🧠:[<think>Okay, so I need to figure out the possible values that the first term, let's call it a₁, of an arithmetic progression can take. The given conditions are that the common difference is 1, and that the sum of the first 2022 terms, S₂₀₂₂, is the smallest among all possible sums Sₙ for any n. That means S₂₀₂₂ has to be less than Sₙ for all other n, not just some of them. Hmm, arithmetic progression with difference 1. Let me recall the formula for the sum of the first n terms of an arithmetic progression.The sum Sₙ is given by Sₙ = n/2 [2a₁ + (n - 1)d], where d is the common difference. Since the common difference here is 1, the formula simplifies to Sₙ = n/2 [2a₁ + (n - 1)]. Let me write that down:Sₙ = (n/2)(2a₁ + n - 1)So, the problem states that S₂₀₂₂ is the minimal sum among all n. Therefore, for all n ≠ 2022, Sₙ > S₂₀₂₂. Our goal is to find the possible values of a₁.First, let's compute S₂₀₂₂. Plugging n = 2022 into the formula:S₂₀₂₂ = (2022/2)(2a₁ + 2022 - 1) = 1011*(2a₁ + 2021)So S₂₀₂₂ = 1011*(2a₁ + 2021)Now, we need to find a₁ such that this sum is smaller than the sum for any other n. Let's think about how Sₙ behaves as a function of n. For an arithmetic progression, the sum Sₙ can be a quadratic function in terms of n. Let's see:Sₙ = (n/2)(2a₁ + n - 1) = (n/2)(n + (2a₁ - 1)) = (n² + (2a₁ - 1)n)/2So, Sₙ is a quadratic function of n: Sₙ = (1/2)n² + (a₁ - 0.5)nTherefore, the sum Sₙ is a quadratic function in n with coefficient (1/2) for n². Since the coefficient of n² is positive (1/2), the parabola opens upwards. That means the function Sₙ has a minimum value at its vertex. The vertex of a quadratic function an² + bn + c is at n = -b/(2a). Here, a = 1/2, b = (a₁ - 0.5). Therefore, the vertex is at:n = -(a₁ - 0.5)/(2*(1/2)) = -(a₁ - 0.5)/1 = 0.5 - a₁But wait, the vertex here gives the value of n where the sum Sₙ is minimized. However, n must be a positive integer because it's the number of terms. So, if the vertex is at n = 0.5 - a₁, then the minimal sum occurs at the integer n closest to 0.5 - a₁. But in our case, it's given that the minimal sum occurs at n = 2022. Therefore, the vertex of the parabola must be such that the closest integer to 0.5 - a₁ is 2022. Therefore, 0.5 - a₁ must be approximately 2022, but since n must be an integer, we need 2021.5 < 0.5 - a₁ < 2022.5.Wait, hold on, let me check that again. If the vertex is at n = 0.5 - a₁, but since the coefficient of n² is positive, the parabola opens upwards, so the minimum is at n = 0.5 - a₁. But since n must be a positive integer, the minimal sum will occur at the integer closest to the vertex. Therefore, to have Sₙ minimized at n = 2022, the vertex must lie in the interval [2021.5, 2022.5). Because if the vertex is at 2022, the closest integer is 2022; if it's halfway between 2021 and 2022 (2021.5), then both 2021 and 2022 are equally close, but since we need S₂₀₂₂ to be strictly less than all other Sₙ, the vertex must not be exactly halfway. Wait, actually, if the vertex is exactly at 2021.5, then S₂₀₂₂ and S₂₀₂₁ would be equal? Let me verify.If the vertex is at 2021.5, then S₂₀₂₁ and S₂₀₂₂ would be equidistant from the vertex. Since the parabola is symmetric, the sums would be the same at n = 2021 and n = 2022? Wait, no. Wait, the sum at n = 2021.5 is the vertex, but n has to be integer. So the sum at n = 2021 and n = 2022 would be symmetric around the vertex. If the vertex is at 2021.5, then S₂₀₂₁ = S₂₀₂₂? Let me test that with a concrete example.Suppose a₁ is such that the vertex is at 2021.5. Then, the formula for the vertex is 0.5 - a₁ = 2021.5, so a₁ = 0.5 - 2021.5 = -2021. Let's compute S₂₀₂₁ and S₂₀₂₂ with a₁ = -2021.Compute S₂₀₂₁: (2021/2)(2*(-2021) + 2021 - 1) = (2021/2)(-4042 + 2020) = (2021/2)(-2022) = -2021*1011Compute S₂₀₂₂: (2022/2)(2*(-2021) + 2022 - 1) = 1011*(-4042 + 2021) = 1011*(-2021)Wait, so S₂₀₂₁ = -2021*1011 and S₂₀₂₂ = -2021*1011. So they are equal. Therefore, if the vertex is exactly at 2021.5, then S₂₀₂₁ = S₂₀₂₂, which contradicts the condition that S₂₀₂₂ is the smallest. Therefore, the vertex must lie in the interval such that S₂₀₂₂ is strictly less than its neighbors. Therefore, the vertex must be closer to 2022 than to 2021 or 2023. So, the vertex must lie in the interval (2021.5, 2022.5). Therefore, n must be 2022 when the vertex is in that interval.But the vertex is given by n = 0.5 - a₁. Therefore:2021.5 < 0.5 - a₁ < 2022.5Subtract 0.5 from all parts:2021 < -a₁ < 2022Multiply all parts by -1, which reverses the inequalities:-2021 > a₁ > -2022So, -2022 < a₁ < -2021Therefore, a₁ must be greater than -2022 and less than -2021. Since a₁ is the first term of an arithmetic progression with common difference 1, the terms are a₁, a₁ + 1, a₁ + 2, etc. But the problem doesn't specify any restrictions on a₁ being an integer or not. Wait, the problem says it's an arithmetic progression with common difference 1, but a₁ could be any real number unless specified otherwise. The problem doesn't state that the terms are integers, so a₁ can be any real number. Therefore, the possible values of a₁ are real numbers between -2022 and -2021, not including the endpoints.But let me confirm this. Let's see:If the vertex is at n = 0.5 - a₁, then to have the minimal sum at n = 2022, we need 0.5 - a₁ to be between 2021.5 and 2022.5. So:2021.5 < 0.5 - a₁ < 2022.5Subtract 0.5 from all sides:2021 < -a₁ < 2022Multiply by -1 (remembering to reverse inequalities):-2021 > a₁ > -2022Which is equivalent to:-2022 < a₁ < -2021Yes, that's correct. Therefore, the first term a₁ must lie in the open interval (-2022, -2021). So, possible values for a₁ are all real numbers between -2022 and -2021, not including the endpoints.But let me test with a concrete example. Suppose a₁ is -2021.5, which is in the middle of the interval. Then the vertex would be at n = 0.5 - (-2021.5) = 0.5 + 2021.5 = 2022. So exactly at 2022. So S₂₀₂₂ would be the minimum. But in this case, the vertex is at n = 2022, so the sum is minimized exactly at n = 2022, right?Wait, but if the vertex is at an integer n, then the sum is minimized exactly there. But if the vertex is not at an integer, then the minimum is at the nearest integer. So in the case where a₁ is in (-2022, -2021), then the vertex is in (2021.5, 2022.5), so the nearest integer is 2022. Therefore, if a₁ is exactly -2021.5, the vertex is at 2022, so S₂₀₂₂ is the minimum. If a₁ is greater than -2021.5, say a₁ = -2021.3, then vertex is 0.5 - (-2021.3) = 2021.8, which is between 2021.5 and 2022.5, so the closest integer is 2022. Similarly, if a₁ is less than -2021.5, say a₁ = -2021.7, then vertex is 0.5 - (-2021.7) = 2022.2, which is still between 2021.5 and 2022.5, so closest integer is 2022.Wait, but actually, if the vertex is at 2021.8, which is closer to 2022 than 2021, so the minimum is at 2022. Similarly, if the vertex is at 2022.2, which is still closer to 2022 than 2023. Therefore, as long as the vertex is between 2021.5 and 2022.5, the minimum is at 2022. Therefore, the first term a₁ must satisfy 0.5 - a₁ ∈ (2021.5, 2022.5), which gives a₁ ∈ (-2022, -2021). Therefore, the answer is that the first term a₁ must be greater than -2022 and less than -2021. So, in interval notation, a₁ ∈ (-2022, -2021).But let me check with a specific example. Let's take a₁ = -2021.5. Then, the sum Sₙ would be (n/2)(2*(-2021.5) + n - 1) = (n/2)(-4043 + n - 1) = (n/2)(n - 4044). Let's compute S₂₀₂₁ and S₂₀₂₂.S₂₀₂₁ = (2021/2)(2021 - 4044) = (2021/2)(-2023) = -2021*2023/2S₂₀₂₂ = (2022/2)(2022 - 4044) = 1011*(-2022) = -2022*1011Wait, let's compute these:First, compute S₂₀₂₂ when a₁ = -2021.5:S₂₀₂₂ = 1011*(2*(-2021.5) + 2021) = 1011*(-4043 + 2021) = 1011*(-2022) = -2022*1011Similarly, S₂₀₂₁ = 2021/2*(2*(-2021.5) + 2020) = 2021/2*(-4043 + 2020) = 2021/2*(-2023) = -2021*2023/2Compare S₂₀₂₁ and S₂₀₂₂:Compute S₂₀₂₁: -2021*2023/2 ≈ -2021*1011.5 ≈ -2021*1011 -2021*0.5Compute S₂₀₂₂: -2022*1011 = -2021*1011 -1011So, S₂₀₂₁ = -2021*1011 - 1010.5S₂₀₂₂ = -2021*1011 - 1011Therefore, S₂₀₂₂ is more negative than S₂₀₂₁, which means S₂₀₂₂ is smaller. Similarly, if we compute S₂₀₂₃:S₂₀₂₃ = (2023/2)(2*(-2021.5) + 2022) = (2023/2)(-4043 + 2022) = (2023/2)(-2021) = -2023*2021/2 ≈ -2021*1011.5 ≈ -2021*1011 -2021*0.5Compare to S₂₀₂₂: -2021*1011 -1011. So, S₂₀₂₃ is -2021*1011 -1010.5, which is greater (less negative) than S₂₀₂₂. Therefore, S₂₀₂₂ is indeed the smallest. So even when a₁ = -2021.5, which is exactly at the midpoint of the interval (-2022, -2021), S₂₀₂₂ is the minimum. Wait, but earlier, when a₁ was -2021.5, the vertex is at n = 2022, so Sₙ is minimized exactly at n = 2022, so S₂₀₂₂ is the minimum, which is consistent. Therefore, even if the vertex is exactly at an integer, the sum is minimized there, and since the problem states that S₂₀₂₂ is the smallest, that's acceptable.But in the problem statement, it says "the smallest among all Sₙ (less than the sum of the first n terms for any other value of n)". So S₂₀₂₂ must be strictly less than all other Sₙ. So even if the vertex is exactly at 2022, then S₂₀₂₂ is the minimal, and for all other n, Sₙ > S₂₀₂₂. But in that case, if the vertex is exactly at 2022, then the sums on either side increase as you move away from 2022, so S₂₀₂₁ and S₂₀₂₃ would both be greater than S₂₀₂₂. Therefore, even if a₁ is exactly -2021.5, S₂₀₂₂ is the minimum, so that's acceptable.Wait, but in the previous example with a₁ = -2021.5, we saw that S₂₀₂₂ is indeed less than S₂₀₂₁ and S₂₀₂₃, right? Because S₂₀₂₂ was -2022*1011, and S₂₀₂₁ was approximately -2021*1011 -1010.5, which is -2021*1011 is about -2021*1000 -2021*11 = -2,021,000 -22,231 = -2,043,231, minus 1010.5 is -2,044,241.5. Similarly, S₂₀₂₂ is -2022*1011 = -2022*1000 -2022*11 = -2,022,000 -22,242 = -2,044,242. So S₂₀₂₂ is -2,044,242, S₂₀₂₁ is -2,044,241.5. Wait, hold on, this is interesting. S₂₀₂₁ is actually greater (less negative) than S₂₀₂₂. Wait, how is that possible? Let me recalculate.Wait, S₂₀₂₁ when a₁ = -2021.5 is:S₂₀₂₁ = (2021/2)(2*(-2021.5) + 2020) = (2021/2)(-4043 + 2020) = (2021/2)(-2023) = (-2021*2023)/2Similarly, S₂₀₂₂ is (2022/2)(2*(-2021.5) + 2021) = 1011*(-4043 + 2021) = 1011*(-2022) = (-2022*1011)Let me compute both:2021*2023 = 2021*(2000 + 23) = 2021*2000 + 2021*23 = 4,042,000 + 46,483 = 4,088,483Therefore, S₂₀₂₁ = -4,088,483 / 2 ≈ -2,044,241.5S₂₀₂₂ = -2022*1011. Let's compute 2022*1000 = 2,022,000; 2022*11 = 22,242. So total is 2,022,000 + 22,242 = 2,044,242. Therefore, S₂₀₂₂ = -2,044,242Therefore, S₂₀₂₂ = -2,044,242 is less than S₂₀₂₁ = -2,044,241.5. So S₂₀₂₂ is indeed smaller. Similarly, S₂₀₂₃ would be:S₂₀₂₃ = (2023/2)(2*(-2021.5) + 2022) = (2023/2)(-4043 + 2022) = (2023/2)(-2021) = (-2023*2021)/22023*2021 = (2022 +1)(2022 -1) = 2022² -1 = 4,088,484 -1 = 4,088,483. Therefore, S₂₀₂₃ = -4,088,483 /2 = -2,044,241.5, same as S₂₀₂₁. So S₂₀₂₁ and S₂₀₂₃ are both -2,044,241.5, which is greater than S₂₀₂₂'s -2,044,242. Therefore, even when a₁ = -2021.5 (the midpoint), S₂₀₂₂ is indeed the smallest. So even if the vertex is exactly at n=2022, the sum at n=2022 is still less than at neighboring n. Therefore, the interval for a₁ is including the exact point where the vertex is at n=2022. But wait, in the initial analysis, we had 2021.5 < 0.5 - a₁ < 2022.5 leading to -2022 < a₁ < -2021. But when a₁ = -2021.5, 0.5 - a₁ = 0.5 - (-2021.5) = 2022, which is the exact integer. So that's included in the interval. But in the inequality 2021.5 < 0.5 - a₁ < 2022.5, 0.5 - a₁ =2022 is allowed? Wait, 2021.5 < 2022 < 2022.5? No, 2022 is equal to 2022, which is not less than 2022.5. Wait, actually, the original inequalities were:To have n=2022 as the closest integer to the vertex, the vertex must lie in [2021.5, 2022.5). If it's equal to 2022.5, then the closest integer would be 2023. Therefore, to ensure that the vertex is strictly closer to 2022 than to 2021 or 2023, we need 2021.5 < 0.5 - a₁ < 2022.5. So this interval includes 2022 as the exact value. Wait, 0.5 - a₁ can be exactly 2022, which would correspond to a₁ = 0.5 - 2022 = -2021.5. So in that case, the vertex is exactly at 2022, so Sₙ is minimized at n=2022. Therefore, the interval for a₁ is -2022 < a₁ ≤ -2021.5 and -2021.5 ≤ a₁ < -2021? Wait, no, wait:From the inequality:2021.5 < 0.5 - a₁ < 2022.5Subtract 0.5:2021 < -a₁ < 2022Multiply by -1 and reverse inequalities:-2021 > a₁ > -2022So, the interval for a₁ is (-2022, -2021). So open interval, not including the endpoints. Therefore, when a₁ approaches -2022 from above, 0.5 - a₁ approaches 0.5 - (-2022) = 2022.5, which is excluded. Similarly, when a₁ approaches -2021 from below, 0.5 - a₁ approaches 0.5 - (-2021) = 2021.5, which is also excluded. Therefore, the interval is open, so a₁ must be strictly between -2022 and -2021. However, in the previous example, when a₁ was exactly -2021.5, which is inside the interval (-2022, -2021), the vertex is at n=2022, and S₂₀₂₂ is still the minimal. Therefore, even though the vertex is exactly at n=2022, it's included in the interval. Wait, but when a₁ = -2021.5, 0.5 - a₁ = 0.5 - (-2021.5) = 2022, which is exactly in the middle of the interval [2021.5, 2022.5). Wait, but according to the original inequality, it's 2021.5 < 0.5 - a₁ < 2022.5. So 2022 is included in the interval. Therefore, the original interval for 0.5 - a₁ is (2021.5, 2022.5). Therefore, 0.5 - a₁ can be exactly 2022, which would mean a₁ = -2021.5. So in that case, the inequalities should actually be 2021.5 ≤ 0.5 - a₁ ≤ 2022.5? But no, because when 0.5 - a₁ is exactly 2021.5, then the vertex is at 2021.5, which would make n=2021 and n=2022 have equal sums? Wait, no. Wait, if the vertex is at 2021.5, then S₂₀₂₁ and S₂₀₂₂ would be equidistant from the vertex. But in reality, when the vertex is at 2021.5, then the minimum sum is at n=2021.5, but since n must be integer, which of the two is smaller? Let's check with a₁ = -2021.5 + ε, where ε is a small positive number. Wait, maybe this is getting too complicated.Alternatively, perhaps the correct interval is [-2022, -2021), but no, because when a₁ approaches -2022, 0.5 - a₁ approaches 2022.5, which would make the closest integer 2023, hence the minimal sum at n=2023, which is not what we want. Similarly, when a₁ approaches -2021, 0.5 - a₁ approaches 2021.5, which would make the closest integers 2021 and 2022. But if a₁ is approaching -2021 from below, then 0.5 - a₁ is approaching 2021.5 from above. Wait, 0.5 - a₁: if a₁ approaches -2021 from below, that is, a₁ approaches -2021 from the left, then -a₁ approaches 2021 from above, so 0.5 - a₁ approaches 2021 + 0.5 = 2021.5. Therefore, if a₁ is just less than -2021, say a₁ = -2021 - ε for very small ε > 0, then 0.5 - a₁ = 0.5 - (-2021 - ε) = 2021.5 + ε. Therefore, 0.5 - a₁ is just above 2021.5, so the closest integer is 2022. Therefore, even as a₁ approaches -2021 from below, the minimal sum is still at n=2022.Wait, but if a₁ is exactly -2021, then 0.5 - a₁ = 0.5 - (-2021) = 2021.5, so the vertex is at 2021.5. Then, S₂₀₂₁ and S₂₀₂₂ would be equidistant from the vertex, but would they be equal?Let's check with a₁ = -2021.Compute S₂₀₂₁: (2021/2)(2*(-2021) + 2020) = (2021/2)(-4042 + 2020) = (2021/2)(-2022) = -2021*1011Compute S₂₀₂₂: (2022/2)(2*(-2021) + 2021) = 1011*(-4042 + 2021) = 1011*(-2021)So S₂₀₂₁ = -2021*1011 and S₂₀₂₂ = -2021*1011. So they are equal. Therefore, if a₁ = -2021, then S₂₀₂₁ = S₂₀₂₂, which violates the condition that S₂₀₂₂ is the smallest. Therefore, a₁ cannot be -2021. Similarly, if a₁ is -2022, then 0.5 - a₁ = 0.5 - (-2022) = 2022.5, which is the midpoint between 2022 and 2023. Then, S₂₀₂₂ and S₂₀₂₃ would be equal. Let's check:a₁ = -2022S₂₀₂₂ = (2022/2)(2*(-2022) + 2021) = 1011*(-4044 + 2021) = 1011*(-2023)S₂₀₂₃ = (2023/2)(2*(-2022) + 2022) = (2023/2)(-4044 + 2022) = (2023/2)(-2022) = -2023*1011Therefore, S₂₀₂₂ = -2023*1011 and S₂₀₂₃ = -2023*1011. So they are equal. Hence, a₁ cannot be -2022 either. Therefore, the interval for a₁ must exclude -2022 and -2021. Thus, a₁ must be strictly between -2022 and -2021. So the open interval (-2022, -2021).Therefore, the possible values of the first term a₁ are all real numbers between -2022 and -2021, not including the endpoints. So, in boxed notation, we can write that a₁ ∈ (-2022, -2021), or using inequalities, -2022 < a₁ < -2021.Just to recap: Since Sₙ is a quadratic function of n opening upwards, its minimum is at the vertex n = 0.5 - a₁. For S₂₀₂₂ to be the smallest sum, this vertex must lie in the interval (2021.5, 2022.5), leading to the requirement that 0.5 - a₁ is in that interval, which solves to a₁ ∈ (-2022, -2021). Therefore, the first term must be in that open interval.Final AnswerThe first term of the progression can take any value in the interval boxed{(-2022, -2021)}.</think>Given an arithmetic progression ((a_n)) with a common difference of 1, we need to determine the values of the first term (a_1) such that (S_{2022}) is the smallest among all sums (S_n).The sum (S_n) of the first (n) terms of an arithmetic progression is given by:[S_n = frac{n}{2} [2a_1 + (n - 1)]]We need (S_{2022}) to be the minimum. This sum is a quadratic function in (n):[S_n = frac{1}{2}n^2 + left(a_1 - frac{1}{2}right)n]The vertex of this quadratic function, which gives the minimum value, occurs at:[n = frac{1}{2} - a_1]For (S_{2022}) to be the minimum, the vertex must lie in the interval ((2021.5, 2022.5)). This ensures that the closest integer to the vertex is 2022. Therefore, we need:[2021.5 < frac{1}{2} - a_1 < 2022.5]Solving these inequalities:1. (2021.5 < frac{1}{2} - a_1)2. (frac{1}{2} - a_1 < 2022.5)Subtracting 0.5 from all parts:1. (2021 < -a_1)2. (-a_1 < 2022)Multiplying by -1 and reversing the inequalities:1. (-2021 > a_1)2. (a_1 > -2022)Combining these, we get:[-2022 < a_1 < -2021]Thus, the first term (a_1) must lie in the open interval ((-2022, -2021)).[boxed{(-2022, -2021)}],<think>Okay, let's see. The problem says that we have an arithmetic progression (aₙ) with a common difference of 1. So, that means each term is 1 more than the previous one. For example, if the first term is a₁, then the second term is a₁ + 1, the third is a₁ + 2, and so on. Got that.Now, it says that S₂₀₂₂ is the smallest among all Sₙ, meaning the sum of the first 2022 terms is less than the sum of the first n terms for any other n. We need to find the possible values that the first term a₁ can take. Hmm, interesting.Let me recall the formula for the sum of the first n terms of an arithmetic progression. The sum Sₙ is given by Sₙ = n/2 [2a₁ + (n - 1)d], where d is the common difference. Since d = 1 here, the formula simplifies to Sₙ = n/2 [2a₁ + (n - 1) * 1] = n/2 [2a₁ + n - 1]. So, Sₙ = (n/2)(2a₁ + n - 1).We need S₂₀₂₂ to be the minimum among all Sₙ. So, the sum Sₙ as a function of n must attain its minimum at n = 2022. Therefore, we can model Sₙ as a quadratic function in terms of n and find the conditions under which its minimum occurs at n = 2022.Wait, let's check that. If we treat n as a continuous variable, then Sₙ is a quadratic function in n. The minimum of this quadratic function occurs at a particular n, and we want this n to be 2022. But since n is actually a positive integer, the minimum might not exactly occur at 2022, but we need S₂₀₂₂ to be the smallest among all integer n. So, perhaps the vertex of the parabola representing Sₙ as a function of n should be near n = 2022. Specifically, the vertex should be such that the closest integer to it is 2022. Therefore, the vertex n-coordinate should be in the interval (2021.5, 2022.5). Hmm, maybe that's the way to approach it.Let me write Sₙ as a quadratic function of n. Let's expand the sum formula:Sₙ = (n/2)(2a₁ + n - 1) = (n/2)(n + 2a₁ - 1) = (1/2)n² + (a₁ - 0.5)n.So, in terms of n, Sₙ is a quadratic function: S(n) = (1/2)n² + (a₁ - 0.5)n. Since the coefficient of n² is positive (1/2 > 0), the parabola opens upwards, so the minimum occurs at the vertex. The vertex of a quadratic function an² + bn + c is at n = -b/(2a). In this case, a = 1/2 and b = (a₁ - 0.5). Therefore, the vertex is at n = - (a₁ - 0.5)/(2*(1/2)) = - (a₁ - 0.5)/1 = -a₁ + 0.5.Wait, that seems odd. Wait, let's check the formula again. For S(n) = (1/2)n² + (a₁ - 0.5)n, the coefficients are:a = 1/2 (coefficient of n²)b = (a₁ - 0.5) (coefficient of n)Thus, vertex at n = -b/(2a) = -(a₁ - 0.5)/(2*(1/2)) = -(a₁ - 0.5)/1 = -a₁ + 0.5.But n is a positive integer, so how can the vertex be at n = -a₁ + 0.5? That would require that -a₁ + 0.5 must be a positive number. So, -a₁ + 0.5 > 0 => a₁ < 0.5. Hmm. Wait, but if a₁ is the first term of the arithmetic progression, which could be any real number, right? Since the problem doesn't specify that the terms are integers or anything. So, a₁ can be any real number.But the vertex is at n = -a₁ + 0.5. For this vertex to correspond to the minimum point of the sum Sₙ, which is a parabola opening upwards, the minimal value is at n = -a₁ + 0.5. However, since n must be a positive integer, the minimal sum Sₙ will occur at the integer n closest to the vertex. Therefore, for S₂₀₂₂ to be the minimal sum, the vertex must be in the interval [2022 - 0.5, 2022 + 0.5). So, the vertex n-coordinate must lie between 2021.5 and 2022.5. Hence:2021.5 ≤ -a₁ + 0.5 < 2022.5Solving for a₁:2021.5 ≤ -a₁ + 0.5 < 2022.5Subtract 0.5 from all parts:2021 ≤ -a₁ < 2022Multiply all parts by -1 and reverse inequalities:-2021 ≥ a₁ > -2022Which can be rewritten as:-2022 < a₁ ≤ -2021Therefore, the first term a₁ must be in the interval (-2022, -2021]. Since a₁ can take any real value in this interval, the possible values of a₁ are all real numbers greater than -2022 and less than or equal to -2021.Wait, but let me verify this conclusion. Let's take an example. Suppose a₁ is -2021.5. Then the vertex is at n = -(-2021.5) + 0.5 = 2021.5 + 0.5 = 2022. So, the vertex is exactly at n = 2022, so the minimum occurs at n = 2022.If a₁ is slightly less than -2021, say a₁ = -2021.1, then the vertex is at n = -(-2021.1) + 0.5 = 2021.1 + 0.5 = 2021.6. So, the vertex is at 2021.6, which is between 2021.5 and 2022.5? Wait, 2021.6 is between 2021.5 and 2022.5? Wait, 2021.5 is 2021.5, 2022.5 is 2022.5. So, 2021.6 is less than 2022.5, but greater than 2021.5. So, according to our initial interval, if the vertex is between 2021.5 and 2022.5, then the closest integer is 2022, right? So, if the vertex is at 2021.6, then the nearest integer is 2022. Similarly, if the vertex is at 2022.4, the nearest integer is 2022. So, as long as the vertex is in [2021.5, 2022.5), the closest integer is 2022, so the minimal Sₙ occurs at n = 2022.Therefore, the interval for the vertex n is [2021.5, 2022.5). Then, the corresponding a₁ is given by:n_vertex = -a₁ + 0.5So,2021.5 ≤ n_vertex < 2022.5Substitute n_vertex:2021.5 ≤ -a₁ + 0.5 < 2022.5Subtract 0.5:2021 ≤ -a₁ < 2022Multiply by -1 and reverse inequalities:-2021 ≥ a₁ > -2022Which is equivalent to:-2022 < a₁ ≤ -2021Therefore, a₁ must be in (-2022, -2021].But let's check the endpoints. If a₁ is exactly -2021, then the vertex is at n = -(-2021) + 0.5 = 2021 + 0.5 = 2021.5. So, the vertex is at 2021.5. Then, the minimum sum would be at n = 2021 or n = 2022. But since the vertex is exactly halfway between 2021 and 2022, we need to check which of S₂₀₂₁ and S₂₀₂₂ is smaller.Wait, when the vertex is exactly at 2021.5, then the function is minimized at that point, but since n must be integer, both n=2021 and n=2022 would give the same minimal value? Wait, no, that's not necessarily the case. Let's compute Sₙ when a₁ = -2021.So, if a₁ = -2021, then Sₙ = (1/2)n² + (a₁ - 0.5)n = (1/2)n² + (-2021 - 0.5)n = (1/2)n² - 2021.5n.Compute S₂₀₂₁ and S₂₀₂₂:S₂₀₂₁ = (1/2)(2021)^2 - 2021.5*2021= (2021^2)/2 - (2021.5)(2021)Similarly, S₂₀₂₂ = (1/2)(2022)^2 - 2021.5*2022Let's compute S₂₀₂₁:First term: (2021^2)/2 = (2021*2021)/2Second term: (2021.5)(2021) = (2021 + 0.5)(2021) = 2021^2 + 0.5*2021Therefore, S₂₀₂₁ = (2021^2)/2 - (2021^2 + 0.5*2021) = - (2021^2)/2 - 0.5*2021Similarly, compute S₂₀₂₂:First term: (2022^2)/2Second term: (2021.5)(2022) = (2021 + 0.5)(2022) = 2021*2022 + 0.5*2022Therefore, S₂₀₂₂ = (2022^2)/2 - (2021*2022 + 0.5*2022) = (2022^2)/2 - 2021*2022 - 0.5*2022Let's simplify S₂₀₂₂:= (2022*(2022))/2 - 2021*2022 - 0.5*2022= (2022*1011) - 2021*2022 - 1011= 2022*(1011 - 2021) - 1011= 2022*(-1010) - 1011= -2022*1010 - 1011Similarly, S₂₀₂₁:= - (2021^2)/2 - 0.5*2021Wait, this is getting complicated. Maybe instead of plugging in the numbers, we can compute the difference S₂₀₂₂ - S₂₀₂₁ and see if it's positive or negative.Compute S₂₀₂₂ - S₂₀₂₁:= [ (1/2)(2022)^2 - 2021.5*2022 ] - [ (1/2)(2021)^2 - 2021.5*2021 ]= (1/2)(2022^2 - 2021^2) - 2021.5*(2022 - 2021)= (1/2)(2022 - 2021)(2022 + 2021) - 2021.5*(1)= (1/2)(1)(4043) - 2021.5= (4043)/2 - 2021.5= 2021.5 - 2021.5 = 0So, when a₁ = -2021, the sums S₂₀₂₁ and S₂₀₂₂ are equal. Therefore, both n=2021 and n=2022 give the same sum. But the problem states that S₂₀₂₂ is the smallest among all Sₙ. So, if a₁ = -2021, then S₂₀₂₁ = S₂₀₂₂, which means S₂₀₂₂ is not strictly smaller than S₂₀₂₁. Therefore, a₁ cannot be equal to -2021, because in that case, there exists another n (namely 2021) where the sum is the same. Therefore, the first term must be greater than -2022 and less than -2021. So, the interval is -2022 < a₁ < -2021. But wait, the initial calculation said -2022 < a₁ ≤ -2021, but the endpoint at a₁ = -2021 is invalid because the sums at n=2021 and n=2022 are equal. Therefore, a₁ must be strictly less than -2021 and greater than -2022. So, the correct interval is (-2022, -2021).Wait, but let's check another endpoint. Suppose a₁ approaches -2022 from the right. Let's say a₁ = -2022 + ε, where ε is a very small positive number. Then, the vertex n is:n_vertex = -a₁ + 0.5 = -(-2022 + ε) + 0.5 = 2022 - ε + 0.5 = 2022.5 - ε.As ε approaches 0 from the positive side, n_vertex approaches 2022.5. So, the vertex is approaching 2022.5 from the left. Therefore, the closest integer to n_vertex is 2022 when ε is between 0 and 1. So, if a₁ is just above -2022, like -2022 + 0.1, then n_vertex is 2022.5 - 0.1 = 2022.4, which is closer to 2022 than to 2023. Therefore, S₂₀₂₂ would be the minimal sum. Similarly, if a₁ is just below -2021, like -2021 - 0.1, then n_vertex = -(-2021 - 0.1) + 0.5 = 2021 + 0.1 + 0.5 = 2021.6, which is closer to 2022 than to 2021. So, S₂₀₂₂ is still the minimal.But if a₁ = -2022, then n_vertex = -(-2022) + 0.5 = 2022 + 0.5 = 2022.5. But then, the vertex is exactly at 2022.5. So, the sum Sₙ would be minimized at both n=2022 and n=2023, because 2022.5 is midway between them. Let's check that. If a₁ = -2022, then Sₙ = (1/2)n² + (-2022 - 0.5)n = (1/2)n² - 2022.5n.Compute S₂₀₂₂ and S₂₀₂₃:S₂₀₂₂ = (1/2)(2022)^2 - 2022.5*2022= (2022*2022)/2 - 2022.5*2022= 2022*(1011) - 2022.5*2022= 2022*(1011 - 2022.5)= 2022*(-1011.5)Similarly, S₂₀₂₃ = (1/2)(2023)^2 - 2022.5*2023= (2023*2023)/2 - 2022.5*2023= 2023*(1011.5) - 2022.5*2023= 2023*(1011.5 - 2022.5)= 2023*(-1011)Comparing S₂₀₂₂ = 2022*(-1011.5) and S₂₀₂₃ = 2023*(-1011). Since both are negative numbers, the one with the smaller absolute value is larger. Let's compute their magnitudes:|S₂₀₂₂| = 2022*1011.5|S₂₀₂₃| = 2023*1011Compute the ratio:|S₂₀₂₂| / |S₂₀₂₃| = (2022*1011.5)/(2023*1011) = (2022/2023)*(1011.5/1011) ≈ (slightly less than 1)*(1.000495) ≈ slightly less than 1.000495. So, |S₂₀₂₂| is slightly less than |S₂₀₂₃|, which means S₂₀₂₂ is slightly more negative than S₂₀₂₃, so S₂₀₂₂ is smaller than S₂₀₂₃.Wait, but if S₂₀₂₂ is more negative, that means it's smaller. So, S₂₀₂₂ < S₂₀₂₃. However, S₂₀₂₂ is also compared to S₂₀₂₁. Let's compute S₂₀₂₁ when a₁ = -2022:S₂₀₂₁ = (1/2)(2021)^2 - 2022.5*2021= (2021^2)/2 - 2022.5*2021= 2021*(2021/2 - 2022.5)= 2021*(1010.5 - 2022.5)= 2021*(-1012)Compare to S₂₀₂₂ = 2022*(-1011.5). Let's compute their actual values:S₂₀₂₁ = -2021*1012 = - (2021*1000 + 2021*12) = -2,021,000 - 24,252 = -2,045,252S₂₀₂₂ = -2022*1011.5 = -2022*(1000 + 11.5) = -2022*1000 - 2022*11.5 = -2,022,000 - 23,253 = -2,045,253Wait, so S₂₀₂₁ = -2,045,252 and S₂₀₂₂ = -2,045,253. Therefore, S₂₀₂₂ is smaller (more negative) than S₂₀₂₁. So, when a₁ = -2022, S₂₀₂₂ is still smaller than both S₂₀₂₁ and S₂₀₂₃, so S₂₀₂₂ is the minimal sum. Wait, but according to our previous interval, when a₁ = -2022, the vertex is at n = 2022.5, but the sums at n=2022 and n=2023 are such that S₂₀₂₂ < S₂₀₂₃, and S₂₀₂₂ < S₂₀₂₁. Therefore, even though the vertex is at 2022.5, the minimal sum is still at n=2022. Therefore, maybe a₁ can be equal to -2022?But wait, in our initial calculation, we had:2021.5 ≤ n_vertex < 2022.5But when a₁ = -2022, n_vertex = 2022.5, which is not less than 2022.5. So, the interval was supposed to be up to but not including 2022.5. Hence, when a₁ = -2022, the vertex is exactly at 2022.5, which is outside the interval [2021.5, 2022.5). Therefore, a₁ = -2022 is not included. However, in reality, even when a₁ = -2022, the minimal sum is still at n=2022, as shown by the calculations. So, perhaps the initial interval was wrong?Wait, perhaps the issue is that even if the vertex is at n=2022.5, the minimal sum is still at the integer n closest to the vertex. But since the vertex is exactly halfway between 2022 and 2023, the sum at n=2022 and n=2023 would need to be compared. However, in this specific case, due to the arithmetic progression's properties, the sum at n=2022 is actually smaller than at n=2023, even though the vertex is at 2022.5. So, maybe even when the vertex is at 2022.5, S₂₀₂₂ is still the minimum. Therefore, maybe a₁ can be as low as -2022, but when a₁ = -2022, the vertex is at 2022.5, but S₂₀₂₂ is still the minimal sum. So, perhaps the initial interval should include a₁ = -2022? Let's double-check.If a₁ = -2022, then Sₙ = (1/2)n² + (-2022 - 0.5)n = (1/2)n² - 2022.5n.The derivative with respect to n (treating n as continuous) is dS/dn = n - 2022.5. Setting derivative to zero gives n = 2022.5. So, the minimum is at n = 2022.5. Since n must be an integer, the minimal sum occurs at either n=2022 or n=2023. To determine which one is smaller, compute S(2022.5) and see the behavior. But since the function is a parabola opening upwards, the sums decrease until n=2022.5 and then increase. Therefore, Sₙ is decreasing for n < 2022.5 and increasing for n > 2022.5. Therefore, the minimal integer n would be the one where the function starts increasing. So, between n=2022 and n=2023, since the minimum is at 2022.5, the sum at n=2022 is decreasing up to that point, and then increasing after. Therefore, S₂₀₂₂ is the last term before the minimum, and S₂₀₂₃ is the first term after the minimum. Wait, but in reality, when n increases, the terms of the arithmetic progression are a₁ + (n-1)*1 = a₁ + n -1. Since the common difference is 1, the terms are increasing by 1 each time. But the sum Sₙ is the sum of these terms. If a₁ is negative enough, the terms might start negative and become positive at some point. So, the sum Sₙ would decrease until the terms become positive, and then start increasing. Wait, but if the common difference is positive (1), then once the terms become positive, they stay positive. So, the sum Sₙ would have a minimum when the terms transition from negative to positive.Wait, maybe another way to think about this: the sum Sₙ is minimized when the terms transition from negative to positive. Because adding positive terms would start increasing the sum, while adding negative terms would decrease the sum. So, the minimal sum occurs when you add all the negative terms and the first non-negative term. Wait, but in an arithmetic progression with common difference 1, starting at a₁, the terms are a₁, a₁ +1, a₁ +2, ..., a₁ + (n -1). So, the terms are negative until a₁ + (k -1) < 0. So, the last negative term is when a₁ + (k -1) < 0. Solving for k: k < -a₁ +1. So, the number of negative terms is floor(-a₁). Then, the sum would be minimized around the time when you stop adding negative terms and start adding positive terms.But in this problem, we have a common difference of 1, so the terms increase by 1 each time. If the first term is a₁, then the terms will be negative until a₁ + (n -1) < 0. So, n < -a₁ +1. Therefore, the number of negative terms is floor(-a₁). Then, the minimal sum occurs around n ≈ -a₁. Hmm, but how does this relate to the sum formula?Alternatively, maybe the minimal sum occurs when the average of the first n terms is closest to zero. The average of the first n terms is Sₙ/n = (1/2)(2a₁ + n -1). So, setting this average to zero gives 2a₁ + n -1 = 0 => n = -2a₁ +1. Therefore, the average is zero when n = -2a₁ +1. So, if n is an integer near -2a₁ +1, then the average is near zero, which might correspond to the minimal sum.Wait, but Sₙ is a quadratic function, so the minimal sum occurs at the vertex n = -a₁ + 0.5 as we derived earlier. Hmm, but according to the average approach, the average is zero at n = -2a₁ +1. There's a discrepancy here. Let me check.Wait, the average of the first n terms is (Sₙ)/n = (1/2)(2a₁ + n -1). So, if we set this equal to zero, we get 2a₁ + n -1 = 0 => n = -2a₁ +1. However, the vertex of Sₙ is at n = -a₁ + 0.5. So, these are different. Therefore, my previous thought process was conflicting. Let me resolve this.The minimal sum Sₙ occurs at the vertex n = -a₁ + 0.5. However, the average being zero occurs at n = -2a₁ +1. These are different. So, perhaps when the average crosses zero, the sum is minimized? Wait, no. The sum being minimized is a separate condition. Let's take an example. Suppose a₁ = -10. Then the arithmetic progression is -10, -9, -8, ..., and so on. The sum Sₙ will decrease until the terms become positive. The minimal sum occurs when we add the last negative term. For example, if a₁ = -10, the terms are negative until n = 10, since a₁ + (n -1) = -10 + 9 = -1 (n=10), and n=11 term is 0. So, the sum S₁₀ is the sum of all negative terms: -10 -9 -8 ... -1 = -55. Then S₁₁ = -55 + 0 = -55, S₁₂ = -55 + 1 = -54, etc. So, the minimal sum occurs at n=10 and n=11. So, in this case, the minimal sum occurs at n where the terms transition from negative to non-negative. So, n = -a₁ (if a₁ is an integer). If a₁ is not an integer, say a₁ = -10.5, then the terms are -10.5, -9.5, ..., up to n=10: term is -10.5 +9 = -1.5, n=11: term is -10.5 +10 = -0.5, n=12: term is -10.5 +11 = 0.5. So, the terms are negative until n=12. So, the minimal sum would occur at n=11 or n=12. Calculating the sum up to n=11: sum of -10.5, -9.5, ..., -0.5. That's 11 terms. The average is (-10.5 + (-0.5))/2 = -11/2 = -5.5. So, sum is 11*(-5.5) = -60.5. Then, sum up to n=12: -60.5 + 0.5 = -60.0. So, the minimal sum is at n=11. So, when a₁ = -10.5, the minimal sum is at n=11, which is approximately -a₁ +0.5 = 10.5 +0.5=11. So, this matches the vertex formula. So, the vertex at n=11, and indeed, the minimal sum is at n=11.So, in this case, the minimal sum occurs at n=ceil(-a₁ -0.5). Wait, maybe not. Let me check with a₁ = -10.5, which gives vertex at n=11. So, the vertex is exactly at 11, so n=11 is the minimal. When a₁ = -10.6, the vertex is at n = -(-10.6) +0.5 = 10.6 +0.5=11.1. So, the minimal sum is at n=11, since 11.1 is closer to 11 than 12? Wait, 11.1 is 0.1 above 11, so the closest integer is 11. Wait, no, 11.1 is between 11 and 12, closer to 11. So, the minimal sum is at n=11. If a₁ = -10.9, vertex at n=10.9 +0.5=11.4, which is still closer to 11 than 12. Wait, 11.4 is 0.4 away from 11 and 0.6 away from 12. So, still n=11. If a₁ = -11, then vertex at n=11 +0.5=11.5, which is equidistant between 11 and 12. Then, as before, the sums at n=11 and n=12 would be equal?Wait, if a₁ = -11, then Sₙ = (1/2)n² + (-11 -0.5)n = (1/2)n² -11.5n.Compute S₁₁ and S₁₂:S₁₁ = (1/2)(121) -11.5*11 = 60.5 -126.5 = -66S₁₂ = (1/2)(144) -11.5*12 =72 -138= -66So, indeed, S₁₁ = S₁₂ when a₁ = -11. Therefore, the minimal sum occurs at both n=11 and n=12. So, similar to the previous example, when a₁ is such that the vertex is at a half-integer, the sums at the two nearest integers are equal. Therefore, in order for S₂₀₂₂ to be strictly smaller than all other Sₙ, the vertex must not be at a half-integer, but must lie strictly within the interval (2022 -0.5, 2022 +0.5), i.e., (2021.5, 2022.5). Therefore, the vertex must not be exactly at 2022.5. Therefore, the initial interval was correct: 2021.5 ≤ n_vertex <2022.5, leading to -2022 <a₁ ≤-2021. However, when a₁ = -2021, n_vertex=2021.5, which would imply the minimal sum occurs at n=2021 and n=2022, but as shown earlier, when a₁ = -2021, S₂₀₂₁ and S₂₀₂₂ are equal, hence S₂₀₂₂ is not strictly smaller. Therefore, a₁ must be greater than -2022 and less than -2021.Wait, but when a₁ = -2022 + ε, where ε is a small positive number, then n_vertex = 2022.5 - ε. If ε approaches 0, n_vertex approaches 2022.5 from the left. So, even if a₁ approaches -2022 from above, n_vertex approaches 2022.5, but never reaches it. Therefore, for all a₁ in (-2022, -2021), n_vertex is in (2021.5, 2022.5), so S₂₀₂₂ is strictly the minimal sum. When a₁ = -2022, n_vertex =2022.5, but in that case, S₂₀₂₂ and S₂₀₂₃ are both minima? Wait, earlier when a₁=-2022, we saw that S₂₀₂₂ was actually smaller than S₂₀₂₁ and S₂₀₂₃. Wait, no:When a₁ = -2022, the sum Sₙ = (1/2)n² -2022.5n.Compute S₂₀₂₂:=0.5*(2022)^2 -2022.5*2022=0.5*(2022)(2022) -2022.5*2022= (1011)(2022) -2022.5*2022=2022*(1011 -2022.5)=2022*(-1011.5)Similarly, S₂₀₂₁:=0.5*(2021)^2 -2022.5*2021=0.5*2021*2021 -2022.5*2021=2021*(0.5*2021 -2022.5)=2021*(1010.5 -2022.5)=2021*(-1012)S₂₀₂₃:=0.5*(2023)^2 -2022.5*2023=0.5*2023*2023 -2022.5*2023=2023*(0.5*2023 -2022.5)=2023*(1011.5 -2022.5)=2023*(-1011)Now, compare these:S₂₀₂₁ = -2021*1012 = -(2021*1012)S₂₀₂₂ = -2022*1011.5S₂₀₂₃ = -2023*1011Calculate numerically:2021*1012: 2021*1000=2,021,000; 2021*12=24,252; total=2,045,2522022*1011.5: 2022*1000=2,022,000; 2022*11.5=23,253; total=2,045,2532023*1011: 2023*1000=2,023,000; 2023*11=22,253; total=2,045,253So,S₂₀₂₁ = -2,045,252S₂₀₂₂ = -2,045,253S₂₀₂₃ = -2,045,253Wait, so S₂₀₂₂ is -2,045,253, which is less than S₂₀₂₁ (-2,045,252) but equal to S₂₀₂₃. Therefore, when a₁ = -2022, S₂₀₂₂ = S₂₀₂₃, both equal to -2,045,253, which is less than S₂₀₂₁. Therefore, the minimal sum occurs at both n=2022 and n=2023. Therefore, S₂₀₂₂ is not strictly smaller than all other Sₙ. Hence, when a₁ = -2022, there exists another n (namely 2023) with the same minimal sum. Therefore, a₁ = -2022 is excluded. Similarly, when a₁ = -2021, S₂₀₂₁ = S₂₀₂₂, so again, there's another n with the same sum. Therefore, to ensure that S₂₀₂₂ is strictly smaller than all other Sₙ, the value of a₁ must lie in the open interval (-2022, -2021).Therefore, the first term a₁ can take any real value between -2022 and -2021, not including the endpoints.So, the answer is that the first term a₁ must satisfy -2022 < a₁ < -2021. In boxed notation, that would be boxed{(-2022, -2021)}.Final AnswerThe first term of the progression can take values in the interval boxed{(-2022, -2021)}.</think>Given an arithmetic progression ((a_n)) with a common difference of 1, we need to determine the values of the first term (a_1) such that (S_{2022}) is the smallest among all (S_n) (the sum of the first (n) terms for any (n)).The sum (S_n) of the first (n) terms of an arithmetic progression is given by:[S_n = frac{n}{2} [2a_1 + (n - 1)]]This simplifies to:[S_n = frac{1}{2}n^2 + left(a_1 - frac{1}{2}right)n]We need (S_{2022}) to be the minimum value. Treating (S_n) as a quadratic function in (n), the vertex of this parabola (which opens upwards) gives the minimum value. The vertex of (S_n = frac{1}{2}n^2 + left(a_1 - frac{1}{2}right)n) occurs at:[n = -frac{b}{2a} = -frac{a_1 - frac{1}{2}}{2 cdot frac{1}{2}} = -a_1 + frac{1}{2}]For (S_{2022}) to be the minimum, the vertex (n = -a_1 + frac{1}{2}) must lie in the interval ([2021.5, 2022.5)). Solving for (a_1):[2021.5 leq -a_1 + frac{1}{2} < 2022.5]Subtracting (frac{1}{2}) from all parts:[2021 leq -a_1 < 2022]Multiplying by (-1) and reversing the inequalities:[-2021 geq a_1 > -2022]This simplifies to:[-2022 < a_1 leq -2021]However, we need (S_{2022}) to be strictly the smallest. Checking the endpoints:- If (a_1 = -2021), the vertex is at (n = 2021.5), leading to equal sums (S_{2021}) and (S_{2022}).- If (a_1 = -2022), the vertex is at (n = 2022.5), leading to equal sums (S_{2022}) and (S_{2023}).Thus, the endpoints are excluded, leading to the interval:[-2022 < a_1 < -2021]The first term (a_1) can take values in the interval (boxed{(-2022, -2021)}).]

🔑:Answer: a_{1} belongs to the interval (-2022; -2021).Solution. Since the difference of the progression is positive, the progression is increasing. Therefore, the described situation is possible if and only if the members of the progression from the first to the 2022nd are negative, and starting from the 2023rd, they are positive. Thus, mathrm{S}_{2022} will be the smallest if and only if a_{2022}<0 and a_{2023}>0. From this, we get the system of inequalitiesleft{begin{array}{l}a_{1}+2021d<0 a_{1}+2022d>0end{array} Leftrightarrow -2022<a_{1}<-2021right..

🧠:[<think>Okay, so I have this problem here where I need to find the ratio of the volume of a convex polyhedron formed by the midpoints of the edges of a tetrahedron to the original tetrahedron's volume. The options are given as fractions, so it's a multiple-choice question. Let me think about how to approach this.First, let's recall that a tetrahedron has four triangular faces, six edges, and four vertices. The problem mentions forming a convex polyhedron by using the midpoints of its edges. Hmm, if we take midpoints of all edges, how many points would that be? Since there are six edges, that would be six midpoints. Now, connecting these midpoints should form a new polyhedron. The question is, what does this polyhedron look like, and how do we compute its volume compared to the original?Maybe I should visualize this. Let me imagine a tetrahedron. Each edge has a midpoint, so if I connect these midpoints, perhaps the new polyhedron is another tetrahedron? But wait, a tetrahedron has four vertices, but here we have six midpoints. So maybe the convex polyhedron formed isn't a tetrahedron but something else. Maybe an octahedron? Because an octahedron has six vertices, right? Let me check: no, a regular octahedron has six vertices and twelve edges. Wait, but here we have six midpoints. If we connect them appropriately, maybe it's a different shape.Alternatively, maybe it's a smaller tetrahedron inside the original one. But with six points, perhaps it's a more complex polyhedron. Wait, another thought: when you connect the midpoints of edges in a tetrahedron, the resulting figure is called the midpoint tetrahedron or maybe the medial polyhedron. But I need to recall or derive its properties.Alternatively, maybe using coordinates would help. Let me assign coordinates to the original tetrahedron to compute the volumes. Let's choose a simple tetrahedron with coordinates that make calculations easier. For example, let me take a regular tetrahedron, but maybe even a simpler one.Wait, but the problem doesn't specify that the tetrahedron is regular, just any tetrahedron. So the ratio should hold for any tetrahedron, not just a regular one. Therefore, I can choose coordinates for a general tetrahedron or perhaps use vector methods.Let me try coordinates. Let's assign coordinates to the original tetrahedron. Let me choose vertices at (0,0,0), (1,0,0), (0,1,0), (0,0,1). This is a standard tetrahedron with volume V1. The volume of this tetrahedron is 1/6, since the volume formula for a tetrahedron with vertices at the origin and (a,0,0), (0,b,0), (0,0,c) is (abc)/6. In this case, a=b=c=1, so 1/6. But maybe I should confirm.Wait, actually, the volume of the tetrahedron with vertices at (0,0,0), (1,0,0), (0,1,0), (0,0,1) is indeed 1/6. So V1 = 1/6. Then, the midpoints of its edges would be the midpoints of each of the six edges. Let's compute those midpoints.First, the edges:1. From (0,0,0) to (1,0,0): midpoint is (0.5, 0, 0)2. From (0,0,0) to (0,1,0): midpoint is (0, 0.5, 0)3. From (0,0,0) to (0,0,1): midpoint is (0, 0, 0.5)4. From (1,0,0) to (0,1,0): midpoint is (0.5, 0.5, 0)5. From (1,0,0) to (0,0,1): midpoint is (0.5, 0, 0.5)6. From (0,1,0) to (0,0,1): midpoint is (0, 0.5, 0.5)So the midpoints are these six points. Now, the convex polyhedron formed by these midpoints. Let me see if I can figure out the shape.Looking at these six points, maybe they form an octahedron? Let me check. An octahedron has eight faces, twelve edges, and six vertices. Wait, no, a regular octahedron has six vertices, twelve edges, and eight triangular faces. But here we have six vertices, which are the midpoints. Let me see if these points form an octahedron.But in 3D space, the convex hull of these six midpoints. Let me try to see how they are connected. Each original edge is connected to two others. Wait, maybe the convex hull is a polyhedron where each midpoint is connected to the midpoints of adjacent edges.Alternatively, perhaps the polyhedron formed is another tetrahedron? But no, since there are six points. Wait, if we take these midpoints, each face of the original tetrahedron will have three midpoints. Maybe the new polyhedron is dual to the original in some way.Alternatively, perhaps the new polyhedron can be considered as the intersection of the original tetrahedron with its dual or something like that. Hmm, this is getting a bit abstract. Maybe I should instead compute the volume directly.But how? Let's think. If I can figure out the coordinates of the convex polyhedron V2, then I can compute its volume.Wait, the convex hull of the six midpoints. Let me list all the midpoints again:1. (0.5, 0, 0)2. (0, 0.5, 0)3. (0, 0, 0.5)4. (0.5, 0.5, 0)5. (0.5, 0, 0.5)6. (0, 0.5, 0.5)Now, let's see. To form the convex hull, we need to find all the convex combinations of these points. But perhaps it's easier to see which points are vertices of the convex hull. All six points are vertices, because none of them can be expressed as a convex combination of the others. For example, (0.5, 0, 0) is not a combination of the others. Similarly for all. So the convex hull is a polyhedron with six vertices.Now, what is this polyhedron? Let's try to see the edges. Each original edge's midpoint is connected to midpoints of adjacent edges. For example, the midpoint of edge (0,0,0)-(1,0,0) is (0.5,0,0). This edge is adjacent to edges (0,0,0)-(0,1,0), (0,0,0)-(0,0,1), and (1,0,0)-(0,1,0), (1,0,0)-(0,0,1). Wait, no. The edge (0,0,0)-(1,0,0) is adjacent in the tetrahedron to the faces that include this edge. Each edge in a tetrahedron is part of two faces.Wait, maybe this is getting too complicated. Let me think of each midpoint connected to others. For example, (0.5, 0, 0) is connected to which other midpoints? The edges connected to (0.5, 0, 0) in the convex hull would be those midpoints that share a coordinate. Let me see.Looking at the list of midpoints:(0.5, 0, 0), (0, 0.5, 0), (0, 0, 0.5), (0.5, 0.5, 0), (0.5, 0, 0.5), (0, 0.5, 0.5)So, (0.5, 0, 0) can be connected to (0.5, 0.5, 0), (0.5, 0, 0.5), and also maybe (0,0.5,0) or (0,0,0.5)? Wait, no, in the convex hull, two midpoints are connected by an edge if the original edges they are midpoints of are adjacent. Because in the original tetrahedron, two edges are adjacent if they share a common vertex. Therefore, their midpoints would be connected by an edge in the new polyhedron.So, for example, the edge between (0.5, 0, 0) and (0.5, 0.5, 0) exists because the original edges (0,0,0)-(1,0,0) and (1,0,0)-(0,1,0) share the vertex (1,0,0). Therefore, their midpoints are connected in the new polyhedron. Similarly, (0.5, 0, 0) is connected to (0.5, 0, 0.5) because the original edges (0,0,0)-(1,0,0) and (1,0,0)-(0,0,1) share (1,0,0). Also, (0.5, 0, 0) is connected to (0, 0.5, 0) and (0, 0, 0.5) because those midpoints correspond to edges sharing (0,0,0).Wait, perhaps each midpoint is connected to four others? Let's check.Take (0.5, 0, 0):- Connected to (0.5, 0.5, 0) [sharing edge (1,0,0)-(0,1,0)]- Connected to (0.5, 0, 0.5) [sharing edge (1,0,0)-(0,0,1)]- Connected to (0, 0.5, 0) [sharing edge (0,0,0)-(0,1,0)]- Connected to (0, 0, 0.5) [sharing edge (0,0,0)-(0,0,1)]Yes, so each midpoint is connected to four others. So each vertex in the new polyhedron has degree four. That suggests that the polyhedron is an octahedron, since a regular octahedron has six vertices, each of degree four. So this is a octahedron, but is it regular?Wait, in our coordinate system, the midpoints are at (0.5,0,0), (0,0.5,0), (0,0,0.5), (0.5,0.5,0), (0.5,0,0.5), (0,0.5,0.5). Let's check the distances between these points. For example, the distance between (0.5,0,0) and (0,0.5,0) is sqrt((0.5)^2 + (-0.5)^2 + 0) = sqrt(0.25 + 0.25) = sqrt(0.5). Similarly, the distance between (0.5,0,0) and (0.5,0.5,0) is sqrt(0 + (0.5)^2 + 0) = 0.5. So the edges have different lengths. Therefore, the octahedron is not regular. So it's a non-regular octahedron.But regardless of regularity, the volume can be computed. So how do we compute the volume of this octahedron?Alternatively, maybe there's a smarter way than coordinates. Let me recall that when you connect midpoints of edges of a tetrahedron, the resulting octahedron can be divided into smaller tetrahedrons whose volumes relate to the original.Wait, another approach: affine transformations. Since the ratio is the same for any tetrahedron, we can choose a convenient one. Let's take the regular tetrahedron for simplicity. If the ratio is the same regardless of the tetrahedron, computing it for a regular tetrahedron would suffice.But before that, maybe scaling. If we consider the original tetrahedron with coordinates as I did before, which is a right tetrahedron with orthogonal edges, maybe the octahedron's volume can be computed by subtracting parts from the original.Wait, let's think. The convex hull of the midpoints is an octahedron inside the original tetrahedron. How is it positioned? Each midpoint is on an edge of the original tetrahedron, so the octahedron is entirely contained within the original tetrahedron.But the original tetrahedron has volume 1/6. So if we can compute V2, the volume of the octahedron, then we can find the ratio.Alternatively, maybe decompose the octahedron into simpler parts. Let's see. The octahedron can be divided into pyramids with a common square base. Wait, but the octahedron here is not regular. Alternatively, maybe split it into tetrahedrons.Alternatively, use the Cayley-Menger determinant to compute the volume given coordinates. But that might be complicated. Let me see.Alternatively, observe that each face of the octahedron corresponds to a face of the original tetrahedron. Each original face of the tetrahedron is a triangle, and the midpoints of its edges form a smaller triangle. So each face of the octahedron is a triangle, and the octahedron is formed by connecting these midpoints across the original edges.Wait, actually, in 3D, when you take midpoints of edges, the convex hull would consist of these midpoints connected in such a way that each original edge's midpoint is connected to the midpoints of adjacent edges. So perhaps each original face gives rise to a face of the octahedron? But the original tetrahedron has four triangular faces, and the octahedron has eight triangular faces. Hmm, no.Wait, maybe the octahedron has eight triangular faces. Let me check. For the six points we have, each vertex is part of several triangles. For example, (0.5,0,0) is connected to (0.5,0.5,0), (0.5,0,0.5), (0,0.5,0), and (0,0,0.5). Each of these connections forms a triangle. For instance, connecting (0.5,0,0), (0.5,0.5,0), and (0,0.5,0) forms a triangle. Similarly, (0.5,0,0), (0.5,0,0.5), and (0,0,0.5) forms another triangle. So each vertex is part of four triangles, but the total number of faces would be more than eight. Wait, maybe not.Alternatively, perhaps the octahedron is formed by two square pyramids glued together at their square bases. Let me see. If there is a square in the middle, then each pyramid would contribute four triangular faces. So total eight faces. But in our case, is there such a square?Looking at the coordinates, let's check if any four points form a square. For example, (0.5,0,0), (0.5,0.5,0), (0,0.5,0), (0,0,0). Wait, but (0,0,0) isn't one of the midpoints. The midpoints are (0.5,0,0), (0,0.5,0), (0,0,0.5), (0.5,0.5,0), (0.5,0,0.5), (0,0.5,0.5). If I take (0.5,0,0), (0.5,0.5,0), (0,0.5,0), (0,0.5,0.5), these four points form a quadrilateral. Let's check the distances:Between (0.5,0,0) and (0.5,0.5,0): 0.5Between (0.5,0.5,0) and (0,0.5,0): 0.5Between (0,0.5,0) and (0,0.5,0.5): 0.5Between (0,0.5,0.5) and (0.5,0,0): Let's compute the distance. The coordinates are (0,0.5,0.5) and (0.5,0,0). The distance is sqrt((0.5)^2 + (-0.5)^2 + (-0.5)^2) = sqrt(0.25 + 0.25 + 0.25) = sqrt(0.75) ≈ 0.866. So this is a non-planar quadrilateral? Wait, but if these four points are coplanar, then they form a quadrilateral, but the sides are not equal. However, maybe it's a planar quadrilateral. Let me check if they lie on the same plane.Take points A=(0.5,0,0), B=(0.5,0.5,0), C=(0,0.5,0), D=(0,0.5,0.5). Let's see if they are coplanar. To check coplanarity, compute the volume of the tetrahedron formed by these four points. If the volume is zero, they are coplanar.The volume can be computed using the scalar triple product. Let's take vectors AB, AC, AD.Vector AB = (0, 0.5, 0)Vector AC = (-0.5, 0.5, 0)Vector AD = (-0.5, 0.5, 0.5)The scalar triple product is AB ⋅ (AC × AD)First compute AC × AD:AC × AD = |i   j   k|         -0.5 0.5 0         -0.5 0.5 0.5= i*(0.5*0.5 - 0*0.5) - j*(-0.5*0.5 - (-0.5)*0) + k*(-0.5*0.5 - (-0.5)*0.5)= i*(0.25) - j*(-0.25 - 0) + k*(-0.25 + 0.25)= (0.25i + 0.25j + 0k)Then AB ⋅ (0.25i + 0.25j + 0k) = (0, 0.5, 0) ⋅ (0.25, 0.25, 0) = 0*0.25 + 0.5*0.25 + 0*0 = 0.125Since the scalar triple product is 0.125 ≠ 0, the four points are not coplanar. Therefore, that quadrilateral is not planar, so the convex hull must have triangular faces.Hmm, so the convex hull is a polyhedron with triangular faces. Given that each vertex is connected to four others, and there are six vertices, this is indeed an octahedron (which typically has eight faces), but in this case, perhaps it's a non-regular octahedron with triangular faces.But how to compute its volume? Maybe divide it into simpler tetrahedrons.Let me pick a point and connect it to others to form tetrahedrons. For example, take the point (0.5, 0, 0) and see which tetrahedrons it is part of.Alternatively, maybe there's a formula for the volume of the convex hull of the edge midpoints of a tetrahedron. Let me think.Alternatively, consider that the midpoints form a smaller polyhedron whose volume relates to the original by a scaling factor. If we can find the scaling factor, then the volume ratio is the cube of that.But how is the new polyhedron scaled? Let's see. If each edge of the original tetrahedron is connected to midpoints, perhaps the new polyhedron is similar but scaled down. Wait, but the original tetrahedron is not regular, so scaling might not be uniform. But since the problem states the ratio is the same for any tetrahedron, maybe it's a linear transformation.Alternatively, think in terms of vectors. Let the original tetrahedron have vertices at vectors a, b, c, d. The midpoints of the edges are (a+b)/2, (a+c)/2, (a+d)/2, (b+c)/2, (b+d)/2, (c+d)/2. The convex hull of these points is the polyhedron V2.Wait, perhaps the polyhedron V2 is the dual of the original tetrahedron? But duals usually have vertices corresponding to faces, which is not the case here.Alternatively, notice that the centroid of the original tetrahedron is the average of its four vertices: (a + b + c + d)/4. Maybe this centroid is also the centroid of the new polyhedron.Alternatively, another approach: the polyhedron formed by the midpoints is called the "midpoint polyhedron" or "edge midpoints polyhedron" of the tetrahedron. Maybe there's a known volume ratio.Alternatively, use linear algebra. The coordinates of V2 are midpoints, so each coordinate is (a + b)/2 for edges a to b. If we consider the transformation from the original edges to the midpoints, this is equivalent to scaling each edge vector by 1/2, but how does this affect the overall volume?Wait, perhaps the volume can be related through the concept of parallelotopes, but a tetrahedron is a simplex. The volume of a simplex is 1/6 times the scalar triple product of its edge vectors. However, the midpoints would form another simplex?Wait, no. Since there are six midpoints, the convex hull is a polyhedron with six vertices, not a simplex. A simplex in 3D has four vertices. So this is more complex.Alternatively, use the fact that each face of the original tetrahedron is divided into four smaller triangles by the midpoints. Then, perhaps the octahedron is formed by connecting these midpoints. Wait, each original face has three midpoints, forming a medial triangle. The volume between these medial triangles and the original faces might form smaller tetrahedrons that are excluded from V2.Wait, maybe the octahedron divides the original tetrahedron into smaller parts. Let's consider the original tetrahedron divided by the midpoints. Each edge is split into two, and each face is divided into four smaller triangles. Then, the original tetrahedron is divided into smaller tetrahedrons and octahedrons. Maybe the octahedron in the center is V2, and the remaining volume is occupied by four smaller tetrahedrons at each corner.Wait, if that's the case, then we can compute the ratio. Suppose the original tetrahedron is divided into one central octahedron and four smaller tetrahedrons. Then the volume ratio would be V2 / V1 = (V1 - 4V_small) / V1. If we can find the volume of the small tetrahedrons, we can compute the ratio.Let me visualize this. Each original vertex is connected to the midpoints of its three edges. So each original vertex is part of a smaller tetrahedron. For example, take the vertex (0,0,0). The midpoints of the edges from (0,0,0) to (1,0,0), (0,1,0), and (0,0,1) are (0.5,0,0), (0,0.5,0), (0,0,0.5). So connecting (0,0,0) to these midpoints forms a small tetrahedron. Similarly for the other three vertices of the original tetrahedron.Therefore, the original tetrahedron is divided into four small tetrahedrons at the corners and one octahedron in the center. Therefore, V1 = 4V_small + V2. So if we can compute V_small, we can find V2.Let's compute V_small. Take the small tetrahedron at (0,0,0). Its vertices are (0,0,0), (0.5,0,0), (0,0.5,0), (0,0,0.5). The volume of this small tetrahedron is (1/6)* | determinant |.The determinant can be computed by the vectors from (0,0,0) to the other points: (0.5,0,0), (0,0.5,0), (0,0,0.5). The volume is (1/6)* | (0.5)(0.5)(0.5) | = (1/6)*(0.125) = 1/48.Wait, but the original tetrahedron's volume was 1/6. So each small tetrahedron has volume 1/48. Then four of them would have total volume 4*(1/48) = 1/12. Therefore, the remaining volume, which is the octahedron V2, would be V1 - 1/12 = 1/6 - 1/12 = 1/12. Therefore, the ratio V2/V1 = (1/12)/(1/6) = 1/2. Wait, but that would mean the answer is 1/2, option A. But that seems conflicting with some intuition. Let me verify.Wait, wait. Let me check the volume of the small tetrahedron again. The vectors from (0,0,0) are (0.5,0,0), (0,0.5,0), (0,0,0.5). The volume is (1/6)*| scalar triple product |. The scalar triple product is the determinant of the matrix formed by these vectors as columns (or rows). The determinant is 0.5*0.5*0.5 = 0.125. So the volume is 0.125/6 = 1/48. Yes. So each small tetrahedron is 1/48, four of them total 1/12. Then the octahedron's volume is 1/6 - 1/12 = 1/12. Then the ratio is (1/12)/(1/6) = 1/2. So 1/2, which is option A. But this contradicts some other sources I recall where the ratio is 1/2. Wait, maybe I made a mistake.Wait, but let's think again. If the original tetrahedron is divided into four small tetrahedrons and one octahedron, each small tetrahedron has 1/4 the edge length? Wait, no, the edges of the small tetrahedrons are half the length of the original edges. Since the midpoints are at half the edges. So scaling factor is 1/2, so volume scales by (1/2)^3 = 1/8. But the original volume is 1/6, so each small tetrahedron would have volume (1/6)*(1/8) = 1/48. Which matches our previous calculation. Then four of them would be 4/48 = 1/12, so the remaining octahedron is 1/6 - 1/12 = 1/12. Then V2/V1 = 1/12 / 1/6 = 1/2.Therefore, the answer should be 1/2, option A. But wait, I also recall that the volume of the convex hull of the edge midpoints of a tetrahedron is 1/2 of the original volume. However, I also remember that sometimes it's 1/4. Wait, maybe I need to check again.Wait, perhaps my assumption that the original tetrahedron is divided into four small tetrahedrons and one octahedron is incorrect. Let's consider the original tetrahedron divided by connecting midpoints. Each face is divided into four smaller triangles, yes. But in 3D, connecting midpoints of edges would create not only four corner tetrahedrons but also other regions. Maybe the central region is more complex.Wait, perhaps I should use the concept of parallelotopes or barycentric subdivisions, but it's getting too vague. Alternatively, let's use coordinates again. Let me compute the coordinates of the octahedron and compute its volume directly.The six midpoints are:A=(0.5,0,0), B=(0,0.5,0), C=(0,0,0.5), D=(0.5,0.5,0), E=(0.5,0,0.5), F=(0,0.5,0.5)We need to compute the volume of the convex hull of these six points. One way to compute this is to choose a triangulation of the convex hull into tetrahedrons and sum their volumes.First, we need to figure out how these six points form a convex polyhedron. Let's try to decompose it.An octahedron can be divided into two square pyramids with a common square base. Let's see if that's possible here. Let's check if there's a square base in the middle.Looking at points D=(0.5,0.5,0), E=(0.5,0,0.5), F=(0,0.5,0.5), and perhaps another point? Wait, there are only six points. Let's see.Alternatively, perhaps the base is a triangle. Wait, maybe the octahedron can be divided into simpler tetrahedrons.Alternatively, pick a point and connect it to adjacent points. Let's take point A=(0.5,0,0). The adjacent points are B=(0,0.5,0), C=(0,0,0.5), D=(0.5,0.5,0), E=(0.5,0,0.5). So A is connected to B, C, D, E.Let's try to split the octahedron into tetrahedrons involving point A.First, tetrahedron ABDA: points A, B, D, and another point. Wait, no, need to check.Wait, perhaps tetrahedron A, B, D, which is a triangle. But in 3D, to form a tetrahedron, we need a fourth point. Maybe point F=(0,0.5,0.5)? Wait, but is F connected to A? Let's check. F is connected to B and C, but not directly to A. So maybe not.Alternatively, take tetrahedron A, B, C, but those three points are not coplanar. Wait, A=(0.5,0,0), B=(0,0.5,0), C=(0,0,0.5). These three points form a triangle, but adding another point to make a tetrahedron. Let's try adding point D=(0.5,0.5,0). The tetrahedron A, B, C, D.Wait, but D is (0.5,0.5,0). So vectors from A to B, A to C, A to D:AB = (-0.5, 0.5, 0)AC = (-0.5, 0, 0.5)AD = (0, 0.5, 0)The volume is 1/6 * | scalar triple product |.Compute the scalar triple product AB ⋅ (AC × AD).First, compute AC × AD:AC = (-0.5, 0, 0.5)AD = (0, 0.5, 0)Cross product:|i   j   k|-0.5 0   0.50   0.5  0= i*(0*0 - 0.5*0.5) - j*(-0.5*0 - 0.5*0) + k*(-0.5*0.5 - 0*0)= i*(-0.25) - j*(0) + k*(-0.25)= (-0.25i - 0.25k)Then AB ⋅ (-0.25i - 0.25k) = (-0.5, 0.5, 0) ⋅ (-0.25, 0, -0.25) = (-0.5)(-0.25) + 0.5*0 + 0*(-0.25) = 0.125 + 0 + 0 = 0.125So the volume is |0.125| / 6 = 0.125 / 6 ≈ 0.0208333, which is 1/48. Wait, but this is the same as the small tetrahedron's volume. So maybe this tetrahedron is part of the small corner tetrahedron. But we already accounted for those. Wait, but this seems to be part of the octahedron.Hmm, this is getting confusing. Maybe decomposing the octahedron into tetrahedrons is leading to confusion. Let me try another approach. Let's use the shoelace formula or another method for volume computation.Alternatively, use the Cayley-Menger determinant for six points. But that's complicated. Alternatively, note that the octahedron is symmetric and can be split into pyramids.Wait, perhaps consider the octahedron as two square pyramids glued together at their bases. If I can find the height and the area of the base, then compute the volume.But first, check if there is a square base. Let's look for four coplanar points forming a square. Take points D=(0.5,0.5,0), E=(0.5,0,0.5), F=(0,0.5,0.5), and maybe another point. Wait, the other points are A=(0.5,0,0), B=(0,0.5,0), C=(0,0,0.5). Is there a square among these points?Wait, take points D=(0.5,0.5,0), E=(0.5,0,0.5), F=(0,0.5,0.5), and maybe C=(0,0,0.5)? Let me check if they are coplanar.Wait, points D, E, F, and C. Let's check coplanarity. The volume of the tetrahedron D, E, F, C is:Vectors from D: E - D = (0, -0.5, 0.5), F - D = (-0.5, 0, 0.5), C - D = (-0.5, -0.5, 0.5)Compute scalar triple product:(0, -0.5, 0.5) ⋅ [ (-0.5, 0, 0.5) × (-0.5, -0.5, 0.5) ]First compute the cross product:|i   j   k|-0.5 0   0.5-0.5 -0.5 0.5= i*(0*0.5 - 0.5*(-0.5)) - j*(-0.5*0.5 - (-0.5)*0.5) + k*(-0.5*(-0.5) - 0*(-0.5))= i*(0 + 0.25) - j*(-0.25 + 0.25) + k*(0.25 - 0)= (0.25i - 0j + 0.25k)Then the dot product with (0, -0.5, 0.5) is:0*0.25 + (-0.5)*0 + 0.5*0.25 = 0 + 0 + 0.125 = 0.125 ≠ 0So they are not coplanar. Therefore, there's no square face. So my initial idea is incorrect.Alternatively, maybe the octahedron is composed of eight tetrahedrons. But this might not help.Alternatively, use the inclusion-exclusion principle. But I'm not sure.Wait, maybe use the formula for the volume of a polyhedron given its vertices. There is a formula where you can compute the volume by dividing the polyhedron into tetrahedrons with a common vertex.Let's choose point A=(0.5,0,0) as the common vertex. Then, the octahedron can be divided into tetrahedrons that all include A.Looking at the connections from A, it is connected to B, C, D, E. So perhaps the octahedron can be divided into four tetrahedrons: A-B-D-E, but no, that's not a tetrahedron. Wait, maybe:1. Tetrahedron A, B, D, E2. Tetrahedron A, B, C, F3. Etc. Hmm, this is not straightforward.Alternatively, since each edge of the octahedron is part of two tetrahedrons.Alternatively, another method: the volume of the octahedron can be calculated by using the determinant formula if we can find a suitable coordinate system.Let me translate the octahedron so that one of its vertices is at the origin. Let's take point A=(0.5,0,0) and translate all points by (-0.5,0,0). Then the translated points are:A'=(0,0,0)B'=(-0.5,0.5,0)C'=(-0.5,0,0.5)D'=(0,0.5,0)E'=(0,0,0.5)F'=(-0.5,0.5,0.5)Now, compute the volume of the convex hull of these translated points. The volume should remain the same after translation.Now, let's try to compute the volume using the Cayley-Menger determinant for the six points. However, that's quite complex. Alternatively,<think>Okay, so I need to find the ratio of the volume V2 of a convex polyhedron formed by the midpoints of the edges of a tetrahedron to the original tetrahedron's volume V1. The options are given as (A) 1/2, (B) 2/3, (C) 3/4, (D) uncertain. Let's see how to approach this.First, I recall that a tetrahedron has 6 edges. If we take the midpoints of each edge and form a convex polyhedron from them, what does that polyhedron look like? Maybe it's another tetrahedron or a different shape. Wait, a convex polyhedron formed by connecting midpoints... Hmm, perhaps it's an octahedron? Because connecting midpoints of edges in a tetrahedron might form an octahedron. Let me visualize.But actually, an octahedron has 8 faces, right? The tetrahedron has 4 triangular faces. If we connect midpoints of edges, each face of the original tetrahedron would have midpoints connected, forming smaller triangles. But the polyhedron formed by all midpoints might be something else. Alternatively, maybe the convex hull of all the midpoints?Alternatively, maybe it's a smaller tetrahedron inside the original one. But if we take midpoints, maybe it's similar to the original but scaled down. Wait, but scaling would affect the volume by a factor. If it's scaled by 1/2, the volume would be (1/2)^3 = 1/8, which isn't one of the options. So maybe not a scaled tetrahedron.Alternatively, perhaps the formed polyhedron is an octahedron. Let me think. A regular octahedron can be formed by connecting midpoints of a tetrahedron? Wait, no. A regular octahedron has all its edges equal and faces are equilateral triangles. But in a general tetrahedron, which isn't regular, the midpoints would form a non-regular octahedron? Maybe.Alternatively, maybe it's a parallelepiped? But a tetrahedron has 6 edges, so midpoints would be 6 points. A convex polyhedron with 6 vertices. Hmm, 6 vertices can form an octahedron? Wait, no, a regular octahedron has 6 vertices. Wait, actually, a regular octahedron has 6 vertices, 12 edges, and 8 faces. So if the midpoints of the edges of a tetrahedron are 6 points, then their convex hull could be an octahedron. Let me confirm that.Consider a regular tetrahedron for simplicity. Let's place the vertices at (1,1,1), (-1,-1,1), (-1,1,-1), (1,-1,-1). Then the midpoints of the edges would be the averages of these coordinates. Let's compute a few. The midpoint between (1,1,1) and (-1,-1,1) is (0,0,1). Similarly, between (1,1,1) and (-1,1,-1) is (0,1,0). Between (1,1,1) and (1,-1,-1) is (1,0,0). Similarly, the other midpoints would be (0,0,-1), (0,-1,0), (-1,0,0). So those are the six points: (0,0,1), (0,0,-1), (0,1,0), (0,-1,0), (1,0,0), (-1,0,0). These are exactly the vertices of a regular octahedron. So in the case of a regular tetrahedron, the midpoints form a regular octahedron. Therefore, the convex polyhedron V2 is an octahedron when the original is a regular tetrahedron.Now, what's the volume ratio between the octahedron and the original tetrahedron?First, let's compute the volume of the regular tetrahedron. The edge length of the regular tetrahedron I described can be calculated. The distance between (1,1,1) and (-1,-1,1) is sqrt[(-2)^2 + (-2)^2 + 0^2] = sqrt(8) = 2√2. So edge length a = 2√2.The volume of a regular tetrahedron is (a³)/(6√2). Plugging in a = 2√2, we get:( (2√2)^3 ) / (6√2) ) = (16 * 2√2) / (6√2) ) Wait, let me compute (2√2)^3: 8 * (√2)^3 = 8 * 2√2 = 16√2. Then divided by 6√2: 16√2 / 6√2 = 16/6 = 8/3. So the volume V1 of the regular tetrahedron is 8/3.Now, the octahedron formed by those midpoints has vertices at (0,0,1), etc. The regular octahedron with vertices at (±1,0,0), (0,±1,0), (0,0,±1) has a volume of 4/3. Wait, how?The regular octahedron can be divided into two square pyramids with a square base. Each pyramid has a base area of 1*1 = 1 (since the distance between (1,0,0) and (-1,0,0) is 2, but wait, no. Wait, the distance between (1,0,0) and (0,1,0) is sqrt(2), so the edge length is sqrt(2). Wait, maybe I need to compute the volume differently.Alternatively, the regular octahedron can be considered as having edge length sqrt(2) between adjacent vertices. The formula for the volume of a regular octahedron with edge length a is (sqrt(2)/3) * a³. But if our octahedron here has vertices at (±1,0,0), (0,±1,0), (0,0,±1), then the edge length between (1,0,0) and (0,1,0) is sqrt((1-0)^2 + (0-1)^2 + (0-0)^2) = sqrt(2). So edge length a = sqrt(2). Then volume is (sqrt(2)/3)*(sqrt(2))^3 = (sqrt(2)/3)*(2*sqrt(2)) )= (sqrt(2)*2*sqrt(2))/3 = (2*2)/3 = 4/3. So the octahedron has volume 4/3.But the original tetrahedron had volume 8/3. Therefore, the ratio V2/V1 is (4/3)/(8/3) = 1/2. So in this case, the answer would be 1/2, which is option A.But wait, this is for a regular tetrahedron. The problem says "a tetrahedron", not necessarily regular. So does the ratio depend on the shape of the tetrahedron? If it does, then the answer would be D) uncertain. But if the ratio is the same for any tetrahedron, then A) 1/2 is correct.Hmm, so the key question is whether this ratio is the same for all tetrahedrons or not.Alternatively, maybe we can approach this problem using linear algebra or affine transformations.Since any tetrahedron can be transformed into any other tetrahedron via an affine transformation, and affine transformations preserve ratios of volumes (up to a scalar determinant), but the ratio V2/V1 should be preserved under affine transformations.Wait, but affine transformations can scale, rotate, shear, etc., but volumes scale by the determinant. However, if we compute the ratio V2/V1, the scaling factor would cancel out, so the ratio should be invariant under affine transformations. Therefore, if it's 1/2 for a regular tetrahedron, it should be 1/2 for any tetrahedron.Therefore, the answer should be A) 1/2.But let me verify this with another example. Suppose we take a degenerate tetrahedron, but wait, a convex polyhedron formed by midpoints would still be non-degenerate? Wait, if the original tetrahedron is degenerate, then the midpoints might form a degenerate shape, but the problem states it's a convex polyhedron, so the original tetrahedron must be non-degenerate.Alternatively, take a different tetrahedron. Let's take a simpler one. Let's say a tetrahedron with vertices at (0,0,0), (1,0,0), (0,1,0), (0,0,1). The volume of this tetrahedron is 1/6.Now, the midpoints of its edges:Edges are between each pair of vertices. So there are 6 edges:1. (0,0,0)-(1,0,0): midpoint at (0.5, 0, 0)2. (0,0,0)-(0,1,0): midpoint at (0, 0.5, 0)3. (0,0,0)-(0,0,1): midpoint at (0, 0, 0.5)4. (1,0,0)-(0,1,0): midpoint at (0.5, 0.5, 0)5. (1,0,0)-(0,0,1): midpoint at (0.5, 0, 0.5)6. (0,1,0)-(0,0,1): midpoint at (0, 0.5, 0.5)So the midpoints are these six points. Now, what is the convex hull of these six points?Let me try to visualize this. The midpoints include three points along the axes at (0.5,0,0), (0,0.5,0), (0,0,0.5), and three points on the faces: (0.5,0.5,0), (0.5,0,0.5), (0,0.5,0.5). Connecting these points, the convex polyhedron formed is a polyhedron with these six vertices. Let's try to figure out its shape.Perhaps it's an octahedron? Let's see. In 3D, an octahedron has six vertices, eight triangular faces, twelve edges. Let's check how many faces this convex hull has.Looking at the midpoints, the convex hull should connect each of these points. For example, the point (0.5,0,0) is connected to (0.5,0.5,0), (0.5,0,0.5), and maybe others. Similarly, (0,0.5,0) is connected to (0.5,0.5,0), (0,0.5,0.5), etc.Alternatively, perhaps this is a octahedron, but not regular. Let's compute the distances between points. For example, the distance between (0.5,0,0) and (0.5,0.5,0) is 0.5. The distance between (0.5,0,0) and (0.5,0,0.5) is 0.5. Similarly, the distance between (0.5,0,0) and (0,0.5,0) is sqrt((0.5)^2 + (0.5)^2) = sqrt(0.25 + 0.25) = sqrt(0.5) ≈ 0.707.So edges have different lengths, meaning it's not a regular octahedron. But it's still an octahedron. The volume of this octahedron can be calculated.Alternatively, perhaps decompose the octahedron into simpler parts. Let's note that the original tetrahedron has vertices at (0,0,0), (1,0,0), (0,1,0), (0,0,1). The midpoints form the convex hull of the six points we listed. To compute the volume of the convex hull, maybe we can use coordinates.Alternatively, perhaps use the fact that the octahedron is dual to the cube in some way, but maybe that's complicating.Alternatively, think of the convex polyhedron V2 as the intersection of regions or use determinant formulas.Alternatively, notice that the formed octahedron can be divided into smaller tetrahedrons whose volumes we can compute.Alternatively, use the formula for the volume of a polyhedron given its vertices, but that might be time-consuming.Alternatively, observe that the octahedron formed here is the dual of the original tetrahedron in some barycentric subdivision, but I might be overcomplicating.Wait, another approach: In 3D, when you connect the midpoints of the edges of a tetrahedron, the resulting polyhedron is called the centroid octahedron. Its volume is known to be 1/2 of the original tetrahedron's volume regardless of the tetrahedron's shape. Therefore, the ratio V2/V1 is 1/2, so the answer is A.But how do I confirm this without prior knowledge?Alternatively, use vectors and coordinates. Let's assign coordinates to the original tetrahedron. Let the original tetrahedron have vertices at points A, B, C, D. Let the midpoints of the edges be M_AB, M_AC, M_AD, M_BC, M_BD, M_CD. Then, the convex hull of these midpoints forms the polyhedron V2.To compute V2, maybe express the coordinates of these midpoints in terms of the original vertices and find the volume.Alternatively, consider that the polyhedron formed by the midpoints is the dual of the original tetrahedron scaled down.Alternatively, use linear transformations. Suppose we apply an affine transformation that maps the original tetrahedron to a regular one. Since affine transformations preserve ratios of volumes, and midpoints are preserved under affine transformations (since they are linear combinations), then the ratio V2/V1 should be the same as in the regular case. Therefore, since we found in the regular tetrahedron case that the ratio is 1/2, then it should be 1/2 in general.Therefore, the answer is A) 1/2.But just to be thorough, let's compute the volume for the second example I considered earlier, the tetrahedron with vertices at (0,0,0), (1,0,0), (0,1,0), (0,0,1). The original volume V1 is 1/6.Now, compute V2, the volume of the convex hull of the midpoints:The midpoints are:1. M_AB = (0.5, 0, 0)2. M_AC = (0, 0.5, 0)3. M_AD = (0, 0, 0.5)4. M_BC = (0.5, 0.5, 0)5. M_BD = (0.5, 0, 0.5)6. M_CD = (0, 0.5, 0.5)Now, let's try to figure out the shape of V2. Let's consider the connections. Each midpoint is connected to others.For example, M_AB (0.5,0,0) is connected to M_AC (0,0.5,0) via the edge between them? Wait, actually, the convex hull will form a polyhedron where each original edge's midpoint is connected to the midpoints of adjacent edges.Alternatively, think of each face of the original tetrahedron. Each face is a triangle, and the midpoints of its three edges form a smaller triangle. The convex hull of all these midpoints would include these smaller triangles and connect them across the original edges.Alternatively, maybe the polyhedron V2 has 8 triangular faces: each original face contributes a smaller triangle, and each original edge contributes a quadrilateral? Hmm, not sure.Alternatively, use the coordinates to compute the volume.Let's choose a subset of these points to form tetrahedrons within V2 and sum their volumes. But since V2 is a convex polyhedron, maybe it's an octahedron, which can be divided into two square pyramids. Wait, but in this case, it might not be regular.Looking at the coordinates:The points are:(0.5,0,0), (0,0.5,0), (0,0,0.5), (0.5,0.5,0), (0.5,0,0.5), (0,0.5,0.5)Let me see if these points form an octahedron. An octahedron has two vertices along each axis, but in this case, the points are all at midpoints. Alternatively, these points can be considered as the centers of the faces of a cube with edge length 0.5, but that might not be the case.Alternatively, let's pick a point and see its adjacent points. Take (0.5,0,0). The adjacent points would be those connected by edges in the convex hull. The convex hull will connect (0.5,0,0) to other midpoints that are adjacent in the original tetrahedron.In the original tetrahedron, the edge AB is connected to edges AC, AD, BC, BD. Therefore, the midpoint M_AB is connected in V2 to midpoints of edges AC, AD, BC, BD. Wait, in the original tetrahedron, each edge is connected to four other edges. So the midpoint of AB is connected to midpoints of AC, AD, BC, BD. So in the polyhedron V2, each vertex (midpoint) is connected to four others.So each vertex has four edges, which is consistent with an octahedron, as each vertex in an octahedron is part of four edges. Therefore, even in this case, the convex hull is an octahedron, albeit not regular.Now, to compute the volume of this octahedron. Let's try to divide it into simpler components.Looking at the coordinates:We have three points along the axes: (0.5,0,0), (0,0.5,0), (0,0,0.5)And three points on the face centers: (0.5,0.5,0), (0.5,0,0.5), (0,0.5,0.5)Wait, if we consider these points, perhaps the octahedron can be divided into pyramids with bases on each face.Alternatively, consider the octahedron as two square pyramids glued together at their square bases.But in this case, the "square" might not be flat. Let's check the coordinates of the points that could form the base.Take the points (0.5,0.5,0), (0.5,0,0.5), (0,0.5,0.5), (0,0,0.5). Wait, no. Alternatively, maybe the square is formed by (0.5,0.5,0), (0.5,0,0.5), (0,0.5,0.5), and another point. Wait, actually, those three points are in different planes.Alternatively, perhaps use the determinant method for volume.Choose four points and compute the volume of the tetrahedron they form, but since V2 is an octahedron, it's not a tetrahedron. So maybe decompose the octahedron into tetrahedrons.For example, divide the octahedron into five tetrahedrons. Wait, no, usually an octahedron can be divided into two square pyramids, each with a square base. If each pyramid has a square base, then each pyramid can be divided into two tetrahedrons, making a total of four tetrahedrons. But let's try.Take the octahedron with vertices at (0.5,0,0), (0,0.5,0), (0,0,0.5), (0.5,0.5,0), (0.5,0,0.5), (0,0.5,0.5). Let's pick a point as the apex. Let's pick (0.5,0,0) as the apex of a pyramid. The base would be some quadrilateral. Wait, but this might not work. Alternatively, pick a point inside the octahedron and connect to all faces, but that's complicated.Alternatively, use the Cayley-Menger determinant. But that might be complex.Alternatively, note that the octahedron is symmetric, and compute the volume by integrating or using coordinates.Alternatively, consider that the octahedron can be seen as the intersection of certain half-spaces.Alternatively, use coordinates and find the convex hull's volume.But this is getting complicated. Maybe there's a smarter way. Since we know that the ratio is 1/2 for the regular tetrahedron, and affine transformations preserve volume ratios, then in any tetrahedron, the ratio V2/V1 should be 1/2.Alternatively, consider that the polyhedron formed by the midpoints is the dual of the original tetrahedron in some sense, but I might be mixing concepts.Alternatively, think of each face of V2 as a parallelogram. Wait, in the original tetrahedron, each pair of opposite edges (there are no opposite edges in a tetrahedron, actually; each edge is connected to two faces). Wait, perhaps each edge midpoint is connected to three other midpoints.Alternatively, consider that each face of the new polyhedron V2 is a parallelogram. For example, the midpoints of edges AB, AC, BC, and BD form a quadrilateral. Wait, not sure.Alternatively, think in terms of vectors. Let the original tetrahedron have vertices at vectors a, b, c, d. The midpoints of the edges are (a + b)/2, (a + c)/2, (a + d)/2, (b + c)/2, (b + d)/2, (c + d)/2.The convex hull of these midpoints is the polyhedron V2. To compute its volume, express it in terms of the original vectors.Alternatively, use the fact that the midpoints form a polyhedron whose volume relates to the original through a linear transformation.Notice that the coordinates of the midpoints are averages of the original vertices. This is similar to applying a linear transformation to the original edges. If we consider that each midpoint is (a + b)/2, etc., then the coordinates of V2 are (a + b)/2, (a + c)/2, etc. So, the polyhedron V2 is the image of the original edge set under a linear transformation that averages pairs of vertices.But how does this affect volume? Perhaps the volume can be expressed in terms of determinants involving these averaged coordinates.Alternatively, consider that the polyhedron formed by the midpoints can be obtained by cutting off the corners of the original tetrahedron at the midpoints. Each vertex of the original tetrahedron is replaced by a smaller face. This process is called truncation, but truncating at midpoints is a specific type of truncation.The volume removed from each corner can be calculated. Since each corner is a small tetrahedron, whose volume is (1/2)^3 = 1/8 of the original corner's volume. But how many such corners are there?Wait, maybe not exactly. Let's see. If we truncate each vertex of the tetrahedron at the midpoints of the edges, each original vertex is replaced by a new face. The volume removed from each vertex would be a smaller tetrahedron whose edge length is half of the original. Since the original tetrahedron has four vertices, truncating each vertex would remove four small tetrahedrons each with volume 1/8 of the original. But wait, the original tetrahedron's volume is V1. If each small tetrahedron is scaled down by 1/2, their volume is (1/2)^3 * V1_vertex, where V1_vertex is the volume associated with each vertex. However, this approach might not be straightforward because truncating a tetrahedron's vertex doesn't just remove a scaled tetrahedron; the actual volume removed depends on the geometry.Alternatively, another approach: The polyhedron formed by the midpoints, V2, is the dual of the original tetrahedron in terms of midpoints. But maybe that's not helpful.Alternatively, use the concept of parallelotopes or other figures.Wait, let me recall that in 2D, if you connect the midpoints of a quadrilateral, you get a parallelogram whose area is half of the original quadrilateral. But in 3D, connecting midpoints in a tetrahedron might lead to a similar ratio. But 3D is different.Wait, in 2D, connecting midpoints of a convex quadrilateral gives a parallelogram with area 1/2. But in 3D, maybe connecting midpoints of a tetrahedron gives a figure with volume 1/2. That seems similar.Alternatively, consider the following: each edge of the original tetrahedron is split into two segments by its midpoint. The convex hull of the midpoints forms a new polyhedron inside the original tetrahedron. The question is how much volume remains.Alternatively, think of the original tetrahedron divided into smaller tetrahedrons and the polyhedron V2. The midpoints divide the original edges, and perhaps the original tetrahedron is divided into smaller pieces, one of which is V2 and others are smaller tetrahedrons.But this might not be the case. Let's think: connecting all midpoints, we might partition the original tetrahedron into several smaller polyhedrons. Let's see.Suppose we connect all midpoints of edges. Each face of the original tetrahedron is divided into four smaller triangles by connecting midpoints. Then, in 3D, connecting these midpoints would create a smaller tetrahedron at the center and four octahedrons? Not sure.Wait, perhaps the original tetrahedron is divided into the inner octahedron V2 and four smaller tetrahedrons. If that's the case, then the total volume would be V1 = V2 + 4*V_small_tetra. Then, if we can find the ratio between V_small_tetra and V1, we can find V2/V1.Let me try this. Suppose each original edge is split into two, so each edge has length 1/2. If we consider each corner of the original tetrahedron, the part near each vertex is a small tetrahedron formed by connecting the midpoints of the edges meeting at that vertex.Each original vertex is connected to three edges. The midpoints of those three edges form a triangle, and connecting that triangle to the original vertex forms a small tetrahedron. However, when we take the convex hull of all midpoints, we remove those small tetrahedrons. Wait, actually, when we connect all midpoints, the convex hull excludes the original vertices, so the small tetrahedrons near each original vertex are indeed removed. Therefore, the volume V2 is the original volume minus the volumes of those four small tetrahedrons.Now, each small tetrahedron near a vertex has volume. Let's compute the volume of one such small tetrahedron.Take vertex A of the original tetrahedron. The three midpoints connected to A are the midpoints of edges AB, AC, AD. Let’s denote these midpoints as M_AB, M_AC, M_AD. The small tetrahedron near A is formed by the points A, M_AB, M_AC, M_AD.The volume of this small tetrahedron can be computed using coordinates. Let’s place vertex A at the origin (0,0,0) for simplicity. Suppose the original tetrahedron has vertices at A(0,0,0), B(2,0,0), C(0,2,0), D(0,0,2). Then the midpoints would be M_AB(1,0,0), M_AC(0,1,0), M_AD(0,0,1), M_BC(1,1,0), M_BD(1,0,1), M_CD(0,1,1).The volume of the original tetrahedron V1 is (1/6)*|det([B-A, C-A, D-A])| = (1/6)*|det([2,0,0; 0,2,0; 0,0,2])| = (1/6)*8 = 4/3.The small tetrahedron near A is formed by A(0,0,0), M_AB(1,0,0), M_AC(0,1,0), M_AD(0,0,1). The volume of this small tetrahedron is (1/6)*|det([1,0,0; 0,1,0; 0,0,1])| = 1/6.Similarly, each of the four small tetrahedrons near the vertices A, B, C, D has volume 1/6. So the total volume removed is 4*(1/6) = 2/3. Therefore, the volume of V2 is V1 - 2/3 = 4/3 - 2/3 = 2/3. Therefore, the ratio V2/V1 is (2/3)/(4/3) = 1/2.Wait a minute! So in this case, V2 is 1/2 of V1, which matches the previous example. Therefore, this suggests that regardless of the tetrahedron, the ratio is 1/2.But let me verify with the coordinates. In this specific example, the original tetrahedron has volume 4/3, and the octahedron V2 has volume 2/3, which is 1/2. So this supports the answer A) 1/2.Therefore, even in a non-regular tetrahedron, the ratio is preserved. Therefore, the answer is A) 1/2.This method works by considering that removing four small tetrahedrons each with 1/8 the volume of the original? Wait, no, in this case, the small tetrahedrons were each 1/6, and the original volume was 4/3. So each small tetrahedron is 1/6, and 4*(1/6) = 2/3. Then V2 = 4/3 - 2/3 = 2/3. So the ratio is 1/2.But why is the volume of each small tetrahedron 1/6 in this case?Because in this coordinate system, the original tetrahedron has edge lengths of 2, and the small tetrahedrons at each corner are similar to the original but scaled by 1/2. Therefore, the volume scales by (1/2)^3 = 1/8. But the original volume is 4/3, so each small tetrahedron would have volume (4/3)*(1/8) = 1/6. Hence, four of them total to 4/6 = 2/3. So V2 = 4/3 - 2/3 = 2/3, which is half of the original.Therefore, this shows that scaling the tetrahedron, the small tetrahedrons' volumes are 1/8 of the original, but since there are four of them, the total removed volume is 4*(1/8) = 1/2, hence V2 = V1 - 1/2 V1 = 1/2 V1. Wait, this seems conflicting with the previous calculation. Wait, in the coordinate system where the original tetrahedron had edges of length 2, the original volume was 4/3, and each small tetrahedron had volume 1/6. So 4*(1/6) = 2/3, which is half of 4/3. So 2/3 is half of 4/3. Therefore, the ratio is 1/2. Therefore, regardless of the original tetrahedron's volume, removing four small tetrahedrons each with 1/8 the volume of the original would result in V2 = V1 - 4*(1/8 V1) = V1 - 1/2 V1 = 1/2 V1.Wait, but in our example, the original volume was 4/3, and each small tetrahedron had volume 1/6, which is (1/6)/(4/3) = 1/8 of the original. Therefore, each small tetrahedron is 1/8 of V1. Therefore, removing four of them removes 4*(1/8) = 1/2 of V1. Hence, V2 = V1 - 1/2 V1 = 1/2 V1. So the ratio is 1/2.Therefore, this seems to be a general result. Therefore, regardless of the original tetrahedron's shape, connecting the midpoints of its edges forms a polyhedron V2 whose volume is half of the original tetrahedron's volume.Thus, the answer should be A) 1/2.Final Answerboxed{A}</think>To determine the ratio (frac{V_2}{V_1}) where (V_1) is the volume of a tetrahedron and (V_2) is the volume of the convex polyhedron formed by the midpoints of its edges, we start by considering the structure of the polyhedron formed by these midpoints. 1. Regular Tetrahedron Analysis: - For a regular tetrahedron with vertices at ((1,1,1)), ((-1,-1,1)), ((-1,1,-1)), and ((1,-1,-1)), the midpoints of its edges form a regular octahedron. - The volume of the regular tetrahedron (V_1) is (frac{8}{3}). - The volume of the regular octahedron (V_2) formed by the midpoints is (frac{4}{3}). - The ratio (frac{V_2}{V_1} = frac{4/3}{8/3} = frac{1}{2}).2. General Tetrahedron Analysis: - For any tetrahedron, the polyhedron formed by the midpoints of its edges is an octahedron. This result holds regardless of the tetrahedron's shape due to affine invariance. - Affine transformations preserve volume ratios, so the ratio (frac{V_2}{V_1}) remains consistent across all tetrahedrons.3. Volume Calculation Verification: - Considering a tetrahedron with vertices at ((0,0,0)), ((1,0,0)), ((0,1,0)), and ((0,0,1)), the volume (V_1) is (frac{1}{6}). - The midpoints of the edges form an octahedron. The volume (V_2) is calculated by subtracting the volumes of four smaller tetrahedrons (each with (frac{1}{6}) the volume of the original) from (V_1), resulting in (V_2 = frac{1}{6} - 4 times frac{1}{48} = frac{1}{12}). - The ratio (frac{V_2}{V_1} = frac{1/12}{1/6} = frac{1}{2}).Thus, the ratio (frac{V_2}{V_1}) is consistently (frac{1}{2}) for any tetrahedron, leading to the final answer:(boxed{A}),<think>Alright, let's tackle this geometry problem. So, we have a tetrahedron with volume V₁, and another convex polyhedron formed by connecting the midpoints of its edges, which has volume V₂. We need to find the ratio V₂/V₁. The options given are (A) 1/2, (B) 2/3, (C) 3/4, and (D) uncertain. Hmm, okay.First, let me recall what a tetrahedron is. It's a three-dimensional shape with four triangular faces, six edges, and four vertices. The convex polyhedron formed by the midpoints of its edges—so each edge is split into two equal parts, and those midpoints are connected somehow. I need to visualize this. If we take midpoints of all edges and connect them appropriately, what does that shape look like?Maybe it's helpful to think about simpler cases. In two dimensions, if you take a triangle and connect the midpoints of its edges, you get a smaller triangle called the medial triangle, whose area is 1/4 of the original. But here it's three dimensions, so the analogy might not be direct. However, perhaps similar principles apply.Let me try to imagine a tetrahedron. Each edge has a midpoint, so there are six midpoints. Connecting these midpoints—how exactly? Each original edge is connected to other edges. Maybe each midpoint is connected to the midpoints of adjacent edges. Wait, in a tetrahedron, each vertex is connected to three edges. So, for each vertex, the midpoints of those three edges would form a triangle? Maybe.Alternatively, perhaps the convex polyhedron formed by the midpoints is another polyhedron inside the original tetrahedron. Let me think. If we connect midpoints of edges, perhaps the new polyhedron is an octahedron? Because an octahedron has eight faces, which could be formed by the midpoints. Wait, no, an octahedron has six vertices. Wait, but we have six midpoints here—so maybe each midpoint becomes a vertex of the new polyhedron. If there are six vertices, then the polyhedron could be an octahedron. Because a regular octahedron has six vertices, twelve edges, and eight triangular faces. But is the polyhedron formed here regular?Alternatively, perhaps it's a different type of octahedron. Let me recall that in three dimensions, connecting midpoints of edges of a tetrahedron does indeed form an octahedron. Let me check: a tetrahedron has six edges. The midpoints of these edges can be considered as the vertices of a new polyhedron. Each face of the original tetrahedron is a triangle, and each face's midpoints would form a smaller triangle, but maybe each face of the new polyhedron corresponds to a face or an edge of the original.Wait, actually, in the case of the tetrahedron, connecting the midpoints of the edges forms a smaller octahedron inside the original tetrahedron. Let me confirm this. For each original edge, there is a midpoint, and each face of the original tetrahedron has three midpoints. Connecting these midpoints should form a tetrahedron? Wait, no. Wait, each face has three midpoints; connecting them would form a medial triangle on each face, but in three dimensions, these triangles are connected. Hmm. Alternatively, perhaps each edge of the original tetrahedron is replaced by a new vertex at its midpoint, and then these vertices are connected in such a way that forms an octahedron.Alternatively, maybe the convex polyhedron formed is a parallelepiped? Wait, a parallelepiped has six faces, all parallelograms. But an octahedron has triangular faces. Hmm. Maybe I need to think more carefully.Let me consider coordinates. Let's assign coordinates to the original tetrahedron to make things concrete. Let's choose a regular tetrahedron for simplicity, maybe with vertices at (0,0,0), (1,0,0), (0,1,0), and (0,0,1). Wait, but a regular tetrahedron is one where all edges are equal, but assigning coordinates like that might not give equal edges. Let me correct that. The standard regular tetrahedron can have vertices at (1,1,1), (-1,-1,1), (-1,1,-1), and (1,-1,-1), but that might complicate calculations. Alternatively, maybe a simpler coordinate system.Alternatively, let me take a tetrahedron with vertices at A(0,0,0), B(2,0,0), C(0,2,0), D(0,0,2). This is a tetrahedron with edges along the axes, and the edges from A to B, A to C, A to D, B to C, B to D, and C to D. The midpoints of these edges would be:Midpoint of AB: (1,0,0)Midpoint of AC: (0,1,0)Midpoint of AD: (0,0,1)Midpoint of BC: (1,1,0)Midpoint of BD: (1,0,1)Midpoint of CD: (0,1,1)So these six midpoints are the points:M1(1,0,0), M2(0,1,0), M3(0,0,1), M4(1,1,0), M5(1,0,1), M6(0,1,1)Now, we need to form a convex polyhedron from these six points. Let's see how they connect. Each original edge is connected to two others. Let's see if these points form an octahedron. In a regular octahedron, all edges are equal, but here, let's check the distances between these midpoints.For example, the distance between M1(1,0,0) and M2(0,1,0) is sqrt((1-0)^2 + (0-1)^2 + (0-0)^2) = sqrt(1 + 1) = sqrt(2)Distance between M1(1,0,0) and M4(1,1,0) is sqrt((1-1)^2 + (0-1)^2 + (0-0)^2) = sqrt(0 + 1 + 0) = 1Similarly, distance between M1 and M5(1,0,1): sqrt((1-1)^2 + (0-0)^2 + (0-1)^2) = sqrt(0 + 0 + 1) = 1So the distances vary. So it's not a regular octahedron. Hmm. But maybe it's a non-regular octahedron. An octahedron has eight triangular faces, six vertices. Wait, but here we have six vertices. Wait, no, an octahedron has six vertices. Wait, in this case, our convex polyhedron has six vertices, which are the midpoints. So perhaps it's a octahedron. Let's see.But in our case, with six points, perhaps the convex hull is a octahedron. Let's check the structure.Looking at the midpoints:M1(1,0,0), M2(0,1,0), M3(0,0,1), M4(1,1,0), M5(1,0,1), M6(0,1,1)Let's try to visualize the convex hull. For example, M1(1,0,0) is connected to M4(1,1,0), M5(1,0,1), and perhaps others. Let's see.If we consider the edges of the original tetrahedron, each midpoint is connected to two others. Wait, but in the new polyhedron, the edges are not the original edges but the connections between midpoints.Alternatively, perhaps each face of the new polyhedron is a quadrilateral. Wait, but convex polyhedron—so the convex hull of these six points. Let's check the coordinates.Alternatively, perhaps it's helpful to compute the convex hull. The convex hull of these six points would form a polyhedron. Let's see.Looking at the points:(1,0,0), (0,1,0), (0,0,1), (1,1,0), (1,0,1), (0,1,1)These points are all midpoints of edges of the tetrahedron. The convex hull of these points is a polyhedron. Let me see.Each of these points lies on the edges of the original tetrahedron. The convex hull would consist of all convex combinations of these points. Let's see if we can figure out the faces.First, note that the points (1,0,0), (0,1,0), (1,1,0) lie on the base face of the original tetrahedron (the face opposite D). Similarly, (1,0,0), (0,0,1), (1,0,1) lie on the face opposite C, and (0,1,0), (0,0,1), (0,1,1) lie on the face opposite B. Similarly, (1,1,0), (1,0,1), (0,1,1) might form another face.But wait, in three dimensions, connecting these midpoints—perhaps each original face of the tetrahedron contributes a smaller face to the new polyhedron. Wait, each original triangular face has three midpoints, which form a smaller triangle. So, the original four faces would each contribute a triangular face to the new polyhedron. That's four triangular faces. Then, perhaps the remaining faces come from connecting the midpoints across different original faces.Alternatively, in the original tetrahedron, each edge is connected to two faces. So, the midpoints of edges not on the same face might form another face. Let me think. For instance, the midpoints of edges AB, AC, and AD are (1,0,0), (0,1,0), (0,0,1). But these three points form a triangle. Wait, actually, in our coordinate system, the original tetrahedron has vertices at (0,0,0), (2,0,0), (0,2,0), (0,0,2). But I think I scaled down to midpoints at (1,0,0) etc. Maybe the original edge lengths are 2 units. Wait, perhaps the coordinates should be adjusted.Wait, no, if the original tetrahedron is at (0,0,0), (2,0,0), (0,2,0), (0,0,2), then the edges are of length 2, sqrt(8), etc. But maybe this complicates things. Let me instead take edge lengths as 2, so midpoints are at 1 unit. Hmm.Alternatively, maybe I should compute the volume of the convex hull of the six midpoints and compare it to the original tetrahedron.First, let's compute the volume V₁ of the original tetrahedron. If the original tetrahedron has vertices at (0,0,0), (2,0,0), (0,2,0), (0,0,2), then the volume is |(2)(2)(2)|/6 = 8/6 = 4/3. Wait, no, the formula for the volume of a tetrahedron with vertices at (0,0,0), (a,0,0), (0,b,0), (0,0,c) is |abc|/6. So in this case, a=2, b=2, c=2, so volume is 8/6 = 4/3.But in this case, the midpoints would be at (1,0,0), (0,1,0), (0,0,1), (1,1,0), (1,0,1), (0,1,1). So those midpoints. Now, the convex hull of these six points. Let's see if this polyhedron is an octahedron.An octahedron can be defined as the convex hull of six points: (±1,0,0), (0,±1,0), (0,0,±1). But in our case, the points are (1,0,0), (0,1,0), (0,0,1), (1,1,0), (1,0,1), (0,1,1). These are different. The standard octahedron has points along the axes, but our points are at midpoints, which include points like (1,1,0). So perhaps the convex hull is a truncated tetrahedron or something else.Alternatively, maybe the convex hull is a polyhedron composed of six quadrilateral faces and eight triangular faces? Wait, maybe not. Let's try to determine the faces.Looking at the points:(1,0,0), (0,1,0), (0,0,1), (1,1,0), (1,0,1), (0,1,1)Let me check which of these points lie on the same plane. For example, consider the plane x + y + z = 1. Let's test the points:(1,0,0): 1 + 0 + 0 = 1, yes.(0,1,0): 0 + 1 + 0 = 1, yes.(0,0,1): 0 + 0 + 1 = 1, yes.(1,1,0): 1 + 1 + 0 = 2 ≠ 1, so no.(1,0,1): 1 + 0 + 1 = 2 ≠ 1.(0,1,1): 0 + 1 + 1 = 2 ≠ 1.So three points lie on the plane x + y + z = 1, and the other three lie on the plane x + y + z = 2. Wait, but (1,1,0): x + y + z = 2? 1 + 1 + 0 = 2? Wait, no. Wait, (1,1,0) is 1 + 1 + 0 = 2? Wait, no, if the original tetrahedron was scaled with edge length 2, then yes. Wait, if the original edges are from (0,0,0) to (2,0,0), etc., then (1,1,0) would be a midpoint of BC, which is from (2,0,0) to (0,2,0). The midpoint is (1,1,0), yes. Then, in this case, x + y + z for (1,1,0) is 1 + 1 + 0 = 2. Similarly, (1,0,1) is 1 + 0 + 1 = 2, and (0,1,1) is 0 + 1 + 1 = 2. So the three midpoints of the edges connected to the original vertices (A, B, C, D) are on the plane x + y + z = 1, and the midpoints of the edges not connected to the original vertices (i.e., the midpoints of BC, BD, CD) are on the plane x + y + z = 2.Wait, but the original tetrahedron's vertices are at (0,0,0), (2,0,0), (0,2,0), (0,0,2). So the midpoints of edges connected to (0,0,0) (i.e., edges OA, OB, OC, where O is (0,0,0)) are (1,0,0), (0,1,0), (0,0,1). The midpoints of the other edges (BC, BD, CD) are (1,1,0), (1,0,1), (0,1,1). So, these two sets of points lie on parallel planes: x + y + z = 1 and x + y + z = 2. The distance between these two planes is |2 - 1| / sqrt(1^2 + 1^2 + 1^2) = 1 / sqrt(3).So the convex hull of these six points is a sort of prismatoid between these two parallel planes. Each plane has three points. The polyhedron would have two triangular bases (on the planes x + y + z = 1 and x + y + z = 2) and three quadrilateral faces connecting them. Wait, but three points on each plane, so connecting each point on the lower plane to two points on the upper plane? Wait, actually, each point on the lower plane is connected to two points on the upper plane. For example, (1,0,0) on the lower plane is connected to (1,1,0) and (1,0,1) on the upper plane. Similarly, (0,1,0) is connected to (1,1,0) and (0,1,1), and (0,0,1) is connected to (1,0,1) and (0,1,1). So, each connection is a edge between a lower point and two upper points. So the lateral faces are quadrilaterals. So, in total, the polyhedron would have two triangular faces and three quadrilateral faces? Wait, no. Wait, each lateral face is a quadrilateral formed by connecting (1,0,0) to (1,1,0) to (0,1,0) to (0,1,1) to (0,0,1) to (1,0,1) to (1,0,0)? No, that seems more complicated.Wait, perhaps if we have three points on each plane, and each point on the lower plane is connected to two points on the upper plane, then each lateral face is a quadrilateral. So three quadrilaterals connecting the two triangles. So the polyhedron has two triangular faces and three quadrilateral faces, making it a pentahedron? Wait, a pentahedron is a polyhedron with five faces. But two triangles and three quadrilaterals would make five faces. But we started with six points. Wait, no, actually, the convex hull of six points can have more faces. Wait, maybe my analysis is off.Alternatively, perhaps the convex hull is a octahedron. Let's check the number of faces. An octahedron has eight triangular faces. If we compute the convex hull, how many faces does it have?Alternatively, maybe the easiest way is to compute the volume V₂ of this convex hull and then compare it to V₁.Given that the original tetrahedron has volume V₁ = 4/3 (as calculated earlier), let's compute the volume of the convex hull of the six midpoints.Given that the six midpoints are divided into two sets: three on the plane x + y + z = 1 and three on x + y + z = 2. The distance between these planes is 1/sqrt(3). The area of the triangular base on the lower plane: the triangle formed by (1,0,0), (0,1,0), (0,0,1). Let's compute its area.The triangle with vertices (1,0,0), (0,1,0), (0,0,1). The vectors from (1,0,0) are (-1,1,0) and (-1,0,1). The cross product of these vectors is determinant of the matrix:i   j   k-1  1   0-1  0   1Calculating the determinant: i*(1*1 - 0*0) - j*(-1*1 - (-1)*0) + k*(-1*0 - (-1)*1)= i*(1 - 0) - j*(-1 - 0) + k*(0 + 1)= i*1 - j*(-1) + k*1= i + j + kThe magnitude of this cross product is sqrt(1^2 + 1^2 + 1^2) = sqrt(3). Therefore, the area of the triangle is half of this: sqrt(3)/2.Similarly, the area of the upper triangle formed by (1,1,0), (1,0,1), (0,1,1) is the same. Let's verify.Vectors from (1,1,0) to (1,0,1): (0, -1, 1)From (1,1,0) to (0,1,1): (-1, 0, 1)Cross product:i   j   k0  -1   1-1  0   1Determinant: i*(-1*1 - 1*0) - j*(0*1 - (-1)*1) + k*(0*0 - (-1)*(-1))= i*(-1 - 0) - j*(0 + 1) + k*(0 - 1)= -i - j - kMagnitude: sqrt((-1)^2 + (-1)^2 + (-1)^2) = sqrt(3), so area is sqrt(3)/2.So both the upper and lower triangular faces have area sqrt(3)/2. The distance between the planes is 1/sqrt(3). But the volume of a prismatoid (a polyhedron with all vertices lying in two parallel planes) can be computed using the formula:Volume = (h/6) * (A1 + A2 + 2Am)Where h is the distance between the planes, A1 and A2 are the areas of the two bases, and Am is the area of the midsection. However, in this case, since the two bases are triangles and the midsection is a hexagon? Wait, maybe not. Alternatively, if the two bases are triangles and the lateral faces are quadrilaterals, maybe it's a kind of frustum.Alternatively, since the two triangular bases are similar and parallel, and connected by three quadrilaterals, the volume can be computed as the average of the areas of the two bases multiplied by the distance between them. But wait, that's only true if the solid is a prism or a pyramid. For a prismatoid, the formula is more general.The formula for the volume of a prismatoid is:V = (h/6)(B1 + B2 + 4M)Where B1 and B2 are the areas of the two bases, and M is the area of the midsection (the intersection of the prismatoid with a plane midway between the two bases). However, computing M might be complex here.Alternatively, since we know all the coordinates, maybe we can compute the volume using coordinates.We can use the convex hull of the six points. To compute the volume, we can divide the polyhedron into simpler components whose volumes we can calculate and sum them up.Alternatively, perhaps use the Cayley-Menger determinant. But that might be complicated.Alternatively, notice that the polyhedron is a centrally symmetric polyhedron with respect to the center of the original tetrahedron. The center of the original tetrahedron is the average of its vertices. If the original tetrahedron has vertices at (0,0,0), (2,0,0), (0,2,0), (0,0,2), then the centroid is at ( (0 + 2 + 0 + 0)/4, (0 + 0 + 2 + 0)/4, (0 + 0 + 0 + 2)/4 ) = (0.5, 0.5, 0.5). Similarly, the midpoints of the edges are at (1,0,0), (0,1,0), etc., which are symmetric with respect to the centroid (0.5, 0.5, 0.5). For example, (1,0,0) and (0,1,1) are symmetric with respect to the centroid: (1 + 0)/2 = 0.5, (0 + 1)/2 = 0.5, (0 + 1)/2 = 0.5. Similarly for other points.Therefore, the polyhedron formed by the midpoints is centrally symmetric about the centroid of the original tetrahedron. This suggests that the polyhedron might be a octahedron, but scaled down. However, given the coordinates, we saw that it's not regular. But maybe it's a rhombic dodecahedron or another centrally symmetric polyhedron. But perhaps instead of focusing on the type, we can compute the volume.Alternatively, observe that the original tetrahedron can be divided into smaller tetrahedrons and the polyhedron V₂. Maybe the volume V₂ is a certain fraction of V₁.Alternatively, using linear transformations. Since the problem doesn't specify that the tetrahedron is regular, the answer should hold for any tetrahedron. Therefore, it might be helpful to use affine invariance. The ratio of volumes should be the same regardless of the specific tetrahedron.So perhaps using barycentric coordinates or affine transformations.But let's think in terms of the original tetrahedron. If we connect the midpoints of the edges, we create a new polyhedron. Each edge of the original is replaced by a vertex at its midpoint. Then, these midpoints are connected to form the new polyhedron.Alternatively, this process is similar to creating a dual polyhedron, but not exactly. Alternatively, it's a form of a midsphere polyhedron.Wait, another approach: the polyhedron formed by connecting the midpoints is called the "midpoint polyhedron" or "edge midpoints polyhedron." Maybe known properties?Alternatively, consider that each face of the new polyhedron is a parallelogram. Because when you connect midpoints of edges in a tetrahedron, each face of the new polyhedron is formed by midpoints of adjacent edges, forming a parallelogram. Wait, how?For example, take two adjacent edges of the original tetrahedron, say AB and AC. Their midpoints are M1 and M2. Then, the edge connecting M1 and M2 would be part of the new polyhedron. But in three dimensions, these midpoints might form a tetrahedron or another shape.Wait, perhaps each face of the new polyhedron is a quadrilateral. Let's take two adjacent faces of the original tetrahedron. Each face has three midpoints. Connecting these midpoints across adjacent faces would form quadrilaterals. For example, consider the original edge AB, with midpoint M1. The two faces adjacent to AB are ABC and ABD. The midpoints of edges in these faces connected to M1 would form a quadrilateral. Wait, perhaps each original edge gives rise to a quadrilateral in the new polyhedron.Alternatively, think of the polyhedron as composed of several smaller tetrahedrons and octahedrons. For example, when you connect midpoints of edges, you might divide the original tetrahedron into smaller components whose volumes can be summed up.In 3D, splitting a tetrahedron at the midpoints of its edges can create smaller tetrahedrons and octahedrons. Let me recall that in 2D, splitting a triangle at the midpoints creates four smaller triangles, each with 1/4 the area. In 3D, splitting a tetrahedron at midpoints might create smaller pieces. However, the exact decomposition is more complex.Wait, if we connect all midpoints of edges, the original tetrahedron is divided into several smaller polyhedrons. Specifically, the original tetrahedron is divided into the convex polyhedron formed by the midpoints (V₂) and some surrounding smaller tetrahedrons.Alternatively, perhaps the volume of V₂ can be found by subtracting the volumes of the surrounding smaller tetrahedrons from V₁.Let's try that. Suppose that connecting the midpoints divides the original tetrahedron into the central polyhedron V₂ and several smaller tetrahedrons attached to each original vertex.How many smaller tetrahedrons would there be? For a tetrahedron, there are four vertices. Each vertex is connected to three edges. The midpoints of those three edges form a triangle. If we connect that triangle back to the original vertex, we get a smaller tetrahedron at each original vertex.So, four smaller tetrahedrons, each at a corner of the original tetrahedron, and the remaining central part is V₂.Therefore, the volume V₁ = 4 * volume of small tetrahedron + V₂.If we can compute the volume of each small tetrahedron, then we can find V₂.Let's compute the volume of one of these small tetrahedrons. Take, for example, the vertex A(0,0,0) of the original tetrahedron. The midpoints of the edges from A are M1(1,0,0), M2(0,1,0), M3(0,0,1). The small tetrahedron at vertex A is formed by points A, M1, M2, M3.The volume of this small tetrahedron can be calculated using the determinant formula. The vectors from A to M1, M2, M3 are (1,0,0), (0,1,0), (0,0,1). The volume is |(1,0,0) ⋅ [(0,1,0) × (0,0,1)]| / 6. The cross product of (0,1,0) and (0,0,1) is (1,0,0). Then, the dot product with (1,0,0) is 1*1 + 0*0 + 0*0 = 1. So the volume is |1| / 6 = 1/6.Wait, but in our original tetrahedron with vertices at (0,0,0), (2,0,0), (0,2,0), (0,0,2), the edges from A are of length 2, so midpoints are at 1 unit. So the small tetrahedron at A has edges of length 1. The volume of a tetrahedron with edges of length 1 (in this case, a right tetrahedron) is 1/6. Similarly, the other three small tetrahedrons at vertices B, C, D also have volume 1/6 each. So total volume of the four small tetrahedrons is 4 * 1/6 = 4/6 = 2/3.But the original volume V₁ is 4/3 (as calculated earlier). Therefore, the remaining volume V₂ = V₁ - 2/3 = 4/3 - 2/3 = 2/3. Therefore, the ratio V₂/V₁ = (2/3)/(4/3) = 1/2. Wait, that would suggest the answer is 1/2, option A. But wait, this contradicts my initial thought that it might be 1/2 or another ratio.Wait, but let me verify. If the original tetrahedron has volume 4/3, and we subtract four small tetrahedrons each of volume 1/6, getting 4/3 - 4*(1/6) = 4/3 - 2/3 = 2/3. So V₂ is 2/3, which is half of the original volume 4/3. Wait, 2/3 is half of 4/3? No, 4/3 divided by 2 is 2/3. So yes, V₂ = (1/2)V₁. Therefore, the ratio V₂/V₁ is 1/2. So answer A.But wait, I need to confirm that the remaining volume is indeed the convex polyhedron formed by the midpoints. Is the central part V₂ indeed equal to 2/3? Let's visualize.If we have the original tetrahedron, and at each vertex, we cut off a small tetrahedron with volume 1/6, then the remaining central part is an octahedron. Wait, but in our coordinates, the convex hull of the midpoints is a polyhedron with six vertices. An octahedron has six vertices. So perhaps the remaining central part is an octahedron with volume 2/3, which is half of the original tetrahedron's volume 4/3. So the ratio is 1/2.But wait, in general, for any tetrahedron, not just a regular one, does this hold? Because we did the calculation in a specific coordinate system, but the problem states "a tetrahedron" in general.Since the ratio of volumes is preserved under affine transformations, which can stretch or shrink space but preserve ratios, the result should hold for any tetrahedron, not just the regular one or the one we chose.Therefore, the ratio V₂/V₁ is 1/2, so the answer is A.Wait, but let me check again. In our coordinates, the original volume is 4/3, and V₂ is 2/3, so ratio is 1/2. However, another way: when you connect midpoints in a tetrahedron, you're essentially taking the dual or the midpoint figure. In 3D, the volume ratio might not be as straightforward as in 2D. But according to the calculation above, it's 1/2.Alternatively, I recall that in 3D, the figure formed by connecting midpoints is called the centroid octahedron, and its volume is 1/2 of the original tetrahedron. But I need to confirm.Alternatively, another approach: The convex polyhedron formed by the midpoints is called the "midpoint polyhedron," and in a tetrahedron, this is indeed an octahedron whose volume is 1/2 of the original tetrahedron. Therefore, the answer should be 1/2, option A.But wait, let me check with another example. Suppose we have a cube. If we connect midpoints of edges, what do we get? But a cube is a different polyhedron. However, a tetrahedron is different. Wait, perhaps in a cube, connecting midpoints of edges would form a different shape, but in a tetrahedron, as shown, it forms an octahedron with half the volume.Alternatively, let me consider a degenerate tetrahedron where three vertices are colinear. Wait, but the problem states it's a convex polyhedron formed by midpoints, so the original must be a valid tetrahedron.Alternatively, maybe using vectors. Let the original tetrahedron have vertices at vectors a, b, c, d. The midpoints of the edges are (a + b)/2, (a + c)/2, (a + d)/2, (b + c)/2, (b + d)/2, (c + d)/2. The convex hull of these points. To find the volume, express these points in terms of vectors.But maybe a better approach is to use the concept of parallelotope or scaling. Since each midpoint is a average of two vertices, the coordinates of the midpoint polyhedron are linear combinations of the original vertices.Alternatively, think of the midpoint polyhedron as the image of the original tetrahedron under a linear transformation. If we consider that each edge is scaled by 1/2, but in three dimensions, scaling all edges by 1/2 would result in a volume scaled by (1/2)^3 = 1/8. But this is not the case here, since we are not scaling the tetrahedron but forming a new polyhedron from midpoints.Alternatively, perhaps the new polyhedron is related to the original by a homothety transformation. A homothety is a transformation that scales space about a fixed point. If the centroid of the original tetrahedron is the center of scaling, then perhaps the midpoint polyhedron is a homothety image with scale factor -1/2. Wait, because each midpoint is halfway between two vertices, so if you invert through the centroid, it would map each vertex to the midpoint of the opposite edge.Wait, consider that the centroid of the original tetrahedron is G = (a + b + c + d)/4. If we take a homothety with center G and ratio -1/2, then each vertex x is mapped to G - 1/2(x - G) = (3G - x)/2. Let's compute this:Given a vertex a, the image under this homothety would be (3G - a)/2.Substituting G = (a + b + c + d)/4:3G = 3(a + b + c + d)/4So (3G - a)/2 = (3(a + b + c + d)/4 - a)/2 = ( -a + 3b + 3c + 3d )/8 * 2? Wait, let me compute:Wait, (3G - a)/2 = [ 3*(a + b + c + d)/4 - a ] / 2= [ (3a + 3b + 3c + 3d - 4a)/4 ] / 2= [ (-a + 3b + 3c + 3d)/4 ] / 2= (-a + 3b + 3c + 3d)/8Hmm, not sure if this is a midpoint of an edge. Let's check for vertex a:The midpoint of edge ab is (a + b)/2. Compare to (-a + 3b + 3c + 3d)/8. Not obviously the same.Alternatively, perhaps another approach. If the midpoint polyhedron is the dual of the original tetrahedron, but in three dimensions, the dual of a tetrahedron is another tetrahedron. But we have an octahedron here. Wait, the dual of a tetrahedron is indeed a tetrahedron, but in the case of midpoints, we're getting an octahedron. Maybe my earlier thought is incorrect.Alternatively, the process of connecting midpoints of edges is dual to something else. Alternatively, perhaps not dual but a different operation.But given the coordinate calculation earlier where V₂ = 2/3 when V₁ = 4/3, giving a ratio of 1/2, and considering affine invariance, it seems the answer is 1/2. But I need to make sure there's no mistake in the volume calculation.Wait, in the coordinate system where the original,<think>Okay, so I need to find the ratio of the volume V2, which is the volume of a convex polyhedron formed by the midpoints of the edges of a tetrahedron, to the original tetrahedron's volume V1. The options are (A) 1/2, (B) 2/3, (C) 3/4, (D) uncertain. Hmm, let me think.First, let me recall what a tetrahedron is. A tetrahedron has four triangular faces, six edges, and four vertices. Each edge connects two vertices. Now, if I take the midpoints of all these edges and form a convex polyhedron from them, what does that polyhedron look like?I think forming a polyhedron from the midpoints of a tetrahedron's edges... Maybe connecting these midpoints creates a smaller polyhedron inside the original tetrahedron? Or perhaps it's a different shape. Let me visualize it. If I connect all midpoints, maybe it's another tetrahedron? Wait, no, because each original edge's midpoint would be connected, but a tetrahedron has six edges, so there would be six midpoints. Connecting six points might form a different kind of polyhedron. Hmm.Alternatively, maybe it's an octahedron? An octahedron has eight faces, which are triangles. Let me think: a regular octahedron can be formed by connecting the midpoints of a cube's edges, but in this case, it's a tetrahedron. Wait, maybe not. Let me check.Alternatively, perhaps the convex polyhedron formed by the midpoints of the edges is called the "midpoint polyhedron" or something similar. But I don't recall the exact term. Maybe it's a type of dual polyhedron?Alternatively, perhaps we can consider the coordinates of a tetrahedron and compute the volumes. Let me try that approach. Let's assign coordinates to the original tetrahedron. Let me pick a regular tetrahedron for simplicity. If the original tetrahedron is regular, then maybe the calculations will be easier, and since the problem doesn't specify the tetrahedron is irregular, maybe the ratio is the same regardless.Wait, but the answer might be different for non-regular tetrahedrons. Wait, the problem just says "a tetrahedron", so it could be any tetrahedron, not necessarily regular. But the answer options are specific numbers, not "uncertain" unless (D) is the answer. But let me check.But perhaps the ratio is the same for any tetrahedron, so maybe it's a general result. Let me see.First, let's consider a regular tetrahedron. Let me assign coordinates to it. Let's say the vertices are at (0,0,0), (1,0,0), (0,1,0), and (0,0,1). Is that a regular tetrahedron? Wait, no. The edges in that case would have different lengths. For example, the distance from (0,0,0) to (1,0,0) is 1, but from (0,0,0) to (0,0,1) is also 1, but the edges between (1,0,0) and (0,1,0) is sqrt(2), so that's not regular. A regular tetrahedron has all edges equal.Alternatively, let me use coordinates for a regular tetrahedron. One possible coordinate system for a regular tetrahedron is:(1,1,1), (-1,-1,1), (-1,1,-1), (1,-1,-1). These points are all equidistant from each other. Let me check the distance between (1,1,1) and (-1,-1,1): sqrt[(-2)^2 + (-2)^2 + 0^2] = sqrt(8) = 2*sqrt(2). Similarly, between (1,1,1) and (-1,1,-1): sqrt[(-2)^2 + 0^2 + (-2)^2] = sqrt(8) = 2*sqrt(2). So all edges are 2*sqrt(2), so that's a regular tetrahedron.But maybe using coordinates that are simpler. Alternatively, consider a regular tetrahedron with edge length 2, for easier calculations.But perhaps instead of coordinates, think about the properties. The midpoints of the edges of a tetrahedron. Each edge has a midpoint, so there are six midpoints. Now, connecting these midpoints. Let's see: in a tetrahedron, each vertex is connected to three edges. The midpoints of those three edges connected to a vertex would form a smaller triangle near that vertex. But connecting all midpoints... Maybe the polyhedron formed is an octahedron? Because an octahedron has six vertices. Wait, six midpoints would correspond to the six vertices of an octahedron. Hmm.Wait, a regular octahedron has six vertices, twelve edges, and eight triangular faces. If the midpoints of a tetrahedron's edges form an octahedron, that could be possible. Let me check.In a regular tetrahedron, connecting the midpoints of the edges would indeed form a regular octahedron inside it. Let me confirm. Each edge of the original tetrahedron is split into two segments by the midpoint. The octahedron would have vertices at these midpoints. Each face of the octahedron would correspond to a pair of opposite edges in the tetrahedron. Since a tetrahedron has three pairs of opposite edges, each pair forming a "skew" line in 3D space. Wait, but in the regular tetrahedron, the midpoints would form an octahedron. So, if that's the case, then the volume of the octahedron compared to the original tetrahedron?But wait, the problem says "convex polyhedron formed by the midpoints of its edges". So if the original is a regular tetrahedron, then V2 would be the volume of the octahedron formed by those midpoints.But then, what's the volume ratio between the octahedron and the original tetrahedron? Let me compute that.First, let's find the volume of the regular octahedron formed by the midpoints. Let's assume the original regular tetrahedron has edge length a. Then, the edge length of the octahedron would be half the edge length of the tetrahedron? Wait, no. The distance between two midpoints would depend on which edges they belong to.Wait, in a regular tetrahedron, the distance between midpoints of adjacent edges (edges that meet at a vertex) would be half the edge length of the original tetrahedron. Because each original edge is length a, so the midpoint divides it into two segments of length a/2. The distance between midpoints of two edges that meet at a vertex would form a triangle with sides a/2, a/2, and the angle between them. The angle between two edges of a regular tetrahedron is arccos(1/3). Wait, but maybe it's easier to compute coordinates.Let me place the regular tetrahedron with vertices at (1,1,1), (1,-1,-1), (-1,1,-1), (-1,-1,1). These coordinates form a regular tetrahedron centered at the origin. The edge length here can be calculated. Let's compute the distance between (1,1,1) and (1,-1,-1): sqrt[(0)^2 + (-2)^2 + (-2)^2] = sqrt(0 + 4 + 4) = sqrt(8) = 2*sqrt(2). So edge length is 2*sqrt(2). Then the midpoints of the edges would be the average of the coordinates of each pair of vertices.So for example, the midpoint between (1,1,1) and (1,-1,-1) is ((1+1)/2, (1 + (-1))/2, (1 + (-1))/2) = (1, 0, 0). Similarly, midpoint between (1,1,1) and (-1,1,-1) is ((1 + (-1))/2, (1 +1)/2, (1 + (-1))/2) = (0, 1, 0). Midpoint between (1,1,1) and (-1,-1,1) is ((1 + (-1))/2, (1 + (-1))/2, (1 +1)/2) = (0, 0, 1). Similarly, midpoint between (1,-1,-1) and (-1,1,-1) is ((1 + (-1))/2, (-1 +1)/2, (-1 + (-1))/2) = (0, 0, -1). Midpoint between (1,-1,-1) and (-1,-1,1) is ((1 + (-1))/2, (-1 + (-1))/2, (-1 +1)/2) = (0, -1, 0). Midpoint between (-1,1,-1) and (-1,-1,1) is ((-1 + (-1))/2, (1 + (-1))/2, (-1 +1)/2) = (-1, 0, 0).Wait, so the midpoints are (1,0,0), (0,1,0), (0,0,1), (0,0,-1), (0,-1,0), (-1,0,0). These are the six points at the unit axes. Connecting these points forms a regular octahedron centered at the origin with vertices at (±1,0,0), (0,±1,0), (0,0,±1). The edge length of this octahedron is the distance between (1,0,0) and (0,1,0), which is sqrt[(1-0)^2 + (0 - 1)^2 + (0 - 0)^2] = sqrt(1 + 1) = sqrt(2). So the edge length of the octahedron is sqrt(2), and the original tetrahedron had edge length 2*sqrt(2). So the octahedron is scaled down by a factor of 1/2 in edge length.But volume scales with the cube of the linear dimensions. However, in this case, the octahedron is not scaled uniformly from the tetrahedron. Wait, but maybe it's easier to compute the volume directly.The volume of a regular octahedron with edge length a is (sqrt(2)/3) * a^3. For our octahedron here, edge length is sqrt(2), so volume would be (sqrt(2)/3) * (sqrt(2))^3 = (sqrt(2)/3)*(2*sqrt(2)) )= (sqrt(2)*2*sqrt(2))/3 = (2*2)/3 = 4/3.Wait, but let's check the original tetrahedron's volume. The original tetrahedron with vertices at (1,1,1), (1,-1,-1), (-1,1,-1), (-1,-1,1). To compute its volume, we can use the formula for the volume of a tetrahedron given coordinates. The volume is |det(B - A, C - A, D - A)| / 6, where A, B, C, D are the vertices.Let me take A = (1,1,1), B = (1,-1,-1), C = (-1,1,-1), D = (-1,-1,1). Then vectors B - A = (0, -2, -2), C - A = (-2, 0, -2), D - A = (-2, -2, 0). The determinant is:| 0 -2 -2 || -2 0 -2 || -2 -2 0 |Calculating the determinant:0*(0*0 - (-2)*(-2)) - (-2)*(-2*0 - (-2)*(-2)) + (-2)*(-2*(-2) - 0*(-2)).Wait, determinant is calculated as:0*(0*0 - (-2)*(-2)) - (-2)*( (-2)*0 - (-2)*(-2) ) + (-2)*( (-2)*(-2) - 0*(-2) )So first term: 0*(0 - 4) = 0Second term: -(-2)*(0 - 4) = 2*(-4) = -8Third term: (-2)*(4 - 0) = -2*4 = -8So total determinant is 0 -8 -8 = -16. The absolute value is 16, so volume is 16/6 = 8/3.Wait, so original tetrahedron volume V1 is 8/3. The octahedron volume V2 we computed as 4/3. Then the ratio V2/V1 is (4/3)/(8/3) = 1/2. So that would be option (A) 1/2.But wait, in this case, the original tetrahedron was regular, and the convex polyhedron formed by the midpoints is a regular octahedron. So the ratio is 1/2. But the problem doesn't specify that the original tetrahedron is regular, so does this ratio hold for any tetrahedron?Hmm, the answer options include (D) uncertain, which suggests that the ratio might depend on the shape of the original tetrahedron. However, sometimes certain ratios are preserved under affine transformations. Let me recall that volume ratios can be preserved under affine transformations if the construction is linear.But is the ratio V2/V1 preserved for any tetrahedron? Let's check with a different tetrahedron.Let me take a degenerate tetrahedron first, but wait, a convex polyhedron formed by midpoints would require the original tetrahedron to be non-degenerate. Let me pick a different non-regular tetrahedron.Let's take a tetrahedron with vertices at (0,0,0), (2,0,0), (0,2,0), (0,0,2). Let's compute its volume. The volume is |det([2,0,0], [0,2,0], [0,0,2])| /6 = |2*2*2| /6 = 8/6 = 4/3.Now, compute the midpoints of all edges. The edges are:Between (0,0,0) and (2,0,0): midpoint (1,0,0)Between (0,0,0) and (0,2,0): midpoint (0,1,0)Between (0,0,0) and (0,0,2): midpoint (0,0,1)Between (2,0,0) and (0,2,0): midpoint (1,1,0)Between (2,0,0) and (0,0,2): midpoint (1,0,1)Between (0,2,0) and (0,0,2): midpoint (0,1,1)So the midpoints are:(1,0,0), (0,1,0), (0,0,1), (1,1,0), (1,0,1), (0,1,1)Now, what is the convex polyhedron formed by these points? Let's see. Connecting these midpoints. Let's see if this is an octahedron or a different polyhedron.An octahedron has eight triangular faces. Let's see how these points connect. For example, (1,0,0) is connected to (1,1,0), (1,0,1), and maybe others? Wait, in the original tetrahedron, each edge's midpoint is connected to other midpoints. Let's try to figure out the structure.Alternatively, maybe this polyhedron is a parallelepiped? Let's check. The points (1,0,0), (0,1,0), (0,0,1) form three axes. Then (1,1,0), (1,0,1), (0,1,1) are combinations. Wait, if you take the convex hull of these points, you might get a polyhedron with vertices at these six points.Wait, actually, in 3D, the convex hull of these six points is a polyhedron known as a "rhombic dodecahedron"? Hmm, no. Wait, let me think again.Alternatively, maybe it's a polyhedron with eight triangular faces. Let's consider how these midpoints are connected. Each original edge is connected to two others at each vertex, so the midpoints form edges that correspond to the original edges' adjacency.Alternatively, maybe it's easier to compute the volume by considering coordinates.Let me list all the midpoints:1. (1,0,0)2. (0,1,0)3. (0,0,1)4. (1,1,0)5. (1,0,1)6. (0,1,1)Now, to find the convex polyhedron (V2) formed by these points. Let's see what the shape is.If we connect these points, the polyhedron is called a "rectified tetrahedron", which is an octahedron. Wait, in this case, with non-regular tetrahedron, the midpoints would form a non-regular octahedron. Let me check.Wait, in the regular case, it's a regular octahedron, but in a non-regular tetrahedron, it's a general octahedron. Let's see if the volume ratio is still 1/2.Alternatively, perhaps we can use linear algebra to decompose the problem.Given that the original tetrahedron has volume V1, the polyhedron formed by the midpoints can be considered as a combination of smaller tetrahedrons and octahedrons. Wait, maybe there's a formula here.Alternatively, note that when you connect the midpoints of the edges of a tetrahedron, the resulting polyhedron (called the midpoint polyhedron) can be divided into smaller components whose volumes relate to the original.Alternatively, consider that each original edge is replaced by a midpoint, and the new polyhedron is the convex hull of these midpoints. The volume of this convex hull relative to the original can be determined by affine transformations.Wait, another approach: since the midpoints are at the centroids of the edges, the new polyhedron is the image of the original under a linear transformation. Wait, but which transformation?If we think of the midpoints as being the average of two vertices, then perhaps the transformation scales the original tetrahedron by a factor of 1/2. But that might not be accurate. Because each midpoint is an average of two vertices, but the entire set of midpoints might not form a scaled-down version of the original.Wait, but in the regular tetrahedron case, the octahedron formed by the midpoints is indeed scaled down. But in a non-regular tetrahedron, scaling factors could be different along different axes, affecting the volume ratio.Wait, but perhaps the key is that the midpoint polyhedron (octahedron) can be decomposed into smaller tetrahedrons whose total volume is 1/2 of the original.Alternatively, in the regular tetrahedron case, the octahedron's volume is 1/2 of the original tetrahedron. Maybe this holds for any tetrahedron due to affine invariance.Wait, affine transformations can distort shapes but preserve ratios of volumes. So if we can show that the ratio V2/V1 is the same for any tetrahedron under affine transformations, then the ratio is fixed.Since any tetrahedron can be affinely transformed into any other tetrahedron, and affine transformations preserve volume ratios, then if in the regular case the ratio is 1/2, it should be 1/2 for all tetrahedrons.But is that true?Wait, but when we take midpoints of edges, which is a linear operation, then under an affine transformation, the midpoints are preserved. So if we have an affine transformation that maps one tetrahedron to another, then the midpoint polyhedron of the transformed tetrahedron is the transform of the original midpoint polyhedron, hence the volume ratio remains the same.Therefore, since the ratio is 1/2 for the regular tetrahedron, it must be 1/2 for all tetrahedrons.Therefore, the answer should be (A) 1/2.Wait, but let me verify with the non-regular tetrahedron example.Original tetrahedron with vertices at (0,0,0), (2,0,0), (0,2,0), (0,0,2). Volume V1 is 4/3.Midpoints are (1,0,0), (0,1,0), (0,0,1), (1,1,0), (1,0,1), (0,1,1). Now, need to compute the volume of the convex hull of these points.Let me see. The convex hull of these six points is a polyhedron. To compute its volume, perhaps we can decompose it into simpler parts.First, notice that the points (1,0,0), (0,1,0), (0,0,1), (1,1,0), (1,0,1), (0,1,1) can form an octahedron. Let's see how.An octahedron has two pyramids with a square base glued together. But in this case, the octahedron might be irregular.Alternatively, perhaps the convex hull can be divided into pyramids whose volumes can be calculated.Alternatively, use the convex hull's coordinates to compute the volume.Alternatively, use the Cayley-Menger determinant. But that might be complicated.Alternatively, realize that the convex polyhedron formed by these midpoints is a centrally symmetric polyhedron. Each point has a counterpart on the opposite side. For example, (1,0,0) and (0,1,1) are not opposites, but maybe (1,0,0) and (0,1,1) are connected? Wait, not sure.Wait, maybe the convex hull consists of all points where each coordinate is either 0 or 1, except for the originals. Wait, but the points here have coordinates with one coordinate 1 and the others 0, or two coordinates 1 and one 0. Wait, no: (1,0,0), (0,1,0), (0,0,1) are the midpoints of the edges from the origin to the other vertices. Then (1,1,0), (1,0,1), (0,1,1) are the midpoints of the edges connecting the other vertices.So, to compute the volume of the convex hull, perhaps think of it as a polyhedron with vertices at these midpoints. Let me see if I can decompose this into tetrahedrons.Alternatively, note that the convex hull of these six points is an octahedron. Let's confirm. An octahedron has six vertices, twelve edges, eight triangular faces. Let's see if this is the case here.Take the points (1,0,0), (0,1,0), (0,0,1), (1,1,0), (1,0,1), (0,1,1). Each of these points can be connected to form triangles.For example, (1,0,0) connected to (1,1,0) connected to (0,1,0) connected back to (1,0,0) forms a triangle in the plane z=0. Similarly, (1,0,0) connected to (1,0,1) connected to (0,0,1) connected back to (1,0,0) forms a triangle in the plane y=0. Similarly, (0,1,0) connected to (0,1,1) connected to (0,0,1) connected back to (0,1,0) forms a triangle in the plane x=0. Similarly, the points (1,1,0), (1,0,1), (0,1,1), and the other points might form the upper part.Wait, maybe it's better to note that the convex hull of these six points is a polyhedron with eight triangular faces. For example, each face is a triangle connecting three midpoints.Alternatively, since this seems complicated, maybe use the fact that this polyhedron is dual to the original tetrahedron in some way. But I'm not sure.Alternatively, note that the original tetrahedron can be divided into smaller tetrahedrons and octahedrons, and the volume V2 corresponds to the octahedron part.Wait, actually, when you connect the midpoints of a tetrahedron's edges, the resulting polyhedron is called the "medial tetrahedron" or "midpoint tetrahedron", but actually, it's an octahedron. In a general tetrahedron, connecting the midpoints of the edges forms an octahedron whose volume is 1/2 of the original tetrahedron. Is that a known result?Yes, I recall that in any tetrahedron, the midpoints of the edges form an octahedron whose volume is 1/2 of the original tetrahedron. This is because each face of the octahedron corresponds to a pair of opposite edges in the tetrahedron, and the octahedron can be split into four smaller tetrahedrons each similar to the original, but scaled down. Wait, perhaps not.Alternatively, using vector analysis. Let’s denote the original tetrahedron as ABCD. The midpoints of the edges can be labeled as follows:Let M_AB be the midpoint of AB, M_AC of AC, M_AD of AD, M_BC of BC, M_BD of BD, M_CD of CD.The convex hull of these midpoints forms the octahedron. Now, to compute its volume, perhaps express the octahedron in terms of the original tetrahedron.Another approach: the octahedron can be divided into two square pyramids with a common square base. Each pyramid's volume is (base area * height)/3. If we can find the base area and the height, we can compute the volume.But in this case, since the original tetrahedron is not regular, the base might not be a square, but a parallelogram. However, in the general case, the midpoints form a octahedron which is not regular, but still the volume can be computed as 1/2 of the original.Alternatively, consider that each edge of the original tetrahedron is split into two, and the octahedron occupies the central region. The original tetrahedron can be divided into the octahedron and four smaller tetrahedrons at each corner. Each of these smaller tetrahedrons has volume 1/8 of the original, so 4*(1/8) = 1/2, leaving 1/2 for the octahedron.Wait, let me elaborate. Suppose we have a tetrahedron ABCD. If we connect the midpoints of the edges, we create an octahedron inside. Additionally, at each original vertex, there's a small tetrahedron formed by connecting the midpoints. For example, at vertex A, the small tetrahedron is formed by midpoints M_AB, M_AC, M_AD, and another midpoint? Wait, actually, each original vertex is surrounded by three midpoints. Connecting those three midpoints and the centroid?Wait, maybe not. Let me think.If we take the original tetrahedron and connect all midpoints, the octahedron is in the center, and the remaining parts are four smaller tetrahedrons near each original vertex. Each of these smaller tetrahedrons is similar to the original but scaled down by a factor of 1/2. Since volume scales with the cube of the scaling factor, each small tetrahedron would have volume (1/2)^3 = 1/8 of the original. With four such tetrahedrons, total volume would be 4*(1/8) = 1/2. Therefore, the remaining volume, which is the octahedron, would be 1 - 1/2 = 1/2 of the original volume. Thus, V2/V1 = 1/2.This seems like a valid argument. Therefore, regardless of the original tetrahedron's shape, connecting the midpoints of edges creates an octahedron whose volume is half of the original tetrahedron's volume. Hence, the answer should be (A) 1/2.But let me confirm this with the non-regular example.Original volume V1 is 4/3. If the octahedron volume V2 is 2/3, then the ratio is 1/2. Let's compute V2 for the non-regular tetrahedron.The midpoints are (1,0,0), (0,1,0), (0,0,1), (1,1,0), (1,0,1), (0,1,1). Let's try to compute the volume of the convex hull.One way is to note that the convex hull of these points is the intersection of the half-spaces defined by the original tetrahedron's midplanes.Alternatively, consider that the octahedron can be split into pyramids. Let me pick a point and calculate the volume of the pyramid it forms with a base.For example, take the point (1,1,0). The base could be the triangle formed by (1,0,0), (0,1,0), (0,0,1). Wait, but how?Alternatively, use the Cayley-Menger determinant for six points, but that might be too complex.Alternatively, consider that the octahedron can be divided into five tetrahedrons. Let me try to decompose it.First, choose a tetrahedron formed by (1,0,0), (0,1,0), (0,0,1), (1,1,0). But (1,1,0) is coplanar with (1,0,0), (0,1,0), and (0,0,1) is not. Wait, no.Alternatively, let's choose the tetrahedron (1,0,0), (0,1,0), (0,0,1), (1,1,1). Wait, but (1,1,1) is not one of our points. Our points are (1,0,0), (0,1,0), (0,0,1), (1,1,0), (1,0,1), (0,1,1).Wait, perhaps split the octahedron into two square pyramids with a common square base. Let's see if there is a square base.Looking at the points (1,0,0), (0,1,0), (1,1,0), and (0,0,0). Wait, (0,0,0) isn't one of our midpoints. The midpoints are (1,0,0), (0,1,0), (0,0,1), (1,1,0), (1,0,1), (0,1,1).Wait, perhaps the square base is formed by (1,0,0), (0,1,0), (1,0,1), (0,1,1). Is that a square?Distance between (1,0,0) and (0,1,0) is sqrt(2).Distance between (0,1,0) and (0,1,1) is 1.Distance between (0,1,1) and (1,0,1) is sqrt(2).Distance between (1,0,1) and (1,0,0) is 1.So this is a rectangle, but not a square. The sides are sqrt(2) and 1. Therefore, the area is sqrt(2)*1 = sqrt(2).But to form a pyramid, we need a apex. Let's take the point (1,1,0) as apex. Then the volume of this pyramid would be (base area * height)/3.The base is the rectangle with vertices (1,0,0), (0,1,0), (0,1,1), (1,0,1). Wait, but how is that a rectangle? The coordinates are (1,0,0), (0,1,0), (0,1,1), (1,0,1). Connecting these points in order forms a non-planar quadrilateral? Wait, is this a planar figure?Let me check if these four points lie on the same plane. To check planarity, compute the volume of the tetrahedron formed by three vectors from one point. Take point (1,0,0). The vectors to (0,1,0), (0,1,1), and (1,0,1) are (-1,1,0), (-1,1,1), (0,0,1). The scalar triple product is:| -1 -1 0 || 1 1 0 || 0 1 1 |Wait, no. Wait, vectors from (1,0,0) to the other points:To (0,1,0): (-1, 1, 0)To (0,1,1): (-1, 1, 1)To (1,0,1): (0, 0, 1)The scalar triple product is the determinant of the matrix formed by these vectors as columns:| -1 -1 0 || 1 1 0 || 0 1 1 |Calculating determinant:-1*(1*1 - 0*1) - (-1)*(1*1 - 0*0) + 0*(1*1 - 1*0)= -1*(1) - (-1)*(1) + 0*1= -1 + 1 + 0 = 0Since the scalar triple product is zero, the four points are coplanar. Therefore, they form a planar quadrilateral, which is a rectangle? Wait, earlier distances: between (1,0,0) and (0,1,0) is sqrt(2), between (0,1,0) and (0,1,1) is 1, between (0,1,1) and (1,0,1) is sqrt(2), between (1,0,1) and (1,0,0) is 1. So the sides alternate between sqrt(2) and 1, and the angles are 90 degrees?Wait, let's compute the vectors:From (1,0,0) to (0,1,0): (-1,1,0)From (0,1,0) to (0,1,1): (0,0,1)From (0,1,1) to (1,0,1): (1,-1,0)From (1,0,1) to (1,0,0): (0,0,-1)So the sides are (-1,1,0), (0,0,1), (1,-1,0), (0,0,-1). The adjacent vectors are perpendicular:Dot product of (-1,1,0) and (0,0,1) is 0.Dot product of (0,0,1) and (1,-1,0) is 0.Dot product of (1,-1,0) and (0,0,-1) is 0.Dot product of (0,0,-1) and (-1,1,0) is 0.Therefore, the quadrilateral is a rectangle, with sides of length sqrt(2) and 1. So the area is sqrt(2) * 1 = sqrt(2).Now, the apex of the pyramid is (1,1,0). The height of the pyramid is the distance from (1,1,0) to the plane of the base. But since the base is in the plane z=0 to z=1. Wait, the base has points with z-coordinates 0 and 1. Wait, actually, the base is the rectangle with points (1,0,0), (0,1,0), (0,1,1), (1,0,1). The plane equation can be determined.Let me find the equation of the plane. Taking three points: (1,0,0), (0,1,0), (0,1,1).Vectors in the plane: from (1,0,0) to (0,1,0): (-1,1,0)From (1,0,0) to (0,1,1): (-1,1,1)The normal vector is the cross product of these two vectors:(-1,1,0) × (-1,1,1) = determinant:i j k-1 1 0-1 1 1= i*(1*1 - 0*1) - j*(-1*1 - 0*(-1)) + k*(-1*1 - (-1)*1)= i*(1) - j*(-1) + k*(-1 +1)= i + j + 0kSo normal vector is (1,1,0). The plane equation is 1*(x -1) + 1*(y -0) + 0*(z -0) = 0 → x + y -1 = 0. So the plane equation is x + y = 1.Now, the apex (1,1,0) has coordinates (1,1,0). The distance from this point to the plane x + y = 1 is |1 +1 -1| / sqrt(1^2 +1^2 +0^2) = |1| / sqrt(2) = 1/sqrt(2).But height of the pyramid is this distance, so the volume is (base area * height)/3 = (sqrt(2) * 1/sqrt(2))/3 = (1)/3.But since there are two such pyramids (one with apex (1,1,0) and another with apex (0,0,1)? Wait, but all other apexes? Wait, no.Wait, the convex hull of the six midpoints should be an octahedron, which can be divided into two square pyramids. But in this non-regular case, it's divided into two]

🔑:2.A.The volume of the convex polyhedron V_{2} is the hexahedron obtained by cutting off 4 small tetrahedra from the original tetrahedron, i.e.,V_{2}=V_{1}-4 V_{1} timesleft(frac{1}{2}right)^{3}=frac{V_{1}}{2} .

🧠:[<think>Okay, so the problem here is about converting the number 935,000,000 into scientific notation and choosing the correct option from A to D. Let me remember how scientific notation works. From what I recall, scientific notation is a way to write very large or very small numbers more conveniently. It's expressed as a number between 1 and 10 multiplied by a power of 10. So, the general form is a × 10^b, where 1 ≤ a < 10 and b is an integer.Alright, so let's take the given number: 935,000,000. I need to convert this into the form where there's one non-zero digit before the decimal point. Let's break this down. The number 935,000,000 can be written as 935 million. But in scientific notation, we need to have a number between 1 and 10 multiplied by 10 raised to some exponent. Starting with 935,000,000. If I move the decimal point to the left, each move will increase the exponent by 1. The original number is 935,000,000.00, right? The decimal is after the last zero. So to get a number between 1 and 10, I need to move the decimal point eight places to the left. Let's see: moving the decimal one place gives 93,500,000.0, which is still greater than 10. Move another place: 9,350,000.00. Still greater than 10. Wait, maybe I need to count how many places I move until the number is between 1 and 10.Wait, actually, 935,000,000 is the same as 9.35 × 10^8. Let me check that. If I take 9.35 and multiply it by 10^8, that's 9.35 × 100,000,000, which equals 935,000,000. So that seems correct. But let me make sure I didn't skip any steps.Alternatively, maybe the process is to count the number of digits in the original number and subtract one to get the exponent? Let's see. 935,000,000 has 9 digits. If I subtract 1, that gives 8, which would be the exponent. So 9.35 × 10^8, which is option B. But let me verify the other options to be sure.Option A is 0.935 × 10^9. Hmm, that would be 0.935 multiplied by 1,000,000,000, which is 935,000,000. Wait, that also works. But in scientific notation, the coefficient has to be at least 1 and less than 10. So 0.935 is less than 1, which means option A isn't correct scientific notation. So even though 0.935 × 10^9 equals the same number, it's not in proper scientific notation because the leading number isn't between 1 and 10. So A is out.Option C is 93.5 × 10^7. Let's compute that. 93.5 × 10^7 is 93.5 × 10,000,000, which is 935,000,000. Again, that's the correct value, but 93.5 is greater than 10, so it's not between 1 and 10. Therefore, option C is also not correct scientific notation.Option D is 935 × 10^6. Calculating that gives 935 × 1,000,000 = 935,000,000. Same number, but 935 is way larger than 10, so again, not valid scientific notation. So D is out.So the only one that follows the rule of having the coefficient between 1 and 10 is option B: 9.35 × 10^8. Therefore, the correct answer should be B.But wait, just to double-check. Let's convert 935,000,000 step by step. Start with the number: 935,000,000. The decimal is at the end (after the last zero). To get a number between 1 and 10, move the decimal point to after the first non-zero digit. Here, the first digit is 9 (from 935...). So moving the decimal between 9 and 3. So that's 9.35000000. How many places did I move the decimal? Let's count. Starting from the end of 935,000,000:Original number: 935,000,000.0 Move decimal left by 8 places: 9.35000000Yes, moving the decimal 8 places to the left gives 9.35, so the exponent is 8. Therefore, 9.35 × 10^8 is indeed correct, which is option B. So that's the right answer. The other options either have the coefficient outside the 1-10 range or use a different exponent but still get the same numerical value, but they don't adhere to the scientific notation rules.Another way to think about it is that in scientific notation, you want exactly one non-zero digit before the decimal point. So for 935,000,000, that non-zero digit is 9, and the rest (35) come after the decimal, making it 9.35. Then, since we moved the decimal 8 places, the exponent is 8. So 9.35 × 10^8.Just to be thorough, let's check the exponents in each option:A: 0.935 × 10^9. Here, 0.935 is less than 1, which is invalid.B: 9.35 × 10^8. 9.35 is between 1 and 10, and exponent 8. Correct.C: 93.5 × 10^7. 93.5 is greater than 10, invalid.D: 935 × 10^6. 935 is way greater than 10, invalid.Thus, B is the only valid scientific notation. So the answer is B.Final Answerboxed{B}</think>To convert the number 935,000,000 into scientific notation, we need to express it in the form (a times 10^b) where (1 leq a < 10) and (b) is an integer.1. Identify the number: 935,000,000.2. Move the decimal point to create a number between 1 and 10. For 935,000,000, we move the decimal point 8 places to the left, resulting in 9.35.3. Determine the exponent: Moving the decimal point 8 places to the left means the exponent (b) is 8.4. Form the scientific notation: Thus, 935,000,000 becomes (9.35 times 10^8).Checking the options:- Option A: (0.935 times 10^9) is incorrect because 0.935 is not between 1 and 10.- Option B: (9.35 times 10^8) is correct because 9.35 is between 1 and 10, and the exponent 8 is appropriate.- Option C: (93.5 times 10^7) is incorrect because 93.5 is not between 1 and 10.- Option D: (935 times 10^6) is incorrect because 935 is not between 1 and 10.Therefore, the correct answer is (boxed{B}).,<think>Okay, let's see. The question is about expressing the number 935,000,000 in scientific notation. The options are A through D. Hmm, scientific notation, right? Let me recall how that works.From what I remember, scientific notation is a way to write numbers that are too big or too small to be conveniently written in decimal form. It's usually expressed as a number between 1 and 10 multiplied by a power of 10. So, the general form is a × 10^b, where 1 ≤ a < 10 and b is an integer.Alright, so the number given here is 935,000,000. Let me write that out: 935 followed by six zeros. So, 935,000,000. To convert this into scientific notation, I need to move the decimal point so that there's only one non-zero digit before the decimal. Right now, the number is 935,000,000. The decimal is at the end, right? So, 935,000,000. To make it a number between 1 and 10, I should move the decimal point two places to the left. Let me check that. Moving it once: 93,500,000.0. Moving it again: 9,350,000.00. Wait, that's not right. Wait, no, actually, if the original number is 935,000,000, the decimal is after the last zero. To make it between 1 and 10, we need to move the decimal point eight places to the left? Wait, maybe I need to count how many digits there are.Let me count the digits in 935,000,000. 9, 3, 5, followed by six zeros. That's 9 digits. So, in scientific notation, the exponent is one less than the number of digits? Wait, no, the exponent is the number of places you move the decimal point to get from the original number to a number between 1 and 10.Original number: 935,000,000. To get a number between 1 and 10, we need to move the decimal point eight places to the left. Let's see:Start at 935,000,000.0Move decimal one place left: 93,500,000.00Second place: 9,350,000.000Third: 935,000.0000Fourth: 93,500.00000Fifth: 9,350.000000Sixth: 935.0000000Seventh: 93.50000000Eighth: 9.350000000Wait, that can't be. Wait, starting from 935,000,000, which is 935 million. So, moving the decimal point how many places? Let's see. 935 million is 9.35 times 100,000,000, right? Because 100,000,000 is 10^8. So, 9.35 × 10^8. That would make sense. So, moving the decimal point 8 places to the left from 935,000,000 gives 9.35 × 10^8. So that would be option B.But let me check the options again. A is 0.935 × 10^9. Well, 0.935 is less than 1, so that's not correct because in scientific notation, the coefficient should be at least 1 and less than 10. So A is invalid. Option B is 9.35 × 10^8. That fits because 9.35 is between 1 and 10. Then 10^8 is 100,000,000, so 9.35 × 100,000,000 is 935,000,000. That works.Option C is 93.5 × 10^7. 93.5 is more than 10, so that's not correct for scientific notation. Even though 93.5 × 10^7 is equal to 935,000,000, it's not proper scientific notation because the coefficient must be less than 10. Similarly, option D is 935 × 10^6. 935 is way over 10, so that's not correct either. So even though all these options might equal 935,000,000 when multiplied out, only option B is in proper scientific notation.Therefore, the correct answer should be B. Let me just verify once again. Original number: 935,000,000. Move decimal 8 places to the left: 9.35. Multiply by 10^8. Yes, 9.35 × 10^8 = 935,000,000. So B is correct. The other options either have coefficients outside the 1-10 range or use different exponents. So, definitely B.Final Answerboxed{B}</think>To express the number 935,000,000 in scientific notation, we need to convert it into a number between 1 and 10 multiplied by a power of 10. 1. The number 935,000,000 can be written as 935,000,000.0.2. Moving the decimal point 8 places to the left gives us 9.35.3. This adjustment requires multiplying by (10^8) to maintain the original value.Thus, 935,000,000 in scientific notation is (9.35 times 10^8).Checking the options:- Option A: (0.935 times 10^9) is incorrect because the coefficient is less than 1.- Option B: (9.35 times 10^8) is correct as the coefficient is between 1 and 10.- Option C: (93.5 times 10^7) is incorrect because the coefficient is greater than 10.- Option D: (935 times 10^6) is incorrect because the coefficient is much greater than 10.Therefore, the correct answer is (boxed{B}).]

🔑:To express the number 935,000,000 in scientific notation, we start by placing the decimal point after the first non-zero digit, which gives us 9.35. Next, we count the number of places the decimal has moved from its original position at the end of the number to its new position. The decimal point has moved 8 places to the left. Therefore, we express 935,000,000 as 9.35 times 10^{8}.Hence, the correct option is boxed{B}.

🧠:[<think>Okay, so I need to solve this equation where the sum of the integer parts of x divided by n factorial from n=1 to n=10 equals 1001. Hmm, let me break this down. The equation is:[left[frac{x}{1!}right] + left[frac{x}{2!}right] + left[frac{x}{3!}right] + cdots + left[frac{x}{10!}right] = 1001]First, I remember that the brackets here denote the floor function, which means taking the greatest integer less than or equal to the value inside. So, for each term, I have to take x divided by n! and then take the floor of that. Then add them all up from 1! to 10! and set it equal to 1001. The question asks for an integer solution, so x must be an integer.Since factorials grow very quickly, the terms for higher n (like 7! to 10!) are going to be much smaller compared to the lower ones. For example, 10! is 3,628,800, so even if x is a large number, when divided by 10!, it might be a very small fraction, and the floor of that would be zero unless x is really huge. But since the total sum is 1001, which isn't astronomically large, maybe those higher terms (like from 7! up) contribute very little or even zero? Let me check.Suppose x is not extremely large. Let's say x is around 500. Then x/7! = 500/5040 ≈ 0.099, so floor of that is 0. Similarly, x/8! would be even smaller. So maybe starting from 7!, all terms are zero. So maybe the equation simplifies to:floor(x/1!) + floor(x/2!) + floor(x/3!) + floor(x/4!) + floor(x/5!) + floor(x/6!) = 1001But I need to verify if that's the case. Let's test with a larger x. Let's say x is 1000. Then x/7! = 1000/5040 ≈ 0.198, still floor to 0. x/8! = 1000/40320 ≈ 0.024, still 0. Similarly, 9! and 10! are even larger, so those terms would definitely be 0. So yes, for x up to 5039 (since 7! is 5040), x/7! would be less than 1, so floor to 0. Therefore, the terms from 7! to 10! don't contribute anything. So the equation reduces to the sum from n=1 to n=6. Therefore, the equation is:floor(x) + floor(x/2) + floor(x/6) + floor(x/24) + floor(x/120) + floor(x/720) = 1001Wait, hold on. Wait, 1! is 1, so floor(x/1!) is floor(x). 2! is 2, so floor(x/2). 3! is 6, floor(x/6). 4! is 24, floor(x/24). 5! is 120, floor(x/120). 6! is 720, floor(x/720). Then 7! is 5040, which we established would be 0 unless x is over 5040, but if x is over 5040, then the sum would have more terms. But since the total is 1001, which is not that big, perhaps x is less than 5040? Let's check.Suppose x is 720 (which is 6!). Then floor(x/720) = 1. Then floor(x/120) = floor(720/120) = 6, floor(x/24) = 30, floor(x/6) = 120, floor(x/2) = 360, floor(x) = 720. Adding all those up: 720 + 360 + 120 + 30 + 6 + 1 = 1237. Hmm, that's already higher than 1001. So if x=720, the sum is 1237, which is way over. So x must be less than 720.Wait, but then let's check x=500. Then floor(x)=500, floor(250)=250, floor(500/6)=83 (since 500/6≈83.333), floor(500/24)=20 (since 500/24≈20.833), floor(500/120)=4 (500/120≈4.166), floor(500/720)=0. So sum is 500 + 250 + 83 + 20 + 4 + 0 = 857. That's less than 1001. So 500 gives 857. So we need something higher. Let's try x=600. Then floor(600)=600, floor(300)=300, floor(100)=100 (600/6=100), floor(25)=25 (600/24=25), floor(5)=5 (600/120=5), floor(0.833)=0. Sum is 600+300+100+25+5+0=1030. Hmm, that's closer to 1001. So x=600 gives 1030, which is 29 over. So maybe x is a bit less. Let's try x=590.x=590: floor(590)=590, floor(295)=295 (590/2=295), floor(590/6)=floor(98.333)=98, floor(590/24)=floor(24.583)=24, floor(590/120)=floor(4.916)=4, floor(590/720)=0. Sum: 590+295+98+24+4+0=1011. Still over by 10. Let's try x=580.x=580: floor(580)=580, floor(290)=290, floor(580/6)=96.666→96, floor(580/24)=24.166→24, floor(580/120)=4.833→4, floor(0.805)=0. Sum: 580+290+96+24+4=994. Hmm, 994. So that's 7 less than 1001. So the sum increases as x increases. So between x=580 and x=590, the sum goes from 994 to 1011, which is an increase of 17 over an increase of 10 in x. Wait, but x increases by 10, but sum increases by 17. So perhaps the difference is not linear.Wait, but floor functions can have different contributions. Let me check x=585.x=585: floor(585)=585, floor(292.5)=292, floor(585/6)=floor(97.5)=97, floor(585/24)=floor(24.375)=24, floor(585/120)=floor(4.875)=4, floor(585/720)=0. Sum: 585+292+97+24+4=585+292=877, 877+97=974, 974+24=998, 998+4=1002. So 1002. That's very close. Only 1 over. So desired sum is 1001, so maybe x=584.x=584: floor(584)=584, floor(584/2)=292, floor(584/6)=floor(97.333)=97, floor(584/24)=floor(24.333)=24, floor(584/120)=floor(4.866)=4, floor(584/720)=0. Sum: 584+292+97+24+4=584+292=876, 876+97=973, 973+24=997, 997+4=1001. Perfect! So x=584 gives the sum 1001. Wait, so is x=584 the solution?But let me verify once again. Let me compute each term step by step.For x=584:1. [584 / 1!] = 5842. [584 / 2!] = [584 / 2] = 2923. [584 / 3!] = [584 / 6] ≈ 97.333, so floor is 974. [584 / 4!] = [584 / 24] ≈ 24.333, floor is 245. [584 / 5!] = [584 / 120] ≈ 4.866, floor is 46. [584 / 6!] = [584 / 720] ≈ 0.811, floor is 07. [584 / 7!] onwards are all less than 1, so floor is 0.Adding them up: 584 + 292 + 97 + 24 + 4 + 0 + ... + 0 = 584 + 292 is 876, 876 +97 is 973, 973 +24 is 997, 997 +4 is 1001. Yep, that's exactly the target sum. So x=584 seems to work. But let me check if x=583 gives 1000.x=583:1. [583] = 5832. [583/2] = 291 (since 583/2=291.5)3. [583/6] ≈ 97.166, floor 974. [583/24] ≈24.291, floor 245. [583/120] ≈4.858, floor 46. [583/720] ≈0.809, floor 0Sum: 583 + 291 +97 +24 +4 = 583+291=874, 874+97=971, 971+24=995, 995+4=999. Wait, that's 999. So x=583 gives 999. Then x=584 gives 1001. So there is a jump from 999 to 1001 when x increases from 583 to 584. Wait, where is the x=584 gives 1001, but between 583 and 584, is there a missing number? Let's check x=584:As above, sum is 1001. So x=584 is the first integer where the sum reaches 1001, and x=583 is 999, which is 2 less. So that seems like there's a jump of 2. How does that happen? Let me check if I miscalculated.Wait, when x increases from 583 to 584, which terms increase?Term 1: floor(x/1!) increases by 1 (from 583 to 584).Term 2: floor(x/2) when x is even increases by 1. But 584 is even, so 584/2=292, whereas 583/2=291.5, floor to 291. So term 2 increases by 1.Term 3: floor(x/6). 583/6≈97.166, floor 97. 584/6≈97.333, still floor 97. So term 3 doesn't increase.Term 4: floor(x/24). 583/24≈24.291, floor 24. 584/24≈24.333, floor 24. So term 4 doesn't increase.Term 5: floor(x/120). 583/120≈4.858, floor 4. 584/120≈4.866, floor 4. No increase.Term 6: floor(x/720). 584/720≈0.811, floor 0. No increase.So the total increase from x=583 to x=584 is term1 + term2: 1 +1=2. Hence the sum goes from 999 to 1001, so there's no x where the sum is 1000. Therefore, the equation skips 1000 and jumps from 999 to 1001. Therefore, the only integer solution is x=584.But wait, let me check x=584 again. Wait, term 1 is 584, term2 is 292, term3 is 97, term4 is 24, term5 is 4, rest are 0. So sum is 584 + 292 + 97 + 24 + 4 = 584+292=876, 876+97=973, 973+24=997, 997+4=1001. Correct.But let me check x=585 again. Earlier, I thought x=585 gives 1002, which is correct. So x=584 gives 1001, x=585 gives 1002, and between 583 and 584, we have the jump. So is x=584 the unique solution? Wait, the problem says "integer solution", so maybe 584 is the only integer x that satisfies the equation. Because when x increases beyond 584, the sum increases again. So 584 is the only integer that gives 1001. Therefore, the answer is 584.But just to be thorough, let me check another nearby number. Let's take x=584.5, but since x must be integer, we don't need to consider fractions. So yes, x must be integer. Therefore, x=584 is the solution.Wait, but the problem says "integer solution", so x must be integer, and we found that x=584 works, and the next integer x=585 oversteps the sum. Therefore, the solution is x=584. Let me check if there's another x in higher ranges. For example, if x is greater than 720, does that contribute?Suppose x=721. Then floor(x/720)=1. So floor(721/720)=1. Let's compute the sum.x=721:floor(721) = 721floor(721/2)=360 (since 721/2=360.5)floor(721/6)=120 (721/6≈120.166)floor(721/24)=30 (721/24≈30.041)floor(721/120)=6 (721/120≈6.008)floor(721/720)=1floor(721/5040)=0... and higher terms are 0.So sum is 721 + 360 + 120 + 30 + 6 + 1 = 721+360=1081, 1081+120=1201, 1201+30=1231, 1231+6=1237, 1237+1=1238. Which is way over 1001. So x=721 is too big. So clearly, x must be less than 720.Wait, but x=584 is less than 720, so our previous calculation is correct. Hence, the only integer solution is 584. Therefore, the answer is boxed{584}.Final Answerboxed{584}</think>To find the integer solution of the equation [left[frac{x}{1!}right] + left[frac{x}{2!}right] + left[frac{x}{3!}right] + cdots + left[frac{x}{10!}right] = 1001,]we start by noting that the terms for ( n geq 7 ) (i.e., ( 7! ) to ( 10! )) will be very small and likely contribute zero to the sum. Therefore, we focus on the terms from ( n = 1 ) to ( n = 6 ).We need to solve:[leftlfloor frac{x}{1!} rightrfloor + leftlfloor frac{x}{2!} rightrfloor + leftlfloor frac{x}{3!} rightrfloor + leftlfloor frac{x}{4!} rightrfloor + leftlfloor frac{x}{5!} rightrfloor + leftlfloor frac{x}{6!} rightrfloor = 1001.]Testing various values of ( x ):- For ( x = 584 ): - ( leftlfloor frac{584}{1!} rightrfloor = 584 ) - ( leftlfloor frac{584}{2!} rightrfloor = 292 ) - ( leftlfloor frac{584}{6} rightrfloor = 97 ) - ( leftlfloor frac{584}{24} rightrfloor = 24 ) - ( leftlfloor frac{584}{120} rightrfloor = 4 ) - ( leftlfloor frac{584}{720} rightrfloor = 0 )Summing these values:[584 + 292 + 97 + 24 + 4 + 0 = 1001]Checking neighboring values confirms that ( x = 584 ) is the correct solution. Values less than 584 result in sums less than 1001, and values greater than 584 result in sums exceeding 1001.Thus, the integer solution is (boxed{584}).,<think>Okay, so I need to find an integer x such that the sum of the floor functions [x/1!] + [x/2!] + [x/3!] + ... + [x/10!] equals 1001. Hmm, floor functions can be tricky because they involve taking the integer part of a division. Let me start by understanding the problem step by step.First, let me recall that the floor function [a] gives the greatest integer less than or equal to a. For example, [3.7] = 3 and [5] = 5. So, each term in the sum is essentially how many times the factorial divides into x, discarding the remainder. The equation given is:[leftlfloor frac{x}{1!} rightrfloor + leftlfloor frac{x}{2!} rightrfloor + leftlfloor frac{x}{3!} rightrfloor + cdots + leftlfloor frac{x}{10!} rightrfloor = 1001]Since x is an integer, some of these floor terms might simplify. Let's note that for k ≥ x, the term [x/k!] would be 0 because x is less than k! if k! is larger than x. However, since we're going up to 10!, which is 3,628,800, and x is an integer that we need to find such that the sum is 1001, it's possible that x is a relatively small number compared to 10! So, perhaps the higher factorial terms (like from 7! onwards) might contribute 0 or small integers to the sum.Let me first check approximately how large x might be. If I ignore the floor functions for a moment and consider the sum:[frac{x}{1!} + frac{x}{2!} + frac{x}{3!} + cdots + frac{x}{10!}]This is a geometric-like series. Let me compute the sum of 1/n! from n=1 to 10:Sum = 1/1! + 1/2! + 1/3! + ... + 1/10!I remember that the sum of 1/n! from n=0 to infinity is e ≈ 2.71828. So subtracting 1/0! = 1, the sum from n=1 to infinity is e - 1 ≈ 1.71828. However, we are only summing up to 10!, so the sum will be slightly less than e - 1. Let me calculate it:1/1! = 11/2! = 0.51/3! ≈ 0.16666671/4! ≈ 0.04166671/5! ≈ 0.00833331/6! ≈ 0.00138891/7! ≈ 0.00019841/8! ≈ 0.00002481/9! ≈ 0.0000027551/10! ≈ 0.000000275Adding these up:1 + 0.5 = 1.5+0.1666667 ≈ 1.6666667+0.0416667 ≈ 1.7083334+0.0083333 ≈ 1.7166667+0.0013889 ≈ 1.7180556+0.0001984 ≈ 1.718254+0.0000248 ≈ 1.7182788+0.000002755 ≈ 1.718281555+0.000000275 ≈ 1.71828183So the total sum is approximately 1.71828183. Therefore, if I approximate the original sum (with floor functions) by x times this sum, I can get an estimate of x. So:x ≈ 1001 / 1.71828183 ≈ Let's compute that.First, 1.71828183 is approximately e - 1, so 1001 / (e - 1) ≈ 1001 / 1.718281828 ≈ Let's divide 1001 by 1.718281828.Well, 1.718281828 * 580 ≈ 1.71828 * 580 ≈ 580*1 + 580*0.7 + 580*0.01828 ≈ 580 + 406 + 10.6 ≈ 996.6So 1.71828*580 ≈ 996.6, which is less than 1001. The difference is 1001 - 996.6 = 4.4. So to get the remaining 4.4, we divide by 1.71828 ≈ 4.4 / 1.718 ≈ 2.56. So approximately, x ≈ 580 + 2.56 ≈ 582.56. So, roughly, x is around 583. But since we are dealing with floor functions, which reduce the sum, maybe the actual x is a bit higher? Hmm, but this is a rough estimate. Let's note that x is likely around 580-590.But let me check for x=583:Compute the sum [583/1!] + [583/2!] + ... + [583/10!]Compute each term:1! term: 583 /1 = 583, floor is 5832! term: 583 /2 = 291.5, floor is 2913! term: 583 /6 ≈ 97.166..., floor is 974! term: 583 /24 ≈ 24.291..., floor is 245! term: 583 /120 ≈ 4.858..., floor is 46! term: 583 /720 ≈ 0.809..., floor is 07! term: 583 /5040 ≈ 0.115..., floor is 0Similarly, 8! to 10! terms will also be 0.So sum is 583 + 291 + 97 + 24 + 4 + 0 + 0 + 0 + 0 + 0 = Let's add these up:583 + 291 = 874874 + 97 = 971971 + 24 = 995995 + 4 = 999So total sum is 999. But we need 1001. So x=583 gives a sum of 999, which is 2 short. Let's try x=584.Compute each term:1! term: 5842! term: 584 /2 = 292, floor 2923! term: 584 /6 ≈ 97.333..., floor 974! term: 584 /24 ≈ 24.333..., floor 245! term: 584 /120 ≈ 4.866..., floor 46! term: 584 /720 ≈ 0.811..., floor 0Others 0.Sum: 584 + 292 + 97 + 24 + 4 = 584 + 292 = 876; 876 +97=973; 973 +24=997; 997 +4=1001Wait, that's exactly 1001! So x=584 gives the sum 1001. Wait, but let me verify each term carefully:1! term: 584/1 = 584, floor 5842! term: 584/2 = 292, floor 2923! term: 584 divided by 6: 6*97=582, so 584-582=2, so 97 + 2/6, floor 974! term: 584/24. 24*24=576, 584-576=8, so 24 + 8/24, floor 245! term: 584/120. 120*4=480, 584-480=104, so 4 + 104/120, floor 46! term: 584/720. 720 >584, so floor 0Similarly, higher terms are 0.Sum: 584 +292=876, +97=973, +24=997, +4=1001, rest are 0. So total sum is 1001. So x=584 is the solution? Wait, but wait, that seems too straightforward. However, let me check if there's another x that could also satisfy this. For example, maybe x=585? Let's check.x=585:1! term: 5852! term: 585/2=292.5, floor 2923! term: 585/6=97.5, floor 974! term: 585/24≈24.375, floor 245! term: 585/120=4.875, floor 46! term: 585/720≈0.8125, floor 0Sum: 585 +292=877, +97=974, +24=998, +4=1002, which is over by 1. So x=585 gives 1002, which is over. Therefore, x=584 is the only integer solution. Therefore, the answer is 584.But wait, let me check once again, because sometimes when you cross a multiple in the floor function, the sum can decrease or increase. Let me confirm.Wait, for x=584, all terms except maybe the 5! term? Wait, 584/120 is 4.866..., which is still floored to 4. So no, 5! term is 4.Wait, but if x=584. Let me recalculate all terms:1! term: 584/1 = 5842! term: 584/2 = 2923! term: 584/6 = 97.333... → 974! term: 584/24 = 24.333... →245! term:584/120=4.866... →46! term:584/720≈0.811→07! to 10! terms: 0Sum:584+292=876; 876+97=973; 973+24=997; 997+4=1001. Perfect. So x=584 is correct.But let me check with x=583 again, just to be thorough.x=583:1! term:5832! term:291.5→2913! term:583/6≈97.166→974! term:583/24≈24.291→245! term:583/120≈4.858→4Sum:583+291=874; 874+97=971; 971+24=995; 995+4=999. Correct, as before. So 999, which is 2 less than 1001. Then x=584 gives 1001. So x=584 is the solution.Wait, but the problem states "integer solution". So since x=584 is an integer, and gives exactly 1001, that must be the answer. Let me just check if there's a possibility of a lower x that also gives 1001. For example, maybe between 583 and 584, but x must be integer, so no. If x=584 is the first integer that achieves the sum 1001, then that's the only solution. Since x=585 gives 1002, which is over, so there's no other integer. Therefore, the solution is x=584.But let me check with x=584.5 or something, but x must be integer. So yes, x=584 is the only integer. Therefore, the answer is 584.Final AnswerThe integer solution is boxed{584}.</think>To find the integer solution of the equation [leftlfloor frac{x}{1!} rightrfloor + leftlfloor frac{x}{2!} rightrfloor + leftlfloor frac{x}{3!} rightrfloor + cdots + leftlfloor frac{x}{10!} rightrfloor = 1001,]we start by estimating the value of ( x ). The sum of the series ( frac{x}{1!} + frac{x}{2!} + frac{x}{3!} + cdots + frac{x}{10!} ) approximates ( x times (e - 1) approx 1.71828x ). Solving ( 1.71828x approx 1001 ) gives ( x approx 582.56 ). Testing integer values around this estimate:- For ( x = 583 ): [ leftlfloor frac{583}{1!} rightrfloor + leftlfloor frac{583}{2!} rightrfloor + leftlfloor frac{583}{3!} rightrfloor + leftlfloor frac{583}{4!} rightrfloor + leftlfloor frac{583}{5!} rightrfloor = 583 + 291 + 97 + 24 + 4 = 999. ]- For ( x = 584 ): [ leftlfloor frac{584}{1!} rightrfloor + leftlfloor frac{584}{2!} rightrfloor + leftlfloor frac{584}{3!} rightrfloor + leftlfloor frac{584}{4!} rightrfloor + leftlfloor frac{584}{5!} rightrfloor = 584 + 292 + 97 + 24 + 4 = 1001. ]- For ( x = 585 ): [ leftlfloor frac{585}{1!} rightrfloor + leftlfloor frac{585}{2!} rightrfloor + leftlfloor frac{585}{3!} rightrfloor + leftlfloor frac{585}{4!} rightrfloor + leftlfloor frac{585}{5!} rightrfloor = 585 + 292 + 97 + 24 + 4 = 1002. ]Thus, ( x = 584 ) is the integer solution as it results in the sum 1001. Higher terms (from ( 6! ) to ( 10! )) contribute 0 for these values of ( x ).The integer solution is (boxed{584}).]

🔑:Solution: because the solution x of the equation is a natural number, and xx+frac{x}{2}+frac{x}{6}+frac{x}{24}+frac{x}{120}-4 =frac{103}{60} x-4, therefore 583<x<585, x=584,end{array}Verification shows that x=584 is the unique solution of the equation.

🧠:[<think>Okay, so I need to find the value of m such that the minimum value of F(x) = f(g(x)) - g(f(x)) is 1/4. The functions given are f(x) = x² and g(x) = 2x + 2m. Hmm, let me start by understanding what F(x) actually is. First, let's compute f(g(x)). Since f is a square function, f(g(x)) would be [g(x)]². Given that g(x) = 2x + 2m, squaring that gives (2x + 2m)². Let me expand that: (2x)² + 2*(2x)*(2m) + (2m)² = 4x² + 8mx + 4m². Wait, is that right? Let me check. (a + b)² = a² + 2ab + b². So (2x + 2m)² = (2x)² + 2*(2x)*(2m) + (2m)² = 4x² + 8mx + 4m². Yes, that's correct.Next, compute g(f(x)). Since g is a linear function, 2x + 2m, applying it to f(x) would be 2*f(x) + 2m. Since f(x) is x², this becomes 2x² + 2m.Therefore, F(x) = f(g(x)) - g(f(x)) = (4x² + 8mx + 4m²) - (2x² + 2m). Let's simplify this:4x² - 2x² = 2x²8mx remains as is4m² - 2m remains as isSo F(x) = 2x² + 8mx + (4m² - 2m)Now, this is a quadratic function in terms of x. Since the coefficient of x² is 2, which is positive, the parabola opens upwards, meaning the minimum value occurs at the vertex.To find the vertex of a quadratic ax² + bx + c, the x-coordinate is at -b/(2a). Let's compute that:Here, a = 2, b = 8m. Therefore, x = -8m / (2*2) = -8m/4 = -2m. So the x-coordinate of the vertex is -2m.To find the minimum value of F(x), substitute x = -2m back into F(x):F(-2m) = 2*(-2m)² + 8m*(-2m) + (4m² - 2m)Let's compute each term:2*(-2m)² = 2*(4m²) = 8m²8m*(-2m) = -16m²4m² - 2m remains as isAdding these together: 8m² -16m² + 4m² - 2m = (8m² -16m² +4m²) + (-2m) = (-4m²) -2mSo the minimum value of F(x) is -4m² -2m, and according to the problem, this minimum is equal to 1/4. Therefore:-4m² -2m = 1/4Let's rearrange the equation:-4m² -2m -1/4 = 0Multiply both sides by -4 to eliminate the fractions:16m² + 8m + 1 = 0Now, we have a quadratic equation in terms of m: 16m² +8m +1 =0Let me solve for m using the quadratic formula. The quadratic is 16m² +8m +1=0, so a=16, b=8, c=1.Discriminant D = b² -4ac = 64 - 64 = 0. Wait, discriminant is zero? So there is exactly one real root.m = [-b ± sqrt(D)]/(2a) = [-8 ± 0]/32 = -8/32 = -1/4.So m = -1/4. Let me check this.Let me verify the calculation steps again to be sure.First, F(x) = f(g(x)) - g(f(x)).Computed f(g(x)) = (2x + 2m)² = 4x² +8mx +4m².Computed g(f(x)) = 2x² +2m.Therefore, F(x) = 4x² +8mx +4m² -2x² -2m = 2x² +8mx +4m² -2m. Correct.So F(x) is quadratic with a=2, b=8m, c=4m² -2m.Vertex at x = -b/(2a) = -8m/(4) = -2m. Correct.F(-2m) = 2*(-2m)^2 +8m*(-2m) +4m² -2m = 2*4m² -16m² +4m² -2m = 8m² -16m² +4m² -2m = (-4m²) -2m. Correct.Set -4m² -2m = 1/4.Multiply both sides by 4: -16m² -8m =1.Bring all terms to left: -16m² -8m -1=0.Multiply both sides by -1:16m² +8m +1=0.Discriminant D=64 -64=0. So m=(-8)/(2*16)= -8/32=-1/4. Correct.Thus, m= -1/4.Let me just check if substituting m=-1/4 gives the minimum value 1/4.Compute F(x) with m=-1/4:F(x)=2x² +8*(-1/4)x +4*(-1/4)^2 -2*(-1/4)=2x² -2x +4*(1/16) +1/2=2x² -2x +1/4 +1/2=2x² -2x +3/4.Now, the vertex is at x=-2m = -2*(-1/4)=1/2.Compute F(1/2):2*(1/2)^2 -2*(1/2) +3/4 = 2*(1/4) -1 +3/4 = 1/2 -1 +3/4 = (1/2 +3/4) -1 = (5/4) -1 =1/4. Yes, correct.Therefore, m=-1/4 is the correct answer.Final AnswerThe value of ( m ) is boxed{-dfrac{1}{4}}.</think>Given the functions ( f(x) = x^2 ) and ( g(x) = 2x + 2m ), we need to find the value of ( m ) such that the minimum value of ( F(x) = f(g(x)) - g(f(x)) ) is ( frac{1}{4} ).First, we compute ( f(g(x)) ) and ( g(f(x)) ):- ( f(g(x)) = (2x + 2m)^2 = 4x^2 + 8mx + 4m^2 )- ( g(f(x)) = 2x^2 + 2m )Thus, the function ( F(x) ) is:[F(x) = f(g(x)) - g(f(x)) = (4x^2 + 8mx + 4m^2) - (2x^2 + 2m) = 2x^2 + 8mx + 4m^2 - 2m]This is a quadratic function in ( x ) with the form ( F(x) = 2x^2 + 8mx + (4m^2 - 2m) ). The minimum value of a quadratic function ( ax^2 + bx + c ) occurs at ( x = -frac{b}{2a} ). Here, ( a = 2 ) and ( b = 8m ), so the x-coordinate of the vertex is:[x = -frac{8m}{2 cdot 2} = -2m]Substituting ( x = -2m ) back into ( F(x) ):[F(-2m) = 2(-2m)^2 + 8m(-2m) + 4m^2 - 2m = 8m^2 - 16m^2 + 4m^2 - 2m = -4m^2 - 2m]We are given that the minimum value is ( frac{1}{4} ), so:[-4m^2 - 2m = frac{1}{4}]Multiplying through by 4 to clear the fraction:[-16m^2 - 8m = 1 implies 16m^2 + 8m + 1 = 0]Solving this quadratic equation using the quadratic formula ( m = frac{-b pm sqrt{b^2 - 4ac}}{2a} ) where ( a = 16 ), ( b = 8 ), and ( c = 1 ):[m = frac{-8 pm sqrt{64 - 64}}{32} = frac{-8}{32} = -frac{1}{4}]Thus, the value of ( m ) is boxed{-dfrac{1}{4}}.,<think>Okay, let me try to figure this problem out. So, we have two functions, f(x) = x² and g(x) = 2x + 2m. Then there's this composite function F(x) = f(g(x)) - g(f(x)), and we need to find the value of m such that the minimum value of F(x) is 1/4. Hmm, okay. Let's break this down step by step.First, let me make sure I understand what F(x) is. It's the composition of f and g, subtracted by the composition of g and f. So, f(g(x)) means plugging g(x) into f, which would be (g(x))². Similarly, g(f(x)) would be plugging f(x) into g, which is 2*f(x) + 2m. So, let's write that out.f(g(x)) = [g(x)]² = (2x + 2m)². Let me expand that. That's (2x)² + 2*(2x)*(2m) + (2m)² = 4x² + 8mx + 4m².Wait, hold on, actually, expanding (2x + 2m)²: first term squared is (2x)² = 4x², second term is 2*(2x)*(2m) = 8mx, third term is (2m)² = 4m². Yeah, that's right.Then, g(f(x)) = 2*f(x) + 2m = 2x² + 2m.So, F(x) = f(g(x)) - g(f(x)) = [4x² + 8mx + 4m²] - [2x² + 2m] = 4x² + 8mx + 4m² - 2x² - 2m. Combine like terms.4x² - 2x² is 2x². Then 8mx remains. Then 4m² - 2m remains. So, F(x) = 2x² + 8mx + (4m² - 2m). Okay, so that simplifies to a quadratic function in terms of x.Since it's a quadratic function, its graph is a parabola. The coefficient of x² is 2, which is positive, so the parabola opens upwards. Therefore, the minimum value occurs at the vertex. The question states that this minimum value is 1/4, so we need to find m such that the vertex of this parabola has a y-coordinate of 1/4.To find the vertex of a quadratic function ax² + bx + c, the x-coordinate is -b/(2a), and then plug that back into the function to find the y-coordinate. So here, a = 2, b = 8m. Therefore, the x-coordinate of the vertex is -8m/(2*2) = -8m/4 = -2m. Then, plugging x = -2m into F(x) gives the minimum value.Let me compute F(-2m):F(-2m) = 2*(-2m)² + 8m*(-2m) + (4m² - 2m)First term: 2*(4m²) = 8m²Second term: 8m*(-2m) = -16m²Third term: 4m² - 2mSo adding them up: 8m² - 16m² + 4m² - 2m = (8m² -16m² +4m²) + (-2m) = (-4m²) -2mTherefore, the minimum value of F(x) is -4m² -2m. But the problem states that this minimum is 1/4. Wait, but that would mean -4m² -2m = 1/4. But that equation seems problematic because the left side is a quadratic in m, and we can set up the equation and solve for m.So, let's write:-4m² - 2m = 1/4Multiply both sides by 4 to eliminate the fraction:-16m² -8m = 1Bring all terms to one side:-16m² -8m -1 = 0Multiply both sides by -1 to make the coefficient of m² positive:16m² +8m +1 = 0Now, solve for m using the quadratic formula. The quadratic is 16m² +8m +1=0.Discriminant D = b² -4ac = (8)^2 -4*16*1 = 64 -64 = 0Oh, discriminant is zero, so there's one real solution.m = [-b ± sqrt(D)]/(2a) = [-8 ± 0]/32 = -8/32 = -1/4So, m = -1/4. But wait, let me check if this is correct.Wait, let's verify this step-by-step. So, F(x) = 2x² +8mx +4m² -2m. Then, the vertex is at x = -2m, and substituting back gives F(-2m) = 2*(4m²) +8m*(-2m) +4m² -2m = 8m² -16m² +4m² -2m = (8 -16 +4)m² -2m = (-4m²) -2m. So, that's correct. Then setting that equal to 1/4 gives -4m² -2m = 1/4.Multiply both sides by 4: -16m² -8m = 1, so 16m² +8m +1 =0. Which factors as (4m +1)^2 = 0. Because 16m² +8m +1 = (4m)^2 + 2*(4m)*(1) +1^2 = (4m +1)^2. So, yes, (4m +1)^2 =0 => 4m +1=0 => m= -1/4. So, m is -1/4. But let's check if this makes sense.Wait, let's plug m = -1/4 back into the expression for F(x) and see if the minimum is indeed 1/4.First, compute F(x):F(x) = 2x² +8m x +4m² -2mSubstitute m = -1/4:First, 8m = 8*(-1/4) = -24m² = 4*(1/16) = 1/4-2m = -2*(-1/4) = 1/2So, F(x) = 2x² -2x +1/4 +1/2 = 2x² -2x + 3/4Now, the vertex of this quadratic is at x = -b/(2a) = -(-2)/(2*2) = 2/4 = 1/2Then, compute F(1/2):2*(1/2)^2 -2*(1/2) + 3/4 = 2*(1/4) -1 + 3/4 = 1/2 -1 + 3/4 = (1/2 + 3/4) -1 = (5/4) -1 = 1/4Yes, that works. So, the minimum value is indeed 1/4 when m = -1/4. So, that seems correct.But wait, the problem states that the minimum value is 1/4, so m is -1/4. Therefore, the answer should be m = -1/4.But wait, let me check the original problem again. The functions are f(x) = x² and g(x) = 2x + 2m. Then F(x) = f(g(x)) - g(f(x)).We computed f(g(x)) as (2x + 2m)^2 and g(f(x)) as 2x² + 2m. Then F(x) = (2x + 2m)^2 - (2x² + 2m). Expanding that:(4x² +8mx +4m²) - 2x² -2m = 2x² +8mx +4m² -2m. Then, vertex at x = -2m, leading to F(-2m) = -4m² -2m. Then setting that equal to 1/4 gives -4m² -2m = 1/4, leading to 16m² +8m +1=0, which factors to (4m+1)^2=0, so m = -1/4. Yes, seems correct. Then when m = -1/4, the minimum is 1/4. So, answer is m = -1/4.But let me just check once more, maybe I made a mistake in expanding (2x + 2m)^2. Let's confirm:(2x + 2m)^2 = (2x)^2 + 2*(2x)*(2m) + (2m)^2 = 4x² +8mx +4m². Correct.Then g(f(x)) = 2*f(x) +2m = 2x² +2m. Correct.So, F(x) = 4x² +8mx +4m² -2x² -2m = 2x² +8mx +4m² -2m. Correct.Then, vertex at x = -b/(2a) = -8m/(2*2) = -2m. Correct.Then, F(-2m) = 2*(-2m)^2 +8m*(-2m) +4m² -2m = 2*4m² -16m² +4m² -2m = 8m² -16m² +4m² -2m = (-4m²) -2m. Correct.Set that equal to 1/4: -4m² -2m = 1/4. Multiply by 4: -16m² -8m =1. Then 16m² +8m +1=0. Which factors as (4m +1)^2=0. Correct. So, m = -1/4. So, all steps check out. Therefore, m is -1/4.Therefore, the answer is m = -1/4, which in boxed form is boxed{-dfrac{1}{4}}.Final AnswerThe value of ( m ) is boxed{-dfrac{1}{4}}.</think>Given the functions ( f(x) = x^2 ) and ( g(x) = 2x + 2m ), we need to find the value of ( m ) such that the minimum value of ( F(x) = f(g(x)) - g(f(x)) ) is ( frac{1}{4} ).First, we compute the compositions:- ( f(g(x)) = (2x + 2m)^2 = 4x^2 + 8mx + 4m^2 )- ( g(f(x)) = 2x^2 + 2m )Thus, the function ( F(x) ) is:[F(x) = f(g(x)) - g(f(x)) = (4x^2 + 8mx + 4m^2) - (2x^2 + 2m) = 2x^2 + 8mx + 4m^2 - 2m]This is a quadratic function in ( x ) with the form ( F(x) = 2x^2 + 8mx + (4m^2 - 2m) ). The minimum value of a quadratic function ( ax^2 + bx + c ) occurs at ( x = -frac{b}{2a} ). Here, ( a = 2 ) and ( b = 8m ), so the x-coordinate of the vertex is:[x = -frac{8m}{2 cdot 2} = -2m]Substituting ( x = -2m ) into ( F(x) ):[F(-2m) = 2(-2m)^2 + 8m(-2m) + 4m^2 - 2m = 8m^2 - 16m^2 + 4m^2 - 2m = -4m^2 - 2m]We are given that the minimum value is ( frac{1}{4} ), so we set up the equation:[-4m^2 - 2m = frac{1}{4}]Multiplying through by 4 to clear the fraction:[-16m^2 - 8m = 1 implies 16m^2 + 8m + 1 = 0]Solving this quadratic equation using the quadratic formula ( m = frac{-b pm sqrt{b^2 - 4ac}}{2a} ):[m = frac{-8 pm sqrt{64 - 64}}{32} = frac{-8}{32} = -frac{1}{4}]Thus, the value of ( m ) is (boxed{-dfrac{1}{4}}).]

🔑:3. -frac{1}{4}.Notice,begin{array}{l}F(x)=f(g(x))-g(f(x)) =(2 x+2 m)^{2}-left(2 x^{2}+2 mright) =2 x^{2}+8 m x+4 m^{2}-2 m =2(x+2 m)^{2}-4 m^{2}-2 m .end{array}Then -4 m^{2}-2 m=frac{1}{4} Rightarrow m=-frac{1}{4}.

🧠:[<think>Okay, so I need to find the negation of the proposition p: "There exists an x in real numbers such that cos(x) > 1." Let me think about how negations work with quantifiers. First, the original statement is an existential quantifier: "There exists an x in ℝ such that cos(x) > 1." The general rule for negating such statements is that the negation of "There exists x such that P(x)" is "For all x, not P(x)." So, applying that here, the negation should be "For all x in ℝ, cos(x) is not greater than 1." Translating "not greater than 1" into mathematical terms, that would be "cos(x) ≤ 1" because if it's not greater than 1, it must be less than or equal to 1. So putting it all together, the negation should be "For all x in ℝ, cos(x) ≤ 1." Looking at the options provided:A: Says there exists an x where cos(x) ≤ 1. But that's not the negation. The original was about existence of some x where cos(x) > 1. The negation isn't just asserting that there's some x where it's not; it's saying that for all x, it's not. So A is incorrect.B: For all x, cos(x) ≤ 1. That matches what I just thought. But let me check the other options to be sure.C: Exists x with cos(x) < 1. Similar issue as A. The negation requires a universal quantifier, not existential, and the inequality is not strict. Also, even if it were universal, <1 is different from ≤1. So C is not correct.D: For all x, cos(x) < 1. This is a universal statement but uses the strict inequality. However, we know that cos(0) = 1, so cos(x) can equal 1. Therefore, D is false because it excludes cases where cos(x) is exactly 1. Hence, the correct negation must allow for equality.Therefore, B is the correct answer. Let me just double-check: the original statement claims there's some real number x with cosine greater than 1. The negation should state that all real numbers x have cosine less than or equal to 1. Since cosine's maximum value is 1, this makes sense. Yep, B must be the right choice.</think>The original proposition ( p ) is an existential statement: "There exists an ( x in mathbb{R} ) such that ( cos(x) > 1 )." To negate this, we use the rule that the negation of "There exists ( x ) such that ( P(x) )" is "For all ( x ), not ( P(x) )." Applying this, the negation becomes: "For all ( x in mathbb{R} ), ( cos(x) leq 1 )." This eliminates options A and C, which incorrectly use existential quantifiers. Option D is incorrect because it uses a strict inequality ( < 1 ), which would exclude ( cos(x) = 1 ) (e.g., ( x = 0 )). The correct negation must allow ( cos(x) ) to equal 1, making option B the answer.(boxed{B}),<think>Okay, let's see. The problem is about finding the negation of the proposition p: "There exists an x in real numbers such that cos(x) > 1." And there are four options to choose from. Alright, so first, I need to remember how to negate statements with quantifiers. So, the original statement is an existential quantifier: "There exists an x in ℝ such that cos(x) > 1." The negation of an existential statement is a universal statement, right? Because if it's not true that there's at least one thing with a certain property, then it must be true that all things do not have that property. So, the negation should be "For all x in ℝ, cos(x) is not greater than 1." Which translates symbolically to ∀x ∈ ℝ, cos(x) ≤ 1. Let me check the options. Option A says: "There exists an x in ℝ such that cos(x) ≤ 1." Wait, that doesn't seem right. If the original was about existence of an x where cos(x) >1, then the negation shouldn't just switch the inequality but keep the existence quantifier. Because if there exists an x where cos(x) ≤1, that doesn't contradict the original statement. The original could still be true even if some x's have cos(x) ≤1. So Option A is probably not correct.Option B: "For all x in ℝ, cos(x) ≤1." This matches what I thought earlier. Since the negation of ∃x P(x) is ∀x ¬P(x). So changing "there exists" to "for all" and negating the inner condition. So here, P(x) is "cos(x) >1", so ¬P(x) is "cos(x) ≤1". So yes, Option B seems correct.Option C: "There exists an x in ℝ such that cos(x) <1." Again, similar to Option A, but with a strict inequality. But same reasoning as before. Even if some x's have cos(x) <1, there might still be an x with cos(x) >1. So the existence of such an x doesn't negate the original statement. So Option C is also not the correct negation.Option D: "For all x in ℝ, cos(x) <1." Hmm. This is a stronger statement than B. It claims that cosine of every real number is strictly less than 1. But we know that cosine of 0, for example, is exactly 1. So this statement is actually false. But is it the negation of the original? Well, the original was about existence of x where cos(x) >1. The negation should be that all x have cos(x) ≤1. So Option D is too strong because it's saying <1 instead of ≤1. Therefore, Option D is incorrect.So, putting it all together, the correct negation should be Option B: For all x in ℝ, cos(x) ≤1. Let me just verify again. The original proposition p is ∃x∈ℝ, cos(x) >1. The negation ¬p is ∀x∈ℝ, ¬(cos(x) >1), which is equivalent to ∀x∈ℝ, cos(x) ≤1. Yep, that's exactly Option B. Also, considering the truth values. The original proposition p is false because the cosine function never exceeds 1. Therefore, the negation ¬p should be true. And indeed, Option B is a true statement since for all real numbers x, cos(x) is always less than or equal to 1. So that makes sense. Just to make sure, let's check if any other options could be correct. For example, if someone thought the negation was changing the existence to "for all" but messed up the inequality. But in this case, Options A and C are existence claims, which we already determined are not negations. Option D is a universal claim but with the wrong inequality. So yes, Option B is definitely the right answer.Final Answerboxed{B}</think>The original proposition ( p ) is: "There exists an ( x in mathbb{R} ) such that ( cos(x) > 1 )." To find the negation of this proposition, we need to negate both the existential quantifier and the inner condition. The negation of an existential quantifier ((exists)) is a universal quantifier ((forall)), and the negation of the condition ( cos(x) > 1 ) is ( cos(x) leq 1 ). Therefore, the negation of the proposition ( p ) is: "For all ( x in mathbb{R} ), ( cos(x) leq 1 )."Checking the options:- Option A and C are incorrect because they use an existential quantifier instead of a universal quantifier.- Option D is incorrect because it uses a strict inequality ( < 1 ) instead of ( leq 1 ).- Option B is correct because it uses the universal quantifier and the correct inequality ( leq 1 ).Thus, the correct negation is given by Option B.[boxed{B}]]

🔑:Since the negation of an existential proposition is a universal proposition, the negation of proposition p: There exists an x in mathbb{R}, such that cos(x) > 1 is "For all x in mathbb{R}, cos(x) leq 1".Therefore, the correct choice is boxed{text{B}}.