💡Okay, so I have this problem where I need to arrange the numbers 1 through 7 in a circle. The goal is to maximize the number of "good" numbers. A "good" number is one that equals the sum of its two neighbors. Hmm, interesting.First, I need to understand what makes a number "good." If a number is good, it means that the two numbers next to it add up to it. So, for example, if I have numbers a, b, c in a row, and b is good, then b = a + c.Since the numbers are arranged in a circle, every number has exactly two neighbors. So, each number could potentially be good if the sum of its two neighbors equals it. But I know that not all numbers can be good because the numbers are fixed from 1 to 7, and their sums have to fit within that range.Let me think about the possible sums. The smallest number is 1, and the largest is 7. If I want a number to be good, it has to be the sum of two smaller numbers because the numbers are positive integers. So, for example, 7 could be good if its neighbors are 3 and 4 because 3 + 4 = 7. Similarly, 6 could be good if its neighbors are 2 and 4 because 2 + 4 = 6.But wait, if I use 3 and 4 for 7, I can't use them again for another number. So, I have to be careful about how I arrange the numbers to maximize the number of good numbers without overlapping the neighbors.Let me try to list out possible good numbers and their required neighbors:- 7: needs neighbors that add up to 7. Possible pairs: (1,6), (2,5), (3,4)- 6: needs neighbors that add up to 6. Possible pairs: (1,5), (2,4)- 5: needs neighbors that add up to 5. Possible pairs: (1,4), (2,3)- 4: needs neighbors that add up to 4. Possible pairs: (1,3)- 3: needs neighbors that add up to 3. Possible pairs: (1,2)- 2 and 1: can't be good because their neighbors would have to be negative or zero, which isn't possible.So, the potential good numbers are 7, 6, 5, 4, and 3. But I need to see how many of these I can fit into the circle without overlapping neighbors.Let's start by trying to make 7 good. If 7 is good, its neighbors could be 3 and 4. So, let's say the arrangement is ...3,7,4...Now, let's see if 6 can be good. If 6 is good, its neighbors could be 2 and 4. But 4 is already a neighbor of 7, so I can't use it again. Alternatively, 6 could be between 1 and 5. Let's try that. So, maybe ...1,6,5...Now, let's see if 5 can be good. If 5 is good, its neighbors could be 2 and 3. But 3 is already a neighbor of 7, so that's a conflict. Alternatively, 5 could be between 1 and 4, but 4 is already next to 7. Hmm, this is tricky.Maybe I should try a different approach. Let's try to place the numbers in a way that allows multiple good numbers without overlapping neighbors.Suppose I start with 2, then 7, then 5. So, 2,7,5. Now, 7 is good because 2 + 5 = 7. Next, after 5, let's place 6. So, 2,7,5,6. Now, 6 needs to be good, so its neighbors should add up to 6. The neighbors of 6 are 5 and the next number. 5 + x = 6, so x = 1. So, let's place 1 after 6. Now we have 2,7,5,6,1.Next, after 1, let's place 4. So, 2,7,5,6,1,4. Now, 4 needs to be good, so its neighbors should add up to 4. The neighbors are 1 and the next number. 1 + x = 4, so x = 3. Let's place 3 after 4. Now we have 2,7,5,6,1,4,3.Finally, we need to close the circle by connecting back to 2. So, the last number is 3, and its neighbors are 4 and 2. Let's check if 3 is good: 4 + 2 = 6, which is not equal to 3. So, 3 is not good. That's okay.Now, let's check which numbers are good:- 7: neighbors 2 and 5. 2 + 5 = 7. Good.- 6: neighbors 5 and 1. 5 + 1 = 6. Good.- 4: neighbors 1 and 3. 1 + 3 = 4. Good.So, in this arrangement, we have 3 good numbers: 7, 6, and 4.Is it possible to have more than 3 good numbers? Let's try.Suppose I try to make 5 good as well. If 5 is good, its neighbors should add up to 5. Let's see if we can adjust the arrangement.Starting with 2,7,5,6,1,4,3 as before. If I want 5 to be good, its neighbors need to add up to 5. Currently, 5 is between 7 and 6. 7 + 6 = 13, which is not 5. So, 5 is not good here.If I rearrange to make 5 good, maybe place 2 and 3 next to 5. So, 2,5,3. Then 5 is good because 2 + 3 = 5. Now, let's try to build the circle.Starting with 2,5,3. Next, let's place 7. So, 2,5,3,7. Now, 7 needs to be good, so its neighbors should add up to 7. The neighbors are 3 and the next number. 3 + x = 7, so x = 4. Place 4 after 7. Now we have 2,5,3,7,4.Next, after 4, let's place 6. So, 2,5,3,7,4,6. Now, 6 needs to be good, so its neighbors should add up to 6. The neighbors are 4 and the next number. 4 + x = 6, so x = 2. But 2 is already at the beginning. Wait, we can't have duplicates. So, this doesn't work.Alternatively, after 4, place 1. So, 2,5,3,7,4,1. Now, 1 needs to be good, but 1 can't be good because its neighbors would have to be 0 and something, which isn't possible. So, 1 is not good.Then, after 1, place 6. So, 2,5,3,7,4,1,6. Now, 6 needs to be good. Its neighbors are 1 and 2 (since it's a circle). 1 + 2 = 3, which is not equal to 6. So, 6 is not good.In this arrangement, only 5 and 7 are good. That's worse than before.Maybe another arrangement. Let's try making 3 good. If 3 is good, its neighbors should add up to 3. So, neighbors could be 1 and 2. Let's try that.Start with 1,3,2. Now, 3 is good because 1 + 2 = 3. Next, place 7. So, 1,3,2,7. 7 needs to be good, so its neighbors should add up to 7. The neighbors are 2 and the next number. 2 + x = 7, so x = 5. Place 5 after 7. Now we have 1,3,2,7,5.Next, after 5, place 6. So, 1,3,2,7,5,6. 6 needs to be good, so its neighbors should add up to 6. The neighbors are 5 and the next number. 5 + x = 6, so x = 1. Place 1 after 6. Now we have 1,3,2,7,5,6,1. Wait, we already have 1 at the beginning, so this duplicates 1, which isn't allowed.Alternatively, after 5, place 4. So, 1,3,2,7,5,4. Now, 4 needs to be good, so its neighbors should add up to 4. The neighbors are 5 and the next number. 5 + x = 4, which would require x = -1, which isn't possible. So, 4 is not good.Then, after 4, place 6. So, 1,3,2,7,5,4,6. Now, 6 needs to be good. Its neighbors are 4 and 1. 4 + 1 = 5, which is not equal to 6. So, 6 is not good.In this arrangement, only 3 and 7 are good. Again, worse than the previous arrangement.It seems that trying to make more than 3 good numbers leads to conflicts or duplicates. Maybe 3 is the maximum.Wait, let's try another arrangement. Suppose I place 4 in the middle. Let's see.Start with 1,4,3. 4 is good because 1 + 3 = 4. Next, place 7. So, 1,4,3,7. 7 needs to be good, so its neighbors should add up to 7. The neighbors are 3 and the next number. 3 + x = 7, so x = 4. But 4 is already used. So, can't do that.Alternatively, after 3, place 6. So, 1,4,3,6. 6 needs to be good, so its neighbors should add up to 6. The neighbors are 3 and the next number. 3 + x = 6, so x = 3. But 3 is already used. Can't do that.Alternatively, after 3, place 2. So, 1,4,3,2. 2 needs to be good, but 2 can't be good because its neighbors would have to be 0 and something. So, 2 is not good.This isn't working either.Maybe another approach. Let's try to place the largest numbers first since they have more potential to be good.Start with 7. To make 7 good, its neighbors should add up to 7. Let's choose 3 and 4. So, 3,7,4.Now, next to 4, let's place 6. So, 3,7,4,6. 6 needs to be good, so its neighbors should add up to 6. The neighbors are 4 and the next number. 4 + x = 6, so x = 2. Place 2 after 6. Now we have 3,7,4,6,2.Next, after 2, place 5. So, 3,7,4,6,2,5. 5 needs to be good, so its neighbors should add up to 5. The neighbors are 2 and the next number. 2 + x = 5, so x = 3. But 3 is already used. Can't do that.Alternatively, after 2, place 1. So, 3,7,4,6,2,1. 1 can't be good. Then, after 1, place 5. So, 3,7,4,6,2,1,5. Now, 5 needs to be good. Its neighbors are 1 and 3. 1 + 3 = 4, which is not equal to 5. So, 5 is not good.In this arrangement, only 7 and 6 are good. Not better than before.Wait, maybe I can adjust the arrangement to make 5 good. Let's see.Start with 3,7,4,6,2,5,1. Now, check:- 7: neighbors 3 and 4. 3 + 4 = 7. Good.- 6: neighbors 4 and 2. 4 + 2 = 6. Good.- 5: neighbors 2 and 1. 2 + 1 = 3 ≠ 5. Not good.- 4: neighbors 7 and 6. 7 + 6 = 13 ≠ 4. Not good.- 3: neighbors 1 and 7. 1 + 7 = 8 ≠ 3. Not good.- 2: neighbors 6 and 5. 6 + 5 = 11 ≠ 2. Not good.- 1: neighbors 5 and 3. 5 + 3 = 8 ≠ 1. Not good.So, only 7 and 6 are good. Still, 2 good numbers.Hmm, maybe I need to try a different starting point.Let's try starting with 2,7,5. So, 2,7,5. 7 is good because 2 + 5 = 7.Next, after 5, place 6. So, 2,7,5,6. 6 needs to be good, so its neighbors should add up to 6. The neighbors are 5 and the next number. 5 + x = 6, so x = 1. Place 1 after 6. Now we have 2,7,5,6,1.Next, after 1, place 4. So, 2,7,5,6,1,4. 4 needs to be good, so its neighbors should add up to 4. The neighbors are 1 and the next number. 1 + x = 4, so x = 3. Place 3 after 4. Now we have 2,7,5,6,1,4,3.Finally, connect back to 2. So, the circle is 2,7,5,6,1,4,3.Now, check which numbers are good:- 7: neighbors 2 and 5. 2 + 5 = 7. Good.- 6: neighbors 5 and 1. 5 + 1 = 6. Good.- 4: neighbors 1 and 3. 1 + 3 = 4. Good.- 3: neighbors 4 and 2. 4 + 2 = 6 ≠ 3. Not good.- 5: neighbors 7 and 6. 7 + 6 = 13 ≠ 5. Not good.- 1: neighbors 6 and 4. 6 + 4 = 10 ≠ 1. Not good.- 2: neighbors 3 and 7. 3 + 7 = 10 ≠ 2. Not good.So, in this arrangement, we have 3 good numbers: 7, 6, and 4. That's the same as before.Is there a way to get more than 3? Let's try.Suppose I try to make 5 good as well. If 5 is good, its neighbors should add up to 5. Let's see if we can adjust the arrangement.Starting with 2,7,5,6,1,4,3 as before. If I want 5 to be good, its neighbors need to add up to 5. Currently, 5 is between 7 and 6. 7 + 6 = 13, which is not 5. So, 5 is not good here.If I rearrange to make 5 good, maybe place 2 and 3 next to 5. So, 2,5,3. Then 5 is good because 2 + 3 = 5. Now, let's try to build the circle.Starting with 2,5,3. Next, let's place 7. So, 2,5,3,7. Now, 7 needs to be good, so its neighbors should add up to 7. The neighbors are 3 and the next number. 3 + x = 7, so x = 4. Place 4 after 7. Now we have 2,5,3,7,4.Next, after 4, let's place 6. So, 2,5,3,7,4,6. Now, 6 needs to be good, so its neighbors should add up to 6. The neighbors are 4 and the next number. 4 + x = 6, so x = 2. But 2 is already at the beginning. Wait, we can't have duplicates. So, this doesn't work.Alternatively, after 4, place 1. So, 2,5,3,7,4,1. Now, 1 needs to be good, but 1 can't be good because its neighbors would have to be 0 and something, which isn't possible. So, 1 is not good.Then, after 1, place 6. So, 2,5,3,7,4,1,6. Now, 6 needs to be good. Its neighbors are 1 and 2 (since it's a circle). 1 + 2 = 3, which is not equal to 6. So, 6 is not good.In this arrangement, only 5 and 7 are good. That's worse than before.Maybe another arrangement. Let's try making 3 good. If 3 is good, its neighbors should add up to 3. So, neighbors could be 1 and 2. Let's try that.Start with 1,3,2. Now, 3 is good because 1 + 2 = 3. Next, place 7. So, 1,3,2,7. 7 needs to be good, so its neighbors should add up to 7. The neighbors are 2 and the next number. 2 + x = 7, so x = 5. Place 5 after 7. Now we have 1,3,2,7,5.Next, after 5, place 6. So, 1,3,2,7,5,6. 6 needs to be good, so its neighbors should add up to 6. The neighbors are 5 and the next number. 5 + x = 6, so x = 1. Place 1 after 6. Now we have 1,3,2,7,5,6,1. Wait, we already have 1 at the beginning, so this duplicates 1, which isn't allowed.Alternatively, after 5, place 4. So, 1,3,2,7,5,4. Now, 4 needs to be good, so its neighbors should add up to 4. The neighbors are 5 and the next number. 5 + x = 4, which would require x = -1, which isn't possible. So, 4 is not good.Then, after 4, place 6. So, 1,3,2,7,5,4,6. Now, 6 needs to be good. Its neighbors are 4 and 1. 4 + 1 = 5, which is not equal to 6. So, 6 is not good.In this arrangement, only 3 and 7 are good. Again, worse than the previous arrangement.It seems that trying to make more than 3 good numbers leads to conflicts or duplicates. Maybe 3 is the maximum.Wait, let's try another arrangement. Suppose I place 4 in the middle. Let's see.Start with 1,4,3. 4 is good because 1 + 3 = 4. Next, place 7. So, 1,4,3,7. 7 needs to be good, so its neighbors should add up to 7. The neighbors are 3 and the next number. 3 + x = 7, so x = 4. But 4 is already used. So, can't do that.Alternatively, after 3, place 6. So, 1,4,3,6. 6 needs to be good, so its neighbors should add up to 6. The neighbors are 3 and the next number. 3 + x = 6, so x = 3. But 3 is already used. Can't do that.Alternatively, after 3, place 2. So, 1,4,3,2. 2 needs to be good, but 2 can't be good because its neighbors would have to be 0 and something. So, 2 is not good.This isn't working either.Maybe another approach. Let's try to place the largest numbers first since they have more potential to be good.Start with 7. To make 7 good, its neighbors should add up to 7. Let's choose 3 and 4. So, 3,7,4.Now, next to 4, let's place 6. So, 3,7,4,6. 6 needs to be good, so its neighbors should add up to 6. The neighbors are 4 and the next number. 4 + x = 6, so x = 2. Place 2 after 6. Now we have 3,7,4,6,2.Next, after 2, place 5. So, 3,7,4,6,2,5. 5 needs to be good, so its neighbors should add up to 5. The neighbors are 2 and the next number. 2 + x = 5, so x = 3. But 3 is already used. Can't do that.Alternatively, after 2, place 1. So, 3,7,4,6,2,1. 1 can't be good. Then, after 1, place 5. So, 3,7,4,6,2,1,5. Now, 5 needs to be good. Its neighbors are 1 and 3. 1 + 3 = 4, which is not equal to 5. So, 5 is not good.In this arrangement, only 7 and 6 are good. Not better than before.Wait, maybe I can adjust the arrangement to make 5 good. Let's see.Start with 3,7,4,6,2,5,1. Now, check:- 7: neighbors 3 and 4. 3 + 4 = 7. Good.- 6: neighbors 4 and 2. 4 + 2 = 6. Good.- 5: neighbors 2 and 1. 2 + 1 = 3 ≠ 5. Not good.- 4: neighbors 7 and 6. 7 + 6 = 13 ≠ 4. Not good.- 3: neighbors 1 and 7. 1 + 7 = 8 ≠ 3. Not good.- 2: neighbors 6 and 5. 6 + 5 = 11 ≠ 2. Not good.- 1: neighbors 5 and 3. 5 + 3 = 8 ≠ 1. Not good.So, only 7 and 6 are good. Still, 2 good numbers.Hmm, maybe I need to try a different starting point.Let's try starting with 2,7,5. So, 2,7,5. 7 is good because 2 + 5 = 7.Next, after 5, place 6. So, 2,7,5,6. 6 needs to be good, so its neighbors should add up to 6. The neighbors are 5 and the next number. 5 + x = 6, so x = 1. Place 1 after 6. Now we have 2,7,5,6,1.Next, after 1, place 4. So, 2,7,5,6,1,4. 4 needs to be good, so its neighbors should add up to 4. The neighbors are 1 and the next number. 1 + x = 4, so x = 3. Place 3 after 4. Now we have 2,7,5,6,1,4,3.Finally, connect back to 2. So, the circle is 2,7,5,6,1,4,3.Now, check which numbers are good:- 7: neighbors 2 and 5. 2 + 5 = 7. Good.- 6: neighbors 5 and 1. 5 + 1 = 6. Good.- 4: neighbors 1 and 3. 1 + 3 = 4. Good.- 3: neighbors 4 and 2. 4 + 2 = 6 ≠ 3. Not good.- 5: neighbors 7 and 6. 7 + 6 = 13 ≠ 5. Not good.- 1: neighbors 6 and 4. 6 + 4 = 10 ≠ 1. Not good.- 2: neighbors 3 and 7. 3 + 7 = 10 ≠ 2. Not good.So, in this arrangement, we have 3 good numbers: 7, 6, and 4. That's the same as before.I think I've tried multiple arrangements, and the maximum number of good numbers I can get is 3. It seems that getting more than 3 leads to conflicts where the required neighbors are already used or result in duplicates, which isn't allowed.Therefore, the maximum number of good numbers is 3.

💡Alright, so I have this math problem here about hexagons and polygons. It's divided into two parts. Let me try to tackle them one by one.Starting with part (1): "Starting from a vertex of a hexagon, we can draw ____ diagonals, dividing the hexagon into ____ triangles; the sum of the interior angles of a hexagon is ____."Okay, so first, I need to figure out how many diagonals can be drawn from a single vertex in a hexagon. A hexagon has six sides, right? So, if I pick one vertex, how many other vertices can I connect to it with a diagonal?Well, from any vertex in a polygon, you can't draw a diagonal to itself or to its two adjacent vertices because those would just be sides of the polygon, not diagonals. So, for a hexagon, which has six vertices, I subtract three (the vertex itself and the two adjacent ones). That leaves me with three vertices. So, from one vertex, I can draw three diagonals.Next, it says these diagonals divide the hexagon into triangles. If I draw three diagonals from one vertex, how many triangles does that create? Let me visualize a hexagon. If I pick one vertex and draw diagonals to the three non-adjacent vertices, it should split the hexagon into four triangles. Wait, is that right? Let me count: each diagonal creates a new triangle, so starting from one vertex, the first diagonal creates one triangle, the second diagonal creates another, and the third diagonal creates the third. But actually, since it's a hexagon, it should create four triangles in total. Hmm, maybe I need to think about it differently.Wait, no. If I have a hexagon and I draw three diagonals from one vertex, it's splitting the hexagon into four triangles. Yeah, that makes sense because each diagonal adds a triangle, and starting from one vertex, you end up with four triangles. So, that part is four triangles.Now, the sum of the interior angles of a hexagon. I remember there's a formula for the sum of interior angles of a polygon. It's (n-2) multiplied by 180 degrees, where n is the number of sides. So, for a hexagon, n is 6. Let's calculate that: (6-2)*180 = 4*180 = 720 degrees. So, the sum of the interior angles of a hexagon is 720 degrees.Alright, so part (1) seems to be: 3 diagonals, 4 triangles, and 720 degrees.Moving on to part (2): "If the sum of the interior angles of a polygon is 180 degrees less than twice the sum of its exterior angles. Find the number of sides and the sum of the interior angles of this polygon."Okay, let's break this down. First, I know that the sum of the exterior angles of any polygon is always 360 degrees, regardless of the number of sides. That's a key fact. So, twice the sum of the exterior angles would be 2*360 = 720 degrees.Now, the problem says the sum of the interior angles is 180 degrees less than this. So, the sum of the interior angles is 720 - 180 = 540 degrees.Now, I need to find the number of sides of the polygon whose interior angles sum up to 540 degrees. Again, using the formula for the sum of interior angles: (n-2)*180 = sum of interior angles.So, plugging in 540 for the sum: (n-2)*180 = 540.Let's solve for n:Divide both sides by 180: n - 2 = 540 / 180 = 3.So, n = 3 + 2 = 5.Therefore, the polygon has 5 sides, which is a pentagon, and the sum of its interior angles is 540 degrees.Wait, let me double-check that. If n=5, then the sum should be (5-2)*180 = 3*180 = 540. Yep, that matches. So, that seems correct.So, summarizing part (2): the polygon has 5 sides and the sum of its interior angles is 540 degrees.I think that covers both parts of the problem. Let me just recap to make sure I didn't miss anything.For part (1):- Diagonals from one vertex: 3- Number of triangles formed: 4- Sum of interior angles: 720 degreesFor part (2):- Number of sides: 5- Sum of interior angles: 540 degreesEverything seems to add up. I don't think I made any calculation errors, but let me just verify the key steps.In part (1), drawing diagonals from one vertex in a hexagon: 6 vertices total, subtract 3 (itself and two adjacent), so 3 diagonals. That's correct. Dividing into triangles: each diagonal adds a triangle, but in a hexagon, starting from one vertex, you end up with four triangles. That makes sense because the number of triangles formed is equal to the number of sides minus two, which is 6-2=4. Wait, actually, that's the formula for the sum of interior angles, but in terms of triangles, yes, it's four triangles.Sum of interior angles: (6-2)*180=720. Correct.In part (2), sum of exterior angles is 360, twice that is 720, minus 180 is 540. Then, using the formula, (n-2)*180=540, so n=5. That all checks out.I think I'm confident with these answers.

💡Okay, so I've got this problem here about Sonny and his cookies. Let me try to understand what's going on. So, Sonny received 45 boxes of cookies from his friend. That's the starting point. Then, he gave some of those boxes to his brother, sister, and cousin. The question is asking how many boxes are left for Sonny after he gave some away.Alright, let's break this down step by step. First, I need to figure out how many boxes Sonny gave away in total. He gave 12 boxes to his brother, 9 boxes to his sister, and 7 boxes to his cousin. So, to find the total number of boxes he gave away, I need to add those numbers together.Let me write that out: 12 (to his brother) plus 9 (to his sister) plus 7 (to his cousin). So, 12 + 9 is 21, and then 21 + 7 is 28. Okay, so Sonny gave away a total of 28 boxes.Now, he started with 45 boxes and gave away 28. To find out how many boxes are left for him, I need to subtract the number of boxes he gave away from the total number he received. So, that's 45 minus 28.Let me do that subtraction: 45 - 28. Hmm, 45 minus 20 is 25, and then minus 8 more would be 17. So, 45 - 28 equals 17.Wait, let me double-check that to make sure I didn't make a mistake. If I add 28 back to 17, I should get 45. So, 17 + 28 is 45. Yep, that checks out.So, after giving away 28 boxes, Sonny has 17 boxes left for himself. That seems straightforward, but let me think if there's another way to approach this problem to confirm my answer.Another way could be to subtract each amount one by one. Starting with 45, subtract 12 for his brother: 45 - 12 is 33. Then, subtract 9 for his sister: 33 - 9 is 24. Finally, subtract 7 for his cousin: 24 - 7 is 17. Yep, same result.Okay, so both methods give me 17 boxes left for Sonny. That makes me more confident that 17 is the correct answer.I wonder if there's any other detail I might have missed. The problem mentions that Sonny received the cookies from his friend, and then he gave them to his brother, sister, and cousin. It doesn't mention anything about him eating any or giving them to anyone else, so I think it's safe to assume that the only boxes given away are the ones mentioned.Also, the problem specifies "boxes of cookies," so I don't think there's any need to consider individual cookies or anything like that. It's all about the number of boxes.Let me visualize this to make sure I'm not missing anything. Imagine Sonny has a pile of 45 boxes. He takes 12 boxes and gives them to his brother. Now, he has 33 boxes left. Then, he takes 9 boxes from those 33 and gives them to his sister, leaving him with 24 boxes. Finally, he gives 7 boxes to his cousin, which brings the total down to 17 boxes. Yep, that matches what I calculated earlier.I think I've covered all the angles here. The key steps are:1. Determine the total number of boxes given away.2. Subtract that total from the original number of boxes.3. Verify the result by adding back or subtracting step by step.Everything seems to align, so I feel confident that the answer is 17 boxes left for Sonny.

💡Okay, so I have this problem where point A is on a curve defined by parametric equations, and I need to find the maximum distance from A to the origin. Let me try to figure this out step by step.First, the parametric equations are given as:x = sqrt(3) + cos(theta)y = 1 + sin(theta)So, point A has coordinates (sqrt(3) + cos(theta), 1 + sin(theta)). The origin is at (0,0). I need to find the maximum distance between A and the origin as theta varies.I remember that the distance between two points (x, y) and (0,0) is sqrt(x^2 + y^2). So, I can write the distance from A to the origin as:Distance = sqrt[(sqrt(3) + cos(theta))^2 + (1 + sin(theta))^2]Hmm, that looks a bit complicated, but maybe I can simplify it. Let me expand the squares inside the square root.First, expand (sqrt(3) + cos(theta))^2:= (sqrt(3))^2 + 2*sqrt(3)*cos(theta) + cos^2(theta)= 3 + 2*sqrt(3)*cos(theta) + cos^2(theta)Next, expand (1 + sin(theta))^2:= 1^2 + 2*1*sin(theta) + sin^2(theta)= 1 + 2*sin(theta) + sin^2(theta)Now, add these two results together:3 + 2*sqrt(3)*cos(theta) + cos^2(theta) + 1 + 2*sin(theta) + sin^2(theta)Combine like terms:3 + 1 = 4cos^2(theta) + sin^2(theta) = 1 (since cos^2 + sin^2 = 1)So, total becomes:4 + 2*sqrt(3)*cos(theta) + 2*sin(theta) + 1Wait, that's 4 + 1 = 5, so:5 + 2*sqrt(3)*cos(theta) + 2*sin(theta)So, the distance squared is 5 + 2*sqrt(3)*cos(theta) + 2*sin(theta). Therefore, the distance is sqrt(5 + 2*sqrt(3)*cos(theta) + 2*sin(theta)).Now, to find the maximum distance, I need to maximize this expression. Since the square root is a monotonically increasing function, maximizing the inside will maximize the distance.So, let me focus on maximizing the expression inside the square root: 5 + 2*sqrt(3)*cos(theta) + 2*sin(theta).This is a linear combination of sin(theta) and cos(theta). I remember that expressions of the form a*sin(theta) + b*cos(theta) can be rewritten as R*sin(theta + phi), where R = sqrt(a^2 + b^2) and phi is some phase shift.Let me try that. Let me write 2*sqrt(3)*cos(theta) + 2*sin(theta) as R*sin(theta + phi).First, compute R:R = sqrt[(2)^2 + (2*sqrt(3))^2] = sqrt[4 + 12] = sqrt[16] = 4.So, R is 4. Now, let me find phi such that:sin(phi) = (2*sqrt(3))/R = (2*sqrt(3))/4 = sqrt(3)/2cos(phi) = 2/R = 2/4 = 1/2So, sin(phi) = sqrt(3)/2 and cos(phi) = 1/2. That means phi is pi/3, since sin(pi/3) = sqrt(3)/2 and cos(pi/3) = 1/2.Therefore, 2*sqrt(3)*cos(theta) + 2*sin(theta) = 4*sin(theta + pi/3).So, the expression inside the square root becomes:5 + 4*sin(theta + pi/3)Therefore, the distance is sqrt(5 + 4*sin(theta + pi/3)).Now, to maximize this distance, I need to maximize sin(theta + pi/3). The maximum value of sine is 1, so the maximum value inside the square root is 5 + 4*1 = 9.Therefore, the maximum distance is sqrt(9) = 3.Wait, that seems straightforward. Let me double-check my steps.1. Expanded the distance squared correctly: yes, I think so.2. Combined like terms: 3 + 1 = 4, and cos^2 + sin^2 = 1, so 4 + 1 = 5. Then added the other terms: 2*sqrt(3)*cos(theta) + 2*sin(theta). That seems correct.3. Rewrote the linear combination as R*sin(theta + phi): correct approach. Calculated R as sqrt( (2)^2 + (2*sqrt(3))^2 ) = sqrt(4 + 12) = sqrt(16) = 4. That's right.4. Found phi: sin(phi) = (2*sqrt(3))/4 = sqrt(3)/2, cos(phi) = 2/4 = 1/2. So phi = pi/3. Correct.5. Rewrote the expression as 4*sin(theta + pi/3). Then added 5: 5 + 4*sin(theta + pi/3). Correct.6. Since sin(theta + pi/3) has a maximum of 1, the maximum inside the sqrt is 9, so sqrt(9) = 3. That makes sense.Is there another way to verify this? Maybe by considering the curve itself.Looking at the parametric equations:x = sqrt(3) + cos(theta)y = 1 + sin(theta)This is a circle with center at (sqrt(3), 1) and radius 1, because cos(theta) and sin(theta) vary between -1 and 1, so x varies between sqrt(3) - 1 and sqrt(3) + 1, and y varies between 0 and 2.So, the curve is a circle centered at (sqrt(3), 1) with radius 1. The origin is at (0,0). The maximum distance from the origin to any point on this circle would be the distance from the origin to the center of the circle plus the radius.Let me compute the distance from the origin to the center (sqrt(3), 1):Distance = sqrt( (sqrt(3))^2 + 1^2 ) = sqrt(3 + 1) = sqrt(4) = 2.So, the distance from the origin to the center is 2, and the radius of the circle is 1. Therefore, the maximum distance from the origin to any point on the circle is 2 + 1 = 3.That matches the earlier result. So, that's a good confirmation.Alternatively, if I think about the distance formula, it's essentially the same as finding the maximum value of |AO|, which is sqrt(x^2 + y^2). Since the curve is a circle, the maximum distance occurs when the point is diametrically opposite from the origin relative to the center.But in this case, since the center is at (sqrt(3),1), the point on the circle farthest from the origin would be in the direction away from the origin. So, the line connecting the origin to the center of the circle, extended by the radius, gives the farthest point.Calculating that, the vector from the origin to the center is (sqrt(3),1). The unit vector in that direction is (sqrt(3)/2, 1/2), since the distance is 2. So, moving from the center in that direction by the radius (1 unit) gives the farthest point:x = sqrt(3) + (sqrt(3)/2)*1 = sqrt(3) + sqrt(3)/2 = (3/2)*sqrt(3)y = 1 + (1/2)*1 = 1 + 1/2 = 3/2Then, the distance from the origin to this point is sqrt( ( (3/2)*sqrt(3) )^2 + (3/2)^2 ) = sqrt( (27/4) + (9/4) ) = sqrt(36/4) = sqrt(9) = 3.Again, same result. So, that's consistent.Therefore, I'm confident that the maximum distance is 3.**Final Answer**The maximum distance from point A to the origin is boxed{3}.

💡Okay, so I have this problem here: I need to find how many three-digit numbers meet three conditions. First, they shouldn't be divisible by 5. Second, the sum of their digits should be exactly 14. Third, the first digit has to be equal to the last digit. The answer choices are from A to E, with numbers 6, 8, 3, 4, and 5. Hmm, let me think about how to approach this.Alright, let's break it down. A three-digit number can be represented as (overline{abc}), where (a) is the hundreds digit, (b) is the tens digit, and (c) is the units digit. The first condition is that the number isn't divisible by 5. Divisibility by 5 depends on the units digit, right? So if the number ends with 0 or 5, it's divisible by 5. Therefore, to satisfy the first condition, the units digit (c) shouldn't be 0 or 5.The second condition is that the sum of the digits is exactly 14. So, (a + b + c = 14).The third condition is that the first digit is equal to the last digit, which means (a = c). So, combining this with the first condition, since (a = c), the units digit (c) can't be 0 or 5. Therefore, (a) also can't be 0 or 5. But wait, (a) is the hundreds digit, so it can't be 0 anyway because then it wouldn't be a three-digit number. So, (a) can be 1, 2, 3, 4, 6, 7, 8, or 9. But since (a = c), and (c) can't be 0 or 5, (a) is restricted to 1, 2, 3, 4, 6, 7, 8, 9.But let's also remember that the sum of the digits is 14, so (a + b + c = 14). Since (a = c), we can rewrite this as (2a + b = 14). So, (b = 14 - 2a). Now, since (b) is a single digit, it has to be between 0 and 9, inclusive. So, (0 leq b leq 9). Therefore, (14 - 2a) must be between 0 and 9.Let me write that down: (0 leq 14 - 2a leq 9). Let's solve for (a). Starting with the left inequality: (14 - 2a geq 0). So, (14 geq 2a), which means (a leq 7). Then, the right inequality: (14 - 2a leq 9). So, (14 - 9 leq 2a), which is (5 leq 2a), so (a geq 2.5). But (a) has to be an integer, so (a geq 3).So, combining these, (a) must be between 3 and 7, inclusive. But earlier, we saw that (a) can't be 5 because (c) can't be 5. So, possible values for (a) are 3, 4, 6, 7.Let me list them out:1. If (a = 3), then (b = 14 - 2*3 = 14 - 6 = 8). So, the number is 383.2. If (a = 4), then (b = 14 - 2*4 = 14 - 8 = 6). So, the number is 464.3. If (a = 6), then (b = 14 - 2*6 = 14 - 12 = 2). So, the number is 626.4. If (a = 7), then (b = 14 - 2*7 = 14 - 14 = 0). So, the number is 707.Wait, let me check if these numbers are valid. Each of them is a three-digit number, the first digit equals the last digit, their digits sum to 14, and none of them end with 0 or 5. So, 383: ends with 3, which is fine. 464: ends with 4, also fine. 626: ends with 6, good. 707: ends with 7, which is okay. So, all four numbers meet the criteria.But hold on, the answer choices are A)6, B)8, C)3, D)4, E)5. So, according to my calculation, the answer is 4, which is option D.But just to make sure I didn't miss anything, let me double-check. Maybe I missed some values of (a). Let's see, (a) has to be between 3 and 7, inclusive, and can't be 5. So, 3,4,6,7. That's four values. Each gives a unique number, so four numbers in total. Yeah, that seems right.I don't think I made a mistake here. So, I think the answer is D)4.**Final Answer**boxed{D}

💡Okay, so I have this function f(x) = ax² + (3 - 2a)x - 6, and I need to solve two parts. Let me start with part (1).When a = 1, I need to find the values of x where the graph of f(x) is above the x-axis. That means I need to solve f(x) > 0. First, substitute a = 1 into the function:f(x) = 1*x² + (3 - 2*1)x - 6f(x) = x² + (3 - 2)x - 6f(x) = x² + x - 6Now, I need to solve x² + x - 6 > 0. To do this, I can factor the quadratic or find its roots. Let me try factoring first.Looking for two numbers that multiply to -6 and add to 1. Hmm, 3 and -2 work because 3*(-2) = -6 and 3 + (-2) = 1.So, x² + x - 6 factors to (x + 3)(x - 2).Therefore, (x + 3)(x - 2) > 0.To solve this inequality, I can use a sign chart. The critical points are x = -3 and x = 2. These divide the number line into three intervals:1. x < -32. -3 < x < 23. x > 2Now, I'll test each interval:1. For x < -3, let's pick x = -4: (-4 + 3)(-4 - 2) = (-1)(-6) = 6 > 02. For -3 < x < 2, let's pick x = 0: (0 + 3)(0 - 2) = 3*(-2) = -6 < 03. For x > 2, let's pick x = 3: (3 + 3)(3 - 2) = 6*1 = 6 > 0So, the inequality (x + 3)(x - 2) > 0 holds true when x < -3 or x > 2.Therefore, the solution for part (1) is x < -3 or x > 2.Moving on to part (2): When a < 0, find the solution set for f(x) < 0.So, f(x) = ax² + (3 - 2a)x - 6 < 0, and a is negative.Since a is negative, the parabola opens downward. That means the graph will be below the x-axis between its two roots.First, let's find the roots of f(x) = 0.f(x) = ax² + (3 - 2a)x - 6 = 0To solve this quadratic equation, we can use the quadratic formula:x = [-b ± sqrt(b² - 4ac)] / (2a)Where a = a, b = (3 - 2a), and c = -6.Plugging in:x = [-(3 - 2a) ± sqrt((3 - 2a)² - 4*a*(-6))] / (2a)Simplify inside the square root:(3 - 2a)² - 4*a*(-6) = 9 - 12a + 4a² + 24a = 9 + 12a + 4a²So, sqrt(9 + 12a + 4a²) = sqrt((2a + 3)²) = |2a + 3|Since a is negative, let's see if 2a + 3 is positive or negative.If a < 0, then 2a is negative. So, 2a + 3 could be positive or negative depending on the value of a.Let's find when 2a + 3 = 0:2a + 3 = 0 => a = -3/2So, if a > -3/2, then 2a + 3 > 0.If a < -3/2, then 2a + 3 < 0.Therefore, |2a + 3| = 2a + 3 if a >= -3/2, and -(2a + 3) if a < -3/2.But since a < 0 in this problem, we need to consider both cases: a between -3/2 and 0, and a < -3/2.Wait, actually, since a is negative, and we're considering a < 0, the critical point is at a = -3/2.So, let's split into two cases:Case 1: -3/2 < a < 0Case 2: a < -3/2Let's handle each case separately.Case 1: -3/2 < a < 0In this case, 2a + 3 > 0 because a > -3/2, so |2a + 3| = 2a + 3.So, the roots are:x = [-(3 - 2a) ± (2a + 3)] / (2a)Let's compute both roots.First root with +:x = [-(3 - 2a) + (2a + 3)] / (2a)= [-3 + 2a + 2a + 3] / (2a)= [4a] / (2a)= 2Second root with -:x = [-(3 - 2a) - (2a + 3)] / (2a)= [-3 + 2a - 2a - 3] / (2a)= [-6] / (2a)= -3/aSo, the roots are x = 2 and x = -3/a.Since a is negative, -3/a is positive.Now, since the parabola opens downward (a < 0), the graph is below the x-axis between the two roots.Therefore, f(x) < 0 when x is between the smaller root and the larger root.But we need to determine which root is smaller.Since a is negative, let's see:-3/a is positive because a is negative.So, we have two roots: x = 2 and x = -3/a.We need to compare 2 and -3/a.Since a is negative, let's denote a = -k where k > 0.Then, -3/a = -3/(-k) = 3/k.So, we have x = 2 and x = 3/k.We need to see which is larger: 2 or 3/k.Given that -3/2 < a < 0, which is equivalent to 0 < k < 3/2.So, k is between 0 and 1.5.Therefore, 3/k is greater than 2 because:If k = 1, 3/k = 3 > 2If k approaches 0, 3/k approaches infinityIf k approaches 1.5, 3/k approaches 2So, 3/k is always greater than or equal to 2 in this interval.Therefore, the roots are x = 2 and x = 3/k (which is greater than 2).Thus, the interval where f(x) < 0 is between 2 and 3/k, which is 2 < x < -3/a.But wait, since a is negative, -3/a is positive, so the interval is between 2 and -3/a.But since -3/a > 2, the solution is 2 < x < -3/a.Wait, but the parabola opens downward, so f(x) < 0 between the roots.But in this case, the roots are 2 and -3/a, with -3/a > 2.Therefore, f(x) < 0 for x between 2 and -3/a.But wait, the problem says a < 0, so we have to express the solution in terms of a.So, the solution is 2 < x < -3/a.But let me check with a specific value to make sure.Let's take a = -1, which is between -3/2 and 0.Then, f(x) = -1x² + (3 - 2*(-1))x -6 = -x² + 5x -6Set f(x) = 0:-x² +5x -6 =0Multiply both sides by -1:x² -5x +6=0(x-2)(x-3)=0So, roots at x=2 and x=3.Since a = -1 <0, the parabola opens downward, so f(x) <0 outside the roots, but wait, no.Wait, if a is negative, the parabola opens downward, so f(x) <0 between the roots.But in this case, the roots are 2 and 3, so f(x) <0 for 2 <x <3.But according to our earlier solution, it should be 2 <x < -3/a.Since a = -1, -3/a = 3.So, 2 <x <3, which matches.So, in this case, the solution is 2 <x < -3/a.Similarly, if a = -1.2, which is less than -3/2.Wait, no, -1.2 is greater than -3/2 (-1.5). So, let's pick a = -2, which is less than -3/2.Then, f(x) = -2x² + (3 -2*(-2))x -6 = -2x² +7x -6Set f(x)=0:-2x² +7x -6=0Multiply by -1:2x² -7x +6=0Using quadratic formula:x = [7 ± sqrt(49 - 48)] /4 = [7 ±1]/4So, x= (7+1)/4=2 and x=(7-1)/4=1.5So, roots at x=1.5 and x=2.Since a=-2 <0, parabola opens downward, so f(x) <0 between the roots.So, 1.5 <x <2.But according to our earlier solution, when a < -3/2, the solution is x < -3/a or x >2.Wait, but in this case, the solution is between 1.5 and 2.Hmm, seems like my earlier reasoning might have a mistake.Wait, let's go back.In the case where a < -3/2, which is a < -1.5.So, a = -2 is less than -1.5.Then, the roots are x=2 and x=-3/a.Since a = -2, -3/a = -3/(-2)=1.5.So, the roots are x=1.5 and x=2.Since the parabola opens downward, f(x) <0 between the roots, i.e., 1.5 <x <2.But according to my earlier conclusion, when a < -3/2, the solution is x < -3/a or x >2.But in this case, the solution is between 1.5 and 2, not x <1.5 or x >2.Wait, that contradicts.So, perhaps my earlier reasoning was wrong.Let me re-examine.When a <0, the parabola opens downward.So, f(x) <0 between the roots.But the roots are x=2 and x=-3/a.Depending on the value of a, -3/a can be greater than 2 or less than 2.So, when a < -3/2, which is a < -1.5, then -3/a is less than 2 because:If a = -2, -3/a = 1.5 <2If a approaches -infty, -3/a approaches 0 <2So, in this case, the smaller root is -3/a and the larger root is 2.Therefore, f(x) <0 between -3/a and 2.But since a <0, -3/a is positive.So, the solution is -3/a <x <2.But wait, in the case when a = -2, the solution is 1.5 <x <2, which is -3/a <x <2.Similarly, when a = -1, which is between -3/2 and 0, -3/a =3, so the solution is 2 <x <3.Wait, so in general:If a <0, the roots are x=2 and x=-3/a.If -3/a >2, which is when a >-3/2, then the solution is 2 <x < -3/a.If -3/a <2, which is when a < -3/2, then the solution is -3/a <x <2.But wait, when a < -3/2, -3/a is less than 2, so the solution is between -3/a and 2.But since a is negative, -3/a is positive, so the interval is from a positive number less than 2 to 2.But in the case when a = -2, the solution is 1.5 <x <2, which is correct.Similarly, when a approaches -infty, -3/a approaches 0, so the solution approaches 0 <x <2.But wait, when a approaches 0 from the negative side, -3/a approaches -infty, but since a is negative, -3/a is positive and approaches +infty.Wait, no, if a approaches 0 from the negative side, a approaches 0-, so -3/a approaches +infty.So, when a is between -3/2 and 0, -3/a is greater than 2, so the solution is 2 <x < -3/a.When a < -3/2, -3/a is less than 2, so the solution is -3/a <x <2.But wait, in the case when a = -3/2, -3/a = 2, so the quadratic becomes a perfect square.Let me check that.If a = -3/2, then f(x) = (-3/2)x² + (3 - 2*(-3/2))x -6Simplify:= (-3/2)x² + (3 +3)x -6= (-3/2)x² +6x -6Set f(x)=0:(-3/2)x² +6x -6=0Multiply both sides by -2:3x² -12x +12=0Divide by 3:x² -4x +4=0(x-2)^2=0So, double root at x=2.Therefore, when a = -3/2, f(x) =0 only at x=2, and since the parabola opens downward, f(x) <0 everywhere except at x=2.But the problem is to find f(x) <0, so the solution is all real numbers except x=2.So, in summary:When a <0:- If a > -3/2 (i.e., -3/2 <a <0), then -3/a >2, so f(x) <0 for 2 <x < -3/a.- If a = -3/2, f(x) <0 for all x ≠2.- If a < -3/2, then -3/a <2, so f(x) <0 for -3/a <x <2.Therefore, the solution set depends on the value of a.But the problem says "When a <0, find the solution set for f(x) <0."So, we need to express the solution in terms of a.So, the solution set is:If -3/2 <a <0, then x ∈ (2, -3/a)If a = -3/2, then x ∈ ℝ {2}If a < -3/2, then x ∈ (-3/a, 2)But the problem doesn't specify to separate into cases, just to find the solution set when a <0.So, perhaps we can write it as:If a ≠ -3/2, then x is between the smaller root and the larger root, which are 2 and -3/a.But depending on whether -3/a is greater than or less than 2, the interval changes.Alternatively, we can write the solution set as:x ∈ (min(2, -3/a), max(2, -3/a))But since a <0, -3/a is positive.So, depending on whether -3/a is greater than 2 or not, the interval is either (2, -3/a) or (-3/a, 2).But to express it without piecewise, perhaps we can write:x ∈ (min(2, -3/a), max(2, -3/a))But in terms of a, we can note that:If a > -3/2, then -3/a >2, so x ∈ (2, -3/a)If a < -3/2, then -3/a <2, so x ∈ (-3/a, 2)If a = -3/2, x ∈ ℝ {2}Therefore, the solution set is:- If -3/2 <a <0, then x ∈ (2, -3/a)- If a = -3/2, then x ∈ ℝ {2}- If a < -3/2, then x ∈ (-3/a, 2)But the problem asks for the solution set when a <0, so we can express it as:If a <0, then:- If a ≠ -3/2, x is between the smaller root and the larger root, which are 2 and -3/a.- If a = -3/2, x is all real numbers except 2.But perhaps the problem expects a general solution without splitting into cases, but I think it's necessary to consider the cases because the interval changes depending on a.So, to write the solution set:When a <0, the solution set for f(x) <0 is:- If a > -3/2, then x ∈ (2, -3/a)- If a = -3/2, then x ∈ ℝ {2}- If a < -3/2, then x ∈ (-3/a, 2)Therefore, the solution set is:- For -3/2 <a <0: {x | 2 <x < -3/a}- For a = -3/2: {x | x ≠2}- For a < -3/2: {x | -3/a <x <2}But the problem says "When a <0, find the solution set for f(x) <0."So, perhaps we can write it as:If a <0, then f(x) <0 for x between the roots, which are 2 and -3/a. The interval depends on whether -3/a is greater than or less than 2.Therefore, the solution set is:- If -3/2 <a <0, then x ∈ (2, -3/a)- If a = -3/2, then x ∈ ℝ {2}- If a < -3/2, then x ∈ (-3/a, 2)So, to express this concisely, we can write:When a <0, the solution set for f(x) <0 is:- If a ≠ -3/2, then x is between 2 and -3/a, specifically: - If a > -3/2, then 2 <x < -3/a - If a < -3/2, then -3/a <x <2- If a = -3/2, then x ≠2Therefore, the final answer for part (2) is:- If -3/2 <a <0, then x ∈ (2, -3/a)- If a = -3/2, then x ∈ ℝ {2}- If a < -3/2, then x ∈ (-3/a, 2)But since the problem asks for the solution set when a <0, we can present it as:When a <0, the solution set for f(x) <0 is:- If -3/2 <a <0, then x ∈ (2, -3/a)- If a = -3/2, then x ∈ ℝ {2}- If a < -3/2, then x ∈ (-3/a, 2)So, in boxed form, we can write:For part (1): x < -3 or x >2, which is {x | x < -3 or x >2}For part (2):- If -3/2 <a <0, then {x | 2 <x < -3/a}- If a = -3/2, then {x | x ≠2}- If a < -3/2, then {x | -3/a <x <2}But since the problem asks for the solution set when a <0, we can present it as:When a <0, the solution set for f(x) <0 is:- If -3/2 <a <0, then x ∈ (2, -3/a)- If a = -3/2, then x ∈ ℝ {2}- If a < -3/2, then x ∈ (-3/a, 2)So, in conclusion, the solution sets are as above.