🧠:Okay, so the question is whether it's possible to pair all lines on a plane into sets of perpendicular lines. Hmm, let's start by understanding what exactly is being asked here. Pairing all lines into sets of perpendicular lines—does that mean grouping lines such that each group consists of two lines that are perpendicular to each other? And every line on the plane must be part of exactly one such pair?First, I need to recall that in a plane, two lines are perpendicular if they intersect at a 90-degree angle. So, if I have a line L, then any line perpendicular to L must have a slope that's the negative reciprocal of L's slope. For example, if L has a slope of m, then a perpendicular line would have a slope of -1/m. Of course, vertical and horizontal lines are perpendicular since vertical lines have undefined slope and horizontal lines have a slope of 0, but I guess that's a special case.Now, the problem is asking if we can pair every line on the plane with another line such that they are perpendicular, and every line is included in exactly one pair. So, essentially, we need a way to partition the set of all lines in the plane into disjoint pairs, each pair being perpendicular.Let me think about the cardinality of the set of all lines in the plane. The set of lines in a plane is uncountably infinite. Each line can be uniquely identified by its slope and y-intercept (except vertical lines, which can be represented as x = c). So, the set of lines is a two-dimensional space—slope and intercept. Similarly, the set of all perpendicular lines would also be uncountable.But how do we pair them? Let's consider that for each line, there exists a unique perpendicular line. Wait, not exactly. For each line, there are infinitely many lines perpendicular to it, differing by their intercepts. For instance, if I have a line y = mx + b, then all lines of the form y = (-1/m)x + c are perpendicular to it, where c is any real number. So, each line has an infinite number of perpendicular lines. But how can we pair them if each line has infinitely many perpendicular counterparts? That might complicate things because pairing requires a one-to-one correspondence.Alternatively, maybe the problem is considering lines without considering their intercepts—just their slopes. But no, the question is about all lines on the plane, which includes both slope and intercept. So, intercepts matter here.Wait, perhaps the key is to realize that perpendicularity is a relationship that depends only on the slope, not the intercept. So two lines are perpendicular if their slopes are negative reciprocals, regardless of their intercepts. So, for each line with slope m, any line with slope -1/m is perpendicular to it, no matter the intercept. Therefore, for each line, there are infinitely many lines perpendicular to it. But to pair them, each line must be paired with exactly one other line such that they are perpendicular, and every line is in exactly one pair.But since there are infinitely many lines with the same slope (different intercepts), how can we pair them? For example, consider all horizontal lines (slope 0). Each horizontal line can be paired with a vertical line (undefined slope), which is perpendicular. However, there are as many horizontal lines as there are vertical lines (each determined by their y-intercept or x-intercept). Since the number of horizontal lines is uncountably infinite (one for each real number y-intercept), and similarly for vertical lines, perhaps we can pair each horizontal line with a vertical line. That would take care of all horizontal and vertical lines. But then, what about lines with other slopes?Take, for example, lines with slope 1. Their perpendicular counterparts would have slope -1. So, each line y = x + b can be paired with a line y = -x + c. But here, similar to the horizontal and vertical case, each line with slope 1 has infinitely many lines with slope -1 that are perpendicular to it. How do we choose which one to pair it with?If we can find a bijection between the set of lines with slope m and the set of lines with slope -1/m, then perhaps we could pair each line with slope m with a unique line of slope -1/m. Since the intercepts for both sets are real numbers (uncountably infinite), we can establish a bijection between them. For example, for each line y = mx + b, pair it with y = (-1/m)x + b. But wait, does that work? If we pair them based on the same intercept, would that ensure that each line is paired uniquely?Wait, if we take a line y = mx + b, and pair it with y = (-1/m)x + b, then the paired line has the same y-intercept. But does this lead to a unique pairing? Let's check. Suppose m is fixed. Then, for each b, we have a line with slope m and intercept b, paired with a line of slope -1/m and intercept b. Since the mapping from b to b is bijective (it's the identity function), this would pair each line of slope m with a unique line of slope -1/m. But then, what about the lines of slope -1/m? Each of those lines would be paired with a line of slope m, but with the same intercept. So, for example, the line y = (-1/m)x + b is paired with y = mx + b. So, this seems to create a bijection between the two sets of lines (slope m and slope -1/m). Thus, if we do this for every pair of slopes that are negative reciprocals, we could potentially pair all non-vertical and non-horizontal lines.But then, vertical and horizontal lines are special cases. For horizontal lines (slope 0), their perpendiculars are vertical lines (undefined slope). Each horizontal line is y = c, and each vertical line is x = d. The intercepts here are c and d, both real numbers. So, similar to the previous case, we can pair each horizontal line y = c with a vertical line x = c. That would be a bijection since for each real number c, we pair y = c with x = c. This way, every horizontal line is paired with a vertical line and vice versa, with no overlaps.So, putting this all together, perhaps we can pair all lines in the plane as follows:1. For horizontal lines y = c, pair each with vertical line x = c.2. For each slope m ≠ 0, pair each line y = mx + b with the line y = (-1/m)x + b.This would result in every line being paired with a unique perpendicular line, and each line is included in exactly one pair. Let me verify this.Take a line y = mx + b. Its pair is y = (-1/m)x + b. The paired line is indeed perpendicular. Now, take the paired line y = (-1/m)x + b; its pair would be y = mx + b. So, each pair is mutual. Also, since for each m and b, there's exactly one pair, and every line is included. For vertical and horizontal lines, similar logic applies: x = c is paired with y = c, and vice versa. Since all lines are either vertical, horizontal, or have some non-zero, defined slope, this seems to cover all possibilities.Wait, but what about lines with the same slope but different intercepts? For example, take two different lines with slope m: y = mx + b1 and y = mx + b2. Their respective pairs would be y = (-1/m)x + b1 and y = (-1/m)x + b2. Since b1 ≠ b2, these are distinct lines. Therefore, each line with slope m is paired with a unique line with slope -1/m, and since the intercepts are preserved in the pairing, there's no overlap or duplication.Similarly, vertical lines x = c1 and x = c2 (c1 ≠ c2) are paired with horizontal lines y = c1 and y = c2, which are distinct. So, no issues there.This seems to work. Therefore, it's possible to pair all lines on a plane into sets of perpendicular lines by using this method. Each line is paired with another line such that their slopes are negative reciprocals, and their intercepts are the same (or in the case of vertical/horizontal, matching the constant term). This creates a bijection between lines of reciprocal slopes, ensuring every line is included exactly once.But wait, let me check if there's a problem with this approach. Suppose we have a line y = mx + b. Its pair is y = (-1/m)x + b. Now, if we take another line with slope -1/m, say y = (-1/m)x + c. Its pair would be y = mx + c. So, unless c = b, this line is paired with a different line. Therefore, as long as we fix the intercept when pairing, each line is uniquely paired. This seems okay because the intercepts are unique for each line in their respective slope classes.Another potential issue: What if a line is paired with itself? For example, if there exists a line that is perpendicular to itself. But in Euclidean geometry, a line cannot be perpendicular to itself because the angle between a line and itself is 0 degrees, not 90. Therefore, no line is paired with itself, which is good.Also, we need to ensure that every line is included. For vertical lines, each x = c is paired with y = c. For horizontal lines, each y = c is paired with x = c. For lines with other slopes, each y = mx + b is paired with y = (-1/m)x + b. Since every line is either vertical, horizontal, or has some non-zero, defined slope, and we've covered all these cases, every line is indeed included in exactly one pair.Therefore, this method works. Hence, the answer should be yes, it's possible to pair all lines on a plane into sets of perpendicular lines.But wait, let me think again. Is there a problem with the intercepts? For example, pairing y = mx + b with y = (-1/m)x + b. Suppose m is 1. Then y = x + b is paired with y = -x + b. If we take b = 0, then the pair is y = x and y = -x. If b = 1, it's y = x +1 and y = -x +1. So, each of these pairs are distinct and don't interfere with each other. So, seems okay.Another check: Take a line not in the standard form. For example, a vertical line x = 5 is paired with y =5. A horizontal line y =5 is paired with x=5. So, mutual pairing. For a diagonal line, say y = 2x +3, its pair is y = -1/2 x +3. Then, the pair of y = -1/2 x +3 is y = 2x +3. So, mutual.Therefore, this seems to satisfy all conditions. Each line is paired with exactly one perpendicular line, and all lines are covered. Therefore, the answer should be yes.Wait, but what about lines that are already perpendicular? For example, take two perpendicular lines y = x and y = -x. If they have the same intercept (0), then according to our pairing, y = x is paired with y = -x +0, and y = -x is paired with y = x +0. So, they form a pair together. But in this case, they intersect at the origin. However, if two lines have different intercepts, their paired lines would not intersect unless the intercepts are arranged such that they cross. But the problem doesn't require the pairs to intersect at a particular point, just to be perpendicular. So, even if two perpendicular lines are parallel in their intercepts, as long as they have the right slope, they are considered perpendicular regardless of where they are placed on the plane.Wait, no. Wait, two lines are perpendicular if their slopes are negative reciprocals. The actual position (intercept) doesn't affect whether they are perpendicular. They just need to have the correct slopes. So, even if two lines with slopes m and -1/m are parallel in terms of intercepts, they are still considered perpendicular lines, even if they don't intersect. Wait, no! If two lines have slopes m and -1/m, they must intersect somewhere unless they are parallel. But lines with different slopes (not parallel) will always intersect at exactly one point. Wait, but if two lines have slopes m and -1/m, they are definitely not parallel (unless m is 0 or undefined, but we already handled vertical and horizontal lines). Therefore, any two lines with slopes m and -1/m will intersect at exactly one point, and the angle between them will be 90 degrees. Therefore, they are indeed perpendicular, regardless of their intercepts.But in our pairing, we are fixing the intercepts. So, for example, pairing y = mx + b with y = (-1/m)x + b. These two lines will intersect at some point, but the key is that they are perpendicular. The pairing is valid because their slopes are negative reciprocals. Therefore, even though the intercept is the same, the lines are not the same (unless m = -1/m, which would imply m^2 = -1, which is not possible for real numbers). So, the pairing is valid.Therefore, this method successfully pairs every line with a unique perpendicular line, covering all lines on the plane. Hence, the answer is yes.But wait, hold on. Let me consider the case when m = 1. Then, the paired line is slope -1. But in this case, suppose we have the line y = x + b paired with y = -x + b. These two lines intersect at the point where x + b = -x + b => 2x = 0 => x=0, y = b. So, they intersect at (0, b). So, they are indeed perpendicular and intersect at that point. Similarly, for any b, they intersect somewhere, but they are still perpendicular.Another check: Suppose we have a line y = 2x + 5. Its pair is y = -1/2 x + 5. They intersect when 2x +5 = -1/2 x +5 => 2x + 1/2 x = 0 => 5/2 x =0 => x=0, y=5. So, the point is (0,5). The angle between them is 90 degrees because their slopes are negative reciprocals. So, this works.Therefore, the pairing works as intended, and every line is included in exactly one pair. Therefore, it's possible to pair all lines on a plane into sets of perpendicular lines.

✅:Yes, it is possible to pair all lines on a plane into sets of perpendicular lines. Here's the systematic reasoning:1. Key Insight: Perpendicularity depends solely on slopes. Two lines are perpendicular if their slopes ( m ) and ( m' ) satisfy ( m cdot m' = -1 ). Vertical (undefined slope) and horizontal (slope 0) lines are also perpendicular.2. Pairing Strategy: - For non-vertical/non-horizontal lines: For every line ( y = mx + b ), pair it with ( y = -frac{1}{m}x + b ). This ensures a unique perpendicular partner for each line by preserving the intercept ( b ). - For vertical and horizontal lines: Pair each vertical line ( x = c ) with the horizontal line ( y = c ). This bijection covers all vertical and horizontal lines.3. Verification: - Uniqueness: Each line is paired with exactly one perpendicular line. For example, ( y = mx + b ) maps uniquely to ( y = -frac{1}{m}x + b ), and vice versa. - Completeness: Every line is included. Non-vertical/non-horizontal lines are covered by reciprocal slope pairings, while vertical/horizontal lines are matched via intercepts. - No Overlaps: Distinct intercepts ensure no duplication. For instance, ( y = mx + b_1 ) and ( y = mx + b_2 ) (( b_1 neq b_2 )) map to distinct perpendicular lines with different intercepts.4. Conclusion: This systematic pairing satisfies the requirement to group all lines into disjoint perpendicular pairs without omission or repetition.Final Answer: boxed{Yes}

🔑:1. Understanding the problem: - We need to determine whether all lines on a plane can be paired into perpendicular pairs. 2. Analysis of different cases: - Case 1: Non-axis parallel lines - Consider a line that is not parallel to either the (x)-axis or the (y)-axis. - Let's denote this line by ( L ). - The line perpendicular to ( L ) will intersect ( L ) at some point, making a (90^circ) angle. - Since the slope of ( L ) is not zero or undefined, the perpendicular line will cross ( L ) somewhere in the coordinate plane, ensuring that every line intersects at a unique angle that is a right angle. - Case 2: Lines parallel to the coordinate axes - Consider a line that is parallel to the (x)-axis or (y)-axis. - Let's denote the line parallel to the (x)-axis by ( L_x ) and the line parallel to the (y)-axis by ( L_y ). - A line perpendicular to ( L_x ) must be parallel to the (y)-axis, thus making the coordinate axes themselves useful references. Specifically: - For any set of lines parallel to the (x)-axis (( L_x )), its perpendicular companion should be parallel to the (y)-axis (( L_y )). - If there are bisectors involved, the bisectors of the angle between the lines provide a solution: - Suppose (L_x) bisects the first coordinate quadrant. - The perpendicular line will intersect at this bisector when it is oriented parallel to the (y)-axis.3. Conclusion: - Given that non-axis parallel lines naturally intersect perpendicularly and axis-parallel lines can be paired into perpendicular bisectors effectively, we have shown that it is possible to pair all lines on the plane into perpendicular pairs.Thus, it is possible to pair all lines on a plane into perpendicular pairs.[boxed{text{Possible}} ]

🧠:Okay, so I need to find all functions ( f: D rightarrow D ), where ( D ) is the set of real numbers except -1, such that for all ( x, y in D ) with ( x neq 0 ) and ( y neq -x ), the equation[ (f(f(x)) + y) fleft(frac{y}{x}right) + f(f(y)) = x ]holds. Alright, functional equations can be tricky, but let's take it step by step.First, functional equations often require making educated guesses about the form of the function. Common functions to test are linear functions, constant functions, or maybe reciprocal functions. Let me try to see if a linear function might work here. Let's assume ( f(x) = ax + b ), where ( a ) and ( b ) are constants. Then, maybe I can substitute this into the equation and solve for ( a ) and ( b ).But before jumping into that, maybe there's a smarter approach. Let's see if plugging in specific values for ( x ) or ( y ) can simplify the equation and give us some information about ( f ).First, let's try setting ( y = 0 ). Wait, but ( y neq -x ). If I set ( y = 0 ), then the condition ( y neq -x ) becomes ( 0 neq -x ), which implies ( x neq 0 ). However, the original conditions already require ( x neq 0 ). So, setting ( y = 0 ) is allowed here.So, let's set ( y = 0 ):[ (f(f(x)) + 0) fleft(frac{0}{x}right) + f(f(0)) = x ]Simplifying:[ f(f(x)) cdot f(0) + f(f(0)) = x ]That's an interesting equation. Let's denote ( c = f(0) ) and ( d = f(f(0)) ). Then the equation becomes:[ f(f(x)) cdot c + d = x ]So, rearranged:[ c cdot f(f(x)) = x - d ]Therefore,[ f(f(x)) = frac{x - d}{c} ]Hmm, so ( f circ f ) is a linear function if ( c neq 0 ). That might be useful. Let's keep this in mind.Now, perhaps try another substitution. Let's try setting ( x = 1 ). Then, the equation becomes:[ (f(f(1)) + y) fleft(frac{y}{1}right) + f(f(y)) = 1 ]Which simplifies to:[ (f(f(1)) + y) f(y) + f(f(y)) = 1 ]This holds for all ( y neq -1 ). Hmm, this might be useful. Let me denote ( f(f(1)) ) as a constant, say ( k ). So:[ (k + y) f(y) + f(f(y)) = 1 ]But from the previous result when we set ( y = 0 ), we have ( f(f(y)) = frac{y - d}{c} ). Wait, if that's the case, then substituting back into this equation:[ (k + y) f(y) + frac{y - d}{c} = 1 ]So this gives us a relation involving ( f(y) ). Let's see if we can combine this with other equations.Wait, perhaps if I use the expression for ( f(f(x)) ), which is ( frac{x - d}{c} ), then maybe I can express ( f ) in terms of itself. Let's suppose that ( f ) is invertible. But I don't know that yet. Alternatively, maybe ( f ) is linear. Let's test the linear case.Assume ( f(x) = ax + b ). Then, ( f(f(x)) = a(ax + b) + b = a^2 x + ab + b ).From the equation ( f(f(x)) = frac{x - d}{c} ), which we derived earlier, we have:[ a^2 x + ab + b = frac{x - d}{c} ]Comparing coefficients:For the x term: ( a^2 = frac{1}{c} )Constant term: ( ab + b = -frac{d}{c} )But ( c = f(0) = a cdot 0 + b = b ), so ( c = b ). Similarly, ( d = f(f(0)) = f(b) = a cdot b + b ). Therefore, ( d = ab + b = b(a + 1) ).So, substituting ( c = b ) and ( d = b(a + 1) ) into the equations:From the x coefficient: ( a^2 = frac{1}{b} )From the constant term: ( ab + b = -frac{b(a + 1)}{b} = -(a + 1) )Simplify the constant term equation:( ab + b = -a - 1 )Factor left side:( b(a + 1) = -a - 1 )Assuming ( a + 1 neq 0 ), we can divide both sides by ( a + 1 ):( b = -1 )But ( D ) excludes -1, so ( f(x) ) maps into ( D ), which means ( f(x) neq -1 ) for any ( x in D ). If ( f(x) = ax + b ), and ( b = -1 ), then ( f(x) = ax - 1 ). But if ( a = 0 ), then ( f(x) = -1 ), which is not allowed. So ( a neq 0 ). Also, we must have ( ax - 1 neq -1 ), which would require ( ax neq 0 ). But since ( x in D ), which is all real numbers except -1, ( x ) can be zero unless excluded by other conditions. Wait, in the original problem, the functional equation is given for ( x neq 0 ) and ( y neq -x ). But the function ( f ) is defined on ( D ), which is ( mathbb{R} setminus {-1} ). So ( f(x) ) must be defined for all ( x neq -1 ), including ( x = 0 ).But if ( f(0) = -1 ), that's a problem because the codomain ( D ) excludes -1. Therefore, ( f(0) neq -1 ). But according to our previous assumption, if ( b = -1 ), then ( f(0) = -1 ), which is invalid. Therefore, this leads to a contradiction. Hence, our assumption that ( a + 1 neq 0 ) is invalid. Therefore, ( a + 1 = 0 ), so ( a = -1 ).So, ( a = -1 ), then from the x coefficient equation ( a^2 = 1/b ), since ( a = -1 ), we have ( 1 = 1/b implies b = 1 ).Therefore, the function would be ( f(x) = -x + 1 ). Let's check if this works.First, check if ( f(x) = -x + 1 ) maps ( D ) to ( D ). Since ( D = mathbb{R} setminus {-1} ), we need to ensure that ( f(x) neq -1 ) for any ( x in D ). Suppose ( f(x) = -1 ), then:( -x + 1 = -1 implies x = 2 ). But 2 is in ( D ), so ( f(2) = -1 ), which is not allowed. Therefore, this function is invalid because ( f(2) = -1 ), which is excluded from the codomain ( D ). So, this suggests that our assumption of a linear function might not work, or we need to adjust our approach.Wait, maybe there is a mistake here. Let's verify again.If ( f(x) = -x + 1 ), then ( f(2) = -2 + 1 = -1 ), which is not allowed in ( D ). Therefore, this function is invalid. Hence, our assumption that ( f ) is linear leads to a contradiction. Therefore, perhaps ( f ) is not linear. Hmm, so linear functions might not work. Let's consider another approach.Going back to the equation we derived when ( y = 0 ):[ c cdot f(f(x)) + d = x ]Where ( c = f(0) ) and ( d = f(f(0)) ).If we can express ( f(f(x)) ) in terms of ( x ), maybe we can find an expression for ( f ).Suppose that ( f ) is invertible. Then, applying ( f^{-1} ) to both sides of ( f(f(x)) = frac{x - d}{c} ), we get:( f(x) = f^{-1}left( frac{x - d}{c} right) )But this might not help directly unless we know more about ( f ).Alternatively, let's assume that ( f ) is linear but make sure it doesn't map any ( x ) to -1. Wait, perhaps if we set ( f(x) = kx ), a linear function without the constant term. Let's try that.Let ( f(x) = kx ). Then ( f(f(x)) = k(kx) = k^2 x ). Let's substitute into the equation obtained when ( y = 0 ):[ c cdot f(f(x)) + d = x ]Here, ( c = f(0) = k cdot 0 = 0 ). But then, ( c = 0 ), so the equation becomes ( 0 cdot f(f(x)) + d = x implies d = x ). But ( d = f(f(0)) = f(0) = 0 ). Therefore, ( 0 = x ), which must hold for all ( x neq 0 ). This is impossible. Therefore, constant term is necessary, or the function is not purely linear. Hence, linear functions with ( b neq 0 ) might not work either, as seen before.Alternatively, maybe ( f ) is a reciprocal function? Let's try ( f(x) = frac{1}{x} ). But ( f ) is defined on ( D = mathbb{R} setminus {-1} ), so we have to check if ( f(x) neq -1 ). If ( frac{1}{x} = -1 implies x = -1 ), which is excluded from the domain. Therefore, ( f(x) = frac{1}{x} ) maps ( D ) to ( D setminus {0} ), but 0 is not in ( D ), so actually, the codomain would be ( mathbb{R} setminus {-1, 0} ), which is a subset of ( D ). Therefore, ( f(x) = frac{1}{x} ) is not a valid function from ( D ) to ( D ), because, for example, ( f(1) = 1 in D ), but ( f(2) = 1/2 in D ), but actually, the image excludes 0 and -1. Wait, but 0 is not in the codomain. Wait, codomain is ( D ), which is ( mathbb{R} setminus {-1} ). Therefore, 0 is allowed in the codomain. Wait, no, codomain is the set of real numbers excluding -1, so 0 is allowed. So, ( f(x) = 1/x ) maps ( D ) to ( mathbb{R} setminus {-1, 0} ), which is a subset of ( D ). Therefore, as long as 0 is not achieved, but ( 1/x ) can never be 0, so actually, ( f(x) = 1/x ) maps ( D ) to ( mathbb{R} setminus {0} ), but since the codomain is ( D = mathbb{R} setminus {-1} ), and ( mathbb{R} setminus {0} ) is a subset of ( mathbb{R} setminus {-1} ) only if -1 is not in ( mathbb{R} setminus {0} ). Wait, -1 is in ( mathbb{R} setminus {0} ), so ( f(x) = 1/x ) would map some ( x ) to -1. Indeed, if ( x = -1 ), which is not in the domain, so actually, ( f(x) = 1/x ) is defined on ( D ), which is ( mathbb{R} setminus {-1} ), so ( x neq -1 ). Then, ( 1/x = -1 implies x = -1 ), which is excluded. Therefore, ( f(x) = 1/x ) maps ( D ) to ( D ), because ( 1/x neq -1 ) for all ( x in D ), and ( 1/x ) is defined since ( x neq 0 ) is not required here; the domain is ( mathbb{R} setminus {-1} ), so ( x = 0 ) is allowed, but ( f(0) ) would be undefined. Wait, no, hold on. The function ( f: D rightarrow D ), so ( f(x) ) must be defined for all ( x in D ), which includes all real numbers except -1. However, if ( f(x) = 1/x ), then ( f(0) ) is undefined, which is a problem because 0 is in ( D ). Therefore, ( f(x) = 1/x ) is not a valid function here because 0 is in the domain, and ( f(0) ) would be undefined. Therefore, reciprocal function is invalid.Hmm, so maybe another approach. Let's try to find ( f(0) ). From earlier, when we set ( y = 0 ), we have:[ c cdot f(f(x)) + d = x ]But ( c = f(0) ), ( d = f(f(0)) ). If we set ( x = 0 ), wait, but in the original functional equation, ( x neq 0 ). So, we can't set ( x = 0 ) in the equation. However, the function ( f ) is defined for all ( x in D ), including 0. So perhaps we can find ( f(0) ) by other means. Let me think.Alternatively, let's try to plug ( x = 1 ) and some specific ( y ). Let's say ( x = 1 ), and ( y = 1 ), provided ( y neq -x implies 1 neq -1 ), which is true. Then the equation becomes:[ (f(f(1)) + 1) f(1) + f(f(1)) = 1 ]Let me denote ( f(f(1)) = k ), then:[ (k + 1) f(1) + k = 1 ]But we also have from the equation when ( x = 1 ):[ (k + y) f(y) + f(f(y)) = 1 ]Wait, earlier we considered this for general ( y ). Let me see if I can relate this with the equation we just got. If I set ( y = 1 ), then:[ (k + 1) f(1) + k = 1 ]But this equation might help us find ( f(1) ) if we know ( k ), but ( k = f(f(1)) ). So, perhaps this is a recursive equation. Let me write it as:[ (k + 1) f(1) + k = 1 ]But ( k = f(f(1)) ). Let me denote ( a = f(1) ), so ( k = f(a) ). Then, the equation becomes:[ (f(a) + 1) a + f(a) = 1 ]Which simplifies to:[ a f(a) + a + f(a) = 1 ]Factorizing:[ f(a) (a + 1) + a = 1 ]Therefore:[ f(a) = frac{1 - a}{a + 1} ]Provided ( a + 1 neq 0 ). Since ( a = f(1) ), and ( f: D rightarrow D ), ( a neq -1 ). Therefore, ( a + 1 neq 0 ), so this is valid.So, we have ( f(a) = frac{1 - a}{a + 1} ). But ( a = f(1) ), so ( f(f(1)) = frac{1 - f(1)}{f(1) + 1} ). Therefore, ( k = frac{1 - a}{a + 1} ).But from the previous equation when ( y = 0 ), we had ( f(f(x)) = frac{x - d}{c} ). Let's recall that ( c = f(0) ), ( d = f(f(0)) ). Let me try to compute ( f(f(1)) ).If ( f(f(x)) = frac{x - d}{c} ), then setting ( x = 1 ):[ f(f(1)) = frac{1 - d}{c} ]But we also have ( f(f(1)) = frac{1 - a}{a + 1} ). Therefore:[ frac{1 - d}{c} = frac{1 - a}{a + 1} ]But ( a = f(1) ), ( c = f(0) ), ( d = f(f(0)) ).This seems a bit convoluted. Maybe we need another substitution. Let's try setting ( x = y ). Wait, but we have to make sure ( y neq -x ), so if ( x = y ), then ( x neq -x implies x neq 0 ). Which is already satisfied since ( x neq 0 ) is given. So, set ( x = y ):[ (f(f(x)) + x) fleft( frac{x}{x} right) + f(f(x)) = x ]Simplify:[ (f(f(x)) + x) f(1) + f(f(x)) = x ]Combine terms:[ f(f(x)) (f(1) + 1) + x f(1) = x ]Rearranged:[ f(f(x)) (f(1) + 1) = x - x f(1) ][ f(f(x)) = frac{x (1 - f(1))}{f(1) + 1} ]But from the earlier result when ( y = 0 ), we had ( f(f(x)) = frac{x - d}{c} ). Therefore, equating the two expressions:[ frac{x - d}{c} = frac{x (1 - f(1))}{f(1) + 1} ]For this equality to hold for all ( x neq 0 ), the coefficients of ( x ) and the constants must be equal. Therefore:Coefficient of ( x ):[ frac{1}{c} = frac{1 - f(1)}{f(1) + 1} ]Constant term:[ frac{-d}{c} = 0 implies d = 0 ]So, from the constant term, ( d = 0 ). Recall that ( d = f(f(0)) ), so ( f(f(0)) = 0 ).From the coefficient equation:[ frac{1}{c} = frac{1 - f(1)}{f(1) + 1} ]But ( c = f(0) ), so:[ frac{1}{f(0)} = frac{1 - f(1)}{f(1) + 1} ]Let me note that ( a = f(1) ), so:[ frac{1}{c} = frac{1 - a}{a + 1} implies c = frac{a + 1}{1 - a} ]But ( c = f(0) ), so:[ f(0) = frac{a + 1}{1 - a} ]Additionally, from the equation when we set ( y = 1 ):[ f(a) = frac{1 - a}{a + 1} ]But ( a = f(1) ), and ( f(a) = f(f(1)) = k ), which we also found to be ( frac{1 - d}{c} ). But ( d = 0 ), so:[ f(f(1)) = frac{1 - 0}{c} = frac{1}{c} ]But from earlier, ( f(a) = frac{1 - a}{a + 1} ), so:[ frac{1}{c} = frac{1 - a}{a + 1} ]But we already established that ( c = frac{a + 1}{1 - a} ), so substituting:[ frac{1}{c} = frac{1 - a}{a + 1} implies frac{1 - a}{a + 1} = frac{1 - a}{a + 1} ]Which is an identity, so no new information here.So far, we have:1. ( f(f(x)) = frac{x}{c} ) (since ( d = 0 )), so ( f(f(x)) = frac{x}{c} ).2. ( c = f(0) = frac{a + 1}{1 - a} ), where ( a = f(1) ).3. ( f(a) = frac{1 - a}{a + 1} ).Let me try to find ( a ). Let's denote ( a = f(1) ). Then, ( f(a) = frac{1 - a}{a + 1} ).But also, from the expression ( f(f(x)) = frac{x}{c} ), we can substitute ( x = 1 ):[ f(f(1)) = frac{1}{c} implies f(a) = frac{1}{c} ]But from point 3, ( f(a) = frac{1 - a}{a + 1} ), and from point 2, ( c = frac{a + 1}{1 - a} ). Therefore,[ frac{1 - a}{a + 1} = frac{1}{c} = frac{1 - a}{a + 1} ]Again, this is consistent but doesn't give new information.Perhaps we need another substitution. Let me try to find ( f(0) ). Since ( d = f(f(0)) = 0 ), we have ( f(f(0)) = 0 ). Let ( f(0) = c ), so ( f(c) = 0 ). Therefore, ( f(c) = 0 ).Now, let's plug ( x = c ) into the original equation. Wait, ( x ) must be in ( D ), so ( c neq -1 ). Since ( c = f(0) ), and ( f: D rightarrow D ), ( c neq -1 ). So ( x = c ) is allowed as long as ( c neq -1 ).Let me set ( x = c ). Then, for any ( y neq -c ), the equation becomes:[ (f(f(c)) + y) fleft( frac{y}{c} right) + f(f(y)) = c ]But ( f(c) = 0 ), so ( f(f(c)) = f(0) = c ). Therefore:[ (c + y) fleft( frac{y}{c} right) + f(f(y)) = c ]But from earlier, ( f(f(y)) = frac{y}{c} ). Therefore:[ (c + y) fleft( frac{y}{c} right) + frac{y}{c} = c ]Let me denote ( z = frac{y}{c} implies y = c z ). Then, substituting:[ (c + c z) f(z) + frac{c z}{c} = c ]Simplify:[ c (1 + z) f(z) + z = c ]Divide both sides by ( c ) (assuming ( c neq 0 )):[ (1 + z) f(z) + frac{z}{c} = 1 ]So,[ (1 + z) f(z) = 1 - frac{z}{c} ]Therefore,[ f(z) = frac{1 - frac{z}{c}}{1 + z} ]Which simplifies to:[ f(z) = frac{c - z}{c (1 + z)} ]Wait, this is interesting. So for any ( z neq -1 ), which is consistent with the domain ( D ). Therefore, this suggests that ( f(z) = frac{c - z}{c (1 + z)} ). But this must hold for all ( z in D ), i.e., for all ( z neq -1 ).But this expression for ( f(z) ) must be consistent with our previous findings, such as ( f(1) = a ), ( f(0) = c ), etc.Let's check ( z = 0 ):[ f(0) = frac{c - 0}{c (1 + 0)} = frac{c}{c} = 1 ]But ( f(0) = c ), so this implies ( c = 1 ).Ah! So ( c = 1 ). Therefore, substituting back into the expression for ( f(z) ):[ f(z) = frac{1 - z}{1 cdot (1 + z)} = frac{1 - z}{1 + z} ]Therefore, ( f(z) = frac{1 - z}{1 + z} ).Now, let's verify if this function satisfies all the conditions.First, check if ( f ) maps ( D ) to ( D ). ( D = mathbb{R} setminus {-1} ). So, for ( z neq -1 ), ( 1 + z neq 0 implies z neq -1 ), which is already in the domain. Then, ( f(z) = frac{1 - z}{1 + z} ). We need to ensure ( f(z) neq -1 ).Suppose ( frac{1 - z}{1 + z} = -1 ). Then:( 1 - z = - (1 + z) implies 1 - z = -1 - z implies 1 = -1 ), which is impossible. Therefore, ( f(z) neq -1 ) for any ( z in D ). Therefore, ( f ) maps ( D ) to ( D ).Good. Now, let's check if this function satisfies the original functional equation.So, the function is ( f(x) = frac{1 - x}{1 + x} ).First, compute ( f(f(x)) ):Let ( f(x) = frac{1 - x}{1 + x} ), then( f(f(x)) = fleft( frac{1 - x}{1 + x} right) = frac{1 - frac{1 - x}{1 + x}}{1 + frac{1 - x}{1 + x}} )Simplify numerator:( 1 - frac{1 - x}{1 + x} = frac{(1 + x) - (1 - x)}{1 + x} = frac{2x}{1 + x} )Denominator:( 1 + frac{1 - x}{1 + x} = frac{(1 + x) + (1 - x)}{1 + x} = frac{2}{1 + x} )Therefore,( f(f(x)) = frac{2x / (1 + x)}{2 / (1 + x)} = x )So, ( f(f(x)) = x ). That's interesting. Therefore, ( f ) is an involution, meaning ( f(f(x)) = x ).Now, substitute back into the original equation:[ (f(f(x)) + y) fleft( frac{y}{x} right) + f(f(y)) = x ]Since ( f(f(x)) = x ) and ( f(f(y)) = y ), this simplifies to:[ (x + y) fleft( frac{y}{x} right) + y = x ]Therefore, we need to check if:[ (x + y) fleft( frac{y}{x} right) + y = x ]Substitute ( fleft( frac{y}{x} right) = frac{1 - frac{y}{x}}{1 + frac{y}{x}} = frac{x - y}{x + y} )Therefore, the left-hand side becomes:[ (x + y) cdot frac{x - y}{x + y} + y = (x - y) + y = x ]Which equals the right-hand side. Therefore, the equation holds.Therefore, the function ( f(x) = frac{1 - x}{1 + x} ) satisfies the given functional equation.Now, we need to check if there are other possible functions.Earlier, we derived that ( f(z) = frac{c - z}{c (1 + z)} ) with ( c = 1 ). Therefore, the only solution is ( f(x) = frac{1 - x}{1 + x} ).To ensure there are no other solutions, let's review the steps. We assumed ( f ) had a particular form based on substitutions and derived the function. Each substitution led us to the conclusion that ( c = 1 ), and hence the function is uniquely determined. Since all steps were necessary and consistent, this should be the only solution.Final AnswerThe function is boxed{dfrac{1 - x}{1 + x}}.

✅:To find all functions ( f: D rightarrow D ) such that for all ( x, y in D ) with ( x neq 0 ) and ( y neq -x ), the equation[ (f(f(x)) + y) fleft(frac{y}{x}right) + f(f(y)) = x ]holds, we proceed as follows:1. Substitute ( y = 0 ): [ f(f(x)) cdot f(0) + f(f(0)) = x ] Let ( c = f(0) ) and ( d = f(f(0)) ). This simplifies to: [ c cdot f(f(x)) + d = x implies f(f(x)) = frac{x - d}{c} ]2. Substitute ( x = 1 ) and ( y = 1 ): [ (f(f(1)) + 1) f(1) + f(f(1)) = 1 ] Let ( a = f(1) ). This simplifies to: [ (f(a) + 1) a + f(a) = 1 implies f(a) = frac{1 - a}{a + 1} ]3. Substitute ( x = y ): [ f(f(x)) (f(1) + 1) + x f(1) = x ] Simplifies to: [ f(f(x)) = frac{x (1 - f(1))}{f(1) + 1} ]4. Equate expressions for ( f(f(x)) ): From the earlier result and the equation derived by setting ( y = 0 ), we find: [ frac{x - d}{c} = frac{x (1 - f(1))}{f(1) + 1} ] This gives ( d = 0 ) and ( c = frac{a + 1}{1 - a} ).5. Determine ( f(0) ): From ( f(f(0)) = 0 ) and ( f(0) = c ), we find ( c = 1 ).6. Derive the function: Substituting ( c = 1 ) into the expression for ( f(z) ), we get: [ f(z) = frac{1 - z}{1 + z} ]7. Verify the function: Check that ( f(f(x)) = x ) and substitute back into the original equation to confirm it holds.Thus, the function satisfying the given functional equation is:[boxed{dfrac{1 - x}{1 + x}}]

🔑:To solve for the function ( f ), we need to satisfy the given functional equation[ (f(f(x)) + y) fleft( frac{y}{x} right) + f(f(y)) = x]for all ( x, y in D ) where ( x neq 0 ) and ( y neq -x ).1. Initial assumption: Let's put ( x = y ). This can simplify the equation significantly. The condition ( x = y neq 0 ) gives us: [ (f(f(x)) + x) fleft( frac{x}{x} right) + f(f(x)) = x ] Simplifying: [ (f(f(x)) + x) f(1) + f(f(x)) = x ]2. Introduce constants: Assume ( f(f(x)) = alpha x ) for some constant ( alpha ). Then, the above equation becomes: [ (alpha x + x) f(1) + alpha x = x ] Factor out ( x ): [ x (alpha f(1) + f(1) + alpha) = x ] For this to hold for any ( x neq 0 ), the coefficient of ( x ) must be 1: [ alpha f(1) + f(1) + alpha = 1 ]3. Analyze ( f(1) ): Try specific value combinations to explore further. Let ( f(1) = k ), we get: [ alpha k + k + alpha = 1 ]4. Case ( alpha neq 0 ): If ( alpha neq 0 ), dividing through by ( alpha ): [ k + frac{k}{alpha} + 1 = frac{1}{alpha} ]5. Special values of ( alpha ): - Examine ( alpha = 1 ): [ alpha = 1 implies f(f(x)) = x ] - Check equation: Plugging in ( f(f(x)) ): [ (x + y) fleft(frac{y}{x}right) + y = x ] Since ( f(f(x)) = x ): [ y = - ( x+ y ) f left( frac{y}{x} right) ] - Setting ( y =1 ): ( f'(x) neq 0): ( f(1) = k) and ...8. Final functional form:If results fail previous steps:6. - Try ( alpha = 0 implies contradiction ): contradiction shows nonzero explore reveals.Continuing deeper verification: Let ( P(x, y) rightarrow ...) evaluation paired transform: derive ( valid domain ):7 clarify insight for range and deduced follow.Conclude: [boxed{ f(x) = frac{1-x}{1+x} equiv satisfied conditions checked. validity holds )] Verification completed total consistency ensures function mapped correct }

🧠:Okay, let me try to figure out this division problem. The question says that the quotient is five times smaller than the dividend and seven times larger than the divisor. Hmm, I need to find numbers such that when you divide the dividend by the divisor, you get a quotient, and that quotient has these specific relationships to the dividend and the divisor.First, let's recall the basic division formula: Dividend ÷ Divisor = Quotient. So, Dividend = Divisor × Quotient. That might be useful here.The problem states two things:1. The quotient is five times smaller than the dividend. So, if the dividend is, say, D and the quotient is Q, then D = 5Q? Wait, "five times smaller" can be a bit ambiguous. Sometimes people say "times smaller" to mean divided by that number. So if something is five times smaller than another, it means it's the original divided by five. So, in this case, Q = D / 5. That makes sense because if the quotient is five times smaller than the dividend, then the dividend is five times the quotient.2. The quotient is seven times larger than the divisor. Again, similar phrasing. If the divisor is d, then Q = 7d. Because if the quotient is seven times larger than the divisor, then it's seven multiplied by the divisor.So from these two relationships:Q = D / 5 (from the first condition)Q = 7d (from the second condition)But we also know from the division algorithm that D = d × Q. Let me write down these equations:1. Q = D / 52. Q = 7d3. D = d × QSo we have three equations here. Let's substitute Q from the first equation into the third equation. So D = d × (D / 5). Let's write that:D = d × (D / 5)Multiplying both sides by 5 to eliminate the denominator:5D = d × DWait, that would give 5D = dD. If I divide both sides by D (assuming D ≠ 0, which makes sense because you can't divide by zero), then 5 = d. So d = 5?Wait, that's interesting. So the divisor is 5. Then from the second equation, Q = 7d = 7 × 5 = 35. Then from the first equation, Q = D / 5, so D = 5 × Q = 5 × 35 = 175.Let me check if this works. Dividend is 175, divisor is 5, quotient would be 175 ÷ 5 = 35. Then check the conditions:- The quotient (35) should be five times smaller than the dividend (175). 175 / 5 = 35, which matches.- The quotient (35) should be seven times larger than the divisor (5). 5 × 7 = 35, which also matches.So that seems to work. But let me make sure I didn't make any mistakes here. Let's go through the steps again.Starting with the three equations:1. Q = D / 52. Q = 7d3. D = d × QSubstituting equation 1 into equation 3: D = d × (D / 5). Then, solving for d:Multiply both sides by 5: 5D = dD. Then divide both sides by D (assuming D ≠ 0): 5 = d. So d = 5. Then Q = 7d = 35. Then D = 5 × 35 = 175. So yes, that works.Alternatively, let's try plugging in Q from equation 2 into equation 1. From equation 2: Q = 7d. Plug into equation 1: 7d = D / 5 → D = 35d. Then from equation 3: D = d × Q = d × 7d = 7d². So D is both 35d and 7d². Therefore:35d = 7d² → 35d - 7d² = 0 → 7d(5 - d) = 0. So either d = 0 or d = 5. But d can't be 0 because divisor can't be zero. So d = 5. Then Q = 7×5 = 35, D = 35×5 = 175. Same result. So that checks out.Therefore, the division example is 175 ÷ 5 = 35, where 35 is five times smaller than 175 and seven times larger than 5.Wait, just to make sure there's no other possible solution. Suppose "five times smaller" was interpreted differently. Sometimes people get confused with "times smaller." For example, if something is five times smaller than X, does that mean X - 5X = -4X? But that doesn't make sense here because quotient can't be negative if we're talking about positive numbers. So the correct interpretation is that the quotient is D divided by 5. Similarly, "seven times larger" would mean multiplied by seven, so Q = 7d. So the initial interpretation seems correct.Another way to check: If I use different numbers, say, suppose the quotient is 10. Then according to the first condition, the dividend would be 5×10=50. According to the second condition, the divisor would be 10/7 ≈ 1.428. Then dividing 50 by 1.428 would give approximately 35, which doesn't match the quotient of 10. So that's inconsistent. Hence, the only consistent solution is when d=5, Q=35, D=175.Yes, I think that's solid. Let me just present the example clearly.

✅:Step-by-Step Explanation:1. Define Variables: - Let the dividend be ( D ), the divisor be ( d ), and the quotient be ( Q ).2. Translate Conditions into Equations: - "Quotient is five times smaller than the dividend" → ( Q = frac{D}{5} ). - "Quotient is seven times larger than the divisor" → ( Q = 7d ). - Division relationship: ( D = d times Q ).3. Substitute ( Q ) from the First Equation into the Third: - ( D = d times frac{D}{5} ). - Multiply both sides by 5: ( 5D = dD ). - Divide by ( D ): ( 5 = d ).4. Find ( Q ) and ( D ): - ( d = 5 ). - Substitute ( d ) into ( Q = 7d ): ( Q = 35 ). - Substitute ( Q ) into ( Q = frac{D}{5} ): ( D = 175 ).5. Verify: - ( 175 ÷ 5 = 35 ). - ( 35 times 5 = 175 ) (checks quotient is 1/5 of dividend). - ( 35 ÷ 7 = 5 ) (checks quotient is 7 times divisor).Final Answer:The division example is ( 175 ÷ 5 = 35 ). Here, the quotient ( 35 ) is five times smaller than the dividend ( 175 ) and seven times larger than the divisor ( 5 ).

🔑:: 1. We are given that the quotient is five times smaller than the dividend. - Let the dividend be ( D ) and the divisor be ( d ). - We can express the quotient as ( frac{D}{d} ).2. According to the problem, the quotient is ( frac{1}{5} ) of the dividend: [ frac{D}{d} = frac{D}{5} ]3. From this, we can infer that: [ d = 5 ]4. Next, we know the quotient is seven times larger than the divisor: - Substitute the value of the divisor into the equation for the quotient: [ frac{D}{d} = 7 cdot d ] [ frac{D}{5} = 7 cdot 5 ]5. Simplify the right-hand side: [ frac{D}{5} = 35 ]6. To find the dividend ( D ), multiply both sides by 5: [ D = 5 cdot 35 ] [ D = 175 ]7. We now verify the solution: - Dividend: ( D = 175 ) - Divisor: ( d = 5 ) - Quotient: ( frac{D}{d} = frac{175}{5} = 35 )Conclusion: The example given fits all the given conditions, hence the division is [boxed{175 div 5 = 35}]

🧠:Okay, so I need to prove that this expression is greater than or equal to 9. Let me start by understanding the problem. The inequality given is:[ frac{1}{cos^2 alpha} + frac{1}{sin^2 alpha sin^2 beta cos^2 beta} geq 9.]Given that both α and β are between 0 and π/2. Hmm, all the trigonometric functions here will be positive since sine and cosine are positive in the first quadrant. That's good, so we don't have to worry about flipping inequalities or anything like that.First, I need to recall some trigonometric identities or inequalities that might help here. The problem mentions proving an inequality, so maybe the AM-GM inequality could be useful? That's a common tool for such proofs. Let me remember that the AM-GM inequality states that for non-negative real numbers, the arithmetic mean is greater than or equal to the geometric mean. So, maybe I can apply AM-GM to the terms in the given expression.But let me look at the structure of the expression. The first term is 1/cos²α, and the second term is 1/(sin²α sin²β cos²β). Both terms are reciprocals of squared trigonometric functions. So, maybe if I can express these terms in a way that I can apply AM-GM with multiple variables. Let me see.Alternatively, maybe I can simplify the second term. Let's consider sin²β cos²β. There's an identity here: sin(2β) = 2 sinβ cosβ, so sin²β cos²β = (sin2β / 2)². Therefore, sin²β cos²β = (sin²2β)/4. Therefore, the second term becomes 1/(sin²α * (sin²2β)/4) = 4/(sin²α sin²2β). So the entire expression can be rewritten as:1/cos²α + 4/(sin²α sin²2β)Hmm, maybe that's helpful. Now, the variables are α and β, but β is inside sin2β. Let me see if I can find a substitution or fix one variable first.Alternatively, since the inequality must hold for all α and β in the given ranges, perhaps we can find the minimum value of the expression. If the minimum is 9, then the inequality is proven.So, maybe I can set up the problem as a minimization problem. Let me denote f(α, β) = 1/cos²α + 4/(sin²α sin²2β). We need to find the minimum of f(α, β) over 0 < α < π/2 and 0 < β < π/2.To find the minimum, perhaps take partial derivatives with respect to α and β, set them to zero, and solve. But calculus might get complicated here. Maybe there's a smarter way using inequalities.Alternatively, notice that sin²2β is involved. Let me see: sin2β varies between 0 and 1 when β is between 0 and π/4, and then back to 0 at π/2. But sin²2β has a maximum of 1 when β = π/4, and minimum approaching 0 as β approaches 0 or π/2. However, since β is between 0 and π/2, sin2β is positive, so sin²2β is between 0 and 1.But in the denominator, sin²2β is in the denominator, so the term 4/(sin²α sin²2β) will be minimized when sin²2β is maximized, which is at β = π/4. Similarly, the term 1/cos²α is minimized when cos²α is maximized, which is when α approaches 0. But the problem is that we need to find the minimum of the sum, so there's a trade-off here. If α is small, then cosα is close to 1, so 1/cos²α is small, but sinα is small, so the second term becomes large. Conversely, if α is large (close to π/2), then 1/cos²α becomes large, but sinα is close to 1, so the second term might be manageable. Similarly for β: if β is π/4, sin2β is 1, so the second term is minimized. If β is different, then sin2β is smaller, so the second term is larger.Therefore, perhaps the minimum occurs when β is π/4, which minimizes the second term. Let me check that. Suppose β = π/4. Then sin2β = sin(π/2) = 1, so the expression becomes:1/cos²α + 4/(sin²α * 1) = 1/cos²α + 4/sin²α.Now, we can write this as sec²α + 4 csc²α.We need to find the minimum of this expression with respect to α in (0, π/2). Let's see if we can apply AM-GM here. Let me recall that sec²α = 1 + tan²α, and csc²α = 1 + cot²α, but maybe that's not helpful.Alternatively, write everything in terms of sinα and cosalpha. Let me consider variables x = sinα and y = cosα. Then x² + y² = 1. The expression becomes:1/y² + 4/x².So, we need to minimize 1/y² + 4/x² with x² + y² = 1 and x, y > 0. That's a constrained optimization problem. Maybe use Lagrange multipliers here. Let me set up the Lagrangian:L = 1/y² + 4/x² - λ(x² + y² - 1).Take partial derivatives:dL/dx = -8/x³ - 2λx = 0,dL/dy = -2/y³ - 2λy = 0,dL/dλ = x² + y² - 1 = 0.So from the first equation: -8/x³ - 2λx = 0 => -8/x³ = 2λx => λ = -4/x⁴.From the second equation: -2/y³ - 2λy = 0 => -2/y³ = 2λy => λ = -1/y⁴.Therefore, setting the two expressions for λ equal:-4/x⁴ = -1/y⁴ => 4/x⁴ = 1/y⁴ => y⁴ = x⁴/4 => y = x / √2.But since x and y are positive, we can write y = x / √2. Now, substituting into the constraint x² + y² = 1:x² + (x² / 2) = 1 => (3x²)/2 = 1 => x² = 2/3 => x = √(2/3) => y = √(2/3)/√2 = √(1/3) = 1/√3.Therefore, the minimum occurs at sinα = √(2/3) and cosα = 1/√3. Therefore, α = arcsin(√(2/3)) ≈ 54.7 degrees. Then, plugging back into the expression:1/y² + 4/x² = 1/(1/3) + 4/(2/3) = 3 + 4*(3/2) = 3 + 6 = 9.So, the minimum is indeed 9 when β = π/4 and α = arcsin(√(2/3)). Therefore, the original inequality holds, and equality is achieved at those values.But wait, this was under the assumption that β = π/4. But does this give the global minimum? Because if β is different, then the second term in the original expression would be larger, so the sum would be larger. Therefore, the minimum occurs when β is chosen to minimize the second term, which is at β = π/4, and then given that β, the minimum over α is 9. Therefore, the overall minimum is 9.Alternatively, let's think of it as a two-step optimization: for any α, the second term is minimized when β = π/4, and given that, then the minimum over α is 9. Therefore, the inequality holds.But let me verify this reasoning. Suppose we fix α, then the second term 4/(sin²α sin²2β) is minimized when sin²2β is maximized, which is 1, achieved at β=π/4. Therefore, for each α, the minimal value of the expression is 1/cos²α + 4/sin²α. Then, we need to find the minimum of 1/cos²α + 4/sin²α over α. As we did earlier, and found that the minimum is 9.Therefore, the minimal value of the original expression is indeed 9, achieved when β=π/4 and α=arcsin(√(2/3)). Hence, the inequality holds.Alternatively, is there another approach using inequalities directly?Let me try using the AM-GM inequality on the two terms. But there are two terms, and AM-GM would state that (A + B)/2 >= sqrt(AB). But then A + B >= 2 sqrt(AB). So if we can show that 2 sqrt(AB) >=9, but that would require AB >= (9/2)^2=81/4. However, let me compute AB:A = 1/cos²α,B = 4/(sin²α sin²2β).So AB = 4/(cos²α sin²α sin²2β).But sin²2β = 4 sin²β cos²β, so AB = 4/(cos²α sin²α * 4 sin²β cos²β) = 1/(cos²α sin²α sin²β cos²β).Not sure if that helps. Alternatively, perhaps split the terms into more terms to apply AM-GM with more variables.Wait, the original expression is:1/cos²α + 1/(sin²α sin²β cos²β).But in the second term, sin²β cos²β is similar to (sinβ cosβ)^2. Let me see, maybe express the second term as [1/(sinα sinβ cosβ)]². So the expression is [1/cosα]^2 + [1/(sinα sinβ cosβ)]^2.But how can we relate these terms?Alternatively, consider writing the entire expression as the sum of squares and use Cauchy-Schwarz inequality. But Cauchy-Schwarz usually gives an upper bound, but we need a lower bound.Alternatively, maybe use the Cauchy-Schwarz in the form (a1^2 + a2^2)(b1^2 + b2^2) >= (a1b1 + a2b2)^2. But not sure how to apply here.Alternatively, perhaps use substitution. Let me set x = cosα and y = sinβ cosβ. Then, since 0 < α < π/2, 0 < x < 1, and 0 < β < π/2, so y = sinβ cosβ = (sin2β)/2, so 0 < y < 1/2. Then, the original expression becomes:1/x² + 1/( (1 - x²) y² )We need to show that 1/x² + 1/( (1 - x²) y² ) >=9.But maybe this substitution complicates things. Alternatively, set t = sinβ cosβ, so t = (sin2β)/2, so t is between 0 and 1/2, as above. Then, the second term is 1/( sin²α * t² ). So the expression is 1/cos²α + 1/( sin²α t² ). Let me denote u = cosα, so sinα = sqrt(1 - u²). Then the expression becomes:1/u² + 1/( (1 - u²) t² )We need to minimize this over 0 < u < 1 and 0 < t < 1/2.But perhaps first minimize over t. For fixed u, the term 1/( (1 - u²) t² ) is minimized when t is maximized. Since t = sinβ cosβ, which is maximized at β = π/4, so t_max = 1/2. Therefore, for fixed u, the minimal value of the second term is 1/( (1 - u²) (1/2)^2 ) = 4/(1 - u²). Therefore, the minimal expression for each u is 1/u² + 4/(1 - u²). Now, we need to minimize 1/u² + 4/(1 - u²) over 0 < u < 1.This brings us back to the same expression as before, leading to the minimum of 9. Therefore, this confirms that the minimal value is indeed 9.Alternatively, let's try using AM-GM on multiple variables. Let's split the terms into more components. Suppose we express the original expression as a sum of terms such that their product is a constant, allowing us to apply AM-GM.Let me see: 1/cos²α can be written as 1 + tan²α. But maybe not helpful. Alternatively, let's consider writing the original expression as:1/cos²α + 1/(sin²α sin²β cos²β) = sec²α + 1/(sin²α sin²β cos²β).But how to split this into terms for AM-GM?Alternatively, think of the second term as [1/(sinα sinβ cosβ)]². Then the expression is [secα]^2 + [1/(sinα sinβ cosβ)]^2. If we can write this as the sum of squares, maybe relate them via some identity?Alternatively, consider using the AM-GM inequality on multiple terms. For example, if we can express the expression as the sum of 9 terms (since the right-hand side is 9), each term being 1, and then apply AM-GM. But this seems vague.Wait, AM-GM with 9 terms: if we can split the left-hand side into 9 terms such that their product is 1, then their sum would be at least 9. Let me try that.But first, the left-hand side is two terms. It's hard to split into 9 terms. Alternatively, maybe split each term into multiple terms.For example, 1/cos²α can be written as 1/cos²α = sec²α. Let's split this into, say, four terms: a + a + a + a, where a = (1/(4 cos²α)). Then, similarly, split the second term into five terms. Wait, but 4 + 5 = 9. Let's try that.Let me denote A = 1/cos²α and B = 1/(sin²α sin²β cos²β). Then, we can write A = 4*(1/(4 cos²α)) and B = 5*(1/(5 sin²α sin²β cos²β)). But that seems arbitrary. Alternatively, if we split A into m terms and B into n terms such that m + n = 9, then apply AM-GM. However, the problem is to choose m and n such that the product of all terms is a constant.Alternatively, if we split A into x terms each equal to 1/(x cos²α) and B into y terms each equal to 1/(y sin²α sin²β cos²β), such that x + y = 9.Then by AM-GM:(A + B)/9 >= ( (1/(x cos²α))^x * (1/(y sin²α sin²β cos²β))^y )^(1/9)But then we need the right-hand side to be greater than or equal to 1, so that (A + B)/9 >= 1 => A + B >=9. Therefore, we need:(1/(x cos²α))^x * (1/(y sin²α sin²β cos²β))^y >=1But simplifying:(1/(x^x y^y)) * (1/(cos²α)^x sin²α^y sin²β^y cos²β^y) ) >=1Which can be written as:1/(x^x y^y) * 1/( cos^{2x}α sin^{2y}α sin^{2y}β cos^{2y}β ) >=1Hmm, this seems complicated. For this inequality to hold for all α and β, the expression must be independent of α and β, which is not the case here unless the exponents for the trigonometric functions cancel out. Let's check the exponents:For cosα: exponent is 2xFor sinα: exponent is 2ySimilarly, for β: exponent in sinβ is 2y and cosβ is 2y.To make the expression independent of α and β, we need the exponents to be zero. That is:For α: 2x + 2y =0, but x and y are positive numbers, so this is impossible. Therefore, this approach is not feasible.Alternatively, perhaps use weighted AM-GM. Let me recall that for positive real numbers a and b, and positive weights p and q with p + q =1, we have pa + qb >= a^p b^q. But I need to check the exact statement.Wait, the weighted AM-GM states that for non-negative real numbers a and b, and positive weights p and q such that p + q =1, then pa + qb >= a^p b^q. But here, if we want to split into 9 terms, perhaps with equal weights. Alternatively, maybe this is not the right approach.Alternatively, let's consider using the Cauchy-Schwarz inequality. For example, (a^2 + b^2)(c^2 + d^2) >= (ac + bd)^2. But how to apply this here?Alternatively, think of the two terms as two vectors. Let me see:Let me consider vectors u = (1/cosα, 1/(sinα sinβ cosβ)) and v = (1, 2). Then, by Cauchy-Schwarz:( (1/cosα)^2 + (1/(sinα sinβ cosβ))^2 )(1^2 + 2^2) >= (1/cosα *1 + 1/(sinα sinβ cosβ)*2)^2.But the left-hand side would be our expression multiplied by 5, and the right-hand side is (1/cosα + 2/(sinα sinβ cosβ))^2. Not sure if this helps. Maybe not directly.Alternatively, maybe use Hölder's inequality. Hölder's inequality generalizes Cauchy-Schwarz and can relate sums with products. But I need to recall Hölder's inequality.Hölder's inequality states that for sequences of non-negative real numbers (a_i), (b_i), ..., (z_i) and exponents p, q, ..., r such that 1/p + 1/q + ... + 1/r =1, then the product of the sums is greater than or equal to the sum of the products. Not sure how to apply here.Alternatively, think of the original inequality and try to use substitution. Let me set t = tanα. Then, since 0 < α < π/2, t >0. Then, cosα = 1/sqrt(1 + t²), sinα = t/sqrt(1 + t²). Therefore, 1/cos²α = 1 + t².Similarly, the second term is 1/( sin²α sin²β cos²β ) = 1/( (t²/(1 + t²)) sin²β cos²β ) = (1 + t²)/( t² sin²β cos²β )So the original expression becomes:(1 + t²) + (1 + t²)/( t² sin²β cos²β )Factor out (1 + t²):(1 + t²) [1 + 1/( t² sin²β cos²β ) ]Hmm, not sure if this helps. Let me denote s = sinβ cosβ. Then s = (sin2β)/2, so 0 < s ≤ 1/2. Then the expression becomes:(1 + t²) [1 + 1/( t² s² ) ]So we have:(1 + t²) + (1 + t²)/( t² s² )But maybe this is more complicated. Alternatively, set variables u = t s, then u = t s = tanα * sinβ cosβ. Not sure.Alternatively, fix t and s as variables. Let me consider s = sinβ cosβ. Then, s is in (0, 1/2]. So the expression becomes:(1 + t²) + (1 + t²)/( t² s² ) = (1 + t²)(1 + 1/( t² s² )).Let me denote this as f(t, s) = (1 + t²)(1 + 1/( t² s² )). We need to find the minimum of f(t, s) over t >0 and 0 < s ≤1/2.First, for fixed t, the expression is minimized when s is maximized, i.e., s =1/2. Therefore, for each t, the minimal value is (1 + t²)(1 + 1/( t² (1/2)^2 )) = (1 + t²)(1 + 4/(t² )).So now, we need to minimize (1 + t²)(1 + 4/t² ) over t >0.Let's compute this:(1 + t²)(1 + 4/t² ) = (1)(1) + 1*(4/t²) + t²*1 + t²*(4/t²) = 1 + 4/t² + t² +4 = 5 + t² +4/t².So the expression simplifies to 5 + t² + 4/t². Now, we need to minimize 5 + t² + 4/t² over t >0. Let's set f(t) = t² + 4/t² +5. Take derivative: f’(t) = 2t - 8/t³. Set to zero: 2t -8/t³ =0 => 2t =8/t³ => t^4 =4 => t=√2.Therefore, the minimum occurs at t=√2. Then f(√2) = (√2)^2 +4/( (√2)^2 ) +5 = 2 +4/2 +5=2 +2 +5=9.Therefore, the minimal value is 9, achieved when t=√2 and s=1/2. Translating back, t=tanα=√2, so α=arctan(√2), and s=sinβ cosβ=1/2, which occurs when β=π/4. So this confirms our earlier result.Therefore, the original inequality holds with equality when α=arctan(√2) and β=π/4.Thus, through multiple approaches—using substitution, calculus, and optimization—we've shown that the minimum value of the expression is indeed 9, hence proving the inequality.Final Answerboxed{9}

✅:Given (0 < alpha < frac{pi}{2}) and (0 < beta < frac{pi}{2}), we need to prove that:[frac{1}{cos^2 alpha} + frac{1}{sin^2 alpha sin^2 beta cos^2 beta} geq 9.]First, we rewrite the second term using the identity (sin^2 beta cos^2 beta = left(frac{sin 2beta}{2}right)^2):[frac{1}{sin^2 alpha sin^2 beta cos^2 beta} = frac{4}{sin^2 alpha sin^2 2beta}.]Thus, the expression becomes:[frac{1}{cos^2 alpha} + frac{4}{sin^2 alpha sin^2 2beta}.]We note that (sin 2beta) is maximized when (beta = frac{pi}{4}), giving (sin 2beta = 1). Substituting (beta = frac{pi}{4}), the expression simplifies to:[frac{1}{cos^2 alpha} + frac{4}{sin^2 alpha}.]Next, we need to minimize this expression. Let (x = cos alpha) and (y = sin alpha), where (x^2 + y^2 = 1). The expression becomes:[frac{1}{x^2} + frac{4}{y^2}.]Using the method of Lagrange multipliers to minimize (frac{1}{x^2} + frac{4}{y^2}) under the constraint (x^2 + y^2 = 1), we find that the minimum occurs at (x = frac{1}{sqrt{3}}) and (y = sqrt{frac{2}{3}}). Substituting these values back, we get:[frac{1}{left(frac{1}{sqrt{3}}right)^2} + frac{4}{left(sqrt{frac{2}{3}}right)^2} = 3 + 6 = 9.]Thus, the minimum value of the original expression is 9, achieved when (alpha = arcsinleft(sqrt{frac{2}{3}}right)) and (beta = frac{pi}{4}). Therefore, the inequality holds:[boxed{9}]

🔑: 1. We start with the given inequality: [ frac{1}{cos^2 alpha} + frac{1}{sin^2 alpha sin^2 beta cos^2 beta} geqslant 9 ]2. Rewriting the second term: [ frac{1}{sin^2 alpha sin^2 beta cos^2 beta} ]3. Using the identity ( sin^2 2beta = 4 sin^2 beta cos^2 beta ), we get: [ frac{1}{sin^2 alpha sin^2 beta cos^2 beta} = frac{4}{sin^2 alpha sin^2 2 beta} ]4. Therefore, we can express the inequality as: [ frac{1}{cos^2 alpha} + frac{4}{sin^2 alpha sin^2 2 beta} ]5. Since (sin^2 2 beta leq 1) for all ( 0 < beta < frac{pi}{2} ), we can use this bound: [ frac{1}{cos^2 alpha} + frac{4}{sin^2 alpha} geqslant frac{1}{cos^2 alpha} + frac{4}{sin^2 alpha} ]6. Applying the Cauchy-Schwarz inequality, which states: [ frac{a^2}{x} + frac{b^2}{y} geq frac{(a+b)^2}{x+y} ] with (a = 1), (b = 2), (x = cos^2 alpha), (y = sin^2 alpha), we proceed: [ frac{1}{cos^2 alpha} + frac{4}{sin^2 alpha} geq frac{(1+2)^2}{cos^2 alpha + sin^2 alpha} ]7. Simplifying: [ frac{(1+2)^2}{cos^2 alpha + sin^2 alpha} = frac{3^2}{1} = 9 ]8. The above inequality turns to equality when: [ sin beta = cos beta text{ and } cos^2 alpha = frac{sin^2 alpha}{2} ]9. Therefore: [ sin^2 beta = cos^2 beta = frac{1}{2}, cos^2 alpha = frac{1}{3}, sin^2 alpha = frac{2}{3} ]10. Thus, we have shown that: [ frac{1}{cos^2 alpha} + frac{1}{sin^2 alpha sin^2 beta cos^2 beta} geqslant 9 ]Conclusion.[boxed{9}]

🧠:Alright, let's tackle this problem step by step. So, we have a function ( f(n) ) defined for all positive integers ( n ), taking nonnegative integer values. The key property is that for any two positive integers ( m ) and ( n ), the expression ( f(m+n) - f(m) - f(n) ) is either 0 or 1. Additionally, we have specific conditions: ( f(2) = 0 ), ( f(3) > 0 ), and ( f(9999) = 3333 ). We need to determine ( f(1982) ).First, let me try to understand the functional equation. The equation ( f(m+n) - f(m) - f(n) in {0, 1} ) suggests that the function is almost additive, but there's a possible "error" term of 1. In other words, when you split ( m + n ) into ( m ) and ( n ), the value of the function at ( m + n ) is either exactly the sum of the values at ( m ) and ( n ), or one more than that sum. This kind of reminds me of subadditive or superadditive functions, but here the difference is bounded by 1. Maybe it's something similar to a function that's close to linear but with some adjustments.Given that ( f(2) = 0 ), let's note that. Also, ( f(3) ) is positive, so ( f(3) geq 1 ). The big condition is ( f(9999) = 3333 ). Since 9999 is a large number, maybe the function has some linear behavior asymptotically? If ( f(n) ) is approximately ( frac{n}{3} ) for large ( n ), since ( 9999 / 3 = 3333 ). Hmm, that might be a hint. Let's explore that.First, let's consider small values of ( n ). We know ( f(2) = 0 ), and ( f(3) geq 1 ). Let's see if we can compute ( f(1) ). Wait, the problem states ( f(n) ) is defined for all positive integers, so ( n geq 1 ). But there's no information given about ( f(1) ). Maybe we need to deduce it.Suppose we try to express ( f(n) ) in terms of smaller values. For example, ( f(4) ). Let's compute ( f(4) ). Let's split 4 as 2 + 2. Then, according to the functional equation:( f(4) - f(2) - f(2) = 0 ) or 1.Since ( f(2) = 0 ), this simplifies to ( f(4) - 0 - 0 = f(4) in {0, 1} ). But wait, ( f(4) ) has to be a nonnegative integer, so it's either 0 or 1. However, we need to check if 0 is possible. If ( f(4) = 0 ), then how does that affect other values? Let's see. Let's try to compute ( f(5) ).Splitting 5 as 4 + 1. Then:( f(5) - f(4) - f(1) in {0, 1} ).But we don't know ( f(1) ). Alternatively, split 5 as 2 + 3:( f(5) - f(2) - f(3) in {0, 1} ).Since ( f(2) = 0 ), this becomes ( f(5) - f(3) in {0, 1} ), so ( f(5) in {f(3), f(3) + 1} ).But without knowing ( f(3) ), this isn't immediately helpful. Let's think of another approach.Alternatively, maybe the function ( f(n) ) is related to the floor of ( n/3 ), but adjusted somehow? Let's check:If ( f(n) = lfloor frac{n}{3} rfloor ), then ( f(3) = 1 ), which is positive, as required. ( f(2) = 0 ), which matches. Let's check the functional equation. For example, take ( m = 2 ), ( n = 2 ), then ( f(4) - f(2) - f(2) = 1 - 0 - 0 = 1 ), which is allowed. Similarly, ( m = 3 ), ( n = 3 ): ( f(6) - f(3) - f(3) = 2 - 1 - 1 = 0 ), which is okay. But let's check another case: ( m = 1 ), ( n = 1 ). However, we don't know ( f(1) ). Wait, maybe this approach is not directly applicable.Alternatively, perhaps ( f(n) = lfloor frac{n + k}{3} rfloor ) for some constant k. But since ( f(2) = 0 ), then ( lfloor frac{2 + k}{3} rfloor = 0 ). So, ( 2 + k < 3 ), meaning ( k < 1 ). Since k has to be an integer (assuming k is an integer), then k <= 0. If k=0, then ( f(n) = lfloor frac{n}{3} rfloor ). Let's test this for ( f(3) = 1 ), which is positive, and for ( f(9999) ), which would be ( lfloor 9999 / 3 rfloor = 3333 ), which matches. So this seems promising. However, we need to check if the functional equation holds.Wait, let's see. If ( f(n) = lfloor n/3 rfloor ), then for any m, n, ( f(m + n) - f(m) - f(n) ) should be either 0 or 1. Let's verify this.Take m and n. Then, ( f(m + n) = lfloor (m + n)/3 rfloor ), and ( f(m) + f(n) = lfloor m/3 rfloor + lfloor n/3 rfloor ). The difference between ( lfloor (m + n)/3 rfloor ) and ( lfloor m/3 rfloor + lfloor n/3 rfloor ) can be 0 or 1, depending on the remainders of m and n modulo 3.If we denote m = 3a + b, n = 3c + d, where b, d ∈ {0, 1, 2}, then m + n = 3(a + c) + (b + d). Then, ( lfloor (m + n)/3 rfloor = a + c + lfloor (b + d)/3 rfloor ). Since ( lfloor (b + d)/3 rfloor ) can be 0 or 1 (since b and d are at most 2, their sum is at most 4). If the sum is less than 3, it's 0; if it's 3 or 4, it's 1. Therefore, the difference ( lfloor (m + n)/3 rfloor - (lfloor m/3 rfloor + lfloor n/3 rfloor) = lfloor (b + d)/3 rfloor ), which is either 0 or 1. Therefore, this function satisfies the condition. So, this suggests that ( f(n) = lfloor n/3 rfloor ).But wait, let's check the given conditions. If ( f(n) = lfloor n/3 rfloor ), then ( f(2) = 0 ), which is correct. ( f(3) = 1 > 0 ), which is good. And ( f(9999) = lfloor 9999 /3 rfloor = 3333 ), which matches. Therefore, this function seems to satisfy all the given conditions. If that's the case, then ( f(1982) = lfloor 1982 /3 rfloor ).Calculating 1982 divided by 3: 3*660 = 1980, so 1982 = 3*660 + 2, hence floor(1982/3) = 660. So, ( f(1982) = 660 ). But wait, hold on. Let's confirm if this function indeed satisfies the functional equation for all m and n.Earlier, we saw that when m and n are split into their quotient and remainder modulo 3, the difference is at most 1. However, in our problem statement, the functional equation allows the difference to be exactly 0 or 1, which aligns with the floor function's behavior. Therefore, it seems that ( f(n) = lfloor n/3 rfloor ) is the function in question.But let me verify with a few small examples. Let's take m = 1, n = 1. If f(1) = floor(1/3) = 0. Then f(2) - f(1) - f(1) = 0 - 0 - 0 = 0, which is allowed. But wait, f(2) is given as 0, which is consistent. Then, f(3) = 1. Let's check m=1, n=2: f(3) - f(1) - f(2) = 1 - 0 - 0 = 1, which is allowed. Similarly, m=2, n=2: f(4) - f(2) - f(2) = floor(4/3) - 0 - 0 = 1 - 0 = 1, which is okay. m=3, n=3: f(6) - f(3) - f(3) = 2 - 1 -1 = 0, allowed. m=4, n=5: f(9) - f(4) - f(5) = 3 -1 -1=1, which is allowed. Seems okay.But wait, the problem statement says that the function takes nonnegative integer values, which is satisfied here. So, if this function satisfies all the given conditions, then the answer is 660. However, I need to make sure that there isn't another function that also satisfies the conditions but gives a different value for f(1982). Is this the only possible function?Suppose there is another function f that differs from floor(n/3) at some points but still satisfies the conditions. For example, maybe at some n, f(n) = floor(n/3) + 1, but adjusted such that the functional equation holds. Let's see if that's possible.Suppose f(3) = 1 (as per floor function), but maybe f(4) could be 1 or 2? Wait, according to the functional equation, for m = 2 and n = 2, f(4) - f(2) - f(2) must be 0 or 1. Since f(2) = 0, f(4) must be 0 or 1. However, if we follow the floor function, f(4) = 1. But suppose f(4) = 0. Is that possible?If f(4) = 0, then let's check for m = 3, n = 1: f(4) - f(3) - f(1) = 0 - 1 - f(1) must be 0 or 1. Therefore, -1 - f(1) must be in {0, 1}. But since f(1) is a nonnegative integer, -1 - f(1) is at most -1, which is not in {0, 1}. This is a contradiction. Therefore, f(4) cannot be 0. Hence, f(4) must be 1, which matches the floor function. So, perhaps the function is uniquely determined as floor(n/3).Wait, let's check another value. Suppose f(5). If we split 5 as 2 + 3, then f(5) - f(2) - f(3) = f(5) - 0 -1 must be 0 or 1. Hence, f(5) is 1 or 2. According to the floor function, f(5) = 1. But could it be 2? Let's check if that's possible.If f(5) = 2, then let's take m = 3 and n = 2: f(5) - f(3) - f(2) = 2 -1 -0 =1, which is allowed. Then, splitting 5 as 4 +1: f(5) - f(4) - f(1) = 2 -1 - f(1) ∈ {0,1}. Therefore, 2 -1 - f(1) =1 - f(1) ∈ {0,1}. Hence, 1 - f(1) >=0, so f(1) <=1. Also, since f(1) is nonnegative integer, f(1) can be 0 or 1.If f(1) =1, then 1 -1 =0, which is allowed. If f(1)=0, then 1 -0=1, which is allowed. So, depending on f(1), f(5) could be 2. However, in the floor function, f(1)=0, f(5)=1. But here, maybe f(1) can be 1? Let's check if f(1)=1 is possible.If f(1)=1, then let's compute f(2). Since f(2) is given as 0. Now, m=1, n=1: f(2) - f(1) - f(1) =0 -1 -1 =-2, which is not in {0,1}. Therefore, this is invalid. Therefore, f(1) cannot be 1. Hence, f(1) must be 0. Therefore, f(5) must satisfy 1 -0 =1, so f(5)=1 + f(4) +0 or 1. Wait, no, earlier, when splitting 5 as 4 +1: f(5) - f(4) - f(1) = f(5) -1 -0 ∈ {0,1}. Therefore, f(5) must be 1 or 2. But if f(5) is 1, then floor(5/3)=1, which is okay. If f(5)=2, then we need to check if that affects other values.Wait, but if f(1)=0, then m=1, n=1: f(2) - f(1)-f(1)=0 -0 -0=0, which is okay. If f(5)=2, let's see how that affects higher numbers. For example, m=5, n=1: f(6) - f(5) -f(1) =f(6) -2 -0 ∈{0,1}, so f(6) is 2 or 3. But according to floor(6/3)=2. If f(6)=3, then splitting 6 as 3+3: f(6) - f(3) -f(3)=3 -1 -1=1, which is allowed. But if f(6)=3, then m=6, n=1: f(7)-f(6)-f(1)=f(7)-3 -0 ∈{0,1}, so f(7)=3 or 4. Floor(7/3)=2, but here f(7) would be 3 or 4. This contradicts the floor function.Hence, allowing f(5)=2 leads to inconsistencies with the floor function behavior. Therefore, f(5) must be 1. Therefore, the function must follow the floor(n/3) function. Therefore, f(n)=floor(n/3) is the only possible function that satisfies all conditions. Therefore, f(1982)=floor(1982/3)=660. However, 1982 divided by 3 is 660.666..., so floor of that is 660. Therefore, the answer is 660.Wait, but before finalizing, let me check 1982 divided by 3: 3*600=1800, 1982-1800=182; 3*60=180, 182-180=2. So 600+60=660, remainder 2. Therefore, floor(1982/3)=660. Correct.But let me confirm once again if there isn't another possible function. Suppose, for example, that f(n) = floor(n/3) + k(n), where k(n) is some function that sometimes adds 1, but such that the functional equation is still satisfied. For instance, maybe at certain points, the function is 1 more, but in a way that the difference f(m + n) - f(m) - f(n) doesn't exceed 1.However, given the constraint at n=9999, which is exactly 3333, which is floor(9999/3), then if we have any additions elsewhere, they would have to cancel out over the sum. But 9999 is a very large number, so if there were any accumulated additions, it's unlikely that they would exactly cancel out to give 3333. Therefore, it's probable that the function is exactly floor(n/3).Another angle: Let's consider that the functional equation allows f(m + n) to be either f(m) + f(n) or f(m) + f(n) + 1. So, over the course of building up n, the function can sometimes add an extra 1 when you add two numbers. However, given that f(9999) = 3333, which is exactly 9999/3, this suggests that over the sum of 9999, all those possible extra 1's must have canceled out, i.e., the total number of extra 1's added must be zero. But since 9999 is a multiple of 3 (wait, 9999 divided by 3 is 3333, so yes, it is a multiple of 3). If the function is linear, then f(n) = n/3 for n divisible by 3, and floor(n/3) otherwise. But since n must be integer, f(n) must be integer. Therefore, for n divisible by 3, f(n) = n/3, and for others, floor(n/3). Which is exactly the floor function. Therefore, the function must be floor(n/3).Therefore, f(1982) = floor(1982/3) = 660.Final Answerboxed{660}

✅:The function ( f(n) ) is defined for all positive integers ( n ) and takes nonnegative integer values. For all ( m, n ), the equation ( f(m+n) - f(m) - f(n) in {0, 1} ) holds. The given conditions are ( f(2) = 0 ), ( f(3) > 0 ), and ( f(9999) = 3333 ).To determine ( f(1982) ), we consider the functional equation and the given conditions. The key insight is that the function ( f(n) ) behaves similarly to the floor function divided by 3. We verify this by checking small values and the given conditions:1. Verification of Small Values: - ( f(2) = 0 ) matches ( lfloor 2/3 rfloor = 0 ). - ( f(3) geq 1 ) matches ( lfloor 3/3 rfloor = 1 ).2. Functional Equation Check: - For any ( m ) and ( n ), splitting ( m ) and ( n ) into their remainders modulo 3 shows that the difference ( f(m+n) - f(m) - f(n) ) is either 0 or 1, consistent with the floor function.3. Large Value Verification: - ( f(9999) = 3333 ) matches ( lfloor 9999/3 rfloor = 3333 ).Given these verifications, we conclude that ( f(n) = lfloor n/3 rfloor ) satisfies all conditions. Therefore, calculating ( f(1982) ):[f(1982) = leftlfloor frac{1982}{3} rightrfloor = leftlfloor 660.666ldots rightrfloor = 660]Thus, the final answer is (boxed{660}).

🔑:1. Let us first explore the given conditions of the function ( f(n) ): - ( f(m+n) - f(m) - f(n) = 0 ) or ( 1 ) - ( f(2) = 0 ) - ( f(3) > 0 ) - ( f(9999) = 3333 )2. To find ( f(n) ), we start by using the condition ( f(m+n) - f(m) - f(n) = 0 ) or ( 1 ). We apply this to ( m = n = 1 ): [ f(2) - 2f(1) = 0 quad text{or} quad 1 ] Given ( f(2) = 0 ), we have: [ 0 - 2f(1) = 0 quad text{or} quad 1 ] So: [ 2f(1) = 0 implies f(1) = 0 ]3. Consider ( f(3) ): [ f(3) > 0, text{ and } f(2+1) - f(2) - f(1) = f(3) - 0 - 0 = f(3) ] Since ( f(3) ) must be 1 from the condition ( f(3) > 0 ), then: [ 0 < f(3) leq 1 ] Thus: [ f(3) = 1 ]4. Now, we use these insights to deduce properties of ( f ). Consider ( f(3n) ): [ f(3n) geq n ] We proceed by induction on ( n ) to show that ( f(3n) = n ) for all ( n leq 3333 ): - Base case: ( n = 1 ) [ f(3 cdot 1) = f(3) = 1 ] which satisfies ( f(3) = 1 ). - Induction step: Assume ( f(3k) = k ) for ( k leq n ): [ f(3(n+1)) = f(3n+3) geq f(3n) + f(3) = n + 1 ] By the induction hypothesis, we know ( f(3n) = n ), thus ( f(3n+3) geq n+1 ).5. Given ( f(9999) = 3333 ): [ f(3n) = n quad text{ for } n = 3333 ]6. By the conditions ( 3f(n) leq f(3n) leq 3f(n) + 2 ): - For ( n leq 3333 ): [ f(n) = leftlfloor frac{f(3n)}{3} rightrfloor = leftlfloor frac{n}{3} rightrfloor ]7. Thus: [ f(1982) = leftlfloor frac{1982}{3} rightrfloor = leftlfloor 660.6667 rightrfloor = 660 ] Conclusion:(boxed{660})

🧠:Okay, let's try to figure out how many matches C won. Hmm, so there are four players: A, B, C, and D. Each plays against every other person once, right? So, first, I need to figure out how many total matches there are. Since each person plays three matches (against the other three), but that would count each match twice (once for each player). So the total number of matches should be 4 choose 2, which is 6 matches in total. Yeah, that makes sense because each pair plays once.Now, the results are given: A won 3 matches, B won 1, D didn't win any. We need to find out how many C won. Let's list out all the possible matches.Each player plays three matches:- A plays against B, C, D- B plays against A, C, D- C plays against A, B, D- D plays against A, B, CSo the matches are: A vs B, A vs C, A vs D, B vs C, B vs D, and C vs D.Now, A has won 3 matches. That means A must have beaten B, C, and D. Because those are the only three matches A played. So A won all three of their matches. That makes sense because if A won 3, that's the maximum possible.Then B has only won 1 match. But B plays against A, C, and D. Since A already beat B (because A won all three), B's one win must be against either C or D. But wait, D didn't win any matches, right? So D lost all their matches. So if B played against D, then B must have beaten D because D can't win. Wait, but if D lost all their matches, that means D lost to A, B, and C. Wait, but B has only one win. So B's only win is against D? But if D lost all their matches, then when B played D, B must have won. But that would mean B's one win is against D. Then B lost to A and C. Wait, but B plays C, right? So if B lost to C, then C would have a win against B. Let me check.So let's map out the matches:1. A vs B: A wins2. A vs C: A wins3. A vs D: A wins4. B vs C: ?5. B vs D: B wins (since D can't win)6. C vs D: C wins (since D can't win)So from the above, A has three wins (matches 1,2,3). B has one win (match 5). D has zero wins. So then the remaining matches are B vs C (match 4) and C vs D (match 6). Wait, but C vs D must have been won by C because D can't win. So match 6 is C's win. Then match 4 is B vs C. If B only has one win, which is against D, then B must have lost to C. So match 4 is a win for C. So that would mean C has two wins: against B and against D. But wait, let's tally up the total wins.Total matches: 6. Each match has one winner, so total wins should be 6. Let's check:A: 3 winsB: 1 winC: 2 wins (from matches 4 and 6)D: 0 winsTotal: 3+1+2+0=6. Perfect, that adds up. So C must have won 2 matches. Therefore, the answer should be 2.Wait, but let me double-check. Let's look at each player's matches.Player A: played B, C, D. Won all three. So A's record is 3-0.Player B: played A, C, D. Lost to A, lost to C (since B only has one win), and beat D. So B's record is 1-2.Player D: played A, B, C. Lost all three. So 0-3.Player C: played A, B, D. Lost to A, beat B, beat D. So C's record is 2-1. So yes, C has two wins. That makes sense. So the answer is 2.

✅:C won 2 matches.Step-by-Step Explanation:1. Total Matches: With 4 players, each plays 3 matches. Total matches = C(4,2) = 6.2. A's Matches: A won all 3 matches (vs B, C, D).3. B's Matches: B's only win is against D (since D lost all matches), so B lost to A and C.4. D's Matches: D lost all 3 matches (vs A, B, C).5. C's Matches: - Lost to A (as A won all). - Defeated B (since B only beat D). - Defeated D (D lost all). Thus, C has 2 wins.Final Answer: C won (boxed{2}) matches.

🔑:We have four players, ( text{A} ), ( text{B} ), ( text{C} ), and ( text{D} ), competing in a round-robin table tennis tournament, where each player plays with every other player exactly once, and there are no drawn matches. Let's denote the players as follows:- ( text{A} ) is 甲- ( text{B} ) is 乙- ( text{C} ) is 丙- ( text{D} ) is 丁Given the outcomes of the matches:- ( text{A} ) (甲) won 3 matches,- ( text{B} ) (乙) won 1 match,- ( text{D} ) (丁) did not win any match. Step-by-Step Calculation:1. Total number of matches: Since each player plays with every other player exactly once, the total number of unique matches among four players is calculated by the combination formula ( binom{4}{2} ): [ binom{4}{2} = frac{4!}{2!(4-2)!} = frac{4 times 3}{2 times 1} = 6 ]2. Sum of wins: Since each match must have a winner, the total number of wins is equal to the number of matches played: [ text{Total wins} = 6 ]3. Known wins: We already have the number of wins for players ( text{A} ), ( text{B} ), and ( text{D} ): [ text{Wins by A} + text{Wins by B} + text{Wins by D} = 3 + 1 + 0 = 4 ]4. Calculating ( text{C} )'s (丙) wins: Since the total number of wins must add up to the total number of matches (6), we can find the number of wins for ( text{C} ) by subtracting the known wins from the total number of matches: [ text{Wins by C} = 6 - 4 = 2 ] Conclusion:( text{C} ) (丙) won 2 matches.[boxed{2}]