🧠:Okay, so there's this problem about two symmetrical dice, and we need to figure out if we can number their faces such that when we roll them, the sum can take on all values from 1 to 36, each with equal probability. Hmm, that seems interesting. Let me start breaking this down.First, let me recall what standard dice are like. A regular die has numbers 1 through 6, and when you roll two, the sums range from 2 to 12. Each sum has a different probability because there are multiple combinations that can result in the same sum. For example, a sum of 7 has the highest number of combinations (6 ways), so it's the most probable. The probabilities decrease as you move away from 7 towards the extremes (2 and 12). So in standard dice, the sums don't have equal probabilities.Now, this problem is asking the opposite: can we design two six-sided dice (symmetrical, so each die has the same numbers, right?) such that every sum from 1 to 36 is equally likely? Wait, 36? That's a big range. Let me check. If each die has 6 faces, then the number of possible outcomes when rolling two dice is 6*6 = 36. So, if we want 36 different sums, each occurring exactly once, that would require each sum from 1 to 36 to appear exactly once in the 36 possible outcomes. Is that the question?Wait, hold on. If we have two dice, each with numbers arranged such that every possible sum from 1 to 36 is achievable, and each sum occurs exactly once. That would mean that for each pair of numbers on the two dice, their sum is unique. Therefore, the total number of unique sums would need to be 36, which is exactly the number of possible outcomes. Therefore, each sum must be achieved exactly once. So, the question is whether such a numbering exists for two six-sided dice.But standard dice have numbers from 1 to 6, and their sums go up to 12. Here, we need sums up to 36. So, obviously, the numbers on each die must be larger. Let me think: if we need the maximum sum to be 36, then the maximum numbers on each die must add up to 36. Since the dice are symmetrical, does that mean both dice have the same numbers? Wait, the problem says "two symmetrical dice." Hmm, does symmetrical mean that each die is symmetrical in itself (i.e., opposite faces add up to the same number), or that the two dice are identical? The wording is a bit ambiguous.Wait, the problem says "two symmetrical dice." So, maybe each die is symmetrical, meaning that opposite faces add up to the same number. For example, in a standard die, opposite faces add up to 7. So, symmetrical here might refer to that property. But the problem is asking about writing numbers on their faces in such a way that the sum of the points when rolling takes on values 1, 2, ..., 36 with equal probabilities. So, the key is whether such numbering is possible, regardless of symmetry, but the dice themselves are symmetrical. Wait, maybe the dice are not required to be identical, just each die individually is symmetrical. Hmm.Wait, maybe I need to clarify. Let me check the original question again: "There are two symmetrical dice. Is it possible to write certain numbers on their faces in such a way that the sum of the points when rolling takes on values 1, 2, …, 36 with equal probabilities?" So, "symmetrical dice" probably refers to each die being symmetrical, i.e., opposite faces have numbers that add up to the same value. For standard dice, that's 7. So, perhaps here, each die must have opposite faces adding to a constant, but the numbers can be different as long as they follow that symmetry.Alternatively, maybe "symmetrical" here refers to the two dice being identical. But that seems less likely. Because if the dice were identical, then the maximum sum would be twice the maximum face value. If they are identical and each has numbers up to 18, then the maximum sum would be 36. But having numbers up to 18 on each die, but each die being symmetrical. But in that case, each die would need to have pairs of numbers that add up to, say, 19 (if following the standard die's symmetry of 1-6, 2-5, 3-4, adding to 7). Wait, but 18 is the maximum, so opposite faces would add up to 19? But 18 +1 = 19. So, perhaps if the numbers on each die are 1-18 arranged such that opposite faces add up to 19. Then, the two dice would each have numbers from 1 to 18, with opposite faces adding to 19, and when rolled, the sums could go from 2 (1+1) to 36 (18+18). But the problem is asking for sums from 1 to 36. Wait, sum of 1 is impossible with two dice, each having at least 1. So, maybe there's a misinterpretation here.Wait, hold on. The problem says "the sum of the points when rolling takes on values 1, 2, …, 36 with equal probabilities". But if each die has a minimum of 1, the minimum sum would be 2. So, sum of 1 is impossible. Therefore, perhaps there's a translation issue or a misstatement in the problem. Alternatively, maybe the dice can have zero or negative numbers? But dice typically have positive integers. The problem says "points", which usually are positive integers.Hmm, maybe the problem is mistyped? Or perhaps "symmetrical" allows for some other interpretation. Alternatively, maybe the dice aren't required to have numbers starting at 1? For example, if one die has numbers from 0 to 5 and the other from 1 to 6, then the minimum sum would be 1. But the problem says "symmetrical dice", which might require that each die's numbers are arranged symmetrically. For instance, if a die has 0,1,2,3,4,5, then its opposite faces would add up to 5 (0+5, 1+4, 2+3). But I don't know if that's considered symmetrical. The problem doesn't specify, but in standard terms, symmetry in dice usually refers to opposite faces adding to the same total.So, if we allow zero on a die, then perhaps we can get a minimum sum of 0+1=1? But the problem is about two symmetrical dice. If both dice are symmetrical, but one has numbers starting at 0, then perhaps that's allowed. But I need to check if that's permissible.Wait, the problem says "write certain numbers on their faces". It doesn't specify that the numbers have to start at 1 or be positive. So maybe zero is allowed. If that's the case, then perhaps one die has numbers from 0 to 5, and the other from 1 to 6. But then each die must be symmetrical. For the first die with 0 to 5, the opposite faces would need to add up to the same value. So 0 and 5, 1 and 4, 2 and 3. That would add up to 5. So, the die would have pairs (0,5), (1,4), (2,3). Similarly, the other die, if numbered 1 to 6, would have pairs adding to 7, as standard. Then, the minimum sum would be 0+1=1 and the maximum sum would be 5+6=11. But that only gives sums from 1 to 11, which is way less than 36. So that doesn't help.Alternatively, if we use larger numbers. Let's think. If we need the maximum sum to be 36, then the maximum face values on the dice must be 36 when added together. So, if the two dice are A and B, then max(A) + max(B) = 36. If the dice are symmetrical, meaning that each die's opposite faces add up to a constant, say S for die A and T for die B.So, for die A, if it's symmetrical, then the numbers on opposite faces add up to S. Similarly, die B's opposite faces add up to T. The numbers on each die must be arranged such that opposite faces sum to S and T respectively.To get the maximum sum of 36, the maximum number on each die would depend on S and T. For example, if die A has a maximum face value of a, then its opposite face would be S - a. So, the maximum number on die A is either a or S - a, whichever is larger. Similarly for die B. Therefore, to have the maximum sum of 36, we need a + b = 36, where a is the maximum face value of die A and b is the maximum face value of die B.But if die A is symmetrical, then the numbers on die A are pairs that add up to S. So, the maximum number on die A would be either S/2 if S is even, or floor(S/2) if S is odd. Wait, no. For example, on a standard die with S=7, the numbers are 1,2,3,4,5,6. The maximum is 6, which is S - 1. So, if S is the sum of opposite faces, then the numbers on die A would be three pairs of numbers adding up to S. So, the maximum number on die A would be S - minimum number. So, if the minimum number on die A is m, then the maximum is S - m.Similarly for die B. So, in order to have maximum sum 36, (S_A - m_A) + (S_B - m_B) = 36, where S_A and S_B are the sums of opposite faces for dice A and B, and m_A, m_B are the minimum numbers on each die.But this is getting a bit complicated. Let's think differently. We need two dice, each with 6 faces, numbers arranged such that opposite faces sum to a constant (symmetrical). The sum of the two numbers rolled must produce each number from 1 to 36 exactly once. So, there are 36 possible outcomes, each corresponding to a unique sum from 1 to 36. Therefore, each possible pair of numbers on the two dice must produce a distinct sum. Therefore, the set of sums must be exactly the numbers 1 through 36, each appearing once. So, the total number of sums is 36, which is exactly the number of possible outcomes (6*6). Therefore, each outcome must result in a unique sum. Therefore, the function f(a,b) = a + b must be injective over the domain of face pairs (a,b). Therefore, the numbers on the two dice must be such that all pairwise sums are distinct.Therefore, the problem reduces to: Can we find two sets of numbers A and B, each of size 6, such that all sums a + b for a in A and b in B are distinct, and cover the numbers 1 through 36? Additionally, each set A and B must be symmetrical, meaning that for each number x in A, there is a number S_A - x in A (similarly for B). So, each die is a set of three pairs of numbers that add up to S_A and S_B respectively.So, first, the key is that the sums must be all distinct. This is similar to the concept of "orthogonal" or "Sidon sequences", where all pairwise sums are unique. But in this case, we have two different sets, so it's a bit different. The problem is to find two sets A and B such that the sumset A + B covers 1 to 36 exactly once, with each set being symmetrical.Given that, let's consider the properties required. Since each die has 6 numbers, and the sums must cover 1 to 36, each die must have numbers such that their pairwise sums can reach up to 36. So, the maximum numbers on each die must add to 36, and the minimum numbers must add to 1. But if the minimum sum is 1, then the minimum numbers on each die must be 0 and 1, or both 0.5, but since we need integers, the minimum numbers must be 0 and 1. But dice typically have positive integers, but the problem doesn't specify, so maybe 0 is allowed. Wait, but the problem says "points", which might imply positive integers. Hmm.If we have to use positive integers, then the minimum sum is 2 (1+1), which contradicts the problem's requirement of sums from 1. Therefore, perhaps the problem has a typo, or "points" can be zero. If we allow zero, then one die could have 0, and the other die could have 1 as the minimum, leading to a sum of 1. So, maybe that's acceptable.Assuming that zero is allowed, let's proceed. So, die A could have numbers including 0, and die B could have numbers starting from 1. But each die must be symmetrical. So, for die A, if it has a 0, then its opposite face must be S_A - 0 = S_A. So, the maximum number on die A is S_A. Similarly, die B, if it has 1 as the minimum, then its opposite face is S_B - 1. So, the maximum number on die B is S_B - 1.To get the minimum sum of 1, we need 0 (from die A) + 1 (from die B) = 1. To get the maximum sum of 36, we need S_A (max of die A) + (S_B - 1) (max of die B) = 36.So, S_A + S_B - 1 = 36 => S_A + S_B = 37.Also, since die A has numbers 0 and S_A, and three pairs adding to S_A. Similarly, die B has 1 and S_B - 1, and three pairs adding to S_B.But die A has three pairs: (0, S_A), (a, S_A - a), (b, S_A - b). Similarly, die B: (1, S_B - 1), (c, S_B - c), (d, S_B - d).The numbers on each die must be distinct, right? Because each face has a unique number. So, die A's numbers are 0, S_A, a, S_A - a, b, S_A - b. Similarly for die B.Now, the key is that all sums of die A and die B must be unique and cover 1 to 36. Let's think about how the sums can be arranged.Each number on die A can be paired with each number on die B, so we need 36 unique sums. The smallest sum is 0 + 1 = 1, and the largest sum is S_A + (S_B - 1) = 36.Given that S_A + S_B = 37, so S_A = 37 - S_B.So, we need to choose S_A and S_B such that the numbers on each die are integers, and all sums are unique.Let me try to pick S_A and S_B. Let's suppose S_A and S_B are such that the numbers on each die are integers. Let's try S_A = 18 and S_B = 19. Then, die A would have numbers: 0, 18, a, 18 - a, b, 18 - b. Die B would have numbers: 1, 18, c, 19 - c, d, 19 - d.Wait, die B's sum is S_B = 19, so numbers are (1, 18), (c, 19 - c), (d, 19 - d). Wait, 1 + 18 = 19, so that's correct.But then, the maximum sum would be 18 + 18 = 36. Okay, that works. So, die A's maximum is 18, die B's maximum is 18 (since 19 - 1 = 18). So, 18 + 18 = 36.But die A has 0, which when added to die B's 1 gives 1. So, that's the minimum sum.Now, we need to assign the numbers on die A and die B such that all sums from 1 to 36 are covered exactly once.Let me attempt to construct such dice.First, die A: symmetrical with sum 18. So, the pairs are (0,18), (a, 18 - a), (b, 18 - b). Similarly, die B: symmetrical with sum 19. The pairs are (1,18), (c, 19 - c), (d, 19 - d).So, die A's numbers: 0, 18, a, 18 - a, b, 18 - b.Die B's numbers: 1, 18, c, 19 - c, d, 19 - d.Now, we need to choose a, b, c, d such that all sums are unique from 1 to 36.First, let's note that die A has 0, 18, a, 18 - a, b, 18 - b. To ensure all numbers are distinct and positive (except 0). So, a and b must be between 1 and 17, and distinct from each other and from their complements.Similarly, die B has 1, 18, c, 19 - c, d, 19 - d. c and d must be between 2 and 17 (since 19 - c must be at least 2, so c ≤ 17), and distinct from each other and their complements.This is quite a complex setup. Let's try to pick numbers step by step.Starting with die A. Let's pick a and b such that all numbers on die A are distinct. Let's say a = 1, then 18 - a = 17. Then, b can be 2, so 18 - b = 16. So die A would be {0, 18, 1, 17, 2, 16}. Let's check if these are all distinct: 0,1,2,16,17,18. Yes, distinct.Now die B: needs to be {1, 18, c, 19 - c, d, 19 - d}. Let's try to choose c and d such that the numbers are distinct. Let's pick c = 2, then 19 - c = 17. Then d = 3, so 19 - d = 16. So die B would be {1,18,2,17,3,16}. Wait, but die B already has 1 and 18. Then adding 2,17,3,16. But die B would then have 1,2,3,16,17,18. All distinct. Let's check:Numbers on die B: 1,18,2,17,3,16. Yes, all distinct.Now, let's check the sums between die A and die B.Die A: 0,1,2,16,17,18Die B: 1,2,3,16,17,18Now, let's compute all possible sums:0+1=10+2=20+3=30+16=160+17=170+18=181+1=21+2=31+3=41+16=171+17=181+18=192+1=32+2=42+3=52+16=182+17=192+18=2016+1=1716+2=1816+3=1916+16=3216+17=3316+18=3417+1=1817+2=1917+3=2017+16=3317+17=3417+18=3518+1=1918+2=2018+3=2118+16=3418+17=3518+18=36Now, let's list all sums and check if they cover 1 to 36 uniquely.Looking at the sums:1,2,3,16,17,18 (from 0)Then from 1:2,3,4,17,18,19From 2:3,4,5,18,19,20From 16:17,18,19,32,33,34From 17:18,19,20,33,34,35From 18:19,20,21,34,35,36Now, compiling all sums:1,2,3,16,17,18,2,3,4,17,18,19,3,4,5,18,19,20,17,18,19,32,33,34,18,19,20,33,34,35,19,20,21,34,35,36.Clearly, there are many duplicates. For example, sum 2 appears twice (0+2 and 1+1). Similarly, sum 3 appears multiple times. So, this attempt doesn't work. The sums are not unique.Therefore, this choice of numbers for die A and die B doesn't satisfy the requirement. So, maybe the approach of setting S_A=18 and S_B=19 is not the right way.Alternatively, perhaps we need different values for S_A and S_B. Let's think.Suppose we want the numbers on die A to be spaced out such that when added to die B's numbers, the sums cover all numbers from 1 to 36. To do this, die A could be designed with numbers that are multiples of 6, and die B with numbers 0-5, but that might not work since we need symmetry.Wait, another idea: If die A has numbers 0, 6, 12, 18, 24, 30 and die B has numbers 1, 2, 3, 4, 5, 6. Then, the sums would range from 1 (0+1) to 36 (30+6). Each sum would be unique because each number in die A is a multiple of 6, and die B has numbers 1-6. So, for each a in A and b in B, a + b would be unique since the multiple of 6 plus 1-6 gives distinct remainders modulo 6. However, die A here is not symmetrical. Because symmetrical dice require that opposite faces add to the same sum. So, for die A, the numbers 0,6,12,18,24,30 – if opposite faces add to 30 (0+30, 6+24, 12+18), then die A would be symmetrical. Similarly, die B with numbers 1,2,3,4,5,6 – standard die with opposite faces adding to 7. Then, die B is symmetrical. So, in this case, die A is symmetrical with opposite faces adding to 30, and die B is symmetrical with opposite faces adding to 7.Now, let's check the sums. Die A: 0,6,12,18,24,30. Die B:1,2,3,4,5,6. Then, the sums would be:0+1=10+2=20+3=30+4=40+5=50+6=66+1=76+2=86+3=96+4=106+5=116+6=1212+1=1312+2=1412+3=1512+4=1612+5=1712+6=1818+1=1918+2=2018+3=2118+4=2218+5=2318+6=2424+1=2524+2=2624+3=2724+4=2824+5=2924+6=3030+1=3130+2=3230+3=3330+4=3430+5=3530+6=36So, the sums are 1 to 36, each appearing exactly once! Wow, that works. So, die A is 0,6,12,18,24,30 with opposite faces adding to 30 (0-30, 6-24, 12-18), and die B is the standard 1-6 die with opposite faces adding to 7. Then, their sums produce each number from 1 to 36 exactly once. Therefore, this satisfies the problem's condition.But wait, the problem states "two symmetrical dice". In this solution, die A is symmetrical (opposite faces add to 30) and die B is symmetrical (opposite faces add to 7). So, yes, both dice are symmetrical. However, die A includes 0, which might be considered non-standard, but the problem didn't specify that numbers must start from 1. It just said "points", but if points can be zero, then this works.However, the problem says "the sum of the points when rolling takes on values 1, 2, ..., 36 with equal probabilities". Each sum occurs exactly once, so each has probability 1/36, which is equal. Therefore, this construction works. Therefore, the answer is yes, it is possible.But wait, let me check if there's any restriction I missed. The problem says "two symmetrical dice". In my solution, die A has numbers 0,6,12,18,24,30, which are symmetrical around 15 (the mid-point between 0 and 30). Each pair of opposite faces adds to 30. Die B is a standard die with opposite faces adding to 7. So, both dice are individually symmetrical.Therefore, this seems to satisfy all the problem's conditions. Therefore, the answer is yes.But let me verify the sums again to be sure. Listing all sums:0+1=10+2=20+3=30+4=40+5=50+6=66+1=76+2=86+3=96+4=106+5=116+6=1212+1=1312+2=1412+3=1512+4=1612+5=1712+6=1818+1=1918+2=2018+3=2118+4=2218+5=2318+6=2424+1=2524+2=2624+3=2724+4=2824+5=2924+6=3030+1=3130+2=3230+3=3330+4=3430+5=3530+6=36Yes, each sum from 1 to 36 is present exactly once. Therefore, this numbering works. Therefore, it is indeed possible.But wait, another thought: the problem says "symmetrical dice". In some contexts, symmetrical dice might mean that the numbers are arranged symmetrically around the center, but perhaps also that the numbering is the same on both dice? But no, in this solution, the two dice are different: one is non-standard with zeros and multiples of six, the other is standard. But they are both symmetrical individually, so I think this is acceptable.Alternatively, maybe the problem requires the two dice to be identical? If that's the case, then this solution wouldn't work because die A and die B are different. However, the problem states "two symmetrical dice", not "two identical symmetrical dice". Therefore, as long as each die is symmetrical individually, it's acceptable.Therefore, the answer is yes, and an example is die A with faces 0,6,12,18,24,30 (symmetrical with opposite faces summing to 30) and die B with faces 1,2,3,4,5,6 (symmetrical with opposite faces summing to 7).But let me check if there's a different approach where both dice have the same numbering. Suppose both dice are identical and symmetrical. Then, each die would have numbers a, b, c, S - a, S - b, S - c, where S is the sum for symmetry. Then, the maximum sum would be 2*(S - min(a,b,c)). To reach 36, we need 2*(S - min) = 36 => S - min = 18. If min is 1, then S = 19. So, each die would have numbers 1, b, c, 18, 19 - b, 19 - c. Then, the maximum sum would be 18 + 18 = 36. But then, the sums would range from 2 (1+1) to 36. But the problem requires sums from 1 to 36. Therefore, this approach wouldn't work because the minimum sum is 2. Hence, if the dice must be identical, it's impossible. However, since the problem allows two different symmetrical dice, the previous solution is valid.Therefore, the answer is yes, it is possible. One such way is to have one die with numbers 0, 6, 12, 18, 24, 30 (symmetrical with opposite faces adding to 30) and the other die with numbers 1, 2, 3, 4, 5, 6 (symmetrical with opposite faces adding to 7). When rolled, their sums produce each integer from 1 to 36 exactly once, each with equal probability.

✅:To determine if two symmetrical dice can be numbered such that their sums range from 1 to 36 with equal probabilities, we analyze the requirements:1. Symmetrical Dice: Each die must have opposite faces summing to a constant (S for die A, T for die B).2. Sum Coverage: The 36 possible outcomes (6×6) must produce sums from 1 to 36, each exactly once. Key Insight:- For sums to cover 1–36 uniquely, the dice must be designed so that every pair of numbers yields a distinct sum. This resembles a Sidon sequence in two dimensions.- A standard die (1–6) can't achieve this, but using non-traditional numbers allows unique sums. Construction:1. Die A: Use numbers 0, 6, 12, 18, 24, 30 with opposite faces summing to 30 (symmetrical pairs: 0–30, 6–24, 12–18).2. Die B: Use a standard die 1, 2, 3, 4, 5, 6 with opposite faces summing to 7 (symmetrical pairs: 1–6, 2–5, 3–4). Verification:- Minimum Sum: (0 + 1 = 1)- Maximum Sum: (30 + 6 = 36)- All Sums: Pairing die A and die B produces every integer from 1 to 36 exactly once (verified below). Summation Table:- Die A (0, 6, 12, 18, 24, 30) + Die B (1, 2, 3, 4, 5, 6) yields: ``` 0+1=1, 0+2=2, ..., 0+6=6, 6+1=7, 6+2=8, ..., 6+6=12, 12+1=13, 12+2=14, ..., 12+6=18, 18+1=19, ..., 18+6=24, 24+1=25, ..., 24+6=30, 30+1=31, ..., 30+6=36. ``` All sums from 1 to 36 appear exactly once. Conclusion:Yes, it is possible. The required dice are:- Die 1: (0, 6, 12, 18, 24, 30) (opposite faces sum to 30).- Die 2: (1, 2, 3, 4, 5, 6) (standard symmetry with pairs summing to 7).Final Answer: boxed{Yes}

🔑:To solve the problem, we need to show that there exists a way to label two dice such that the sums of the numbers on the faces can be any of the numbers from 1 to 36 equally likely. We will demonstrate such a constructive example.1. Labeling the First Die: - We assign the numbers 1, 2, 3, 4, 5, and 6 to the faces of the first die. Therefore, the possible outcomes when rolling this die are: ( {1, 2, 3, 4, 5, 6} ).2. Labeling the Second Die: - We assign the numbers 0, 6, 12, 18, 24, and 30 to the faces of the second die. Therefore, the possible outcomes when rolling this die are: ( {0, 6, 12, 18, 24, 30} ).3. Examining the Possible Sums: - When these dice are rolled, the sum of the numbers on the resulted faces ranges from the minimum sum (1 + 0 = 1) to the maximum sum (6 + 30 = 36). - Therefore, the possible sums are given by ( (a + b) ) for every combination where ( a ) is a number from the first die and ( b ) is a number from the second die.4. Ensuring All Sums from 1 to 36: - By carefully analyzing, we can pair each number ( k ) (for ( k = 1, 2, 3, cdots, 36 )) with a unique sum ( (a + b) ): - For example: - Sum (1): (1 + 0) - Sum (2): (2 + 0) - Sum (6+1): (5 + 0) - Sum (7): (1+6) - Sum (12): (1+12) - And so forth up to the largest sum (36): (6 + 30).5. Conclusion: - With these labels, every number from 1 to 36 can be formed as a unique sum of one number from the first die and one number from the second die. - Given the uniform distribution of the numbers on both dice, each sum from 1 to 36 has an equal likelihood.Thus, it is possible to label the two dice such that the sums range from 1 to 36 with equal probabilities, confirming that the solution is correct.[boxed{text{Yes, it is possible}}]

🧠:Okay, so I need to find the maximum possible value of |a| + |b| + |c| given that the quadratic polynomial ax² + bx + c doesn't exceed 1 in absolute value on the interval [-1, 1]. The answer is supposed to be 3, but I need to figure out how to get there. Let me start by understanding the problem.First, the polynomial is quadratic, so it's of degree 2. The coefficients are a, b, c, and we need to maximize the sum of their absolute values. The constraint is that for all x in [-1, 1], |ax² + bx + c| ≤ 1. So we need to find the maximum of |a| + |b| + |c| under this condition.Hmm, how do I approach this? Maybe I can think about specific points in the interval [-1, 1] where the polynomial reaches its maximum or minimum. Since it's a quadratic, it's a parabola, so it will have a vertex. Depending on the coefficient a, it could open upwards or downwards.But wait, the problem is about the absolute value of the polynomial being bounded by 1 on the entire interval. So the maximum of |ax² + bx + c| on [-1, 1] is at most 1. So maybe I can use some known inequalities or optimization techniques with constraints.Alternatively, maybe there's a known result about polynomials bounded on an interval and their coefficients. I recall something called Chebyshev polynomials, which have minimal maximum norm on [-1, 1] among polynomials with leading coefficient 1. But here, we're dealing with a quadratic, so maybe Chebyshev polynomials of degree 2 could be related?Chebyshev polynomials are defined as T_n(x) = cos(n arccos x). For n=2, T_2(x) = 2x² - 1. This polynomial oscillates between -1 and 1 at three points in [-1, 1]: x = -1, x = 0, and x = 1. Let me check that. At x = -1, T_2(-1) = 2(-1)^2 -1 = 2 -1 =1. At x=0, T_2(0) = -1. At x=1, it's 1 again. So yes, it alternates between 1 and -1 three times. This is the Chebyshev polynomial of degree 2, which has the minimal maximum norm on [-1, 1] for leading coefficient 1. But how does this relate to our problem?In our case, we have a quadratic polynomial ax² + bx + c. The Chebyshev polynomial T_2(x) is 2x² - 1. If we scale it, maybe we can get some relation. Let's see. If we have a quadratic polynomial that attains its maximum absolute value at three points with alternating signs, like the Chebyshev polynomial, then maybe such a polynomial would be extremal in some sense. Since we need the maximum of |a| + |b| + |c|, perhaps scaling the Chebyshev polynomial could give us a candidate for the maximum.But let's compute |a| + |b| + |c| for T_2(x). For T_2(x) = 2x² -1, the coefficients are a=2, b=0, c=-1. So |a| + |b| + |c| = |2| + |0| + |-1| = 3. So that's exactly the value we're supposed to get. So maybe this is the maximal sum. But is this indeed the maximum?Wait, but the problem states that the polynomial is bounded by 1 in absolute value on [-1,1], so if we take T_2(x) scaled down by a factor, but here the original T_2(x) has maximum 1 and minimum -1. So if we take the same coefficients, then |ax² + bx + c| ≤1 on [-1,1], and the sum |a| + |b| + |c| is 3. But maybe there's another polynomial where the sum is larger? How can we be sure?Alternatively, maybe we can use Lagrange multipliers to maximize |a| + |b| + |c| under the constraint that |ax² + bx + c| ≤1 for all x in [-1,1]. But Lagrange multipliers can be tricky here because it's an infinite set of constraints (for each x in [-1,1]). However, for polynomials, sometimes it's possible to consider critical points where the maximum is achieved. For a quadratic, the maximum on [-1,1] can occur either at the endpoints or at the critical point inside the interval.So let's first find the critical point of the quadratic polynomial ax² + bx + c. The derivative is 2ax + b, so the critical point is at x = -b/(2a). Now, if this critical point lies within [-1,1], then the extremum (maximum or minimum) occurs there; otherwise, the extrema are at x = -1 and x = 1.Therefore, to ensure |ax² + bx + c| ≤1 on [-1,1], we must have that the polynomial evaluated at x = -1, x = 1, and x = -b/(2a) (if it's in the interval) is within [-1,1].So maybe we can write the constraints as:| a(1)^2 + b(1) + c | ≤1,| a(-1)^2 + b(-1) + c | ≤1,and | a(-b/(2a))² + b(-b/(2a)) + c | ≤1.Simplifying these:At x=1: |a + b + c| ≤1,At x=-1: |a - b + c| ≤1,At critical point x = -b/(2a): |c - b²/(4a)| ≤1, assuming that -1 ≤ -b/(2a) ≤1.But this is getting a bit complicated. Also, note that we need to maximize |a| + |b| + |c|. Since the problem involves absolute values, maybe we can consider different cases based on the signs of a, b, c. But that might be too tedious.Alternatively, perhaps we can use the fact that the maximum of |a| + |b| + |c| occurs when the polynomial touches 1 or -1 at several points in [-1,1], similar to the Chebyshev polynomial. The Chebyshev polynomial alternates between 1 and -1 exactly n+1 times (for degree n), which is the maximum number of alternations. For quadratic, that would be 3 points. So maybe such a polynomial would maximize the coefficients.Wait, in the case of T_2(x) = 2x² -1, it does touch 1 at x=-1, x=1, and -1 at x=0. So three points. And as we saw, the sum |a| + |b| + |c| is 3 here. So maybe that's the maximum.But how to prove that no other polynomial can have a higher sum?Suppose we have another quadratic polynomial f(x) = ax² + bx + c with |f(x)| ≤1 on [-1,1]. We need to show that |a| + |b| + |c| ≤3.Alternatively, maybe we can bound |a|, |b|, |c| individually and then sum them up. Let's see.First, let's evaluate f at x=1: f(1) = a + b + c.Similarly, at x=-1: f(-1) = a - b + c.At x=0: f(0) = c.So from these three points, we have:|c| ≤1,|a + b + c| ≤1,|a - b + c| ≤1.But also, if the critical point is inside [-1,1], then the value at x = -b/(2a) is f(-b/(2a)) = c - b²/(4a). So |c - b²/(4a)| ≤1.Hmm, but how can we use these inequalities to bound |a| + |b| + |c|?Let me try adding the inequalities for x=1 and x=-1:|a + b + c| + |a - b + c| ≤2.But by the triangle inequality, |a + b + c| + |a - b + c| ≥ |(a + b + c) + (a - b + c)| = |2a + 2c| = 2|a + c|.So 2|a + c| ≤2, hence |a + c| ≤1.Similarly, if we subtract the inequalities, maybe?Alternatively, let's consider c. We know |c| ≤1. So |c| is at most 1.Also, from |a + c + b| ≤1 and |a + c - b| ≤1, we can think of these as two inequalities that bound b. Let me denote s = a + c. Then we have |s + b| ≤1 and |s - b| ≤1. Then, the maximum of |b| is when these inequalities are tight. For a given s, |b| ≤1 - |s|. But since |s| = |a + c| ≤1, then |b| ≤1 - |a + c|. But I'm not sure how helpful this is.Alternatively, maybe we can use some norm comparisons. The sum |a| + |b| + |c| is the l1 norm of the coefficient vector, and we want to maximize it given that the l∞ norm of the polynomial on [-1,1] is bounded by 1. There might be some inequality relating these norms.But I don't recall a direct inequality for this case. Maybe we can use the fact that the coefficients can be related to the values of the polynomial at certain points.Alternatively, think of the polynomial in terms of the basis of Legendre polynomials or Chebyshev polynomials, but I'm not sure.Wait, another approach: use Lagrange interpolation. Since a quadratic polynomial is determined by its values at three points. If we take three points in [-1,1], say -1, 0, 1, then we can express the polynomial in terms of its values at these points.Let me try that. Let’s denote f(-1) = y1, f(0) = y2, f(1) = y3. Then the polynomial can be written as:f(x) = y1 * [(x)(x - 1)/2] + y2 * [(x + 1)(x - 1)/(-1)] + y3 * [(x + 1)(x)/2]Wait, let me check that. The Lagrange interpolation polynomial through points x=-1, 0, 1.The Lagrange basis polynomials are:L1(x) for x=-1: [(x - 0)(x - 1)] / [(-1 -0)(-1 -1)] = [x(x -1)] / [(-1)(-2)] = [x(x -1)] / 2L2(x) for x=0: [(x +1)(x -1)] / [(0 +1)(0 -1)] = [(x +1)(x -1)] / (-1)L3(x) for x=1: [(x +1)(x -0)] / [(1 +1)(1 -0)] = [(x +1)x] / 2Therefore, f(x) = y1 * [x(x -1)/2] + y2 * [-(x +1)(x -1)] + y3 * [x(x +1)/2]Simplify each term:First term: y1*(x² -x)/2Second term: -y2*(x² -1)Third term: y3*(x² +x)/2Combine all terms:= (y1/2 + y3/2 - y2)x² + (-y1/2 + y3/2)x + y2Therefore, coefficients:a = (y1 + y3)/2 - y2b = (-y1 + y3)/2c = y2Given that |y1| ≤1, |y2| ≤1, |y3| ≤1, since |f(x)| ≤1 at x=-1,0,1.But wait, actually, even if the maximum is attained at other points, we still know that |y1|, |y2|, |y3| ≤1 because the polynomial doesn't exceed 1 in absolute value anywhere on [-1,1].Therefore, we can write a, b, c in terms of y1, y2, y3, which are each bounded by 1 in absolute value. Then, |a| + |b| + |c| can be expressed in terms of |(y1 + y3)/2 - y2| + |(-y1 + y3)/2| + |y2|.Our goal is to maximize this expression over all y1, y2, y3 in [-1,1].So let's denote:Let’s set variables as y1, y2, y3 each in [-1,1].Compute S = |(y1 + y3)/2 - y2| + |(-y1 + y3)/2| + |y2|.We need to maximize S.This seems complicated, but maybe we can consider possible combinations of y1, y2, y3 to maximize S.Alternatively, set y1, y3 to extremes (either 1 or -1) and adjust y2 to maximize the terms. Let's try some examples.First, take y1 = 1, y3 = 1, and y2 = -1.Compute:a = (1 +1)/2 - (-1) = 1 +1 = 2b = (-1 +1)/2 = 0c = -1Then |a| + |b| + |c| = 2 +0 +1 = 3. Which matches the Chebyshev case.Another case: y1=1, y3=1, y2=1.Then a = (1 +1)/2 -1 = 1 -1 =0b = (-1 +1)/2 =0c=1Sum: 0 +0 +1=1. Not good.Another case: y1=1, y3=-1, y2=0.a=(1 + (-1))/2 -0 =0 -0=0b=(-1 + (-1))/2= (-2)/2=-1c=0Sum:0 +1 +0=1. Still low.Another case: y1=1, y3=1, y2=0.a=(1 +1)/2 -0=1b=(-1 +1)/2=0c=0Sum:1 +0 +0=1.Not better.Wait, maybe if y1 and y3 are 1, but y2 is different. Let's see.Suppose y1=1, y3=1, y2= -1. Then, as above, sum is 3. If y2= -1, then |a| + |b| + |c| =3. That seems like the maximum.Alternatively, maybe y1=1, y3=-1, y2= something.Wait, let's compute S:S = |(y1 + y3)/2 - y2| + |(-y1 + y3)/2| + |y2|.Let’s consider different possibilities:Case 1: y1 =1, y3=1. Then:S = |(1 +1)/2 - y2| + |(-1 +1)/2| + |y2| = |1 - y2| + 0 + |y2| = |1 - y2| + |y2|.To maximize |1 - y2| + |y2|, note that for y2 in [-1,1], let's consider different intervals.If y2 ≥0:|1 - y2| + |y2| = (1 - y2) + y2 =1.If y2 <0:|1 - y2| + |y2| = (1 - y2) - y2 =1 -2y2. Since y2 <0, the maximum occurs at the minimal y2, which is y2 = -1, giving 1 -2*(-1)=3.So in this case, if y1=1, y3=1, y2=-1, then S=3. That's the maximum here.Case 2: y1=1, y3=-1.Then S = |(1 -1)/2 - y2| + |(-1 -1)/2| + |y2| = |0 - y2| + |-1| + |y2| = |y2| +1 + |y2| =1 + 2|y2|.The maximum of this is when |y2|=1, so S=1+2*1=3.So if y1=1, y3=-1, y2=1 or y2=-1:Wait, y2=1:S=1 +2*1=3. But let's check the coefficients:a=(1 + (-1))/2 - y2 =0 - y2 = -y2If y2=1, then a=-1b=(-1 + (-1))/2= (-2)/2=-1c=y2=1So coefficients a=-1, b=-1, c=1.Sum: |-1| + |-1| + |1| =1 +1 +1=3.Similarly, if y2=-1:a= -y2=1b=(-1 + (-1))/2= -1c=y2=-1Sum:1 +1 +1=3. Same result.So in this case, also, we can get S=3.Case 3: y1=-1, y3=-1, y2=1.Then S= |(-1 + (-1))/2 -1| + |(1 + (-1))/2| + |1| = | -1 -1 | + 0 +1=2 +1=3.Similarly, coefficients:a=(-1 + (-1))/2 -1= -1 -1= -2b=(1 + (-1))/2=0c=1Sum:2 +0 +1=3.Same as the previous cases.So in these cases, S=3. So that's consistent.But can we get a higher S? Let's see.Case 4: y1=1, y3=0, y2=-1.Compute S:|(1 +0)/2 - (-1)| + |(-1 +0)/2| + |-1| = |0.5 +1| + | -0.5 | +1 =1.5 +0.5 +1=3. Still 3.Coefficients:a=(1 +0)/2 - (-1)=0.5 +1=1.5b=(-1 +0)/2= -0.5c=-1Sum:1.5 +0.5 +1=3.Same.Another case: y1=1, y3=0, y2=1.S= |(1 +0)/2 -1| + |(-1 +0)/2| + |1| = |0.5 -1| + 0.5 +1=0.5 +0.5 +1=2.Not better.So seems like maximum S=3.Alternatively, take y1=1, y3=1, y2=-1, which gives S=3. Similarly, other combinations. So from these cases, the maximum seems to be 3. Therefore, the answer is 3.But to confirm, is there a case where S could be higher? Let's see.Suppose we take y1, y3 at other points. For example, y1=1, y3=1, y2=-1. We already saw that gives S=3. Similarly, other sign combinations. If all y1, y3, y2 are set to 1 or -1 in some combination, the maximum sum S is 3.Another angle: suppose we model this as an optimization problem. Let's consider variables a, b, c with the constraints |ax² + bx + c| ≤1 for all x in [-1,1]. We need to maximize |a| + |b| + |c|.To handle the absolute values, we can consider linear programming techniques, but since the constraints are quadratic, it's not straightforward. However, note that for x in [-1,1], the maximum of |ax² + bx + c| occurs at one of the endpoints or the critical point. So we can convert the problem into a set of inequalities:1. |a + b + c| ≤1,2. |a - b + c| ≤1,3. |c - b²/(4a)| ≤1 (if the critical point x=-b/(2a) is in [-1,1]).But this third constraint complicates things because it's nonlinear. However, maybe the maximum of |a| + |b| + |c| occurs when the critical point is within the interval, so we need to consider that.Alternatively, if we suppose that the maximum occurs when all three points x=-1,0,1 are points where the polynomial touches 1 or -1 alternately, then we have the Chebyshev case, which gives sum 3. If we can show that any deviation from this would lead to a lower sum, then 3 is indeed the maximum.Alternatively, consider that the sum |a| + |b| + |c| is maximized when each of |a|, |b|, |c| is as large as possible. But they are constrained by the polynomial's maximum on [-1,1]. For example, c is the value at x=0, so |c| ≤1. Similarly, a + b + c and a - b + c are bounded by 1. From these, can we find upper bounds on |a|, |b|?Let’s denote:From x=1: |a + b + c| ≤1,From x=-1: |a - b + c| ≤1,From x=0: |c| ≤1.Let’s add and subtract the first two inequalities.Let’s set s1 = a + b + c,s2 = a - b + c.Then, adding s1 + s2 = 2a + 2c,Subtracting s1 - s2 = 2b.So,|s1 + s2| = |2a + 2c| ≤ |s1| + |s2| ≤2,Thus, |a + c| ≤1.Similarly,|2b| = |s1 - s2| ≤ |s1| + |s2| ≤2,Thus, |b| ≤1.But we already know |c| ≤1. So individually, |a + c| ≤1, |b| ≤1, |c| ≤1. But how does this help in finding |a| + |b| + |c|?Note that |a| = |(a + c) - c| ≤ |a + c| + |c| ≤1 +1=2. But this is a loose bound. Similarly, we can have:|a| + |b| + |c| ≤ |a + c| + |c| + |b| ≤1 +1 +1=3.Wait, this is interesting. If |a + c| ≤1, |b| ≤1, |c| ≤1, then |a| + |b| + |c| ≤ |a + c| + |c| + |b| ≤1 +1 +1=3.Therefore, the sum |a| + |b| + |c| is bounded by 3. So the maximum possible is 3. And this is achieved when |a + c|=1, |b|=1, and |c|=1. Which is possible only if:|a + c| =1,|b|=1,|c|=1.For example, take c=1, then a +1 must satisfy |a +1|=1 ⇒ a=0 or a=-2.If a=0, then b=1 or -1. Then |a| + |b| + |c|=0 +1 +1=2 <3.If a=-2, then |a +c|= |-2 +1|=1. Then b=1 or -1. So coefficients a=-2, b=1, c=1. Let's check if this satisfies the original condition.Compute the polynomial: -2x² + x +1.Evaluate at x=1: -2 +1 +1=0, which is within the bound.At x=-1: -2 + (-1) +1=-2, absolute value is 2 >1. So this doesn't satisfy the condition.So this combination doesn't work. Therefore, just because |a + c|=1, |b|=1, |c|=1, it doesn't mean that the polynomial satisfies the condition for all x in [-1,1]. So that approach isn't valid.But the earlier example using Chebyshev's polynomial gives a valid case where the sum is 3. So maybe even though the above approach gives an upper bound of 3, it's achievable by the Chebyshev polynomial. Hence, the maximum is indeed 3.Another way to see this is that the Chebyshev polynomial T_2(x) = 2x² -1 has coefficients a=2, b=0, c=-1. So |a| + |b| + |c|=3. And we know that |T_2(x)| ≤1 on [-1,1]. So this is a valid case where the sum is 3.Moreover, any other polynomial with a higher sum would violate the bound on the polynomial's maximum. Hence, 3 is indeed the maximum.Therefore, after considering different cases, using Lagrange interpolation, and applying bounds based on constraints at specific points, the maximum possible value of |a| + |b| + |c| is 3.

✅:The maximum possible value of ( |a| + |b| + |c| ) is achieved by the Chebyshev polynomial ( T_2(x) = 2x^2 - 1 ), which satisfies ( |T_2(x)| leq 1 ) on ([-1, 1]). For this polynomial, ( |a| + |b| + |c| = |2| + |0| + |-1| = 3 ). Rigorous analysis confirms no other quadratic polynomial under the given constraint yields a higher sum.Answer: (boxed{3})

🔑: Problem:Given that the values of the quadratic polynomial ( ax^2 + bx + c ) on the interval ([-1, 1]) do not exceed 1 in absolute value, find the maximum possible value of the sum ( |a| + |b| + |c| ).Let's proceed step-by-step to find the maximum possible value of the sum ( |a| + |b| + |c| ). 1. Substitute key points: We start by substituting the key points ( x = 0, 1, -1 ) into the polynomial ( ax^2 + bx + c ): - For ( x = 0 ): [ ax^2 + bx + c = c implies -1 leq c leq 1 ] - For ( x = 1 ): [ ax^2 + bx + c = a + b + c implies -1 leq a + b + c leq 1 ] - For ( x = -1 ): [ ax^2 + bx + c = a - b + c implies -1 leq a - b + c leq 1 ]2. Combine inequalities: To simplify, we combine the three inequalities as follows: - Summing the inequalities for ( x = 1 ) and ( x = -1 ): [ (a + b + c) + (a - b + c) = 2a + 2c implies -2 leq 2a + 2c leq 2 ] [ -1 leq a + c leq 1 ] - Subtracting the inequalities for ( x = 1 ) and ( x = -1 ): [ (a + b + c) - (a - b + c) = 2b implies -2 leq 2b leq 2 ] [ -1 leq b leq 1 ]3. Summarize ranges: Consequently, we have the following: [ -1 leq c leq 1 ] [ -1 leq b leq 1 ] [ -1 leq a + c leq 1 ] From here, we subtract the independent inequality involving ( c ) to get the bounds on ( a ): [ -1 leq a + c leq 1 ] [ -1 leq c leq 1 ] [ implies -2 leq a leq 2 ]4. Consider all combinations of signs: We now examine combinations of signs to find the maximum ( |a| + |b| + |c| ). Notice the symmetry of the problem regarding ( a ): - If ( b, c geq 0 ), then (|a| + |b| + |c|) could be given by: [ |a| + |b| + |c| = a + b + c implies a + b + c leq 1 ] - If ( b < 0, c geq 0 ), then (|a| + |b| + |c|) becomes: [ |a| + |b| + |c| = a - b + c implies a - b + c leq 1 ] - If ( b geq 0, c < 0 ), then (|a| + |b| + |c|) can be: [ |a| + |b| + |c| = a + b - c = (a + b + c) - 2c leq 1 + 2 = 3 ] - If ( b, c < 0 ), the calculation yields: [ |a| + |b| + |c| = a - b - c = (a - b + c) - 2c leq 1 + 2 = 3 ]5. Verify the maximum value: We can verify that the maximum ( |a| + |b| + |c| = 3 ). For example, consider the polynomial ( f(x) = 2x^2 - 1 ). The polynomial achieves its critical values within the interval: - ( x = 0 longrightarrow f(0) = -1 ) - ( x = 1 longrightarrow f(1) = 2(1) - 1 = 1 ) - ( x = -1 longrightarrow f(-1) = 2 - 1 = 1 )Therefore, the polynomial ( f(x) = 2x^2 - 1 ) fits the given constraints and achieves the sum ( |a| + |b| + |c| = 3 ). Conclusion:The maximum possible value of the sum ( |a| + |b| + |c| ) is:[boxed{3}]

🧠:Okay, so I need to find the maximum value of the function y = 3x + 4√(1 - x²). Hmm, let's see. I remember that when you have functions involving square roots, especially with variables inside them, you might need to use calculus to find maxima or minima. But maybe there's another way, like using trigonometric substitution or something? Let me think.First, let me understand the domain of the function. The square root √(1 - x²) requires that 1 - x² ≥ 0. So x² ≤ 1, which means x is between -1 and 1, inclusive. So the domain is [-1, 1]. Got that.Now, the function is y = 3x + 4√(1 - x²). To find its maximum value on the interval [-1, 1], I can use calculus. Find the derivative, set it equal to zero, and solve for x. Then check the critical points and endpoints.Let me start by computing the derivative of y with respect to x. The derivative of 3x is 3. For the second term, 4√(1 - x²), which is 4*(1 - x²)^(1/2). Using the chain rule, the derivative would be 4*(1/2)*(1 - x²)^(-1/2)*(-2x). Let me compute that step by step.First, derivative of (1 - x²)^(1/2) is (1/2)(1 - x²)^(-1/2)*(-2x) = (-x)/(√(1 - x²)). Then multiply by 4, so the derivative of the second term is 4*(-x)/√(1 - x²) = -4x/√(1 - x²).Therefore, the derivative of y is dy/dx = 3 - 4x/√(1 - x²). To find critical points, set dy/dx = 0:3 - (4x)/√(1 - x²) = 0So, 3 = (4x)/√(1 - x²)Let me solve for x. Let's rearrange:Multiply both sides by √(1 - x²):3√(1 - x²) = 4xNow, square both sides to eliminate the square root:[3√(1 - x²)]² = (4x)²9(1 - x²) = 16x²Expand left side: 9 - 9x² = 16x²Bring all terms to one side: 9 = 16x² + 9x² => 9 = 25x²Therefore, x² = 9/25 => x = ±3/5.But wait, the original equation after taking derivatives had a square root, so we need to check if these solutions are valid in the original equation (since squaring both sides can introduce extraneous solutions).First, let's check x = 3/5. Let's plug into 3√(1 - x²) = 4x:Left side: 3√(1 - (9/25)) = 3√(16/25) = 3*(4/5) = 12/5.Right side: 4*(3/5) = 12/5. Equal. So x=3/5 is valid.Now, x = -3/5. Let's check:Left side: 3√(1 - (9/25)) = 3*(4/5) = 12/5.Right side: 4*(-3/5) = -12/5. So 12/5 ≠ -12/5. Therefore, x=-3/5 is extraneous. So only x=3/5 is a critical point.Therefore, critical points are x=3/5 and also endpoints x=-1 and x=1. Wait, but when we found x=3/5, that's the only critical point. The endpoints are -1 and 1. Let's compute the function at these three points.First, at x=3/5:y = 3*(3/5) + 4√(1 - (9/25)) = 9/5 + 4*(√(16/25)) = 9/5 + 4*(4/5) = 9/5 + 16/5 = 25/5 = 5.At x=1:y = 3*1 + 4√(1 - 1) = 3 + 0 = 3.At x=-1:y = 3*(-1) + 4√(1 - 1) = -3 + 0 = -3.So the maximum value is 5 at x=3/5.But wait, let me make sure I didn't make a mistake. Let me verify the derivative steps again.Original function: y = 3x + 4√(1 - x²)Derivative: dy/dx = 3 + 4*( derivative of √(1 - x²) )Derivative of √(1 - x²) is (1/(2√(1 - x²)))*(-2x) = -x/√(1 - x²). So dy/dx = 3 + 4*(-x/√(1 - x²)) = 3 - (4x)/√(1 - x²). Correct.Setting that equal to zero: 3 - (4x)/√(1 - x²) = 0 => 4x/√(1 - x²) = 3 => 4x = 3√(1 - x²). Then squaring both sides: 16x² = 9(1 - x²) => 16x² = 9 - 9x² => 25x² = 9 => x² = 9/25 => x=±3/5. Then checking validity, x=-3/5 invalid. Correct.So critical point at x=3/5. Then function value there is 5. The endpoints at x=1 and x=-1 give 3 and -3. So yes, maximum is 5. Seems right.Alternatively, maybe there's another method without calculus? Let's think. For functions like a x + b√(1 - x²), sometimes trigonometric substitution works. Let me try that.Let x = sinθ, since √(1 - x²) becomes cosθ. So substitute x = sinθ, where θ ∈ [-π/2, π/2] to cover x from -1 to 1.Then y = 3 sinθ + 4 cosθ.Now, the expression 3 sinθ + 4 cosθ can be written as R sin(θ + φ), where R is the amplitude. The maximum value would be R, since the maximum of sine function is 1.Alternatively, using the formula: a sinθ + b cosθ = √(a² + b²) sin(θ + φ), where φ = arctan(b/a) or something like that. Wait, let me recall.Actually, the identity is a sinθ + b cosθ = √(a² + b²) sin(θ + φ), where φ = arctan(b/a) if we write it as sinθ * a + cosθ * b. Wait, maybe it's better to write it as:Let’s consider R = √(3² + 4²) = 5. Then 3 sinθ + 4 cosθ = 5*( (3/5) sinθ + (4/5) cosθ ) = 5 sin(θ + α), where α is the angle such that cosα = 3/5 and sinα = 4/5. Because (3/5)^2 + (4/5)^2 = 1.So, the maximum value of 5 sin(θ + α) is 5*1 = 5. Therefore, the maximum value of y is 5. That's the same answer as before. So this confirms the result.Therefore, regardless of the method, calculus or trigonometric substitution, the maximum is 5.Wait, but let me check with θ. If we have y = 3 sinθ + 4 cosθ, then maximum is indeed √(3² +4²)=5. So yeah, that's a standard result. So this gives another way to see it.So the maximum value is 5. Therefore, the answer is 5.But let me also think if there's another approach, maybe using Cauchy-Schwarz inequality or something. Let's see.We have y = 3x + 4√(1 - x²). Let's denote vector u = (3, 4) and vector v = (x, √(1 - x²)). Then by Cauchy-Schwarz inequality, the dot product u · v ≤ ||u|| ||v||.Compute ||u|| = √(3² + 4²) = 5. Compute ||v|| = √(x² + (1 - x²)) = √1 = 1. Therefore, u · v ≤ 5*1 =5. Therefore, maximum value of y is 5. Equality holds when vectors are scalar multiples, i.e., when (x, √(1 - x²)) = k*(3,4). So x = 3k and √(1 - x²) =4k. Then x=3k, so substitute into the second equation: √(1 -9k²)=4k. Square both sides: 1 -9k²=16k² => 1=25k² => k=1/5. Therefore, x=3*(1/5)=3/5. So that's the same critical point as before. Therefore, equality holds at x=3/5, and the maximum value is 5.Wow, that's another method. So regardless of the approach, calculus, trigonometric substitution, or Cauchy-Schwarz, we get the same maximum value of 5. So that's reassuring.Therefore, the maximum value of the function is 5, achieved at x=3/5.I think that's thorough. Let me check if there's any possible mistake. For example, in the derivative step, did I compute it correctly? Let's verify again:y = 3x + 4√(1 - x²)dy/dx = 3 + 4*(1/(2√(1 - x²)))*(-2x) = 3 - (4x)/√(1 - x²). Correct.Set to zero, solve for x. Then x=3/5. Compute y at that point: 3*(3/5) +4*sqrt(1 -9/25)=9/5 +4*(4/5)=25/5=5. Correct.Endpoints give lower values. So all checks out. I think that's solid.Final AnswerThe maximum value of the function is boxed{5}.

✅:To find the maximum value of the function ( y = 3x + 4 sqrt{1 - x^2} ), we start by noting the domain of the function is ([-1, 1]). Using Calculus1. Compute the derivative: [ frac{dy}{dx} = 3 - frac{4x}{sqrt{1 - x^2}} ]2. Set the derivative to zero and solve for ( x ): [ 3 - frac{4x}{sqrt{1 - x^2}} = 0 implies 3sqrt{1 - x^2} = 4x ] Squaring both sides: [ 9(1 - x^2) = 16x^2 implies 9 = 25x^2 implies x^2 = frac{9}{25} implies x = pm frac{3}{5} ]3. Check validity: - For ( x = frac{3}{5} ), the equation holds. - For ( x = -frac{3}{5} ), it does not hold. Thus, the critical point is ( x = frac{3}{5} ).4. Evaluate the function at critical points and endpoints: - At ( x = frac{3}{5} ): [ y = 3 left( frac{3}{5} right) + 4 sqrt{1 - left( frac{3}{5} right)^2} = frac{9}{5} + frac{16}{5} = 5 ] - At ( x = 1 ): [ y = 3(1) + 4 sqrt{1 - 1^2} = 3 ] - At ( x = -1 ): [ y = 3(-1) + 4 sqrt{1 - (-1)^2} = -3 ] Using Trigonometric SubstitutionSubstitute ( x = sin theta ):[y = 3 sin theta + 4 cos theta]This can be written as ( R sin(theta + phi) ) where ( R = sqrt{3^2 + 4^2} = 5 ). The maximum value is ( 5 ). Using Cauchy-Schwarz Inequality[y = 3x + 4 sqrt{1 - x^2} leq sqrt{(3^2 + 4^2)(x^2 + (1 - x^2))} = 5]Equality holds when ( x = frac{3}{5} ). ConclusionThe maximum value of the function is (boxed{5}).

🔑:1. Determine the domain of the function y = 3x + 4sqrt{1 - x^2}: Since the square root function sqrt{1 - x^2} must be non-negative, we need: [ 1 - x^2 geq 0 implies -1 leq x leq 1 ] The domain is: [ D(f) = {x mid -1 leq x leq 1} ]2. Find the maximum of the function y = 3x + 4sqrt{1 - x^2}: Consider the function y = 3x + 4sqrt{1 - x^2} and its derivatives. We use the First Derivative Test to find the critical points. Compute the derivative of y with respect to x: [ frac{dy}{dx} = 3 - frac{4x}{sqrt{1 - x^2}} ] Set the derivative equal to zero to find the critical points: [ 3 - frac{4x}{sqrt{1 - x^2}} = 0 implies 3 = frac{4x}{sqrt{1 - x^2}} ] Solving the equation for x: [ 3sqrt{1 - x^2} = 4x implies 9(1 - x^2) = 16x^2 implies 9 - 9x^2 = 16x^2 implies 25x^2 = 9 implies x^2 = frac{9}{25} implies x = pm frac{3}{5} ] So, the critical points are x = frac{3}{5} and x = -frac{3}{5}. 3. Evaluate the function at the endpoints and the critical points: - At x = frac{3}{5}: [ yleft(frac{3}{5}right) = 3left(frac{3}{5}right) + 4sqrt{1 - left(frac{3}{5}right)^2} = frac{9}{5} + 4sqrt{1 - frac{9}{25}} = frac{9}{5} + 4sqrt{frac{16}{25}} = frac{9}{5} + 4 cdot frac{4}{5} = frac{9}{5} + frac{16}{5} = 5 ] - At x = -frac{3}{5}: [ yleft(-frac{3}{5}right) = 3left(-frac{3}{5}right) + 4sqrt{1 - left(-frac{3}{5}right)^2} = -frac{9}{5} + 4sqrt{1 - frac{9}{25}} = -frac{9}{5} + 4 cdot frac{4}{5} = -frac{9}{5} + frac{16}{5} = frac{7}{5} ] - At x = 1: [ y(1) = 3(1) + 4sqrt{1 - 1^2} = 3 + 4 cdot 0 = 3 ] - At x = -1: [ y(-1) = 3(-1) + 4sqrt{1 - (-1)^2} = -3 + 4 cdot 0 = -3 ]4. Conclude: Comparing all the values obtained, [ yleft(frac{3}{5}right) = 5, quad yleft(-frac{3}{5}right) = frac{7}{5}, quad y(1) = 3, quad y(-1) = -3 ] Therefore, the maximum value of y is achieved at x = frac{3}{5} and it is: [ boxed{5} ]

🧠:Okay, let's see. So the problem is about a trainee cook who peeled two buckets of unpeeled potatoes in an hour, and 25% of the potatoes went into peels. The question is, how long did it take him to collect exactly one bucket of peeled potatoes?Hmm, first, I need to understand the problem step by step. Let me break it down. The cook starts with two buckets of unpeeled potatoes. After peeling, 25% by volume (I assume) becomes peels, and the remaining 75% is the peeled potatoes. So, from two buckets of unpeeled potatoes, he ends up with 75% of two buckets as peeled potatoes. That would be 1.5 buckets of peeled potatoes, right? Because 25% of two buckets is 0.5 buckets, so subtracting that gives 1.5 buckets.But he did all that in one hour. So, he peeled two buckets of unpeeled potatoes in an hour, resulting in 1.5 buckets of peeled potatoes. Now, the question is asking for the time it took him to collect exactly one bucket of peeled potatoes.Wait, so is the rate of peeling constant? I guess so. Since he took an hour to peel two buckets of unpeeled potatoes, which results in 1.5 buckets of peeled potatoes. So his rate of producing peeled potatoes is 1.5 buckets per hour. Therefore, to get one bucket, he would need less time. Let's think in terms of rates.First, let's confirm the rates. If two buckets of unpeeled potatoes become 1.5 buckets of peeled potatoes in one hour, then the rate at which he's producing peeled potatoes is 1.5 buckets per hour. Therefore, time to get one bucket would be 1 divided by 1.5 hours, which is 2/3 of an hour, or 40 minutes. But wait, is this correct?Wait, maybe not. Because when he peels potatoes, the amount of peeled potatoes depends on the amount of unpeeled potatoes he processes. So maybe we need to model this as peeling a certain number of buckets per hour, considering that 25% is lost as peels.Alternatively, let's think of the relationship between unpeeled and peeled potatoes. For every bucket of unpeeled potatoes, after peeling, you get 0.75 buckets of peeled potatoes. So, the conversion factor is 0.75. Therefore, to get one bucket of peeled potatoes, how many buckets of unpeeled potatoes does he need to peel?Let me set up the equation. Let’s denote the amount of unpeeled potatoes needed to get one bucket of peeled potatoes as x. Then, 0.75x = 1 bucket. Solving for x gives x = 1 / 0.75 = 4/3 ≈ 1.333 buckets of unpeeled potatoes.So, he needs to peel 4/3 buckets of unpeeled potatoes to get 1 bucket of peeled potatoes. Now, how long does it take him to peel 4/3 buckets of unpeeled potatoes?Given that he can peel 2 buckets of unpeeled potatoes in 1 hour, his peeling rate is 2 buckets per hour. Therefore, the time required to peel 4/3 buckets would be (4/3) / 2 = (4/3) * (1/2) = 2/3 hours, which is 40 minutes. So, the answer would be 40 minutes.But let me verify this again because it's crucial to ensure that this is correct. Let's cross-check.If he peels 2 buckets of unpeeled potatoes in 1 hour, that's 2 buckets/hour. Therefore, peeling 4/3 buckets would take (4/3) / 2 = 2/3 hours, which is 40 minutes. During that time, he processes 4/3 buckets of unpeeled potatoes, which results in 4/3 * 0.75 = 1 bucket of peeled potatoes. That matches the requirement. So, 40 minutes seems correct.Alternatively, another way to think about it: in one hour, he produces 1.5 buckets of peeled potatoes. So, the rate is 1.5 buckets per hour. To get 1 bucket, time = 1 / 1.5 = 2/3 hours, which is 40 minutes. Same result. So both methods confirm that.But wait, why did we get the same answer using both approaches? Because peeling rate in terms of unpeeled potatoes and the conversion to peeled potatoes are directly related. So, both perspectives lead to the same time. That's reassuring.But let me make sure there's no misunderstanding in the problem statement. The problem says he took two buckets of unpeeled potatoes and peeled everything in an hour. Meanwhile, 25% of the potatoes went into peels. So, 25% by volume? So, the total peeled potatoes are 75% of the original volume. So, two buckets become 1.5 buckets. So, that's correct.Therefore, to get one bucket of peeled potatoes, he needs to process enough unpeeled potatoes such that 75% of that amount equals 1 bucket. So, as before, 1 / 0.75 = 4/3 buckets of unpeeled. Then, time is (4/3) divided by his peeling rate of 2 buckets per hour.Alternatively, if his peeling rate is 2 buckets of unpeeled per hour, then the rate of producing peeled potatoes is 2 * 0.75 = 1.5 buckets per hour, so time to get 1 bucket is 1 / 1.5 = 2/3 hours.Either way, it's 40 minutes. Therefore, the answer should be 40 minutes, or 2/3 of an hour.Is there any chance I missed something here? Let me consider another angle. Suppose the 25% is by weight, but the problem mentions buckets, which are volume measures. So, we can safely assume it's by volume. Also, the problem doesn't mention any compaction or expansion, just that 25% becomes peels.Alternatively, maybe the problem is interpreted differently. For example, maybe 25% of the potatoes by count went into peels, but that would complicate things since potatoes vary in size, but the problem uses buckets, which is volume. So, the 25% is likely by volume.Another check: if he peels two buckets in an hour, and 25% is lost, then per hour he produces 1.5 buckets. So, the rate is 1.5 buckets per hour, so time for one bucket is 1 / 1.5 hours = 40 minutes. So, that's straightforward.Therefore, I think the answer is 40 minutes. There's no conflicting information here. The key is understanding the relationship between unpeeled and peeled potatoes and translating that into the required time based on the given rate.Wait, but maybe there's another way someone could misinterpret it. For example, thinking that 25% of the peeled potatoes become peels, but that would be a different problem. However, the problem states "25% of the potatoes went into peels," which likely refers to the original unpeeled potatoes. So, 25% of the original volume is peels, 75% is peeled potatoes. So, that interpretation is correct.Another possible confusion: is the 25% by weight or volume? But since buckets are a volume measure, and the problem doesn't specify otherwise, it's safe to assume volume.Therefore, all checks point to 40 minutes as the correct answer.Final AnswerIt took him boxed{dfrac{2}{3}} hours, which is equivalent to 40 minutes.

✅:The trainee cook peeled two buckets of unpeeled potatoes in one hour, with 25% of the potatoes going into peels. This means 75% of the volume remains as peeled potatoes. 1. Conversion from unpeeled to peeled potatoes: - For every bucket of unpeeled potatoes, 75% (or 0.75 buckets) becomes peeled potatoes. - Therefore, peeling two buckets of unpeeled potatoes results in (2 times 0.75 = 1.5) buckets of peeled potatoes.2. Rate of peeling: - The cook peels 2 buckets of unpeeled potatoes per hour. - This results in a rate of (1.5) buckets of peeled potatoes per hour.3. Time to collect one bucket of peeled potatoes: - To find the time required to collect 1 bucket of peeled potatoes, we use the rate of peeled potato production. - The time (t) is given by ( frac{1 text{ bucket}}{1.5 text{ buckets per hour}} = frac{2}{3} text{ hours} ).4. Verification: - The cook needs to peel ( frac{4}{3} ) buckets of unpeeled potatoes to get 1 bucket of peeled potatoes (since ( frac{1}{0.75} = frac{4}{3} )). - At a rate of 2 buckets of unpeeled potatoes per hour, the time required is ( frac{4}{3} div 2 = frac{2}{3} text{ hours} ).Thus, the time it took for him to collect exactly one bucket of peeled potatoes is boxed{dfrac{2}{3}} hours, which is equivalent to 40 minutes.

🔑:To determine the time taken by the trainee cook to get exactly one bucket of peeled potatoes, follow the detailed steps below:1. Identify the amount of uncleaned potatoes in two buckets: - Let the total amount of uncleaned potatoes be (2) buckets.2. Calculate the proportion of potatoes that remains after peeling: - Since (25%) of the potatoes are lost during peeling, this means (75%) of the potatoes remain. - Hence, for (2) buckets of uncleaned potatoes, the amount of peeled potatoes is calculated as: [ 2 text{ buckets} times 0.75 = 1.5 text{ buckets of peeled potatoes} ]3. Determine the amount of peeled potatoes per hour: - It takes (1) hour to peel (2) buckets of uncleaned potatoes, resulting in (1.5) buckets of peeled potatoes. - Thus, the rate at which the trainee cook peels potatoes is: [ 1.5 text{ buckets per hour} ]4. Calculate the time taken to peel exactly one bucket of potatoes: - To find out how long it takes to get exactly one bucket of peeled potatoes, use the proportion of the peeled potatoes rate. [ text{Time} = frac{1 text{ bucket}}{1.5 text{ buckets per hour}} = frac{1}{1.5} text{ hours} = frac{2}{3} text{ hours} ]5. Convert the time to minutes: - Since (1) hour equals (60) minutes, convert (frac{2}{3} text{ hours}) to minutes. [ frac{2}{3} text{ hours} times 60 text{ minutes per hour} = 40 text{ minutes} ] Conclusion:[boxed{40 text{ minutes}}]

🧠:Okay, so I need to find the remainder when the sum of 1⁵ + 2⁵ + 3⁵ + ... + 2007⁵ is divided by 5. Hmm, let's see. I remember that when dealing with remainders, modular arithmetic can be really helpful. Since we're dealing with division by 5, maybe each term modulo 5 can be calculated first, then sum them up, and take modulo 5 again at the end. That should give the remainder. Let me try to structure this step by step.First, perhaps I can find a pattern or formula for the sum of fifth powers modulo 5. But fifth powers... Hmm. Wait, maybe instead of calculating each term individually, there's a periodicity in the residues when numbers are raised to powers modulo 5. Because modulo operations cycle, right? For example, the residues of n⁵ mod 5 might repeat every certain number of terms. Let me check that.Let's recall Fermat's little theorem. It states that if p is a prime number, then for any integer a not divisible by p, a^(p-1) ≡ 1 mod p. Here, 5 is a prime, so for any a not divisible by 5, a^4 ≡ 1 mod 5. Then, a^5 ≡ a mod 5. Wait, that's interesting! So if a is not divisible by 5, a^5 mod 5 is the same as a mod 5. But if a is divisible by 5, then a^5 ≡ 0 mod 5. So in general, for any integer a, a^5 ≡ a mod 5. Is that correct?Let me verify with some examples. Take a = 2. 2⁵ = 32, 32 mod 5 is 2, which is the same as 2 mod 5. Similarly, 3⁵ = 243, 243 mod 5 is 3, same as 3 mod 5. For a = 5, 5⁵ is 3125, which mod 5 is 0, which is the same as 5 mod 5. So yes, it seems that a^5 ≡ a mod 5 for any integer a. That's a crucial observation!Therefore, instead of summing all the fifth powers and then taking modulo 5, we can just sum each term mod 5. So the sum 1⁵ + 2⁵ + ... + 2007⁵ mod 5 is equivalent to (1 + 2 + 3 + ... + 2007) mod 5. That simplifies the problem significantly! Now, I just need to compute the sum of the first 2007 natural numbers and then find its remainder when divided by 5.The formula for the sum of the first n natural numbers is n(n + 1)/2. Let me apply that here. So sum = 2007 * 2008 / 2. Let me compute that. First, 2007 divided by 2 is 1003.5, but since we're dealing with integers, 2007 * 2008 is 2007 multiplied by 2008, and then divided by 2. Let me compute 2007 * 2008 first.Alternatively, maybe compute 2007*2008/2 modulo 5 directly. That might be easier. Because if I can compute each component modulo 5 first, the calculations will be simpler.So, let's compute 2007 mod 5 and 2008 mod 5. 2007 divided by 5: 5*401 = 2005, so 2007 - 2005 = 2. Therefore, 2007 ≡ 2 mod 5. Similarly, 2008 = 2005 + 3, so 2008 ≡ 3 mod 5.Therefore, the sum modulo 5 is (2007 * 2008 / 2) mod 5 ≡ (2 * 3 / 2) mod 5. Let's compute this step by step. 2 * 3 = 6. Then, 6 / 2 = 3. So 3 mod 5 is 3. Therefore, is the sum modulo 5 equal to 3? Wait, that seems straightforward, but let me check if this is correct.Wait a second, division in modular arithmetic isn't as straightforward. When we divide by 2 modulo 5, we have to multiply by the modular inverse of 2 modulo 5. The inverse of 2 modulo 5 is 3 because 2*3=6≡1 mod5. So 6 divided by 2 is equivalent to 6 * 3 mod5, which is 18 mod5=3. So yes, the calculation is correct. Therefore, the sum is congruent to 3 mod5. Therefore, the remainder is 3.But wait, let me make sure that this is correct. Let me check the steps again. First, we used the fact that a^5 ≡ a mod5 for any integer a, so the sum of fifth powers is congruent to the sum of the numbers themselves mod5. Then, the sum from 1 to 2007 is 2007*2008/2. Then, we computed 2007≡2, 2008≡3 mod5. So 2*3=6, 6 divided by 2 is 3 mod5. So the sum is congruent to 3 mod5. Therefore, the remainder is 3. Hmm. That seems correct. But let me verify with a smaller example.Take a smaller sum, say 1⁵ + 2⁵ + 3⁵ + 4⁵ + 5⁵. Let's compute this sum mod5. According to the formula, this should be equivalent to (1 + 2 + 3 + 4 + 5) mod5. Compute the sum: 1+2+3+4+5=15. 15 mod5=0. Now compute the fifth powers:1⁵=1 mod5=12⁵=32 mod5=23⁵=243 mod5=34⁵=1024 mod5=4 (since 1024/5=204*5 + 4)5⁵=3125 mod5=0So sum is 1+2+3+4+0=10 mod5=0. Which matches the result. So the formula works here.Another example: 1⁵ + 2⁵ =1 +32=33 mod5=3. Alternatively, sum is 1+2=3 mod5=3. Which also matches. So the method is correct.Therefore, applying this to the original problem, the sum 1⁵ + ... +2007⁵ mod5 is equivalent to (1 + 2 + ... +2007) mod5=3. Therefore, the remainder is 3. So I think the answer is 3.Wait, but just to be thorough, let me compute the sum from 1 to 2007 step by step. The formula is n(n+1)/2. So n=2007. So 2007*2008/2. Let me compute 2007*2008 first. 2007*2008. Hmm, 2007*(2000 +8)=2007*2000 +2007*8=4,014,000 +16,056=4,030,056. Then divide by 2: 4,030,056 /2=2,015,028. Now, compute 2,015,028 mod5. To find this, we can note that 10≡0 mod5, so we can just look at the last digit. 2,015,028 ends with 8. 8 mod5=3. Therefore, 2,015,028 mod5=3. Which confirms the earlier result. Therefore, the remainder is 3. So yes, the answer is 3.But just to make sure, let's think if there's any possibility that the initial step where we use a^5≡a mod5 is flawed. Let me check for a few numbers:a=0: 0⁵=0≡0 mod5. Correct.a=1:1⁵=1≡1 mod5.a=2:32≡2 mod5. Correct.a=3:243≡3 mod5. Correct.a=4:4⁵=1024≡4 mod5 (since 1024-4=1020, which is divisible by 5). Correct.a=5:5⁵=3125≡0 mod5. Correct.So for all residues mod5 (0,1,2,3,4), the congruence a⁵≡a mod5 holds. Therefore, by the properties of modular arithmetic, it's valid for all integers a. Therefore, the reduction of the sum of fifth powers to the sum of the numbers themselves mod5 is correct.Therefore, the answer is 3. I think that's solid.Final AnswerThe remainder is boxed{3}.

✅:To find the remainder when the sum (1^5 + 2^5 + 3^5 + cdots + 2007^5) is divided by 5, we can use modular arithmetic. By Fermat's Little Theorem, for any integer (a), (a^5 equiv a pmod{5}). This allows us to simplify the sum of fifth powers modulo 5 to the sum of the integers themselves modulo 5. Therefore, we need to compute the sum (1 + 2 + 3 + cdots + 2007 mod 5).The sum of the first (n) natural numbers is given by the formula (frac{n(n+1)}{2}). For (n = 2007):[text{Sum} = frac{2007 times 2008}{2}]We compute the modulo 5 of 2007 and 2008 first:- (2007 equiv 2 pmod{5})- (2008 equiv 3 pmod{5})Thus, the product modulo 5 is:[2007 times 2008 equiv 2 times 3 = 6 pmod{5}]Dividing by 2 in modular arithmetic involves multiplying by the modular inverse of 2 modulo 5. The inverse of 2 modulo 5 is 3 because (2 times 3 equiv 1 pmod{5}). Therefore:[frac{6}{2} equiv 6 times 3 equiv 18 equiv 3 pmod{5}]Hence, the remainder when the sum (1^5 + 2^5 + 3^5 + cdots + 2007^5) is divided by 5 is (boxed{3}).

🔑:1. Applying Fermat's Little Theorem: According to Fermat's Little Theorem, for any integer ( a ) and a prime ( p ), if ( p ) does not divide ( a ), then: [ a^{p-1} equiv 1 pmod{p} ] When applied to ( p = 5 ), this leads to: [ a^5 equiv a pmod{5} ] This congruence holds true for any integer ( a ).2. Simplifying each term in the sum: Given the expression ( 1^5 + 2^5 + 3^5 + cdots + 2007^5 ), using the result from Fermat's Little Theorem, we can simplify: [ 1^5 equiv 1 pmod{5},quad 2^5 equiv 2 pmod{5},quad 3^5 equiv 3 pmod{5},quad ldots,quad 2007^5 equiv 2007 pmod{5} ]3. Sum of the simplified terms: Consequently, the original sum simplifies to: [ 1 + 2 + 3 + cdots + 2007 pmod{5} ]4. Calculating the sum modulo 5: To find the sum modulo 5, we first use the formula for the sum of the first ( n ) positive integers: [ 1 + 2 + 3 + cdots + n = frac{n(n+1)}{2} ] For ( n = 2007 ): [ 1 + 2 + 3 + cdots + 2007 = frac{2007 times 2008}{2} = 2007 times 1004 ]5. Finding the product modulo 5: We need to find the remainder of ( 2007 times 1004 ) modulo 5. Note that: [ 2007 equiv 2 pmod{5} quad (text{since } 2007 = 5 times 401 + 2) ] and [ 1004 equiv 4 pmod{5} quad (text{since } 1004 = 5 times 200 + 4) ] Therefore, the product modulo 5 is: [ 2007 times 1004 equiv 2 times 4 equiv 8 equiv 3 pmod{5} ] Conclusion:Thus, the remainder when the sum ( 1^{5} + 2^{5} + 3^{5} + cdots + 2007^{5} ) is divided by 5 is:[boxed{3}]

🧠:Okay, so I need to prove that if you have a point on the external bisector of an angle, and a transversal passes through this point, cutting segments a and b from the sides of the angle or their extensions, then the absolute value of (1/a - 1/b) is constant, meaning it doesn't depend on where the transversal is. Then, I also have to formulate and prove the converse theorem. Hmm, let me start by understanding the problem step by step.First, let me recall what an external angle bisector is. For a given angle, say angle formed by two lines OA and OB, the internal bisector is the line that splits the angle into two equal parts towards the interior of the angle. The external bisector, on the other hand, would split the angle into two equal parts but towards the exterior. So, if the original angle is less than 180 degrees, the external bisector would be the line such that the angle between OA and the external bisector is equal to the angle between OB and the external bisector, but on the outside of the original angle.Now, the point is on this external bisector. Let's call this point P. Then, a transversal passes through P and intersects the sides of the angle (OA and OB) or their extensions, creating segments of lengths a and b. The claim is that |1/a - 1/b| is constant regardless of how the transversal is positioned (as long as it passes through P). So, even if we rotate the transversal around P, changing a and b, this expression remains the same.I need to prove this. Let me visualize it. Imagine angle AOB with external bisector OP. Point P is somewhere along OP. A transversal through P intersects OA at some point A' and OB (or its extension) at B', with PA' = a and PB' = b. Wait, but actually, the problem says "cuts segments a and b from the sides of the angle or their extensions." So, maybe a is the length from the vertex O to A' and b is from O to B'? Or is it the length from where the transversal intersects OA and OB (or extensions) to P? The problem statement says "cuts segments a and b from the sides of the angle or their extensions." Hmm, maybe the segments are the lengths between the intersection points on the sides and the transversal. So, if the transversal cuts OA at A' and OB at B', then a is the length OA' and b is OB'? Or is it PA' and PB'?Wait, the problem states "cuts segments a and b from the sides of the angle or their extensions." So perhaps a and b are the lengths from the vertex O to the points where the transversal intersects the sides OA and OB (or their extensions). Let me check this.Alternatively, it might be the lengths from the point P to the sides. But the problem says "cuts segments a and b from the sides of the angle or their extensions." So, if a transversal cuts the sides OA and OB (or their extensions) at two points, then a and b are the lengths of the segments from those intersection points to somewhere. Since the transversal passes through P, which is on the external bisector, maybe a and b are the distances from P to the sides? But that seems like it would be fixed if P is fixed. But the problem says the transversal passes through P and cuts segments a and b from the sides. So maybe a and b are the lengths along the transversal from P to the points where it intersects the sides OA and OB (or their extensions). So, for example, if the transversal intersects OA at A' and OB at B', then a is PA' and b is PB'. Then, as the transversal rotates around P, PA' and PB' would change, but |1/a - 1/b| remains constant.Alternatively, if the segments a and b are OA' and OB', then OA' and OB' would vary as the transversal moves, and the expression |1/a - 1/b| is constant. So I need to clarify this.Looking back at the problem: "a transversal passing through this point, which cuts segments a and b from the sides of the angle or their extensions". The wording "cuts segments... from the sides" suggests that a and b are the lengths of the segments on the sides OA and OB (or their extensions) that are cut by the transversal. So, if the transversal intersects OA at A' and OB at B', then a = OA' and b = OB', regardless of the direction of the transversal. Then, the claim is that |1/a - 1/b| is constant, regardless of the transversal's position through P.Alternatively, if a and b are the lengths along the transversal from P to the sides, then PA' = a and PB' = b, but in that case, if P is fixed, rotating the transversal would change a and b. But the problem states that the point is on the external bisector and the transversal passes through this point, cutting segments a and b from the sides. So maybe the segments are OA' and OB', the distances from the vertex to the intersection points. That seems more plausible.Alternatively, perhaps a and b are the lengths from the intersection points on the sides to the point P? That is, PA' = a and PB' = b. Then, as the transversal rotates around P, PA' and PB' change, but |1/a - 1/b| remains constant.I need to figure out which interpretation is correct. Let me think about similar problems. There's a theorem called the Angle Bisector Theorem which relates the ratio of the adjacent sides to the ratio of the segments created by the bisector. But here, it's the external bisector, and the transversal is arbitrary through a fixed point on the bisector.Wait, perhaps using coordinates would help. Let me try to model this.Let's set up a coordinate system. Let the vertex of the angle be at the origin O(0,0). Let one side of the angle be along the positive x-axis, OA, and the other side OB making an angle 2θ with OA (since we are dealing with the external bisector, which would bisect the external angle, so if the internal angle is α, the external angle is π - α, and the external bisector would be at θ from OA, where θ is (π - α)/2. Wait, perhaps it's better to define the angle explicitly.Suppose the angle at O is 2θ, so the internal bisector would split it into two angles of θ each, and the external bisector would split the external angle (which is π - 2θ) into two angles of (π - 2θ)/2 each. But maybe working in coordinates is better.Let me consider angle AOB with OA along the x-axis and OB making an angle of 2θ with OA. Then, the internal bisector would be the line y = tan(θ)x, and the external bisector would be the line y = tan(θ + π/2)x, but wait, that might not be right. Wait, the external bisector is the bisector of the angle supplementary to angle AOB. If angle AOB is 2θ, then the external angle is π - 2θ, so the external bisector would be at an angle of θ + π/2 from OA? Hmm, maybe not. Wait, if the internal bisector is at angle θ from OA, then the external bisector would be at angle θ from the opposite direction? Wait, perhaps it's easier to use coordinate geometry.Let me take OA as the positive x-axis, and let OB be a line making an angle of 2θ with OA. So the angle between OA and OB is 2θ. The internal bisector would be the line that splits 2θ into θ and θ, so its direction is θ from OA. The external bisector would split the angle outside, which is π - 2θ, into two equal angles of (π - 2θ)/2. So, the external bisector would be at an angle of θ + (π - 2θ)/2 = π/2 - θ/2 from OA? Wait, maybe not. Alternatively, the external bisector is perpendicular to the internal bisector if the angle is 90 degrees, but in general, for an angle of 2θ, the internal and external bisectors are separated by π/2.Wait, perhaps another approach. Let me consider the angle between OA and OB as 2θ, so the external angle is π - 2θ. The external bisector is the line that divides this external angle into two equal parts, each of (π - 2θ)/2. So, starting from OA, if we go in the external direction (i.e., the reflex angle), the external bisector would be at an angle of π - (π - 2θ)/2 = π/2 + θ from OA. Wait, perhaps this is getting too confusing. Let me just assign coordinates.Let’s set OA along the positive x-axis, O at (0,0). Let angle AOB be 2θ, so point B is in the plane such that line OB makes an angle of 2θ with OA. The internal bisector is the line making angle θ with OA. The external bisector would be the line that bisects the angle outside the original angle. So, if the original angle is 2θ, the external angle is 2π - 2θ (the reflex angle), but actually, the external angle is considered as the angle formed by one side and the extension of the other side. Wait, maybe that's a better way. If you extend OA beyond O, forming a straight line, then the external angle between OB and the extension of OA is π - 2θ. The external bisector would bisect this angle into two angles of (π - 2θ)/2 each. So, from point O, the external bisector would be a line that is at an angle of π - (π - 2θ)/2 = π/2 + θ from OA? Hmm, maybe.Alternatively, perhaps it's better to assign specific coordinates. Let me take θ as 30 degrees for simplicity, so the angle between OA and OB is 60 degrees. Then, the external bisector would bisect the angle formed by OB and the extension of OA beyond O, which would be 180 - 60 = 120 degrees, so the external bisector would be at 60 degrees from OB towards the extended OA. Wait, maybe. Alternatively, if OA is along the positive x-axis, and OB is at 60 degrees above OA, then the external bisector would be in the opposite direction. Maybe this is getting too complicated without a diagram.Alternatively, maybe using vector geometry or coordinate geometry. Let me place point O at the origin. Let’s suppose OA is along the positive x-axis, and OB is in the upper half-plane making an angle of 2θ with OA. The external bisector would then be the line that bisects the angle between OA and the opposite direction of OB. Wait, perhaps. Alternatively, if we consider the external angle as the angle between one side and the extension of the other side, then the external bisector can be constructed. For example, if OA is along the positive x-axis, and OB is at an angle 2θ from OA, then extending OA beyond O to the negative x-axis, the external angle between OB and the extended OA is π - 2θ. The external bisector would bisect this angle into two angles of (π - 2θ)/2 each. Therefore, the external bisector would be a line starting at O and making an angle of (π - 2θ)/2 with the extended OA (i.e., the negative x-axis). So, from the negative x-axis, going towards OB, the external bisector is at (π - 2θ)/2. Therefore, in standard coordinates, the external bisector would be at an angle of π - (π - 2θ)/2 = (π + 2θ)/2 from the positive x-axis. So, angle (π + 2θ)/2.Alternatively, if θ is the angle of the internal bisector, then the external bisector would be perpendicular to the internal bisector if the original angle is 90 degrees. But perhaps this is not necessary. Let me try to define coordinates.Let’s define OA as the positive x-axis, O at (0,0), and OB as the line making an angle 2θ with OA. The external bisector can be found as follows: The external angle is π - 2θ, so the external bisector is at an angle of (π - 2θ)/2 from each side forming the external angle. The external angle is formed by OB and the extension of OA beyond O (i.e., the negative x-axis). Therefore, the external bisector is a line that is at an angle of (π - 2θ)/2 from both OB and the negative x-axis.Therefore, starting from the negative x-axis, the external bisector is at an angle of (π - 2θ)/2 towards OB. Therefore, in standard position (from the positive x-axis), the external bisector is at an angle of π - (π - 2θ)/2 = (π + 2θ)/2. So, the external bisector has direction (π + 2θ)/2 from the positive x-axis.So, for example, if the original angle 2θ is 60 degrees (θ = 30 degrees), then the external bisector is at (π + π/3)/2 = (4π/3)/2 = 2π/3 radians, which is 120 degrees from the positive x-axis. So, in this case, the external bisector is along the 120-degree direction.Therefore, the external bisector is a line starting at O and going in the direction of (π + 2θ)/2.Now, we have a point P on this external bisector. Let's parametrize point P. Since it's on the external bisector, its coordinates can be written as P(t) = t*(cos((π + 2θ)/2), sin((π + 2θ)/2)), where t is a scalar (positive if P is on the external bisector beyond O, negative if in the opposite direction, but probably we can take t > 0 for simplicity). Wait, but since the external bisector is a line, the point P can be anywhere along it. So, depending on the position of P, t varies.Now, a transversal passes through P and cuts segments a and b from the sides OA and OB (or their extensions). Let me consider a line passing through P(t) that intersects OA and OB (or their extensions). Let me parametrize the transversal.Let’s denote the transversal as a line passing through P(t) with some slope m. This line will intersect OA (the x-axis) and OB (the line making an angle 2θ with OA). Let me find the coordinates of these intersection points.First, OA is the x-axis (y=0). The transversal passes through P(t) = (t*cos((π + 2θ)/2), t*sin((π + 2θ)/2)). Let me compute cos((π + 2θ)/2) and sin((π + 2θ)/2). Using trigonometric identities:cos((π + 2θ)/2) = cos(π/2 + θ) = -sinθsin((π + 2θ)/2) = sin(π/2 + θ) = cosθTherefore, coordinates of P(t) are (-t sinθ, t cosθ). Hmm, interesting. So regardless of θ, the coordinates are (-t sinθ, t cosθ). Let's verify with θ = 30 degrees (π/6 radians). Then, cos((π + π/3)/2) = cos(2π/3) = -1/2, sin(2π/3) = √3/2. So, P(t) would be (-t*(1/2), t*(√3/2)), which is equivalent to t*(-1/2, √3/2), which is indeed 120 degrees from the positive x-axis. So, correct.So, point P(t) has coordinates (-t sinθ, t cosθ). Now, a transversal through P(t) can be represented parametrically. Let's denote the direction of the transversal by an angle φ, but maybe using parametric equations is better.Suppose the transversal has a slope m, so its equation is y - t cosθ = m(x + t sinθ). We can find where this line intersects OA (y=0) and OB.First, intersection with OA (y=0):0 - t cosθ = m(x + t sinθ)=> x = (-t cosθ / m) - t sinθSo, the intersection point A' is at ( (-t cosθ / m) - t sinθ, 0 )Similarly, intersection with OB. Let's find the equation of OB. Since OB makes an angle 2θ with OA (the x-axis), its slope is tan(2θ). Therefore, the equation of OB is y = tan(2θ) x.Find the intersection of the transversal with OB:Set y = tan(2θ)x into the transversal's equation:tan(2θ)x - t cosθ = m(x + t sinθ)Solve for x:tan(2θ)x - m x = m t sinθ + t cosθx (tan(2θ) - m) = t (m sinθ + cosθ)Therefore,x = [ t (m sinθ + cosθ) ] / (tan(2θ) - m )Then, y = tan(2θ) x = [ t tan(2θ) (m sinθ + cosθ) ] / (tan(2θ) - m )Therefore, the intersection point B' is at ( [ t (m sinθ + cosθ) ] / (tan(2θ) - m ), [ t tan(2θ) (m sinθ + cosθ) ] / (tan(2θ) - m ) )Now, we need to find the lengths a and b. The problem states that the transversal cuts segments a and b from the sides OA and OB or their extensions. The question is, are a and b the distances from O to A' and O to B', or from P to A' and P to B'? The problem says "cuts segments a and b from the sides of the angle or their extensions". So, if the transversal intersects OA at A', then the segment cut from OA is OA', and similarly, the segment cut from OB is OB'. However, if the transversal intersects the extension of OA beyond O, then OA' would be negative, depending on the direction.But the problem mentions "segments a and b", so perhaps we take the absolute lengths, regardless of direction. Alternatively, a and b could be signed lengths, but since the expression is |1/a - 1/b|, the absolute value would take care of any sign issues. However, since the problem mentions "segments", it's more natural to take a and b as positive lengths. Therefore, OA' and OB' would be the lengths from O to the intersection points on OA and OB (or their extensions), with a = |OA'| and b = |OB'|. However, given that the transversal passes through P, which is on the external bisector, the intersections could be on the extensions of OA and OB.But let's proceed with the coordinates. Let's compute OA' and OB'.First, OA' is the x-coordinate of A', which we found as (-t cosθ / m) - t sinθ. Since OA is along the x-axis, OA' is just the x-coordinate. However, depending on the slope m, this could be positive or negative. If the intersection is on OA (the positive x-axis), then OA' is positive. If it's on the extension beyond O, then OA' is negative. Similarly, OB' is the distance from O to B', which can be computed using the coordinates of B'.The distance OB' is sqrt(x_B'^2 + y_B'^2). But since B' is on OB, which is a straight line from O, the distance OB' can be expressed as the scalar multiple along the direction of OB. Alternatively, if OB is parametrized as (r cos2θ, r sin2θ), then the distance OB' is r. But in our case, the coordinates of B' are ( [ t (m sinθ + cosθ) ] / (tan2θ - m ), [ t tan2θ (m sinθ + cosθ) ] / (tan2θ - m ) )Therefore, the distance OB' is sqrt( x_B'^2 + y_B'^2 ) = sqrt( [ t (m sinθ + cosθ) / (tan2θ - m ) ]^2 [1 + tan^2 2θ ] )But 1 + tan^2 2θ = sec^2 2θ, so sqrt becomes [ t (m sinθ + cosθ) / (tan2θ - m ) ] * sec2θTherefore, OB' = [ t (m sinθ + cosθ) / (tan2θ - m ) ] * sec2θSimplify tan2θ = sin2θ / cos2θ, so sec2θ = 1 / cos2θ.Therefore, OB' = [ t (m sinθ + cosθ) / ( (sin2θ / cos2θ ) - m ) ] * (1 / cos2θ ) = [ t (m sinθ + cosθ) / ( (sin2θ - m cos2θ ) / cos2θ ) ] * (1 / cos2θ )Simplify denominator: (sin2θ - m cos2θ ) / cos2θSo, dividing by that is multiplying by cos2θ / (sin2θ - m cos2θ )Thus, OB' = t (m sinθ + cosθ ) * [ cos2θ / (sin2θ - m cos2θ ) ] * (1 / cos2θ ) = t (m sinθ + cosθ ) / (sin2θ - m cos2θ )Similarly, OA' is the x-coordinate of A', which is (-t cosθ / m ) - t sinθ. Let's factor out -t:OA' = -t ( cosθ / m + sinθ ) = -t ( (cosθ + m sinθ ) / m )But OA' is the distance from O to A', so depending on the sign, we take the absolute value. However, since we are dealing with segments cut from OA or its extension, the length a is |OA'|. So, a = | -t ( (cosθ + m sinθ ) / m ) | = t | (cosθ + m sinθ ) / m |. Similarly, b = |OB'| = | t (m sinθ + cosθ ) / (sin2θ - m cos2θ ) |.But note that in the expression for OB', m sinθ + cosθ is the same as in OA', but in the denominator for OB', we have sin2θ - m cos2θ.But let's note that sin2θ = 2 sinθ cosθ and cos2θ = cos^2θ - sin^2θ.Therefore, sin2θ - m cos2θ = 2 sinθ cosθ - m (cos^2θ - sin^2θ )But perhaps instead of expanding, we can relate a and b.We have:a = t | (cosθ + m sinθ ) / m | = t | (cosθ / m + sinθ ) |b = t | (m sinθ + cosθ ) / (sin2θ - m cos2θ ) | = t | (cosθ + m sinθ ) / (sin2θ - m cos2θ ) |Notice that the numerator of b is the same as the numerator inside the absolute value of a, except a has an extra division by m. Wait, actually, in a:From OA', a = | -t ( (cosθ + m sinθ ) / m ) | = t | (cosθ + m sinθ ) / m | = t | (cosθ / m + sinθ ) |But in b, we have t | (cosθ + m sinθ ) / (sin2θ - m cos2θ ) |.So, if we denote k = cosθ + m sinθ, then a = t |k / m| and b = t |k / (sin2θ - m cos2θ )|.Therefore, 1/a - 1/b = (1 / (t |k / m| )) - (1 / (t |k / (sin2θ - m cos2θ )| )) = ( |m| / (t |k| ) ) - ( |sin2θ - m cos2θ | / (t |k| ) )Therefore, |1/a - 1/b| = (1 / (t |k| )) | |m| - |sin2θ - m cos2θ | |.But this seems complicated. Maybe there's a better approach. Let's see if we can find a relationship between a and b.Alternatively, perhaps using similar triangles or the properties of the external bisector.Wait, since point P is on the external bisector, maybe there's a relation of distances from P to the sides of the angle. Wait, in general, for a point on the angle bisector, the distances to the sides are equal. But here, it's the external bisector. For the external bisector, the ratio of the distances from any point on the external bisector to the sides of the angle is constant. Wait, actually, for the internal bisector, the distances are equal. For the external bisector, the ratio of distances is equal to the ratio of the adjacent sides. Wait, no, the Angle Bisector Theorem states that the internal bisector divides the opposite side in the ratio of the adjacent sides. Similarly, the external bisector also divides the opposite side externally in the same ratio.But perhaps in this case, since P is on the external bisector, the ratio of the distances from P to the sides OA and OB is constant. Let me recall that for a point on the external bisector, the ratio of the distances to the two sides is equal to the ratio of the lengths of the sides. Wait, but in this case, the original angle is between OA and OB, which are both sides of the angle. Wait, perhaps not. Wait, if the external bisector is the bisector of the external angle, then the ratio of the distances from any point on the external bisector to the two sides OA and OB is equal to the ratio of the lengths of OA and OB. But if OA and OB are just the sides of the angle (not necessarily of any specific length), then the ratio is 1, since the external bisector is equidistant in terms of angle, but maybe not in terms of Euclidean distance. Wait, maybe not.Alternatively, perhaps we can use the formula for the distance from a point to a line. Let's compute the distance from P(t) to OA and to OB.Since OA is the x-axis, the distance from P(t) to OA is simply the y-coordinate of P(t), which is t cosθ.The distance from P(t) to OB. The line OB has equation y = tan2θ x. The distance from point P(t) = (-t sinθ, t cosθ) to line OB is |tan2θ*(-t sinθ) - t cosθ + 0| / sqrt(tan^2 2θ + 1) )Wait, the formula for the distance from a point (x0, y0) to the line ax + by + c = 0 is |ax0 + by0 + c| / sqrt(a^2 + b^2). So, first, let's write the equation of OB in standard form. OB is y = tan2θ x, which can be rewritten as tan2θ x - y = 0. Therefore, a = tan2θ, b = -1, c = 0.Thus, the distance from P(t) to OB is |tan2θ*(-t sinθ) - 1*(t cosθ) + 0| / sqrt(tan^2 2θ + 1 )= | -t sinθ tan2θ - t cosθ | / sqrt(tan^2 2θ + 1 )Factor out -t:= | -t ( sinθ tan2θ + cosθ ) | / sqrt(tan^2 2θ + 1 )= t | sinθ tan2θ + cosθ | / sqrt(tan^2 2θ + 1 )Note that sqrt(tan^2 2θ + 1 ) = sec2θ, so denominator is sec2θ.Thus, distance from P(t) to OB is t | sinθ tan2θ + cosθ | / sec2θ = t | sinθ tan2θ + cosθ | * cos2θSimplify sinθ tan2θ:tan2θ = sin2θ / cos2θ = 2 sinθ cosθ / cos2θTherefore, sinθ tan2θ = sinθ * 2 sinθ cosθ / cos2θ = 2 sin^2θ cosθ / cos2θTherefore, sinθ tan2θ + cosθ = (2 sin^2θ cosθ / cos2θ ) + cosθ = cosθ ( 2 sin^2θ / cos2θ + 1 )But 2 sin^2θ = 1 - cos2θ, so substituting:= cosθ ( (1 - cos2θ)/cos2θ + 1 ) = cosθ ( (1 - cos2θ + cos2θ ) / cos2θ ) = cosθ (1 / cos2θ )Therefore, sinθ tan2θ + cosθ = cosθ / cos2θTherefore, the distance from P(t) to OB is t | cosθ / cos2θ | * cos2θ = t | cosθ |. Since θ is between 0 and π/2, cosθ is positive, so distance is t cosθ.Wait, that's interesting. The distance from P(t) to OA is t cosθ, and the distance from P(t) to OB is also t cosθ. So, the distances from P to both sides OA and OB are equal? That seems like the property of the angle bisector. Wait, but we are talking about the external bisector here. Wait, but according to this calculation, the distances are equal. That suggests that the external bisector is also equidistant from both sides, which is a property of the internal bisector. Wait, perhaps there's a mistake here.Wait, no, the internal bisector is equidistant to both sides. The external bisector, however, is equidistant in terms of angles but not necessarily in Euclidean distance. But according to the calculation above, the distances from P(t) to OA and OB are both t cosθ. So that suggests that the external bisector is also equidistant to both sides, which seems contradictory. Wait, perhaps the error comes from how we defined the external bisector.Wait, in our coordinate system, point P(t) is at (-t sinθ, t cosθ). The distance to OA (y=0) is indeed t cosθ. The distance to OB, which is the line y = tan2θ x, calculated as t cosθ. So, according to this, any point on the external bisector is equidistant to both sides OA and OB. But that's the property of the internal bisector. Hmm, so perhaps there's a confusion here between internal and external bisectors.Wait, but in reality, the external bisector should not be equidistant to both sides. Wait, maybe my coordinate system is flawed. Let me check again.Wait, if OA is the x-axis, and OB is at angle 2θ, then the internal bisector would be the line that splits angle AOB into two angles of θ each. The external bisector would split the external angle, which is the angle between OB and the extension of OA beyond O, which is π - 2θ. The external bisector would be the line that splits this π - 2θ angle into two equal angles. However, in this case, the external bisector is not equidistant to OA and OB, but the internal bisector is.But according to the calculation above, the distance from P(t) to both OA and OB is t cosθ. This suggests that the external bisector, as defined, is also equidistant to OA and OB, which is not correct. Therefore, there must be an error in the calculation.Wait, let's recast the problem. Let me take θ = 30 degrees, so 2θ = 60 degrees. Then, OA is along the x-axis, OB is at 60 degrees. The internal bisector is at 30 degrees, and the external bisector is bisecting the external angle of 120 degrees (180 - 60), so the external bisector is at 60 degrees from the negative x-axis, which is 120 degrees from the positive x-axis. Then, point P(t) is at (-t sinθ, t cosθ). For θ = 30 degrees, sinθ = 1/2, cosθ = √3/2. So, P(t) is at (-t/2, (t√3)/2 ). Now, the distance from P(t) to OA (y=0) is indeed (t√3)/2. The distance from P(t) to OB: line OB is at 60 degrees. The distance from P(t) to OB should be different. Wait, let's compute it.Line OB: passing through the origin at 60 degrees, equation y = tan60° x = √3 x. The distance from point (-t/2, (t√3)/2 ) to line √3 x - y = 0 is |√3*(-t/2) - (t√3)/2 + 0| / sqrt( (√3)^2 + (-1)^2 ) = | - (√3 t)/2 - (√3 t)/2 | / 2 = | -√3 t | / 2 = (√3 t)/2. Which is the same as the distance to OA. So in this case, the distance from P(t) to both OA and OB is the same. So, even though P(t) is on the external bisector, it's equidistant to both sides. That seems contradictory to my initial thought. Wait, but maybe in this specific case, when the external bisector is constructed, it is equidistant to both sides. Wait, no, that's supposed to be a property of the internal bisector. So perhaps the external bisector in this case also has that property? Or maybe my understanding is wrong.Wait, according to the Angle Bisector Theorem, the internal bisector divides the opposite side in the ratio of the adjacent sides. The external bisector divides it externally in the same ratio. However, regarding distances, for any point on the internal bisector, the distances to both sides are equal. For the external bisector, perhaps the ratio of the distances is equal to the ratio of the sides, but if the sides are of equal length, then the ratio is 1. Wait, but in our case, OA and OB are just sides of an angle, not necessarily of any specific length. So, if the angle is 2θ, then OA and OB are of arbitrary length. But in our coordinate system, OA is along the x-axis, and OB is a line at angle 2θ. So, OA and OB are infinite lines.But the calculation shows that the distance from P(t) to both OA and OB is the same, which is t cosθ. That suggests that the external bisector is also a locus of points equidistant from OA and OB. But that contradicts the standard geometric understanding, where only the internal bisector has that property. So, something is wrong here.Wait, perhaps the confusion arises because in this case, when we talk about external bisector, in the case of a straight angle, the external and internal bisectors coincide? No, in this case, the angle is 60 degrees, so external angle is 120 degrees, and the external bisector is at 60 degrees from each side forming the external angle, which are OB and the extension of OA. Wait, but in our coordinate system, the external bisector was calculated to be equidistant to OA and OB, which seems incorrect.Wait, maybe the error is in the definition of external bisector. Let me check another source. The external angle bisector is the bisector of the angle formed by one side and the extension of the other side. For angle AOB, the external bisector would be the line that bisects the angle between OB and the extension of OA beyond O. The set of points equidistant to OA and OB is the internal bisector. The external bisector, however, consists of points where the ratio of distances to OA and OB is equal to the ratio of the lengths of OA and OB. If OA and OB are considered to be of equal length (since they are sides of an angle, not actual segments), then the ratio is 1, and hence the external bisector would also consist of points equidistant to OA and OB. But in that case, the internal and external bisectors would both be equidistant, which is not the case.Wait, this is conflicting. Let me think differently. Suppose OA and OB are two lines forming an angle. The internal bisector is the locus of points equidistant from OA and OB. The external bisector is another line, different from the internal bisector, which also has some relation to the distances from OA and OB. Wait, but according to the calculation above, points on the external bisector are equidistant to OA and OB. So perhaps both internal and external bisectors consist of points equidistant to OA and OB. But that contradicts standard geometry. Wait, no, actually, no. The set of points equidistant to two lines is both the internal and external bisectors. Wait, but for two intersecting lines, the set of points equidistant to both lines is both angle bisectors (internal and external). Because for any two intersecting lines, there are two angle bisectors, each dividing one pair of vertical angles. Therefore, both the internal and external bisectors are equidistant to the two lines. Therefore, in that case, both bisectors are loci of points equidistant to OA and OB.Wait, that makes sense. Because if you have two intersecting lines, they form two pairs of vertical angles. Each pair has its own bisector. Both bisectors are equidistant to the original two lines. Therefore, both internal and external bisectors are equidistant. Therefore, in our problem, since P is on the external bisector, it is equidistant to OA and OB, and the distance is t cosθ.Therefore, coming back to the problem. The transversal passes through P, which is equidistant to OA and OB. The transversal cuts segments a and b from OA and OB (or their extensions). We need to show that |1/a - 1/b| is constant.Given that a and b are the lengths from O to the intersection points on OA and OB (or their extensions). Let's suppose OA' = a and OB' = b. Then, using the coordinates we found earlier:OA' = x_A' = (-t cosθ / m ) - t sinθOB' = [ t (m sinθ + cosθ ) ] / (sin2θ - m cos2θ )But we need to relate these expressions to show that |1/a - 1/b| is constant.Alternatively, let's express m in terms of a or b and substitute.Alternatively, perhaps express 1/a - 1/b.First, let's compute 1/a - 1/b.Given a = |OA'| = | (-t cosθ / m ) - t sinθ |. But OA' is a length, so we can take absolute value.But let's suppose that the intersection is on the extension of OA beyond O, so OA' is negative. However, since a is a length, we take a = |OA'|.But to simplify, let's consider the case where the transversal intersects OA and OB beyond O, i.e., on the extensions. Wait, but depending on the slope m, the transversal could intersect OA on the positive side or the negative side. Similarly for OB.But perhaps instead of absolute values, we can work with directed segments, keeping track of signs. Let's try that.Let’s define a and b as directed lengths. So, if the intersection is on the OA side beyond O, a is positive, otherwise negative. Similarly for b. Then, the expression 1/a - 1/b would account for the signs, and the absolute value would give the magnitude.Alternatively, perhaps there's a relation between a and b due to P being on the external bisector.Wait, since P is equidistant to OA and OB, and the transversal passes through P, maybe there's a harmonic division or something related to reciprocal lengths.Alternatively, let's use areas. The area of triangle OA'P is (1/2)*a*d, where d is the distance from P to OA, which is t cosθ. Similarly, the area of triangle OB'P is (1/2)*b*d, where d is the distance from P to OB, which is also t cosθ. But since P is on both triangles, maybe the areas relate to a and b. However, since the transversal passes through P, the areas would depend on the angles.Alternatively, use coordinates to express a and b, then compute 1/a - 1/b.From earlier, we have:a = OA' = (-t cosθ / m ) - t sinθ = -t (cosθ / m + sinθ )Similarly, b = OB' = t (m sinθ + cosθ ) / (sin2θ - m cos2θ )Let’s compute 1/a - 1/b.1/a = 1 / [ -t (cosθ / m + sinθ ) ] = -1 / [ t ( cosθ / m + sinθ ) ]1/b = [ sin2θ - m cos2θ ] / [ t (m sinθ + cosθ ) ]Therefore, 1/a - 1/b = -1 / [ t ( cosθ / m + sinθ ) ] - [ sin2θ - m cos2θ ] / [ t (m sinθ + cosθ ) ]Factor out 1/t:= (1/t)[ -1 / ( cosθ / m + sinθ ) - ( sin2θ - m cos2θ ) / (m sinθ + cosθ ) ]Let’s combine the terms inside the brackets:Let’s denote the first term as T1 = -1 / ( cosθ / m + sinθ )Second term T2 = - ( sin2θ - m cos2θ ) / (m sinθ + cosθ )So, the expression is (1/t)(T1 + T2)Compute T1:T1 = -1 / ( (cosθ + m sinθ ) / m ) ) = -m / (cosθ + m sinθ )T2 = - ( sin2θ - m cos2θ ) / (m sinθ + cosθ )Therefore, T1 + T2 = [ -m / (cosθ + m sinθ ) ] + [ - ( sin2θ - m cos2θ ) / (cosθ + m sinθ ) ]= [ -m - sin2θ + m cos2θ ] / (cosθ + m sinθ )Factor numerator:= [ -m - sin2θ + m cos2θ ] = -m(1 - cos2θ ) - sin2θNote that 1 - cos2θ = 2 sin²θ, so:= -m(2 sin²θ ) - sin2θ = -2 m sin²θ - 2 sinθ cosθFactor out -2 sinθ:= -2 sinθ ( m sinθ + cosθ )Therefore, T1 + T2 = [ -2 sinθ ( m sinθ + cosθ ) ] / ( cosθ + m sinθ ) = -2 sinθTherefore, the expression becomes (1/t)( -2 sinθ ) = -2 sinθ / tTherefore, |1/a - 1/b| = | -2 sinθ / t | = 2 sinθ / tSince θ and t are constants (θ is half the angle of the external bisector, and t is the distance from O to P along the external bisector), this expression 2 sinθ / t is a constant, independent of m, which determines the slope of the transversal. Therefore, |1/a - 1/b| is constant, not depending on the position of the transversal.Therefore, the proof is complete. The key was to express a and b in terms of the slope m, compute 1/a - 1/b, and find that all terms involving m cancel out, leaving a constant dependent only on θ and t, which are fixed for a given point P on the external bisector.Now, for the converse theorem: If for a point P and any transversal through P cutting segments a and b from the sides of an angle or their extensions, the quantity |1/a - 1/b| is constant, then P lies on the external bisector of the angle.To prove the converse, assume that |1/a - 1/b| is constant for all transversals through P. We need to show that P must lie on the external bisector.Suppose that P is not on the external bisector. Then, the distances from P to OA and OB are not equal (since only points on the internal/external bisectors are equidistant to the sides). However, in our previous calculation, we saw that if P is equidistant to OA and OB (i.e., on one of the bisectors), then |1/a - 1/b| is constant. Conversely, if |1/a - 1/b| is constant for all transversals through P, then the expression derived earlier must hold, which required that the distances from P to OA and OB are equal (since otherwise, the expression would depend on m). Therefore, if the distances were unequal, then the term involving m would not cancel out, and |1/a - 1/b| would depend on m, contradicting the assumption. Hence, P must be equidistant from OA and OB, meaning it lies on one of the angle bisectors. However, the internal bisector would require a different sign in the expression (without absolute value), so considering the external bisector specifically, P must lie on the external bisector.Alternatively, more rigorously, suppose that for any transversal through P cutting segments a and b, |1/a - 1/b| is constant. We can use coordinate geometry again. Assume the same coordinate system as before, with O at the origin, OA on the x-axis, and OB at angle 2θ. Let P be at some point (h, k). For the quantity |1/a - 1/b| to be constant regardless of the slope m of the transversal, the earlier computation shows that this is only possible if the distances from P to OA and OB are equal, which places P on one of the bisectors. Since the problem is about the external bisector, the converse would state that if |1/a - 1/b| is constant, then P is on the external bisector. However, technically, it could be on either the internal or external bisector, since both are equidistant. But the difference would be in the sign. Since the problem uses the external bisector, perhaps the constant's sign determines which bisector it is. However, since the problem uses the absolute value, both bisectors would give the same absolute value. Therefore, the converse theorem would state that if |1/a - 1/b| is constant for all transversals through P, then P lies on one of the angle bisectors (internal or external). However, the original theorem specifies the external bisector. Therefore, to make the converse precise, we might need to adjust the statement to specify the external bisector, possibly by considering the sign in the expression without absolute value. But given the problem mentions the external bisector, the converse should probably be formulated accordingly.Therefore, the converse theorem would be: If a point P is such that for every transversal through P cutting segments a and b from the sides of an angle or their extensions, the quantity |1/a - 1/b| is constant, then P lies on the external bisector of the angle.To prove this, assume |1/a - 1/b| = C for some constant C and all transversals through P. Using the previous calculation, we found that this implies 2 sinθ / t = C, where t is the distance along the external bisector. Therefore, sinθ and t must be constants determined by the position of P. Hence, P must lie on the external bisector such that this relation holds. If P were not on the external bisector, the expression would depend on m, contradicting the constancy. Hence, P must lie on the external bisector.Final Answerboxed{left|frac{1}{a} - frac{1}{b}right| text{ is constant and the converse holds as stated.}}

✅:Given a point on the external bisector of an angle and a transversal passing through this point, which cuts segments (a) and (b) from the sides of the angle or their extensions, we need to prove that (left|frac{1}{a} - frac{1}{b}right|) does not depend on the position of the transversal. We also need to formulate and prove the converse theorem. Proof1. Setup and Coordinates: - Consider an angle ( angle AOB ) with vertex ( O ) and sides ( OA ) and ( OB ). - The external bisector of ( angle AOB ) is considered. A point ( P ) on this external bisector is parametrized as ( P(t) = (-t sin theta, t cos theta) ) in a coordinate system where ( O ) is the origin, ( OA ) is along the x-axis, and ( OB ) makes an angle ( 2theta ) with ( OA ).2. Intersection Points: - A transversal through ( P ) intersects ( OA ) (or its extension) at ( A' ) and ( OB ) (or its extension) at ( B' ). - The coordinates of ( A' ) and ( B' ) are derived using the equation of the transversal line passing through ( P ).3. Segments (a) and (b): - The lengths ( a ) and ( b ) are the distances from ( O ) to ( A' ) and ( O ) to ( B' ), respectively. - Using coordinate geometry, these lengths are expressed in terms of the slope ( m ) of the transversal.4. Expression for ( left|frac{1}{a} - frac{1}{b}right| ): - After calculating ( a ) and ( b ) in terms of ( m ), the expression ( left|frac{1}{a} - frac{1}{b}right| ) is simplified. - Key steps involve trigonometric identities and algebraic manipulation, leading to the result that the expression simplifies to ( frac{2 sin theta}{t} ), which is a constant independent of ( m ). Converse TheoremStatement: If for a point ( P ) and any transversal through ( P ) cutting segments ( a ) and ( b ) from the sides of an angle or their extensions, the quantity ( left|frac{1}{a} - frac{1}{b}right| ) is constant, then ( P ) lies on the external bisector of the angle.Proof:- Assume ( left|frac{1}{a} - frac{1}{b}right| ) is constant for all transversals through ( P ).- Using the derived expression, this constancy implies that the distances from ( P ) to the sides ( OA ) and ( OB ) must be equal.- Equality of distances from ( P ) to both sides of the angle implies ( P ) lies on one of the angle bisectors. Given the problem context and the use of absolute value, ( P ) must specifically lie on the external bisector. Final Answer[boxed{left|frac{1}{a} - frac{1}{b}right| text{ is constant and the converse holds as stated.}}]

🔑:To solve this problem, we first need to understand the indications from problem № 270, as the solution hints that it is related. The problem deals with a point on the external angle bisector of an angle with a line intersecting the sides of the angle at segments a and b, and we aim to show that left|frac{1}{a}-frac{1}{b}right| is independent of the position of the secant line.# Step 1: Set Up the ProblemLet’s denote the given angle as angle XAY and the external bisector as line l. Assume that there is a point P on the external bisector l, and a secant line through P intersecting the sides AX and AY at points B and C, respectively.# Step 2: Use Similar TrianglesSince P lies on the external bisector of the angle angle XAY:- triangle ABP and triangle ACP are similar by the Angle-Angle (AA) criterion for similarity (as they share angle angle BAP or angle CAP and have corresponding angles at A).# Step 3: Relationship with AnglesSince triangle ABP sim triangle ACP, we have:[frac{AP}{AP} = frac{AB}{AC} = frac{PB}{PC}]However, since P is on the external bisector, there’s an additional property related to the segments cut off by any secant through P. We know that:[angle BAX = angle CAY]Thus, the triangle similarity tells us that the lengths AB and AC are proportional under the angle bisector theorem modified for external bisectors.From the Angle-Bisector Theorem, we have:[frac{AB}{BX} = frac{AC}{CY}]But since the bisector is external and we consider segments forming a secant, the essence of a and b is:[a = AB + BX quad text{and} quad b = AC + CY]# Step 4: Deriving left|frac{1}{a} - frac{1}{b}right|Since AB and AC are segments proportional to BX and CY, let's substitute:[a = k_1 cdot AB quad text{and} quad b = k_2 cdot AC]with certain constants of proportionality, k_1 and k_2. From the similarity relations, it follows that:[frac{1}{a} = frac{1}{k_1 cdot AB}quad text{and}quad frac{1}{b} = frac{1}{k_2 cdot AC}]Thus,[left|frac{1}{a} - frac{1}{b}right| = left|frac{1}{k_1 cdot AB} - frac{1}{k_2 cdot AC}right|]By cross multiplication and relations obtained from the angle bisector properties,[left| frac{AC}{k_2 cdot AB cdot AC} - frac{AB}{k_1 cdot AB cdot AC} right|= left| frac{k_2}{k_1 cdot k_2 cdot k_2} - frac{k_1}{k_1 cdot k_1 cdot k_2} right|= frac{|k_2 - k_1|}{k_1 cdot k_2}]Thus it is consistent and does not depend on the particular position of the secant lines as constants k_1 and k_2 serve the equal proportions. Conclusion:[boxed{left|frac{1}{a} - frac{1}{b}right| text{ is independent of the position of the secant line}}]