💡Okay, so I have this problem with three circles, P, Q, and R, each with radius 2. Circles P and Q are tangent to each other, and circle R is tangent to the midpoint of the line segment PQ. I need to find the area inside circle R but outside circles P and Q. Hmm, let me try to visualize this.First, since circles P and Q are tangent to each other, the distance between their centers must be equal to the sum of their radii. Since each has a radius of 2, the distance between centers P and Q is 2 + 2 = 4 units. So, if I imagine circles P and Q sitting next to each other, touching at a single point, the line segment PQ is 4 units long.Now, circle R is tangent to the midpoint of PQ. The midpoint of PQ would be exactly halfway between P and Q, so that's 2 units from each center. Since circle R has a radius of 2, and it's tangent to this midpoint, the distance from the center of R to the midpoint must be equal to the radius of R, which is 2 units. So, the center of R is 2 units away from the midpoint of PQ.Let me try to sketch this mentally. If I place circle P at coordinates (0,0) and circle Q at (4,0), then the midpoint of PQ is at (2,0). Circle R is tangent to this midpoint, so its center must be somewhere 2 units away from (2,0). Since it's a circle, it could be either above or below the line PQ. I think it's above because if it were below, it might interfere with circles P and Q, but the problem doesn't specify, so I'll assume it's above for simplicity. So, the center of R is at (2,2).Now, I need to find the area inside circle R but outside circles P and Q. So, essentially, I need to calculate the area of circle R and subtract the areas where R overlaps with P and Q.First, let's find the area of circle R. Since the radius is 2, the area is πr² = π*(2)² = 4π.Next, I need to find the overlapping areas between R and P, and R and Q. Since the setup is symmetric, the overlapping areas with P and Q should be the same. So, I can calculate one and double it.To find the overlapping area between R and P, I can use the formula for the area of intersection between two circles. The formula is:Area = r² cos⁻¹(d²/(2r²)) - (d/2)√(4r² - d²)where r is the radius of the circles and d is the distance between their centers.Wait, but in this case, both circles have the same radius, which is 2, and the distance between their centers is the distance between (0,0) and (2,2). Let me calculate that distance.Using the distance formula: √[(2-0)² + (2-0)²] = √[4 + 4] = √8 = 2√2.So, d = 2√2, and r = 2.Plugging into the formula:Area = 2² cos⁻¹((2√2)²/(2*2²)) - (2√2/2)√(4*2² - (2√2)²)Simplify step by step.First, calculate (2√2)² = 8.Then, 2*2² = 8.So, cos⁻¹(8/8) = cos⁻¹(1) = 0 radians.Wait, that can't be right. If the distance between centers is 2√2, which is approximately 2.828, and each radius is 2, so the circles are overlapping because 2 + 2 = 4 > 2√2 ≈ 2.828. So, they do overlap, but the formula is giving me 0? That doesn't make sense.Wait, maybe I misapplied the formula. Let me check the formula again.The area of intersection between two circles of radius r separated by distance d is:2r² cos⁻¹(d/(2r)) - (d/2)√(4r² - d²)Ah, I see. I missed the division by 2r inside the arccos. So, it's cos⁻¹(d/(2r)).So, let's recalculate.d = 2√2, r = 2.So, d/(2r) = (2√2)/(2*2) = √2/2 ≈ 0.7071.cos⁻¹(√2/2) is π/4 radians.So, the first part is 2*(2)² * (π/4) = 8*(π/4) = 2π.The second part is (d/2)*√(4r² - d²).d = 2√2, so d/2 = √2.4r² = 4*(2)² = 16.d² = (2√2)² = 8.So, √(16 - 8) = √8 = 2√2.So, the second part is √2 * 2√2 = (√2)*(2√2) = 2*(√2*√2) = 2*2 = 4.Putting it all together, the area of intersection is 2π - 4.So, the overlapping area between R and P is 2π - 4. Since the overlapping area with Q is the same, the total overlapping area is 2*(2π - 4) = 4π - 8.But wait, hold on. The area of circle R is 4π. If I subtract the overlapping areas, which are 4π - 8, that would give me 4π - (4π - 8) = 8. But 8 is just a number, not in terms of π. That seems off because the answer choices are all multiples of π.Hmm, maybe I made a mistake in interpreting the overlapping areas. Let me think again.Wait, actually, the overlapping area between R and P is 2π - 4, as calculated. So, subtracting this from the area of R would give the area inside R but outside P. Similarly, subtracting the overlapping area with Q would give the area inside R but outside Q. But since these overlapping areas are separate, I need to subtract both from the area of R.But wait, if I subtract both overlapping areas, I might be subtracting the intersection of R, P, and Q twice. Is there an overlap where all three circles intersect? Let me see.Given that circle R is centered at (2,2), and circles P and Q are at (0,0) and (4,0), respectively. The distance from R to P is 2√2, and similarly to Q is 2√2. The distance from P to Q is 4. So, the three circles form a sort of triangle, but since R is only tangent to the midpoint of PQ, it doesn't overlap with both P and Q in the same region. So, the overlapping areas between R and P, and R and Q, are distinct and don't overlap with each other. Therefore, I can safely subtract both overlapping areas from the area of R without worrying about double-counting.So, the area inside R but outside P and Q would be:Area of R - (Area of overlap with P + Area of overlap with Q) = 4π - (2π - 4 + 2π - 4) = 4π - (4π - 8) = 4π - 4π + 8 = 8.But 8 is not one of the answer choices. The answer choices are 2π, 4π, 6π, π, and 8π. So, I must have made a mistake somewhere.Wait, maybe I misapplied the formula. Let me double-check the area of intersection between two circles.The formula is:Area = 2r² cos⁻¹(d/(2r)) - (d/2)√(4r² - d²)Where r is the radius of both circles, and d is the distance between their centers.In this case, r = 2, d = 2√2.So, plugging in:Area = 2*(2)² * cos⁻¹((2√2)/(2*2)) - (2√2/2)*√(4*(2)² - (2√2)²)Simplify:= 8 * cos⁻¹(√2/2) - √2 * √(16 - 8)= 8*(π/4) - √2*√8= 2π - √2*(2√2)= 2π - (2*2)= 2π - 4So, that part seems correct. So, each overlapping area is 2π - 4, and two of them would be 4π - 8.But subtracting that from the area of R (4π) gives 8, which isn't an answer choice. Hmm.Wait, maybe I need to consider that the overlapping regions are lens-shaped, and when I subtract them from R, I'm left with the area inside R but outside both P and Q. But perhaps the regions I'm subtracting are not entirely within R? Or maybe I need to consider that the overlapping areas are actually smaller.Alternatively, maybe I should approach this problem differently. Since circle R is tangent to the midpoint of PQ, and PQ is 4 units long, the midpoint is 2 units from both P and Q. Circle R has a radius of 2, so its center is 2 units above the midpoint, making the distance from R's center to P and Q equal to 2√2, as calculated before.Perhaps instead of calculating the overlapping areas, I can find the area inside R that doesn't overlap with P or Q by considering the segments of R that are outside P and Q.Since the distance between R and P is 2√2, which is greater than the radius of P (which is 2), the circles intersect. The area inside R but outside P would be the area of R minus the overlapping lens-shaped area with P. Similarly for Q.But since the overlapping areas are the same for both P and Q, I can calculate one and double it.Wait, but earlier I tried that and got 8, which isn't an answer choice. Maybe I need to think about the geometry differently.Let me consider the triangle formed by the centers of P, Q, and R. The centers are at (0,0), (4,0), and (2,2). This forms a triangle with sides 4, 2√2, and 2√2. So, it's an isosceles triangle with two sides equal to 2√2 and the base equal to 4.The area inside R but outside P and Q would be the area of R minus the areas of the two lens-shaped regions where R overlaps with P and Q.Alternatively, maybe I can calculate the area of R that is above the line PQ, since the overlapping regions are below that line.Wait, circle R is centered at (2,2), so the line PQ is along the x-axis from (0,0) to (4,0). The area inside R but above the x-axis is a semicircle, which has area 2π. But I need to subtract the areas where this semicircle overlaps with P and Q.But wait, the semicircle is entirely above the x-axis, while circles P and Q are below. So, actually, the area inside R above the x-axis doesn't overlap with P and Q at all. Therefore, the area inside R but outside P and Q is just the area of the semicircle, which is 2π.Wait, that makes sense. Because circle R is centered at (2,2), and its radius is 2, so the bottom of R touches the midpoint of PQ at (2,0). The rest of R is above the x-axis, and since P and Q are below the x-axis (except for their tangent point at (2,0)), the area above the x-axis in R doesn't overlap with P or Q.Therefore, the area inside R but outside P and Q is the area of the upper half of R, which is a semicircle with area (1/2)*π*(2)^2 = 2π.So, the answer should be 2π, which is option A.Wait, but earlier I thought the overlapping areas were 2π - 4 each, but that led to a contradiction. Maybe I was overcomplicating it by considering the entire overlapping areas, but actually, the area inside R above the x-axis doesn't overlap with P or Q at all, so it's simply the area of the semicircle.Yes, that makes sense. So, the correct answer is 2π.**Final Answer**boxed{A}

💡First, I need to simplify the expression ((6x^{2}y - 2xy^{2}) div 2xy).I can start by distributing the division across the subtraction inside the parentheses:[(6x^{2}y - 2xy^{2}) div 2xy = 6x^{2}y div 2xy - 2xy^{2} div 2xy]Next, I'll simplify each term separately.For the first term:[6x^{2}y div 2xy = frac{6x^{2}y}{2xy} = frac{6}{2} cdot frac{x^{2}}{x} cdot frac{y}{y} = 3x]For the second term:[-2xy^{2} div 2xy = frac{-2xy^{2}}{2xy} = frac{-2}{2} cdot frac{x}{x} cdot frac{y^{2}}{y} = -y]Combining the simplified terms gives the final result:[3x - y]

💡First, I need to determine the total distance the trains need to cover to be completely clear of each other. This is the sum of their lengths.Next, I'll calculate the relative speed at which the trains are moving apart since they're traveling in opposite directions. I'll add their speeds together.After obtaining the relative speed in kilometers per hour, I'll convert it to meters per second to match the units of the distance.Finally, I'll use the formula Time = Distance / Speed to find out how long it takes for the trains to be completely clear of each other.

💡Okay, so I have this problem here about pulling a stump out of the ground using a lever. It says that the force required varies inversely with the length of the lever. Hmm, okay, inverse variation. I remember that means if one thing gets bigger, the other gets smaller, right? So, if the lever is longer, the force needed should be less, and if the lever is shorter, you need more force. That makes sense because a longer lever gives you more mechanical advantage.Alright, the problem gives me that a 10-foot lever requires 180 pounds of force. And they want to know how much force is needed if I use a 25-foot lever instead. Okay, so I need to find the force when the lever is longer. Since it's an inverse relationship, I think I can set up a proportion or something like that.Let me recall the formula for inverse variation. If two quantities, say F and L, vary inversely, then F times L equals some constant k. So, F * L = k. That means if I know F and L for one situation, I can find k, and then use that k to find F or L in another situation.So, for the first situation, F1 is 180 pounds and L1 is 10 feet. Let me plug those into the formula to find k. That would be 180 * 10 = k. So, 180 times 10 is 1800. Therefore, k is 1800. Okay, got that.Now, for the second situation, I need to find F2 when L2 is 25 feet. Using the same formula, F2 * L2 = k. So, F2 * 25 = 1800. To find F2, I need to divide both sides by 25. So, F2 = 1800 / 25.Let me do that division. 1800 divided by 25. Hmm, 25 times 70 is 1750, and 25 times 72 is 1800. Wait, no, 25 times 72 is actually 1800 because 25 times 70 is 1750 and 25 times 2 is 50, so 1750 + 50 is 1800. So, F2 is 72 pounds.Wait, let me double-check that. If I have a 25-foot lever, and the force is 72 pounds, then 72 times 25 should be 1800. Let me multiply 72 by 25. 70 times 25 is 1750, and 2 times 25 is 50, so 1750 + 50 is 1800. Yep, that checks out.So, using a longer lever reduces the force needed, which makes sense because you're distributing the effort over a longer distance. It's like how a longer wrench makes it easier to loosen a bolt because you have more leverage.I think I've got it. The force required with a 25-foot lever is 72 pounds.

💡Okay, so I have this problem where I need to find the maximum number of terms in a finite sequence of real numbers. The conditions are that the sum of any seven successive terms is negative, and the sum of any eleven successive terms is positive. Hmm, that's interesting. Let me try to break this down step by step.First, I'll denote the sequence as ( x_1, x_2, x_3, ldots, x_n ), where ( n ) is the number of terms we need to find. The problem states that for any seven consecutive terms, their sum is negative. So, for example, ( x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 < 0 ), and similarly for any other set of seven consecutive terms. Similarly, the sum of any eleven consecutive terms is positive. So, ( x_1 + x_2 + ldots + x_{11} > 0 ), and this holds for any eleven consecutive terms in the sequence.My goal is to find the maximum possible value of ( n ). Let me think about how to approach this. Maybe I can use these inequalities to derive some relationships between the terms or find a contradiction if ( n ) is too large.Let me consider the case where ( n = 14 ). If I take the first seven terms and the next seven terms, their sums are both negative. So, ( x_1 + x_2 + ldots + x_7 < 0 ) and ( x_8 + x_9 + ldots + x_{14} < 0 ). If I add these two inequalities, I get ( x_1 + x_2 + ldots + x_{14} < 0 ).But wait, the problem also says that the sum of any eleven consecutive terms is positive. So, if I take terms from ( x_4 ) to ( x_{14} ), that's eleven terms, and their sum should be positive. So, ( x_4 + x_5 + ldots + x_{14} > 0 ).Now, if I subtract the sum of ( x_4 ) to ( x_{14} ) from the sum of all fourteen terms, I get ( x_1 + x_2 + x_3 < 0 ). That's because ( (x_1 + x_2 + ldots + x_{14}) - (x_4 + x_5 + ldots + x_{14}) = x_1 + x_2 + x_3 ).So, from this, I can conclude that ( x_1 + x_2 + x_3 < 0 ). Interesting. Now, let's see if I can find another relationship. Maybe by shifting the window of seven terms.Consider the sum ( x_5 + x_6 + ldots + x_{11} < 0 ). Adding this to the sum of the first seven terms, ( x_1 + x_2 + ldots + x_7 < 0 ), gives ( x_1 + x_2 + ldots + x_{11} < 0 ). But wait, the problem states that the sum of any eleven consecutive terms is positive. So, this leads to a contradiction because ( x_1 + x_2 + ldots + x_{11} ) should be positive, but we just got that it's negative.Hmm, so this suggests that ( n ) cannot be 14 because it leads to a contradiction. Therefore, ( n ) must be less than 14. Wait, but I'm not sure if this is the right way to approach it. Maybe I need to consider more terms or a different approach.Let me try another angle. Suppose the sequence has ( n ) terms. Then, for any ( i ) such that ( 1 leq i leq n - 6 ), the sum ( x_i + x_{i+1} + ldots + x_{i+6} < 0 ). Similarly, for any ( j ) such that ( 1 leq j leq n - 10 ), the sum ( x_j + x_{j+1} + ldots + x_{j+10} > 0 ).I wonder if I can use these inequalities to bound ( n ). Maybe by considering overlapping sums or something like that.Let me try to write down some inequalities. For example, consider the sum of the first seven terms: ( S_7 = x_1 + x_2 + ldots + x_7 < 0 ). The sum of the next seven terms: ( S'_7 = x_8 + x_9 + ldots + x_{14} < 0 ). Adding these gives ( S_7 + S'_7 = x_1 + x_2 + ldots + x_{14} < 0 ).But as before, the sum of terms from ( x_4 ) to ( x_{14} ) is positive, so ( x_4 + x_5 + ldots + x_{14} > 0 ). Subtracting this from the total sum ( x_1 + x_2 + ldots + x_{14} ), we get ( x_1 + x_2 + x_3 < 0 ).Similarly, if I consider the sum of terms from ( x_5 ) to ( x_{11} ), which is less than zero, and add it to the sum of the first seven terms, I get ( x_1 + x_2 + ldots + x_{11} < 0 ), which contradicts the condition that the sum of any eleven consecutive terms is positive.So, this suggests that having ( n = 14 ) is impossible. Therefore, ( n ) must be less than 14. But wait, maybe ( n = 13 ) is possible? Let me check.If ( n = 13 ), then the sum of the first seven terms is negative, and the sum of the last seven terms (from ( x_7 ) to ( x_{13} )) is also negative. Adding these gives ( x_1 + x_2 + ldots + x_{13} < 0 ). However, the sum of terms from ( x_3 ) to ( x_{13} ) is positive, so ( x_3 + x_4 + ldots + x_{13} > 0 ). Subtracting this from the total sum gives ( x_1 + x_2 < 0 ).Similarly, considering the sum of terms from ( x_5 ) to ( x_{11} ), which is negative, and adding it to the sum of the first seven terms, we get ( x_1 + x_2 + ldots + x_{11} < 0 ), which again contradicts the condition that the sum of any eleven consecutive terms is positive.Hmm, so ( n = 13 ) also leads to a contradiction. Maybe I need to go lower. Let's try ( n = 12 ).For ( n = 12 ), the sum of the first seven terms is negative, and the sum of the last seven terms (from ( x_6 ) to ( x_{12} )) is also negative. Adding these gives ( x_1 + x_2 + ldots + x_{12} < 0 ). The sum of terms from ( x_2 ) to ( x_{12} ) is positive, so ( x_2 + x_3 + ldots + x_{12} > 0 ). Subtracting this from the total sum gives ( x_1 < 0 ).Similarly, considering the sum of terms from ( x_5 ) to ( x_{11} ), which is negative, and adding it to the sum of the first seven terms, we get ( x_1 + x_2 + ldots + x_{11} < 0 ), which again contradicts the condition that the sum of any eleven consecutive terms is positive.Wait a minute, this seems to be a pattern. Every time I try to set ( n ) to a certain value, I end up with a contradiction because the sum of eleven terms is supposed to be positive, but my inequalities suggest it's negative.Maybe I need to approach this differently. Perhaps instead of trying specific values, I can find a general relationship or bound on ( n ).Let me consider the total sum of the sequence. If I denote ( S = x_1 + x_2 + ldots + x_n ), then depending on ( n ), I can express ( S ) in terms of the sums of seven and eleven terms.But I'm not sure how to proceed with that. Maybe I can use the fact that the sum of any seven consecutive terms is negative and the sum of any eleven consecutive terms is positive to create a system of inequalities.Alternatively, perhaps I can use the concept of overlapping sums. For example, if I have overlapping blocks of seven and eleven terms, their sums have different signs, which might give me some constraints on the terms themselves.Wait, another idea: if I consider the sum of the first eleven terms, it's positive. Then, the sum of the first seven terms is negative. So, the sum of the remaining four terms (from ( x_8 ) to ( x_{11} )) must be positive enough to make the total sum of eleven terms positive. Similarly, the sum of the next seven terms (from ( x_5 ) to ( x_{11} )) is negative, which means the sum of ( x_5 ) to ( x_7 ) must be negative, but ( x_8 ) to ( x_{11} ) is positive.This seems a bit convoluted, but maybe I can formalize it.Let me denote ( S_k = x_k + x_{k+1} + ldots + x_{k+6} < 0 ) for any ( k ) such that ( k + 6 leq n ).Similarly, ( T_k = x_k + x_{k+1} + ldots + x_{k+10} > 0 ) for any ( k ) such that ( k + 10 leq n ).Now, if I take ( T_1 = x_1 + x_2 + ldots + x_{11} > 0 ), and ( S_1 = x_1 + x_2 + ldots + x_7 < 0 ). Then, ( T_1 - S_1 = x_8 + x_9 + x_{10} + x_{11} > 0 ).Similarly, ( S_5 = x_5 + x_6 + ldots + x_{11} < 0 ). So, ( T_1 - S_5 = x_1 + x_2 + x_3 + x_4 > 0 ).So, from ( T_1 - S_1 > 0 ) and ( T_1 - S_5 > 0 ), I can get some information about the sums of the first four terms and the last four terms of the eleven-term block.But how does this help me find ( n )? Maybe by considering more such relationships.Let me try to generalize. For any ( k ), ( T_k - S_k = x_{k+7} + x_{k+8} + x_{k+9} + x_{k+10} > 0 ). Similarly, ( T_k - S_{k+4} = x_1 + x_2 + x_3 + x_4 > 0 ).Wait, maybe I can create a chain of such inequalities. For example, starting from ( T_1 ), I can express the sum of the last four terms as positive. Then, considering ( T_2 ), the sum of terms from ( x_2 ) to ( x_{12} ) is positive, and so on.But I'm not sure if this is leading me anywhere. Maybe I need to think about the maximum number of terms before the contradictions become unavoidable.From my earlier attempts, I saw that ( n = 14 ) leads to a contradiction because the sum of the first eleven terms would have to be both positive and negative. Similarly, ( n = 13 ) and ( n = 12 ) also lead to contradictions.Wait, maybe I need to consider that the maximum ( n ) is 16. Let me check that.If ( n = 16 ), then I can have multiple overlapping blocks of seven and eleven terms. Let me see if I can construct such a sequence without contradictions.Suppose I set the first seven terms to sum to a negative number, say -1. Then, the next seven terms (from ( x_8 ) to ( x_{14} )) also sum to -1. Adding these gives the total sum of the first fourteen terms as -2. However, the sum of terms from ( x_4 ) to ( x_{14} ) is positive, say +1. Then, subtracting this from the total sum gives ( x_1 + x_2 + x_3 = -3 ).Similarly, considering the sum of terms from ( x_5 ) to ( x_{11} ), which is negative, and adding it to the sum of the first seven terms, we get ( x_1 + x_2 + ldots + x_{11} < 0 ), which contradicts the condition that the sum of any eleven consecutive terms is positive.Wait, so even ( n = 16 ) leads to a contradiction? Or maybe I'm not constructing the sequence correctly.Perhaps I need to distribute the negative and positive sums in a way that avoids this contradiction. Maybe by having some terms positive and some negative in a specific pattern.Let me try to construct a sequence where the sum of any seven consecutive terms is negative, but the sum of any eleven consecutive terms is positive. Maybe by alternating blocks of positive and negative terms.For example, suppose I have a block of three positive terms, followed by four negative terms, and repeat this pattern. Let me see if this works.Let me denote the blocks as ( A = [a, a, a, -b, -b, -b, -b] ), where ( a ) and ( b ) are positive numbers. The sum of seven terms in this block is ( 3a - 4b < 0 ), so ( 3a < 4b ).Now, the sum of eleven consecutive terms would cover one full block ( A ) and part of the next block. Let's see: ( A + ) first three terms of the next block ( A ). So, the sum would be ( (3a - 4b) + 3a = 6a - 4b ). We need this to be positive, so ( 6a > 4b ) or ( 3a > 2b ).But from the seven-term sum condition, we have ( 3a < 4b ). So, combining these, ( 2b < 3a < 4b ). This is possible if ( 2b < 4b ), which is always true since ( b > 0 ).So, for example, let me choose ( a = 2 ) and ( b = 3 ). Then, ( 3a = 6 ) and ( 4b = 12 ), so ( 6 < 12 ), satisfying the seven-term condition. For the eleven-term sum, ( 6a = 12 ) and ( 4b = 12 ), so ( 12 - 12 = 0 ), which is not positive. Hmm, that's not good.Wait, maybe I need to adjust the values. Let me choose ( a = 3 ) and ( b = 2 ). Then, ( 3a = 9 ) and ( 4b = 8 ), so ( 9 < 8 ) is false. That doesn't work.Wait, maybe I need a different ratio. Let me set ( a = 4 ) and ( b = 3 ). Then, ( 3a = 12 ) and ( 4b = 12 ), so ( 12 < 12 ) is false.Hmm, maybe this approach isn't working. Perhaps I need a different pattern or more careful selection of ( a ) and ( b ).Alternatively, maybe I can have the sequence where the first three terms are positive, the next four are negative, and then repeat. Let me try that.Let me define the sequence as follows:- ( x_1 = a )- ( x_2 = a )- ( x_3 = a )- ( x_4 = -b )- ( x_5 = -b )- ( x_6 = -b )- ( x_7 = -b )- ( x_8 = a )- ( x_9 = a )- ( x_{10} = a )- ( x_{11} = -b )- ( x_{12} = -b )- ( x_{13} = -b )- ( x_{14} = -b )- ( x_{15} = a )- ( x_{16} = a )Now, let's check the sums.First, the sum of any seven consecutive terms:- ( x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 = 3a - 4b < 0 )- ( x_2 + x_3 + x_4 + x_5 + x_6 + x_7 + x_8 = 3a - 4b + a = 4a - 4b < 0 )- Similarly, all other seven-term sums will be either ( 3a - 4b ) or ( 4a - 4b ), both of which need to be negative.So, ( 3a < 4b ) and ( 4a < 4b ) which simplifies to ( a < b ).Now, the sum of any eleven consecutive terms:- ( x_1 + x_2 + ldots + x_{11} = 3a - 4b + 3a - b = 6a - 5b > 0 )- ( x_2 + x_3 + ldots + x_{12} = 3a - 4b + a - b = 4a - 5b > 0 )- ( x_3 + x_4 + ldots + x_{13} = a - 4b - b = a - 5b > 0 ) Wait, this can't be right because ( a < b ), so ( a - 5b ) would be negative.Hmm, that's a problem. The sum of terms from ( x_3 ) to ( x_{13} ) is ( a - 5b ), which is negative, contradicting the condition that it should be positive.So, my construction is flawed. Maybe I need a different pattern or adjust the values of ( a ) and ( b ) more carefully.Alternatively, perhaps I can have overlapping blocks where the positive sums from the eleven-term blocks compensate for the negative sums in the seven-term blocks.Wait, maybe I can set up a system of inequalities based on the sums.Let me denote the sum of the first seven terms as ( S_7 = x_1 + x_2 + ldots + x_7 < 0 ).The sum of the first eleven terms is ( T_{11} = x_1 + x_2 + ldots + x_{11} > 0 ).Similarly, the sum of terms from ( x_5 ) to ( x_{11} ) is ( S'_7 = x_5 + x_6 + ldots + x_{11} < 0 ).So, ( T_{11} = S_7 + x_8 + x_9 + x_{10} + x_{11} > 0 ).But ( S'_7 = x_5 + x_6 + x_7 + x_8 + x_9 + x_{10} + x_{11} < 0 ).So, ( T_{11} = S_7 + (x_8 + x_9 + x_{10} + x_{11}) > 0 ).But ( S'_7 = (x_5 + x_6 + x_7) + (x_8 + x_9 + x_{10} + x_{11}) < 0 ).Let me denote ( A = x_1 + x_2 + x_3 + x_4 ), ( B = x_5 + x_6 + x_7 ), ( C = x_8 + x_9 + x_{10} + x_{11} ).Then, ( S_7 = A + B < 0 ), ( T_{11} = A + B + C > 0 ), and ( S'_7 = B + C < 0 ).From ( S_7 = A + B < 0 ) and ( T_{11} = A + B + C > 0 ), we can write ( C > - (A + B) ).From ( S'_7 = B + C < 0 ), we have ( C < -B ).Combining these, ( - (A + B) < C < -B ).This implies ( - (A + B) < -B ), which simplifies to ( -A - B < -B ), so ( -A < 0 ), meaning ( A > 0 ).So, ( A = x_1 + x_2 + x_3 + x_4 > 0 ).Similarly, from ( S_7 = A + B < 0 ) and ( A > 0 ), it follows that ( B < -A ).So, ( B = x_5 + x_6 + x_7 < -A ).Now, let's consider the next set of sums. The sum of terms from ( x_2 ) to ( x_8 ) is ( S''_7 = x_2 + x_3 + x_4 + x_5 + x_6 + x_7 + x_8 < 0 ).This can be written as ( (x_2 + x_3 + x_4) + (x_5 + x_6 + x_7) + x_8 < 0 ).Let me denote ( D = x_2 + x_3 + x_4 ), ( B = x_5 + x_6 + x_7 ), and ( E = x_8 ).So, ( S''_7 = D + B + E < 0 ).But we know ( A = x_1 + x_2 + x_3 + x_4 > 0 ), so ( D = x_2 + x_3 + x_4 = A - x_1 ).Since ( A > 0 ), ( D ) could be positive or negative depending on ( x_1 ).But let's see, from ( S_7 = A + B < 0 ) and ( A > 0 ), ( B < -A ).So, ( B ) is quite negative.Now, considering ( S''_7 = D + B + E < 0 ), and knowing ( B < -A ), we can write ( D + E < A ).But ( D = A - x_1 ), so ( (A - x_1) + E < A ), which simplifies to ( -x_1 + E < 0 ), so ( E < x_1 ).So, ( x_8 < x_1 ).Similarly, I can continue this process for other sums, but it's getting quite involved.Maybe instead of trying to construct the sequence, I can use these inequalities to find a bound on ( n ).From the earlier steps, I saw that ( n = 14 ) leads to a contradiction because the sum of the first eleven terms would have to be both positive and negative. Similarly, ( n = 13 ), ( 12 ), and even ( 16 ) might lead to contradictions.Wait, but earlier I thought ( n = 16 ) might be possible, but my construction led to a contradiction. Maybe the maximum ( n ) is actually 16, but I need to find a way to construct the sequence without contradictions.Alternatively, perhaps the maximum ( n ) is 16, as per some known results or similar problems.Wait, I recall that in similar problems involving overlapping sums, the maximum length is often related to the sum of the two window sizes minus one, but I'm not sure if that applies here.Alternatively, maybe the maximum ( n ) is 16 because beyond that, the contradictions become unavoidable.Let me try to think differently. Suppose the sequence has ( n ) terms. Then, the sum of the first seven terms is negative, and the sum of the last seven terms is also negative. Adding these gives the sum of the first ( n - 7 ) terms plus the last seven terms is negative. But the sum of the entire sequence can be expressed in terms of these.Wait, maybe I can use induction or some recursive approach, but I'm not sure.Alternatively, perhaps I can model this as a system of linear inequalities and find the maximum ( n ) such that the system is consistent.But that might be too abstract. Maybe I can consider the differences between the sums.Let me define ( S_k = x_k + x_{k+1} + ldots + x_{k+6} < 0 ) for ( k = 1, 2, ldots, n - 6 ).Similarly, ( T_k = x_k + x_{k+1} + ldots + x_{k+10} > 0 ) for ( k = 1, 2, ldots, n - 10 ).Now, consider the difference ( T_k - S_k = x_{k+7} + x_{k+8} + x_{k+9} + x_{k+10} > 0 ).Similarly, ( T_{k+1} - S_{k+1} = x_{k+8} + x_{k+9} + x_{k+10} + x_{k+11} > 0 ).Subtracting these two, we get ( (T_{k+1} - S_{k+1}) - (T_k - S_k) = x_{k+11} - x_{k+7} > 0 ).So, ( x_{k+11} > x_{k+7} ).This suggests that each term is greater than the term four positions before it. So, ( x_{k+11} > x_{k+7} ) for all ( k ) such that ( k + 11 leq n ).This is an interesting recursive relationship. It implies that the sequence is increasing every four terms. So, ( x_5 > x_1 ), ( x_6 > x_2 ), ( x_7 > x_3 ), ( x_8 > x_4 ), and so on.But wait, if the sequence is increasing every four terms, but the sum of any seven consecutive terms is negative, that might impose a limit on how much it can increase.Let me try to formalize this. Suppose ( x_{k+4} > x_k ) for all ( k ). Then, the sequence is strictly increasing every four terms.But if the sequence is increasing, the later terms are larger than the earlier ones. However, the sum of any seven consecutive terms is negative, which suggests that the negative terms must dominate in some way.This seems contradictory because if the sequence is increasing, the later terms are larger, but the sum of seven terms is negative, implying that the earlier terms are more negative.Wait, maybe the sequence alternates between negative and positive in a way that the sums of seven terms are negative, but the sums of eleven terms are positive.But I'm not sure how to reconcile this with the increasing nature every four terms.Alternatively, perhaps the sequence has a periodic pattern where the negative sums of seven terms are offset by positive sums of eleven terms.Wait, maybe I can consider the sequence as a combination of overlapping negative and positive blocks.Let me try to think of the sequence as having a repeating pattern where every seven terms sum to a negative number, but every eleven terms sum to a positive number.Perhaps the sequence has a period of 16, where the first seven terms are negative, the next four are positive, and so on, but I'm not sure.Alternatively, maybe the maximum ( n ) is 16 because beyond that, the overlapping sums would force the sequence to have both positive and negative sums in a way that's impossible.Wait, I think I've heard of a similar problem before where the maximum length is 16. Let me try to recall.In some problems involving overlapping sums with different signs, the maximum length is often the sum of the two window sizes minus one, which in this case would be ( 7 + 11 - 1 = 17 ). But since we're getting contradictions at 14, maybe it's less.Wait, but in my earlier attempt, ( n = 16 ) led to a contradiction when I tried to construct the sequence. Maybe the maximum ( n ) is actually 16, but I need to find a specific construction that avoids the contradiction.Alternatively, perhaps the maximum ( n ) is 16 because beyond that, the overlapping sums would force the sequence to have both positive and negative sums in a way that's impossible.Wait, let me try to think differently. Suppose the sequence has ( n = 16 ) terms. Then, I can have multiple overlapping blocks of seven and eleven terms.Let me try to construct the sequence such that the sum of any seven consecutive terms is negative, but the sum of any eleven consecutive terms is positive.One way to do this is to have the first three terms positive, the next four negative, and repeat this pattern. Let me try that.Let me define the sequence as follows:- ( x_1 = a )- ( x_2 = a )- ( x_3 = a )- ( x_4 = -b )- ( x_5 = -b )- ( x_6 = -b )- ( x_7 = -b )- ( x_8 = a )- ( x_9 = a )- ( x_{10} = a )- ( x_{11} = -b )- ( x_{12} = -b )- ( x_{13} = -b )- ( x_{14} = -b )- ( x_{15} = a )- ( x_{16} = a )Now, let's check the sums.First, the sum of any seven consecutive terms:- ( x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 = 3a - 4b < 0 )- ( x_2 + x_3 + x_4 + x_5 + x_6 + x_7 + x_8 = 3a - 4b + a = 4a - 4b < 0 )- Similarly, all other seven-term sums will be either ( 3a - 4b ) or ( 4a - 4b ), both of which need to be negative.So, ( 3a < 4b ) and ( 4a < 4b ) which simplifies to ( a < b ).Now, the sum of any eleven consecutive terms:- ( x_1 + x_2 + ldots + x_{11} = 3a - 4b + 3a - b = 6a - 5b > 0 )- ( x_2 + x_3 + ldots + x_{12} = 3a - 4b + a - b = 4a - 5b > 0 )- ( x_3 + x_4 + ldots + x_{13} = a - 4b - b = a - 5b > 0 ) Wait, this can't be right because ( a < b ), so ( a - 5b ) would be negative.Hmm, that's a problem. The sum of terms from ( x_3 ) to ( x_{13} ) is ( a - 5b ), which is negative, contradicting the condition that it should be positive.So, my construction is flawed. Maybe I need a different pattern or adjust the values of ( a ) and ( b ) more carefully.Alternatively, perhaps I can have overlapping blocks where the positive sums from the eleven-term blocks compensate for the negative sums in the seven-term blocks.Wait, maybe I can set up a system of inequalities based on the sums.Let me denote the sum of the first seven terms as ( S_7 = x_1 + x_2 + ldots + x_7 < 0 ).The sum of the first eleven terms is ( T_{11} = x_1 + x_2 + ldots + x_{11} > 0 ).Similarly, the sum of terms from ( x_5 ) to ( x_{11} ) is ( S'_7 = x_5 + x_6 + ldots + x_{11} < 0 ).So, ( T_{11} = S_7 + x_8 + x_9 + x_{10} + x_{11} > 0 ).But ( S'_7 = x_5 + x_6 + x_7 + x_8 + x_9 + x_{10} + x_{11} < 0 ).Let me denote ( A = x_1 + x_2 + x_3 + x_4 ), ( B = x_5 + x_6 + x_7 ), ( C = x_8 + x_9 + x_{10} + x_{11} ).Then, ( S_7 = A + B < 0 ), ( T_{11} = A + B + C > 0 ), and ( S'_7 = B + C < 0 ).From ( S_7 = A + B < 0 ) and ( T_{11} = A + B + C > 0 ), we can write ( C > - (A + B) ).From ( S'_7 = B + C < 0 ), we have ( C < -B ).Combining these, ( - (A + B) < C < -B ).This implies ( - (A + B) < -B ), which simplifies to ( -A - B < -B ), so ( -A < 0 ), meaning ( A > 0 ).So, ( A = x_1 + x_2 + x_3 + x_4 > 0 ).Similarly, from ( S_7 = A + B < 0 ) and ( A > 0 ), it follows that ( B < -A ).So, ( B = x_5 + x_6 + x_7 < -A ).Now, let's consider the next set of sums. The sum of terms from ( x_2 ) to ( x_8 ) is ( S''_7 = x_2 + x_3 + x_4 + x_5 + x_6 + x_7 + x_8 < 0 ).This can be written as ( (x_2 + x_3 + x_4) + (x_5 + x_6 + x_7) + x_8 < 0 ).Let me denote ( D = x_2 + x_3 + x_4 ), ( B = x_5 + x_6 + x_7 ), and ( E = x_8 ).So, ( S''_7 = D + B + E < 0 ).But we know ( A = x_1 + x_2 + x_3 + x_4 > 0 ), so ( D = x_2 + x_3 + x_4 = A - x_1 ).Since ( A > 0 ), ( D ) could be positive or negative depending on ( x_1 ).But let's see, from ( S_7 = A + B < 0 ) and ( A > 0 ), ( B < -A ).So, ( B ) is quite negative.Now, considering ( S''_7 = D + B + E < 0 ), and knowing ( B < -A ), we can write ( D + E < A ).But ( D = A - x_1 ), so ( (A - x_1) + E < A ), which simplifies to ( -x_1 + E < 0 ), so ( E < x_1 ).So, ( x_8 < x_1 ).Similarly, I can continue this process for other sums, but it's getting quite involved.Maybe instead of trying to construct the sequence, I can use these inequalities to find a bound on ( n ).From the earlier steps, I saw that ( n = 14 ) leads to a contradiction because the sum of the first eleven terms would have to be both positive and negative. Similarly, ( n = 13 ), ( 12 ), and even ( 16 ) might lead to contradictions.Wait, but earlier I thought ( n = 16 ) might be possible, but my construction led to a contradiction. Maybe the maximum ( n ) is actually 16, but I need to find a way to construct the sequence without contradictions.Alternatively, perhaps the maximum ( n ) is 16 because beyond that, the contradictions become unavoidable.Wait, I think I've heard of a similar problem before where the maximum length is 16. Let me try to recall.In some problems involving overlapping sums with different signs, the maximum length is often the sum of the two window sizes minus one, which in this case would be ( 7 + 11 - 1 = 17 ). But since we're getting contradictions at 14, maybe it's less.Wait, but in my earlier attempt, ( n = 16 ) led to a contradiction when I tried to construct the sequence. Maybe the maximum ( n ) is actually 16, but I need to find a specific construction that avoids the contradiction.Alternatively, perhaps the maximum ( n ) is 16 because beyond that, the overlapping sums would force the sequence to have both positive and negative sums in a way that's impossible.After considering all these possibilities and trying different constructions, I think the maximum number of terms in the sequence is 16. This is because beyond 16, the overlapping sums of seven and eleven terms would lead to unavoidable contradictions, making it impossible to satisfy both conditions simultaneously.

💡Okay, so I have this problem here: α is an angle in the second quadrant, and sin α + cos α equals √3 over 3. I need to find the value of cos 2α. Hmm, let me think about how to approach this.First, I remember that when dealing with expressions like sin α + cos α, squaring both sides can sometimes help because it relates to the Pythagorean identity. So, let me try that.Starting with sin α + cos α = √3 / 3. If I square both sides, I get:(sin α + cos α)^2 = (√3 / 3)^2Expanding the left side, using the formula (a + b)^2 = a^2 + 2ab + b^2, so:sin²α + 2 sin α cos α + cos²α = (√3 / 3)^2Simplify the right side:(√3)^2 is 3, and 3 squared is 9, so it becomes 3/9, which simplifies to 1/3.So, sin²α + 2 sin α cos α + cos²α = 1/3I know that sin²α + cos²α = 1, so substituting that in:1 + 2 sin α cos α = 1/3Now, subtract 1 from both sides:2 sin α cos α = 1/3 - 1Which is 2 sin α cos α = -2/3Divide both sides by 2:sin α cos α = -1/3Wait, but I need to find cos 2α. I remember there's a double angle formula for cosine that might help here. Let me recall the formulas:cos 2α can be expressed in terms of sin²α and cos²α, or in terms of sin α cos α. The formula is:cos 2α = cos²α - sin²αAlternatively, I also know that:cos 2α = 1 - 2 sin²αorcos 2α = 2 cos²α - 1But another formula is:cos 2α = (cos α - sin α)(cos α + sin α)Wait, that might be useful because I have sin α + cos α given, which is √3 / 3. If I can find cos α - sin α, then I can multiply them together to get cos 2α.So, let me try that approach.Let me denote:A = sin α + cos α = √3 / 3B = cos α - sin αThen, cos 2α = A * BSo, if I can find B, then I can find cos 2α.To find B, let's square B:B² = (cos α - sin α)^2 = cos²α - 2 sin α cos α + sin²αAgain, using the Pythagorean identity, cos²α + sin²α = 1, so:B² = 1 - 2 sin α cos αBut from earlier, we found that sin α cos α = -1/3So, substituting that in:B² = 1 - 2*(-1/3) = 1 + 2/3 = 5/3Therefore, B = sqrt(5/3) or -sqrt(5/3)But since α is in the second quadrant, let's think about the signs of sin α and cos α.In the second quadrant, sin α is positive and cos α is negative. So, cos α - sin α would be negative minus positive, which is negative. Therefore, B is negative.So, B = -sqrt(5/3)But let me write sqrt(5/3) as sqrt(15)/3 for simplicity.So, B = -sqrt(15)/3Now, cos 2α = A * B = (√3 / 3) * (-sqrt(15)/3)Multiply the numerators and denominators:Numerator: √3 * (-sqrt(15)) = -sqrt(45) = -3 sqrt(5)Denominator: 3 * 3 = 9So, cos 2α = (-3 sqrt(5)) / 9 = -sqrt(5)/3Wait, but let me double-check my steps to make sure I didn't make a mistake.Starting from sin α + cos α = √3 / 3, squared to get 1 + 2 sin α cos α = 1/3, leading to sin α cos α = -1/3. Then, using the identity for cos 2α as (sin α + cos α)(cos α - sin α), which is A * B. Then, finding B by squaring and taking square roots, considering the quadrant to determine the sign.Yes, that seems correct. So, cos 2α is -sqrt(5)/3.Looking at the options, that's option A.