💡First, I need to find the vectors AB and AC using the given points A, B, and C.For vector AB, I subtract the coordinates of point A from point B:AB = (3 - 2, -2 - (-1), 1 - 1) = (1, -1, 0).For vector AC, I subtract the coordinates of point A from point C:AC = (0 - 2, 1 - (-1), -1 - 1) = (-2, 2, -2).Next, I calculate the dot product of vectors AB and AC:AB · AC = (1)(-2) + (-1)(2) + (0)(-2) = -2 - 2 + 0 = -4.Then, I find the magnitude of vector AC:|AC| = √((-2)² + 2² + (-2)²) = √(4 + 4 + 4) = √12 = 2√3.Finally, I determine the magnitude of the projection vector of AB onto AC by dividing the absolute value of the dot product by the magnitude of AC:|Projection| = |AB · AC| / |AC| = |-4| / 2√3 = 4 / 2√3 = 2√3 / 3.Therefore, the magnitude of the projection vector is 2√3 divided by 3.

💡Alright, so I have this problem where there are ten coins, each with different weights, and I need to figure out if I can identify the heaviest and the lightest coin using just 13 weighings on a balance scale. Hmm, okay, let's break this down.First, I know that a balance scale can compare two groups of coins and tell me which side is heavier, lighter, or if they're equal. Since all the coins have different weights, I don't have to worry about equal weights in my comparisons. That simplifies things a bit.Now, my goal is to find both the heaviest and the lightest coin. If I think about it, finding the heaviest and the lightest separately might take a lot of weighings, but maybe there's a smarter way to do it together. I remember something about tournaments in sorting algorithms where you can find the maximum and minimum with fewer comparisons than sorting the entire list.Let me recall: in a tournament method, to find the maximum, you can have a series of pairwise comparisons, and the winner of each comparison moves on to the next round until you find the maximum. Similarly, for the minimum, you can do the same. But if I do both simultaneously, maybe I can save some weighings.Wait, how many weighings would that take? For ten coins, if I pair them up, I can have five weighings in the first round to find the heavier and lighter in each pair. Then, in the next round, I can take the winners (heaviest from each pair) and compare them to find the overall heaviest. Similarly, take the losers (lightest from each pair) and compare them to find the overall lightest.But hold on, that would be five weighings in the first round, then two more rounds for the heaviest and two more for the lightest, totaling nine weighings. That's way under 13. Maybe I'm missing something.Oh, right, because in reality, after the first round, the heaviest and lightest are already identified in their respective groups, but to confirm them, I might need additional weighings. Also, I need to ensure that I haven't missed any potential candidates for the heaviest or lightest.Let me think differently. Maybe I can use a method where I divide the coins into groups and compare them, narrowing down the candidates step by step. For example, divide the ten coins into two groups of five and compare them. The heavier group will contain the heaviest coin, and the lighter group will contain the lightest coin. Then, I can focus on those groups separately.So, first weighing: compare group A (coins 1-5) with group B (coins 6-10). Suppose group A is heavier. That means the heaviest coin is in group A, and the lightest is in group B. Now, I need to find the heaviest in group A and the lightest in group B.To find the heaviest in group A, I can use a tournament method within group A. Similarly, to find the lightest in group B, I can use a tournament method within group B.For group A (5 coins), to find the heaviest, I can have two rounds: first, compare coins 1 vs 2, 3 vs 4, and 5 sits out. That's two weighings. Then, take the winners of those and compare them, plus the one that sat out. That's three weighings total for group A.Similarly, for group B (5 coins), to find the lightest, I can do the same: compare 6 vs 7, 8 vs 9, and 10 sits out. Two weighings. Then, take the losers of those and compare them, plus the one that sat out. Three weighings total for group B.So, in total, I've used 1 (initial group comparison) + 3 (heaviest in group A) + 3 (lightest in group B) = 7 weighings. That's way under 13. Maybe I need to account for the fact that I might have to do more comparisons to ensure I haven't missed any candidates.Wait, perhaps I need to consider that the initial group comparison might not always give a clear heavier or lighter group, or maybe I need to do more detailed comparisons within the groups. Also, I might need to verify my findings to ensure accuracy.Alternatively, maybe I should think about the problem in terms of information theory. Each weighing gives me some information, and I need enough information to distinguish between all possible outcomes. There are 10 coins, so there are 10! possible orderings. But I only need to identify two specific coins, the heaviest and the lightest.The number of possible pairs for heaviest and lightest is 10 choices for the heaviest and 9 remaining choices for the lightest, so 90 possibilities. Each weighing can have three outcomes, so the number of possible outcomes with 13 weighings is 3^13, which is much larger than 90. That suggests that 13 weighings should be more than enough.But wait, that's a very rough estimate. The actual number of possible outcomes isn't just 90 because the weighings are dependent on each other, and the information gained from each weighing affects the next. So, maybe 13 weighings are sufficient, but I need a concrete method to achieve it.Let me try to outline a step-by-step approach:1. Divide the ten coins into two groups of five: Group A (coins 1-5) and Group B (coins 6-10).2. Weigh Group A against Group B. - If Group A is heavier, then the heaviest coin is in Group A, and the lightest is in Group B. - If Group B is heavier, then the heaviest coin is in Group B, and the lightest is in Group A. - If they're equal, which they can't be since all coins have different weights, but just in case, we can proceed similarly.3. Suppose Group A is heavier. Now, focus on Group A to find the heaviest and Group B to find the lightest.4. Within Group A (coins 1-5): a. Weigh coin 1 vs coin 2. b. Weigh coin 3 vs coin 4. c. The heavier of these two weighings is compared with the fifth coin to find the heaviest. d. This takes three weighings.5. Within Group B (coins 6-10): a. Weigh coin 6 vs coin 7. b. Weigh coin 8 vs coin 9. c. The lighter of these two weighings is compared with the tenth coin to find the lightest. d. This takes three weighings.6. So, total weighings: 1 (initial) + 3 (heaviest) + 3 (lightest) = 7 weighings.But the problem allows for 13 weighings, which is more than enough. Maybe I need to account for the fact that I might have to do additional comparisons to ensure that the heaviest and lightest are correctly identified, especially if there are any discrepancies or if the initial group comparison doesn't give a clear result.Alternatively, perhaps I'm overcomplicating it. If I can find the heaviest and lightest in 7 weighings, then 13 weighings should definitely be possible. Maybe the question is testing whether it's possible within 13, and since 7 is less than 13, the answer is yes.But to be thorough, let me consider another approach. Maybe using a sorting algorithm like merge sort or quick sort, but adapted for finding just the max and min.In merge sort, to find the max and min, you can do it in n log n time, but in terms of weighings, it might be similar. However, since we're dealing with a balance scale, it's more about comparisons rather than operations.Alternatively, think about it as a decision tree. Each internal node represents a weighing, and each leaf represents a possible outcome (i.e., which coin is the heaviest and which is the lightest). The depth of the tree corresponds to the number of weighings needed.The number of leaves needed is 10 * 9 = 90, as there are 90 possible pairs for heaviest and lightest. The maximum number of leaves in a ternary tree of depth 13 is 3^13, which is much larger than 90, so theoretically, it's possible.But again, this is a theoretical upper bound. The actual method needs to be constructed.Maybe a better way is to use a method where we progressively narrow down the candidates for heaviest and lightest.Start by comparing coins in pairs:1. Weigh coin 1 vs coin 2. Suppose coin 1 is heavier.2. Weigh coin 3 vs coin 4. Suppose coin 3 is heavier.3. Weigh coin 5 vs coin 6. Suppose coin 5 is heavier.4. Weigh coin 7 vs coin 8. Suppose coin 7 is heavier.5. Weigh coin 9 vs coin 10. Suppose coin 9 is heavier.Now, from these five weighings, we have the winners: 1, 3, 5, 7, 9.Next, find the heaviest among these winners:6. Weigh 1 vs 3. Suppose 1 is heavier.7. Weigh 5 vs 7. Suppose 5 is heavier.8. Weigh 1 vs 5. Suppose 1 is heavier.So, coin 1 is the heaviest.Now, to find the lightest, we need to consider the losers from the initial pairings and the losers from the subsequent comparisons.From the initial pairings:- Coin 2 lost to 1.- Coin 4 lost to 3.- Coin 6 lost to 5.- Coin 8 lost to 7.- Coin 10 lost to 9.Additionally, in the second round:- Coin 3 lost to 1.- Coin 7 lost to 5.So, the potential lightest candidates are 2, 4, 6, 8, 10, 3, 7.Wait, that's seven coins. Maybe I need to track all the losers.Alternatively, to find the lightest, I can take the losers from each initial pairing and then find the lightest among them.From initial pairings:- Losers: 2, 4, 6, 8, 10.Now, find the lightest among these:9. Weigh 2 vs 4. Suppose 2 is lighter.10. Weigh 6 vs 8. Suppose 6 is lighter.11. Weigh 2 vs 6. Suppose 2 is lighter.12. Weigh 2 vs 10. Suppose 2 is lighter.So, coin 2 is the lightest.But wait, I've used 12 weighings so far. I have one more weighing left.But in this method, I already found the heaviest and lightest in 12 weighings. So, 13 weighings would definitely be sufficient.However, this method might not always work because the initial pairings could have different outcomes, and the losers might not necessarily include the lightest coin. For example, if the lightest coin was in a group that was compared later, it might not have been identified.Alternatively, maybe I need to ensure that all coins are compared in a way that the lightest and heaviest are definitely identified.Another approach is to use a method where you maintain two candidates for the heaviest and lightest and update them as you go.Start by assuming the first coin is both the heaviest and lightest.Then, for each subsequent coin, compare it with the current heaviest and lightest, updating them accordingly.But since we're using a balance scale, we can't compare three coins at once, so this might not be straightforward.Alternatively, use a method where you compare coins in a way that each comparison provides information about both the heaviest and lightest.For example, compare coin 1 vs coin 2:- If coin 1 is heavier, then coin 1 is a candidate for heaviest, and coin 2 is a candidate for lightest.- If coin 2 is heavier, then coin 2 is a candidate for heaviest, and coin 1 is a candidate for lightest.Then, compare coin 3 vs coin 4:- Similarly, determine candidates for heaviest and lightest.Continue this process for all coins, keeping track of the current heaviest and lightest candidates.After all initial comparisons, you'll have a set of candidates for heaviest and lightest, which you can then compare among themselves to find the overall heaviest and lightest.This method might take more weighings, but it ensures that all coins are considered.Given that, let's estimate the number of weighings:- For ten coins, if we compare them in pairs, that's five weighings.- Then, to find the heaviest among the five winners, that's log2(5) ≈ 3 weighings.- Similarly, to find the lightest among the five losers, that's another 3 weighings.Total: 5 + 3 + 3 = 11 weighings.Again, under 13. So, it seems that 13 weighings are more than enough.But to be safe, let's consider that in the worst case, we might need to do additional comparisons to confirm the heaviest and lightest, especially if there are any ties or if the initial comparisons don't give a clear path.Alternatively, maybe I need to account for the fact that the balance scale can only compare two coins at a time, and I need to structure the weighings in a way that maximizes the information gained each time.In summary, based on these different approaches, it seems that identifying the heaviest and lightest coin among ten can be done in fewer than 13 weighings. Therefore, 13 weighings should definitely be sufficient.I think the key takeaway is that there are efficient methods to find both the heaviest and lightest coins without having to sort all of them, and these methods require fewer than 13 weighings. Therefore, the answer should be yes, it's possible.

💡Okay, so I have this inequality to solve: 3x + 2 < 7 - 2x. Hmm, I need to find the greatest integer that satisfies this. Alright, let me think about how to approach this step by step.First, I remember that solving inequalities is somewhat similar to solving equations, but I have to be careful with the inequality sign, especially if I ever need to multiply or divide by a negative number, which can flip the inequality. But in this case, I don't think that will happen, so maybe I don't have to worry about that right now.So, the inequality is 3x + 2 < 7 - 2x. I think the first step is to get all the x terms on one side and the constants on the other. To do that, I can add 2x to both sides to eliminate the -2x on the right. Let me write that down:3x + 2 + 2x < 7 - 2x + 2xSimplifying both sides, that becomes:5x + 2 < 7Okay, now I have 5x + 2 < 7. Next, I need to isolate the term with x, which is 5x. To do that, I'll subtract 2 from both sides:5x + 2 - 2 < 7 - 2That simplifies to:5x < 5Now, to solve for x, I need to divide both sides by 5:5x / 5 < 5 / 5Which gives me:x < 1Alright, so the solution to the inequality is x < 1. But the question asks for the greatest integer that satisfies this inequality. So, I need to find the largest integer that is less than 1.Thinking about integers, the integers less than 1 are 0, -1, -2, and so on. The greatest of these is 0 because 0 is larger than -1, -2, etc. So, 0 is the largest integer that is still less than 1.Let me double-check to make sure I didn't make any mistakes. If I plug x = 0 back into the original inequality:3(0) + 2 < 7 - 2(0)That simplifies to:0 + 2 < 7 - 0Which is:2 < 7That's true. Now, what if I tried the next integer, which is 1? Plugging x = 1 into the original inequality:3(1) + 2 < 7 - 2(1)That gives:3 + 2 < 7 - 2Which is:5 < 5Wait, 5 is not less than 5; it's equal. So, x = 1 doesn't satisfy the inequality because the inequality is strict (it's less than, not less than or equal to). Therefore, 1 is not a solution.So, confirming again, the greatest integer less than 1 is indeed 0. I think that's the correct answer.Just to recap the steps:1. Start with the inequality: 3x + 2 < 7 - 2x2. Add 2x to both sides: 5x + 2 < 73. Subtract 2 from both sides: 5x < 54. Divide both sides by 5: x < 15. Identify the greatest integer less than 1, which is 0.Everything seems to check out. I don't think I missed anything here. It's a straightforward inequality, and the steps are pretty basic, but it's always good to verify each step to avoid any mistakes.Another thing I could consider is graphing the inequality on a number line to visualize where x lies. If I draw a number line, with 0, 1, 2, etc., and mark the point x = 1, the solution x < 1 would be all the numbers to the left of 1. The greatest integer to the left of 1 is 0, which reinforces my earlier conclusion.I might also think about what happens if I choose a number slightly less than 1, like 0.9. Plugging that into the original inequality:3(0.9) + 2 < 7 - 2(0.9)Calculating:2.7 + 2 < 7 - 1.8Which is:4.7 < 5.2That's true. But since we're looking for integers, 0.9 isn't an integer, so the next lower integer is 0, which we've already confirmed works.In summary, after carefully solving the inequality and verifying each step, I'm confident that the greatest integer satisfying 3x + 2 < 7 - 2x is 0.

💡First, I need to find the value of ( x ) such that ( k(x) = 3 ). Given that ( k(x) = 4x - 7 ), I set up the equation:[4x - 7 = 3]Solving for ( x ):[4x = 10 x = frac{10}{4} = 2.5]Now that I have ( x = 2.5 ), I substitute this value into the expression for ( s(k(x)) ):[s(k(2.5)) = (2.5)^2 - 2(2.5) + 5]Calculating each term:[(2.5)^2 = 6.25 2(2.5) = 5]Putting it all together:[s(k(2.5)) = 6.25 - 5 + 5 = 6.25]Therefore, ( s(3) = 6.25 ).

💡Okay, so I have this triangle ABC, and the sides opposite angles A, B, and C are a, b, and c respectively. The problem gives me an equation: a cos B + b cos A = (√3)/3 * c tan B. I need to find the measure of angle B first, and then, if b is 2, find the maximum area of triangle ABC.Alright, starting with part (1). Hmm, I remember in triangles, there are laws like the Law of Sines and the Law of Cosines that relate sides and angles. Maybe I can use those here.Given the equation: a cos B + b cos A = (√3)/3 * c tan B.I think I can express a and b in terms of the sine of their opposite angles using the Law of Sines. The Law of Sines says that a/sin A = b/sin B = c/sin C = 2R, where R is the radius of the circumcircle. So, a = 2R sin A, b = 2R sin B, c = 2R sin C.Let me substitute these into the equation:(2R sin A) cos B + (2R sin B) cos A = (√3)/3 * (2R sin C) tan B.Okay, so I can factor out 2R from both sides:2R [sin A cos B + sin B cos A] = (√3)/3 * 2R sin C tan B.Since 2R is on both sides and R is not zero, I can divide both sides by 2R:sin A cos B + sin B cos A = (√3)/3 sin C tan B.Hmm, I also remember that sin A cos B + sin B cos A is equal to sin(A + B). Because sin(A + B) = sin A cos B + cos A sin B. So that simplifies the left side:sin(A + B) = (√3)/3 sin C tan B.But in a triangle, A + B + C = π, so A + B = π - C. Therefore, sin(A + B) = sin(π - C) = sin C. So substituting that in:sin C = (√3)/3 sin C tan B.Now, sin C is on both sides. Since C is an angle in a triangle, it's between 0 and π, so sin C is not zero. Therefore, I can divide both sides by sin C:1 = (√3)/3 tan B.So, tan B = 3/√3 = √3.Therefore, B is an angle whose tangent is √3. The principal value is 60 degrees because tan(60°) = √3. Since in a triangle, angles are between 0 and 180 degrees, and tan is positive in the first and third quadrants, but in a triangle, angles are less than 180, so B must be 60 degrees.So, part (1) is solved: B is 60 degrees.Now, moving on to part (2). Given that b = 2, find the maximum area of triangle ABC.First, since we know angle B is 60 degrees, and side b is opposite angle B, which is 2. So, using the Law of Sines, we can relate sides a, c, and the other angles.Law of Sines: a/sin A = b/sin B = c/sin C.Given that b = 2 and B = 60°, sin B = sin 60° = √3/2. So, 2 / (√3/2) = 4/√3. Therefore, a = (4/√3) sin A, and c = (4/√3) sin C.The area of triangle ABC can be given by (1/2)ab sin C, but I think another formula might be more useful here. Alternatively, the area can be expressed as (1/2)ac sin B, since we know angle B.Let me write that down: Area = (1/2) * a * c * sin B.Since B is 60°, sin B = √3/2. So, Area = (1/2) * a * c * (√3/2) = (√3/4) * a * c.So, to maximize the area, I need to maximize the product a * c.Given that a = (4/√3) sin A and c = (4/√3) sin C, so a * c = (16/3) sin A sin C.Therefore, Area = (√3/4) * (16/3) sin A sin C = (4√3/3) sin A sin C.So, I need to maximize sin A sin C.But in triangle ABC, A + B + C = π, so A + C = π - B = π - π/3 = 2π/3. So, A + C = 120°, or 2π/3 radians.Let me denote A = x, so C = 2π/3 - x. So, sin A sin C = sin x sin(2π/3 - x).I can use the identity for sin x sin y: sin x sin y = [cos(x - y) - cos(x + y)] / 2.So, sin x sin(2π/3 - x) = [cos(x - (2π/3 - x)) - cos(x + (2π/3 - x))]/2.Simplify the arguments:First term: cos(2x - 2π/3)Second term: cos(2π/3)So, sin x sin(2π/3 - x) = [cos(2x - 2π/3) - cos(2π/3)] / 2.We know that cos(2π/3) = -1/2.So, sin x sin(2π/3 - x) = [cos(2x - 2π/3) - (-1/2)] / 2 = [cos(2x - 2π/3) + 1/2] / 2 = (1/2) cos(2x - 2π/3) + 1/4.So, sin A sin C = (1/2) cos(2x - 2π/3) + 1/4.To maximize sin A sin C, we need to maximize this expression.The maximum value of cos(theta) is 1, so the maximum of (1/2) cos(theta) + 1/4 is (1/2)(1) + 1/4 = 3/4.Therefore, the maximum of sin A sin C is 3/4.Therefore, the maximum area is (4√3/3) * (3/4) = √3.So, the maximum area is √3.Wait, let me double-check that.So, Area = (√3/4) * a * c.We expressed a * c as (16/3) sin A sin C, so Area = (√3/4) * (16/3) sin A sin C = (4√3/3) sin A sin C.Then, sin A sin C is maximized at 3/4, so Area_max = (4√3/3)*(3/4) = √3.Yes, that seems correct.Alternatively, maybe I can approach this using the Law of Cosines.Given that angle B is 60°, and side b = 2. So, by the Law of Cosines:b² = a² + c² - 2ac cos B.So, 4 = a² + c² - 2ac cos 60°.Since cos 60° = 0.5, so 4 = a² + c² - 2ac*(1/2) = a² + c² - ac.So, 4 = a² + c² - ac.We can rearrange this as a² + c² - ac = 4.We need to maximize the area, which is (√3/4)ac.So, we need to maximize ac given that a² + c² - ac = 4.Let me denote S = ac. Then, from the equation:a² + c² = S + 4.But we also know that a² + c² ≥ 2ac by the AM-GM inequality. So, S + 4 ≥ 2S, which implies that 4 ≥ S, so ac ≤ 4.Therefore, the maximum value of ac is 4.Thus, the maximum area is (√3/4)*4 = √3.So, same result. That confirms it.Therefore, the maximum area is √3.**Final Answer**(1) The measure of angle ( B ) is (boxed{60^circ}).(2) The maximum area of ( triangle ABC ) is (boxed{sqrt{3}}).

💡First, determine the work rates of Amit and Ananthu. Amit can complete the work in 15 days, so his work rate is ( frac{1}{15} ) of the work per day. Ananthu can complete the work in 45 days, so his work rate is ( frac{1}{45} ) of the work per day.Next, calculate the amount of work Amit completes in the first 3 days. This is done by multiplying his work rate by the number of days he worked:[3 times frac{1}{15} = frac{3}{15} = frac{1}{5}]Subtract the work completed by Amit from the total work to find the remaining work:[1 - frac{1}{5} = frac{4}{5}]Now, determine how many days Ananthu needs to complete the remaining ( frac{4}{5} ) of the work. This is done by dividing the remaining work by Ananthu's work rate:[frac{frac{4}{5}}{frac{1}{45}} = frac{4}{5} times 45 = 36 text{ days}]Finally, add the days Amit worked to the days Ananthu worked to find the total time taken to complete the work:[3 text{ days} + 36 text{ days} = 39 text{ days}]