💡Okay, so I need to find the coefficient of ( x^5 ) in the expansion of ( (x^2 + x + 1)^5 ). Hmm, that sounds like a problem where I can use the binomial theorem or maybe the multinomial theorem. Let me think about how to approach this.First, I remember that the binomial theorem is for expressions with two terms, like ( (a + b)^n ), but here we have three terms: ( x^2 ), ( x ), and ( 1 ). So maybe the multinomial theorem is more appropriate here. The multinomial theorem generalizes the binomial theorem to expressions with more than two terms.The multinomial theorem states that:[(a_1 + a_2 + dots + a_k)^n = sum_{n_1 + n_2 + dots + n_k = n} frac{n!}{n_1! n_2! dots n_k!} a_1^{n_1} a_2^{n_2} dots a_k^{n_k}]In our case, ( a_1 = x^2 ), ( a_2 = x ), and ( a_3 = 1 ), and ( n = 5 ). So, we need to find all combinations of exponents ( n_1 ), ( n_2 ), and ( n_3 ) such that ( n_1 + n_2 + n_3 = 5 ) and the term ( (x^2)^{n_1} cdot x^{n_2} cdot 1^{n_3} ) results in ( x^5 ).Let me write that out:[(x^2)^{n_1} cdot x^{n_2} cdot 1^{n_3} = x^{2n_1 + n_2}]We need this to equal ( x^5 ), so:[2n_1 + n_2 = 5]And we also have the constraint:[n_1 + n_2 + n_3 = 5]So, we need to find all non-negative integer solutions ( (n_1, n_2, n_3) ) to these equations.Let me solve ( 2n_1 + n_2 = 5 ) for possible values of ( n_1 ) and ( n_2 ).Starting with ( n_1 = 0 ):[2(0) + n_2 = 5 Rightarrow n_2 = 5]Then, ( n_3 = 5 - 0 - 5 = 0 ). So, one solution is ( (0, 5, 0) ).Next, ( n_1 = 1 ):[2(1) + n_2 = 5 Rightarrow n_2 = 3]Then, ( n_3 = 5 - 1 - 3 = 1 ). So, another solution is ( (1, 3, 1) ).Next, ( n_1 = 2 ):[2(2) + n_2 = 5 Rightarrow n_2 = 1]Then, ( n_3 = 5 - 2 - 1 = 2 ). So, another solution is ( (2, 1, 2) ).If I try ( n_1 = 3 ):[2(3) + n_2 = 5 Rightarrow n_2 = -1]That's not possible since exponents can't be negative. So, we stop here.So, the valid solutions are:1. ( (n_1, n_2, n_3) = (0, 5, 0) )2. ( (1, 3, 1) )3. ( (2, 1, 2) )Now, for each of these solutions, I need to compute the corresponding multinomial coefficients and sum them up to get the total coefficient for ( x^5 ).The multinomial coefficient for each term is:[frac{5!}{n_1! n_2! n_3!}]So, let's compute each one.1. For ( (0, 5, 0) ):[frac{5!}{0! 5! 0!} = frac{120}{1 times 120 times 1} = 1]2. For ( (1, 3, 1) ):[frac{5!}{1! 3! 1!} = frac{120}{1 times 6 times 1} = 20]3. For ( (2, 1, 2) ):[frac{5!}{2! 1! 2!} = frac{120}{2 times 1 times 2} = 30]Now, adding these up:[1 + 20 + 30 = 51]So, the coefficient of ( x^5 ) is 51.Wait, let me double-check to make sure I didn't miss any solutions or make a calculation error.Starting with ( n_1 = 0 ), ( n_2 = 5 ), ( n_3 = 0 ): that seems correct.Then ( n_1 = 1 ), ( n_2 = 3 ), ( n_3 = 1 ): correct.Then ( n_1 = 2 ), ( n_2 = 1 ), ( n_3 = 2 ): correct.And ( n_1 = 3 ) gives negative ( n_2 ), so it's invalid.Calculations for the coefficients:1. ( 5! / (0!5!0!) = 1 ): correct.2. ( 5! / (1!3!1!) = 120 / (1*6*1) = 20 ): correct.3. ( 5! / (2!1!2!) = 120 / (2*1*2) = 30 ): correct.Adding them: 1 + 20 + 30 = 51: correct.So, I think 51 is the right answer. Let me see if there's another way to approach this problem to confirm.Alternatively, I could expand ( (x^2 + x + 1)^5 ) step by step, but that might take longer. Maybe I can use generating functions or another combinatorial approach.Another thought: since ( (x^2 + x + 1) ) is a polynomial, raising it to the 5th power will involve convolving the coefficients multiple times. But that might be tedious.Alternatively, I could think of this as a trinomial expansion, but I think the multinomial approach I took earlier is solid.Wait, another idea: maybe using generating functions, the coefficient of ( x^5 ) can be found by considering the number of ways to pick terms from each of the five factors ( (x^2 + x + 1) ) such that their exponents add up to 5.Each factor contributes either ( x^2 ), ( x ), or ( 1 ). So, over five factors, the exponents contributed will be the sum of the exponents from each factor.So, to get ( x^5 ), we need to choose, from each of the five factors, either ( x^2 ), ( x ), or ( 1 ), such that the total exponent is 5.Let me model this as a problem of counting the number of sequences of five terms, each term being 2, 1, or 0, such that their sum is 5.So, each term in the product corresponds to choosing either 2, 1, or 0, and we need the sum of these five choices to be 5.So, the number of such sequences is the coefficient we're looking for.This is similar to solving the equation:[a_1 + a_2 + a_3 + a_4 + a_5 = 5]where each ( a_i in {0, 1, 2} ).So, how many solutions are there to this equation?This is equivalent to finding the number of non-negative integer solutions with each ( a_i leq 2 ).This is a classic stars and bars problem with restrictions.The formula for the number of solutions is given by inclusion-exclusion.The total number of solutions without restrictions is ( C(5 + 5 - 1, 5 - 1) = C(9,4) = 126 ).But we have restrictions: each ( a_i leq 2 ).So, we need to subtract the cases where one or more ( a_i geq 3 ).Using inclusion-exclusion:Number of solutions = Total solutions - solutions where at least one ( a_i geq 3 ) + solutions where at least two ( a_i geq 3 ) - ... and so on.But since 5 is the total sum, and each ( a_i leq 2 ), the maximum any single ( a_i ) can be is 2. So, if we have any ( a_i geq 3 ), it would require the sum to be at least 3, but since our total is 5, it's possible for one ( a_i ) to be 3 or 4 or 5, but not more than that.Wait, actually, in our case, each ( a_i ) can be at most 2, so any ( a_i geq 3 ) is invalid.So, let's compute the number of solutions where at least one ( a_i geq 3 ).Let me denote ( A_j ) as the set of solutions where ( a_j geq 3 ). We need to compute ( |A_1 cup A_2 cup A_3 cup A_4 cup A_5| ).By inclusion-exclusion:[|A_1 cup A_2 cup A_3 cup A_4 cup A_5| = sum |A_j| - sum |A_j cap A_k| + sum |A_j cap A_k cap A_l| - dots]But since the total sum is 5, and each ( a_i geq 3 ), the intersections where two or more ( a_i geq 3 ) would require the sum to be at least 6, which is more than 5. So, those intersections are empty.Therefore, the number of solutions where at least one ( a_i geq 3 ) is just ( 5 times ) the number of solutions where a specific ( a_j geq 3 ).So, let's compute ( |A_j| ). For a specific ( a_j geq 3 ), let ( a_j' = a_j - 3 ), so ( a_j' geq 0 ). Then, the equation becomes:[a_j' + 3 + a_1 + a_2 + dots + a_{j-1} + a_{j+1} + dots + a_5 = 5]Simplifying:[a_j' + a_1 + a_2 + dots + a_{j-1} + a_{j+1} + dots + a_5 = 2]The number of non-negative integer solutions to this is ( C(2 + 4, 4) = C(6,4) = 15 ).Since there are 5 choices for ( j ), the total number of solutions where at least one ( a_i geq 3 ) is ( 5 times 15 = 75 ).But wait, this can't be right because the total number of solutions without restrictions is 126, and subtracting 75 would give 51, which matches our earlier result. So, the number of valid solutions is 126 - 75 = 51.Therefore, the coefficient is 51.Wait, that's the same answer as before. So, this method also gives 51.Hmm, that's reassuring. So, both methods give me 51, which is option A.Let me just think if there's another way to approach this, maybe using generating functions or recursion.Alternatively, I could consider that ( (x^2 + x + 1)^5 ) is the same as ( (1 + x + x^2)^5 ), which is a generating function for the number of ways to roll a certain number with five dice, each die having faces 0, 1, 2. So, the coefficient of ( x^5 ) would be the number of ways to get a sum of 5 with five dice, each die showing 0, 1, or 2.But that's essentially the same as the combinatorial approach I took earlier, so it would also result in 51.Alternatively, I could use dynamic programming to compute the coefficients step by step, but that might be overkill for this problem.Another idea: since ( (x^2 + x + 1) ) is a geometric series, maybe I can express it in a different form and then raise it to the 5th power.Wait, ( x^2 + x + 1 = frac{x^3 - 1}{x - 1} ) when ( x neq 1 ). So, ( (x^2 + x + 1)^5 = left( frac{x^3 - 1}{x - 1} right)^5 ).But I'm not sure if that helps directly in finding the coefficient of ( x^5 ). Maybe expanding this would complicate things further.Alternatively, I could use the fact that ( (x^2 + x + 1)^5 = (x^2 + x + 1)(x^2 + x + 1)^4 ), and then recursively compute the coefficients. But that might not be more efficient.Wait, another thought: maybe using the binomial theorem twice. Let me try that.Let me write ( (x^2 + x + 1)^5 ) as ( ((x^2 + x) + 1)^5 ). Then, using the binomial theorem, this becomes:[sum_{k=0}^{5} C(5, k) (x^2 + x)^k (1)^{5 - k}]So, each term is ( C(5, k) (x^2 + x)^k ).Now, I need to find the coefficient of ( x^5 ) in this expansion. So, for each ( k ), I need to find the coefficient of ( x^5 ) in ( (x^2 + x)^k ), multiply it by ( C(5, k) ), and sum over all ( k ).So, let's consider ( (x^2 + x)^k ). This can be expanded as:[sum_{m=0}^{k} C(k, m) (x^2)^{k - m} x^m = sum_{m=0}^{k} C(k, m) x^{2(k - m) + m} = sum_{m=0}^{k} C(k, m) x^{2k - m}]We need the exponent ( 2k - m = 5 ). So, ( m = 2k - 5 ).Since ( m ) must be between 0 and ( k ), we have:[0 leq 2k - 5 leq k]Which simplifies to:[5 leq 2k leq k + 5]Wait, that seems a bit confusing. Let me solve for ( k ).From ( m = 2k - 5 geq 0 ):[2k - 5 geq 0 Rightarrow k geq frac{5}{2} Rightarrow k geq 3]And from ( m = 2k - 5 leq k ):[2k - 5 leq k Rightarrow k leq 5]So, ( k ) can be 3, 4, or 5.Therefore, for each ( k = 3, 4, 5 ), we can find the corresponding ( m ) and compute the coefficient.Let's compute each case:1. For ( k = 3 ): - ( m = 2(3) - 5 = 6 - 5 = 1 ) - Coefficient in ( (x^2 + x)^3 ) is ( C(3, 1) = 3 ) - Multiply by ( C(5, 3) = 10 ) - Contribution: ( 10 times 3 = 30 )2. For ( k = 4 ): - ( m = 2(4) - 5 = 8 - 5 = 3 ) - Coefficient in ( (x^2 + x)^4 ) is ( C(4, 3) = 4 ) - Multiply by ( C(5, 4) = 5 ) - Contribution: ( 5 times 4 = 20 )3. For ( k = 5 ): - ( m = 2(5) - 5 = 10 - 5 = 5 ) - Coefficient in ( (x^2 + x)^5 ) is ( C(5, 5) = 1 ) - Multiply by ( C(5, 5) = 1 ) - Contribution: ( 1 times 1 = 1 )Adding these contributions:[30 + 20 + 1 = 51]So, again, the coefficient is 51.This method also confirms the result. It seems like 51 is definitely the correct answer.Just to make sure, let me think if there's any other perspective or method that could be used here.Another approach might be to use generating functions and consider the convolution of coefficients. Since ( (x^2 + x + 1)^5 ) is the product of five such polynomials, the coefficient of ( x^5 ) can be found by convolving the coefficients five times.But that would involve a lot of computation, and I think the methods I've already used are sufficient.Alternatively, I could use the fact that the coefficient of ( x^k ) in ( (x^2 + x + 1)^n ) is equal to the number of ways to write ( k ) as the sum of ( n ) numbers, each of which is 0, 1, or 2. Which is essentially the same as the combinatorial approach.In this case, ( n = 5 ) and ( k = 5 ). So, the number of ways to write 5 as the sum of five numbers, each 0, 1, or 2.This is similar to distributing 5 indistinguishable balls into 5 distinguishable boxes, each box holding at most 2 balls.The formula for this is the same as the number of non-negative integer solutions to ( a_1 + a_2 + a_3 + a_4 + a_5 = 5 ) with ( 0 leq a_i leq 2 ), which we already calculated as 51.So, all these different methods converge to the same answer, 51.Therefore, I'm confident that the coefficient of ( x^5 ) in the expansion of ( (x^2 + x + 1)^5 ) is 51, which corresponds to option A.**Final Answer**boxed{A}

💡Okay, so I need to prove that every integer can be written as (x^2 + y^2 - z^2) where (x), (y), and (z) are integers. Hmm, that sounds interesting. Let me think about how to approach this.First, I know that squares of integers are always non-negative, so (x^2), (y^2), and (z^2) are all non-negative. But since we're subtracting (z^2), the expression (x^2 + y^2 - z^2) can potentially be negative or positive, which is good because we need to represent all integers, both positive and negative.Maybe I can start by considering specific cases, like even and odd integers, and see if I can find a pattern or a way to express them in the desired form.Let me try with some small integers first to get a feel for it.Take the integer 0. Can I write 0 as (x^2 + y^2 - z^2)? Sure, if I set (x = 0), (y = 0), and (z = 0), then (0 + 0 - 0 = 0). That works.How about 1? Let's see. If I set (x = 1), (y = 0), and (z = 0), then (1 + 0 - 0 = 1). Perfect.What about 2? Hmm, let's see. If I set (x = 1), (y = 1), and (z = 0), then (1 + 1 - 0 = 2). That works too.How about a negative number, say -1? Let me try (x = 0), (y = 0), and (z = 1). Then (0 + 0 - 1 = -1). Nice, that works.Okay, so small numbers seem manageable. But I need to prove this for all integers, not just specific ones.Maybe I can find a general expression or formula that can represent any integer in this form. Let me think about how to manipulate the equation (n = x^2 + y^2 - z^2) to express (x), (y), and (z) in terms of (n).Alternatively, perhaps I can express (n) as a difference of squares. I know that (a^2 - b^2 = (a - b)(a + b)), so maybe I can use that identity somehow.Wait, in the equation (n = x^2 + y^2 - z^2), I can rearrange it as (n = (x^2 + y^2) - z^2). So, if I can write (n + z^2 = x^2 + y^2), then I can think of it as representing (n + z^2) as a sum of two squares.But I'm not sure if that helps directly. Maybe I need a different approach.Let me consider specific forms for (x), (y), and (z). For example, if I set (x = 0), then the equation becomes (n = y^2 - z^2). So, (n = (y - z)(y + z)). That means (n) can be expressed as a product of two integers of the same parity (since (y - z) and (y + z) are both even or both odd).But not all integers can be expressed as a product of two integers of the same parity. For example, if (n) is odd, then (y - z) and (y + z) must both be odd, which is possible. If (n) is even, then (y - z) and (y + z) must both be even, which is also possible because their product would be divisible by 4. Wait, but if (n) is even but not divisible by 4, like 2, then (n = 2) cannot be expressed as a product of two even numbers because 2 is not divisible by 4. Hmm, so setting (x = 0) might not work for all even integers.So, maybe I need to allow (x) to be non-zero. Let me try setting (x = 1). Then the equation becomes (n = 1 + y^2 - z^2), or (n - 1 = y^2 - z^2). So, similar to before, (n - 1) needs to be expressible as a difference of squares.If (n) is odd, then (n - 1) is even. Let me see if an even number can be expressed as a difference of squares. For example, take (n = 3), then (n - 1 = 2). Can 2 be expressed as (y^2 - z^2)? Let's see: (y^2 - z^2 = (y - z)(y + z)). So, 2 can be factored as 1 * 2. Therefore, (y - z = 1) and (y + z = 2). Solving these, we get (y = frac{1 + 2}{2} = 1.5) and (z = frac{2 - 1}{2} = 0.5). But these are not integers. Hmm, so that doesn't work.Wait, maybe I need to choose different factors. 2 can also be factored as 2 * 1, but that would lead to the same non-integer solutions. So, perhaps 2 cannot be expressed as a difference of squares of integers. That means setting (x = 1) might not help for (n = 3).Hmm, maybe I need a different approach. Let me think about parametrizing (x), (y), and (z) in terms of (n).Suppose I fix (x) and (y) in terms of (n) and then solve for (z). Let's say (x = k) and (y = k) for some integer (k). Then the equation becomes (n = k^2 + k^2 - z^2), which simplifies to (n = 2k^2 - z^2). Rearranging, (z^2 = 2k^2 - n). For (z) to be an integer, (2k^2 - n) must be a perfect square.But I don't know if this is always possible. Maybe I need a different parametrization.Alternatively, perhaps I can express (n) as (x^2 + y^2 - z^2) by choosing (x), (y), and (z) such that (x^2 + y^2 = n + z^2). So, for any integer (n), I need to find integers (x), (y), and (z) such that (x^2 + y^2 = n + z^2).This seems similar to expressing (n + z^2) as a sum of two squares. I know that not all integers can be expressed as a sum of two squares, but perhaps by adjusting (z), I can make (n + z^2) expressible as a sum of two squares.Wait, maybe I can choose (z) such that (n + z^2) is a sum of two squares. For example, if I set (z = 0), then (n) must be a sum of two squares. But not all integers are sums of two squares. However, if I allow (z) to vary, maybe I can adjust it to make (n + z^2) a sum of two squares.Let me try with (n = 7). Can I find (x), (y), and (z) such that (7 = x^2 + y^2 - z^2)? Let's see. If I set (z = 1), then (x^2 + y^2 = 8). The possible squares less than 8 are 0, 1, 4. So, 0 + 8 is not possible, 1 + 7 is not possible, 4 + 4 = 8. So, (x = 2), (y = 2), (z = 1). Then (2^2 + 2^2 - 1^2 = 4 + 4 - 1 = 7). Perfect.How about (n = 6)? Let me try (z = 1). Then (x^2 + y^2 = 7). The squares less than 7 are 0, 1, 4. So, 0 + 7 is not possible, 1 + 6 is not possible, 4 + 3 is not possible. So, no solution with (z = 1). Try (z = 2). Then (x^2 + y^2 = 6 + 4 = 10). Squares less than 10: 0, 1, 4, 9. 1 + 9 = 10, so (x = 1), (y = 3), (z = 2). Then (1^2 + 3^2 - 2^2 = 1 + 9 - 4 = 6). Perfect.Okay, so by adjusting (z), I can find solutions for specific (n). But I need a general proof that this is always possible.Maybe I can use the fact that any integer can be expressed as a difference of two squares, but with an adjustment for the third square.Wait, I recall that any integer can be expressed as a difference of two squares, but that requires the integer to be either odd or divisible by 4. But in our case, we have three squares, so maybe that gives more flexibility.Let me think about the equation (n = x^2 + y^2 - z^2). I can rearrange it as (n + z^2 = x^2 + y^2). So, for any integer (n), I need to find integers (x), (y), and (z) such that (x^2 + y^2 = n + z^2).Now, I know that every integer can be expressed as a sum of four squares, but here we have three squares. Maybe I can use some properties of quadratic forms.Alternatively, perhaps I can choose (x) and (y) in terms of (n) and (z). For example, set (x = z + a) and (y = z + b) for some integers (a) and (b). Then, substituting into the equation:(n = (z + a)^2 + (z + b)^2 - z^2)Expanding:(n = z^2 + 2az + a^2 + z^2 + 2bz + b^2 - z^2)Simplify:(n = z^2 + 2az + a^2 + 2bz + b^2)Combine like terms:(n = z^2 + 2z(a + b) + (a^2 + b^2))Hmm, this seems a bit complicated. Maybe I can choose (a) and (b) such that the coefficients of (z) cancel out or simplify.Alternatively, perhaps I can set (a = 1) and (b = 0) to simplify. Then:(n = z^2 + 2z(1 + 0) + (1 + 0))So,(n = z^2 + 2z + 1)Which is,(n = (z + 1)^2)But this only works if (n) is a perfect square, which it's not necessarily. So, that approach might not help.Maybe I need a different substitution. Let me try setting (x = y). Then the equation becomes:(n = 2x^2 - z^2)So, (z^2 = 2x^2 - n). For (z) to be an integer, (2x^2 - n) must be a perfect square. Let me see if I can choose (x) such that this is possible.Suppose I set (x = k), then (z^2 = 2k^2 - n). I need (2k^2 - n) to be a perfect square. Let me denote (m^2 = 2k^2 - n), so (n = 2k^2 - m^2).This is similar to expressing (n) as a difference of two squares, but scaled by 2. I'm not sure if this helps directly, but maybe I can choose (k) and (m) such that this holds.Alternatively, perhaps I can express (n) as (x^2 + y^2 - z^2) by choosing (x), (y), and (z) in a way that leverages the flexibility of three variables.Let me consider the parity of (n). If (n) is even, maybe I can express it differently than if (n) is odd.Suppose (n) is even. Let me set (x = 1), then the equation becomes (n = 1 + y^2 - z^2), so (n - 1 = y^2 - z^2). Now, (n - 1) is odd. I know that any odd integer can be expressed as a difference of two squares because (a^2 - b^2 = (a - b)(a + b)), and if the product is odd, both factors must be odd, which is possible.For example, let (n - 1 = k), where (k) is odd. Then, (k = (m + 1)^2 - m^2 = 2m + 1). So, (m = frac{k - 1}{2}). Therefore, (y = m + 1) and (z = m). Thus, (y = frac{k - 1}{2} + 1 = frac{k + 1}{2}) and (z = frac{k - 1}{2}).Since (k = n - 1), we have (y = frac{n}{2}) and (z = frac{n - 2}{2}). But wait, (n) is even, so (n = 2m), so (y = m + frac{1}{2}) and (z = m - frac{1}{2}), which are not integers. Hmm, that's a problem.Wait, maybe I need to adjust my approach. Let me set (x = 1), then (n - 1 = y^2 - z^2). Since (n - 1) is odd, I can write it as ( (y - z)(y + z) ). Let me set (y - z = 1) and (y + z = n - 1). Then, solving these equations:(y - z = 1)(y + z = n - 1)Adding both equations:(2y = n)So, (y = frac{n}{2})Subtracting the first equation from the second:(2z = n - 2)So, (z = frac{n - 2}{2})But since (n) is even, (n = 2k), so (y = k) and (z = k - 1). Therefore, (y) and (z) are integers. So, for even (n), setting (x = 1), (y = frac{n}{2}), and (z = frac{n}{2} - 1) works.For example, let (n = 4). Then (x = 1), (y = 2), (z = 1). So, (1^2 + 2^2 - 1^2 = 1 + 4 - 1 = 4). Perfect.Now, let's consider when (n) is odd. Let me set (x = 0). Then the equation becomes (n = y^2 - z^2). Since (n) is odd, (y^2 - z^2) must be odd. As I mentioned earlier, this is possible because (y) and (z) must be of opposite parity, so their squares will differ by an odd number.Let me write (n = (y - z)(y + z)). Since (n) is odd, both (y - z) and (y + z) must be odd. Let me set (y - z = 1) and (y + z = n). Then, solving these:Adding both equations:(2y = n + 1)So, (y = frac{n + 1}{2})Subtracting the first equation from the second:(2z = n - 1)So, (z = frac{n - 1}{2})Since (n) is odd, both (y) and (z) are integers. Therefore, for odd (n), setting (x = 0), (y = frac{n + 1}{2}), and (z = frac{n - 1}{2}) works.For example, let (n = 3). Then (x = 0), (y = 2), (z = 1). So, (0^2 + 2^2 - 1^2 = 0 + 4 - 1 = 3). Perfect.So, putting it all together, for any integer (n):- If (n) is even, set (x = 1), (y = frac{n}{2}), and (z = frac{n}{2} - 1).- If (n) is odd, set (x = 0), (y = frac{n + 1}{2}), and (z = frac{n - 1}{2}).This shows that every integer can indeed be expressed in the form (x^2 + y^2 - z^2) with integers (x), (y), and (z).I think this approach covers all integers, both positive and negative, because the expressions for (x), (y), and (z) work regardless of the sign of (n). For negative integers, the same logic applies because the squares will adjust accordingly.Let me test with a negative integer, say (n = -1). Since (-1) is odd, set (x = 0), (y = frac{-1 + 1}{2} = 0), and (z = frac{-1 - 1}{2} = -1). So, (0^2 + 0^2 - (-1)^2 = 0 + 0 - 1 = -1). Perfect.Another test with (n = -2), which is even. Set (x = 1), (y = frac{-2}{2} = -1), and (z = frac{-2}{2} - 1 = -2). So, (1^2 + (-1)^2 - (-2)^2 = 1 + 1 - 4 = -2). Perfect.Okay, I think this method works for all integers. By choosing (x) appropriately based on the parity of (n), and then solving for (y) and (z), we can always find integers (x), (y), and (z) such that (n = x^2 + y^2 - z^2).So, to summarize:- For even (n), set (x = 1), (y = frac{n}{2}), and (z = frac{n}{2} - 1).- For odd (n), set (x = 0), (y = frac{n + 1}{2}), and (z = frac{n - 1}{2}).This covers all integers, proving that every integer can be expressed in the desired form.

💡Okay, so I have this problem about a convex quadrilateral ABCD with perpendicular diagonals intersecting at point O. They give me AO = 2 and OC = 3, so the diagonal AC is 5 units long. There's a point K on side BC such that BK:KC = 1:2, and triangle AKD is equilateral. I need to find the area of triangle AKD.Hmm, let me start by drawing a diagram to visualize the problem. Since the diagonals are perpendicular, that might help in using some properties of perpendicular diagonals in quadrilaterals. I remember that in such cases, the area of the quadrilateral can be found using the formula (AC * BD)/2. But I'm not sure if that's directly useful here.Given that AO = 2 and OC = 3, the ratio of AO to OC is 2:3. Point K divides BC in the ratio 1:2, so BK:KC = 1:2. Maybe I can use coordinate geometry here. Let me assign coordinates to the points to make it easier.Let's place point O at the origin (0,0). Since diagonals are perpendicular, I can let AC lie along the x-axis and BD lie along the y-axis. So, point A is at (-2,0) because AO = 2, and point C is at (3,0) because OC = 3. Now, points B and D will be somewhere along the y-axis. Let me denote point B as (0,b) and point D as (0,d). Since the diagonals intersect at O, which is the origin, and they are perpendicular, this setup should work.Now, point K is on BC such that BK:KC = 1:2. Let me find the coordinates of point K. Since B is (0,b) and C is (3,0), the coordinates of K can be found using the section formula. If BK:KC = 1:2, then K divides BC in the ratio 1:2. So, the coordinates of K will be:K_x = (1*3 + 2*0)/(1+2) = 3/3 = 1K_y = (1*0 + 2*b)/(1+2) = (2b)/3So, K is at (1, 2b/3).Now, triangle AKD is equilateral. So, all sides AK, KD, and AD are equal. Let's find the coordinates of points A, K, and D:A: (-2,0)K: (1, 2b/3)D: (0,d)So, I need the distances AK, KD, and AD to be equal.First, let's compute AK:AK = sqrt[(1 - (-2))^2 + (2b/3 - 0)^2] = sqrt[(3)^2 + (2b/3)^2] = sqrt[9 + (4b²)/9]Next, compute KD:KD = sqrt[(0 - 1)^2 + (d - 2b/3)^2] = sqrt[1 + (d - 2b/3)^2]AD is the distance between A and D:AD = sqrt[(0 - (-2))^2 + (d - 0)^2] = sqrt[4 + d²]Since triangle AKD is equilateral, AK = KD = AD.So, set AK = AD:sqrt[9 + (4b²)/9] = sqrt[4 + d²]Square both sides:9 + (4b²)/9 = 4 + d²Simplify:(4b²)/9 = d² - 5Similarly, set KD = AD:sqrt[1 + (d - 2b/3)^2] = sqrt[4 + d²]Square both sides:1 + (d - 2b/3)^2 = 4 + d²Expand the left side:1 + d² - (4b/3)d + (4b²)/9 = 4 + d²Subtract d² from both sides:1 - (4b/3)d + (4b²)/9 = 4Simplify:- (4b/3)d + (4b²)/9 = 3Multiply both sides by 9 to eliminate denominators:-12b d + 4b² = 27So, we have two equations:1) (4b²)/9 = d² - 52) -12b d + 4b² = 27Let me rewrite equation 1:d² = (4b²)/9 + 5Now, substitute d² into equation 2. But equation 2 is linear in d, so maybe express d from equation 1 and substitute into equation 2.Wait, equation 1 gives d² in terms of b², but equation 2 is linear in d. Maybe express d from equation 1 as sqrt[(4b²)/9 + 5], but that might complicate things. Alternatively, let me see if I can express d from equation 2 in terms of b.From equation 2:-12b d + 4b² = 27Let me solve for d:-12b d = 27 - 4b²d = (4b² - 27)/(12b)So, d = (4b² - 27)/(12b)Now, substitute this into equation 1:d² = (4b²)/9 + 5So,[(4b² - 27)/(12b)]² = (4b²)/9 + 5Let me compute the left side:[(4b² - 27)^2]/(144b²) = (4b²)/9 + 5Multiply both sides by 144b² to eliminate denominators:(4b² - 27)^2 = 144b²*(4b²/9 + 5)Simplify the right side:144b²*(4b²/9) + 144b²*5 = 64b⁴ + 720b²Left side:(4b² - 27)^2 = 16b⁴ - 216b² + 729So, we have:16b⁴ - 216b² + 729 = 64b⁴ + 720b²Bring all terms to one side:16b⁴ - 216b² + 729 - 64b⁴ - 720b² = 0Combine like terms:(16b⁴ - 64b⁴) + (-216b² - 720b²) + 729 = 0-48b⁴ - 936b² + 729 = 0Multiply both sides by -1:48b⁴ + 936b² - 729 = 0Let me divide all terms by 3 to simplify:16b⁴ + 312b² - 243 = 0Let me set y = b², so the equation becomes:16y² + 312y - 243 = 0This is a quadratic in y. Let's solve for y using quadratic formula:y = [-312 ± sqrt(312² - 4*16*(-243))]/(2*16)Compute discriminant:D = 312² + 4*16*243Calculate 312²:312*312: 300²=90000, 12²=144, 2*300*12=7200, so (300+12)²=90000 + 7200 + 144=973444*16*243=64*243=64*(200+43)=64*200=12800 + 64*43=2752=12800+2752=15552So, D=97344 + 15552=112896sqrt(112896): Let's see, 336²=112896 because 300²=90000, 36²=1296, 2*300*36=21600, so (300+36)²=90000+21600+1296=112896. So sqrt(D)=336Thus,y = [-312 ± 336]/32Compute both possibilities:First solution:y = (-312 + 336)/32 = 24/32 = 3/4Second solution:y = (-312 - 336)/32 = (-648)/32 = -20.25But y = b² cannot be negative, so y = 3/4Thus, b² = 3/4, so b = sqrt(3)/2 or b = -sqrt(3)/2Since the quadrilateral is convex and diagonals intersect at O, which is the origin, and point B is (0,b). Depending on the orientation, b could be positive or negative, but since AO is 2 and OC is 3, and diagonals are perpendicular, I think b should be positive. So, b = sqrt(3)/2Now, compute d from earlier:d = (4b² - 27)/(12b)We have b² = 3/4, so 4b² = 3Thus,d = (3 - 27)/(12b) = (-24)/(12b) = (-2)/bSince b = sqrt(3)/2,d = (-2)/(sqrt(3)/2) = (-4)/sqrt(3) = (-4 sqrt(3))/3So, d = -4 sqrt(3)/3So, point D is at (0, -4 sqrt(3)/3)Now, let me compute the coordinates of points:A: (-2, 0)K: (1, 2b/3) = (1, 2*(sqrt(3)/2)/3) = (1, sqrt(3)/3)D: (0, -4 sqrt(3)/3)Now, let's compute the lengths AK, KD, and AD to confirm they are equal.First, AK:Distance between A(-2,0) and K(1, sqrt(3)/3):sqrt[(1 - (-2))² + (sqrt(3)/3 - 0)²] = sqrt[(3)² + (sqrt(3)/3)²] = sqrt[9 + (3)/9] = sqrt[9 + 1/3] = sqrt[28/3] = (2 sqrt(21))/3Wait, that doesn't seem right. Let me compute it again.Wait, (sqrt(3)/3)^2 = (3)/9 = 1/3. So, 9 + 1/3 = 28/3. So sqrt(28/3) = (2 sqrt(21))/3? Wait, sqrt(28/3) = sqrt(84)/3 = (2 sqrt(21))/3. Yes, that's correct.Next, KD:Distance between K(1, sqrt(3)/3) and D(0, -4 sqrt(3)/3):sqrt[(0 - 1)^2 + (-4 sqrt(3)/3 - sqrt(3)/3)^2] = sqrt[1 + (-5 sqrt(3)/3)^2] = sqrt[1 + (25*3)/9] = sqrt[1 + 75/9] = sqrt[1 + 25/3] = sqrt[(3 + 25)/3] = sqrt[28/3] = same as AK, which is (2 sqrt(21))/3Finally, AD:Distance between A(-2,0) and D(0, -4 sqrt(3)/3):sqrt[(0 - (-2))² + (-4 sqrt(3)/3 - 0)^2] = sqrt[4 + (16*3)/9] = sqrt[4 + 48/9] = sqrt[4 + 16/3] = sqrt[(12 + 16)/3] = sqrt[28/3] = same as before.So, all sides are equal, which confirms that triangle AKD is equilateral.Now, to find the area of triangle AKD. Since it's equilateral, the area can be calculated using the formula:Area = (sqrt(3)/4) * (side length)^2We have the side length as sqrt(28/3). So,Area = (sqrt(3)/4) * (28/3) = (sqrt(3)/4) * (28/3) = (7 sqrt(3))/3Wait, but let me compute it step by step:First, side length squared is (sqrt(28/3))² = 28/3So,Area = (sqrt(3)/4) * (28/3) = (28 sqrt(3))/12 = (7 sqrt(3))/3Yes, that's correct.Alternatively, I can compute the area using coordinates. The coordinates of A, K, D are:A(-2,0), K(1, sqrt(3)/3), D(0, -4 sqrt(3)/3)Using the shoelace formula:Area = (1/2)|sum(x_i y_{i+1} - x_{i+1} y_i)|Order of points: A, K, D, ACompute terms:x_A y_K - x_K y_A = (-2)*(sqrt(3)/3) - 1*0 = (-2 sqrt(3))/3 - 0 = (-2 sqrt(3))/3x_K y_D - x_D y_K = 1*(-4 sqrt(3)/3) - 0*(sqrt(3)/3) = (-4 sqrt(3))/3 - 0 = (-4 sqrt(3))/3x_D y_A - x_A y_D = 0*0 - (-2)*(-4 sqrt(3)/3) = 0 - (8 sqrt(3)/3) = (-8 sqrt(3))/3Sum these terms:(-2 sqrt(3))/3 + (-4 sqrt(3))/3 + (-8 sqrt(3))/3 = (-14 sqrt(3))/3Take absolute value and multiply by 1/2:Area = (1/2)*|(-14 sqrt(3))/3| = (1/2)*(14 sqrt(3)/3) = (7 sqrt(3))/3Same result as before. So, the area is (7 sqrt(3))/3.Wait, but the problem says to put the answer in a box, so I think that's the final answer.**Final Answer**The area of triangle (AKD) is boxed{dfrac{7sqrt{3}}{3}}.

💡Okay, so I have this problem here: "6 packs of DVDs can be bought with 120 dollars. How much does each pack cost?" Alright, let's see. I need to find out the cost per pack of DVDs. Hmm, okay, so if 6 packs cost 120 dollars in total, then each pack should cost less than 120 dollars, right? Probably, I need to divide the total cost by the number of packs to find the cost per pack.Wait, let me think. If I have 6 packs and I spent 120 dollars, then each pack would be 120 divided by 6. Let me write that down: 120 ÷ 6. Okay, what's 120 divided by 6? Well, 6 times 20 is 120, so 120 divided by 6 is 20. So, each pack costs 20 dollars.But wait, let me double-check to make sure I didn't make a mistake. If each pack is 20 dollars, then 6 packs would be 6 times 20, which is 120 dollars. Yeah, that matches the total amount given. So, it seems like 20 dollars per pack is the right answer.Is there another way to think about this? Maybe using fractions or something? Let's see. The total cost is 120 dollars for 6 packs, so the cost per pack is 120/6 dollars. Simplifying that, 120 divided by 6 is 20. Yep, same result.I guess another way to look at it is by breaking down the numbers. 120 divided by 6 can be thought of as dividing both the numerator and the denominator by 6. So, 120 divided by 6 is 20, and 6 divided by 6 is 1. So, it's 20 dollars per 1 pack, which is 20 dollars per pack.Wait, what if I think about it in terms of unit cost? Unit cost is the cost per unit, which in this case is per pack. So, unit cost equals total cost divided by the number of units. That's exactly what I did earlier: 120 divided by 6 equals 20. So, the unit cost is 20 dollars per pack.Is there any chance I could have misread the problem? Let me check again. It says 6 packs cost 120 dollars. So, yeah, I think I interpreted it correctly. It's not like 6 packs cost 120 dollars each or something like that. No, it's 6 packs for 120 dollars total.Maybe I can visualize it. Imagine I have 120 dollars and I want to buy 6 packs of DVDs. If I split the 120 dollars equally among the 6 packs, each pack would get 20 dollars. So, each pack costs 20 dollars.Another way to think about division is repeated subtraction. So, if I start with 120 dollars and subtract 6 dollars repeatedly until I reach zero, how many times would I subtract? Wait, no, that's not exactly right. Actually, division is about finding how many times a number fits into another number. So, how many times does 6 fit into 120? It fits 20 times because 6 times 20 is 120.I think I'm overcomplicating it, but it's good to make sure I understand it thoroughly. So, to recap: total cost is 120 dollars for 6 packs. To find the cost per pack, divide the total cost by the number of packs. 120 divided by 6 is 20. Therefore, each pack costs 20 dollars.Just to be absolutely sure, let's do a quick multiplication check. If each pack is 20 dollars, then 6 packs would be 6 times 20, which is 120 dollars. That matches the total cost given in the problem. So, everything checks out.I don't think I need to consider any other factors here, like taxes or discounts, because the problem doesn't mention anything like that. It's a straightforward division problem. So, I'm confident that the cost per pack is 20 dollars.

💡First, I need to determine how far ahead the coyote is when Darrel starts chasing it. Since the coyote has been traveling at 15 miles per hour for 1 hour, it is 15 miles ahead.Next, I'll calculate the relative speed between Darrel and the coyote. Darrel is traveling at 30 miles per hour, and the coyote is moving at 15 miles per hour. The relative speed is the difference between their speeds, which is 15 miles per hour.Finally, to find out how long it will take Darrel to catch up, I'll divide the distance he needs to cover (15 miles) by the relative speed (15 miles per hour). This gives me the time it will take for Darrel to catch the coyote.

💡Okay, so I have this geometry problem here about triangle PQR. It's a right-angled triangle with the right angle at Q. They tell me that PQ is 4 cm, and QR is some length that I need to find. There's a point X that moves along PQ, and from X, they draw a line parallel to QR, which intersects PR at Y. Then, from Y, they draw another line parallel to PQ, which intersects QR at Z. The least possible length of XZ is given as 3.2 cm. My goal is to find the length of QR.Alright, let me try to visualize this. Triangle PQR is right-angled at Q, so PQ and QR are the legs, and PR is the hypotenuse. Point X is somewhere on PQ, and from X, I draw a line parallel to QR, which meets PR at Y. Then from Y, I draw a line parallel to PQ, which meets QR at Z. So, XZ is the segment connecting X and Z, and we're told that the minimal length of XZ is 3.2 cm.Hmm, since the lines are drawn parallel to the sides, I think this forms some similar triangles or maybe a rectangle? Let me think. If I draw a line from X parallel to QR, which is one of the legs, that should be a horizontal line if I imagine QR as vertical. Then, from Y, drawing a line parallel to PQ, which is another leg, that would be a vertical line. So, connecting X to Y and Y to Z, it seems like we're forming a rectangle or a parallelogram.Wait, if XY is parallel to QR and YZ is parallel to PQ, then XY is perpendicular to YZ because PQ and QR are perpendicular. So, actually, the figure formed is a rectangle. That means XY is equal in length to QR, and YZ is equal in length to PQ. But the problem mentions XZ, which is the diagonal of this rectangle. So, XZ is the diagonal of a rectangle with sides XY and YZ. Since XY is parallel to QR, XY should be proportional to QR based on where X is on PQ.Let me denote some variables. Let's let the length of QR be 'a' cm. Since PQ is 4 cm, and QR is 'a' cm, the hypotenuse PR can be found using the Pythagorean theorem: PR = sqrt(PQ² + QR²) = sqrt(16 + a²).Now, point X is a variable point on PQ. Let's denote the distance from P to X as 'x' cm. So, XQ would be (4 - x) cm. Since XY is parallel to QR, triangle PXQ is similar to triangle PQR. Wait, is that correct? Let me see.Actually, since XY is parallel to QR, triangle PXQ is similar to triangle PQR by the Basic Proportionality Theorem (also known as Thales' theorem). So, the ratio of similarity would be the ratio of PX to PQ, which is x/4.Therefore, the length of XY, which is parallel to QR, would be (x/4)*QR = (x/4)*a. Similarly, the length of YZ, which is parallel to PQ, would be (x/4)*PQ = (x/4)*4 = x cm.But wait, earlier I thought that XY is equal to QR, but that doesn't seem right because XY is a segment inside the triangle, not the entire side. So, actually, XY is a scaled-down version of QR based on the position of X on PQ.So, if XY = (x/4)*a, and YZ = x, then the rectangle formed has sides of length (x/4)*a and x. Therefore, the diagonal XZ can be found using the Pythagorean theorem: XZ = sqrt[(x/4 * a)² + x²].We are told that the minimal length of XZ is 3.2 cm. So, we need to find the value of 'a' such that the minimal value of sqrt[(x/4 * a)² + x²] is 3.2 cm.To find the minimal value, we can consider the expression inside the square root: f(x) = (x²/16)*a² + x². We can find the minimum of this function with respect to x.Let me write f(x) = (a²/16)x² + x² = x²(a²/16 + 1). To find the minimum, we can take the derivative of f(x) with respect to x and set it to zero.But wait, f(x) is a quadratic function in terms of x², which is a parabola opening upwards. Therefore, the minimum occurs at the vertex. However, since x is a length along PQ, it can vary from 0 to 4 cm. So, the minimum of f(x) occurs either at the vertex or at the endpoints.But let's compute the derivative to be thorough. f'(x) = 2x(a²/16 + 1). Setting f'(x) = 0 gives x = 0. But x=0 would mean point X coincides with P, and then Y would coincide with P as well, making Z coincide with Q. Then, XZ would be PQ, which is 4 cm. But we are told the minimal XZ is 3.2 cm, which is less than 4 cm. So, the minimum must occur somewhere inside the interval (0,4).Wait, but f(x) is x²(a²/16 + 1), which is minimized when x is as small as possible, but x cannot be zero because then XZ becomes 4 cm. So, maybe my approach is incorrect.Alternatively, perhaps I need to consider the rectangle formed by XY and YZ, and find the minimal diagonal XZ. Since XZ is the diagonal, its length depends on both XY and YZ. Maybe I can express XZ in terms of x and a, then find the minimum with respect to x.Let me write XZ² = (XY)² + (YZ)² = (x*a/4)² + x² = x²(a²/16 + 1). So, XZ = x*sqrt(a²/16 + 1). To minimize XZ, we need to minimize x*sqrt(a²/16 + 1). Since sqrt(a²/16 + 1) is a constant with respect to x, the minimal XZ occurs when x is minimal. But x cannot be zero because then XZ would be zero, but in reality, when x approaches zero, X approaches P, and Z approaches Q, making XZ approach PQ, which is 4 cm. Wait, that contradicts the given minimal XZ of 3.2 cm.I think I'm missing something here. Maybe the minimal XZ doesn't occur when x is minimal or maximal, but somewhere in between. Let me consider the coordinates approach.Let's place triangle PQR on a coordinate system with point Q at the origin (0,0), PQ along the y-axis, and QR along the x-axis. So, point P is at (0,4), point Q is at (0,0), and point R is at (a,0). Then, point X is somewhere on PQ, so its coordinates are (0, x), where x ranges from 0 to 4.Now, the line through X parallel to QR (which is along the x-axis) would be a horizontal line. So, the equation of this line is y = x. Wait, no, because QR is along the x-axis, so a line parallel to QR would also be horizontal, meaning constant y-coordinate. Since X is at (0, x), the line through X parallel to QR is y = x. Wait, that can't be right because QR is along the x-axis, so its slope is zero. Therefore, the line through X parallel to QR would have the same slope, which is zero, so it's a horizontal line. Therefore, the equation is y = x-coordinate of X, which is x. Wait, no, the y-coordinate of X is x, so the horizontal line is y = x. But that would mean the line is y = x, which is a diagonal line, not horizontal. That doesn't make sense.Wait, I think I made a mistake. If QR is along the x-axis from (0,0) to (a,0), then a line parallel to QR would also be horizontal, meaning it has a slope of zero. Therefore, the line through X(0, x) parallel to QR is y = x. Wait, no, that's not correct. The line through X(0, x) with slope zero is y = x. Wait, no, slope zero means y is constant. So, the line is y = x, but that's not correct because x is the y-coordinate of X. Wait, I'm getting confused.Let me clarify. If QR is along the x-axis, then a line parallel to QR would also be horizontal, meaning it's of the form y = k, where k is a constant. Since X is at (0, x), the horizontal line through X is y = x. Wait, that would mean the line is y = x, but that's a diagonal line, not horizontal. I think I'm mixing up variables here.Let me denote the coordinates more clearly. Let me say point P is at (0,4), Q at (0,0), and R at (a,0). Then, point X is on PQ, so its coordinates are (0, t), where t ranges from 0 to 4. The line through X parallel to QR (which is along the x-axis) would be a horizontal line, so its equation is y = t.This line intersects PR at Y. Let's find the coordinates of Y. The line PR goes from P(0,4) to R(a,0). Its equation can be found using the two-point form. The slope of PR is (0 - 4)/(a - 0) = -4/a. So, the equation of PR is y - 4 = (-4/a)(x - 0), which simplifies to y = (-4/a)x + 4.Now, the horizontal line y = t intersects PR at Y. So, setting y = t in the equation of PR, we get t = (-4/a)x + 4. Solving for x, we have x = (4 - t)*(a/4) = a*(4 - t)/4.Therefore, the coordinates of Y are (a*(4 - t)/4, t).Now, from Y, we draw a line parallel to PQ. Since PQ is along the y-axis, a line parallel to PQ would be vertical. So, the line through Y parallel to PQ is a vertical line passing through Y, which has the equation x = a*(4 - t)/4.This vertical line intersects QR at Z. Since QR is along the x-axis from (0,0) to (a,0), the point Z must be at (a*(4 - t)/4, 0).Therefore, the coordinates of Z are (a*(4 - t)/4, 0).Now, we need to find the length of XZ. Point X is at (0, t), and point Z is at (a*(4 - t)/4, 0). So, the distance between X and Z is:XZ = sqrt[(a*(4 - t)/4 - 0)^2 + (0 - t)^2]= sqrt[(a²*(4 - t)²)/16 + t²]We are told that the minimal length of XZ is 3.2 cm. So, we need to find the value of 'a' such that the minimum of sqrt[(a²*(4 - t)²)/16 + t²] is 3.2.Let me denote f(t) = (a²*(4 - t)²)/16 + t². To find the minimum of f(t), we can take the derivative with respect to t and set it to zero.f'(t) = (a²/16)*2*(4 - t)*(-1) + 2t= (-a²/8)*(4 - t) + 2tSet f'(t) = 0:(-a²/8)*(4 - t) + 2t = 0Multiply both sides by 8 to eliminate the denominator:-a²*(4 - t) + 16t = 0-4a² + a² t + 16t = 0t(a² + 16) = 4a²t = (4a²)/(a² + 16)So, the value of t that minimizes f(t) is t = (4a²)/(a² + 16).Now, substitute this value of t back into f(t):f(t) = (a²*(4 - t)²)/16 + t²First, compute (4 - t):4 - t = 4 - (4a²)/(a² + 16) = [4(a² + 16) - 4a²]/(a² + 16) = [4a² + 64 - 4a²]/(a² + 16) = 64/(a² + 16)So, (4 - t)² = (64/(a² + 16))² = 4096/(a² + 16)²Now, compute (a²*(4 - t)²)/16:= a² * 4096/(a² + 16)² / 16= (a² * 4096)/(16(a² + 16)²)= (a² * 256)/(a² + 16)²Next, compute t²:t² = [(4a²)/(a² + 16)]² = 16a⁴/(a² + 16)²So, f(t) = (a² * 256)/(a² + 16)² + 16a⁴/(a² + 16)²= [256a² + 16a⁴]/(a² + 16)²= 16a²(16 + a²)/(a² + 16)²= 16a²/(a² + 16)Therefore, f(t) = 16a²/(a² + 16)Since f(t) is the square of XZ, we have XZ² = 16a²/(a² + 16)Given that the minimal XZ is 3.2 cm, we have:sqrt(16a²/(a² + 16)) = 3.2Square both sides:16a²/(a² + 16) = (3.2)² = 10.24Now, solve for 'a':16a² = 10.24(a² + 16)16a² = 10.24a² + 163.8416a² - 10.24a² = 163.845.76a² = 163.84a² = 163.84 / 5.76a² = 28.444...a = sqrt(28.444...)a ≈ 5.333...Wait, 28.444... is 28 and 4/9, which is 256/9. So, a² = 256/9, so a = 16/3 ≈ 5.333 cm.Therefore, the length of QR is 16/3 cm, which is approximately 5.333 cm.But let me double-check the calculations to make sure I didn't make a mistake.Starting from f(t) = 16a²/(a² + 16) = (3.2)² = 10.24So, 16a² = 10.24(a² + 16)16a² = 10.24a² + 163.8416a² - 10.24a² = 163.845.76a² = 163.84a² = 163.84 / 5.76Let me compute 163.84 ÷ 5.76:First, 5.76 × 28 = 161.28Subtract: 163.84 - 161.28 = 2.56Now, 5.76 × 0.444... ≈ 2.56Because 5.76 × 0.4 = 2.3045.76 × 0.444 = 5.76 × (4/9) = 2.56So, 5.76 × (28 + 4/9) = 163.84Therefore, a² = 28 + 4/9 = 256/9So, a = sqrt(256/9) = 16/3 ≈ 5.333 cmYes, that seems correct.Therefore, the length of QR is 16/3 cm, which is approximately 5.333 cm.