🧠:[<think>Okay, so I need to find the equation of a line that passes through the point P(5,6), and its x-intercept is twice its y-intercept. Hmm, let's start by recalling some basic concepts about lines and intercepts. First, the equation of a line can be written in slope-intercept form, which is y = mx + b, where m is the slope and b is the y-intercept. But here, they mentioned intercepts, so maybe the intercept form of a line would be more useful here. The intercept form is x/a + y/b = 1, where a is the x-intercept and b is the y-intercept. The problem states that the x-intercept is twice the y-intercept. So, if I let the y-intercept be b, then the x-intercept a should be 2b. So substituting a = 2b into the intercept form, the equation becomes x/(2b) + y/b = 1. But the line also passes through the point (5,6). So I can substitute x = 5 and y = 6 into the equation to solve for b. Let's do that. Substituting into x/(2b) + y/b = 1:5/(2b) + 6/b = 1Let me simplify this equation. First, let's find a common denominator for the two terms on the left side. The denominators are 2b and b, so the common denominator is 2b. So rewriting 6/b as 12/(2b), the equation becomes:5/(2b) + 12/(2b) = 1Adding those fractions together:(5 + 12)/(2b) = 117/(2b) = 1Now, solving for b. Multiply both sides by 2b:17 = 2bThen divide both sides by 2:b = 17/2So the y-intercept is 17/2, which is 8.5. Then the x-intercept is twice that, so a = 2*(17/2) = 17. Therefore, the equation of the line in intercept form is x/17 + y/(17/2) = 1. But maybe they want the equation in standard form or slope-intercept form. Let me convert it to slope-intercept form to check. Alternatively, since I know the intercepts, I can find the slope. The x-intercept is (17, 0) and the y-intercept is (0, 17/2). The slope m is (y2 - y1)/(x2 - x1). So using the two intercepts:m = (17/2 - 0)/(0 - 17) = (17/2)/(-17) = -1/2So the slope is -1/2. Then using the slope-intercept form with the y-intercept b = 17/2:y = (-1/2)x + 17/2Alternatively, let's check if this line passes through (5,6). Plugging in x=5:y = (-1/2)(5) + 17/2 = (-5/2) + 17/2 = (12/2) = 6. Yes, that works. But maybe another way to approach this problem is using the two-intercept form. Let me verify again.If a line has x-intercept a and y-intercept b, then the equation is x/a + y/b = 1. Given that a = 2b, so substituting that in:x/(2b) + y/b = 1Then plug in (5,6):5/(2b) + 6/b = 1Multiply both sides by 2b to eliminate denominators:5 + 12 = 2b17 = 2b => b=17/2 as before.So that's consistent. So the equation in intercept form is x/17 + y/(17/2) = 1. To convert this into standard form (Ax + By = C), multiply both sides by 17:x + 2y = 17Wait, let's check that:Starting from x/17 + y/(17/2) = 1. Multiply each term by 17:17*(x/17) + 17*(y/(17/2)) = 17*1Simplify:x + (17)*(2/17)y = 17x + 2y = 17Yes, that's correct. So the standard form is x + 2y = 17.Alternatively, if we write this in slope-intercept form, solving for y:2y = -x +17y = (-1/2)x + 17/2Which matches the earlier result. So both forms are correct. Depending on how the answer is expected, either form is acceptable, but maybe the problem expects one or the other. Since they didn't specify, perhaps either is okay, but standard form is also common. Let me check both.Alternatively, maybe they want it in the form like Ax + By + C =0? Then x + 2y -17=0. But in the problem statement, it's better to see what's required. The problem says "Find the equation of this line." Typically, equations of lines can be written in any form, but perhaps slope-intercept or standard. Let's check both.But wait, let me just verify once again.Given that the line passes through (5,6) and the intercepts satisfy a=2b. We found that the equation is x + 2y =17. Let's confirm the intercepts.For the x-intercept, set y=0: x +0 =17 => x=17. That's a=17. For the y-intercept, set x=0: 0 +2y=17 => y=17/2. So y-intercept is 17/2. Which is 8.5, and indeed 17 is twice 8.5. So that's correct. Also, plugging in (5,6): 5 +2*6 =5 +12=17. Correct. So the equation x + 2y =17 is correct.Alternatively, in slope-intercept form, y = (-1/2)x + 17/2. Either is correct. The problem didn't specify the form, so maybe both are acceptable, but perhaps they want it in a specific form. Let me check the problem again. It says "Find the equation of this line." So maybe either is okay, but perhaps slope-intercept is more common unless specified otherwise. However, in some cases, standard form is preferred. Since both forms are correct, but maybe the answer expects one of them. Let me check if my calculation is correct once more.Alternatively, maybe I made a mistake in the calculation. Let me redo the steps.Start with intercept form: x/a + y/b =1. Given a=2b. So x/(2b) + y/b=1. Plugging in (5,6):5/(2b) +6/b=1. Let me write 6/b as 12/(2b). So 5/(2b) +12/(2b)=17/(2b)=1. So 17/(2b)=1 => 2b=17 => b=17/2. Then a=2b=17. So equation is x/17 + y/(17/2)=1. Multiply both sides by 17: x + 2y=17. Correct. So that's standard form, which is a valid equation. Therefore, the answer is x + 2y=17. Alternatively, y = (-1/2)x +17/2. Both are correct, but likely the problem expects standard form. Let me check.Alternatively, the problem might accept any correct form, but since the user hasn't specified, perhaps write both? Wait, no, the instructions just say "put your final answer within boxed{}". So probably need to pick one. Let me check standard form versus slope-intercept. Since slope-intercept is more unique (because standard form can have multiples), but in this case, the equation x + 2y =17 is in simplest terms with integer coefficients. So that's acceptable. Alternatively, the slope-intercept form is also simplified. But let me check with the original problem statement. The problem gives coordinates as integers, and the intercepts are a=17 and b=17/2, so fractions are involved. So if the answer is in slope-intercept, fractions are okay. If in standard form, x +2y=17 is also acceptable. Either way is correct. Let me check if the problem is from a source that prefers one form. Since it's unspecified, maybe present it in standard form. But the user might prefer slope-intercept. Alternatively, the problem might expect slope-intercept. Wait, the original thought process derived both, so perhaps present both? But in the final answer, only one is needed.Alternatively, since the user hasn't specified, and the problem doesn't indicate a preferred form, either is acceptable. However, in many math problems, the standard form with integer coefficients is preferred unless slope-intercept is specifically requested. But in this case, since the intercepts are given in terms that lead to fractions in slope-intercept, but integers in standard form. So x +2y=17 is cleaner.But let me check again if x +2y=17 is correct. Plug in (5,6): 5 +12=17. Correct. x-intercept at (17,0). y-intercept at (0,17/2). All correct. So the equation is correct. Therefore, the answer is x +2y=17.Alternatively, if written as y = -1/2 x + 17/2, that's also correct. So the problem might accept either. But since standard form is more straightforward here, perhaps present that. Wait, but in the problem statement, when they mention the x-intercept is twice the y-intercept, maybe they want an equation without fractions. But x +2y=17 is already with integer coefficients. So that's good. Alternatively, maybe they want it in the form Ax + By = C where A, B, C are integers with no common factors, which x +2y=17 satisfies. So yes, that's a valid answer.So after going through this step-by-step, I think the correct equation is x + 2y =17.Final AnswerThe equation of the line is boxed{x + 2y = 17}.</think>Given a line that passes through the point ( P(5, 6) ) and has an x-intercept that is twice its y-intercept, we start by using the intercept form of a line equation, which is (frac{x}{a} + frac{y}{b} = 1). Here, ( a ) is the x-intercept and ( b ) is the y-intercept. Since the x-intercept is twice the y-intercept, we let ( a = 2b ).Substituting ( a = 2b ) into the intercept form, we get:[frac{x}{2b} + frac{y}{b} = 1]Next, we substitute the coordinates of point ( P(5, 6) ) into this equation:[frac{5}{2b} + frac{6}{b} = 1]To solve for ( b ), we first find a common denominator:[frac{5}{2b} + frac{12}{2b} = 1]Combining the fractions:[frac{17}{2b} = 1]Solving for ( b ):[17 = 2b implies b = frac{17}{2}]Thus, the x-intercept ( a ) is:[a = 2b = 2 times frac{17}{2} = 17]Substituting ( a = 17 ) and ( b = frac{17}{2} ) back into the intercept form equation, we get:[frac{x}{17} + frac{y}{frac{17}{2}} = 1]Multiplying through by 17 to clear the denominators:[x + 2y = 17]Verifying this equation with the point ( P(5, 6) ):[5 + 2 times 6 = 5 + 12 = 17]Thus, the equation of the line is (boxed{x + 2y = 17}).,<think>Okay, let's see. I have to find the equation of a line that passes through the point P(5,6). Also, the x-intercept of this line is twice its y-intercept. Hmm, intercepts... Let me recall what intercepts are. The x-intercept is where the line crosses the x-axis, so the y-coordinate there is 0. Similarly, the y-intercept is where the line crosses the y-axis, so the x-coordinate there is 0. The problem says the x-intercept is twice the y-intercept. Let me denote the y-intercept as some number, maybe 'b'. Then the x-intercept would be twice that, so 2b. But wait, in standard terms, the x-intercept is usually denoted as 'a' and the y-intercept as 'b'. So maybe the x-intercept is 2 times the y-intercept. So if the y-intercept is b, then the x-intercept is 2b. Alternatively, I can write the equation of the line in intercept form. The intercept form of a line is x/a + y/b = 1, where a is the x-intercept and b is the y-intercept. Since the x-intercept is twice the y-intercept, that means a = 2b. So substituting that into the intercept form equation, we get x/(2b) + y/b = 1. Maybe I can work with that. Alternatively, I can use the slope-intercept form of the line, which is y = mx + b, where m is the slope and b is the y-intercept. But if I use this form, I need to relate the intercepts. Let's see. The x-intercept is the value of x when y is 0. So if I set y = 0, then 0 = mx + b, so x = -b/m. So the x-intercept is -b/m. According to the problem, the x-intercept is twice the y-intercept. So -b/m = 2b. Hmm, that's an equation I can solve for m in terms of b. Let's try that. Starting with -b/m = 2b. Let's divide both sides by b, assuming b ≠ 0. Then we get -1/m = 2. So solving for m, we get m = -1/2. So the slope is -1/2. Then the equation of the line would be y = (-1/2)x + b. Since the line passes through the point (5,6), I can substitute x = 5 and y = 6 into the equation to find b. Let's do that. Substituting: 6 = (-1/2)(5) + b. Calculating (-1/2)(5) gives -5/2. So 6 = -5/2 + b. To solve for b, add 5/2 to both sides. 6 + 5/2 = b. Converting 6 to halves, that's 12/2 + 5/2 = 17/2. So b = 17/2. Therefore, the equation of the line is y = (-1/2)x + 17/2. But wait, let me check this. If the y-intercept is 17/2, then the x-intercept should be twice that, which is 17. Let's verify. The x-intercept is when y=0. So set y=0 in the equation: 0 = (-1/2)x + 17/2. Solving for x: (1/2)x = 17/2 => x = 17. That's correct. So the x-intercept is 17, which is twice the y-intercept of 17/2. So 17 = 2*(17/2) = 17, which checks out. Also, does the line pass through (5,6)? Let's check. Plugging x=5 into the equation: y = (-1/2)(5) + 17/2 = (-5/2) + 17/2 = (12/2) = 6. Yes, that works. Alternatively, maybe I should try the intercept form method to see if I get the same result. The intercept form is x/a + y/b = 1. Given a = 2b, so substituting, we have x/(2b) + y/b = 1. Since the line passes through (5,6), plugging in x=5 and y=6: 5/(2b) + 6/b = 1. Let's solve for b. First, multiply both sides by 2b to eliminate denominators. That gives 5 + 12 = 2b. So 17 = 2b => b = 17/2. Then a = 2b = 17. So the equation is x/17 + y/(17/2) = 1. To convert this to standard form, multiply both sides by 17: x + 2y = 17. Then rearranged, that's x + 2y = 17. Let's check if this is the same as the previous equation. From the slope-intercept form: y = (-1/2)x + 17/2. Multiplying both sides by 2: 2y = -x + 17. Then bringing x to the left: x + 2y = 17. Yes, that's the same as the intercept form method. So both methods give the same equation, which is good. So the equation of the line is x + 2y = 17, or in slope-intercept form, y = (-1/2)x + 17/2. The problem doesn't specify the form, so either is acceptable, but maybe they prefer standard form. Let me check if there's a more simplified version. Alternatively, standard form is usually written as Ax + By = C, where A, B, C are integers and A is positive. So x + 2y = 17 is already in standard form. So either form is okay. But perhaps the problem expects slope-intercept form? Let me check the original problem again. It just says "Find the equation of this line." So both forms are correct, but maybe writing it in slope-intercept or standard form. Since both methods give x + 2y = 17, that might be the answer. But let me check once more. If someone uses the point-slope form. Wait, maybe not necessary since we already have two methods confirming the same result. But just to be thorough. Suppose I use the two intercepts to find the slope. The x-intercept is (17, 0) and the y-intercept is (0, 17/2). The slope m is (0 - 17/2)/(17 - 0) = (-17/2)/17 = -1/2. Then using point-slope form with point (0, 17/2): y - 17/2 = -1/2(x - 0), which simplifies to y = -1/2x + 17/2. Same result. Alternatively, using the x-intercept (17,0) in point-slope: y - 0 = -1/2(x - 17), so y = -1/2x + 17/2. Also same. Therefore, the equation is correct. So the answer is y = -1/2 x + 17/2 or x + 2y = 17. Either form is acceptable. Since the problem doesn't specify the form, but in most cases, unless specified, either form is okay, but maybe standard form is preferable. Let me see how the question is phrased. "Find the equation of this line." So it's acceptable as both. But perhaps they want it in the form similar to y = mx + b. But since we have fractions, maybe standard form is better. Let me check if x + 2y = 17 is the standard form. Yes. Alternatively, if written in slope-intercept form: y = -1/2 x + 8.5, but 17/2 is 8.5. However, fractional coefficients are often left as fractions. So perhaps writing y = (-1/2)x + 17/2 is more precise. But maybe the problem expects standard form with integer coefficients. Let me check the process again. Wait, if I use the intercepts (17,0) and (0,17/2), the equation can be written in standard form as x/17 + y/(17/2) = 1. Multiplying through by 17 gives x + 2y = 17. So this is the standard form. Alternatively, if they want it in slope-intercept, it's y = -1/2 x + 17/2. Both are correct, but unless specified, standard form is often preferred. But since the problem mentions intercepts, perhaps intercept form is acceptable. But intercept form is x/a + y/b = 1. But they might want it in the most simplified version. But the problem says "Find the equation of this line," so either answer is correct. However, in most math problems like this, standard form or slope-intercept are both acceptable. Let me check some examples. But in the answer, the user instructions say "put your final answer within boxed{}". So perhaps the standard form is more likely. Wait, let me see, in the initial problem statement, the point is given as (5,6). Maybe the answer is expected in Ax + By = C form. Let me check if x + 2y =17 passes through (5,6). 5 + 2*6 = 5 +12=17. Yes, correct. So yes, x + 2y =17 is the equation. Alternatively, if written as y = -1/2x +17/2. Either way. But to confirm, perhaps the answer is in standard form. Let me check both. But maybe the problem expects slope-intercept. However, since intercepts were mentioned, standard form might be better. Hmm. Wait, in the problem, they mention x-intercept and y-intercept. So intercept form is x/a + y/b =1. But converting that to standard form, x + 2y =17. So either way. Alternatively, perhaps the problem expects the answer in the form similar to how intercepts are given, but with integers. Since x + 2y =17 is simple. Alternatively, if written as y = -1/2 x + 17/2, that's also acceptable. But to make sure, perhaps present both? Wait, no, the problem asks for the equation of the line, so either is correct. But in textbooks, sometimes standard form is preferred when dealing with intercepts. Alternatively, perhaps the problem expects slope-intercept. But in any case, both equations are correct. However, since in the problem statement, the x-intercept is twice the y-intercept, which gives a relationship between a and b. So, in standard form, the equation is x + 2y =17. Therefore, I think the answer is x + 2y =17. So boxed, it would be boxed{x + 2y = 17}. Alternatively, if they prefer slope-intercept, then boxed{y = -frac{1}{2}x + frac{17}{2}}. But given that the problem mentions intercepts, I think standard form is more appropriate here. So I will go with boxed{x + 2y = 17}.Final AnswerThe equation of the line is boxed{x + 2y = 17}.</think>Given a line that passes through the point ( P(5, 6) ) and whose x-intercept is twice its y-intercept, we need to find the equation of this line.1. Identify the intercepts: - Let the y-intercept be ( b ). - Then, the x-intercept is ( 2b ).2. Use the intercept form of the line equation: - The intercept form is ( frac{x}{a} + frac{y}{b} = 1 ), where ( a ) is the x-intercept and ( b ) is the y-intercept. - Substituting ( a = 2b ), the equation becomes ( frac{x}{2b} + frac{y}{b} = 1 ).3. Substitute the point ( P(5, 6) ) into the equation: - Substitute ( x = 5 ) and ( y = 6 ): [ frac{5}{2b} + frac{6}{b} = 1 ] - Multiply through by ( 2b ) to clear the denominators: [ 5 + 12 = 2b implies 17 = 2b implies b = frac{17}{2} ] - Therefore, the x-intercept ( a = 2b = 17 ).4. Convert the intercept form to standard form: - The intercept form equation becomes ( frac{x}{17} + frac{y}{frac{17}{2}} = 1 ). - Multiplying through by 17 to clear the denominators: [ x + 2y = 17 ]5. Verification: - Check if the line passes through ( P(5, 6) ): [ 5 + 2 times 6 = 5 + 12 = 17 ] - Check the intercepts: - x-intercept: ( 17 ) - y-intercept: ( frac{17}{2} )Thus, the equation of the line is (boxed{x + 2y = 17}).]

🔑:Let's denote the x-intercept as (2a, 0) and the y-intercept as (0, a). Since both these points, along with P(5,6), lie on the same line, we can find the equation of this line by using the two-point form of a linear equation.We'll first use the points (0, a) and P(5,6) to find the slope of the line, m. The slope m is given bym = frac{6 - a}{5 - 0} = frac{6 - a}{5}.Since (2a, 0) also lies on this line, we can use the slope calculated above and use it in the point-slope form of the equation of a line:y - 0 = m(x - 2a),which simplifies toy = frac{6 - a}{5}(x - 2a).To find the value of a, we substitute (0, a) into the equation, giving us:a = frac{6 - a}{5}(0 - 2a).Simplifying, we find a = -frac{2a(6 - a)}{5}.Multiplying each side by 5 to clear the fraction, we have:5a = -2a(6 - a).Expanding the right side, we obtain:5a = -12a + 2a^2.Rearranging all terms to one side to solve for a:2a^2 - 17a = 0.Factoring a out:a(2a - 17) = 0.Setting each factor equal to zero gives us the potential solutions for a:a = 0quad text{or}quad 2a - 17 = 0.Since the y-intercept cannot be zero (as that would contradict the initial condition), we discard a = 0 and solve 2a - 17 = 0:2a = 17,a = frac{17}{2}.Using a = frac{17}{2}, the slope of the line is given by:m = frac{6 - frac{17}{2}}{5},m = frac{12 - 17}{10},m = frac{-5}{10},m = -frac{1}{2}.The equation of the line can now be written using point (0, frac{17}{2}) and the slope m = -frac{1}{2}:y - frac{17}{2} = -frac{1}{2}(x - 0),y = -frac{1}{2}x + frac{17}{2}.Multiplying each term by 2 to clear the fraction, we get the standard form of the line:2y = -x + 17,x + 2y - 17 = 0.So the equation of the line is boxed{x+2y-17=0}.

🧠:[<think>Okay, so here's this problem about Mario and the moving sidewalk. Let me try to figure it out step by step. Hmm, let's see. The problem says there's a moving sidewalk from Point A to Point B. When the sidewalk is turned off, Mario takes 90 seconds to walk from A to B. Then, when he just stands on the moving sidewalk without walking, it takes him 45 seconds to get from A to B. The question is asking how long it would take him to walk from A to B along the moving sidewalk when it's turned on. Alright, so basically, Mario can walk on his own, the sidewalk moves on its own, and when he walks on the moving sidewalk, their speeds add up. So we need to find the combined speed and then calculate the time it takes. But let's break it down more formally.First, let's denote some variables. Let's say the distance from Point A to Point B is D. Then, Mario's walking speed is V_m, and the speed of the moving sidewalk is V_s. The problem is asking for the time T it takes Mario to walk from A to B when both his walking and the sidewalk's movement are contributing.We know two scenarios:1. When the sidewalk is off, Mario takes 90 seconds. So his walking speed alone covers distance D in 90 seconds.2. When Mario is not walking (just standing on the sidewalk), the sidewalk takes 45 seconds to carry him from A to B.We need to find the time when both are moving together. Since their speeds will add up, the combined speed is V_m + V_s. Then, the time T should be D divided by (V_m + V_s).So let's write the equations.First, from the first scenario: Mario's walking speed V_m = D / 90. Because speed is distance divided by time.Second scenario: Sidewalk's speed V_s = D / 45. Again, speed is distance over time.Then, when both are moving, the combined speed is V_m + V_s = D/90 + D/45. Let's compute that.But hold on, let me check the units. Since D is in distance (let's say meters), and time is in seconds, speed is meters per second. Adding them together gives total speed. Then, time would be D divided by total speed.So, let's compute V_m + V_s:V_m = D / 90V_s = D / 45So, V_m + V_s = D/90 + D/45. Let's factor out D:= D (1/90 + 1/45)Now, 1/90 + 1/45. Since 1/45 is equal to 2/90, so:= D (1/90 + 2/90) = D (3/90) = D (1/30)So, combined speed is D/30 per second.Therefore, the time T to cover distance D at speed D/30 is:T = D / (D/30) = 30 seconds.Wait, so the answer is 30 seconds? Hmm, that seems logical because if walking alone takes 90 seconds and the sidewalk alone takes 45 seconds, then combining their speeds should result in a time less than both. Since 30 seconds is indeed less than 45 seconds, that seems reasonable.But let me check my steps again to be sure.First, define D as the distance between A and B. Correct.Then, Mario's walking speed: V_m = D / 90. That's correct because speed equals distance over time.Sidewalk speed: V_s = D / 45. Also correct.Combined speed: V_m + V_s = D/90 + D/45. Yes, because when moving in the same direction, their speeds add.Adding those fractions: 1/90 + 1/45. Since 1/45 is 2/90, so total 3/90 = 1/30. So combined speed is D/30. Therefore, time is D / (D/30) = 30 seconds. All steps check out.Alternatively, we can think of it in terms of rates. Mario's rate is 1/90 per second (of the distance D), and the sidewalk's rate is 1/45 per second. When combined, their rates add up: 1/90 + 1/45 = 1/30 per second. Therefore, time is 1 / (1/30) = 30 seconds. That also works.Alternatively, using algebra without assuming D. Let’s say the distance D is the same in both cases. So Mario's walking speed is V_m = D / 90, sidewalk speed V_s = D / 45. Then together, speed V = V_m + V_s = D/90 + D/45. As before, this is D*(1/90 + 2/90) = D*(3/90) = D/30. Therefore, time is D / (D/30) = 30. So same answer.Alternatively, maybe use the harmonic mean? Wait, no, harmonic mean is for when you go the same distance at different speeds. Here, it's different. The rates are additive. So the time should be reciprocal of the sum of reciprocals? Wait, no, that's when dealing with parallel resistors or something. Wait, maybe let's think of it as work rates. If Mario can complete 1 walk in 90 seconds, his work rate is 1/90. The sidewalk can complete 1 walk in 45 seconds, its work rate is 1/45. Together, their combined work rate is 1/90 + 1/45 = 1/30. So time is 1 / (1/30) = 30. Yep, same answer. So that's another way to see it.So, all methods lead to 30 seconds. Therefore, the answer is 30 seconds. So, 30 seconds is the time it takes Mario to walk from A to B on the moving sidewalk.Wait, but just to ensure, let's think of it in terms of real numbers. Let's assume the distance is, say, 90 meters. Then, Mario's walking speed is 90 meters / 90 seconds = 1 m/s. The sidewalk's speed is 90 meters / 45 seconds = 2 m/s. Then, when Mario walks on the sidewalk, his speed is 1 + 2 = 3 m/s. Therefore, time is 90 meters / 3 m/s = 30 seconds. Yep, that works. So concrete numbers also confirm the answer. Therefore, 30 seconds is correct.Final AnswerThe time it takes Mario to walk from Point A to Point B along the moving sidewalk is boxed{30} seconds.</think>Let ( D ) be the distance from Point ( A ) to Point ( B ). 1. When the sidewalk is turned off, Mario's walking speed ( V_m ) is given by: [ V_m = frac{D}{90 text{ seconds}} ]2. When Mario is not walking, the speed of the moving sidewalk ( V_s ) is given by: [ V_s = frac{D}{45 text{ seconds}} ]3. When Mario walks on the moving sidewalk, their speeds add up. The combined speed ( V ) is: [ V = V_m + V_s = frac{D}{90} + frac{D}{45} ]4. Simplifying the combined speed: [ V = frac{D}{90} + frac{D}{45} = frac{D}{90} + frac{2D}{90} = frac{3D}{90} = frac{D}{30} ]5. The time ( T ) it takes to cover distance ( D ) at the combined speed ( frac{D}{30} ) is: [ T = frac{D}{frac{D}{30}} = 30 text{ seconds} ]Thus, the time it takes Mario to walk from Point ( A ) to Point ( B ) along the moving sidewalk is (boxed{30}) seconds.,<think>Okay, let's see. So the problem is about Mario walking on a moving sidewalk from point A to point B. There are two scenarios given: one where the sidewalk is turned off, and he takes 90 seconds to walk from A to B. Another where he just stands on the moving sidewalk and it takes him 45 seconds. The question is, how long does it take him if he walks on the moving sidewalk while it's moving?Hmm, I think this is a relative speed problem. When two things are moving in the same direction, their speeds add up. So Mario's walking speed plus the sidewalk's speed should give the combined speed when he's walking on the moving sidewalk. Then, the time taken would be the distance divided by that combined speed.Let me start by defining some variables. Let me call the distance from A to B as D. Let Mario's walking speed be Vm, and the speed of the sidewalk be Vs. First scenario: sidewalk is off, so Mario's speed is just Vm. He takes 90 seconds to cover distance D. So, distance equals speed multiplied by time. So D = Vm * 90.Second scenario: Mario is not walking, just standing on the moving sidewalk. His speed is then Vs. He takes 45 seconds. So, D = Vs * 45.Since both these scenarios cover the same distance D, I can set these two equations equal to each other. So Vm * 90 = Vs * 45. Let me note that down.Vm * 90 = Vs * 45So if I solve for one variable in terms of the other, let's divide both sides by 45:2 * Vm = VsSo the speed of the sidewalk is twice Mario's walking speed. Interesting.Third scenario: Mario walks on the moving sidewalk. So his effective speed would be Vm + Vs. Then the time taken T is equal to distance D divided by (Vm + Vs).But we know D from the first equation: D = Vm * 90. So substituting that into the time equation:T = (Vm * 90) / (Vm + Vs)But since Vs = 2 Vm from the earlier equation, substitute that in:T = (Vm * 90) / (Vm + 2 Vm) = (Vm * 90) / (3 Vm) = 90 / 3 = 30.So the time taken should be 30 seconds. Wait, that seems straightforward, but let me check again.Wait, if the sidewalk is twice as fast as Mario, so when Mario is walking on it, their speeds add up. So his total speed is three times his normal walking speed? Wait, if the sidewalk is twice as fast as Mario, so Vs = 2 Vm, so combined speed is Vm + 2 Vm = 3 Vm. Then, since the original time without the sidewalk was 90 seconds, which is at speed Vm. So at 3 Vm, the time should be 90 / 3 = 30 seconds. That makes sense.Alternatively, using the other distance equation. D = Vs * 45. But Vs = 2 Vm, so D = 2 Vm * 45 = 90 Vm. Which is the same as D = Vm * 90, so that checks out.Therefore, when moving on the sidewalk, the combined speed is 3 Vm, so time is D / (3 Vm) = (90 Vm) / (3 Vm) = 30 seconds. Yes, that seems right.Wait, but let me think again. Sometimes when you have these problems, you can also approach them by considering the rates. For example, Mario's walking rate is 1/90 per second (of the distance per second), and the sidewalk's rate is 1/45 per second. Then, when combined, their rates add up. So 1/90 + 1/45 = 1/90 + 2/90 = 3/90 = 1/30 per second. Therefore, the time taken is 1 / (1/30) = 30 seconds. That's another way to get the same answer. That seems to confirm it.Wait, actually, this is the same calculation as before, just phrased differently. Instead of speed, they talk about rates. But yeah, the result is the same. So, adding the rates gives the combined rate, and the reciprocal gives the time. So either way, 30 seconds.But just to make sure, let's plug numbers. Suppose the distance D is, let's say, 90 meters. Then Mario's walking speed Vm is 90 meters / 90 seconds = 1 m/s. Then the sidewalk's speed Vs is 2 * Vm = 2 m/s. Then, when he walks on the sidewalk, his speed is 1 + 2 = 3 m/s. Time taken would be 90 meters / 3 m/s = 30 seconds. Perfect, that works.Alternatively, if D is 45 meters. Then the sidewalk's speed is 45 meters / 45 seconds = 1 m/s. Then Mario's walking speed would be D / 90 = 45 / 90 = 0.5 m/s. Wait, but according to the first equation, Vm = D / 90. So if D is 45, Vm is 0.5 m/s. Then Vs would be 2 * Vm = 1 m/s. Then combined speed is 0.5 + 1 = 1.5 m/s. Time taken is 45 / 1.5 = 30 seconds. Again, same result. So regardless of the actual distance, the time is 30 seconds.Therefore, the answer should be 30 seconds. So boxed, it's boxed{30}Final Answerboxed{30}</think>Let ( D ) be the distance from Point ( A ) to Point ( B ), ( V_m ) be Mario's walking speed, and ( V_s ) be the speed of the moving sidewalk.1. When the sidewalk is turned off, Mario takes 90 seconds to walk from ( A ) to ( B ): [ D = V_m times 90 ]2. When Mario is carried by the moving sidewalk without walking, he takes 45 seconds: [ D = V_s times 45 ]Equating the two expressions for ( D ): [ V_m times 90 = V_s times 45 ] [ 2 V_m = V_s ]3. When Mario walks on the moving sidewalk, their speeds add up. The combined speed is ( V_m + V_s ). Substituting ( V_s = 2 V_m ): [ text{Combined speed} = V_m + 2 V_m = 3 V_m ]The time ( T ) taken to cover distance ( D ) with the combined speed is: [ T = frac{D}{3 V_m} ]Using ( D = V_m times 90 ): [ T = frac{V_m times 90}{3 V_m} = frac{90}{3} = 30 ]Thus, the time it takes Mario to walk from ( A ) to ( B ) along the moving sidewalk is (boxed{30}) seconds.]

🔑:Suppose that the distance from Point A to Point B is d mathrm{~m}, that Mario's speed is w mathrm{~m} / mathrm{s}, and that the speed of the moving sidewalk is v mathrm{~m} / mathrm{s}.From the given information, frac{d}{w}=45 and frac{d}{v}=90.The amount of the time, in seconds, that it takes Mario to walk along the sidewalk when it is moving is frac{d}{v+w}, since his resulting speed is the sum of his walking speed and the speed of the sidewalk.Here,frac{d}{v+w}=frac{1}{frac{v+w}{d}}=frac{1}{frac{v}{d}+frac{w}{d}}=frac{1}{frac{1}{90}+frac{1}{45}}=frac{1}{frac{3}{90}}=30Therefore, it takes Mario 30 seconds to walk from Point A to Point B with the sidewalk moving.ANSWER: 30 seconds

🧠:[<think>Okay, let's see. The problem is: we have 12 sticks of the same length, and we need to cut them into smaller sticks so that we can form 13 equal triangles. Each small stick has to be a side of one of these triangles. Hmm, interesting. Let me try to break this down step by step.First, all the original sticks are the same length. Let's call the length of each original stick L. So we have 12L total length of sticks. Now, after cutting them into smaller sticks, we need to have enough sticks to form 13 equal triangles. Since each triangle has three sides, we need a total of 13*3 = 39 sticks, right? But wait, each stick is a side of one triangle, so each small stick is used exactly once in one triangle. That means we need exactly 39 small sticks after cutting. However, we start with 12 sticks. So how do we get from 12 to 39? By cutting the original sticks into smaller pieces. Each time we cut a stick, we increase the number of sticks by one. So if we start with 12 sticks and want to end up with 39, we need to make 39 - 12 = 27 cuts. But each original stick can be cut multiple times. Wait, but maybe I need to verify this approach.Alternatively, perhaps we can think in terms of the total length. The total length of all the small sticks must equal the total length of the original sticks, which is 12L. Each triangle is equal, so all 13 triangles are congruent. Therefore, each triangle must have sides of lengths a, b, c, and all 13 triangles will have the same a, b, c. Therefore, the total number of each side length needed would be 13 of each. Wait, no. Each triangle has three sides, which could be of different lengths. But if the triangles are equal, then each triangle must have the same set of side lengths. So for example, if each triangle is equilateral, then we need 13*3 = 39 sticks of the same length. But if the triangles are not equilateral, say they are all scalene, then each triangle would have three different side lengths, so we need 13 sticks of each length. Wait, but in that case, each triangle would require a, b, c, so each side length a, b, c would need to be present 13 times. So total sticks would be 13*3 = 39, each length a, b, c repeated 13 times. Therefore, if the triangles are not equilateral, we need three different lengths, each occurring 13 times, totaling 39 sticks. So in that case, the problem would require cutting the original 12 sticks into 39 smaller sticks, with three different lengths, each length appearing 13 times.Alternatively, if the triangles are equilateral, then all 39 sticks must be of the same length. Then, the total length required would be 39 * s, where s is the length of each small stick. Since the original total length is 12L, we have 39s = 12L, so s = (12/39)L = (4/13)L. So each original stick of length L would need to be cut into 13/4 pieces? But 13/4 is 3.25, which is not an integer, which is a problem. Since we can't cut a stick into a fraction of a piece, this approach might not work. Therefore, equilateral triangles might not be feasible.Therefore, maybe the triangles are not equilateral. Let's think about triangles with sides of different lengths. Suppose each triangle is, say, a certain type of triangle with sides a, b, c. Then, we need 13 copies of a, 13 copies of b, and 13 copies of c. Therefore, the total length needed is 13(a + b + c). But the total length available is 12L. Therefore, 13(a + b + c) = 12L. Therefore, the sum of the sides of each triangle must be 12L / 13. So each triangle has a perimeter of 12L / 13. But each triangle must have sides such that a + b + c = 12L / 13. However, we also need to ensure that the sides satisfy the triangle inequality. So for any triangle, a + b > c, a + c > b, and b + c > a. Therefore, each side must be less than half the perimeter. Since the perimeter of each triangle is 12L/13, each side must be less than 6L/13. Hmm, so each of a, b, c must be less than 6L/13. Also, the sum a + b + c = 12L/13. So the sides have to add up to that, and each less than 6L/13. Let's think of possible integer ratios or something.But perhaps the problem doesn't require the triangles to have sides of different lengths. Maybe they can be isoceles or something. Wait, but the problem says "equal triangles", which I think means congruent triangles. So they have to be exact copies. So each triangle must have the same side lengths. So, in any case, we need 13 triangles each with sides a, b, c. So 13 of each a, b, c. So total sticks: 13*3 = 39. So 39 sticks in total, each of length a, b, or c, with 13 of each. Therefore, the total length required is 13(a + b + c) = 12L. Therefore, a + b + c = 12L /13. So each triangle's perimeter is 12L/13.So the key is to figure out how to cut the 12 original sticks into 39 smaller sticks of three different lengths a, b, c, each repeated 13 times, such that a + b + c = 12L /13. Then, each original stick of length L must be cut into pieces whose total length is L, and the combination of pieces across all 12 sticks gives exactly 13 of each a, b, c.Alternatively, maybe the triangles could be equilateral, but as we saw earlier, that leads to fractional sticks. So perhaps we need to find lengths a, b, c such that they can be obtained by cutting the original sticks in integer multiples. Let's think.Suppose each original stick is cut into n pieces. Then, the total number of sticks would be 12n. We need 39 sticks, so 12n = 39 => n = 39/12 = 13/4 = 3.25. Not an integer. So that's a problem. Therefore, perhaps the original sticks are not all cut into the same number of pieces. Some sticks might be cut into 3 pieces, some into 4, to get a total of 39. Let's check: 12 sticks. If some are cut into 3 and some into 4. Let x sticks be cut into 3 pieces, and y sticks into 4 pieces. Then, total pieces: 3x + 4y = 39. And x + y =12. So substituting y =12 -x into the first equation: 3x +4(12 -x) =39 => 3x +48 -4x=39 => -x +48=39 => -x= -9 =>x=9. So y=3. So 9 sticks are cut into 3 pieces each, and 3 sticks are cut into 4 pieces each. Then total pieces: 9*3 +3*4=27 +12=39. That works. So, that's possible. So in this case, each of the 9 sticks is divided into 3 parts, and each of the 3 sticks is divided into 4 parts.Now, but we need each of those pieces to be of lengths a, b, or c, such that there are 13 of each. So, how can we arrange the cuts so that the total number of each length is 13. Let me think.Each original stick is either cut into 3 pieces or 4 pieces. The sum of the lengths of the pieces from each original stick must equal L. So for the sticks cut into 3 pieces, each of those pieces must add up to L. Similarly, for the ones cut into 4 pieces.But we need 13 of each length a, b, c. So, across all sticks, each length must appear exactly 13 times. Let's denote the three lengths as a, b, c. Then, in each original stick, depending on how it's cut, it can contribute some combination of a, b, c.For example, a stick cut into 3 pieces could have all three pieces of different lengths (a, b, c), or two of one and one of another, or all three the same. Similarly, a stick cut into 4 pieces could have various combinations. But we need the total counts to be 13 for each a, b, c. So perhaps each stick cut into 3 pieces contributes one a, one b, one c. Then, 9 sticks cut into 3 pieces would contribute 9 a, 9 b, 9 c. Similarly, the 3 sticks cut into 4 pieces need to contribute the remaining 4 a, 4 b, 4 c (since 13 total of each). Wait, 13 -9 =4. So if each of the 3 sticks cut into 4 pieces contributes 4/3 of a, b, c? Wait, that's impossible because you can't have a fraction of a stick. Hmm. Alternatively, maybe the sticks cut into 4 pieces contribute a different combination. Let's think.Alternatively, maybe the sticks cut into 4 pieces each contribute one a, one b, and two c's. Then, each such stick would contribute 1a +1b +2c. Then, three sticks would contribute 3a +3b +6c. Then, combined with the 9 sticks cut into 3 pieces each contributing 1a +1b +1c, which gives 9a +9b +9c. Total would be 12a +12b +15c. But we need 13a +13b +13c. So that's not matching. So this approach might not work.Alternatively, maybe the sticks cut into 3 pieces contribute two of one length and one of another. For example, if each of the 9 sticks cut into 3 pieces is split into two a's and one b, then total contribution would be 18a +9b. Then, the 3 sticks cut into 4 pieces need to contribute (13-18)a, (13-9)b, and 13c. But 13 -18 is negative, which is impossible. So that's not feasible.Alternatively, perhaps the sticks cut into 3 pieces each contribute three different lengths, and the sticks cut into 4 pieces each contribute three of one length and one of another. Wait, this is getting complicated. Maybe we need to consider that each original stick is divided into equal-length pieces. But then, if we divide a stick into three equal parts, each part is L/3. If we divide a stick into four equal parts, each part is L/4. But then, if we use sticks of length L/3 and L/4, how do we form triangles? Each triangle needs three sides. But if the triangles are made up of these sticks, they would need to satisfy the triangle inequality. For example, if a triangle is made with three L/3 sticks, that's an equilateral triangle. Similarly, three L/4 sticks. But since we need 13 triangles, maybe we can mix.Wait, but the problem says "each of the smaller sticks is a side of one of these triangles". So each small stick is used as a side in exactly one triangle. So the total number of sticks is 39, which is 13*3. So each triangle uses three sticks, each of which is a side. So perhaps all triangles are made from sticks of the same length, but different triangles could use different lengths. But the problem states "13 equal triangles", so all triangles must be congruent. So all 13 triangles must have the same side lengths. Therefore, each triangle must have sides of lengths a, b, c (maybe equal, maybe not). So each triangle is congruent, so all have the same a, b, c. Therefore, we need 13 copies of each a, b, c.Therefore, the problem reduces to: cut 12 sticks of length L into 39 smaller sticks, with 13 of each length a, b, c, such that a + b + c = 12L /13 (since total length is 12L, and each triangle has perimeter 12L/13). Additionally, the lengths a, b, c must satisfy the triangle inequality.So now, the challenge is to find such lengths a, b, c and a way to partition the original sticks into pieces of these lengths such that the counts work out.Let me think about possible lengths. Suppose we let each original stick be cut into pieces of lengths that add up to L. For example, if we can divide some sticks into three pieces and others into four, as per the earlier calculation (9 sticks into 3 pieces, 3 sticks into 4 pieces), then each of those pieces must be a, b, or c, such that the total number of a, b, c is 13 each.Let me think of an example. Suppose a, b, c are such that 3a + 4b = L. Wait, but this is just a random thought. Maybe not. Let's think numerically. Let me assign a numerical value to L for easier calculation. Let's say L =13 units. Then the total length is 12*13=156 units. Each triangle has a perimeter of 12 units (since 156 /13 =12). So each triangle has sides a, b, c such that a + b + c =12. Also, each side must be less than 6 (triangle inequality). So, for example, possible sides could be 3,4,5. Because 3+4+5=12, and each side is less than 6 (wait, 5 is less than 6? 3+4>5, 3+5>4, 4+5>3, so yes, 3,4,5 is a valid triangle with perimeter 12. So if we use 13 triangles each with sides 3,4,5, then we need 13 sticks of 3, 13 of 4, 13 of 5. Total sticks 39. Total length 13*(3+4+5)=13*12=156, which matches the total length of original sticks (12*13=156). So this works in terms of total length.Now, the question is: can we cut the original 12 sticks (each of length 13) into 39 sticks of lengths 3,4,5, with 13 of each? So each original stick of length 13 must be divided into some combination of 3,4,5 that sum to 13. Let's see.How can we break a stick of length 13 into pieces of 3,4,5? Let's find combinations where the sum is 13.Possible combinations:- 5 +5 +3 =13 (three pieces: two 5s and one 3)- 5 +4 +4 =13 (one 5 and two 4s)- 4 +4 +5 = same as above- 3 + 3 + 7, but 7 is not allowedWait, but the only possible combinations using 3,4,5 are the first two. Let's check:1. 5 +5 +3=13. That's three pieces: two 5s and one 3.2. 5 +4 +4=13. That's three pieces: one 5 and two 4s.3. 4 +3 +3 +3=13. That's four pieces: one 4 and three 3s. Wait, 4+3+3+3=13. Yes, that's a valid combination with four pieces.4. 5 +3 +5= same as the first one.So there are different ways. For example, cutting into three pieces with combinations 5,5,3 or 5,4,4. Or cutting into four pieces like 4,3,3,3.Now, we need to cut the 12 sticks into these combinations such that the total number of 3s, 4s, and 5s is 13 each.Wait, but 13 of each. So total needed: 13 threes, 13 fours, 13 fives.Let me think about how many of each combination we need.Suppose we use x sticks cut into [5,5,3], y sticks cut into [5,4,4], and z sticks cut into [4,3,3,3]. Each of these uses different numbers of 3s,4s,5s.Each [5,5,3] contributes two 5s and one 3. Each [5,4,4] contributes one 5 and two 4s. Each [4,3,3,3] contributes one 4 and three 3s.We need:For 5s: 2x + y =13For 4s: 2y + z =13For 3s: x +3z =13And the total number of sticks used:Each [5,5,3] is one original stick, contributing three pieces.Each [5,4,4] is one original stick, contributing three pieces.Each [4,3,3,3] is one original stick, contributing four pieces.But we have 12 original sticks. So x + y + z =12.So we have four equations:1. 2x + y =13 (from 5s)2. 2y + z =13 (from 4s)3. x +3z =13 (from 3s)4. x + y + z =12 (total sticks)Let me try to solve these equations.From equation 1: y =13 -2xFrom equation 2: z=13 -2y=13 -2(13 -2x)=13 -26 +4x=4x -13From equation 3: x +3z=13. Substitute z=4x-13:x +3(4x -13)=13x +12x -39=1313x -39=1313x=52x=4Then y=13 -2x=13 -8=5z=4x -13=16 -13=3Check equation 4: x + y + z=4 +5 +3=12, which matches.So the solution is x=4, y=5, z=3.So we need to cut:- 4 sticks into [5,5,3]- 5 sticks into [5,4,4]- 3 sticks into [4,3,3,3]Let's check the counts:5s: 2x + y =2*4 +5=8 +5=13 ✔️4s: 2y + z=2*5 +3=10 +3=13 ✔️3s: x +3z=4 +3*3=4 +9=13 ✔️Perfect. So this works.Therefore, if we take 4 original sticks and cut each into two 5s and one 3, 5 original sticks into one 5 and two 4s, and 3 original sticks into one 4 and three 3s, we will get exactly 13 of each length. Then, we can form 13 triangles each with sides 3,4,5. Since 3,4,5 is a valid triangle, this works.Therefore, the answer is to cut the sticks such that you have 13 sticks each of lengths 3, 4, and 5 (in the original problem, the total length of each original stick is 13 units). Then, each triangle is a 3-4-5 triangle, which is a right-angled triangle.But wait, in the problem statement, the original sticks are of the same length, but the actual length isn't specified. However, in my example, I set L=13 for simplicity. But if the original sticks are of length L, then the sides would be 3k, 4k, 5k, where k is a scaling factor such that 3k +4k +5k=12L/13. Wait, let's check:The perimeter of each triangle is 3k +4k +5k=12k. The total perimeter for 13 triangles is13*12k=156k. This must equal the total length of all sticks, which is12L. Therefore,156k=12L =>k=12L/156= L/13. Therefore, the lengths are 3*(L/13)=3L/13, 4L/13, and5L/13. So each original stick of length L must be cut into pieces of lengths 3L/13,4L/13,5L/13 in the combinations mentioned earlier.But practically, how do we ensure that these lengths are possible? For example, if L is 13 units, then as per my earlier example, it works. But if L is arbitrary, then we need to be able to divide each stick into these fractional lengths, which is possible in theory but requires that the original stick can be divided accordingly. Since the problem doesn't specify that the lengths have to be integers or any specific units, just that we need to form triangles, it should be acceptable.Therefore, the solution is:Cut 4 original sticks into two pieces of length 5L/13 and one piece of length 3L/13 each.Cut 5 original sticks into one piece of length 5L/13 and two pieces of length 4L/13 each.Cut 3 original sticks into one piece of length4L/13 and three pieces of length3L/13 each.This way, we have 13 pieces of each length, and each triangle (3L/13,4L/13,5L/13) is a valid right-angled triangle.Therefore, the answer is to cut the sticks into lengths of 3L/13,4L/13, and5L/13 in the specified manner.</think>To solve the problem, we need to cut 12 sticks of length ( L ) into smaller sticks such that we can form 13 congruent triangles. Each triangle must have sides of lengths ( frac{3L}{13} ), ( frac{4L}{13} ), and ( frac{5L}{13} ), which form a right-angled triangle. Here's how to cut the sticks:1. Cut 4 sticks into two pieces of ( frac{5L}{13} ) and one piece of ( frac{3L}{13} ). - Total pieces per stick: 3 - Total pieces: ( 4 times 3 = 12 ) - Contribution: ( 8 times frac{5L}{13} ), ( 4 times frac{3L}{13} )2. Cut 5 sticks into one piece of ( frac{5L}{13} ) and two pieces of ( frac{4L}{13} ). - Total pieces per stick: 3 - Total pieces: ( 5 times 3 = 15 ) - Contribution: ( 5 times frac{5L}{13} ), ( 10 times frac{4L}{13} )3. Cut 3 sticks into one piece of ( frac{4L}{13} ) and three pieces of ( frac{3L}{13} ). - Total pieces per stick: 4 - Total pieces: ( 3 times 4 = 12 ) - Contribution: ( 3 times frac{4L}{13} ), ( 9 times frac{3L}{13} )Totals After Cutting:- ( frac{5L}{13} ): ( 8 + 5 = 13 ) sticks - ( frac{4L}{13} ): ( 10 + 3 = 13 ) sticks - ( frac{3L}{13} ): ( 4 + 9 = 13 ) sticks Each triangle uses one stick of each length. Thus, with 13 of each length, you can form 13 congruent right-angled triangles with sides ( frac{3L}{13} ), ( frac{4L}{13} ), and ( frac{5L}{13} ).Answer: Cut 4 sticks into two ( frac{5L}{13} ) and one ( frac{3L}{13} ), 5 sticks into one ( frac{5L}{13} ) and two ( frac{4L}{13} ), and 3 sticks into one ( frac{4L}{13} ) and three ( frac{3L}{13} ). This yields 13 triangles with sides (boxed{frac{3L}{13}}), (boxed{frac{4L}{13}}), and (boxed{frac{5L}{13}}).,<think>Okay, let's see. The problem is to take 12 sticks of the same length and cut them into smaller sticks so that we can form 13 equal triangles, with each smaller stick being a side of one of these triangles. Hmm, so each triangle has to be the same size and shape, right? So they must be congruent triangles. Each triangle will have three sides, each made from a smaller stick. And all the triangles are equal, so all their corresponding sides must be the same length. Therefore, each triangle must have three sticks of specific lengths, and all 13 triangles will use sticks of those same lengths.First, let's note that each original stick is going to be cut into some number of smaller sticks. The total number of sticks after cutting will depend on how we cut each original stick. But the key is that all the small sticks must be used as sides of the 13 triangles. Since each triangle has three sides, the total number of sticks needed is 13 * 3 = 39. So we need to end up with 39 smaller sticks after cutting the original 12 sticks. Each original stick can be cut into multiple smaller sticks. If we cut each original stick into k pieces, then the total number of sticks would be 12k. We need 12k = 39. But 39 divided by 12 is 3.25, which isn't an integer. So that suggests that we can't cut each stick into the same number of pieces. Instead, we need to cut some sticks into 3 pieces and others into 4 pieces so that the total number of pieces is 39. Let's check: if x sticks are cut into 3 pieces and y sticks are cut into 4 pieces, then 3x + 4y = 39, and x + y = 12. Solving these equations: from the second equation, x = 12 - y. Substitute into the first equation: 3(12 - y) + 4y = 39 => 36 - 3y + 4y = 39 => y = 3. Therefore, x = 9. So 9 sticks would be cut into 3 pieces each, and 3 sticks would be cut into 4 pieces each, resulting in 9*3 + 3*4 = 27 + 12 = 39 sticks. That works.So the cutting part is feasible. But now we need to make sure that the lengths of these 39 sticks can form 13 congruent triangles. Since all triangles are congruent, each triangle must have sides of lengths a, b, c. So we need 13 sets of (a, b, c). That means we need 13 sticks of length a, 13 sticks of length b, and 13 sticks of length c. Therefore, the 39 sticks must consist of 13 of each of three different lengths. So the total number of sticks of each length must be 13. Thus, we need 13 sticks of length a, 13 of length b, and 13 of length c. However, these lengths a, b, c must be such that the original 12 sticks, when cut into pieces of these lengths, sum up correctly.Let’s denote the original length of each stick as L. When we cut a stick into pieces, the sum of the lengths of the pieces must equal L. So for each original stick, the sum of the lengths of the pieces it's cut into must be L. So if a stick is cut into 3 pieces, say lengths x, y, z, then x + y + z = L. Similarly, if cut into 4 pieces, the sum of those four lengths is L.But we need all the pieces to be either a, b, or c. So each piece from every original stick must be of length a, b, or c. Therefore, each original stick is partitioned into some combination of a, b, c. So the problem reduces to determining how to partition the original 12 sticks into pieces of lengths a, b, c such that the total number of pieces of each length is 13. Also, the triangles must be congruent, so a, b, c must satisfy the triangle inequality. Since all triangles are congruent, each triangle must be a triangle with sides a, b, c. Therefore, a + b > c, a + c > b, and b + c > a.So the problem is two-fold: first, determine the lengths a, b, c such that 13 of each are needed, and then determine how to cut the original 12 sticks (each of length L) into pieces of a, b, c such that the total number of each piece is 13. Also, the sum of the pieces from each original stick must be L.Since all original sticks are the same length, let's assume that each original stick is of length 1 unit for simplicity. Then the total length of all sticks is 12 units. The total length required for all the small sticks is 13*(a + b + c). Therefore, 12 = 13*(a + b + c). So a + b + c = 12/13. So each triangle has a perimeter of 12/13, and each triangle uses a, b, c which sum to 12/13. But we also need to have that each original stick is cut into pieces whose lengths sum to 1 (since each original stick is length 1). So each original stick is divided into segments of a, b, and/or c such that their sum is 1. So each stick's cutting must consist of some combination of a, b, c that adds up to 1. Now, we need to find a, b, c such that:1. a + b + c = 12/13.2. The total number of each of a, b, c is 13.3. Each original stick (length 1) is divided into pieces of a, b, c whose total is 1.4. The triangle inequalities hold: a + b > c, a + c > b, b + c > a.This seems complicated, but maybe there's a way to simplify. Let's think of possible values for a, b, c. Since each triangle must be congruent, perhaps the triangles are equilateral? If so, a = b = c. Then each triangle would have three sides of length (12/13)/3 = 4/13. But then all sticks would need to be cut into lengths of 4/13. However, we need 13*3 = 39 sticks of length 4/13. The total length would be 39*(4/13) = 12, which works. However, each original stick must be cut into pieces of length 4/13. Each original stick is length 1, so 1/(4/13) = 13/4 = 3.25. But you can't cut a stick into 3.25 pieces. So that's impossible. Therefore, equilateral triangles won't work. So the triangles must be of different side lengths.Alternatively, maybe two sides are equal. Isosceles triangles. For example, a = b ≠ c. Then 2a + c = 12/13, and the triangle inequalities would require a + a > c, so 2a > c, and a + c > a, which is c > 0. So with 2a > c and 2a + c = 12/13. Then we could solve for a and c. But we also need to ensure that we can cut the original sticks into combinations of a, a, c such that each original stick sums to 1, and we get 13 of each a and c.But this might be complicated. Let's consider that we need 13 sticks of each length a, b, c. So each original stick must contribute some number of a's, b's, and c's. Let’s suppose that each original stick is cut into pieces in such a way that they can contribute to the total count of a, b, c.Given that each stick can be cut into 3 or 4 pieces (as per earlier, 9 sticks cut into 3 pieces and 3 sticks into 4 pieces), but perhaps the number of pieces per stick can vary as long as the total is 39. Wait, but earlier we found that 9 sticks cut into 3 and 3 sticks into 4 gives 39. So that's the way to get 39 sticks. So each of the 9 sticks cut into 3 pieces will contribute 3 segments (maybe a combination of a, b, c), and each of the 3 sticks cut into 4 pieces will contribute 4 segments.But since each segment must be either a, b, or c, and each original stick must sum to 1, we need to figure out how to combine a, b, c such that their sum is 1, and the total number of each across all sticks is 13.This feels like a system of equations. Let’s denote:Let’s let x_i be the number of sticks cut into i pieces. Wait, but earlier we found that 9 sticks are cut into 3 pieces and 3 into 4. So each of those 9 sticks will have 3 pieces, each of the 3 sticks will have 4 pieces. Each of these pieces is a, b, or c.Let’s suppose that in the 9 sticks that are cut into 3 pieces, each such stick is cut into, say, combinations like a, a, b or a, c, c, etc., such that a + a + b = 1 or a + c + c = 1. Similarly, the sticks cut into 4 pieces would have combinations like a, a, a, a (if a = 1/4), but since a must be part of a triangle with other lengths, that might not work.Alternatively, maybe all sticks, whether cut into 3 or 4 pieces, have a combination of a, b, c such that the sum is 1. But how?Alternatively, maybe some sticks are cut into three equal pieces and others into four equal pieces. But if a stick is cut into three equal pieces, each of length 1/3, but 1/3 is a possible side length. However, then we would have 9 sticks each contributing three 1/3-length pieces, totaling 27 pieces, and 3 sticks each cut into four 1/4-length pieces, totaling 12 pieces. But then we have 27 pieces of 1/3 and 12 pieces of 1/4. But we need 13 of each of three different lengths, which doesn't match here. So that approach won't work.Alternatively, perhaps all the sticks are cut into a combination of two different lengths. For example, each stick is cut into two pieces of length a and b. But then total pieces would be 24, which is less than 39. Not helpful.Wait, but the key is that we need three different lengths (a, b, c), each appearing 13 times. So the total number of sticks is 39. So we need to distribute these 39 sticks across the 12 original sticks, with each original stick being partitioned into 3 or 4 smaller sticks. The total partitions are 9 sticks into 3 pieces (27 pieces) and 3 sticks into 4 pieces (12 pieces), totaling 39.So each of the 9 sticks contributes 3 pieces, and each of the 3 sticks contributes 4 pieces. Each piece is either a, b, or c. The total count of a, b, c must each be 13. Therefore, the sum over all sticks of the number of a's, b's, and c's in each stick must equal 13 for each.Let’s think in terms of equations. Let’s define variables for the number of a's, b's, c's in each original stick. For the 9 sticks cut into 3 pieces, each stick can have some combination of a, b, c adding up to 1. Let’s denote that for each of these 9 sticks, maybe some have 3 a's, some have 2 a's and 1 b, etc. Similarly, for the 3 sticks cut into 4 pieces, each must have a combination of a, b, c adding to 1, with 4 pieces.But this seems too vague. Maybe there's a better way. Let's think about the total number of each piece:Total a's: 13Total b's: 13Total c's: 13Total pieces: 39.Each of the 9 sticks cut into 3 pieces contributes 3 pieces, and each of the 3 sticks cut into 4 pieces contributes 4. Let’s denote the following:For the 9 sticks (cut into 3):Let’s say that each of these sticks contributes x_a a's, x_b b's, x_c c's per stick, such that x_a + x_b + x_c = 3, and x_a*a + x_b*b + x_c*c = 1.Similarly, for the 3 sticks (cut into 4):Each contributes y_a a's, y_b b's, y_c c's, with y_a + y_b + y_c = 4, and y_a*a + y_b*b + y_c*c = 1.Then, the total number of a's is 9*x_a + 3*y_a = 13Similarly, total b's: 9*x_b + 3*y_b = 13Total c's: 9*x_c + 3*y_c = 13So we have three equations:9x_a + 3y_a = 13 ...(1)9x_b + 3y_b = 13 ...(2)9x_c + 3y_c = 13 ...(3)But since x_a + x_b + x_c = 3 for the 3-piece sticks, and y_a + y_b + y_c = 4 for the 4-piece sticks.Additionally, for each 3-piece stick:x_a*a + x_b*b + x_c*c = 1Similarly, for each 4-piece stick:y_a*a + y_b*b + y_c*c = 1This seems quite a system to solve. Let's see if we can find integer solutions for x_a, x_b, x_c and y_a, y_b, y_c such that equations (1), (2), (3) hold.Looking at equation (1): 9x_a + 3y_a = 13. Let's divide both sides by 3: 3x_a + y_a = 13/3. But 13/3 is not an integer, which is a problem because x_a and y_a must be integers (number of a's per stick). This suggests that perhaps our initial assumption that each stick cut into 3 or 4 pieces may not work because we can't have fractional counts. Wait, but x_a and y_a are per-stick counts. For example, each of the 9 sticks could contribute different numbers of a's, b's, c's. So maybe x_a and y_a are averages?Wait, no. Each of the 9 sticks is cut into 3 pieces. Let’s suppose that among these 9 sticks, some have a certain number of a's, b's, c's. For example, maybe some sticks have 1 a, 1 b, 1 c, and others have different combinations. Similarly, the 3 sticks cut into 4 pieces might have different distributions. So the variables x_a, x_b, x_c could vary per stick, but equations (1), (2), (3) are totals across all sticks. Therefore, perhaps instead of assuming each stick has the same number of a's, b's, c's, we need to allow variation.But this complicates things. Let me think.Alternatively, maybe all the 3-piece sticks have the same combination, and all the 4-piece sticks have the same combination. Let’s try that.Suppose all 9 sticks cut into 3 pieces each have the same combination: x_a, x_b, x_c. Then total a's from these would be 9x_a. Similarly, the 3 sticks cut into 4 pieces each have y_a, y_b, y_c, so total a's from these would be 3y_a. Then total a's: 9x_a + 3y_a = 13. Similarly for b and c.Also, x_a + x_b + x_c = 3, and y_a + y_b + y_c = 4.Let’s try to solve 9x_a + 3y_a = 13. Let’s divide both sides by 3: 3x_a + y_a = 13/3 ≈ 4.333. Since x_a and y_a must be integers, this is impossible. Similarly for the other equations. Therefore, this approach of uniform combinations per stick type (3-piece or 4-piece) doesn't work. Therefore, we need different combinations among the sticks.Therefore, some of the 3-piece sticks have different numbers of a's, b's, c's, and similarly for the 4-piece sticks. For example, maybe some 3-piece sticks contribute 2 a's and 1 b, and others contribute 1 a and 2 c's, etc., such that the total counts add up to 13 each.This seems like a Diophantine equation problem. Let’s think about how to distribute the counts.Let’s start with the a's. We need 13 a's in total. These come from the 9 sticks cut into 3 pieces and the 3 sticks cut into 4 pieces. Let’s denote the number of a's in the 3-piece sticks as A1 and in the 4-piece sticks as A2. Then A1 + A2 = 13. Similarly for B1 + B2 = 13 and C1 + C2 = 13.Each 3-piece stick can contribute 0, 1, 2, or 3 a's. Similarly for the 4-piece sticks. But since there are 9 sticks in the 3-piece group and 3 sticks in the 4-piece group, the maximum number of a's from the 3-piece group is 9*3=27 and from the 4-piece group is 3*4=12. But we need only 13 of each.This seems feasible. Let's try to find integer solutions. For A1 + A2 = 13. Let's suppose that in the 3-piece sticks, each stick can contribute 1 a. Then A1 = 9*1 = 9. Then A2 = 13 - 9 = 4. So the 4-piece sticks need to contribute 4 a's. Since there are 3 sticks, each can contribute 1 or 2 a's. For example, one stick contributes 2 a's and two sticks contribute 1 a each: 2 +1 +1 =4. That works.Similarly, for B's and C's. Let's try this.Assume each 3-piece stick has 1 a, 1 b, and 1 c. Then each 3-piece stick contributes 1 of each. Then total a's from 3-piece sticks: 9*1=9. Similarly, total b's and c's: 9 each. Then from the 4-piece sticks, we need 13 -9=4 of each. So each 4-piece stick needs to contribute (4/3) of each, which is impossible. So that approach fails.Alternatively, let's vary the distribution.Suppose that for the a's: 9 sticks (3-piece) contribute 1 a each, so A1=9. Then A2=4. So the 3 sticks (4-piece) need to contribute 4 a's. So 4 a's over 3 sticks: possible if one stick has 2 a's and two sticks have 1 a each. Similarly for b's and c's.But how do we ensure that each stick's pieces sum to 1?If we have a stick (3-piece) with 1 a, 1 b, 1 c, then a + b + c =1. If another stick (4-piece) has 2 a's, 1 b, 1 c, then 2a + b + c =1. Wait, but if a + b + c =1 (from the 3-piece stick), then 2a + b + c = a + (a + b + c) = a + 1. Therefore, 2a + b + c = a +1. So this would be equal to 1 + a. But the stick's total length is 1, so a +1 =1 implies a=0, which is impossible. Therefore, this approach is invalid.Hmm. So perhaps the combinations in the 3-piece and 4-piece sticks have to be such that their equations are consistent.Alternatively, maybe all the sticks are cut into the same combination of a, b, c. But we saw earlier that 9 sticks cut into 3 pieces each would require 3x_a + y_a =13/3, which isn't possible. So perhaps not.Wait, let's think differently. Suppose the triangles are such that two sides are of one length and the third is another. For example, let's say each triangle is isosceles with sides a, a, b. Then we need 13*2=26 a's and 13 b's. But the problem states that each of the smaller sticks is a side of one of the triangles. So maybe this isn't allowed since each triangle has sides a, a, b, but we need each small stick to be a side of one triangle. But if we have 26 a's and 13 b's, that would total 39 sticks. Then we need to cut the original 12 sticks into 26 a's and 13 b's. Let's see. Then total length would be 26a +13b =12. Also, for the triangles, a + a + b =12/13 (since each triangle's perimeter is 12/13). So 2a + b =12/13. Therefore, from total length: 26a +13b =12. Let's substitute b =12/13 -2a into the total length equation:26a +13*(12/13 -2a) =1226a + (156/13 -26a) =1226a +12 -26a =1212=12. So this holds for any a? Wait, that's interesting. So the total length equation is automatically satisfied if 2a + b =12/13. Therefore, we have infinite solutions? Wait, but how?Actually, if we set each triangle as isosceles with sides a, a, b, then 2a + b =12/13 (perimeter). The total number of a's is 26 and b's is13. Therefore, the total length used is26a +13b=13*(2a +b)=13*(12/13)=12, which matches the total original length. So this works for any a and b such that 2a +b=12/13. But we also need to make sure that the triangle inequality holds: a +a >b, so 2a >b. But since 2a +b=12/13, then 2a >b implies 2a > (12/13 -2a), so 4a >12/13 => a>3/13. Similarly, since b=12/13 -2a, and b>0, so 12/13 -2a >0 =>a<6/13. So a must be between 3/13 and6/13.Now, we need to choose a and b such that we can cut the original sticks into pieces of a and b. So each original stick (length1) is cut into some number of a's and b's, summing to1. So for each original stick, we can have different combinations of a's and b's such that the sum is1.But since we need 26 a's and13 b's, and there are12 original sticks. Let's figure out how to distribute these.Each original stick can be cut into, for example:- 3 pieces: two a's and one b. Then 2a +b=1. But since 2a +b=12/13≈0.923, which is less than1. So this doesn't work. Hmm.Wait, the problem is that the length of each original stick is1, but the sum of pieces in each stick must be1. However, in this case, if each triangle's perimeter is12/13≈0.923, but each original stick is length1. So we need to cut each original stick into pieces that sum to1, but the sum of a's and b's in each stick must be1. But with 2a +b=12/13. So for example, if we have a stick cut into one triangle's worth of pieces: two a's and one b. Then 2a +b=12/13. But then the total length of that stick's pieces is12/13, which is less than1. Therefore, this is impossible. Therefore, we cannot have a stick cut into exactly one triangle's worth of pieces. Instead, each stick must be cut into multiple triangles' worth of pieces. But each triangle requires two a's and one b, so multiple triangles would need multiple a's and b's. However, since we need exactly26 a's and13 b's, which is exactly 13 triangles, each requiring two a's and one b. So perhaps each original stick is cut into pieces that correspond to multiple triangles. For example, if a stick is cut into two a's and one b, that's one triangle. But since the stick is length1, and the sum of those pieces would be12/13, which is less than1. Therefore, we need to have more pieces per stick.Alternatively, if a stick is cut into four a's and two b's, which would correspond to two triangles (each needing two a's and one b). But then the sum would be4a +2b=2*(2a +b)=2*(12/13)=24/13≈1.846, which exceeds1. So that's no good.Wait, this seems like a dead end. Because any combination of a's and b's that forms whole triangles would either sum to less than1 or more than1. Therefore, perhaps this approach with isosceles triangles doesn't work.Alternatively, maybe each stick is cut into a combination of a's and b's that don't correspond to whole triangles. For example, a stick could have some a's and some b's that add up to1, but those a's and b's are parts of different triangles. However, this complicates the problem because we need to ensure that all a's and b's are used in triangles, but they can come from different sticks.But this seems possible. For instance, if we have sticks that are cut into two a's and one b, and other sticks cut into one a and three b's, such that when combined, the total a's and b's sum up to26 and13 respectively.Let’s try this. Suppose some sticks are cut into two a's and one b (total length 2a +b=12/13), but this sum needs to be1. Therefore, 2a +b=1. But earlier, we have 2a +b=12/13≈0.923, which is less than1. Therefore, this doesn't work. So this approach is invalid.Alternatively, maybe we need to introduce a third length, c. Let's get back to the original problem where each triangle has sides a, b, c. Then each triangle's perimeter is a +b +c=12/13. The total number of each side is13. So we need13 a's,13 b's,13 c's. Total length13a +13b +13c=13*(a +b +c)=13*(12/13)=12, which matches.Now, we need to cut each original stick (length1) into pieces of a, b, c such that the sum per stick is1, and the total pieces are13 of each. Each original stick can be cut into 3 or4 pieces as per the earlier calculation (9 sticks into3 pieces,3 sticks into4 pieces).Let’s think of possible combinations. For example, if some sticks are cut into three pieces: a, b, c. Then a +b +c=1. If we have nine sticks each cut into a, b, c, then each stick contributes one a, one b, one c. Then total from these nine sticks:9 a's,9 b's,9 c's. Then we need four more of each from the remaining three sticks (which are cut into four pieces each). So each of these three sticks must contribute (4/3) a's, (4/3) b's, and (4/3) c's, which is impossible since we can't have fractions of a piece.Alternatively, maybe the three sticks are cut into four pieces each, with combinations that give the remaining four a's, four b's, and four c's. For example, each of the three sticks could have one a, one b, and two c's. Then each stick contributes1 a,1 b,2 c's, summing to1*a +1*b +2*c. The total from three sticks would be3 a's,3 b's,6 c's. But we need4 a's,4 b's,4 c's. So this gives us total a's:9 +3=12, which is one short. Similarly for b's:9 +3=12, and c's:9 +6=15, which is over. So this doesn't work.Alternatively, two of the three sticks could have two a's, one b, one c (sum 2a +b +c=1), and one stick could have0 a's, two b's, two c's (sum0*a +2b +2c=1). Then total a's from the three sticks:2*2 +0=4. Total b's:2*1 +2=4. Total c's:2*1 +2=4. So this works. Let's check:Original nine sticks: each has1 a,1 b,1 c. So nine each. The three sticks:Two sticks with2 a's,1 b,1 c: each contributes2a +1b +1c=1. Total for two sticks:4 a's,2 b's,2 c's.One stick with0 a's,2 b's,2 c's:0a +2b +2c=1. Total:0a,2b,2c.Total from three sticks:4 a's,4 b's,4 c's. Combined with nine sticks: total a's=9+4=13, b's=9+4=13, c's=9+4=13. Perfect.Now, we need to ensure that the equations for the sums hold:For the nine sticks: a +b +c=1.For the two sticks cut into four pieces:2a +b +c=1.For the one stick cut into four pieces:2b +2c=1.So we have three equations:1. a +b +c=12. 2a +b +c=13. 2b +2c=1Let’s solve these equations.From equation1 and equation2:Equation2 - equation1: (2a +b +c) - (a +b +c)=1 -1 =>a=0. But if a=0, then from equation1, 0 +b +c=1 =>b +c=1.But equation3:2b +2c=1 =>b +c=0.5. But from equation1, b +c=1. Contradiction. Therefore, this system has no solution.Hmm, that's a problem. So this particular combination doesn't work. Maybe another combination?Let’s try different combinations. Suppose the three sticks are cut as follows:One stick has3 a's and1 c:3a +c=1.One stick has3 b's and1 c:3b +c=1.One stick has1 a,1 b,2 c's:a +b +2c=1.Then total a's from three sticks:3 +0 +1=4.Total b's:0 +3 +1=4.Total c's:1 +1 +2=4.Combined with nine sticks (1a,1b,1c each): total a=9+4=13, b=9+4=13, c=9+4=13. Good.Now, the equations are:For the nine sticks: a +b +c=1.For the first three-piece stick:3a +c=1.For the second:3b +c=1.For the third:a +b +2c=1.Let’s solve these equations.From the first equation: a +b +c=1.From the second:3a +c=1.From the third:3b +c=1.From the fourth:a +b +2c=1.Let’s subtract the first equation from the fourth equation:(a +b +2c) - (a +b +c)=1 -1 =>c=0. But c=0 is impossible. Therefore, contradiction again. Hmm.Alternatively, maybe two sticks have three of one kind and one of another, and the third stick has a different combination.Alternatively, let's think of another approach. Since we need to have each original stick sum to1, and we need a +b +c=12/13≈0.923. Wait, but if each original stick is cut into pieces that sum to1, which is longer than12/13. So perhaps the pieces in each stick come from more than one triangle. For example, a stick might have pieces that are parts of two different triangles. For instance, a stick could have two a's and one b, where the two a's are sides from two different triangles, and the b is a side from another triangle. But since each triangle must have exactly one a, one b, and one c, this complicates things.Wait, but each triangle has sides a, b, c, so each triangle must have one of each. Therefore, each a must pair with a b and a c. So if a stick has multiple a's, each a must be paired with a b and a c from other sticks. This seems complicated, but maybe manageable.Alternatively, consider that each original stick is cut into one a, one b, and one c, plus an extra piece. Wait, but the total number of pieces would then be4, which is possible for the three sticks that are cut into4 pieces. For example, each of those three sticks could be cut into a, b, c, and another piece. But then we have an additional piece, which would be a fourth length, which is not allowed because we can only have a, b, c.Alternatively, maybe the three sticks cut into four pieces are cut into two a's, one b, and one c. Then total from three sticks:6 a's,3 b's,3 c's. Combined with nine sticks each contributing1 a,1 b,1 c: total a's=9+6=15, which exceeds13. So that's no good.This is quite challenging. Maybe there's a different way to approach the problem. Let's think about the fact that all triangles must be congruent, so their sides are in a fixed ratio. Let’s assume that the triangles are similar to each other, but scaled. Wait, but they must be congruent, so they have to be the same size. Therefore, the sides must be fixed lengths a, b, c.Given that a + b + c =12/13, and that each original stick (length1) is cut into pieces summing to1. Let’s assume that we can cut each original stick into several segments that are multiples of a, b, or c.Wait, perhaps we can have sticks that are cut into multiple copies of a, b, or c. For example, if a stick is cut into three pieces of length a, then 3a=1 =>a=1/3. But then, since a +b +c=12/13≈0.923, so b +c=12/13 -1/3≈0.923 -0.333≈0.590. But then the triangle inequality requires a +b >c and a +c >b. If a=1/3≈0.333, then b +c≈0.590. Let’s say b and c are both≈0.295. Then a +b≈0.333+0.295=0.628 >c≈0.295. Similarly for a +c >b. And b +c≈0.590 >a≈0.333. So that works.But in this case, we need13 a's,13 b's,13 c's. If we have some sticks cut into three a's (each a=1/3), then each such stick contributes three a's. If we have 13 a's, then we need 13/3≈4.33 sticks cut into three a's. Which is impossible. So that approach won't work.Alternatively, if some sticks are cut into two a's and one something else. For example, suppose a stick is cut into two a's and one b. Then 2a +b=1. If another stick is cut into two c's and one b, then 2c +b=1. And others are cut into a, b, c. But this is getting too vague.Perhaps the key is to use the fact that 13 is a prime number and 12 is one less. Maybe we need to have each original stick contribute one of each side, plus an extra piece. For example, if we have 12 original sticks, and each contributes one a, one b, one c, plus an extra a, then we would have 13 a's, 12 b's, 12 c's. But we need 13 of each. So maybe one stick contributes an extra b and c instead. This is getting convoluted.Alternatively, think of the problem as a combinatorial optimization. We need 13 of each side, and each original stick can be partitioned into 3 or 4 pieces. Let’s try to find a combination where some sticks contribute more a's, others more b's, and others more c's.Suppose we have:- Four original sticks each cut into three pieces: two a's and one b. So per stick:2a +1b=1. Total from four sticks:8a +4b.- Four original sticks each cut into three pieces: two b's and one c. Per stick:2b +1c=1. Total:8b +4c.- Four original sticks each cut into three pieces: two c's and one a. Per stick:2c +1a=1. Total:8c +4a.Total a's:4 from the last group. But wait, no. Let's recast:Wait, each of the first four sticks gives two a's and one b. So a's:4*2=8, b's:4*1=4.Each of the next four sticks gives two b's and one c. So b's:4*2=8, c's:4*1=4.Each of the last four sticks gives two c's and one a. So c's:4*2=8, a's:4*1=4.Total a's:8 +4=12.Total b's:4 +8=12.Total c's:4 +8=12.But we need13 of each. So we're short by1 for each. If we then have three sticks cut into four pieces each, perhaps each contributing one a, one b, one c, and one extra piece. But each stick must sum to1. If we have three sticks each cut into four pieces: one a, one b, one c, and one d. But we can't have a fourth length d. So this isn't allowed.Alternatively, the three sticks could be cut into four pieces of existing lengths. For example, each of the three sticks is cut into one a, one b, and two c's. Then per stick:1a +1b +2c=1. Total from three sticks:3a +3b +6c. Adding to previous totals: a=12 +3=15, b=12 +3=15, c=12 +6=18. Too much.Alternatively, each of the three sticks is cut into one a, one b, one c, and one piece that sums with the others to1. But again, introducing a fourth length.This approach isn't working.Maybe instead, let's use the fact that we need 13 of each, and 13 is a prime number. The key might be that each original stick contributes one of each side, plus an extra piece every now and then. For example, 12 original sticks each contribute one a, one b, one c (totaling12 of each), and then one additional stick contributes an extra a, b, c. But we only have12 original sticks. Wait, no, we have12 original sticks, and need to produce13 of each. So perhaps each original stick contributes one a, one b, one c, and then one of them contributes an extra a, b, c. But we need to split the cutting such that one stick contributes more.Wait, but each original stick must be cut into3 or4 pieces. If12 original sticks each are cut into3 pieces, we get36 pieces. But we need39. So three sticks must be cut into4 pieces. So the initial idea of9 sticks cut into3 and3 sticks into4 is necessary.If we have9 sticks cut into3 pieces and3 sticks into4 pieces.Now, suppose that among the9 sticks, each is cut into one a, one b, one c. So total from these:9a,9b,9c.Then the3 sticks cut into4 pieces need to contribute4a,4b,4c to reach13 each. So each of the3 sticks must contribute (4/3)a, (4/3)b, (4/3)c. But that's not possible. However, if each of these three sticks is cut into1 a,1 b,1 c, and1 more piece. But that more piece would have to be a, b, or c. Let’s say each of the three sticks is cut into1 a,1 b,1 c, and1 a. Then each contributes2a +1b +1c=1. So total from three sticks:6a +3b +3c. Combined with the nine sticks: total a=9+6=15, b=9+3=12, c=9+3=12. Still missing1 b and1 c.Alternatively, if two of the three sticks are cut into2a +1b +1c=1, and one stick is cut into2b +1a +1c=1. Then total a's:9 +2*2 +1=14, b's:9 +2*1 +2=13, c's:9 +3=12. Still missing1 c.Alternatively, maybe two sticks contribute an extra a and one stick contributes an extra c. So two sticks with2a +1b +1c=1 and one stick with2c +1a +1b=1. Then total a's:9 +2*2 +1=14, b's:9 +2*1 +1=12, c's:9 +2*1 +2=13. Now we have14 a's,12 b's,13 c's. Still missing1 b.This is not working. Maybe a different distribution.Alternatively, if each of the three 4-piece sticks is cut into1 a,1 b, and2 c's. Then each stick contributes1a +1b +2c=1. Total from three sticks:3a +3b +6c. Combined with nine sticks (1a +1b +1c each): total a=9+3=12, b=9+3=12, c=9+6=15. Need13 each. Still missing1 a,1 b, and have2 extra c's.Not helpful.Alternatively, each of the three 4-piece sticks is cut into1 a,2 b,1 c. So each stick is1a +2b +1c=1. Total from three sticks:3a +6b +3c. Combined with nine sticks: total a=12, b=15, c=12. Still missing1 a and1 c.Hmm. This is tricky. The problem seems to require a careful balance. Maybe we need to use different combinations in the 3-piece and 4-piece sticks to get the exact counts.Let’s try another approach. Let's assume that all the sticks are cut into three pieces, but that would give36 pieces. We need39, so we need three more pieces. Therefore, three sticks need to be cut into four pieces instead of three. So now, we have nine sticks cut into three pieces and three sticks cut into four pieces.Suppose that each of the nine 3-piece sticks is cut into1 a,1 b, and1 c. So total from these:9a,9b,9c. Then, the three 4-piece sticks need to contribute4a,4b,4c. So each of these three sticks must contribute (4/3)a, (4/3)b, (4/3)c. But this isn't possible. However, if we allow that the three sticks have different combinations:For example, one 4-piece stick contributes2a,1b,1c (sum 2a +b +c=1).Another contributes1a,2b,1c (sum a +2b +c=1).Another contributes1a,1b,2c (sum a +b +2c=1).Then total from these three sticks:a's:2+1+1=4b's:1+2+1=4c's:1+1+2=4Combined with the nine sticks: total a=9+4=13, b=9+4=13, c=9+4=13. Perfect.Now, we need to check if these equations are consistent.The equations for the three 4-piece sticks are:1. 2a +b +c=12. a +2b +c=13. a +b +2c=1And for the 3-piece sticks:4. a +b +c=1Let’s solve these equations.From equation4: a +b +c=1.Subtract equation4 from equation1:(2a +b +c) - (a +b +c)=1 -1 =>a=0. But then from equation4,0 +b +c=1 =>b +c=1.From equation3:0 +b +2c=1 =>b +2c=1.Subtract equation4 from this: (b +2c) - (b +c)=1 -1 =>c=0. But then b=1. But then from equation2:0 +2*1 +0=2=1, which is false. Contradiction.Therefore, no solution.This is the same problem as before. The equations are dependent and lead to a contradiction. Therefore, this combination is invalid.Hmm. Is there a way to set the equations so that they are consistent?Let’s think. Suppose that all four equations:1. 2a +b +c=12. a +2b +c=13. a +b +2c=14. a +b +c=1If we subtract equation4 from equations1,2,3, we get:From1-4: a=0From2-4: b=0From3-4: c=0Which implies a=b=c=0, which is impossible. Therefore, no solution.Therefore, this approach doesn't work.Perhaps we need to consider different combinations for the 4-piece sticks that don't all involve adding an extra piece of one type. For example, one stick could have3a +1b=1, another3b +1c=1, another3c +1a=1. Let's see:Then total a's from three sticks:3a (from first) +0 +1a=4a.Similarly, b's:1b +3b +0=4b.c's:0 +1c +3c=4c.Combined with nine sticks (1a +1b +1c): total a=9+4=13, b=9+4=13, c=9+4=13.But we need to check if the equations hold:For the first 4-piece stick:3a +1b=1Second:3b +1c=1Third:3c +1a=1Also, for the 3-piece sticks:a +b +c=1.Now, we have four equations:1. 3a +b=12. 3b +c=13. 3c +a=14. a +b +c=1Let’s solve these equations.From equation1: b=1 -3aFrom equation2: c=1 -3b=1 -3*(1 -3a)=1 -3 +9a= -2 +9aFrom equation3: 3c +a=1. Substitute c from above:3*(-2 +9a) +a=1 =>-6 +27a +a=1 =>28a=7 =>a=7/28=1/4.Then b=1 -3*(1/4)=1 -3/4=1/4.c= -2 +9*(1/4)= -2 +9/4= (-8/4 +9/4)=1/4.Check equation4: a +b +c=1/4 +1/4 +1/4=3/4≠1. Contradiction. So this doesn't work.Another dead end.Perhaps the solution requires a different approach altogether. Let’s think outside the box. What if the triangles are arranged such that some sticks are shared between triangles? But the problem states that each small stick is a side of one of the triangles. So no sharing.Alternatively, maybe the triangles are not all separate. But the problem says "form 13 equal triangles", so they need to be 13 distinct triangles.Alternatively, think about the fact that 12 sticks are cut into 39 pieces, which is 3*13. So each triangle is made of three sticks. The total number of triangles is13, each requiring three sticks. So the 39 pieces form 13 triangles. Each triangle has three sticks of lengths a, b, c. Therefore, to form 13 congruent triangles, each with sides a, b, c, we need13 of each stick a, b, c.Now, the challenge is to cut the original12 sticks into these39 pieces such that each original stick is cut into pieces whose lengths add up to1 (original stick length), and the total number of each piece is13.Assuming each original stick is length1, the total length of all pieces is12.The total length required is13*(a +b +c)=12, so a +b +c=12/13.We need to find a way to partition each original stick into3 or4 pieces of lengths a, b, c such that their sum is1, and the total count of each is13.One possible way is to use a repeating pattern that allows for the counts to add up.For example, suppose we cut each original stick into two a's and one b. Then 2a +b=1. If we do this for multiple sticks, but we need to balance the counts of a, b, c. However, this only gives a and b, not c.Alternatively, use a combination of different cuts. For example:- For four original sticks, cut each into three pieces: a, a, b (sum 2a +b=1). This gives eight a's and four b's.- For another four original sticks, cut each into three pieces: b, b, c (sum 2b +c=1). This gives eight b's and four c's.- For the last four original sticks, cut each into three pieces: c, c, a (sum 2c +a=1). This gives eight c's and four a's.Total a's:8 +4=12Total b's:4 +8=12Total c's:4 +8=12Plus the three remaining sticks, which need to be cut into four pieces each to reach39 total. Each of these three sticks needs to contribute1 a,1 b,1 c, and an extra piece. Let's say each contributes1 a,1 b,1 c, and1 a. Then each such stick has2a +1b +1c=1. Total from three sticks:6a +3b +3c. Adding to previous totals: a=12 +6=18, b=12 +3=15, c=12 +3=15. Too many.Alternatively, each of the three sticks contributes1 a,1 b,1 c, and1 d (but we can't have d).This seems impossible. Another approach is needed.Wait, perhaps the solution involves using the same length in multiple sticks. For example, if a= b= c=4/13. Then each triangle would be equilateral with sides4/13. But as we saw earlier, cutting a stick into three pieces of4/13 would sum to12/13, which is less than1. Similarly, cutting into four pieces of4/13 would sum to16/13, which is more than1. So this doesn't work. Therefore, equilateral triangles are out.Another idea: perhaps use right-angled triangles. For example, a=3/13, b=4/13, c=5/13. Then a +b +c=12/13, which matches. Then each triangle is a 3-4-5 triangle scaled down by1/13. Check triangle inequality:3 +4 >5, etc., which holds.Total length of all sticks:13*(3/13 +4/13 +5/13)=13*(12/13)=12, which matches.Now, need to cut the original12 sticks into pieces of3/13,4/13,5/13 such that each original stick's pieces sum to1.Each original stick of length1 can be cut into combinations of3/13,4/13,5/13. Let’s find how.We need to find non-negative integers x, y, z such that (3/13)x + (4/13)y + (5/13)z =1. Multiply both sides by13:3x +4y +5z=13.We need to solve for non-negative integers x, y, z.Possible combinations:Let’s try z=0:3x +4y=13Solutions:x= (13 -4y)/3. We need 13 -4y divisible by3.Trying y=1:13-4=9 → x=3. So (x,y,z)=(3,1,0)y=4:13-16=-3 → x=-1 invalid.y=0:13-0=13 → x=13/3 invalid.So only y=1 gives a solution.Next, z=1:3x +4y +5=13 ⇒3x +4y=8.Solutions:x=(8-4y)/3.y=0:8/3 invalid.y=2: (8-8)/3=0 → x=0. So (0,2,1).y=1: (8-4)/3=4/3 invalid.y=3: (8-12)/3 negative.So only y=2, x=0.Next, z=2:3x +4y +10=13 ⇒3x +4y=3.Solutions:x=(3-4y)/3.y=0: x=1.y=1: (3-4)/3=-1/3 invalid.So (1,0,2).z=3:3x +4y +15=13 →3x +4y=-2 invalid.So possible combinations for a stick:Either (3,1,0), (0,2,1), or (1,0,2).These are the possible ways to cut a stick into3/13,4/13,5/13 pieces.Now, let's check the number of pieces for each combination:- (3,1,0):3 pieces of3/13,1 piece of4/13. Total pieces:4.- (0,2,1):2 pieces of4/13,1 piece of5/13. Total pieces:3.- (1,0,2):1 piece of3/13,2 pieces of5/13. Total pieces:3.Wait, but our earlier calculation said that to get39 pieces, we need9 sticks cut into3 and3 sticks into4. So the combination (3,1,0) gives4 pieces, which would be part of the3 sticks cut into4. The combinations (0,2,1) and (1,0,2) give3 pieces, part of the9 sticks.Now, let's calculate how many of each combination we need to get13 of each piece.Let’s denote:Let’s use variables:Let’s let:- m be the number of sticks cut into (3,1,0) (4 pieces:3a +1b)- n be the number of sticks cut into (0,2,1) (3 pieces:2b +1c)- p be the number of sticks cut into (1,0,2) (3 pieces:1a +2c)We need to solve for m, n, p such that:Total a's:3m +0n +1p =13Total b's:1m +2n +0p =13Total c's:0m +1n +2p =13Also, since m sticks are cut into4 pieces, and n + p sticks are cut into3 pieces, and total sticks is12:m +n +p=12But also, the total number of sticks cut into4 pieces is m, and into3 pieces is n +p=12 -m. But from earlier, we need9 sticks cut into3 and3 cut into4. Therefore, m=3 and n +p=9.So m=3. Therefore:From a's:3*3 +0n +1p=9 +p=13 ⇒p=4From b's:1*3 +2n +0p=3 +2n=13 ⇒2n=10 ⇒n=5From c's:0*3 +1*5 +2*4=5 +8=13. Checks out.Therefore, we need:- m=3 sticks cut into (3a +1b)=3*(3/13) +1*(4/13)=13/13=1. Good.- n=5 sticks cut into (0a +2b +1c)=0 +2*(4/13) +1*(5/13)=13/13=1. Good.- p=4 sticks cut into (1a +2c)=1*(3/13) +2*(5/13)=13/13=1. Good.Total sticks:3 +5 +4=12. Correct.Total pieces:- 3 sticks with4 pieces:3*4=12- 5 sticks with3 pieces:5*3=15-4 sticks with3 pieces:4*3=12Total pieces:12 +15 +12=39. Correct.Now, check the counts:a's:3*3 +0*5 +1*4=9 +0 +4=13b's:1*3 +2*5 +0*4=3 +10 +0=13c's:0*3 +1*5 +2*4=0 +5 +8=13Perfect. All counts match.Therefore, the solution is to cut the sticks as follows:- Cut3 sticks into four pieces: three pieces of length3/13 and one piece of length4/13.- Cut5 sticks into three pieces: two pieces of length4/13 and one piece of length5/13.- Cut4 sticks into three pieces: one piece of length3/13 and two pieces of length5/13.This way, we end up with13 pieces of each length3/13,4/13,5/13, which can then form13 congruent right-angled triangles with sides3/13,4/13,5/13.Therefore, the answer is to cut the sticks such that you have13 pieces of lengths3/13,4/13, and5/13 each, and then form13 triangles with those sides.</think>To solve the problem of cutting 12 sticks to form 13 congruent triangles, follow these steps:1. Determine Side Lengths: Each triangle must have sides (a = frac{3}{13}), (b = frac{4}{13}), and (c = frac{5}{13}) (a 3-4-5 triangle scaled by (frac{1}{13})). This ensures the perimeter of each triangle is (frac{12}{13}), matching the total length available.2. Cutting the Sticks: - 3 sticks are each cut into four pieces: three pieces of (frac{3}{13}) and one piece of (frac{4}{13}). - 5 sticks are each cut into three pieces: two pieces of (frac{4}{13}) and one piece of (frac{5}{13}). - 4 sticks are each cut into three pieces: one piece of (frac{3}{13}) and two pieces of (frac{5}{13}).3. Verification: - Total pieces: (3 times 4 + 5 times 3 + 4 times 3 = 39) (13 of each length). - Each original stick's total length remains 1 after cutting.Answer: Cut 3 sticks into three (frac{3}{13}) and one (frac{4}{13}), 5 sticks into two (frac{4}{13}) and one (frac{5}{13}), and 4 sticks into one (frac{3}{13}) and two (frac{5}{13}). This yields 13 triangles with sides (boxed{frac{3}{13}}), (boxed{frac{4}{13}}), and (boxed{frac{5}{13}}).]

🔑:We need to form triangles with sides in the ratio of 3: 4: 5.## SolutionLet's assume the length of each stick is 13. We will cut three sticks into segments of length 3,3,3,4; four sticks into segments of length 3, 5, 5; and five sticks into segments of length 4, 4, 5. As a result, we will have 13 sticks of length 3, 13 sticks of length 4, and 13 sticks of length 5. From these, we can form 13 equal triangles with sides 3,4,5.Send a comment

🧠:[<think>Okay, so I need to solve the equation arctg(1/x) + arctg(1/y) = arctg(1/10) where x and y are integers. Hmm, arctangent equations can sometimes be tricky, but maybe I can use the tangent addition formula here. Let me recall that formula. The tangent of the sum of two angles is (tan α + tan β) / (1 - tan α tan β). So if I let α = arctg(1/x) and β = arctg(1/y), then tan α = 1/x and tan β = 1/y. Then, the equation becomes arctg(1/x) + arctg(1/y) = arctg(1/10). Taking the tangent of both sides might help. Let me try that.So, tan(arctg(1/x) + arctg(1/y)) = tan(arctg(1/10)). The right side simplifies to 1/10. The left side, using the tangent addition formula, should be (tan(arctg(1/x)) + tan(arctg(1/y))) / (1 - tan(arctg(1/x)) tan(arctg(1/y))). Which simplifies to (1/x + 1/y) / (1 - (1/x)(1/y)) = 1/10.So, ( (1/x + 1/y) ) / (1 - 1/(xy)) = 1/10. Let me write that equation out:( (y + x) / (xy) ) / ( (xy - 1) / (xy) ) ) = 1/10.Since the denominators are both over xy, when I divide them, the xy cancels out. So, (x + y) / (xy - 1) = 1/10.Cross-multiplying gives 10(x + y) = xy - 1. Let's rearrange that equation:xy - 10x - 10y - 1 = 0.Hmm, this is a quadratic Diophantine equation. Maybe I can factor this or rearrange terms to make it easier. Let me add 100 to both sides to complete the square:xy - 10x - 10y + 100 - 101 = 0.Wait, let's see: xy -10x -10y -1 = (x -10)(y -10) - 101 = 0. Let me check that expansion. (x -10)(y -10) = xy -10x -10y + 100. So, if we subtract 101, we get xy -10x -10y -1. So, yes, (x -10)(y -10) = 101.Therefore, the equation becomes (x -10)(y -10) = 101. Now, since 101 is a prime number, the integer pairs (x -10, y -10) must be the pairs of divisors of 101. The divisors of 101 are (1,101), (101,1), (-1,-101), (-101,-1). Therefore, possible solutions:1. x -10 = 1 and y -10 = 101 => x =11, y=1112. x -10 =101 and y -10 =1 => x=111, y=113. x -10 = -1 and y -10 = -101 => x=9, y=-914. x -10 = -101 and y -10 = -1 => x=-91, y=9So these are four possible solutions. Now, I need to check if these solutions satisfy the original equation because sometimes when dealing with trigonometric equations, especially when taking inverses, there can be issues with the principal values or the quadrants.First, let's check solution 1: x=11, y=111. Then arctg(1/11) + arctg(1/111). Let me compute this. Since both 1/11 and 1/111 are positive, the sum should be in the first quadrant, which matches the right-hand side arctg(1/10) which is also in the first quadrant. So this seems okay. Let me check with the tangent addition formula. The sum should be arctg( (1/11 + 1/111)/(1 - (1/11)(1/111)) ). Calculating numerator: (122/1221) denominator: (1 - 1/1221) = 1220/1221. So the total is (122/1221) / (1220/1221) = 122/1220 = 61/610 = 1/10. So yes, this works.Solution 2: x=111, y=11. Same reasoning, just swapped. So arctg(1/111) + arctg(1/11) = arctg(1/10). Which is the same as above, so this works as well.Now solution 3: x=9, y=-91. Let's compute arctg(1/9) + arctg(-1/91). Since y=-91, 1/y = -1/91. So we have arctg(1/9) + arctg(-1/91). Let's check if this equals arctg(1/10). The tangent of the sum would be (1/9 - 1/91)/(1 + (1/9)(1/91)) = ( (91 -9)/819 ) / (1 + 1/819 ) = (82/819) / (820/819 ) = 82/820 = 41/410 = 1/10. So yes, the tangent is 1/10. However, we need to check the angle. Since arctg(1/9) is positive (in first quadrant), and arctg(-1/91) is negative (in fourth quadrant). So the sum could be in the fourth or first quadrant depending on the magnitudes. However, arctg(1/9) is approximately 6.34 degrees, arctg(-1/91) is approximately -0.63 degrees. So the sum is approximately 5.71 degrees, which is positive. However, arctg(1/10) is approximately 5.71 degrees. So the arctg function will return the principal value, which is correct here. So this solution is valid.Similarly, solution 4: x=-91, y=9. arctg(-1/91) + arctg(1/9). Same as solution 3, just swapped. The sum is the same, so this is also valid.Wait, but does arctg(-1/91) + arctg(1/9) equal arctg(1/10)? Let me verify. The tangent of the sum is (-1/91 + 1/9)/(1 + (-1/91)(1/9)) = ( (-9 + 91)/819 ) / (1 - 1/819 ) = (82/819) / (818/819) = 82/818 = 41/409 ≈ 0.10024, which is approximately 1/10 (0.1). Wait, 41/409 is approximately 0.10024, which is slightly more than 0.1. Hmm, but 41/409 is exactly 41/(400+9) = 41/409. Wait, but 41*10=410, which is 1 more than 409, so 41/409 = (410 -1)/409 = 10/10 - 1/409 ≈ 1 - 0.00244, which is not 1/10. Wait, that can't be. Wait, no, my calculation is wrong.Wait, let me recalculate:For solution 3: x=9, y=-91. Then 1/x = 1/9, 1/y = -1/91.So tan(α + β) = (1/9 + (-1/91)) / (1 - (1/9)(-1/91)) = ( (91 - 9)/ (9*91) ) / (1 + 1/(9*91) )Calculate numerator: 82 / 819Denominator: 1 + 1/819 = 820 / 819So tan(α + β) = (82/819) / (820/819) ) = 82 / 820 = 41 / 410 = 1 / 10. So exactly 1/10. So the angle whose tangent is 1/10 is indeed arctg(1/10). So even though one angle is positive and one is negative, their sum still has a tangent of 1/10. The actual sum would be arctg(1/9) - arctg(1/91). Let's check numerically:arctg(1/9) ≈ 0.11066 radians (≈6.34 degrees)arctg(1/91) ≈0.01099 radians (≈0.63 degrees)So arctg(1/9) - arctg(1/91) ≈0.09967 radians, which is approximately 5.71 degrees. arctg(1/10) is approximately 0.0997 radians (≈5.71 degrees). So yes, it is correct.Similarly, for solution 4, x=-91, y=9, it's arctg(-1/91) + arctg(1/9), which is the same as arctg(1/9) - arctg(1/91), which is the same as solution 3, so it works.Therefore, all four solutions are valid. Therefore, the integer solutions are (11,111), (111,11), (9,-91), (-91,9).Wait, but the problem says "In the set of integers", so both x and y have to be integers. The solutions we found are indeed integer pairs. Let me check if I didn't miss any divisors. The divisors of 101 are ±1 and ±101. So the four pairs we considered are all possible factorizations. Therefore, there are no other integer solutions. So these four are the only solutions.Hence, the solutions are (11, 111), (111, 11), (9, -91), (-91, 9).Final AnswerThe integer solutions are boxed{(11, 111)}, boxed{(111, 11)}, boxed{(9, -91)}, and boxed{(-91, 9)}.</think>To solve the equation (operatorname{arctg} frac{1}{x} + operatorname{arctg} frac{1}{y} = operatorname{arctg} frac{1}{10}) in the set of integers, we start by using the tangent addition formula. Let (alpha = operatorname{arctg} frac{1}{x}) and (beta = operatorname{arctg} frac{1}{y}). Then, we have:[tan(alpha + beta) = frac{tan alpha + tan beta}{1 - tan alpha tan beta}]Substituting (tan alpha = frac{1}{x}) and (tan beta = frac{1}{y}), we get:[frac{frac{1}{x} + frac{1}{y}}{1 - frac{1}{xy}} = frac{1}{10}]Simplifying this equation:[frac{frac{x + y}{xy}}{frac{xy - 1}{xy}} = frac{1}{10} implies frac{x + y}{xy - 1} = frac{1}{10}]Cross-multiplying gives:[10(x + y) = xy - 1 implies xy - 10x - 10y - 1 = 0]Rewriting this equation by completing the square:[(x - 10)(y - 10) = 101]Since 101 is a prime number, the integer pairs ((x - 10, y - 10)) must be the divisors of 101. The divisors are (pm 1) and (pm 101), leading to the pairs:1. (x - 10 = 1) and (y - 10 = 101) ⟹ (x = 11), (y = 111)2. (x - 10 = 101) and (y - 10 = 1) ⟹ (x = 111), (y = 11)3. (x - 10 = -1) and (y - 10 = -101) ⟹ (x = 9), (y = -91)4. (x - 10 = -101) and (y - 10 = -1) ⟹ (x = -91), (y = 9)Each solution is verified to satisfy the original equation. Therefore, the integer solutions are:[boxed{(11, 111)}, boxed{(111, 11)}, boxed{(9, -91)}, boxed{(-91, 9)}],<think>Okay, so I need to solve this equation involving arctangent functions: arctg(1/x) + arctg(1/y) = arctg(1/10). Hmm, arctg is the inverse tangent function, right? So maybe I can use some trigonometric identities here. Let me recall that if we have arctg(a) + arctg(b) = arctg(c), then there's a formula for that. I remember that tan(arctg(a) + arctg(b)) = (a + b)/(1 - ab), as long as 1 - ab ≠ 0. But since the equation is set equal to arctg(1/10), maybe I can take the tangent of both sides to simplify. Let me try that.Taking tangent of both sides:tan(arctg(1/x) + arctg(1/y)) = tan(arctg(1/10)).On the right side, tan(arctg(1/10)) is just 1/10. On the left side, using the tangent addition formula:tan(arctg(1/x) + arctg(1/y)) = [tan(arctg(1/x)) + tan(arctg(1/y))] / [1 - tan(arctg(1/x)) * tan(arctg(1/y))].Which simplifies to:(1/x + 1/y) / (1 - (1/x)(1/y)) = 1/10.So now the equation is:( (1/x + 1/y) ) / (1 - 1/(xy)) = 1/10.Let me write that as:( (x + y)/xy ) / ( (xy - 1)/xy ) = 1/10.Simplify numerator divided by denominator:(x + y)/xy * xy/(xy - 1) = (x + y)/(xy - 1) = 1/10.So that simplifies to:(x + y)/(xy - 1) = 1/10.Cross-multiplying:10(x + y) = xy - 1.Bring all terms to one side:xy - 10x - 10y - 1 = 0.Hmm, now I have a diophantine equation: xy -10x -10y -1 =0. Need to find integer solutions x and y.This seems like a quadratic in two variables. Maybe I can rearrange terms to make it factorable. Let me try adding 100 to both sides:xy -10x -10y + 100 -1 = 99.So, xy -10x -10y +100 = 99.The left side can be factored as (x -10)(y -10) = 99.Ah, perfect! That's a common technique, completing the rectangle or something. So, by adding and subtracting 100, I can factor the left-hand side.So, (x -10)(y -10) = 99.Now, since x and y are integers, we need integer pairs (a, b) such that a*b = 99, where a = x -10 and b = y -10.So, the integer divisors of 99 are: ±1, ±3, ±9, ±11, ±33, ±99.Therefore, possible pairs (a, b):(1, 99), (3, 33), (9, 11), (-1, -99), (-3, -33), (-9, -11), and also reversed pairs since multiplication is commutative.So, for each pair (a, b), x = a +10, y = b +10.So let's list all possible pairs:1. (1, 99): x = 11, y = 1092. (3, 33): x =13, y=433. (9,11): x=19, y=214. (-1, -99): x=9, y=-895. (-3, -33): x=7, y=-236. (-9, -11): x=1, y=-1And the reversed pairs:7. (99,1): x=109, y=118. (33,3): x=43, y=139. (11,9): x=21, y=1910. (-99, -1): x=-89, y=911. (-33, -3): x=-23, y=712. (-11, -9): x=-1, y=1So these are all possible solutions. Now, but wait, do we need to check if these solutions actually satisfy the original equation? Because sometimes when dealing with inverse trigonometric functions and taking tangents, we might introduce extraneous solutions or lose some information due to the periodicity of tangent.So maybe some of these solutions don't work. Let me verify a few.First, take x=11, y=109.Compute arctg(1/11) + arctg(1/109). Let's see.Using the formula: arctg(a) + arctg(b) = arctg((a + b)/(1 - ab)), but since we know that (1/11 + 1/109)/(1 - (1/11)(1/109)) = (120/1199)/(1 - 1/1199) = (120/1199)/(1198/1199) = 120/1198 = 60/599 ≈ 0.100167, which is approximately 1/10 (0.1). But 60/599 is not exactly 1/10. Wait, but 10(x + y) = 10*(11 + 109)=10*120=1200, and xy -1=11*109 -1=1199 -1=1198. So 1200=1198? Wait, that can't be. Wait, hold on, earlier steps: after cross-multiplying, we had 10(x + y) = xy -1. So 10*(11 +109)=10*120=1200, and 11*109 -1=1199 -1=1198. 1200 ≠ 1198. So something's wrong here. Wait, but we derived that (x -10)(y -10)=99, so (11 -10)(109 -10)=1*99=99, which is correct. So why is the equation not satisfied?Wait, but we did cross-multiply to get 10(x + y)=xy -1. But if x=11, y=109, then 10*(11 +109)=10*120=1200 and xy -1=11*109 -1=1199 -1=1198. So 1200≠1198. That's a problem. So that suggests that maybe there's an error in the logic.Wait, but how come (x -10)(y -10)=99 gives us solutions that don't satisfy the original equation? That must mean that when we took the tangent of both sides, we might have introduced extraneous solutions or perhaps lost some information about the quadrant.Wait, let's recall that arctg(1/x) and arctg(1/y) are angles. Since x and y are integers, 1/x and 1/y can be positive or negative. The arctangent function returns values between -π/2 and π/2. But when we add two arctangent terms, the result could be in a different quadrant, and taking the tangent might not capture the correct angle. So even though the tangent values are equal, the actual angles might differ by a multiple of π. But since the original equation is equality of arctg(1/x) + arctg(1/y) to arctg(1/10), which is between -π/2 and π/2, the sum must also lie in that interval. However, the sum of two arctangent terms could be outside of (-π/2, π/2). Wait, actually, no. Each arctg(1/x) is between -π/2 and π/2, but if x is positive, then 1/x is positive, so arctg(1/x) is between 0 and π/2. If x is negative, arctg(1/x) is between -π/2 and 0. So the sum of two such angles would be between -π and π, but arctg(1/10) is between -π/2 and π/2. So perhaps some solutions where the sum is actually in a different quadrant but has the same tangent value.So, for example, if arctg(1/x) + arctg(1/y) = arctg(1/10) + kπ for some integer k. But since the left side is between -π and π, and the right side is between -π/2 and π/2, the only possible k is 0. So maybe the equation is only valid when the sum is in (-π/2, π/2). But we need to check whether the sum is actually in that interval.Alternatively, maybe some of the solutions lead to a negative argument for the arctg, which complicates things. Let's check.Take the solution x=11, y=109. Then 1/x and 1/y are both positive, so arctg(1/11) and arctg(1/109) are both positive angles. Their sum is a positive angle less than π/2 + π/2 = π, but arctg(1/10) is less than π/2, so the sum must actually be less than π/2. Wait, but is that the case?Wait, let's compute arctg(1/11) + arctg(1/109). Since 1/11 ≈0.0909 and 1/109≈0.00917. Their arctangent values are approximately arctg(0.0909)≈0.0907 radians and arctg(0.00917)≈0.00917 radians. Sum≈0.0999 radians. arctg(1/10)=arctg(0.1)≈0.0997 radians. So they are very close. But is 0.0999 ≈0.0997? The slight difference is due to the approximation, but perhaps they are equal? Wait, but according to our algebra, the equation (x + y)/(xy -1)=1/10. For x=11 and y=109, we have (11 +109)/(11*109 -1)=120/1198≈0.100167, which is not exactly 1/10=0.1, which suggests that this solution does not satisfy the original equation. But why?But we derived (x -10)(y -10)=99 from the original equation. So there's a contradiction here. That must mean that when we took the tangent of both sides, we assumed that the sum of arctg(1/x) + arctg(1/y) equals arctg(1/10), but in reality, their tangents are equal, but the angles might differ by π. Wait, but if the sum of the two arctg is equal to arctg(1/10) + kπ for some integer k. But since arctg(1/x) and arctg(1/y) are each between -π/2 and π/2, their sum is between -π and π. arctg(1/10) is between -π/2 and π/2. So possible k values are -1,0,1. But to have arctg(1/x) + arctg(1/y) = arctg(1/10) + kπ, and given that the original problem is in integers, maybe positive integers? Wait, the problem says "In the set of integers", so x and y can be positive or negative. But if x or y is zero, 1/x or 1/y is undefined. So x and y must be non-zero integers.But perhaps some of the solutions have x or y negative. Let's check the solutions.Looking at the possible solutions:1. (11, 109): both positive2. (13,43): both positive3. (19,21): both positive4. (9, -89): x=9, y=-895. (7, -23): x=7, y=-236. (1, -1): x=1, y=-17. (109,11): same as 18. (43,13): same as 29. (21,19): same as 310. (-89,9): x=-89, y=911. (-23,7): x=-23, y=712. (-1,1): x=-1, y=1So solutions with x and y positive: 1,2,3,7,8,9. Solutions with one positive and one negative: 4,5,6,10,11,12. Let's check if positive solutions work.Take x=11, y=109. Let's compute arctg(1/11) + arctg(1/109). As I approximated before, this is roughly 0.0907 + 0.00917 ≈0.0999 radians. arctg(1/10)=arctg(0.1)≈0.0997 radians. So they are very close but not exactly equal. Wait, but maybe due to the approximation. Let's compute more precisely.But how can we check if arctg(1/11) + arctg(1/109) equals arctg(1/10)?Using the tangent addition formula:tan(arctg(1/11) + arctg(1/109)) = (1/11 + 1/109)/(1 - (1/11)(1/109)) = (120/1199)/(1 - 1/1199) = (120/1199)/(1198/1199) = 120/1198 = 60/599 ≈0.100167.But tan(arctg(1/10)) is 1/10=0.1. So 60/599 ≈0.100167 is not equal to 1/10=0.1. So the left side is not equal to the right side. Therefore, even though (11,109) satisfies the equation (x -10)(y -10)=99, it does not satisfy the original trigonometric equation. Therefore, this must be an extraneous solution introduced when we took the tangent of both sides.Hmm, so the problem arises because the formula tan(A + B) = (tan A + tan B)/(1 - tan A tan B) assumes that A + B ≠ π/2 + kπ. But even more importantly, when we take the arctg of both sides, the equality of tangents does not necessarily imply equality of angles, because the arctg function only returns principal values. So even if tan(alpha) = tan(beta), alpha and beta can differ by a multiple of π. However, in our case, since the original equation is alpha + beta = gamma, where gamma is arctg(1/10), which is between -π/2 and π/2. But alpha and beta are arctg(1/x) and arctg(1/y), each between -π/2 and π/2. Therefore, their sum is between -π and π. Therefore, if the sum is equal to gamma, which is between -π/2 and π/2, then the difference between the left and right sides can't be a multiple of π. Wait, but tan(alpha + beta) = tan(gamma), so alpha + beta = gamma + kπ. But since gamma is between -π/2 and π/2, and alpha + beta is between -π and π, then k can be -1,0,1.But in order for alpha + beta = gamma + kπ, if k=0, then alpha + beta = gamma. If k=1, then alpha + beta = gamma + π. But alpha + beta would have to be greater than π/2 (if gamma is positive) but since alpha and beta are each less than π/2 in absolute value, their sum can't exceed π. Similarly, if k=-1, then alpha + beta = gamma - π, which would be less than -π/2 - π = -3π/2, but alpha + beta is greater than -π, so that's impossible. Therefore, possible k=0 or k=1 or k=-1. But given the ranges, k=0 is the only possible, unless alpha and beta are both negative. Wait, but if x and y are positive, then alpha and beta are positive, so their sum is positive, so gamma must be positive. If x or y is negative, then alpha or beta could be negative.This is getting complicated. Maybe we need to check which solutions actually satisfy the original equation.So let's check each solution step by step.First, take solution (11,109):Calculate LHS = arctg(1/11) + arctg(1/109). Compute tan(LHS):(1/11 + 1/109)/(1 - 1/(11*109)) = (120/1199)/(1198/1199) = 120/1198 ≈0.100167.RHS is arctg(1/10), tan(RHS)=1/10=0.1. So tan(LHS) ≈0.100167≠0.1. Therefore, LHS ≠ RHS. So this solution is extraneous. So discard (11,109).Similarly, check (13,43):tan(LHS) = (1/13 + 1/43)/(1 - 1/(13*43)) = (56/559)/(558/559) =56/558≈0.10036. Again, not equal to 0.1. So extraneous.Next, (19,21):(1/19 +1/21)/(1 -1/(19*21))= (40/399)/(398/399)=40/398≈0.1005. Again, not 0.1. So extraneous.Similarly, the reversed pairs (109,11), (43,13), (21,19) would give the same results. So none of the positive solutions work. So maybe all the positive solutions are extraneous. That's strange.Now check the solutions where one is positive and one is negative.Take solution (9, -89):x=9, y=-89.Compute arctg(1/9) + arctg(-1/89). Let's compute tan(LHS):(1/9 -1/89)/(1 + (1/9)(1/89)) = (80/801)/(802/801) =80/802≈0.09975. Which is approximately 0.1. Let's compute exactly:80/802=40/401≈0.09975. 1/10=0.1. So very close, but not exactly equal.Wait, but let's check using cross-multiplied equation:10(x + y)=10*(9 + (-89))=10*(-80)=-800.xy -1=9*(-89)-1=-801 -1=-802.So -800 vs -802. Not equal. So equation not satisfied. Therefore, even though (x-10)(y-10)=9*(-99)=-891=99? Wait no, x=9, y=-89. So x-10= -1, y-10=-99. (-1)*(-99)=99. Correct. So (x-10)(y-10)=99, but the equation 10(x+y)=xy -1 is not satisfied. So this solution is extraneous. Hmm.Similarly, take solution (7, -23):x=7, y=-23.Compute 10(x + y)=10*(7 -23)=10*(-16)=-160.xy -1=7*(-23)-1=-161 -1=-162. So -160 vs -162. Not equal.But (x -10)(y -10)= (7 -10)(-23 -10)= (-3)(-33)=99. Correct. But the original equation is not satisfied.Another solution: (1, -1):x=1, y=-1.10(1 + (-1))=10*0=0.xy -1=1*(-1)-1=-1 -1=-2. So 0 vs -2. Not equal.But (1 -10)(-1 -10)=(-9)(-11)=99. Correct.So all these solutions satisfy (x -10)(y -10)=99 but do not satisfy 10(x + y)=xy -1. Wait, but how is that possible? Because we derived (x -10)(y -10)=99 from 10(x + y)=xy -1. So there's a contradiction here.Wait, let me check the algebra again.Starting from:( (1/x + 1/y) ) / (1 - (1/x)(1/y)) = 1/10.Multiply numerator and denominator by xy:( (y + x) ) / (xy -1 ) =1/10.Therefore:(x + y)/(xy -1)=1/10.Cross-multiplying:10(x + y)=xy -1.So that's correct. Then moving all terms to left:xy -10x -10y -1=0.Then adding 100 to both sides:xy -10x -10y +99=99.Factor as (x -10)(y -10)=99.Wait, no: (x -10)(y -10)=xy -10x -10y +100. So if we have xy -10x -10y -1=0, then xy -10x -10y=1. Therefore, (x -10)(y -10)=xy -10x -10y +100=1 +100=101. Wait, wait, that contradicts previous steps. Wait, what's wrong here.Wait, let's redo the step.Original equation after cross-multiplying:xy -10x -10y -1=0.Then, to complete the rectangle, add 100 to both sides:xy -10x -10y -1 +100=99.Thus, xy -10x -10y +99=99.But xy -10x -10y +100= (x -10)(y -10). Therefore,(x -10)(y -10)=99 +1=100? Wait, no. Wait:Wait, (x -10)(y -10)=xy -10x -10y +100.But in the equation xy -10x -10y -1=0, we have xy -10x -10y=1.Therefore, (x -10)(y -10)=xy -10x -10y +100=1 +100=101.But that contradicts the previous step where we added 100 to both sides. Wait, so I must have made a mistake in the factoring step.Wait, let's do it step by step:Starting from 10(x + y) = xy -1.Rewriting:xy -10x -10y -1=0.Now, to complete the rectangle, we can add 100 to both sides:xy -10x -10y -1 +100=99.Thus, xy -10x -10y +99=99.But xy -10x -10y +100= (x -10)(y -10).Therefore, (x -10)(y -10)=99 +1=100? No.Wait, if xy -10x -10y +100= (x -10)(y -10). Then, from xy -10x -10y +99=99,We have xy -10x -10y +99=99.Thus, xy -10x -10y=0.But original equation was xy -10x -10y=1.So something's wrong here. There is a mistake in the algebra.Wait, let's go back. The original equation after cross-multiplying is:10(x + y) = xy -1.Bring all terms to left side:xy -10x -10y -1=0.Now, to factor this, the technique is to add and subtract a constant to complete the rectangle. The standard form is xy + ax + by + c =0. To complete the rectangle, we can write (x + b)(y + a) = ab - c.Wait, let me recall that. Alternatively, for the equation xy + ax + by + c =0, we can rewrite it as (x + b)(y + a) = ab - c.In our case, equation is xy -10x -10y -1=0. So it's xy + (-10)x + (-10)y + (-1)=0. Comparing with xy + ax + by + c =0, here a=-10, b=-10, c=-1.Then, according to the formula, (x + b)(y + a) = ab - c. So:(x + (-10))(y + (-10)) = (-10)(-10) - (-1) =100 +1=101.Therefore, (x -10)(y -10)=101. Wait, but that contradicts our previous step where we thought it was 99.Wait, this is conflicting with my initial steps. Therefore, I must have made a mistake earlier. Let me check again.Wait, the standard method for equations of the form xy + ax + by + c =0 is to add ab to both sides:xy + ax + by + c + ab =ab.Then, factor the left side as (x + b)(y + a)=ab -c.In our case, equation is xy -10x -10y -1=0. So a=-10, b=-10, c=-1.Add ab=(-10)(-10)=100 to both sides:xy -10x -10y -1 +100=100 -1=99.Left side: xy -10x -10y +99= (x -10)(y -10).Wait, but according to the formula, it should be (x + b)(y + a) = (x -10)(y -10). So yes, equals to 99. Therefore, (x -10)(y -10)=99.But earlier when I tried to check the solution (11,109), which gives (1)(99)=99, the equation 10(x + y)=xy -1 gives 10*(120)=1199 -1=1198, which is 1200=1198, which is false. So there is a contradiction here.Wait, but if we use the standard method, (x -10)(y -10)=99, then substituting x=11, y=109 gives (1)(99)=99, which is correct, but plugging into the original equation doesn't hold. Therefore, there must be a mistake in the derivation.Wait, but step by step:Original equation:arctg(1/x) + arctg(1/y) = arctg(1/10).Take tangent of both sides:tan(arctg(1/x) + arctg(1/y)) = tan(arctg(1/10)).Left side becomes (1/x +1/y)/(1 -1/(xy)) =1/10.Therefore, (x + y)/(xy -1)=1/10.Cross-multiplying:10(x + y)=xy -1.Rearranged:xy -10x -10y -1=0.Then, adding 100 to both sides:xy -10x -10y +99=99.Factor left side as (x -10)(y -10)=99.Therefore, this derivation is correct. Then why do the solutions not satisfy the original equation?Wait, let's check with x= -1, y=1.Plug into original equation:arctg(1/(-1)) + arctg(1/1)= arctg(-1) + arctg(1)= -π/4 + π/4=0. But arctg(1/10) is arctg(0.1)≈0.0997. So 0≠0.0997. Therefore, this solution doesn't work. Similarly, x= -23, y=7: compute arctg(-1/23) + arctg(1/7). Let's compute tan of the sum:(-1/23 +1/7)/(1 + (1/23)(1/7))= (16/161)/(162/161)=16/162=8/81≈0.0988. Which is close to 0.1, but not exactly.But according to the equation 10(x + y)=xy -1. For x=-23, y=7: 10*(-23 +7)=10*(-16)=-160. xy -1=(-23)(7)-1=-161 -1=-162. So -160≠-162. Therefore, the equation is not satisfied. So even though (x -10)(y -10)=99 is satisfied, the original equation is not.This suggests that the derivation is missing something. Maybe the step where we take tangent of both sides is not reversible? Because tan(alpha)=tan(beta) implies alpha=beta +kπ, but when we derived (x + y)/(xy -1)=1/10, we have to ensure that alpha + beta=gamma +kπ, but since the original equation is alpha + beta=gamma, which is in a specific range, perhaps some solutions correspond to k≠0 and thus are invalid.Therefore, to resolve this, we need to check not only that (x -10)(y -10)=99, but also that 10(x + y)=xy -1. Wait, but we derived (x -10)(y -10)=99 from 10(x + y)=xy -1. So they should be equivalent. But in reality, when we check the solutions, they don't satisfy 10(x + y)=xy -1. How is that possible?Wait, maybe a calculation error when substituting.Take solution x=11, y=109:10(x + y)=10*(120)=1200.xy -1=11*109 -1=1199 -1=1198.1200≠1198.But according to the equation, (x -10)(y -10)=99. 11-10=1, 109-10=99. 1*99=99. Correct.But according to 10(x + y)=xy -1: 10(x + y)=1200, xy -1=1198. So they differ by 2. How? There must be an error in the derivation.Wait, let's re-derive the equation step by step.Original equation:arctg(1/x) + arctg(1/y) = arctg(1/10).Take tangent of both sides:tan(arctg(1/x) + arctg(1/y)) = tan(arctg(1/10)).Left side:[tan(arctg(1/x)) + tan(arctg(1/y))]/[1 - tan(arctg(1/x))tan(arctg(1/y))] = (1/x + 1/y)/(1 - (1/x)(1/y)) = (x + y)/[xy -1].Right side:tan(arctg(1/10))=1/10.Therefore:(x + y)/(xy -1) =1/10.Cross-multiplying:10(x + y) = xy -1.Therefore:xy -10x -10y -1=0.So far, so good.Then, to factor, add 100 to both sides:xy -10x -10y -1 +100=99.Thus:xy -10x -10y +99=99.But xy -10x -10y +100=(x -10)(y -10).Therefore:(x -10)(y -10) -1=99.Wait, no. Wait, xy -10x -10y +100= (x -10)(y -10). Therefore, from xy -10x -10y +99=99, we have:(x -10)(y -10) -1=99.Thus:(x -10)(y -10)=100.But this contradicts previous steps. Wait, I'm getting confused.Wait, let's do it carefully:Starting from xy -10x -10y -1=0.We want to write this as (x -10)(y -10)=something.Expand (x -10)(y -10)=xy -10x -10y +100.Therefore, xy -10x -10y = (x -10)(y -10) -100.But from the equation, xy -10x -10y -1=0 ⇒ xy -10x -10y=1.Therefore, (x -10)(y -10) -100=1.Thus, (x -10)(y -10)=101.But this contradicts the previous result where we added 100 to both sides. So where is the mistake?Earlier, when we had:xy -10x -10y -1=0.Then, added 100 to both sides:xy -10x -10y +99=99.But xy -10x -10y +100= (x -10)(y -10). Therefore, (x -10)(y -10) =99 +1=100. So (x -10)(y -10)=100.Wait, but that contradicts the previous derivation where we arrived at (x -10)(y -10)=101.Wait, this is very confusing. Let's check with numbers. Take x=11, y=109.(x -10)(y -10)=1*99=99. But according to the equation, it should be 100 or 101?This suggests that there's a fundamental error in the algebra steps. Let me start over.From the original equation:10(x + y) = xy -1.Rearrange:xy -10x -10y -1=0.Let me attempt to factor this equation. To factor expressions of the form xy + ax + by + c=0, the trick is to add a constant to both sides to make it factorable.Let me consider the equation:xy -10x -10y -1=0.Add 100 to both sides:xy -10x -10y +99=99.Now, can the left side be factored?Let me write it as:x(y -10) -10(y -10) + something= ?Wait, x(y -10) -10(y -10)= (x -10)(y -10). Then:(x -10)(y -10)= xy -10x -10y +100.But in our equation, we have xy -10x -10y +99=99. So:(x -10)(y -10) -1=99.Therefore:(x -10)(y -10)=100.Ah! So actually, (x -10)(y -10)=100. Not 99. Therefore, earlier steps were wrong. That's the mistake. When I added 100 to both sides, I should have:xy -10x -10y -1 +100=99 ⇒ xy -10x -10y +99=99.But xy -10x -10y +99=(x -10)(y -10) -1. Therefore:(x -10)(y -10) -1=99 ⇒ (x -10)(y -10)=100.Therefore, the correct factoring is (x -10)(y -10)=100, not 99. So my previous factorization was incorrect due to an error in the algebra. That explains why the solutions didn't satisfy the original equation.This was the critical mistake. So now, we have to solve (x -10)(y -10)=100. So the previous analysis was incorrect because of this miscalculation.Therefore, the correct equation is (x -10)(y -10)=100. Therefore, the integer solutions are pairs (a, b) such that a*b=100, where a=x -10 and b=y -10.The integer divisors of 100 are ±1, ±2, ±4, ±5, ±10, ±20, ±25, ±50, ±100.So possible pairs (a, b):(1,100),(2,50),(4,25),(5,20),(10,10),(-1,-100),(-2,-50),(-4,-25),(-5,-20),(-10,-10), and all their permutations.Therefore, the corresponding (x,y) pairs are:For positive divisors:1. (1,100): x=11, y=1102. (2,50): x=12, y=603. (4,25):x=14, y=354. (5,20):x=15, y=305. (10,10):x=20, y=206. (100,1):x=110, y=117. (50,2):x=60, y=128. (25,4):x=35, y=149. (20,5):x=30, y=15For negative divisors:10. (-1,-100):x=9, y=-9011. (-2,-50):x=8, y=-4012. (-4,-25):x=6, y=-1513. (-5,-20):x=5, y=-1014. (-10,-10):x=0, y=0 (but x and y can't be zero)15. (-100,-1):x=-90, y=916. (-50,-2):x=-40, y=817. (-25,-4):x=-15, y=618. (-20,-5):x=-10, y=5But we have to exclude pairs where x or y is zero. So pair 14 is invalid. Similarly, check others.So let's list all possible valid (x,y):Positive divisors:1. (11,110)2. (12,60)3. (14,35)4. (15,30)5. (20,20)6. (110,11)7. (60,12)8. (35,14)9. (30,15)Negative divisors:10. (9,-90)11. (8,-40)12. (6,-15)13. (5,-10)14. (-90,9)15. (-40,8)16. (-15,6)17. (-10,5)Now, let's verify if these solutions satisfy the original equation.Take first solution (11,110):Check 10(x + y)=10*(11 +110)=10*121=1210.xy -1=11*110 -1=1210 -1=1209.1210 vs 1209. Not equal. Hmm. Wait, but according to the equation, they should be equal. What's wrong here?Wait, we have (x -10)(y -10)=100. For x=11, y=110: (1)(100)=100. Correct.But 10(x + y)=1210, xy -1=1209. So they differ by 1. This inconsistency suggests another mistake.Wait, no. Wait, if we have (x -10)(y -10)=100, then expanding:xy -10x -10y +100=100 ⇒ xy -10x -10y=0. But original equation is xy -10x -10y -1=0. So xy -10x -10y=1.But according to factoring, we have xy -10x -10y=0. Therefore, contradiction. So how did this happen?Wait, there must be a mistake in the factoring.Original equation: xy -10x -10y -1=0.We added 100 to both sides:xy -10x -10y -1 +100=99 ⇒ xy -10x -10y +99=99.But (x -10)(y -10)=xy -10x -10y +100.So, from the equation xy -10x -10y +99=99 ⇒ (x -10)(y -10) -1=99 ⇒ (x -10)(y -10)=100.Therefore, the correct equation is (x -10)(y -10)=100.But expanding (x -10)(y -10)=100 gives:xy -10x -10y +100=100 ⇒ xy -10x -10y=0.But original equation is xy -10x -10y -1=0 ⇒ xy -10x -10y=1.Contradiction. Therefore, this suggests that there is no solution? But we derived (x -10)(y -10)=100 from the equation, which implies that xy -10x -10y=0, conflicting with original equation's xy -10x -10y=1. Therefore, the only possible conclusion is that there is no solution, which contradicts the factoring approach.This is very confusing. It seems like there's a paradox here.Wait, going back to the original trigonometric equation:arctg(1/x) + arctg(1/y) = arctg(1/10).We took the tangent of both sides and arrived at:(x + y)/(xy -1) =1/10 ⇒ 10(x + y)=xy -1 ⇒ xy -10x -10y -1=0.Then, attempting to factor led to a contradiction. So perhaps there is no integer solution. But this contradicts the earlier step where we found factors. Therefore, there must be a miscalculation.Wait, the key is that when we added 100 to both sides:xy -10x -10y -1 +100=99 ⇒ xy -10x -10y +99=99.But (x -10)(y -10)=xy -10x -10y +100.Therefore, from xy -10x -10y +99=99 ⇒ (x -10)(y -10) -1=99 ⇒ (x -10)(y -10)=100.But expanding this, we get:xy -10x -10y +100=100 ⇒ xy -10x -10y=0.But original equation is xy -10x -10y=1. Therefore, 0=1, which is impossible.This means that there is no solution. Therefore, the original equation has no integer solutions.But the problem states "In the set of integers, solve the equation", implying that there is a solution. So where is the mistake?Wait, maybe when we took the tangent of both sides, we introduced an inconsistency. Let me check the original trigonometric equation.Suppose there are no solutions, which is possible. Let's test with x=20, y=20. Plug into original equation:arctg(1/20) + arctg(1/20). What's the sum?Using tan(alpha + beta)= (2/20)/(1 -1/400)= (1/10)/(399/400)=400/3990≈0.10025. Which is close to 1/10=0.1, but not exactly. Therefore, the sum is not equal to arctg(1/10). So even though (20,20) satisfies (x -10)(y -10)=100, it doesn't satisfy the original equation. Therefore, there are no solutions.But the problem asks to solve the equation in integers. So is there no solution? But the problem wouldn't ask to solve an equation with no solution. Therefore, I must have made a mistake in the initial steps.Wait, let's check the tangent addition formula again.We have tan(A + B) = (tan A + tan B)/(1 - tan A tan B). But this formula is valid when A + B ≠ π/2 + kπ. If A + B = π/2 + kπ, the tangent is undefined. But in our case, the result is 1/10, which is defined. So the formula should hold. But when we derived (x + y)/(xy -1)=1/10, leading to xy -10x -10y -1=0, which leads to a contradiction when factoring. Therefore, there must be no solution.But why does the factoring suggest solutions? Because when we forced the equation to factor, we introduced an inconsistency. Therefore, the conclusion is that there are no integer solutions to the equation. However, the problem asks to solve it, so this can't be. Therefore, I must have made a mistake.Wait, let's try specific values. Suppose x and y are positive integers. Let's assume x ≤ y. Try small values.Let me try x=11. Then the equation is:arctg(1/11) + arctg(1/y)= arctg(1/10).Using the tangent addition formula:(1/11 + 1/y)/(1 -1/(11y))=1/10.Cross-multiplying:10(1/11 +1/y)=1 -1/(11y).Multiply both sides by 11y:10(y +11)=11y -1.Simplify:10y +110=11y -1.Therefore, y=111.But check x=11,y=111:10(x + y)=10*(122)=1220.xy -1=11*111 -1=1221 -1=1220. So 1220=1220. This satisfies the equation. But why isn't this solution listed in our previous factoring?Because according to factoring (x -10)(y -10)=100. For x=11,y=111: (1)(101)=101≠100. So this solution is not in the list.Wait, but according to our previous factoring, it's impossible. But here we have a valid solution. Therefore, our factoring method was incorrect.This is a contradiction. Let's see:With x=11,y=111:arctg(1/11) + arctg(1/111)= arctg(1/10).Check tan(arctg(1/11)+arctg(1/111))= (1/11 +1/111)/(1 -1/(11*111))= (122/1221)/(1220/1221)=122/1220=61/610≈0.09998≈0.1.Which is very close to 1/10. And according to the equation, 10(x + y)=10*(122)=1220 and xy -1=1221 -1=1220. Therefore, equality holds. So (11,111) is a solution.But according to the factoring, (x -10)(y -10)=100. For x=11,y=111: (1)(101)=101≠100. So where is the mistake?This suggests that the factoring approach is incorrect. Let's check the algebra again.Original equation:10(x + y)=xy -1.Rewriting:xy -10x -10y -1=0.Let me attempt to solve for y in terms of x:xy -10x -10y =1.xy -10y=10x +1.y(x -10)=10x +1.Therefore:y=(10x +1)/(x -10).Since y must be an integer, (x -10) must divide (10x +1). So x -10 divides 10x +1.Let me write 10x +1=10(x -10)+101. Therefore:10x +1=10(x -10)+101.Therefore, y=(10(x -10) +101)/(x -10)=10 +101/(x -10).Therefore, for y to be integer, (x -10) must divide 101.101 is a prime number. Therefore, the divisors of 101 are ±1, ±101.Thus, x -10=1 ⇒x=11, y=10 +101/1=111.x -10=-1⇒x=9, y=10 +101/(-1)=10 -101=-91.x -10=101⇒x=111, y=10 +101/101=10 +1=11.x -10=-101⇒x=-91, y=10 +101/(-101)=10 -1=9.Therefore, the integer solutions are:(11,111), (9,-91), (111,11), (-91,9).These are the only solutions where x and y are integers.Therefore, the correct integer solutions are:x=11,y=111; x=9,y=-91; x=111,y=11; x=-91,y=9.This makes sense because when we derived y=(10x +1)/(x -10), we see that only when x -10 divides 101, which is prime, so only four solutions.Therefore, the factoring approach earlier was wrong. The correct solutions are these four.Therefore, the mistake was in the initial factoring approach where I tried to add 100 to both sides and factor, which led to an incorrect result. Instead, solving for y in terms of x and using divisibility gives the correct solutions.Therefore, the integer solutions are:(11,111), (9,-91), (111,11), (-91,9).Let's verify one more solution, say x=9,y=-91.Check 10(x + y)=10*(9 -91)=10*(-82)=-820.xy -1=9*(-91)-1=-819 -1=-820. So equality holds.Check the original trigonometric equation:arctg(1/9) + arctg(-1/91)= arctg(1/10).Compute tan(arctg(1/9) + arctg(-1/91)):(1/9 -1/91)/(1 + (1/9)(-1/91))= (82/819)/(818/819)=82/818=41/409≈0.1002.Which is approximately 1/10=0.1. The exact value is 41/409≈0.1002, which is slightly more than 0.1, but the equation is exact because 10(x + y)=xy -1 holds exactly.However, since tan(alpha)=1/10, alpha=arctg(1/10). But tan(beta)=41/409≈0.1002. However, since 41/409= (10(x + y))/ (xy -1)= (-820)/(-820)=1/10. Wait, no. Wait, 10(x + y)=xy -1=-820. So 10(x + y)=xy -1.But the tangent of the sum is (1/9 -1/91)/(1 + (1/9)(-1/91))= (82/819)/(818/819)=82/818=41/409≈0.1002. But 1/10=0.1. So these are not equal. Therefore, this suggests that even though 10(x + y)=xy -1 holds, the original equation does not. Therefore, this solution is also extraneous.Wait, this is perplexing. If 10(x + y)=xy -1, then the equation (x + y)/(xy -1)=1/10 holds, which was derived from taking the tangent of both sides. However, if the tangent of the sum equals 1/10, but the sum itself is not equal to arctg(1/10), but instead differs by a multiple of π. But since the tangent is periodic with period π, this could be the case.However, given that x and y are integers, and the values of arctg(1/x) and arctg(1/y) are in (-π/2, π/2), their sum can be in (-π, π). Therefore, if the sum is equal to arctg(1/10) + kπ, but since arctg(1/10) is approximately 0.0997 radians, which is in (-π/2, π/2), the only possible k is 0, because for k=1, arctg(1/10) + π would be > π/2, which the sum cannot reach if x and y are positive. For negative x or y, the sum could be negative, but adding π would make it positive.This suggests that the solutions with positive x and y might actually satisfy the original equation, whereas those with negative values might not.Take x=11,y=111.Compute arctg(1/11) + arctg(1/111).Using precise calculation:arctg(1/11) ≈0.0906598872 radians.arctg(1/111)≈0.0089997293 radians.Sum≈0.0996596165 radians.arctg(1/10)≈0.0996686525 radians.The difference is about 0.000009 radians, which is negligible and likely due to calculator precision. Therefore, these are essentially equal. Therefore, the solution (11,111) is valid.Similarly, x=111,y=11 gives the same result.For negative solutions, take x=9,y=-91:arctg(1/9) + arctg(-1/91)= arctg(1/9) - arctg(1/91).Compute this:arctg(1/9)≈0.1106572212 radians.arctg(1/91)≈0.0109948583 radians.Therefore, difference≈0.0996623629 radians≈arctg(1/10). So this also holds.Therefore, even though the tangent values differ slightly in decimal approximation, in exact terms, they are equal. Therefore, all four solutions are valid.Therefore, the integer solutions are:(11,111), (111,11), (9,-91), (-91,9).But the problem didn't specify that x and y must be positive, so all four are valid.Therefore, the final answer is these four pairs.Final AnswerThe integer solutions are boxed{(11, 111)}, boxed{(111, 11)}, boxed{(9, -91)}, and boxed{(-91, 9)}.</think>To solve the equation (operatorname{arctg} frac{1}{x} + operatorname{arctg} frac{1}{y} = operatorname{arctg} frac{1}{10}) in the set of integers, we start by taking the tangent of both sides. Using the tangent addition formula, we get:[tanleft(operatorname{arctg} frac{1}{x} + operatorname{arctg} frac{1}{y}right) = tanleft(operatorname{arctg} frac{1}{10}right)]This simplifies to:[frac{frac{1}{x} + frac{1}{y}}{1 - frac{1}{xy}} = frac{1}{10}]Rewriting this equation, we have:[frac{frac{x + y}{xy}}{1 - frac{1}{xy}} = frac{1}{10}]Simplifying further:[frac{x + y}{xy - 1} = frac{1}{10}]Cross-multiplying gives:[10(x + y) = xy - 1]Rearranging terms, we get:[xy - 10x - 10y - 1 = 0]To solve this equation, we rewrite it as:[y = frac{10x + 1}{x - 10}]For (y) to be an integer, (x - 10) must divide (10x + 1). Factoring 101 (a prime number), we find the divisors of 101 are (pm 1, pm 101). This leads to the solutions:- (x - 10 = 1 implies x = 11) and (y = 111)- (x - 10 = -1 implies x = 9) and (y = -91)- (x - 10 = 101 implies x = 111) and (y = 11)- (x - 10 = -101 implies x = -91) and (y = 9)Verifying these solutions, we find they satisfy the original equation. Thus, the integer solutions are:[boxed{(11, 111)}, boxed{(111, 11)}, boxed{(9, -91)}, boxed{(-91, 9)}]]

🔑:Solution. If we use that operatorname{tg}(alpha+beta)=frac{operatorname{tg} alpha+operatorname{tg} beta}{1-operatorname{tg} alpha operatorname{tg} beta} and operatorname{tg}(operatorname{arctg} alpha)=alpha, for every alpha in(-infty,+infty), operatorname{tg} alpha operatorname{tg} beta neq 1, we get that the given equation is equivalent in turn to the equationsoperatorname{tg}left(operatorname{arctg} frac{1}{1}+operatorname{arctg} frac{1}{y}right)=operatorname{tg}left(operatorname{arctg} frac{1}{10}right)begin{aligned}& frac{frac{1}{x}+frac{1}{y}}{1-frac{1}{x} cdot frac{1}{y}}=frac{1}{10} & (x-10)(y-10)=101end{aligned}Since 101 is a prime number, from the last equation we get the systemsbegin{array}{ll}left{begin{array}{l}x-10=1 y-10=101end{array}right. & left{begin{array}{l}x-10=-1 y-10=-101end{array}right. left{begin{array}{l}x-10=101 y-10=1end{array}right. & left{begin{array}{l}x-10=-101 y-10=-1end{array}right.end{array}whose solutions are the ordered pairs (11,111),(9,-91),(111,11),(-91,9), respectively. Finally, the solution is any ordered pair (x, y) in{(11,111),(9,-91),(111,11),(-91,9)}.

🧠:[<think>Okay, let's try to solve this problem step by step. First, I need to understand what's being asked. We have two sets, M and N, which are defined based on certain conditions involving the function φ(x) and another function f(x). We need to find the intersection of M and N, which means the values of m that satisfy both conditions.Starting with set M: M is the set of all m such that for every x in [0, π/2], φ(x) < 0. The function φ(x) is given by sin²x + m·cosx - 2m. So, we need to find all m values where this expression is negative for all x in that interval.Next, set N: N is the set of all m such that for every x in [0, π/2], f(φ(x)) < 0. Here, f(x) is an odd function defined on all real numbers, and when x < 0, f(x) is monotonically increasing with f(-1) = 0. Since f is odd, that means f(-x) = -f(x). Also, since it's increasing for x < 0, then for x > 0, it must be increasing as well because the odd function property reflects the behavior about the origin. Wait, actually, if f(x) is odd and increasing on x < 0, then for x > 0, it's also increasing. Let me verify that.If f is odd, then f(-x) = -f(x). If f is increasing on x < 0, then for a < b < 0, f(a) < f(b). Translating to x > 0, let's take 0 < c < d, then -d < -c < 0. Since f is increasing on the negative side, f(-d) < f(-c). But since f is odd, f(-d) = -f(d) and f(-c) = -f(c). Therefore, -f(d) < -f(c), which implies f(d) > f(c). So, f is increasing on x > 0 as well. Therefore, f is monotonically increasing on all of R. Wait, but the problem only states that when x < 0, f(x) is monotonically increasing. But since it's odd, maybe the behavior on x > 0 is determined. Let's check.Given that f is odd and increasing on x < 0. Let’s suppose x1 < x2 < 0, then f(x1) < f(x2). For x > 0, take x1' = -x1 and x2' = -x2 where x1' > x2' > 0. Then f(x1') = -f(-x1') = -f(x1) and f(x2') = -f(-x2') = -f(x2). Since x1 < x2 < 0, f(x1) < f(x2). Therefore, -f(x1) > -f(x2), meaning f(x1') > f(x2'). So, on the positive side, when x increases, f(x) decreases. Wait, that contradicts my previous thought. Let me clarify.Wait, if f is odd and increasing on x < 0, then on x > 0, it must be increasing as well? Or decreasing?Wait, let's take a concrete example. Let’s say f(x) is odd and increasing on x < 0. For example, f(x) = x. Then f(x) is odd and increasing everywhere. But if f(x) is, say, -x^3. Then f is odd. Let's check if it's increasing on x < 0. The derivative is -3x², which is negative for all x ≠ 0. So actually, f(x) = -x³ is decreasing everywhere. Wait, but in the problem, they say f(x) is monotonically increasing when x < 0. So perhaps the function is increasing on x < 0, but since it's odd, on x > 0, does it have to be increasing or decreasing?Wait, take f(x) = x. Then it's odd and increasing everywhere. So in that case, both sides are increasing.Alternatively, suppose f(x) is increasing on x < 0. Let's pick x1 < x2 < 0, so f(x1) < f(x2). Then for x > 0, take two points, x1' = -x1, x2' = -x2. Since x1 < x2 < 0, then x1' > x2' > 0. Then f(x1') = -f(-x1') = -f(x1) and f(x2') = -f(x2). Since f(x1) < f(x2), multiplying by -1 reverses the inequality: -f(x1) > -f(x2). Therefore, f(x1') > f(x2'). So, when moving from x2' to x1' (which is increasing x), f(x) decreases. Therefore, f is decreasing on x > 0. Therefore, if f is odd and increasing on x < 0, it is decreasing on x > 0. That's an important point. So f is increasing on x < 0 and decreasing on x > 0. Moreover, f(0) = 0 because it's odd.Also, given that f(-1) = 0. Since f is odd, f(1) = -f(-1) = 0. So, f(-1) = 0, f(1) = 0. But f is increasing on x < 0, so as x approaches 0 from the left, f(x) increases towards f(0) = 0. Similarly, on x > 0, f(x) decreases from f(0) = 0 to f(1) = 0. Wait, but if f is decreasing on x > 0 and f(1) = 0, then for x > 1, does f(x) continue decreasing? But the problem states f is defined on R, so maybe it's decreasing on x > 0, but only that interval is considered here. Hmm, but the problem says "when x < 0, f(x) is monotonically increasing". So we know about x < 0, and being odd gives us the rest. But perhaps we can sketch the graph.Since f is odd, increasing on x < 0, so as x approaches 0 from the left, f(x) approaches 0. Then on x > 0, since it's odd, the function is decreasing. Also, f(-1) = 0, so f(1) = -f(-1) = 0. So f(-1) = 0, and since it's increasing on x < 0, for x < -1, f(x) is less than 0? Wait, when x approaches -infty, if f is increasing on x < 0, then as x decreases (goes to -infty), f(x) tends to -infty? Because if it's increasing, then as x becomes more negative, f(x) becomes more negative? Wait, no. Wait, if f is increasing on x < 0, then when x increases towards 0, f(x) increases. Therefore, as x approaches -infty, since it's increasing, f(x) tends to -infty. So, for example, f(x) could be like x, which is increasing everywhere, but for x < 0, f(x) is negative and increasing towards 0. But in the problem, f(-1) = 0. So, if f is increasing on x < 0 and f(-1) = 0, then for x < -1, f(x) < 0, and for -1 < x < 0, f(x) > 0. But since f is odd, for 0 < x < 1, f(x) = -f(-x). Since for -x in (-1,0), f(-x) is positive, so f(x) is negative. And at x = 1, f(1) = 0. For x > 1, since f is decreasing, f(x) continues to decrease from 0, so f(x) becomes negative again? Wait, no. If f is decreasing on x > 0, then starting from x = 0, f(0) = 0, then as x increases to 1, f(x) decreases from 0 to f(1) = 0. Wait, that can't be. If f(1) = 0, and f is decreasing on x > 0, then between x = 0 and x = 1, f(x) must be decreasing from 0 to 0? That would mean f(x) is constant zero on x > 0, which contradicts being decreasing. Therefore, perhaps my previous conclusion is wrong.Wait, maybe there's a mistake here. Let's re-examine. Since f is odd, f(1) = -f(-1) = 0. So f(1) = 0, f(-1) = 0. Now, for x < 0, f is monotonically increasing. So, let's consider x < -1: as x increases from -infty to -1, f(x) increases from -infty to f(-1) = 0. Then, between x = -1 and 0, since f is still increasing, but f(-1) = 0, so from x = -1 to x = 0, f(x) increases from 0 to f(0) = 0. Wait, that can't be. If f is increasing on x < 0, then moving from x = -infty to x = 0, f(x) increases from -infty to f(0) = 0. But f(-1) = 0. So, how does that work? If f(-1) = 0, and f is increasing on x < 0, then for x < -1, f(x) < f(-1) = 0. Then from x = -1 to x = 0, since f is increasing, f(x) increases from 0 to 0? That would mean f(x) is 0 on [-1, 0]. But that contradicts being increasing unless it's constant. But if it's constant on [-1, 0], then it's not strictly increasing. Hmm, maybe the problem says "monotonically increasing", which allows for non-strict. So, if f is monotonically increasing on x < 0, then f(x) can be constant on some intervals. However, f(-1) = 0. So, perhaps f(x) is 0 for x in [-1, 0]. But since f is odd, then for x in [0,1], f(x) = -f(-x) = -0 = 0. Then, for x > 1, since f is decreasing (from the previous logic), but f(1) = 0, so f(x) would have to be less than 0 for x > 1? But that seems inconsistent.Wait, perhaps there is a different way. Let's consider that f is increasing on x < 0, so for x < 0, if a < b, then f(a) ≤ f(b). Given that f(-1) = 0, then for x < -1, f(x) ≤ f(-1) = 0, and for -1 ≤ x < 0, f(x) ≥ f(-1) = 0. Since f is odd, for x > 0, f(x) = -f(-x). Therefore, for 0 < x < 1, since -x is between -1 and 0, f(-x) ≥ 0, so f(x) = -f(-x) ≤ 0. For x > 1, -x < -1, so f(-x) ≤ 0, hence f(x) = -f(-x) ≥ 0. Also, since f is decreasing on x > 0 (as we saw earlier), then as x increases beyond 1, f(x) decreases from 0 to ... Wait, but f(x) for x > 1 is non-negative, and if it's decreasing, then as x increases, f(x) decreases, approaching some limit. But since f is odd, if x approaches infinity, f(x) would have to mirror the behavior as x approaches -infty. If f(x) approaches -infty as x approaches -infty, then as x approaches +infty, f(x) approaches +infty? But that contradicts being decreasing on x > 0. Hmm, maybe my earlier analysis is missing something.Wait, perhaps f(x) is increasing on x < 0, but we don't have information about x > 0 except that it's odd. Wait, the problem states: "when x < 0, f(x) is monotonically increasing". It doesn't say anything about x > 0, other than f is odd. So perhaps on x > 0, f(x) is determined by being odd, but its monotonicity isn't specified? But from the oddness, we can infer something.Wait, let's get back. Maybe we don't need to know the exact behavior of f on x > 0 because in set N, we are dealing with f(φ(x)) < 0 for all x in [0, π/2]. Since φ(x) is sin²x + m·cosx - 2m, and we are considering x in [0, π/2], so cosx ranges from 0 to 1. Let's see, φ(x) is sin²x + m cosx - 2m. Let me analyze φ(x):First, sin²x can be written as 1 - cos²x, so φ(x) = 1 - cos²x + m cosx - 2m. Let's set t = cosx, where x ∈ [0, π/2], so t ∈ [0, 1]. Then φ(x) becomes 1 - t² + m t - 2m = -t² + m t + (1 - 2m). So φ is a quadratic in t: φ(t) = -t² + m t + (1 - 2m), t ∈ [0,1].Therefore, the problem reduces to analyzing this quadratic in t over [0,1]. For set M, we need φ(t) < 0 for all t ∈ [0,1]. For set N, we need f(φ(t)) < 0 for all t ∈ [0,1]. Since f is odd, and we need to figure out when f(φ(t)) < 0. Let's remember that f is odd, so f(-y) = -f(y). Also, we know that f is monotonically increasing on x < 0, with f(-1) = 0. Therefore, for x < 0, as x increases (towards 0), f(x) increases. Since f(-1) = 0, then for x < -1, f(x) < f(-1) = 0, and for -1 < x < 0, f(x) > f(-1) = 0. But since f is increasing on x < 0, f(x) < 0 for x < -1 and f(x) > 0 for -1 < x < 0.Given that, if we have f(φ(t)) < 0, then φ(t) must be in a region where f is negative. Since f is negative for x < -1 and positive for -1 < x < 0. Wait, but for x < -1, f(x) is negative, and for x between -1 and 0, f(x) is positive. Therefore, f(φ(t)) < 0 if and only if φ(t) < -1. Because if φ(t) is between -1 and 0, then f(φ(t)) is positive. If φ(t) < -1, then f(φ(t)) is negative. If φ(t) > 0, then since f is odd, f(φ(t)) = -f(-φ(t)), but we need to check the sign. Wait, if φ(t) > 0, then f(φ(t)) = -f(-φ(t)). Since φ(t) > 0, -φ(t) < 0. If -φ(t) < -1, then f(-φ(t)) < 0, so f(φ(t)) = -f(-φ(t)) > 0. If -1 < -φ(t) < 0, then f(-φ(t)) > 0, so f(φ(t)) = -f(-φ(t)) < 0. Wait, this is getting confusing.Wait, let's think step by step. To have f(φ(t)) < 0, what does that imply? Since f is odd:If φ(t) > 0, then f(φ(t)) = -f(-φ(t)). For φ(t) > 0, -φ(t) < 0. Then, if -φ(t) is in (-1, 0), i.e., φ(t) in (0,1), then f(-φ(t)) is positive (since f is increasing on x < 0 and f(-1) = 0, so for x between -1 and 0, f(x) is positive). Therefore, f(φ(t)) = -f(-φ(t)) would be negative. If -φ(t) < -1, i.e., φ(t) > 1, then f(-φ(t)) is negative (since for x < -1, f(x) is increasing and f(-1)=0, so x < -1 implies f(x) < 0). Therefore, f(φ(t)) = -f(-φ(t)) would be positive. Therefore, if φ(t) > 1, f(φ(t)) is positive. If φ(t) is between 0 and 1, f(φ(t)) is negative. If φ(t) is between -1 and 0, f(φ(t)) is positive (since f is positive there). If φ(t) < -1, f(φ(t)) is negative.Therefore, f(φ(t)) < 0 in two cases: φ(t) ∈ (0,1) or φ(t) < -1.But set N is defined as all m such that for all x ∈ [0, π/2], f(φ(x)) < 0. Therefore, for all t ∈ [0,1], φ(t) must be either in (0,1) or in (-∞, -1). But since φ(t) is a continuous function of t, and t ranges from 0 to 1, unless φ(t) is entirely in (0,1) or entirely in (-∞, -1), but if φ(t) crosses between these regions, then f(φ(t)) would not be always negative.Wait, but can φ(t) be in both (0,1) and (-∞, -1) over the interval [0,1]? It's a quadratic function, so it can have at most two roots. Let's consider the quadratic φ(t) = -t² + m t + (1 - 2m). Let's analyze its maximum and minimum. Since the coefficient of t² is -1, the parabola opens downward. Therefore, it has a maximum at vertex. The vertex is at t = -b/(2a) = -m/(2*(-1)) = m/2. So the maximum occurs at t = m/2.Since t ∈ [0,1], the maximum is either at t = m/2 if m/2 ∈ [0,1], or at t=0 or t=1 otherwise. So, if m/2 is within [0,1], i.e., m ∈ [0,2], then the maximum is at t = m/2. Otherwise, for m < 0, maximum at t=0, and for m > 2, maximum at t=1.Therefore, to ensure φ(t) < 0 for all t ∈ [0,1] (which is set M), we need the maximum of φ(t) to be less than 0. Since the parabola opens downward, the maximum is at the vertex. So, if m ∈ [0,2], then maximum at t = m/2. Let's compute φ(m/2):φ(m/2) = -( (m/2)^2 ) + m*(m/2) + (1 - 2m) = -m²/4 + m²/2 + 1 - 2m = ( -m²/4 + 2m²/4 ) + 1 - 2m = (m²/4) + 1 - 2m.Set this maximum to be less than 0:m²/4 + 1 - 2m < 0Multiply both sides by 4:m² + 4 - 8m < 0m² -8m +4 <0Solve quadratic inequality:Discriminant: 64 - 16 = 48Roots: [8 ± sqrt(48)] / 2 = [8 ± 4*sqrt(3)] / 2 = 4 ± 2*sqrt(3)Approximate values: sqrt(3) ≈ 1.732, so 2*sqrt(3) ≈ 3.464. Therefore, roots at 4 - 3.464 ≈ 0.536 and 4 + 3.464 ≈ 7.464. So the quadratic is less than 0 between 0.536 and 7.464. But since m ∈ [0,2] in this case, the inequality m² -8m +4 <0 holds for m ∈ (4 - 2*sqrt(3), 4 + 2*sqrt(3)) ≈ (0.536, 7.464). But since m is in [0,2], the valid interval here is m ∈ (0.536, 2). Therefore, if m ∈ (0.536, 2), then the maximum at t = m/2 is negative, so φ(t) < 0 for all t ∈ [0,1].If m is not in [0,2], then the maximum is at t=0 or t=1.For m < 0, maximum at t=0. φ(0) = -0 + 0 +1 -2m = 1 -2m. Since m <0, 1 -2m >1. So φ(0) >1, which is not less than 0. Therefore, for m <0, φ(t) is not always negative.For m >2, maximum at t=1. φ(1) = -1 + m*1 +1 -2m = -1 + m +1 -2m = -m. For m >2, -m < -2 <0. So φ(1) = -m <0. However, since the parabola opens downward, with maximum at t=1, but φ(1) is negative. Then we need to check the minimum at t=0. Wait, but the parabola opens downward, so the minimum would be at the ends. Wait, actually, if the maximum is at t=1 and φ(1) is negative, then the entire parabola is below zero? Wait, no. If the parabola opens downward, and the maximum is at t=1, but φ(1) is negative, then the entire parabola is negative. So, if m >2, then φ(t) <0 for all t ∈ [0,1]. Let me check φ(0) when m >2:φ(0) = 1 -2m. If m >2, 1 -2m <1 -4 = -3 <0. So both φ(0) and φ(1) are negative, and since the parabola opens downward, the entire function is below zero. Therefore, for m >2, φ(t) <0 for all t ∈ [0,1].Wait, but in the previous analysis, when m ∈ (4 -2√3, 2), which is approximately (0.536,2), φ(t) is negative for all t. And for m >2, φ(t) is also negative for all t. So set M is m ∈ (4 -2√3, ∞). Wait, but wait, when m approaches infinity, φ(t) = -t² + m t -2m +1. For large m, φ(t) ≈ m(t - 2). Since t ∈ [0,1], t -2 ∈ [-2, -1], so m(t -2) is large negative if m is positive. Therefore, φ(t) ≈ -t² + m(t - 2) +1. For m large positive, φ(t) ≈ m(t - 2), which is negative for all t ∈ [0,1], since t -2 is negative. So yes, as m increases beyond 2, φ(t) becomes more negative. Therefore, set M is m > 4 -2√3. Wait, but 4 -2√3 ≈0.536, so set M is m > 4 -2√3.Wait, but we need to check m >2 as well. Let me confirm:For m >2:φ(t) = -t² + m t +1 -2m. Let's check at t=0: φ(0) =1 -2m <1 -4= -3 <0. At t=1: φ(1)= -1 + m +1 -2m = -m <0. The vertex is at t = m/2 >1, which is outside the interval [0,1], so the maximum in [0,1] is at t=1, which is -m <0. Therefore, φ(t) is decreasing on [0,1] (since the vertex is at t=m/2 >1), so φ(t) is decreasing from φ(0)=1-2m to φ(1)=-m. Both endpoints are negative, and since it's decreasing, the entire function is negative. Therefore, for m >2, φ(t) <0 for all t ∈ [0,1].For m =2:φ(t)= -t² +2t +1 -4= -t² +2t -3. The discriminant is 4 -12= -8 <0, so it's always negative? Let's check φ(1)= -1 +2 -3= -2 <0, φ(0)= -3 <0. The vertex is at t=1, which is the maximum. So yes, for m=2, φ(t) <0.Therefore, set M is m >4 -2√3 (≈0.536). Therefore, M = (4 -2√3, ∞).Wait, but when m is exactly 4 -2√3, what happens? Let's compute φ(t) at m=4 -2√3. At the vertex t=m/2=(4 -2√3)/2=2 -√3≈2 -1.732≈0.268, which is in [0,1]. Then φ(m/2)=0 as that's the boundary case. So, for m=4 -2√3, the maximum of φ(t) is 0. Therefore, φ(t) ≤0 for all t ∈ [0,1], but since the question requires φ(x) <0, strict inequality, so m=4 -2√3 is excluded. Therefore, set M is m >4 -2√3.Now, moving to set N: N is the set of m such that for all x ∈ [0, π/2], f(φ(x)) <0. As established earlier, f(φ(x)) <0 when φ(x) ∈ (0,1) or φ(x) < -1. Therefore, for f(φ(x)) <0 for all x ∈ [0, π/2], we need that for all t ∈ [0,1], φ(t) ∈ (0,1) ∪ (-∞, -1). But φ(t) is a continuous function. If φ(t) attains values in both (0,1) and (-∞, -1), then there must be some t where φ(t) is in between, which would violate the condition. Therefore, φ(t) must lie entirely in (0,1) or entirely in (-∞, -1) for all t ∈ [0,1].But φ(t) cannot lie entirely in (-∞, -1) because φ(t) is a quadratic function. Let's see: if φ(t) < -1 for all t ∈ [0,1], then that would imply φ(t) +1 <0, so -t² +m t +1 -2m +1 <0 → -t² +m t +2 -2m <0. But this is another quadratic. Maybe it's complicated. Alternatively, since φ(t) < -1 for all t ∈ [0,1], which is a stricter condition than φ(t) <0, so such m would be a subset of M. However, we need to see when φ(t) is either in (0,1) or in (-∞, -1).But actually, since φ(t) is continuous, to have φ(t) in (0,1) for all t or in (-∞, -1) for all t. Let's check:Case 1: φ(t) ∈ (0,1) for all t ∈ [0,1].Case 2: φ(t) ∈ (-∞, -1) for all t ∈ [0,1].But φ(t) is a quadratic function. For φ(t) to be in (0,1) for all t, we need 0 < φ(t) <1 for all t. Similarly, for φ(t) < -1 for all t.Let's analyze Case 2 first: φ(t) < -1 for all t ∈ [0,1]. Then, φ(t) = -t² +m t +1 -2m < -1. Rearranged: -t² +m t +1 -2m +1 <0 → -t² +m t +2 -2m <0. Let's call this ψ(t) = -t² +m t +2 -2m. For ψ(t) <0 for all t ∈ [0,1]. The maximum of ψ(t) occurs at t = m/2. If m/2 ∈ [0,1], then maximum at t=m/2. Otherwise, at t=0 or t=1.Compute ψ(m/2) when m ∈ [0,2]:ψ(m/2) = - (m/2)^2 + m*(m/2) +2 -2m = -m²/4 + m²/2 +2 -2m = m²/4 +2 -2m. Set this <0:m²/4 +2 -2m <0 → m² -8m +8 <0. Solve m² -8m +8 =0:Discriminant: 64 -32=32. Roots: (8 ± √32)/2= (8 ±4√2)/2=4 ±2√2≈4±2.828≈6.828 and 1.172. Therefore, inequality m² -8m +8 <0 holds for m ∈(4 -2√2,4 +2√2). But since m ∈[0,2] in this case, the overlapping interval is (4 -2√2,2). 4 -2√2≈4 -2.828≈1.172. Therefore, for m ∈(1.172,2), ψ(m/2) <0. But ψ(t) also needs to be less than 0 at t=0 and t=1. ψ(0)=0 +0 +2 -2m=2 -2m. For ψ(0) <0 →2 -2m <0 →m >1. Similarly, ψ(1)= -1 +m +2 -2m=1 -m <0 →m >1. Therefore, combining these, for m >1, ψ(0) and ψ(1) are negative. If m ∈(4 -2√2,2), approximately (1.172,2), then the maximum ψ(m/2) is negative, so ψ(t) <0 for all t. Therefore, φ(t) < -1 for all t ∈[0,1] when m ∈(4 -2√2,2).Wait, but 4 -2√2≈1.172, which is greater than 1. Therefore, when m >1.172, φ(t) < -1 for all t ∈[0,1].For m >2, as before, φ(t) = -t² +m t +1 -2m. At m=2, φ(t)= -t² +2t -3. Let's check φ(t) at m=3: φ(t)= -t² +3t +1 -6= -t² +3t -5. This is a downward parabola. At t=0: -5 < -1; at t=1: -1 +3 -5= -3 < -1. Vertex at t=3/2=1.5, which is outside [0,1]. Therefore, the maximum in [0,1] is at t=1, which is -3 < -1. So φ(t) is decreasing on [0,1], starting at -5 and going to -3, both less than -1. So for m >2, φ(t) < -1 for all t ∈ [0,1]. Wait, but φ(t)= -t² +m t +1 -2m. Let's check at t=0: φ(0)=1 -2m. For m >2, 1 -2m <1 -4= -3 < -1. At t=1: φ(1)= -1 +m +1 -2m= -m. For m >2, -m < -2 < -1. Since the parabola opens downward, the maximum in [0,1] is at t=1 (since vertex at m/2 >1), so φ(t) is decreasing from φ(0)=1 -2m to φ(1)= -m, both less than -1. Therefore, for m >2, φ(t) < -1 for all t ∈[0,1].Therefore, combining the intervals, for φ(t) < -1 for all t ∈[0,1], m must be in (4 -2√2, ∞). But wait, when m >2, φ(t) < -1, as shown. When m ∈(4 -2√2,2), which is approximately (1.172,2), φ(t) < -1 for all t. Therefore, the set where φ(t) < -1 is m ∈ (4 -2√2, ∞). Therefore, set N is m ∈ (4 -2√2, ∞). But wait, we have to also consider Case 1 where φ(t) ∈ (0,1) for all t ∈ [0,1]. Let's check if that's possible.Case 1: φ(t) ∈ (0,1) for all t ∈ [0,1]. For this, we need 0 < φ(t) <1 for all t. So, φ(t) >0 and φ(t) <1.First, φ(t) >0 for all t ∈ [0,1]. φ(t) is a downward parabola, so it's positive everywhere only if the minimum is positive. But since it's a downward parabola, it has a maximum, so the minimum occurs at the endpoints. Wait, no. For a downward parabola, the minimums are at the endpoints. So, to have φ(t) >0 for all t ∈ [0,1], we need φ(0) >0 and φ(1) >0, and the minimum in between is positive. Wait, no, since it's a downward parabola, the minimum would actually be at the endpoints. Wait, for a downward parabola, the maximum is at the vertex, and the minima are at the endpoints. So, if the minimum at both endpoints is positive, then the entire function is positive. Therefore, to have φ(t) >0 for all t ∈ [0,1], we need φ(0) >0 and φ(1) >0.φ(0) =1 -2m >0 → m <0.5φ(1)= -1 +m +1 -2m = -m >0 → m <0Therefore, φ(t) >0 for all t ∈ [0,1] only if m <0. But we also need φ(t) <1 for all t ∈ [0,1].But φ(t) <1 for all t ∈ [0,1]. Since φ(t) is a downward parabola, its maximum is at the vertex. So, compute the maximum of φ(t):If m ∈ [0,2], maximum at t=m/2. φ(m/2)=m²/4 +1 -2m. Set this <1:m²/4 +1 -2m <1 →m²/4 -2m <0 →m² -8m <0 →m(m -8) <0. Therefore, m ∈(0,8). But since m ∈[0,2], this holds for m ∈(0,2). But we also need φ(t) >0, which requires m <0. But m ∈(0,2) contradicts m <0. Therefore, there is no overlap. Hence, Case 1 where φ(t) ∈ (0,1) for all t is only possible if m <0, but φ(t) <1 is automatically satisfied if φ(t) >0, but since m <0, φ(0)=1 -2m >1, which contradicts φ(t) <1. Therefore, there is no m for which φ(t) ∈ (0,1) for all t ∈ [0,1]. Therefore, set N corresponds to m ∈ (4 -2√2, ∞).Wait, but earlier we saw that when φ(t) is between 0 and1, f(φ(t)) is negative. However, for φ(t) to be between0 and1 for all t, we need m <0.5 and m <0, which is m <0, but φ(0)=1 -2m >1 when m <0, which contradicts φ(t) <1. Therefore, there's no m where φ(t) ∈ (0,1) for all t ∈ [0,1]. Therefore, set N must correspond to φ(t) < -1 for all t ∈ [0,1], which is m ∈ (4 -2√2, ∞). Therefore, set N is (4 -2√2, ∞). But wait, earlier analysis showed that for m ∈ (4 -2√2,2), φ(t) < -1, and for m >2, φ(t) < -1. Therefore, N is m >4 -2√2.Now, set M is m >4 -2√3≈0.536, and set N is m >4 -2√2≈1.172. Therefore, the intersection M ∩ N is m >4 -2√2, since 4 -2√2 >4 -2√3. Therefore, M ∩ N = N = (4 -2√2, ∞). But we need to confirm.Wait, 4 -2√3≈4-3.464≈0.536; 4 -2√2≈4-2.828≈1.172. So yes, 4 -2√2 is greater than4 -2√3. Therefore, M is (0.536, ∞), N is (1.172, ∞), so their intersection is (1.172, ∞), which is N. But let me verify.But wait, maybe I made a mistake in defining set N. Set N is the set of m such that for all x ∈ [0, π/2], f(φ(x)) <0. As per previous analysis, f(φ(x)) <0 requires that φ(x) ∈ (0,1) or φ(x) < -1. But since there are no m for which φ(x) ∈ (0,1) for all x, set N is the set of m for which φ(x) < -1 for all x. Therefore, N = (4 -2√2, ∞). But M is the set where φ(x) <0 for all x, which is (4 -2√3, ∞). So the intersection is the overlapping part, which is (4 -2√2, ∞), since 4 -2√2 >4 -2√3. Hence, M ∩ N = (4 -2√2, ∞).But to confirm, let's check with exact values.First, compute 4 -2√3 and 4 -2√2:√3≈1.732, so 2√3≈3.464; 4 -3.464≈0.536.√2≈1.414, so 2√2≈2.828; 4 -2.828≈1.172.Therefore, M is (0.536, ∞), N is (1.172, ∞). Therefore, their intersection is (1.172, ∞), which is N.But the problem asks for M ∩ N, which would be the values of m that are in both sets. Since N is a subset of M (as 1.172 >0.536), then M ∩ N = N = (4 -2√2, ∞).But wait, is N indeed a subset of M? For m ∈ N, which requires φ(x) < -1 <0, so yes, φ(x) <0, hence m ∈ M. Therefore, N ⊂ M, so M ∩ N = N.Therefore, the answer is (4 -2√2, ∞). But the problem might expect an exact form, so in terms of radicals.But let me re-express 4 -2√2 as 2*(2 -√2). However, in the problem statement, the answer should be presented in boxed notation. Therefore, the final answer is boxed{(4 - 2sqrt{2}, +infty)}.Final Answerboxed{(4 - 2sqrt{2}, +infty)}</think>Given that ( f(x) ) is an odd function defined on ( mathbb{R} ), and when ( x < 0 ), ( f(x) ) is monotonically increasing with ( f(-1) = 0 ). We need to find the intersection of sets ( M ) and ( N ) where:- ( M = left{ m mid forall x in left[ 0, frac{pi}{2} right], varphi(x) < 0 right} )- ( N = left{ m mid forall x in left[ 0, frac{pi}{2} right], f(varphi(x)) < 0 right} )The function ( varphi(x) = sin^2 x + m cos x - 2m ) can be rewritten in terms of ( t = cos x ) (where ( t in [0, 1] )) as ( varphi(t) = -t^2 + mt + (1 - 2m) ). Analysis for Set ( M )To ensure ( varphi(t) < 0 ) for all ( t in [0, 1] ):- The quadratic ( varphi(t) ) opens downward (coefficient of ( t^2 ) is -1).- The maximum value of ( varphi(t) ) must be less than 0. The vertex of the parabola is at ( t = frac{m}{2} ).Solving for ( varphi(t) < 0 ):- For ( m in (4 - 2sqrt{3}, infty) ), the quadratic ( varphi(t) ) is negative for all ( t in [0, 1] ). Analysis for Set ( N )To ensure ( f(varphi(x)) < 0 ) for all ( x in [0, frac{pi}{2}] ):- ( f ) is odd, increasing on ( x < 0 ), and ( f(-1) = 0 ).- ( f(varphi(x)) < 0 ) if ( varphi(x) < -1 ) (since ( f ) is negative for ( x < -1 ) and positive for ( -1 < x < 0 )).Solving for ( varphi(t) < -1 ):- The quadratic ( varphi(t) ) must be less than -1 for all ( t in [0, 1] ).- For ( m in (4 - 2sqrt{2}, infty) ), the quadratic ( varphi(t) ) is less than -1 for all ( t in [0, 1] ). Intersection ( M cap N )- Set ( M ) is ( (4 - 2sqrt{3}, infty) )- Set ( N ) is ( (4 - 2sqrt{2}, infty) )- The intersection ( M cap N ) is ( (4 - 2sqrt{2}, infty) ) since ( 4 - 2sqrt{2} > 4 - 2sqrt{3} ).[boxed{(4 - 2sqrt{2}, +infty)}],<think>Okay, let's try to tackle this problem. It seems a bit involved, but I'll take it step by step.First, let's parse the problem statement again to make sure I understand all parts.We have an odd function ( f(x) ) defined on all real numbers. When ( x < 0 ), ( f(x) ) is monotonically increasing, and we know that ( f(-1) = 0 ). Then there's this function ( varphi(x) = sin^2 x + m cdot cos x - 2m ). Sets ( M ) and ( N ) are defined as:- ( M = { m mid forall x in [0, frac{pi}{2}], varphi(x) < 0 } )- ( N = { m mid forall x in [0, frac{pi}{2}], f(varphi(x)) < 0 } )We need to find ( M cap N ), which is the set of all ( m ) that satisfy both conditions.Alright, let's break this down.Understanding Set M:First, ( M ) is the set of all ( m ) such that for every ( x ) in ( [0, pi/2] ), ( varphi(x) < 0 ). So we need to find all ( m ) for which the expression ( sin^2 x + m cos x - 2m ) is negative throughout the interval ( [0, pi/2] ).Understanding Set N:Set ( N ) is a bit trickier. It's the set of all ( m ) such that for every ( x in [0, pi/2] ), ( f(varphi(x)) < 0 ). Since ( f ) is an odd function, we know that ( f(-y) = -f(y) ). Also, ( f ) is monotonically increasing when ( x < 0 ), and ( f(-1) = 0 ). Let's recall properties of odd functions:- ( f(0) = 0 ) because ( f(-0) = -f(0) ) implies ( f(0) = 0 ).- Since ( f ) is increasing on ( x < 0 ), then for ( x < 0 ), as ( x ) increases (approaches 0), ( f(x) ) increases. Since ( f(-1) = 0 ) and ( f ) is increasing on ( x < 0 ), for ( x < -1 ), ( f(x) ) would be less than 0, and for ( -1 < x < 0 ), ( f(x) ) would be greater than 0. Wait, let me verify that.Wait, if ( f ) is monotonically increasing on ( x < 0 ), then as ( x ) increases towards 0 from the left, ( f(x) ) increases. Given ( f(-1) = 0 ), then for ( x < -1 ), since ( x ) is more negative, and ( f ) is increasing, ( f(x) ) would be less than ( f(-1) = 0 ). For ( -1 < x < 0 ), ( x ) is greater than -1 but still negative, so ( f(x) ) would be greater than ( f(-1) = 0 ). Therefore:- If ( x < -1 ), ( f(x) < 0 )- If ( -1 < x < 0 ), ( f(x) > 0 )- At ( x = -1 ), ( f(-1) = 0 )- At ( x = 0 ), ( f(0) = 0 )Therefore, ( f(y) < 0 ) if and only if ( y < -1 ), because for ( y < -1 ), ( f(y) < 0 ), and for ( -1 < y < 0 ), ( f(y) > 0 ). At ( y = -1 ), ( f(y) = 0 ), and at ( y = 0 ), ( f(y) = 0 ).Wait, but since ( f ) is odd, ( f(-y) = -f(y) ). So if ( y > 0 ), then ( f(y) = -f(-y) ). But we don't have information about ( f ) when ( x > 0 ). However, since ( f ) is odd, once we know its behavior on ( x < 0 ), we can determine it on ( x > 0 ). Let me confirm.Since ( f ) is odd and increasing on ( x < 0 ), then on ( x > 0 ), ( f ) must be increasing as well? Wait, not necessarily. Let's think:If ( f ) is odd, then ( f'(x) = f'(-x) ) if it's differentiable. Wait, actually, the derivative of an odd function is even. So if ( f ) is increasing on ( x < 0 ), then its derivative is positive there, so the derivative on ( x > 0 ) would also be positive, meaning ( f ) is increasing on ( x > 0 ) as well. Hmm, but we weren't told that ( f ) is differentiable. Hmm.But the problem says only that when ( x < 0 ), ( f(x) ) is monotonically increasing. For ( x > 0 ), since it's odd, we can say that ( f(x) = -f(-x) ). So if ( x > 0 ), then ( -x < 0 ), and since ( f(-x) ) is increasing as ( -x ) increases (i.e., as ( x ) decreases), then ( f(x) = -f(-x) ). Therefore, if ( f(-x) ) is increasing as ( -x ) increases, then ( f(-x) ) is decreasing as ( x ) increases. Therefore, ( f(x) = -f(-x) ) would be increasing as ( x ) increases, because the negative of a decreasing function is increasing.Therefore, ( f(x) ) is increasing on ( x < 0 ) and increasing on ( x > 0 ). But at ( x = 0 ), since ( f(0) = 0 ).But perhaps we don't need the behavior on ( x > 0 ). Let's focus back on the problem. For ( N ), we have ( f(varphi(x)) < 0 ) for all ( x in [0, pi/2] ).Given that ( f(y) < 0 ) when ( y < -1 ), and ( f(y) > 0 ) when ( -1 < y < 0 ), and ( f(-1) = 0 ), ( f(0) = 0 ).Therefore, ( f(varphi(x)) < 0 ) if and only if ( varphi(x) < -1 ). Because when ( varphi(x) < -1 ), then ( f(varphi(x)) < 0 ). If ( -1 < varphi(x) < 0 ), then ( f(varphi(x)) > 0 ). If ( varphi(x) = -1 ), then ( f(varphi(x)) = 0 ). Similarly, if ( varphi(x) geq 0 ), since ( f(y) ) is odd, but we don't know its behavior for ( y > 0 ), but in the interval ( [0, pi/2] ), ( varphi(x) ) could potentially be positive. Wait, but in set ( M ), we require ( varphi(x) < 0 ). So maybe in set ( N ), even if ( varphi(x) ) is positive, we have to consider ( f(varphi(x)) ). Hmm.Wait, but ( M ) and ( N ) are separate sets. So ( N ) is defined as all ( m ) such that for all ( x in [0, pi/2] ), ( f(varphi(x)) < 0 ). So regardless of whether ( varphi(x) ) is positive or negative, as long as ( f(varphi(x)) < 0 ). But since ( f ) is odd, if ( varphi(x) > 0 ), then ( f(varphi(x)) = -f(-varphi(x)) ). But we don't know how ( f ) behaves on positive numbers except that it's odd. However, since ( f ) is increasing on ( x < 0 ), and given that it's odd, for ( x > 0 ), ( f ) would be increasing as well. Let's verify:Suppose ( x_1 > x_2 > 0 ). Then ( -x_1 < -x_2 < 0 ). Since ( f ) is increasing on ( x < 0 ), ( f(-x_1) < f(-x_2) ). Therefore, ( f(x_1) = -f(-x_1) ) and ( f(x_2) = -f(-x_2) ). Since ( f(-x_1) < f(-x_2) ), multiplying by -1 reverses the inequality: ( -f(-x_1) > -f(-x_2) ), so ( f(x_1) > f(x_2) ). Therefore, ( f ) is increasing on ( x > 0 ).Therefore, ( f ) is increasing on the entire real line except possibly at 0? Wait, but at 0, ( f(0) = 0 ). So, if ( f ) is increasing on ( x < 0 ) and increasing on ( x > 0 ), but with a jump at 0? For example, if ( x ) approaches 0 from the left, ( f(x) ) approaches 0 from below, and from the right, approaches 0 from above? Wait, but since ( f ) is odd, if it's increasing on both sides, then yes. For instance, ( f(x) = x ) is odd, increasing everywhere, and ( f(0) = 0 ).But in our case, the given information is that ( f(x) ) is increasing on ( x < 0 ), and ( f(-1) = 0 ). Since ( f ) is odd, then ( f(1) = -f(-1) = 0 ). So at ( x = 1 ), ( f(1) = 0 ). But ( since f is increasing on ( x > 0 ), then for ( x > 1 ), ( f(x) > 0 ), and for ( 0 < x < 1 ), ( f(x) < 0 )? Wait, no. Wait, if ( f ) is increasing on ( x > 0 ), then as ( x ) increases, ( f(x) ) increases. Since ( f(1) = 0 ), then for ( x > 1 ), ( f(x) > 0 ), and for ( 0 < x < 1 ), since it's increasing, ( f(x) < f(1) = 0 ). Therefore, on ( x > 0 ):- ( 0 < x < 1 ): ( f(x) < 0 )- ( x = 1 ): ( f(x) = 0 )- ( x > 1 ): ( f(x) > 0 )Similarly, on ( x < 0 ):- ( x < -1 ): ( f(x) < 0 )- ( x = -1 ): ( f(x) = 0 )- ( -1 < x < 0 ): ( f(x) > 0 )And at ( x = 0 ), ( f(0) = 0 ).Therefore, compiling this:- For ( x < -1 ), ( f(x) < 0 )- For ( -1 < x < 0 ), ( f(x) > 0 )- For ( 0 < x < 1 ), ( f(x) < 0 )- For ( x > 1 ), ( f(x) > 0 )- At ( x = -1, 0, 1 ), ( f(x) = 0 )Therefore, ( f(y) < 0 ) when ( y in (-infty, -1) cup (0, 1) ), and ( f(y) > 0 ) when ( y in (-1, 0) cup (1, infty) ), with zeros at ( y = -1, 0, 1 ).But in our problem, ( varphi(x) ) is a function defined on ( [0, pi/2] ). So for ( x in [0, pi/2] ), ( varphi(x) = sin^2 x + m cos x - 2m ). Let's analyze the range of ( varphi(x) ).First, let's understand ( varphi(x) ):( varphi(x) = sin^2 x + m cos x - 2m )We can rewrite ( sin^2 x ) as ( 1 - cos^2 x ), so:( varphi(x) = 1 - cos^2 x + m cos x - 2m )Simplify:( varphi(x) = -cos^2 x + m cos x + (1 - 2m) )That's a quadratic in ( cos x ). Let's denote ( t = cos x ). Since ( x in [0, pi/2] ), ( t in [0, 1] ).Therefore, ( varphi(x) = -t^2 + m t + (1 - 2m) ), where ( t in [0, 1] ).So, ( varphi ) as a function of ( t ) is ( varphi(t) = -t^2 + m t + (1 - 2m) ).Therefore, analyzing ( varphi(t) ) over ( t in [0, 1] ).This seems useful. Let's proceed.Analyzing Set M:Set ( M ) requires ( varphi(x) < 0 ) for all ( x in [0, pi/2] ). Which is equivalent to ( varphi(t) < 0 ) for all ( t in [0, 1] ).So, we need ( -t^2 + m t + (1 - 2m) < 0 ) for all ( t in [0, 1] ).Let's rearrange the inequality:( -t^2 + m t + 1 - 2m < 0 )Multiply both sides by -1 (remember to reverse inequality):( t^2 - m t - 1 + 2m > 0 )But maybe it's better to keep it as is and analyze:( -t^2 + m t + 1 - 2m < 0 )Which is:( -t^2 + m t + (1 - 2m) < 0 )Let's think of this quadratic in terms of ( t ). Let's denote ( varphi(t) = -t^2 + m t + (1 - 2m) ).We need ( varphi(t) < 0 ) for all ( t in [0, 1] ).Since it's a quadratic function in ( t ), opening downward (coefficient of ( t^2 ) is -1). Therefore, the maximum occurs at the vertex.The vertex of ( varphi(t) ) is at ( t = -b/(2a) ). Here, ( a = -1 ), ( b = m ), so vertex at ( t = -m/(2*(-1)) = m/2 ).So the maximum of ( varphi(t) ) is at ( t = m/2 ).But since ( t in [0, 1] ), the maximum could be at ( t = m/2 ) if ( m/2 in [0, 1] ), i.e., ( m in [0, 2] ). Otherwise, the maximum is at the endpoint.So, to ensure ( varphi(t) < 0 ) for all ( t in [0, 1] ), the maximum of ( varphi(t) ) in [0,1] must be less than 0.Therefore, we need to check the maximum of ( varphi(t) ) over [0,1] and set it less than 0.Case 1: If the vertex ( t = m/2 ) is in [0,1], i.e., ( 0 leq m leq 2 ), then the maximum is at ( t = m/2 ).Compute ( varphi(m/2) ):( varphi(m/2) = - (m/2)^2 + m*(m/2) + (1 - 2m) )Simplify:( -m²/4 + m²/2 + 1 - 2m )Combine terms:( (-m²/4 + 2m²/4) + 1 - 2m = (m²/4) + 1 - 2m )Set this maximum < 0:( m²/4 + 1 - 2m < 0 )Multiply both sides by 4:( m² + 4 - 8m < 0 )( m² - 8m + 4 < 0 )Solve the quadratic inequality:The quadratic equation ( m² - 8m + 4 = 0 ).Discriminant: ( 64 - 16 = 48 )Roots: ( [8 ± √48]/2 = [8 ± 4√3]/2 = 4 ± 2√3 ).So, ( 4 - 2√3 ≈ 4 - 3.464 ≈ 0.536 )and ( 4 + 2√3 ≈ 7.464 )Therefore, the inequality ( m² - 8m + 4 < 0 ) holds for ( m in (4 - 2√3, 4 + 2√3) ).But in this case, we're considering ( m in [0, 2] ).Therefore, the intersection is ( m in (4 - 2√3, 2] ).But 4 - 2√3 ≈ 0.536, so (0.536, 2]. However, this is only when the maximum is at the vertex. So in this case, if m is between 0 and 2, the maximum is at t = m/2. So we need that maximum <0. Therefore, m must be in (4 - 2√3, 2]. But 4 - 2√3 ≈ 0.536, so m must be greater than approximately 0.536 and less than or equal to 2.Case 2: If the vertex is outside [0,1], i.e., m < 0 or m > 2, then the maximum is at the endpoints.So, let's check the endpoints:At t = 0:( varphi(0) = -0 + 0 + (1 - 2m) = 1 - 2m )At t = 1:( varphi(1) = -1 + m*1 + (1 - 2m) = -1 + m + 1 - 2m = -m )So, if m < 0 or m > 2, the maximum of φ(t) over [0,1] is the maximum of φ(0) = 1 - 2m and φ(1) = -m.We need both φ(0) < 0 and φ(1) < 0.So:1. For m < 0:- φ(0) = 1 - 2m. If m < 0, then 1 - 2m > 1 (since m is negative, subtracting a negative is adding). So φ(0) > 1, which is positive. Therefore, φ(0) < 0 is not satisfied. Therefore, no solutions for m < 0.2. For m > 2:- φ(0) = 1 - 2m. If m > 2, then 1 - 2m < 1 - 4 = -3 < 0.- φ(1) = -m. If m > 2, then -m < -2 < 0.Therefore, both φ(0) and φ(1) are negative. However, since the vertex is at t = m/2 > 1 (since m > 2), the quadratic is decreasing on [0,1] (since it opens downward, and vertex is to the right of 1). Therefore, φ(t) is decreasing from φ(0) = 1 - 2m to φ(1) = -m. Since both endpoints are negative, and the function is decreasing, the entire function is negative on [0,1]. Therefore, for m > 2, φ(t) < 0 for all t ∈ [0,1].However, in Case 1, when m ∈ (4 - 2√3, 2], the maximum at the vertex is negative. Therefore, combining both cases:Set M is m > 4 - 2√3. Because:- For m > 2, φ(t) < 0 for all t.- For 4 - 2√3 < m ≤ 2, φ(t) < 0 for all t.- For m ≤ 4 - 2√3, the maximum at the vertex would be ≥0, so φ(t) is not always negative.But wait, 4 - 2√3 is approximately 0.536, so in Case 1, m between 0 and 2, the maximum is at t = m/2, and we need m in (4 - 2√3, 2]. So for m between approximately 0.536 and 2, the maximum is negative, and for m > 2, all values are negative. Therefore, set M is m > 4 - 2√3.But let's confirm with the endpoints:Wait, when m = 4 - 2√3, the maximum at the vertex is exactly 0. Therefore, M is m > 4 - 2√3.But the problem says M is the set where φ(x) < 0 for all x in [0, π/2]. So we must have φ(t) < 0 for all t ∈ [0,1]. Therefore, M is m > 4 - 2√3. Because for m = 4 - 2√3, the maximum is 0, which is not less than 0, so m must be strictly greater.Wait, but let's check when m = 4 - 2√3:The maximum at the vertex is 0, so φ(t) = 0 at t = m/2 = (4 - 2√3)/2 = 2 - √3 ≈ 2 - 1.732 ≈ 0.267. So at t ≈ 0.267, φ(t) = 0. Therefore, φ(x) is not strictly less than 0 everywhere, hence m = 4 - 2√3 is excluded. Therefore, M is m > 4 - 2√3.Similarly, for m > 2, we already have φ(t) < 0 for all t. Therefore, set M is ( m > 4 - 2sqrt{3} ).But wait, 4 - 2√3 is approximately 0.536, so M is (4 - 2√3, ∞). But wait, when m approaches infinity, φ(t) = -t² + m t + 1 - 2m. As m becomes very large, the dominant terms are m t - 2m = m(t - 2). Since t ∈ [0,1], t - 2 is negative, so for large m, φ(t) ≈ m(t - 2) which is negative. So for m approaching infinity, φ(t) is negative. Therefore, M is m > 4 - 2√3. So M is the interval ( (4 - 2sqrt{3}, +infty) ).Wait, but when m = 3, for example, φ(t) = -t² + 3t + 1 - 6 = -t² + 3t -5. Let's check at t = 0: -5 < 0, at t = 1: -1 + 3 -5 = -3 < 0. The vertex is at t = 3/2, which is outside [0,1], so maximum at t=1. So φ(t) is decreasing on [0,1], so yes, all values are negative. So m > 4 - 2√3 is correct.Thus, set M is ( m > 4 - 2sqrt{3} ).Now, Analyzing Set N:Set N is defined as all m such that for all x ∈ [0, π/2], f(φ(x)) < 0.From our earlier analysis, f(y) < 0 if and only if y ∈ (-∞, -1) ∪ (0, 1). So f(φ(x)) < 0 when φ(x) ∈ (-∞, -1) ∪ (0, 1).But we need this to hold for all x ∈ [0, π/2]. Therefore, for all x in [0, π/2], φ(x) must be in (-∞, -1) ∪ (0, 1).But φ(x) is a continuous function (as composition of continuous functions), so its image over [0, π/2] is an interval. Therefore, to have φ(x) ∈ (-∞, -1) ∪ (0, 1) for all x, the entire image of φ(x) must lie either in (-∞, -1) or in (0, 1).But since φ(x) is continuous, its image is connected. Therefore, either:1. The entire image is in (-∞, -1), or2. The entire image is in (0, 1).But since for some m, φ(x) can have parts in both intervals? Wait, but we need for all x, φ(x) must be in either (-∞, -1) or (0,1). But if for some x, φ(x) is in (-∞, -1), and for others in (0,1), then the function φ(x) would have to cross the interval [-1, 0], but since f(φ(x)) needs to be <0 for all x, φ(x) cannot be in [-1, 0] because:From earlier:- If φ(x) ∈ (-∞, -1), then f(φ(x)) < 0.- If φ(x) ∈ (0, 1), then f(φ(x)) < 0 (since for 0 < y < 1, f(y) < 0).But if φ(x) ∈ [-1, 0], then f(φ(x)) ≥ 0 (since for -1 ≤ y < 0, f(y) ≥ 0). Therefore, to have f(φ(x)) < 0 for all x ∈ [0, π/2], we must have φ(x) ∈ (-∞, -1) ∪ (0, 1) for all x, and φ(x) cannot enter [-1, 0].Therefore, the image of φ(x) must be entirely in (-∞, -1) or entirely in (0, 1). Because if part of the image is in (-∞, -1) and part in (0, 1), then by continuity, φ(x) would have to cross the interval [-1, 0], which would make f(φ(x)) ≥ 0 for those x where φ(x) ∈ [-1, 0], violating the condition.Therefore, set N requires that either:1. For all x ∈ [0, π/2], φ(x) < -1, or2. For all x ∈ [0, π/2], 0 < φ(x) < 1.But wait, in set M, we require φ(x) < 0 for all x, which would correspond to case 1 here. But N is a different set. However, N could include m where φ(x) is in (0,1) for all x. But φ(x) is defined as sin²x + m cos x - 2m.But let's check if φ(x) can be in (0,1) for all x ∈ [0, π/2].But first, let's note that set N is separate from set M. So N includes all m where either φ(x) < -1 for all x, or 0 < φ(x) < 1 for all x. Because in both cases, f(φ(x)) < 0.Therefore, set N is the union of two sets:- N1 = { m | ∀x ∈ [0, π/2], φ(x) < -1 }- N2 = { m | ∀x ∈ [0, π/2], 0 < φ(x) < 1 }Therefore, N = N1 ∪ N2.So we need to find both N1 and N2 and then take their union.Finding N1:N1 is the set of m where φ(x) < -1 for all x ∈ [0, π/2].Which is equivalent to φ(t) < -1 for all t ∈ [0,1].So, similar to set M but stricter. So φ(t) < -1 for all t ∈ [0,1].Again, φ(t) = -t² + m t + 1 - 2m < -1.Therefore:- t² + m t + 1 - 2m < -1Wait, let's write the inequality:φ(t) = -t² + m t + 1 - 2m < -1Adding 1 to both sides:- t² + m t + 1 - 2m + 1 < 0Wait, no:Wait, φ(t) < -1So:- t² + m t + 1 - 2m < -1Wait, φ(t) = -t² + m t + 1 - 2m < -1Therefore:- t² + m t + 1 - 2m < -1Add 1 to both sides:- t² + m t - 2m < -2Multiply both sides by -1 (reverse inequality):t² - m t + 2m > 2So:t² - m t + 2m - 2 > 0 for all t ∈ [0,1].So, we need t² - m t + 2m - 2 > 0 for all t ∈ [0,1].Let's denote this quadratic as ψ(t) = t² - m t + 2m - 2.We need ψ(t) > 0 for all t ∈ [0,1].ψ(t) is a quadratic in t, opening upwards (coefficient of t² is 1). Therefore, the minimum occurs at the vertex.Vertex at t = m/(2*1) = m/2.So, depending on where the vertex is relative to [0,1], the minimum is either at the vertex or at the endpoints.Case 1: Vertex t = m/2 ∈ [0,1], i.e., m ∈ [0,2].Compute ψ(m/2):ψ(m/2) = (m/2)² - m*(m/2) + 2m - 2= m²/4 - m²/2 + 2m - 2= -m²/4 + 2m - 2Set this minimum > 0:- m²/4 + 2m - 2 > 0Multiply by -4 (reverse inequality):m² - 8m + 8 < 0Solve m² -8m +8 =0:Discriminant: 64 -32=32Roots: [8 ± √32]/2 = [8 ± 4√2]/2 = 4 ± 2√2 ≈ 4 ± 2.828Thus, roots ≈ 1.172 and 6.828.So inequality m² -8m +8 <0 holds for m ∈ (4 - 2√2, 4 + 2√2) ≈ (1.172, 6.828)But in this case, m ∈ [0,2]. So intersection is m ∈ (4 - 2√2, 2].But 4 - 2√2 ≈ 1.172, so m ∈ (1.172, 2].Therefore, for m ∈ (1.172, 2], the minimum at the vertex is >0, so ψ(t) >0 for all t ∈ [0,1].Case 2: Vertex outside [0,1], i.e., m <0 or m >2.If m <0, vertex at t = m/2 <0. Then ψ(t) is increasing on [0,1], so minimum at t=0.ψ(0) = 0 -0 + 2m -2 = 2m -2 >0So 2m -2 >0 → m >1. But m <0 here, contradiction. Therefore, no solutions.If m >2, vertex at t = m/2 >1. Then ψ(t) is decreasing on [0,1], minimum at t=1.ψ(1) = 1 - m + 2m -2 = (1 -2) + (-m + 2m) = -1 + m >0 → m >1.But m >2, so this holds. Therefore, ψ(1) = m -1 >0 for m >2. Since m >2, m -1 >1 >0.Therefore, for m >2, ψ(t) >0 for all t ∈ [0,1].Therefore, combining cases:N1 is m ∈ (4 - 2√2, 2] ∪ (2, ∞) = m >4 - 2√2.But 4 - 2√2 ≈1.172.Therefore, N1 = (4 - 2√2, ∞).Finding N2:N2 is the set of m where for all x ∈ [0, π/2], 0 < φ(x) <1.Which is equivalent to 0 < φ(t) <1 for all t ∈ [0,1].So, we need two inequalities:1. φ(t) >0 for all t ∈ [0,1]2. φ(t) <1 for all t ∈ [0,1]Let's tackle them one by one.First, φ(t) >0 for all t ∈ [0,1]:- t² + m t +1 -2m >0Wait, φ(t) = -t² + m t +1 -2m >0So:- t² + m t +1 -2m >0Which is:- t² + m t + (1 - 2m) >0This is similar to the previous analysis but reversed.Again, since it's a quadratic in t opening downward (coefficient of t² is -1). The maximum is at t = m/2.Thus, to have φ(t) >0 for all t ∈ [0,1], the minimum of φ(t) over [0,1] must be >0.But since it's opening downward, the minimum occurs at the endpoints.Therefore, check φ(0) and φ(1):φ(0) =1 -2m >0 → 1 -2m >0 → m <1/2φ(1) = -1 +m +1 -2m = -m >0 → -m >0 → m <0Therefore, to have φ(t) >0 for all t ∈ [0,1], we need both φ(0) >0 and φ(1) >0.Which gives m <0.But if m <0, let's check the function φ(t) = -t² + m t +1 -2m.Since m <0, let's see:At t=0, φ(0) =1 -2m. Since m <0, -2m is positive, so φ(0) =1 + |2m| >1 >0.At t=1, φ(1) = -1 +m +1 -2m = -m. Since m <0, -m >0.Since the quadratic opens downward, and the maximum is at t = m/2 <0 (since m <0). Therefore, the maximum is outside the interval [0,1], and the function is decreasing on [0,1]. Therefore, φ(t) is decreasing from φ(0) >0 to φ(1) >0, hence φ(t) >0 for all t ∈ [0,1].Therefore, for m <0, φ(t) >0 for all t ∈ [0,1].Now, the second part of N2: φ(t) <1 for all t ∈ [0,1].Which is:- t² + m t +1 -2m <1Simplify:- t² + m t -2m <0So:- t² + m t -2m <0 → multiply by -1:t² - m t + 2m >0We need t² - m t + 2m >0 for all t ∈ [0,1].Wait, this is similar to ψ(t) in N1.But here, the inequality is t² - m t + 2m >0 for all t ∈ [0,1]. Let's analyze this quadratic.Let’s denote this as ψ'(t) = t² - m t + 2m.Need ψ'(t) >0 for all t ∈ [0,1].This is a quadratic in t, opening upwards (coefficient of t² is 1). Therefore, the minimum occurs at the vertex t = m/2.Case 1: Vertex t = m/2 ∈ [0,1], i.e., m ∈ [0,2].Compute ψ'(m/2):= (m/2)² - m*(m/2) + 2m= m²/4 - m²/2 + 2m= -m²/4 + 2mSet this minimum >0:- m²/4 + 2m >0Multiply by -4 (reverse inequality):m² -8m <0 → m(m -8) <0Thus, m ∈ (0,8)But since m ∈ [0,2] here, intersection is m ∈ (0,2].Case 2: Vertex outside [0,1], i.e., m <0 or m >2.If m <0, vertex at t = m/2 <0. Then ψ'(t) is increasing on [0,1]. Therefore, minimum at t=0.ψ'(0) =0 -0 +2m =2m >0 → m >0. But m <0, so no solutions.If m >2, vertex at t =m/2 >1. Then ψ'(t) is decreasing on [0,1]. Minimum at t=1.ψ'(1) =1 -m +2m =1 +m >0. Since m >2, 1 +m >3 >0. Therefore, ψ'(t) >0 for all t ∈ [0,1].Therefore, combining cases:For ψ'(t) >0 for all t ∈ [0,1], we need:- m ∈ (0,2] ∪ (2, ∞) → m >0.But wait, in Case 1, when m ∈ (0,2], we need ψ'(m/2) >0, which led to m ∈ (0,8). But since m ∈ [0,2], it's m ∈ (0,2]. For m >2, it's always true.Therefore, ψ'(t) >0 for all t ∈ [0,1] when m >0.But for m >0, we already have φ(t) >0 for all t ∈ [0,1] only if m <0. Wait, this is conflicting.Wait, N2 requires both φ(t) >0 and φ(t) <1 for all t ∈ [0,1].From earlier, φ(t) >0 for all t ∈ [0,1] only if m <0.But when m <0, we need ψ'(t) = t² -m t + 2m >0 for all t ∈ [0,1].But if m <0, let's check ψ'(t) = t² -m t + 2m.Since m <0, -m t is positive, and 2m is negative.At t=0: ψ'(0) =0 +0 +2m =2m <0 (since m <0). Therefore, ψ'(0) <0, which violates the requirement ψ'(t) >0 for all t ∈ [0,1].Therefore, when m <0, φ(t) <1 is not satisfied because at t=0, ψ'(0) =2m <0.Thus, there is a contradiction here. Therefore, N2 is empty.Wait, let's go back. N2 requires 0 < φ(t) <1 for all t ∈ [0,1].But φ(t) >0 requires m <0, but for φ(t) <1, which translates to ψ'(t) >0 (t² - m t + 2m >0), but when m <0, ψ'(0) =2m <0, which violates ψ'(t) >0. Therefore, there are no m such that 0 < φ(t) <1 for all t ∈ [0,1].Hence, N2 is empty.Therefore, set N is just N1, which is m >4 - 2√2.Wait, but let me verify again.If N2 is empty, then N = N1 = {m | φ(x) < -1 for all x ∈ [0, π/2]} = (4 - 2√2, ∞).But earlier, we found that N1 is m >4 - 2√2.But let me check with m =3, which is in N1.φ(t) = -t² +3t +1 -6 = -t² +3t -5.At t=0: -5 < -1, at t=1: -1 +3 -5 =-3 < -1. The vertex is at t=3/2, outside [0,1], so it's decreasing on [0,1]. Thus, φ(t) < -1 for all t, so f(φ(t)) <0. So yes, m=3 is in N.For m=4 -2√2 +ε (where ε is small positive), φ(t) < -1 for all t. Therefore, N is m >4 -2√2.Therefore, Set N is also (4 - 2√2, ∞).Wait, but 4 - 2√2 ≈1.172, and set M was found to be (4 -2√3, ∞) ≈(0.536, ∞). Therefore, M is a larger interval than N.Therefore, M ∩ N would be the overlap, which is N, since N is a subset of M.But let's verify:Set M is m >4 -2√3 ≈0.536.Set N is m >4 -2√2 ≈1.172.Since 4 -2√2 >4 -2√3 (because √2 ≈1.414, √3≈1.732, so 2√2≈2.828, 2√3≈3.464; thus, 4 -2√2≈1.172, 4 -2√3≈0.536), therefore, N is a subset of M. Hence, their intersection is N.But let's confirm with numbers:Take m=1.5, which is greater than 4 -2√3≈0.536 and greater than 4 -2√2≈1.172.So m=1.5 is in both M and N.Take m=1.0, which is greater than 0.536 but less than 1.172. So m=1.0 is in M but not in N. Therefore, M ∩ N is m >4 -2√2.Therefore, the final answer is ( (4 - 2sqrt{2}, +infty) ).But let's check with the exact values.Given that:4 -2√3 ≈4 -3.464≈0.5364 -2√2≈4 -2.828≈1.172Therefore, M is (0.536, ∞), N is (1.172, ∞). So their intersection is (1.172, ∞), which is N.Therefore, M ∩ N = N = (4 -2√2, ∞).But wait, according to the problem statement, set N is defined using f(φ(x)) <0 for all x. Since N1 requires φ(x) < -1, which is a subset of φ(x) <0, so N1 is a subset of M. And N2 is empty. Therefore, N is a subset of M, hence M ∩ N = N.Therefore, the answer should be ( (4 - 2sqrt{2}, +infty) ).But let's check with exact expressions:4 - 2√2 is equal to 2*(2 -√2). But perhaps it's better to leave it as 4 -2√2.Therefore, boxed answer: boxed{(4 - 2sqrt{2}, +infty)}Final Answerboxed{(4 - 2sqrt{2}, +infty)}</think>Given that ( f(x) ) is an odd function defined on ( mathbb{R} ), and when ( x < 0 ), ( f(x) ) is monotonically increasing with ( f(-1) = 0 ). The function ( varphi(x) = sin^2 x + m cos x - 2m ). We need to find ( M cap N ) where ( M = { m mid forall x in [0, frac{pi}{2}], varphi(x) < 0 } ) and ( N = { m mid forall x in [0, frac{pi}{2}], f(varphi(x)) < 0 } ).1. Analyzing Set ( M ): - ( varphi(x) = -cos^2 x + m cos x + (1 - 2m) ). - Convert to quadratic in ( t = cos x ): ( varphi(t) = -t^2 + mt + (1 - 2m) ). - Require ( varphi(t) < 0 ) for all ( t in [0, 1] ). - The maximum value of ( varphi(t) ) occurs at ( t = m/2 ). - For ( varphi(t) < 0 ), the maximum value must be less than 0. Solving, we find ( M = (4 - 2sqrt{3}, +infty) ).2. Analyzing Set ( N ): - ( f(varphi(x)) < 0 ) implies ( varphi(x) in (-infty, -1) cup (0, 1) ). - For ( varphi(x) < -1 ) for all ( x ), we need ( -t^2 + mt + (1 - 2m) < -1 ). - Convert to quadratic inequality: ( t^2 - mt + 2m - 2 > 0 ). - Solving, we find ( N = (4 - 2sqrt{2}, +infty) ).3. Intersection ( M cap N ): - ( M = (4 - 2sqrt{3}, +infty) ) and ( N = (4 - 2sqrt{2}, +infty) ). - Since ( 4 - 2sqrt{2} > 4 - 2sqrt{3} ), the intersection is ( N ).Thus, the final answer is (boxed{(4 - 2sqrt{2}, +infty)}).]

🔑:Question 9, Answer: Since f(x) is an odd function defined on R, and when x>0, f(x) is also monotonically increasing, and f(1)=0.According to the definition: M cap N=left{m midright. for forall x inleft[0, frac{pi}{2}right], varphi(x)>0Leftrightarrow(2-cos x) m>2-cos ^{2} xLeftrightarrow m>max _{x inleft[0, frac{pi}{2}right]} frac{2-cos ^{2} x}{2-cos x}=max _{t in[0,1]}left{4-left[(2-t)+frac{2}{2-t}right]right}=4-2 sqrt{2}In summary, we know M cap N=(4-2 sqrt{2},+infty).

🧠:[<think>Okay, so I need to find the value of k such that the intersection of sets A and B is a singleton, meaning they intersect at exactly one point. Let me start by understanding what each set represents.Set A is defined as all points (x, y) such that x² + y² = 2(x + y). Hmm, this looks like the equation of a circle. Let me rewrite it to get it into the standard form. Starting with x² + y² = 2x + 2y. If I move the terms on the right to the left, I get x² - 2x + y² - 2y = 0. To complete the squares for both x and y terms:For x: Take x² - 2x, complete the square by adding (2/2)² = 1.For y: Take y² - 2y, complete the square by adding (2/2)² = 1.So adding those to both sides:x² - 2x + 1 + y² - 2y + 1 = 0 + 1 + 1Which simplifies to:(x - 1)² + (y - 1)² = 2.Okay, so set A is a circle centered at (1, 1) with radius √2.Set B is defined as all points (x, y) such that kx - y + k + 3 ≥ 0. That's a linear inequality. Let me rearrange it to get a better sense:kx - y + k + 3 ≥ 0 => y ≤ kx + k + 3.So set B is the region on or below the line y = kx + k + 3. We need A ∩ B to be a singleton, meaning the line y = kx + k + 3 is tangent to the circle (x - 1)² + (y - 1)² = 2. Because if the line is tangent to the circle, they intersect at exactly one point, which would make the intersection a singleton set.Therefore, the problem reduces to finding the value of k such that the line y = kx + k + 3 is tangent to the circle centered at (1,1) with radius √2.To find the condition for tangency between a line and a circle, the distance from the center of the circle to the line must equal the radius of the circle.The formula for the distance from a point (h, m) to the line Ax + By + C = 0 is |Ah + Bm + C| / sqrt(A² + B²). First, let's write the line in standard form. The line is y = kx + k + 3. Let's rearrange that:kx - y + k + 3 = 0.So A = k, B = -1, C = k + 3.The center of the circle is (1, 1). Plugging into the distance formula:Distance = |k*1 + (-1)*1 + k + 3| / sqrt(k² + (-1)²) = |k - 1 + k + 3| / sqrt(k² + 1)Simplify numerator:|2k + 2| / sqrt(k² + 1)This distance must equal the radius √2. So:|2k + 2| / sqrt(k² + 1) = √2Let me square both sides to eliminate the absolute value and the square root:(2k + 2)² / (k² + 1) = 2Multiply both sides by (k² + 1):(2k + 2)² = 2(k² + 1)Expand the left side:(2k + 2)² = 4k² + 8k + 4So:4k² + 8k + 4 = 2k² + 2Bring all terms to left side:4k² + 8k + 4 - 2k² - 2 = 0 => 2k² + 8k + 2 = 0Divide both sides by 2:k² + 4k + 1 = 0Now, solve for k using quadratic formula:k = [-4 ± sqrt(16 - 4*1*1)] / 2 = [-4 ± sqrt(12)] / 2 = [-4 ± 2*sqrt(3)] / 2 = -2 ± sqrt(3)Wait, so k can be -2 + sqrt(3) or -2 - sqrt(3). But we need to check if these solutions are valid, because when we squared both sides, we might have introduced extraneous solutions. However, since the original equation was |2k + 2| = √2 * sqrt(k² + 1), squaring both sides doesn't introduce extraneous solutions here because both sides are non-negative. Therefore, both solutions are valid.But let me verify by plugging back into the original distance equation.First, take k = -2 + sqrt(3):Compute numerator |2k + 2|:2*(-2 + sqrt(3)) + 2 = -4 + 2*sqrt(3) + 2 = -2 + 2*sqrt(3)Denominator sqrt(k² + 1):Compute k²: (-2 + sqrt(3))² = 4 - 4*sqrt(3) + 3 = 7 - 4*sqrt(3)So sqrt(7 - 4*sqrt(3) + 1) = sqrt(8 - 4*sqrt(3))Hmm, let's see if this equals √2 * sqrt(k² + 1). Wait, actually, we can check if the original equation holds.But maybe it's better to compute the left side and the right side:Left side: |2k + 2| = |-2 + 2*sqrt(3)| = | -2 + 3.464 | ≈ |1.464| ≈ 1.464Right side: √2 * sqrt(k² + 1) = √2 * sqrt( (7 - 4*sqrt(3)) + 1 ) = √2 * sqrt(8 - 4*sqrt(3)).Compute sqrt(8 - 4*sqrt(3)). Let me rationalize or see if it's a perfect square.Let me denote sqrt(8 - 4*sqrt(3)) = sqrt(a) - sqrt(b). Then squaring both sides: 8 - 4*sqrt(3) = a + b - 2*sqrt(ab). Comparing, we have:a + b = 8-2*sqrt(ab) = -4*sqrt(3) => sqrt(ab) = 2*sqrt(3) => ab = 12So solving a + b =8 and ab =12. The solutions are a and b roots of x² -8x +12=0, which factors as (x-6)(x-2)=0. So a=6, b=2. Hence sqrt(8 -4√3)=sqrt(6) - sqrt(2). Therefore:sqrt(8 -4√3) = sqrt(6) - sqrt(2)Thus, right side: √2*(sqrt(6) - sqrt(2)) = √2*sqrt(6) - √2*sqrt(2) = sqrt(12) - 2 = 2*sqrt(3) - 2 ≈ 3.464 - 2 ≈ 1.464. Which matches the left side. So this solution is valid.Now check k = -2 - sqrt(3):Compute |2k + 2| = |2*(-2 - sqrt(3)) + 2| = |-4 - 2*sqrt(3) + 2| = |-2 - 2*sqrt(3)| = | - (2 + 2*sqrt(3)) | = 2 + 2*sqrt(3)Denominator sqrt(k² +1):k² = (-2 - sqrt(3))² = 4 + 4*sqrt(3) + 3 = 7 + 4*sqrt(3)sqrt(k² +1) = sqrt(8 + 4*sqrt(3))Similarly, sqrt(8 +4*sqrt(3)) can be written as sqrt(a) + sqrt(b). Then squaring: 8 +4√3 = a + b + 2*sqrt(ab). So:a + b =82*sqrt(ab)=4√3 => sqrt(ab)=2√3 => ab=12Same as before, a=6, b=2. Therefore sqrt(8 +4√3)=sqrt(6)+sqrt(2)Thus, right side: √2*(sqrt(6) + sqrt(2)) = √2*sqrt(6) + √2*sqrt(2) = sqrt(12) + 2 = 2*sqrt(3) + 2 ≈ 3.464 + 2 ≈ 5.464Compare with left side |2k +2| = 2 + 2*sqrt(3) ≈ 2 + 3.464 ≈ 5.464. So that's equal. Therefore, both solutions are valid.So the values of k are -2 + sqrt(3) and -2 - sqrt(3). But the problem says "the value of k is", implying a unique answer. Wait, maybe both are possible? But let's check if both lines are on different sides of the circle.Wait, the original inequality is kx - y +k +3 >=0, which is y <= kx +k +3. So the region is below the line. So the line is the upper boundary. For the intersection to be a singleton, the line must be tangent to the circle from above. However, depending on the slope k, the line could be tangent from different sides. But perhaps both solutions are valid. Let me visualize.The circle is centered at (1,1), radius √2. Let's see the lines y = kx +k +3. For k = -2 + sqrt(3) ≈ -2 + 1.732 ≈ -0.268. So the line would have a slight negative slope. Let's compute the y-intercept: k +3 ≈ -0.268 +3 ≈ 2.732. So the line is y ≈ -0.268x + 2.732. This line would be tangent to the circle from above.For k = -2 - sqrt(3) ≈ -3.732. Then the line would have a steeper negative slope. The y-intercept is k +3 ≈ -3.732 +3 ≈ -0.732. So the line is y ≈ -3.732x -0.732. This line would be tangent to the circle from below.But since the region B is y <= kx +k +3, so for the line with k ≈ -0.268, the region is below the tangent line. Since the line is tangent from above, the region B would include all points below the line, which would intersect the circle only at the tangent point. Similarly, for k ≈ -3.732, the line is tangent from below, but the region B is below the line. However, if the line is tangent from below, the region B (being below the line) would include the entire circle except the tangent point? Wait, no.Wait, if the line is tangent from below, then the region B is y <= line. Since the line is tangent from below, the entire circle except the tangent point is above the line. So the intersection A ∩ B would be just the tangent point. So both cases would result in a singleton set. Therefore, both k values are valid.But the problem says "the value of k is", so maybe both solutions are acceptable. However, the problem might be expecting both answers. Let me check again.Wait, in Chinese problems sometimes even if there are two solutions, they might require both. Let me check the problem statement again:"6. Let k be a real number, in the Cartesian coordinate system xOy there are two point sets A = {(x, y) | x² + y² = 2(x + y)} and B = {(x, y) | kx - y +k +3 ≥ 0}. If A ∩ B is a singleton set, then the value of k is ___"The problem says "the value of k is", but since there are two solutions, they might need both. But maybe in the original problem, only one is possible due to some constraints. Wait, let's think again.Wait, the region B is kx - y +k +3 ≥0, which is y ≤ kx +k +3. So if the line is tangent from above, the region B is below the line. The circle is centered at (1,1). If the line is tangent from above, the region B includes the tangent point and all below. Since the circle is above and below, the intersection would only be the tangent point.If the line is tangent from below, the region B is all below the line. The tangent point is the only point on the line, and since the line is below the circle except at that point, the intersection is that single point. So both cases are possible.Therefore, there are two values of k. But let's check the problem again. It says "the value of k is". Maybe the answer is both. However, in Chinese textbooks, sometimes problems like this might have two answers. Let me check calculations again.Wait, but when we solved the quadratic equation, we got two solutions: k = -2 ± sqrt(3). Both satisfy the tangency condition. So the answer should be both k = -2 + sqrt(3) and k = -2 - sqrt(3). But the problem says "the value of k is". Maybe in the original problem, there is a specific answer, but since in my calculation there are two, perhaps both are needed. But let me check once more.Alternatively, perhaps there is a miscalculation. Let me go through steps again.Original equation of circle: (x -1)^2 + (y -1)^2 = 2.Equation of line: y = kx +k +3.Distance from center (1,1) to line should be equal to radius sqrt(2).Line in standard form: kx - y +k +3 = 0.Distance formula: |k*1 -1*1 +k +3| / sqrt(k^2 +1) = |2k +2| / sqrt(k^2 +1) = sqrt(2).Squaring: (2k +2)^2 = 2(k^2 +1) => 4k^2 +8k +4 = 2k^2 +2 => 2k^2 +8k +2=0 => k^2 +4k +1=0 => k = [-4 ± sqrt(16 -4)]/2 = [-4 ± sqrt(12)]/2 = [-4 ± 2*sqrt(3)]/2 = -2 ± sqrt(3). So that's correct.Therefore, the two solutions are k = -2 + sqrt(3) and k = -2 - sqrt(3). So unless there's a restriction on k being positive or something, which the problem doesn't specify, both are valid. Therefore, the answer should be both.But the problem says "the value of k is", so maybe the answer is presented as two values. However, in the format requested, the user said "put your final answer within boxed{}". If there are two answers, we need to box both. But perhaps in the original problem, due to translation or something, maybe only one is expected. Wait, maybe I made a mistake in interpreting the inequality.Wait, the region B is kx - y +k +3 ≥ 0, which is y ≤ kx +k +3. So for the line being tangent from above, the region below the line would intersect the circle at only the tangent point. For the line being tangent from below, the region below the line would include all the circle below the line, but since the line is tangent, it only touches at one point. Wait, no. If the line is tangent from below, the circle is above the line except at the tangent point. So the region B, which is below the line, would only include the tangent point from the circle. So both cases are possible. Hence, two values.Therefore, the answer is both k = -2 + sqrt(3) and k = -2 - sqrt(3). So box both answers.But in Chinese textbooks, sometimes problems expect multiple answers in one box separated by commas. Let me check the original problem statement again:"If A ∩ B is a singleton set, then the value of k is ______"Given that, and since there are two solutions, the answer should be both. However, maybe the problem expects both answers. So I need to write both values.But let me check the problem statement once again to confirm there are no constraints on k. It just says "k is a real number", so both solutions are valid.Thus, the values are k = -2 ± sqrt(3). So I can write this as boxed{-2 + sqrt{3}} and boxed{-2 - sqrt{3}}. But if they want both in one box, maybe boxed{-2 pm sqrt{3}}.Given that the problem is translated from Chinese, and sometimes in such contexts, they might present the answer as boxed{-2 pm sqrt{3}} to denote both solutions. Alternatively, if the answer requires two separate boxes, but the user instruction says "put your final answer within boxed{}", implying a single box. Hmm. Maybe the question expects both values. Alternatively, maybe I made a mistake in the sign.Wait, let me verify with a specific example. Take k = -2 + sqrt(3) ≈ -0.268. The line is y ≈ -0.268x + (-0.268 +3) ≈ -0.268x +2.732. Let's compute the distance from (1,1) to this line:| -0.268*1 -1 + (-0.268) +3 | / sqrt( (-0.268)^2 +1 ) = | -0.268 -1 -0.268 +3 | / sqrt(0.0718 +1 ) = |1.464| / sqrt(1.0718) ≈ 1.464 / 1.035 ≈ 1.414 ≈ sqrt(2). Which is correct.Similarly for k = -2 - sqrt(3) ≈ -3.732. The line is y ≈ -3.732x + (-3.732 +3) ≈ -3.732x -0.732. Distance from (1,1):| -3.732*1 -1 + (-3.732) +3 | / sqrt( (-3.732)^2 +1 ) ≈ | -3.732 -1 -3.732 +3 | / sqrt(13.928 +1 ) ≈ | -5.464 | / sqrt(14.928) ≈5.464 / 3.863 ≈1.414 ≈ sqrt(2). Correct.Therefore, both solutions are valid. Therefore, the answer is k = -2 ± sqrt(3). So in boxed form, that would be boxed{-2 pm sqrt{3}}.Final AnswerThe value of ( k ) is boxed{-2 pm sqrt{3}}.</think>Given the problem, we need to find the value of ( k ) such that the intersection of sets ( A ) and ( B ) is a singleton set. Set ( A ) is defined as the circle ( (x - 1)^2 + (y - 1)^2 = 2 ), which is centered at ( (1, 1) ) with radius ( sqrt{2} ). Set ( B ) is the region below the line ( y = kx + k + 3 ). For ( A cap B ) to be a singleton set, the line must be tangent to the circle.To find the tangent condition, we calculate the distance from the center of the circle ( (1, 1) ) to the line ( kx - y + k + 3 = 0 ) and set it equal to the radius ( sqrt{2} ). The distance formula gives:[frac{|k cdot 1 - 1 cdot 1 + k + 3|}{sqrt{k^2 + 1}} = sqrt{2}]Simplifying the numerator:[|2k + 2| = sqrt{2} sqrt{k^2 + 1}]Squaring both sides:[(2k + 2)^2 = 2(k^2 + 1)]Expanding and simplifying:[4k^2 + 8k + 4 = 2k^2 + 2 implies 2k^2 + 8k + 2 = 0 implies k^2 + 4k + 1 = 0]Solving this quadratic equation using the quadratic formula:[k = frac{-4 pm sqrt{16 - 4}}{2} = frac{-4 pm 2sqrt{3}}{2} = -2 pm sqrt{3}]Both solutions ( k = -2 + sqrt{3} ) and ( k = -2 - sqrt{3} ) are valid as they satisfy the tangency condition. Therefore, the value of ( k ) is:[boxed{-2 pm sqrt{3}}],<think>Okay, so I have this problem here where I need to find the value of k such that the intersection of sets A and B is a singleton set. That means they touch at exactly one point, right? So set A is defined by the equation x² + y² = 2(x + y), and set B is defined by the inequality kx - y + k + 3 ≥ 0. Hmm, okay. Let me start by understanding what each set represents.First, set A: x² + y² = 2(x + y). That looks like a circle equation. Let me rewrite it to standard form. If I move the 2x and 2y to the left side, it becomes x² - 2x + y² - 2y = 0. To complete the squares for both x and y terms:For the x terms: x² - 2x. Take half of -2, which is -1, square it to get 1. So, x² - 2x + 1 = (x - 1)².Similarly, for the y terms: y² - 2y. Half of -2 is -1, square it to 1. So, y² - 2y + 1 = (y - 1)².Therefore, adding these squares to both sides:x² - 2x + 1 + y² - 2y + 1 = 0 + 1 + 1 → (x - 1)² + (y - 1)² = 2.So, set A is a circle with center at (1, 1) and radius √2. Got that.Now, set B: kx - y + k + 3 ≥ 0. That's a linear inequality. Let me rearrange it to the form y ≤ mx + b or something similar. So, moving y to the other side:kx + k + 3 ≥ y → y ≤ kx + k + 3.Wait, but inequalities can be a bit tricky here. The set B is all the points (x, y) such that y ≤ kx + k + 3. So it's the region below the line y = kx + k + 3. But the problem states that A ∩ B is a singleton set, which would mean that the line y = kx + k + 3 is tangent to the circle A. Because if the line is tangent, then there's exactly one point where the line touches the circle, so the intersection would be that single point. But since B is the region below the line, the intersection being a singleton would imply that the line is tangent and all other points of the circle are above the line. Wait, is that correct?Wait, if B is the region where y ≤ kx + k + 3, then the intersection A ∩ B would consist of all points on the circle A that are also below or on the line. If the line is tangent to the circle from below, then the only point of intersection would be the tangent point. But if the line cuts through the circle, there would be more points in the intersection. So in order for A ∩ B to be a singleton, the line must be tangent to the circle, and the circle must lie entirely above the line except for the tangent point. Therefore, the distance from the center of the circle to the line must be equal to the radius, and the line must be below the circle except at the tangent point.Alternatively, maybe if the line is above the circle, but the inequality is y ≤ line, so the intersection would be points on the circle below the line. So if the line is tangent from above, then the only point in the intersection would be the tangent point, but the rest of the circle is below the line? Wait, no. If the line is tangent from above, then the region y ≤ line would include all the circle except the tangent point. Wait, no, maybe.Wait, this is confusing. Let me think again.Set A is the circle (x - 1)² + (y - 1)² = 2. Set B is all points below or on the line y = kx + k + 3. The intersection A ∩ B is the set of points on the circle that are also below or on the line. For this intersection to be a singleton, there must be exactly one point on the circle that is below or on the line. But if the line is tangent to the circle from above, then all other points on the circle would be below the line. Wait, that's possible. Wait, if the line is tangent to the circle from above, then the tangent point is the only point on the line, and all other points on the circle are below the line. Therefore, A ∩ B would be the entire circle except for the tangent point? Wait, no. If the line is tangent from above, then the tangent point is on the line, and all other points of the circle are below the line. Therefore, the intersection A ∩ B would be the entire circle, because all points of the circle are below or on the line. But the tangent point is on the line. Wait, but if the line is tangent from above, then the tangent point is the highest point of the circle relative to the line, right? So if the line is above the circle, touching at one point, then all other points of the circle are below the line, hence the intersection A ∩ B would be the entire circle. But the problem says A ∩ B is a singleton set, meaning only one point. So that can't be the case.Alternatively, if the line is tangent from below, then the tangent point is the only point of the circle that is on or below the line. Wait, because the rest of the circle would be above the line. So in that case, A ∩ B would be just the tangent point, which is a singleton set. That makes sense. So, if the line is tangent to the circle from below, then the intersection would only be that tangent point. Therefore, we need to find k such that the line y = kx + k + 3 is tangent to the circle (x - 1)² + (y - 1)² = 2, and the tangent occurs from below, meaning the line is below the circle except at the tangent point.Therefore, the key is to find the values of k for which the distance from the center of the circle (1, 1) to the line y = kx + k + 3 is equal to the radius √2.Wait, but the formula for the distance from a point (x₀, y₀) to the line ax + by + c = 0 is |ax₀ + by₀ + c| / sqrt(a² + b²). Let's rewrite the line equation in standard form.The line is y = kx + k + 3. Let's rearrange it to kx - y + (k + 3) = 0. So, a = k, b = -1, c = k + 3.Then, the distance from (1, 1) to this line is |k*1 + (-1)*1 + k + 3| / sqrt(k² + (-1)²) = |k - 1 + k + 3| / sqrt(k² + 1) = |2k + 2| / sqrt(k² + 1).This distance must be equal to the radius √2. So:|2k + 2| / sqrt(k² + 1) = √2.Let's square both sides to eliminate the absolute value and the square root:(2k + 2)² / (k² + 1) = 2.Multiply both sides by (k² + 1):(2k + 2)² = 2(k² + 1).Expand the left side:4k² + 8k + 4 = 2k² + 2.Bring all terms to one side:4k² + 8k + 4 - 2k² - 2 = 0 → 2k² + 8k + 2 = 0.Divide both sides by 2:k² + 4k + 1 = 0.Now, solve this quadratic equation:k = [-4 ± sqrt(16 - 4*1*1)] / 2 = [-4 ± sqrt(12)] / 2 = [-4 ± 2*sqrt(3)] / 2 = -2 ± sqrt(3).So, the solutions are k = -2 + sqrt(3) and k = -2 - sqrt(3).But wait, we need to check if these values of k actually result in the line being tangent from below the circle. Because even though the distance is equal to the radius, the line could be tangent from above or below. Since we want A ∩ B to be a singleton set, which requires the line to be tangent from below, we need to ensure that the line is indeed below the circle except at the tangent point.How can we check which of these k values correspond to the line being below the circle?Alternatively, maybe both solutions are valid, but the problem states that k is a real number, so both are possible. Wait, but perhaps when solving the distance equation, both possibilities are there. Let me verify.Suppose k = -2 + sqrt(3). Let's compute the line equation: y = kx + k + 3. Substitute k:y = (-2 + sqrt(3))x + (-2 + sqrt(3)) + 3 = (-2 + sqrt(3))x + 1 + sqrt(3).Similarly, for k = -2 - sqrt(3):y = (-2 - sqrt(3))x + (-2 - sqrt(3)) + 3 = (-2 - sqrt(3))x + 1 - sqrt(3).Now, let's see the positions of these lines relative to the circle.Alternatively, since we have the distance equal to the radius, the line is tangent. The direction of the tangent depends on the slope. But how do we know if the line is above or below? Hmm.Alternatively, maybe both solutions are valid, but depending on the slope, the line could be tangent from above or below. However, since the set B is defined by y ≤ kx + k + 3, the intersection A ∩ B will be a singleton only if the line is tangent from below, such that the rest of the circle is above the line. Therefore, the tangent point is the only point on the circle that is on or below the line.Alternatively, if the line is tangent from above, then the entire circle except the tangent point would be below the line, so A ∩ B would be the entire circle except that point, which is not a singleton.Therefore, we need to ensure that the line is tangent from below. To determine that, we can check the y-coordinate of the tangent point. If the tangent point is the lowest point on the circle relative to the line, then the line is below the circle.Alternatively, perhaps we can check the position of the center relative to the line. The center is (1,1). If the line is below the center, then the tangent point would be the closest point on the circle to the line. Wait, but the distance from the center to the line is equal to the radius, so the line is tangent. The direction depends on the slope. Wait, maybe not. Let me think.Alternatively, the line being tangent to the circle, regardless of the direction, but for the intersection A ∩ B to be singleton, the line must touch the circle at exactly one point, and all other points of the circle lie above the line. Therefore, if the line is tangent and the center of the circle is above the line, then the entire circle except the tangent point is above the line.So, let's check if the center (1,1) is above the line y = kx + k + 3.For each k, compute the value of the line at x=1:y = k*1 + k + 3 = 2k + 3.Compare to the y-coordinate of the center, which is 1.If 2k + 3 < 1, then the center is above the line.So 2k + 3 < 1 → 2k < -2 → k < -1.Looking at the solutions k = -2 + sqrt(3) ≈ -2 + 1.732 ≈ -0.268, which is greater than -1. So 2k + 3 ≈ 2*(-0.268) + 3 ≈ -0.536 + 3 ≈ 2.464, which is greater than 1. Therefore, the center (1,1) is below the line. So if the line is tangent, but the center is below the line, then the entire circle would be below the line, except the tangent point. Wait, no. Wait, if the center is below the line, then the tangent line is above the center. Since the distance from the center to the line is equal to the radius, which is √2. So, if the center is below the line, the tangent point would be in the direction towards the line. Therefore, the circle would cross the line? Wait, no, because it's tangent. If the distance from the center to the line is equal to the radius, then the line is tangent. If the center is below the line, the tangent point is above the center. Then, the circle lies both below and above the line? Wait, no. If the center is a distance of √2 below the line, and the radius is √2, then the circle would just touch the line at one point above the center. Then, the rest of the circle is below the line. Wait, but the center is (1,1). If the line is above the center, and the distance is √2, then the tangent point is on the line, and the rest of the circle is below the line. Therefore, the intersection A ∩ B (points on the circle below or on the line) would be the entire circle, since all points are below the line except the tangent point. But the tangent point is on the line, so it's included in B. Therefore, A ∩ B would be the entire circle. Wait, that can't be. So if the line is tangent above the center, then the circle is mostly below the line, with one point on the line. So all points of the circle are in B (since they are below or on the line), so A ∩ B would be the entire circle. But the problem states that A ∩ B is a singleton, which would mean only one point. Therefore, this case is not possible.Wait, maybe my reasoning is flawed. Let me try again. If the line is tangent to the circle from above, then the tangent point is the highest point of the circle. Then, all other points are below the line, so the intersection A ∩ B would be all the points of the circle, because they are all below or on the line. Therefore, that can't be a singleton set. On the other hand, if the line is tangent from below, then the tangent point is the lowest point of the circle, and all other points are above the line. Therefore, the intersection A ∩ B would be just the tangent point, which is a singleton. So in this case, the line is tangent from below, meaning the center is above the line. Therefore, to have the center above the line, the y-coordinate of the center (1) must be greater than the y-coordinate of the line at x=1. Wait, let's see. The line at x=1 is y = k*1 + k + 3 = 2k + 3. So, if the center (1,1) is above the line, then 1 > 2k + 3 → 2k + 3 < 1 → 2k < -2 → k < -1. So, if k < -1, then the center is above the line. So the line is below the center. Then, the tangent point would be the closest point on the circle to the line, which is in the direction away from the center. Wait, no. If the center is above the line, then the closest point on the circle to the line would be in the direction towards the line. But the distance is equal to the radius, so the line is tangent. Therefore, the circle would touch the line at one point below the center. Therefore, the rest of the circle is above the line. Therefore, the intersection A ∩ B, which is the set of points on the circle that are below or on the line, would be just the tangent point. Therefore, that's the case we want.Therefore, the line must be tangent from below, so the center is above the line. Therefore, the value of k must satisfy k < -1. Let's check our solutions. The solutions were k = -2 ± sqrt(3). Let's approximate sqrt(3) ≈ 1.732. So k = -2 + 1.732 ≈ -0.268, and k = -2 - 1.732 ≈ -3.732. So between the two, k ≈ -0.268 is greater than -1, and k ≈ -3.732 is less than -1. Therefore, the solution k = -2 - sqrt(3) satisfies k < -1, so the center is above the line, and the line is tangent from below. Therefore, this is the correct solution. The other solution, k = -2 + sqrt(3), would result in the center being below the line, leading to the entire circle being in B except the tangent point, but then the intersection would be the entire circle, which is not a singleton. Therefore, only k = -2 - sqrt(3) is valid.But let me verify this with an example. Suppose k = -3.732 (approximately). Then the line is y = -3.732x - 3.732 + 3 = -3.732x - 0.732. So this line has a steep negative slope. Let's see where the center (1,1) is relative to the line. Plugging x=1 into the line equation: y = -3.732*1 -0.732 ≈ -4.464. The center is at (1,1), which is above the line (since 1 > -4.464). The distance from the center to the line is √2, so the line is tangent to the circle. Therefore, the tangent point is the only point on the circle that is on or below the line. Hence, A ∩ B is a singleton. Conversely, for k ≈ -0.268, the line is y = (-0.268)x + (-0.268) + 3 ≈ -0.268x + 2.732. Plugging x=1, the line at x=1 is y ≈ -0.268 + 2.732 ≈ 2.464. The center (1,1) is below this line. Therefore, the distance from the center to the line is √2, so the line is tangent to the circle. However, since the center is below the line, the tangent point is above the center, and the rest of the circle is above the line as well. Wait, no. If the line is tangent above the center, then part of the circle is above the line and part below? Wait, no. If the line is tangent to the circle, then it touches at exactly one point. If the center is below the line, then the tangent point is the closest point on the circle to the line, which would be above the center. Therefore, the rest of the circle is above the tangent point. Therefore, all other points on the circle would be above the line, so A ∩ B would only include the tangent point. Wait, but that contradicts my previous reasoning.Wait, this is confusing. Let me take a specific example. Suppose we have a circle centered at (1,1) with radius √2. Let's take k = -2 + sqrt(3) ≈ -0.268. Then the line is y ≈ -0.268x + 2.732. Let's see where this line is. At x=1, it's approximately 2.464. So the center (1,1) is below this line. The line is above the center. The distance from the center to the line is √2. So, the line is tangent to the circle. The tangent point would be in the direction from the center towards the line, which is upwards. So the tangent point is the highest point on the circle relative to the line. Then, all other points on the circle are below the line. Therefore, the intersection A ∩ B would be all the points on the circle except the tangent point, because they are below the line. But the tangent point is on the line, so it's included. Therefore, A ∩ B would be the entire circle, which is not a singleton. Wait, that can't be right.Wait, if the line is tangent above the center, then the tangent point is the only point on the circle that is on the line, and all other points are below the line. Therefore, the set B includes all points below or on the line, so all points of the circle are in B, hence A ∩ B is the entire circle. Which contradicts the problem's requirement of being a singleton. Therefore, only when the line is tangent below the center, so that the tangent point is the only point on the circle that is in B. Therefore, the correct k is the one where the center is above the line, so k = -2 - sqrt(3). Let me check again. For k = -2 - sqrt(3) ≈ -3.732, the line is y ≈ -3.732x - 0.732. At x=1, y ≈ -4.464. The center (1,1) is above the line. The distance from the center to the line is √2, so the line is tangent. The tangent point is the closest point on the circle to the line, which is below the center. Therefore, all other points on the circle are above the line, so A ∩ B is just the tangent point. Hence, this is the correct solution.Therefore, the answer is k = -2 - sqrt(3).Final AnswerThe value of ( k ) is (boxed{-2 - sqrt{3}}).</think>Given the problem, we need to find the value of ( k ) such that the intersection of sets ( A ) and ( B ) is a singleton set. Set ( A ) is defined by the equation ( x^2 + y^2 = 2(x + y) ), which represents a circle. Rewriting this equation in standard form, we get:[(x - 1)^2 + (y - 1)^2 = 2]This circle has center ((1, 1)) and radius (sqrt{2}).Set ( B ) is defined by the inequality ( kx - y + k + 3 geq 0 ), which represents the region below or on the line ( y = kx + k + 3 ).For ( A cap B ) to be a singleton set, the line ( y = kx + k + 3 ) must be tangent to the circle. The distance from the center of the circle ((1, 1)) to the line must be equal to the radius (sqrt{2}).Rewriting the line equation in standard form ( kx - y + k + 3 = 0 ), we use the distance formula from a point to a line:[frac{|k cdot 1 - 1 + k + 3|}{sqrt{k^2 + 1}} = sqrt{2}]Simplifying the numerator:[|2k + 2| = sqrt{2} sqrt{k^2 + 1}]Squaring both sides:[(2k + 2)^2 = 2(k^2 + 1)]Expanding and simplifying:[4k^2 + 8k + 4 = 2k^2 + 2 implies 2k^2 + 8k + 2 = 0 implies k^2 + 4k + 1 = 0]Solving this quadratic equation:[k = frac{-4 pm sqrt{16 - 4}}{2} = frac{-4 pm 2sqrt{3}}{2} = -2 pm sqrt{3}]We need to determine which value of ( k ) results in the line being tangent from below the circle. Checking the position of the center ((1, 1)) relative to the line for each ( k ):- For ( k = -2 + sqrt{3} ), the line is above the center.- For ( k = -2 - sqrt{3} ), the line is below the center.Since the intersection ( A cap B ) must be a singleton set, the line must be tangent from below, ensuring the center is above the line. Thus, the correct value is ( k = -2 - sqrt{3} ).[boxed{-2 - sqrt{3}}]]

🔑:The point set A represents the circle odot C:(x-1)^{2}+(y-1)^{2}=2, and the point set B represents the region below the line y=k(x+1)+3. Since the line is tangent to odot C, the distance from the center C(1,1) to the line isd=frac{|2 k+2|}{sqrt{k^{2}+1}}=sqrt{2} Rightarrow k^{2}+4 k+1=0 Rightarrow k=frac{-4 pm 2 sqrt{3}}{2}=-2 pm sqrt{3} text {. }Combining the graph, we can see that the value of k is -2-sqrt{3}.