💡Okay, so I came across this interesting geometry problem: "Does there exist a hexahedron (six-faced polyhedron) other than a cube, in which all faces are congruent rhombuses?" At first glance, it seems pretty straightforward, but I want to make sure I understand all the nuances before jumping to any conclusions.First, let me recall what a hexahedron is. A hexahedron is a polyhedron with six faces. The most common one is a cube, which is a regular hexahedron because all its faces are squares, which are special types of rhombuses. But the question is asking if there's another type of hexahedron, not a cube, where all six faces are congruent rhombuses. So, essentially, it's asking if there's a shape with six rhombus faces that isn't a cube.I know that a rhombus is a quadrilateral with all sides equal in length, but unlike a square, its angles aren't necessarily 90 degrees. So, if all faces are congruent rhombuses, each face must have the same side lengths and the same angles. That means the hexahedron would have to be made up of six identical rhombuses.I remember that in three-dimensional geometry, there are shapes called parallelepipeds, which are like skewed cubes or rectangles. A cube is a special case of a parallelepiped where all angles are 90 degrees. If I can think of a parallelepiped where all the faces are rhombuses, that might be the hexahedron we're looking for.So, let's think about a rhombic parallelepiped. If I have three vectors defining the edges of the parallelepiped, and each face is a rhombus, that means each pair of vectors must have the same length. Wait, no, actually, in a rhombic parallelepiped, the faces are rhombuses, but the edges don't necessarily have to be equal in length. Hmm, I need to clarify that.In a rhombic parallelepiped, all faces are rhombuses, but the edges can have different lengths. However, in our case, we need all faces to be congruent rhombuses. That means not only are all faces rhombuses, but they are identical in shape and size. So, each rhombus must have the same side lengths and the same angles.I think this might lead us to a specific type of parallelepiped where all edges are equal, but the angles between the edges are not 90 degrees. That would make all the faces rhombuses, and if the angles are consistent, they could all be congruent.Let me try to visualize this. Imagine taking a cube and then squeezing it along one of its space diagonals. This would distort the cube into a shape where all the faces remain rhombuses, but they are no longer squares. If I can adjust the angles appropriately, maybe all the rhombuses can be congruent.But wait, if I squeeze a cube along a space diagonal, the faces that were originally squares would become rhombuses, but are they congruent? Let me think. When you squeeze a cube along a space diagonal, the faces that were adjacent to the direction of squeezing would become rhombuses with angles different from the original 90 degrees. However, I'm not sure if all six faces would remain congruent after such a transformation.Maybe another approach is needed. Let's consider the properties of a hexahedron with all faces congruent rhombuses. Each face is a rhombus, so each face has four edges of equal length. Since all faces are congruent, all edges of the hexahedron must be of equal length. That means the hexahedron would have all edges equal, but the angles between the edges could vary.Wait, if all edges are equal, then it's a type of equilateral hexahedron. But is it necessarily a cube? No, because the angles between the edges can differ. For example, a regular tetrahedron has all edges equal, but it's a tetrahedron, not a hexahedron. So, maybe there's a way to construct a hexahedron with all edges equal and all faces congruent rhombuses.I think this might be related to the concept of a rhombohedron. A rhombohedron is a three-dimensional figure with six faces, each of which is a rhombus. It's essentially a parallelepiped with all edges equal. So, a rhombohedron is a hexahedron with all faces congruent rhombuses. That seems to fit the description.But is a rhombohedron the same as a cube? No, because in a rhombohedron, the angles between the edges can be different from 90 degrees. If all angles are 90 degrees, then it's a cube. But if the angles are different, it's a different shape. So, a rhombohedron is a hexahedron with all faces congruent rhombuses, and it's not necessarily a cube.Therefore, the answer should be yes, such a hexahedron exists, and it's called a rhombohedron. It's a three-dimensional figure with six congruent rhombic faces, and it's distinct from a cube because its angles aren't necessarily right angles.But let me double-check to make sure I'm not missing something. I should consider if all faces being congruent rhombuses necessarily implies that the hexahedron is a rhombohedron. I think so because a rhombohedron is defined by having six congruent rhombic faces. So, yes, it fits the criteria.Another way to think about it is by considering the properties of a rhombus. A rhombus has opposite angles equal, and the sum of any two adjacent angles is 180 degrees. In a rhombohedron, each face is a rhombus, and all faces are congruent, meaning all the rhombuses have the same side lengths and angles. This uniformity allows the shape to close up into a hexahedron without being a cube.I also recall that a rhombohedron can be formed by deforming a cube along one of its space diagonals, which changes the angles between the edges but keeps all the edges equal and all the faces as rhombuses. This transformation preserves the congruency of the faces because the scaling is uniform along the direction of the space diagonal.So, to summarize, a rhombohedron is a hexahedron with six congruent rhombic faces, and it's distinct from a cube because its angles aren't 90 degrees. Therefore, the answer to the question is yes, such a hexahedron exists.I think I've covered the main points. I considered the definition of a hexahedron, thought about rhombuses and their properties, recalled the concept of a rhombohedron, and verified that it fits the criteria of having six congruent rhombic faces without being a cube. I also considered the geometric transformation from a cube to a rhombohedron, which helps visualize why such a shape is possible.One thing I might have overlooked is whether there are other types of hexahedrons with congruent rhombic faces that aren't rhombohedrons. But from what I know, the rhombohedron is the primary example of a hexahedron with all faces congruent rhombuses. There might be other more complex polyhedrons, but the rhombohedron is the most straightforward and well-known example.Another consideration is whether the rhombohedron is convex or not. I believe it is convex because all its dihedral angles are less than 180 degrees, which is a property of convex polyhedrons. If it were concave, it might not fit the typical definition of a hexahedron, but in this case, it's convex.In conclusion, after thinking through the definitions, properties, and possible transformations, I'm confident that a rhombohedron is the hexahedron we're looking for. It's not a cube, but it has all faces as congruent rhombuses. Therefore, the answer is yes.

💡Alright, so I've got this problem here with two curves, C₁ and C₂. I need to find their general equations and then figure out the value of m based on the minimum distance condition. Let me try to break this down step by step.Starting with part (1): Finding the general equation of curve C₁. The parametric equations are given as x = t and y = m + t, where t is the parameter and m is a real number. Hmm, okay, so if x = t, then t is just x. So, substituting t into the equation for y, we get y = m + x. That seems straightforward. So, rearranging that, the general equation should be y - x = m, or x - y + m = 0. Yeah, that makes sense. So, that's the general equation for C₁.Now, moving on to curve C₂. Its polar equation is given as ρ² = 3 / (3 - 2cos²θ). I need to convert this into a Cartesian coordinate equation. I remember that in polar coordinates, ρ² = x² + y², and cosθ = x / ρ. So, let's try substituting these into the equation.Starting with ρ² = 3 / (3 - 2cos²θ). Let's rewrite the denominator:3 - 2cos²θ = 3 - 2(x² / ρ²). So, substituting back into the equation:ρ² = 3 / [3 - 2(x² / ρ²)]. Let's multiply both sides by the denominator to eliminate the fraction:ρ² * [3 - 2(x² / ρ²)] = 3Expanding the left side:3ρ² - 2x² = 3But since ρ² = x² + y², substitute that in:3(x² + y²) - 2x² = 3Simplify:3x² + 3y² - 2x² = 3Which simplifies to:x² + 3y² = 3Divide both sides by 3 to get it into standard form:x²/3 + y² = 1Okay, so that's the Cartesian equation for C₂. It looks like an ellipse centered at the origin with semi-major axis √3 along the x-axis and semi-minor axis 1 along the y-axis.Wait, but the original polar equation was given with ρ², so I should check if there are any restrictions on θ. The problem mentions taking the origin as the pole and the positive x-axis as the polar axis, so θ is in [0, π]. Hmm, does that affect the Cartesian equation? I think it just means that we're considering the upper half of the ellipse because θ ranges from 0 to π, which covers the upper half-plane. So, in Cartesian terms, y would be non-negative. So, the equation is x²/3 + y² = 1 with y ≥ 0. Got it.Alright, so part (1) seems done. Now, moving on to part (2). We need to find the value of m such that the minimum distance from any point P on C₂ to the curve C₁ is 2√2.First, let's recall that the distance from a point (x₀, y₀) to the line ax + by + c = 0 is given by |ax₀ + by₀ + c| / √(a² + b²). In our case, the line C₁ is x - y + m = 0, so a = 1, b = -1, c = m. So, the distance from a point P(x, y) on C₂ to C₁ is |x - y + m| / √(1 + 1) = |x - y + m| / √2.We need to find the minimum value of this distance as P moves along C₂, and set that minimum equal to 2√2. Then solve for m.But since C₂ is an ellipse, we can parametrize it. Let me use the parametric equations for an ellipse. For x²/3 + y² = 1, the parametric equations are x = √3 cosθ, y = sinθ, where θ is in [0, π] as per the polar equation's θ range.So, substituting these into the distance formula:Distance d = |√3 cosθ - sinθ + m| / √2We need to find the minimum of d as θ varies in [0, π], and set that minimum equal to 2√2.So, let's write d as:d(θ) = |√3 cosθ - sinθ + m| / √2To find the minimum of d(θ), we can consider the expression inside the absolute value:Let me denote f(θ) = √3 cosθ - sinθ + mSo, d(θ) = |f(θ)| / √2We need to find the minimum of |f(θ)| over θ in [0, π], and set it equal to 2√2 * √2 = 4? Wait, no. Wait, if d(θ) = |f(θ)| / √2, and we want the minimum d(θ) = 2√2, then |f(θ)| = 2√2 * √2 = 4. Wait, hold on.Wait, no. If d(θ) = |f(θ)| / √2, and we want the minimum d(θ) to be 2√2, then:Minimum |f(θ)| / √2 = 2√2Multiply both sides by √2:Minimum |f(θ)| = 4So, the minimum of |f(θ)| over θ in [0, π] is 4. Therefore, we need to find m such that the minimum of |√3 cosθ - sinθ + m| is 4.Hmm, okay. So, let's analyze f(θ) = √3 cosθ - sinθ + m.I can write √3 cosθ - sinθ as a single sinusoidal function. Let's recall that A cosθ + B sinθ can be written as C cos(θ + φ), where C = √(A² + B²), and tanφ = B/A.In our case, A = √3, B = -1. So, C = √( (√3)² + (-1)² ) = √(3 + 1) = √4 = 2.And tanφ = B/A = (-1)/√3, so φ = -30° or 330°, which is equivalent to -π/6 radians.So, √3 cosθ - sinθ = 2 cos(θ + π/6)Therefore, f(θ) = 2 cos(θ + π/6) + mSo, the expression inside the absolute value is 2 cos(θ + π/6) + m.Therefore, |2 cos(θ + π/6) + m|.We need the minimum of this expression over θ in [0, π] to be 4.So, let's analyze the range of 2 cos(θ + π/6) as θ varies in [0, π].First, θ is in [0, π], so θ + π/6 is in [π/6, 7π/6].The cosine function over [π/6, 7π/6] has a maximum at θ + π/6 = 0 (which is not in our interval), and the minimum at θ + π/6 = π, which is 7π/6 - π/6 = π. Wait, no.Wait, θ + π/6 ranges from π/6 to 7π/6. The maximum value of cos occurs at θ + π/6 = 0, but 0 is not in [π/6, 7π/6]. The maximum in this interval is at θ + π/6 = π/6, which is cos(π/6) = √3/2. The minimum occurs at θ + π/6 = π, which is cos(π) = -1.So, cos(θ + π/6) ranges from -1 to √3/2 in this interval.Therefore, 2 cos(θ + π/6) ranges from -2 to √3.So, 2 cos(θ + π/6) + m ranges from m - 2 to m + √3.Therefore, the expression inside the absolute value, |2 cos(θ + π/6) + m|, will have its minimum value depending on where the interval [m - 2, m + √3] is relative to zero.We need the minimum of |2 cos(θ + π/6) + m| to be 4. So, let's consider the possible cases.Case 1: The entire interval [m - 2, m + √3] is above 4. That is, m - 2 > 4, which implies m > 6. Then, the minimum |2 cos(θ + π/6) + m| would be m - 2, which we set equal to 4. So, m - 2 = 4 => m = 6.Case 2: The entire interval [m - 2, m + √3] is below -4. That is, m + √3 < -4, which implies m < -4 - √3. Then, the minimum |2 cos(θ + π/6) + m| would be |m + √3|, which we set equal to 4. So, |m + √3| = 4 => m + √3 = -4 (since m < -4 - √3, m + √3 is negative) => m = -4 - √3.Case 3: The interval [m - 2, m + √3] crosses zero. In this case, the minimum |2 cos(θ + π/6) + m| would be zero, which is less than 4. But we need the minimum to be 4, so this case is not possible.Therefore, the only possible solutions are m = 6 and m = -4 - √3.Let me double-check this reasoning.If m = 6, then f(θ) = 2 cos(θ + π/6) + 6. The minimum value of f(θ) is 6 - 2 = 4, so |f(θ)| minimum is 4, which satisfies the condition.If m = -4 - √3, then f(θ) = 2 cos(θ + π/6) - 4 - √3. The maximum value of f(θ) is -4 - √3 + √3 = -4, so |f(θ)| minimum is 4, which also satisfies the condition.Yes, that seems correct. So, the values of m are 6 and -4 - √3.**Final Answer**The value of the real number ( m ) is (boxed{-4 - sqrt{3}}) or (boxed{6}).

💡First, recall that in an arithmetic sequence, each term is obtained by adding a constant difference ( d ) to the previous term. Therefore, the general form of the ( n )-th term is:[a_n = a_1 + (n-1)d]Given the equations:[a_2 + a_6 = 8][a_3 + a_4 = 3]Substitute the general form into these equations:[(a_1 + d) + (a_1 + 5d) = 8 quad Rightarrow quad 2a_1 + 6d = 8][(a_1 + 2d) + (a_1 + 3d) = 3 quad Rightarrow quad 2a_1 + 5d = 3]Now, subtract the second equation from the first to eliminate ( a_1 ):[(2a_1 + 6d) - (2a_1 + 5d) = 8 - 3 quad Rightarrow quad d = 5]Thus, the common difference ( d ) is 5.

💡First, I need to determine the average of one fourth of the class, given that the overall class average is 84 and the average of the remaining three fourths is 80.I'll start by letting ( N ) represent the total number of students in the class and ( A ) represent the average of one fourth of the class.Since one fourth of the class has an average of ( A ), the remaining three fourths have an average of 80.Next, I'll calculate the total sum of grades for the entire class by adding the sum of the grades of one fourth of the class and the sum of the grades of the remaining three fourths.For one fourth of the class:[frac{N}{4} times A]For three fourths of the class:[frac{3N}{4} times 80]The total sum of grades for the entire class is:[frac{N}{4} times A + frac{3N}{4} times 80]Since the overall average is 84, the total sum of grades can also be expressed as:[N times 84]Setting the two expressions for the total sum of grades equal to each other:[frac{N}{4} times A + frac{3N}{4} times 80 = N times 84]I can simplify this equation by dividing both sides by ( N ):[frac{1}{4} times A + frac{3}{4} times 80 = 84]Now, solving for ( A ):[frac{1}{4} times A = 84 - frac{3}{4} times 80][frac{1}{4} times A = 84 - 60][frac{1}{4} times A = 24][A = 24 times 4][A = 96]Therefore, the average of one fourth of the class is 96.

💡Alright, so I have this problem about probability and statistics. It's about selecting students for volunteer services. Let me try to understand it step by step.First, the community has 8 college students: 5 males and 3 females. They need to select 3 students for volunteer services. There are two parts to this problem.**Part 1: Finding P(B|A)**Okay, so event A is "female student A is selected," and event B is "male student B is selected." I need to find the conditional probability P(B|A), which is the probability that male student B is selected given that female student A is already selected.Hmm, conditional probability. I remember the formula for conditional probability is:[ P(B|A) = frac{P(A text{ and } B)}{P(A)} ]So, I need to find both P(A) and P(A and B).Let's start with P(A). This is the probability that female student A is selected. There are 8 students, and we're selecting 3. So, the total number of ways to select 3 students out of 8 is:[ C(8, 3) = frac{8!}{3!(8-3)!} = 56 ]Now, the number of ways to select 3 students where A is included. If A is already selected, we need to choose 2 more students from the remaining 7. So:[ C(7, 2) = frac{7!}{2!(7-2)!} = 21 ]Therefore, P(A) is:[ P(A) = frac{21}{56} = frac{3}{8} ]Okay, got P(A). Now, P(A and B) is the probability that both A and B are selected. So, if both A and B are selected, we need to choose 1 more student from the remaining 6 (since 8 - 2 = 6). So:[ C(6, 1) = 6 ]Therefore, P(A and B) is:[ P(A text{ and } B) = frac{6}{56} = frac{3}{28} ]Now, plug these into the conditional probability formula:[ P(B|A) = frac{frac{3}{28}}{frac{3}{8}} = frac{3}{28} times frac{8}{3} = frac{8}{28} = frac{2}{7} ]So, the probability that male student B is selected given that female student A is selected is 2/7.**Part 2: Probability Distribution and Expected Value of X**Now, X is the number of male students among the selected 3 students. I need to find the probability distribution of X and its expected value.First, let's identify the possible values of X. Since we're selecting 3 students, and there are 5 males, X can be 0, 1, 2, or 3.To find the probability distribution, I need to calculate P(X=0), P(X=1), P(X=2), and P(X=3).The general formula for the probability of selecting k males out of 3 is:[ P(X=k) = frac{C(5, k) times C(3, 3 - k)}{C(8, 3)} ]Where:- C(5, k) is the number of ways to choose k males from 5.- C(3, 3 - k) is the number of ways to choose the remaining (3 - k) females from 3.- C(8, 3) is the total number of ways to choose 3 students from 8.Let's compute each probability.**For X=0:**[ P(X=0) = frac{C(5, 0) times C(3, 3)}{C(8, 3)} = frac{1 times 1}{56} = frac{1}{56} ]**For X=1:**[ P(X=1) = frac{C(5, 1) times C(3, 2)}{C(8, 3)} = frac{5 times 3}{56} = frac{15}{56} ]**For X=2:**[ P(X=2) = frac{C(5, 2) times C(3, 1)}{C(8, 3)} = frac{10 times 3}{56} = frac{30}{56} = frac{15}{28} ]**For X=3:**[ P(X=3) = frac{C(5, 3) times C(3, 0)}{C(8, 3)} = frac{10 times 1}{56} = frac{10}{56} = frac{5}{28} ]So, summarizing the probability distribution:- P(X=0) = 1/56- P(X=1) = 15/56- P(X=2) = 15/28- P(X=3) = 5/28Now, to find the expected value E(X), I can use the formula:[ E(X) = sum [x times P(X=x)] ]Plugging in the values:[ E(X) = 0 times frac{1}{56} + 1 times frac{15}{56} + 2 times frac{15}{28} + 3 times frac{5}{28} ]Let's compute each term:- 0 × 1/56 = 0- 1 × 15/56 = 15/56- 2 × 15/28 = 30/28 = 15/14- 3 × 5/28 = 15/28Now, sum them up:[ E(X) = 0 + frac{15}{56} + frac{15}{14} + frac{15}{28} ]To add these fractions, they need a common denominator. The least common denominator for 56, 14, and 28 is 56.Convert each fraction:- 15/56 remains as is.- 15/14 = (15 × 4)/(14 × 4) = 60/56- 15/28 = (15 × 2)/(28 × 2) = 30/56Now, add them:[ E(X) = frac{15}{56} + frac{60}{56} + frac{30}{56} = frac{105}{56} ]Simplify:[ frac{105}{56} = frac{15}{8} ]So, the expected value E(X) is 15/8.**Summary:**- The conditional probability P(B|A) is 2/7.- The probability distribution of X is: - P(X=0) = 1/56 - P(X=1) = 15/56 - P(X=2) = 15/28 - P(X=3) = 5/28- The expected value E(X) is 15/8.I think that covers both parts of the problem. I should double-check my calculations to make sure I didn't make any mistakes.For part 1, I used the conditional probability formula correctly, and the calculations seem right. For part 2, I used the hypergeometric distribution approach, which is appropriate for selecting without replacement. The expected value calculation also seems correct, as I converted all fractions to have a common denominator before adding.Everything checks out. I feel confident with these answers.

💡Okay, so I have this problem here about sets M and P. Let me try to understand what it's asking. First, set M is defined as all real numbers x such that x is less than or equal to 1. So, M = {x | x ≤ 1}. That means M includes every number from negative infinity up to and including 1. Then, set P is defined as all real numbers x such that x is greater than t. So, P = {x | x > t}. This set P includes every number greater than t, but not t itself. The problem says that the empty set is a proper subset of the intersection of M and P. In mathematical terms, that's written as ∅ ⊊ (M ∩ P). I know that a proper subset means that the empty set is contained within M ∩ P, but M ∩ P isn't just the empty set. So, this tells me that M ∩ P must contain at least one element. So, M ∩ P is the set of all x that are both less than or equal to 1 and greater than t. In other words, M ∩ P = {x | t < x ≤ 1}. For this set to not be empty, there must be some x that satisfies both conditions. Let me visualize this. If I draw a number line, M goes from negative infinity up to 1, and P starts just above t and goes to positive infinity. The intersection of these two sets would be the overlap between them. For there to be an overlap, t must be less than 1. If t were equal to 1, then P would start just above 1, and M ends at 1, so there would be no overlap. Similarly, if t were greater than 1, P would start even higher, so the overlap would still be empty. Wait, but the problem says ∅ is a proper subset, which means M ∩ P is not empty. So, t must be less than 1. That way, there are numbers between t and 1 that are in both M and P. Let me test this with some numbers. If t is 0, which is less than 1, then P is {x | x > 0}, and M is {x | x ≤ 1}. Their intersection would be {x | 0 < x ≤ 1}, which is definitely not empty. What if t is exactly 1? Then P is {x | x > 1}, and M is {x | x ≤ 1}. Their intersection would be empty because there's no number that is both greater than 1 and less than or equal to 1. So, t can't be equal to 1. What if t is greater than 1, say t = 2? Then P is {x | x > 2}, and M is still {x | x ≤ 1}. Their intersection would be empty because there's no number that is both greater than 2 and less than or equal to 1. So, t can't be greater than 1 either. Therefore, the only possibility is that t is less than 1. That way, there is some overlap between M and P, making their intersection non-empty. So, looking back at the options: A: t > 1 – This would make the intersection empty, which isn't allowed. B: t ≥ 1 – This includes t = 1 and t > 1, both of which result in an empty intersection. C: t < 1 – This is what we determined is necessary for the intersection to be non-empty. D: t ≤ 1 – This includes t = 1, which we saw makes the intersection empty. Therefore, the correct answer must be C: t < 1.