🧠:Okay, so I need to find all functions ( f: mathbb{N}^{*} rightarrow mathbb{N}^{*} ) such that for every positive integer ( m ) and ( n ), the expression ( m^2 + f(n) ) divides ( m f(m) + n ). Hmm, let's start by understanding the problem. First, let's parse the divisibility condition. The notation ( a mid b ) means that ( a ) divides ( b ), so ( m^2 + f(n) ) must be a divisor of ( m f(m) + n ). That implies there exists some integer ( k ) such that ( m f(m) + n = k(m^2 + f(n)) ). Since all variables are positive integers, ( k ) must also be a positive integer. My goal is to find all such functions ( f ). Maybe I can start by plugging in some specific values for ( m ) and ( n ) to see if I can find a pattern or deduce the form of ( f ).Let me try setting ( m = 1 ). Then the condition becomes:[1^2 + f(n) mid 1 cdot f(1) + n]So, ( 1 + f(n) ) divides ( f(1) + n ). Let's denote ( f(1) = c ), where ( c ) is a positive integer. Then, for each ( n ), ( 1 + f(n) ) divides ( c + n ). This means that ( 1 + f(n) ) is a divisor of ( c + n ). Since ( 1 + f(n) ) must be a positive integer greater than 1 (because ( f(n) ) is at least 1), the divisor must be at least 2. So for each ( n ), there exists some integer ( k_n ) such that ( c + n = k_n (1 + f(n)) ). Let's rearrange this:[c + n = k_n (1 + f(n)) implies 1 + f(n) = frac{c + n}{k_n}]Since ( 1 + f(n) ) must be a positive integer, ( k_n ) must be a positive divisor of ( c + n ). Also, ( k_n leq c + n ) because ( 1 + f(n) geq 2 ), so ( frac{c + n}{k_n} geq 2 implies k_n leq frac{c + n}{2} ).But since ( f(n) = frac{c + n}{k_n} - 1 ), and ( f(n) ) must be a positive integer, we have that ( frac{c + n}{k_n} ) must be an integer greater than 1, which implies ( k_n ) divides ( c + n ). This seems a bit abstract. Maybe I can find some constraints on ( c ). Let's consider specific values of ( n ).Take ( n = 1 ). Then:[1 + f(1) mid c + 1 implies 1 + c mid c + 1]Which is always true because ( 1 + c ) divides itself. So no new information here.Take ( n = 2 ):[1 + f(2) mid c + 2]So ( 1 + f(2) ) is a divisor of ( c + 2 ). Similarly, for ( n = 3 ):[1 + f(3) mid c + 3]And so on. This suggests that ( 1 + f(n) ) must divide ( c + n ) for each ( n ). Therefore, ( 1 + f(n) leq c + n ), which implies ( f(n) leq c + n - 1 ). But since ( c ) is a constant (f(1)), this gives a linear upper bound on ( f(n) ). However, we need more constraints. Let's see if we can find another condition by fixing ( n ) and varying ( m ), or vice versa.Alternatively, maybe set ( n = 1 ). Then the original condition becomes for all ( m ):[m^2 + f(1) mid m f(m) + 1]Which is:[m^2 + c mid m f(m) + 1]Hmm, so ( m^2 + c ) divides ( m f(m) + 1 ). Let's see if we can use this to find more information about ( f(m) ).For this to hold for all ( m ), the left side ( m^2 + c ) must be less than or equal to the right side ( m f(m) + 1 ), unless the right side is a multiple of the left side. But since ( m^2 ) grows quadratically and ( m f(m) ) grows linearly with ( f(m) ), unless ( f(m) ) is linear or faster, the left side could be larger. Wait, but both sides are positive integers, so the divisibility requires that ( m f(m) + 1 geq m^2 + c ). Wait, no, that's not necessarily true. Divisibility doesn't require the divisor to be smaller, but in practice, for the quotient to be a positive integer, if ( m^2 + c > m f(m) + 1 ), then the quotient would be 0 or negative, which is impossible. Therefore, we must have ( m^2 + c leq m f(m) + 1 ). So, rearranged:[m f(m) geq m^2 + c - 1 implies f(m) geq m + frac{c - 1}{m}]Since ( f(m) ) is a positive integer, and ( frac{c - 1}{m} ) must be such that the right side is an integer. But ( c ) is a constant. Let's take ( m = 1 ):For ( m = 1 ), the inequality becomes:[f(1) geq 1 + (c - 1)/1 implies c geq 1 + c - 1 implies c geq c]Which is okay, equality holds. For ( m = 2 ):[f(2) geq 2 + frac{c - 1}{2}]Since ( f(2) ) must be an integer, ( frac{c - 1}{2} ) must be an integer or a half-integer. Wait, but ( c ) is a positive integer (since ( f(1) = c )), so ( c - 1 ) is an integer. Therefore, ( frac{c - 1}{2} ) is either integer or half-integer. But ( f(2) ) has to be an integer, so ( 2 + frac{c - 1}{2} ) must be an integer. Let's denote ( c - 1 = 2k ) or ( c - 1 = 2k + 1 ). If ( c - 1 ) is even, then ( frac{c -1}{2} ) is integer. If odd, then it's a half-integer. However, ( f(2) ) must be an integer, so if ( c -1 ) is odd, then ( 2 + frac{c -1}{2} = 2 + k + 0.5 = k + 2.5 ), which is not an integer. Therefore, ( c -1 ) must be even, so ( c ) is odd. Therefore, ( c ) must be odd. Let's write ( c = 2k + 1 ), where ( k ) is a non-negative integer. Then ( c -1 = 2k ), so for ( m = 2 ):[f(2) geq 2 + frac{2k}{2} = 2 + k]So ( f(2) geq 2 + k ). But ( k = frac{c -1}{2} ), so ( f(2) geq 2 + frac{c -1}{2} ). But we need to relate this back to the original problem. Let's think about possible values for ( c ). If ( c = 1 ), which is allowed since ( c ) is a positive integer, then ( c -1 = 0 ), so ( k = 0 ). Then, for ( m = 2 ), ( f(2) geq 2 + 0 = 2 ). Similarly, for ( c = 3 ), ( k = 1 ), so ( f(2) geq 2 + 1 = 3 ), and so on.But how does this help us? Maybe we can test small values of ( c ). Let's try ( c = 1 ). Then ( f(1) = 1 ). For ( m = 1 ), the divisibility condition is ( 1 + f(n) mid 1 + n ). So ( 1 + f(n) ) divides ( n + 1 ). Therefore, ( 1 + f(n) ) is a divisor of ( n + 1 ), which implies ( 1 + f(n) leq n + 1 implies f(n) leq n ). Also, since ( f(n) geq 1 ), we have ( 1 leq f(n) leq n ).But also, from the earlier condition when ( c = 1 ), for each ( m ), ( m^2 + 1 ) divides ( m f(m) + 1 ). Let's check for ( m = 2 ):Left side: ( 4 + 1 = 5 )Right side: ( 2 f(2) + 1 )So 5 divides ( 2 f(2) + 1 ). Therefore, ( 2 f(2) + 1 equiv 0 mod 5 implies 2 f(2) equiv -1 mod 5 implies 2 f(2) equiv 4 mod 5 implies f(2) equiv 2 mod 5 ). Since ( f(2) leq 2 ) (from ( f(n) leq n )), then ( f(2) = 2 ).Similarly, check ( m = 3 ):Left side: ( 9 + 1 = 10 )Right side: ( 3 f(3) + 1 )So 10 divides ( 3 f(3) + 1 ). Therefore, ( 3 f(3) + 1 equiv 0 mod 10 implies 3 f(3) equiv -1 mod 10 implies 3 f(3) equiv 9 mod 10 implies f(3) equiv 3 mod 10 ). But ( f(3) leq 3 ), so ( f(3) = 3 ).Continuing, ( m = 4 ):Left side: ( 16 + 1 = 17 )Right side: ( 4 f(4) + 1 )Thus, 17 divides ( 4 f(4) + 1 ). Therefore, ( 4 f(4) + 1 = 17 k ). Since ( f(4) leq 4 ), let's compute possible values:If ( f(4) = 1 ): ( 4 + 1 = 5 ), not divisible by 17.f(4) = 2: 8 + 1 = 9, nope.f(4) = 3: 12 + 1 =13, nope.f(4)=4:16 +1=17, which is divisible by 17. So f(4)=4.So far, for c=1, f(n)=n seems to work. Let's check m=5:Left side:25 +1=26. Right side:5 f(5) +1. If f(5)=5, then 25 +1=26, which divides 26. So yes, 26 divides 26. So f(5)=5 works.So maybe f(n)=n is a solution when c=1. Let's verify for general m. If f(n)=n, then for all m,n:m² + n divides m f(m) + n = m² + n. So m² + n divides itself, which is always true. Therefore, f(n)=n is a solution.Wait, so that's trivial. So identity function is a solution. Is that the only solution?Wait, let's check if c=1 is the only possibility. Suppose c=3. Then f(1)=3. Then for each n, 1 + f(n) divides 3 + n. So similar to before, 1 + f(n) divides n +3. So f(n) <= n +3 -1= n +2. But maybe we can check m=1, n=1: 1 +3 divides 3 +1=4. So 4 divides 4, which is okay. For n=2: 1 + f(2) divides 5. So possible divisors of 5 are 1,5. But 1 + f(2) must be at least 2, so 5. Thus, 1 + f(2)=5 => f(2)=4.For m=2, the original condition becomes:2² + f(n) divides 2 f(2) + n. Since f(2)=4, this becomes 4 + f(n) divides 8 + n. So for all n, 4 + f(n) divides n +8.Similarly, when n=2: 4 +4=8 divides 8 +2=10. But 8 does not divide 10. Contradiction. Therefore, c=3 is invalid. So c=1 is the only possibility?Wait, that was a mistake. If c=3, then for n=2, 1 + f(2) divides 3 +2=5. So 1 + f(2) must be 5, hence f(2)=4. Then, for m=2 and n=2, the divisibility condition is 4 +4=8 divides 2*4 +2=10, which is false. So c=3 is invalid.Similarly, if we try c=5, but that would lead to similar contradictions. Therefore, c=1 is the only possible value. Therefore, f(n)=n is the only solution?Wait, but perhaps there are other functions when c=1. Because even if c=1, the condition is that 1 + f(n) divides n +1, so f(n) can be any function such that 1 + f(n) divides n +1. But earlier, when we considered m=1, n=2, etc., we saw that f(n)=n is the only possibility. Let me check.If c=1, then for each n, 1 + f(n) divides n +1. So 1 + f(n) must be a divisor of n +1. So possible values of 1 + f(n) are the divisors of n +1. Since 1 + f(n) ≥ 2, the divisors are from 2 up to n +1. But f(n) must be a positive integer, so f(n) = d -1, where d is a divisor of n +1 and d ≥ 2. So f(n) can be any such value. However, in the original problem, the function f must satisfy the condition for all m and n. So even if for each n, f(n) is chosen such that 1 + f(n) divides n +1, we still have to ensure that for all m and n, m² + f(n) divides m f(m) + n.But when we tested with f(n)=n, it worked. If we choose a different function for c=1, would it still work?Suppose, for example, take n=2. Then n +1=3, so possible divisors d=3. Therefore, 1 + f(2)=3 => f(2)=2. Which is the same as identity function. For n=3, n+1=4, divisors are 2,4. So f(3) can be 1 or 3. But if we choose f(3)=1, let's see if that works.Wait, let's try f(n) defined as follows: f(n) = n for all n except f(3)=1. Then check the original condition for m=3 and n=3:Left side: 3² + f(3) =9 +1=10.Right side:3 f(3) +3=3*1 +3=6.But 10 divides 6? No, 10 does not divide 6. Therefore, this function doesn't work. Therefore, even if for some n we choose f(n) different from n, it may violate the condition for some m.Therefore, perhaps f(n) must equal n for all n. Let's check another possibility. Suppose for n=4, if we set f(4)=1 instead of 4. Then for m=4, check:Left side:16 +1=17.Right side:4 f(4) +4=4*1 +4=8.17 divides 8? No. So that's invalid. Hence, f(n) must be such that m² + f(n) divides m f(m) +n for all m,n. If we set f(m)=m for all m, then m² +n divides m² +n, which is true. So identity function works.But are there other functions? Let's check.Suppose there exists another function. Let's assume f(n) = k*n for some constant k. Let's see if that's possible.Assume f(n) = k n. Then, the condition becomes:m² + k n divides m*(k m) +n =k m² +n.So m² + k n divides k m² +n. Let's see when this can be true.Suppose k=1: then m² +n divides m² +n, which is true.If k=2: Then m² +2n divides 2m² +n. Let's check for m=1:Left side:1 +2n. Right side:2 +n. So 1 +2n divides 2 +n. Let's take n=1: 3 divides 3: okay. n=2:5 divides 4: No. So k=2 doesn't work.k=0 is invalid since f(n) must be positive integers.Alternatively, suppose f(n) =n +c for some constant c. Let's check.Then m² +n +c divides m*(m +c) +n =m² +c m +n.So m² +n +c divides m² +c m +n. The difference between them is c m -c =c(m -1). For divisibility, m² +n +c must divide c(m -1). But since m² +n +c is at least m² +1 +c (since n ≥1), and c(m -1) is linear in m, for large m, the left side is quadratic and the right side is linear, so it's impossible unless c=0, which is invalid. Therefore, f(n)=n +c is not a solution unless c=0, which is the identity function.Therefore, linear functions other than identity don't work. What about constant functions? Suppose f(n) =c for all n.Then the condition becomes m² +c divides m c +n for all m,n.But let's fix m and vary n. For example, take m=1:1 +c divides c +n for all n. But as n can be any positive integer, the left side must divide c +n for all n. However, 1 +c must divide (c +n) for all n. Let's set n =k(1 +c) -c for some k, but since n can be arbitrary, unless 1 +c divides all integers, which is only possible if 1 +c=1, so c=0, which is invalid. Therefore, constant functions are impossible.Hence, the only linear solution is the identity function. What about affine functions f(n) =a n +b? Let's try with a=1 and b=0, which is identity. a≠1?Suppose f(n)=2n +1. Let's test for m=1:1 +2n +1=2n +2 divides 1*(2*1 +1) +n=3 +n. So 2n +2 divides n +3. For n=1:4 divides4: okay. n=2:6 divides5: no. So invalid.Alternatively, maybe f(n) = floor(n/2) +1? Probably not, since the divisibility condition is quite strict.Alternatively, suppose f(n) =n for even n, and something else for odd n. Let's test. For example, let f(n)=n when n even, and f(n)=1 when n odd.Check m=1, n=1:1 +1=2 divides1*1 +1=2: okay.n=2:1 +2=3 divides1*2 +2=4: 3 divides4? No. So invalid.Therefore, deviating from identity function even slightly causes problems. Alternatively, maybe f(n) =n for all n except some specific n, but as we saw earlier, changing even one value can break the condition for some m and n.Alternatively, perhaps there's another function. Let's consider f(n) =n +k where k divides something. Wait, maybe not.Alternatively, suppose f is a linear function, but with different coefficients. Wait, we saw that f(n)=kn only works for k=1.Alternatively, maybe f(n) =n^s for some exponent s. Let's try s=2. Then:m² +n² divides m*m² +n =m³ +n.But for m=2 and n=1:4 +1=5 divides8 +1=9: 5 doesn't divide9. So no.Alternatively, s=0.5? Not allowed since f(n) must be integer.Alternatively, maybe f(n) is a constant multiple of n. Wait, we tried that.Alternatively, let's go back to the original condition. For all m,n:m² +f(n) divides m f(m) +n.If I can write this as:m f(m) +n ≡0 mod(m² +f(n)).Perhaps for fixed m, varying n, or vice versa.Suppose we fix n and vary m. Let's pick n=1. Then the condition is:m² +f(1) divides m f(m) +1.We saw that for f(1)=1, f(m)=m works. Suppose there's another solution. Let's suppose that for some m, f(m)≠m. Let's see if that's possible.Suppose for m=2, f(2)=2 +k for some k≥1. Then, for n=1, the condition is:4 +1=5 divides2 f(2) +1=2(2 +k)+1=5 +2k.So 5 divides5 +2k ⇒ 5 divides2k ⇒5 dividesk. Let k=5t. Then f(2)=2 +5t. But since f(2) must also satisfy the condition from m=1, n=2:1 +f(2) divides1 +2=3. So 1 +f(2)=3 ⇒f(2)=2. But if t=0, then k=0 ⇒f(2)=2. If t≥1, then f(2)=2 +5t, which would imply 1 +f(2)=3 +5t divides3, but 3 +5t divides3 only if 3 +5t ≤3 ⇒t=0. Hence, f(2)=2. Therefore, even if we try to set f(2) higher, the condition from m=1, n=2 forces it back to 2. So f(2)=2.Similarly, suppose for m=3, let's assume f(3)=3 +k. For n=1:9 +1=10 divides3 f(3)+1=3(3 +k)+1=10 +3k. Thus,10 divides10 +3k ⇒10 divides3k ⇒10 dividesk. Let k=10t. Then f(3)=3 +10t. But from m=1, n=3:1 +f(3) divides1 +3=4. So 1 +f(3)=4 ⇒f(3)=3. Hence, t=0. So f(3)=3.Therefore, this pattern suggests that for each m, f(m)=m is forced by the combination of the conditions when n=1 and varying m. Because when we set n=1, the divisibility condition forces f(m) to be m, otherwise, the divisibility fails unless f(m)=m. Let's verify this.Assume for some m, f(m) ≠m. Then, from the divisibility condition with n=1:m² +c divides m f(m) +1. But if f(m) >m, then m f(m) +1 >m² +c (since c=1), so m f(m) +1 must be a multiple of m² +1. Similarly, if f(m) <m, then m f(m) +1 could be less than m² +1, but divisibility would require m f(m) +1 ≥m² +1, implying m f(m) ≥m² ⇒f(m) ≥m. Contradiction. Hence, f(m) cannot be less than m. If f(m) >m, then m f(m) +1 must be a multiple of m² +1.Suppose f(m)=m +k for some k ≥1. Then:m f(m) +1 =m(m +k) +1 =m² +k m +1.This must be divisible by m² +1. Therefore:m² +1 divides m² +k m +1 ⇒ m² +1 divides k m.Since m² +1 and m are coprime (because any common divisor d must divide m² +1 and m, so d divides1), therefore m² +1 divides k. But m² +1 >k unless k is very large. However, since k is a positive integer, this is only possible if k is a multiple of m² +1. Therefore, f(m)=m +t(m² +1), where t is a positive integer. But then, for m=1, f(1)=1 +t(1 +1)=1 +2t. But from earlier analysis, when m=1, n=2:1 +f(2) divides1 +2=3. If f(2) is of the form2 +t(4 +1)=2 +5t, but we already saw that this forces t=0. Similarly, for other m, t must be zero. Hence, the only possibility is f(m)=m for all m.Therefore, through these steps, it seems that the only function satisfying the condition is the identity function f(n)=n.To confirm, let's check another case. Take m=2, n=3:Left side:4 +3=7Right side:2*2 +3=77 divides7: yes.Another example, m=3, n=4:Left side:9 +4=13Right side:3*3 +4=1313 divides13: yes.Therefore, all tests confirm that f(n)=n is a solution. Since our analysis didn't reveal any other possible functions and all deviations led to contradictions, it's likely that the identity function is the only solution.Final AnswerThe only such function is boxed{f(n) = n}.

✅:To find all functions ( f: mathbb{N}^{*} rightarrow mathbb{N}^{*} ) such that for all ( m, n in mathbb{N}^{*} ), we have ( m^{2} + f(n) mid m f(m) + n ), we start by analyzing specific values of ( m ) and ( n ).1. Setting ( m = 1 ): [ 1 + f(n) mid f(1) + n ] Let ( f(1) = c ). Then ( 1 + f(n) ) must divide ( c + n ). This implies ( f(n) leq c + n - 1 ).2. Setting ( n = 1 ): [ m^2 + c mid m f(m) + 1 ] For this to hold for all ( m ), we find that ( f(m) geq m ). Further analysis shows that ( f(m) = m ) is necessary to satisfy the divisibility condition.3. Verification: - For ( f(n) = n ), the condition ( m^2 + n mid m^2 + n ) is trivially true. - Testing other forms of ( f(n) ) (constant functions, linear functions with different coefficients, etc.) leads to contradictions with the divisibility condition.4. Conclusion: The identity function ( f(n) = n ) is the only solution that satisfies the given condition for all ( m, n in mathbb{N}^{*} ).[boxed{f(n) = n}]

🔑:# Step 1:Let's start by substituting ( m = n ) in the given functional equation:[ m^2 + f(n) mid m f(m) + n ]Which becomes:[ n^2 + f(n) mid n f(n) + n ]# Step 2:Next, we substitute ( n ) by a prime number ( p ). If ( p ) does not divide ( f(p) ), then ( p^{2} + f(p) ) is coprime with ( p ). Hence:[ p^{2} + f(p) mid f(p) + 1 ]Since ( f(p) + 1 < f(p) + p^2 ), it must necessarily follow that ( p mid f(p) ) for all prime numbers ( p ). This means ( f(p) = kp ) for some integer ( k ).# Step 3:Now, set ( m = 1 ) in the given equation:[ 1 + f(n) mid f(1) + n ]From the above, we get:[ 1 + f(n) leq f(1) + n ]which implies:[ f(n) leq f(1) + n - 1 ]# Step 4:Notice that if we choose ( p ) as large prime, from the previous deduction ( p mid f(p) ) and it follows that ( f(p) = p ). Consequently, as ( p ) grows larger, the inequality ( f(n) leq n ) and the nature of ( n ) being prime forces ( f(n) = n ) for prime ( n ).# Step 5:Now, consider any ( m in mathbb{N}^* ). For any prime ( p ) large enough, we have:[ m^2 + p mid mf(m) + p ]From this, it implies:[ m^2 + p leq mf(m) + p ]giving us:[ m^2 leq mf(m) ]This results in:[ f(m) geq m ]# Step 6:Let us focus now on:[ p^2 + f(n) mid p^2 + n ]for large primes ( p ):[ f(n) leq n ]Combining the two previous results, namely ( f(m) geq m ) and ( f(n) leq n ), we conclude:[ f(n) = n ]for all ( n in mathbb{N}^* ).Thus, the only function that satisfies the given functional equation is:[ boxed{f(n) = n text{ for all } n in mathbb{N}^{*}} ]

🧠:Okay, so I need to prove that the length of segment CD is determined by a and r, and then find the locus of points C. Hmm, let me start by visualizing the problem. There's a segment AB with length a. At points A and B, they're drawing lines that are perpendicular to each other and also perpendicular to AB. Then, points C and D are on these lines, respectively. The key condition is that when you draw segment CD and find where it intersects the plane through the midpoint O of AB (which is also perpendicular to AB), that intersection point is at a distance r from O. The goal is to show CD's length depends only on a and r, and then find the path that point C can take.First, let me set up a coordinate system to model this. Let's place point O, the midpoint of AB, at the origin (0, 0, 0). Then, since AB is a segment of length a, points A and B can be at (-a/2, 0, 0) and (a/2, 0, 0), respectively. The segment AB lies along the x-axis.Now, from points A and B, we need to draw lines that are perpendicular to AB and to each other. Since AB is along the x-axis, a line perpendicular to AB at A would be in the y-z plane. Similarly, at B, a line perpendicular to AB would also be in the y-z plane. But they also need to be perpendicular to each other. Wait, how exactly are these lines oriented?If the lines at A and B are both perpendicular to AB, then they each lie in their own planes perpendicular to AB. But the lines themselves must also be perpendicular to each other. So, maybe the line through A is along the y-axis direction, and the line through B is along the z-axis direction? Or perhaps one is along y and the other along z, but rotated such that they are perpendicular in 3D space. Hmm.Wait, actually, in 3D space, two lines can be perpendicular even if they are not in the same plane. So, if we have a line through A that's in the y-direction and a line through B that's in the z-direction, then those lines are perpendicular to each other because their direction vectors (0,1,0) and (0,0,1) are perpendicular. But also, each of these lines is perpendicular to AB, which is along the x-axis. That works. So, point C can be on the line through A in the y-direction, and point D can be on the line through B in the z-direction. Alternatively, maybe the lines are in other directions, but as long as they are both perpendicular to AB and to each other.Wait, the problem says "lines perpendicular to each other and to the segment AB are drawn." So, each line is perpendicular to AB and also perpendicular to each other. So, the lines at A and B must each be perpendicular to AB (so their direction vectors are perpendicular to AB's direction vector, which is along x-axis). Also, the lines at A and B must be perpendicular to each other. So, their direction vectors must be perpendicular.Therefore, let's define the lines as follows: line through A has direction vector (0, 1, 0), so it's along the y-axis, and line through B has direction vector (0, 0, 1), so it's along the z-axis. These are both perpendicular to AB (which is along x-axis) and also perpendicular to each other since their direction vectors are perpendicular. So, points C and D can be parametrized as:Point C: A + t*(0,1,0) = (-a/2, t, 0)Point D: B + s*(0,0,1) = (a/2, 0, s)But wait, is that correct? Wait, the lines are drawn from A and B, perpendicular to AB and to each other. So, if the line through A is along the y-axis, then moving along that line would be changing the y-coordinate. Similarly, the line through B along the z-axis changes the z-coordinate. So, points C and D can be parametrized as C(-a/2, y, 0) and D(a/2, 0, z), where y and z are parameters. Then, the segment CD connects these two points.Now, the plane passing through the midpoint O (which is the origin) and perpendicular to AB. Since AB is along the x-axis, the plane perpendicular to AB through O is the y-z plane (x=0). So, the intersection point of CD with the plane x=0. Let's find where CD intersects the plane x=0.Parametrize segment CD. Let's write parametric equations for CD. Starting at point C(-a/2, y, 0) going to D(a/2, 0, z). Let parameter λ go from 0 to 1.So, x-coordinate: -a/2 + λ(a) (since from -a/2 to a/2 is a distance of a)y-coordinate: y - λ y (since from y to 0)z-coordinate: 0 + λ z (since from 0 to z)Wait, let me check. The parametric equations can be written as:x(λ) = -a/2 + λ*(a/2 - (-a/2)) = -a/2 + λ*(a)y(λ) = y + λ*(0 - y) = y*(1 - λ)z(λ) = 0 + λ*(z - 0) = λ zSo, yes, that's correct.We need to find the value of λ where x(λ) = 0 (intersection with the plane x=0).Set x(λ) = -a/2 + a λ = 0.Solving for λ:a λ = a/2λ = 1/2.So, at λ = 1/2, the segment CD intersects the plane x=0. The coordinates at this point are:x=0,y = y*(1 - 1/2) = y/2,z = (1/2) z.So, the intersection point is (0, y/2, z/2). The distance from O (which is at (0,0,0)) to this point is sqrt( (y/2)^2 + (z/2)^2 ) = (1/2) sqrt(y² + z²). According to the problem, this distance is equal to r. Therefore:(1/2) sqrt(y² + z²) = rMultiply both sides by 2:sqrt(y² + z²) = 2rSquare both sides:y² + z² = (2r)^2 = 4r²So, this is the relation between y and z for points C and D such that the intersection point is at distance r from O.Now, we need to find the length of CD. Let's compute CD's length in terms of y and z.Point C is (-a/2, y, 0), Point D is (a/2, 0, z).The distance CD is sqrt[ (a/2 - (-a/2))² + (0 - y)^2 + (z - 0)^2 ]Compute each component:x: a/2 - (-a/2) = ay: 0 - y = -yz: z - 0 = zTherefore, CD = sqrt(a² + y² + z²)But from earlier, we have y² + z² = 4r². Substitute that into CD's length:CD = sqrt(a² + 4r²)Therefore, CD is determined solely by a and r, as required. So that's the first part done.Now, for the locus of points C. Point C is moving on the line through A perpendicular to AB and to the other line. Wait, point C is on the line through A which is along the y-axis (as per our coordinate system). So, in our setup, point C has coordinates (-a/2, y, 0). But we also have the condition that when connected to D (which is (a/2, 0, z)), the intersection point is at distance r from O. Which gives us y² + z² = 4r².But how does this relate to the locus of C? Since point D is determined once point C is chosen, or vice versa. So, for each position of C along its line, there's a corresponding D on its line such that y² + z² = 4r². Therefore, for the locus of C, we need to see how y can vary while satisfying this condition with z.But in our parametrization, for each C(-a/2, y, 0), there exists a D(a/2, 0, z) such that y² + z² = 4r². But z is dependent on y. Therefore, for each y, z can be sqrt(4r² - y²), but z can be positive or negative. However, since D is on the line through B, which in our coordinate system was along the z-axis, so z can be any real number. Therefore, for each y such that |y| ≤ 2r, there is a corresponding z = ±sqrt(4r² - y²). But does this affect the locus of C?Wait, point C is only moving along the line through A (the y-axis in our coordinates). However, the existence of a D such that CD intersects the plane at distance r requires that y² + z² = 4r². But z is determined by y. However, z is the coordinate of D along its line (the z-axis through B). Therefore, for a given C with coordinate y, there exists a D with z such that z = ±sqrt(4r² - y²). But how does this affect the locus of C?Wait, the problem states that points C and D are chosen respectively on their lines such that the intersection point is at distance r from O. Therefore, for each possible position of C on its line, there might be a corresponding D on its line that satisfies this condition. However, in our setup, if we fix C at (-a/2, y, 0), then D must be at (a/2, 0, z) where z = ±sqrt(4r² - y²). Therefore, as long as |y| ≤ 2r, such a D exists. However, if |y| > 2r, then there is no real z that satisfies the equation, so such a D does not exist. Therefore, the possible positions of C are those where |y| ≤ 2r. Therefore, the locus of points C is the set of all points on the line through A (which is the y-axis at x = -a/2, z = 0) with y ranging from -2r to 2r. Therefore, it's a line segment from (-a/2, -2r, 0) to (-a/2, 2r, 0). But wait, is that correct?Wait, but in 3D space, the line through A is infinite, but due to the condition y² + z² = 4r² and z being dependent on y, but in our case, z is determined once y is chosen. However, point C is only moving along the y-axis at A, so z-coordinate of C is always 0. The z in the equation is the coordinate of D. So, for point C, its y-coordinate can range between -2r and 2r so that z is real. Therefore, the locus of C is the line segment on the line through A (along y-axis) from y = -2r to y = 2r. So, the locus is a line segment of length 4r centered at A, but shifted along the y-axis.But wait, in the problem statement, it's said that through endpoints A and B, lines perpendicular to each other and to AB are drawn. So, in our setup, the line through A is along y-axis, and line through B is along z-axis. Therefore, C is on the y-axis line, D is on the z-axis line. For each C on the y-axis line (with y between -2r and 2r), there is a D on the z-axis line such that CD intersects the plane at distance r. Therefore, the locus of C is the portion of the line through A (perpendicular to AB) where the y-coordinate satisfies |y| ≤ 2r. Therefore, the locus is a line segment of length 4r on the line through A, perpendicular to AB.But wait, is there a different interpretation? Maybe the problem allows for the lines through A and B to be any lines perpendicular to AB and to each other, not necessarily aligned with y and z axes. If the lines through A and B are arbitrary as long as they're perpendicular to each other and to AB, then perhaps the locus of C is a circle or something else.Wait, but in my coordinate system, I fixed the lines through A and B to be along the y and z axes. But if the lines can be any pair of perpendicular lines (each perpendicular to AB), then the locus might be different. However, the problem states: "lines perpendicular to each other and to the segment AB are drawn." So, each line is perpendicular to AB, and the two lines are also perpendicular to each other. In 3D space, two lines that are both perpendicular to AB (which is along x-axis) can be oriented in any direction in the y-z plane, as long as they are perpendicular to each other.Therefore, perhaps my initial assumption of aligning them with y and z axes was just a convenient choice, but the problem might be more general. However, since the problem states that the lines are drawn through A and B, respectively, and are perpendicular to each other and to AB, then the specific orientation of these lines might affect the locus.Wait, but in 3D space, if two lines are both perpendicular to AB (x-axis) and also perpendicular to each other, then they must form a right angle in the plane perpendicular to AB. So, regardless of their specific orientation, as long as they are orthogonal, we can choose coordinates such that one is along y and the other along z. Because any two perpendicular vectors in the y-z plane can be rotated into the y and z axes. Therefore, without loss of generality, we can assume the lines through A and B are along y and z directions.Therefore, the previous analysis holds: the locus of C is the line segment on the line through A (along y-axis) from y = -2r to y = 2r. Therefore, it's a line segment of length 4r centered at A.But wait, that seems too simplistic. Alternatively, maybe the locus is a circle or a cylinder. Wait, let me double-check.Wait, in our coordinate system, point C is on the line through A along y-axis, so its coordinates are (-a/2, y, 0). But according to the condition, when connected to D(a/2, 0, z), the intersection point is at distance r from O. We found that y² + z² = 4r². However, point C is only parameterized by y, and z is dependent on y. But in this case, z is the coordinate of D, not of C. So, for each y (of C), z (of D) is determined. Therefore, the position of C is restricted only by the existence of such a D. Since z must be real, y² ≤ 4r², so |y| ≤ 2r. Therefore, point C can move along the line through A (the y-axis) between y = -2r and y = 2r. Therefore, the locus is the line segment from (-a/2, -2r, 0) to (-a/2, 2r, 0). Similarly, if the line through A was in a different direction, say some other direction perpendicular to AB, then the locus would be a line segment in that direction. But in our case, since we fixed the line through A as the y-axis, the locus is a line segment along the y-axis.But the problem says "Find the locus of points C." So, if the lines through A and B can be any pair of perpendicular lines (each perpendicular to AB), then the locus would depend on the orientation. But the problem states "lines perpendicular to each other and to the segment AB are drawn." So, these lines are fixed once chosen. The problem doesn't specify that the lines can rotate; they are drawn through A and B, perpendicular to each other and to AB. Therefore, once drawn, the lines are fixed, and points C and D move along these lines. Therefore, the locus of C is a line segment on the line through A, with length 4r, centered at A.But wait, in our coordinate system, the line through A is along the y-axis, so moving from y = -2r to y = 2r gives a segment of length 4r. But in the problem statement, the length from A to C would be |y - 0| since A is at (-a/2, 0, 0). Wait, point C is at (-a/2, y, 0), so the distance from A to C is |y|. So, if y ranges from -2r to 2r, then the distance from A to C ranges from 0 to 2r? Wait, no. If A is at (-a/2, 0, 0), and C is at (-a/2, y, 0), then the distance AC is sqrt( (0)^2 + (y - 0)^2 + (0)^2 ) = |y|. Therefore, if y ranges from -2r to 2r, the distance from A to C ranges from 0 to 2r, but since y can be negative, the locus is the line segment from 2r above A to 2r below A along the y-axis. So, total length 4r.But the problem is in 3D space, but the locus of C is confined to a line segment on the line through A. However, wait, perhaps there's another way to interpret the problem where the lines through A and B can be any lines perpendicular to AB and to each other, not fixed as y and z axes. In that case, the locus might be different.Alternatively, maybe the lines through A and B are both perpendicular to AB but can rotate around AB as long as they remain perpendicular to each other. Then, the locus of C would be a circle or a cylinder. Hmm, this is a critical point.Wait, the problem states: "through the endpoints A and B of a segment of length a, lines perpendicular to each other and to the segment AB are drawn." The phrasing suggests that for each endpoint, a line is drawn, and these two lines are perpendicular to each other and to AB. So, once these lines are drawn, points C and D are chosen on them. Therefore, the lines are fixed; they don't rotate. Therefore, once the lines are drawn (perpendicular to each other and to AB), then C and D move along these fixed lines.Therefore, in this case, the locus of C is a subset of the fixed line through A. So, if the line through A is fixed, then as per previous analysis, the locus is the segment on that line where |y| ≤ 2r. But if the problem allows the lines to be arbitrary (i.e., any pair of perpendicular lines through A and B, each perpendicular to AB), then the locus could vary depending on the orientation.But since the problem doesn't specify any particular orientation, just that the lines are perpendicular to each other and to AB, we might need to consider all possible such lines. However, in 3D space, all such configurations are isometric; that is, we can rotate the coordinate system so that the lines through A and B are along y and z axes. Therefore, without loss of generality, the locus would be a line segment of length 4r on any such line through A. However, if the lines through A and B can be in any orientation (rotated around AB), then the locus of C would trace a circle or another shape.Wait, but if the lines through A and B are fixed once chosen, then the locus is a line segment. But if the problem allows the lines to rotate while maintaining their perpendicularity to AB and to each other, then the locus of C could be a circle. However, the problem says "lines perpendicular to each other and to the segment AB are drawn." The use of past tense "drawn" suggests that these lines are fixed once drawn. Therefore, points C and D are moving along fixed lines. Therefore, the locus of C is a line segment on the fixed line through A.But let me check the problem statement again: "Through the endpoints A and B of a segment of length a, lines perpendicular to each other and to the segment AB are drawn. On these lines, points C and D are chosen respectively such that the intersection point of the segment CD with the plane passing through the midpoint O of the segment AB and perpendicular to it is at a distance r from point O."The key part is "lines... are drawn." So, once drawn, points C and D are chosen on these lines. Therefore, the lines are fixed. Therefore, the locus of C is a subset of the fixed line through A. Therefore, the answer is a line segment of length 4r on the line through A, which is perpendicular to AB and to the line through B.But in my previous analysis, with the coordinate system, the locus is the line segment from (-a/2, -2r, 0) to (-a/2, 2r, 0). So, in terms of the problem's geometry, this is a line segment on the line through A, perpendicular to AB and to the line through B, extending 2r on either side of A.Wait, but hold on. If the line through A is perpendicular to AB and to the line through B, but the line through B is also perpendicular to AB and to the line through A. In our coordinate system, the line through A is along the y-axis, and the line through B is along the z-axis. So, they are both perpendicular to AB (x-axis) and to each other. Therefore, the locus of C is along the y-axis line through A, between y = -2r and y = 2r.But in the problem statement, is there any restriction on the position of C and D? For example, can C be on either side of A? In our coordinate system, yes, because y can be positive or negative. Therefore, the locus is a straight line segment symmetric about A, extending 2r in both directions along the line through A (which is perpendicular to AB and to the line through B).Alternatively, if the line through A was not along the y-axis but some other direction perpendicular to AB and to the line through B, then the locus would be a line segment in that direction. However, due to the problem's symmetry, regardless of the orientation of the lines (as long as they are perpendicular to each other and to AB), the locus should be a line segment of length 4r along the line through A.Wait, but 4r? If from -2r to 2r, that's a length of 4r. Yes.Therefore, the locus of points C is a line segment of length 4r, centered at A, lying along the line through A that is perpendicular to AB and to the line through B.But the problem says "Find the locus of points C." So, in mathematical terms, if we consider the line through A as a one-dimensional subspace, the locus is the set of all points on this line at a distance up to 2r from A. Therefore, the locus is a line segment of length 4r, symmetric around A.But wait, in our coordinate system, A is at (-a/2, 0, 0), and the line through A is the y-axis. Therefore, the points C are from (-a/2, -2r, 0) to (-a/2, 2r, 0). Therefore, this is a line segment along the y-axis direction, centered at A, with length 4r. So, in terms of locus, it's a straight line segment.But the problem might be expecting a different answer, perhaps a circle or something else. Let me think again.Wait, no. Since C is constrained to move along a fixed line (the one through A, perpendicular to AB and to the other line), its locus is inherently a subset of that line. Therefore, unless there's a different interpretation where the lines through A and B can rotate, the locus is a line segment. However, the problem states the lines are drawn through A and B, meaning they are fixed once established. Therefore, points C and D are moving along these fixed lines, so the locus of C is a line segment on the fixed line through A.Therefore, the final answer should be that the length of CD is sqrt(a² + 4r²), and the locus of C is a line segment of length 4r along the line through A perpendicular to AB and to the line through B.But let me check once more. Suppose instead of aligning the lines with y and z axes, we choose a different coordinate system where the line through A is along some other direction perpendicular to AB. For example, suppose the line through A is along a vector (0, cosθ, sinθ) and the line through B is along (0, -sinθ, cosθ), ensuring they are perpendicular. Then, points C and D would have coordinates depending on θ, but the relation y² + z² = 4r² would still hold in some rotated coordinate system. However, since the lines are fixed, θ is fixed, so the locus of C is still a line segment along the line through A, with length 4r. So regardless of the orientation of the lines (as long as they're perpendicular), the locus is a line segment.Therefore, the answer is:The length of CD is √(a² + 4r²), and the locus of points C is a line segment of length 4r on the line through A perpendicular to AB and the line through B, centered at A.But the problem says "Find the locus of points C." In the problem's statement, the lines are drawn through A and B, perpendicular to each other and to AB. Then, points C and D are on these lines. So, since the lines are fixed, the locus of C is just the set of all possible positions of C on its line such that there exists a D on the other line fulfilling the intersection condition. As we saw, this restricts C to a segment of length 4r on its line.But in the problem statement, it's possible that the lines through A and B are not aligned with coordinate axes, but since they are perpendicular to each other and to AB, their direction vectors form an orthogonal basis in the plane perpendicular to AB. Therefore, regardless of their specific orientation, the locus of C would be a line segment on its respective line.Thus, after thorough analysis, I conclude:1. The length of segment CD is √(a² + 4r²).2. The locus of points C is a line segment of length 4r, located on the line through A that is perpendicular to both AB and the line through B, extending 2r on either side of A.But to express this more formally, in mathematical terms, considering the line through A is one of the two mutually perpendicular lines, the locus is the set of all points on that line at a distance up to 2r from A. Therefore, it's a line segment with length 4r.Alternatively, if the problem expects a geometric shape like a circle, maybe there's a misunderstanding. Wait, let's think differently.Suppose that instead of the lines through A and B being fixed, we consider all possible such lines (i.e., all possible pairs of perpendicular lines through A and B, each perpendicular to AB). Then, for each such pair of lines, points C and D are chosen such that CD intersects the plane at distance r from O. In this case, the locus of C would be the union of all possible line segments from the previous case, which might form a circle or a cylinder.But the problem states: "Through the endpoints A and B... lines perpendicular to each other and to the segment AB are drawn." The use of "are drawn" suggests a single pair of lines is drawn, not all possible pairs. Therefore, the lines are fixed once drawn, hence the locus of C is a line segment on that fixed line.Therefore, the correct answer is that CD has length √(a² + 4r²), and the locus of C is a line segment of length 4r along the line through A perpendicular to AB and the other line.Final AnswerThe length of segment ( CD ) is (boxed{sqrt{a^2 + 4r^2}}) and the locus of points ( C ) is a line segment of length (4r) on the line through ( A ) perpendicular to both ( AB ) and the line through ( B ).

✅:To solve the problem, we start by setting up a coordinate system with the midpoint ( O ) of segment ( AB ) at the origin ((0, 0, 0)). Points ( A ) and ( B ) are at ((-a/2, 0, 0)) and ((a/2, 0, 0)) respectively. Lines perpendicular to ( AB ) and each other are drawn through ( A ) and ( B ). We assume the line through ( A ) is along the y-axis and the line through ( B ) is along the z-axis. Points ( C ) and ( D ) are parametrized as ( C(-a/2, y, 0) ) and ( D(a/2, 0, z) ).The segment ( CD ) intersects the plane ( x = 0 ) (the y-z plane) at a point whose distance from ( O ) is ( r ). Parametrizing ( CD ), we find the intersection occurs at ( lambda = 1/2 ), giving coordinates ((0, y/2, z/2)). The distance from this point to ( O ) is (sqrt{(y/2)^2 + (z/2)^2} = r), leading to ( y^2 + z^2 = 4r^2 ).The length of ( CD ) is calculated as:[CD = sqrt{(a/2 - (-a/2))^2 + (0 - y)^2 + (z - 0)^2} = sqrt{a^2 + y^2 + z^2}]Substituting ( y^2 + z^2 = 4r^2 ), we get:[CD = sqrt{a^2 + 4r^2}]For the locus of ( C ), since ( y ) must satisfy ( y^2 leq 4r^2 ), the coordinates of ( C ) are ((-a/2, y, 0)) with ( |y| leq 2r ). This describes a line segment on the line through ( A ) perpendicular to ( AB ), extending from ((-a/2, -2r, 0)) to ((-a/2, 2r, 0)), thus having length (4r). Final AnswerThe length of segment ( CD ) is (boxed{sqrt{a^2 + 4r^2}}) and the locus of points ( C ) is a line segment of length (4r) on the line through ( A ) perpendicular to both ( AB ) and the line through ( B ).

🔑:1. Given points ( A ) and ( B ) are the endpoints of a segment of length ( a ). Let's denote the midpoint of ( overline{AB} ) as ( O ).2. Perpendicular lines are drawn through points ( A ) and ( B ) to segment ( overline{AB} ) and points ( C ) and ( D ) are located on these lines such that the line segment ( overline{CD} ) intersects the plane passing through midpoint ( O ) orthogonally at a distance ( r ) from ( O ).3. The perpendicular lines from ( A ) and ( B ) intersect the given plane at points ( E ). Let ( C^{prime} ) and ( D^{prime} ) be the projections of points ( C ) and ( D ) onto the plane that contains ( overline{AB} ) and point ( E ). From symmetry and the problem setup, we have: - ( C C^{prime} = D D^{prime} = frac{a}{2} ) - ( angle CEC^{prime} = angle DED^{prime} ) 4. Point ( E ) is determined to be the median of the right triangles ( triangle COC^{prime} ) and ( triangle DOD^{prime} ), hence: [ O E = 2 r ]5. Considering the right triangle ( C^{prime} O D^{prime} ), where ( O E ) is the median, we express the length ( overline{C^{prime} D^{prime}} ) using the distance formula applied in right-angled triange ( COC^{prime} ): [ C^{prime} D^{prime} = 2 O E = 2 r ] Now, substituting the given values: [ C D = 2 sqrt{C^{prime} C^2 + C^{prime} E^2} = 2sqrt{left(frac{a}{2}right)^2 + r^2} = 2 sqrt{frac{a^2}{4} + r^2} = sqrt{a^2 + 4r^2} ]6. The set of points ( C ) form a line segment of length ( 4r ) centered at point ( A ). Conclusion:Thus, the length of the segment ( overline{CD} ) can be expressed as:[boxed{sqrt{a^2 + 4r^2}}]

🧠:Okay, so I need to figure out two things here: first, how many different k-element subsets of the numbers from 1 to k² can be obtained using this selection method where you pick a number, delete its row and column, and repeat until you have k numbers. Second, I need to find the sum of the numbers in each of these subsets. Let me start by understanding the problem step by step.First, let's visualize the matrix. For a given k, the matrix is a k x k grid filled with numbers from 1 to k² arranged row-wise. So the first row is 1 to k, the second row starts at k+1 and goes to 2k, and so on until the last row, which has numbers from (k-1)k +1 to k². For example, if k=3, the matrix would be:1 2 34 5 67 8 9Now, the selection process: when you pick a number, you delete its entire row and column, so you can't pick another number from that row or column again. This means that each subsequent selection has to be from a smaller matrix. For instance, if I pick 5 from the 3x3 matrix, I delete the second row and second column, leaving me with:1 37 9Then I pick another number, say 1, delete its row and column, leaving only 9 as the last number. So the subset would be {5, 1, 9}. Alternatively, if I first pick 1, delete the first row and first column, leaving:5 68 9Then pick 5, delete its row and column, leaving 9. So the subset is {1, 5, 9}. Wait, that's the same as before but in a different order. But since subsets are unordered, maybe these are considered the same? Wait, the problem says "k different numbers" forming a subset, so the order of selection doesn't matter. Therefore, regardless of the order in which we pick the numbers, the subset is determined by the positions selected. Since each time we pick a number, we remove its row and column, each subsequent pick must be from a different row and column. So essentially, the selection is equivalent to choosing one number from each row and each column, which is exactly a permutation matrix. That is, each subset corresponds to selecting exactly one element from each row and each column, which is like a permutation.Wait, let me check. Suppose k=2. The matrix is:1 23 4If I pick 1 first, then delete the first row and column, leaving only 4. So the subset is {1,4}. Alternatively, if I pick 2 first, then delete first row and second column, leaving 3. So subset {2,3}. Similarly, if I pick 3 first, delete second row and first column, leaving 2. Subset {3,2}. And picking 4 first leaves 1. So subsets {4,1}. But subsets are unordered, so {1,4} and {4,1} are the same. Similarly, {2,3} and {3,2} are the same. So there are two distinct subsets: {1,4} and {2,3}. But wait, in the case of k=2, how many permutation matrices are there? There are 2 permutations: the identity permutation and the transposition. Each corresponds to a different subset. But in terms of subsets, since order doesn't matter, the two permutation matrices correspond to the same two subsets. So the number of subsets is k! for a k x k matrix? Wait, for k=2, k! is 2, which matches. For k=3, it would be 6 subsets. Wait, but let's check with k=3.Take the 3x3 matrix. If we pick one number from each row and column, that's 3! = 6 subsets. Each subset corresponds to a permutation. For example, the permutation (1,2,3) would correspond to selecting 1 from the first row, 5 from the second, 9 from the third. The permutation (1,3,2) would correspond to 1, 6, 8. Wait, let's see:Permutation (1,2,3): row 1 column 1 (1), row 2 column 2 (5), row 3 column 3 (9). Sum is 1+5+9=15.Permutation (1,3,2): row 1 column 1 (1), row 2 column 3 (6), row 3 column 2 (8). Sum is 1+6+8=15.Wait, same sum. Similarly, permutation (2,1,3): row 1 column 2 (2), row 2 column 1 (4), row 3 column 3 (9). Sum is 2+4+9=15.Permutation (2,3,1): row 1 column 2 (2), row 2 column 3 (6), row 3 column 1 (7). Sum is 2+6+7=15.Permutation (3,1,2): row 1 column 3 (3), row 2 column 1 (4), row 3 column 2 (8). Sum is 3+4+8=15.Permutation (3,2,1): row 1 column 3 (3), row 2 column 2 (5), row 3 column 1 (7). Sum is 3+5+7=15.Wow, all subsets sum to 15. That's interesting. So for k=3, regardless of the permutation, the sum is 15. Similarly, for k=2, let's check:Subset {1,4}: sum 5.Subset {2,3}: sum 5. So they both sum to 5. Which is 1+4=5 and 2+3=5. So for k=2, sum is 5. For k=1, trivially 1.So perhaps for any k, the sum is the same for all subsets? If that's the case, then the answer to the second part is that the sum is fixed, regardless of the subset. Then the first part would be the number of such subsets, which is k!.Wait, but let's confirm with k=4. Let's see if all subsets sum to the same value. Let's take a 4x4 matrix:1 2 3 45 6 7 89 10 11 1213 14 15 16Let's pick a permutation. Say, identity permutation: 1,6,11,16. Sum is 1+6+11+16=34.Another permutation: 4,5,10,15. Let's see: row 1 column 4 (4), row 2 column 1 (5), row 3 column 2 (10), row 4 column 3 (15). Sum: 4+5+10+15=34.Another permutation: 2,5,12,15. Wait, row 1 column 2 (2), row 2 column 1 (5), row 3 column 4 (12), row 4 column 3 (15). Sum: 2+5+12+15=34.Another permutation: 3,8,9,14. Row 1 col3=3, row2 col4=8, row3 col1=9, row4 col2=14. Sum: 3+8+9+14=34.Hmm, all these permutations give the same sum. So it seems that for any k, the sum of the selected numbers is constant, regardless of the permutation. Therefore, the answer to the second question is that the sum is fixed, and the first question is that the number of subsets is k!.But let's formalize this. Why is the sum the same for any permutation? Let's think about the matrix structure.Each element in the matrix can be written as a function of its row and column. Let's index the rows from i=1 to k and columns from j=1 to k. Then the element in row i and column j is (i-1)k + j. For example, in row 1, column 1: (1-1)k +1=1. In row 2, column 3: (2-1)k +3= k +3. Wait, that seems correct.Wait, let's check with k=3: row 2, column 3 would be (2-1)*3 +3=3+3=6, which is correct. Similarly, row 3, column 1: (3-1)*3 +1=6+1=7. Correct.So general formula: element a_ij = (i-1)k + j.Now, if we select one element from each row and each column, which is a permutation σ of the columns, then the selected elements are a_1σ(1), a_2σ(2), ..., a_kσ(k).So the sum S = Σ_{i=1 to k} [(i-1)k + σ(i)] = Σ_{i=1}^k (i-1)k + Σ_{i=1}^k σ(i).First term: Σ_{i=1}^k (i-1)k = k Σ_{i=1}^k (i-1) = k Σ_{m=0}^{k-1} m = k*(k-1)k/2 = k²(k-1)/2.Second term: Σ σ(i). Since σ is a permutation of {1,2,...,k}, the sum Σ σ(i) is Σ_{j=1}^k j = k(k+1)/2.Therefore, total sum S = k²(k-1)/2 + k(k+1)/2 = [k³ -k² +k² +k]/2 = [k³ +k]/2 = k(k² +1)/2.Therefore, regardless of the permutation σ, the sum S is always k(k² +1)/2. That explains why for k=2, sum is 2*(4 +1)/2=5, k=3: 3*(9 +1)/2=15, k=4:4*(16 +1)/2=34, which matches our earlier examples.So the sum is fixed. Therefore, the answer to the second question is that each subset sums to k(k² +1)/2.Now, for the first question: the number of subsets. Since each subset corresponds to a permutation of the columns (each permutation corresponds to selecting one element from each row and column), the number of such subsets is k!.But wait, in the k=2 case, the subsets are {1,4} and {2,3}, which are two subsets, and 2! =2, which matches. For k=3, 6 subsets, which matches 3!.However, wait, in the problem statement, it says "different k-element subsets". So even though the selection process can be done in different orders, the resulting subset is the same if it contains the same elements. But in our analysis, each permutation corresponds to a different subset, but in reality, when you select elements in different orders, the subset is the same. Wait no, the subset is determined by the elements selected, regardless of the order. So even if you pick the elements in different orders, the subset is the same. However, in our permutation matrix approach, each permutation corresponds to a unique subset, because each permutation selects different elements. Wait, for example, in the k=3 case, permutation (1,2,3) gives {1,5,9}, permutation (1,3,2) gives {1,6,8}, which are different subsets. Therefore, each permutation gives a distinct subset. Therefore, the number of subsets is indeed k!.But let's verify with k=3. The subsets would be:1,5,91,6,82,4,92,6,73,4,83,5,7Yes, these are 6 distinct subsets, each corresponding to a permutation. So indeed, the number is k!.Therefore, the answers should be:Number of subsets: k!Sum of each subset: k(k² +1)/2.But let's check if there's any catch here. The problem says "select any number, then delete all numbers in the same row and column... continue until k numbers have been selected". So the process is equivalent to selecting one element from each row and column, but in sequence, removing the options as you go. But regardless of the order, the final subset must consist of one element from each row and column, hence forming a permutation. Therefore, the number of such subsets is exactly the number of permutations, k!.But wait, the problem is phrased as "how many different k-element subsets... can be obtained using this method". So it's not the number of ways to select them, but the number of distinct subsets. Since each permutation leads to a unique subset (as different permutations lead to different element selections), the number of subsets is k!.However, I need to make sure that there aren't two different permutations leading to the same subset. Suppose we have two different permutations σ and τ such that for all i, a_iσ(i) = a_iτ(i). But since a_iσ(i) = (i-1)k + σ(i), and similarly for τ, then (i-1)k + σ(i) = (i-1)k + τ(i) implies σ(i) = τ(i) for all i. Therefore, different permutations must correspond to different subsets. Hence, the number of subsets is exactly k!.Therefore, the answers are k! and k(k² +1)/2.But wait, let me confirm with k=1. For k=1, the matrix is just [1]. The number of subsets is 1! =1, and the sum is 1(1 +1)/2=1, which is correct.Another example, k=2. As before, two subsets {1,4} and {2,3}, sum is 5, which is 2(4 +1)/2=5.Therefore, the reasoning seems solid.Thus, the answers are:Number of subsets: k!Sum of each subset: k(k² +1)/2Final AnswerThe number of different subsets is boxed{k!} and the sum of the numbers in each subset is boxed{dfrac{k(k^2 + 1)}{2}}.

✅:To determine the number of different ( k )-element subsets of ({1, 2, cdots, k^2}) that can be obtained using the described method and the sum of the numbers in each subset, we analyze the problem as follows:1. Matrix Structure and Selection Process: - The numbers (1, 2, ldots, k^2) are arranged in a (k times k) matrix where each element at row (i) and column (j) is given by ((i-1)k + j). - Selecting a number and deleting its row and column ensures that each subsequent selection is from a different row and column. This process is equivalent to selecting one number from each row and each column, which corresponds to a permutation of the columns.2. Number of Subsets: - Each permutation of the columns corresponds to a unique subset of numbers. The number of permutations of (k) elements is (k!), which is also the number of distinct (k)-element subsets.3. Sum of Each Subset: - The element in row (i) and column (j) is ((i-1)k + j). For a permutation (sigma) of the columns, the sum of the selected elements is: [ sum_{i=1}^{k} left[(i-1)k + sigma(i)right] ] - This sum can be split into two parts: - The sum of ((i-1)k) for each row (i), which is (k sum_{i=1}^{k} (i-1) = k cdot frac{k(k-1)}{2} = frac{k^2(k-1)}{2}). - The sum of (sigma(i)) for each column (j), which is the sum of the first (k) natural numbers, (frac{k(k+1)}{2}). - Combining these parts, the total sum is: [ frac{k^2(k-1)}{2} + frac{k(k+1)}{2} = frac{k(k^2 + 1)}{2} ] Final AnswerThe number of different ( k )-element subsets is (boxed{k!}) and the sum of the numbers in each subset is (boxed{dfrac{k(k^2 + 1)}{2}}).

🔑:1. Understanding the problem: Given a matrix of natural numbers from 1 to k^2 arranged in the following manner: [ begin{array}{cccc} 1 & 2 & 3 & cdots & k k+1 & k+2 & k+3 & cdots & 2k vdots & vdots & vdots & & vdots (k-1)k+1 & (k-1)k+2 & (k-1)k+3 & cdots & k^2 end{array} ] Remove a number and all other numbers in its row and column. Repeat from the remaining numbers until you extract a total of k numbers.2. Observation: After selecting a number and removing its row and column, each successive choice must come from a different row and different column than all previous choices. Thus, the k numbers will come from distinct rows and columns.3. Counting the possibilities: To count the number of distinct sets of k numbers one can form, arrange them in the natural sequence of rows. Denote their column indices as j_1, j_2, ..., j_k. Each j_i must be unique. Hence, these column indices are a permutation of {1, 2, ldots, k}. Therefore, the number of ways to choose these k numbers is the number of permutations of k distinct elements: [ k! = A_k^k ]4. Calculating the sum of these k numbers: Each number in the i^{th} row and j^{th} column is of the form (i-1)k + j. Summing up these numbers for any permutation j_1, j_2, ldots, j_k: [ sum_{i=1}^k left((i-1)k + j_iright) = sum_{i=1}^k (i-1)k + sum_{i=1}^k j_i ] We know: [ sum_{i=1}^k (i-1)k = k(0 + 1 + 2 + cdots + (k-1)) = k left(frac{(k-1)k}{2}right) = frac{k^2(k-1)}{2} ] Since j_1, j_2, ..., j_k is a permutation of {1, 2, ldots, k}, we also have: [ sum_{i=1}^k j_i = 1 + 2 + 3 + cdots + k = frac{k(k+1)}{2} ] Adding these sums together: [ sum_{i=1}^k left((i-1)k + j_iright) = frac{k^2(k-1)}{2} + frac{k(k+1)}{2} = frac{k^3 - k^2 + k^2 + k}{2} = frac{k^3 + k}{2} = frac{k(k^2 + 1)}{2} ] Conclusion:The number of different sets of k numbers is k!, and the sum of the k numbers in each set is:[boxed{frac{k(k^2 + 1)}{2}}]

🧠:Alright, let's try to tackle this problem. So, the numbers from 2 to 30 are written on a board. We need to mark all of them, but there's a rule: for one ruble, we can mark any number. However, once a number is marked, we can freely mark its divisors and multiples without paying. The goal is to find the minimal number of rubles needed to mark all numbers. Hmm, okay, interesting.First, let me make sure I understand the problem correctly. We start with all numbers from 2 to 30 unmarked. Each time we pay a ruble, we can choose a number to mark, and then we get to mark all of its divisors and multiples for free. So the strategy here is probably to pick numbers that allow us to cover as many other numbers as possible through their divisors and multiples. That way, each ruble we spend gives us maximum coverage, minimizing the total cost.So, the key here is to select numbers such that each chosen number covers the most uncovered numbers. But there's a catch: once a number is marked (either by paying a ruble or as a result of being a divisor/multiple of a paid number), we can't use it again to mark other numbers unless we pay for it again. Wait, no—if a number is already marked, can we still use it to mark its divisors and multiples? Let me check the problem statement again."If any number is already marked, you can freely mark its divisors and multiples." Wait, does that mean that if a number is marked (whether by paying or by being a divisor/multiple), then you can mark its divisors and multiples for free? Or is it only when you pay for a number that you can mark its divisors and multiples?Hmm, the wording says: "For one ruble, you may mark any number. If any number is already marked, you can freely mark its divisors and multiples." So, perhaps once a number is marked (by paying or otherwise), then you can use that marked number to mark its divisors and multiples. So, once a number is marked, you can mark its divisors and multiples without paying. So, maybe it's a chain reaction. For example, if I pay to mark 6, then I can mark all divisors of 6 (2, 3) and multiples of 6 (12, 18, 24, 30) for free. Then, since 2 and 3 are now marked, I can also mark their divisors and multiples. But 2's divisors are only 1, which isn't on the board, and multiples of 2 (4, 6, 8, ..., 30). But 6 is already marked, but 4, 8, etc., might not be. Wait, but once 2 is marked, can I mark all multiples of 2 for free? Similarly for 3.Wait, this seems a bit ambiguous. Let me parse the problem again: "For one ruble, you may mark any number. If any number is already marked, you can freely mark its divisors and multiples." So, maybe whenever a number is marked (whether by paying or by being marked through a previous operation), you can mark its divisors and multiples without paying. So, each time you mark a number, you can then mark its divisors and multiples. So, it's a two-step process: pay to mark a number, then mark its divisors and multiples. Then, if those divisors and multiples are marked, you can mark their divisors and multiples as well. So, it's like a chain reaction. So, for example, if I pay to mark 6, then I can mark 2, 3, 12, 18, 24, 30. Then, since 2 is now marked, I can mark all divisors of 2 (which is 1, not on the board) and multiples of 2 (4, 6, 8, etc.). But 6 is already marked, so maybe 4, 8, 10, etc., get marked. Similarly, 3 is marked, so multiples of 3 (6, 9, 12, etc.) get marked. But 6 and 12 are already marked, so 9, 15, 18, 21, etc., would be marked.Wait, this is getting complicated. So, perhaps when you mark a number, you trigger a chain where all its divisors and multiples get marked, and then those divisors and multiples trigger their own divisors and multiples to be marked, and so on, until no more numbers can be marked. So, in effect, paying for a number allows you to mark its entire connected component in some divisor/multiple graph. Hmm, but maybe not. Let's think step by step.Suppose you pay for a number, say 6. Then you mark 6. Then, you can mark all divisors and multiples of 6. Divisors of 6 are 2 and 3. Multiples of 6 in the range are 12, 18, 24, 30. So, those get marked. Now, since 2 is marked, you can mark all divisors and multiples of 2. Divisors of 2: 1 (not on the board). Multiples of 2: 4, 6, 8, 10, ..., 30. But 6 is already marked, but 4, 8, 10, etc., would now be marked. Similarly, for 3: divisors of 3 are 1 (not on the board), multiples are 6, 9, 12, 15, 18, 21, 24, 27, 30. So, 9, 15, 21, 27 get marked now. Then, moving on, 4 is marked (from multiple of 2), so divisors of 4 are 2, 4 (already marked), multiples are 8, 12, 16, 20, 24, 28. So, 8, 16, 20, 28 get marked. Then, 8's multiples: 16, 24, etc., but those are already covered. Continuing this way, all even numbers would get marked through 2, and multiples of 3 would get marked through 3. So, paying for 6 might result in marking a large portion of the numbers. But maybe not all. For example, numbers like 5, 7, 11, etc., which are primes not covered by 2 or 3, would not be marked. So, if you pay for 6, you get 2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30. Wait, but 5 is still not marked. So, perhaps 5 and 7, 11, etc., need to be paid for separately.Alternatively, maybe if I pay for a number, say 5, then I can mark 5 and its multiples: 10, 15, 20, 25, 30. But divisors of 5 are only 1 and 5. So, only 5 and its multiples. Then, if I pay for 5, I get 5,10,15,20,25,30. But 10,15,20,25,30 might already be marked through other numbers. But if they aren't, paying for 5 would mark those. Similarly, if I pay for 7, I get 7,14,21,28. Etc.So, perhaps the key is to cover all the primes, because primes can't be obtained as multiples of smaller numbers (except 1, which isn't on the board). Therefore, each prime number would need to be either paid for directly or covered by paying for a multiple of it. But since primes have no divisors other than 1 and themselves, the only way to cover them is to either pay for them directly or pay for one of their multiples.For example, take prime number 7. If I pay for 14 (which is a multiple of 7), then when I pay for 14, I can mark 14, its divisors (which include 2 and 7), and its multiples (28). So, paying for 14 would mark 2,7,14,28. Therefore, instead of paying for 7 directly, I could pay for 14 and get 7 for free. Similarly, for prime 5: if I pay for 10, which is a multiple of 5, then paying for 10 would mark 10, its divisors (2,5) and multiples (20, 30). So, paying for 10 gives me 2,5,10,20,30. But 2 might already be covered by another number.Therefore, the strategy might be to cover primes by paying for their composite multiples, which can also cover other numbers. However, this might require some optimization. For instance, paying for composite numbers that are multiples of several primes might be more efficient.Alternatively, paying for the primes themselves might be cheaper because their multiples are more numbers. Let's see. Let's list all primes between 2 and 30: 2,3,5,7,11,13,17,19,23,29. There are 10 primes here. If we paid for each prime, then by paying 10 rubles, we could mark each prime and then their multiples. But maybe we can do better by paying for some composite numbers that cover multiple primes.For example, if we pay for 6 (which is 2*3), then we can mark 2 and 3, as well as their multiples. Similarly, paying for 10 (2*5) can mark 2 and 5. But if we have already paid for 6, then 2 and 3 are already marked, so paying for 10 would only give us 5 and its multiples. Hmm, but if we pay for 6 and 10, then 2,3,5 are covered. Then, their multiples would be covered as well.But maybe we can do even better by paying for numbers that are products of multiple primes, so that each payment covers several primes. For example, paying for 30 (which is 2*3*5) would allow us to mark 2,3,5, and their multiples. But 30 is a multiple of 2,3,5. If we pay for 30, then divisors of 30 are 2,3,5,6,10,15,30. So, paying for 30 would mark 30, then we can mark its divisors (2,3,5,6,10,15) and multiples (but 30 is the maximum number, so no multiples beyond 30). So, paying for 30 would mark 2,3,5,6,10,15,30. Then, through those numbers, we can mark their multiples. For example, 2's multiples are 4,6,8,...,30. But since 2 is marked, we can mark all even numbers. Similarly, 3's multiples: 6,9,12,...,30. 5's multiples: 10,15,20,25,30. So, by paying for 30, we can get a lot of numbers marked. Let's see:Paying for 30:- Mark 30.- Then mark divisors: 2,3,5,6,10,15,30.- Then, from 2: mark all multiples (4,6,8,...,30). But 6,10, etc., are already marked.- From 3: mark multiples 6,9,12,...,30.- From 5: mark multiples 10,15,20,25,30.- From 6: mark divisors (2,3,6) and multiples (12,18,24,30). Already covered.- From 10: mark divisors (2,5,10) and multiples (20,30). Already covered.- From 15: divisors (3,5,15) and multiples (30). Already covered.So, paying for 30 would result in all numbers divisible by 2,3,5 being marked. That includes all even numbers, all multiples of 3, and all multiples of 5. But numbers not divisible by 2,3,5 would still be unmarked. These are the primes 7,11,13,17,19,23,29 and composites like 49 (which isn't on the board), 77, etc., but within 2-30, the numbers not divisible by 2,3,5 are:7,11,13,17,19,23,29 (primes) and composite numbers like 7*7=49 (too big), 7*11=77 (too big), etc. Wait, actually, within 2-30, the numbers not divisible by 2,3,5 are:7, 11, 13, 17, 19, 23, 29, and also composite numbers like 7*7=49 (not in range), so actually, only the primes 7,11,13,17,19,23,29 and the number 1 (which isn't on the board). So, numbers like 7,11,13,17,19,23,29 are primes not covered by 2,3,5. Additionally, numbers like 21=3*7, but 21 is divisible by 3, which is already covered if we paid for 30 (since 3 is marked). Wait, 21 is divisible by 3 and 7. Since 3 is already marked (from paying for 30), then 21 would be marked as a multiple of 3. Similarly, 25 is covered as a multiple of 5. 27 is covered as a multiple of 3. 28 is covered as a multiple of 2. 14 is covered as a multiple of 2. Wait, but 7 itself is not covered. So, numbers divisible by 7 but not by 2,3,5 would be 7, 49 (which isn't there). So, 7 is a prime not covered. Similarly, 11,13, etc.Therefore, after paying for 30, we still need to cover primes 7,11,13,17,19,23,29. Each of these primes can't be obtained as multiples of 2,3,5, so they need to be covered either by paying for them directly or by paying for one of their multiples. For example, 7 can be covered by paying for 7 itself or paying for 14, 21, 28. However, 14 is a multiple of 2 and 7; 21 is a multiple of 3 and 7; 28 is a multiple of 2 and 7. But since 2 and 3 are already marked (from paying for 30), paying for 14 would allow us to mark 7 (as a divisor of 14), and then 7's multiples (14,21,28). Similarly, paying for 21 would mark 7 (as a divisor) and 21's multiples (42, which is out of range). But 21 is already marked as a multiple of 3. Wait, but 7 isn't marked yet. So, if we pay for 14, which is a multiple of 2 and 7, since 2 is already marked, paying for 14 would mark 14, and then we can mark its divisors (2 and 7) and multiples (28). Since 2 is already marked, the new thing here is 7 and 28. So, paying for 14 would give us 7 and 28. Similarly, paying for 28 would mark 28, and its divisors (2,4,7,14,28). Since 2 and 4 are already marked (from 30 and 2's multiples), paying for 28 would give us 7,14,28. But 14 is a multiple of 2 and 7, so if we pay for 14 or 28, we can get 7 covered.Similarly, for prime 11: to cover 11, we can pay for 11 itself or 22 (multiple of 2 and 11). Since 2 is already marked, paying for 22 would mark 22 and its divisors (2 and 11), so 11 gets marked. Then 11's multiples (22, 33 which is out of range). So paying for 22 would mark 11 and 22. Similarly for 13: paying for 26 (multiple of 2 and 13), which would mark 13 and 26. For 17: paying for 34 (out of range), so must pay for 17. Similarly, 19: pay for 38 (out of range), so pay for 19. 23: pay for 46 (out of range), so pay for 23. 29: pay for 58 (out of range), so pay for 29.Therefore, after paying for 30, we need to cover 7,11,13,17,19,23,29. To cover each of these, we can either pay for the prime itself or pay for their multiple within the range. Let's see:- 7: can pay for 14, 21, 28. If we pay for 14, we get 7 and 14 (and 28). But 14 is already covered as a multiple of 2. Wait, but 14 is a multiple of 2, which is already marked, so is 14 already marked? Let's see. After paying for 30, we have 2,3,5 marked. Then, from 2, all multiples (even numbers) are marked, so 4,6,8,...,30. Therefore, 14 is already marked as a multiple of 2. Therefore, paying for 14 would be redundant because 14 is already marked. Wait, but the problem says that if a number is already marked, you can freely mark its divisors and multiples. So, if 14 is already marked (as a multiple of 2), then you can mark its divisors (2 and 7) and multiples (28). But 2 is already marked, so 7 and 28 would be marked. Therefore, if 14 is already marked, you can mark 7 and 28 for free. But wait, is that how it works?Wait, the process is: when you pay for a number, you mark it, and then you mark its divisors and multiples. But if a number is already marked (without paying), can you still mark its divisors and multiples? The problem states: "If any number is already marked, you can freely mark its divisors and multiples." So, once a number is marked (either by paying or by being marked through another operation), you can use it to mark its divisors and multiples. So, if 14 is marked because it's a multiple of 2 (which was marked when paying for 30), then since 14 is marked, you can mark its divisors (2 and 7) and multiples (28). However, 2 is already marked, so 7 and 28 would be marked. Therefore, if 14 is already marked, then you can get 7 and 28 for free. Therefore, you don't need to pay for 7 or 28; they would be automatically marked once 14 is marked. But since 14 is already marked (as a multiple of 2), then 7 and 28 should already be marked.Wait, hold on. Let's think through this step by step. After paying for 30, we have 30 marked. Then, its divisors (2,3,5,6,10,15,30) are marked. Then, since 2 is marked, all multiples of 2 (4,6,8,...,30) are marked. That includes 14, 22, 26, 28. So, 14,22,26,28 are marked because they are multiples of 2. Then, since these numbers are marked, we can mark their divisors and multiples. For example, 14 is marked, so we can mark its divisors (2,7) and multiples (28). Since 2 is already marked, marking 7 and 28. Similarly, 22 is marked, so we can mark 11 and 22's multiples (44, which is out of range). So, 11 would be marked. Similarly, 26 is marked, so divisors 2 and 13 are marked, so 13 gets marked. 28 is marked, divisors 2,4,7,14,28, so 7 gets marked. Therefore, after paying for 30, since all even numbers are marked (including 14,22,26,28), their divisors include 7,11,13. Therefore, 7,11,13 are marked through their even multiples. Therefore, after paying for 30, we automatically get 7,11,13 marked through 14,22,26. Then, their multiples would be marked as well.But wait, 7's multiples are 14,21,28. 14 and 28 are already marked. 21 is a multiple of 3 (which is marked), so 21 is already marked. Similarly, 11's multiples are 22 and 33. 22 is already marked. 13's multiples are 26 and 39. 26 is already marked. So, actually, paying for 30 might automatically cover 7,11,13 through their even multiples. Therefore, after paying for 30, the remaining primes to cover are 17,19,23,29. Let's verify:Numbers 17,19,23,29 are primes. Their multiples within 2-30 are:- 17: 34 (out of range)- 19: 38 (out of range)- 23: 46 (out of range)- 29: 58 (out of range)Therefore, there are no multiples of these primes within the board except themselves. Therefore, the only way to mark them is to either pay for them directly or pay for a number that is a multiple of them. But since their multiples are beyond 30, paying for them is the only option.Wait, but 17 is prime, and it's on the board. If we can mark 17 by paying for it, but maybe we can mark it through some other number. Wait, 17's divisors are 1 and 17. Since 1 is not on the board, the only way to mark 17 is to either pay for 17 or pay for a multiple of 17. But the multiples are 34, 51, etc., which are not on the board. So, we have to pay for 17. Similarly for 19,23,29.Therefore, after paying for 30, we still need to pay for 17,19,23,29. That's four more rubles. So, total rubles would be 1 (for 30) + 4 (for 17,19,23,29) = 5 rubles.Wait, but let's check if this actually works. Let's go step by step:1. Pay for 30: - Mark 30. - Mark divisors of 30: 2,3,5,6,10,15,30. - Mark multiples of 30: none beyond 30. Then, from 2: - Mark all multiples of 2: 4,6,8,10,12,...,30. So all even numbers. From 3: - Mark all multiples of 3: 6,9,12,15,18,21,24,27,30. From 5: - Mark all multiples of 5:10,15,20,25,30. So, after paying for 30, all numbers divisible by 2,3,5 are marked. Then, through the marked numbers like 14 (even), which allows us to mark 7 and 28. Similarly, 22 (even) allows marking 11, 26 allows marking 13. So, primes 7,11,13 are marked via their even multiples. Then, their multiples (like 21=3*7, which is already covered by 3; 28=2*14, etc.) are already covered. So, the remaining primes are 17,19,23,29.Therefore, total rubles: 1 (for 30) + 4 (for 17,19,23,29) = 5.But wait, is there a way to cover some of these remaining primes by paying for a composite number that covers multiple primes? For example, is there a number that is a multiple of two of these primes? Let's see:- 17 is prime, and the next multiple is 34.- 19 is prime, multiple is 38.- 23 is prime, multiple is 46.- 29 is prime, multiple is 58.All of these are outside the range. Therefore, there is no composite number within 2-30 that is a multiple of any two of these primes. Therefore, each of these primes must be paid for individually. Therefore, we need to pay for each of them separately.Therefore, total rubles would be 5.But wait, let's check if paying for 30 is the optimal choice. Maybe paying for a different number could cover more primes.Alternatively, instead of paying for 30, maybe paying for other composite numbers can cover more primes.For example, paying for 6 (covers 2,3), paying for 10 (covers 2,5), paying for 14 (covers 2,7), paying for 22 (covers 2,11), paying for 26 (covers 2,13), and then paying for 17,19,23,29. But this seems like more rubles.Wait, let's compute:If we pay for 6: covers 2,3, and their multiples.Then, paying for 10: covers 2,5, and their multiples. But 2 is already covered.Paying for 14: covers 2,7, and their multiples. But 2 is already covered.Paying for 22: covers 2,11, and their multiples. 2 is covered.Paying for 26: covers 2,13, and their multiples. 2 is covered.Then, paying for 17,19,23,29. So, total rubles: 6 (6,10,14,22,26,17,19,23,29) – wait, no, that's 9 payments. That's worse than paying for 30 and four primes.Alternatively, maybe paying for 6, 10, 14, 22, 26, 17,19,23,29. That's 9 rubles, which is worse than 5. So paying for 30 is better.Alternatively, paying for 2,3,5,7,11,13,17,19,23,29. That's 10 rubles. Worse.So, paying for 30 and then 17,19,23,29 is 5 rubles. But maybe there is a smarter way.Wait, but after paying for 30, we also cover numbers like 25 (multiple of 5) and others. But the key is whether paying for 30 covers 7,11,13 through their even multiples. Let's verify this.After paying for 30:- 30 is marked.- Divisors of 30: 2,3,5,6,10,15,30 are marked.- Then, multiples of 2: all even numbers (4,6,8,...,30).- Among these even numbers are 14,22,26,28.- Each of these even numbers, when marked, allows marking their divisors. For example, 14 is marked (as a multiple of 2), so we can mark divisors of 14: 2 and 7. 2 is already marked, so 7 is marked.- Similarly, 22 is marked (multiple of 2), so divisors 2 and 11. 11 is marked.- 26 is marked (multiple of 2), divisors 2 and 13. 13 is marked.- 28 is marked (multiple of 2), divisors 2 and 7. 7 is already marked.Therefore, after paying for 30, when we process the even numbers 14,22,26,28, we can mark their prime divisors 7,11,13. Therefore, 7,11,13 are marked for free.So, after paying for 30, the only primes left are 17,19,23,29. These primes are not divisible by 2,3,5, and their multiples are beyond 30, so we have to pay for them individually. Hence, 4 more rubles.Thus, total rubles: 1 + 4 = 5.But wait, let's check if there's another composite number that can cover two of these primes. For example, 17*19=323, which is way beyond 30. Similarly, other combinations are too big. So, no, there's no composite number within 2-30 that can cover two of these primes. Therefore, each must be paid for separately.Therefore, the minimal number of rubles is 5.But let's check if this is correct. Let's simulate the entire process.Step 1: Pay for 30.- Mark 30.- Mark divisors of 30: 2,3,5,6,10,15,30.- Now, since 2 is marked, mark all multiples of 2: 4,6,8,...,30.- Since 3 is marked, mark all multiples of 3: 6,9,12,15,18,21,24,27,30.- Since 5 is marked, mark all multiples of 5:10,15,20,25,30.- Now, looking at marked numbers: All even numbers (2,4,6,...,30), all multiples of 3 (6,9,12,...,30), all multiples of 5 (10,15,20,25,30).- Now, check if 7 is marked: 7 is not a multiple of 2,3,5. However, 14 is a multiple of 2, so 14 is marked. Then, since 14 is marked, we can mark its divisors (2 and 7). 2 is already marked, so 7 is marked.- Similarly, 22 is marked (even), so mark divisors 2 and 11. 11 is marked.- 26 is marked (even), so divisors 2 and 13. 13 is marked.- 28 is marked (even), divisors 2 and 7. 7 is already marked.- Therefore, 7,11,13 are marked through their even multiples.- Now, the remaining numbers are: Primes:17,19,23,29. Composite numbers: All others are covered by 2,3,5,7,11,13. For example, 21=3*7 (covered by 3), 25=5*5 (covered by 5), 27=3*9 (covered by 3), etc.- Therefore, the only numbers left are 17,19,23,29. Each must be paid for individually.Step 2: Pay for 17.- Mark 17. Since it's a prime, no divisors except 1 and itself. Multiples are 34, which is out of range.Step 3: Pay for 19.- Mark 19. Similarly, no multiples within range.Step 4: Pay for 23.- Mark 23.Step 5: Pay for 29.- Mark 29.Therefore, total rubles: 1 (30) + 4 (17,19,23,29) = 5.Is there a way to cover any of these remaining primes with a composite number? Let's check if any composite number in the range 2-30 has a divisor among 17,19,23,29. For example:- 17: only multiple is 34 (out of range).- 19: multiple is 38 (out of range).- 23: multiple is 46 (out of range).- 29: multiple is 58 (out of range).Therefore, no composite numbers in the range are multiples of these primes. Therefore, they must be paid for individually.Therefore, 5 rubles is the minimal number.But wait, let's consider another approach. What if instead of paying for 30, we pay for numbers that cover multiple primes more efficiently. For example, paying for numbers like 6, 10, 14, 22, 26, etc., to cover 2,3,5,7,11,13, and then pay for 17,19,23,29. Let's calculate the total.Paying for 6: covers 2,3.Paying for 10: covers 2,5.Paying for 14: covers 2,7.Paying for 22: covers 2,11.Paying for 26: covers 2,13.Then paying for 17,19,23,29.Total rubles: 5 (6,10,14,22,26) +4 (17,19,23,29) =9. Which is worse than 5.Alternatively, maybe paying for 2,3,5,7,11,13,17,19,23,29. That's 10 rubles. Worse.Alternatively, paying for 30 (covers 2,3,5,7,11,13) as before, then 17,19,23,29. 5 rubles.Alternatively, paying for 15 (covers 3,5) and 10 (covers 2,5). Then 14 (covers 2,7), 22 (covers 2,11), 26 (covers 2,13). Then 17,19,23,29. Total: 6 rubles. Still worse.Alternatively, is there a way to cover some of the remaining primes (17,19,23,29) via their own multiples? For example, if we pay for 17, can we get anything else? No, because multiples of 17 are 34, which is out of range. Same for the others.Therefore, the conclusion is that paying for 30 covers the primes 2,3,5, and through their multiples, covers 7,11,13. Then, the remaining primes 17,19,23,29 must be paid for individually. Total rubles:5.But wait, let's check if 7,11,13 are indeed covered when we pay for 30.When we pay for 30, we mark 30 and its divisors (2,3,5,6,10,15). Then, from 2, we mark all even numbers, which includes 14,22,26,28. Then, since 14,22,26,28 are marked, their divisors (including 7,11,13) are marked. Therefore, yes, 7,11,13 are marked through their even multiples. So, that works.Thus, the minimal number of rubles is 5.But let's verify with another example. Suppose instead of paying for 30, we paid for 2. Then, mark 2 and all even numbers. Then, from even numbers like 6,10,14, etc., we can mark their divisors. For example, 6 is even, so marked. Divisors of 6:2,3. So, 3 is marked. Then, multiples of 3:9,12,15,...,30. Then, 10 is even, divisors 2,5. So 5 is marked. Multiples of 5:10,15,20,...,30. Similarly, 14 is even, divisors 2,7. So 7 is marked. Then multiples of 7:14,21,28. 21 is marked via 3 (if 3 is marked). 28 is even. Then, 22 is even, divisors 2,11. So 11 is marked. 26 is even, divisors 2,13. So 13 is marked. So, paying for 2 would cover 2,3,5,7,11,13 through their even multiples. Then, the remaining primes would be 17,19,23,29. So total rubles: 1 (for 2) +4 (for 17,19,23,29) =5. Same as before.So, paying for 2 or paying for 30 both result in 5 rubles. Is there any difference? Let's see.If we pay for 2:- Mark 2 and all even numbers.- Through even numbers, mark 3,5,7,11,13.- Then pay for 17,19,23,29.Total rubles:5.If we pay for 30:- Mark 30 and its divisors (2,3,5,6,10,15,30).- Then mark all multiples of 2,3,5.- Which includes even numbers and multiples of 3 and 5.- Then, through even numbers, mark 7,11,13.- Then pay for 17,19,23,29.Total rubles:5.So, either way, paying for a single number (2 or 30) that triggers a chain reaction covering multiple primes, then paying for the remaining four primes. Therefore, minimal rubles is 5.But wait, what if we pay for a different number that covers more?For example, paying for 6 (covers 2,3). Then, all multiples of 2 and 3 are marked. This would cover even numbers and multiples of 3. Then, through multiples like 6,12,18, etc., but we still need to cover 5,7,11,13,17,19,23,29.To cover 5, we need to pay for 5 or a multiple like 10. If we pay for 10, that covers 2,5. Then, multiples of 5 (10,15,20,25,30). But 2 is already covered. Then, paying for 10 gives us 5. Then, through 10's multiples, we get 10,15,20,25,30. So, paying for 10 after paying for 6 would cover 5.Then, similarly, to cover 7,11,13, we'd need to pay for their even multiples (14,22,26) or pay for them directly. If we pay for 14,22,26, that's three more rubles. Then, pay for 17,19,23,29: four rubles. Total rubles: 1 (6) +1 (10) +3 (14,22,26) +4 (17,19,23,29) =9. Worse than 5.Alternatively, paying for 6, 10, 14, 22, 26, and then 17,19,23,29: total of 9.Therefore, clearly, paying for 2 or 30 is better.Alternatively, paying for 2 and then paying for 3,5,7,11,13,17,19,23,29. But that's 10 rubles.No, that's worse.Alternatively, paying for 2, then through even numbers mark 3,5,7,11,13. Then paying for 17,19,23,29. Total 5 rubles.Same as before.So, the key is that paying for 2 or 30 allows you to cover multiple primes through their multiples, and then you only have to cover the remaining primes.But in both cases, the total is 5. Therefore, the minimal number of rubles is 5.But let's check if there is another way. For example, paying for a number that covers multiple of the remaining primes. For instance, paying for a number like 17*2=34, but it's out of range. Similarly, 17+ something.No, there's no such number. Therefore, we need to pay for each of the remaining primes.Therefore, the answer should be 5.However, let me check once again to make sure I'm not missing anything.Numbers from 2 to 30.If we pay for 2:- Mark 2, all even numbers (4,6,...,30).- From even numbers like 6: divisors 2,3. So 3 is marked. Multiples of 3:6,9,12,...,30.- From 10: divisors 2,5. So 5 is marked. Multiples of 5:10,15,20,25,30.- From 14: divisors 2,7. So 7 is marked. Multiples of 7:14,21,28.- From 22: divisors 2,11. So 11 is marked. Multiples of 11:22,33.- From 26: divisors 2,13. So 13 is marked. Multiples of 13:26,39.- So, after paying for 2, we get 3,5,7,11,13 marked for free.- Then, remaining primes:17,19,23,29.Total rubles:1 (for 2) +4=5.Same as paying for 30.Therefore, regardless of whether we pay for 2 or 30, we end up needing to pay for four more primes.Therefore, the minimal number of rubles is 5.I think this is correct. However, let's check if there's a number that covers two of the remaining primes. For example, is there a number between 2-30 that is a multiple of two of the primes 17,19,23,29? Let's see:- 17*19=323- 17*23=391- 17*29=493- 19*23=437- 19*29=551- 23*29=667All these are way above 30. Therefore, there are no composite numbers within 2-30 that are multiples of two of these primes. Therefore, each must be paid for individually.Therefore, the minimal number of rubles is indeed 5.

✅:To determine the minimal number of rubles required to mark all numbers from 2 to 30, we analyze the optimal strategy:1. Key Insight: Numbers can be marked by paying for them directly or through their divisors/multiples. Primes must be covered either by paying for them or via their multiples. Composite numbers can often be covered through their prime factors.2. Strategic Payment: - Paying for 30 marks it and its divisors (2, 3, 5, 6, 10, 15, 30). This triggers a chain reaction: - Multiples of 2 (all even numbers), multiples of 3, and multiples of 5 are marked. - Even multiples like 14, 22, 26, and 28 allow marking their prime divisors (7, 11, 13) for free.3. Remaining Primes: - After paying for 30, the uncovered primes are 17, 19, 23, 29. These primes have no multiples within 2–30, so each must be paid for individually.4. Total Cost: - 1 ruble for 30. - 4 rubles for 17, 19, 23, and 29.Final Answer: The minimal number of rubles required is (boxed{5}).

🔑:We need to mark all the numbers from 2 to 30 on the board. To do this cost-effectively, we follow these steps:1. Identify Large Prime Numbers: Note that we need to mark prime numbers greater than 15 first because they are not divisible nor are multiples of any numbers already on the board. These prime numbers within our range (17, 19, 23, 29) must be marked independently, costing 1 ruble each. [ text{Cost so far: } 4 text{ rubles} ]2. Mark the Smallest Prime Number (2): Next, we mark the smallest prime number, which is 2. [ text{Cost so far: } 4 + 1 = 5 text{ rubles} ]3. Mark Even Numbers: - Once 2 is marked, we can freely mark all multiples of 2 without additional cost. This includes all even numbers from 4 to 30.4. Marking Remaining Odd Numbers ≤ 15: - Next, we consider the odd numbers less than or equal to 15. Each of these numbers, say ( n ), will have ( 2n ) already marked since ( 2n ) is an even number. Since ( 2n ) is already marked, ( n ) can be marked without additional cost.5. Marking Special Numbers: - Observe specific numbers like 21, 25, and 27: - 21 is divisible by 3 (marked as part of previous steps, 2×3=6). - 25 is divisible by 5 (as multiples of marked 5). - 27 is divisible by 3 (as it's a multiple of marked 3).By marking these significant divisors and multiples, we cover all numbers. No further costs are incurred beyond the initial ones. ConclusionThus, the minimum amount of rubles needed to mark all numbers on the board from 2 to 30 is ( boxed{5} ).

🧠:Okay, so I need to find the minimum number k of subsets A₁, A₂, ..., A_k of the set X = {1, 2, ..., n} such that the family of subsets is both covering and separating. Let me first make sure I understand what covering and separating mean in this context.A covering family means that the union of all subsets A₁ through A_k is equal to X. So every element in X is included in at least one of the subsets. That part seems straightforward.A separating family, on the other hand, requires that for any pair of distinct elements i and j in X, there is at least one subset A_m in the family that contains exactly one of i or j. So for any i ≠ j, either i is in A_m and j is not, or j is in A_m and i is not. This ensures that each pair of elements can be distinguished or separated by at least one subset in the family.So the problem is asking for the smallest k such that there exists a family of k subsets that both covers X and separates every pair of elements in X.Hmm. Let me think about how to approach this. Maybe I can start with small values of n and see if I can find a pattern or come up with a general formula.Let's start with n = 1. Then X = {1}. For the family to be covering, we need at least one subset containing 1. Since there's only one element, there are no pairs to separate, so the separating condition is trivially satisfied. So k = 1 suffices. But n=1 is a trivial case.n = 2: X = {1, 2}. To cover both elements, the union of subsets must include both 1 and 2. The separating condition requires that for the pair (1,2), there is a subset that contains exactly one of them. So we need at least one subset that contains 1 but not 2, or 2 but not 1. But we also need the union to cover both 1 and 2. Let's see:If we have two subsets: A₁ = {1}, A₂ = {2}. Then the union is {1,2}, so covering is satisfied. For separating, the pair (1,2): A₁ contains exactly 1, and A₂ contains exactly 2. So both subsets can separate them. Wait, but the definition says "there is a subset in the family that contains exactly one of i or j". So for 1 and 2, A₁ contains exactly one (1) and A₂ contains exactly one (2). So actually, either A₁ or A₂ can separate 1 and 2. Wait, but in this case, each subset separates them. Wait, but maybe even a single subset can separate them if it contains exactly one. So if we have A₁ = {1}, then 1 is in A₁ and 2 is not, so that subset alone separates 1 and 2. But then the union would be {1}, which doesn't cover 2. So we need another subset {2}. Therefore, k=2. But wait, is there a way to do it with k=1? If we have A₁ = {1,2}, then the union is covered, but it doesn't separate 1 and 2 because the subset contains both. So no. So for n=2, k=2.n=3: X = {1,2,3}. Let's try to find the minimal k. Let's consider subsets. To cover all elements, each element must be in at least one subset. To separate all pairs: (1,2), (1,3), (2,3). Let's try k=2.Suppose we use two subsets. Let's see:Case 1: A₁ = {1,2}, A₂ = {2,3}. The union is {1,2,3}, so covering is okay. Now check separation:- For pair (1,2): A₁ contains both, A₂ contains 2 but not 1. So in A₂, does it contain exactly one? No, because A₂ contains 2 and 3. Wait, 1 is not in A₂, so 2 is in A₂ but 1 is not. So A₂ separates 1 and 2. Similarly, for pair (1,3): Check each subset. A₁ contains 1 but not 3; A₂ contains 3 but not 1. So either A₁ or A₂ can separate them. Wait, actually, in A₁, 1 is there and 3 is not; in A₂, 3 is there and 1 is not. So both subsets can be used to separate. But according to the problem, we need at least one subset that contains exactly one. For pair (1,3), both subsets contain exactly one: A₁ has 1, A₂ has 3. So either of them works. Then for pair (2,3): A₁ contains 2 but not 3, and A₂ contains both 2 and 3. So A₁ can separate 2 and 3? Wait, in A₁, 2 is present and 3 is not. So yes, A₁ separates 2 and 3. Similarly, in A₂, both are present. So actually, all pairs can be separated with these two subsets. Wait, so maybe k=2 works for n=3?Wait, but let me check again. Pair (1,2): A₂ contains 2, not 1. So A₂ can separate them. Pair (1,3): A₁ contains 1, A₂ contains 3. So either subset can separate. Pair (2,3): A₁ contains 2, not 3. So A₁ can separate. So all pairs are separated. And the union is all three elements. So k=2 works for n=3? That seems possible.Wait, but maybe there's a mistake here. Let me take a different example. Suppose A₁ = {1,2}, A₂ = {1,3}. Then union is covered. Now check separation:- (1,2): A₂ contains 1 but not 2? No, A₂ contains 1 and 3. So in A₂, 1 is present, 2 is not. So A₂ separates 1 and 2. Wait, but 1 is in A₂, 2 is not. So yes, A₂ separates them. Similarly, (1,3): A₁ contains 1 but not 3. So A₁ separates 1 and 3. (2,3): Both subsets: A₁ contains 2, A₂ contains 3. So in A₁, 2 is present and 3 is not; in A₂, 3 is present and 2 is not. So either subset can separate 2 and 3. So this also works. So k=2 for n=3. Hmm. So maybe k=2 suffices for n=3.Wait, but I thought maybe there's a lower bound that k has to be at least log₂n or something, but here for n=3, k=2 which is log₂3 ≈ 1.58, so rounded up to 2. Hmm. Maybe that's the case. Let me check n=4.n=4: X={1,2,3,4}. Let's see if k=3 works. Let's try to construct such subsets.Alternatively, think in terms of binary representation. If we assign each element a binary string of length k, and each subset corresponds to a bit position where the subset contains all elements with a 1 in that position. Then the separating condition is equivalent to any two elements differing in at least one bit, which is the same as all codes being distinct. The covering condition is that every element is included in at least one subset, which corresponds to every binary string having at least one 1. Wait, but in that case, the minimal k would be the smallest number such that n ≤ 2^k -1, since each element is assigned a unique non-zero binary string. Wait, but that would be for the case where each subset corresponds to a bit, and the codes are unique. However, in our problem, the separating condition is slightly different: for any two elements, there exists a subset that contains exactly one of them. This is equivalent to the characteristic vectors of the elements (where each vector is a 0-1 string indicating membership in each subset) having Hamming distance at least 1. Wait, but actually, more precisely, for any two elements, there is at least one coordinate (subset) where their membership differs. Which is exactly the definition of the characteristic vectors being distinct. Wait, but if two elements are in exactly the same subsets, then they can't be separated. So the separating family requires that all characteristic vectors are distinct. But also, covering requires that every element is in at least one subset, so each characteristic vector has at least one 1.Therefore, the problem reduces to finding the minimal k such that there are n distinct binary vectors of length k, each with at least one 1. The number of such vectors is 2^k -1 (since each vector has to be non-zero). Therefore, we need 2^k -1 ≥ n, so k ≥ log₂(n+1). Therefore, the minimal k is the ceiling of log₂(n+1).Wait, but let me check this with the previous examples.For n=2: log₂(3) ≈ 1.58, so ceiling is 2. Which matches our earlier result.For n=3: log₂(4) = 2, ceiling is 2. Which also matches.For n=4: log₂(5) ≈ 2.32, ceiling is 3. Let's see if k=3 works for n=4. The characteristic vectors would be 3-bit non-zero vectors, of which there are 7. So we can assign each of the 4 elements a distinct 3-bit vector with at least one 1. For example:1: 0012: 0103: 0114: 100Wait, but wait, 011 is already covered by 01 and 11? Wait, no. Each element is assigned a unique 3-bit vector. Then the subsets correspond to each bit position. So subset A₁ corresponds to the first bit, A₂ to the second, A₃ to the third. Then each element's membership in the subsets is determined by their bits. For example, element 1 (001) is in A₃ only. Element 2 (010) is in A₂ only. Element 3 (011) is in A₂ and A₃. Element 4 (100) is in A₁. Then check covering: every element is in at least one subset, yes. Check separating: for any two elements, they have different vectors, so there must be at least one bit where they differ. Therefore, in that bit's subset, one is present and the other is not. So yes, this should work. So k=3 for n=4. Which matches the ceiling(log₂(n+1)) formula.Wait, so this seems like a general solution. Therefore, the minimal k is the smallest integer such that 2^k -1 ≥ n, so k = ceiling(log₂(n+1)).But wait, let me confirm with n=1. For n=1, log₂(2) = 1, so k=1. Correct. n=2: log₂3 ≈1.58, k=2. Correct. So seems this is the case.But wait, the problem states that the family must be both covering and separating. The covering requires that every element is in at least one subset, which is equivalent to each characteristic vector having at least one 1. The separating requires that any two elements have different characteristic vectors, which is equivalent to the vectors being distinct. Therefore, the total number of available vectors is 2^k -1 (since each vector must have at least one 1). Therefore, to have n distinct vectors, we need 2^k -1 ≥ n. Therefore, k ≥ log₂(n+1). Since k must be an integer, the minimal k is the ceiling of log₂(n+1). Therefore, the answer is ⌈log₂(n+1)⌉.But let me check this with n=5. log₂(6) ≈ 2.58, so k=3. Then 2^3 -1=7 ≥5. So yes, k=3 works. Similarly, for n=7, log₂(8)=3, so k=3. For n=8, log₂(9)≈3.17, so k=4. So 2^4 -1=15 ≥8. Correct.Therefore, the minimal k is the smallest integer k such that 2^k ≥n+1, which is equivalent to k=⌈log₂(n+1)⌉.But let me double-check if this is indeed correct, or if there's a different approach.Another way to think about it is that each element can be associated with a unique non-empty subset of the k subsets, but since each element is in some subsets, the mapping is injective. Wait, no, because subsets can overlap. The characteristic vectors must be unique and non-zero.Alternatively, the problem is equivalent to finding a binary code of length k, with n codewords, each with Hamming weight at least 1, and pairwise Hamming distance at least 1. The minimal such k is indeed the minimal k such that 2^k -1 ≥n, since the maximum number of codewords with Hamming weight at least 1 is 2^k -1. Therefore, we need k ≥ log₂(n+1).Therefore, the minimal k is ⌈log₂(n+1)⌉.But wait, let's see if there's another perspective. For example, in covering, each element is in at least one subset. For separating, each pair is separated by at least one subset.Alternatively, think of each subset as a test that partitions the set X into two parts: those in the subset and those not in it. The covering condition requires that every element is in at least one subset. The separating condition requires that for any two elements, there is at least one test (subset) where they are separated (one in, one out). This is similar to building a binary identification system where each element is identified by its responses to the k tests.This is analogous to a binary code where each element has a unique code word (hence the separating) and each code word has at least one '1' (hence covering). Therefore, the minimal k is the smallest number such that the number of possible non-zero code words is at least n, which is exactly 2^k -1 ≥n. Therefore, k=⌈log₂(n+1)⌉.Therefore, the answer should be the ceiling of log₂(n+1). So, for example, if n=7, then log₂(8)=3, so k=3. If n=8, log₂(9)≈3.17, so k=4.But let me check with n=4 again. According to the formula, k=3. Let's see if we can actually construct such a family.Using k=3, the characteristic vectors are 3-bit non-zero vectors. Assign each element a unique vector:1: 0012: 0103: 0114: 100Now, the subsets are A₁, A₂, A₃, where A_i corresponds to the i-th bit. So:A₁ = {4} (since only element 4 has the first bit set)A₂ = {2,3} (elements with second bit set)A₃ = {1,3} (elements with third bit set)Check covering: All elements are in at least one subset. 1 is in A₃, 2 in A₂, 3 in A₂ and A₃, 4 in A₁. Good.Check separating:- 1 and 2: A₂ contains 2, not 1; A₃ contains 1, not 2. So both subsets separate them.- 1 and 3: A₂ contains 3, not 1. So A₂ separates them.- 1 and 4: A₁ contains 4, not 1; A₃ contains 1, not 4. So multiple subsets separate.- 2 and 3: A₃ contains 3, not 2. So A₃ separates them.- 2 and 4: A₁ contains 4, not 2; A₂ contains 2, not 4. So multiple subsets separate.- 3 and 4: A₁ contains 4, not 3; A₂ contains 3, not 4. So multiple subsets separate.All pairs are separated. So yes, k=3 works for n=4, which matches the formula.Another way to think is using information theory: to distinguish n elements, you need log₂n bits. But here, since each "bit" corresponds to a subset, and each element must be in at least one subset (so no all-zero code word), the number of required bits is log₂(n+1). Wait, but why n+1?Because with k bits, you can have 2^k code words, but excluding the all-zero code word, you have 2^k -1 code words. Therefore, to cover n elements, you need 2^k -1 ≥n. Hence, k ≥ log₂(n+1).Therefore, the minimal k is indeed the ceiling of log₂(n+1).Thus, the answer is ⌈log₂(n+1)⌉.But let me check another example, say n=5.log₂(6)≈2.58, so k=3. Let's see:Characteristic vectors for 5 elements:001, 010, 011, 100, 101.Check if these are all unique and non-zero.Subsets:A₁: 100, 101 → elements 4 and 5.A₂: 010, 011 → elements 2 and 3.A₃: 001, 011, 101 → elements 1,3,5.Wait, but element 3 is in A₂ and A₃, element 5 is in A₁ and A₃.Covering: All elements are in at least one subset. Good.Separating:Check pair 1 and 2: A₂ contains 2, not 1; A₃ contains 1, not 2.Pair 1 and 3: A₂ contains 3, not 1.Pair 1 and 4: A₁ contains 4, not 1.Pair 1 and 5: A₁ contains 5, not 1; but wait, element 5 is in A₁ and A₃. Wait, element 1 is in A₃ only. Wait, in subset A₁, 5 is present, 1 is not. So A₁ separates 1 and 5.Pair 2 and 3: A₃ contains 3, not 2.Pair 2 and 4: A₁ contains 4, not 2.Pair 2 and 5: A₁ contains 5, not 2.Pair 3 and 4: A₁ contains 4, not 3.Pair 3 and 5: A₁ contains 5, not 3; A₂ contains 3, not 5. Either works.Pair 4 and 5: A₃ contains 5, not 4. So A₃ separates them.All pairs are separated. So k=3 works for n=5.Therefore, the formula holds. Therefore, the minimal k is ⌈log₂(n+1)⌉.But let me see if there's any possible way to get a lower k. Suppose for n=4, could we do it with k=2? Let's see.For k=2, we have subsets A₁ and A₂. The possible non-zero characteristic vectors are 01, 10, 11. So three vectors. But n=4, so we need four distinct vectors. Impossible. So k=2 is insufficient. Therefore, k=3 is necessary.Similarly, for n=3, using k=2 gives us three non-zero vectors: 01, 10, 11. Wait, but 01, 10, 11: three vectors. So n=3 is possible with k=2, which matches our earlier example.Therefore, the formula seems consistent.Hence, the minimal k is the smallest integer k such that 2^k -1 ≥n, which is equivalent to k=⌈log₂(n+1)⌉.Therefore, the answer is the ceiling of log base 2 of (n+1).Expressed in LaTeX, that's boxed{lceil log_2 (n+1) rceil}.Final AnswerThe minimum ( k ) is (boxed{lceil log_2 (n+1) rceil}).

✅:To find the minimum ( k ) such that there exists a family of subsets ( A_1, A_2, ldots, A_k ) of ( X = {1, 2, ldots, n} ) that is both covering and separating, we need to consider the following:1. Covering: The union of all subsets ( A_1, A_2, ldots, A_k ) must be ( X ). This means every element in ( X ) must be included in at least one subset.2. Separating: For any pair of distinct elements ( i ) and ( j ) in ( X ), there must be at least one subset that contains exactly one of ( i ) or ( j ).To solve this, we can model the problem using binary representation. Each element in ( X ) can be associated with a binary vector (characteristic vector) of length ( k ), where each bit indicates membership in a corresponding subset. The covering condition requires that each vector has at least one '1', and the separating condition requires that all vectors are distinct.The number of distinct non-zero binary vectors of length ( k ) is ( 2^k - 1 ). Therefore, we need ( 2^k - 1 geq n ). Solving for ( k ), we get:[ k geq lceil log_2 (n + 1) rceil ]This is because ( lceil log_2 (n + 1) rceil ) is the smallest integer ( k ) such that ( 2^k - 1 ) is at least ( n ).Thus, the minimum ( k ) such that there exists a family of subsets that is both covering and separating is:[ boxed{lceil log_2 (n+1) rceil} ]

🔑:1. Define the problem context: We need to cover the set X = {1, 2, ldots, n} using subsets A_1, A_2, ldots, A_k such that every pair of elements {i, j} from X can be found in at least one of these subsets. This family of subsets is said to be separable if no two columns (representing the elements of each subset) are identical.2. Binary representation of subsets: Treat each subset A_1, A_2, ldots, A_k as a row of a k times n matrix where each element of the matrix is either 0 or 1. Each column of the matrix represents whether an element i from X is in a particular subset A_j. 3. Conditions for covering: For the subsets to be covering, every pair {i, j} in X must appear together in at least one subset. Therefore, each column in the matrix must have at least one 1.4. Conditions for separable subsets: For the subsets to be separable, no two columns can be exactly the same. This means each binary vector (each column) must be unique, and non-trivial in the sense that it is not entirely zeros.5. Maximum number of unique k-length binary vectors: Since each element in the matrix is either 0 or 1, the total number of different columns (binary vectors of length k) that can be formed (excluding the all-zero vector) is 2^k - 1. This is the maximum number of unique binary vectors if we avoid all-zero vectors.6. Inequality to cover n elements: To cover n elements with unique vectors: [ 2^k - 1 geq n ] Solving for k, we get: [ 2^k geq n + 1 implies k geq log_2(n + 1) ]7. Ceiling function to ensure integer k: Since k needs to be an integer, we use the ceiling function: [ k geq lceil log_2(n + 1) rceil ]8. Verification of the bound: We can confirm this bound is tight by checking that k - 1 is insufficient: [ 2^{k-1} < n + 1 < 2^k ] This also implies: [ boxed{k = leftlceil log_2(n + 1) rightrceil} ]9. Conclusion: The smallest integer k that ensures the subsets are both covering and separable is: [ boxed{k = leftlceil log_2(n + 1) rightrceil} ]

🧠:Alright, so I need to solve this geometry problem where a triangle has one side of length √2, and the two angles adjacent to that side are 75° and 60°. The question is asking for the length of the segment connecting the bases of the altitudes dropped from the vertices of these angles. Hmm, let me visualize this first.First, let me sketch the triangle in my mind. Let's denote the triangle as triangle ABC, where side BC is √2. The angles at B and C are 75° and 60°, respectively. Wait, actually, the problem says the angles adjacent to the side √2 are 75° and 60°. So if the side is between the two angles, that would mean the two angles adjacent to it are at the two ends of the side. So maybe the side is between the two angles. Let me confirm: in a triangle, the angles adjacent to a side are the two angles at each end of that side. So if the side is, say, BC, then the angles at B and C are adjacent to side BC. So in this case, side BC is √2, angle at B is 75°, angle at C is 60°, which would make angle at A equal to 180° - 75° - 60° = 45°. Is that right? Wait, yes, the sum of angles in a triangle is 180°, so angle A must be 45°.So triangle ABC has BC = √2, angle B = 75°, angle C = 60°, angle A = 45°. Now, we need to drop altitudes from the vertices of the angles adjacent to BC, which are B and C. Wait, the problem says "the segment connecting the bases of the altitudes dropped from the vertices of these angles." The vertices of the angles adjacent to side BC are B and C themselves. So we need to drop altitudes from B and C onto the opposite sides, and then find the distance between the bases of these two altitudes.Wait, in a triangle, an altitude from a vertex is perpendicular to the opposite side. So altitude from B would be dropped onto side AC, and the altitude from C would be dropped onto side AB. Then the bases of these altitudes are points on AC and AB respectively. The segment connecting these two bases is the one we need to find. Hmm, okay. Let me make sure I got that right.Alternatively, maybe the problem is referring to dropping the altitudes from B and C onto the side BC? Wait, but if we drop an altitude from B, since B is already on BC, the altitude from B to BC would just be the point B itself. That doesn't make sense. Similarly for C. So no, the altitude from a vertex is to the opposite side. So from B, the altitude is to side AC, and from C, the altitude is to side AB. Then the bases of these altitudes are two points on AC and AB, and the segment between them is what we need to compute.Alright, so how do I find the coordinates of these bases? Maybe coordinate geometry would help here. Let me assign coordinates to the triangle to make this easier.Let me place point B at the origin (0,0). Then side BC is of length √2. Let me orient the triangle such that side BC is along the x-axis, so point C is at (√2, 0). Then point A is somewhere in the plane, forming angles of 75° at B and 60° at C. Wait, but angle at B is 75°, angle at C is 60°, so angle at A is 45°. Let me confirm that.Yes, 180 - 75 - 60 = 45. So angle at A is 45°. Now, to find the coordinates of point A, we can use the Law of Sines or the Law of Cosines. Let's see. Maybe Law of Sines would be helpful here.Law of Sines states that a / sin A = b / sin B = c / sin C. Here, sides are opposite the angles. So in triangle ABC:- Side BC is opposite angle A, which is 45°, so BC = a = √2.- Side AC is opposite angle B, which is 75°, so AC = b.- Side AB is opposite angle C, which is 60°, so AB = c.So by Law of Sines:a / sin A = b / sin B = c / sin CTherefore,√2 / sin 45° = b / sin 75° = c / sin 60°Compute sin 45°, which is √2/2.So √2 / (√2/2) = 2. Therefore, the common ratio is 2.Thus, b = 2 * sin 75°, and c = 2 * sin 60°.Compute sin 75° and sin 60°.Sin 75° is sin(45° + 30°) = sin45 cos30 + cos45 sin30 = (√2/2)(√3/2) + (√2/2)(1/2) = √6/4 + √2/4 = (√6 + √2)/4.Sin 60° is √3/2.Therefore,b = 2 * (√6 + √2)/4 = (√6 + √2)/2 ≈ value, but we can keep it exact.c = 2 * (√3/2) = √3.So sides AC = (√6 + √2)/2 and AB = √3.Now, let's find coordinates of point A. Since we've placed point B at (0,0) and point C at (√2, 0), we need to find coordinates (x,y) such that distance from A to B is c = √3, distance from A to C is b = (√6 + √2)/2, and the angles at B and C are 75° and 60° respectively.Alternatively, maybe it's easier to use coordinate geometry by setting up the triangle with BC on the x-axis and computing coordinates based on angles.Alternatively, let's use vectors or trigonometry to find coordinates of A.Since angle at B is 75°, and side BC is along the x-axis from B(0,0) to C(√2,0), then point A can be located by moving at an angle of 75° from point B and some distance. Wait, but the length of AB is c = √3. Wait, AB is the side opposite angle C (60°), which we found to be √3.So from point B(0,0), moving at an angle of 75° (since angle at B is 75°, the angle between BA and BC is 75°), the coordinates of A can be determined.Wait, maybe better to parametrize.Let me consider that from point B(0,0), side BA has length c = √3, and makes an angle of 75° with BC (which is along the x-axis). Therefore, coordinates of A can be found using trigonometry:x-coordinate: BA * cos(75°) = √3 * cos(75°)y-coordinate: BA * sin(75°) = √3 * sin(75°)Similarly, from point C(√2,0), angle at C is 60°, so the side CA makes an angle of 180° - 60° = 120° with the positive x-axis. Wait, is that correct?Alternatively, since angle at C is 60°, the angle between side CB and CA is 60°. Since CB is along the x-axis from C to B (leftwards), then the angle between CA and the x-axis is 180° - 60° = 120° from the positive x-axis.But perhaps this is getting complicated. Maybe better to use coordinates based on BA and CA.But maybe there's a better approach. Let me think.Alternatively, using coordinates:Let me place point B at (0,0), point C at (√2,0). Then point A is somewhere in the plane. The coordinates of A must satisfy the distances AB = √3 and AC = (√6 + √2)/2.Alternatively, since AB = √3 and angle at B is 75°, we can compute coordinates of A.From point B(0,0), moving at an angle of 75° above the x-axis (since angle between BA and BC is 75°), for a distance of √3. Therefore, coordinates of A would be:A_x = √3 * cos(75°)A_y = √3 * sin(75°)Similarly, from point C(√2,0), angle at C is 60°, so the angle between CA and CB is 60°. Since CB is along the negative x-axis from C, the direction of CA is 60° above the negative x-axis, which is 180° - 60° = 120° from the positive x-axis. Therefore, coordinates of A can also be expressed as:A_x = √2 + AC * cos(120°)A_y = 0 + AC * sin(120°)Since AC = (√6 + √2)/2, so:A_x = √2 + [(√6 + √2)/2] * cos(120°)A_y = [(√6 + √2)/2] * sin(120°)But these two expressions for A_x and A_y must be equal. Let me verify that.First, compute cos(75°) and sin(75°):cos(75°) = cos(45° + 30°) = cos45 cos30 - sin45 sin30 = (√2/2)(√3/2) - (√2/2)(1/2) = √6/4 - √2/4 = (√6 - √2)/4sin(75°) = sin(45° + 30°) = sin45 cos30 + cos45 sin30 = (√2/2)(√3/2) + (√2/2)(1/2) = √6/4 + √2/4 = (√6 + √2)/4Therefore, coordinates from point B:A_x = √3 * (√6 - √2)/4A_y = √3 * (√6 + √2)/4Compute A_x:√3*(√6 - √2)/4 = (√18 - √6)/4 = (3√2 - √6)/4Similarly, A_y:√3*(√6 + √2)/4 = (√18 + √6)/4 = (3√2 + √6)/4Now, compute coordinates from point C:First, cos(120°) = cos(180° - 60°) = -cos60° = -1/2sin(120°) = sin(180° - 60°) = sin60° = √3/2Therefore,A_x = √2 + [(√6 + √2)/2] * (-1/2) = √2 - (√6 + √2)/4 = (4√2 - √6 - √2)/4 = (3√2 - √6)/4A_y = [(√6 + √2)/2] * (√3/2) = (√6 + √2)√3 /4Let's compute A_y:(√6 + √2)√3 = √6*√3 + √2*√3 = √18 + √6 = 3√2 + √6Therefore, A_y = (3√2 + √6)/4So both methods give the same coordinates for A: ( (3√2 - √6)/4 , (3√2 + √6)/4 ). Great, so coordinates of A are established.Now, we need to find the bases of the altitudes from B and C.First, the altitude from B to AC. Let's call the foot of this altitude as D. Similarly, the altitude from C to AB, foot as E. Then, we need to find the distance between points D and E.So first, find coordinates of D and E.To find D: foot of altitude from B(0,0) to side AC.First, equation of side AC. Points A( (3√2 - √6)/4 , (3√2 + √6)/4 ) and C(√2, 0).Compute the slope of AC.Slope m_AC = [0 - (3√2 + √6)/4] / [√2 - (3√2 - √6)/4]First, compute numerator:- (3√2 + √6)/4Denominator:√2 - (3√2 - √6)/4 = (4√2 - 3√2 + √6)/4 = (√2 + √6)/4Therefore, slope m_AC = [ - (3√2 + √6)/4 ] / [ (√2 + √6)/4 ] = - (3√2 + √6) / (√2 + √6)Simplify this expression:Multiply numerator and denominator by (√2 - √6) to rationalize denominator.Denominator: (√2 + √6)(√2 - √6) = 2 - 6 = -4Numerator: - (3√2 + √6)(√2 - √6) = - [ 3√2*√2 - 3√2*√6 + √6*√2 - √6*√6 ]Compute term by term:3√2*√2 = 3*2 = 6-3√2*√6 = -3*√12 = -3*2√3 = -6√3√6*√2 = √12 = 2√3-√6*√6 = -6Therefore, numerator inside the brackets:6 -6√3 + 2√3 -6 = (6 -6) + (-6√3 + 2√3) = 0 -4√3 = -4√3Therefore, numerator is - (-4√3) = 4√3So slope m_AC = 4√3 / (-4) = -√3Therefore, the equation of AC is y - y1 = m_AC (x - x1). Let's use point C(√2, 0):y - 0 = -√3(x - √2)So equation of AC: y = -√3 x + √3*√2 = -√3 x + √6Now, we need the foot of the altitude from B(0,0) to AC. The altitude is perpendicular to AC, so its slope is the negative reciprocal of m_AC. Since m_AC = -√3, the slope of altitude BD is 1/√3.Equation of altitude BD: passes through B(0,0), slope 1/√3: y = (1/√3)xFind intersection point D between BD and AC.Set equations equal:(1/√3)x = -√3 x + √6Multiply both sides by √3 to eliminate denominator:x = -3x + √6*√3x + 3x = √184x = 3√2x = (3√2)/4Then y = (1/√3)*(3√2)/4 = (3√2)/(4√3) = (3√6)/12 = √6/4Therefore, coordinates of D are ( (3√2)/4 , √6/4 )Now, find foot of altitude from C(√2,0) to AB. Let's call this point E.First, find equation of AB. Points A( (3√2 - √6)/4 , (3√2 + √6)/4 ) and B(0,0). Let's compute the slope of AB.Slope m_AB = [ (3√2 + √6)/4 - 0 ] / [ (3√2 - √6)/4 - 0 ] = (3√2 + √6)/4 / (3√2 - √6)/4 = (3√2 + √6)/(3√2 - √6)Simplify this expression by rationalizing denominator:Multiply numerator and denominator by (3√2 + √6):Numerator: (3√2 + √6)^2Denominator: (3√2)^2 - (√6)^2 = 18 - 6 = 12Compute numerator:(3√2)^2 + 2*(3√2)*(√6) + (√6)^2 = 9*2 + 2*3*√12 + 6 = 18 + 6*2√3 + 6 = 24 + 12√3Therefore, m_AB = (24 + 12√3)/12 = 2 + √3So slope of AB is 2 + √3. Therefore, equation of AB is y = (2 + √3)x.Now, altitude from C(√2,0) to AB is perpendicular to AB. Therefore, slope of altitude CE is negative reciprocal of (2 + √3), which is -1/(2 + √3). Rationalize denominator:Multiply numerator and denominator by (2 - √3):-1*(2 - √3)/[(2 + √3)(2 - √3)] = -(2 - √3)/(4 - 3) = -(2 - √3)/1 = -2 + √3Therefore, slope of CE is √3 - 2.Equation of altitude CE: passes through C(√2,0), slope (√3 - 2):y - 0 = (√3 - 2)(x - √2)So y = (√3 - 2)(x - √2)Find intersection point E between CE and AB.Equation of AB: y = (2 + √3)xEquation of CE: y = (√3 - 2)(x - √2)Set equal:(2 + √3)x = (√3 - 2)(x - √2)Expand right-hand side:(√3 - 2)x - (√3 - 2)√2Bring all terms to left-hand side:(2 + √3)x - (√3 - 2)x + (√3 - 2)√2 = 0Factor x:[ (2 + √3) - (√3 - 2) ]x + (√3 - 2)√2 = 0Compute coefficient of x:2 + √3 - √3 + 2 = 4Thus,4x + (√3 - 2)√2 = 0Solve for x:4x = - (√3 - 2)√2x = - (√3 - 2)√2 / 4 = (2 - √3)√2 / 4Simplify:x = (2√2 - √3*√2)/4 = (2√2 - √6)/4Then y = (2 + √3)x = (2 + √3)*(2√2 - √6)/4Multiply numerator:(2 + √3)(2√2 - √6) = 2*2√2 - 2*√6 + √3*2√2 - √3*√6Compute term by term:2*2√2 = 4√2-2*√6 = -2√6√3*2√2 = 2√6-√3*√6 = -√18 = -3√2Therefore, total:4√2 -2√6 + 2√6 -3√2 = (4√2 -3√2) + (-2√6 +2√6) = √2 + 0 = √2Therefore, y = √2 /4So coordinates of E are ( (2√2 - √6)/4 , √2/4 )Now, we have coordinates of D( (3√2)/4 , √6/4 ) and E( (2√2 - √6)/4 , √2/4 )Need to find the distance between D and E.Use distance formula:Distance DE = sqrt[ (x_D - x_E)^2 + (y_D - y_E)^2 ]Compute x_D - x_E:(3√2/4) - ( (2√2 - √6)/4 ) = (3√2 -2√2 + √6)/4 = (√2 + √6)/4Similarly, y_D - y_E:(√6/4 - √2/4 ) = (√6 - √2)/4Therefore, DE = sqrt[ ( (√2 + √6)/4 )^2 + ( (√6 - √2)/4 )^2 ]Compute squares:First term: ( (√2 + √6)^2 ) / 16 = (2 + 2√12 +6)/16 = (8 + 4√3)/16 = (2 + √3)/4Second term: ( (√6 - √2)^2 ) / 16 = (6 - 2√12 +2)/16 = (8 - 4√3)/16 = (2 - √3)/4Therefore, DE = sqrt[ (2 + √3)/4 + (2 - √3)/4 ) ] = sqrt[ (4/4) ] = sqrt[1] = 1Wait, that's surprisingly simple. So despite all the complex coordinates, the distance DE is 1.But let me verify this calculation step by step.First, compute (√2 + √6)^2:(√2)^2 + 2*√2*√6 + (√6)^2 = 2 + 2√12 +6 = 2 + 6 + 2*2√3 = 8 + 4√3Similarly, (√6 - √2)^2:(√6)^2 - 2*√6*√2 + (√2)^2 =6 - 2√12 +2= 8 -4√3So then:[ (8 +4√3) + (8 -4√3) ] / 16 = (16)/16 =1Therefore sqrt(1) =1.So yes, DE=1.Wow, so after a long process, the answer is 1. That seems clean, so probably correct. Let me check if there's an alternative way to see this.Alternatively, perhaps using trigonometry in the triangle without coordinates. Let me think.Given triangle ABC with BC=√2, angles at B=75°, C=60°, angle at A=45°.We drop altitudes from B and C to their opposite sides, getting D on AC and E on AB.We need DE.Alternatively, maybe using the formula for the distance between two feet of altitudes. Not sure, but perhaps properties of orthocentric system or something.Alternatively, recognizing that in triangle ABC, the two altitudes from B and C, their feet D and E, maybe triangle ADE has some properties? Not sure.Alternatively, perhaps in coordinate geometry, once we found coordinates of D and E, compute DE distance. Which we did and got 1.Alternatively, maybe using vectors. Let's see.But since coordinate geometry gave DE=1, and all steps check out, I think that's the answer.Therefore, the segment connecting the bases of the altitudes is 1.Final Answerboxed{1}

✅:Given a triangle with side ( BC = sqrt{2} ) and adjacent angles ( angle B = 75^circ ) and ( angle C = 60^circ ), we need to find the segment connecting the bases of the altitudes dropped from vertices ( B ) and ( C ).1. Determine the third angle: [ angle A = 180^circ - 75^circ - 60^circ = 45^circ ]2. Use the Law of Sines to find the lengths of sides ( AB ) and ( AC ): [ frac{sqrt{2}}{sin 45^circ} = frac{AB}{sin 60^circ} = frac{AC}{sin 75^circ} ] [ frac{sqrt{2}}{frac{sqrt{2}}{2}} = 2 implies AB = 2 sin 60^circ = sqrt{3}, quad AC = 2 sin 75^circ = frac{sqrt{6} + sqrt{2}}{2} ]3. Assign coordinates to the vertices: - ( B(0, 0) ) - ( C(sqrt{2}, 0) ) - ( A left( frac{3sqrt{2} - sqrt{6}}{4}, frac{3sqrt{2} + sqrt{6}}{4} right) )4. Find the equations of the altitudes: - Altitude from ( B ) to ( AC ): - Slope of ( AC ) is ( -sqrt{3} ) - Equation of ( AC ): ( y = -sqrt{3}x + sqrt{6} ) - Equation of altitude from ( B ): ( y = frac{1}{sqrt{3}}x ) - Intersection ( D left( frac{3sqrt{2}}{4}, frac{sqrt{6}}{4} right) ) - Altitude from ( C ) to ( AB ): - Slope of ( AB ) is ( 2 + sqrt{3} ) - Equation of ( AB ): ( y = (2 + sqrt{3})x ) - Equation of altitude from ( C ): ( y = (sqrt{3} - 2)(x - sqrt{2}) ) - Intersection ( E left( frac{2sqrt{2} - sqrt{6}}{4}, frac{sqrt{2}}{4} right) )5. Calculate the distance ( DE ): [ DE = sqrt{left( frac{3sqrt{2}}{4} - frac{2sqrt{2} - sqrt{6}}{4} right)^2 + left( frac{sqrt{6}}{4} - frac{sqrt{2}}{4} right)^2} ] [ = sqrt{left( frac{sqrt{2} + sqrt{6}}{4} right)^2 + left( frac{sqrt{6} - sqrt{2}}{4} right)^2} ] [ = sqrt{frac{(8 + 4sqrt{3}) + (8 - 4sqrt{3})}{16}} = sqrt{frac{16}{16}} = sqrt{1} = 1 ]The segment connecting the bases of the altitudes is (boxed{1}).

🔑:1. Establish the Given Data: We are given a triangle ( triangle ABC ) with the following: - ( BC = sqrt{2} ) - ( angle B = 75^circ ) - ( angle C = 60^circ )2. Determine the Remaining Angle: Since the sum of the angles in a triangle is ( 180^circ ): [ angle A = 180^circ - angle B - angle C = 180^circ - 75^circ - 60^circ = 45^circ ]3. Use of Heights in the Triangle: Let ( BM ) and ( CN ) be the altitudes from vertices ( B ) and ( C ) respectively to side ( AC ) and side ( AB ).4. Identifying Similar Triangles: Notice that triangle ( AMN ) is nested within ( triangle ABC ) where ( M ) and ( N ) are the feet of the perpendiculars from ( B ) and ( C ) respectively.5. Triangle Similarity Argument: Triangles ( triangle AMN ) and ( triangle ABC ) are similar with similarity ratio ( k ) given by the cosine of angle ( A ), because both triangles share angle ( A ): [ k = cos 45^circ = frac{1}{sqrt{2}} ]6. Calculating the Length of ( MN ): Since ( MN ) corresponds to side ( BC ) of the larger triangle ( triangle ABC ), the relationship is given by similarity ratio: [ MN = BC cdot cos 45^circ ] Substitute the known values: [ MN = sqrt{2} cdot frac{1}{sqrt{2}} = 1 ] Conclusion:The length of the segment ( MN ) that connects the bases of the altitudes is therefore:[boxed{1}]