💡First, I need to find the number for which 30% of it is equal to 25% of 50.I'll start by calculating 25% of 50. To do this, I'll convert 25% to a decimal, which is 0.25, and multiply it by 50.Next, I'll set up an equation where 30% of the unknown number (let's call it x) equals the result from the first step. Converting 30% to a decimal gives me 0.30, so the equation becomes 0.30x = 12.5.To solve for x, I'll divide both sides of the equation by 0.30. This will isolate x and give me its value.After performing the division, I'll have the value of x, which is the number I'm looking for.

💡Okay, so I'm trying to understand this problem about a function ( f(x) ) with specific properties. Let me break it down step by step.First, the function ( f(x) ) has two main properties:1. ( f(0) = 0 ) and ( f(1) = 1 ).2. There are infinitely many positive integers ( n ) such that if the equation [ f(x_1) + f(x_2) + cdots + f(x_s) = n ] has integer solutions ( x_1, x_2, ldots, x_s ), then the number of terms ( s ) must be at least ( 2^k - 1 ).Hmm, okay. So, the function starts at 0 when ( x = 0 ) and is 1 when ( x = 1 ). The second property is a bit more complex. It says that for infinitely many ( n ), if you can express ( n ) as a sum of ( f(x) ) values, you need at least ( 2^k - 1 ) terms. I'm not entirely sure what ( k ) refers to here, but maybe it's a parameter related to the function or the problem.Let me think about the second property more carefully. It mentions that for infinitely many ( n ), the minimal number of terms ( s ) needed to express ( n ) as a sum of ( f(x) )s is at least ( 2^k - 1 ). So, this is about the additive basis of the function ( f(x) ). It's saying that the function isn't too "dense" in the integers because you need a significant number of terms to reach certain integers ( n ).I wonder if this relates to something like the binary representation of numbers. Since ( 2^k - 1 ) is one less than a power of two, which is the maximum number you can represent with ( k ) bits. Maybe the function ( f(x) ) has something to do with binary expansions or something similar.Wait, the problem also mentions that ( f(x) ) is an integer polynomial. That might be important. So, ( f(x) ) is a polynomial that takes integer values for integer inputs. That could mean it's something like a binomial coefficient or a combination of such terms.Let me consider an example. Suppose ( f(x) ) is the binomial coefficient ( binom{x}{i} ) for some ( i ). These are integer-valued polynomials, right? For example, ( binom{x}{1} = x ), ( binom{x}{2} = frac{x(x-1)}{2} ), and so on. These are all integer-valued for integer ( x ).But in the problem, ( f(x) ) is a sum of such binomial coefficients multiplied by powers of 2. Specifically, the function is given as ( f(x) = sum_{i=1}^{k} binom{x}{i} cdot (-2)^{i-1} ). Hmm, that seems a bit complicated, but let's try to unpack it.First, let's compute ( f(x) ) for small values of ( x ) to see if it makes sense. Let's take ( k = 2 ) as a simple case.For ( k = 2 ):- ( f(x) = binom{x}{1} cdot (-2)^{0} + binom{x}{2} cdot (-2)^{1} )- Simplify: ( f(x) = x cdot 1 + frac{x(x-1)}{2} cdot (-2) )- Which is ( f(x) = x - x(x - 1) )- Simplify further: ( f(x) = x - x^2 + x = -x^2 + 2x )Let's compute ( f(0) = 0 ), which is correct. ( f(1) = -1 + 2 = 1 ), which is also correct. ( f(2) = -4 + 4 = 0 ). Hmm, interesting. ( f(3) = -9 + 6 = -3 ). Wait, negative values? But the problem didn't specify that ( f(x) ) has to be non-negative. So, maybe that's okay.But in the problem statement, we're considering sums of ( f(x) ) to reach positive integers ( n ). If ( f(x) ) can take negative values, that complicates things because you could potentially subtract to reach ( n ). But the problem says "integer solutions," so maybe negative ( x ) are allowed? Or perhaps ( x ) are non-negative integers?Wait, the problem says "integer solutions ( x_1, x_2, ldots, x_s )", so ( x ) can be any integer, positive or negative. That might be important.But let's go back to the function ( f(x) ). For ( k = 2 ), we have ( f(x) = -x^2 + 2x ). So, ( f(x) ) can be positive or negative depending on ( x ). For ( x = 0 ), it's 0; ( x = 1 ), it's 1; ( x = 2 ), it's 0; ( x = 3 ), it's -3; ( x = -1 ), it's -1 - 2 = -3; and so on.Now, let's think about the second property. For infinitely many ( n ), the equation ( f(x_1) + f(x_2) + cdots + f(x_s) = n ) requires at least ( 2^k - 1 ) terms. For ( k = 2 ), that would be ( 3 ) terms. So, for infinitely many ( n ), you can't express ( n ) as the sum of fewer than 3 ( f(x) )s.But wait, ( f(x) ) can take both positive and negative values. So, maybe you can combine positive and negative terms to reach certain ( n ). But the problem says "integer solutions," so negative ( x ) are allowed, but does that mean negative ( f(x) ) can be used to subtract?Hmm, this is getting a bit confusing. Maybe I should think about the general form of ( f(x) ). The function is given as ( f(x) = sum_{i=1}^{k} binom{x}{i} cdot (-2)^{i-1} ). Let's see if this can be simplified or expressed in another way.I recall that binomial coefficients can be related to generating functions. Maybe if I consider the generating function for ( f(x) ), I can find a pattern or a closed-form expression.The generating function for ( binom{x}{i} ) is ( (1 + t)^x ). So, the generating function for ( f(x) ) would be ( sum_{i=1}^{k} (-2)^{i-1} binom{x}{i} t^i ). Hmm, not sure if that helps directly.Wait, maybe I can write ( f(x) ) as a combination of lower-degree polynomials. Since each ( binom{x}{i} ) is a polynomial of degree ( i ), the sum ( f(x) ) will be a polynomial of degree ( k ).Let me try to compute ( f(x) ) for a general ( k ). For example, for ( k = 3 ):- ( f(x) = binom{x}{1} cdot (-2)^0 + binom{x}{2} cdot (-2)^1 + binom{x}{3} cdot (-2)^2 )- Simplify: ( f(x) = x - 2 cdot frac{x(x-1)}{2} + 4 cdot frac{x(x-1)(x-2)}{6} )- Which is ( f(x) = x - x(x - 1) + frac{2}{3}x(x - 1)(x - 2) )- Simplify further: ( f(x) = x - x^2 + x + frac{2}{3}x^3 - 2x^2 + frac{4}{3}x )- Combine like terms: ( f(x) = frac{2}{3}x^3 - 3x^2 + frac{8}{3}x )Hmm, this is getting messy with fractions. Maybe I should consider a different approach.Wait, the problem mentions that ( f(x) ) is an integer polynomial, but not necessarily with integer coefficients. So, even though the coefficients might not be integers, the function evaluated at integer ( x ) gives integers. That makes sense because binomial coefficients are integers when ( x ) is an integer.But how does this relate to the second property? The key seems to be in the modular arithmetic part. The problem mentions that for ( x geq k ), the expression ( 2f(x) - 1 ) is congruent to ( -(-1)^x ) modulo ( 2^k ). Let me try to understand that.So, for ( x geq k ), we have:[ 2f(x) - 1 equiv -(-1)^x pmod{2^k} ]This implies that ( 2f(x) equiv 1 - (-1)^x pmod{2^k} ). Therefore, ( f(x) equiv frac{1 - (-1)^x}{2} pmod{2^{k-1}} ).Wait, that seems like a useful congruence. So, depending on whether ( x ) is even or odd, ( f(x) ) is congruent to either 0 or 1 modulo ( 2^{k-1} ).If ( x ) is even, ( (-1)^x = 1 ), so ( f(x) equiv 0 pmod{2^{k-1}} ).If ( x ) is odd, ( (-1)^x = -1 ), so ( f(x) equiv 1 pmod{2^{k-1}} ).Interesting. So, for ( x geq k ), ( f(x) ) is either 0 or 1 modulo ( 2^{k-1} ). That might help in analyzing the sum ( f(x_1) + f(x_2) + cdots + f(x_s) ).Now, let's consider the equation ( f(x_1) + f(x_2) + cdots + f(x_s) = n ). If ( n equiv -1 pmod{2^k} ), then we have:[ f(x_1) + f(x_2) + cdots + f(x_s) equiv 2^k - 1 pmod{2^k} ]From the earlier congruence, each ( f(x_i) ) is either 0 or 1 modulo ( 2^{k-1} ). So, when we sum them up modulo ( 2^k ), each term contributes either 0 or 1 modulo ( 2^{k-1} ).Wait, but the sum needs to be ( 2^k - 1 ) modulo ( 2^k ). That is, the sum is congruent to ( -1 ) modulo ( 2^k ). So, how can we get such a sum?Each ( f(x_i) ) is either 0 or 1 modulo ( 2^{k-1} ). So, if we have ( s ) terms, each contributing either 0 or 1 modulo ( 2^{k-1} ), the total sum modulo ( 2^{k-1} ) would be the number of terms that are congruent to 1 modulo ( 2^{k-1} ).But we need the sum to be ( 2^k - 1 ) modulo ( 2^k ). Let's see what ( 2^k - 1 ) is modulo ( 2^{k-1} ). Since ( 2^k = 2 cdot 2^{k-1} ), ( 2^k - 1 = 2 cdot 2^{k-1} - 1 equiv -1 pmod{2^{k-1}} ).So, the sum modulo ( 2^{k-1} ) is ( -1 ), which is equivalent to ( 2^{k-1} - 1 ). Therefore, the number of terms contributing 1 modulo ( 2^{k-1} ) must be ( 2^{k-1} - 1 ).But each such term is a ( f(x_i) ) that is 1 modulo ( 2^{k-1} ). From the earlier analysis, ( f(x) equiv 1 pmod{2^{k-1}} ) only when ( x ) is odd and ( x geq k ). So, each ( x_i ) must be odd and at least ( k ).But wait, if ( x_i ) is odd and at least ( k ), then ( f(x_i) ) is 1 modulo ( 2^{k-1} ), but what about the higher bits? Since ( f(x_i) ) is a polynomial, it can have higher values, but modulo ( 2^k ), it's either 0 or 1.Wait, no. The earlier congruence was modulo ( 2^k ), right? So, ( 2f(x) - 1 equiv -(-1)^x pmod{2^k} ), which implies ( f(x) equiv frac{1 - (-1)^x}{2} pmod{2^{k-1}} ).So, if ( x ) is odd, ( f(x) equiv 1 pmod{2^{k-1}} ), and if ( x ) is even, ( f(x) equiv 0 pmod{2^{k-1}} ).Therefore, when summing ( f(x_i) ), each odd ( x_i ) contributes 1 modulo ( 2^{k-1} ), and each even ( x_i ) contributes 0. So, to get a sum of ( 2^k - 1 ) modulo ( 2^k ), which is ( -1 ) modulo ( 2^k ), we need the sum modulo ( 2^{k-1} ) to be ( 2^{k-1} - 1 ).This means we need exactly ( 2^{k-1} - 1 ) terms where ( x_i ) is odd and at least ( k ). But each such term contributes 1 modulo ( 2^{k-1} ), so the total number of such terms must be ( 2^{k-1} - 1 ).However, each of these terms is a single ( f(x_i) ), so to get the required sum, we need at least ( 2^{k-1} - 1 ) terms. But the problem states that ( s geq 2^k - 1 ). Wait, that seems like a discrepancy.Wait, maybe I made a mistake in the modular arithmetic. Let me double-check.We have ( 2f(x) - 1 equiv -(-1)^x pmod{2^k} ). So, if ( x ) is even, ( (-1)^x = 1 ), so ( 2f(x) - 1 equiv -1 pmod{2^k} ), which implies ( 2f(x) equiv 0 pmod{2^k} ), so ( f(x) equiv 0 pmod{2^{k-1}} ).If ( x ) is odd, ( (-1)^x = -1 ), so ( 2f(x) - 1 equiv 1 pmod{2^k} ), which implies ( 2f(x) equiv 2 pmod{2^k} ), so ( f(x) equiv 1 pmod{2^{k-1}} ).So, for even ( x ), ( f(x) equiv 0 pmod{2^{k-1}} ), and for odd ( x ), ( f(x) equiv 1 pmod{2^{k-1}} ).Now, when we sum ( s ) terms ( f(x_i) ), the sum modulo ( 2^{k-1} ) is equal to the number of odd ( x_i )s modulo ( 2^{k-1} ). Because each odd ( x_i ) contributes 1, and each even ( x_i ) contributes 0.We need the total sum to be ( 2^k - 1 ) modulo ( 2^k ), which is equivalent to ( -1 ) modulo ( 2^k ). But modulo ( 2^{k-1} ), ( -1 ) is equivalent to ( 2^{k-1} - 1 ).Therefore, the number of odd ( x_i )s must be congruent to ( 2^{k-1} - 1 ) modulo ( 2^{k-1} ). Since we're dealing with counts, the minimal number of odd ( x_i )s needed is ( 2^{k-1} - 1 ).But each odd ( x_i ) is a single term contributing 1 modulo ( 2^{k-1} ). So, to get a total of ( 2^{k-1} - 1 ) modulo ( 2^{k-1} ), we need exactly ( 2^{k-1} - 1 ) odd terms.However, the problem states that ( s geq 2^k - 1 ). Wait, that's a much larger number. So, maybe my reasoning is missing something.Perhaps I need to consider higher bits beyond ( 2^{k-1} ). Each ( f(x_i) ) is not just 0 or 1 modulo ( 2^{k-1} ), but it can have higher contributions modulo ( 2^k ).Wait, let's think about the actual value of ( f(x) ) modulo ( 2^k ). From the earlier equation:[ 2f(x) - 1 equiv -(-1)^x pmod{2^k} ]So, rearranging:[ 2f(x) equiv 1 - (-1)^x pmod{2^k} ][ f(x) equiv frac{1 - (-1)^x}{2} pmod{2^{k-1}} ]But modulo ( 2^k ), ( f(x) ) can be either ( 0 ) or ( 1 ) depending on ( x )'s parity, but scaled appropriately.Wait, maybe I should consider the binary representation of ( f(x) ). Since ( f(x) ) is designed such that each bit beyond the ( (k-1) )-th bit is determined by the lower bits.But I'm getting a bit stuck here. Let me try a different approach.Suppose we have ( n equiv -1 pmod{2^k} ). Then, ( n = 2^k m - 1 ) for some integer ( m ). We need to express ( n ) as a sum of ( f(x_i) ).From the earlier analysis, each ( f(x_i) ) is either 0 or 1 modulo ( 2^{k-1} ). So, to get a sum of ( 2^k - 1 ), which is ( 2 cdot 2^{k-1} - 1 ), we need to have contributions from both the lower ( 2^{k-1} ) bits and the higher bits.Wait, maybe each ( f(x_i) ) can contribute up to ( 2^{k-1} ) in the sum. So, to reach ( 2^k - 1 ), which is ( 2 cdot 2^{k-1} - 1 ), we need at least ( 2^{k} - 1 ) terms because each term can contribute at most 1 in the higher bits.But I'm not entirely sure. Let me think about it differently.If each ( f(x_i) ) is either 0 or 1 modulo ( 2^{k-1} ), then to get a sum of ( 2^k - 1 ), which is ( 2 cdot 2^{k-1} - 1 ), we need to have contributions from both the lower and higher bits.Specifically, the sum modulo ( 2^{k} ) is ( -1 ), which means that the sum has a carryover into the ( 2^k ) bit. To achieve this, we need enough terms such that their contributions modulo ( 2^{k} ) sum up to ( -1 ).But how does this relate to the number of terms ( s )?Wait, each term ( f(x_i) ) can contribute either 0 or 1 modulo ( 2^{k-1} ), but when considering modulo ( 2^k ), each term can contribute either 0 or 1 in the lower ( 2^{k-1} ) bits, and potentially more in higher bits.But since we're summing modulo ( 2^k ), the higher bits beyond ( 2^k ) don't affect the result. So, the key is that each term can contribute at most 1 in the lower ( 2^{k-1} ) bits, and potentially 0 or more in higher bits, but since we're working modulo ( 2^k ), higher bits wrap around.Wait, maybe I'm overcomplicating it. Let's think about the minimal number of terms needed to reach a sum of ( 2^k - 1 ) modulo ( 2^k ).Each term ( f(x_i) ) can contribute either 0 or 1 modulo ( 2^{k-1} ). To get a total of ( 2^k - 1 ) modulo ( 2^k ), which is ( -1 ), we need the sum of the lower ( 2^{k-1} ) bits to be ( 2^{k-1} - 1 ), and the higher bits to contribute an additional ( 2^{k-1} ).But how does that translate to the number of terms?Wait, each term can contribute at most 1 to the lower ( 2^{k-1} ) bits. So, to get ( 2^{k-1} - 1 ) in the lower bits, we need at least ( 2^{k-1} - 1 ) terms. But since we also need to account for the higher bits contributing ( 2^{k-1} ), which is another ( 2^{k-1} ) terms, the total number of terms needed is ( 2^{k-1} - 1 + 2^{k-1} = 2^k - 1 ).Ah, that makes sense! So, to get the required sum modulo ( 2^k ), we need ( 2^{k-1} - 1 ) terms contributing to the lower bits and ( 2^{k-1} ) terms contributing to the higher bits, totaling ( 2^k - 1 ) terms.Therefore, for ( n equiv -1 pmod{2^k} ), the minimal number of terms ( s ) needed is ( 2^k - 1 ). Since there are infinitely many such ( n ) (for example, ( n = 2^k - 1, 2^{k+1} - 1, 2^{k+2} - 1, ldots )), the function ( f(x) ) satisfies the second property.So, putting it all together, the function ( f(x) ) is constructed in such a way that it forces the sum to require at least ( 2^k - 1 ) terms for infinitely many ( n ). This is achieved by leveraging the properties of binomial coefficients and modular arithmetic to control the contributions of each term in the sum.I think I've got a handle on it now. The key was understanding how the function ( f(x) ) behaves modulo powers of 2 and how that constrains the number of terms needed to reach certain sums.

💡Okay, so I have this problem about calculating the average donation per person in a company during a disaster relief effort. Let me try to understand what's given and figure out how to approach it.First, the problem states that 1/10 of the people donated 200 yuan each. Then, 3/4 of the people donated 100 yuan each. The remaining people donated 50 yuan each. I need to find the average donation per person.Hmm, averages usually involve adding up all the donations and then dividing by the number of people. But since the problem is given in fractions of the total number of people, maybe I can work with fractions directly without knowing the exact number of people.Let me denote the total number of people in the company as N. That might make it easier to calculate each group's contribution.So, if there are N people:- 1/10 of N donated 200 yuan each. That would be (1/10)*N people donating 200 yuan.- 3/4 of N donated 100 yuan each. That's (3/4)*N people donating 100 yuan.- The remaining people donated 50 yuan each. I need to figure out what fraction that is.Wait, the remaining people would be N minus the ones who donated 200 and 100 yuan. So, that's N - (1/10)*N - (3/4)*N. Let me compute that.First, let's find a common denominator for 1/10 and 3/4. The denominators are 10 and 4. The least common multiple of 10 and 4 is 20. So, converting both fractions to have a denominator of 20:1/10 = 2/203/4 = 15/20So, adding those together: 2/20 + 15/20 = 17/20Therefore, the remaining fraction is 1 - 17/20 = 3/20So, 3/20 of N donated 50 yuan each.Now, to find the total donation, I can calculate each group's total contribution and then add them up.Total donation from the first group: (1/10)*N * 200 yuanTotal donation from the second group: (3/4)*N * 100 yuanTotal donation from the third group: (3/20)*N * 50 yuanLet me compute each of these:First group: (1/10)*N * 200 = (200/10)*N = 20*NSecond group: (3/4)*N * 100 = (300/4)*N = 75*NThird group: (3/20)*N * 50 = (150/20)*N = 7.5*NNow, adding all these together:Total donation = 20*N + 75*N + 7.5*N = (20 + 75 + 7.5)*N = 102.5*NSo, the total donation is 102.5*N yuan.But the average donation per person is total donation divided by the number of people, which is N.So, average donation = (102.5*N) / N = 102.5 yuanTherefore, the average donation per person is 102.5 yuan.Wait, let me double-check my calculations to make sure I didn't make any mistakes.Starting with the fractions:1/10 + 3/4 + 3/20Convert all to 20ths:1/10 = 2/203/4 = 15/203/20 = 3/20Adding them up: 2 + 15 + 3 = 20/20 = 1, which checks out.Calculating each group's total:First group: (1/10)*N * 200 = 20*NSecond group: (3/4)*N * 100 = 75*NThird group: (3/20)*N * 50 = 7.5*NAdding them: 20 + 75 + 7.5 = 102.5Yes, that seems correct.So, the average donation per person is indeed 102.5 yuan.I think that's the right approach. I considered all the groups, converted their fractions to a common denominator to ensure the remaining fraction was calculated correctly, then multiplied each group's donation amount by their respective fractions of the total number of people, summed them up, and then divided by the total number of people to get the average. Everything adds up, so I feel confident about this answer.

💡Okay, so I have this problem here: If the square root of (x minus 9) is meaningful within the real number range, then what is the range of real number x? Hmm, let me think about this step by step.First off, I remember that when dealing with square roots, the expression inside the square root has to be non-negative. That's because you can't take the square root of a negative number and get a real number. So, for the square root of (x - 9) to be meaningful, the part inside the square root, which is (x - 9), must be greater than or equal to zero.Alright, so I need to set up an inequality where (x - 9) is greater than or equal to zero. Let me write that down:x - 9 ≥ 0Now, to solve for x, I just need to isolate x on one side of the inequality. I can do that by adding 9 to both sides. Let's see:x - 9 + 9 ≥ 0 + 9Simplifying that, the -9 and +9 on the left side cancel each other out, leaving me with:x ≥ 9So, this tells me that x has to be greater than or equal to 9 for the square root of (x - 9) to be a real number. That makes sense because if x were less than 9, then (x - 9) would be negative, and the square root of a negative number isn't a real number—it's a complex number, which isn't what we're dealing with here.Let me double-check my reasoning. If x is exactly 9, then (x - 9) is 0, and the square root of 0 is 0, which is perfectly fine. If x is greater than 9, say 10, then (10 - 9) is 1, and the square root of 1 is 1, which is also fine. What if x is less than 9, like 8? Then (8 - 9) is -1, and the square root of -1 isn't a real number. So, my inequality x ≥ 9 seems correct.Is there any other consideration I need to make here? Well, sometimes with square roots, people worry about other things like domain restrictions or if there are any other operations involved, but in this case, it's just a straightforward square root. So, I think I've covered all the bases.To sum it up, the expression √(x - 9) is only meaningful in the real number system when the value inside the square root is zero or positive. That happens when x is greater than or equal to 9. So, the range of x is all real numbers from 9 upwards, including 9 itself.I don't think I've missed anything here. It seems pretty straightforward. Maybe I can visualize it on a number line to make sure. If I draw a number line, with 9 marked on it, and then shade all the numbers to the right of 9, including 9, that should represent all the possible x values that make √(x - 9) meaningful. Yep, that looks right.Another way to think about it is to consider the graph of the function y = √(x - 9). The graph of a square root function starts at the point where the expression inside the square root is zero, which in this case is at x = 9, y = 0. From there, it increases as x increases. So, the domain of this function, which is the set of all possible x-values, starts at 9 and goes to infinity. That matches what I found earlier.I guess I could also test some values to confirm. Let's try x = 9: √(9 - 9) = √0 = 0, which is fine. x = 10: √(10 - 9) = √1 = 1, also fine. x = 15: √(15 - 9) = √6, which is approximately 2.45, still a real number. Now, if I try x = 8: √(8 - 9) = √(-1), which isn't a real number. So, that confirms that x needs to be at least 9.Is there a way this could be more complicated? Maybe if there were other operations or if the square root was part of a larger expression, but in this case, it's isolated. So, I think my conclusion is solid.In summary, the key takeaway is that for any square root expression √(expression), the expression inside must be greater than or equal to zero for the square root to be real. Applying that to √(x - 9), we set x - 9 ≥ 0 and solve for x, resulting in x ≥ 9. This gives us the range of x as all real numbers from 9 onwards.I feel confident about this answer now. It's always good to go through the reasoning step by step, especially when you're just starting out with these concepts. It helps solidify the understanding and ensures that there are no gaps in your knowledge.

💡First, calculate the fraction of medals won by Team USA: ( frac{23}{120} approx 0.1917 ).Next, convert each answer choice to its decimal equivalent:[frac{1}{4} = 0.25, quad frac{1}{5} = 0.2, quad frac{1}{6} approx 0.1667, quad frac{1}{7} approx 0.1429, quad frac{1}{8} = 0.125]Compare these decimals with ( 0.1917 ) to find the closest match. The closest value is ( 0.2 ), which corresponds to ( frac{1}{5} ).

💡Alright, so I have this problem about a right triangle ABC with legs AC = a and CB = b. I need to find two things: a) The side length of the largest square that can fit inside the triangle with one vertex at C.b) The dimensions of the largest rectangle that can fit inside the triangle with one vertex at C.Okay, let me start with part a). I remember that in a right triangle, you can fit a square such that one of its sides lies along the legs of the triangle. The square will touch the hypotenuse as well. I think the key here is to use similar triangles because the square will create smaller similar triangles within the original triangle.So, let me visualize this. Triangle ABC is right-angled at C. If I place a square inside it with one vertex at C, the square will have sides along AC and CB. Let's denote the side length of the square as x. The square will touch the hypotenuse AB at some point, say D.Now, the square divides the original triangle into two smaller triangles and the square itself. These smaller triangles should be similar to the original triangle ABC because all the angles will be the same. Let me denote the triangle above the square as ADE and the triangle to the side of the square as BDF. Both of these should be similar to ABC.Since triangle ADE is similar to ABC, the ratio of their corresponding sides should be equal. The side AC of triangle ABC is 'a', and the corresponding side of triangle ADE would be (a - x). Similarly, the side CB of triangle ABC is 'b', and the corresponding side of triangle BDF would be (b - x).Wait, is that correct? Let me think again. If the square has side length x, then along AC, the remaining length after the square would be (a - x), and along CB, it would be (b - x). But actually, the triangles formed might not both have sides (a - x) and (b - x). Maybe I need to consider the proportions differently.Let me consider the coordinates. Let's place point C at the origin (0,0), point A at (0,a), and point B at (b,0). The hypotenuse AB would then be the line connecting (0,a) to (b,0). The equation of this hypotenuse can be found using the two-point form.The slope of AB is (0 - a)/(b - 0) = -a/b. So, the equation of AB is y = (-a/b)x + a.Now, if I place a square with side length x inside the triangle, one corner at C (0,0), and extending along the x and y axes, the square will have corners at (0,0), (x,0), (0,x), and (x,x). The point (x,x) must lie on the hypotenuse AB.So, plugging (x,x) into the equation of AB: x = (-a/b)x + a.Let me solve for x:x = (-a/b)x + aBring the term with x to the left:x + (a/b)x = aFactor out x:x(1 + a/b) = aCombine the terms:x((b + a)/b) = aMultiply both sides by b:x(a + b) = a*bSo,x = (a*b)/(a + b)Okay, that seems to make sense. So, the side length of the square is (a*b)/(a + b). That's the answer for part a).Now, moving on to part b). I need to find the dimensions of the largest rectangle that can fit inside the triangle with one vertex at C.Similar to the square, the rectangle will have sides along AC and CB. Let me denote the length along AC as y and the length along CB as z. So, the rectangle has dimensions y and z.Again, the rectangle will touch the hypotenuse AB at some point. Let's denote this point as (z, y). Since this point lies on AB, it must satisfy the equation of AB, which is y = (-a/b)z + a.So, we have the equation:y = (-a/b)z + aOur goal is to maximize the area of the rectangle, which is A = y*z.But since y is expressed in terms of z, we can substitute:A = z*(-a/b*z + a) = (-a/b)z^2 + a*zThis is a quadratic equation in terms of z, and it opens downward, so the maximum occurs at the vertex.The vertex of a quadratic equation ax^2 + bx + c is at x = -b/(2a). In this case, the quadratic is (-a/b)z^2 + a*z, so the coefficient of z^2 is (-a/b) and the coefficient of z is a.So, the value of z at maximum area is:z = -a/(2*(-a/b)) = -a / (-2a/b) = (a)/(2a/b) = b/2So, z = b/2.Then, plugging back into the equation for y:y = (-a/b)*(b/2) + a = (-a/2) + a = a/2So, y = a/2 and z = b/2.Therefore, the dimensions of the rectangle are a/2 and b/2.Wait, that seems too straightforward. Let me verify.If I take a rectangle with sides a/2 and b/2, its area would be (a/2)*(b/2) = ab/4.Is this indeed the maximum area? Let me see.Suppose I take a different rectangle, say with z = b/3, then y would be (-a/b)*(b/3) + a = (-a/3) + a = 2a/3. The area would be (2a/3)*(b/3) = 2ab/9 ≈ 0.222ab, which is less than ab/4 ≈ 0.25ab.Similarly, if I take z = b, then y would be (-a/b)*b + a = -a + a = 0, which gives area 0.If I take z = 0, then y = a, area 0.So, indeed, the maximum area occurs at z = b/2, y = a/2, giving area ab/4.Therefore, the dimensions of the largest rectangle are a/2 and b/2.So, summarizing:a) The side length of the square is (a*b)/(a + b).b) The dimensions of the rectangle are a/2 and b/2.I think that's it.