💡Alright, so I have this problem where I need to find the largest possible value of the expression ( ab + bc + cd + da ) where ( a, b, c, d ) are the numbers 1, 2, 3, 4 in some order. Hmm, okay, let's see how I can approach this.First, I need to understand the expression ( ab + bc + cd + da ). It looks like it's a combination of products of consecutive variables. Maybe there's a way to factor this or find a pattern that can help me simplify it.Let me write down the expression again: ( ab + bc + cd + da ). Hmm, I notice that each term is a product of two variables, and each variable is multiplied by two others. For example, ( a ) is multiplied by ( b ) and ( d ), ( b ) is multiplied by ( a ) and ( c ), and so on. Maybe I can factor this expression somehow.Let me try to factor it. If I look at ( ab + bc + cd + da ), I can group terms differently. Let's see:( ab + bc + cd + da = b(a + c) + d(a + c) = (a + c)(b + d) ).Oh, that's interesting! So the expression can be factored into ( (a + c)(b + d) ). That simplifies things a bit because now instead of dealing with four variables, I can think of it as the product of two sums.So, now my goal is to maximize ( (a + c)(b + d) ) where ( a, b, c, d ) are 1, 2, 3, 4 in some order. Since all four numbers are used, ( a + c + b + d = 1 + 2 + 3 + 4 = 10 ). So, ( (a + c) + (b + d) = 10 ).Now, I need to maximize the product ( (a + c)(b + d) ) given that their sum is 10. I remember from math class that for two numbers with a fixed sum, their product is maximized when the numbers are equal or as close to equal as possible. This is related to the Arithmetic Mean-Geometric Mean Inequality (AM-GM).So, if I can make ( a + c ) and ( b + d ) as close to each other as possible, their product will be the largest. Since their sum is 10, the closest they can be is 5 and 5. But wait, can I actually get both ( a + c ) and ( b + d ) equal to 5?Let me check. The numbers I have are 1, 2, 3, 4. Let's see if I can split them into two pairs that each add up to 5.Possible pairs:- 1 and 4: 1 + 4 = 5- 2 and 3: 2 + 3 = 5Yes! So, if I pair 1 with 4 and 2 with 3, both pairs add up to 5. Therefore, ( (a + c) = 5 ) and ( (b + d) = 5 ), so their product is ( 5 times 5 = 25 ).Is this the maximum possible? Let me verify by trying other combinations.Suppose I pair 1 with 3 and 2 with 4. Then, ( a + c = 4 ) and ( b + d = 6 ). Their product is ( 4 times 6 = 24 ), which is less than 25.Another combination: pairing 1 with 2 and 3 with 4. Then, ( a + c = 3 ) and ( b + d = 7 ). Their product is ( 3 times 7 = 21 ), which is even less.How about pairing 1 with 4 and 2 with 3, which we already did, giving 25. Is there any other way to pair them?Wait, let me think. Maybe if I assign the numbers differently, not just pairing the smallest with the largest. But since all numbers are used, the pairs have to be two numbers each, so the only possible pairings are the ones I considered.Therefore, the maximum product is indeed 25 when ( a + c = 5 ) and ( b + d = 5 ).But just to be thorough, let me list all possible pairings and their products:1. (1,4) and (2,3): ( 5 times 5 = 25 )2. (1,3) and (2,4): ( 4 times 6 = 24 )3. (1,2) and (3,4): ( 3 times 7 = 21 )4. (1,4) and (3,2): Same as the first, just reordered, still 255. (2,3) and (1,4): Same as the first, 256. Etc.So, the maximum is definitely 25.But wait, let me make sure that when I factor the original expression, I didn't make any mistakes. The original expression was ( ab + bc + cd + da ). When I factored it, I got ( (a + c)(b + d) ). Let me check that:( (a + c)(b + d) = ab + ad + cb + cd ). Hmm, that's ( ab + ad + cb + cd ), which is the same as ( ab + bc + cd + da ). Yes, because ( ad ) is the same as ( da ). So, my factoring was correct.Therefore, the maximum value of ( ab + bc + cd + da ) is 25.Just to wrap up, the key steps were:1. Recognize that the expression can be factored into ( (a + c)(b + d) ).2. Note that the sum ( a + b + c + d = 10 ).3. Use the AM-GM principle to realize that the product is maximized when the two sums are equal.4. Verify that such a pairing is possible with the given numbers.5. Confirm by checking other possible pairings that 25 is indeed the maximum.So, I think I'm confident that 25 is the correct answer.

💡Okay, so I have this problem about a geometric sequence. All the terms are positive, and I know that the product of the first term and the fifth term is 16. I need to find the sum of the logarithms (base 2) of each term from a1 to a5. Hmm, let's break this down step by step.First, I remember that in a geometric sequence, each term is multiplied by a common ratio, r. So, the nth term can be written as a_n = a1 * r^(n-1). That means a1 is just a1, a2 is a1*r, a3 is a1*r^2, a4 is a1*r^3, and a5 is a1*r^4.The problem says that a1 * a5 = 16. Let me write that out using the formula for a5. So, a1 * (a1 * r^4) = 16. That simplifies to a1^2 * r^4 = 16. Hmm, okay, so (a1 * r^2)^2 = 16. Taking the square root of both sides, I get a1 * r^2 = 4. Wait, that's interesting because a1 * r^2 is actually a3. So, a3 = 4. That's a useful piece of information.Now, the problem asks for the sum of the logarithms: log2(a1) + log2(a2) + log2(a3) + log2(a4) + log2(a5). I remember that the sum of logs is the log of the product. So, this sum is equal to log2(a1 * a2 * a3 * a4 * a5). That might be easier to handle.Let me write out the product a1 * a2 * a3 * a4 * a5. Using the terms from the geometric sequence:a1 = a1a2 = a1 * ra3 = a1 * r^2a4 = a1 * r^3a5 = a1 * r^4Multiplying them all together: a1 * (a1 * r) * (a1 * r^2) * (a1 * r^3) * (a1 * r^4). Let's compute this step by step.First, multiply all the a1 terms: a1 * a1 * a1 * a1 * a1 = a1^5.Next, multiply all the r terms: r * r^2 * r^3 * r^4. Adding the exponents: 1 + 2 + 3 + 4 = 10, so that's r^10.So, the entire product is a1^5 * r^10. But from earlier, I know that a1 * r^2 = 4, which is a3. So, a1^5 * r^10 can be written as (a1 * r^2)^5. Since a1 * r^2 is 4, this becomes 4^5.Calculating 4^5: 4 * 4 = 16, 16 * 4 = 64, 64 * 4 = 256, 256 * 4 = 1024. Wait, that seems too big. Let me check: 4^1=4, 4^2=16, 4^3=64, 4^4=256, 4^5=1024. Yeah, that's correct.So, the product a1 * a2 * a3 * a4 * a5 is 1024. Therefore, log2(1024) is the value we're looking for.I know that 2^10 = 1024, so log2(1024) = 10. That means the sum of the logarithms is 10.Wait, let me make sure I didn't skip any steps or make a mistake. So, starting from a1 * a5 = 16, I found that a3 = 4. Then, the product of all five terms is (a3)^5, which is 4^5 = 1024. Taking log base 2 of 1024 is indeed 10 because 2^10 = 1024. Yeah, that seems right.I don't think I made any errors here. The key was recognizing that in a geometric sequence, the product of terms equidistant from the ends is equal, and that the middle term squared is equal to the product of the first and fifth terms. Then, using logarithm properties to simplify the sum to a single logarithm of the product. Everything checks out.

💡First, I need to determine the speed of the train. When the train crosses a signal pole, it covers its own length in 18 seconds.Next, I'll calculate the speed using the formula:Speed = Distance / TimeSpeed = 300 meters / 18 secondsSpeed ≈ 16.67 meters per secondNow, when the train crosses the platform, it covers the combined length of the train and the platform in 39 seconds.Let ( P ) be the length of the platform. The total distance covered is:Total Distance = 300 meters + ( P )Using the speed calculated earlier:16.67 meters/second = (300 meters + ( P )) / 39 secondsSolving for ( P ):( P = 16.67 times 39 - 300 )( P ≈ 650.13 - 300 )( P ≈ 350.13 ) metersTherefore, the length of the platform is approximately 350.13 meters.

💡Alright, so I'm trying to figure out this probability problem where four points are chosen uniformly at random on a circle, and I need to find the probability that no three of these points form an obtuse triangle with the circle's center. Hmm, okay, let's break this down step by step.First, let me recall that when you have a triangle with the center of the circle, the type of triangle (acute, right, or obtuse) depends on the angles subtended at the center. Specifically, if the angle at the center is greater than 90 degrees (π/2 radians), the triangle is obtuse. So, for a triangle formed by two points on the circle and the center, if the arc between those two points is greater than π/2, the triangle is obtuse.Now, since we're dealing with four points, there are multiple triangles we can form by choosing any three points. Each triangle will have the center as one vertex, and two of the four points as the other two vertices. So, for four points, there are C(4,3) = 4 triangles to consider. Each of these triangles could potentially be obtuse if the arc between the two points is greater than π/2.The problem asks for the probability that none of these four triangles are obtuse. That means, for every pair of points among the four, the arc between them must be less than or equal to π/2. So, essentially, all four points must be arranged on the circle such that no two points are more than π/2 apart from each other along the circumference.Wait, hold on. If all four points must be within π/2 of each other, that seems really restrictive. Because if you have four points on a circle, each pair must be within π/2. That would mean all four points lie within a semicircle of π/2, right? But wait, a semicircle is π radians, so π/2 is a quarter of the circle. So, all four points must lie within a quarter-circle? That seems too strict because if you randomly place four points on a circle, the chance that all four lie within any particular quarter-circle is quite low.But maybe I'm misunderstanding something. Let me think again. The condition is that no three points form an obtuse triangle with the center. So, for each triangle, the arc between any two points must be less than or equal to π/2. But does that mean that for every pair of points, the arc between them is less than or equal to π/2? Or is it sufficient that for each triangle, at least one of the arcs is less than or equal to π/2?Wait, no, the triangle is obtuse if the angle at the center is obtuse, which happens when the arc between the two points is greater than π/2. So, for each triangle, if the arc between any two points is greater than π/2, then that triangle is obtuse. Therefore, to have no obtuse triangles, all arcs between any two points must be less than or equal to π/2.But if all arcs between any two points are less than or equal to π/2, that would mean that all four points lie within a semicircle of π/2? Wait, no, a semicircle is π radians, so π/2 is a quarter-circle. So, all four points must lie within a quarter-circle? That seems too strict because the probability of four random points lying within a quarter-circle is (1/4)^3 = 1/64, which is quite low. But I'm not sure if that's the correct approach.Wait, maybe I'm overcomplicating it. Let me think about the problem differently. Instead of considering all four points, maybe I can fix one point and then consider the positions of the other three points relative to it.Let's fix one point, say A, at angle 0 on the circle. Then, the other three points, B, C, and D, are randomly distributed around the circle. For no three points to form an obtuse triangle with the center, the arcs between A and B, A and C, A and D, as well as between B and C, B and D, and C and D must all be less than or equal to π/2.But wait, if I fix point A at 0, then points B, C, and D must all lie within the arc from 0 to π/2. Because if any of them lie beyond π/2 from A, then the arc between A and that point would be greater than π/2, making the triangle obtuse. So, all three points B, C, and D must lie within the arc from 0 to π/2.But then, what about the arcs between B, C, and D? If all three are within π/2 of each other, then the maximum arc between any two of them would also be less than or equal to π/2. So, that seems to satisfy the condition.Therefore, the probability that all four points lie within a π/2 arc is the same as the probability that all three points B, C, and D lie within a π/2 arc starting at A. Since the circle is continuous and we can fix A anywhere, the probability is the same regardless of where A is fixed.The probability that n points lie within a particular arc of length θ on a circle is n*(θ/(2π))^{n-1}. Wait, is that correct? Let me recall. For n points on a circle, the probability that all lie within some arc of length θ is n*(θ/(2π))^{n-1}, assuming θ ≤ 2π.But in our case, θ is π/2, and n is 3 (since we fixed one point). So, the probability would be 3*(π/2 / 2π)^{3-1} = 3*(1/4)^2 = 3/16.But wait, is that the correct formula? I think the formula is actually n*(θ/(2π))^{n-1} when θ ≤ 2π. So, for θ = π/2, n = 3, it would be 3*(1/4)^2 = 3/16.But I'm not entirely sure if this formula applies here. Let me think about it differently. The probability that all three points lie within a particular arc of length π/2 is (π/2 / 2π)^3 = (1/4)^3 = 1/64. But since the arc can start anywhere, we have to multiply by the number of possible positions, which is 4 (since the circle is divided into four equal arcs of π/2 each). Wait, no, that's not quite right because the arc can start anywhere, not just at multiples of π/2.Actually, the correct way to calculate the probability that all n points lie within some arc of length θ is n*(θ/(2π))^{n-1}, as long as θ ≤ 2π. So, for θ = π/2 and n = 3, it's 3*(1/4)^2 = 3/16.But wait, I think I might be mixing up the formula. Let me look it up in my mind. The probability that all n points lie within any arc of length θ is n*(θ/(2π))^{n-1} when θ ≤ 2π. So, yes, for θ = π/2 and n = 3, it's 3*(1/4)^2 = 3/16.But in our case, we fixed one point, so we're only considering the other three points. So, the probability that all three lie within a π/2 arc starting at the fixed point is (π/2 / 2π)^3 = (1/4)^3 = 1/64. But since the arc can start anywhere, we have to multiply by the number of possible starting positions, which is 4 (since the circle is 2π, and π/2 is a quarter of that). So, 4*(1/64) = 1/16.Wait, that doesn't seem right. Because if we fix one point, the probability that the other three lie within a π/2 arc starting at that point is (1/4)^3 = 1/64. But the probability that all three lie within any π/2 arc is higher because the arc can be anywhere. So, the correct formula should account for the fact that the arc can be anywhere on the circle.I think the correct formula is n*(θ/(2π))^{n-1} for the probability that all n points lie within some arc of length θ. So, for n = 3 and θ = π/2, it's 3*(1/4)^2 = 3/16.But wait, in our case, we fixed one point, so n = 3. So, the probability that all three lie within a π/2 arc is 3*(1/4)^2 = 3/16.But I'm still a bit confused because I thought fixing a point would change the calculation. Maybe I need to approach it differently.Let me consider the circle as a circumference of 1 instead of 2π to simplify calculations. So, the total circumference is 1, and π/2 corresponds to 1/4 of the circumference.If we fix one point at position 0, the other three points are uniformly distributed between 0 and 1. The probability that all three points lie within an interval of length 1/4 is the same as the probability that the maximum of the three points minus the minimum is less than or equal to 1/4.But since we fixed one point at 0, the other three points must lie within [0, 1/4] to satisfy the condition. The probability that all three points lie within [0, 1/4] is (1/4)^3 = 1/64.But this is only for the specific interval starting at 0. However, the arc can start anywhere on the circle, so we have to consider all possible positions of the interval.Wait, no. Since we fixed one point at 0, the interval must start at 0 to include all three points. Otherwise, if the interval starts elsewhere, it might not include the fixed point at 0. So, actually, the interval must start at 0, and the other three points must lie within [0, 1/4].Therefore, the probability is (1/4)^3 = 1/64.But that seems too low. Maybe I'm missing something.Alternatively, perhaps I should consider the circle without fixing any points. The probability that all four points lie within some arc of length π/2 is 4*(π/2 / 2π)^3 = 4*(1/4)^3 = 4/64 = 1/16.Wait, that seems more reasonable. So, the probability that all four points lie within a π/2 arc is 1/16.But wait, I thought earlier that the formula was n*(θ/(2π))^{n-1}, which for n=4 and θ=π/2 would be 4*(1/4)^3 = 1/16. So, that matches.Therefore, the probability that all four points lie within a π/2 arc is 1/16.But does that mean that the probability that no three points form an obtuse triangle with the center is 1/16?Wait, no, because the condition is that no three points form an obtuse triangle, which is equivalent to all arcs between any two points being less than or equal to π/2. But if all four points lie within a π/2 arc, then the maximum arc between any two points is less than or equal to π/2, which satisfies the condition.Therefore, the probability that no three points form an obtuse triangle with the center is equal to the probability that all four points lie within a π/2 arc, which is 1/16.But wait, I'm not sure if that's the case. Let me think again.If all four points lie within a π/2 arc, then indeed, the maximum arc between any two points is less than or equal to π/2, so no triangle formed by any three points and the center will be obtuse.However, is the converse true? If no three points form an obtuse triangle, does that necessarily mean that all four points lie within a π/2 arc?Suppose that the four points do not all lie within a π/2 arc. Then, there exists at least one pair of points with an arc greater than π/2 between them. Therefore, the triangle formed by those two points and the center would be obtuse. Hence, the condition that no three points form an obtuse triangle is equivalent to all four points lying within a π/2 arc.Therefore, the probability we're seeking is indeed the probability that all four points lie within a π/2 arc.So, using the formula for the probability that n points lie within some arc of length θ on a circle, which is n*(θ/(2π))^{n-1} for θ ≤ 2π.In our case, n=4 and θ=π/2, so the probability is 4*(π/2 / 2π)^3 = 4*(1/4)^3 = 4/64 = 1/16.Therefore, the probability that no three of the four points form an obtuse triangle with the center is 1/16.But wait, I'm a bit unsure because I remember that for n points on a circle, the probability that all lie within some semicircle is n/(2^{n-1}}). So, for n=4, it's 4/8 = 1/2. But in our case, θ is π/2, which is a quarter-circle, not a semicircle.So, the formula for the probability that all n points lie within an arc of length θ is n*(θ/(2π))^{n-1} when θ ≤ 2π.Therefore, for θ=π/2 and n=4, it's 4*(1/4)^3 = 1/16.Yes, that seems correct.But let me double-check by considering the problem differently. Suppose we fix one point, say A, at angle 0. Then, the other three points, B, C, and D, must lie within the arc from 0 to π/2. The probability that each of these points lies within this arc is 1/4, so the probability that all three lie within this arc is (1/4)^3 = 1/64.However, the arc can start anywhere on the circle, not just at 0. So, we have to consider all possible starting positions of the arc. Since the circle is continuous, the probability is actually the expected value over all possible positions.But wait, since we fixed point A at 0, the arc must start at 0 to include A. Therefore, the probability is indeed 1/64.But that contradicts the earlier result of 1/16. So, which one is correct?I think the confusion arises from whether we fix a point or not. If we fix a point, the probability that the other three lie within a π/2 arc starting at that point is (1/4)^3 = 1/64. However, if we don't fix any point, the probability that all four points lie within some π/2 arc is higher because the arc can be anywhere.The formula n*(θ/(2π))^{n-1} accounts for the fact that the arc can be anywhere on the circle. So, for n=4 and θ=π/2, it's 4*(1/4)^3 = 1/16.Therefore, the correct probability is 1/16.But wait, let me think about it another way. Suppose we don't fix any point. The probability that all four points lie within a π/2 arc is the same as the expected value of the minimum arc covering all four points being less than or equal to π/2.To calculate this, we can use the formula for the probability that the maximum arc between consecutive points is less than or equal to θ. But I'm not sure about that.Alternatively, we can use the inclusion-exclusion principle. The probability that all four points lie within some π/2 arc is equal to the sum over all possible arcs of length π/2 of the probability that all four points lie within that arc, minus the overlaps, and so on.But that seems complicated. Instead, the formula n*(θ/(2π))^{n-1} is a known result for this kind of problem when θ ≤ 2π.So, for n=4 and θ=π/2, the probability is 4*(1/4)^3 = 1/16.Therefore, I think the correct answer is 1/16.But wait, I'm still a bit unsure because when I fixed one point, the probability was 1/64, but when considering all possible positions, it's 1/16. So, which one is it?I think the key is that when we fix one point, we're effectively reducing the problem to three points, and the probability that all three lie within a π/2 arc starting at the fixed point is 1/64. However, since the arc can start anywhere, we have to consider all possible positions, which is why the formula gives 1/16.Therefore, the correct probability is 1/16.But wait, let me check with n=3. If we have three points, the probability that all three lie within a π/2 arc is 3*(1/4)^2 = 3/16. If we fix one point, the probability that the other two lie within a π/2 arc starting at that point is (1/4)^2 = 1/16. But since the arc can start anywhere, the probability is 3/16.So, for n=4, it's 4*(1/4)^3 = 1/16.Yes, that seems consistent.Therefore, the probability that no three of the four points form an obtuse triangle with the center is 1/16.

💡To evaluate the expression (sqrt{7 + 4sqrt{3}} - sqrt{7 - 4sqrt{3}}), we can follow these steps:1. **Let the expression be equal to (x):** [ x = sqrt{7 + 4sqrt{3}} - sqrt{7 - 4sqrt{3}} ]2. **Square both sides to eliminate the square roots:** [ x^2 = left(sqrt{7 + 4sqrt{3}} - sqrt{7 - 4sqrt{3}}right)^2 ] Expanding the right-hand side: [ x^2 = (7 + 4sqrt{3}) + (7 - 4sqrt{3}) - 2sqrt{(7 + 4sqrt{3})(7 - 4sqrt{3})} ] Simplify the terms: [ x^2 = 14 - 2sqrt{49 - 48} = 14 - 2sqrt{1} = 14 - 2 = 12 ]3. **Take the square root of both sides to solve for (x):** [ x = sqrt{12} = 2sqrt{3} ]Thus, the value of the expression is (2sqrt{3}).

💡First, I need to determine the speed of the stream based on the information provided about Sushi's upstream and downstream journeys.I'll start by defining the variables:- Let ( V ) represent Sushi's rowing speed in still water (in km/h).- Let ( S ) represent the speed of the stream (in km/h).When Sushi is rowing upstream, her effective speed is reduced by the stream's speed, so her upstream speed is ( V - S ). Conversely, when she is rowing downstream, her effective speed is increased by the stream's speed, making her downstream speed ( V + S ).Given that Sushi rows 27 km upstream and 81 km downstream, each taking 9 hours, I can set up two equations using the formula ( text{Distance} = text{Speed} times text{Time} ).For the upstream journey:[27 = (V - S) times 9]Simplifying this equation:[3 = V - S quad text{(1)}]For the downstream journey:[81 = (V + S) times 9]Simplifying this equation:[9 = V + S quad text{(2)}]Now, I have a system of two equations:[begin{cases}3 = V - S 9 = V + Send{cases}]To solve for ( V ) and ( S ), I'll add the two equations together to eliminate ( S ):[3 + 9 = (V - S) + (V + S)][12 = 2V][V = 6 , text{km/h}]With ( V ) known, I can substitute it back into equation (1) to find ( S ):[3 = 6 - S][S = 6 - 3][S = 3 , text{km/h}]Therefore, the speed of the stream is 3 km/h.