💡First, I need to convert 4 hours and 35 minutes into total minutes. Since there are 60 minutes in an hour, I multiply 4 hours by 60 to get 240 minutes. Then, I add the additional 35 minutes to obtain the total of 275 minutes.Next, to convert 4 tons and 35 kilograms into kilograms, I recognize that 1 ton is equal to 1000 kilograms. Therefore, I multiply 4 tons by 1000 to get 4000 kilograms. Adding the 35 kilograms gives a total of 4035 kilograms.For converting 678 seconds into minutes and seconds, I divide 678 by 60, which gives 11 minutes with a remainder of 18 seconds. This means 678 seconds is equivalent to 11 minutes and 18 seconds.Finally, to convert 120,000 square meters into hectares, I use the conversion factor that 1 hectare equals 10,000 square meters. Dividing 120,000 by 10,000 results in 12 hectares.

💡Alright, so I've got this problem here where they've defined a new operation "⊕" for real numbers. The rules are: if a is greater than or equal to b, then a ⊕ b equals a. But if a is less than b, then a ⊕ b equals b squared. Interesting. I need to find the maximum value of the function f(x) = (1 ⊕ x) * x - (2 ⊕ x) where x is between -2 and 2. The options are A: -1, B: 1, C: 6, D: 12.Okay, let's break this down. First, I need to understand how the operation ⊕ works. So, for any two numbers a and b, if a is bigger or equal to b, then a ⊕ b is just a. But if a is smaller than b, then it's b squared. Got it.Now, let's look at the function f(x). It's composed of two parts: (1 ⊕ x) multiplied by x, and then subtracting (2 ⊕ x). So, f(x) = (1 ⊕ x) * x - (2 ⊕ x). I think the key here is to figure out what 1 ⊕ x and 2 ⊕ x are for different ranges of x, and then compute f(x) accordingly.Let me consider different intervals for x. Since x is in [-2, 2], I can split this interval into parts where 1 ⊕ x and 2 ⊕ x change their behavior based on the definition of ⊕.First, let's consider 1 ⊕ x. The operation depends on whether 1 is greater than or equal to x or not. So, if x is less than or equal to 1, then 1 ⊕ x = 1. But if x is greater than 1, then 1 ⊕ x = x squared. Similarly, for 2 ⊕ x, since 2 is a larger number, we need to see when 2 is greater than or equal to x. Since x is up to 2, when x is less than or equal to 2, 2 ⊕ x = 2. But wait, actually, 2 is always greater than or equal to x in the interval [-2, 2], except when x is greater than 2, which isn't the case here. So, 2 ⊕ x is always 2 in this interval.Wait, hold on. Let me think again. For 2 ⊕ x, since 2 is the first operand, and x is the second. So, if 2 >= x, then 2 ⊕ x = 2. But if 2 < x, then 2 ⊕ x = x squared. But in our interval, x is up to 2, so 2 is always greater than or equal to x. Therefore, 2 ⊕ x is always 2. So, that simplifies things a bit.So, f(x) = (1 ⊕ x) * x - 2. Now, let's focus on 1 ⊕ x. As I thought earlier, if x <= 1, then 1 ⊕ x = 1. If x > 1, then 1 ⊕ x = x squared.Therefore, we can split the function f(x) into two cases:1. When x <= 1: f(x) = 1 * x - 2 = x - 22. When x > 1: f(x) = (x^2) * x - 2 = x^3 - 2So, now we have two different expressions for f(x) depending on whether x is less than or equal to 1 or greater than 1. Our interval is from -2 to 2, so we can analyze f(x) in these two sub-intervals: [-2, 1] and (1, 2].Let's first analyze the interval [-2, 1]. Here, f(x) = x - 2. This is a linear function with a slope of 1, so it's increasing. Therefore, its maximum in this interval will occur at the right endpoint, which is x = 1. Plugging in x = 1, f(1) = 1 - 2 = -1.Now, let's look at the interval (1, 2]. Here, f(x) = x^3 - 2. This is a cubic function, which is also increasing because the derivative f’(x) = 3x^2 is always positive for x ≠ 0. So, in the interval (1, 2], f(x) is increasing. Therefore, its maximum occurs at the right endpoint, x = 2. Plugging in x = 2, f(2) = 8 - 2 = 6.So, comparing the maximum values from both intervals: in [-2, 1], the maximum is -1, and in (1, 2], the maximum is 6. Therefore, the overall maximum value of f(x) on the interval [-2, 2] is 6.Wait, but let me double-check to make sure I didn't make any mistakes. For x in [-2, 1], f(x) = x - 2. At x = -2, f(-2) = -2 - 2 = -4. At x = 1, f(1) = 1 - 2 = -1. So, yes, it's increasing from -4 to -1, so the maximum is indeed -1 in that interval.For x in (1, 2], f(x) = x^3 - 2. At x = 1, f(1) = 1 - 2 = -1. At x = 2, f(2) = 8 - 2 = 6. Since it's increasing, the maximum is 6. So, yes, 6 is the maximum value.Therefore, the answer should be C: 6.**Final Answer**boxed{C}

💡Okay, so I need to construct a quadratic equation with rational coefficients and a leading coefficient of 1, and it should have √5 - 2 as one of its roots. Hmm, let me think about how to approach this.First, I remember that if a quadratic equation has rational coefficients and one of its roots is irrational, like √5 - 2, then the other root must be its conjugate. I think the conjugate in this case would be -√5 - 2. Wait, is that right? Let me double-check. The conjugate of √5 - 2 should involve changing the sign of the radical part, so yes, it would be -√5 - 2. Okay, that makes sense.Now, if I have both roots, I can use Vieta's formulas to find the sum and product of the roots, which will help me construct the quadratic equation. Vieta's formulas say that for a quadratic equation ax² + bx + c = 0, the sum of the roots is -b/a, and the product is c/a. Since the leading coefficient is 1, a = 1, so the sum of the roots will be -b, and the product will be c.Let me calculate the sum of the roots first. So, (√5 - 2) + (-√5 - 2). Let's compute that step by step. The √5 and -√5 will cancel each other out, right? So, √5 - √5 is 0. Then, -2 - 2 is -4. So, the sum of the roots is -4. That means -b = -4, so b = 4.Next, I need to find the product of the roots. That's (√5 - 2) multiplied by (-√5 - 2). Hmm, let me expand this using the distributive property. So, (√5)(-√5) + (√5)(-2) + (-2)(-√5) + (-2)(-2). Let's compute each term:1. (√5)(-√5) = - (√5)² = -52. (√5)(-2) = -2√53. (-2)(-√5) = 2√54. (-2)(-2) = 4Now, let's add all these terms together: -5 - 2√5 + 2√5 + 4. The -2√5 and +2√5 cancel each other out, leaving -5 + 4, which is -1. Wait, that's not right because earlier I thought the product was 1. Did I make a mistake?Let me check my multiplication again. Maybe I should use the formula (a - b)(a + b) = a² - b², but in this case, it's not exactly that because the signs are different. Let me try another approach. Let me consider (√5 - 2)(-√5 - 2). Let me factor out a negative sign from the second term: (√5 - 2)(-1)(√5 + 2) = - (√5 - 2)(√5 + 2). Now, (√5 - 2)(√5 + 2) is (√5)² - (2)² = 5 - 4 = 1. So, the product becomes -1. Hmm, so the product is -1, not 1 as I initially thought. That changes things.Wait, so if the product of the roots is -1, then c = -1. So, putting it all together, the quadratic equation would be x² + 4x - 1 = 0. But wait, earlier I thought the product was 1, but now I'm getting -1. Did I make a mistake in my initial calculation?Let me go back to the original multiplication: (√5 - 2)(-√5 - 2). Let's do it term by term again without factoring out the negative sign. First term: √5 * -√5 = -5. Second term: √5 * -2 = -2√5. Third term: -2 * -√5 = +2√5. Fourth term: -2 * -2 = +4. Now, adding them up: -5 -2√5 + 2√5 +4. The -2√5 and +2√5 cancel out, leaving -5 +4 = -1. So, yes, the product is indeed -1.Wait, so that means the product of the roots is -1, so c = -1. Therefore, the quadratic equation should be x² + 4x -1 = 0. But earlier, when I thought the product was 1, I concluded x² +4x +1=0. So, which one is correct?Let me verify by plugging in the root √5 -2 into the equation x² +4x -1=0. Let's compute (√5 -2)² +4(√5 -2) -1. First, (√5 -2)² = (√5)² - 2*√5*2 + 2² = 5 -4√5 +4 = 9 -4√5. Then, 4(√5 -2) =4√5 -8. Adding these together: (9 -4√5) + (4√5 -8) =9 -4√5 +4√5 -8 =1. Then, subtracting 1 gives 0. So, yes, it satisfies the equation. Therefore, the correct quadratic equation is x² +4x -1=0.Wait, but earlier I thought the product was 1, but now it's -1. So, I must have made a mistake in my initial reasoning when I thought the product was 1. It's actually -1. So, the correct quadratic equation is x² +4x -1=0.But let me double-check everything again to make sure I didn't make any mistakes. Let's start from the beginning.Given that √5 -2 is a root, and the coefficients are rational, the other root must be its conjugate, which is -√5 -2. So, the two roots are √5 -2 and -√5 -2.Sum of roots: (√5 -2) + (-√5 -2) = √5 -√5 -2 -2 = -4. So, sum is -4, which means -b/a = -4, so b=4.Product of roots: (√5 -2)(-√5 -2). Let's compute this carefully.First, expand the product:(√5)(-√5) = -5(√5)(-2) = -2√5(-2)(-√5) = +2√5(-2)(-2) = +4Now, add all these terms: -5 -2√5 +2√5 +4.The -2√5 and +2√5 cancel out, leaving -5 +4 = -1.So, product is -1, which means c/a = -1. Since a=1, c=-1.Therefore, the quadratic equation is x² +4x -1=0.Wait, but earlier I thought the product was 1, but now it's -1. So, I must have made a mistake in my initial calculation. Therefore, the correct equation is x² +4x -1=0.Let me verify by plugging in the root √5 -2 into this equation.Compute (√5 -2)² +4(√5 -2) -1.First, (√5 -2)² = (√5)^2 - 2*√5*2 + 2^2 =5 -4√5 +4=9-4√5.Then, 4(√5 -2)=4√5 -8.Adding these together: (9-4√5)+(4√5 -8)=9-4√5+4√5-8=1.Subtracting 1: 1-1=0. So, yes, it satisfies the equation.Therefore, the correct quadratic equation is x² +4x -1=0.Wait, but in the initial problem, the user wrote that the product was 1, leading to x² +4x +1=0. But according to my calculations, the product is -1, so the equation should be x² +4x -1=0.I think the initial answer was incorrect because the product was miscalculated. So, the correct quadratic equation is x² +4x -1=0.

💡Okay, so I have this problem: I need to determine all integers ( m geq 2 ) for which there exists an integer ( n geq 1 ) such that ( gcd(m, n) = d ) and ( gcd(m, 4n + 1) = 1 ). Hmm, that sounds a bit tricky, but let me try to break it down step by step.First, let me understand the problem. I need to find all ( m ) such that there's some ( n ) where the greatest common divisor of ( m ) and ( n ) is ( d ), and the greatest common divisor of ( m ) and ( 4n + 1 ) is 1. So, ( m ) and ( n ) share a common divisor ( d ), but ( m ) and ( 4n + 1 ) are coprime.Wait, the problem says "for which there exists an integer ( n geq 1 )", so ( d ) must be a divisor of ( m ), right? Because ( gcd(m, n) = d ) implies that ( d ) divides both ( m ) and ( n ). So, ( d ) is a divisor of ( m ). But the problem doesn't specify ( d ); it just says "there exists an integer ( n geq 1 ) with ( gcd(m, n) = d ) and ( gcd(m, 4n + 1) = 1 )". Hmm, so ( d ) is given? Or is it part of the problem?Wait, no, the problem is asking for all ( m geq 2 ) such that there exists some ( n geq 1 ) with ( gcd(m, n) = d ) and ( gcd(m, 4n + 1) = 1 ). So, ( d ) is a specific divisor of ( m ), but the problem doesn't fix ( d ); it just says "there exists an integer ( n )" with those two gcd conditions. So, actually, ( d ) is determined by ( m ) and ( n ). So, for each ( m ), we need to find if there's an ( n ) such that ( gcd(m, n) = d ) and ( gcd(m, 4n + 1) = 1 ). So, ( d ) is dependent on ( m ) and ( n ).Wait, maybe I need to rephrase. The problem is asking for all ( m geq 2 ) such that there exists an ( n geq 1 ) where ( gcd(m, n) = d ) and ( gcd(m, 4n + 1) = 1 ). So, ( d ) is the gcd of ( m ) and ( n ), and we need ( m ) and ( 4n + 1 ) to be coprime.So, perhaps I can write ( m = d cdot k ) and ( n = d cdot l ) where ( gcd(k, l) = 1 ). Because if ( gcd(m, n) = d ), then ( m = d cdot k ) and ( n = d cdot l ) with ( gcd(k, l) = 1 ). That makes sense.So substituting back, ( gcd(m, 4n + 1) = gcd(dk, 4dl + 1) ). Since ( dk ) and ( 4dl + 1 ) are involved, let's compute this gcd.( gcd(dk, 4dl + 1) ). Since ( dk ) and ( 4dl ) share a common factor ( d ), but ( 4dl + 1 ) is one more than a multiple of ( d ). So, ( gcd(dk, 4dl + 1) = gcd(dk, 1) = 1 ), because ( 4dl + 1 ) is congruent to 1 modulo ( d ). Wait, is that right?Wait, ( 4dl + 1 ) modulo ( d ) is ( 1 ), because ( 4dl ) is a multiple of ( d ), so ( 4dl + 1 equiv 1 mod d ). Therefore, ( gcd(dk, 4dl + 1) ) must divide both ( dk ) and ( 4dl + 1 ). But since ( 4dl + 1 equiv 1 mod d ), the gcd can't have any common factors with ( d ). So, the gcd is actually ( gcd(k, 4dl + 1) ).Wait, let me think again. If ( gcd(dk, 4dl + 1) ), since ( d ) and ( 4dl + 1 ) are coprime (as ( 4dl + 1 equiv 1 mod d )), then ( gcd(dk, 4dl + 1) = gcd(k, 4dl + 1) ). So, the condition ( gcd(m, 4n + 1) = 1 ) simplifies to ( gcd(k, 4dl + 1) = 1 ).So, in summary, we have ( m = dk ), ( n = dl ), with ( gcd(k, l) = 1 ), and we need ( gcd(k, 4dl + 1) = 1 ). So, the problem reduces to finding ( m = dk ) such that there exists an ( l ) with ( gcd(k, l) = 1 ) and ( gcd(k, 4dl + 1) = 1 ).Hmm, okay, so perhaps I can think about this in terms of the prime factors of ( k ). Let me consider the prime factors of ( k ). Suppose ( p ) is a prime divisor of ( k ). Then, for ( gcd(k, 4dl + 1) = 1 ), we must have that ( p ) does not divide ( 4dl + 1 ).So, for each prime ( p ) dividing ( k ), we need that ( 4dl + 1 notequiv 0 mod p ). That is, ( 4dl notequiv -1 mod p ). Since ( l ) is such that ( gcd(k, l) = 1 ), ( l ) is coprime to ( k ), so ( l ) is invertible modulo ( p ).Therefore, for each prime ( p ) dividing ( k ), we can choose ( l ) such that ( 4dl notequiv -1 mod p ). Since ( l ) is invertible modulo ( p ), this is equivalent to choosing ( l ) such that ( l notequiv - (4d)^{-1} mod p ).But since ( l ) is arbitrary except for being coprime to ( k ), we can choose ( l ) such that ( l ) is not congruent to ( - (4d)^{-1} mod p ) for any prime ( p ) dividing ( k ). Is that possible?Well, for each prime ( p ) dividing ( k ), the number of forbidden residues modulo ( p ) is 1 (since ( l ) cannot be congruent to ( - (4d)^{-1} mod p )). Since ( l ) is chosen to be coprime to ( k ), which has ( phi(k) ) choices modulo ( k ). But since we're working modulo each prime ( p ), we can use the Chinese Remainder Theorem to find such an ( l ).Wait, maybe I need to think about this more carefully. Let me consider the Chinese Remainder Theorem. If I can find an ( l ) such that for each prime ( p ) dividing ( k ), ( l notequiv - (4d)^{-1} mod p ), then such an ( l ) exists because the forbidden residues are only one per prime, and the Chinese Remainder Theorem allows us to find such an ( l ).But wait, actually, ( l ) must satisfy ( gcd(k, l) = 1 ), so ( l ) must not be divisible by any prime ( p ) dividing ( k ). So, ( l ) is coprime to ( k ), which is equivalent to ( l ) being coprime to each prime ( p ) dividing ( k ).Therefore, for each prime ( p ) dividing ( k ), ( l ) must not be congruent to 0 modulo ( p ), and also not congruent to ( - (4d)^{-1} mod p ). So, for each prime ( p ), there are two forbidden residues: 0 and ( - (4d)^{-1} mod p ). Since ( p geq 2 ), there are at least two residues, so as long as ( p ) is greater than 2, there are more than two residues, so we can choose ( l ) such that it avoids both.But wait, if ( p = 2 ), then modulo 2, the residues are 0 and 1. If ( p = 2 ) divides ( k ), then ( l ) must be odd (since ( l ) must be coprime to ( k )), so ( l equiv 1 mod 2 ). Then, we need to check if ( 4dl + 1 equiv 0 mod 2 ). But ( 4dl ) is even, so ( 4dl + 1 ) is odd, so ( gcd(2, 4dl + 1) = 1 ). So, for ( p = 2 ), it's automatically satisfied because ( 4dl + 1 ) is odd, so ( gcd(2, 4dl + 1) = 1 ). Therefore, for ( p = 2 ), we don't have any additional constraints beyond ( l ) being odd.For odd primes ( p ), we need ( l notequiv - (4d)^{-1} mod p ). Since ( l ) is already coprime to ( p ), ( l ) can be any residue modulo ( p ) except 0 and ( - (4d)^{-1} mod p ). Since ( p geq 3 ), there are at least two non-zero residues, so we can choose ( l ) such that ( l notequiv - (4d)^{-1} mod p ).Therefore, for each prime ( p ) dividing ( k ), we can choose ( l ) such that ( l ) is not congruent to ( - (4d)^{-1} mod p ). Since these are congruence conditions modulo each prime ( p ), by the Chinese Remainder Theorem, there exists an ( l ) satisfying all these conditions simultaneously.Therefore, as long as ( k ) is such that for each prime ( p ) dividing ( k ), ( 4d ) is invertible modulo ( p ), which it is because ( p ) divides ( k ) and ( gcd(k, l) = 1 ), so ( p ) does not divide ( l ), but ( 4d ) may or may not be invertible modulo ( p ). Wait, actually, ( 4d ) is an integer, and if ( p ) divides ( 4d ), then ( 4d equiv 0 mod p ), so ( (4d)^{-1} ) doesn't exist modulo ( p ). Hmm, that complicates things.Wait, so if ( p ) divides ( 4d ), then ( 4d equiv 0 mod p ), so ( 4dl equiv 0 mod p ), so ( 4dl + 1 equiv 1 mod p ). Therefore, ( gcd(p, 4dl + 1) = 1 ), since ( 4dl + 1 equiv 1 mod p ). So, in this case, even if ( p ) divides ( 4d ), ( 4dl + 1 ) is congruent to 1 modulo ( p ), so ( p ) does not divide ( 4dl + 1 ). Therefore, ( gcd(k, 4dl + 1) = 1 ) automatically for primes ( p ) dividing both ( k ) and ( 4d ).Wait, so maybe I don't have to worry about primes dividing both ( k ) and ( 4d ), because in that case, ( 4dl + 1 equiv 1 mod p ), so ( p ) doesn't divide ( 4dl + 1 ). Therefore, the problematic primes are those that divide ( k ) but do not divide ( 4d ). For those primes, ( 4d ) is invertible modulo ( p ), so ( (4d)^{-1} ) exists, and we need to avoid ( l equiv - (4d)^{-1} mod p ).So, to summarize, for each prime ( p ) dividing ( k ), if ( p ) divides ( 4d ), then ( gcd(k, 4dl + 1) = 1 ) automatically. If ( p ) does not divide ( 4d ), then we need to choose ( l ) such that ( l notequiv - (4d)^{-1} mod p ). Since ( l ) is already coprime to ( p ), we can choose such an ( l ) as long as ( p ) is greater than 2, which it is because if ( p = 2 ), we already handled it.Therefore, the conclusion is that for any ( k ), as long as we can choose ( l ) such that ( l ) is coprime to ( k ) and avoids the forbidden residues modulo each prime ( p ) dividing ( k ) and not dividing ( 4d ), then such an ( l ) exists. Therefore, for any ( m = dk ), there exists an ( n = dl ) such that ( gcd(m, n) = d ) and ( gcd(m, 4n + 1) = 1 ).Wait, but does this hold for any ( m )? Or are there restrictions on ( m )?Wait, let's think about specific examples. Let me take ( m = 2 ). Then ( d ) must be 1 or 2. Let's see:Case 1: ( d = 1 ). Then ( m = 2 ), ( n ) must satisfy ( gcd(2, n) = 1 ), so ( n ) is odd. Then ( 4n + 1 ) is ( 4 times text{odd} + 1 = 4k + 1 ), which is 1 mod 4, so ( gcd(2, 4n + 1) = 1 ). So, yes, for ( m = 2 ), ( d = 1 ), and ( n ) odd works.Case 2: ( d = 2 ). Then ( m = 2 ), ( n ) must be even. Let ( n = 2l ). Then ( 4n + 1 = 8l + 1 ). Now, ( gcd(2, 8l + 1) = 1 ), since ( 8l + 1 ) is odd. So, yes, ( d = 2 ) also works with ( n = 2l ).So, for ( m = 2 ), both ( d = 1 ) and ( d = 2 ) are possible, and in both cases, we can find an ( n ) such that the gcd conditions are satisfied.Another example: ( m = 3 ). Let's see:Case 1: ( d = 1 ). Then ( n ) must be coprime to 3. Let's choose ( n = 1 ). Then ( 4n + 1 = 5 ), and ( gcd(3, 5) = 1 ). So, that works.Case 2: ( d = 3 ). Then ( n = 3l ). Then ( 4n + 1 = 12l + 1 ). Now, ( gcd(3, 12l + 1) ). Since ( 12l ) is divisible by 3, ( 12l + 1 equiv 1 mod 3 ), so ( gcd(3, 12l + 1) = 1 ). So, yes, ( d = 3 ) works.So, for ( m = 3 ), both ( d = 1 ) and ( d = 3 ) are possible.Wait, so maybe for any ( m ), both ( d = 1 ) and ( d = m ) are possible? Let me check ( m = 4 ).Case 1: ( d = 1 ). Then ( n ) must be coprime to 4, so ( n ) is odd. Then ( 4n + 1 ) is ( 4 times text{odd} + 1 = 4k + 1 ), which is 1 mod 4, so ( gcd(4, 4n + 1) = 1 ). So, that works.Case 2: ( d = 2 ). Then ( n = 2l ). Then ( 4n + 1 = 8l + 1 ). ( gcd(4, 8l + 1) = 1 ), since ( 8l + 1 ) is odd. So, that works.Case 3: ( d = 4 ). Then ( n = 4l ). Then ( 4n + 1 = 16l + 1 ). ( gcd(4, 16l + 1) = 1 ), since ( 16l + 1 ) is 1 mod 4. So, that works.So, for ( m = 4 ), ( d = 1, 2, 4 ) all work.Wait, so maybe for any ( m ), all divisors ( d ) of ( m ) work? Because in each case, I can choose ( n = dl ) such that ( l ) is coprime to ( k = m/d ), and then ( 4n + 1 ) is congruent to 1 modulo ( d ), so ( gcd(m, 4n + 1) = gcd(dk, 4dl + 1) = gcd(k, 4dl + 1) ). Then, as long as ( l ) is chosen such that ( 4dl + 1 ) is coprime to ( k ), which is possible by the Chinese Remainder Theorem, as I thought earlier.Therefore, perhaps the answer is that for any ( m geq 2 ), there exists such an ( n ), meaning that all integers ( m geq 2 ) satisfy the condition. But wait, that can't be right because the problem is asking to determine all such ( m ), implying that not all ( m ) satisfy it.Wait, maybe I'm missing something. Let me think again. Suppose ( m ) is such that all its prime factors are 2. For example, ( m = 8 ). Let's see:Case 1: ( d = 1 ). Then ( n ) must be odd. Then ( 4n + 1 ) is 1 mod 4, so ( gcd(8, 4n + 1) = 1 ). So, that works.Case 2: ( d = 2 ). Then ( n = 2l ). Then ( 4n + 1 = 8l + 1 ). ( gcd(8, 8l + 1) = 1 ), since ( 8l + 1 ) is 1 mod 8. So, that works.Case 3: ( d = 4 ). Then ( n = 4l ). Then ( 4n + 1 = 16l + 1 ). ( gcd(8, 16l + 1) = 1 ), since ( 16l + 1 ) is 1 mod 8. So, that works.Case 4: ( d = 8 ). Then ( n = 8l ). Then ( 4n + 1 = 32l + 1 ). ( gcd(8, 32l + 1) = 1 ), since ( 32l + 1 ) is 1 mod 8. So, that works.So, even for ( m = 8 ), all divisors ( d ) work.Wait, maybe I need to consider a different ( m ). Let me try ( m = 6 ).Case 1: ( d = 1 ). Then ( n ) must be coprime to 6, so ( n ) is not divisible by 2 or 3. Then ( 4n + 1 ). Let's see, ( gcd(6, 4n + 1) ). Since ( n ) is coprime to 6, ( n ) is 1, 5, 7, 11, etc. Let's take ( n = 1 ). Then ( 4n + 1 = 5 ), ( gcd(6, 5) = 1 ). So, that works.Case 2: ( d = 2 ). Then ( n = 2l ). Then ( 4n + 1 = 8l + 1 ). ( gcd(6, 8l + 1) ). Let's choose ( l = 1 ), so ( 8 + 1 = 9 ). ( gcd(6, 9) = 3 ). Oh, that's not 1. Hmm, so that's a problem.Wait, so ( d = 2 ) might not always work. Let me check again. If ( d = 2 ), ( m = 6 ), ( n = 2l ). Then ( 4n + 1 = 8l + 1 ). We need ( gcd(6, 8l + 1) = 1 ). So, ( 8l + 1 ) must not be divisible by 2 or 3.But ( 8l + 1 ) is always odd, so it's not divisible by 2. However, we need to ensure it's not divisible by 3. So, ( 8l + 1 notequiv 0 mod 3 ). ( 8 equiv 2 mod 3 ), so ( 2l + 1 notequiv 0 mod 3 ). So, ( 2l notequiv -1 mod 3 ), which is ( 2l notequiv 2 mod 3 ), so ( l notequiv 1 mod 3 ).Therefore, if we choose ( l ) such that ( l notequiv 1 mod 3 ), then ( gcd(6, 8l + 1) = 1 ). For example, ( l = 2 ): ( 8*2 + 1 = 17 ), ( gcd(6, 17) = 1 ). So, yes, ( d = 2 ) works if we choose ( l ) appropriately.Wait, so in this case, even though for ( l = 1 ), it didn't work, for ( l = 2 ), it did. So, as long as we can choose ( l ) such that ( l notequiv 1 mod 3 ), then ( d = 2 ) works.Similarly, for ( d = 3 ): ( n = 3l ). Then ( 4n + 1 = 12l + 1 ). ( gcd(6, 12l + 1) ). Since ( 12l + 1 ) is 1 mod 2 and 1 mod 3, so ( gcd(6, 12l + 1) = 1 ). So, that works.For ( d = 6 ): ( n = 6l ). Then ( 4n + 1 = 24l + 1 ). ( gcd(6, 24l + 1) = 1 ), since ( 24l + 1 ) is 1 mod 6. So, that works.So, for ( m = 6 ), all divisors ( d ) work as long as we choose ( l ) appropriately.Wait, so maybe for any ( m ), regardless of its prime factors, as long as we can choose ( l ) such that ( l ) avoids certain residues modulo the prime factors of ( k = m/d ), then such an ( l ) exists. Therefore, the answer is that all integers ( m geq 2 ) satisfy the condition.But wait, let me test ( m = 5 ). Let's see:Case 1: ( d = 1 ). Then ( n ) must be coprime to 5. Let's choose ( n = 1 ). Then ( 4n + 1 = 5 ). ( gcd(5, 5) = 5 ), which is not 1. Oh, that's a problem.Wait, so ( d = 1 ) might not always work. Let me check again. If ( d = 1 ), ( m = 5 ), ( n ) must be coprime to 5. So, ( n ) can be 1, 2, 3, 4, 6, etc. Let's try ( n = 2 ). Then ( 4n + 1 = 9 ). ( gcd(5, 9) = 1 ). So, that works. So, ( d = 1 ) works for ( m = 5 ) if we choose ( n = 2 ).Similarly, ( n = 3 ): ( 4*3 + 1 = 13 ), ( gcd(5, 13) = 1 ). So, that works too. So, as long as ( n ) is chosen such that ( 4n + 1 ) is not a multiple of 5. So, ( 4n + 1 notequiv 0 mod 5 ), which is ( 4n notequiv -1 mod 5 ), so ( n notequiv 1 mod 5 ). So, as long as ( n ) is not congruent to 1 mod 5, it works. Since ( n ) is coprime to 5, ( n ) can be 1, 2, 3, 4 mod 5. So, if ( n equiv 1 mod 5 ), then ( 4n + 1 equiv 0 mod 5 ), which is bad. But if ( n ) is 2, 3, or 4 mod 5, then ( 4n + 1 ) is not 0 mod 5.Therefore, for ( m = 5 ), ( d = 1 ) works as long as ( n ) is not congruent to 1 mod 5. So, we can choose such an ( n ).Wait, so even though for ( n = 1 ), it didn't work, for ( n = 2 ), it did. So, as long as we can choose ( n ) such that ( n ) is coprime to ( m ) and ( 4n + 1 ) is not divisible by any prime divisor of ( m ), then it works.Therefore, the conclusion is that for any ( m geq 2 ), there exists an ( n ) such that ( gcd(m, n) = d ) and ( gcd(m, 4n + 1) = 1 ). Therefore, all integers ( m geq 2 ) satisfy the condition.But wait, the problem says "determine all integers ( m geq 2 ) for which there exists an integer ( n geq 1 ) with ( gcd(m,n)=d ) and ( gcd(m,4n+1)=1 )." So, the answer is all integers ( m geq 2 ).But wait, in the case of ( m = 5 ), ( d = 1 ) works, but what if ( m ) is such that all its prime factors are 2? For example, ( m = 8 ). As I saw earlier, ( d = 1, 2, 4, 8 ) all work. So, even powers of 2 are okay.Wait, another test case: ( m = 7 ). Let's see:Case 1: ( d = 1 ). Then ( n ) must be coprime to 7. Let's choose ( n = 2 ). Then ( 4n + 1 = 9 ), ( gcd(7, 9) = 1 ). So, that works.Case 2: ( d = 7 ). Then ( n = 7l ). Then ( 4n + 1 = 28l + 1 ). ( gcd(7, 28l + 1) = 1 ), since ( 28l + 1 equiv 1 mod 7 ). So, that works.So, ( m = 7 ) works.Wait, so maybe indeed, for any ( m geq 2 ), such an ( n ) exists. Therefore, the answer is all integers ( m geq 2 ).But wait, let me think about ( m = 1 ). Wait, the problem says ( m geq 2 ), so ( m = 1 ) is not considered. So, all ( m geq 2 ) satisfy the condition.But wait, let me think about ( m = 9 ). Let's see:Case 1: ( d = 1 ). Then ( n ) must be coprime to 9. Let's choose ( n = 2 ). Then ( 4n + 1 = 9 ), ( gcd(9, 9) = 9 ), which is not 1. So, that's a problem.Wait, but we can choose another ( n ). Let's try ( n = 1 ). Then ( 4n + 1 = 5 ), ( gcd(9, 5) = 1 ). So, that works. So, ( d = 1 ) works for ( m = 9 ) if we choose ( n = 1 ).Similarly, ( n = 4 ): ( 4*4 + 1 = 17 ), ( gcd(9, 17) = 1 ). So, that works too.So, as long as ( n ) is chosen such that ( 4n + 1 ) is not a multiple of 3. Since ( m = 9 ), which is ( 3^2 ), we need ( 4n + 1 notequiv 0 mod 3 ). ( 4n + 1 equiv n + 1 mod 3 ). So, ( n + 1 notequiv 0 mod 3 ), so ( n notequiv 2 mod 3 ). Since ( n ) is coprime to 9, ( n ) can be 1, 2, 4, 5, 7, 8, etc. So, if ( n equiv 2 mod 3 ), then ( 4n + 1 equiv 0 mod 3 ), which is bad. But if ( n equiv 1 ) or ( 0 mod 3 ), but ( n ) must be coprime to 9, so ( n ) cannot be 0 mod 3. So, ( n ) must be 1 or 2 mod 3. But if ( n equiv 2 mod 3 ), then ( 4n + 1 equiv 0 mod 3 ). Therefore, to avoid that, we need ( n equiv 1 mod 3 ). So, as long as we choose ( n equiv 1 mod 3 ), then ( 4n + 1 equiv 4 + 1 = 5 mod 3 ), which is 2 mod 3, so ( gcd(9, 4n + 1) = 1 ).Therefore, for ( m = 9 ), ( d = 1 ) works if we choose ( n equiv 1 mod 3 ). So, yes, it works.Wait, so even for ( m = 9 ), which is a square of a prime, it works. So, perhaps indeed, for any ( m geq 2 ), such an ( n ) exists.Therefore, the answer is that all integers ( m geq 2 ) satisfy the condition. So, the set of all integers ( m geq 2 ).But wait, let me think about ( m = 15 ). Let's see:Case 1: ( d = 1 ). Then ( n ) must be coprime to 15. Let's choose ( n = 2 ). Then ( 4n + 1 = 9 ), ( gcd(15, 9) = 3 ), which is not 1. So, that's a problem.But we can choose another ( n ). Let's try ( n = 1 ). Then ( 4n + 1 = 5 ), ( gcd(15, 5) = 5 ), which is not 1. Hmm, that's also a problem.Wait, let's try ( n = 4 ). Then ( 4n + 1 = 17 ), ( gcd(15, 17) = 1 ). So, that works. So, ( d = 1 ) works for ( m = 15 ) if we choose ( n = 4 ).Similarly, ( n = 7 ): ( 4*7 + 1 = 29 ), ( gcd(15, 29) = 1 ). So, that works too.So, as long as ( n ) is chosen such that ( 4n + 1 ) is not divisible by 3 or 5. Since ( m = 15 = 3 times 5 ), we need ( 4n + 1 notequiv 0 mod 3 ) and ( 4n + 1 notequiv 0 mod 5 ).For mod 3: ( 4n + 1 equiv n + 1 mod 3 ). So, ( n + 1 notequiv 0 mod 3 ), so ( n notequiv 2 mod 3 ).For mod 5: ( 4n + 1 equiv 4n + 1 mod 5 ). So, ( 4n notequiv -1 mod 5 ), which is ( 4n notequiv 4 mod 5 ), so ( n notequiv 1 mod 5 ).Therefore, as long as ( n ) is coprime to 15 and ( n notequiv 2 mod 3 ) and ( n notequiv 1 mod 5 ), then ( gcd(15, 4n + 1) = 1 ). Since ( n ) is coprime to 15, it's already not divisible by 3 or 5, so ( n ) can be 1, 2, 4, 7, 8, 11, 13, 14, etc. So, choosing ( n = 4 ), which is 1 mod 3 and 4 mod 5, which is not 2 mod 3 or 1 mod 5, works.Therefore, for ( m = 15 ), ( d = 1 ) works as long as ( n ) is chosen appropriately.So, putting it all together, it seems that for any ( m geq 2 ), there exists an ( n ) such that ( gcd(m, n) = d ) and ( gcd(m, 4n + 1) = 1 ). Therefore, the answer is that all integers ( m geq 2 ) satisfy the condition.But wait, let me think about ( m = 1 ). Wait, the problem specifies ( m geq 2 ), so ( m = 1 ) is not considered. So, yes, all ( m geq 2 ) work.Therefore, the final answer is that all integers ( m geq 2 ) satisfy the condition.

💡To determine the value of ( m ) such that the points ( A(2, -3) ), ( B(4, 3) ), and ( C(5, m) ) lie on the same straight line, we can use the concept of slope. The slope between two points ( (x_1, y_1) ) and ( (x_2, y_2) ) is given by the formula:[text{Slope} = frac{y_2 - y_1}{x_2 - x_1}]First, calculate the slope between points ( A ) and ( B ):[text{Slope}_{AB} = frac{3 - (-3)}{4 - 2} = frac{6}{2} = 3]Next, calculate the slope between points ( B ) and ( C ):[text{Slope}_{BC} = frac{m - 3}{5 - 4} = m - 3]Since the points are collinear, their slopes must be equal:[m - 3 = 3]Solving for ( m ):[m = 3 + 3 = 6]

💡Alright, let me try to figure out this geometry problem. It's about an acute-angled triangle ABC, with M being the midpoint of AC. There's a circle Γ passing through B and M, intersecting AB and BC at points P and Q respectively. Then, there's a point T such that BPQT is a parallelogram, and T lies on the circumcircle of triangle ABC. We need to find the possible values of BT/BM.First, let me visualize the triangle ABC. Since it's acute-angled, all angles are less than 90 degrees. M is the midpoint of AC, so AM = MC. The circle Γ passes through B and M, so it intersects AB at P and BC at Q. So, points P and Q are on AB and BC respectively, and Γ passes through B, P, M, and Q.Now, BPQT is a parallelogram. That means BP is parallel to QT and BQ is parallel to PT. Also, BP = QT and BQ = PT. Since T is on the circumcircle of ABC, it must lie somewhere on that circle.I think I need to use properties of circles, parallelograms, and midpoints here. Maybe some similar triangles or power of a point. Let me try to draw this out step by step.First, since M is the midpoint of AC, and Γ passes through M and B, perhaps there's some symmetry or midpoint theorem involved. Also, since BPQT is a parallelogram, the diagonals bisect each other. So, the midpoint of BT should be the same as the midpoint of PQ.Let me denote the midpoint of PQ as S. Then, S is also the midpoint of BT. So, BS = ST and PS = SQ.Since Γ passes through B, P, M, and Q, maybe I can use cyclic quadrilateral properties. For example, angles subtended by the same chord are equal. So, angle BPM should be equal to angle BQM because they both subtend chord BM.Wait, but BPQT is a parallelogram, so angle BPQ is equal to angle BQT. Hmm, not sure if that helps directly.Maybe I can use power of a point. For point A, the power with respect to Γ would be AP * AB = AM * something? Wait, not sure. Let me think.Alternatively, since M is the midpoint, maybe I can use vectors or coordinate geometry. But since it's a ratio, maybe coordinate geometry could simplify things.Let me set up a coordinate system. Let me place point B at the origin (0,0). Let me assume AB is along the x-axis, so point A is at (2a, 0) for some a. Since M is the midpoint of AC, let me denote point C as (2c, 2d), so M would be at (a + c, d).Wait, maybe that's too complicated. Alternatively, let me set coordinates such that B is at (0,0), A is at (2,0), and C is at (0,2). Then M, the midpoint of AC, would be at (1,1).Then, the circle Γ passes through B(0,0) and M(1,1). Let me find the equation of circle Γ. The general equation is x² + y² + Dx + Ey + F = 0. Since it passes through B(0,0), substituting gives F = 0. So equation is x² + y² + Dx + Ey = 0.It also passes through M(1,1), so substituting gives 1 + 1 + D + E = 0 => D + E = -2.Now, the circle intersects AB and BC at P and Q. AB is the x-axis from (0,0) to (2,0). BC is the line from (0,0) to (0,2). So, let me find points P and Q.For point P on AB: y=0. Substitute into circle equation: x² + 0 + Dx + E*0 = 0 => x² + Dx = 0. So, x(x + D) = 0. So, x=0 (which is point B) and x = -D. So, P is at (-D, 0).Similarly, for point Q on BC: x=0. Substitute into circle equation: 0 + y² + D*0 + Ey = 0 => y² + Ey = 0 => y(y + E) = 0. So, y=0 (point B) and y = -E. So, Q is at (0, -E).So, points P(-D, 0) and Q(0, -E). Now, BPQT is a parallelogram. So, vectors BP and BQ should define the sides, and T should be such that BPQT is a parallelogram.Wait, in a parallelogram, the vector from B to P plus the vector from B to Q should give the vector from B to T. So, vector BP is (-D, 0) and vector BQ is (0, -E). So, vector BT would be (-D, -E). Therefore, point T is at (-D, -E).But T is supposed to lie on the circumcircle of triangle ABC. The circumcircle of ABC with points A(2,0), B(0,0), and C(0,2). Let me find the equation of the circumcircle.The circumcircle of ABC can be found using the perpendicular bisectors. The midpoint of AB is (1,0), and the perpendicular bisector is the line y = something. Wait, AB is horizontal, so its perpendicular bisector is vertical through (1,0). Similarly, the midpoint of AC is (1,1), and the perpendicular bisector of AC is the line y = x. Wait, no.Wait, point A(2,0), point C(0,2). The midpoint is (1,1). The slope of AC is (2-0)/(0-2) = -1, so the perpendicular bisector has slope 1. So, equation is y -1 = 1*(x -1), so y = x.Similarly, the perpendicular bisector of AB is the vertical line x=1. So, the intersection of x=1 and y=x is (1,1). So, the circumcircle is centered at (1,1) with radius the distance from (1,1) to A(2,0), which is sqrt((1)^2 + (-1)^2) = sqrt(2). So, equation is (x -1)^2 + (y -1)^2 = 2.Now, point T is at (-D, -E). It must lie on this circumcircle. So, substituting into the equation:(-D -1)^2 + (-E -1)^2 = 2.Expanding: (D +1)^2 + (E +1)^2 = 2.We also know from earlier that D + E = -2.So, we have two equations:1. D + E = -22. (D +1)^2 + (E +1)^2 = 2Let me substitute E = -2 - D into the second equation.So, (D +1)^2 + (-2 - D +1)^2 = 2Simplify:(D +1)^2 + (-1 - D)^2 = 2Note that (-1 - D)^2 = (D +1)^2, so:2*(D +1)^2 = 2Divide both sides by 2:(D +1)^2 = 1Take square roots:D +1 = ±1So, D = 0 or D = -2Case 1: D = 0Then, E = -2 - D = -2So, point P is (-D, 0) = (0,0), which is point B. But P is supposed to be another intersection point on AB, so P can't be B. So, this case is invalid.Case 2: D = -2Then, E = -2 - (-2) = 0So, point P is (-D, 0) = (2,0), which is point A. But P is supposed to be on AB, which is from (0,0) to (2,0). So, P=A is allowed? Wait, but the circle Γ passes through B and M, and intersects AB at P and B. If P=A, then circle Γ passes through A, B, and M. Let me check if that's possible.Points A(2,0), B(0,0), M(1,1). The circle through these three points: Let me compute its equation.Using three points: (2,0), (0,0), (1,1).The general equation is x² + y² + Dx + Ey + F = 0.Substitute (0,0): 0 + 0 + 0 + 0 + F = 0 => F=0.Substitute (2,0): 4 + 0 + 2D + 0 = 0 => 2D = -4 => D = -2.Substitute (1,1): 1 + 1 + (-2) + E = 0 => 2 -2 + E = 0 => E = 0.So, equation is x² + y² -2x = 0. Completing the square: (x -1)^2 + y^2 = 1.So, circle with center (1,0) and radius 1. But point M(1,1) is on this circle, since (1-1)^2 +1^2 =1. So, yes, it works.So, in this case, point P is A(2,0), and point Q is (0, -E) = (0,0), which is point B. But Q is supposed to be another intersection point on BC, so Q=B is not allowed. So, this case is also invalid.Wait, both cases lead to P or Q being B, which is not allowed. Hmm, that's a problem. Maybe my coordinate setup is causing issues.Perhaps I should choose a different coordinate system. Let me try to place point B at (0,0), point A at (2,0), and point C at (0,2c) for some c >0, so that M is at (1, c). Maybe this will make the calculations more general.So, points:A(2,0), B(0,0), C(0,2c), M(1,c).Circle Γ passes through B(0,0) and M(1,c). Let me find its equation.General equation: x² + y² + Dx + Ey + F =0.Passes through B(0,0): 0 +0 +0 +0 + F=0 => F=0.Passes through M(1,c): 1 + c² + D + E c =0 => D + E c = -1 -c².So, equation is x² + y² + Dx + Ey =0.Now, find intersections P on AB and Q on BC.AB is from (0,0) to (2,0), so y=0.Substitute y=0 into circle equation: x² + Dx =0 => x(x + D)=0. So, x=0 (point B) and x=-D. So, P is at (-D, 0).Similarly, BC is from (0,0) to (0,2c), so x=0.Substitute x=0 into circle equation: y² + Ey =0 => y(y + E)=0. So, y=0 (point B) and y=-E. So, Q is at (0, -E).Now, BPQT is a parallelogram. So, as before, vector BP is (-D,0), vector BQ is (0,-E). So, vector BT is (-D, -E). Therefore, point T is at (-D, -E).T lies on the circumcircle of ABC. Let me find the equation of the circumcircle.Points A(2,0), B(0,0), C(0,2c). Let me find the circumcircle.Midpoint of AB is (1,0). The perpendicular bisector is the line perpendicular to AB (which is horizontal) through (1,0). So, it's the vertical line x=1.Midpoint of AC is (1, c). The slope of AC is (2c -0)/(0 -2) = -c. So, the perpendicular bisector has slope 1/c.Equation of perpendicular bisector of AC: passes through (1,c) with slope 1/c.So, equation: y - c = (1/c)(x -1).Intersection with x=1: y - c = (1/c)(0) => y = c.So, the circumcircle is centered at (1, c) with radius equal to the distance from (1,c) to A(2,0):sqrt((2-1)^2 + (0 - c)^2) = sqrt(1 + c²).So, equation is (x -1)^2 + (y - c)^2 = 1 + c².Now, point T(-D, -E) lies on this circle:(-D -1)^2 + (-E - c)^2 = 1 + c².Expanding:(D +1)^2 + (E + c)^2 = 1 + c².Simplify:(D +1)^2 + E² + 2E c + c² = 1 + c².Cancel c²:(D +1)^2 + E² + 2E c =1.From earlier, we have D + E c = -1 -c².Let me write that as D = -1 -c² - E c.Substitute D into the equation:(-1 -c² - E c +1)^2 + E² + 2E c =1.Simplify inside the square:(-c² - E c)^2 + E² + 2E c =1.Expand (-c² - E c)^2:c⁴ + 2c³ E + c² E².So, equation becomes:c⁴ + 2c³ E + c² E² + E² + 2E c =1.Let me collect like terms:c⁴ + 2c³ E + (c² +1) E² + 2c E -1 =0.This seems complicated. Maybe there's a better approach.Alternatively, since T is at (-D, -E), and it's on the circumcircle, maybe we can relate BT and BM.BM is the distance from B(0,0) to M(1,c):BM = sqrt(1² + c²) = sqrt(1 + c²).BT is the distance from B(0,0) to T(-D, -E):BT = sqrt(D² + E²).We need to find BT/BM = sqrt(D² + E²)/sqrt(1 + c²).From earlier, D + E c = -1 -c².Let me denote this as equation (1): D + E c = -1 -c².We also have from the circle equation:(D +1)^2 + (E + c)^2 =1 + c².Expanding:D² + 2D +1 + E² + 2E c + c² =1 + c².Simplify:D² + 2D + E² + 2E c =0.From equation (1): D = -1 -c² - E c.Substitute into D²:(-1 -c² - E c)^2 = (1 + c² + E c)^2 =1 + 2c² + c⁴ + 2E c + 2E c³ + E² c².So, D² =1 + 2c² + c⁴ + 2E c + 2E c³ + E² c².Similarly, 2D = 2*(-1 -c² - E c) = -2 -2c² -2E c.So, D² + 2D = [1 + 2c² + c⁴ + 2E c + 2E c³ + E² c²] + [-2 -2c² -2E c] = (1 -2) + (2c² -2c²) + c⁴ + (2E c -2E c) + 2E c³ + E² c² = -1 + c⁴ + 2E c³ + E² c².Now, D² + 2D + E² + 2E c = (-1 + c⁴ + 2E c³ + E² c²) + E² + 2E c = -1 + c⁴ + 2E c³ + E² c² + E² + 2E c.But from earlier, D² + 2D + E² + 2E c =0.So,-1 + c⁴ + 2E c³ + E² c² + E² + 2E c =0.Let me factor E² terms:E²(c² +1) + 2E(c³ + c) + (c⁴ -1)=0.This is a quadratic in E:E²(c² +1) + 2E(c³ + c) + (c⁴ -1)=0.Let me solve for E:E = [-2(c³ + c) ± sqrt{4(c³ + c)^2 -4(c² +1)(c⁴ -1)}}]/[2(c² +1)].Simplify discriminant:4(c³ + c)^2 -4(c² +1)(c⁴ -1) =4[ (c³ +c)^2 - (c² +1)(c⁴ -1) ].Compute inside:(c³ +c)^2 =c⁶ + 2c⁴ +c².(c² +1)(c⁴ -1)=c⁶ -c² +c⁴ -1.Subtract:(c⁶ + 2c⁴ +c²) - (c⁶ -c² +c⁴ -1) =c⁶ +2c⁴ +c² -c⁶ +c² -c⁴ +1= (2c⁴ -c⁴) + (c² +c²) +1= c⁴ +2c² +1= (c² +1)^2.So, discriminant=4*(c² +1)^2.Thus,E = [-2(c³ +c) ± 2(c² +1)]/[2(c² +1)] = [- (c³ +c) ± (c² +1)]/(c² +1).So, two cases:Case 1: E = [ - (c³ +c) + (c² +1) ]/(c² +1) = [ -c³ -c +c² +1 ]/(c² +1).Factor numerator:- c³ +c² -c +1 = -c²(c -1) -1(c -1) = (-c² -1)(c -1).Wait, let me factor:- c³ +c² -c +1 = -(c³ -c² +c -1) = -(c²(c -1) +1(c -1)) = -(c² +1)(c -1).So, E = [ -(c² +1)(c -1) ] / (c² +1 )= -(c -1).So, E = 1 -c.Case 2: E = [ - (c³ +c) - (c² +1) ]/(c² +1) = [ -c³ -c -c² -1 ]/(c² +1).Factor numerator:- c³ -c² -c -1 = - (c³ +c² +c +1 )= - (c²(c +1) +1(c +1))= - (c² +1)(c +1).Thus, E = [ - (c² +1)(c +1) ] / (c² +1 )= -(c +1).So, E = - (c +1).Now, let's consider both cases.Case 1: E =1 -c.From equation (1): D + E c = -1 -c².So, D + (1 -c)c = -1 -c².Compute:D + c -c² = -1 -c².Thus, D = -1 -c² -c +c² = -1 -c.So, D= -1 -c.Thus, point P is at (-D,0)= (1 +c, 0).Point Q is at (0, -E)= (0, c -1).Now, point T is at (-D, -E)= (1 +c, c -1).We need to ensure that P is on AB and Q is on BC.AB is from (0,0) to (2,0). So, P is at (1 +c,0). Since the triangle is acute, c must be positive, and 1 +c <2, so c <1.Similarly, BC is from (0,0) to (0,2c). Q is at (0, c -1). Since Q must be on BC, c -1 must be between 0 and 2c. So, c -1 >=0 => c >=1, but earlier we had c <1. Contradiction. So, this case is invalid.Case 2: E = - (c +1).From equation (1): D + E c = -1 -c².So, D + (-c -1)c = -1 -c².Compute:D -c² -c = -1 -c².Thus, D = -1 -c² +c² +c = -1 +c.So, D= c -1.Thus, point P is at (-D,0)= (1 -c, 0).Point Q is at (0, -E)= (0, c +1).Now, check if P is on AB and Q is on BC.AB is from (0,0) to (2,0). So, P is at (1 -c,0). Since c >0 and triangle is acute, c must be such that 1 -c >0 => c <1.Similarly, BC is from (0,0) to (0,2c). Q is at (0, c +1). Since c <1, c +1 <2, so Q is on BC.So, this case is valid.Now, point T is at (-D, -E)= (1 -c, c +1).We need to find BT/BM.BT is the distance from B(0,0) to T(1 -c, c +1):BT = sqrt( (1 -c)^2 + (c +1)^2 )= sqrt(1 -2c +c² +c² +2c +1 )= sqrt(2 + 2c² )= sqrt(2(1 +c² )).BM is the distance from B(0,0) to M(1,c):BM = sqrt(1 +c² ).Thus, BT/BM = sqrt(2(1 +c² )) / sqrt(1 +c² )= sqrt(2).So, regardless of c, as long as c <1, the ratio is sqrt(2).Therefore, the only possible value is sqrt(2).