answer：Okay, so I have this geometry problem here. Let me read it again: In triangle ABC, BC is 15 times the square root of 2, and angle C is 45 degrees. The perpendicular bisector of BC intersects BC at D and AC at E. I need to find the length of DE. Hmm, okay.First, let me visualize the triangle. Triangle ABC with BC = 15√2 and angle C = 45°. So, angle at point C is 45 degrees. The sides adjacent to angle C are BC and AC, and the side opposite angle C is AB.Now, the problem mentions the perpendicular bisector of BC. A perpendicular bisector is a line that is perpendicular to BC and passes through its midpoint. So, point D is the midpoint of BC because it's where the perpendicular bisector intersects BC. That makes sense because a bisector splits the segment into two equal parts.Since D is the midpoint, BD equals DC. Given that BC is 15√2, then BD and DC should each be half of that. Let me calculate that: 15√2 divided by 2 is (15/2)√2, which is 7.5√2. So, BD = DC = 7.5√2.Now, the perpendicular bisector intersects BC at D and AC at E. So, DE is a segment from the midpoint D of BC to some point E on AC. I need to find the length of DE.Let me think about the properties of a perpendicular bisector. It's perpendicular to BC, so the line DE is perpendicular to BC. Since BC is a side of the triangle, and DE is perpendicular to it, DE forms a right angle with BC at point D.So, triangle BDE is a right triangle with right angle at D. Wait, but actually, DE is the perpendicular bisector, so it's not just triangle BDE; it's also triangle CDE. But since D is the midpoint, both BD and DC are equal, so triangles BDE and CDE are congruent if DE is the same for both. Hmm, maybe that's not directly helpful.Alternatively, maybe I should consider triangle ABC and see if I can find some other lengths or angles that can help me find DE.Given that angle C is 45°, and we know BC, maybe I can use the Law of Sines or Cosines to find other sides of triangle ABC. Let me recall the Law of Sines: in any triangle, the ratio of the length of a side to the sine of its opposite angle is constant. So, in triangle ABC, we have:BC / sin(A) = AC / sin(B) = AB / sin(C)We know BC is 15√2 and angle C is 45°, so sin(C) is sin(45°) which is √2/2. So, AB / (√2/2) = BC / sin(A). Hmm, but I don't know angles A or B yet.Alternatively, maybe using the Law of Cosines would be better. The Law of Cosines states that c² = a² + b² - 2ab cos(C), where c is the side opposite angle C. In triangle ABC, if I let AB = c, BC = a = 15√2, AC = b, and angle C = 45°, then:AB² = AC² + BC² - 2 * AC * BC * cos(45°)But I don't know AB or AC, so maybe that's not directly helpful either.Wait, perhaps I can assign coordinates to the triangle to make it easier. Let me place point C at the origin (0,0). Since angle C is 45°, and BC is 15√2, I can position points B and A accordingly.Let me assume point C is at (0,0). Let me place point B along the x-axis for simplicity, so point B is at (15√2, 0). Then, point A is somewhere in the plane such that angle at C is 45°. So, the coordinates of point A can be determined based on the angle.Since angle C is 45°, the triangle is either a 45-45-90 triangle or some other configuration. But since we only know one angle and one side, it might not necessarily be a right triangle. Hmm.Alternatively, maybe I can use coordinate geometry. Let me assign coordinates as follows:- Let point C be at (0,0).- Let point B be at (15√2, 0).- Let point A be at (x, y).Since angle at C is 45°, the angle between vectors CB and CA is 45°. Vector CB is (15√2, 0), and vector CA is (x, y). The angle between them is 45°, so the dot product formula can be used:CB · CA = |CB| |CA| cos(45°)Calculating the dot product: (15√2)(x) + (0)(y) = 15√2 xThe magnitudes: |CB| is 15√2, |CA| is √(x² + y²). So,15√2 x = (15√2)(√(x² + y²)) * (√2 / 2)Simplify the right side: 15√2 * √(x² + y²) * √2 / 2 = 15√2 * √(x² + y²) * √2 / 2√2 * √2 is 2, so this becomes 15 * 2 * √(x² + y²) / 2 = 15√(x² + y²)So, 15√2 x = 15√(x² + y²)Divide both sides by 15: √2 x = √(x² + y²)Square both sides: 2x² = x² + y²Simplify: 2x² - x² = y² => x² = y² => y = ±xSince we're dealing with a triangle above the x-axis, y should be positive, so y = x.So, point A lies somewhere along the line y = x. So, coordinates of A are (x, x).Now, we can find the coordinates of point E, which is the intersection of the perpendicular bisector of BC with AC.First, let's find the equation of the perpendicular bisector of BC.Since BC is from (0,0) to (15√2, 0), its midpoint D is at ((15√2)/2, 0). The slope of BC is (0 - 0)/(15√2 - 0) = 0, so it's a horizontal line. Therefore, the perpendicular bisector will be a vertical line passing through D. So, the equation of the perpendicular bisector is x = (15√2)/2.Now, we need to find the equation of AC. Since point A is at (x, x) and point C is at (0,0), the line AC has a slope of (x - 0)/(x - 0) = 1. So, the equation of AC is y = x.Wait, but if AC is y = x, and the perpendicular bisector is x = (15√2)/2, then their intersection point E is at x = (15√2)/2, y = (15√2)/2.So, point E is at ((15√2)/2, (15√2)/2).Now, point D is at ((15√2)/2, 0). So, DE is the distance between D and E.Since both points have the same x-coordinate, the distance DE is simply the difference in y-coordinates. So, DE = |y_E - y_D| = |(15√2)/2 - 0| = (15√2)/2.Wait, but (15√2)/2 is equal to 7.5√2. So, DE is 7.5√2.Wait, that seems straightforward. Let me double-check.We placed point C at (0,0), point B at (15√2, 0). Point A is somewhere on y = x. The perpendicular bisector of BC is x = (15√2)/2, which intersects AC at E. Since AC is y = x, E is at ((15√2)/2, (15√2)/2). D is at ((15√2)/2, 0). So, DE is vertical distance from (15√2)/2, 0 to (15√2)/2, (15√2)/2, which is (15√2)/2.Yes, that seems correct. So, DE is 7.5√2.Alternatively, I can think about triangle CDE. Since DE is perpendicular to BC, and D is the midpoint, triangle CDE is a right triangle with legs CD and DE, and hypotenuse CE.We know CD is 7.5√2, and angle at C is 45°, so triangle CDE is a 45-45-90 triangle. Therefore, DE should be equal to CD, which is 7.5√2. Wait, but that would mean DE = CD, but in reality, in a 45-45-90 triangle, the legs are equal, and the hypotenuse is leg * √2. So, if CD is one leg, and DE is the other leg, then yes, they should be equal. Therefore, DE = CD = 7.5√2.Wait, but in our coordinate system, DE is vertical, and CD is horizontal, so they are perpendicular. So, triangle CDE is indeed a right triangle with legs CD and DE, and angle at D is 90°, angle at C is 45°, so the other angle at E must be 45°, making it an isosceles right triangle. Therefore, DE = CD = 7.5√2.So, both methods confirm that DE is 7.5√2.I think that's the answer. Let me just recap:1. Assign coordinates to the triangle with C at (0,0), B at (15√2, 0), and A somewhere on y = x.2. Find the perpendicular bisector of BC, which is x = (15√2)/2.3. Find the intersection E of this bisector with AC, which is at ((15√2)/2, (15√2)/2).4. Calculate DE as the vertical distance between D ((15√2)/2, 0) and E ((15√2)/2, (15√2)/2), which is (15√2)/2 = 7.5√2.5. Alternatively, recognize that triangle CDE is a 45-45-90 triangle, so DE = CD = 7.5√2.Yes, that all makes sense. I don't see any mistakes in the reasoning. So, DE is 7.5√2.

answer：Okay, so I have this problem where there are 12 people standing equally spaced around a circle. Each person knows exactly 3 others: the two next to them and the second person clockwise from themselves. I need to figure out how many ways they can split into 6 couples where each couple knows each other. The answer choices are from 2 to 6, so it's a relatively small number, which suggests that the solution might involve some combinatorial patterns or symmetries.First, I should probably visualize the setup. Imagine a clock face with 12 people instead of numbers. Each person is connected to their immediate neighbors and the person two spots away clockwise. So, for example, person 1 knows person 2, person 12, and person 3. Person 2 knows person 1, person 3, and person 4, and so on.Since each person knows exactly three others, the graph formed by their relationships is a 3-regular graph (each vertex has degree 3). Specifically, this graph is known as the "Pappus graph," which is a well-known 3-regular graph with 12 vertices. However, I might not need to know that term for this problem, but it's good to recognize the structure.Now, the problem is asking for the number of ways to partition these 12 people into 6 couples such that each couple consists of people who know each other. In graph theory terms, this is equivalent to finding the number of perfect matchings in the graph. A perfect matching is a set of edges where every vertex is included exactly once, and no two edges share a vertex.Given that each person has three connections, the number of perfect matchings isn't immediately obvious. It might be helpful to consider symmetries or specific properties of the graph. Since the people are arranged in a circle, the graph is symmetric, which might mean that certain matchings can be rotated or reflected to form others, but I need to be careful not to overcount.Let me try to approach this step by step. First, fix a person, say person 1. Person 1 can be paired with one of three people: person 2, person 12, or person 3. Let's consider each case:1. **Case 1: Person 1 is paired with person 2.** - If person 1 is paired with person 2, then person 3 must be paired with someone else. Person 3 knows person 4 and person 1 (but person 1 is already paired), so person 3 can be paired with person 4 or person 5. - **Subcase 1a: Person 3 is paired with person 4.** - Then person 5 must be paired with someone. Person 5 knows person 6 and person 4 (already paired), so person 5 can be paired with person 6 or person 7. - **Subsubcase 1a1: Person 5 is paired with person 6.** - Continuing this pattern, person 7 must be paired with person 8 or person 9. If we continue pairing adjacent people, we end up with the matching: (1,2), (3,4), (5,6), (7,8), (9,10), (11,12). This is one valid matching. - **Subsubcase 1a2: Person 5 is paired with person 7.** - Then person 6 must be paired with person 7 or person 8. But person 7 is already paired with person 5, so person 6 must be paired with person 8. Then person 9 must be paired with person 10 or person 11. If we continue, this leads to a different matching: (1,2), (3,4), (5,7), (6,8), (9,10), (11,12). Wait, but person 11 is only connected to person 10 and person 12. Since person 12 is already paired with person 1, person 11 must be paired with person 10. So this seems valid as well. - **Subcase 1b: Person 3 is paired with person 5.** - Then person 4 must be paired with person 5 or person 6. But person 5 is already paired with person 3, so person 4 must be paired with person 6. Then person 7 must be paired with person 8 or person 9, and so on. This leads to another matching: (1,2), (3,5), (4,6), (7,8), (9,10), (11,12). But wait, person 11 is only connected to person 10 and person 12. Since person 12 is paired with person 1, person 11 must be paired with person 10. So this is another valid matching.2. **Case 2: Person 1 is paired with person 12.** - This is symmetric to Case 1, so it will result in similar matchings. Essentially, it's just a mirror image of the previous case, leading to the same number of valid matchings.3. **Case 3: Person 1 is paired with person 3.** - If person 1 is paired with person 3, then person 2 must be paired with someone else. Person 2 knows person 1 (already paired), person 3 (already paired), and person 4. So person 2 must be paired with person 4. - Then person 5 must be paired with someone. Person 5 knows person 4 (already paired), person 6, and person 7. So person 5 can be paired with person 6 or person 7. - **Subcase 3a: Person 5 is paired with person 6.** - Then person 7 must be paired with person 8 or person 9. If we continue, this leads to the matching: (1,3), (2,4), (5,6), (7,8), (9,10), (11,12). But person 11 is only connected to person 10 and person 12. Since person 12 is paired with person 1, person 11 must be paired with person 10. So this is valid. - **Subcase 3b: Person 5 is paired with person 7.** - Then person 6 must be paired with person 7 or person 8. But person 7 is already paired with person 5, so person 6 must be paired with person 8. Then person 9 must be paired with person 10 or person 11. This leads to another matching: (1,3), (2,4), (5,7), (6,8), (9,10), (11,12). Again, person 11 must be paired with person 10.Wait, I'm starting to see a pattern here. Depending on how I pair the initial people, I can end up with different configurations. However, I need to ensure that all pairings are valid, meaning that each person is only paired with someone they know.But I'm also noticing that some of these cases might lead to the same overall matching when considering the circular symmetry. For example, pairing person 1 with person 2 and then proceeding around the circle might be equivalent to pairing person 1 with person 12 and proceeding in the opposite direction.Let me try to count the distinct matchings:1. **All adjacent pairs**: This is the case where everyone is paired with their immediate neighbor. There are two such matchings: one where each person is paired with the next person clockwise, and one where each person is paired with the previous person clockwise. However, in a circle, these are actually the same matching because it's just a rotation. Wait, no, actually, in terms of pairings, they are distinct because the direction matters. But since the problem doesn't specify direction, maybe they are considered the same. Hmm, I need to clarify.2. **Pairs with one person skipped**: This is where each person is paired with the person two spots away. For example, person 1 with person 3, person 2 with person 4, etc. This creates a different kind of matching. However, in this case, person 12 would be paired with person 2, which is adjacent, not two spots away. Wait, that might not work because person 12's connections are person 11, person 1, and person 2. So if person 12 is paired with person 2, that's valid, but then person 11 must be paired with someone else. This might complicate things.Wait, maybe I need to think in terms of the graph structure. The graph is a combination of two cycles: one being the outer circle (adjacent pairs) and the other being the inner connections (second person clockwise). So, the graph is actually a combination of two 12-cycles and a 6-cycle? No, that's not quite right. Each person is connected to their two neighbors and the person two spots away, so the graph is actually a 3-regular graph with girth 3, known as the Pappus graph.But perhaps I can decompose the graph into perfect matchings. Since the graph is 3-regular and has 12 vertices, it should have a certain number of perfect matchings. I recall that the number of perfect matchings in a 3-regular graph can vary, but for the Pappus graph, it's known to have 3 perfect matchings. Wait, is that correct?Let me check. The Pappus graph is a bipartite graph? No, it's not bipartite because it has odd-length cycles. For example, person 1 connected to person 2, person 3, and person 12. Person 2 connected to person 1, person 3, and person 4. This creates a triangle between person 1, 2, and 3, which is an odd cycle. Therefore, it's not bipartite, so it can't be decomposed into perfect matchings in the same way as a bipartite graph.Wait, but the problem is just asking for the number of perfect matchings, not a decomposition. So, how many perfect matchings does the Pappus graph have? I think it's 3, but I'm not entirely sure. Let me try to reason it out.If I fix person 1, they can be paired with person 2, person 12, or person 3. Let's see how each choice affects the rest of the pairings.1. **Person 1 paired with person 2**: - Then person 3 must be paired with either person 4 or person 5. - If person 3 is paired with person 4, then person 5 must be paired with person 6 or person 7. - If person 5 is paired with person 6, then person 7 must be paired with person 8 or person 9, and so on. This leads to a specific matching. - If person 5 is paired with person 7, then person 6 must be paired with person 8, and so on. This leads to another matching. - If person 3 is paired with person 5, then person 4 must be paired with person 6, and so on. This leads to a third matching.2. **Person 1 paired with person 12**: - This is symmetric to the first case, so it doesn't lead to any new matchings beyond what we've already counted.3. **Person 1 paired with person 3**: - Then person 2 must be paired with person 4, and person 12 must be paired with person 11 (since person 12 is connected to person 11, person 1, and person 2. Person 1 is already paired with person 3, and person 2 is paired with person 4, so person 12 must be paired with person 11). - Then person 5 must be paired with person 6 or person 7, leading to similar subcases as before.Wait, but if I consider all these possibilities, it seems like there are more than 3 matchings. However, I might be overcounting because some pairings might lead to the same overall matching when considering the circular symmetry.Alternatively, maybe the graph has exactly 3 perfect matchings due to its structure. I think the Pappus graph is known to have 3 perfect matchings, which correspond to the three different ways of pairing the vertices such that each pair is connected by an edge.Therefore, after considering all the cases and recognizing the structure of the graph, I conclude that there are 3 ways to split the 12 people into 6 couples where each couple knows each other.

answer：Alright, so I've got this problem here: I need to find the sum of the two integers adjacent to 21 when all the positive integer divisors of 294, except 1, are arranged around a circle. The condition is that every pair of adjacent integers must have a common factor greater than 1. Hmm, okay, let's break this down step by step.First, I need to figure out what the divisors of 294 are. To do that, I should probably start by finding the prime factorization of 294. Let me recall how to do that. I know that 294 is an even number because it ends with a 4, so it's divisible by 2. Dividing 294 by 2 gives me 147. Now, 147 is divisible by 3 because 1 + 4 + 7 equals 12, which is divisible by 3. So, 147 divided by 3 is 49. And 49 is 7 squared, so that's 7 times 7. Putting it all together, the prime factors of 294 are 2, 3, and 7, with 7 appearing twice. So, the prime factorization is 2 × 3 × 7².Now, to find all the divisors of 294, I can use the prime factors. The formula for finding the number of divisors is to add one to each of the exponents in the prime factorization and then multiply those together. Here, the exponents are 1 for 2, 1 for 3, and 2 for 7. So, adding one to each gives me 2, 2, and 3. Multiplying those together: 2 × 2 × 3 equals 12. So, there are 12 divisors in total. But the problem says to exclude 1, so we'll have 11 divisors to arrange around the circle.Let me list out all the divisors. Starting from the smallest, excluding 1, we have:- 2- 3- 6 (which is 2 × 3)- 7- 14 (which is 2 × 7)- 21 (which is 3 × 7)- 42 (which is 2 × 3 × 7)- 49 (which is 7²)- 98 (which is 2 × 7²)- 147 (which is 3 × 7²)- 294 (which is 2 × 3 × 7²)Okay, so that's 11 divisors. Now, the task is to arrange these numbers around a circle such that every pair of adjacent numbers has a common factor greater than 1. That means that any two numbers next to each other must share at least one prime factor.Since we're focusing on the number 21, let's think about its factors. 21 is 3 × 7, so any number adjacent to 21 must share either a factor of 3 or 7. Let's list the other numbers and see which ones share a factor with 21.Looking at the list:- 2: Factors are 2. Doesn't share with 21.- 3: Shares factor 3.- 6: Shares factor 3 and 2.- 7: Shares factor 7.- 14: Shares factor 7.- 42: Shares factors 3 and 7.- 49: Shares factor 7.- 98: Shares factor 7.- 147: Shares factor 3 and 7.- 294: Shares factors 2, 3, and 7.So, the numbers that can be adjacent to 21 are: 3, 6, 7, 14, 42, 49, 98, 147, and 294. That's quite a few. But since we're arranging them in a circle, each number will have two neighbors. So, 21 will have two neighbors from this list.Now, the challenge is to arrange all 11 numbers in a circle where each adjacent pair shares a common factor greater than 1. This sounds like forming a cycle where each edge (adjacency) is defined by a common prime factor.One approach is to model this as a graph where each node is a divisor, and edges connect nodes that share a common factor greater than 1. Then, we need to find a Hamiltonian cycle in this graph, which is a cycle that visits every node exactly once.But since we're specifically interested in the neighbors of 21, maybe we don't need to construct the entire cycle. Instead, we can focus on the possible neighbors of 21 and see which ones can fit into the cycle.Looking back at the possible neighbors of 21: 3, 6, 7, 14, 42, 49, 98, 147, 294.But in reality, since we have to arrange all 11 numbers, each number will have two neighbors, so 21 will have two specific neighbors from this list.To figure out which ones, maybe we can look at the structure of the graph. Let's think about the prime factors:- 2: Only connected to numbers that have 2 as a factor: 2, 6, 14, 42, 98, 294.- 3: Connected to numbers with 3: 3, 6, 21, 42, 147, 294.- 7: Connected to numbers with 7: 7, 14, 21, 42, 49, 98, 147, 294.So, 21 is connected to all numbers with 3 or 7. Now, since we need to arrange them in a circle, each number must be connected to two others. Let's see if we can find a pattern or a way to chain these numbers.Maybe starting from 21, let's see what could come next. If we go with 3, then the next number after 3 must share a factor with 3. The options are 6, 21, 42, 147, 294. But 21 is already used, so it could be 6, 42, 147, or 294.Alternatively, if we go from 21 to 7, then the next number must share a factor with 7, which are 14, 42, 49, 98, 147, 294.This seems a bit too broad. Maybe another approach is needed.Perhaps we can look for numbers that only have one connection, meaning they can only be adjacent to one specific number. For example, 49 is 7², so it only has 7 as a factor. Therefore, it can only be adjacent to numbers that have 7 as a factor. Similarly, 98 is 2 × 7², so it has 2 and 7 as factors, meaning it can be adjacent to numbers with 2 or 7.Wait, 49 only has 7 as a factor, so it can only be adjacent to numbers that have 7. Similarly, 147 is 3 × 7², so it has 3 and 7. 294 has 2, 3, and 7.Maybe we can think about the numbers with only one prime factor, like 2, 3, and 7. These are primes, so they can only connect to numbers that share their prime.For example, 2 can only connect to numbers with 2 as a factor: 2, 6, 14, 42, 98, 294.Similarly, 3 can only connect to numbers with 3: 3, 6, 21, 42, 147, 294.7 can connect to numbers with 7: 7, 14, 21, 42, 49, 98, 147, 294.So, primes 2, 3, and 7 are like hubs connecting to multiple numbers.Now, considering that we need to form a circle, each prime hub must be connected in a way that the chain can loop back.But maybe this is getting too abstract. Let's try to construct the circle step by step.Start with 21. It needs two neighbors. Let's pick two numbers from its possible neighbors: 3, 6, 7, 14, 42, 49, 98, 147, 294.Let's say we pick 3 and 7 as neighbors. So, 21 is between 3 and 7.Now, from 3, we need to go to another number that shares a factor with 3. The options are 6, 42, 147, 294.Let's pick 6 next. So, 3 is between 21 and 6.From 6, it shares factors with 2 and 3. So, next could be 2 or another multiple of 3 or 2.If we go to 2, then from 2, we can go to 14, 42, 98, 294.Let's say we go to 14. So, 2 is between 6 and 14.From 14, it shares factors with 2 and 7. So, next could be 42, 49, 98, 294.Let's pick 42. So, 14 is between 2 and 42.From 42, it shares factors with 2, 3, and 7. So, next could be 21, but 21 is already used, or 147, 294.Let's pick 147. So, 42 is between 14 and 147.From 147, it shares factors with 3 and 7. So, next could be 3, but 3 is already used, or 21, which is already used, or 49, 98, 294.Let's pick 49. So, 147 is between 42 and 49.From 49, it shares factors with 7. So, next could be 7, 14, 42, 98, 294.But 7 is already connected to 21, so let's pick 98. So, 49 is between 147 and 98.From 98, it shares factors with 2 and 7. So, next could be 2, which is already used, or 14, which is already used, or 42, which is already used, or 294.Let's pick 294. So, 98 is between 49 and 294.From 294, it shares factors with 2, 3, and 7. So, next could be 2, which is already used, or 3, which is already used, or 42, which is already used, or 147, which is already used, or 21, which is already used.Hmm, that's a problem. We've reached 294, and all its possible connections are already used. So, we're stuck. This means our current arrangement doesn't form a complete circle.Maybe we need to backtrack and choose different connections.Let's go back to 21. Instead of connecting to 3 and 7, maybe connect to 7 and another number.Let's try 21 connected to 7 and 42.So, 21 is between 7 and 42.From 7, it can go to 14, 49, 98, 147, 294.Let's pick 14. So, 7 is between 21 and 14.From 14, it can go to 2, 42, 49, 98, 294.Let's pick 2. So, 14 is between 7 and 2.From 2, it can go to 6, 42, 98, 294.Let's pick 6. So, 2 is between 14 and 6.From 6, it can go to 3, 42, 294.Let's pick 3. So, 6 is between 2 and 3.From 3, it can go to 21, 42, 147, 294.21 is already used, so let's pick 42. So, 3 is between 6 and 42.From 42, it can go to 21, which is already used, or 147, 294.Let's pick 147. So, 42 is between 3 and 147.From 147, it can go to 49, 98, 294.Let's pick 49. So, 147 is between 42 and 49.From 49, it can go to 7, which is already used, or 98, 294.Let's pick 98. So, 49 is between 147 and 98.From 98, it can go to 2, which is already used, or 42, which is already used, or 294.Let's pick 294. So, 98 is between 49 and 294.From 294, it can go to 2, which is already used, or 3, which is already used, or 42, which is already used, or 147, which is already used, or 21, which is already used.Again, stuck. Hmm.Maybe we need a different approach. Perhaps 21 should be connected to 42 and another number.Wait, 42 is connected to both 3 and 7, so it's a good hub.Let me try arranging the circle by starting with 21 connected to 42 and 7.So, 21 is between 42 and 7.From 42, it can go to 3, 6, 14, 294.Let's pick 3. So, 42 is between 21 and 3.From 3, it can go to 6, 21, 42, 147, 294.21 and 42 are already used, so pick 6. So, 3 is between 42 and 6.From 6, it can go to 2, 42, 294.42 is already used, so pick 2. So, 6 is between 3 and 2.From 2, it can go to 14, 42, 98, 294.42 is already used, so pick 14. So, 2 is between 6 and 14.From 14, it can go to 7, 42, 49, 98, 294.7 and 42 are already used, so pick 49. So, 14 is between 2 and 49.From 49, it can go to 7, which is already used, or 98, 294.Pick 98. So, 49 is between 14 and 98.From 98, it can go to 2, which is already used, or 42, which is already used, or 294.Pick 294. So, 98 is between 49 and 294.From 294, it can go to 2, which is already used, or 3, which is already used, or 42, which is already used, or 147, which hasn't been used yet.So, 294 is between 98 and 147.From 147, it can go to 49, which is already used, or 21, which is already used, or 3, which is already used, or 294, which is already used.Wait, 147 is connected to 49 and 294, but 49 is already connected to 14 and 98, and 294 is connected to 98 and 147. So, 147 is only connected to 49 and 294, but we need to connect it to another number.Hmm, seems like we're missing something. Maybe we need to adjust earlier connections.Alternatively, perhaps 147 should be connected to 42 instead of 49.Let me try that.Starting over:21 is between 42 and 7.42 is between 21 and 3.3 is between 42 and 6.6 is between 3 and 2.2 is between 6 and 14.14 is between 2 and 49.49 is between 14 and 98.98 is between 49 and 294.294 is between 98 and 147.147 is between 294 and... Hmm, 147 needs to connect back to someone. It's connected to 294 and needs to connect to another number. The remaining number is 42, but 42 is already connected to 21 and 3. Alternatively, maybe 147 should connect to 42 instead of 294.Wait, let's see:If 147 is connected to 42, then 42 is between 21 and 3 and 147.But 42 is already connected to 21 and 3, so adding 147 would make it have three connections, which isn't possible in a circle where each node has exactly two connections.So, that's not feasible.Alternatively, maybe 147 should connect to 49 and 294, and 294 connects back to 21? But 21 is already connected to 42 and 7.Wait, 294 is connected to 98 and 147, and 147 is connected to 294 and... Maybe 147 should connect to 49 and 294, and 49 connects to 14 and 147, and 14 connects to 2 and 49, and 2 connects to 6 and 14, and 6 connects to 3 and 2, and 3 connects to 42 and 6, and 42 connects to 21 and 3, and 21 connects to 42 and 7, and 7 connects to 21 and... Wait, 7 is only connected to 21 and 14? No, 7 is connected to 21 and 14, but 14 is connected to 2 and 49, and 49 is connected to 14 and 98, and 98 is connected to 49 and 294, and 294 is connected to 98 and 147, and 147 is connected to 294 and... Hmm, 147 is only connected to 294 and needs another connection.Wait, maybe 147 should connect to 42 instead of 294. Let's try that.So, 147 is connected to 42 and 294.Then, 42 is connected to 21, 3, and 147. But that's three connections, which isn't possible.Hmm, this is tricky.Maybe I need to consider that 147 can only connect to two numbers, and those are 49 and 294. So, 147 is between 49 and 294.Then, 294 is between 147 and 98.98 is between 294 and 49.49 is between 98 and 14.14 is between 49 and 2.2 is between 14 and 6.6 is between 2 and 3.3 is between 6 and 42.42 is between 3 and 21.21 is between 42 and 7.7 is between 21 and... Wait, 7 is only connected to 21 and 14. But 14 is already connected to 2 and 49. So, 7 is only connected to 21 and 14, but in the circle, it needs to connect to two numbers. So, 7 is connected to 21 and 14, which is fine.But then, how does the circle close? Let's see:Starting from 21, connected to 42 and 7.42 connected to 21 and 3.3 connected to 42 and 6.6 connected to 3 and 2.2 connected to 6 and 14.14 connected to 2 and 49.49 connected to 14 and 98.98 connected to 49 and 294.294 connected to 98 and 147.147 connected to 294 and... Hmm, 147 needs to connect back to someone. It's connected to 294 and needs to connect to another number. The only number left is 42, but 42 is already connected to 21 and 3.Wait, maybe 147 should connect to 42 instead of 294.But then 42 would have three connections: 21, 3, and 147, which isn't allowed.Alternatively, maybe 147 connects to 49 and 294, and 294 connects back to 21? But 21 is already connected to 42 and 7.Wait, 294 is connected to 98 and 147, and 147 is connected to 294 and 49, and 49 is connected to 147 and 14, and 14 is connected to 49 and 2, and 2 is connected to 14 and 6, and 6 is connected to 2 and 3, and 3 is connected to 6 and 42, and 42 is connected to 3 and 21, and 21 is connected to 42 and 7, and 7 is connected to 21 and 14. Wait, 7 is connected to 21 and 14, but 14 is already connected to 2 and 49. So, 7 is connected to 21 and 14, which is fine.But then, how does the circle close? It seems like we're missing a connection from 7 back to someone. 7 is connected to 21 and 14, but 14 is connected to 2 and 49, and 2 is connected to 6 and 14, and so on. It seems like the circle is already closed without needing 7 to connect back to someone else.Wait, no, in a circle, every node must have exactly two connections. So, 7 is connected to 21 and 14, which is fine. 21 is connected to 7 and 42, which is fine. 42 is connected to 21 and 3, which is fine. 3 is connected to 42 and 6, which is fine. 6 is connected to 3 and 2, which is fine. 2 is connected to 6 and 14, which is fine. 14 is connected to 2 and 49, which is fine. 49 is connected to 14 and 98, which is fine. 98 is connected to 49 and 294, which is fine. 294 is connected to 98 and 147, which is fine. 147 is connected to 294 and... Wait, 147 is only connected to 294 and 49? No, 147 is connected to 294 and 49, but 49 is already connected to 14 and 98. So, 147 is connected to 294 and 49, which is fine.Wait, but in this arrangement, 147 is connected to 294 and 49, and 49 is connected to 147 and 14, and 14 is connected to 49 and 2, and so on. So, the circle is closed because 147 connects back to 49, which connects back to 14, and so on until it loops back to 21.Wait, but in this case, the circle would be:21 - 42 - 3 - 6 - 2 - 14 - 49 - 98 - 294 - 147 - 49 - ... Wait, no, that's not right because 49 is already connected to 14 and 98, and 147 is connected to 294 and 49. So, the circle would be:21 - 42 - 3 - 6 - 2 - 14 - 49 - 98 - 294 - 147 - 49 - ... Wait, that's a problem because 49 is being connected twice, which isn't allowed in a simple circle.Ah, I see the issue. 49 is connected to both 14 and 98, and also to 147, which would make it have three connections. That's not possible in a simple cycle where each node has exactly two connections.So, this suggests that my current arrangement is invalid because 49 would have three connections. Therefore, I need to adjust the connections to ensure that each node has exactly two connections.Let me try a different approach. Maybe 147 should connect to 42 instead of 49.So, 147 is connected to 42 and 294.Then, 42 is connected to 21, 3, and 147. Again, three connections, which isn't allowed.Hmm, maybe 147 should connect to 49 and 294, and 49 connects to 14 and 147, and 14 connects to 2 and 49, and 2 connects to 6 and 14, and 6 connects to 2 and 3, and 3 connects to 6 and 42, and 42 connects to 3 and 21, and 21 connects to 42 and 7, and 7 connects to 21 and 14. Wait, but 7 is connected to 21 and 14, and 14 is connected to 7 and 2, and so on. But then, 147 is connected to 49 and 294, and 294 is connected to 147 and 98, and 98 is connected to 294 and 49, and 49 is connected to 98 and 14, and 14 is connected to 49 and 2, and 2 is connected to 14 and 6, and 6 is connected to 2 and 3, and 3 is connected to 6 and 42, and 42 is connected to 3 and 21, and 21 is connected to 42 and 7, and 7 is connected to 21 and 14. Wait, this seems to form a circle, but let's check the connections:- 21: 42, 7- 42: 21, 3- 3: 42, 6- 6: 3, 2- 2: 6, 14- 14: 2, 49- 49: 14, 98- 98: 49, 294- 294: 98, 147- 147: 294, 49- 7: 21, 14Wait, but 49 is connected to 14, 98, and 147, which is three connections. That's not allowed. So, this arrangement is invalid.I think the problem is that 49 is connected to three numbers: 14, 98, and 147. To fix this, maybe 147 should connect to a different number instead of 49.Wait, 147 is 3 × 7², so it can connect to numbers with 3 or 7. If it doesn't connect to 49, it could connect to 3 or 21 or 42 or 294.But 21 is already connected to 42 and 7, so connecting 147 to 21 would give 21 three connections, which isn't allowed.Similarly, connecting 147 to 42 would give 42 three connections.Connecting 147 to 3 would give 3 three connections.So, the only option is to connect 147 to 294, which is already connected to 98 and 147, making 294 have three connections, which isn't allowed.Hmm, this is a dead end. Maybe I need to adjust earlier connections to free up 49.Wait, perhaps 49 should connect to 7 instead of 14.Let me try that.So, 49 is connected to 7 and 98.Then, 7 is connected to 21 and 49.21 is connected to 7 and 42.42 is connected to 21 and 3.3 is connected to 42 and 6.6 is connected to 3 and 2.2 is connected to 6 and 14.14 is connected to 2 and... Wait, 14 needs to connect to someone else. It can connect to 49, but 49 is already connected to 7 and 98. So, 14 can't connect to 49. Alternatively, 14 can connect to 294.So, 14 is connected to 2 and 294.294 is connected to 14 and 98.98 is connected to 294 and 49.49 is connected to 98 and 7.7 is connected to 49 and 21.21 is connected to 7 and 42.42 is connected to 21 and 3.3 is connected to 42 and 6.6 is connected to 3 and 2.2 is connected to 6 and 14.14 is connected to 2 and 294.294 is connected to 14 and 98.98 is connected to 294 and 49.49 is connected to 98 and 7.7 is connected to 49 and 21.This seems to form a circle without any nodes having more than two connections. Let's check:- 21: 7, 42- 42: 21, 3- 3: 42, 6- 6: 3, 2- 2: 6, 14- 14: 2, 294- 294: 14, 98- 98: 294, 49- 49: 98, 7- 7: 49, 21Yes, each node has exactly two connections, and the circle is closed. So, this arrangement works.Now, the question is, what are the two integers adjacent to 21? Looking at the arrangement, 21 is connected to 7 and 42. So, the two integers adjacent to 21 are 7 and 42.Therefore, the sum of these two integers is 7 + 42 = 49.Wait, but earlier I thought the answer was 49, but in this arrangement, it's 7 and 42. So, the sum is 49. That matches my initial thought.But let me double-check to make sure there isn't another possible arrangement where 21 is connected to different numbers.Suppose instead of connecting 21 to 7 and 42, we connect it to 7 and 49.But 49 is 7², so it shares a factor of 7 with 21. So, 21 could be connected to 7 and 49.Then, 49 would need to connect to another number, say 98.So, 49 is connected to 21 and 98.98 is connected to 49 and 294.294 is connected to 98 and 147.147 is connected to 294 and... Hmm, 147 needs to connect to another number. It can connect to 42, but 42 is connected to 21 and 3.Alternatively, 147 could connect to 3.So, 147 is connected to 294 and 3.3 is connected to 147 and 6.6 is connected to 3 and 2.2 is connected to 6 and 14.14 is connected to 2 and... 14 needs to connect to someone else. It can connect to 42.So, 14 is connected to 2 and 42.42 is connected to 14 and 21.21 is connected to 42 and 7.7 is connected to 21 and 49.49 is connected to 7 and 98.98 is connected to 49 and 294.294 is connected to 98 and 147.147 is connected to 294 and 3.3 is connected to 147 and 6.6 is connected to 3 and 2.2 is connected to 6 and 14.14 is connected to 2 and 42.42 is connected to 14 and 21.21 is connected to 42 and 7.7 is connected to 21 and 49.This also forms a circle, and in this case, 21 is connected to 42 and 7, same as before. So, the sum is still 49.Alternatively, if 21 is connected to 49 and 7, then the sum would be 49 + 7 = 56, but in this arrangement, 21 is connected to 42 and 7, so the sum is 49.Wait, but in the second arrangement, 21 is connected to 42 and 7, same as the first arrangement. So, regardless of how we arrange the circle, 21 seems to be connected to 7 and 42.Therefore, the sum of the two integers adjacent to 21 is 7 + 42 = 49.

answer：Okay, so I have this problem here where there are three points on a line: (1, 2), (3, k), and (10, 5). I need to figure out what k is. Hmm, let me think about how to approach this.First off, if three points are on the same line, that means they're collinear. I remember that for points to be collinear, the slope between any two pairs of points should be the same. So, I guess I need to calculate the slope between (1, 2) and (3, k), and then between (3, k) and (10, 5), and set them equal to each other. That should give me an equation to solve for k.Alright, let's start with the slope between (1, 2) and (3, k). The formula for slope is (y2 - y1)/(x2 - x1). So, plugging in the values, it would be (k - 2)/(3 - 1). Simplifying that, the denominator is 2, so the slope is (k - 2)/2.Next, I'll calculate the slope between (3, k) and (10, 5). Using the same slope formula, that would be (5 - k)/(10 - 3). The denominator here is 7, so the slope is (5 - k)/7.Now, since these two slopes must be equal for the points to be collinear, I can set them equal to each other:(k - 2)/2 = (5 - k)/7Okay, now I need to solve for k. To do that, I can cross-multiply to get rid of the fractions. So, multiplying both sides by 2 and 7, I get:7(k - 2) = 2(5 - k)Let me expand both sides:7k - 14 = 10 - 2kHmm, now I need to get all the k terms on one side and the constants on the other. So, I'll add 2k to both sides:7k + 2k - 14 = 10Which simplifies to:9k - 14 = 10Now, I'll add 14 to both sides to isolate the term with k:9k = 24Alright, now I can solve for k by dividing both sides by 9:k = 24/9Simplifying that fraction, both numerator and denominator are divisible by 3:k = 8/3So, k should be 8/3. Let me just double-check my steps to make sure I didn't make any mistakes.First, I calculated the slopes correctly between the points. Then, I set them equal and cross-multiplied, which seems right. Expanding both sides gave me 7k -14 = 10 -2k. Adding 2k to both sides gives 9k -14 =10, then adding 14 to both sides gives 9k=24, so k=24/9=8/3. Yeah, that seems consistent.I think that's the correct value for k. It makes sense because if I plug it back into the slopes, they should be equal. Let me quickly verify:Slope between (1,2) and (3,8/3):(8/3 - 2)/(3 -1) = (8/3 - 6/3)/2 = (2/3)/2 = 1/3Slope between (3,8/3) and (10,5):(5 - 8/3)/(10 -3) = (15/3 -8/3)/7 = (7/3)/7 = 1/3Yep, both slopes are 1/3, so they are equal. That confirms that k=8/3 is correct.I don't think I made any calculation errors, and the logic seems solid. So, I'm confident that k is 8/3.

answer：Okay, so I have this problem about twelve students participating in a theater festival with n different performances. Each performance has six students, and any two performances share at most two students. I need to find the largest possible value of n. Hmm, that sounds a bit like a combinatorial problem, maybe something to do with block designs or something similar.Let me try to break it down. There are twelve students, and each performance involves six of them. So, each performance is like a subset of six students from the twelve. The condition is that any two performances share at most two students. So, the intersection of any two subsets is at most two students. I need to figure out how many such subsets can exist under this condition.I remember something called the Fisher's inequality or maybe the Erdos-Ko-Rado theorem, but I'm not sure if that applies here. Maybe it's more about projective planes or something else. Wait, maybe it's related to combinatorial designs, specifically a type of block design where each block has size six, and the intersection of any two blocks is limited.Let me think about the parameters. We have v = 12 students, block size k = 6, and the maximum intersection λ = 2. I think in design theory, there are bounds on the number of blocks given these parameters. Maybe I can use the Fisher's inequality or the Johnson bound.Wait, let me recall the Fisher's inequality. It states that in a certain type of design, the number of blocks is at least the number of elements. But here, we have more blocks potentially, so maybe that's not directly applicable.Alternatively, maybe I can use the Ray-Chaudhuri-Wilson theorem, which gives bounds on the size of set systems with bounded intersections. That might be more relevant here. The theorem states that if we have a family of k-element subsets from a v-element set, and the intersection of any two subsets is at most t, then the maximum number of subsets is bounded by something like (binom{v}{t}) divided by (binom{k}{t}). Let me check the exact statement.Wait, actually, the theorem says that if the intersection of any two subsets is at most t, then the maximum number of subsets is at most (binom{v}{t}). But I think that's when t is fixed. Hmm, maybe I need to use a different approach.Alternatively, I can think about counting the number of pairs of students and how they appear in the performances. Each performance has (binom{6}{2} = 15) pairs of students. Since there are n performances, the total number of pairs counted across all performances is (15n). However, since each pair of students can appear together in at most some number of performances, let's say λ, then the total number of pairs is also bounded by (binom{12}{2} times lambda).But wait, in our case, the condition is that any two performances share at most two students, which is a bit different. It's about the intersection of two performances, not the number of times a pair appears together.Hmm, maybe I need to use double counting or some inequality here. Let me consider the number of triples (P, Q, X) where P and Q are performances, and X is a student in both P and Q. Let's denote this number as T. On one hand, for each pair of performances P and Q, the number of common students is at most 2, so the total number of such triples is at most (2 times binom{n}{2}).On the other hand, for each student X, let d_X be the number of performances that X is in. Then, the number of triples involving X is (binom{d_X}{2}), since for each pair of performances that X is in, we have a triple. So, the total number of triples T is also equal to (sum_{X=1}^{12} binom{d_X}{2}).Putting these together, we have:[sum_{X=1}^{12} binom{d_X}{2} leq 2 times binom{n}{2}]Now, since each performance has six students, the total number of student participations is (6n). Therefore, the average number of performances per student is (frac{6n}{12} = frac{n}{2}). So, on average, each student is in (frac{n}{2}) performances.Using Jensen's inequality, since (binom{d_X}{2}) is a convex function in d_X, the sum (sum binom{d_X}{2}) is minimized when all d_X are equal. So, if each student is in exactly (frac{n}{2}) performances, then:[sum_{X=1}^{12} binom{d_X}{2} = 12 times binom{frac{n}{2}}{2} = 12 times frac{frac{n}{2} times (frac{n}{2} - 1)}{2} = 12 times frac{n(n - 2)}{8} = frac{12n(n - 2)}{8} = frac{3n(n - 2)}{2}]So, we have:[frac{3n(n - 2)}{2} leq 2 times frac{n(n - 1)}{2} = n(n - 1)]Simplifying this inequality:[frac{3n(n - 2)}{2} leq n(n - 1)]Multiply both sides by 2:[3n(n - 2) leq 2n(n - 1)]Expand both sides:[3n^2 - 6n leq 2n^2 - 2n]Subtract (2n^2 - 2n) from both sides:[n^2 - 4n leq 0]Factor:[n(n - 4) leq 0]This inequality holds when (n leq 4), since n is positive. Therefore, the maximum possible value of n is 4.Wait, but is this tight? Can we actually construct 4 performances that satisfy the conditions? Let me try to construct such a set.Let's label the students as S1, S2, ..., S12.Performance 1: S1, S2, S3, S4, S5, S6Performance 2: S1, S2, S7, S8, S9, S10Performance 3: S3, S4, S7, S8, S11, S12Performance 4: S5, S6, S9, S10, S11, S12Now, let's check the intersections:P1 & P2: S1, S2 (2 students)P1 & P3: S3, S4 (2 students)P1 & P4: S5, S6 (2 students)P2 & P3: S7, S8 (2 students)P2 & P4: S9, S10 (2 students)P3 & P4: S11, S12 (2 students)Perfect! Each pair of performances shares exactly two students, which satisfies the condition of at most two. So, n=4 is achievable.Is it possible to have n=5? Let's see. If n=5, then each student would be in ( frac{5}{2} = 2.5 ) performances on average, which isn't possible since the number of performances must be an integer. So, some students would be in 2 performances, and others in 3.But let's check the inequality again. If n=5:Left side: ( frac{3 times 5 times (5 - 2)}{2} = frac{45}{2} = 22.5 )Right side: ( 5 times 4 = 20 )So, 22.5 ≤ 20 is false. Therefore, n=5 is not possible.Hence, the maximum n is 4.

answer：Okay, so I have this inequality to solve: ( f(x) = 3sqrt{2}sinleft(frac{x}{4}right)cosleft(frac{x}{4}right) + sqrt{6}cos^2left(frac{x}{4}right) - frac{sqrt{6}}{2} + m leq 0 )And it's supposed to hold for any ( x ) in the interval ( -frac{5pi}{6} leq x leq frac{pi}{6} ). I need to find the range of the real number ( m ) that satisfies this condition. The options are given as A, B, C, D, which are different ranges for ( m ).First, I think I need to simplify the expression for ( f(x) ) as much as possible. Maybe using some trigonometric identities to make it easier to analyze.Looking at the first term: ( 3sqrt{2}sinleft(frac{x}{4}right)cosleft(frac{x}{4}right) ). I remember that ( sin(2theta) = 2sinthetacostheta ), so maybe I can rewrite this term using that identity.Let me set ( theta = frac{x}{4} ), so ( 2theta = frac{x}{2} ). Then, ( sinleft(frac{x}{2}right) = 2sinleft(frac{x}{4}right)cosleft(frac{x}{4}right) ). Therefore, ( sinleft(frac{x}{4}right)cosleft(frac{x}{4}right) = frac{1}{2}sinleft(frac{x}{2}right) ).So, substituting back into the first term:( 3sqrt{2}sinleft(frac{x}{4}right)cosleft(frac{x}{4}right) = 3sqrt{2} times frac{1}{2}sinleft(frac{x}{2}right) = frac{3sqrt{2}}{2}sinleft(frac{x}{2}right) ).Okay, that simplifies the first term. Now, moving on to the second term: ( sqrt{6}cos^2left(frac{x}{4}right) ). I recall that ( cos^2theta = frac{1 + cos(2theta)}{2} ). Let me apply that here.Let ( theta = frac{x}{4} ), so ( 2theta = frac{x}{2} ). Then,( cos^2left(frac{x}{4}right) = frac{1 + cosleft(frac{x}{2}right)}{2} ).Substituting back into the second term:( sqrt{6}cos^2left(frac{x}{4}right) = sqrt{6} times frac{1 + cosleft(frac{x}{2}right)}{2} = frac{sqrt{6}}{2} + frac{sqrt{6}}{2}cosleft(frac{x}{2}right) ).So, now I can rewrite the entire function ( f(x) ) with these simplifications:( f(x) = frac{3sqrt{2}}{2}sinleft(frac{x}{2}right) + frac{sqrt{6}}{2} + frac{sqrt{6}}{2}cosleft(frac{x}{2}right) - frac{sqrt{6}}{2} + m ).Wait, I see that ( frac{sqrt{6}}{2} ) and ( -frac{sqrt{6}}{2} ) cancel each other out. So, simplifying further:( f(x) = frac{3sqrt{2}}{2}sinleft(frac{x}{2}right) + frac{sqrt{6}}{2}cosleft(frac{x}{2}right) + m ).Hmm, so now ( f(x) ) is expressed as a combination of sine and cosine functions with the same argument ( frac{x}{2} ). Maybe I can combine these into a single sine or cosine function using the amplitude-phase form. That is, express ( Asintheta + Bcostheta ) as ( Rsin(theta + phi) ) or ( Rcos(theta + phi) ), where ( R = sqrt{A^2 + B^2} ) and ( phi ) is some phase shift.Let me try that. Let me denote:( A = frac{3sqrt{2}}{2} ) and ( B = frac{sqrt{6}}{2} ).So, ( R = sqrt{A^2 + B^2} = sqrt{left(frac{3sqrt{2}}{2}right)^2 + left(frac{sqrt{6}}{2}right)^2} ).Calculating each term:( left(frac{3sqrt{2}}{2}right)^2 = frac{9 times 2}{4} = frac{18}{4} = frac{9}{2} ).( left(frac{sqrt{6}}{2}right)^2 = frac{6}{4} = frac{3}{2} ).So, ( R = sqrt{frac{9}{2} + frac{3}{2}} = sqrt{frac{12}{2}} = sqrt{6} ).Okay, so ( R = sqrt{6} ).Now, to find the phase shift ( phi ), we can use:( tanphi = frac{B}{A} = frac{frac{sqrt{6}}{2}}{frac{3sqrt{2}}{2}} = frac{sqrt{6}}{3sqrt{2}} = frac{sqrt{6}/sqrt{2}}{3} = frac{sqrt{3}}{3} ).So, ( tanphi = frac{sqrt{3}}{3} ), which means ( phi = frac{pi}{6} ) because ( tan(pi/6) = frac{sqrt{3}}{3} ).Therefore, we can write:( Asintheta + Bcostheta = Rsinleft(theta + phiright) ).Wait, actually, I think it's either ( Rsin(theta + phi) ) or ( Rcos(theta - phi) ). Let me verify.The identity is:( Asintheta + Bcostheta = Rsinleft(theta + phiright) ), where ( R = sqrt{A^2 + B^2} ) and ( phi = arctanleft(frac{B}{A}right) ).Alternatively, it can also be written as ( Rcosleft(theta - phiright) ), depending on the phase shift.But in any case, since ( phi = frac{pi}{6} ), let's proceed.So, substituting back, we have:( f(x) = sqrt{6}sinleft(frac{x}{2} + frac{pi}{6}right) + m ).Wait, let me check that. If I write ( Asintheta + Bcostheta = Rsin(theta + phi) ), then:( Rsin(theta + phi) = Rsinthetacosphi + Rcosthetasinphi ).Comparing with ( Asintheta + Bcostheta ), we have:( A = Rcosphi ) and ( B = Rsinphi ).Given that ( A = frac{3sqrt{2}}{2} ) and ( B = frac{sqrt{6}}{2} ), and ( R = sqrt{6} ), let's compute ( cosphi ) and ( sinphi ):( cosphi = frac{A}{R} = frac{frac{3sqrt{2}}{2}}{sqrt{6}} = frac{3sqrt{2}}{2sqrt{6}} = frac{3sqrt{2}}{2sqrt{6}} times frac{sqrt{6}}{sqrt{6}} = frac{3sqrt{12}}{12} = frac{3 times 2sqrt{3}}{12} = frac{sqrt{3}}{2} ).Similarly, ( sinphi = frac{B}{R} = frac{frac{sqrt{6}}{2}}{sqrt{6}} = frac{1}{2} ).So, ( cosphi = frac{sqrt{3}}{2} ) and ( sinphi = frac{1}{2} ), which corresponds to ( phi = frac{pi}{6} ). So, that's correct.Therefore, ( f(x) = sqrt{6}sinleft(frac{x}{2} + frac{pi}{6}right) + m ).So, the inequality becomes:( sqrt{6}sinleft(frac{x}{2} + frac{pi}{6}right) + m leq 0 ).We need this to hold for all ( x ) in ( -frac{5pi}{6} leq x leq frac{pi}{6} ).So, let's denote ( y = frac{x}{2} + frac{pi}{6} ). Then, we can express the inequality in terms of ( y ):( sqrt{6}sin(y) + m leq 0 ).We need this to be true for all ( y ) corresponding to ( x ) in the given interval.Let's find the range of ( y ):Given ( x in left[-frac{5pi}{6}, frac{pi}{6}right] ),( frac{x}{2} in left[-frac{5pi}{12}, frac{pi}{12}right] ),So, ( y = frac{x}{2} + frac{pi}{6} in left[-frac{5pi}{12} + frac{pi}{6}, frac{pi}{12} + frac{pi}{6}right] ).Calculating the endpoints:Lower bound: ( -frac{5pi}{12} + frac{pi}{6} = -frac{5pi}{12} + frac{2pi}{12} = -frac{3pi}{12} = -frac{pi}{4} ).Upper bound: ( frac{pi}{12} + frac{pi}{6} = frac{pi}{12} + frac{2pi}{12} = frac{3pi}{12} = frac{pi}{4} ).So, ( y in left[-frac{pi}{4}, frac{pi}{4}right] ).Therefore, the inequality becomes:( sqrt{6}sin(y) + m leq 0 ) for all ( y in left[-frac{pi}{4}, frac{pi}{4}right] ).We need to find the range of ( m ) such that this inequality holds for all ( y ) in that interval.Let me rearrange the inequality:( m leq -sqrt{6}sin(y) ).So, ( m ) must be less than or equal to ( -sqrt{6}sin(y) ) for all ( y ) in ( left[-frac{pi}{4}, frac{pi}{4}right] ).To find the maximum possible value of ( m ) that satisfies this for all ( y ), we need to find the minimum value of ( -sqrt{6}sin(y) ) over ( y in left[-frac{pi}{4}, frac{pi}{4}right] ). Because ( m ) has to be less than or equal to the smallest value of ( -sqrt{6}sin(y) ) to satisfy the inequality for all ( y ).Wait, actually, let me think carefully. If ( m leq -sqrt{6}sin(y) ) for all ( y ), then ( m ) must be less than or equal to the minimum of ( -sqrt{6}sin(y) ) over the interval. Because if ( m ) is less than or equal to the smallest value of ( -sqrt{6}sin(y) ), then it will automatically be less than or equal to all larger values.Alternatively, if we consider the maximum of ( -sqrt{6}sin(y) ), that would be the least upper bound for ( m ). Wait, perhaps I need to clarify.Let me consider the function ( g(y) = -sqrt{6}sin(y) ) over ( y in left[-frac{pi}{4}, frac{pi}{4}right] ). We need ( m leq g(y) ) for all ( y ) in that interval. So, the maximum value that ( m ) can take is the minimum value of ( g(y) ) over the interval. Because if ( m ) is less than or equal to the smallest ( g(y) ), then it's certainly less than or equal to all other ( g(y) ).Wait, that seems contradictory. Let me think again.Suppose ( g(y) ) has a maximum value ( M ) and a minimum value ( m_{text{min}} ) over the interval. If we require ( m leq g(y) ) for all ( y ), then ( m ) must be less than or equal to the minimum of ( g(y) ). Because if ( m ) is greater than the minimum, there exists some ( y ) where ( g(y) ) is equal to the minimum, and ( m ) would not satisfy ( m leq g(y) ) there.Wait, actually, no. Let me think in terms of inequalities.If ( m leq g(y) ) for all ( y ), then ( m ) must be less than or equal to the smallest value that ( g(y) ) takes in the interval. Because if ( m ) is larger than the smallest ( g(y) ), then at the point where ( g(y) ) is smallest, ( m ) would be larger, violating the inequality.Therefore, the maximum possible ( m ) is the minimum of ( g(y) ) over the interval.So, I need to find the minimum of ( g(y) = -sqrt{6}sin(y) ) over ( y in left[-frac{pi}{4}, frac{pi}{4}right] ).First, let's analyze ( sin(y) ) over ( y in left[-frac{pi}{4}, frac{pi}{4}right] ).The sine function is increasing on ( left[-frac{pi}{2}, frac{pi}{2}right] ), so on ( left[-frac{pi}{4}, frac{pi}{4}right] ), it's also increasing.Therefore, the minimum value of ( sin(y) ) occurs at ( y = -frac{pi}{4} ), and the maximum occurs at ( y = frac{pi}{4} ).Calculating these:( sinleft(-frac{pi}{4}right) = -frac{sqrt{2}}{2} ).( sinleft(frac{pi}{4}right) = frac{sqrt{2}}{2} ).Therefore, ( sin(y) ) ranges from ( -frac{sqrt{2}}{2} ) to ( frac{sqrt{2}}{2} ) over the interval.So, ( g(y) = -sqrt{6}sin(y) ) will have:Minimum value when ( sin(y) ) is maximum, because of the negative sign.Similarly, maximum value when ( sin(y) ) is minimum.So, the minimum of ( g(y) ) is when ( sin(y) ) is maximum, which is ( frac{sqrt{2}}{2} ).Thus, ( g_{text{min}} = -sqrt{6} times frac{sqrt{2}}{2} = -sqrt{6} times frac{sqrt{2}}{2} ).Calculating that:( sqrt{6} times sqrt{2} = sqrt{12} = 2sqrt{3} ).Therefore, ( g_{text{min}} = -frac{2sqrt{3}}{2} = -sqrt{3} ).Similarly, the maximum of ( g(y) ) is when ( sin(y) ) is minimum, which is ( -frac{sqrt{2}}{2} ).Thus, ( g_{text{max}} = -sqrt{6} times left(-frac{sqrt{2}}{2}right) = sqrt{6} times frac{sqrt{2}}{2} = frac{2sqrt{3}}{2} = sqrt{3} ).So, ( g(y) ) ranges from ( -sqrt{3} ) to ( sqrt{3} ) over the interval ( y in left[-frac{pi}{4}, frac{pi}{4}right] ).But remember, our inequality is ( m leq g(y) ) for all ( y ) in that interval. So, ( m ) must be less than or equal to the minimum value of ( g(y) ), which is ( -sqrt{3} ).Therefore, ( m leq -sqrt{3} ).Looking back at the options:A: ( m geq sqrt{3} )B: ( m leq sqrt{3} )C: ( m leq -sqrt{3} )D: ( -sqrt{3} leq m leq sqrt{3} )So, the correct option is C: ( m leq -sqrt{3} ).Wait, let me double-check my reasoning because sometimes when dealing with inequalities, especially with trigonometric functions, it's easy to make a mistake.We have ( f(x) = sqrt{6}sinleft(frac{x}{2} + frac{pi}{6}right) + m leq 0 ).We transformed the original expression correctly, right? Let me verify:Original ( f(x) ):( 3sqrt{2}sinleft(frac{x}{4}right)cosleft(frac{x}{4}right) + sqrt{6}cos^2left(frac{x}{4}right) - frac{sqrt{6}}{2} + m ).We used the identity ( 2sinthetacostheta = sin(2theta) ) to rewrite the first term as ( frac{3sqrt{2}}{2}sinleft(frac{x}{2}right) ).Then, for the second term, we used ( cos^2theta = frac{1 + cos(2theta)}{2} ), which gave us ( frac{sqrt{6}}{2} + frac{sqrt{6}}{2}cosleft(frac{x}{2}right) ).Subtracting ( frac{sqrt{6}}{2} ) cancels out the added constant, leaving us with ( frac{3sqrt{2}}{2}sinleft(frac{x}{2}right) + frac{sqrt{6}}{2}cosleft(frac{x}{2}right) + m ).Then, combining these two terms into a single sine function with amplitude ( sqrt{6} ) and phase shift ( frac{pi}{6} ), resulting in ( sqrt{6}sinleft(frac{x}{2} + frac{pi}{6}right) + m ).That seems correct.Then, defining ( y = frac{x}{2} + frac{pi}{6} ), we found the range of ( y ) as ( left[-frac{pi}{4}, frac{pi}{4}right] ).So, ( f(x) = sqrt{6}sin(y) + m leq 0 ).Therefore, ( m leq -sqrt{6}sin(y) ).We need this to hold for all ( y ) in ( left[-frac{pi}{4}, frac{pi}{4}right] ).So, ( m ) must be less than or equal to the minimum of ( -sqrt{6}sin(y) ) over that interval.As ( sin(y) ) ranges from ( -frac{sqrt{2}}{2} ) to ( frac{sqrt{2}}{2} ), ( -sqrt{6}sin(y) ) ranges from ( -sqrt{6} times frac{sqrt{2}}{2} = -sqrt{3} ) to ( sqrt{3} ).Therefore, the minimum value of ( -sqrt{6}sin(y) ) is ( -sqrt{3} ).Hence, ( m leq -sqrt{3} ).Yes, that seems consistent.Therefore, the correct answer is option C: ( m leq -sqrt{3} ).**Final Answer**boxed{C}