💡Okay, so I've got this problem about an infinite arithmetic sequence that has a term which is a perfect square and another term which is a perfect cube. I need to prove that this sequence contains a term that is a perfect sixth power. Hmm, interesting. Let me try to unpack this step by step.First, let's recall what an arithmetic sequence is. It's a sequence where each term after the first is obtained by adding a constant difference. So, if the first term is ( a ) and the common difference is ( d ), the sequence looks like ( a, a + d, a + 2d, a + 3d, ldots ). Since it's infinite, we can go on forever adding ( d ).Now, the problem states that this sequence contains a perfect square and a perfect cube. Let's denote the perfect square term as ( x^2 ) and the perfect cube term as ( y^3 ), where ( x ) and ( y ) are integers. So, there exist integers ( m ) and ( n ) such that:[a + md = x^2][a + nd = y^3]Our goal is to show that there exists some integer ( k ) such that:[a + kd = z^6]for some integer ( z ).Hmm, okay. So, we have two equations:1. ( a + md = x^2 )2. ( a + nd = y^3 )I need to somehow combine these to find a term that's a sixth power. Maybe I can subtract these two equations to eliminate ( a ):[(a + nd) - (a + md) = y^3 - x^2]Simplifying, we get:[(n - m)d = y^3 - x^2]So, ( d ) divides ( y^3 - x^2 ). That's interesting. But how does that help me?Wait, maybe I should think about the properties of arithmetic sequences and perfect powers. Since the sequence is arithmetic, the difference ( d ) is constant. If ( d ) is such that it allows both a square and a cube, perhaps ( d ) has some special properties.Let me think about the structure of ( d ). If ( d ) is 1, then the sequence is just all integers starting from ( a ). In that case, it's clear that there are infinitely many squares and cubes, so certainly a sixth power would appear. But the problem doesn't specify that ( d ) is 1, so I need a more general approach.Maybe I can use modular arithmetic. If ( a + md = x^2 ) and ( a + nd = y^3 ), then modulo ( d ), both ( x^2 ) and ( y^3 ) are congruent to ( a ). So:[x^2 equiv a pmod{d}][y^3 equiv a pmod{d}]Therefore, ( x^2 equiv y^3 pmod{d} ). So, ( x^2 ) and ( y^3 ) are congruent modulo ( d ). That might be useful.I wonder if I can find some integer ( k ) such that ( a + kd ) is a sixth power. Let's denote ( z^6 ) as the term we're looking for. So:[a + kd = z^6]We need to find such a ( k ).Maybe I can express ( z^6 ) in terms of ( x^2 ) and ( y^3 ). Since ( z^6 = (z^3)^2 = (z^2)^3 ), it's both a square and a cube. So, perhaps if I can find a term that is both a square and a cube, it would be a sixth power.But how do I ensure that such a term exists in the sequence? Well, the sequence has a term that's a square and another that's a cube. Maybe I can find a term that's a multiple of both, or something like that.Wait, another thought: since ( d ) divides ( y^3 - x^2 ), maybe ( d ) has factors that are common to both ( x^2 ) and ( y^3 ). If I can factor ( d ) into parts that relate to squares and cubes, perhaps I can construct a sixth power.Let me think about the prime factorization of ( d ). Suppose ( d ) has prime factors ( p_1, p_2, ldots, p_n ). Then, ( x^2 ) and ( y^3 ) must align with these factors in some way.Since ( x^2 ) is a square, all exponents in its prime factorization are even. Similarly, ( y^3 ) has exponents that are multiples of 3. So, if ( d ) is such that it allows both these conditions, maybe the exponents in ( d ) can be adjusted to form a sixth power.Wait, sixth powers have exponents that are multiples of 6. So, if I can find a term in the sequence where the exponents of all primes are multiples of 6, that term would be a sixth power.But how do I ensure that such a term exists? Maybe by using the Chinese Remainder Theorem or something similar.Let me try to formalize this. Suppose ( d ) has a prime factorization ( d = p_1^{e_1} p_2^{e_2} ldots p_n^{e_n} ). Then, for each prime ( p_i ), the exponent in ( x^2 ) must be even, and in ( y^3 ) must be a multiple of 3.Therefore, for each prime ( p_i ), the exponent ( e_i ) in ( d ) must satisfy that ( e_i ) is such that ( x^2 ) and ( y^3 ) can be formed. So, perhaps ( e_i ) is such that it allows both even and multiple of 3 exponents when combined with ( a ).Wait, maybe I'm overcomplicating. Let's think about the problem differently. Since the sequence is arithmetic, it's linear. So, if I can find a term that is both a square and a cube, it's a sixth power. So, perhaps I can find a term that is congruent to ( a ) modulo ( d ) and is a sixth power.Let me denote ( z^6 ) as the term we're looking for. So, we need:[z^6 equiv a pmod{d}]But we already know that ( x^2 equiv a pmod{d} ) and ( y^3 equiv a pmod{d} ). So, ( z^6 equiv x^2 pmod{d} ) and ( z^6 equiv y^3 pmod{d} ).Hmm, so ( z^6 ) needs to be congruent to both ( x^2 ) and ( y^3 ) modulo ( d ). Since ( x^2 equiv y^3 pmod{d} ), this is consistent.But how do I ensure that such a ( z ) exists? Maybe by solving the congruence ( z^6 equiv a pmod{d} ). If I can find such a ( z ), then ( z^6 ) is in the sequence.But how do I know that this congruence has a solution? Maybe by using the Chinese Remainder Theorem. If I can solve ( z^6 equiv a pmod{p_i^{e_i}} ) for each prime power ( p_i^{e_i} ) dividing ( d ), then I can combine these solutions to get a solution modulo ( d ).So, let's consider each prime power ( p_i^{e_i} ) dividing ( d ). We need to solve ( z^6 equiv a pmod{p_i^{e_i}} ).But we already know that ( x^2 equiv a pmod{p_i^{e_i}} ) and ( y^3 equiv a pmod{p_i^{e_i}} ). So, ( x^2 equiv y^3 pmod{p_i^{e_i}} ).Therefore, ( z^6 equiv x^2 pmod{p_i^{e_i}} ) and ( z^6 equiv y^3 pmod{p_i^{e_i}} ).Wait, so if ( z^6 equiv x^2 pmod{p_i^{e_i}} ), then ( z^3 equiv x pmod{p_i^{e_i}} ) or ( z^3 equiv -x pmod{p_i^{e_i}} ), depending on whether ( p_i ) is 2 or odd.Similarly, ( z^6 equiv y^3 pmod{p_i^{e_i}} ) implies ( z^2 equiv y pmod{p_i^{e_i}} ) or ( z^2 equiv -y pmod{p_i^{e_i}} ).Hmm, this seems a bit messy. Maybe there's a better approach.Wait, another idea: Since ( x^2 equiv y^3 pmod{d} ), perhaps we can set ( z = x^{3} ) or ( z = y^{2} ) modulo ( d ). Let me see.If I set ( z = x^{3} ), then ( z^6 = x^{18} ). But I don't know if ( x^{18} equiv a pmod{d} ). Similarly, if I set ( z = y^{2} ), then ( z^6 = y^{12} ), and again, I don't know if that's congruent to ( a pmod{d} ).Wait, maybe I can use the fact that ( x^2 equiv y^3 pmod{d} ). So, ( x^2 equiv y^3 pmod{d} ) implies that ( x^6 equiv y^9 pmod{d} ). So, ( x^6 equiv y^9 pmod{d} ). Hmm, but I'm not sure how that helps.Alternatively, perhaps I can find a common multiple of the exponents. Since ( x^2 ) and ( y^3 ) are in the sequence, maybe I can find a term that is a multiple of both 2 and 3, which is 6. So, perhaps ( z^6 ) is the term we're looking for.Wait, but how do I ensure that such a term exists? Maybe by using the fact that the sequence is infinite, so I can always find a term that satisfies the necessary congruence conditions.Let me think about the exponents again. If I can find a ( k ) such that ( a + kd ) is a sixth power, then I'm done. So, perhaps I can set ( z^6 = a + kd ), and solve for ( k ).But ( k ) has to be an integer, so ( z^6 - a ) must be divisible by ( d ). So, ( z^6 equiv a pmod{d} ). So, I need to find a ( z ) such that ( z^6 equiv a pmod{d} ).But how do I know such a ( z ) exists? Well, since ( x^2 equiv a pmod{d} ) and ( y^3 equiv a pmod{d} ), maybe I can use these to construct such a ( z ).Wait, perhaps I can set ( z = x^3 ). Then, ( z^6 = x^{18} ). But ( x^2 equiv a pmod{d} ), so ( x^{18} equiv a^9 pmod{d} ). Hmm, not sure if that helps.Alternatively, set ( z = y^2 ). Then, ( z^6 = y^{12} ). Since ( y^3 equiv a pmod{d} ), ( y^{12} equiv a^4 pmod{d} ). Again, not directly helpful.Wait, maybe I can use the fact that ( x^2 equiv y^3 pmod{d} ). So, ( x^2 equiv y^3 pmod{d} ) implies that ( x^6 equiv y^9 pmod{d} ). So, ( x^6 equiv y^9 pmod{d} ). Hmm, but I'm not sure how that helps.Another approach: Let's consider the exponents in the prime factorization of ( d ). Suppose ( d = p_1^{e_1} p_2^{e_2} ldots p_n^{e_n} ). For each prime ( p_i ), since ( x^2 equiv a pmod{p_i^{e_i}} ) and ( y^3 equiv a pmod{p_i^{e_i}} ), we have ( x^2 equiv y^3 pmod{p_i^{e_i}} ).Therefore, for each ( p_i ), ( x^2 equiv y^3 pmod{p_i^{e_i}} ). So, ( x^6 equiv y^9 pmod{p_i^{e_i}} ). But I'm not sure how that helps.Wait, maybe I can use the fact that ( x^2 equiv y^3 pmod{p_i^{e_i}} ) to find a ( z ) such that ( z^6 equiv a pmod{p_i^{e_i}} ).Let me think about each prime ( p_i ) separately. For each ( p_i ), we have ( x^2 equiv a pmod{p_i^{e_i}} ) and ( y^3 equiv a pmod{p_i^{e_i}} ). So, ( x^2 equiv y^3 pmod{p_i^{e_i}} ).Therefore, ( x^6 equiv y^9 pmod{p_i^{e_i}} ). But ( x^6 ) is a sixth power, and ( y^9 ) is also a sixth power (since 9 is a multiple of 3, and 6 is the least common multiple of 2 and 3). Wait, no, 9 is not a multiple of 6. Hmm.Wait, maybe I can find a ( z ) such that ( z^6 equiv a pmod{p_i^{e_i}} ). Since ( x^2 equiv a pmod{p_i^{e_i}} ), perhaps ( z ) can be expressed in terms of ( x ) and ( y ).Let me try to set ( z = x^k y^m ) for some integers ( k ) and ( m ). Then, ( z^6 = x^{6k} y^{6m} ). Since ( x^2 equiv a pmod{p_i^{e_i}} ) and ( y^3 equiv a pmod{p_i^{e_i}} ), we can write ( x^2 = a + t p_i^{e_i} ) and ( y^3 = a + s p_i^{e_i} ) for some integers ( t ) and ( s ).But I'm not sure if this approach is leading me anywhere.Wait, another idea: Since ( x^2 equiv y^3 pmod{d} ), maybe ( x^3 equiv y^2 pmod{d} ). Is that necessarily true? Let me check.If ( x^2 equiv y^3 pmod{d} ), then raising both sides to the power of 3, we get ( x^6 equiv y^9 pmod{d} ). Similarly, raising both sides to the power of 2, we get ( x^4 equiv y^6 pmod{d} ). Hmm, not sure.Wait, maybe I can use the fact that ( x^2 equiv y^3 pmod{d} ) to write ( x = y^{3/2} ) modulo ( d ). But that's not necessarily an integer exponent, so that might not help.Wait, perhaps I can consider the exponents in terms of the least common multiple. Since 2 and 3 are coprime, their least common multiple is 6. So, maybe if I can find a term that is both a square and a cube, it would be a sixth power.But how do I ensure that such a term exists in the sequence? Well, the sequence has a square and a cube, but they might not coincide. However, since the sequence is infinite, maybe I can find a term that is both.Wait, another approach: Let's consider the sequence modulo ( d ). Since ( a + md equiv x^2 pmod{d} ) and ( a + nd equiv y^3 pmod{d} ), we have ( x^2 equiv y^3 pmod{d} ).Therefore, ( x^6 equiv y^9 pmod{d} ). But ( x^6 ) is a sixth power, and ( y^9 ) is also a sixth power (since 9 is a multiple of 3, and 6 is the least common multiple of 2 and 3). Wait, no, 9 is not a multiple of 6. Hmm.Wait, actually, ( y^9 = (y^3)^3 ), which is a cube, but not necessarily a sixth power. Hmm.Wait, maybe I can use the fact that ( x^2 equiv y^3 pmod{d} ) to write ( x^6 equiv y^9 pmod{d} ), and since ( x^6 ) is a sixth power, perhaps ( y^9 ) can be expressed as a sixth power.Wait, ( y^9 = (y^{3})^3 = (y^{2})^{4.5} ), which isn't an integer exponent. Hmm, not helpful.Wait, maybe I can consider the exponents in terms of the greatest common divisor. Since ( x^2 equiv y^3 pmod{d} ), perhaps the exponents in the prime factorization of ( d ) must satisfy certain conditions.Let me think about a specific prime ( p ) dividing ( d ). Suppose ( p ) divides ( d ). Then, ( x^2 equiv y^3 pmod{p} ). So, either ( p ) divides both ( x ) and ( y ), or it doesn't.If ( p ) doesn't divide ( x ) or ( y ), then ( x ) and ( y ) are invertible modulo ( p ). Therefore, ( x^2 equiv y^3 pmod{p} ) implies that ( x ) is a cube modulo ( p ) and ( y ) is a square modulo ( p ). Hmm, interesting.But I'm not sure how to use this information. Maybe if I can find a ( z ) such that ( z^6 equiv a pmod{p} ), then by the Chinese Remainder Theorem, I can piece together solutions for each prime ( p ) dividing ( d ).Wait, let's try to construct such a ( z ). Since ( x^2 equiv a pmod{p} ) and ( y^3 equiv a pmod{p} ), we have ( x^2 equiv y^3 pmod{p} ). So, ( x^6 equiv y^9 pmod{p} ). But ( x^6 ) is a sixth power, so ( y^9 ) must also be a sixth power modulo ( p ).Wait, but ( y^9 = (y^3)^3 ), and since ( y^3 equiv a pmod{p} ), we have ( y^9 equiv a^3 pmod{p} ). Similarly, ( x^6 equiv (x^2)^3 equiv a^3 pmod{p} ). So, ( x^6 equiv y^9 equiv a^3 pmod{p} ).Therefore, ( a^3 equiv a pmod{p} ), which implies ( a(a^2 - 1) equiv 0 pmod{p} ). So, either ( a equiv 0 pmod{p} ), ( a equiv 1 pmod{p} ), or ( a equiv -1 pmod{p} ).Hmm, interesting. So, for each prime ( p ) dividing ( d ), ( a ) is congruent to 0, 1, or -1 modulo ( p ).Now, if ( a equiv 0 pmod{p} ), then ( p ) divides ( a ). But since ( a + md = x^2 ) and ( a + nd = y^3 ), ( p ) divides both ( x^2 ) and ( y^3 ), so ( p ) divides ( x ) and ( y ). Therefore, ( p^2 ) divides ( x^2 ) and ( p^3 ) divides ( y^3 ). But ( p ) divides ( d ), so ( p ) divides ( x^2 - y^3 ). Wait, but ( x^2 - y^3 = (a + md) - (a + nd) = (m - n)d ). So, ( p ) divides ( (m - n)d ). Since ( p ) divides ( d ), this is consistent.But I'm not sure how this helps me find a sixth power in the sequence.Wait, maybe I can use the fact that ( a equiv 0, 1, ) or ( -1 pmod{p} ) to construct a ( z ) such that ( z^6 equiv a pmod{p} ).Let's consider each case:1. If ( a equiv 0 pmod{p} ), then ( z = 0 pmod{p} ) works because ( 0^6 equiv 0 pmod{p} ).2. If ( a equiv 1 pmod{p} ), then ( z = 1 pmod{p} ) works because ( 1^6 equiv 1 pmod{p} ).3. If ( a equiv -1 pmod{p} ), then ( z = -1 pmod{p} ) works because ( (-1)^6 equiv 1 equiv -1 pmod{p} ) only if ( p = 2 ). Wait, no, ( (-1)^6 = 1 ), which is congruent to 1 modulo any ( p ). So, if ( a equiv -1 pmod{p} ), then ( z^6 equiv 1 pmod{p} ), which is not equal to ( a pmod{p} ) unless ( p = 2 ).Wait, that's a problem. If ( a equiv -1 pmod{p} ), then ( z^6 equiv 1 pmod{p} ), which isn't equal to ( a pmod{p} ) unless ( p = 2 ). So, maybe this approach doesn't work for primes where ( a equiv -1 pmod{p} ).Hmm, maybe I need to adjust my approach. Perhaps instead of trying to solve ( z^6 equiv a pmod{p} ) directly, I can use the fact that ( x^2 equiv a pmod{p} ) and ( y^3 equiv a pmod{p} ) to find a ( z ) such that ( z^6 equiv a pmod{p} ).Let me try setting ( z = x^k y^m ) for some exponents ( k ) and ( m ). Then, ( z^6 = x^{6k} y^{6m} ). Since ( x^2 equiv a pmod{p} ) and ( y^3 equiv a pmod{p} ), we can write ( x^2 = a + t p ) and ( y^3 = a + s p ) for some integers ( t ) and ( s ).But this seems too vague. Maybe I can choose ( k ) and ( m ) such that ( 6k ) is a multiple of 2 and ( 6m ) is a multiple of 3. For example, set ( k = 1 ) and ( m = 1 ). Then, ( z = x y ), and ( z^6 = x^6 y^6 ). But ( x^6 = (x^2)^3 equiv a^3 pmod{p} ) and ( y^6 = (y^3)^2 equiv a^2 pmod{p} ). So, ( z^6 equiv a^3 a^2 = a^5 pmod{p} ). Hmm, not helpful.Wait, maybe I can set ( k = 3 ) and ( m = 2 ). Then, ( z = x^3 y^2 ), and ( z^6 = x^{18} y^{12} ). Since ( x^2 equiv a pmod{p} ), ( x^{18} = (x^2)^9 equiv a^9 pmod{p} ). Similarly, ( y^{12} = (y^3)^4 equiv a^4 pmod{p} ). So, ( z^6 equiv a^9 a^4 = a^{13} pmod{p} ). Still not helpful.Wait, maybe I'm overcomplicating. Let's think about the multiplicative order of ( a ) modulo ( p ). If ( a ) is a quadratic residue and a cubic residue modulo ( p ), then it's a sixth power residue modulo ( p ). Is that true?Yes, actually, if ( a ) is both a square and a cube modulo ( p ), then it's a sixth power modulo ( p ). Because the multiplicative group modulo ( p ) is cyclic, so if an element is both a square and a cube, it's a sixth power.So, since ( x^2 equiv a pmod{p} ) and ( y^3 equiv a pmod{p} ), ( a ) is both a square and a cube modulo ( p ), hence a sixth power modulo ( p ). Therefore, there exists some ( z ) such that ( z^6 equiv a pmod{p} ).Therefore, for each prime ( p ) dividing ( d ), there exists a solution ( z ) to ( z^6 equiv a pmod{p} ). By the Chinese Remainder Theorem, there exists a ( z ) such that ( z^6 equiv a pmod{d} ).Therefore, there exists an integer ( k ) such that ( a + kd = z^6 ). Hence, the sequence contains a term that is a perfect sixth power.Wait, that seems to wrap it up. So, the key idea was that since ( a ) is both a square and a cube modulo each prime dividing ( d ), it must be a sixth power modulo each such prime, and hence a sixth power modulo ( d ) itself. Therefore, such a term exists in the sequence.I think that makes sense. I might have taken a long way around, but I think I got there in the end.

💡Okay, so I need to prove that these four pairs of logical statements are equivalent. That means each pair should have the same truth table, right? Let me start by recalling what each logical connective means.First, for the pair ( p land q ) and ( neg((neg p) lor (neg q)) ). I remember that ( land ) is the "and" operator, which is true only when both ( p ) and ( q ) are true. On the other hand, ( lor ) is the "or" operator, which is true when at least one of the operands is true. The negation ( neg ) just flips the truth value.So, ( neg((neg p) lor (neg q)) ) means "it is not the case that either ( p ) is false or ( q ) is false." Hmm, that sounds like both ( p ) and ( q ) have to be true for this to hold. So, maybe ( p land q ) is equivalent to ( neg((neg p) lor (neg q)) ). I think this is related to De Morgan's Law, which states that ( neg(A lor B) ) is equivalent to ( neg A land neg B ). Wait, but here we have ( neg((neg p) lor (neg q)) ), so applying De Morgan's Law, that would be ( neg(neg p) land neg(neg q) ), which simplifies to ( p land q ). Okay, so that makes sense. So, ( p land q ) is equivalent to ( neg((neg p) lor (neg q)) ).Next, the pair ( p lor q ) and ( neg((neg p) land (neg q)) ). Similar idea here. ( p lor q ) is true when either ( p ) or ( q ) is true. The expression ( neg((neg p) land (neg q)) ) means "it is not the case that both ( p ) is false and ( q ) is false." So, if it's not the case that both are false, then at least one of them must be true. That sounds exactly like ( p lor q ). Again, this seems like De Morgan's Law. Specifically, ( neg(A land B) ) is equivalent to ( neg A lor neg B ). But here, we have ( neg((neg p) land (neg q)) ), which by De Morgan's Law would be ( neg(neg p) lor neg(neg q) ), simplifying to ( p lor q ). So, yes, ( p lor q ) is equivalent to ( neg((neg p) land (neg q)) ).Moving on to the pair ( p rightarrow q ) and ( neg q rightarrow neg p ). This one is a bit trickier. I know that ( p rightarrow q ) is an implication that is false only when ( p ) is true and ( q ) is false. The contrapositive of ( p rightarrow q ) is ( neg q rightarrow neg p ), and I remember that an implication is logically equivalent to its contrapositive. So, if ( p rightarrow q ) is true, then ( neg q rightarrow neg p ) should also be true, and vice versa. Let me verify this with a truth table.For ( p rightarrow q ):- When ( p ) is true and ( q ) is true, ( p rightarrow q ) is true.- When ( p ) is true and ( q ) is false, ( p rightarrow q ) is false.- When ( p ) is false and ( q ) is true, ( p rightarrow q ) is true.- When ( p ) is false and ( q ) is false, ( p rightarrow q ) is true.For ( neg q rightarrow neg p ):- When ( q ) is true (so ( neg q ) is false) and ( p ) is true (so ( neg p ) is false), ( neg q rightarrow neg p ) is true.- When ( q ) is true (so ( neg q ) is false) and ( p ) is false (so ( neg p ) is true), ( neg q rightarrow neg p ) is true.- When ( q ) is false (so ( neg q ) is true) and ( p ) is true (so ( neg p ) is false), ( neg q rightarrow neg p ) is false.- When ( q ) is false (so ( neg q ) is true) and ( p ) is false (so ( neg p ) is true), ( neg q rightarrow neg p ) is true.Comparing these, the truth values match exactly. So, ( p rightarrow q ) is equivalent to ( neg q rightarrow neg p ).Lastly, the pair ( p equiv q ) and ( (p rightarrow q) land (q rightarrow p) ). I know that ( p equiv q ) is the biconditional, which is true when ( p ) and ( q ) have the same truth values. The expression ( (p rightarrow q) land (q rightarrow p) ) means both ( p rightarrow q ) and ( q rightarrow p ) are true. If both implications are true, then ( p ) and ( q ) must have the same truth values, which is exactly what ( p equiv q ) means. Let me check this with a truth table.For ( p equiv q ):- When both ( p ) and ( q ) are true, ( p equiv q ) is true.- When ( p ) is true and ( q ) is false, ( p equiv q ) is false.- When ( p ) is false and ( q ) is true, ( p equiv q ) is false.- When both ( p ) and ( q ) are false, ( p equiv q ) is true.For ( (p rightarrow q) land (q rightarrow p) ):- When both ( p rightarrow q ) and ( q rightarrow p ) are true, the conjunction is true.- Otherwise, it's false.Looking at the truth values, they match exactly. So, ( p equiv q ) is equivalent to ( (p rightarrow q) land (q rightarrow p) ).In summary, by applying logical equivalences like De Morgan's Law and understanding the contrapositive, as well as constructing truth tables, I've shown that each pair of statements is equivalent.

💡Okay, so I have this geometry problem here about triangle ABC where AB is 10 units, AC is 12 units, and BC is 8 units. I need to find the length of BI, where I is the incenter of the triangle. Hmm, okay, let me think about how to approach this.First, I remember that the incenter is the point where the angle bisectors of the triangle meet, and it's also the center of the incircle. The incircle touches each side of the triangle at a point, and these points are called the points of tangency. In this case, the incircle touches BC at D, AC at E, and AB at F. So, D, E, and F are the points where the incircle touches the sides BC, AC, and AB respectively.I think I need to find some lengths related to these points of tangency. Maybe I can use the properties of tangents from a point to a circle being equal in length. So, if I let AE = AF = x, BD = BF = y, and CD = CE = z, then I can set up some equations based on the given side lengths.Given that AB = 10, which is equal to AF + BF, so that would be x + y = 10. Similarly, AC = 12, which is equal to AE + CE, so x + z = 12. And BC = 8, which is equal to BD + CD, so y + z = 8. Now I have three equations:1. x + y = 102. x + z = 123. y + z = 8I can solve these equations to find the values of x, y, and z. Let me add all three equations together: (x + y) + (x + z) + (y + z) = 10 + 12 + 8. That simplifies to 2x + 2y + 2z = 30. Dividing both sides by 2, I get x + y + z = 15.Now, if I subtract the second equation (x + z = 12) from this sum, I get (x + y + z) - (x + z) = 15 - 12, which simplifies to y = 3. So, BD = BF = y = 3 units. That's one value found.Next, I can find x by plugging y back into the first equation: x + 3 = 10, so x = 7. Similarly, I can find z by plugging y into the third equation: 3 + z = 8, so z = 5. So, AE = AF = 7, BD = BF = 3, and CD = CE = 5.Now, I need to find the length of BI. I remember that the inradius can be found using the formula r = K/s, where K is the area of the triangle and s is the semi-perimeter. Let me calculate the semi-perimeter first.The semi-perimeter s is (AB + BC + AC)/2 = (10 + 8 + 12)/2 = 30/2 = 15. Okay, so s = 15.Next, I need to find the area K of triangle ABC. I can use Heron's formula for this, which is K = sqrt[s(s - AB)(s - BC)(s - AC)]. Plugging in the values, we get K = sqrt[15(15 - 10)(15 - 8)(15 - 12)] = sqrt[15 * 5 * 7 * 3]. Let me compute that: 15 * 5 is 75, 75 * 7 is 525, and 525 * 3 is 1575. So, K = sqrt[1575].Wait, can I simplify sqrt[1575]? Let me factor 1575. 1575 divided by 25 is 63, so 1575 = 25 * 63. 63 is 9 * 7, so 1575 = 25 * 9 * 7. Therefore, sqrt[1575] = sqrt[25 * 9 * 7] = 5 * 3 * sqrt[7] = 15sqrt[7]. So, the area K is 15sqrt[7].Now, the inradius r is K/s = (15sqrt[7])/15 = sqrt[7]. So, r = sqrt[7]. That's the radius of the incircle.Now, to find BI, which is the distance from vertex B to the incenter I. I think I can use the formula for the length of the angle bisector or maybe use coordinates. But perhaps there's a simpler way using right triangles.I recall that in a triangle, the distance from a vertex to the incenter can be found using the formula: sqrt[r^2 + (s - side)^2], where r is the inradius and s is the semi-perimeter, and 'side' is the side opposite the vertex. Wait, let me think if that's correct.Alternatively, I remember that in the right triangle formed by the inradius, the distance from the vertex to the point of tangency, and the segment from the vertex to the incenter. So, in this case, if I consider triangle BDI, where D is the point where the incircle touches BC, then BD is 3, DI is the inradius r = sqrt[7], and BI is the hypotenuse.So, using the Pythagorean theorem, BI^2 = BD^2 + DI^2. Plugging in the values, BI^2 = 3^2 + (sqrt[7])^2 = 9 + 7 = 16. Therefore, BI = sqrt[16] = 4.Wait, that seems straightforward. Let me verify if this approach is correct. So, triangle BDI is a right triangle because the radius is perpendicular to the tangent at the point of contact. So, DI is perpendicular to BC, making triangle BDI a right-angled triangle at D. Therefore, BI is the hypotenuse, and BD and DI are the legs. So, yes, using Pythagoras here is valid.So, BI = sqrt[BD^2 + DI^2] = sqrt[3^2 + (sqrt[7})^2] = sqrt[9 + 7] = sqrt[16] = 4. That makes sense.Alternatively, I could have used coordinates to solve this problem. Let me try that approach to confirm the result.Let me place triangle ABC on a coordinate system. Let me put point B at (0, 0), point C at (8, 0), and point A somewhere in the plane. I need to find the coordinates of A such that AB = 10 and AC = 12.So, let me denote point A as (x, y). Then, the distance from A to B is sqrt[(x - 0)^2 + (y - 0)^2] = sqrt[x^2 + y^2] = 10. So, x^2 + y^2 = 100.Similarly, the distance from A to C is sqrt[(x - 8)^2 + (y - 0)^2] = sqrt[(x - 8)^2 + y^2] = 12. So, (x - 8)^2 + y^2 = 144.Now, I can set up these two equations:1. x^2 + y^2 = 1002. (x - 8)^2 + y^2 = 144Subtracting equation 1 from equation 2:(x - 8)^2 + y^2 - x^2 - y^2 = 144 - 100Expanding (x - 8)^2: x^2 - 16x + 64 + y^2 - x^2 - y^2 = 44Simplify: -16x + 64 = 44So, -16x = 44 - 64 = -20Therefore, x = (-20)/(-16) = 20/16 = 5/4 = 1.25Now, plug x = 5/4 into equation 1 to find y:(5/4)^2 + y^2 = 10025/16 + y^2 = 100y^2 = 100 - 25/16 = (1600/16 - 25/16) = 1575/16Therefore, y = sqrt(1575/16) = (sqrt(1575))/4 = (15sqrt(7))/4So, point A is at (5/4, (15sqrt(7))/4).Now, the incenter I can be found using the formula:I = (aA + bB + cC)/(a + b + c)Wait, no, that's the formula for the centroid. The incenter has a different formula. The coordinates of the incenter are given by ( (a*x_A + b*x_B + c*x_C)/perimeter, (a*y_A + b*y_B + c*y_C)/perimeter ), where a, b, c are the lengths of the sides opposite to vertices A, B, C respectively.Wait, in triangle ABC, side a is opposite vertex A, which is BC = 8, side b is opposite vertex B, which is AC = 12, and side c is opposite vertex C, which is AB = 10.So, the incenter coordinates would be:x = (a*x_A + b*x_B + c*x_C)/(a + b + c)y = (a*y_A + b*y_B + c*y_C)/(a + b + c)Plugging in the values:a = BC = 8, b = AC = 12, c = AB = 10x_A = 5/4, y_A = (15sqrt(7))/4x_B = 0, y_B = 0x_C = 8, y_C = 0So,x = (8*(5/4) + 12*0 + 10*8)/(8 + 12 + 10) = (10 + 0 + 80)/30 = 90/30 = 3y = (8*(15sqrt(7)/4) + 12*0 + 10*0)/30 = (30sqrt(7) + 0 + 0)/30 = sqrt(7)So, the incenter I is at (3, sqrt(7)).Now, point B is at (0, 0), so the distance BI is the distance between (0, 0) and (3, sqrt(7)).Using the distance formula:BI = sqrt[(3 - 0)^2 + (sqrt(7) - 0)^2] = sqrt[9 + 7] = sqrt[16] = 4.So, that confirms the earlier result. Therefore, BI is 4 units long.Alternatively, I could have used trigonometric methods or other properties, but both the coordinate geometry approach and the right triangle approach gave me the same answer, so I'm confident that BI = 4.**Final Answer**The length of ( BI ) is (boxed{4}).

💡Okay, so I have this problem about an arithmetic sequence. Let me try to understand what it's asking. First, it says that the arithmetic sequence {a_n} has a common difference d, which is not zero. The first term a_1 is 2. Then, it mentions that a_1, a_3, and a_9 form a geometric sequence. Hmm, okay, so I need to use this information to find the general formula for the arithmetic sequence. That's part (I). Then, part (II) is about finding the sum of the first n terms of another sequence, which is {2^{a_n} - 1}. Starting with part (I). Let me recall what an arithmetic sequence is. An arithmetic sequence is a sequence where each term after the first is obtained by adding a constant difference. So, the nth term is given by a_n = a_1 + (n - 1)d. Here, a_1 is 2, so a_n = 2 + (n - 1)d. Now, the problem says that a_1, a_3, a_9 form a geometric sequence. A geometric sequence is a sequence where each term after the first is obtained by multiplying the previous term by a constant ratio. So, if a_1, a_3, a_9 are in geometric progression, then the ratio between a_3 and a_1 should be the same as the ratio between a_9 and a_3. Mathematically, this means that (a_3)^2 = a_1 * a_9. That's a property of geometric sequences: the square of the middle term is equal to the product of the first and the third terms. So, let's write down what a_3 and a_9 are in terms of d. a_3 is the third term of the arithmetic sequence. So, a_3 = a_1 + 2d = 2 + 2d. Similarly, a_9 is the ninth term. So, a_9 = a_1 + 8d = 2 + 8d. Now, according to the geometric sequence property, (a_3)^2 = a_1 * a_9. Let's plug in the expressions we have:(2 + 2d)^2 = 2 * (2 + 8d)Let me expand the left side: (2 + 2d)^2 = 4 + 8d + 4d^2The right side is 2*(2 + 8d) = 4 + 16dSo, setting them equal:4 + 8d + 4d^2 = 4 + 16dHmm, let's subtract 4 from both sides to simplify:8d + 4d^2 = 16dNow, subtract 16d from both sides:4d^2 + 8d - 16d = 0Which simplifies to:4d^2 - 8d = 0I can factor out a 4d:4d(d - 2) = 0So, the solutions are d = 0 or d = 2. But the problem states that d ≠ 0, so we discard d = 0 and take d = 2.Therefore, the common difference d is 2. Now, let's write the general formula for the arithmetic sequence. Since a_n = a_1 + (n - 1)d, plugging in a_1 = 2 and d = 2:a_n = 2 + (n - 1)*2 = 2 + 2n - 2 = 2nSo, the general formula is a_n = 2n. That seems straightforward.Wait, let me double-check. If a_n = 2n, then a_1 = 2*1 = 2, which matches. a_3 = 2*3 = 6, and a_9 = 2*9 = 18. Let's check if 2, 6, 18 form a geometric sequence. The ratio between a_3 and a_1 is 6/2 = 3, and the ratio between a_9 and a_3 is 18/6 = 3. So yes, they have the same ratio, which confirms it's a geometric sequence. Great, that seems correct.So, part (I) is done, and the general formula is a_n = 2n.Moving on to part (II). We need to find the sum of the first n terms of the sequence {2^{a_n} - 1}. First, let's understand what this sequence is. The nth term of this sequence is 2^{a_n} - 1. Since we already found a_n = 2n, then each term is 2^{2n} - 1. Simplify 2^{2n}: that's the same as (2^2)^n = 4^n. So, each term is 4^n - 1. Therefore, the sequence is 4^1 - 1, 4^2 - 1, 4^3 - 1, ..., 4^n - 1. So, the sum S_n is the sum from k=1 to n of (4^k - 1). We can split this sum into two separate sums:S_n = sum_{k=1}^n 4^k - sum_{k=1}^n 1The first sum is a geometric series with first term 4 and common ratio 4. The second sum is just adding 1 n times, which is n.Let's compute each sum separately.First, the sum of the geometric series: sum_{k=1}^n 4^k. The formula for the sum of a geometric series is S = a*(r^n - 1)/(r - 1), where a is the first term, r is the common ratio, and n is the number of terms.Here, a = 4, r = 4, so:sum_{k=1}^n 4^k = 4*(4^n - 1)/(4 - 1) = (4^{n+1} - 4)/3Wait, let me check that. The formula is a*(r^n - 1)/(r - 1). So, plugging in a=4, r=4:sum = 4*(4^n - 1)/(4 - 1) = 4*(4^n - 1)/3 = (4^{n+1} - 4)/3Yes, that's correct.Then, the second sum is just n.So, putting it together:S_n = (4^{n+1} - 4)/3 - nAlternatively, we can write 4^{n+1} as 4*4^n, so:S_n = (4*4^n - 4)/3 - n = (4^{n+1} - 4)/3 - nBut perhaps it's better to factor out the 4:S_n = (4(4^n - 1))/3 - n = (4/3)(4^n - 1) - nYes, that seems a cleaner way to write it.Let me verify this formula with a small n to make sure.Let's take n=1. Then, the sequence {2^{a_n} - 1} is {2^{2*1} - 1} = {4 - 1} = {3}. So, S_1 = 3.Plugging into our formula:(4/3)(4^1 - 1) - 1 = (4/3)(4 - 1) - 1 = (4/3)(3) - 1 = 4 - 1 = 3. Correct.Now, n=2. The terms are 3 and 2^{4} - 1 = 16 - 1 = 15. So, S_2 = 3 + 15 = 18.Using the formula:(4/3)(4^2 - 1) - 2 = (4/3)(16 - 1) - 2 = (4/3)(15) - 2 = 20 - 2 = 18. Correct.Another test, n=3. The terms are 3, 15, 2^{6} -1 = 64 -1 = 63. So, S_3 = 3 + 15 + 63 = 81.Using the formula:(4/3)(4^3 - 1) - 3 = (4/3)(64 - 1) - 3 = (4/3)(63) - 3 = 84 - 3 = 81. Correct.Okay, so the formula seems to work for n=1, 2, 3. So, I think it's correct.Therefore, the sum S_n is (4/3)(4^n - 1) - n.So, summarizing:(I) The general formula for the arithmetic sequence is a_n = 2n.(II) The sum of the first n terms of the sequence {2^{a_n} - 1} is S_n = (4/3)(4^n - 1) - n.**Final Answer**(I) The general formula is boxed{a_n = 2n}.(II) The sum of the first ( n ) terms is boxed{S_n = dfrac{4}{3}(4^n - 1) - n}.

💡Alright, so I have this problem with two parts. Let me try to tackle them one by one. Starting with part (1): I need to discuss the monotonicity of the function f(x) = ln(x+1) - (ax² + x)/(x+1)² when 1 < a ≤ 2. Monotonicity refers to whether the function is increasing or decreasing over certain intervals. To find that, I remember I need to look at the first derivative of f(x).So, let's find f'(x). The function f(x) has two parts: ln(x+1) and the rational function (ax² + x)/(x+1)². I'll differentiate each part separately.First, the derivative of ln(x+1) is straightforward. It's 1/(x+1).Now, for the second part, (ax² + x)/(x+1)². I'll use the quotient rule here. The quotient rule is (num’ * den - num * den’)/den². Let me denote the numerator as u = ax² + x and the denominator as v = (x+1)².So, u’ = 2ax + 1, and v’ = 2(x+1).Applying the quotient rule: (u’v - uv’) / v².Plugging in the values:[(2ax + 1)(x+1)² - (ax² + x)(2(x+1))] / (x+1)^4.Hmm, that looks a bit complicated. Maybe I can factor out some terms to simplify.Let me factor out (x+1) from the numerator:(x+1)[(2ax + 1)(x+1) - 2(ax² + x)] / (x+1)^4.Simplify the denominator: (x+1)^4 becomes (x+1)^3 after canceling one (x+1) from the numerator.So now, the numerator inside the brackets is:(2ax + 1)(x+1) - 2(ax² + x).Let me expand (2ax + 1)(x+1):2ax(x) + 2ax(1) + 1(x) + 1(1) = 2ax² + 2ax + x + 1.Now subtract 2(ax² + x):2ax² + 2ax + x + 1 - 2ax² - 2x.Simplify term by term:2ax² - 2ax² = 0.2ax + x - 2x = 2ax - x.And the constant term is +1.So the numerator becomes (2ax - x + 1).Putting it all together, the derivative of the second part is:(x+1)(2ax - x + 1) / (x+1)^3.Simplify (x+1) in numerator and denominator:(2ax - x + 1) / (x+1)^2.So, the derivative of the second part is (2ax - x + 1)/(x+1)^2.Therefore, the derivative of f(x) is:f'(x) = 1/(x+1) - (2ax - x + 1)/(x+1)^2.To combine these terms, I'll get a common denominator, which is (x+1)^2.So, f'(x) = (x+1)/(x+1)^2 - (2ax - x + 1)/(x+1)^2.Combine the numerators:[x + 1 - 2ax + x - 1] / (x+1)^2.Simplify the numerator:x + 1 - 2ax + x - 1.Combine like terms:x + x = 2x.1 - 1 = 0.So, 2x - 2ax = 2x(1 - a).Therefore, f'(x) = [2x(1 - a)] / (x+1)^2.Wait, that seems different from what I thought earlier. Let me double-check my steps.Wait, when I expanded (2ax + 1)(x+1), I think I made a mistake. Let me redo that.(2ax + 1)(x + 1) = 2ax*x + 2ax*1 + 1*x + 1*1 = 2ax² + 2ax + x + 1.Then subtract 2(ax² + x):2ax² + 2ax + x + 1 - 2ax² - 2x.Simplify:2ax² - 2ax² = 0.2ax + x - 2x = 2ax - x.Constant term: +1.So numerator is 2ax - x + 1.So, f'(x) is [2ax - x + 1]/(x+1)^2.Wait, but earlier I thought f'(x) was [x(x - 2a + 3)]/(x+1)^2. Hmm, maybe I made a mistake in simplifying.Wait, let me go back to the derivative:f'(x) = 1/(x+1) - (2ax - x + 1)/(x+1)^2.To combine these, I need to express 1/(x+1) as (x+1)/(x+1)^2.So, f'(x) = (x+1)/(x+1)^2 - (2ax - x + 1)/(x+1)^2.Combine numerators:(x + 1) - (2ax - x + 1) = x + 1 - 2ax + x - 1.Simplify:x + x = 2x.1 - 1 = 0.So, 2x - 2ax = 2x(1 - a).Therefore, f'(x) = [2x(1 - a)] / (x+1)^2.Wait, that's different from the initial thought. So, f'(x) = 2x(1 - a)/(x+1)^2.But in the initial problem, the user mentioned f'(x) = x(x - 2a + 3)/(x+1)^2. So, perhaps I made a mistake in the differentiation.Let me try a different approach. Maybe I should use the chain rule or product rule instead of the quotient rule.Wait, the function is f(x) = ln(x+1) - (ax² + x)/(x+1)^2.Let me denote the second term as (ax² + x)/(x+1)^2. Maybe I can write this as (ax² + x)(x+1)^{-2} and use the product rule.So, derivative of (ax² + x) is 2ax + 1.Derivative of (x+1)^{-2} is -2(x+1)^{-3}.So, using product rule:(2ax + 1)(x+1)^{-2} + (ax² + x)(-2)(x+1)^{-3}.Factor out (x+1)^{-3}:(x+1)^{-3} [ (2ax + 1)(x+1) - 2(ax² + x) ].Now, expand (2ax + 1)(x+1):2ax*x + 2ax*1 + 1*x + 1*1 = 2ax² + 2ax + x + 1.Subtract 2(ax² + x):2ax² + 2ax + x + 1 - 2ax² - 2x.Simplify:2ax² - 2ax² = 0.2ax + x - 2x = 2ax - x.Constant term: +1.So, numerator is 2ax - x + 1.Therefore, derivative of the second term is (2ax - x + 1)/(x+1)^3.Thus, f'(x) = 1/(x+1) - (2ax - x + 1)/(x+1)^3.To combine these, express 1/(x+1) as (x+1)^2/(x+1)^3.So, f'(x) = (x+1)^2/(x+1)^3 - (2ax - x + 1)/(x+1)^3.Combine numerators:(x+1)^2 - (2ax - x + 1).Expand (x+1)^2: x² + 2x + 1.Subtract (2ax - x + 1):x² + 2x + 1 - 2ax + x - 1.Simplify:x² + (2x + x) + (1 - 1) - 2ax.Which is x² + 3x - 2ax.Factor x: x(x + 3 - 2a).So, numerator is x(x + 3 - 2a).Therefore, f'(x) = x(x + 3 - 2a)/(x+1)^3.Wait, that's different from my previous result. So, f'(x) = x(x + 3 - 2a)/(x+1)^3.Hmm, okay, so f'(x) = x(x - 2a + 3)/(x+1)^3.Wait, that's similar to what the user mentioned, except the denominator is (x+1)^3 instead of (x+1)^2. Maybe I made a mistake in the exponent.Wait, in the initial differentiation, I had (x+1)^{-3} as a factor, so when I combine everything, the denominator is (x+1)^3.But in the user's initial problem, the derivative was given as f'(x) = x(x - 2a + 3)/(x+1)^2. So, perhaps there was a miscalculation.Alternatively, maybe I should consider the original function again.Wait, f(x) = ln(x+1) - (ax² + x)/(x+1)^2.Let me compute f'(x) again carefully.Derivative of ln(x+1) is 1/(x+1).Derivative of (ax² + x)/(x+1)^2:Let me use the quotient rule: [ (2ax + 1)(x+1)^2 - (ax² + x)(2)(x+1) ] / (x+1)^4.Factor out (x+1):[ (2ax + 1)(x+1) - 2(ax² + x) ] / (x+1)^3.Now, expand (2ax + 1)(x+1):2ax*x + 2ax*1 + 1*x + 1*1 = 2ax² + 2ax + x + 1.Subtract 2(ax² + x):2ax² + 2ax + x + 1 - 2ax² - 2x.Simplify:2ax² - 2ax² = 0.2ax + x - 2x = 2ax - x.Constant term: +1.So, numerator is 2ax - x + 1.Thus, derivative of the second term is (2ax - x + 1)/(x+1)^3.Therefore, f'(x) = 1/(x+1) - (2ax - x + 1)/(x+1)^3.Express 1/(x+1) as (x+1)^2/(x+1)^3:f'(x) = (x+1)^2/(x+1)^3 - (2ax - x + 1)/(x+1)^3.Combine numerators:(x+1)^2 - (2ax - x + 1).Expand (x+1)^2: x² + 2x + 1.Subtract (2ax - x + 1):x² + 2x + 1 - 2ax + x - 1.Simplify:x² + (2x + x) + (1 - 1) - 2ax.Which is x² + 3x - 2ax.Factor x: x(x + 3 - 2a).So, numerator is x(x + 3 - 2a).Therefore, f'(x) = x(x + 3 - 2a)/(x+1)^3.So, f'(x) = x(x - 2a + 3)/(x+1)^3.Wait, so the denominator is (x+1)^3, not (x+1)^2 as the user mentioned. So, perhaps there was a typo in the initial problem statement.Anyway, proceeding with f'(x) = x(x - 2a + 3)/(x+1)^3.Now, to discuss the monotonicity, we need to analyze the sign of f'(x).The denominator (x+1)^3 is always positive for x > -1, since x+1 > 0.So, the sign of f'(x) depends on the numerator: x(x - 2a + 3).So, let's set the numerator equal to zero:x(x - 2a + 3) = 0.Solutions are x = 0 and x = 2a - 3.Given that 1 < a ≤ 2, let's compute 2a - 3:For a = 1, 2a - 3 = -1.For a = 2, 2a - 3 = 1.So, when 1 < a ≤ 2, 2a - 3 ranges from -1 to 1.But since x > -1, the critical points are x = 0 and x = 2a - 3, which is between -1 and 1.So, depending on the value of a, 2a - 3 can be less than 0 or greater than 0.Let me consider different cases:Case 1: 1 < a < 3/2.Then, 2a - 3 < 0, because 2*(3/2) - 3 = 0. So, for a < 3/2, 2a - 3 < 0.Thus, the critical points are x = 0 and x = 2a - 3 < 0.But since x > -1, the relevant critical points are x = 0 and x = 2a - 3 (which is between -1 and 0).So, the intervals to consider are:(-1, 2a - 3), (2a - 3, 0), and (0, ∞).Now, let's test the sign of f'(x) in each interval.For x in (-1, 2a - 3):x is negative, and x - 2a + 3 is also negative because x < 2a - 3.So, x(x - 2a + 3) is positive (negative * negative).Thus, f'(x) > 0 in (-1, 2a - 3).For x in (2a - 3, 0):x is negative, and x - 2a + 3 is positive because x > 2a - 3.So, x(x - 2a + 3) is negative (negative * positive).Thus, f'(x) < 0 in (2a - 3, 0).For x in (0, ∞):x is positive, and x - 2a + 3 is positive because x > 0 and 2a - 3 < 0 for a < 3/2.Wait, no. Wait, 2a - 3 is negative, so x - 2a + 3 = x + (3 - 2a). Since 3 - 2a > 0 (because a < 3/2), and x > 0, so x + (3 - 2a) > 0.Thus, x(x - 2a + 3) is positive (positive * positive).So, f'(x) > 0 in (0, ∞).Therefore, for 1 < a < 3/2, f(x) is increasing on (-1, 2a - 3), decreasing on (2a - 3, 0), and increasing again on (0, ∞).Case 2: a = 3/2.Then, 2a - 3 = 0.So, the critical points coincide at x = 0.Thus, the numerator becomes x(x - 0) = x², which is always non-negative.So, f'(x) = x²/(x+1)^3, which is always non-negative for x > -1.Thus, f(x) is monotonically increasing on (-1, ∞).Case 3: a > 3/2.Wait, but the problem states 1 < a ≤ 2, so a can be greater than 3/2 but up to 2.So, for 3/2 < a ≤ 2, 2a - 3 is positive.Thus, the critical points are x = 0 and x = 2a - 3 > 0.So, the intervals are:(-1, 0), (0, 2a - 3), and (2a - 3, ∞).Testing the sign of f'(x):For x in (-1, 0):x is negative, x - 2a + 3 is positive because x > -1 and 2a - 3 > 0, so x - 2a + 3 = x + (3 - 2a). Since x > -1 and 3 - 2a < 0 (because a > 3/2), but x + (3 - 2a) could be positive or negative.Wait, let me think again.Wait, x is in (-1, 0), and 2a - 3 is positive.So, x - 2a + 3 = x + (3 - 2a).Since 3 - 2a is negative (because a > 3/2), so x + (3 - 2a) is x minus a positive number.But x is between -1 and 0, so x + (3 - 2a) could be positive or negative.Wait, let's compute 3 - 2a:For a = 2, 3 - 2a = 3 - 4 = -1.So, x + (3 - 2a) = x -1.Since x is in (-1, 0), x -1 is in (-2, -1), which is negative.Thus, x - 2a + 3 is negative in (-1, 0).So, x is negative, x - 2a + 3 is negative.Thus, x(x - 2a + 3) is positive (negative * negative).Therefore, f'(x) > 0 in (-1, 0).For x in (0, 2a - 3):x is positive, x - 2a + 3 is negative because x < 2a - 3.So, x(x - 2a + 3) is negative (positive * negative).Thus, f'(x) < 0 in (0, 2a - 3).For x in (2a - 3, ∞):x is positive, x - 2a + 3 is positive because x > 2a - 3.Thus, x(x - 2a + 3) is positive.So, f'(x) > 0 in (2a - 3, ∞).Therefore, for a > 3/2, f(x) is increasing on (-1, 0), decreasing on (0, 2a - 3), and increasing again on (2a - 3, ∞).Putting it all together:When 1 < a < 3/2, f(x) is increasing on (-1, 2a - 3) and (0, ∞), decreasing on (2a - 3, 0).When a = 3/2, f(x) is monotonically increasing on (-1, ∞).When 3/2 < a ≤ 2, f(x) is increasing on (-1, 0) and (2a - 3, ∞), decreasing on (0, 2a - 3).Okay, that seems consistent.Now, moving on to part (2): Find the maximum value of g(x) = x ln(1 + 1/x) + (1/x) ln(1 + x) when x > 0.First, let me write down g(x):g(x) = x ln(1 + 1/x) + (1/x) ln(1 + x).I notice that g(x) is symmetric in a way. Let me see if g(x) = g(1/x).Let me substitute x with 1/x:g(1/x) = (1/x) ln(1 + x) + x ln(1 + 1/(1/x)).Simplify ln(1 + 1/(1/x)) = ln(1 + x).So, g(1/x) = (1/x) ln(1 + x) + x ln(1 + x).Which is the same as g(x) = x ln(1 + 1/x) + (1/x) ln(1 + x).Wait, no, because g(1/x) = (1/x) ln(1 + x) + x ln(1 + x), which is different from g(x).Wait, actually, let me compute g(1/x):g(1/x) = (1/x) ln(1 + x) + x ln(1 + 1/(1/x)).Simplify 1/(1/x) = x, so ln(1 + x).Thus, g(1/x) = (1/x) ln(1 + x) + x ln(1 + x).Which is the same as g(x) = x ln(1 + 1/x) + (1/x) ln(1 + x).Wait, no, because in g(x), the first term is x ln(1 + 1/x) and the second term is (1/x) ln(1 + x). In g(1/x), it's (1/x) ln(1 + x) + x ln(1 + x). So, they are not the same unless x ln(1 + 1/x) = x ln(1 + x), which is not true.Wait, maybe I made a mistake. Let me re-express g(x):g(x) = x ln(1 + 1/x) + (1/x) ln(1 + x).Let me write 1 + 1/x as (x + 1)/x, so ln(1 + 1/x) = ln((x + 1)/x) = ln(x + 1) - ln x.Similarly, ln(1 + x) is just ln(x + 1).So, g(x) = x [ln(x + 1) - ln x] + (1/x) ln(x + 1).Simplify:g(x) = x ln(x + 1) - x ln x + (1/x) ln(x + 1).Combine like terms:g(x) = (x + 1/x) ln(x + 1) - x ln x.Hmm, interesting. Maybe this form is easier to work with.Now, to find the maximum value of g(x) for x > 0.I can consider taking the derivative of g(x) and setting it to zero to find critical points.So, let's compute g'(x).First, write g(x) as:g(x) = (x + 1/x) ln(x + 1) - x ln x.Compute derivative term by term.First term: (x + 1/x) ln(x + 1).Use product rule: derivative of (x + 1/x) times ln(x + 1) plus (x + 1/x) times derivative of ln(x + 1).Derivative of (x + 1/x) is 1 - 1/x².Derivative of ln(x + 1) is 1/(x + 1).So, first term derivative:[1 - 1/x²] ln(x + 1) + (x + 1/x) * [1/(x + 1)].Simplify the second part: (x + 1/x)/(x + 1) = [x(x + 1) + 1]/(x(x + 1)).Wait, no, let me compute it step by step.(x + 1/x) * [1/(x + 1)] = [x + 1/x]/(x + 1).Let me write x + 1/x as (x² + 1)/x.So, [ (x² + 1)/x ] / (x + 1) = (x² + 1)/(x(x + 1)).Now, the second term in g(x) is -x ln x.Derivative of -x ln x is - [ln x + x*(1/x)] = - ln x - 1.Putting it all together, g'(x) is:[1 - 1/x²] ln(x + 1) + (x² + 1)/(x(x + 1)) - ln x - 1.Simplify term by term.First term: [1 - 1/x²] ln(x + 1).Second term: (x² + 1)/(x(x + 1)).Third term: - ln x.Fourth term: -1.Let me see if I can combine these terms.First, let's look at the second term: (x² + 1)/(x(x + 1)).We can write this as (x² + 1)/(x² + x).Alternatively, maybe split the fraction:(x² + 1)/(x(x + 1)) = [x² + x - x + 1]/(x(x + 1)) = [x(x + 1) - (x - 1)]/(x(x + 1)).But that might not help much.Alternatively, perform polynomial division:Divide x² + 1 by x(x + 1) = x² + x.So, x² + 1 divided by x² + x is 1 with a remainder of -x + 1.Thus, (x² + 1)/(x² + x) = 1 + (-x + 1)/(x² + x).But not sure if that helps.Alternatively, let me write it as:(x² + 1)/(x(x + 1)) = (x² + 1)/(x² + x) = 1 + (1 - x)/(x² + x).Hmm, not sure.Alternatively, maybe leave it as is for now.So, g'(x) = [1 - 1/x²] ln(x + 1) + (x² + 1)/(x(x + 1)) - ln x - 1.Let me see if I can combine the logarithmic terms.We have [1 - 1/x²] ln(x + 1) - ln x.Let me factor out ln(x + 1):ln(x + 1) [1 - 1/x²] - ln x.Hmm, not sure.Alternatively, let me consider substituting t = x, and see if I can find symmetry or other properties.Wait, earlier I noticed that g(x) is not symmetric, but maybe I can consider substitution y = 1/x.Let me define y = 1/x, so x = 1/y.Then, g(x) = x ln(1 + 1/x) + (1/x) ln(1 + x).Substitute x = 1/y:g(1/y) = (1/y) ln(1 + y) + y ln(1 + 1/y).Which is the same as g(y) = y ln(1 + 1/y) + (1/y) ln(1 + y).So, g(1/y) = g(y).Thus, g(x) is symmetric in the sense that g(x) = g(1/x).Therefore, the function is symmetric around x = 1 in some sense.This suggests that the maximum might occur at x = 1, but I need to verify.Alternatively, since g(x) = g(1/x), the function is symmetric with respect to x and 1/x, so the maximum on (0, ∞) should be the same as the maximum on (0, 1], because for x > 1, it's equivalent to 1/x < 1.Therefore, to find the maximum, I can consider x in (0, 1], and find the maximum there, which will be the same as the maximum on (0, ∞).So, let's restrict our attention to x in (0, 1].Now, let's compute g'(x) on (0, 1].We have g'(x) = [1 - 1/x²] ln(x + 1) + (x² + 1)/(x(x + 1)) - ln x - 1.Let me see if I can simplify this expression.First, note that x is in (0, 1], so x + 1 is in (1, 2].Let me compute each term:1. [1 - 1/x²] ln(x + 1):Since x ≤ 1, 1/x² ≥ 1, so 1 - 1/x² ≤ 0. Thus, this term is non-positive.2. (x² + 1)/(x(x + 1)):This is always positive because numerator and denominator are positive.3. - ln x:Since x ≤ 1, ln x ≤ 0, so - ln x ≥ 0.4. -1:Negative.So, overall, g'(x) is a combination of positive and negative terms.To analyze the sign of g'(x), perhaps I can consider the behavior as x approaches 0+ and as x approaches 1.As x approaches 0+:- [1 - 1/x²] ln(x + 1) ≈ [ -1/x² ] * ln(1) = 0, but actually, ln(x + 1) ~ x, so [1 - 1/x²] * x ≈ -x/x² = -1/x, which tends to -∞.- (x² + 1)/(x(x + 1)) ≈ (1)/(x*1) = 1/x, which tends to +∞.- - ln x tends to +∞.- -1 is negligible.So, overall, as x approaches 0+, g'(x) tends to (-∞) + (+∞) + (+∞) - 1. It's unclear which term dominates.Similarly, as x approaches 1:Compute g'(1):[1 - 1/1²] ln(2) + (1 + 1)/(1*2) - ln 1 - 1.Simplify:(0) ln(2) + (2)/(2) - 0 - 1 = 0 + 1 - 0 - 1 = 0.So, g'(1) = 0.Now, let's check the behavior of g'(x) near x = 1.Let me compute g'(x) for x slightly less than 1, say x = 0.9.Compute each term:1. [1 - 1/(0.9)^2] ln(1.9):1 - 1/0.81 ≈ 1 - 1.2345 ≈ -0.2345.ln(1.9) ≈ 0.6419.So, term ≈ -0.2345 * 0.6419 ≈ -0.1513.2. (0.81 + 1)/(0.9*1.9) ≈ 1.81 / (1.71) ≈ 1.058.3. - ln(0.9) ≈ -(-0.1054) ≈ 0.1054.4. -1.So, total g'(0.9) ≈ -0.1513 + 1.058 + 0.1054 - 1 ≈ (-0.1513 - 1) + (1.058 + 0.1054) ≈ (-1.1513) + 1.1634 ≈ 0.0121.So, g'(0.9) ≈ 0.0121 > 0.Similarly, for x = 0.5:Compute each term:1. [1 - 1/(0.5)^2] ln(1.5) = [1 - 4] ln(1.5) = (-3)(0.4055) ≈ -1.2165.2. (0.25 + 1)/(0.5*1.5) = 1.25 / 0.75 ≈ 1.6667.3. - ln(0.5) ≈ -(-0.6931) ≈ 0.6931.4. -1.Total g'(0.5) ≈ -1.2165 + 1.6667 + 0.6931 - 1 ≈ (-1.2165 - 1) + (1.6667 + 0.6931) ≈ (-2.2165) + 2.3598 ≈ 0.1433 > 0.Wait, so at x = 0.5, g'(x) is positive.At x approaching 0+, g'(x) tends to ?Wait, earlier I thought it was indeterminate, but let me compute the limit as x approaches 0+ of g'(x).Compute limit as x→0+ of g'(x):g'(x) = [1 - 1/x²] ln(x + 1) + (x² + 1)/(x(x + 1)) - ln x - 1.Let me compute each term:1. [1 - 1/x²] ln(x + 1):As x→0+, ln(x + 1) ~ x - x²/2 + x³/3 - ... ~ x.So, [1 - 1/x²] * x ≈ [ -1/x² ] * x = -1/x, which tends to -∞.2. (x² + 1)/(x(x + 1)) ≈ (1)/(x*1) = 1/x, which tends to +∞.3. - ln x tends to +∞.4. -1 is negligible.So, we have -∞ + ∞ + ∞ - 1. It's an indeterminate form.To resolve this, let's combine the terms:[1 - 1/x²] ln(x + 1) + (x² + 1)/(x(x + 1)) - ln x - 1.Let me write it as:[1 - 1/x²] ln(x + 1) - ln x + (x² + 1)/(x(x + 1)) - 1.Now, let's consider the first two terms:[1 - 1/x²] ln(x + 1) - ln x.Let me factor out ln(x + 1):ln(x + 1) [1 - 1/x²] - ln x.But not sure.Alternatively, let me write 1 - 1/x² = (x² - 1)/x².So, [ (x² - 1)/x² ] ln(x + 1) - ln x.= [ (x² - 1) ln(x + 1) ] / x² - ln x.Now, as x→0+, x² - 1 ≈ -1, ln(x + 1) ≈ x, so numerator ≈ -1 * x = -x.Thus, [ (x² - 1) ln(x + 1) ] / x² ≈ (-x)/x² = -1/x, which tends to -∞.So, the first two terms together tend to -∞.The other terms: (x² + 1)/(x(x + 1)) - 1.As x→0+, (x² + 1)/(x(x + 1)) ≈ 1/x, so 1/x - 1 ≈ 1/x, which tends to +∞.Thus, overall, we have (-∞) + (+∞). It's still indeterminate.To resolve this, let me combine all terms:g'(x) = [1 - 1/x²] ln(x + 1) + (x² + 1)/(x(x + 1)) - ln x - 1.Let me write it as:[1 - 1/x²] ln(x + 1) - ln x + (x² + 1)/(x(x + 1)) - 1.Now, let me consider the limit as x→0+.Let me denote A = [1 - 1/x²] ln(x + 1), B = - ln x, C = (x² + 1)/(x(x + 1)) - 1.Compute limit of A + B + C.As x→0+:A ≈ [ -1/x² ] * x = -1/x.B = - ln x → +∞.C = (x² + 1)/(x(x + 1)) - 1 ≈ (1)/(x) - 1.So, C ≈ 1/x - 1.Thus, A + B + C ≈ (-1/x) + (+∞) + (1/x - 1) = (-1/x + 1/x) + (+∞) - 1 = 0 + ∞ - 1 = ∞.Wait, but that contradicts earlier thoughts. Wait, let me re-express:A ≈ -1/x, B ≈ +∞, C ≈ 1/x - 1.So, A + B + C ≈ (-1/x + 1/x) + ∞ - 1 = 0 + ∞ - 1 = ∞.Thus, the limit as x→0+ of g'(x) is +∞.Wait, but earlier when I computed g'(0.5) and g'(0.9), I got positive values, and as x approaches 0+, g'(x) tends to +∞. At x=1, g'(1)=0.So, the derivative starts at +∞ when x approaches 0+, decreases, and reaches 0 at x=1.Thus, g'(x) is positive on (0, 1), and g'(1)=0.Therefore, g(x) is increasing on (0, 1], and since g(x) is symmetric around x=1, the maximum occurs at x=1.Thus, the maximum value of g(x) is g(1).Compute g(1):g(1) = 1 * ln(1 + 1/1) + (1/1) * ln(1 + 1) = ln(2) + ln(2) = 2 ln(2).Therefore, the maximum value is 2 ln 2.So, the final answer is 2 ln 2.

💡Okay, so I have this problem here: In triangle ABC, angle A is an interior angle, and it's given that sin A + cos A equals 12/25. I need to figure out what kind of triangle ABC is, and the options are acute, obtuse, isosceles right, or isosceles. Hmm, let me think about how to approach this.First, I remember that in trigonometry, when you have expressions like sin A + cos A, sometimes squaring both sides can help because it relates to the Pythagorean identity. So maybe I can square both sides of the equation to simplify it.Let me write that down:(sin A + cos A)^2 = (12/25)^2Expanding the left side, I get:sin²A + 2 sin A cos A + cos²A = 144/625I know that sin²A + cos²A equals 1, so substituting that in:1 + 2 sin A cos A = 144/625Now, subtracting 1 from both sides:2 sin A cos A = 144/625 - 1Calculating the right side:144/625 - 1 is the same as 144/625 - 625/625, which equals (144 - 625)/625 = (-481)/625So, 2 sin A cos A = -481/625Dividing both sides by 2:sin A cos A = -481/1250Hmm, so sin A cos A is negative. I remember that sin A and cos A can be positive or negative depending on the quadrant where angle A is located. Since A is an interior angle of a triangle, it must be between 0 and 180 degrees, or 0 and π radians.In the first quadrant (0 to 90 degrees), both sin A and cos A are positive, so their product would be positive. In the second quadrant (90 to 180 degrees), sin A is positive and cos A is negative, so their product would be negative. Since sin A cos A is negative here, that means angle A must be in the second quadrant. Therefore, angle A is obtuse, meaning it's greater than 90 degrees.If one angle in a triangle is obtuse, the triangle itself is classified as an obtuse triangle. So, triangle ABC must be an obtuse triangle.Let me double-check my steps to make sure I didn't make any mistakes. I squared both sides correctly, used the Pythagorean identity, and solved for sin A cos A. The negative result indicates that angle A is in the second quadrant, making it obtuse. That seems right.I also considered the other options: acute triangle would mean all angles are less than 90 degrees, which isn't the case here. Isosceles right triangle would require a 90-degree angle and two equal sides, but we don't have information about the sides or other angles. Isosceles triangle just means two sides are equal, but again, we don't have information about the sides. So, the most accurate classification is an obtuse triangle.Yeah, I think I'm confident with that answer.**Final Answer**boxed{B}