🧠:<think>Okay, so I have this problem about athletes represented by prime numbers, and I need to figure out two things. First, find a set of 10 distinct primes that add up to 200. Second, determine the target number T that maximizes the number of subsets of these primes that sum to T. Hmm, let's start with the first part.1. Finding 10 distinct primes summing to 200.Alright, so I need to find 10 unique prime numbers whose total is 200. Primes are numbers greater than 1 that have no divisors other than 1 and themselves. The primes start from 2, 3, 5, 7, 11, 13, and so on.First thought: The smallest 10 primes. Let me list them:2, 3, 5, 7, 11, 13, 17, 19, 23, 29.Let me add these up:2 + 3 = 55 + 5 = 1010 + 7 = 1717 + 11 = 2828 + 13 = 4141 + 17 = 5858 + 19 = 7777 + 23 = 100100 + 29 = 129.Hmm, that's only 129. We need 200. So the sum is way too low. So, we need to replace some of the smaller primes with larger ones to increase the total.But wait, primes can't be too large, otherwise, the sum might overshoot 200. Let's think about the average. 200 divided by 10 is 20. So, on average, each prime should be around 20. So, primes around 17, 19, 23, etc.But since primes are unique, we need to pick 10 distinct ones. Let me think about how to approach this.One strategy is to start with the smallest primes and then replace the smallest ones with larger primes until the total reaches 200.So, starting with the first 10 primes: sum is 129. We need an additional 71.So, we can try replacing the smallest prime, which is 2, with a larger prime. Let's see:If we remove 2 and add a prime p, the new sum becomes 129 - 2 + p = 127 + p. We need 127 + p = 200, so p = 73. So, replacing 2 with 73 would make the sum 200. But wait, is 73 a prime? Yes, 73 is a prime number.But let's check if that's the only replacement needed. So, the set would be: 3, 5, 7, 11, 13, 17, 19, 23, 29, 73. Let me add these:3 + 5 = 88 + 7 = 1515 + 11 = 2626 + 13 = 3939 + 17 = 5656 + 19 = 7575 + 23 = 9898 + 29 = 127127 + 73 = 200.Yes, that works. So, one possible set is replacing 2 with 73. But wait, is 73 the only option? Let's see if we can replace another prime instead.Alternatively, maybe we can replace multiple primes to get a more balanced set. For example, replacing 2 and 3 with larger primes.Let me try replacing 2 and 3 with two larger primes. The total sum would be 129 - 2 - 3 + p + q = 124 + p + q. We need 124 + p + q = 200, so p + q = 76. We need two distinct primes that sum to 76.Looking for primes p and q such that p + q = 76. Let's see:76 is even, so by Goldbach's conjecture, it can be expressed as the sum of two primes. Let's find such pairs.Starting from the smallest prime:76 - 3 = 73, which is prime. So, 3 and 73.But 3 is already in our original set, so we can't use 3 again since all primes must be distinct. So, next:76 - 5 = 71, which is prime. So, 5 and 71.But 5 is already in the set. Next:76 - 7 = 69, not prime.76 - 11 = 65, not prime.76 - 13 = 63, not prime.76 - 17 = 59, which is prime. So, 17 and 59.17 is already in the set. Next:76 - 19 = 57, not prime.76 - 23 = 53, which is prime. 23 is in the set. Next:76 - 29 = 47, which is prime. 29 is in the set.76 - 31 = 45, not prime.76 - 37 = 39, not prime.76 - 41 = 35, not prime.76 - 43 = 33, not prime.76 - 47 = 29, which is prime, but 29 is already in the set.So, the only possible pairs are (3,73), (5,71), (17,59), (23,53), (29,47). All of these include primes already in the original set except for 73, 71, 59, 53, 47.So, if we replace 2 and 3 with 71 and 5, but 5 is already there. Wait, no, we're replacing 2 and 3, so we can add 71 and another prime. Wait, no, we're replacing two primes, so we need two new primes.Wait, perhaps replacing 2 and 3 with 71 and 5 is not possible because 5 is already in the set. So, maybe we need to replace 2 and another prime.Alternatively, maybe replacing 2 and 5 with 71 and something else.Wait, this is getting complicated. Maybe it's better to just stick with replacing 2 with 73, as that gives a valid set without overlapping primes.So, one possible set is: 3, 5, 7, 11, 13, 17, 19, 23, 29, 73.But let me check if there are other possible sets. Maybe instead of replacing 2, we can replace another small prime with a larger one.For example, replacing 3 with a larger prime. Let's see:Original sum is 129. If we remove 3 and add p, the new sum is 129 - 3 + p = 126 + p. We need 126 + p = 200, so p = 74. But 74 is not a prime. Next prime after 74 is 75, which is not prime, 76 not prime, 77 not prime, 78 no, 79 is prime. So p = 79.So, replacing 3 with 79. Then the set would be: 2, 5, 7, 11, 13, 17, 19, 23, 29, 79.Let me add these:2 + 5 = 77 + 7 = 1414 + 11 = 2525 + 13 = 3838 + 17 = 5555 + 19 = 7474 + 23 = 9797 + 29 = 126126 + 79 = 205. Oh, that's over 200. So, that doesn't work. So, replacing 3 with 79 overshoots.Wait, maybe I miscalculated. Let me add them again:2 + 5 = 77 + 7 = 1414 + 11 = 2525 + 13 = 3838 + 17 = 5555 + 19 = 7474 + 23 = 9797 + 29 = 126126 + 79 = 205. Yes, that's correct. So, that's too much.So, p needs to be 74, which isn't prime. So, no solution here.Alternatively, maybe replacing 5 with a larger prime.Original sum 129. Remove 5, add p: 129 -5 + p = 124 + p = 200 => p=76. Not prime. Next prime is 77, not prime, 79 is prime. So p=79.So, replacing 5 with 79. The set becomes: 2, 3, 7, 11, 13, 17, 19, 23, 29, 79.Let's add:2 + 3 = 55 + 7 = 1212 + 11 = 2323 + 13 = 3636 + 17 = 5353 + 19 = 7272 + 23 = 9595 + 29 = 124124 + 79 = 203. Still over.Hmm, so replacing 5 with 79 gives 203, which is still over. Maybe we need a smaller prime.Wait, p needs to be 76, which isn't prime. So, no solution here either.Alternatively, maybe replacing two small primes with two larger ones. For example, replacing 2 and 3 with 73 and 71. Let's see:Original sum 129. Remove 2 and 3: 129 -2 -3 = 124. Add 71 and 73: 124 +71 +73 = 268. That's way over 200. So, that's not good.Alternatively, maybe replacing 2 and 5 with 73 and something else. Let's see:Remove 2 and 5: 129 -2 -5 = 122. We need to add two primes p and q such that p + q = 200 - 122 = 78.Looking for two primes that sum to 78.78 is even, so possible pairs:78 - 5 = 73, which is prime. So, 5 and 73. But 5 is already in the set.78 - 7 = 71, which is prime. So, 7 and 71. 7 is in the set.78 - 11 = 67, prime. 11 is in the set.78 - 13 = 65, not prime.78 - 17 = 61, prime. 17 is in the set.78 - 19 = 59, prime. 19 is in the set.78 - 23 = 55, not prime.78 - 29 = 49, not prime.78 - 31 = 47, prime. 31 is not in the set, 47 is not in the set.So, 31 and 47. Both are primes not in the original set. So, replacing 2 and 5 with 31 and 47.Let's check the sum:Original sum 129 -2 -5 = 122. Add 31 +47 = 78. 122 +78=200. Perfect.So, the new set would be: 3, 7, 11, 13, 17, 19, 23, 29, 31, 47.Let me verify the sum:3 +7=1010+11=2121+13=3434+17=5151+19=7070+23=9393+29=122122+31=153153+47=200. Yes, that works.So, another possible set is: 3,7,11,13,17,19,23,29,31,47.So, there are multiple possible sets. The first one I found was replacing 2 with 73, giving 3,5,7,11,13,17,19,23,29,73.Another set is replacing 2 and 5 with 31 and 47, giving 3,7,11,13,17,19,23,29,31,47.Are there more? Let's see.Alternatively, maybe replacing 2 and 7 with two primes.Original sum 129. Remove 2 and7: 129-2-7=120. Need two primes p and q such that p + q=80.Looking for primes summing to 80.80 is even, so possible pairs:80-3=77, not prime.80-5=75, not prime.80-7=73, prime. So, 7 and73. But 7 is already in the set.80-11=69, not prime.80-13=67, prime. 13 is in the set.80-17=63, not prime.80-19=61, prime. 19 is in the set.80-23=57, not prime.80-29=51, not prime.80-31=49, not prime.80-37=43, prime. 37 is not in the set, 43 is not in the set.So, 37 and43. Both primes not in the original set.So, replacing 2 and7 with37 and43.Sum: 129-2-7=120. Add37+43=80. 120+80=200.So, the set becomes:3,5,11,13,17,19,23,29,37,43.Let me check the sum:3+5=88+11=1919+13=3232+17=4949+19=6868+23=9191+29=120120+37=157157+43=200. Yes, that works.So, another possible set is:3,5,11,13,17,19,23,29,37,43.So, there are multiple solutions. The key is to replace the smallest primes with larger ones in such a way that the total sum reaches 200.Another approach: Maybe replacing 2,3,5 with three larger primes.Original sum 129. Remove 2,3,5: 129-2-3-5=129-10=119. Need three primes p,q,r such that p+q+r=81.Looking for three primes summing to81.This might be more complex, but let's try.Looking for primes around 27 each.Let me think: 23, 29, 29. But duplicates not allowed.Alternatively, 19, 23, 37. 19+23=42, 42+37=79. Not enough.19+23+37=79. Close.19+23+39=81, but 39 not prime.19+29+33=81, 33 not prime.23+29+29=81, duplicates.23+29+29=81, same.Wait, maybe 17+23+41=81. 17+23=40, 40+41=81. Yes, 17,23,41.But 17 and23 are already in the set. So, can't use them again.Alternatively, 13+29+39=81, 39 not prime.11+29+41=81. 11 is in the set, 29 is in the set, 41 is not.So, 11,29,41. But 11 and29 are already in the set.Alternatively, 7+37+37=81, duplicates.7+31+43=81. 7 is in the set, 31 and43 are not.So, replacing 2,3,5 with7,31,43. But 7 is already in the set, so can't do that.Alternatively, 5+31+45=81, 45 not prime.Hmm, this is tricky. Maybe it's better to stick with replacing two primes at a time.So, in summary, there are multiple possible sets. The key is to replace the smallest primes with larger ones until the total sum reaches 200.So, for part 1, one possible set is:3,5,7,11,13,17,19,23,29,73.Another set is:3,7,11,13,17,19,23,29,31,47.And another set is:3,5,11,13,17,19,23,29,37,43.So, the answer isn't unique, but these are possible sets.2. Finding T that maximizes S(T), the number of subsets summing to T.Now, for the second part, we need to find the target T such that the number of subsets of these 10 primes that sum to T is maximized.This is similar to the subset sum problem, where we want to find the T with the maximum number of subsets.In general, for a set of numbers, the number of subsets that sum to a particular T tends to be maximized around the middle of the possible sum range. This is because the number of subsets increases as you approach the middle, due to combinatorial reasons.The total sum of all primes is 200, so the possible subset sums range from 0 (empty set) to 200 (all primes). The number of subsets that sum to T is maximized around T=100, because that's the middle point.However, since all primes are positive integers, and we're dealing with distinct primes, the distribution might be slightly different. But generally, the maximum number of subsets should be around half the total sum.But let's think more carefully.In the subset sum problem, the number of subsets that sum to T is maximized when T is around half the total sum, especially when the numbers are all positive and distinct. This is due to the symmetry of the problem. For each subset that sums to T, there is a complementary subset that sums to total - T. So, the number of subsets for T and total - T are equal. Therefore, the maximum number of subsets should occur at T = total / 2, which is 100 in this case.But wait, is that always true? It depends on the specific set of numbers. For example, if the set has many small numbers, the distribution might be skewed.In our case, the primes are all distinct and range from small to large. The smallest primes are 3,5,7, etc., and the largest could be 73 or similar.Given that, the number of subsets that sum to 100 should be the highest.But let's verify.Alternatively, maybe the maximum occurs slightly above or below 100, depending on the distribution of primes.But without knowing the exact set, it's hard to say. However, since the problem doesn't specify a particular set, just that the sum is 200, we can assume that the maximum occurs at T=100.But wait, in the first part, we have multiple possible sets. Each set might have a different T that maximizes S(T). But the problem says "the mental fortitude of these athletes is challenged by a sequence of mathematical tasks." So, it's referring to the same set of primes from part 1.But since part 1 has multiple possible sets, each with different distributions, the T that maximizes S(T) might vary.However, the problem asks to "find the value of T such that S(T) is maximized." It doesn't specify which set, so perhaps we need to consider the general case, or maybe all sets have the same T.But given that the total sum is 200, and the maximum number of subsets occurs around 100, I think T=100 is the answer.But let me think again.In the subset sum problem, the number of subsets is symmetric around the total sum divided by 2. So, S(T) = S(200 - T). Therefore, the maximum must occur at T=100, because it's the center of symmetry.Therefore, regardless of the specific set, as long as the total sum is 200, the number of subsets is maximized at T=100.Wait, but is that always true? For example, if the set has all numbers even, then the subset sums would also be even, so the maximum might be at 100, but if the set has a mix, it might not.But in our case, primes are mostly odd, except for 2. But in our sets from part 1, 2 is not included in the first set I found (3,5,7,11,13,17,19,23,29,73). So, all primes are odd. Therefore, the subset sums will be:- Even if the subset has an even number of odd primes.- Odd if the subset has an odd number of odd primes.But since all primes are odd, the subset sums will alternate between even and odd depending on the subset size.Wait, no. The sum of an even number of odd numbers is even, and the sum of an odd number of odd numbers is odd.Therefore, the possible subset sums will be either even or odd, depending on the number of primes in the subset.But since the total sum is 200, which is even, the number of subsets that sum to T and 200 - T are equal.But the maximum number of subsets should still occur at T=100, because it's the midpoint.However, since 100 is even, and the subset sums can be both even and odd, but in our case, all primes are odd, so the subset sums will be even if the subset has an even number of primes, and odd if it has an odd number.Therefore, T=100 is even, so it can be achieved by subsets with an even number of primes.But does that affect the maximum?Wait, actually, the number of subsets that sum to T=100 might be less than the number of subsets that sum to T=101, because 101 is odd, and there might be more subsets with an odd number of primes.But no, because the number of subsets is symmetric around 100. For every subset that sums to T, there is a complementary subset that sums to 200 - T. Therefore, the number of subsets for T and 200 - T are equal.But since 100 is the midpoint, it's the only T where S(T) is equal to itself (since 200 - 100 = 100). So, the maximum must occur at T=100.Wait, but if the total sum is even, then the number of subsets that sum to 100 is equal to the number of subsets that sum to 100, which is itself. So, it's the peak.But in reality, the number of subsets might be higher for T=100 than for other Ts, but I'm not entirely sure.Alternatively, maybe the maximum occurs at T=100, but I need to confirm.Let me think about a smaller example. Suppose we have primes 3,5,7. Total sum=15.Possible subset sums:0,3,5,7,8(3+5),10(3+7),12(5+7),15(3+5+7).So, S(0)=1, S(3)=1, S(5)=1, S(7)=1, S(8)=1, S(10)=1, S(12)=1, S(15)=1.Wait, that's not right. Wait, no, actually, for each subset, the sum is unique? No, that's not correct.Wait, in this case, the subset sums are:- {}:0- {3}:3- {5}:5- {7}:7- {3,5}:8- {3,7}:10- {5,7}:12- {3,5,7}:15So, each subset sum is unique, so S(T)=1 for all T except 0 and 15, which have S(T)=1 as well.But that's because the primes are small and unique. So, in this case, the maximum S(T)=1.But if we have more primes, the number of subsets increases, and some sums might overlap.Wait, let's take another example with more primes.Suppose primes are 3,5,7,11. Total sum=26.Possible subset sums:0,3,5,7,11,8(3+5),10(3+7),14(3+11),12(5+7),16(5+11),18(7+11),20(3+5+7),21(3+5+11),23(3+7+11),25(5+7+11),26(3+5+7+11).So, S(T)=1 for each T except:- T=0:1- T=3:1- T=5:1- T=7:1- T=8:1- T=10:1- T=11:1- T=12:1- T=14:1- T=16:1- T=18:1- T=20:1- T=21:1- T=23:1- T=25:1- T=26:1So, again, each T has only one subset. So, S(T)=1 for all T.But that's because the primes are small and unique, leading to unique subset sums.But in our problem, we have 10 primes, which is a larger set, so the number of subsets is 2^10=1024. However, many of these subsets will have the same sum.Therefore, the number of subsets S(T) will vary, with some Ts having many subsets and others few.In such cases, the number of subsets tends to peak around the middle of the possible sum range.Given that, and considering the symmetry, the maximum number of subsets should occur at T=100.But let me think about another example.Suppose we have primes 2,3,5,7. Total sum=17.Possible subset sums:0,2,3,5,7,5(2+3),7(2+5),9(2+7),8(3+5),10(3+7),12(5+7),12(2+3+7),14(2+5+7),15(3+5+7),17(2+3+5+7).So, here, T=12 appears twice: {5,7} and {2,3,7}.Similarly, T=7 appears twice: {7} and {2,5}.So, in this case, the maximum S(T)=2 occurs at T=7 and T=12.But the total sum is 17, so the midpoint is 8.5. So, T=8 and T=9 are around there.But in this case, T=7 and T=12 have the maximum S(T)=2.So, the maximum doesn't necessarily occur exactly at the midpoint, but near it.Similarly, in our problem, with 10 primes summing to 200, the maximum S(T) might occur near 100, but not exactly at 100.But without knowing the exact distribution of primes, it's hard to say.However, given that the primes are all odd (except possibly 2, but in our sets from part 1, 2 is excluded), the subset sums will alternate between even and odd.But since the total sum is even (200), the number of subsets that sum to T and 200 - T are equal.Therefore, the maximum must occur at T=100, because it's the only T where S(T) is equal to itself, and due to the symmetry, it's likely the peak.But wait, in the small example with primes 2,3,5,7, the maximum occurred at T=7 and T=12, which are equidistant from the midpoint (8.5). So, in that case, the maximum occurs at two points symmetric around the midpoint.Similarly, in our problem, the maximum might occur at T=100, but perhaps also at T=100 ±k for some k, but since 100 is the midpoint, it's the most probable candidate.Alternatively, maybe the maximum occurs just below or above 100, but given the symmetry, it's likely at 100.Therefore, I think T=100 is the answer.But let me think again.In the subset sum problem, the number of subsets is maximized around the mean, which is total_sum / 2.In our case, total_sum=200, so mean=100.Therefore, T=100 is the value where S(T) is maximized.Yes, that makes sense.So, the answer is T=100.Final Answer1. One possible set of prime numbers is boxed{3, 5, 7, 11, 13, 17, 19, 23, 29, 73}.2. The value of ( T ) that maximizes ( S(T) ) is boxed{100}.</think>

🧠:<think>Okay, so I have this problem about a playground shaped like an irregular hexagon, and I need to calculate its area using the Shoelace theorem. Then, I also need to figure out the total length of a safety net that extends 1 meter outward from every point along the perimeter. Hmm, let's take this step by step.First, for the area calculation. I remember the Shoelace theorem is a method to find the area of a polygon when you know the coordinates of its vertices. It's called the Shoelace theorem because when you write down the coordinates and multiply them in a certain way, it looks like lacing a shoe. The formula is something like taking the sum of the products of each coordinate and the next one, subtracting the sum of the products going the other way, and then taking half the absolute value of that. Let me write that down to make sure I have it right.The formula is:Area = (1/2) * |sum from i=1 to n of (x_i * y_{i+1} - x_{i+1} * y_i)|Where (x_{n+1}, y_{n+1}) is the same as (x_1, y_1) to close the polygon.So, the vertices given are (0,0), (4,0), (6,3), (4,5), (2,5), and (0,3). Let me list them in order and make sure they are either clockwise or counterclockwise. Looking at the coordinates, starting at (0,0), moving to (4,0), then up to (6,3), then to (4,5), then to (2,5), then to (0,3), and back to (0,0). That seems to form a hexagon without crossing over itself, so it should be fine.I think I should write down the coordinates in order and then apply the Shoelace formula. Let me set up a table to compute the terms x_i * y_{i+1} and x_{i+1} * y_i for each pair of consecutive vertices.So, the vertices in order are:1. (0,0)2. (4,0)3. (6,3)4. (4,5)5. (2,5)6. (0,3)7. Back to (0,0) to close the polygon.Let me compute each term step by step.First, compute x_i * y_{i+1} for each i:1. i=1: x1=0, y2=0. So, 0*0=02. i=2: x2=4, y3=3. So, 4*3=123. i=3: x3=6, y4=5. So, 6*5=304. i=4: x4=4, y5=5. So, 4*5=205. i=5: x5=2, y6=3. So, 2*3=66. i=6: x6=0, y7=0. So, 0*0=0Now, sum all these up: 0 + 12 + 30 + 20 + 6 + 0 = 68Next, compute x_{i+1} * y_i for each i:1. i=1: x2=4, y1=0. So, 4*0=02. i=2: x3=6, y2=0. So, 6*0=03. i=3: x4=4, y3=3. So, 4*3=124. i=4: x5=2, y4=5. So, 2*5=105. i=5: x6=0, y5=5. So, 0*5=06. i=6: x7=0, y6=3. So, 0*3=0Sum these up: 0 + 0 + 12 + 10 + 0 + 0 = 22Now, subtract the second sum from the first sum: 68 - 22 = 46Take half the absolute value: (1/2)*|46| = 23So, the area of the playground is 23 square units. Wait, the coordinates are given in meters, I assume, so the area is 23 square meters? Hmm, that seems a bit small for a playground, but maybe it's a small one. Let me double-check my calculations.Looking back at the first sum: 0 + 12 + 30 + 20 + 6 + 0. That adds up to 68. Correct.Second sum: 0 + 0 + 12 + 10 + 0 + 0. That adds up to 22. Correct.68 - 22 is 46, half of that is 23. Hmm, okay, maybe it's correct. The shape is an irregular hexagon, so maybe it's not too big.Alright, so the area is 23 square meters.Now, moving on to the second part: determining the total length of the safety net required. The net needs to cover the entire hexagonal playground and extend 1 meter outward from every point along the perimeter. So, essentially, the safety net is a buffer zone around the playground, 1 meter wide.Wait, but the problem says the net covers the entire playground and extends 1 meter outward from every point along the perimeter. So, does that mean the net is a larger polygon that is offset outward by 1 meter from the original hexagon? Or is it a perimeter that is extended by 1 meter in some way?Wait, the question says: "the safety net covers the entire hexagonal playground and extends 1 meter outward from every point along the perimeter." Hmm, so it's not just a simple buffer; it's a net that covers the playground and extends 1 meter beyond the perimeter. So, perhaps the net is a combination of the original perimeter and an additional 1 meter around it? Or maybe it's a perimeter that is increased by 1 meter on each side?Wait, actually, the problem says: "the total length of the safety net required, accounting for the extension." So, the perimeter is the original perimeter, but extended outward by 1 meter. Hmm, but how does that affect the length?Wait, when you offset a polygon outward by a certain distance, the perimeter increases by twice the distance times the number of sides, but actually, no, that's not quite right. Because when you offset a polygon, especially a convex polygon, the perimeter increases by 2πr, where r is the offset distance, but for a polygon, it's a bit different because the corners add extra length.Wait, actually, for a polygon, when you offset it outward by a distance d, the perimeter increases by 2πd if it's a circle, but for a polygon, it's more complicated because each corner adds a certain amount of length.Wait, but in this case, the playground is a hexagon, which is a polygon with straight sides. So, if we offset each side outward by 1 meter, the perimeter will increase by the length added at each corner.Wait, actually, when you offset a polygon outward by a distance d, the new perimeter is equal to the original perimeter plus 2πd. But wait, that's only for a circle. For a polygon, the perimeter increases by 2πd minus the original perimeter times something? Hmm, maybe I'm overcomplicating.Alternatively, perhaps the safety net is not just a simple offset polygon, but a buffer zone around the playground, which would form a sort of rounded shape, but since the playground is a polygon, the buffer would have rounded corners. However, the problem says to consider the perimeter as the piecewise linear segments connecting the vertices. So, maybe the net is just the original perimeter plus an additional 1 meter around each side?Wait, no, the problem says the net covers the entire playground and extends 1 meter outward from every point along the perimeter. So, it's not just adding a strip around the playground, but the net is a larger shape that includes the playground and extends 1 meter beyond each point on the perimeter.Wait, but how is that different from just the perimeter? Or is it that the net is a perimeter that is 1 meter away from the original perimeter?Wait, perhaps the net is a closed loop that is 1 meter outside the original playground's perimeter. So, the net would form a larger polygon, each side of which is parallel to the original polygon's sides but offset outward by 1 meter.But in that case, the perimeter of the larger polygon would be equal to the original perimeter plus some additional length due to the corners.Wait, actually, when you offset a polygon outward by a distance d, the perimeter increases by 2πd. But is that accurate?Wait, let me think. For a convex polygon, when you offset it outward by a distance d, each corner effectively becomes a circular arc with radius d, and the total length added is equal to the sum of the arc lengths at each corner. Since each corner is a convex angle, the total added length is equal to the sum over all corners of (π - θ_i) * d, where θ_i is the internal angle at corner i.But for a polygon, the sum of (π - θ_i) over all corners is equal to 2π, because the sum of internal angles is (n-2)π, so the sum of (π - θ_i) is nπ - (n-2)π = 2π.Therefore, the total added perimeter is 2πd.So, in this case, d is 1 meter, so the total added perimeter is 2π*1 = 2π meters.Therefore, the total length of the safety net would be the original perimeter plus 2π meters.Wait, but is that correct? Let me verify.Suppose we have a square with side length a. The perimeter is 4a. If we offset it outward by 1 meter, the new perimeter would be the original perimeter plus 2π*1, because each corner adds a quarter-circle, and four quarter-circles make a full circle, so 2π*1.But wait, actually, for a square, each corner is a right angle, so when you offset outward, each corner becomes a quarter-circle with radius 1 meter. So, each corner adds a length of (π/2)*1, and four corners add up to 2π. So, the total perimeter becomes 4a + 2π.Similarly, for any convex polygon, the perimeter after offsetting outward by d is equal to the original perimeter plus 2πd.Therefore, in our case, the playground is a convex hexagon? Wait, is it convex?Looking at the coordinates: (0,0), (4,0), (6,3), (4,5), (2,5), (0,3). Let me plot these points mentally.Starting at (0,0), moving to (4,0) – that's a straight line along the x-axis. Then to (6,3) – that's a diagonal up. Then to (4,5) – that's a diagonal up and to the left. Then to (2,5) – that's straight left. Then to (0,3) – that's diagonal down and left. Then back to (0,0). Hmm, does this form a convex hexagon?Wait, convex polygons have all interior angles less than 180 degrees. Let me see if any of the vertices are reflex (greater than 180 degrees). Looking at the point (6,3): the angle there. From (4,0) to (6,3) to (4,5). The turn is from southeast to northwest, so the internal angle is probably less than 180. Similarly, at (4,5): from (6,3) to (4,5) to (2,5). That's a straight line? Wait, (6,3) to (4,5) is a line with slope (5-3)/(4-6) = 2/(-2) = -1. Then from (4,5) to (2,5) is a horizontal line to the left. So, the angle at (4,5) is a right angle, which is 90 degrees, so convex.At (2,5): from (4,5) to (2,5) to (0,3). The slope from (2,5) to (0,3) is (3-5)/(0-2) = (-2)/(-2) = 1. So, the angle there is also convex. At (0,3): from (2,5) to (0,3) to (0,0). The slope from (0,3) to (0,0) is undefined (vertical line). So, the angle at (0,3) is between a line with slope 1 and a vertical line, which is less than 180 degrees. Similarly, at (0,0): from (0,3) to (0,0) to (4,0). That's a right angle again. And at (4,0): from (0,0) to (4,0) to (6,3). The slope from (4,0) to (6,3) is (3-0)/(6-4)= 3/2. So, the angle there is also convex.Therefore, the hexagon is convex. So, when we offset it outward by 1 meter, the perimeter increases by 2π meters.Therefore, the total length of the safety net is the original perimeter plus 2π meters.So, first, I need to compute the original perimeter of the hexagon.The perimeter is the sum of the lengths of all sides. So, I need to compute the distance between each pair of consecutive vertices.The vertices are:1. (0,0)2. (4,0)3. (6,3)4. (4,5)5. (2,5)6. (0,3)7. Back to (0,0)So, let's compute each side's length.Side 1: from (0,0) to (4,0). That's along the x-axis, so distance is |4-0| = 4 meters.Side 2: from (4,0) to (6,3). Use the distance formula: sqrt[(6-4)^2 + (3-0)^2] = sqrt[4 + 9] = sqrt[13] ≈ 3.6055 meters.Side 3: from (6,3) to (4,5). Distance: sqrt[(4-6)^2 + (5-3)^2] = sqrt[4 + 4] = sqrt[8] ≈ 2.8284 meters.Side 4: from (4,5) to (2,5). That's along the y=5 line, so distance is |4-2| = 2 meters.Side 5: from (2,5) to (0,3). Distance: sqrt[(0-2)^2 + (3-5)^2] = sqrt[4 + 4] = sqrt[8] ≈ 2.8284 meters.Side 6: from (0,3) to (0,0). That's along the y-axis, so distance is |3-0| = 3 meters.Now, let's sum these up:Side 1: 4Side 2: sqrt(13) ≈ 3.6055Side 3: sqrt(8) ≈ 2.8284Side 4: 2Side 5: sqrt(8) ≈ 2.8284Side 6: 3Adding them together:4 + 3.6055 + 2.8284 + 2 + 2.8284 + 3Let me compute step by step:4 + 3.6055 = 7.60557.6055 + 2.8284 = 10.433910.4339 + 2 = 12.433912.4339 + 2.8284 = 15.262315.2623 + 3 = 18.2623 meters.So, the original perimeter is approximately 18.2623 meters.But wait, let me compute it more accurately without approximating the square roots.Compute each side exactly:Side 1: 4Side 2: sqrt(13)Side 3: sqrt(8) = 2*sqrt(2)Side 4: 2Side 5: sqrt(8) = 2*sqrt(2)Side 6: 3So, total perimeter P = 4 + sqrt(13) + 2*sqrt(2) + 2 + 2*sqrt(2) + 3Combine like terms:Constants: 4 + 2 + 3 = 9sqrt(13): sqrt(13)sqrt(2) terms: 2*sqrt(2) + 2*sqrt(2) = 4*sqrt(2)So, P = 9 + sqrt(13) + 4*sqrt(2)That's the exact perimeter.So, the original perimeter is 9 + sqrt(13) + 4*sqrt(2) meters.Now, when we offset the polygon outward by 1 meter, the perimeter increases by 2π meters, as per the earlier reasoning.Therefore, the total length of the safety net is P + 2π = 9 + sqrt(13) + 4*sqrt(2) + 2π meters.But let me make sure that this is correct.Wait, is the perimeter of the offset polygon equal to the original perimeter plus 2π*d? For a convex polygon, yes, because the offset adds a circular buffer around each vertex, effectively adding a semicircle at each vertex, but since each vertex is a corner, the total added length is 2π*d.Wait, actually, for each vertex, when you offset outward, the corner is replaced by a circular arc with radius d, and the angle of the arc is equal to the external angle at that vertex.For a convex polygon, the external angles sum up to 2π radians. Therefore, the total length added by all the arcs is 2π*d.Therefore, the perimeter of the offset polygon is the original perimeter plus 2π*d.In this case, d=1, so the total perimeter is P + 2π.Therefore, the total length of the safety net is (9 + sqrt(13) + 4*sqrt(2)) + 2π meters.So, that's the exact value.Alternatively, if we need a numerical approximation, we can compute it:sqrt(13) ≈ 3.6055sqrt(2) ≈ 1.4142So, 4*sqrt(2) ≈ 5.6568Therefore, 9 + 3.6055 + 5.6568 ≈ 9 + 3.6055 = 12.6055 + 5.6568 ≈ 18.2623Then, 2π ≈ 6.2832So, total length ≈ 18.2623 + 6.2832 ≈ 24.5455 meters.But since the problem asks for the exact value, we should present it in terms of sqrt and π.Therefore, the total length is 9 + sqrt(13) + 4*sqrt(2) + 2π meters.Wait, but let me double-check if the perimeter increases by exactly 2π meters. Because when you offset a polygon, the increase in perimeter is 2π*d only if the polygon is convex and the offset is such that the buffer doesn't cause any overlaps or self-intersections.In our case, since the playground is a convex hexagon, and we're offsetting outward by 1 meter, which is a small distance relative to the size of the playground, I think it's safe to assume that the perimeter increases by 2π meters.Alternatively, another way to think about it is that the buffer zone around the polygon is akin to a Minkowski sum with a disk of radius 1. The perimeter of the Minkowski sum of a convex polygon and a disk is equal to the perimeter of the polygon plus 2π times the radius of the disk.Yes, that's a theorem in geometry. So, that confirms it.Therefore, the total length of the safety net is the original perimeter plus 2π meters.So, putting it all together, the exact area is 23 square meters, and the exact length of the safety net is 9 + sqrt(13) + 4*sqrt(2) + 2π meters.Wait, but let me confirm the original perimeter calculation once more.Original perimeter:Sides:1. (0,0) to (4,0): 42. (4,0) to (6,3): sqrt[(6-4)^2 + (3-0)^2] = sqrt[4 + 9] = sqrt(13)3. (6,3) to (4,5): sqrt[(4-6)^2 + (5-3)^2] = sqrt[4 + 4] = sqrt(8) = 2*sqrt(2)4. (4,5) to (2,5): 25. (2,5) to (0,3): sqrt[(0-2)^2 + (3-5)^2] = sqrt[4 + 4] = sqrt(8) = 2*sqrt(2)6. (0,3) to (0,0): 3So, adding up: 4 + sqrt(13) + 2*sqrt(2) + 2 + 2*sqrt(2) + 3Combine constants: 4 + 2 + 3 = 9Combine sqrt(13): sqrt(13)Combine sqrt(2): 2*sqrt(2) + 2*sqrt(2) = 4*sqrt(2)So, total perimeter: 9 + sqrt(13) + 4*sqrt(2). Correct.Therefore, the safety net length is 9 + sqrt(13) + 4*sqrt(2) + 2π.So, summarizing:1. Area using Shoelace theorem: 23 m²2. Safety net length: 9 + sqrt(13) + 4*sqrt(2) + 2π metersI think that's it. Let me just make sure I didn't make any calculation errors.For the area, the Shoelace formula gave me 23. Let me recount:First sum: 0 + 12 + 30 + 20 + 6 + 0 = 68Second sum: 0 + 0 + 12 + 10 + 0 + 0 = 22Difference: 68 - 22 = 46Half of that: 23. Correct.For the perimeter, each side was calculated correctly, and the total is 9 + sqrt(13) + 4*sqrt(2). Then, adding 2π for the offset. Correct.So, I think I'm confident with these answers.Final Answer1. The exact area of the playground is boxed{23} square meters.2. The total length of the safety net required is boxed{9 + sqrt{13} + 4sqrt{2} + 2pi} meters.</think>

🧠:<think>Alright, so I have this integral to solve: ( I = int_0^1 e^{x^2} cdot sin(pi x) , dx ). Hmm, that looks pretty challenging. I remember that integrals involving exponentials and trigonometric functions can sometimes be tricky, especially when they're multiplied together. Let me think about the techniques I know for integrating such functions.First, I recall that integration by parts is a common method for products of functions. The formula is ( int u , dv = uv - int v , du ). Maybe I can set ( u = sin(pi x) ) and ( dv = e^{x^2} dx ). But wait, if I do that, then I need to find ( du ) and ( v ). The derivative of ( sin(pi x) ) is ( pi cos(pi x) ), so ( du = pi cos(pi x) dx ). But integrating ( dv = e^{x^2} dx ) to get ( v ) is problematic because ( int e^{x^2} dx ) doesn't have an elementary antiderivative. That's a dead end.Maybe I should try switching the roles. Let ( u = e^{x^2} ) and ( dv = sin(pi x) dx ). Then, ( du = 2x e^{x^2} dx ) and ( v = -frac{1}{pi} cos(pi x) ). Plugging into integration by parts, I get:[I = uv|_0^1 - int_0^1 v , du = left[ -frac{1}{pi} e^{x^2} cos(pi x) right]_0^1 + frac{2}{pi} int_0^1 x e^{x^2} cos(pi x) dx]Okay, so now I have another integral to solve: ( int_0^1 x e^{x^2} cos(pi x) dx ). Hmm, that doesn't look much simpler. Maybe I can apply integration by parts again on this new integral. Let me set ( u = x e^{x^2} ) and ( dv = cos(pi x) dx ). Then, ( du = (e^{x^2} + 2x^2 e^{x^2}) dx ) and ( v = frac{1}{pi} sin(pi x) ).Applying integration by parts again:[int x e^{x^2} cos(pi x) dx = frac{x e^{x^2}}{pi} sin(pi x) - int frac{1}{pi} sin(pi x) (e^{x^2} + 2x^2 e^{x^2}) dx]Simplifying, that becomes:[frac{x e^{x^2}}{pi} sin(pi x) - frac{1}{pi} int e^{x^2} sin(pi x) dx - frac{2}{pi} int x^2 e^{x^2} sin(pi x) dx]Wait a second, the first integral here is our original integral ( I ), and the second integral is similar but with an extra ( x^2 ) term. This seems to be getting more complicated rather than simpler. Maybe integration by parts isn't the right approach here.Let me think about other methods. Sometimes, integrals involving ( e^{x^2} ) can be expressed in terms of the error function, but I don't think that applies here because of the sine term. Alternatively, perhaps a series expansion could help. I know that both ( e^{x^2} ) and ( sin(pi x) ) can be expressed as power series.The Taylor series for ( e^{x^2} ) is ( sum_{n=0}^{infty} frac{x^{2n}}{n!} ), and the Taylor series for ( sin(pi x) ) is ( sum_{m=0}^{infty} (-1)^m frac{(pi x)^{2m+1}}{(2m+1)!} ). Maybe I can multiply these two series together and integrate term by term.So, multiplying the two series:[e^{x^2} sin(pi x) = left( sum_{n=0}^{infty} frac{x^{2n}}{n!} right) left( sum_{m=0}^{infty} (-1)^m frac{(pi x)^{2m+1}}{(2m+1)!} right )]Multiplying these gives a double sum:[sum_{n=0}^{infty} sum_{m=0}^{infty} (-1)^m frac{pi^{2m+1} x^{2n + 2m + 1}}{n! (2m+1)!}]Now, integrating term by term from 0 to 1:[I = int_0^1 sum_{n=0}^{infty} sum_{m=0}^{infty} (-1)^m frac{pi^{2m+1} x^{2n + 2m + 1}}{n! (2m+1)!} dx = sum_{n=0}^{infty} sum_{m=0}^{infty} (-1)^m frac{pi^{2m+1}}{n! (2m+1)!} int_0^1 x^{2n + 2m + 1} dx]The integral of ( x^{k} ) from 0 to 1 is ( frac{1}{k+1} ), so:[I = sum_{n=0}^{infty} sum_{m=0}^{infty} (-1)^m frac{pi^{2m+1}}{n! (2m+1)! (2n + 2m + 2)}]Hmm, that's a double series. It might converge, but it's not exactly giving me an exact value in a closed form. Maybe I can change the order of summation or find a way to express it more compactly. Alternatively, perhaps recognizing that this integral doesn't have an elementary antiderivative, so the best we can do is express it as a series.Alternatively, maybe using complex analysis or special functions. I recall that integrals involving ( e^{x^2} ) can sometimes be related to the error function, but with the sine term, it might be more complicated. Let me see if I can express ( sin(pi x) ) in terms of complex exponentials.We know that ( sin(pi x) = frac{e^{ipi x} - e^{-ipi x}}{2i} ). So, substituting that into the integral:[I = frac{1}{2i} int_0^1 e^{x^2} (e^{ipi x} - e^{-ipi x}) dx = frac{1}{2i} left( int_0^1 e^{x^2 + ipi x} dx - int_0^1 e^{x^2 - ipi x} dx right )]Hmm, so now we have two integrals of the form ( int e^{x^2 + a x} dx ). I remember that integrals of ( e^{x^2} ) don't have elementary forms, but perhaps we can express them in terms of the error function. The integral ( int e^{x^2 + a x} dx ) can be related to the imaginary error function or something similar.Let me recall the formula for ( int e^{x^2 + a x} dx ). I think it can be expressed as ( e^{a^2/4} sqrt{pi} text{erfi}left( x + frac{a}{2} right ) / 2 ), but I'm not entirely sure. Let me check the differentiation:Let ( f(x) = e^{x^2 + a x} ). Then, ( f'(x) = (2x + a) e^{x^2 + a x} ). Hmm, not directly helpful. Alternatively, perhaps completing the square in the exponent.For the exponent ( x^2 + a x ), complete the square:( x^2 + a x = (x + a/2)^2 - a^2/4 ). So,[int e^{x^2 + a x} dx = e^{a^2/4} int e^{(x + a/2)^2} dx]But ( int e^{(x + a/2)^2} dx ) is related to the error function. Specifically,[int e^{(x + a/2)^2} dx = sqrt{pi} text{erfi}(x + a/2) / 2]Wait, actually, the integral of ( e^{t^2} ) is ( sqrt{pi} text{erfi}(t) / 2 ). So, substituting back,[int e^{x^2 + a x} dx = e^{a^2/4} cdot sqrt{pi} text{erfi}(x + a/2) / 2 + C]Therefore, applying this to our integrals:For ( int_0^1 e^{x^2 + ipi x} dx ), let ( a = ipi ). Then,[int_0^1 e^{x^2 + ipi x} dx = e^{(ipi)^2 / 4} cdot frac{sqrt{pi}}{2} left[ text{erfi}left(1 + frac{ipi}{2}right) - text{erfi}left(0 + frac{ipi}{2}right) right ]]Similarly, for ( int_0^1 e^{x^2 - ipi x} dx ), let ( a = -ipi ):[int_0^1 e^{x^2 - ipi x} dx = e^{(-ipi)^2 / 4} cdot frac{sqrt{pi}}{2} left[ text{erfi}left(1 - frac{ipi}{2}right) - text{erfi}left(0 - frac{ipi}{2}right) right ]]Simplify the exponents:( (ipi)^2 = -pi^2 ), so ( e^{(ipi)^2 / 4} = e^{-pi^2 / 4} ). Similarly, ( e^{(-ipi)^2 / 4} = e^{-pi^2 / 4} ).So, both integrals have a factor of ( e^{-pi^2 / 4} ). Therefore, putting it all together:[I = frac{1}{2i} e^{-pi^2 / 4} cdot frac{sqrt{pi}}{2} left[ text{erfi}left(1 + frac{ipi}{2}right) - text{erfi}left(frac{ipi}{2}right) - text{erfi}left(1 - frac{ipi}{2}right) + text{erfi}left(-frac{ipi}{2}right) right ]]Hmm, that's a bit complicated, but perhaps we can simplify it further. I know that the error function is an odd function, so ( text{erfi}(-z) = -text{erfi}(z) ). Therefore, ( text{erfi}left(-frac{ipi}{2}right) = -text{erfi}left(frac{ipi}{2}right) ).So, substituting that in:[I = frac{sqrt{pi}}{4i} e^{-pi^2 / 4} left[ text{erfi}left(1 + frac{ipi}{2}right) - text{erfi}left(frac{ipi}{2}right) - text{erfi}left(1 - frac{ipi}{2}right) + text{erfi}left(frac{ipi}{2}right) right ]]Notice that the ( - text{erfi}left(frac{ipi}{2}right) ) and ( + text{erfi}left(frac{ipi}{2}right) ) cancel each other out. So, we're left with:[I = frac{sqrt{pi}}{4i} e^{-pi^2 / 4} left[ text{erfi}left(1 + frac{ipi}{2}right) - text{erfi}left(1 - frac{ipi}{2}right) right ]]Now, let's consider ( text{erfi}(z) ) for complex arguments. I recall that ( text{erfi}(z) = -i text{erf}(iz) ), where ( text{erf}(z) ) is the error function. So, substituting this in:[text{erfi}left(1 + frac{ipi}{2}right) = -i text{erf}left( i left(1 + frac{ipi}{2}right) right ) = -i text{erf}left( -frac{pi}{2} + i right )][text{erfi}left(1 - frac{ipi}{2}right) = -i text{erf}left( i left(1 - frac{ipi}{2}right) right ) = -i text{erf}left( frac{pi}{2} + i right )]So, plugging these back into the expression for ( I ):[I = frac{sqrt{pi}}{4i} e^{-pi^2 / 4} left[ -i text{erf}left( -frac{pi}{2} + i right ) + i text{erf}left( frac{pi}{2} + i right ) right ]]Simplify the constants:The ( -i ) and ( +i ) inside the brackets factor out:[I = frac{sqrt{pi}}{4i} e^{-pi^2 / 4} cdot i left[ - text{erf}left( -frac{pi}{2} + i right ) + text{erf}left( frac{pi}{2} + i right ) right ]]The ( i ) in the numerator and denominator cancels out:[I = frac{sqrt{pi}}{4} e^{-pi^2 / 4} left[ - text{erf}left( -frac{pi}{2} + i right ) + text{erf}left( frac{pi}{2} + i right ) right ]]Now, using the property of the error function that ( text{erf}(-z) = -text{erf}(z) ), we can rewrite ( text{erf}left( -frac{pi}{2} + i right ) = -text{erf}left( frac{pi}{2} - i right ) ). So,[I = frac{sqrt{pi}}{4} e^{-pi^2 / 4} left[ text{erf}left( frac{pi}{2} - i right ) + text{erf}left( frac{pi}{2} + i right ) right ]]Hmm, that's still a bit complex, but maybe we can relate this to hypergeometric functions or something else. Alternatively, perhaps recognizing that this integral doesn't have a closed-form expression in terms of elementary functions, and the best we can do is express it in terms of the error function with complex arguments.Alternatively, maybe using numerical methods to approximate the integral. However, since the problem asks for the exact value, perhaps the integral is intended to be expressed in terms of the error function as we've done here.So, putting it all together, the exact value of the integral is:[I = frac{sqrt{pi}}{4} e^{-pi^2 / 4} left[ text{erf}left( frac{pi}{2} - i right ) + text{erf}left( frac{pi}{2} + i right ) right ]]But I'm not entirely sure if this is the most simplified form or if there's a way to express this more elegantly. Maybe using properties of the error function or recognizing some symmetry. Alternatively, perhaps the integral can be expressed in terms of sine and cosine integrals, but I'm not sure.Wait, another thought: sometimes integrals involving ( e^{x^2} ) and trigonometric functions can be related to the imaginary error function or the Dawson function. Let me recall that the Dawson function is defined as ( F(x) = e^{-x^2} int_0^x e^{t^2} dt ), which is related to the error function.But in our case, we have ( e^{x^2} ) multiplied by sine, so maybe integrating by parts in a different way or using a different substitution. Alternatively, perhaps using the method of stationary phase or other asymptotic methods, but that might be overcomplicating things.Given that the integral doesn't seem to have an elementary antiderivative, and after trying integration by parts and series expansion, the most precise exact expression we can get is in terms of the error function with complex arguments. So, I think that's the answer we have to settle with.Moving on to the linear algebra problem. We have a 3x3 matrix A:[A = begin{pmatrix}2 & -1 & 0 -1 & 2 & -1 0 & -1 & 2end{pmatrix}]We need to find the eigenvalues and eigenvectors.First, let's find the eigenvalues. The eigenvalues are the solutions to the characteristic equation ( det(A - lambda I) = 0 ).So, let's compute the determinant of ( A - lambda I ):[det begin{pmatrix}2 - lambda & -1 & 0 -1 & 2 - lambda & -1 0 & -1 & 2 - lambdaend{pmatrix} = 0]This is a tridiagonal matrix, and I remember that for such matrices, especially symmetric ones, there are sometimes patterns or known eigenvalues.But let's compute the determinant step by step. The determinant of a 3x3 matrix can be computed using the rule of Sarrus or expansion by minors. Let's expand along the first row.The determinant is:[(2 - lambda) cdot det begin{pmatrix}2 - lambda & -1 -1 & 2 - lambdaend{pmatrix} - (-1) cdot det begin{pmatrix}-1 & -1 0 & 2 - lambdaend{pmatrix} + 0 cdot det(...)]Simplify each minor:First minor: ( det begin{pmatrix} 2 - lambda & -1 -1 & 2 - lambda end{pmatrix} = (2 - lambda)^2 - (-1)(-1) = (2 - lambda)^2 - 1 )Second minor: ( det begin{pmatrix} -1 & -1 0 & 2 - lambda end{pmatrix} = (-1)(2 - lambda) - (-1)(0) = - (2 - lambda) )Third term is zero, so we can ignore it.Putting it all together:[(2 - lambda)[(2 - lambda)^2 - 1] + 1 cdot [ - (2 - lambda) ] = 0]Factor out ( (2 - lambda) ):[(2 - lambda) left[ (2 - lambda)^2 - 1 - 1 right ] = 0]Wait, no. Let's compute it step by step:First term: ( (2 - lambda)[(2 - lambda)^2 - 1] )Second term: ( +1 cdot [ - (2 - lambda) ] = - (2 - lambda) )So, the equation is:[(2 - lambda)[(2 - lambda)^2 - 1] - (2 - lambda) = 0]Factor out ( (2 - lambda) ):[(2 - lambda) left[ (2 - lambda)^2 - 1 - 1 right ] = 0][(2 - lambda) left[ (2 - lambda)^2 - 2 right ] = 0]So, set each factor equal to zero:1. ( 2 - lambda = 0 ) => ( lambda = 2 )2. ( (2 - lambda)^2 - 2 = 0 ) => ( (2 - lambda)^2 = 2 ) => ( 2 - lambda = pm sqrt{2} ) => ( lambda = 2 mp sqrt{2} )Therefore, the eigenvalues are ( lambda_1 = 2 ), ( lambda_2 = 2 - sqrt{2} ), and ( lambda_3 = 2 + sqrt{2} ).Wait, hold on. Let me verify the determinant calculation again because I might have made a mistake in the expansion.Original determinant:[(2 - lambda) cdot [(2 - lambda)^2 - 1] - (-1) cdot [ -1 cdot (2 - lambda) - (-1) cdot 0 ] + 0]Wait, no, the second minor was ( det begin{pmatrix} -1 & -1 0 & 2 - lambda end{pmatrix} ), which is ( (-1)(2 - lambda) - (-1)(0) = - (2 - lambda) ). So, the second term is ( -1 cdot (- (2 - lambda)) = (2 - lambda) ).So, the determinant equation becomes:[(2 - lambda)[(2 - lambda)^2 - 1] + (2 - lambda) = 0]Factor out ( (2 - lambda) ):[(2 - lambda) left[ (2 - lambda)^2 - 1 + 1 right ] = 0][(2 - lambda) left[ (2 - lambda)^2 right ] = 0]Wait, that can't be right because that would imply ( (2 - lambda)^3 = 0 ), which would mean a triple eigenvalue at 2, but that contradicts the earlier result. I must have made a mistake in the expansion.Let me recompute the determinant step by step.The determinant is:[(2 - lambda) cdot det begin{pmatrix} 2 - lambda & -1 -1 & 2 - lambda end{pmatrix} - (-1) cdot det begin{pmatrix} -1 & -1 0 & 2 - lambda end{pmatrix} + 0 cdot det(...)]Compute each minor:First minor: ( (2 - lambda)(2 - lambda) - (-1)(-1) = (2 - lambda)^2 - 1 )Second minor: ( (-1)(2 - lambda) - (-1)(0) = - (2 - lambda) )So, the determinant is:[(2 - lambda)[(2 - lambda)^2 - 1] - (-1)[ - (2 - lambda) ] = (2 - lambda)[(2 - lambda)^2 - 1] - (2 - lambda)]Factor out ( (2 - lambda) ):[(2 - lambda) left[ (2 - lambda)^2 - 1 - 1 right ] = (2 - lambda) left[ (2 - lambda)^2 - 2 right ] = 0]So, that gives:Either ( 2 - lambda = 0 ) => ( lambda = 2 ), or ( (2 - lambda)^2 - 2 = 0 ) => ( (2 - lambda)^2 = 2 ) => ( 2 - lambda = pm sqrt{2} ) => ( lambda = 2 mp sqrt{2} )Therefore, the eigenvalues are ( lambda_1 = 2 ), ( lambda_2 = 2 - sqrt{2} ), and ( lambda_3 = 2 + sqrt{2} ). That seems correct.Now, let's find the eigenvectors for each eigenvalue.Starting with ( lambda = 2 ):We need to solve ( (A - 2I) mathbf{x} = 0 ). Compute ( A - 2I ):[A - 2I = begin{pmatrix}0 & -1 & 0 -1 & 0 & -1 0 & -1 & 0end{pmatrix}]This matrix has a lot of symmetry. Let's write the system of equations:1. ( 0 cdot x_1 - 1 cdot x_2 + 0 cdot x_3 = 0 ) => ( -x_2 = 0 ) => ( x_2 = 0 )2. ( -1 cdot x_1 + 0 cdot x_2 - 1 cdot x_3 = 0 ) => ( -x_1 - x_3 = 0 ) => ( x_1 = -x_3 )3. ( 0 cdot x_1 - 1 cdot x_2 + 0 cdot x_3 = 0 ) => ( -x_2 = 0 ) => ( x_2 = 0 )So, from equations 1 and 3, ( x_2 = 0 ). From equation 2, ( x_1 = -x_3 ). Let ( x_3 = t ), then ( x_1 = -t ). So, the eigenvectors are of the form ( begin{pmatrix} -t 0 t end{pmatrix} = t begin{pmatrix} -1 0 1 end{pmatrix} ). So, an eigenvector is ( begin{pmatrix} -1 0 1 end{pmatrix} ).But wait, let's check if this satisfies all equations. Plugging into equation 2: ( -(-1) - 1 = 1 - 1 = 0 ), which works. So, yes, that's correct.However, since the matrix is symmetric, we should have orthogonal eigenvectors. Let me see if this eigenvector is orthogonal to the others we'll find. But since we're just finding one eigenvector for ( lambda = 2 ), we can proceed.Next, for ( lambda = 2 - sqrt{2} ):Compute ( A - (2 - sqrt{2})I ):[A - (2 - sqrt{2})I = begin{pmatrix}2 - (2 - sqrt{2}) & -1 & 0 -1 & 2 - (2 - sqrt{2}) & -1 0 & -1 & 2 - (2 - sqrt{2})end{pmatrix} = begin{pmatrix}sqrt{2} & -1 & 0 -1 & sqrt{2} & -1 0 & -1 & sqrt{2}end{pmatrix}]Now, we need to solve ( (A - (2 - sqrt{2})I) mathbf{x} = 0 ). Let's write the system:1. ( sqrt{2} x_1 - x_2 = 0 ) => ( x_2 = sqrt{2} x_1 )2. ( -x_1 + sqrt{2} x_2 - x_3 = 0 )3. ( -x_2 + sqrt{2} x_3 = 0 ) => ( x_2 = sqrt{2} x_3 )From equation 1: ( x_2 = sqrt{2} x_1 )From equation 3: ( x_2 = sqrt{2} x_3 ) => ( sqrt{2} x_1 = sqrt{2} x_3 ) => ( x_1 = x_3 )Let ( x_1 = t ), then ( x_3 = t ) and ( x_2 = sqrt{2} t ).Now, plug into equation 2:( -t + sqrt{2} (sqrt{2} t) - t = -t + 2 t - t = 0 ). So, it works.Therefore, the eigenvector is ( begin{pmatrix} t sqrt{2} t t end{pmatrix} = t begin{pmatrix} 1 sqrt{2} 1 end{pmatrix} ).Similarly, for ( lambda = 2 + sqrt{2} ):Compute ( A - (2 + sqrt{2})I ):[A - (2 + sqrt{2})I = begin{pmatrix}2 - (2 + sqrt{2}) & -1 & 0 -1 & 2 - (2 + sqrt{2}) & -1 0 & -1 & 2 - (2 + sqrt{2})end{pmatrix} = begin{pmatrix}- sqrt{2} & -1 & 0 -1 & - sqrt{2} & -1 0 & -1 & - sqrt{2}end{pmatrix}]Solve ( (A - (2 + sqrt{2})I) mathbf{x} = 0 ):1. ( -sqrt{2} x_1 - x_2 = 0 ) => ( x_2 = -sqrt{2} x_1 )2. ( -x_1 - sqrt{2} x_2 - x_3 = 0 )3. ( -x_2 - sqrt{2} x_3 = 0 ) => ( x_2 = -sqrt{2} x_3 )From equation 1: ( x_2 = -sqrt{2} x_1 )From equation 3: ( x_2 = -sqrt{2} x_3 ) => ( -sqrt{2} x_1 = -sqrt{2} x_3 ) => ( x_1 = x_3 )Let ( x_1 = t ), then ( x_3 = t ) and ( x_2 = -sqrt{2} t ).Plug into equation 2:( -t - sqrt{2} (-sqrt{2} t) - t = -t + 2 t - t = 0 ). So, it works.Therefore, the eigenvector is ( begin{pmatrix} t -sqrt{2} t t end{pmatrix} = t begin{pmatrix} 1 -sqrt{2} 1 end{pmatrix} ).So, summarizing the eigenvalues and eigenvectors:- Eigenvalue ( lambda = 2 ): Eigenvector ( begin{pmatrix} -1 0 1 end{pmatrix} )- Eigenvalue ( lambda = 2 - sqrt{2} ): Eigenvector ( begin{pmatrix} 1 sqrt{2} 1 end{pmatrix} )- Eigenvalue ( lambda = 2 + sqrt{2} ): Eigenvector ( begin{pmatrix} 1 -sqrt{2} 1 end{pmatrix} )It's also worth noting that these eigenvectors are orthogonal to each other, which makes sense because the matrix A is symmetric, and symmetric matrices have orthogonal eigenvectors corresponding to distinct eigenvalues.Let me verify the orthogonality:Take eigenvectors for ( lambda = 2 ) and ( lambda = 2 - sqrt{2} ):Dot product: ( (-1)(1) + (0)(sqrt{2}) + (1)(1) = -1 + 0 + 1 = 0 ). So, they are orthogonal.Similarly, eigenvectors for ( lambda = 2 ) and ( lambda = 2 + sqrt{2} ):Dot product: ( (-1)(1) + (0)(-sqrt{2}) + (1)(1) = -1 + 0 + 1 = 0 ). Orthogonal.Eigenvectors for ( lambda = 2 - sqrt{2} ) and ( lambda = 2 + sqrt{2} ):Dot product: ( (1)(1) + (sqrt{2})(-sqrt{2}) + (1)(1) = 1 - 2 + 1 = 0 ). Orthogonal.Good, so all eigenvectors are orthogonal, which is consistent with the properties of symmetric matrices.Therefore, we've successfully found the eigenvalues and corresponding eigenvectors for matrix A.</think>

🧠:<think>First, I need to determine the total reduction in carbon emissions over five years from installing solar panels. The city plans to install solar panels on 15,000 square meters of public building rooftops, and each square meter reduces emissions by 0.2 metric tons per year. I'll calculate the annual reduction by multiplying the area by the reduction rate: 15,000 square meters * 0.2 metric tons/square meter/year = 3,000 metric tons per year. Over five years, this amounts to 3,000 metric tons/year * 5 years = 15,000 metric tons.Next, I'll calculate the reduction from expanding urban green spaces. The city is converting 100 hectares, and each hectare absorbs 0.5 metric tons of CO2 per year. The annual reduction from green spaces is 100 hectares * 0.5 metric tons/hectare/year = 50 metric tons per year. Over five years, this is 50 metric tons/year * 5 years = 250 metric tons.Now, I'll find the total reduction from both initiatives: 15,000 metric tons + 250 metric tons = 15,250 metric tons.The city's current annual emissions are 1,200,000 metric tons, so a 30% reduction target is 0.3 * 1,200,000 metric tons = 360,000 metric tons over five years.Finally, I'll determine how much more reduction is needed by subtracting the achieved reduction from the target: 360,000 metric tons - 15,250 metric tons = 344,750 metric tons. The city needs an additional 344,750 metric tons reduction over five years to meet the target.</think>