💡Okay, so I have this problem where I need to figure out the integers that make the sequence of operations correct inside some circles. I'm not entirely sure what the exact setup is, but I think it involves placing numbers in circles and performing operations between them to get a correct sequence. Maybe it's something like a number chain where each circle has a number and an operation, and the result should flow correctly from one to the next.Let me try to visualize it. If there are circles with numbers and operations between them, perhaps it's similar to a number puzzle where each operation depends on the previous number. For example, if I have a circle with a number, then an operation like addition or subtraction, and then another circle with a number, and so on. The goal is to find the numbers that make the entire sequence of operations valid or correct.Maybe the operations are given, and I need to find the numbers that fit into the circles to satisfy those operations. For instance, if the first circle has a number, and then there's an operation like "+2," the next circle should have the result of that operation. Then, from there, another operation leads to the next number, and so on. If that's the case, I need to work backward or forward to find the numbers that fit.Alternatively, it could be that the operations are not given, and I need to determine both the numbers and the operations. That might be more complex, but I'll assume for now that the operations are provided, and I just need to find the numbers.Let me think of a simple example. Suppose there are three circles in a row: Circle A, Circle B, and Circle C. Between A and B, there's an operation, say "+3," and between B and C, another operation, say "×2." If I know the starting number in Circle A, I can find the numbers in B and C by applying the operations. But if I don't know the starting number, I might need more information or constraints to find it.Wait, the problem says "inside the circles, write the integers that make the sequence of operations correct." So maybe it's about finding the numbers that, when the operations are applied in sequence, result in a correct or valid outcome, perhaps a specific target number or a consistent sequence.Let's consider another approach. Maybe it's a loop where the operations cycle back to the starting number. For example, if I have numbers in circles connected by operations, and after applying all operations, I end up back at the starting number. That could be a way to set up equations to solve for the numbers.Suppose there are two circles, A and B. If I start at A, apply an operation to get B, then apply another operation to get back to A. That would create a system of equations that I can solve to find the values of A and B.Alternatively, if there are more circles, say three or four, the problem becomes more complex but follows a similar principle. Each operation connects one circle to the next, and the sequence should be consistent.Let me try to formalize this. Suppose I have circles labeled C1, C2, C3, ..., Cn. Each circle Ci has a number xi. Between each pair of circles, there's an operation, say op1, op2, ..., op(n-1). The sequence of operations should transform x1 into x2, x2 into x3, and so on, until xn is obtained.If I know the operations, I can express each xi in terms of xi+1. For example, if op1 is addition, then x2 = x1 + a, where a is some integer. Similarly, if op2 is multiplication, then x3 = x2 × b, and so on.But since the problem is about finding the integers that make the sequence correct, I might need to set up equations based on the operations and solve for the unknowns.Let's consider a specific example to make it concrete. Suppose there are two circles, C1 and C2, with an operation between them. If the operation is "+3," then x2 = x1 + 3. If I know one of the numbers, I can find the other. But if I don't know either, I need more information.Maybe the sequence loops back, so applying the operations brings me back to the starting number. For instance, if I have two circles with operations that, when applied in sequence, return to the original number. That would give me an equation to solve.Suppose the operations are "+3" and "×2." Starting from x1, I add 3 to get x2, then multiply by 2 to get back to x1. So, x1 = (x1 + 3) × 2. Solving this equation: x1 = 2x1 + 6 → -x1 = 6 → x1 = -6. Then x2 = -6 + 3 = -3. So the numbers would be -6 and -3.That seems to work. Let me check: Starting at -6, add 3 to get -3, then multiply by 2 to get -6 again. Yes, it loops back correctly.But in the problem, there might be more circles or different operations. Without the exact diagram, it's hard to be precise, but this approach can be generalized.Another example: suppose three circles with operations "+2," "×3," and "-4." Starting from x1, add 2 to get x2, multiply by 3 to get x3, then subtract 4 to get back to x1. So, x1 = ((x1 + 2) × 3) - 4. Solving: x1 = 3x1 + 6 - 4 → x1 = 3x1 + 2 → -2x1 = 2 → x1 = -1. Then x2 = -1 + 2 = 1, x3 = 1 × 3 = 3. Check: 3 - 4 = -1, which loops back. So the numbers are -1, 1, and 3.This method works for any number of circles and operations, as long as the operations are given and the sequence loops back to the starting number.If the operations are not given, it becomes a different problem. I would need to determine both the numbers and the operations, which is more complex. But since the problem mentions "the sequence of operations," I think the operations are provided, and I just need to find the numbers.In summary, the strategy is:1. Identify the operations between the circles.2. Set up equations based on the operations and the requirement that the sequence is correct (possibly looping back to the start).3. Solve the system of equations to find the integers for each circle.Without the exact operations or the number of circles, I can't provide specific numbers, but this approach should work for any such problem.

💡First, I need to determine the number of leaves shed in the first week. The tree sheds ( frac{2}{5} ) of its 1000 leaves.Next, I'll calculate the remaining leaves after the first week by subtracting the shed leaves from the initial total.For the second week, the tree sheds a certain percentage ( P% ) of the remaining leaves. I'll express the number of leaves shed in the second week in terms of ( P% ).In the third week, the tree sheds ( frac{3}{4} ) times the number of leaves shed in the second week. I'll express this as a function of ( P% ).Given that the tree hasn't shed 90 leaves by the third week, I'll set up an inequality to represent this condition.Finally, I'll solve the inequality to find the maximum percentage ( P% ) that satisfies the condition.

💡Alright, so I have this geometry problem here about triangle ABC. Let me try to visualize it first. There's triangle ABC, and CD is an angle bisector of angle C. That means CD splits angle C into two equal parts. Then, AB is extended to a point E, and angle EBC is a right angle, which is 90 degrees. I need to find the relationship between angle BAE and the angles inside triangle ABC. The options are given, so I need to figure out which one is correct.First, let me label the triangle. Let's say point A is at the top, B is at the bottom left, and C is at the bottom right. So, CD is the angle bisector from point C to side AB. That means CD divides angle C into two equal angles. Let's call each of these angles x. So, angle BCD is x and angle ACD is also x.Now, AB is extended to point E, making angle EBC a right angle. So, if I extend AB beyond point B to point E, then angle EBC is 90 degrees. That means triangle EBC is a right-angled triangle at point B.I need to find angle BAE. So, angle BAE is the angle at point A between BA and EA. Since EA is an extension of AB, angle BAE is essentially the external angle at point A when we extend AB to E.Let me recall some geometry concepts. In a triangle, the sum of the internal angles is 180 degrees. Also, when a side of a triangle is extended, the external angle is equal to the sum of the two opposite internal angles. That might be useful here.So, in triangle ABC, the sum of angles at A, B, and C is 180 degrees. Let's denote angle BAC as angle A, angle ABC as angle B, and angle ACB as angle C. Since CD is the angle bisector, angle BCD is half of angle C, which is x. So, angle C is 2x.Now, looking at triangle EBC, which is right-angled at B. So, angle EBC is 90 degrees, and angle BEC is the remaining angle. Since the sum of angles in a triangle is 180 degrees, angle BEC would be 180 - 90 - angle BCE. But angle BCE is the same as angle BCD, which is x. So, angle BEC is 180 - 90 - x, which simplifies to 90 - x degrees.Now, looking back at angle BAE. Since EA is an extension of AB, angle BAE is an external angle at point A. According to the external angle theorem, angle BAE should be equal to the sum of the two non-adjacent internal angles of triangle ABC. In this case, those angles would be angle ABC (angle B) and angle ACB (angle C). But wait, angle C is 2x, and angle B is something else.Wait, maybe I need to approach this differently. Let's consider triangle ABE. In triangle ABE, we have angle EBC as 90 degrees, and angle BAE is what we're trying to find. Maybe I can express angle BAE in terms of the angles in triangle ABC.Alternatively, let's consider the straight line at point B. Since AB is extended to E, the angles around point B should add up to 180 degrees. So, angle ABC plus angle EBC should be 180 degrees. But angle EBC is 90 degrees, so angle ABC plus 90 degrees equals 180 degrees. Therefore, angle ABC is 90 degrees. Wait, that can't be right because triangle ABC is just a general triangle, not necessarily right-angled.Hmm, I think I made a mistake there. Let me correct that. If AB is extended to E, then angle EBC is an external angle at point B. The sum of the internal angle at B (angle ABC) and the external angle EBC should be 180 degrees because they are supplementary. Since angle EBC is 90 degrees, angle ABC must be 90 degrees as well. But that would mean triangle ABC has a right angle at B, which isn't specified in the problem. So, I must have misunderstood something.Wait, no. Actually, angle EBC is the angle between BE and BC, not between BE and BA. So, angle EBC is 90 degrees, but angle ABC is the angle between BA and BC. Therefore, they are not supplementary. Instead, angle EBC is an external angle related to triangle ABC.Let me try to draw this out mentally. Point E is on the extension of AB beyond B. So, starting from A, going through B to E. Then, angle EBC is the angle at point B between BE and BC. Since BE is a straight line extension of AB, angle EBC is formed by BE and BC. This angle is given as 90 degrees.Now, in triangle EBC, we have angle EBC = 90 degrees, angle BCE = x (since CD is the angle bisector), and angle BEC is the remaining angle. So, angle BEC = 180 - 90 - x = 90 - x degrees.Now, looking at triangle ABE, we have point E, and we're interested in angle BAE. Let's see if we can relate angle BAE to the angles in triangle ABC.In triangle ABE, the sum of angles is 180 degrees. So, angle BAE + angle ABE + angle BEA = 180 degrees. We know angle BEA is angle BEC, which is 90 - x degrees. Angle ABE is the same as angle ABC, which we can denote as angle B. So, angle BAE + angle B + (90 - x) = 180 degrees.Therefore, angle BAE = 180 - angle B - (90 - x) = 90 - angle B + x.Now, in triangle ABC, the sum of angles is 180 degrees: angle A + angle B + angle C = 180 degrees. We know angle C is 2x, so angle A + angle B + 2x = 180 degrees. Therefore, angle A = 180 - angle B - 2x.But we have angle BAE = 90 - angle B + x. Let's see if we can express this in terms of angle A or angle C.From angle A = 180 - angle B - 2x, we can solve for angle B: angle B = 180 - angle A - 2x.Substituting this into angle BAE: angle BAE = 90 - (180 - angle A - 2x) + x = 90 - 180 + angle A + 2x + x = -90 + angle A + 3x.Hmm, that doesn't seem to directly relate to the given options. Maybe I need a different approach.Let's consider the external angle at point A. Angle BAE is an external angle for triangle ABC at point A. According to the external angle theorem, angle BAE should be equal to the sum of the two non-adjacent internal angles, which are angle ABC and angle ACB.So, angle BAE = angle ABC + angle ACB.But angle ACB is angle C, which is 2x, and angle ABC is angle B. So, angle BAE = angle B + 2x.But from triangle ABC, angle A + angle B + 2x = 180 degrees, so angle B = 180 - angle A - 2x.Substituting back, angle BAE = (180 - angle A - 2x) + 2x = 180 - angle A.But angle A is angle BAC, so angle BAE = 180 - angle BAC.Wait, that doesn't seem to match any of the options directly. Let me check my reasoning.Alternatively, maybe I should consider triangle EBC. In triangle EBC, angle EBC = 90 degrees, angle BCE = x, so angle BEC = 90 - x degrees.Now, looking at triangle ABE, angle BAE + angle ABE + angle BEA = 180 degrees.We have angle BEA = angle BEC = 90 - x degrees.Angle ABE is the same as angle ABC, which is angle B.So, angle BAE + angle B + (90 - x) = 180 degrees.Therefore, angle BAE = 180 - angle B - (90 - x) = 90 - angle B + x.From triangle ABC, angle A + angle B + 2x = 180 degrees, so angle A = 180 - angle B - 2x.Substituting angle B = 180 - angle A - 2x into angle BAE:angle BAE = 90 - (180 - angle A - 2x) + x = 90 - 180 + angle A + 2x + x = -90 + angle A + 3x.This still doesn't seem to align with the options. Maybe I need to express angle BAE in terms of angle C and angle BCD.Since angle C = 2x, and angle BCD = x, then angle BAE = angle C + 2 * angle BCD.Wait, let's see:angle BAE = 90 - angle B + x.But angle B = 180 - angle A - 2x.So, angle BAE = 90 - (180 - angle A - 2x) + x = 90 - 180 + angle A + 2x + x = -90 + angle A + 3x.But angle A = 180 - angle B - 2x, so substituting back:angle BAE = -90 + (180 - angle B - 2x) + 3x = -90 + 180 - angle B + x = 90 - angle B + x.This seems circular. Maybe I need to approach it differently.Let me consider the straight line at point B. The sum of angles around point B should be 180 degrees. So, angle ABC + angle EBC = 180 degrees. But angle EBC is 90 degrees, so angle ABC = 90 degrees. Wait, that would mean triangle ABC has a right angle at B, but that's not given in the problem. So, this must be incorrect.Wait, no. Angle EBC is not the same as angle ABC. Angle EBC is the angle between BE and BC, while angle ABC is the angle between BA and BC. So, they are different angles. Therefore, angle ABC + angle EBC = 180 degrees is not correct.Instead, angle ABC and angle EBC are adjacent angles forming a linear pair along BC. Therefore, angle ABC + angle EBC = 180 degrees. Since angle EBC is 90 degrees, angle ABC = 90 degrees. But this contradicts unless triangle ABC is right-angled at B, which isn't specified.This suggests that my initial assumption might be wrong. Maybe angle EBC is not adjacent to angle ABC in that way. Let me clarify.Point E is on the extension of AB beyond B. So, starting from A, going through B to E. Then, angle EBC is the angle at point B between BE and BC. Since BE is a straight line extension of AB, angle EBC is an external angle related to triangle ABC.In triangle ABC, angle ABC is the internal angle at B between BA and BC. Angle EBC is the external angle at B between BE and BC. Therefore, angle ABC + angle EBC = 180 degrees because they are supplementary.Given that angle EBC is 90 degrees, angle ABC = 180 - 90 = 90 degrees. So, angle ABC is 90 degrees. Therefore, triangle ABC is right-angled at B.Wait, but the problem doesn't specify that triangle ABC is right-angled. It only says that angle EBC is a right angle. So, does that mean triangle ABC is right-angled at B? It seems so because angle ABC is 90 degrees.If that's the case, then triangle ABC is a right-angled triangle at B, with angle ABC = 90 degrees, angle BAC = angle A, and angle ACB = angle C = 2x, since CD is the angle bisector.Given that, in triangle ABC, angle A + 90 + 2x = 180, so angle A = 90 - 2x.Now, angle BAE is the external angle at A when AB is extended to E. According to the external angle theorem, angle BAE should be equal to the sum of the two non-adjacent internal angles, which are angle ABC and angle ACB.So, angle BAE = angle ABC + angle ACB = 90 + 2x.But angle ACB is angle C = 2x, and angle BCD is x. So, angle BAE = angle ACB + 2 * angle BCD.Wait, no. angle BAE = angle ACB + angle ABC, which is 2x + 90 degrees. But angle ABC is 90 degrees, so angle BAE = 90 + 2x.But angle ACB is 2x, and angle BCD is x. So, angle BAE = angle ACB + 2 * angle BCD.Wait, angle BAE = 90 + 2x, which is angle ACB (2x) plus 90 degrees. But 90 degrees is angle ABC, not angle BCD.I think I'm getting confused here. Let me try to express angle BAE in terms of angle ACB and angle BCD.Since angle ACB = 2x and angle BCD = x, then angle BAE = angle ACB + angle BCD.But from earlier, angle BAE = 90 + 2x, which is angle ACB (2x) plus 90 degrees. But 90 degrees is angle ABC, not angle BCD.Wait, maybe I need to express angle BAE differently. Since angle BAE is an external angle at A, it's equal to the sum of the two opposite internal angles, which are angle ABC and angle ACB.So, angle BAE = angle ABC + angle ACB = 90 + 2x.But angle ACB is 2x, and angle BCD is x. So, angle BAE = angle ACB + 2 * angle BCD.Wait, because angle ACB is 2x, and angle BCD is x, so 2x = angle ACB, and x = angle BCD. Therefore, angle BAE = angle ACB + 2 * angle BCD.Yes, that makes sense. Because angle BAE = 90 + 2x, and angle ACB = 2x, angle BCD = x, so angle BAE = angle ACB + 2 * angle BCD.Therefore, the correct answer should be option (B): angle BAE = angle ACB + 2 * angle BCD.

💡Alright, so I've got this problem here with four functions, and I need to figure out which ones are monotonically increasing in the interval from 0 to pi over 2. Monotonically increasing means that as x increases, the function's value doesn't decrease—it either stays the same or goes up. So, I need to check each function one by one to see if their derivatives are positive throughout that interval.Let me start with the first function: y equals sine x plus cosine x. To find out if it's increasing, I should find its derivative. The derivative of sine x is cosine x, and the derivative of cosine x is negative sine x. So, putting that together, the derivative of y is cosine x minus sine x. Now, I need to see if this derivative is positive between 0 and pi over 2.At x equals 0, cosine 0 is 1, and sine 0 is 0, so the derivative is 1 - 0, which is 1. That's positive. At x equals pi over 2, cosine pi over 2 is 0, and sine pi over 2 is 1, so the derivative is 0 - 1, which is -1. That's negative. Hmm, so the derivative starts positive and ends negative. That means the function is increasing at first but then starts decreasing. So, it's not monotonically increasing throughout the entire interval. Maybe it's increasing up to a certain point and then decreasing. So, function 1 is not monotonically increasing in the whole interval.Moving on to function 2: y equals sine x minus cosine x. Let's find its derivative. The derivative of sine x is cosine x, and the derivative of negative cosine x is sine x. So, the derivative of y is cosine x plus sine x. Now, let's check the sign of this derivative in the interval from 0 to pi over 2.At x equals 0, cosine 0 is 1, and sine 0 is 0, so the derivative is 1 + 0, which is 1—positive. At x equals pi over 2, cosine pi over 2 is 0, and sine pi over 2 is 1, so the derivative is 0 + 1, which is 1—still positive. In between, both sine x and cosine x are positive, so their sum is definitely positive. So, the derivative is always positive in this interval, meaning function 2 is monotonically increasing throughout the interval.Function 3 is y equals sine x times cosine x. To find the derivative, I'll use the product rule. The derivative of sine x is cosine x, and the derivative of cosine x is negative sine x. So, the derivative of y is cosine x times cosine x plus sine x times negative sine x. That simplifies to cosine squared x minus sine squared x. I remember that cosine squared x minus sine squared x is equal to cosine 2x. So, the derivative is cosine 2x.Now, let's analyze cosine 2x in the interval from 0 to pi over 2. When x is 0, 2x is 0, and cosine 0 is 1—positive. When x is pi over 4, 2x is pi over 2, and cosine pi over 2 is 0. When x is pi over 2, 2x is pi, and cosine pi is -1—negative. So, the derivative starts positive, becomes zero at pi over 4, and then negative after that. That means the function is increasing up to pi over 4 and then decreasing after that. Therefore, it's not monotonically increasing throughout the entire interval.Finally, function 4 is y equals sine x over cosine x, which is the same as tangent x. The derivative of tangent x is secant squared x, which is 1 over cosine squared x. Since cosine x is positive in the interval from 0 to pi over 2, cosine squared x is also positive, and thus secant squared x is positive. So, the derivative is always positive in this interval, meaning function 4 is monotonically increasing throughout the interval.So, summarizing my findings:- Function 1: Not monotonically increasing.- Function 2: Monotonically increasing.- Function 3: Not monotonically increasing.- Function 4: Monotonically increasing.Therefore, the functions that are monotonically increasing in the interval from 0 to pi over 2 are functions 2 and 4.

💡Okay, so I have this problem where I need to prove that the sum of sines of some angles is less than or equal to n times the sine of π divided by n. The conditions are that the sum of all these angles is π, and each angle is between 0 and π. Hmm, interesting.Let me write down what I know:Given:- ( sum_{i=1}^{n} x_{i} = pi )- ( x_{i} in [0, pi] ) for each i from 1 to n.Need to prove:- ( sum_{i=1}^{n} sin x_{i} leq n sin frac{pi}{n} )Alright, so I need to show that the sum of sines of these angles is maximized when all the angles are equal, right? Because if all the angles are equal, each would be ( frac{pi}{n} ), and then the sum would be ( n sin frac{pi}{n} ). So, maybe I can use some inequality that tells me that the sine function is concave or convex, and then apply Jensen's inequality?Wait, let me recall. Jensen's inequality applies to convex or concave functions. If a function is concave, then the inequality goes one way, and if it's convex, it goes the other way. So, what is the concavity of the sine function on the interval [0, π]?The second derivative of sine is negative sine. So, ( f''(x) = -sin x ). On [0, π], sine is non-negative, so ( f''(x) leq 0 ). That means sine is concave on [0, π]. Therefore, Jensen's inequality for concave functions states that:( frac{1}{n} sum_{i=1}^{n} f(x_i) leq fleft( frac{1}{n} sum_{i=1}^{n} x_i right) )Multiplying both sides by n, we get:( sum_{i=1}^{n} f(x_i) leq n fleft( frac{1}{n} sum_{i=1}^{n} x_i right) )In our case, ( f(x) = sin x ), and ( sum_{i=1}^{n} x_i = pi ), so ( frac{1}{n} sum_{i=1}^{n} x_i = frac{pi}{n} ). Therefore, substituting into Jensen's inequality:( sum_{i=1}^{n} sin x_i leq n sin left( frac{pi}{n} right) )Which is exactly what we needed to prove!Wait, but hold on. Did I just use Jensen's inequality directly? Let me double-check if all the conditions are satisfied.First, sine is concave on [0, π], which we established because the second derivative is negative there. Second, all ( x_i ) are within [0, π], so they lie within the interval where sine is concave. Third, the weights in Jensen's inequality are equal because we're dealing with an average, which is the case here since each ( x_i ) is treated equally.So, yes, it seems like applying Jensen's inequality directly gives us the desired result. That was straightforward.But just to make sure I'm not missing anything, let me think about another approach, maybe using Lagrange multipliers or some other optimization technique.Suppose I want to maximize ( sum_{i=1}^{n} sin x_i ) subject to the constraint ( sum_{i=1}^{n} x_i = pi ) and ( 0 leq x_i leq pi ). Using Lagrange multipliers, I can set up the function:( L = sum_{i=1}^{n} sin x_i - lambda left( sum_{i=1}^{n} x_i - pi right) )Taking partial derivatives with respect to each ( x_i ):( frac{partial L}{partial x_i} = cos x_i - lambda = 0 )So, ( cos x_i = lambda ) for all i. That suggests that all ( x_i ) must be equal because their cosines are equal. Since ( x_i in [0, pi] ), the only way for all cosines to be equal is if all ( x_i ) are equal. Therefore, each ( x_i = frac{pi}{n} ).Substituting back, the maximum sum of sines is ( n sin frac{pi}{n} ), which confirms the result.Hmm, so both approaches lead me to the same conclusion. That gives me more confidence that the inequality is indeed true and that the maximum occurs when all angles are equal.Is there another way to think about this? Maybe using the concavity of sine and some sort of rearrangement inequality?Well, since sine is concave on [0, π], the maximum of the sum ( sum sin x_i ) under the constraint ( sum x_i = pi ) occurs when the variables are as equal as possible. This is a common theme in optimization problems involving concave functions—equality often gives extremal values.Alternatively, I could consider using the method of Lagrange multipliers with inequality constraints, but since the maximum occurs at the interior point where all ( x_i ) are equal, the Lagrange multiplier method suffices.Wait, but what if some ( x_i ) are at the boundaries, like 0 or π? Would that affect the maximum?If one of the ( x_i ) is π, then ( sin x_i = 0 ), which might actually decrease the sum. Similarly, if one ( x_i ) is 0, ( sin x_i = 0 ) as well. So, having any ( x_i ) at the boundaries doesn't help in increasing the sum; in fact, it might decrease it. Therefore, the maximum should indeed occur when all ( x_i ) are in the interior of the interval, specifically equal to each other.Let me test this with a small n, say n=2.For n=2, the sum of angles is π, so each angle is π/2. Then, ( sin(pi/2) + sin(pi/2) = 1 + 1 = 2 ). On the other hand, ( 2 sin(pi/2) = 2 * 1 = 2 ). So equality holds.What if I take different angles? Suppose one angle is π and the other is 0. Then, ( sin(pi) + sin(0) = 0 + 0 = 0 ), which is less than 2. If I take one angle as 2π/3 and the other as π/3, then ( sin(2π/3) + sin(π/3) = sqrt{3}/2 + sqrt{3}/2 = sqrt{3} approx 1.732 ), which is still less than 2.So, for n=2, the maximum occurs when both angles are equal, as expected.Let me try n=3. The maximum should be ( 3 sin(pi/3) = 3 * (sqrt{3}/2) approx 2.598 ).If I take all angles equal to π/3, then each sine is √3/2, so the sum is 3*(√3/2) ≈ 2.598.If I take one angle as π and the other two as 0, the sum is 0 + 0 + 0 = 0, which is less.If I take one angle as 2π/3 and the other two as π/6, then the sum is ( sin(2π/3) + 2 sin(π/6) = sqrt{3}/2 + 2*(1/2) = sqrt{3}/2 + 1 ≈ 0.866 + 1 = 1.866 ), which is still less than 2.598.So again, the maximum occurs when all angles are equal.This seems consistent. Therefore, it's reasonable to conclude that the maximum of the sum of sines occurs when all angles are equal, given the constraint that their sum is π.Another thought: maybe using the Cauchy-Schwarz inequality? Although I'm not sure if that would directly apply here.Wait, let's see. The Cauchy-Schwarz inequality states that ( (sum a_i b_i)^2 leq (sum a_i^2)(sum b_i^2) ). I'm not sure how to apply this directly to the sum of sines.Alternatively, maybe using the AM-GM inequality? But AM-GM relates the arithmetic mean and geometric mean, which might not directly help with sines.Alternatively, perhaps using the concavity of sine to argue that the sum is maximized when variables are equal. Since sine is concave, the sum is maximized when the variables are equal, which is a standard result in optimization under concave functions.Yes, that seems to be the case. So, given that sine is concave on [0, π], the sum ( sum sin x_i ) is maximized when all ( x_i ) are equal, given the constraint on their sum.Therefore, the maximum value is ( n sin(pi/n) ), which is exactly what we needed to prove.I think I've covered multiple approaches: Jensen's inequality, Lagrange multipliers, testing with small n, and considering the concavity of sine. All of them lead to the same conclusion. So, I feel confident that the inequality holds and that the maximum is achieved when all angles are equal.**Final Answer**boxed{sum_{i=1}^{n} sin x_{i} leq n sin frac{pi}{n}}

💡Okay, so I have this problem: I need to figure out if there are 100 natural numbers, let's call them (a_1, a_2, a_3, ldots, a_{100}), such that each number is bigger than the one before it, meaning (a_1 < a_2 < a_3 < ldots < a_{100}). And not only that, but the greatest common divisors (GCDs) of each pair of consecutive numbers should be strictly decreasing. That is, (gcd(a_1, a_2) > gcd(a_2, a_3) > ldots > gcd(a_{99}, a_{100})).Hmm, okay. So, first, I need to understand what this means. The GCD of two numbers is the largest number that divides both of them without leaving a remainder. So, if I have two numbers, their GCD is a measure of how "similar" they are in terms of factors. If the GCDs are strictly decreasing, each consecutive pair should share a smaller and smaller common factor.Let me think about how to construct such a sequence. Maybe I can start by considering some simple examples with smaller numbers and see if I can spot a pattern or a method that could be extended to 100 numbers.Let's start with a smaller case, say 3 numbers: (a_1, a_2, a_3). I need (a_1 < a_2 < a_3) and (gcd(a_1, a_2) > gcd(a_2, a_3)). How can I choose these numbers?One idea is to have (a_1) and (a_2) share a large common factor, and then (a_2) and (a_3) share a smaller common factor. For example, let me pick (a_1 = 6), (a_2 = 12), and (a_3 = 15). Then, (gcd(6, 12) = 6), and (gcd(12, 15) = 3). So, 6 > 3, which satisfies the condition. Also, 6 < 12 < 15.Okay, that works for 3 numbers. Let's try to extend this to 4 numbers. So, I need (a_1 < a_2 < a_3 < a_4) with (gcd(a_1, a_2) > gcd(a_2, a_3) > gcd(a_3, a_4)).Using the previous example, let's see. If (a_1 = 6), (a_2 = 12), (a_3 = 15), then what should (a_4) be? The GCD of (a_3) and (a_4) needs to be less than 3. So, maybe pick (a_4 = 16). Then, (gcd(15, 16) = 1), which is less than 3. So, the sequence would be 6, 12, 15, 16 with GCDs 6, 3, 1. That works.But wait, 16 is only 1 more than 15. Is there a better way? Maybe pick numbers that are multiples of smaller primes? Let me think.Alternatively, maybe I can use powers of 2 or something. Let's try that. Suppose (a_1 = 2), (a_2 = 4), (a_3 = 8), (a_4 = 16). Then, all the GCDs would be 2, 4, 8, which are increasing, not decreasing. That's the opposite of what I want.Hmm, so powers of 2 won't work because their GCDs increase. Maybe I need to use numbers with decreasing common factors.Wait, in my first example, I used 6, 12, 15, 16. The GCDs were 6, 3, 1. So, each time, the GCD was halved or something. Maybe I can use a similar approach for more numbers.Let me try to create a sequence where each consecutive pair has a GCD that is half of the previous one. So, starting with a high GCD and then each time, the GCD is divided by 2.But wait, GCDs have to be integers, so I can't just divide by 2 every time unless the GCD is even. So, maybe I can use powers of 2 for the GCDs.Wait, but in my first example, the GCDs went from 6 to 3 to 1. So, that's not exactly powers of 2, but it's decreasing by a factor of 2 each time.Alternatively, maybe I can use a sequence where each pair shares a common factor that is one less than the previous. For example, starting with GCD 100, then 99, then 98, etc., down to 1. But that might be complicated because I need to ensure that each number is larger than the previous and that the GCDs actually decrease by 1 each time.Wait, but if I try to do that, I might run into issues where the numbers aren't increasing properly or the GCDs don't behave as expected.Maybe another approach is to use prime numbers. If I can arrange the numbers such that each consecutive pair shares a unique prime factor, and those primes are decreasing. For example, the first pair shares a large prime, the next pair shares a smaller prime, and so on.But primes are only divisible by 1 and themselves, so the GCD of two primes is 1 unless they are the same prime. So, if I use distinct primes for each pair, the GCDs would all be 1, which isn't helpful. So, that approach might not work.Wait, unless I use multiples of primes. For example, if I have (a_1 = p_1 times p_2), (a_2 = p_2 times p_3), (a_3 = p_3 times p_4), and so on, where (p_1 > p_2 > p_3 > ldots) are primes. Then, the GCD of (a_1) and (a_2) would be (p_2), the GCD of (a_2) and (a_3) would be (p_3), and so on. So, if I arrange the primes in decreasing order, the GCDs would also decrease.That sounds promising. Let me try to formalize this.Let me define the sequence as follows: Let (p_1, p_2, ldots, p_{101}) be distinct prime numbers in decreasing order, so (p_1 > p_2 > ldots > p_{101}). Then, define each (a_i) as (a_i = p_i times p_{i+1}). So, (a_1 = p_1 p_2), (a_2 = p_2 p_3), (a_3 = p_3 p_4), and so on up to (a_{100} = p_{100} p_{101}).Now, let's check the GCDs. For each consecutive pair, (gcd(a_i, a_{i+1}) = gcd(p_i p_{i+1}, p_{i+1} p_{i+2})). Since (p_i), (p_{i+1}), and (p_{i+2}) are distinct primes, the GCD is just (p_{i+1}). Therefore, the GCDs are (p_2, p_3, ldots, p_{101}), which are in decreasing order because the primes are arranged in decreasing order.Also, we need to ensure that the sequence (a_1, a_2, ldots, a_{100}) is strictly increasing. Let's check that. Since (p_1 > p_2 > ldots > p_{101}), each (a_i = p_i p_{i+1}) is the product of two primes. Let's see if (a_i < a_{i+1}).We have (a_i = p_i p_{i+1}) and (a_{i+1} = p_{i+1} p_{i+2}). To check if (a_i < a_{i+1}), we need (p_i p_{i+1} < p_{i+1} p_{i+2}). Dividing both sides by (p_{i+1}) (which is positive), we get (p_i < p_{i+2}). But since the primes are in decreasing order, (p_i > p_{i+1} > p_{i+2}), so (p_i > p_{i+2}). Therefore, (p_i p_{i+1} > p_{i+1} p_{i+2}), which means (a_i > a_{i+1}). Oh no, that's the opposite of what we want.So, this approach results in a decreasing sequence (a_i), which is not acceptable because we need (a_1 < a_2 < ldots < a_{100}). Hmm, that's a problem.Maybe I need to adjust the way I construct the sequence. Perhaps instead of multiplying two consecutive primes, I can multiply a prime with a larger number to ensure the sequence increases.Wait, let's think differently. Maybe instead of using just two primes, I can use more factors to make sure the numbers increase while still controlling the GCDs.Another idea: Let's use powers of 2 for the GCDs. Since powers of 2 are easy to handle and decrease by dividing by 2 each time. So, if I can make the GCDs be (2^{99}, 2^{98}, ldots, 2^0 = 1), that would satisfy the strictly decreasing condition.To do this, I can construct the sequence such that each (a_i) is a multiple of (2^{100 - i}), but not a multiple of (2^{101 - i}). That way, the GCD of (a_i) and (a_{i+1}) would be (2^{100 - i}), which decreases by a factor of 2 each time.Let me try to formalize this. Let (a_i = 2^{100} - 2^{100 - i}). Wait, let's see:For (i = 1), (a_1 = 2^{100} - 2^{99} = 2^{99}(2 - 1) = 2^{99}).For (i = 2), (a_2 = 2^{100} - 2^{98} = 2^{98}(2^2 - 1) = 2^{98} times 3).For (i = 3), (a_3 = 2^{100} - 2^{97} = 2^{97}(2^3 - 1) = 2^{97} times 7).And so on, until (a_{100} = 2^{100} - 2^{0} = 2^{100} - 1).Now, let's check the GCDs. For each (i), (gcd(a_i, a_{i+1}) = gcd(2^{100} - 2^{100 - i}, 2^{100} - 2^{100 - (i+1)})).Simplifying, (gcd(2^{100} - 2^{100 - i}, 2^{100} - 2^{99 - i})).Using the property that (gcd(a, b) = gcd(a, b - a)), we can subtract the two terms:(gcd(2^{100} - 2^{100 - i}, 2^{100} - 2^{99 - i} - (2^{100} - 2^{100 - i})) = gcd(2^{100} - 2^{100 - i}, 2^{100 - i} - 2^{99 - i})).Simplifying the second term: (2^{100 - i} - 2^{99 - i} = 2^{99 - i}(2 - 1) = 2^{99 - i}).So now, (gcd(2^{100} - 2^{100 - i}, 2^{99 - i})).But (2^{100} - 2^{100 - i} = 2^{100 - i}(2^i - 1)). So, we have (gcd(2^{100 - i}(2^i - 1), 2^{99 - i})).Since (2^{100 - i}) and (2^{99 - i}) are powers of 2, their GCD is (2^{99 - i}). Also, (2^i - 1) is an odd number, so it doesn't contribute any factors of 2. Therefore, the GCD is (2^{99 - i}).So, the GCDs are (2^{99}, 2^{98}, ldots, 2^0 = 1), which is a strictly decreasing sequence of GCDs.Now, let's check if the sequence (a_1, a_2, ldots, a_{100}) is strictly increasing. Since each (a_i = 2^{100} - 2^{100 - i}), as (i) increases, (2^{100 - i}) decreases, so (a_i) increases. For example:- (a_1 = 2^{100} - 2^{99} = 2^{99})- (a_2 = 2^{100} - 2^{98} = 2^{98} times 3)- (a_3 = 2^{100} - 2^{97} = 2^{97} times 7)- ...- (a_{100} = 2^{100} - 1)Each subsequent (a_i) is larger than the previous one because we're subtracting smaller powers of 2 from (2^{100}). So, the sequence is indeed strictly increasing.Therefore, this construction works. We have a sequence of 100 natural numbers where each number is larger than the previous one, and the GCDs of consecutive pairs are strictly decreasing.Another way to think about this is to use the concept of geometric sequences or arithmetic sequences, but in this case, the construction using powers of 2 and subtracting smaller powers seems to fit perfectly.I can also consider another approach using prime numbers in a different way. For example, if I can find a sequence where each pair shares a unique prime factor, and those primes are arranged in decreasing order, then the GCDs would also decrease. However, as I saw earlier, simply multiplying consecutive primes doesn't work because the sequence ends up decreasing. But maybe if I adjust the construction, I can make it work.Suppose I define each (a_i) as (p_i times q_i), where (p_i) are primes in decreasing order, and (q_i) are chosen such that (q_i) is greater than (p_i) and ensures that (a_i < a_{i+1}). For example, let me try:Let (p_1 = 101), (p_2 = 97), (p_3 = 89), ..., down to (p_{100}), which would be a very small prime, maybe 2 or something. Then, choose (q_i) such that (q_i > p_i) and (a_i = p_i times q_i) is increasing.But I need to ensure that (gcd(a_i, a_{i+1}) = p_{i+1}), which would be decreasing since the primes are decreasing. However, this might not be straightforward because (a_i) and (a_{i+1}) would share (p_{i+1}) only if (p_{i+1}) divides both (a_i) and (a_{i+1}). But (a_i = p_i times q_i) and (a_{i+1} = p_{i+1} times q_{i+1}). For (p_{i+1}) to divide (a_i), (p_{i+1}) must divide (p_i) or (q_i). Since (p_i) and (p_{i+1}) are distinct primes, (p_{i+1}) cannot divide (p_i), so it must divide (q_i). Therefore, (q_i) must be a multiple of (p_{i+1}).So, if I set (q_i = k_i times p_{i+1}), where (k_i) is an integer greater than 1 (to ensure (q_i > p_i)), then (a_i = p_i times k_i times p_{i+1}). Then, (a_{i+1} = p_{i+1} times k_{i+1} times p_{i+2}).Now, (gcd(a_i, a_{i+1}) = gcd(p_i k_i p_{i+1}, p_{i+1} k_{i+1} p_{i+2})). Since (p_i), (p_{i+1}), and (p_{i+2}) are distinct primes, the GCD is (p_{i+1}), which is decreasing as (i) increases.Also, to ensure (a_i < a_{i+1}), we need (p_i k_i p_{i+1} < p_{i+1} k_{i+1} p_{i+2}). Dividing both sides by (p_{i+1}), we get (p_i k_i < k_{i+1} p_{i+2}). Since (p_i > p_{i+1} > p_{i+2}) and (k_i, k_{i+1} geq 2), it's possible to choose (k_{i+1}) large enough to satisfy this inequality.For example, let's choose (k_i = 2) for all (i). Then, (a_i = 2 p_i p_{i+1}). Now, check if (a_i < a_{i+1}):(2 p_i p_{i+1} < 2 p_{i+1} p_{i+2})Divide both sides by (2 p_{i+1}):(p_i < p_{i+2})But since (p_i > p_{i+1} > p_{i+2}), this inequality (p_i < p_{i+2}) is not true. So, (k_i = 2) is not sufficient. We need to choose (k_{i+1}) such that (k_{i+1} > frac{p_i k_i}{p_{i+2}}).Given that (p_i > p_{i+1} > p_{i+2}), and (k_i geq 2), we can choose (k_{i+1}) to be, say, ( lceil frac{p_i k_i}{p_{i+2}} rceil + 1 ) to ensure the inequality holds.This approach is more involved, but it shows that with careful selection of the multipliers (k_i), we can construct such a sequence. However, it's more complex than the previous method using powers of 2, which worked seamlessly.Therefore, the construction using powers of 2 seems more straightforward and guarantees the required properties without additional complications.In summary, by defining each (a_i) as (2^{100} - 2^{100 - i}), we ensure that the sequence is strictly increasing and that the GCDs of consecutive pairs are strictly decreasing from (2^{99}) down to 1. This satisfies all the conditions of the problem.Another way to think about it is to recognize that the GCD of two numbers can be controlled by their prime factorization. By carefully choosing the numbers to have specific common factors, we can manipulate the GCDs to decrease as needed. The key insight here is that using powers of 2 allows us to have a predictable and easily controlled GCD structure, ensuring that each consecutive pair has a GCD that is exactly half of the previous one, leading to a strictly decreasing sequence of GCDs.It's also worth noting that while the powers of 2 approach works, there might be other constructions as well. For example, using different bases or different sequences of numbers that allow for controlled GCDs. However, the powers of 2 method is particularly elegant because it leverages the properties of exponents and the binary system to create a clear and simple structure.In conclusion, through careful construction and understanding of how GCDs behave with respect to number factorizations, it is indeed possible to create a sequence of 100 natural numbers that is strictly increasing and has strictly decreasing GCDs for each consecutive pair. The example using powers of 2 provides a clear and effective method to achieve this.