💡First, calculate the annual cost of the former apartment:- **Monthly Rent**: 750 sq ft * 2/sq ft = 1,500- **Annual Rent**: 1,500 * 12 = 18,000Next, calculate the annual rent for the new apartment:- **First Six Months Rent**: 2,800 * 6 = 16,800- **Last Six Months Rent**: 2,800 * 1.05 = 2,940; 2,940 * 6 = 17,640- **Total Annual Rent**: 16,800 + 17,640 = 34,440Calculate the annual utilities cost:- **Winter Months (3 months)**: 200 * 3 = 600- **Other Months (9 months)**: 150 * 9 = 1,350- **Total Annual Utilities**: 600 + 1,350 = 1,950Adjust for prorated rent:- **First Month (20 days)**: (2,800 / 30) * 20 = 1,866.67- **Last Month (15 days)**: (2,940 / 30) * 15 = 1,470- **Total Prorated Rent**: 1,866.67 + 1,470 = 3,336.67Adjust the total annual rent:- **Adjusted Annual Rent**: 34,440 - (2,800 + 2,940 - 3,336.67) = 29,036.67Calculate the total annual cost for the new apartment:- **Total Cost**: 29,036.67 + 1,950 = 30,986.67Determine John's share:- **John's Share**: 30,986.67 / 2 = 15,493.34Finally, calculate the savings:- **Savings**: 18,000 - 15,493.34 = 2,506.66

💡Alright, so I'm trying to tackle this problem where we have a set X partitioned into three subsets A, B, and C. The condition given is that if you take one number from each of A, B, and C, the largest number among them doesn't equal the sum of the other two. And we need to prove that the sizes of A, B, and C can't all be equal.Hmm, okay. Let me start by understanding the problem better. A partition means that every element of X is in exactly one of A, B, or C. So, A, B, and C are disjoint and their union is X. The condition is about the relationship between numbers selected from each set. Specifically, if I pick any a from A, b from B, and c from C, then the largest of these three numbers shouldn't be equal to the sum of the other two.So, for example, if a is the largest, then a shouldn't be equal to b + c. Similarly, if b is the largest, then b shouldn't be equal to a + c, and the same for c. This seems like a condition that prevents certain kinds of arithmetic progressions or specific number combinations within the sets.Now, the goal is to show that the sizes of A, B, and C can't all be the same. That is, |A| ≠ |B| or |B| ≠ |C| or |A| ≠ |C|. So, we need to show that at least two of these sets must have different sizes.Let me think about how to approach this. Maybe I can assume, for the sake of contradiction, that |A| = |B| = |C| and then show that this leads to a violation of the given condition.Suppose |A| = |B| = |C|. Let's denote the size of each set as n. So, |A| = |B| = |C| = n.Now, since A, B, and C are partitions of X, the total number of elements in X is 3n. So, X has 3n elements.Now, let's think about the numbers in these sets. Since we're dealing with numbers, I assume they are integers, but the problem doesn't specify. Maybe they are just elements of some set with an order, but perhaps it's safer to assume they are integers.If they are integers, then we can order them. Let's say the smallest number in X is 1, and the largest is 3n. So, X = {1, 2, 3, ..., 3n}.Now, if A, B, and C each have n elements, they must each contain some subset of these numbers.Given the condition that in any selection of one number from each set, the largest doesn't equal the sum of the other two, we need to ensure that for any a ∈ A, b ∈ B, c ∈ C, the largest among a, b, c is not equal to the sum of the other two.So, for example, if a is the largest, then a ≠ b + c. Similarly for b and c.This seems similar to avoiding solutions to the equation a = b + c, b = a + c, or c = a + b, depending on which is the largest.So, if we can show that when |A| = |B| = |C|, such a condition cannot be satisfied, then we have our contradiction.Alternatively, maybe we can construct a specific case where if |A| = |B| = |C|, then there must exist some a, b, c such that the largest equals the sum of the other two.Let me try to think of a specific example. Suppose n = 1. Then, A, B, and C each have one element. Let's say A = {1}, B = {2}, C = {3}. Then, taking one from each set, we have 1, 2, 3. The largest is 3, which is equal to 1 + 2. So, this violates the condition. Hence, for n = 1, the condition fails.Wait, but n = 1 is trivial. Maybe for larger n, it's possible. Let's try n = 2.Suppose A = {1, 4}, B = {2, 5}, C = {3, 6}. Now, let's check all combinations.Take 1 from A, 2 from B, 3 from C. Largest is 3, which is not equal to 1 + 2 = 3. Wait, that's equal. So, that violates the condition.Hmm, so in this case, even with n = 2, it's possible to have such a combination.Wait, maybe I need to choose the sets differently. Maybe if I spread out the numbers more.Let me try A = {1, 5}, B = {2, 6}, C = {3, 7}. Now, check combinations.1, 2, 3: largest is 3, which is equal to 1 + 2. Again, violates the condition.Hmm, maybe it's unavoidable when |A| = |B| = |C|.Alternatively, maybe if I choose A, B, C in a way that avoids such sums.But it's tricky. Maybe I need a more general approach.Let me think about the numbers in A, B, and C. Since they are partitions, every number from 1 to 3n is in exactly one of A, B, or C.Now, suppose we have |A| = |B| = |C| = n.Let me consider the numbers 1, 2, ..., 3n.If I can show that there must exist some a ∈ A, b ∈ B, c ∈ C such that one of them is the sum of the other two, then we have a contradiction.Alternatively, maybe I can use the pigeonhole principle or some combinatorial argument.Wait, another idea: if the sets are too "dense", then sums will occur. Maybe if they are equally sized, their density forces such sums.Alternatively, think about the minimal elements.Suppose 1 is in A. Then, 2 could be in B or C. Suppose 2 is in B. Then, 3 could be in C. Then, 1 + 2 = 3, which would be in C. So, that's a problem because 3 is in C, and 1 is in A, 2 is in B, so 3 = 1 + 2, which violates the condition.Wait, so if 1 is in A, 2 in B, then 3 must not be in C. But since 3 has to be in one of A, B, or C, it can't be in C. So, it must be in A or B.If 3 is in A, then A has 1 and 3. Then, 4 could be in B or C. Suppose 4 is in B. Then, 1 + 3 = 4, which is in B. So, that's a problem because 4 is in B, and 1 and 3 are in A, so 4 = 1 + 3, which would violate the condition if we pick 1 from A, 4 from B, and any from C.Wait, but the condition is that when you pick one from each set, the largest is not the sum of the other two. So, if 4 is in B, and 1 and 3 are in A, then picking 1 from A, 4 from B, and any from C, the largest is 4, which is equal to 1 + 3, which are both in A. But wait, the condition is about one from each set. So, in this case, 1 is from A, 4 is from B, and c is from C. The largest is 4, which is equal to 1 + 3, but 3 is also in A. So, does that violate the condition? Wait, the condition is that the largest is not equal to the sum of the other two, regardless of which set they are in.Wait, no, the condition is that when you take one from each set, the largest doesn't equal the sum of the other two. So, in this case, if 4 is in B, and 1 is in A, and 3 is in A, then if I pick 1 from A, 4 from B, and any from C, say c. Then, the largest is 4, which is equal to 1 + 3, but 3 is not in C, it's in A. So, does that matter? Wait, the condition is about the sum of the other two numbers, regardless of their sets. So, if 4 = 1 + 3, and 1 is in A, 3 is in A, and 4 is in B, then when you pick 1 from A, 4 from B, and any from C, the largest is 4, which is equal to 1 + 3, but 3 is not in C. So, does that violate the condition?Wait, the condition is that when you take one from each set, the largest doesn't equal the sum of the other two. So, in this case, the other two are 1 (from A) and c (from C). So, 4 ≠ 1 + c. So, as long as c ≠ 3, it's okay. But 3 is in A, so c can't be 3. So, if 3 is in A, then c can't be 3, so c must be something else. So, maybe 4 is safe.Wait, this is getting complicated. Maybe I need a different approach.Let me think about the minimal elements. Suppose 1 is in A. Then, 2 could be in B or C. Suppose 2 is in B. Then, 3 could be in C. But then, 1 + 2 = 3, which is in C. So, that's a problem because if I pick 1 from A, 2 from B, and 3 from C, the largest is 3, which equals 1 + 2. So, that violates the condition.Therefore, 3 cannot be in C. So, 3 must be in A or B. If 3 is in A, then A has 1 and 3. Then, 4 could be in B or C. Suppose 4 is in B. Then, 1 + 3 = 4, which is in B. So, if I pick 1 from A, 4 from B, and any from C, the largest is 4, which equals 1 + 3, but 3 is in A. So, does that violate the condition? Wait, the condition is about the sum of the other two, which are 1 (from A) and c (from C). So, 4 ≠ 1 + c. So, as long as c ≠ 3, it's okay. But 3 is in A, so c can't be 3. So, 4 is safe.But then, 5 could be in C. Then, 2 + 3 = 5. So, if 5 is in C, then picking 2 from B, 3 from A, and 5 from C, the largest is 5, which equals 2 + 3. That violates the condition.Therefore, 5 cannot be in C. So, 5 must be in A or B. If 5 is in A, then A has 1, 3, 5. Then, 6 could be in B or C. Suppose 6 is in B. Then, 1 + 5 = 6, which is in B. So, picking 1 from A, 6 from B, and any from C, the largest is 6, which equals 1 + 5. But 5 is in A, so as long as c ≠ 5, it's okay. But 5 is in A, so c can't be 5. So, 6 is safe.But then, 7 could be in C. Then, 2 + 5 = 7. So, if 7 is in C, then picking 2 from B, 5 from A, and 7 from C, the largest is 7, which equals 2 + 5. That violates the condition.Therefore, 7 cannot be in C. So, 7 must be in A or B. If 7 is in A, then A has 1, 3, 5, 7. Then, 8 could be in B or C. Suppose 8 is in B. Then, 1 + 7 = 8, which is in B. So, picking 1 from A, 8 from B, and any from C, the largest is 8, which equals 1 + 7. But 7 is in A, so as long as c ≠ 7, it's okay. But 7 is in A, so c can't be 7. So, 8 is safe.But then, 9 could be in C. Then, 2 + 7 = 9. So, if 9 is in C, then picking 2 from B, 7 from A, and 9 from C, the largest is 9, which equals 2 + 7. That violates the condition.This seems to be a pattern. Every time we add a new number, if we place it in C, it creates a sum with previous numbers in A and B, violating the condition. Therefore, to avoid this, we have to place these numbers in A or B instead.But if we keep doing this, A and B will grow larger, while C remains small. However, we assumed that |A| = |B| = |C| = n. So, if we keep having to place numbers in A or B to avoid the condition, C will have fewer elements, which contradicts the assumption that |C| = n.Therefore, our initial assumption that |A| = |B| = |C| must be false.Wait, let me make sure I'm not missing something. Maybe there's a way to arrange the numbers so that C doesn't have these problematic sums. But from the pattern above, every time we try to place a number in C, it creates a sum with previous numbers in A and B, which violates the condition. Therefore, to maintain the condition, we have to keep placing numbers in A or B, which would make C smaller than A and B, contradicting the assumption that all three have equal size.So, this suggests that it's impossible for |A|, |B|, and |C| to all be equal under the given condition.Another way to think about it is using the concept of sum-free sets. A sum-free set is a set of numbers such that no two elements add up to another element. In this problem, we're dealing with three sets, and the condition is similar to a sum-free condition across the sets.If A, B, and C were all sum-free within themselves, that might help, but the condition is stronger because it involves elements from different sets. Specifically, we need that for any a ∈ A, b ∈ B, c ∈ C, the largest is not the sum of the other two.This seems to impose a stricter condition than just being sum-free within each set. Therefore, it's more restrictive, making it harder to have equal-sized sets.Moreover, considering the minimal elements and how sums propagate, it's clear that trying to maintain equal sizes leads to unavoidable sums that violate the condition.Therefore, the conclusion is that |A|, |B|, and |C| cannot all be equal.

💡Okay, so I have these two equations to solve, and I need to figure out for which values of ( c ) they have solutions. If solutions exist, I also need to determine how many there are and what they are. Let me take them one by one.Starting with part (a): ( x + leftlfloor frac{x}{2} rightrfloor = c ).Hmm, the floor function can be tricky, but I remember that ( leftlfloor frac{x}{2} rightrfloor ) essentially divides the real line into intervals where it's constant. For example, when ( x ) is between 0 and 2, ( leftlfloor frac{x}{2} rightrfloor = 0 ); between 2 and 4, it's 1, and so on. So maybe I can break this problem down into intervals based on these steps.Let me consider an interval ( [2k, 2k + 2) ) where ( k ) is an integer. In this interval, ( leftlfloor frac{x}{2} rightrfloor = k ). So the equation becomes ( x + k = c ), which simplifies to ( x = c - k ). But since ( x ) has to be in ( [2k, 2k + 2) ), ( c - k ) must lie within this interval. That means ( 2k leq c - k < 2k + 2 ).Let me solve these inequalities for ( c ). The first inequality: ( 2k leq c - k ) implies ( 3k leq c ). The second inequality: ( c - k < 2k + 2 ) implies ( c < 3k + 2 ). So combining these, ( 3k leq c < 3k + 2 ).This tells me that for each integer ( k ), ( c ) must lie in the interval ( [3k, 3k + 2) ) for the equation to have a solution in ( [2k, 2k + 2) ). But wait, what about the interval ( [2k + 2, 2k + 4) )? In this interval, ( leftlfloor frac{x}{2} rightrfloor = k + 1 ), so the equation becomes ( x + (k + 1) = c ), which gives ( x = c - (k + 1) ). Again, ( x ) must be in ( [2k + 2, 2k + 4) ), so ( 2k + 2 leq c - (k + 1) < 2k + 4 ).Solving these inequalities: The first one, ( 2k + 2 leq c - (k + 1) ), simplifies to ( 3k + 3 leq c ). The second inequality, ( c - (k + 1) < 2k + 4 ), simplifies to ( c < 3k + 5 ). So combining these, ( 3k + 3 leq c < 3k + 5 ).Putting it all together, for each ( k ), ( c ) must lie in either ( [3k, 3k + 2) ) or ( [3k + 3, 3k + 5) ). But wait, there's a gap between ( 3k + 2 ) and ( 3k + 3 ). That means if ( c ) is in ( [3k + 2, 3k + 3) ), there's no solution. So the equation has no solution when ( c ) is congruent to 2 modulo 3, right?So, for part (a), the equation ( x + leftlfloor frac{x}{2} rightrfloor = c ) has a solution if and only if ( c ) is not congruent to 2 modulo 3. And when it does have a solution, there's exactly one solution in each interval where ( c ) is valid.Now, moving on to part (b): ( x - leftlfloor frac{x}{2} rightrfloor = c ).Again, I'll consider the same intervals ( [2k, 2k + 2) ) and ( [2k + 2, 2k + 4) ). In the first interval, ( leftlfloor frac{x}{2} rightrfloor = k ), so the equation becomes ( x - k = c ), which gives ( x = c + k ). Since ( x ) is in ( [2k, 2k + 2) ), ( c + k ) must lie within this interval. So ( 2k leq c + k < 2k + 2 ), which simplifies to ( k leq c < k + 2 ).Similarly, in the interval ( [2k + 2, 2k + 4) ), ( leftlfloor frac{x}{2} rightrfloor = k + 1 ), so the equation becomes ( x - (k + 1) = c ), which gives ( x = c + k + 1 ). Since ( x ) is in ( [2k + 2, 2k + 4) ), ( c + k + 1 ) must lie within this interval. So ( 2k + 2 leq c + k + 1 < 2k + 4 ), which simplifies to ( k + 1 leq c < k + 3 ).Putting these together, for each ( k ), ( c ) must lie in either ( [k, k + 2) ) or ( [k + 1, k + 3) ). But wait, these intervals overlap. For example, ( [k, k + 2) ) and ( [k + 1, k + 3) ) overlap between ( k + 1 ) and ( k + 2 ). So actually, for any real number ( c ), there are always two solutions: one in each interval.Therefore, for part (b), the equation ( x - leftlfloor frac{x}{2} rightrfloor = c ) has solutions for all real numbers ( c ), and there are exactly two solutions for each ( c ).Wait, let me double-check that. If ( c ) is an integer, say ( c = k ), then in the first interval ( [2k, 2k + 2) ), ( x = c + k = 2k ), which is the left endpoint. In the second interval ( [2k + 2, 2k + 4) ), ( x = c + k + 1 = 2k + 1 ), but wait, ( 2k + 1 ) is not in ( [2k + 2, 2k + 4) ). Hmm, so maybe when ( c ) is an integer, there's only one solution? Or does it wrap around?Wait, no. If ( c = k ), then in the first interval, ( x = k + k = 2k ), which is valid. In the second interval, ( x = k + (k + 1) = 2k + 1 ), but ( 2k + 1 ) is not in ( [2k + 2, 2k + 4) ). So actually, for ( c = k ), there's only one solution. Similarly, for ( c = k + 2 ), let's see: in the first interval, ( x = (k + 2) + k = 2k + 2 ), which is the left endpoint of the second interval. In the second interval, ( x = (k + 2) + (k + 1) = 2k + 3 ), which is within ( [2k + 2, 2k + 4) ). So actually, for ( c = k + 2 ), there are two solutions: ( 2k + 2 ) and ( 2k + 3 ).Wait, so maybe when ( c ) is an integer, there's only one solution, but when ( c ) is not an integer, there are two solutions. Hmm, but the problem says "if solutions exist, how many are there". So maybe I need to specify that for each ( c ), there are either one or two solutions depending on whether ( c ) is an integer or not.But actually, looking back, when ( c ) is an integer, say ( c = k ), then in the first interval, ( x = 2k ) is a solution, and in the second interval, ( x = 2k + 1 ) is not a solution because it's not in the interval. So only one solution. Similarly, for ( c = k + 1 ), which is not an integer, there are two solutions: one in each interval.Wait, but ( c ) can be any real number, not necessarily an integer. So if ( c ) is not an integer, then in both intervals, there will be a solution. If ( c ) is an integer, then only one solution exists because the other solution would fall outside the interval.But actually, let me think again. If ( c ) is an integer, say ( c = k ), then in the first interval, ( x = 2k ) is a solution, and in the second interval, ( x = 2k + 1 ) is not a solution because it's not in ( [2k + 2, 2k + 4) ). So only one solution. Similarly, if ( c = k + 1 ), which is not an integer, then in the first interval, ( x = (k + 1) + k = 2k + 1 ), which is in ( [2k, 2k + 2) ) only if ( 2k leq 2k + 1 < 2k + 2 ), which is true. In the second interval, ( x = (k + 1) + (k + 1) = 2k + 2 ), which is in ( [2k + 2, 2k + 4) ). So two solutions.Wait, but ( 2k + 2 ) is the left endpoint of the second interval, so it's included. So yes, two solutions. So in general, for any ( c ), whether integer or not, there are two solutions except when ( c ) is an integer, in which case there's only one solution.But wait, that contradicts my earlier thought. Let me check with specific numbers. Let's say ( c = 1 ). Then in the first interval ( [0, 2) ), ( x = 1 + 0 = 1 ), which is valid. In the second interval ( [2, 4) ), ( x = 1 + 1 = 2 ), which is the left endpoint, so it's valid. So two solutions: 1 and 2. Wait, but 2 is in the second interval, so it's valid. So actually, even when ( c ) is an integer, there are two solutions: one at the endpoint of the first interval and one at the endpoint of the second interval.Wait, but in the first interval, ( x = c + k ), and in the second interval, ( x = c + k + 1 ). So if ( c = k ), then ( x = 2k ) in the first interval, and ( x = 2k + 1 ) in the second interval. But ( 2k + 1 ) is not in ( [2k + 2, 2k + 4) ). So actually, only one solution exists when ( c = k ).Wait, I'm getting confused. Let me take a specific example. Let ( c = 1 ). Then in the first interval ( [0, 2) ), ( x = 1 + 0 = 1 ), which is valid. In the second interval ( [2, 4) ), ( x = 1 + 1 = 2 ), which is the left endpoint, so it's valid. So two solutions: 1 and 2.But if ( c = 2 ), then in the first interval ( [0, 2) ), ( x = 2 + 0 = 2 ), which is the right endpoint, so it's not included. In the second interval ( [2, 4) ), ( x = 2 + 1 = 3 ), which is valid. So only one solution: 3.Wait, so when ( c ) is an integer, there's only one solution because the solution in the first interval falls at the endpoint, which is not included. But when ( c ) is not an integer, both solutions are valid.So, to generalize, for any real number ( c ), there are two solutions except when ( c ) is an integer, in which case there's only one solution.But the problem says "if solutions exist, how many are there". So I think the answer is that for any real number ( c ), there are two solutions, except when ( c ) is an integer, in which case there's only one solution.Wait, but in my earlier example with ( c = 1 ), which is an integer, there were two solutions: 1 and 2. But 2 is the left endpoint of the second interval, so it's included. So maybe even when ( c ) is an integer, there are two solutions.Wait, I'm getting conflicting results. Let me clarify.When ( c ) is an integer, say ( c = k ), then in the first interval ( [2k, 2k + 2) ), ( x = k + k = 2k ), which is the left endpoint, so it's included. In the second interval ( [2k + 2, 2k + 4) ), ( x = k + (k + 1) = 2k + 1 ), which is not in ( [2k + 2, 2k + 4) ). So only one solution: ( x = 2k ).But wait, in the example with ( c = 1 ), ( x = 1 ) is in ( [0, 2) ), and ( x = 2 ) is in ( [2, 4) ). So both are valid because ( x = 2 ) is the left endpoint of the second interval. So in that case, even though ( c = 1 ) is an integer, there are two solutions.Wait, so maybe my earlier conclusion was wrong. Let me think again.If ( c ) is an integer, say ( c = k ), then in the first interval ( [2k, 2k + 2) ), ( x = k + k = 2k ), which is valid. In the second interval ( [2k + 2, 2k + 4) ), ( x = k + (k + 1) = 2k + 1 ), which is not in ( [2k + 2, 2k + 4) ). So only one solution: ( x = 2k ).But in the case of ( c = 1 ), ( k = 0 ), so ( x = 0 + 0 = 0 ) in the first interval ( [0, 2) ), but ( x = 1 ) is also a solution because ( x = 1 + 0 = 1 ). Wait, no, that's not right. If ( c = 1 ), then in the first interval, ( x = 1 + 0 = 1 ), which is valid. In the second interval, ( x = 1 + 1 = 2 ), which is the left endpoint, so it's valid. So two solutions: 1 and 2.Wait, so maybe when ( c ) is an integer, there are two solutions because the second solution is at the left endpoint of the next interval. So in general, for any real ( c ), there are two solutions: one in the first interval and one in the second interval. Even when ( c ) is an integer, the second solution is at the left endpoint of the next interval, which is included.So, in conclusion, for part (b), the equation ( x - leftlfloor frac{x}{2} rightrfloor = c ) has two solutions for any real number ( c ). The solutions are ( x = c + lfloor c rfloor ) and ( x = c + lfloor c rfloor - 1 ).Wait, let me verify that. If ( c = 1 ), then ( lfloor c rfloor = 1 ), so the solutions would be ( x = 1 + 1 = 2 ) and ( x = 1 + 1 - 1 = 1 ). That matches my earlier example. If ( c = 1.5 ), then ( lfloor c rfloor = 1 ), so the solutions are ( x = 1.5 + 1 = 2.5 ) and ( x = 1.5 + 1 - 1 = 1.5 ). Checking these: ( 2.5 - lfloor 2.5/2 rfloor = 2.5 - 1 = 1.5 ), and ( 1.5 - lfloor 1.5/2 rfloor = 1.5 - 0 = 1.5 ). So both solutions work.Therefore, for part (b), the equation has two solutions for any real ( c ), given by ( x = c + lfloor c rfloor ) and ( x = c + lfloor c rfloor - 1 ).So, summarizing:For part (a), the equation ( x + leftlfloor frac{x}{2} rightrfloor = c ) has solutions if and only if ( c ) is not congruent to 2 modulo 3. When it does have solutions, there is exactly one solution, which is ( x = c - lfloor frac{c}{3} rfloor ).For part (b), the equation ( x - leftlfloor frac{x}{2} rightrfloor = c ) has two solutions for any real number ( c ), which are ( x = c + lfloor c rfloor ) and ( x = c + lfloor c rfloor - 1 ).I think that's it. I should probably write this up clearly.

💡Okay, so I have this problem where the ratio of the interior angles of a triangle is 1:5:6, and the longest side has a length of 12. I need to find the length of the altitude (height) of the triangle that is perpendicular to the longest side. Hmm, let me try to figure this out step by step.First, I know that the sum of the interior angles in any triangle is 180 degrees. The ratio given is 1:5:6, so I can let the measures of the angles be 1x, 5x, and 6x. Adding them up: 1x + 5x + 6x = 12x. So, 12x = 180 degrees. Solving for x, I divide both sides by 12: x = 180/12 = 15 degrees. Therefore, the angles are 15 degrees, 75 degrees, and 90 degrees. Wait, 90 degrees? That means this is a right-angled triangle!Alright, so it's a right-angled triangle with angles 15°, 75°, and 90°. The longest side in a triangle is always opposite the largest angle, which in this case is the 90° angle. So, the longest side is the hypotenuse, and it's given as 12 units. Now, I need to find the altitude drawn to this hypotenuse.I remember that in a right-angled triangle, the altitude to the hypotenuse relates to the lengths of the other two sides. There's a formula that connects the area of the triangle with the hypotenuse and the altitude. The area can be calculated in two ways: one using the two legs as base and height, and the other using the hypotenuse and the altitude.Let me denote the legs as 'a' and 'b', and the hypotenuse as 'c' (which is 12). The area is (1/2)*a*b. It's also equal to (1/2)*c*h, where h is the altitude to the hypotenuse. So, setting these equal: (1/2)*a*b = (1/2)*c*h. Simplifying, a*b = c*h. Therefore, h = (a*b)/c.So, if I can find the lengths of the other two sides, I can calculate h. Since it's a 15-75-90 triangle, I might need to use trigonometric ratios to find the sides. Let me recall the sine and cosine of 15° and 75°. Alternatively, maybe there's a ratio for the sides in such a triangle.I remember that in a 15-75-90 triangle, the sides are in the ratio 1 : (2 + √3) : (√6 + √2). Wait, is that correct? Let me think. Alternatively, perhaps it's better to use the sine of 15° and 75° to find the sides.Since the hypotenuse is 12, the side opposite the 15° angle can be found using sine: a = 12*sin(15°). Similarly, the side opposite the 75° angle is b = 12*sin(75°). Let me calculate these.First, sin(15°). I know that sin(15°) is sin(45° - 30°). Using the sine subtraction formula: sin(A - B) = sinA cosB - cosA sinB. So, sin(15°) = sin(45°)cos(30°) - cos(45°)sin(30°). Plugging in the values: sin(45°) = √2/2, cos(30°) = √3/2, cos(45°) = √2/2, sin(30°) = 1/2. So, sin(15°) = (√2/2)(√3/2) - (√2/2)(1/2) = (√6/4) - (√2/4) = (√6 - √2)/4.Similarly, sin(75°) is sin(45° + 30°). Using the sine addition formula: sin(A + B) = sinA cosB + cosA sinB. So, sin(75°) = sin(45°)cos(30°) + cos(45°)sin(30°). Plugging in the values: sin(45°) = √2/2, cos(30°) = √3/2, cos(45°) = √2/2, sin(30°) = 1/2. So, sin(75°) = (√2/2)(√3/2) + (√2/2)(1/2) = (√6/4) + (√2/4) = (√6 + √2)/4.Therefore, the sides opposite 15° and 75° are:a = 12*sin(15°) = 12*(√6 - √2)/4 = 3*(√6 - √2).b = 12*sin(75°) = 12*(√6 + √2)/4 = 3*(√6 + √2).So, now I have the two legs: a = 3*(√6 - √2) and b = 3*(√6 + √2). Now, I can compute the product a*b.Let me compute a*b:a*b = [3*(√6 - √2)] * [3*(√6 + √2)] = 9*( (√6)^2 - (√2)^2 ) = 9*(6 - 2) = 9*4 = 36.So, a*b = 36. Earlier, I had h = (a*b)/c. Since c = 12, h = 36/12 = 3.So, the altitude to the hypotenuse is 3 units long.Wait, let me just verify this another way to make sure I didn't make a mistake. Another way to find the altitude is to use the formula: h = (a*b)/c, which I did. Alternatively, I could use the area formula.The area of the triangle is (1/2)*a*b = (1/2)*36 = 18. The area is also (1/2)*c*h = (1/2)*12*h = 6h. Setting these equal: 6h = 18, so h = 3. Yep, that checks out.So, the length of the altitude perpendicular to the longest side is 3.**Final Answer**The length of the altitude is boxed{3}.

💡Okay, so I have this problem about complex numbers forming a geometric sequence and then transforming them into another set of complex numbers. I need to figure out what shape these transformed points make in the complex plane. Hmm, let me try to break this down step by step.First, the problem says there are n complex numbers z₁, z₂, ..., zₙ that form a geometric sequence with a common ratio q, and q is not equal to ±1. Also, the absolute value of z₁ is not 1. Then, each wₖ is defined as zₖ + 1/zₖ + h, where h is a real number. I need to figure out what kind of figure the points w₁, w₂, ..., wₙ form.Alright, let's start by recalling what a geometric sequence of complex numbers means. In a geometric sequence, each term is the previous term multiplied by a common ratio. So, z₂ = z₁ * q, z₃ = z₂ * q = z₁ * q², and so on. So, in general, zₖ = z₁ * q^(k-1).Now, since we're dealing with complex numbers, it might be helpful to express z₁ in polar form. Let me write z₁ as r * (cos α + i sin α), where r is the modulus of z₁, and α is the argument. Since |z₁| ≠ 1, r ≠ 1. Also, since q ≠ ±1, the angle θ (which is the argument of q) isn't 0 or π, so it's a non-trivial rotation.Next, the common ratio q can also be expressed in polar form. Let me write q as cos θ + i sin θ, where θ is the argument of q. So, q is a complex number on the unit circle, making the sequence zₖ rotate by θ each time.So, zₖ = z₁ * q^(k-1) = r * (cos α + i sin α) * (cos θ + i sin θ)^(k-1). Using De Moivre's theorem, this becomes r * [cos(α + (k-1)θ) + i sin(α + (k-1)θ)]. So, each zₖ is a complex number with modulus r and argument α + (k-1)θ.Now, let's look at wₖ = zₖ + 1/zₖ + h. Let's first compute 1/zₖ. Since zₖ is r * [cos(α + (k-1)θ) + i sin(α + (k-1)θ)], its reciprocal is (1/r) * [cos(α + (k-1)θ) - i sin(α + (k-1)θ)], because the reciprocal of a complex number in polar form is the reciprocal of the modulus and the negative of the argument.So, 1/zₖ = (1/r) * [cos(α + (k-1)θ) - i sin(α + (k-1)θ)]. Therefore, zₖ + 1/zₖ is:zₖ + 1/zₖ = r * [cos(α + (k-1)θ) + i sin(α + (k-1)θ)] + (1/r) * [cos(α + (k-1)θ) - i sin(α + (k-1)θ)].Let me combine these terms. The real parts are r cos(α + (k-1)θ) + (1/r) cos(α + (k-1)θ), and the imaginary parts are r sin(α + (k-1)θ) - (1/r) sin(α + (k-1)θ).So, factoring out the cosine and sine terms, we get:zₖ + 1/zₖ = [r + 1/r] cos(α + (k-1)θ) + i [r - 1/r] sin(α + (k-1)θ).Therefore, wₖ = zₖ + 1/zₖ + h = [r + 1/r] cos(α + (k-1)θ) + h + i [r - 1/r] sin(α + (k-1)θ).Let me denote wₖ as x + yi, where x and y are real numbers. So, we have:x = [r + 1/r] cos(α + (k-1)θ) + h,y = [r - 1/r] sin(α + (k-1)θ).Now, I need to eliminate the angle α + (k-1)θ to find a relationship between x and y. Let me denote φ = α + (k-1)θ for simplicity. Then, we have:x - h = [r + 1/r] cos φ,y = [r - 1/r] sin φ.Now, I can write these as:cos φ = (x - h) / [r + 1/r],sin φ = y / [r - 1/r].Since cos² φ + sin² φ = 1, substituting the expressions above gives:[(x - h) / (r + 1/r)]² + [y / (r - 1/r)]² = 1.This is the equation of an ellipse centered at (h, 0) with semi-major axis a = r + 1/r and semi-minor axis b = |r - 1/r|.Wait, but let me check the semi-major and semi-minor axes. Since r ≠ 1, either r > 1 or 0 < r < 1. If r > 1, then r + 1/r > |r - 1/r|, so the major axis is along the x-axis. If 0 < r < 1, then r + 1/r is still greater than |r - 1/r| because 1/r > r, so again, the major axis is along the x-axis.So, regardless of whether r is greater than 1 or less than 1, the major axis is along the x-axis, and the minor axis is along the y-axis. The lengths of the semi-axes are a = r + 1/r and b = |r - 1/r|.But wait, the problem statement says h is a known real number. So, the ellipse is centered at (h, 0) on the real axis.Now, let's compute the distance between the foci of the ellipse. For an ellipse, the distance between the foci is 2c, where c = sqrt(a² - b²).Let me compute a² - b²:a² = (r + 1/r)² = r² + 2 + 1/r²,b² = (r - 1/r)² = r² - 2 + 1/r².So, a² - b² = (r² + 2 + 1/r²) - (r² - 2 + 1/r²) = 4.Therefore, c = sqrt(4) = 2.So, the distance between the foci is 2c = 4. That's interesting. So, regardless of the value of r (as long as r ≠ 1), the distance between the foci is always 4.Therefore, all the points w₁, w₂, ..., wₙ lie on an ellipse centered at h on the real axis, with major axis length 4 and minor axis length depending on r.Wait, but the problem says "what kind of figure do the points representing w₁, w₂, ..., wₙ form in the complex plane?" So, it's an ellipse.But let me make sure I didn't make a mistake. Let me recap:1. Expressed z₁ in polar form.2. Expressed q in polar form.3. Expressed zₖ as a geometric sequence.4. Calculated 1/zₖ.5. Summed zₖ + 1/zₖ.6. Expressed wₖ as x + yi, leading to parametric equations in terms of cos φ and sin φ.7. Eliminated φ to get the ellipse equation.8. Calculated semi-axes and foci distance.Everything seems to check out. So, the points w₁, w₂, ..., wₙ lie on an ellipse centered at h with major axis length 4.Wait, but the problem didn't specify whether the ellipse is the entire set or just a part of it. Since zₖ are in a geometric sequence, φ = α + (k-1)θ, so as k increases, φ increases by θ each time. So, the points wₖ are equally spaced in angle around the ellipse, but since it's a geometric sequence, they might not cover the entire ellipse unless θ is such that the points are dense on the ellipse. But regardless, they lie on an ellipse.So, the figure formed is an ellipse with major axis length 4, centered at h on the real axis.I think that's the conclusion.

💡Okay, so I have this problem with two parts. Let me try to tackle them one by one. First, part (1): Solve the inequality f(x) ≥ (m + n)x, where f(x) is given as |x - 1| - |x + 1|. Also, it's given that m + n = 7. Hmm, okay. So, I need to solve |x - 1| - |x + 1| ≥ 7x. I remember that absolute value functions can be tricky because they change their behavior depending on the value of x. So, I think I should break this down into different cases based on the critical points where the expressions inside the absolute values change sign. The expressions inside the absolute values are (x - 1) and (x + 1), so they change sign at x = 1 and x = -1, respectively. So, I'll consider three intervals: x ≤ -1, -1 < x < 1, and x ≥ 1. Case 1: x ≤ -1. In this case, both (x - 1) and (x + 1) are negative or zero. So, |x - 1| = -(x - 1) = -x + 1, and |x + 1| = -(x + 1) = -x - 1. Therefore, f(x) becomes (-x + 1) - (-x - 1) = (-x + 1) + x + 1 = 2. So, the inequality becomes 2 ≥ 7x. Solving for x, we get x ≤ 2/7. But wait, in this case, x is already ≤ -1, which is certainly ≤ 2/7. So, all x ≤ -1 satisfy the inequality.Case 2: -1 < x < 1. In this interval, (x + 1) is positive, but (x - 1) is negative. So, |x - 1| = -(x - 1) = -x + 1, and |x + 1| = x + 1. Therefore, f(x) becomes (-x + 1) - (x + 1) = -x + 1 - x - 1 = -2x. So, the inequality is -2x ≥ 7x. Adding 2x to both sides, we get 0 ≥ 9x, which simplifies to x ≤ 0. But in this case, x is between -1 and 1, so the solution here is -1 < x ≤ 0.Case 3: x ≥ 1. Here, both (x - 1) and (x + 1) are positive. So, |x - 1| = x - 1, and |x + 1| = x + 1. Therefore, f(x) becomes (x - 1) - (x + 1) = x - 1 - x - 1 = -2. So, the inequality is -2 ≥ 7x. Solving for x, we get x ≤ -2/7. But in this case, x is ≥ 1, which doesn't overlap with x ≤ -2/7. So, there's no solution in this interval.Putting it all together, the solutions are x ≤ -1 and -1 < x ≤ 0. Combining these, we get x ≤ 0. So, the solution set is all real numbers x such that x ≤ 0.Alright, that seems to make sense. Let me just double-check. For x ≤ -1, f(x) is 2, and 2 ≥ 7x is true because 7x is negative. For -1 < x ≤ 0, f(x) is -2x, which is non-negative, and -2x ≥ 7x implies x ≤ 0, which is consistent. For x ≥ 1, f(x) is -2, which is less than 7x (since x is positive), so no solution there. Yep, that seems right.Now, moving on to part (2). It says: Let max{a, b} be defined as a if a ≥ b, else b. We need to find the minimum value of F = max{|x² - 4y + m|, |y² - 2x + n|}. Given that m + n = 7, so n = 7 - m. Maybe we can express everything in terms of m or n. But I'm not sure yet. Let me see.So, F is the maximum of two absolute values: |x² - 4y + m| and |y² - 2x + n|. We need to find the minimum possible value of F over all real numbers x and y. Hmm, this seems a bit more complex. Maybe I can use some inequality properties or optimization techniques. Let me think.First, since F is the maximum of two expressions, both of which are absolute values, F is at least as large as each of them. So, F ≥ |x² - 4y + m| and F ≥ |y² - 2x + n|. If I add these two inequalities, I get 2F ≥ |x² - 4y + m| + |y² - 2x + n|. Maybe I can find a lower bound for the sum on the right-hand side.I recall the triangle inequality, which says that |a| + |b| ≥ |a + b|. So, perhaps I can apply that here. Let me try:|x² - 4y + m| + |y² - 2x + n| ≥ |(x² - 4y + m) + (y² - 2x + n)|. Simplify the expression inside the absolute value:(x² + y²) + (-4y - 2x) + (m + n). We know that m + n = 7, so this becomes:x² + y² - 2x - 4y + 7.Hmm, that looks like a quadratic in x and y. Maybe I can complete the squares to find its minimum value.Let's rewrite x² - 2x as (x - 1)² - 1, and y² - 4y as (y - 2)² - 4. So, substituting back:(x - 1)² - 1 + (y - 2)² - 4 + 7 = (x - 1)² + (y - 2)² + 2.So, the expression inside the absolute value is (x - 1)² + (y - 2)² + 2, which is always positive because squares are non-negative and we're adding 2. Therefore, the absolute value is just the expression itself.Thus, |x² - 4y + m| + |y² - 2x + n| ≥ (x - 1)² + (y - 2)² + 2.Since (x - 1)² + (y - 2)² is always non-negative, the minimum value of the right-hand side is when both squares are zero, i.e., when x = 1 and y = 2. Then, the expression becomes 0 + 0 + 2 = 2.Therefore, |x² - 4y + m| + |y² - 2x + n| ≥ 2.Going back to the inequality 2F ≥ |x² - 4y + m| + |y² - 2x + n|, we have 2F ≥ 2, so F ≥ 1.So, the minimum possible value of F is at least 1. Now, we need to check if this bound is achievable, i.e., if there exist x and y such that F = 1.To achieve F = 1, both |x² - 4y + m| ≤ 1 and |y² - 2x + n| ≤ 1 must hold, and at least one of them must be equal to 1. But since we are taking the maximum, if both are ≤ 1, then F is ≤ 1. But we also have F ≥ 1, so F must be exactly 1.So, we need to find x and y such that both |x² - 4y + m| ≤ 1 and |y² - 2x + n| ≤ 1, and at least one of them is equal to 1.But wait, earlier, we found that the sum |x² - 4y + m| + |y² - 2x + n| is minimized when x = 1 and y = 2, giving the sum as 2. So, if we set x = 1 and y = 2, let's compute F.Compute |x² - 4y + m|: x = 1, y = 2, so 1 - 8 + m = m - 7. Since m + n = 7, m = 7 - n. Wait, but without knowing m or n individually, I can't compute this directly. Hmm, maybe I need to express m in terms of n or vice versa.Wait, but m + n = 7, so n = 7 - m. So, let's compute both expressions:First expression: |1 - 8 + m| = |m - 7|.Second expression: |4 - 2 + n| = |2 + n|.But since n = 7 - m, the second expression becomes |2 + 7 - m| = |9 - m|.So, F = max{|m - 7|, |9 - m|}.We need to find m such that both |m - 7| ≤ 1 and |9 - m| ≤ 1. Let's solve these inequalities.For |m - 7| ≤ 1: 6 ≤ m ≤ 8.For |9 - m| ≤ 1: 8 ≤ m ≤ 10.The intersection of these intervals is m = 8.So, if m = 8, then n = 7 - 8 = -1.Let me check:First expression: |1 - 8 + 8| = |1| = 1.Second expression: |4 - 2 + (-1)| = |1| = 1.So, both expressions equal 1, so F = max{1, 1} = 1.Therefore, when x = 1, y = 2, m = 8, n = -1, F = 1. So, the minimum value of F is indeed 1.Wait, but in the problem statement, m and n are given such that m + n = 7, but they don't specify any constraints on m and n individually. So, m can be 8, which makes n = -1, which is acceptable because m and n are real numbers. So, yes, this is achievable.Therefore, the minimum value of F is 1.Let me just recap to make sure I didn't miss anything. We transformed the sum of the absolute values into a sum involving squares, found the minimum of that sum, which gave us a lower bound for 2F, leading to F ≥ 1. Then, by setting x = 1, y = 2, and choosing m and n accordingly, we achieved F = 1. So, that's the minimum.I think that makes sense. I don't see any mistakes in the reasoning. So, I'm confident that the minimum value of F is 1.**Final Answer**(1) The solution set is boxed{(-infty, 0]}.(2) The minimum value of ( F ) is boxed{1}.