly, if we have a square with three points inside, but again, the inner points would be too close to the square points or to each other.Alternatively, suppose we have two separate regular triangles, one rotated with respect to the other. For example, two equilateral triangles inscribed in the same circle, rotated by 30 degrees relative to each other. This would give six points on the circle, each 30 degrees apart. But then, the distance between adjacent points would be 2R sin(15°) ≈ 0.517R, which is way too small. So, that doesn't work.Alternatively, three points forming an equilateral triangle on the circle, and three other points forming another equilateral triangle rotated by 60 degrees. But then, the distance between points from different triangles might be less than R. For example, the angle between a point from the first triangle and a point from the second triangle could be 30 degrees, leading to a distance of 2R sin(15°) ≈ 0.517R, again too small.Therefore, such configurations don't work. It seems that the only way to have all distances at least R with maximum modulus R is the regular hexagon plus center.Another thought: maybe if we don't place all the circle points on the same circle, but vary their radii slightly. For example, some points slightly inside and some slightly outside, but keeping the maximum modulus R. However, if a point is slightly inside, then its distance to the origin is less than R, but we need all points to have modulus at most R. If a point is slightly outside, then its modulus would exceed R, which is not allowed. Therefore, all points must be on or inside the circle of radius R. So, varying radii isn't helpful unless we can place points outside, which we can't.Alternatively, placing points in three dimensions? But the problem is about complex numbers, which are in 2D. So, no.Alternatively, arranging points in some non-symmetric way. For example, clusters of points. But if points are clustered, the distances within the cluster would be too small.Therefore, after considering various possibilities, the regular hexagon plus center seems to be the only configuration that allows seven points with the given condition. All other configurations either result in fewer points or violate the minimum distance requirement.Therefore, I conclude that the maximum possible n is 7, and the sets achieving this are those consisting of the origin and the six vertices of a regular hexagon centered at the origin with radius equal to the maximum modulus R.Final AnswerThe maximum possible ( n ) is (boxed{7}), achieved by the origin together with the six vertices of a regular hexagon centered at the origin.</think>Given a set of distinct complex numbers ( z_i ) for ( i = 1, 2, ldots, n ) that satisfy the inequality:[min_{i neq j} |z_i - z_j| geq max_{i} |z_i|,]we need to find the maximum possible ( n ) and characterize all such sets.First, denote ( R = max_i |z_i| ). The condition requires that the minimum distance between any two distinct complex numbers ( z_i ) and ( z_j ) is at least ( R ). Considering the geometric interpretation, all points must lie within or on a circle of radius ( R ) centered at the origin, and each pair of points must be at least ( R ) apart. Key steps and reasoning:1. Equally Spaced Points on a Circle: If all points are equally spaced on a circle of radius ( R ), the chord length between adjacent points is ( 2R sin(pi/n) ). For this distance to be at least ( R ), we need ( sin(pi/n) geq 1/2 ), leading to ( n leq 6 ). However, this configuration alone gives ( n = 6 ).2. Including the Origin: Adding the origin as a seventh point, we check distances. The distance from the origin to any other point is ( R ), and the distance between adjacent points on the circle (now six points) is also ( R ). This configuration satisfies the minimum distance requirement, leading to ( n = 7 ).3. Verification of Other Configurations: Considering other configurations (e.g., multiple concentric circles, different polygons, or interior points) shows that they either fail the distance requirement or do not maximize ( n ). Points inside the circle would be too close to the origin or other points on the circle.4. Conclusion: The optimal configuration is one point at the origin and six points forming a regular hexagon on the circle of radius ( R ). This configuration ensures all distances are at least ( R ) and the maximum modulus is ( R ).Thus, the maximum possible ( n ) is (boxed{7}), achieved by the origin together with the six vertices of a regular hexagon centered at the origin.,<think>Okay, so I have this problem here about complex numbers. Let me try to understand it step by step. The problem states:Given a set of distinct complex numbers ( z_i ) for ( i = 1, 2, ldots, n ), they satisfy the inequality[min_{i neq j} |z_i - z_j| geq max_i |z_i|]And we need to find the maximum possible value of ( n ) and describe all such sets that achieve this maximum.Hmm. So, first, let me parse this condition. The minimum distance between any two distinct complex numbers in the set is at least as large as the maximum modulus (distance from the origin) of any number in the set. Interesting. So the closest any two points are to each other is not smaller than the farthest any point is from the origin.This seems like a packing problem, maybe? Like, how many points can you place in the complex plane (which is essentially the plane (mathbb{R}^2)) such that the smallest distance between any two points is at least equal to the largest distance from the origin. So, the points have to be at least a certain distance apart, and that certain distance is exactly the radius of the circle that contains all the points.Wait, if the maximum modulus is, say, ( R ), then all points lie within or on the circle of radius ( R ). But the minimum distance between any two points is at least ( R ). So, we need to place points inside a circle of radius ( R ), each at least ( R ) apart from each other. So, how many such points can we fit?I remember that in packing problems, the densest packing in 2D is hexagonal, but here we're not concerned with density per se, but with the number of points that can fit in a circle such that each pair is at least a certain distance apart. This is more like a circle packing problem where each circle has radius ( R/2 ), since if two points are at least ( R ) apart, their circles of radius ( R/2 ) around them don't overlap. But the entire configuration has to fit within a circle of radius ( R ).Wait, let me think again. If each point is at least ( R ) apart from each other, then the centers of the circles (each with radius ( R/2 )) must be at least ( R ) apart, which would mean that the circles don't overlap. So the problem reduces to packing non-overlapping circles of radius ( R/2 ) inside a circle of radius ( R + R/2 = 3R/2 )? Wait, no, that might not be the right way.Wait, actually, the original problem says that all points ( z_i ) lie within the circle of radius ( R ) (since ( max |z_i| = R )), and each pair of points is at least ( R ) apart. So, the distance between any two points is at least ( R ), and all points are within a circle of radius ( R ).So, this is like placing points inside a circle of radius ( R ), each pair separated by at least ( R ). How many points can we fit?Let me try small values of ( n ):- For ( n = 1 ): trivial, just one point anywhere inside the circle.- For ( n = 2 ): place two points on the boundary of the circle, diametrically opposed. The distance between them is ( 2R ), which is greater than ( R ), so that's okay.- For ( n = 3 ): Now, if we try to place three points. To maximize the minimal distance, we would place them in an equilateral triangle. The side length of the equilateral triangle inscribed in a circle of radius ( R ) is ( sqrt{3}R ). Wait, no. Wait, the side length ( s ) of an equilateral triangle inscribed in a circle of radius ( R ) is ( s = R sqrt{3} approx 1.732 R ). But our condition requires that the minimal distance between any two points is at least ( R ). So, ( R sqrt{3} approx 1.732 R ) is greater than ( R ), so that satisfies the condition. Therefore, three points can be placed as an equilateral triangle on the boundary of the circle, each pair separated by ( sqrt{3} R ), which is greater than ( R ), so the minimal distance here is actually ( sqrt{3} R ), which is more than ( R ). Wait, but the problem states that the minimal distance has to be at least equal to the maximum modulus. Wait, but the maximum modulus here is ( R ). So, if we can have three points with all pairwise distances at least ( R ), then that's okay.But actually, in the equilateral triangle case, all distances are ( sqrt{3} R ), which is larger than ( R ). So that works. But maybe we can even fit more points?Wait, let's see. Let me consider four points. If we try to place four points on the circle of radius ( R ), forming a square. The side length of a square inscribed in a circle of radius ( R ) is ( sqrt{2} R approx 1.414 R ). So, the minimal distance between adjacent points is ( sqrt{2} R ), which is more than ( R ). But wait, adjacent points on a square have distance ( sqrt{2} R ), but the diagonal of the square is ( 2 R ). So, all pairs of points have distances either ( sqrt{2} R ) or ( 2 R ). The minimal distance is ( sqrt{2} R approx 1.414 R ), which is greater than ( R ). Therefore, four points can be placed on the circle, each at least ( sqrt{2} R ) apart, which is more than ( R ), satisfying the condition.Wait a second, but hold on. If the problem states that the minimal distance between any two points is at least equal to the maximum modulus ( R ), then even if the minimal distance is greater than ( R ), that's still okay. So, in that case, four points on a square would satisfy the condition, since their minimal distance is ( sqrt{2} R geq R ).But wait, the problem says "min_{i≠j} |z_i - z_j| ≥ max_i |z_i|". So, the minimal distance is at least the maximum modulus, which is R. So, if we can have four points in the circle of radius R where each pair is at least R apart, then n=4 is possible.But can we actually place four points in the circle of radius R with all pairwise distances at least R?Wait, let's compute for a square inscribed in a circle of radius R. The side length s of the square is given by s = sqrt(2) * R. Wait, no. Wait, if the square is inscribed in a circle of radius R, then the diagonal of the square is 2R. The diagonal of a square is s*sqrt(2), so s*sqrt(2) = 2R => s = 2R / sqrt(2) = sqrt(2) R. So, the side length is sqrt(2) R ≈ 1.414 R. Therefore, adjacent vertices are sqrt(2) R apart, which is greater than R. So, all pairs of points (adjacent and diagonal) are either sqrt(2) R or 2 R apart. Therefore, the minimal distance is sqrt(2) R ≈ 1.414 R, which is greater than R. So, yes, four points on a square satisfy the condition. So n=4 is possible.Similarly, for five points: if we place five points on a regular pentagon inscribed in the circle of radius R. The side length of a regular pentagon inscribed in a circle of radius R is 2 R sin(π/5) ≈ 2 R * 0.5878 ≈ 1.1756 R. So, the distance between adjacent points is approximately 1.1756 R, which is still greater than R. Therefore, a regular pentagon would have minimal distance ~1.175 R > R, so that works. Wait, so then even n=5 is possible?Wait, this seems to contradict my initial intuition. Let me check. Let's compute the chord length for a regular polygon with n sides inscribed in a circle of radius R.The chord length (distance between adjacent vertices) is given by ( 2 R sin(pi/n) ). So, for the minimal distance to be at least R, we need:( 2 R sin(pi/n) geq R )Divide both sides by R:( 2 sin(pi/n) geq 1 )Thus,( sin(pi/n) geq 1/2 )Which implies:( pi/n geq pi/6 ) (since sin(π/6) = 1/2)Therefore,( n leq 6 )Wait, but this is for adjacent vertices. So, in a regular polygon with n sides inscribed in a circle, the minimal distance between two vertices is the side length, which is ( 2 R sin(pi/n) ). Therefore, setting this greater or equal to R gives ( 2 sin(pi/n) geq 1 implies sin(pi/n) geq 1/2 implies pi/n geq pi/6 implies n leq 6 ).Therefore, for a regular n-gon inscribed in a circle of radius R, the minimal distance is at least R when n ≤ 6. For n=6, the side length is exactly ( 2 R sin(pi/6) = 2 R * 1/2 = R ). So, a regular hexagon inscribed in the circle of radius R has adjacent vertices exactly R apart. Therefore, the minimal distance is R, which satisfies the condition.For n=7, the regular heptagon would have side length ( 2 R sin(pi/7) approx 2 R * 0.4338 ≈ 0.8676 R ), which is less than R, so it doesn't satisfy the condition. Therefore, if we place seven points regularly on the circle, the minimal distance would be less than R, which violates the condition.But wait, this is only considering regular polygons. Maybe there is a non-regular configuration with more points where the minimal distance is still at least R?Alternatively, perhaps we can place points not all on the boundary. For example, maybe placing some points inside the circle as well. But if we have points inside the circle, their distance to other points might be less than R. Wait, but the maximum modulus is R, so all points are within or on the circle of radius R. If we have a point at the origin, then its distance to any other point is at least the modulus of that other point, which is at most R. But the minimal distance in this case would be the modulus of the other points, but since they can be as close as R (if on the boundary). Wait, but the minimal distance between the origin and a boundary point is R, but the minimal distance between two boundary points could be larger. Wait, but if we have the origin and some points on the boundary, then the distance from the origin to a boundary point is exactly R. The distance between two boundary points could be as small as 0 if they are close, but in our case, we need all pairs to have distance at least R.So, if we include the origin, then all other points must be at distance at least R from the origin (since the origin is one of the points, and the distance from the origin to any other point is their modulus, which must be at least R. But since all points are within the circle of radius R, their modulus is exactly R. Therefore, if we include the origin, all other points must lie on the boundary of the circle.But wait, if we include the origin, then the origin is one point, and all other points must lie on the circle of radius R. Then, the distance from the origin to any other point is R, which is exactly the minimal distance. However, the distance between two points on the circle must also be at least R. So, if we have the origin and k points on the circle, then we need that the minimal distance between any two points on the circle is also at least R. Therefore, the minimal distance between the points on the circle is the same as if we had a circle packing with minimal distance R. Therefore, how many points can we place on the circle of radius R with each pair at least R apart?Again, using the chord length formula: the chord length between two points on a circle of radius R separated by angle θ is ( 2 R sin(theta/2) ). To have this chord length ≥ R, we need:( 2 R sin(theta/2) geq R implies sin(theta/2) geq 1/2 implies theta/2 geq pi/6 implies theta geq pi/3 ).Therefore, the angle between any two adjacent points (if placed regularly) must be at least π/3, which allows at most 6 points (since 2π / (π/3) = 6). Therefore, if we have the origin and 6 points on the circle, each separated by 60 degrees, then the distance between any two points on the circle is exactly R (since chord length is 2 R sin(π/6) = R). However, the distance from the origin to each point is R. So, in this case, the minimal distance is R (between the origin and any other point, as well as between adjacent points on the circle). Therefore, if we have 7 points: the origin and 6 points on the circle spaced 60 degrees apart, then all pairs have distance at least R. Wait, but between the origin and the circle points, it's R, and between circle points, it's R. So, the minimal distance is R, which meets the condition. Therefore, n=7 is possible? But wait, wait, let me check.Wait, the problem states that the set of complex numbers must be distinct. If we place the origin and 6 points on the circle, each separated by 60 degrees, then the distance between the origin and each of the 6 points is R, and the distance between any two adjacent points on the circle is R. But the distance between non-adjacent points on the circle would be larger. For example, the distance between two points separated by 120 degrees would be ( 2 R sin(60^circ) = 2 R (sqrt{3}/2) = sqrt{3} R approx 1.732 R ). So, the minimal distance is R (between the origin and the circle points, and between adjacent circle points). Therefore, all pairs are at least R apart. So, in this configuration, n=7.Wait, but is this correct? Let me count the distances:- Between origin and each of the 6 circle points: distance R.- Between adjacent circle points: distance R.- Between non-adjacent circle points: greater than R.Therefore, all pairs are at least R apart. Therefore, this configuration satisfies the condition with n=7.But earlier, when considering regular polygons without the origin, we had n=6 as the maximal regular polygon. But adding the origin gives n=7. But is this allowed? The problem states that the complex numbers are distinct. The origin is distinct from the circle points, so that's okay.Wait, but then could we add more points?Suppose we try n=8. Let's see. Suppose we place two points at the origin? No, the problem says the complex numbers are distinct, so we can't have multiple points at the same location. So, only one point at the origin. Then, if we have 7 points: origin + 6 on the circle. To add an 8th point, where could we place it?If we try to place another point on the circle, between two existing points. But the existing points are spaced 60 degrees apart. If we add a point between two adjacent points, the angle between them becomes 30 degrees. The chord length between the new point and its neighbors would be ( 2 R sin(15^circ) approx 2 R * 0.2588 ≈ 0.5176 R ), which is less than R. Therefore, that distance is too small. Alternatively, if we place the 8th point somewhere inside the circle. But then the distance from the 8th point to the origin is less than R, right? Because the origin is already one of the points, and the new point is inside the circle, so its modulus is less than R. Therefore, the distance between the new point and the origin is less than R, which violates the condition. Therefore, we cannot add any points inside the circle. If we try to place the 8th point on the circle, but not at the existing 6 points. But then, as shown, inserting a point between two existing points would create a chord of length less than R. Therefore, it's not possible. Therefore, n=7 seems possible. But wait, but let me check again: the problem says "distinct complex numbers", so we can have one at the origin and six on the circle. Then all distances are at least R, so this would satisfy the condition. Therefore, n=7 is possible.Wait, but I'm confused because in some references I recall that the maximum number is 5 or 6. Maybe I'm missing something here.Wait, let me check again the distances in the configuration with the origin and six points on the circle. Let's take specific coordinates. Let the origin be z0 = 0. Then, place six points on the circle of radius R at angles 0°, 60°, 120°, 180°, 240°, 300°. Let's compute the distance between z0 and any z_i: that's R, which is good. The distance between z1 (at 0°) and z2 (60°) is 2R sin(30°) = R, since the angle between them is 60°, so half the angle is 30°, and the chord length is 2R sin(30°) = R. Similarly, the distance between z1 and z3 (120°) is 2R sin(60°) ≈ 1.732 R. So that's greater than R. The distance between z1 and z4 (180°) is 2R, which is greater than R. So, all the distances between circle points are either R, ~1.732 R, or 2 R. Therefore, the minimal distance is R. So, this configuration satisfies the condition. Therefore, n=7 is possible. So, why do I have a conflicting thought?Wait, but maybe the problem says "min_{i≠j} |z_i - z_j| ≥ max_i |z_i|". So, in this configuration, the minimal distance is R, and the maximum modulus is R (since all points are either 0 or modulus R). Therefore, R ≥ R, which satisfies the equality. So, this is allowed.But then why in some problems similar to this, the maximum number is 6? Maybe I need to check the exact problem statement again.Wait, let me verify. The problem says:"min_{i≠j} |z_i - z_j| ≥ max_i |z_i|"So, the minimal distance between any two points is at least the maximum modulus. In the configuration with the origin and six circle points, the minimal distance is R, and the maximum modulus is R. Therefore, R ≥ R, which satisfies the condition. Therefore, n=7 is possible. Hmm.But is there a configuration with n=8?Suppose we try to place two points at the origin. No, the points must be distinct. If we place seven points on the circle without the origin. Wait, but then the minimal distance between the points is R (as in a regular hexagon), but then the maximum modulus is R, so that's okay. Wait, but in a regular hexagon, the distance between adjacent points is R, so the minimal distance is R. So, a regular hexagon (6 points) would have all pairs either R or 2R apart, so minimal distance R. Therefore, in that case, n=6 is possible. But adding the origin gives n=7. Therefore, 7 is possible. So, maybe the answer is 7? But I need to confirm whether there's a way to get more than 7.Alternatively, maybe there's a configuration where you have more points. Let me think. Suppose we have a point at the origin, and then arrange points in two concentric circles. Wait, but the maximum modulus is R, so all points must be within or on the circle of radius R. Suppose we have a point at the origin, and some points on an inner circle and some on the outer circle. But the distance from the origin to any other point is their modulus. So, if points are on an inner circle, say radius r < R, then their modulus is r, which is less than R. But the minimal distance between two points on the inner circle would be 2r sin(theta/2), where theta is the angle between them. But since their distance has to be at least R, which is greater than r. For example, if we have two points on the inner circle, separated by 180 degrees, the distance is 2r. To have 2r ≥ R, we need r ≥ R/2. But if we place points on both inner and outer circles, the minimal distance could be between an inner and outer point. The minimal distance between an inner point (radius r) and an outer point (radius R) would be at least R - r. To have R - r ≥ R, which would require -r ≥ 0, which is impossible. Therefore, the minimal distance between an inner and outer point is at least R - r. But we need this to be ≥ R, which can't happen. Therefore, if we have points inside the circle of radius R, their distances to the outer points would be less than R. Therefore, such configurations are impossible. Therefore, all points must lie on the boundary (modulus R) or at the origin. But if we have the origin, then the distance from the origin to any boundary point is R, which is okay. However, if we have another point inside, then the distance from that point to the origin is less than R, which violates the condition. Therefore, all points except possibly the origin must lie on the boundary. And the origin can be included, but then all other points must lie on the boundary, and their mutual distances must also be at least R.So, in the case where we include the origin, the remaining points must lie on the circle of radius R, with each pair at least R apart. How many points can we place on the circle with each pair at least R apart? As we saw earlier, in a regular hexagon, the adjacent points are exactly R apart. So, six points on the circle, each adjacent pair R apart. Therefore, if we include the origin, we can have 1 + 6 = 7 points.But if we don't include the origin, how many points can we place on the circle with minimal distance R? In a regular hexagon, 6 points. If we try to place 7 points on the circle without the origin, each pair must be at least R apart. But the regular heptagon has a smaller chord length. However, maybe an irregular configuration allows 7 points on the circle with all pairs at least R apart? Is that possible?Wait, let's compute the angular separation needed. For two points on a circle of radius R, the chord length is 2R sin(theta/2), where theta is the angle between them. So, to have chord length ≥ R, we need sin(theta/2) ≥ 1/2, which gives theta/2 ≥ 30°, so theta ≥ 60°. Therefore, each pair of points must be at least 60° apart. Therefore, on the circle, how many points can we place such that each pair is at least 60° apart? The maximum number is 6, since 360° / 60° = 6. So, even irregularly, you can't place more than 6 points on the circle with each pair at least 60° apart. Therefore, on the circle, maximum 6 points. Therefore, if you don't include the origin, n=6 is the maximum.But if you include the origin, then you can have 1 + 6 = 7 points, since the origin is 60° apart from all the circle points (in terms of angular distance?), but actually, the distance from the origin to any circle point is R, which is allowed, and the circle points are each 60° apart, so their chord lengths are R. Therefore, 7 points.Therefore, the maximum n is 7.Wait, but is there a configuration with more than 7 points? Let's see.Suppose we try to place two layers: the origin, six points on a circle of radius R, and another six points on a smaller circle. But as I thought before, the distance from the smaller circle points to the origin would be less than R, which is not allowed. Alternatively, place points on two concentric circles with radius R and something else, but then distances between points on different circles would be problematic.Alternatively, place points in some other configuration not on a circle. For example, maybe some points on the circle and some inside, but arranged such that all distances are at least R. But as mentioned, any point inside would have a distance to the origin less than R, unless it's on the circle. So, all non-origin points must be on the circle.Alternatively, maybe a configuration with multiple points on the circle and multiple points not at the origin but somewhere else. Wait, but any other point not at the origin would have modulus ≤ R. Suppose we have two points inside the circle, not at the origin. Then their distance to each other could be less than R, unless they are placed appropriately. But if their distance is at least R, but their modulus is also ≤ R. For example, place two points on the circle and two points inside such that all distances are at least R. It's complicated.Wait, let's consider trying to place two points inside the circle. Suppose we have two points inside the circle, each at a distance of, say, R/2 from the origin. Then the distance between them could be as much as R (if diametrically opposed), but they need to be at least R apart. So, to have two points inside the circle at distance R apart, they need to be on the boundary of a circle of radius R/2, separated by 180 degrees. But then their distance is R. So, placing two points at (R/2, 0) and (-R/2, 0). Then, their distance is R. So, modulus of each is R/2. Then, the minimal distance between them is R, which is equal to the maximum modulus (which is R/2)? Wait, no. The maximum modulus in this case is R/2, so the minimal distance is R, which is greater than R/2. Therefore, this would satisfy the condition. But then we can add more points?Wait, but in this case, the maximum modulus is R/2, but the minimal distance is R. So, according to the problem statement, the minimal distance must be at least the maximum modulus. Here, R ≥ R/2, which is true. But the problem says that the set must satisfy min_{i≠j} |z_i - z_j| ≥ max_i |z_i|. So, in this case, it's satisfied. But in this case, the maximum modulus is R/2, so we could maybe have more points with modulus up to R/2, but placed in such a way that their mutual distances are at least R/2. Wait, but the problem allows the maximum modulus to be as large as possible, but here the maximum modulus is R/2. Wait, but the problem is to find the maximum possible n given that the set satisfies the condition. So, if we can have more points by reducing the maximum modulus, but the problem doesn't fix the maximum modulus. Wait, but the problem says that for the set, the minimal distance is at least the maximum modulus. So, if we have a set where the maximum modulus is smaller, then the required minimal distance is also smaller. But the question is to find the maximum possible n over all such sets. Therefore, perhaps some sets have smaller maximum modulus, allowing more points with smaller minimal distances. But the problem is to find the maximum possible n, regardless of the specific maximum modulus. Wait, no. Wait, the problem says "Find the maximum possible n and for this n all sets satisfying the condition of the problem."Therefore, we need the largest possible n such that there exists a set of n distinct complex numbers where the minimal distance between any two is at least the maximum modulus of any in the set. So, the maximum modulus could be any value, but the minimal distance must be at least that value.But perhaps the maximum n is achieved when the minimal distance is exactly equal to the maximum modulus. Otherwise, if the minimal distance is larger, you could perhaps scale down the configuration to make the maximum modulus equal to the minimal distance, allowing possibly more points. Wait, that might be a direction.Suppose we have a configuration where the minimal distance is greater than the maximum modulus. Then, we can scale down all points by a factor to make the maximum modulus equal to the minimal distance. For example, if the minimal distance is d and the maximum modulus is R, with d > R, then scaling all points by R/d would make the new maximum modulus R*(R/d) = R^2/d, and the new minimal distance d*(R/d) = R. But then the new maximum modulus is R^2/d, which is less than R, since d > R. Therefore, the ratio of minimal distance to maximum modulus becomes R / (R^2/d) = d/R > 1. Wait, but this might not necessarily allow more points. Maybe not. Alternatively, maybe not helpful.Alternatively, perhaps the maximum n is achieved when the minimal distance equals the maximum modulus. Because if you have a configuration where the minimal distance is strictly greater than the maximum modulus, you could potentially add more points by reducing the minimal distance. Wait, but how?Alternatively, think of it in terms of scaling. Suppose you have a configuration with n points where the minimal distance is d and the maximum modulus is R, with d > R. Then, you can scale the entire configuration by a factor of R/d, resulting in a configuration where the maximum modulus is R*(R/d) and the minimal distance is d*(R/d) = R. So, the new configuration has maximum modulus R^2/d < R and minimal distance R. But since R^2/d < R, this configuration's maximum modulus is smaller. But since the problem allows any maximum modulus, as long as the minimal distance is at least that. Therefore, scaling doesn't help in increasing n, because scaling down reduces the modulus but keeps the minimal distance proportionally. However, the problem is to find the maximum n regardless of the modulus. So, perhaps the maximum n is achieved when the minimal distance equals the maximum modulus.Therefore, to find the maximum n, we can assume that the minimal distance equals the maximum modulus. So, let's set R = d. Then, all points lie within a circle of radius R, and each pair is at least R apart. So, the problem reduces to: place as many points as possible within a circle of radius R such that each pair is at least R apart. What's the maximum number of points?As we saw earlier, placing one point at the origin and six points on the circle gives seven points. But is there a way to place more?Wait, here's a thought. Suppose we place three points forming an equilateral triangle with side length R. The circumradius of an equilateral triangle is ( frac{R}{sqrt{3}} approx 0.577 R ). So, all three points lie within a circle of radius ~0.577 R. Then, the maximum modulus is ~0.577 R, and the minimal distance is R. Therefore, this satisfies the condition, but with a much smaller maximum modulus. However, since the problem allows the maximum modulus to be anything (as long as the minimal distance is at least that), this configuration is valid. But in this case, we have only three points. But if we can find a configuration with more points by having them cluster in a smaller circle with minimal distance R, but their maximum modulus is small. But since the problem is to find the maximum n over all possible such configurations, regardless of how small the maximum modulus is, as long as the minimal distance is at least the maximum modulus.Wait, but this seems counterintuitive. If we can make the maximum modulus as small as we want, then we can place an arbitrary number of points within that small circle, as long as their mutual distances are at least the maximum modulus. But if the maximum modulus tends to zero, the number of points you can fit in a circle of radius tending to zero with mutual distances at least that radius would... Wait, actually, if you have a circle of radius r, and you want to place points inside it with mutual distances at least r, then how many points can you fit? For small r, the circle is small, but the required distance between points is also small. Wait, but in terms of packing, in a circle of radius r, the number of points you can place with mutual distances at least r is similar to packing circles of radius r/2 within a circle of radius r + r/2 = 3r/2. Wait, but this is a different problem. Wait, no. If the points are centers of circles of radius r/2, then the distance between centers must be at least r. Therefore, the problem reduces to packing circles of radius r/2 with centers within a circle of radius R = r. So, the centers must lie within a circle of radius r, and each center must be at least r apart from each other. Therefore, how many circles of radius r/2 can be packed into a circle of radius r + r/2 = 3r/2. Wait, no, that's the covering radius. Wait, maybe this is a different approach.Alternatively, if we have points in a circle of radius r, each pair at least distance r apart. So, for example, in a circle of radius r, the maximum number of points with mutual distances at least r. For small r, how many points can we fit? Actually, in the plane, the maximum number of points you can place in a circle of radius r with each pair at least distance r apart is similar to packing non-overlapping disks of radius r/2 into a larger disk of radius 3r/2. The number of such disks is known in disk packing problems. However, I don't remember the exact numbers, but for small r, the number can be large. Wait, but this seems not the case. Wait, actually, if the points are in a circle of radius r, then the distance from the origin to any point is at most r. But if two points are at distance at least r apart, then they can't both be too close to each other. Wait, but if you have points in a circle of radius r, their mutual distance can be up to 2r. But the problem requires their mutual distance to be at least r. So, how many points can you place in a circle of radius r such that each pair is at least r apart?This is equivalent to placing points in a circle where each point is the center of a circle of radius r/2, and all such circles are non-overlapping and fit within a circle of radius r + r/2 = 3r/2. However, this seems like a different problem. Wait, actually, if the original points are within radius r, and each is surrounded by a circle of radius r/2, then those circles must lie within a circle of radius r + r/2 = 3r/2. But the problem is different because we are required to have the centers within radius r, and the circles of radius r/2 around them not overlapping. The number of such circles is equivalent to the number of non-overlapping circles of radius r/2 that can fit within a circle of radius 3r/2. The exact number is known, but I might not recall it.But in our problem, the maximum modulus is r, and the minimal distance between points is at least r. Therefore, it's equivalent to packing points within a circle of radius r with mutual distance at least r. The question is, what's the maximum number of such points?But in our previous consideration, when we set r = R, we saw that we can place up to 7 points (origin + 6 on the circle). But if we allow r to be smaller, can we place more points? For example, if r is very small, say approaching zero, then in theory, you could place an infinite number of points within a circle of radius r, each separated by distance at least r. But that's not true because as r approaches zero, the area remains proportional to r², and the number of points you can pack with mutual distance at least r is roughly proportional to (Area)/(π(r/2)^2) = (πr²)/(π(r²/4)) = 4. So, regardless of how small r is, you can only fit a constant number of points? Wait, no. Wait, if you have a circle of radius r, and you need to place points inside it such that each pair is at least r apart. How many points can you fit?Well, the maximal number is related to the kissing number in 2D, which is 6. The kissing number is the number of non-overlapping spheres of radius r that can touch another sphere of radius r. In 2D, the kissing number is 6. So, in our case, if we have a circle of radius r, and we want to place points inside it such that each pair is at least r apart. The maximal number is 7: one at the center and six around it, each at distance r from the center and adjacent ones at distance r from each other. But wait, in this case, the center point is at distance r from the surrounding points, but the surrounding points are distance r apart from each other. So, this configuration is exactly the same as the one we considered earlier with n=7. Therefore, even if we scale the configuration down, we can't get more than 7 points. Because if we place more than 6 points around the center, the angular separation becomes less than 60 degrees, leading to chord lengths less than r.Therefore, regardless of the radius r, the maximal number of points in a circle of radius r with mutual distances at least r is 7. Therefore, the maximum n is 7. Therefore, the answer is n=7, achieved by the configuration with one point at the origin and six points forming a regular hexagon on the circle of radius R, where R is the maximal modulus (distance from origin to the hexagon points) and the minimal distance (between the origin and the hexagon points, and between adjacent hexagon points) is R.But wait, but if we can have configurations with smaller maximum modulus, why can't we have more points? For instance, if we have a configuration where the maximum modulus is 1, and the minimal distance is 1. If we scale this configuration down, making the maximum modulus 0.5 and minimal distance 0.5, then in terms of geometry, it's similar, just scaled. Therefore, the number of points remains the same. So, regardless of scaling, the maximal number is 7. Therefore, the answer is n=7.But I need to verify this. Let's check online if possible. Wait, since I can't actually browse the web, but based on my knowledge, in disk packing problems, the maximal number of points within a circle of radius R with mutual distances at least R is 7. This configuration is called the "hexagonal packing" with a center point. Yes, this is a known result. For example, the kissing number in 2D is 6, which is the number of points you can place around a central point all touching it. But if you allow the central point, then you get 7 points. Therefore, the maximal n is 7.Therefore, the answer should be n=7, and the configurations are those consisting of the origin and the six vertices of a regular hexagon inscribed in a circle of radius R, with R equal to the distance between any two adjacent points (which is also the distance from the origin to each vertex).But let me just try to see if there's any other configuration with n=7. For example, could we have seven points on the circle without the origin? But as we saw earlier, placing seven points on the circle with each pair at least R apart requires each pair to be at least 60° apart, which would only allow six points. Seven points would require angles less than 60°, leading to chord lengths less than R. Therefore, it's not possible. Therefore, the only way to get seven points is by including the origin.Therefore, the maximum possible n is 7, achieved by the configuration with one point at the origin and six points forming a regular hexagon on the circle of radius equal to the distance from the origin to each of those six points.Thus, the answer is boxed{7}, and the sets are those consisting of zero (the origin) and the sixth roots of unity scaled by a common radius R, along with the origin.Final AnswerThe maximum possible ( n ) is boxed{7}, achieved by the origin and the six vertices of a regular hexagon centered at the origin.</think>Given a set of distinct complex numbers ( z_i ) for ( i = 1, 2, ldots, n ) that satisfy the inequality[min_{i neq j} |z_i - z_j| geq max_i |z_i|,]we need to find the maximum possible value of ( n ) and describe the configurations that achieve this maximum.1. Understanding the Condition: - The minimum distance between any two distinct complex numbers ( z_i ) and ( z_j ) must be at least as large as the maximum modulus of any ( z_i ). - This implies that all points lie within or on a circle of radius ( R ) (where ( R = max_i |z_i| )), and each pair of points is at least ( R ) apart.2. Considering Regular Polygons: - For ( n = 6 ), a regular hexagon inscribed in a circle of radius ( R ) has each pair of adjacent points separated by ( R ), satisfying the condition. - However, adding the origin (0) as a seventh point allows the configuration where the origin is at distance ( R ) from each of the six points on the circle, and the six points on the circle are each separated by ( R ).3. Verification of Distances: - The distance from the origin to any point on the circle is ( R ). - The distance between any two adjacent points on the circle (forming a regular hexagon) is ( R ). - Non-adjacent points on the circle are farther apart than ( R ), ensuring all pairwise distances are at least ( R ).4. Conclusion: - The configuration with the origin and six points forming a regular hexagon on the circle of radius ( R ) satisfies the given condition. - This configuration achieves the maximum ( n = 7 ).Thus, the maximum possible ( n ) is boxed{7}, achieved by the origin and the six vertices of a regular hexagon centered at the origin.]

🔑:275. Let left|z_{m}right|=max _{i}left|z_{i}right|. Then on the complex plane, all points z_{i} will be located inside the circle R of radius left|z_{m}right| centered at the point z=0. Clearly, 6 points z_{i}, located on the circumference of R and forming a regular hexagon, together with the point z_{1}=0 give n=7. If, however, n geqslant 7 and some point z neq 0 is not on the circumference of R, then by applying the cosine theorem, we will get that left|z_{i}-z_{j}right|0 and 0 leqslant delta<frac{pi}{3}.[P. Kornya, A. M. M., 76, 91 (January 1969).]

🧠:[<think>Okay, let's tackle this problem. It's about a regular truncated quadrilateral pyramid, also known as a frustum. The upper base has a side length of 1, and the lower base has a side length of 7. A specific plane cuts through edge B₁C₁ and is perpendicular to the plane AD₁C, dividing the frustum into two parts of equal volume. We need to find the volume of the original pyramid before truncation.First, let's recall some basics. A regular pyramid has a square base and lateral edges that meet at a single apex point. When truncated, the resulting frustum has two square bases, the original larger base and a smaller upper base. The volume of a frustum is given by (1/3) * height * (area of lower base + area of upper base + sqrt(area lower * area upper)). But wait, the problem mentions dividing the pyramid into two parts of equal volume. However, the plane is cutting the frustum, not the original pyramid. Hmm, maybe the problem is actually referring to the original pyramid's volume? Let me check again.Wait, the problem states: "a regular truncated quadrilateral pyramid... a plane passing through the edge B₁C₁ perpendicular to the plane AD₁C divides the pyramid into two parts of equal volume. Find the volume of the pyramid." So it's the original pyramid's volume we need to find, even though the truncation is given. Wait, maybe it's a frustum, but the plane divides the entire original pyramid into two parts of equal volume. Wait, the problem is a bit confusing. Let me parse it again."In a regular truncated quadrilateral pyramid with lateral edges AA₁, BB₁, CC₁, DD₁, the side of the upper base A₁B₁C₁D₁ is 1, and the side of the lower base is 7. A plane passing through the edge B₁C₁ perpendicular to the plane AD₁C divides the pyramid into two parts of equal volume. Find the volume of the pyramid."So the pyramid in question is the truncated one (the frustum), but the problem says "the volume of the pyramid". Wait, maybe "pyramid" here refers to the original pyramid before truncation? Because a frustum is technically not a pyramid. So perhaps the problem is that we have a frustum (truncated pyramid) and a plane cutting it into two parts of equal volume. But the question is asking for the volume of the original pyramid. That makes more sense. So we need to find the volume of the original pyramid, given that when we truncate it to form a frustum with upper base side 1 and lower base 7, a certain plane divides the frustum into two equal volumes.Alternatively, maybe the problem is in the truncated pyramid itself. The wording is a bit unclear. Wait, the problem says "the volume of the pyramid". If it's a truncated pyramid, it's a frustum, and its volume is different from the original pyramid. But the problem says "the pyramid", so maybe they are referring to the original pyramid before truncation? Let me check once more.The problem starts by describing a regular truncated quadrilateral pyramid (i.e., frustum) with lateral edges AA₁, BB₁, etc. The upper base has side 1, lower base 7. Then a plane passes through edge B₁C₁, is perpendicular to plane AD₁C, and divides the pyramid (the frustum?) into two parts of equal volume. The question is to find the volume of the pyramid. Hmm.Wait, maybe "pyramid" here actually refers to the frustum? But a frustum is not a pyramid. So maybe the problem is a translation issue. The original term might have been "frustum", but it's translated as "truncated pyramid". In that case, the volume of the frustum is being referred to. But the problem says "the pyramid", not "the truncated pyramid". So perhaps they are referring to the original pyramid before truncation? But then, how does the truncation relate to the original pyramid?Alternatively, maybe the problem is set in such a way that the frustum (the truncated pyramid) is considered as a part of the original pyramid. The plane divides the original pyramid into two parts of equal volume, one of which is the frustum. Wait, but the problem states "the plane ... divides the pyramid into two parts of equal volume". So if the pyramid is the original pyramid, the plane divides it into two parts, one of which is the frustum? But the frustum is obtained by truncating the pyramid with a plane parallel to the base. However, the plane in the problem is passing through B₁C₁ and is perpendicular to plane AD₁C, which is likely not parallel to the base, so the frustum mentioned here is different from the usual one.This is getting confusing. Let me try to visualize the figure.We have a regular quadrilateral pyramid, truncated. So the original pyramid has a square base ABCD with side 7, and the truncation plane cuts off the top, leaving a smaller square A₁B₁C₁D₁ with side 1. The lateral edges are AA₁, BB₁, CC₁, DD₁. So the truncation plane is parallel to the base, as in a regular frustum.Now, a plane passes through edge B₁C₁ (which is part of the upper base) and is perpendicular to the plane AD₁C. This plane divides the frustum into two parts of equal volume. We need to find the volume of the original pyramid.So, the key steps here are:1. Determine the height of the original pyramid, which will allow us to compute its volume.2. Use the given condition that the plane through B₁C₁ perpendicular to AD₁C divides the frustum into two equal volumes to set up an equation involving the height.But how do we relate the position of the cutting plane to the volume? Maybe we need to find the height of the frustum, which is the distance between the two bases, and relate it to the original pyramid's height. Since the frustum is part of the original pyramid, the height of the frustum is the difference between the original pyramid's height and the height of the smaller, similar pyramid that was cut off.Given that the original pyramid has base side 7 and the truncated top has base side 1, the ratio of similarity is 1:7. Therefore, the heights are in the same ratio. Let h be the height of the original pyramid. Then the height of the smaller pyramid (the part that was cut off) is h * (1/7), so the height of the frustum is h - h/7 = (6h)/7.But how does the plane through B₁C₁ and perpendicular to AD₁C come into play? Let's think about the plane AD₁C. Points A, D₁, and C. Point A is a vertex of the lower base, D₁ is a vertex of the upper base, and C is a vertex of the lower base. So plane AD₁C is a diagonal plane cutting through the frustum.First, let's find some coordinates to model this. Let's place the pyramid in a coordinate system. Let the center of the lower base ABCD be at the origin (0,0,0), and the pyramid's apex at (0,0,h). Since it's a regular pyramid, all edges are congruent. The lower base ABCD is a square with side 7, so the coordinates of the lower base vertices can be:A: (-3.5, -3.5, 0)B: (3.5, -3.5, 0)C: (3.5, 3.5, 0)D: (-3.5, 3.5, 0)But wait, if the side length is 7, then the distance from the center to each vertex along the x and y axes is 7/2. So that's correct.The upper base A₁B₁C₁D₁ is a square with side length 1. Since the truncation is similar, the upper base is centered at (0,0,k), where k is the height from the base to the truncation plane. The coordinates of the upper base vertices would be scaled by a factor of 1/7. So:A₁: (-0.5, -0.5, k)B₁: (0.5, -0.5, k)C₁: (0.5, 0.5, k)D₁: (-0.5, 0.5, k)But since the upper base is similar and parallel, the scaling factor is 1/7, so k would be h - (6h/7) = h/7. Wait, no. If the original pyramid has height h, then the truncation plane is at height k from the base, creating a smaller pyramid with height h - k. The similarity ratio is (h - k)/h = 1/7, so h - k = h/7, so k = 6h/7. Therefore, the upper base is located at height k = 6h/7. So the coordinates of the upper base vertices are scaled by 1/7 and shifted up to z = 6h/7.Therefore:A₁: (-3.5*(1/7), -3.5*(1/7), 6h/7) = (-0.5, -0.5, 6h/7)Similarly for B₁, C₁, D₁.So, coordinates:Lower base ABCD:A: (-3.5, -3.5, 0)B: (3.5, -3.5, 0)C: (3.5, 3.5, 0)D: (-3.5, 3.5, 0)Upper base A₁B₁C₁D₁:A₁: (-0.5, -0.5, 6h/7)B₁: (0.5, -0.5, 6h/7)C₁: (0.5, 0.5, 6h/7)D₁: (-0.5, 0.5, 6h/7)Now, the plane in question passes through edge B₁C₁ and is perpendicular to the plane AD₁C. Let's first understand the plane AD₁C.Points A, D₁, C.Point A: (-3.5, -3.5, 0)Point D₁: (-0.5, 0.5, 6h/7)Point C: (3.5, 3.5, 0)So, three points: A, D₁, C.We can find the equation of plane AD₁C.First, compute two vectors in the plane:Vector AD₁: D₁ - A = (-0.5 - (-3.5), 0.5 - (-3.5), 6h/7 - 0) = (3, 4, 6h/7)Vector AC: C - A = (3.5 - (-3.5), 3.5 - (-3.5), 0 - 0) = (7, 7, 0)Now, the normal vector to plane AD₁C is the cross product of AD₁ and AC.Compute AD₁ × AC:|i j k||3 4 6h/7||7 7 0 |= i*(4*0 - 6h/7*7) - j*(3*0 - 6h/7*7) + k*(3*7 - 4*7)= i*(-6h) - j*(-6h) + k*(21 - 28)= (-6h)i + 6hj -7kSo the normal vector is (-6h, 6h, -7)Therefore, the equation of plane AD₁C can be written as:-6h(x - (-3.5)) + 6h(y - (-3.5)) -7(z - 0) = 0Simplify:-6h(x + 3.5) + 6h(y + 3.5) -7z = 0Expand:-6h x - 21h + 6h y + 21h -7z = 0Simplify terms:-6h x + 6h y -7z = 0Divide both sides by h (assuming h ≠ 0):-6x + 6y - (7/h) z = 0Alternatively, we can write the plane equation as:6(x - y) + (7/h) z = 0But maybe we can leave it as -6h x + 6h y -7z = 0.Now, the plane we are interested in passes through edge B₁C₁ and is perpendicular to plane AD₁C. Let's denote the desired plane as Π.Since plane Π is perpendicular to plane AD₁C, their normal vectors are also perpendicular. The normal vector of plane AD₁C is (-6h, 6h, -7). The normal vector of plane Π must be perpendicular to this.Also, plane Π contains edge B₁C₁. Let's parametrize edge B₁C₁. Points B₁: (0.5, -0.5, 6h/7) and C₁: (0.5, 0.5, 6h/7). So edge B₁C₁ is a horizontal line segment in the upper base, along y from -0.5 to 0.5, x=0.5, z=6h/7.Since plane Π contains B₁C₁, it must contain the line B₁C₁. So, any point on B₁C₁ is (0.5, t, 6h/7) where t ranges from -0.5 to 0.5.To find the equation of plane Π, we need another point or direction. But plane Π is also perpendicular to plane AD₁C. The normal vectors of the two planes must be perpendicular.Let n1 be the normal vector of plane AD₁C: (-6h, 6h, -7).Let n2 be the normal vector of plane Π. Then n1 · n2 = 0.Additionally, plane Π contains the line B₁C₁. Let's find two direction vectors in plane Π. One direction vector is along B₁C₁: from B₁ to C₁: (0, 1, 0). Another direction vector can be found by taking a point on B₁C₁ and another point in plane Π.Alternatively, since the plane Π contains B₁C₁ and is perpendicular to plane AD₁C, the line of intersection of planes Π and AD₁C must be perpendicular to B₁C₁. Hmm, maybe this is getting too abstract.Alternatively, let's find the normal vector n2 of plane Π.Since plane Π contains the line B₁C₁, which has direction vector (0,1,0). Also, plane Π is perpendicular to plane AD₁C, so the normal vector of Π (n2) is perpendicular to the normal vector of AD₁C (n1) and also lies in plane Π. Wait, no. If two planes are perpendicular, their normal vectors are also perpendicular. So n1 · n2 = 0.Therefore, n2 must be orthogonal to n1. Additionally, plane Π contains the line B₁C₁, so the direction vector of B₁C₁, which is (0,1,0), must lie in plane Π. Therefore, the normal vector n2 of plane Π must be orthogonal to both (0,1,0) and n1.Wait, no. The normal vector n2 is orthogonal to all direction vectors in plane Π. Since direction vector (0,1,0) is in Π, then n2 must be orthogonal to (0,1,0). Also, n2 is orthogonal to n1.Therefore, n2 is orthogonal to both (0,1,0) and n1. Therefore, n2 is parallel to the cross product of (0,1,0) and n1.Compute cross product of (0,1,0) and n1 = (-6h, 6h, -7):|i j k||0 1 0||-6h 6h -7|= i*(1*(-7) - 0*6h) - j*(0*(-7) - (-6h)*0) + k*(0*6h - (-6h)*1)= i*(-7) - j*(0) + k*(6h)= (-7, 0, 6h)Therefore, n2 is parallel to (-7, 0, 6h). So the normal vector of plane Π can be taken as (-7, 0, 6h).Therefore, the equation of plane Π is:-7(x - 0.5) + 0*(y - (-0.5)) + 6h(z - 6h/7) = 0Wait, but we need a point on the plane. Since plane Π contains edge B₁C₁, we can use point B₁ (0.5, -0.5, 6h/7) to plug into the equation.So general equation: -7(x - 0.5) + 0*(y + 0.5) + 6h(z - 6h/7) = 0Simplify:-7x + 3.5 + 6h z - 6h*(6h/7) = 0-7x + 3.5 + 6h z - (36h²)/7 = 0Multiply all terms by 7 to eliminate denominators:-49x + 24.5 + 42h z - 36h² = 0But maybe we can keep it as:-7x + 6h z + 3.5 - (36h²)/7 = 0But perhaps it's better not to plug in the point yet. Let me check again.The normal vector is (-7, 0, 6h). So the plane equation is:-7(x - x0) + 0*(y - y0) + 6h(z - z0) = 0We need a point (x0, y0, z0) on the plane. Since the plane contains B₁C₁, take point B₁ (0.5, -0.5, 6h/7):-7(x - 0.5) + 0*(y + 0.5) + 6h(z - 6h/7) = 0Expanding:-7x + 3.5 + 6h z - 6h*(6h/7) = 0Combine constants:-7x + 6h z + 3.5 - (36h²)/7 = 0So the equation of plane Π is:-7x + 6h z + 3.5 - (36h²)/7 = 0Alternatively, multiply through by 7:-49x + 42h z + 24.5 - 36h² = 0But maybe it's better to write in terms of x, y, z:-7x + 6h z = (36h²)/7 - 3.5But perhaps we can leave it as is for now.Now, the plane Π divides the frustum into two parts of equal volume. We need to compute the volume of one part and set it equal to half the volume of the frustum.But first, let's recall that the volume of the frustum is (1/3) * H * (S1 + S2 + sqrt(S1 S2)), where H is the height of the frustum, S1 is the area of the lower base, S2 the area of the upper base.Given that the side lengths are 7 and 1, so areas are 49 and 1. The height H of the frustum is 6h/7 (as computed earlier: original pyramid height h, smaller pyramid height h/7, so frustum height h - h/7 = 6h/7).Therefore, the volume of the frustum is:V_frustum = (1/3) * (6h/7) * (49 + 1 + sqrt(49*1)) = (1/3)*(6h/7)*(50 + 7) = (1/3)*(6h/7)*57 = (6h/7)*(19) = 114h/7.But the problem states that plane Π divides the frustum into two parts of equal volume, so each part has volume 57h/7.But we need to verify if this is indeed the case. However, the plane is cutting through the frustum, but the cutting plane is not necessarily horizontal or parallel to the bases. Therefore, the resulting volumes are not simply portions of the frustum but more complex shapes.Therefore, perhaps we need to compute the volume of one part created by plane Π and set it equal to half of V_frustum.To compute this volume, we might need to find the intersection of plane Π with the frustum, which would form a polygon, and then compute the volume using integration or decomposition into simpler shapes.Alternatively, since the original frustum is part of the pyramid, maybe we can consider extending the frustum to the original pyramid and use similarity ratios.But this might get complicated. Let's see.Alternatively, we can model the frustum as the region between z=0 (lower base) and z=6h/7 (upper base), with the cross-sectional area at height z being a square with side length decreasing from 7 to 1 as z increases from 0 to 6h/7.The plane Π cuts through this frustum. To find the volume on one side of Π, we can integrate over the frustum, but taking into account the intersection with the plane.Alternatively, perhaps we can use coordinates to model the frustum and the plane, then compute the volume.Given that the frustum is between z=0 and z=6h/7, with cross-sectional square side length at height z given by s(z) = 7 - (7 -1)*(z)/(6h/7) = 7 - 6*(7z)/(6h) = 7 - (7z)/h.Wait, let's check:At z=0, s(0)=7; at z=6h/7, s(6h/7)=1. So linear decrease.Thus, s(z) = 7 - (7 -1)*(z)/(6h/7) = 7 - 6*(7z)/(6h) = 7 - (7z)/h. Yes, that's correct.So, at height z, the square has side length 7 - (7z)/h.Therefore, the coordinates of the four corners at height z are (±(7 - 7z/h)/2, ±(7 - 7z/h)/2, z). Wait, but in our coordinate system earlier, the lower base at z=0 is from -3.5 to 3.5 in x and y, which is ±7/2. Similarly, upper base at z=6h/7 is from -0.5 to 0.5, which is ±1/2.So, indeed, the side length at height z is 7 - (7z)/h. Therefore, the half-length in x and y is (7 - 7z/h)/2.Therefore, the frustum can be parametrized as:x ∈ [-(7 - 7z/h)/2, (7 - 7z/h)/2]y ∈ [-(7 - 7z/h)/2, (7 - 7z/h)/2]z ∈ [0, 6h/7]But plane Π is given by -7x + 6h z + 3.5 - (36h²)/7 = 0. Let's solve for x in terms of z:-7x + 6h z = (36h²)/7 - 3.5=> x = (6h z - (36h²)/7 + 3.5)/7Simplify:x = (6h z)/7 - (36h²)/49 + 0.5So x = (6h z)/7 - (36h²)/49 + 0.5This is the x-coordinate of the plane Π at height z.But in the frustum, the x ranges from -(7 - 7z/h)/2 to (7 - 7z/h)/2. Therefore, the intersection of plane Π with the frustum will occur where x is within this range.To find the region of the frustum that is on one side of the plane (say, x ≥ (6h z)/7 - (36h²)/49 + 0.5), we need to set up the limits for integration.But integrating this might be complicated. Alternatively, perhaps we can decompose the frustum into smaller parts or use the method of slicing.Alternatively, use the fact that the plane divides the frustum into two equal volumes. Therefore, the average height of the frustum with respect to the plane Π must be such that the volume on either side is equal.Alternatively, think of the frustum as a part of the original pyramid. If we can model the original pyramid, then maybe the plane Π intersects the original pyramid in some way, but the problem states that the plane divides the frustum into two parts of equal volume, so the plane is entirely within the frustum.Alternatively, use coordinates and find the intersection of plane Π with the lateral edges or faces of the frustum.Wait, the plane Π contains edge B₁C₁. Since edge B₁C₁ is part of the upper base, the plane Π cuts through the frustum along B₁C₁ and exits somewhere on the lower base or lateral faces.Wait, let's check where plane Π intersects the lower base. The lower base is at z=0. Let's substitute z=0 into the plane equation:-7x + 6h*0 + 3.5 - (36h²)/7 = 0=> -7x + 3.5 - (36h²)/7 = 0=> x = (3.5 - (36h²)/7)/7 = 0.5 - (36h²)/49But the lower base's x ranges from -3.5 to 3.5. So for the plane to intersect the lower base within this range, x must be between -3.5 and 3.5.So x = 0.5 - (36h²)/49 must be within [-3.5, 3.5]. Therefore:-3.5 ≤ 0.5 - (36h²)/49 ≤ 3.5Subtract 0.5:-4 ≤ - (36h²)/49 ≤ 3Multiply by -1 (reverse inequalities):4 ≥ (36h²)/49 ≥ -3But (36h²)/49 is always non-negative, so the right inequality (≥ -3) is always true. The left inequality: (36h²)/49 ≤ 4 => h² ≤ (4*49)/36 = 49/9 => h ≤ 7/3 ≈ 2.333...So h must be less than or equal to 7/3. But since the frustum's height is 6h/7, which must be positive, h > 0.But we don't know h yet; h is the original pyramid's height which we need to find.Alternatively, maybe the plane intersects the lateral faces. Let's see.The frustum has four lateral faces: each a trapezoid. For example, face AA₁B₁B is a trapezoid between points A, A₁, B₁, B.Plane Π contains edge B₁C₁ and extends downward. It might intersect other lateral edges or the lower base.Alternatively, since we have the equation of plane Π, we can find its intersection with the lateral edges.The lateral edges are AA₁, BB₁, CC₁, DD₁. Let's check if plane Π intersects any of these edges besides B₁C₁.Take edge AA₁: from A(-3.5, -3.5, 0) to A₁(-0.5, -0.5, 6h/7). Parametrize this edge as:x = -3.5 + 3ty = -3.5 + 3tz = 0 + (6h/7)tfor t ∈ [0,1]Plug into plane equation Π: -7x + 6h z + 3.5 - (36h²)/7 = 0Substitute:-7(-3.5 + 3t) + 6h*(6h/7 t) + 3.5 - (36h²)/7 = 0Compute term by term:-7*(-3.5) = 24.5-7*3t = -21t6h*(6h/7 t) = (36h²)/7 tSo total:24.5 -21t + (36h²)/7 t + 3.5 - (36h²)/7 = 0Combine constants:24.5 + 3.5 = 28Combine terms with t:(-21t + (36h²)/7 t) = t*(-21 + (36h²)/7)Constants minus (36h²)/7:28 - (36h²)/7 = 0So equation becomes:28 + t*(-21 + (36h²)/7) - (36h²)/7 = 0Wait, let me re-express:24.5 -21t + (36h²)/7 t + 3.5 - (36h²)/7 = 0Combine like terms:(24.5 + 3.5) + (-21t + (36h²)/7 t) - (36h²)/7 = 028 + t*(-21 + (36h²)/7) - (36h²)/7 = 0Rearranged:28 - (36h²)/7 + t*(-21 + (36h²)/7) = 0Solve for t:t*(-21 + (36h²)/7) = (36h²)/7 - 28t = [(36h²)/7 - 28] / [ -21 + (36h²)/7 ]Factor numerator and denominator:Numerator: (36h² - 196)/7Denominator: (-147 + 36h²)/7Thus,t = (36h² - 196)/(-147 + 36h²) = (36h² - 196)/(36h² - 147) = [36h² - 196]/[36h² - 147]This simplifies to:t = [36h² - 196]/[36h² - 147] = [ (36h² - 147) - 49 ] / (36h² - 147) ) = 1 - 49/(36h² - 147)But this might not be helpful. However, we need to check if this t is within [0,1].So for t to be in [0,1], the numerator and denominator must have the same sign.Case 1: 36h² - 147 > 0 => h² > 147/36 ≈ 4.0833 => h > sqrt(147/36) ≈ 2.0412.But earlier we had h ≤ 7/3 ≈ 2.333. So if h is between approximately 2.0412 and 2.333, denominator is positive. Then numerator is 36h² - 196. Let's see when numerator is positive:36h² - 196 > 0 => h² > 196/36 ≈ 5.444 => h > sqrt(196/36) ≈ 2.357. But 2.357 is greater than 7/3 ≈2.333, which was our upper bound. Therefore, numerator is negative when denominator is positive. Therefore, t would be negative, which is outside [0,1]. Therefore, no intersection on edge AA₁.Case 2: 36h² - 147 < 0 => h² < 147/36 ≈4.0833. Then denominator is negative. Numerator 36h² - 196 is also negative (since 36h² < 147 <196). Thus, t = positive/positive = positive. Let's compute t:t = (36h² - 196)/(36h² - 147)Since both numerator and denominator are negative, t is positive. But we need t ≤1.So,(36h² - 196)/(36h² - 147) ≤1Multiply both sides by (36h² - 147), which is negative, so inequality flips:36h² - 196 ≥ 36h² - 147Subtract 36h²:-196 ≥ -147Which is false. Therefore, no solution. Therefore, t >1 in this case. Therefore, plane Π does not intersect edge AA₁.Therefore, plane Π intersects edge AA₁ only if t is between 0 and1, but our calculations show it does not. So plane Π does not intersect edge AA₁.Similarly, check intersection with edge DD₁:Edge DD₁: from D(-3.5, 3.5, 0) to D₁(-0.5, 0.5, 6h/7). Parametrize as:x = -3.5 + 3ty = 3.5 - 3tz = 0 + (6h/7)t, t ∈ [0,1]Plug into plane equation:-7x + 6h z + 3.5 - (36h²)/7 = 0Substitute:-7*(-3.5 +3t) +6h*(6h/7 t) +3.5 -36h²/7=0Calculate:24.5 -21t + (36h²/7)t +3.5 -36h²/7=0Same as edge AA₁:28 -21t + (36h²/7)t -36h²/7=0Which is the same equation as before, leading to the same t. Therefore, same result: no intersection within t ∈[0,1].Now check edge BB₁: from B(3.5, -3.5, 0) to B₁(0.5, -0.5, 6h/7). Parametrize as:x=3.5 -3ty=-3.5 +3tz=0 + (6h/7)t, t ∈[0,1]Plug into plane equation:-7x +6h z +3.5 -36h²/7=0Substitute:-7*(3.5 -3t) +6h*(6h/7 t) +3.5 -36h²/7=0Compute:-24.5 +21t + (36h²/7)t +3.5 -36h²/7=0Combine constants:-24.5 +3.5 = -21Combine terms:21t + (36h²/7)t -36h²/7=21So:t*(21 +36h²/7) =21 +36h²/7Thus,t= (21 +36h²/7)/(21 +36h²/7)=1Therefore, t=1 is on edge BB₁, which is point B₁. But B₁ is already on plane Π, so the intersection is at B₁, which is expected since plane Π contains edge B₁C₁.Similarly, check edge CC₁: from C(3.5, 3.5, 0) to C₁(0.5, 0.5, 6h/7). Parametrize as:x=3.5 -3ty=3.5 -3tz=0 + (6h/7)t, t ∈[0,1]Plug into plane equation:-7x +6h z +3.5 -36h²/7=0Substitute:-7*(3.5 -3t) +6h*(6h/7 t) +3.5 -36h²/7=0Compute:-24.5 +21t + (36h²/7)t +3.5 -36h²/7=0Same as edge BB₁:-21 +21t + (36h²/7)t -36h²/7=0Which gives:21t + (36h²/7)t =21 +36h²/7t*(21 +36h²/7)=21 +36h²/7Thus, t=1, which is point C₁, already on the plane.Therefore, plane Π intersects the lateral edges only at B₁ and C₁, which are on the upper base.Additionally, we need to check intersection with the lateral faces. For example, face AA₁B₁B. Since plane Π contains edge B₁C₁ and is perpendicular to plane AD₁C, it may intersect other parts of the frustum.Alternatively, since the plane contains edge B₁C₁ and is perpendicular to plane AD₁C, the intersection with the frustum is likely a quadrilateral formed by edge B₁C₁ and another edge from the lower base.Alternatively, maybe the intersection is a triangle or another polygon. To determine this, let's check where plane Π intersects the lower base.Earlier, we found that at z=0, plane Π intersects the lower base at x=0.5 - (36h²)/49. Let's denote this x-coordinate as x0 = 0.5 - (36h²)/49.Given that the lower base at z=0 has x ranging from -3.5 to 3.5, x0 must be within this interval. As previously calculated, this requires h ≤7/3.So, the line of intersection of plane Π with the lower base is at x=x0, z=0, y can be anything? Wait, no. Wait, plane Π intersects the lower base (which is the square ABCD at z=0) along a line. Since the plane equation at z=0 is -7x + 3.5 - (36h²)/7 =0, solving for x gives x= (3.5 - (36h²)/7)/7=0.5 - (36h²)/49. But this is a constant x, so the intersection line is x= x0, z=0, with y ranging from -3.5 to 3.5? Wait, no.Wait, the lower base is the square ABCD at z=0, with x and y ranging from -3.5 to 3.5. The plane Π intersects this square at x= x0, but for a plane to intersect a square in 3D space, it could be a line. However, since the plane Π has a normal vector with y-component zero, the intersection with the lower base (z=0) is indeed a vertical line x= x0, y arbitrary, but within the square. But in reality, the square is bounded in y from -3.5 to 3.5, so the intersection is the line segment from (x0, -3.5, 0) to (x0, 3.5, 0).But in reality, the plane Π contains edge B₁C₁ and is passing through the lower base at x=x0. Therefore, the intersection of plane Π with the frustum is a quadrilateral: edge B₁C₁ on the upper base, and the line segment from (x0, -3.5, 0) to (x0, 3.5, 0) on the lower base, connected by two other edges along the intersection of Π with the lateral faces.But wait, plane Π contains edge B₁C₁, which is from (0.5, -0.5, 6h/7) to (0.5, 0.5, 6h/7). The intersection with the lower base is the line x=x0, z=0, y from -3.5 to 3.5. But how does the plane connect B₁C₁ to this line?Actually, the intersection of plane Π with the frustum is a polygon. Let's find the exact intersection points.Plane Π intersects the lateral face BB₁C₁C. Let's parametrize this face.Face BB₁C₁C is a trapezoid with vertices B(3.5, -3.5, 0), B₁(0.5, -0.5, 6h/7), C₁(0.5, 0.5, 6h/7), C(3.5, 3.5, 0).To find the intersection of plane Π with this face, we can parametrize the face and find where it meets the plane.Parametrize the face as two triangles or use bilinear coordinates. Alternatively, note that the face consists of the line segments BB₁, B₁C₁, C₁C, and CB.But plane Π already contains B₁C₁, so the intersection on this face is B₁C₁ itself.Now, check intersection with face AA₁D₁D. Wait, plane Π's normal vector is (-7,0,6h). The plane Π might intersect other faces.Wait, let's check intersection with face AA₁B₁B. This face is a trapezoid with vertices A(-3.5, -3.5, 0), A₁(-0.5, -0.5, 6h/7), B₁(0.5, -0.5, 6h/7), B(3.5, -3.5, 0).Parametrize this face. Let's consider a point on this face. Points on this face can be parametrized with parameters s and t, but it's complicated. Alternatively, find the intersection of plane Π with the edges of this face.We already checked edges AA₁ and BB₁; plane Π doesn't intersect them except at B₁. Therefore, the intersection with face AA₁B₁B is only at B₁.Similarly, check intersection with face DD₁C₁C. This face has vertices D(-3.5, 3.5, 0), D₁(-0.5, 0.5, 6h/7), C₁(0.5, 0.5, 6h/7), C(3.5, 3.5, 0). Plane Π contains edge B₁C₁, which is not part of this face. To check intersection with this face, we need to see if plane Π intersects any edges of this face.The edges are DD₁, D₁C₁, C₁C, and CD. Plane Π contains C₁, which is part of edge D₁C₁ and C₁C. But edge D₁C₁ is from D₁ to C₁, both at z=6h/7. Since plane Π contains C₁, but D₁ is not on Π. Let's check if plane Π intersects edge D₁C₁.Edge D₁C₁ is from D₁(-0.5, 0.5, 6h/7) to C₁(0.5, 0.5, 6h/7). Parametrize as x = -0.5 + t(1), y=0.5, z=6h/7, t∈[0,1].Plug into plane equation:-7x +6h z +3.5 -36h²/7=0Substitute:-7*(-0.5 + t) +6h*(6h/7) +3.5 -36h²/7=03.5 -7t + (36h²)/7 +3.5 -36h²/7=0Simplify:7 -7t=0 => t=1Thus, intersection at t=1, which is point C₁, already on the plane.Similarly, edge C₁C: from C₁(0.5,0.5,6h/7) to C(3.5,3.5,0). Parametrize as:x=0.5 +3ty=0.5 +3tz=6h/7 -6h/7 t, t ∈[0,1]Plug into plane equation:-7x +6h z +3.5 -36h²/7=0Substitute:-7*(0.5 +3t) +6h*(6h/7 -6h/7 t) +3.5 -36h²/7=0Calculate term by term:-3.5 -21t + (36h²/7 -36h²/7 t) +3.5 -36h²/7=0Simplify:(-3.5 +3.5) + (-21t) + (36h²/7 -36h²/7 t) -36h²/7=0This simplifies to:-21t -36h²/7 t=0Factor:t*(-21 -36h²/7)=0Therefore, t=0. But t=0 corresponds to point C₁, which is on the plane. Therefore, plane Π intersects edge C₁C only at C₁.Similarly, check edge CD: from C(3.5,3.5,0) to D(-3.5,3.5,0). Parametrize as x=3.5 -7t, y=3.5, z=0, t ∈[0,1]Plug into plane equation:-7x +6h*0 +3.5 -36h²/7=0=> -7*(3.5 -7t) +0 +3.5 -36h²/7=0=> -24.5 +49t +3.5 -36h²/7=0=> -21 +49t -36h²/7=0=>49t=21 +36h²/7=>t=(21 +36h²/7)/49=(3 + 36h²/49)/7But for t ∈[0,1], the value (3 +36h²/49)/7 must be between 0 and1.Compute:Minimum when h=0: t=3/7≈0.428Maximum when h=7/3: t=(3 +36*(49/9)/49)/7=(3 +4)/7=1Therefore, t ranges from 3/7 to1 as h ranges from0 to7/3.Therefore, plane Π intersects edge CD at point:x=3.5 -7t, y=3.5, z=0Where t=(3 +36h²/49)/7So coordinates:x=3.5 -7*(3 +36h²/49)/7=3.5 - (3 +36h²/49)=0.5 -36h²/49Which matches our previous result for x0=0.5 -36h²/49.Thus, the intersection point on edge CD is (x0,3.5,0), where x0=0.5 -36h²/49.Similarly, check intersection with edge DA: from D(-3.5,3.5,0) to A(-3.5,-3.5,0). Parametrize as x=-3.5, y=3.5 -7t, z=0, t ∈[0,1]Plug into plane equation:-7*(-3.5) +6h*0 +3.5 -36h²/7=024.5 +0 +3.5 -36h²/7=0=>28 -36h²/7=0=>36h²/7=28=>h²=28*7/36=196/36=49/9=>h=7/3≈2.333...But h=7/3 is the upper bound we found earlier. Therefore, if h=7/3, plane Π intersects edge DA at all points since the equation is satisfied for any y when x=-3.5 and z=0. But this would only happen if the entire edge DA lies on plane Π, which would require the plane equation to be satisfied for all y when x=-3.5 and z=0. But substituting x=-3.5, z=0 into plane equation:-7*(-3.5) +6h*0 +3.5 -36h²/7=24.5 +0 +3.5 -36h²/7=28 -36h²/7=0Which is only true when h=7/3. Therefore, if h=7/3, plane Π contains edge DA. Otherwise, it does not intersect edge DA.But we don't know h yet. However, according to the problem statement, the plane Π divides the frustum into two parts of equal volume. So if h=7/3, plane Π would contain edge DA and B₁C₁, making the division symmetric? Not sure.But perhaps we need to use the condition that the volume split is equal. To compute the volume of one part, perhaps we can consider the frustum divided by plane Π into two parts: one part is a polyhedron bounded by B₁C₁, the intersection line on the lower base, and parts of the lateral faces.This seems complex, but maybe we can use the principle of similar figures or coordinate integration.Alternatively, consider that the frustum's volume is 114h/7, and each part must be 57h/7. The plane Π divides it into two parts, and we need to find h such that this condition holds.Let me consider setting up an integral to compute the volume of one part.The frustum can be described in coordinates as follows:At any height z between 0 and 6h/7, the cross-section is a square with side length s(z) =7 -7z/h.The plane Π is given by x = (6h z)/7 -36h²/49 +0.5.We need to compute the volume of the region where x ≥ (6h z)/7 -36h²/49 +0.5.This is equivalent to integrating over the frustum the area where x ≥ x0(z), with x0(z) = (6h z)/7 -36h²/49 +0.5.At each height z, the cross-sectional square has side length s(z) =7 -7z/h, centered at x=0, y=0. The leftmost x is -s(z)/2, and the rightmost x is s(z)/2.The plane Π intersects this cross-section at x=x0(z). If x0(z) is within [-s(z)/2, s(z)/2], then the area to the right of x0(z) is a rectangle from x0(z) to s(z)/2, with y ranging from -s(z)/2 to s(z)/2.Therefore, the area at height z is [s(z)/2 - x0(z)] * s(z).Integrate this from z=0 to z=6h/7 and set equal to 57h/7.Wait, but let's verify:The cross-sectional area at height z is a square with side s(z). The plane slices this square at x=x0(z). The area to the right of x=x0(z) is a rectangle with width (s(z)/2 - x0(z)) and height s(z). Wait, no. In 2D, slicing a square at x=x0, the area to the right is a rectangle with width (s(z)/2 - x0(z)) and height s(z). But actually, the cross-section is a square in x and y. If we slice at x=x0(z), the area to the right is the area of the square where x ≥ x0(z). Which is a rectangle from x=x0(z) to x=s(z)/2, with y from -s(z)/2 to s(z)/2. Therefore, area is (s(z)/2 - x0(z)) * s(z).Therefore, the volume is the integral from z=0 to z=6h/7 of (s(z)/2 - x0(z)) * s(z) dz.Set this equal to 57h/7.Let's write this integral.First, express s(z) =7 -7z/h.x0(z) = (6h z)/7 -36h²/49 +0.5.Compute (s(z)/2 - x0(z)) * s(z):= [ (7 -7z/h)/2 - (6h z)/7 +36h²/49 -0.5 ] * (7 -7z/h)Simplify term by term:First, expand (7 -7z/h)/2:= 7/2 - (7z)/(2h)Then subtract x0(z):= 7/2 -7z/(2h) -6h z/7 +36h²/49 -0.5Combine constants:7/2 -0.5 =3.5 -0.5=3Combine terms with z:-7z/(2h) -6h z/7Combine terms with h²:36h²/49So expression becomes:3 - z*(7/(2h) +6h/7) +36h²/49Multiply by s(z)=7 -7z/h:Volume = ∫[0 to 6h/7] [3 - z*(7/(2h) +6h/7) +36h²/49]*(7 -7z/h) dzThis integral looks complicated, but perhaps we can expand it and integrate term by term.Let me denote:A =3 +36h²/49B = - (7/(2h) +6h/7)So the integrand becomes (A + Bz)(7 -7z/h)Expand:=7A +7Bz -7A z/h -7B z²/hThen integrate term by term:Integral from 0 to6h/7 of [7A +7Bz -7A z/h -7B z²/h] dzCompute each term:1. ∫7A dz =7A z from0 to6h/7=7A*(6h/7)=6A h2. ∫7Bz dz=7B*(z²/2) from0 to6h/7=7B*( (36h²/49)/2 )=7B*(18h²/49)= (126B h²)/49= (18B h²)/73. ∫-7A z/h dz= -7A/h * ∫z dz= -7A/h*(z²/2) from0 to6h/7= -7A/h*(36h²/49 /2)= -7A/h*(18h²/49)= - (126A h)/49= - (18A h)/74. ∫-7B z²/h dz= -7B/h * ∫z² dz= -7B/h*(z³/3) from0 to6h/7= -7B/h*( (216h³)/343 /3)= -7B/h*(72h³/343)= - (504B h²)/343= - (72B h²)/49Therefore, total integral:6A h + (18B h²)/7 - (18A h)/7 - (72B h²)/49Combine like terms:A terms:6A h -18A h/7= (42A h -18A h)/7=24A h/7B terms:(18B h²)/7 -72B h²/49= (126B h² -72B h²)/49=54B h²/49Thus, Volume=24A h/7 +54B h²/49Now substitute back A=3 +36h²/49 and B= - (7/(2h) +6h/7):First compute 24A h/7:24*(3 +36h²/49)*h/7=24h/7*(3 +36h²/49)= (72h)/7 + (864h³)/343Then compute54B h²/49:54*(-7/(2h) -6h/7)*h²/49=54*(-7/(2h) *h² -6h/7*h²)/49=54*(-7h/2 -6h³/7)/49=54*(-7h/2 -6h³/7)/49Simplify:=54*(-7h/2 -6h³/7)/49=54*(- (49h +12h³)/14 )/49=54*(-49h -12h³)/(14*49)=54*(-49h -12h³)/(686)= (-54*(49h +12h³))/686Simplify numerator and denominator:54 and686: 54=2*27, 686=2*343=2*7³. So divide numerator and denominator by2:(-27*(49h +12h³))/343Therefore, Volume= (72h/7 +864h³/343) -27*(49h +12h³)/343Convert 72h/7 to denominator343:72h/7= (72h*49)/343= 3528h/343864h³/343 remains.So Volume=3528h/343 +864h³/343 -27*(49h +12h³)/343Factor out1/343:[3528h +864h³ -1323h -324h³]/343Combine like terms:h terms:3528h -1323h=2205hh³ terms:864h³ -324h³=540h³Thus, Volume=(2205h +540h³)/343Factor numerator:2205=45*49=45*7²540=540=54*10=6*9*10=2*3³*5But perhaps factor common terms:2205h +540h³=45*49h +540h³=45*49h +540h³=45h(49) +540h³=45h*49 +540h³=45h*49 +540h³=45h*49 +12*45h³=45h(49 +12h²)Thus, Volume=45h(49 +12h²)/343Set this equal to half the frustum's volume, which is57h/7:45h(49 +12h²)/343 =57h/7Cancel h from both sides (h≠0):45(49 +12h²)/343 =57/7Multiply both sides by343:45(49 +12h²)=57/7 *343Calculate 343/7=49, so:45(49 +12h²)=57*49Divide both sides by49:45(1 +12h²/49)=57Divide both sides by45:1 +12h²/49=57/45=19/15Subtract1:12h²/49=19/15 -1=4/15Multiply both sides by49/12:h²= (4/15)*(49/12)= (196)/(180)=49/45Thus, h=7/√45=7/(3√5)=7√5/15Therefore, h=7√5/15Now, the volume of the original pyramid is (1/3)*base area*height.Base area is7^2=49.Volume= (1/3)*49*(7√5/15)= (343√5)/45But let's check:Wait, original pyramid volume is (1/3)*base area*height=(1/3)*49*(7√5/15)= (343√5)/45.But let me verify the calculation:h=7√5/15Volume= (1/3)*49*(7√5/15)=49*7√5/(3*15)=343√5/45.Yes.But let's check if this is the answer. However, we need to ensure that we didn't make any mistakes in the integral setup and computation.Wait, let me recap:We set up the integral for the volume of the region x ≥x0(z) in the frustum, which is sliced by plane Π. The integral was set equal to half the frustum's volume.Computed the integral and found h=7√5/15. Then computed the original pyramid's volume as343√5/45.But let's confirm the frustum's volume with h=7√5/15:Height of frustum H=6h/7=6*(7√5/15)/7=6√5/15=2√5/5.Volume of frustum=(1/3)*H*(49 +1 +sqrt(49*1))=(1/3)*(2√5/5)*(50 +7)=(1/3)*(2√5/5)*57=(114√5)/15= (38√5)/5.Half of this volume is19√5/5.Now, compute the integral result with h=7√5/15:Volume=(2205h +540h³)/343.Plugging h=7√5/15:2205h=2205*(7√5/15)=2205/15*7√5=147*7√5=1029√5540h³=540*(343*5√5)/(3375)= wait, compute h³:h³=(7√5/15)^3=343*5^(3/2)/3375=343*5√5/3375Thus, 540h³=540*(343*5√5)/3375=540*343*5√5/(3375)= (540/3375)*343*5√5= (540/675)*343*√5= (0.8)*343√5=274.4√5Wait, maybe this is not the best way.Alternatively, compute h=7√5/15≈7*2.236/15≈15.652/15≈1.043.Then h³≈1.043³≈1.134.So 2205h≈2205*1.043≈2298.015540h³≈540*1.134≈612.36Sum≈2298.015+612.36≈2910.375Divide by343≈2910.375/343≈8.485.Which is approximately equal to19√5/5≈19*2.236/5≈42.484/5≈8.4968. Close enough, considering rounding errors. So the calculation is correct.Therefore, the original pyramid's volume is343√5/45.But let's simplify this:343=7³,45=9*5.343/45 cannot be simplified, so volume is343√5/45.But perhaps rationalizing or writing it differently?Alternatively, 343/45 is approximately7.622, but exact form is fine.Therefore, the answer is343√5⁄45, which can be written asboxed{dfrac{343sqrt{5}}{45}}.Final Answerboxed{dfrac{343sqrt{5}}{45}}</think>Given a regular truncated quadrilateral pyramid (frustum) with an upper base side length of 1 and a lower base side length of 7, we need to find the volume of the original pyramid. A plane passing through edge ( B_1C_1 ) and perpendicular to plane ( AD_1C ) divides the frustum into two parts of equal volume.1. Coordinates and Plane Equations: - The lower base vertices are ( A(-3.5, -3.5, 0) ), ( B(3.5, -3.5, 0) ), ( C(3.5, 3.5, 0) ), ( D(-3.5, 3.5, 0) ). - The upper base vertices are ( A_1(-0.5, -0.5, 6h/7) ), ( B_1(0.5, -0.5, 6h/7) ), ( C_1(0.5, 0.5, 6h/7) ), ( D_1(-0.5, 0.5, 6h/7) ).2. Plane AD₁C: - Vectors ( overrightarrow{AD_1} = (3, 4, 6h/7) ) and ( overrightarrow{AC} = (7, 7, 0) ). - Normal vector ( overrightarrow{n_1} = (-6h, 6h, -7) ).3. Plane Π through B₁C₁: - Normal vector ( overrightarrow{n_2} = (-7, 0, 6h) ). - Equation of plane Π: ( -7x + 6hz + 3.5 - frac{36h^2}{7} = 0 ).4. Volume Calculation: - The volume of the frustum is split equally by plane Π. - The height of the frustum is ( frac{6h}{7} ). - The volume of the frustum is ( frac{114h}{7} ), and half of this volume is ( frac{57h}{7} ).5. Integral Setup and Solving: - The integrand for the volume of one part is computed and solved for ( h ). - Solving the integral and equating to half the frustum's volume gives ( h = frac{7sqrt{5}}{15} ).6. Original Pyramid Volume: - The volume of the original pyramid is ( frac{1}{3} times 49 times frac{7sqrt{5}}{15} = frac{343sqrt{5}}{45} ).The final answer is:[boxed{dfrac{343sqrt{5}}{45}}],<think>Okay, so I need to find the volume of a regular truncated quadrilateral pyramid (also called a frustum). The upper base has a side length of 1, and the lower base has a side length of 7. There's also a plane passing through edge B₁C₁ that's perpendicular to the plane AD₁C, which splits the pyramid into two parts with equal volume. The answer is supposed to be the total volume of the pyramid, right?First, let me recall that the volume of a frustum is given by the formula:[ V = frac{1}{3} h (S_1 + S_2 + sqrt{S_1 S_2}) ]where ( S_1 ) and ( S_2 ) are the areas of the two bases, and h is the height. But in this problem, the height isn't given directly. However, the information about the plane cutting through B₁C₁ and being perpendicular to AD₁C might help determine the height or some other necessary parameter. Since the problem states that this plane divides the frustum into two equal volumes, I need to use this condition to find h, and then compute the volume.But wait, the problem mentions a "regular truncated quadrilateral pyramid". Regular usually means that the base is a regular polygon, and the pyramid is right (i.e., the apex is directly above the center of the base). However, since it's truncated, the upper base is a smaller regular quadrilateral (square) parallel to the lower base. The lateral edges are AA₁, BB₁, CC₁, DD₁. So each lateral edge connects a vertex of the lower base to the corresponding vertex of the upper base.Given that the sides of the upper and lower bases are 1 and 7, respectively. So the lower base is a square with side 7, upper base a square with side 1.Now, the plane passes through edge B₁C₁ and is perpendicular to the plane AD₁C. This plane divides the frustum into two parts of equal volume. We need to use this condition to find the height, and then compute the volume.First, maybe I should visualize the frustum. Let me denote the original pyramid before truncation. Let’s say the original pyramid has base ABCD with side 7, and when we cut it with a plane parallel to the base, we get the upper base A₁B₁C₁D₁ with side 1. The height of the frustum is h, and the height of the original pyramid would be H = h + h', where h' is the height from the upper base to the apex.Since the pyramids are similar, the ratio of the sides is 1/7, so the ratio of heights is also 1/7. Therefore, h' = H / 7, and the height of the frustum is h = H - h' = H - H/7 = (6/7) H. Thus, H = (7/6) h. But maybe I need to express things in terms of h, the height of the frustum.Alternatively, perhaps it's better to model the frustum in coordinates.Let me set up a coordinate system. Let’s place the lower base ABCD on the xy-plane, centered at the origin for simplicity. So, since it's a square with side 7, the coordinates can be:A = (-7/2, -7/2, 0)B = (7/2, -7/2, 0)C = (7/2, 7/2, 0)D = (-7/2, 7/2, 0)Then the upper base A₁B₁C₁D₁ is a square with side 1, parallel to the base, and shifted up along the z-axis by height h. The coordinates of the upper base would be scaled by the ratio 1/7 and centered at the same center (since it's a regular truncated pyramid). Therefore, their coordinates are:A₁ = (-1/2, -1/2, h)B₁ = (1/2, -1/2, h)C₁ = (1/2, 1/2, h)D₁ = (-1/2, 1/2, h)Wait, but actually, if the original pyramid had height H, then the truncation occurs at height H - h from the base, so the scaling factor would be (H - h)/H = 1/7, so H - h = H/7 => h = (6/7) H. Hence, h = 6H/7. So the upper base is at height h_truncated = H - h = H - 6H/7 = H/7. Wait, maybe I need to clarify this.Alternatively, perhaps the frustum's height is h, and the original pyramid's total height is H. Then the upper base is a similar pyramid with height H - h. The ratio of similarity is (H - h)/H, which equals the ratio of the sides, 1/7. So:(H - h)/H = 1/7 => 1 - h/H = 1/7 => h/H = 6/7 => H = (7/6) h.So, the original pyramid's height is 7/6 times the frustum's height.But I need to find the volume of the frustum. The volume of the frustum is the difference between the original pyramid and the smaller pyramid that was removed. The original volume is (1/3) * (7)^2 * H, and the removed pyramid's volume is (1/3) * (1)^2 * (H - h). Therefore, the frustum's volume is:V = (1/3) * 49 * H - (1/3) * 1 * (H - h) = (1/3)(49H - (H - h)) = (1/3)(48H + h)But since H = (7/6) h, substitute:V = (1/3)(48*(7/6 h) + h) = (1/3)(56 h + h) = (1/3)(57 h) = 19 h.Alternatively, using the frustum volume formula:V = (1/3) h (S1 + S2 + sqrt(S1 S2)) = (1/3) h (49 + 1 + sqrt(49*1)) = (1/3) h (49 + 1 + 7) = (1/3) h (57) = 19 h.So that's consistent. Therefore, the volume of the frustum is 19 h. So if we can find h, then we can find the volume. The problem is to find the volume, so we need to determine h using the condition about the plane.The plane passes through edge B₁C₁ and is perpendicular to the plane AD₁C. It divides the frustum into two parts of equal volume. So, this plane is cutting the frustum into two regions, each with volume 19 h / 2. So, the problem is to use the geometric condition given to set up an equation that allows us to solve for h.First, let's try to find the equation of the plane passing through B₁C₁ and perpendicular to plane AD₁C.To do this, we need coordinates of the points involved. Let me define coordinates as follows:Let’s place the lower base ABCD on the xy-plane with coordinates:- A: (-7/2, -7/2, 0)- B: (7/2, -7/2, 0)- C: (7/2, 7/2, 0)- D: (-7/2, 7/2, 0)Upper base A₁B₁C₁D₁ is a square of side 1, centered at the origin (since it's a regular truncation), at height h. So:- A₁: (-1/2, -1/2, h)- B₁: (1/2, -1/2, h)- C₁: (1/2, 1/2, h)- D₁: (-1/2, 1/2, h)Now, the plane in question passes through edge B₁C₁. Edge B₁C₁ goes from (1/2, -1/2, h) to (1/2, 1/2, h). So, this edge is a vertical line segment in the upper base, along x = 1/2, z = h, from y = -1/2 to y = 1/2.Wait, in the upper base, which is a square of side 1, the coordinates are correct. So, B₁ is (1/2, -1/2, h), and C₁ is (1/2, 1/2, h). So the edge B₁C₁ is along the line x = 1/2, z = h, varying y from -1/2 to 1/2.Now, the plane passes through this entire edge. So, any plane containing B₁C₁ must contain all points along that edge. Additionally, the plane is perpendicular to the plane AD₁C. So, we need to find a plane that contains B₁C₁ and is perpendicular to plane AD₁C.First, let's find the equation of plane AD₁C.Points A, D₁, and C:- A: (-7/2, -7/2, 0)- D₁: (-1/2, 1/2, h)- C: (7/2, 7/2, 0)So, three points: A(-7/2, -7/2, 0), D₁(-1/2, 1/2, h), C(7/2, 7/2, 0)To find the equation of plane AD₁C, we can compute two vectors in the plane and find the normal vector.Vector AD₁: D₁ - A = [(-1/2 + 7/2), (1/2 + 7/2), h - 0] = (3, 4, h)Vector AC: C - A = [(7/2 + 7/2), (7/2 + 7/2), 0 - 0] = (7, 7, 0)Now, the normal vector to plane AD₁C is the cross product of AD₁ and AC.Compute cross product AD₁ × AC:|i    j    k||3    4    h||7    7    0|= i*(4*0 - h*7) - j*(3*0 - h*7) + k*(3*7 - 4*7)= i*(-7h) - j*(-7h) + k*(21 - 28)= (-7h, 7h, -7)So the normal vector is (-7h, 7h, -7). We can simplify by dividing by -7: (h, -h, 1). So the normal vector is (h, -h, 1). Therefore, the equation of plane AD₁C is:h(x - x_A) - h(y - y_A) + 1(z - z_A) = 0Using point A(-7/2, -7/2, 0):h(x + 7/2) - h(y + 7/2) + 1(z - 0) = 0Simplify:h x + (7h)/2 - h y - (7h)/2 + z = 0Simplify terms:h x - h y + z = 0So the equation of plane AD₁C is h x - h y + z = 0.Now, the plane we are interested in passes through edge B₁C₁ and is perpendicular to plane AD₁C. For two planes to be perpendicular, their normal vectors must be perpendicular.The normal vector of plane AD₁C is (h, -h, 1). Let’s denote the normal vector of our desired plane (the one through B₁C₁) as n = (a, b, c). The dot product of the two normals must be zero:(h)(a) + (-h)(b) + (1)(c) = 0=> h a - h b + c = 0.Additionally, the desired plane contains edge B₁C₁. Edge B₁C₁ is the line segment from B₁(1/2, -1/2, h) to C₁(1/2, 1/2, h). So, the plane must contain all points with x = 1/2, z = h, and y varying between -1/2 and 1/2. Therefore, the plane's equation must satisfy x = 1/2 and z = h for all points on B₁C₁. However, the plane is not necessarily vertical; it could extend beyond B₁C₁.Wait, but actually, the plane contains the entire line B₁C₁. So, parametrizing the line B₁C₁: all points (1/2, y, h) where y ranges from -1/2 to 1/2. So, the plane must contain this line. Therefore, the equation of the plane can be determined by this line and another point not on the line, but since it's a plane in space, we need another direction.But we also know that the plane is perpendicular to plane AD₁C. So, the two planes are perpendicular, so their normals are perpendicular. Let’s see.Alternatively, the plane contains the line B₁C₁ and is perpendicular to plane AD₁C. Therefore, the direction vector of line B₁C₁ is along the y-axis (since x and z are fixed). The direction vector is (0, 1, 0). The normal vector of plane AD₁C is (h, -h, 1). The plane we are seeking must contain line B₁C₁ and be perpendicular to plane AD₁C. So, perhaps the normal vector of our desired plane can be found by taking the cross product of the direction vector of B₁C₁ and the normal vector of plane AD₁C.Wait, if the desired plane is perpendicular to plane AD₁C, then their normals are perpendicular. Alternatively, if the plane contains line B₁C₁, which is in the direction (0,1,0), and is perpendicular to plane AD₁C, then the normal vector of the desired plane should be parallel to the cross product of the direction vector of B₁C₁ and the normal vector of AD₁C.Let me recall that if two planes are perpendicular, their normals are also perpendicular. So the normal vector of the desired plane must be perpendicular to the normal vector of plane AD₁C. So, the dot product of the two normals must be zero.But also, the desired plane contains the line B₁C₁. Let me think. Let me try to find the equation of the desired plane.Given that the plane contains the line B₁C₁, which is x = 1/2, z = h. So any point on this line is (1/2, y, h) where y ∈ [-1/2, 1/2]. Additionally, the plane is perpendicular to plane AD₁C. So, the two planes' normals are perpendicular.Let’s denote the normal vector of the desired plane as n = (a, b, c). Then, as before:h a - h b + c = 0.Also, since the plane contains the line x = 1/2, z = h, we can plug this into the plane equation. The general equation of a plane is a x + b y + c z + d = 0. Since the line x = 1/2, z = h lies on the plane, substituting x = 1/2, z = h into the equation gives:a*(1/2) + b y + c*h + d = 0 for all y. Therefore, the coefficient of y must be zero, so b = 0. Also, the remaining terms must be zero for all y, so:( a/2 + c h + d ) + 0*y = 0 => a/2 + c h + d = 0.But since the plane contains the entire line, this equation must hold for any y, which is already satisfied by b = 0, and the rest must be a constant. However, since (1/2, y, h) must satisfy the plane equation for all y, the equation reduces to a*(1/2) + 0*y + c*h + d = 0. Therefore, this is a single equation involving a, c, d. But we need more conditions to find the plane.Additionally, the plane must be perpendicular to plane AD₁C, whose normal is (h, -h, 1). So, the normal vectors must be perpendicular:h a - h b + c = 0. But since we found b = 0, this simplifies to h a + c = 0 => c = -h a.So, substituting c = -h a into the equation from the line:a*(1/2) + (-h a)*h + d = 0 => (a/2) - h² a + d = 0 => d = (h² - 1/2) a.Therefore, the plane equation is:a x + 0*y + (-h a) z + (h² - 1/2) a = 0.We can factor out a (assuming a ≠ 0):a [x - h z + (h² - 1/2)] = 0.Since a ≠ 0, the equation simplifies to:x - h z + (h² - 1/2) = 0.Therefore, the equation of the desired plane is:x - h z + (h² - 1/2) = 0.Alternatively, rearranged:x - h z = 1/2 - h².So, that's the plane passing through edge B₁C₁ and perpendicular to plane AD₁C.Now, this plane divides the frustum into two parts of equal volume. So, the volume of the part on one side of the plane is 19h / 2.We need to compute the volume of one of the parts and set it equal to 19h / 2, then solve for h. But how do we compute the volume cut by this plane?First, perhaps we need to find the intersection of this plane with the frustum. The frustum has two bases (lower ABCD, upper A₁B₁C₁D₁) and four trapezoidal faces.The plane passes through edge B₁C₁ of the upper base, so it will intersect some other parts of the frustum. Since it's a plane cutting through the frustum, the intersection will be a polygon, possibly a quadrilateral or a triangle, and the volume on one side of the plane will be a portion of the frustum.But to compute this volume, maybe we can use integration or some geometric decomposition. However, given that it's a plane cutting through the frustum, perhaps we can model the frustum as a portion of the original pyramid and find where the plane cuts the original pyramid.But wait, since the frustum is part of the pyramid, maybe extending the lateral edges to the apex would help. Let me recall that in a frustum, if we know the side lengths of the bases, we can find the height h of the frustum and the height H of the original pyramid.But we already saw that H = (7/6) h. However, maybe this plane cuts the original pyramid, not just the frustum. Wait, the problem states that the plane divides the pyramid into two parts of equal volume. Wait, hold on, the problem says "the pyramid" but it's a truncated pyramid (frustum). Wait, the original problem says: "a regular truncated quadrilateral pyramid... divides the pyramid into two parts of equal volume." So, perhaps they refer to the frustum as the pyramid here. But in the problem statement, it's called a truncated pyramid, which is a frustum.Wait, maybe the problem is actually referring to the original pyramid before truncation? But no, the truncation is given with the upper base A₁B₁C₁D₁ of side 1 and lower base ABCD of side 7. So the frustum is the truncated pyramid, and the problem is about this frustum. However, the wording says "the pyramid" which might be confusing. Wait, let me check the original problem again:"In a regular truncated quadrilateral pyramid with lateral edges AA₁, BB₁, CC₁, DD₁, the side of the upper base A₁B₁C₁D₁ is 1, and the side of the lower base is 7. A plane passing through the edge B₁C₁ perpendicular to the plane AD₁C divides the pyramid into two parts of equal volume. Find the volume of the pyramid."Wait, "the pyramid" here refers to the truncated pyramid (the frustum). So the frustum's volume is to be found, given that a certain plane divides it into two equal volumes. So, the total volume is 19h, and we need to find h such that the plane cuts it into two parts each of volume 19h / 2.So, how do we compute the volume cut by this plane?Alternatively, maybe it's easier to consider the entire original pyramid, compute the volume ratio, but since the frustum is part of the pyramid, maybe we need to model the intersection of the cutting plane with the original pyramid? Hmm, not sure.Alternatively, perhaps parametrize the frustum and set up an integral.Alternatively, find the intersection of the cutting plane with the lateral edges or the lower base, and then decompose the frustum into simpler solids whose volumes can be calculated.Let me try to find the intersection of the cutting plane with the frustum. The plane passes through B₁C₁ and is given by the equation x - h z = 1/2 - h².We need to find where else this plane intersects the frustum. The frustum has the lower base ABCD on the xy-plane (z=0) and the upper base at z=h. The lateral edges are AA₁, BB₁, CC₁, DD₁.So, the plane passes through B₁C₁ (which is at z=h). Let's check if it intersects any of the other edges or the lower base.First, check intersection with the lower base (z=0). Substitute z=0 into the plane equation:x - h*0 = 1/2 - h² => x = 1/2 - h².So, on the lower base (z=0), the plane intersects the line x = 1/2 - h². But the lower base is a square from x = -7/2 to x = 7/2, and y from -7/2 to 7/2. The line x = 1/2 - h² on the lower base z=0 will intersect the boundary of the lower base at y = ±7/2, provided that x = 1/2 - h² is within [-7/2, 7/2]. Since h is positive, h² is positive. Therefore, 1/2 - h² must be between -7/2 and 7/2.Assuming h is such that this is true, which it should be for a valid frustum.Similarly, check intersection with lateral edges. For example, take edge AA₁: from A(-7/2, -7/2, 0) to A₁(-1/2, -1/2, h). Parametrize this edge as:x = -7/2 + t*(6/2) = -7/2 + 3ty = -7/2 + t*(6/2) = -7/2 + 3tz = 0 + t*hfor t ∈ [0,1]Plug into plane equation x - h z = 1/2 - h²:(-7/2 + 3t) - h*(h t) = 1/2 - h²Simplify:-7/2 + 3t - h² t = 1/2 - h²Bring all terms to left:-7/2 + 3t - h² t - 1/2 + h² = 0Combine constants:-8/2 + 3t - h² t + h² = 0 => -4 + 3t - h² t + h² = 0Factor t:t*(3 - h²) + (h² - 4) = 0Solve for t:t = (4 - h²)/(3 - h²)If t ∈ [0,1], then the plane intersects edge AA₁ at this t value. Similarly, we can check for other edges.Similarly, check edge BB₁: from B(7/2, -7/2, 0) to B₁(1/2, -1/2, h). Parametrize:x = 7/2 - 3ty = -7/2 + 3tz = 0 + h tPlug into plane equation:x - h z = 7/2 - 3t - h*(h t) = 7/2 - 3t - h² tSet equal to 1/2 - h²:7/2 - 3t - h² t = 1/2 - h²Bring all terms to left:7/2 - 3t - h² t - 1/2 + h² = 0 => 6/2 - 3t - h² t + h² = 0 => 3 - 3t - h² t + h² = 0Factor t:-3t - h² t + (3 + h²) = 0 => t*(-3 - h²) + (3 + h²) = 0 => t = (3 + h²)/(3 + h²) = 1So, t=1, which is point B₁. So the plane passes through B₁, which we already knew. Similarly, check edge CC₁: from C(7/2, 7/2, 0) to C₁(1/2, 1/2, h). Parametrize:x = 7/2 - 3ty = 7/2 - 3tz = 0 + h tPlug into plane equation:x - h z = 7/2 - 3t - h*(h t) = 7/2 - 3t - h² tSet equal to 1/2 - h²:7/2 - 3t - h² t = 1/2 - h²Bring all terms to left:7/2 - 3t - h² t -1/2 + h² = 0 => 6/2 - 3t - h² t + h² = 0 => 3 -3t - h² t + h² = 0Same as edge BB₁, leads to t=1, which is point C₁. So the plane passes through C₁ as expected.Edge DD₁: from D(-7/2, 7/2, 0) to D₁(-1/2, 1/2, h). Parametrize:x = -7/2 + 3ty = 7/2 - 3tz = 0 + h tPlug into plane equation:x - h z = (-7/2 + 3t) - h*(h t) = -7/2 + 3t - h² tSet equal to 1/2 - h²:-7/2 + 3t - h² t = 1/2 - h²Bring terms to left:-7/2 + 3t - h² t -1/2 + h² = 0 => -8/2 + 3t - h² t + h² = 0 => -4 + 3t - h² t + h² = 0Factor t:t*(3 - h²) + (h² -4) =0 => t = (4 - h²)/(3 - h²)Same as edge AA₁. So, if (4 - h²)/(3 - h²) ∈ [0,1], then the plane intersects edge DD₁ at this parameter.So, the plane intersects edges AA₁ and DD₁ at t = (4 - h²)/(3 - h²), provided that this value is between 0 and 1.Let’s check when t ∈ [0,1]:Case 1: 3 - h² > 0 => h² < 3Then, numerator 4 - h² must be between 0 and 3 - h².Since 4 - h² > 0 => h² <4, which is already satisfied if h² <3.So t = (4 - h²)/(3 - h²). Since h² <3, denominator positive.Then, t = [4 - h²]/[3 - h²] = 1 + [1]/[3 - h²]. Since 3 - h² >0, this is greater than 1. Therefore, t >1, which is outside the parameter range [0,1]. Therefore, no intersection with edges AA₁ and DD₁.Case 2: 3 - h² <0 => h² >3Then, t = (4 - h²)/(3 - h²) = [negative]/[negative] = (h² -4)/(h² -3)Which is positive if h² >4, and negative if 3 <h² <4.But in this case, edges AA₁ and DD₁ are parametrized with t ∈ [0,1]. So, if h² >4, then t = (h² -4)/(h² -3). Let's see if this is between 0 and1.(h² -4)/(h² -3) = 1 - 1/(h² -3). Since h² >4, h² -3 >1, so 1/(h² -3) <1, so 1 - something <1, and greater than 0. Therefore, t ∈ (0,1) only if h² >4.Wait, but the frustum's height h must be such that the original pyramid's height is H =7h/6. The original pyramid's height must be greater than the frustum's height, so H >h => 7h/6 >h => 7/6 >1, which is true. So h can be any positive value. But if h² >4, then h >2. So if h >2, then the plane intersects edges AA₁ and DD₁ at t=(h² -4)/(h² -3). If h² <3, then the intersection is at t>1, which is outside, so plane doesn't intersect those edges. If 3 <h² <4, the t would be negative, which is also outside. So only when h² >4 (h>2) does the plane intersect edges AA₁ and DD₁ within the frustum.But in this problem, we don't know h yet. So maybe h could be greater than 2 or not. We have to see.Additionally, the plane intersects the lower base at x =1/2 -h², z=0. So on the lower base, at x =1/2 -h², which is a vertical line (since z=0). Depending on the value of x, this line could intersect the lower base's edges.The lower base ABCD has edges from (-7/2, -7/2,0) to (7/2, -7/2,0) (edge AB), then to (7/2,7/2,0) (edge BC), etc.The line x=1/2 -h² on the lower base will intersect either the front edge (AB or CD) or the back edges (AD or BC), depending on the x value.For example, if x=1/2 -h² is between -7/2 and7/2, which it is as long as |1/2 -h²| ≤7/2. Since h is positive, 1/2 -h² ≥ -7/2 => h² ≤7/2 +1/2=4. So h² ≤4 => h≤2. So if h≤2, then x=1/2 -h² ≥1/2 -4= -7/2, which is the leftmost x-coordinate. If h>2, then x=1/2 -h² < -7/2, which is outside the lower base.Wait, but the lower base's x ranges from -7/2 to7/2. So x=1/2 -h² must be within this interval.Therefore:-7/2 ≤1/2 -h² ≤7/2Subtract 1/2:-7/2 -1/2 ≤ -h² ≤7/2 -1/2=> -4 ≤ -h² ≤3Multiply by -1 (reversing inequalities):4 ≥ h² ≥ -3But h² is always non-negative, so the right inequality is always true. Therefore, the condition reduces to h² ≤4 => h≤2.Therefore, if h≤2, then the plane intersects the lower base at x=1/2 -h², otherwise, if h>2, the intersection is outside the lower base.Therefore, depending on h, the intersection of the plane with the frustum is different:- If h ≤2: Plane intersects the lower base at x=1/2 -h², and does not intersect lateral edges AA₁ and DD₁.- If h >2: Plane intersects lateral edges AA₁ and DD₁ at t=(h² -4)/(h² -3) and does not intersect the lower base.Therefore, the shape of the intersection is:- For h ≤2: A quadrilateral formed by edge B₁C₁ and the intersection line on the lower base.- For h >2: A triangle formed by edge B₁C₁ and the intersection points on edges AA₁ and DD₁.But since we don't know h, we have to consider both cases.However, since the problem states that the plane divides the frustum into two equal volumes, and given the complexity, it's possible that h ≤2, so that the intersection is a quadrilateral, making the volume calculation more straightforward.But we need to confirm.Let me assume h ≤2 first.Case 1: h ≤2.Then, the plane intersects the lower base at x =1/2 -h². Let's find the coordinates of this intersection.On the lower base z=0, x=1/2 -h², and y can be anything, but within the square. However, the line x=1/2 -h² on the lower base will intersect the edges of the lower base. Let's see:The lower base has edges AB, BC, CD, DA.The line x=1/2 -h² on z=0 will intersect edges DA and BC if x=1/2 -h² is between -7/2 and7/2. Since h ≤2, h² ≤4, so x=1/2 -4= -3.5= -7/2. Wait, if h=2, x=1/2 -4= -7/2 +1/2? Wait, 1/2 -h² where h=2: 1/2 -4= -7/2 +1/2? Wait, no. Let's compute x when h=2:x=1/2 -4= -3.5= -7/2. But the lower base's x ranges from -7/2 to7/2. So when h=2, x=-7/2, which is the left edge (edge DA: x=-7/2, y from -7/2 to7/2). When h=0, x=1/2 -0=1/2, which is inside the lower base. So for h from0 to2, x=1/2 -h² goes from1/2 to -7/2. So it moves from the right half of the lower base to the left edge.Therefore, the line x=1/2 -h² on the lower base (z=0) intersects edge DA when x=-7/2 (i.e., when h=2) and edge BC when x=7/2, but x=1/2 -h² can't reach7/2 unless h² is negative, which is impossible. So, for h <2, x=1/2 -h² is greater than -7/2, so it intersects the lower base within the front or back edges? Wait, no. The lower base is a square. The line x=1/2 -h² on z=0 is a vertical line. It will intersect the top and bottom edges of the lower base if it's within x ∈ [-7/2,7/2]. The top and bottom edges are y=7/2 and y=-7/2.Wait, no. The lower base is on the xy-plane. The edges are AB (y=-7/2), BC (x=7/2), CD (y=7/2), and DA (x=-7/2). So a vertical line x=1/2 -h² will intersect the top edge CD (y=7/2) and the bottom edge AB (y=-7/2), provided that x is within [-7/2,7/2]. So for h ≤2, x=1/2 -h² ∈ [-7/2,1/2]. Therefore, the vertical line x=1/2 -h² intersects the lower base at two points: (1/2 -h², -7/2,0) and (1/2 -h²,7/2,0). But these points are on edges AB and CD if x=1/2 -h² is between -7/2 and7/2.Wait, edge AB is from (-7/2,-7/2,0) to (7/2,-7/2,0), so along y=-7/2. Similarly, edge CD is from (-7/2,7/2,0) to (7/2,7/2,0), along y=7/2. So a vertical line x=1/2 -h² on the lower base (z=0) will intersect edge AB at (1/2 -h², -7/2,0) and edge CD at (1/2 -h²,7/2,0), provided that x=1/2 -h² is between -7/2 and7/2, which is true for h² ≤4 (h ≤2).Therefore, the intersection of the cutting plane with the frustum when h ≤2 is a quadrilateral with vertices:1. B₁(1/2, -1/2, h)2. C₁(1/2, 1/2, h)3. (1/2 -h²,7/2,0)4. (1/2 -h², -7/2,0)So this quadrilateral connects B₁ to C₁ to (1/2 -h²,7/2,0) to (1/2 -h², -7/2,0) back to B₁. Wait, not exactly, because the plane passes through B₁C₁ and the line on the lower base. So the intersection should be a polygon connecting B₁, C₁, (1/2 -h²,7/2,0), and (1/2 -h², -7/2,0). But actually, since the plane passes through B₁C₁ and cuts the lower base at x=1/2 -h², the intersection is a quadrilateral with vertices at B₁, C₁, (1/2 -h²,7/2,0), and (1/2 -h², -7/2,0). But this seems a bit non-planar? Wait, no. The plane is defined by those four points. Since B₁ and C₁ are on the upper base, and the other two points are on the lower base. Wait, but the plane passes through B₁C₁ and intersects the lower base along the vertical line x=1/2 -h². So yes, the intersection is a quadrilateral with those four vertices.Therefore, the portion of the frustum that is cut off by the plane is a kind of prismatoid with two triangular faces and two trapezoidal faces? Maybe not. It's a four-sided figure with two edges on the upper base, two edges on the lower base, connected by the plane.Alternatively, the volume cut by the plane can be computed as the sum of two pyramidal volumes or using the formula for the volume of a prismatoid.Alternatively, maybe using coordinates to set up integrals. Let's attempt that.The frustum can be parametrized as a linear interpolation between the lower base at z=0 and upper base at z=h. For any z between0 andh, the cross-section is a square with side length decreasing from7 to1 linearly. The side length at height z is 7 - (7 -1)(z/h) =7 -6(z/h).So, the coordinates can be parametrized as:x = [ - (7/2 - (6/2)(z/h)) , 7/2 - (6/2)(z/h) ] in x-direction,y = [ - (7/2 - (6/2)(z/h)) ,7/2 - (6/2)(z/h) ] in y-direction,with z from0 to h.Alternatively, more precisely, the side length at height z is 7 -6(z/h), so half the side length is (7 -6(z/h))/2. Therefore, at height z, the x ranges from - (7/2 -3(z/h)) to7/2 -3(z/h), same for y.But integrating this might be complicated. Alternatively, perhaps we can use the fact that the volume ratio is 1/2 and set up an equation.The cutting plane divides the frustum into two parts. Let's denote the part that includes the upper base A₁B₁C₁D₁ as V1 and the other part as V2. The problem states that V1 = V2 = 19h / 2.But how to compute V1?V1 is the region of the frustum on one side of the plane. Since the plane passes through B₁C₁ and extends downward to intersect the lower base at x=1/2 -h², the volume V1 is bounded by the upper base A₁B₁C₁D₁, the cutting plane, and part of the lateral faces.Alternatively, maybe it's easier to consider the entire frustum and use the method of slicing. Since the plane is oblique, it's challenging.Alternatively, using the concept of the volume of a wedge.Alternatively, maybe use coordinates to define the region V1 and integrate.The plane equation is x -h z =1/2 -h². Let's solve for x:x = h z +1/2 -h².Therefore, in the frustum, which spans z from0 toh, and x from- (7/2 -3(z/h)) to7/2 -3(z/h), similarly for y.The region V1 is the set of points in the frustum where x ≥h z +1/2 -h².Therefore, the volume V1 can be computed as the integral over z from0 to h of the area of the cross-section at height z where x ≥h z +1/2 -h².At each height z, the cross-section is a square with side length s(z) =7 -6(z/h). The square extends from x=-s(z)/2 to x= s(z)/2 and similarly for y.The plane x =h z +1/2 -h² cuts this square. The area to the right of x =h z +1/2 -h² is a rectangle or trapezoid, depending on whether the line x =h z +1/2 -h² is within the square.For each z, the value x = h z +1/2 -h². Let's see when this x is within the square's x-range, which is from -s(z)/2 to s(z)/2, where s(z)=7 -6(z/h).So, x must satisfy:- (7 -6(z/h))/2 ≤ h z +1/2 -h² ≤ (7 -6(z/h))/2But we are interested in the region x ≥h z +1/2 -h². So, the cross-sectional area at height z is the area of the part of the square where x ≥h z +1/2 -h².This area depends on where the plane cuts the square. For each z, we can compute the limits.Let’s define x0(z) = h z +1/2 -h². The cross-sectional area A(z) is:If x0(z) ≤ -s(z)/2: A(z) =0 (entire cross-section is to the right)If x0(z) ≥s(z)/2: A(z)=0 (entire cross-section is to the left)If -s(z)/2 <x0(z) <s(z)/2: A(z) = (s(z)/2 -x0(z)) * s(z) (since it's a rectangle in the square)Wait, in the square cross-section at height z, which spans x from -s(z)/2 to s(z)/2 and y from -s(z)/2 to s(z)/2. The plane x=x0(z) cuts the square. The area to the right of x=x0(z) is a rectangle with width (s(z)/2 -x0(z)) and height s(z). So, A(z) = (s(z)/2 -x0(z)) * s(z) when x0(z) is within the square. Otherwise, A(z) is 0 or the entire area.But we need to check for each z whether x0(z) is within the square.So, x0(z) =h z +1/2 -h².Given that z ∈ [0,h], let's see the behavior of x0(z):At z=0: x0(0)=1/2 -h².At z=h: x0(h)=h² +1/2 -h²=1/2.So x0(z) increases linearly from1/2 -h² to1/2 as z goes from0 toh.The square at height z has x from -s(z)/2 tos(z)/2, where s(z)=7 -6(z/h).Therefore, the limits of x0(z):At z=0: x0=1/2 -h². The square's x-range is -7/2 to7/2. So, x0=1/2 -h² must be compared to -7/2 and7/2.But we already considered h ≤2, so x0(0)=1/2 -h² ≥1/2 -4= -3.5= -7/2. So at z=0, x0 is within the square's x-range. At z=h, x0=1/2, and the square's x-range at z=h is -1/2 to1/2 (since s(h)=1). So at z=h, x0=1/2 is equal to the upper limit of x, so the area at z=h is zero.Therefore, for z ∈ [0,h], x0(z) ranges from1/2 -h² (which is ≥ -7/2) to1/2. However, as z increases, the square's x-range decreases from7 to1, so the upper limit of x decreases from7/2 to1/2.We need to find for which z, x0(z) ≤s(z)/2.s(z)/2=(7 -6(z/h))/2.Therefore, x0(z) ≤s(z)/2:h z +1/2 -h² ≤(7 -6(z/h))/2Multiply both sides by 2:2 h z +1 -2h² ≤7 -6(z/h)Bring all terms to left:2 h z +1 -2h² -7 +6(z/h) ≤0Simplify:2 h z -2h² -6 +6(z/h) ≤0Multiply through by h to eliminate denominator (h >0):2 h² z -2h³ -6h +6z ≤0But this seems messy. Maybe it's better to solve for z.Let’s rearrange the inequality:h z +1/2 -h² ≤(7 -6(z/h))/2Multiply both sides by2:2 h z +1 -2 h² ≤7 -6(z/h)Let’s collect terms with z on left and constants on right:2 h z +6(z/h) ≤7 -1 +2 h²=> z(2 h +6/h) ≤6 +2 h²Thus,z ≤ (6 +2 h²)/(2 h +6/h)Multiply numerator and denominator by h:z ≤ (6 h +2 h³)/(2 h² +6)Factor numerator and denominator:Numerator: 2 h³ +6 h =2 h (h² +3)Denominator:2 h² +6=2(h² +3)Thus,z ≤ [2 h (h² +3)]/[2 (h² +3)] ]= hWhich is always true since z ∈ [0,h]. Therefore, the inequality x0(z) ≤s(z)/2 holds for all z ∈ [0,h]. Therefore, for all z ∈ [0,h], the cross-sectional area A(z) is the area to the right of x=x0(z), which is:A(z) = (s(z)/2 -x0(z)) * s(z)Because the cross-section at height z is a square of side s(z), so the width in x-direction is from x0(z) to s(z)/2, and the height in y-direction is full s(z).But wait, no. The cross-section is a square, so for each x ≥x0(z), y ranges from -s(z)/2 to s(z)/2. Therefore, the area is:A(z) = [s(z)/2 -x0(z)] * s(z), provided x0(z) ≤s(z)/2.But we just saw that this is true for all z. Therefore, the cross-sectional area is:A(z) = [ (7 -6(z/h))/2 - (h z +1/2 -h²) ] * (7 -6(z/h))Simplify inside the brackets:= [7/2 -3(z/h) -h z -1/2 +h² ] * (7 -6(z/h))Simplify constants and terms:7/2 -1/2 =3,-3(z/h) -h z = -z(3/h +h),So:A(z) = [3 -z(3/h +h) +h²] * (7 -6(z/h))Let me check this:Yes:Original expression:[ (7/2 -3(z/h)) - (h z +1/2 -h²) ]=7/2 -3(z/h) -h z -1/2 +h²= (7/2 -1/2) + (-3(z/h) -h z) +h²=3 -z(3/h +h) +h²Yes.Therefore,A(z)= [3 +h² - z(3/h +h)] * (7 -6(z/h))But this seems complicated. Let's see if we can simplify.Let’s denote t = z/h, where t ∈[0,1]. Then z= t h.Substitute into A(z):A(t)= [3 +h² - t h (3/h +h)] * (7 -6 t )Simplify inside:3 +h² -t h*(3/h +h)=3 +h² -t*(3 +h²)= (3 +h²)(1 -t) +3 -3(1 -t) ??? Wait, no.Let me compute:=3 +h² -t*(3 +h²)= (3 +h²)(1 -t) +3 -3(1 -t) ???Wait, perhaps factor:= (3 +h²)(1 -t)Wait, no:Wait, 3 +h² -t*(3 +h²) = (3 +h²)(1 -t)Yes, correct.Therefore,A(t) = (3 +h²)(1 -t) * (7 -6 t )Therefore, the volume V1 is the integral of A(t) over z from0 toh, which in terms of t is:V1= ∫₀^h A(z) dz = ∫₀¹ A(t) * h dt = h ∫₀¹ (3 +h²)(1 -t)(7 -6t) dtFactor out constants:V1= h (3 +h²) ∫₀¹ (1 -t)(7 -6t) dtCompute the integral:Expand (1 -t)(7 -6t)=7(1 -t) -6t(1 -t)=7 -7t -6t +6t²=7 -13t +6t²Integrate from0 to1:∫₀¹ (7 -13t +6t²) dt=7t - (13/2)t² +2t³ |₀¹=7 -13/2 +2= (14/2 -13/2 +4/2)=5/2Therefore,V1= h (3 +h²)*(5/2)= (5/2) h (3 +h²)But according to the problem, V1=19h /2.Therefore,(5/2) h (3 +h²)= (19/2) hCancel h (assuming h≠0):(5/2)(3 +h²)=19/2Multiply both sides by2:5(3 +h²)=1915 +5h²=195h²=4h²=4/5 => h=2/√5Since h>0.Therefore, the height h is2/√5. Therefore, the volume of the frustum is19h=19*(2/√5)=38/√5. However, rationalizing the denominator, this is (38√5)/5.But the problem might expect an answer in a simplified radical form. So, 38√5 /5.But let me check my steps to ensure I didn't make a mistake.First, the cross-sectional area A(z) was derived as [3 +h² - z(3/h +h)]*(7 -6(z/h)). Then substituting t =z/h, got A(t)= (3 +h²)(1 -t)*(7 -6t). Then integrated over t from0 to1, giving the integral equal to5/2, hence V1=(5/2)h(3 +h²). Set equal to19h/2, solved for h²=4/5, h=2/√5.Thus, volume is19*(2/√5)=38/√5= (38√5)/5.Yes, that seems correct.Therefore, the volume of the pyramid (frustum) is38√5 /5.But let me verify the integral computation:∫₀¹ (7 -13t +6t²) dt=7*1 -13/2*(1)^2 +6*(1)^3 /3 -0=7 -13/2 +2=7 -6.5 +2=2.5=5/2. Correct.Then yes, the integral is5/2. So the computation is correct.Therefore, the answer is38√5 /5, which is7.6√5. But as a fraction,38 and5 are coprime, so it's simplest form.But let me check if the assumption that h ≤2 is valid. Since h=2/√5≈0.894, which is less than2, so our initial assumption holds. Therefore, the case where h ≤2 is valid, and the intersection is a quadrilateral, and the computation is correct.Therefore, the volume of the pyramid is38√5 /5.But wait, let me confirm whether the frustum's volume formula is correct. Earlier, we had V=19h. With h=2/√5, V=19*(2/√5)=38/√5= (38√5)/5. Yes.Alternatively, using the formula for the frustum:V=(1/3)h(S1 +S2 +sqrt(S1S2))=(1/3)h(49 +1 +7)=(1/3)h(57)=19h. Yes, so 19h=38√5 /5.Therefore, the answer is38√5 /5.But let me check with the original problem statement. It says "Find the volume of the pyramid." Even though it's a frustum, the problem refers to it as a pyramid. However, in some contexts, a truncated pyramid can still be called a pyramid, but the volume would be that of the frustum. Given the computations, and the conditions leading to h=2/√5, the volume is38√5 /5. Therefore, the answer should be boxed as such.Final AnswerThe volume of the pyramid is boxed{dfrac{38sqrt{5}}{5}}.</think>To find the volume of the regular truncated quadrilateral pyramid (frustum) with given conditions, we start by noting the side lengths of the upper and lower bases are 1 and 7, respectively. The volume of a frustum is given by:[ V = frac{1}{3} h (S_1 + S_2 + sqrt{S_1 S_2}) ]where ( S_1 = 1^2 = 1 ) and ( S_2 = 7^2 = 49 ). The height ( h ) of the frustum needs to be determined using the condition that a plane passing through edge ( B_1C_1 ) and perpendicular to the plane ( AD_1C ) divides the frustum into two equal volumes.1. Equation of Plane AD₁C: - Points: ( A(-7/2, -7/2, 0) ), ( D_1(-1/2, 1/2, h) ), ( C(7/2, 7/2, 0) ) - Vectors: ( overrightarrow{AD_1} = (3, 4, h) ), ( overrightarrow{AC} = (7, 7, 0) ) - Normal vector: ( (-7h, 7h, -7) ) - Equation: ( h x - h y + z = 0 )2. Equation of Plane Passing Through B₁C₁ and Perpendicular to AD₁C: - Plane contains line ( B_1C_1 ) (points ( (1/2, -1/2, h) ) to ( (1/2, 1/2, h) )) - Normal vector condition: ( h a - h b + c = 0 ) - Derived equation: ( x - h z = 1/2 - h^2 )3. Volume Calculation: - The plane divides the frustum into two equal volumes. - Using integration, the volume of one part is found to be ( frac{5}{2} h (3 + h^2) ). - Setting this equal to half the total volume ( frac{19h}{2} ), solving for ( h ): [ frac{5}{2} h (3 + h^2) = frac{19h}{2} implies 5(3 + h^2) = 19 implies h^2 = frac{4}{5} implies h = frac{2}{sqrt{5}} ]4. Volume of the Frustum: - Substitute ( h = frac{2}{sqrt{5}} ) into the volume formula: [ V = 19h = 19 times frac{2}{sqrt{5}} = frac{38sqrt{5}}{5} ]The volume of the pyramid is boxed{dfrac{38sqrt{5}}{5}}.]

🔑:100. Let the plane passing through B_{1} C_{1} intersect A B and D C at points K and L (Fig. 12). By the condition, the volumes of the polyhedra A K L D A_{1} B_{1} C_{1} D_{1} and K B C L B_{1} C_{1} are equal. Applying the Simpson's formula (Problem 15) to them, denoting |A K|=|D L|=a. Since the heights of these polyhedra are equal, we get the equation for a:7 a+1+4 frac{(a+1)}{2} cdot frac{(7+1)}{2}=(7-a) 7+4 frac{(7-a)}{2} cdot frac{(7+1)}{2}from which a=frac{16}{5}.Let the height of the pyramid be h. Introduce a coordinate system, taking its origin at the center of A B C D, with the x and y axes parallel to A B and B C. The points A, C, and D_{1} will have coordinates left(-frac{7}{2},-frac{7}{2}, 0right),left(frac{7}{2}, frac{7}{2}, 0right),Fig. 11.Fig. 12.left(-frac{1}{2}, frac{1}{2}, hright). It is not difficult to find the equation of the plane A C D_{1} ; h x - h y + z = 0. The plane K L C_{1} B_{1} will have the equation 10 h x - 8 z + 3 h = 0. The normal vector to the first plane is boldsymbol{n}(h, -h, 1), and to the second is boldsymbol{m}(10 h, 0, -8). The condition of their perpendicularity gives 10 h^{2} - 8 = 0, h = frac{2 sqrt{5}}{5}. The volume of the pyramid is frac{38 sqrt{5}}{5}.

🧠:[<think>Okay, let's see. I need to calculate this expression: 2013 divided by 2.2 multiplied by the quantity (4/3 minus 8/15). Hmm, alright. Let me break it down step by step. First, maybe I should handle the parentheses part because of the order of operations—PEMDAS, right? Parentheses first, then division and multiplication from left to right.So, the expression inside the parentheses is (4/3 - 8/15). To subtract these fractions, they need a common denominator. The denominators are 3 and 15. The least common denominator here is 15. Let me convert 4/3 to fifteenths. 4/3 is the same as (4*5)/(3*5) which is 20/15. Then subtract 8/15 from that. So, 20/15 - 8/15 = (20 - 8)/15 = 12/15. Hmm, 12/15 can be simplified by dividing numerator and denominator by 3. That gives 4/5. So, the part in the parentheses simplifies to 4/5. Okay, that seems right.Now the expression is 2013 divided by 2.2 multiplied by 4/5. Let me write that as 2013 ÷ 2.2 × (4/5). Since division and multiplication are at the same level in order of operations, I should go from left to right. So first divide 2013 by 2.2, then multiply the result by 4/5.Let me compute 2013 ÷ 2.2 first. Dividing by a decimal. Maybe I can convert 2.2 to a fraction to make it easier. 2.2 is the same as 11/5. So dividing by 11/5 is the same as multiplying by 5/11. Therefore, 2013 ÷ (11/5) = 2013 × (5/11). Let me compute that. 2013 multiplied by 5, then divided by 11.First, 2013 × 5. Let's do that step by step. 2000 × 5 = 10,000. 13 × 5 = 65. So, 10,000 + 65 = 10,065. Then divide that by 11. Hmm, 10,065 ÷ 11. Let me see. 11 × 915 = 10,065 because 11 × 900 = 9900, and 11 × 15 = 165. 9900 + 165 = 10,065. So, 10,065 ÷ 11 = 915. So, the result of 2013 ÷ 2.2 is 915. Okay, that seems straightforward.Now, multiply that result by 4/5. So, 915 × (4/5). Let me compute this. Multiplying 915 by 4 first, then divide by 5. 915 × 4. Let's do 900 × 4 = 3,600 and 15 × 4 = 60. So, 3,600 + 60 = 3,660. Then divide that by 5. 3,660 ÷ 5. Let's see, 3,600 ÷ 5 = 720, and 60 ÷ 5 = 12. So, 720 + 12 = 732. Therefore, 915 × (4/5) is 732. So, putting it all together: 2013 ÷ 2.2 × (4/3 - 8/15) = 915 × 4/5 = 732. Let me double-check each step to make sure I didn't make a mistake.First, the parentheses: 4/3 - 8/15. Convert to 20/15 - 8/15 = 12/15 = 4/5. That looks good. Then 2013 ÷ 2.2. Converting 2.2 to 11/5, so 2013 × 5/11. 2013 × 5 = 10,065, and 10,065 ÷ 11 = 915. Yes, because 11 × 900 = 9900, and 11 × 15 = 165. 9900 + 165 = 10,065. So that's correct. Then 915 × 4/5: 915 × 4 = 3,660, divided by 5 is 732. Correct again. Alternatively, maybe I can check using decimal conversions instead of fractions. Let's try another way. Starting with 2013 ÷ 2.2. 2.2 is equal to 11/5, so 2013 ÷ 2.2 is 2013 ÷ (11/5) = 2013 × (5/11) which is the same as before. Alternatively, if I do it as decimal division: 2.2 goes into 2013 how many times?First, 2.2 × 900 = 1980. Subtract 1980 from 2013, which gives 33. Then 2.2 goes into 33 how many times? 2.2 × 15 = 33. So total is 900 + 15 = 915. So that confirms the division part is correct. Then multiplying 915 by 4/5. 915 divided by 5 is 183. Then 183 × 4 = 732. Yep, same result. So that checks out. So, I think 732 is the correct answer. Let me just verify once more the original problem: 2013 divided by 2.2 times (4/3 minus 8/15). All steps seem correct, so I think that's the right answer.Final Answerboxed{732}</think>First, simplify the expression inside the parentheses:[frac{4}{3} - frac{8}{15} = frac{20}{15} - frac{8}{15} = frac{12}{15} = frac{4}{5}]Next, perform the division and multiplication from left to right. Convert the division by 2.2 into multiplication by the reciprocal:[2013 div 2.2 = 2013 div frac{11}{5} = 2013 times frac{5}{11} = frac{2013 times 5}{11} = frac{10065}{11} = 915]Then, multiply the result by (frac{4}{5}):[915 times frac{4}{5} = frac{915 times 4}{5} = frac{3660}{5} = 732]Thus, the final answer is:[boxed{732}],<think>Okay, let's tackle this problem step by step. The question is to calculate 2013 divided by 2.2 multiplied by the quantity (4/3 minus 8/15). Alright, first, I need to remember the order of operations, which is parentheses first, then multiplication and division from left to right. So, let's break it down.First, focus on the part inside the parentheses: (4/3 - 8/15). To subtract these fractions, they need a common denominator. The denominators are 3 and 15, so 15 is a common denominator. Let me convert 4/3 to fifteenths. 4/3 is equivalent to (4*5)/(3*5) = 20/15. Then, subtract 8/15 from that: 20/15 - 8/15 = (20 - 8)/15 = 12/15. Hmm, 12/15 can be simplified, right? Both numerator and denominator are divisible by 3. So 12 ÷ 3 = 4 and 15 ÷ 3 = 5, which reduces to 4/5. Okay, so the parentheses simplify to 4/5. Good.Now, the expression is 2013 ÷ 2.2 × 4/5. Let's handle the division and multiplication from left to right. So first, 2013 divided by 2.2. Let me compute that. Dividing by a decimal can be tricky, so maybe convert 2.2 to a fraction. 2.2 is the same as 11/5, right? Because 2.2 = 2 + 0.2 = 2 + 1/5 = 11/5. So, dividing by 11/5 is the same as multiplying by 5/11. Therefore, 2013 ÷ (11/5) = 2013 × (5/11). Let me compute 2013 × 5 first. 2013 × 5: 2000 × 5 = 10,000, and 13 × 5 = 65, so total is 10,000 + 65 = 10,065. Then divide by 11. So 10,065 ÷ 11. Let me do this division. 11 × 914 = 10,054 (because 11 × 900 = 9900, 11 × 14 = 154; 9900 + 154 = 10,054). Then 10,065 - 10,054 = 11. So that's 914 + 1 = 915. So 10,065 ÷ 11 = 915. Wait, really? Let me check. 11 × 915: 11×900=9900, 11×15=165, so 9900+165=10,065. Yes, correct. So 2013 ÷ 2.2 equals 915. That seems straightforward.Now, the next step is to multiply this result by 4/5. So 915 × 4/5. Let me compute that. 915 divided by 5 is 183, then multiplied by 4. 183 × 4: 100 × 4 = 400, 80 × 4 = 320, 3 × 4 = 12. So 400 + 320 = 720, plus 12 is 732. So 915 × 4/5 = 732. Therefore, the entire expression simplifies to 732.Wait, let me verify each step again to make sure I didn't make any mistakes. Starting with the parentheses: 4/3 - 8/15. Convert 4/3 to 20/15, subtract 8/15 gives 12/15, which simplifies to 4/5. That seems correct. Then 2013 divided by 2.2. Convert 2.2 to 11/5, so dividing by 11/5 is multiplying by 5/11. 2013 × 5 = 10,065. Divided by 11 gives 915. Correct. Then 915 multiplied by 4/5: 915 ÷ 5 = 183, 183 × 4 = 732. All steps check out. So the final answer should be 732. Alternatively, maybe I can approach the calculation differently to cross-verify. Let me compute everything as decimals to see if I get the same result. Let's see. Starting with the parentheses: 4/3 is approximately 1.3333..., and 8/15 is approximately 0.5333... So subtracting those: 1.3333 - 0.5333 = 0.8. Which is 4/5 as a decimal, so that part checks out. Then 2013 divided by 2.2. Let's do that division as decimals. 2013 ÷ 2.2. Let's compute 2013 ÷ 2.2.First, 2.2 goes into 2013 how many times? Let's compute 2013 ÷ 2.2. Multiply numerator and denominator by 10 to eliminate the decimal: 20130 ÷ 22. Let's divide 20130 by 22. 22 × 900 = 19,800. Subtract that from 20,130: 20,130 - 19,800 = 330. Now, 22 × 15 = 330. So 900 + 15 = 915. So 20130 ÷ 22 = 915. So again, same result. Then multiply by 0.8 (which is 4/5). 915 × 0.8. 915 × 0.8: 900 × 0.8 = 720, 15 × 0.8 = 12, so 720 + 12 = 732. Yep, same answer. So both methods give 732. That gives me more confidence that the answer is correct.Alternatively, maybe another way: combine all operations into fractions. Let's see. Original expression: 2013 ÷ 2.2 × (4/3 - 8/15). As above, the parentheses become 4/5. So it's 2013 ÷ (11/5) × 4/5. Converting division to multiplication by reciprocal: 2013 × (5/11) × (4/5). The 5 in the numerator and denominator cancels out, leaving 2013 × (4/11). Then compute 2013 × 4 ÷ 11. 2013 × 4 = 8052. Then 8052 ÷ 11. Let's compute that. 11 × 732 = 8052 (since 700 × 11 = 7700, 32 × 11 = 352; 7700 + 352 = 8052). So 8052 ÷ 11 = 732. Again, same result. So three different methods all lead to 732, which must be the correct answer. I think that's thorough enough. Can't see any mistakes here.</think>The problem requires calculating (2013 div 2.2 times left(frac{4}{3} - frac{8}{15}right)).1. Simplify the parentheses: [ frac{4}{3} - frac{8}{15} = frac{20}{15} - frac{8}{15} = frac{12}{15} = frac{4}{5} ]2. Convert division to multiplication by reciprocal: [ 2013 div 2.2 = 2013 div frac{11}{5} = 2013 times frac{5}{11} ] Calculate (2013 times 5 = 10065), then divide by 11: [ frac{10065}{11} = 915 ]3. Multiply by (frac{4}{5}): [ 915 times frac{4}{5} = frac{915 times 4}{5} = frac{3660}{5} = 732 ]All steps confirm the final result is (boxed{732}).]

🔑:【Solution】Solve: 2013 div 2.2 timesleft(frac{4}{3}-frac{8}{15}right)begin{array}{l}=2013 div frac{11}{5} times frac{12}{15} =2013 times frac{5}{11} times frac{12}{15} =732end{array}Therefore, the answer is: 732 .