💡Okay, so I have this problem here where there are n elements, labeled a1, a2, up to an. These elements are organized into n pairs, which are called P1, P2, up to Pn. The condition given is that if two pairs Pi and Pj share exactly one element, then that shared element must be either ai or aj, meaning that (ai, aj) is one of the pairs. I need to prove that every element is in exactly two of these pairs.Hmm, let me try to understand this better. So, we have n elements and n pairs. Each pair consists of two elements, so the total number of element occurrences in all pairs is 2n. If every element is in exactly two pairs, then the total number of element occurrences would be 2n, which matches. So, that seems consistent.But wait, the problem is not just about the total count. It's about the condition that if two pairs share exactly one element, then that shared element must be one of the specific elements ai or aj. So, this seems like a specific kind of pairing structure.Let me try to think of a small example. Suppose n=3. Then we have three elements: a1, a2, a3. We need to form three pairs. But wait, with three elements, how can we form three pairs? Each pair has two elements, so three pairs would require six element occurrences, but we only have three elements. That would mean each element is in two pairs. So, in this case, each element is in exactly two pairs, which satisfies the condition.But let's check the condition. If two pairs share exactly one element, then that element must be either ai or aj. Let's see. If we have pairs (a1, a2), (a1, a3), and (a2, a3). Now, let's check the condition. Take pairs (a1, a2) and (a1, a3). They share a1, which is ai where i=1. Similarly, (a1, a2) and (a2, a3) share a2, which is aj where j=2. And (a1, a3) and (a2, a3) share a3, which is ak where k=3. So, the condition is satisfied, and each element is in exactly two pairs.Okay, that works for n=3. Let's try n=4. So, we have four elements: a1, a2, a3, a4. We need to form four pairs. Each pair has two elements, so total element occurrences are eight. If each element is in exactly two pairs, that's 4 elements * 2 = 8, which matches.Let me try to construct such pairs. Maybe (a1, a2), (a1, a3), (a2, a4), (a3, a4). Let's check the condition. Take pairs (a1, a2) and (a1, a3). They share a1, which is ai where i=1. Good. Pairs (a1, a2) and (a2, a4) share a2, which is aj where j=2. Good. Pairs (a1, a2) and (a3, a4) share no elements, which is fine because the condition only applies when they share exactly one element. Similarly, pairs (a1, a3) and (a2, a4) share no elements. Pairs (a1, a3) and (a3, a4) share a3, which is ak where k=3. Good. Pairs (a2, a4) and (a3, a4) share a4, which is al where l=4. Good.So, in this case, each element is in exactly two pairs, and the condition is satisfied. So, it seems to hold for n=4 as well.But wait, is this always the case? Let me think about n=2. We have two elements, a1 and a2. We need to form two pairs. But with two elements, we can only form one pair: (a1, a2). But we need two pairs, which is impossible because we only have two elements. So, n=2 doesn't make sense here because we can't form two pairs with two elements. So, maybe n needs to be at least 3.But the problem statement says n elements organized into n pairs. So, for n=2, it's impossible, but for n>=3, it's possible. So, maybe the problem assumes n>=3.Anyway, back to the general case. I need to prove that every element is in exactly two pairs. So, let's think about how to approach this.Maybe I can use double counting or some combinatorial argument. Let's consider the number of pairs and the number of elements.We have n pairs, each containing two elements, so total element occurrences are 2n. If each element is in exactly two pairs, then total element occurrences are 2n, which matches. So, that suggests that if we can show that no element is in more than two pairs and no element is in fewer than two pairs, then we are done.So, first, let's show that no element is in more than two pairs. Suppose, for contradiction, that some element, say a1, is in three or more pairs. Let's say a1 is in pairs P1, P2, and P3. Now, according to the condition, if two pairs share exactly one element, that element must be ai or aj for some i,j.But if a1 is in P1, P2, and P3, then consider pairs P1 and P2. They share a1, which is a1, so that's fine. Similarly, P1 and P3 share a1, which is a1, so that's fine. But what about P2 and P3? They also share a1, which is a1. But according to the condition, if two pairs share exactly one element, that element must be ai or aj. So, in this case, the shared element is a1, which is ai where i=1. So, that's okay.Wait, but does this lead to a contradiction? Maybe not directly. Let me think differently.If a1 is in three pairs, say P1, P2, P3, then each of these pairs must contain a1 and another element. Let's say P1 = (a1, a2), P2 = (a1, a3), P3 = (a1, a4). Now, consider pairs P1 and P2. They share a1, which is a1, so that's fine. Similarly, P1 and P3 share a1, which is a1, fine. P2 and P3 share a1, which is a1, fine.But now, what about other pairs? For example, if there is a pair (a2, a3), then P1 and (a2, a3) share a2, which is a2, so that's fine. Similarly, P2 and (a2, a3) share a3, which is a3, fine. But wait, in this case, a2 is in P1 and (a2, a3), so a2 is in two pairs. Similarly, a3 is in P2 and (a2, a3), so a3 is in two pairs. a4 is only in P3, so a4 is in one pair. But we need each element to be in exactly two pairs, so a4 is only in one pair, which violates the condition.Wait, but in this case, a4 is only in P3, which is (a1, a4). So, a4 is in one pair, which is not allowed because we need each element to be in exactly two pairs. So, this leads to a contradiction because a4 is only in one pair.Therefore, our assumption that a1 is in three pairs leads to a contradiction because it forces another element to be in only one pair. Hence, no element can be in more than two pairs.Now, let's show that no element is in fewer than two pairs. Suppose, for contradiction, that some element, say a1, is in only one pair. Then, that pair is, say, (a1, a2). Now, consider another pair that contains a2. Since a2 must be in two pairs, it must be in another pair, say (a2, a3). Now, consider the pair (a1, a2) and (a2, a3). They share a2, which is a2, so that's fine.But what about the pair (a1, a2) and any other pair that doesn't contain a1 or a2? For example, suppose there is a pair (a3, a4). Then, (a1, a2) and (a3, a4) share no elements, which is fine. But what about the pair (a1, a3)? If such a pair exists, then (a1, a2) and (a1, a3) share a1, which is a1, so that's fine. But if a1 is only in one pair, then (a1, a3) cannot exist because a1 is only in (a1, a2). So, in this case, a1 is only in one pair, which might not necessarily lead to a contradiction, but let's see.Wait, but if a1 is only in one pair, then the total number of element occurrences is 2n. If a1 is in one pair, then the remaining 2n - 2 element occurrences must be distributed among the other n-1 elements. So, the average number of pairs per element for the remaining elements is (2n - 2)/(n - 1) = 2. So, on average, each of the remaining elements is in two pairs. But since a1 is only in one pair, some element must be in more than two pairs to compensate, which we already showed is impossible.Therefore, our assumption that a1 is in only one pair leads to a contradiction because it forces another element to be in more than two pairs, which is impossible. Hence, no element can be in fewer than two pairs.Thus, every element must be in exactly two pairs.

💡Okay, so I have this problem where I need to find the value of ( c ) in the ordered 6-tuple ( (a, b, c, d, e, f) ) that minimizes the sum ( a + b + c + d + e + f ), given the equation:[a + 2b + 6c + 30d + 210e + 2310f = 2^{15}]First, I need to understand the problem. It's about finding non-negative integers ( a, b, c, d, e, f ) such that when you plug them into the equation above, it equals ( 2^{15} ), which is 32768. The goal is to find the tuple where the sum ( a + b + c + d + e + f ) is as small as possible, and specifically, determine what ( c ) is in that tuple.I remember that when trying to minimize the sum of variables subject to a linear equation, a common strategy is to use the greedy algorithm. This means starting with the largest coefficients and assigning as much as possible to them to reduce the total number of variables needed.So, let's list the coefficients in order from largest to smallest:- 2310 (for ( f ))- 210 (for ( e ))- 30 (for ( d ))- 6 (for ( c ))- 2 (for ( b ))- 1 (for ( a ))Starting with the largest coefficient, 2310. I need to find the maximum number of times 2310 can fit into 32768.Calculating ( 32768 div 2310 ):- 2310 times 14 is 32340.- 2310 times 15 is 34650, which is too big because it exceeds 32768.So, ( f = 14 ). Subtracting 32340 from 32768 gives a remainder of 428.Now, moving to the next largest coefficient, which is 210 for ( e ). Let's see how many times 210 can fit into 428.Calculating ( 428 div 210 ):- 210 times 2 is 420.- 210 times 3 is 630, which is too big.So, ( e = 2 ). Subtracting 420 from 428 leaves a remainder of 8.Next, the coefficient for ( d ) is 30. Let's see how many times 30 can fit into 8. Since 30 is larger than 8, ( d = 0 ).Moving on to the coefficient for ( c ), which is 6. How many times does 6 fit into 8?Calculating ( 8 div 6 ):- 6 times 1 is 6.- 6 times 2 is 12, which is too big.So, ( c = 1 ). Subtracting 6 from 8 leaves a remainder of 2.Now, the coefficient for ( b ) is 2. How many times does 2 fit into 2?Calculating ( 2 div 2 ):- 2 times 1 is 2.So, ( b = 1 ). Subtracting 2 from 2 leaves a remainder of 0.Finally, the coefficient for ( a ) is 1. Since the remainder is 0, ( a = 0 ).Putting it all together, the tuple is ( (a, b, c, d, e, f) = (0, 1, 1, 0, 2, 14) ).To verify, let's plug these back into the original equation:[0 + 2(1) + 6(1) + 30(0) + 210(2) + 2310(14)]Calculating each term:- ( 0 )- ( 2 times 1 = 2 )- ( 6 times 1 = 6 )- ( 30 times 0 = 0 )- ( 210 times 2 = 420 )- ( 2310 times 14 = 32340 )Adding them up:[0 + 2 + 6 + 0 + 420 + 32340 = 32768]Which matches ( 2^{15} ), so that's correct.Now, checking the sum ( a + b + c + d + e + f = 0 + 1 + 1 + 0 + 2 + 14 = 18 ).I should make sure that this is indeed the minimal sum. Let's see if there's a way to get a smaller sum by adjusting the variables.For example, if I reduce ( f ) by 1, making ( f = 13 ), then the total contribution from ( f ) would be ( 2310 times 13 = 30030 ). The remaining amount would be ( 32768 - 30030 = 2738 ).Then, moving to ( e ): ( 2738 div 210 approx 13.04 ), so ( e = 13 ). Then, ( 210 times 13 = 2730 ), leaving a remainder of 8.Then, ( d = 0 ), ( c = 1 ), ( b = 1 ), ( a = 0 ). So the tuple would be ( (0, 1, 1, 0, 13, 13) ), with a sum of ( 0 + 1 + 1 + 0 + 13 + 13 = 28 ), which is larger than 18. So that's worse.Alternatively, if I try to adjust ( e ) differently, but I think the initial approach gives the minimal sum because we're using the largest coefficients as much as possible, which reduces the number of variables needed.Therefore, the value of ( c ) in the tuple that minimizes the sum is 1.**Final Answer**boxed{1}

💡Okay, so I have this problem where there are four natural numbers a, b, c, d. They are pairwise coprime, which means that any two of them share no common divisors other than 1. They satisfy the equation ab + cd = ac - 10bd. I need to prove that among these four numbers, there are three such that one is the sum of the other two.Alright, let's start by writing down the given equation:ab + cd = ac - 10bdHmm, maybe I can rearrange this equation to see if I can find a relationship between the variables. Let me try moving all terms to one side:ab + cd - ac + 10bd = 0Hmm, let's see. Maybe I can factor some terms here. Let me group terms with similar variables:ab - ac + cd + 10bd = 0Looking at ab - ac, I can factor out an a:a(b - c) + cd + 10bd = 0Now, looking at cd + 10bd, I can factor out a d:a(b - c) + d(c + 10b) = 0So, now the equation is:a(b - c) + d(c + 10b) = 0Hmm, that's interesting. Let me write this as:a(b - c) = -d(c + 10b)So, a(b - c) is equal to negative d times (c + 10b). Since a, b, c, d are natural numbers, the left side is a times (b - c), and the right side is negative d times (c + 10b). Wait, since a, b, c, d are natural numbers, and they are pairwise coprime, that might help. Let me think about the implications of this equation.First, since a, b, c, d are pairwise coprime, a and d must be coprime as well. Similarly, a and (c + 10b) must be coprime because a is coprime with b and c. Similarly, d is coprime with (c + 10b) because d is coprime with b and c.Wait, is that necessarily true? Let me check. If a is coprime with b and c, then a is coprime with any linear combination of b and c, right? So, a is coprime with (c + 10b). Similarly, d is coprime with b and c, so d is coprime with (c + 10b) as well.So, from the equation a(b - c) = -d(c + 10b), since a and (c + 10b) are coprime, a must divide (b - c). Similarly, d must divide (b - c) as well because d is on the right side.Wait, but a and d are coprime, so if both a and d divide (b - c), then their product ad must divide (b - c). But since a and d are coprime, ad divides (b - c). But since a and d are positive integers, this would mean that (b - c) is a multiple of ad.But (b - c) is equal to (b - c), which is an integer. So, let me write:b - c = k * ad, where k is some integer.But since b and c are natural numbers, (b - c) could be positive or negative. However, since a, d are positive, k must be an integer such that b - c is positive or negative accordingly.Wait, but let's think about the original equation:a(b - c) = -d(c + 10b)So, a(b - c) is equal to negative d(c + 10b). Since a, b, c, d are natural numbers, the right side is negative because of the negative sign. Therefore, the left side must also be negative, which implies that (b - c) is negative. So, b - c < 0, which means c > b.So, c is greater than b. Therefore, b - c is negative, so k must be negative as well because a and d are positive.So, let me write:b - c = -k * ad, where k is a positive integer.Therefore, c - b = k * ad.So, c = b + k * ad.Alright, so now I have c expressed in terms of b, a, d, and k.Now, let's substitute this back into the equation.From the equation:a(b - c) = -d(c + 10b)We have c = b + k * ad, so let's substitute c:a(b - (b + k * ad)) = -d((b + k * ad) + 10b)Simplify the left side:a(b - b - k * ad) = a(-k * ad) = -a^2 d kSimplify the right side:-d(b + k * ad + 10b) = -d(11b + k * ad) = -11b d - k * a d^2So, now we have:-a^2 d k = -11b d - k * a d^2Let me multiply both sides by -1 to make it positive:a^2 d k = 11b d + k * a d^2Now, let's factor out d on both sides:d(a^2 k) = d(11b + k a d)Since d is a natural number and non-zero, we can divide both sides by d:a^2 k = 11b + k a dNow, let's rearrange this equation:a^2 k - k a d = 11bFactor out k a:k a (a - d) = 11bSo, now we have:k a (a - d) = 11bHmm, interesting. So, 11b is equal to k a (a - d). Since 11 is a prime number, this suggests that 11 divides the right-hand side. Since a, b, c, d are pairwise coprime, and 11 is prime, 11 must divide one of the factors on the right-hand side.So, 11 divides either k, a, or (a - d). Let's consider each case.Case 1: 11 divides k.If 11 divides k, then we can write k = 11 m, where m is a positive integer.Then, substituting back into the equation:11 m a (a - d) = 11bDivide both sides by 11:m a (a - d) = bSo, b = m a (a - d)But since a, b, c, d are pairwise coprime, and b is expressed as m a (a - d), this suggests that a must divide b. But since a and b are coprime, the only way this can happen is if a = 1.So, a = 1.Then, b = m * 1 * (1 - d) = m (1 - d)But b is a natural number, so m (1 - d) must be positive. Therefore, 1 - d > 0, which implies d < 1. But d is a natural number, so d must be 0, which is not possible because d is a natural number (assuming natural numbers start at 1). Therefore, this case leads to a contradiction.Therefore, Case 1 is not possible.Case 2: 11 divides a.If 11 divides a, then a = 11 n, where n is a positive integer.Substituting back into the equation:k * 11 n (11 n - d) = 11bDivide both sides by 11:k n (11 n - d) = bSo, b = k n (11 n - d)Again, since a, b, c, d are pairwise coprime, and a = 11 n, b is expressed as k n (11 n - d). Since a and b are coprime, n must be 1, because otherwise, n would be a common divisor of a and b.So, n = 1.Therefore, a = 11 * 1 = 11.Then, b = k * 1 * (11 * 1 - d) = k (11 - d)Since b is a natural number, 11 - d must be positive, so d < 11.Also, since d is a natural number, d can be from 1 to 10.But let's remember that a, b, c, d are pairwise coprime. Since a = 11, which is prime, b must not be a multiple of 11. But b = k (11 - d). Since 11 - d is less than 11, and k is a positive integer, b is a multiple of k times (11 - d). However, since a = 11 and b must be coprime with a, (11 - d) must be such that it doesn't introduce any common factors with 11. Since 11 is prime, as long as (11 - d) is not 11, which it isn't because d is at least 1, so (11 - d) is between 1 and 10, which are all coprime with 11. Therefore, b is k times (11 - d), and since a = 11, b must be coprime with 11, so k must be 1. Otherwise, if k > 1, b would have factors other than 11, but since a = 11, b must be coprime with 11, so k must be 1.Therefore, k = 1.Thus, b = 11 - d.So, b = 11 - d.Therefore, d = 11 - b.Since d is a natural number, and b is a natural number less than 11 (since d = 11 - b must be positive), b can be from 1 to 10.Now, let's recall that c = b + k * a * d.Earlier, we had c = b + k * a * d, and in this case, k = 1, a = 11, d = 11 - b.So, c = b + 1 * 11 * (11 - b) = b + 11(11 - b) = b + 121 - 11b = 121 - 10bSo, c = 121 - 10bNow, since c must be a natural number, 121 - 10b must be positive, so 10b < 121, which implies b < 12.1. Since b is a natural number less than 11, as we saw earlier, this is satisfied.Now, let's check if c is positive:c = 121 - 10b > 0Since b is at least 1, c = 121 - 10(1) = 111, which is positive.Similarly, for b = 10, c = 121 - 100 = 21, which is still positive.So, c is positive for all b from 1 to 10.Now, let's check the pairwise coprimality.We have a = 11, b, c = 121 - 10b, d = 11 - b.We need to ensure that a, b, c, d are pairwise coprime.First, a = 11 is prime, so it's coprime with b, c, d as long as b, c, d are not multiples of 11. Since b is from 1 to 10, it's not a multiple of 11. c = 121 - 10b. Let's see if c can be a multiple of 11.c = 121 - 10b121 is 11^2, so 121 ≡ 0 mod 11.10b ≡ (-1)b mod 11.So, c ≡ 0 - (-1)b ≡ b mod 11.So, c ≡ b mod 11.Since b is from 1 to 10, c ≡ b mod 11, which is non-zero, so c is not a multiple of 11. Therefore, a and c are coprime.Similarly, d = 11 - b. Since b is from 1 to 10, d is from 1 to 10, so d is not a multiple of 11. Therefore, a and d are coprime.Now, let's check pairwise between b, c, d.We need to ensure that b and c are coprime, b and d are coprime, and c and d are coprime.First, b and d: d = 11 - b. So, gcd(b, d) = gcd(b, 11 - b) = gcd(b, 11). Since b is from 1 to 10, and 11 is prime, gcd(b, 11) = 1. Therefore, b and d are coprime.Next, b and c: c = 121 - 10b. Let's see if b and c are coprime.We have c = 121 - 10b.Let's compute gcd(b, c) = gcd(b, 121 - 10b).Using the Euclidean algorithm:gcd(b, 121 - 10b) = gcd(b, 121 - 10b - (-10)*b) = gcd(b, 121)Since 121 is 11^2, and b is from 1 to 10, which are coprime with 11, so gcd(b, 121) = 1. Therefore, gcd(b, c) = 1. So, b and c are coprime.Finally, c and d: c = 121 - 10b, d = 11 - b.Let's compute gcd(c, d) = gcd(121 - 10b, 11 - b).Let me denote d = 11 - b, so b = 11 - d.Substitute into c:c = 121 - 10(11 - d) = 121 - 110 + 10d = 11 + 10dSo, c = 11 + 10d.Therefore, gcd(c, d) = gcd(11 + 10d, d).Using the Euclidean algorithm:gcd(11 + 10d, d) = gcd(d, 11 + 10d - 10*d) = gcd(d, 11)Since d is from 1 to 10, and 11 is prime, gcd(d, 11) = 1. Therefore, gcd(c, d) = 1. So, c and d are coprime.Therefore, in this case, all four numbers a, b, c, d are pairwise coprime.Now, let's summarize what we have:a = 11b = 11 - dc = 121 - 10b = 11 + 10dd = 11 - bSo, we have expressions for all variables in terms of b or d.Now, the problem states that among a, b, c, d, there are three numbers such that one is the sum of the other two.Let me check the possible triples:1. a, b, c: Is one of them the sum of the other two?a = 11, b, c = 121 - 10bCheck if a = b + c: 11 = b + (121 - 10b) => 11 = 121 - 9b => 9b = 110 => b = 110/9, which is not an integer. So, no.Check if b = a + c: b = 11 + (121 - 10b) => b = 132 - 10b => 11b = 132 => b = 12. But b is from 1 to 10, so no.Check if c = a + b: 121 - 10b = 11 + b => 121 - 11 = 11b => 110 = 11b => b = 10.So, if b = 10, then c = 121 - 100 = 21, and a = 11, b = 10, c = 21.Indeed, 21 = 11 + 10. So, c = a + b.Therefore, in this case, when b = 10, c is the sum of a and b.Similarly, let's check other triples.2. a, b, d: a = 11, b, d = 11 - bCheck if a = b + d: 11 = b + (11 - b) => 11 = 11, which is always true. So, a = b + d.Wait, that's interesting. So, regardless of the value of b, a = 11 is equal to b + d, since d = 11 - b.So, a = b + d.Therefore, in this case, a is the sum of b and d.So, regardless of the value of b, as long as d = 11 - b, a = b + d.Therefore, in this case, a is the sum of b and d.So, in this case, we have two possibilities:Either c = a + b when b = 10, or a = b + d for any b.But wait, when b = 10, d = 1, so a = 11 = 10 + 1, which is also true.So, in that case, both a = b + d and c = a + b hold.But in other cases, only a = b + d holds.So, regardless, in this case, we have three numbers where one is the sum of the other two: a, b, d with a = b + d.Therefore, in this case, the conclusion holds.Case 3: 11 divides (a - d).If 11 divides (a - d), then a - d = 11 m, where m is a positive integer.So, a = d + 11 m.Substituting back into the equation:k a (a - d) = 11bBut a - d = 11 m, so:k a (11 m) = 11bDivide both sides by 11:k a m = bSo, b = k a mBut since a, b, c, d are pairwise coprime, and b = k a m, this implies that a must divide b. But since a and b are coprime, the only way this can happen is if a = 1.So, a = 1.Then, a = d + 11 m => 1 = d + 11 mBut d is a natural number, so d ≥ 1, and m is a positive integer, so 11 m ≥ 11. Therefore, 1 = d + 11 m ≥ 1 + 11 = 12, which is impossible. Therefore, this case leads to a contradiction.Therefore, Case 3 is not possible.So, the only possible case is Case 2, where 11 divides a, leading to a = 11, and we have a = b + d, or c = a + b when b = 10.Therefore, in all cases, among the four numbers a, b, c, d, there are three numbers such that one is the sum of the other two.Hence, the proof is complete.

💡First, I'll denote the student's marks in Mathematics as ( M ), in Physics as ( P ), and in Chemistry as ( C ).From the problem, I know that the score in Chemistry is 20 marks more than in Physics. This gives me the equation:[ C = P + 20 ]Next, the average marks scored in Mathematics and Chemistry together is 40. To find the total marks in these two subjects, I'll use the average formula:[ frac{M + C}{2} = 40 ]Multiplying both sides by 2 to eliminate the denominator:[ M + C = 80 ]Now, I'll substitute the expression for ( C ) from the first equation into this equation:[ M + (P + 20) = 80 ]Simplifying this:[ M + P + 20 = 80 ]Finally, I'll solve for the total marks in Mathematics and Physics:[ M + P = 80 - 20 ][ M + P = 60 ]So, the total marks obtained by the student in Mathematics and Physics is 60.

💡First, I need to calculate the mean of the vertical drops. I will add all the vertical drop values together and then divide by the number of roller coasters.Next, I will determine the median by arranging the vertical drop values in ascending order. Since there are six values, the median will be the average of the third and fourth values.Finally, I will find the positive difference between the mean and the median.

💡Okay, so I have this problem where I need to find a point P on the line l: 3x - y - 1 = 0 such that the difference between the distances from P to point A(4,1) and P to point B(0,4) is maximized. Hmm, that sounds a bit tricky, but let me try to break it down.First, I remember that when dealing with distances and lines, sometimes reflecting points can help. I think the idea is to use the reflection of one of the points across the line l to simplify the problem. But I'm not entirely sure how that works, so maybe I should look it up or recall some geometry concepts.Wait, I think I remember something about the shortest path involving reflections. Like, if you have a point and you want the shortest path that reflects off a line, you can reflect the point across the line and then find the straight line distance. Maybe that's related here?So, if I reflect point B across the line l, I can get a new point B'. Then, the distance from P to B would be the same as the distance from P to B' because P is on the line l. That might help because then I can express |PA| - |PB| as |PA| - |PB'|, which might be easier to maximize.Alright, let's try that. I need to find the reflection of point B(0,4) across the line l: 3x - y - 1 = 0. How do I find the reflection of a point across a line?I think the formula for reflecting a point (x, y) across the line ax + by + c = 0 is:x' = x - 2a(ax + by + c)/(a² + b²)y' = y - 2b(ax + by + c)/(a² + b²)Wait, is that right? Let me double-check. I think it's something like that, but maybe I should derive it.Alternatively, I can use the fact that the reflection point B' will satisfy two conditions: the midpoint between B and B' lies on the line l, and the line segment BB' is perpendicular to l.Okay, let's use that approach. Let me denote B' as (x', y'). The midpoint M between B(0,4) and B'(x', y') is ((0 + x')/2, (4 + y')/2). This midpoint must lie on the line l: 3x - y - 1 = 0. So plugging in M into l:3*(x'/2) - (4 + y')/2 - 1 = 0Simplify that:(3x'/2) - (4 + y')/2 - 1 = 0Multiply both sides by 2 to eliminate denominators:3x' - (4 + y') - 2 = 0Simplify:3x' - y' - 6 = 0So that's one equation: 3x' - y' = 6.Now, the line segment BB' is perpendicular to l. The slope of l is given by the equation 3x - y - 1 = 0, which can be rewritten as y = 3x - 1, so the slope is 3. Therefore, the slope of BB' should be the negative reciprocal, which is -1/3.The slope of BB' is (y' - 4)/(x' - 0) = (y' - 4)/x' = -1/3.So, (y' - 4)/x' = -1/3Multiply both sides by x':y' - 4 = (-1/3)x'So, y' = (-1/3)x' + 4Now, we have two equations:1. 3x' - y' = 62. y' = (-1/3)x' + 4Let's substitute equation 2 into equation 1:3x' - [(-1/3)x' + 4] = 6Simplify:3x' + (1/3)x' - 4 = 6Multiply through by 3 to eliminate fractions:9x' + x' - 12 = 18Combine like terms:10x' - 12 = 18Add 12 to both sides:10x' = 30Divide by 10:x' = 3Now, plug x' = 3 back into equation 2:y' = (-1/3)(3) + 4 = -1 + 4 = 3So, the reflection point B' is (3, 3). Okay, that wasn't too bad.Now, I need to find the point P on line l such that |PA| - |PB| is maximized. But since |PB| = |PB'| for any P on l, this becomes |PA| - |PB'|. So, I need to maximize |PA| - |PB'|.Hmm, how do I maximize the difference of two distances? I remember that the maximum difference occurs when P is aligned in a certain way with A and B'. Maybe when P is on the line connecting A and B'?Wait, actually, if I think about it, |PA| - |PB'| is maximized when P is as far as possible from A and as close as possible to B'. But since P has to lie on line l, I need to find the point on l that lies on the extension of AB' beyond B', such that |PA| - |PB'| is maximized.Alternatively, maybe it's when P is the intersection point of line AB' with line l. Because beyond that point, P would be moving away from A and towards B', but since we're subtracting |PB'|, maybe the maximum occurs at the intersection.Let me think. If I draw line AB', the point where it intersects l would be the point P where |PA| - |PB'| is maximized. Because beyond that point, moving P further along l would decrease |PA| and increase |PB'|, thus decreasing the difference.Wait, actually, if P is on AB', then |PA| - |PB'| would be equal to |AB'|, but since P is constrained to be on l, maybe the maximum occurs at the intersection point.Alternatively, perhaps the maximum is achieved when P is such that the line PA - PB' is aligned in a certain way.Wait, maybe I can use the triangle inequality here. The difference |PA| - |PB'| is less than or equal to |AB'|. So, the maximum possible value of |PA| - |PB'| is |AB'|, which occurs when P is on the line segment AB' extended beyond B'.But since P has to lie on l, the maximum occurs when P is the intersection point of AB' and l.So, let's find the equation of line AB' and then find its intersection with l.Point A is (4,1) and point B' is (3,3). So, the slope of AB' is (3 - 1)/(3 - 4) = 2/(-1) = -2.So, the equation of AB' is y - 1 = -2(x - 4). Let's write that out:y - 1 = -2x + 8So, y = -2x + 9Now, we need to find the intersection of this line with l: 3x - y - 1 = 0.Substitute y from AB' into l:3x - (-2x + 9) - 1 = 0Simplify:3x + 2x - 9 - 1 = 05x - 10 = 05x = 10x = 2Now, plug x = 2 back into the equation of AB':y = -2(2) + 9 = -4 + 9 = 5So, the intersection point P is (2,5).Therefore, the point P on line l that maximizes |PA| - |PB| is (2,5).Wait, let me double-check my steps to make sure I didn't make a mistake.1. Reflected B(0,4) across l:3x - y -1=0 to get B'(3,3). That seems correct.2. Found the equation of AB' as y = -2x + 9. That also seems correct.3. Found intersection with l:3x - y -1=0 by substituting y = -2x +9 into it, leading to x=2, y=5. That seems correct.So, I think my answer is correct. The coordinates of point P are (2,5).**Final Answer**The coordinates of point ( P ) are boxed{(2,5)}.