💡First, I need to find the distance between point A with coordinates (1, 1, 2) and point B with coordinates (2, 3, 4) in a three-dimensional Cartesian coordinate system.I recall that the distance formula for two points in space is given by:[|AB| = sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}]Next, I'll identify the coordinates for both points:- Point A: ( (x_1, y_1, z_1) = (1, 1, 2) )- Point B: ( (x_2, y_2, z_2) = (2, 3, 4) )Now, I'll calculate the differences in each coordinate:- ( x_2 - x_1 = 2 - 1 = 1 )- ( y_2 - y_1 = 3 - 1 = 2 )- ( z_2 - z_1 = 4 - 2 = 2 )Then, I'll square each of these differences:- ( (1)^2 = 1 )- ( (2)^2 = 4 )- ( (2)^2 = 4 )Adding these squared differences together:[1 + 4 + 4 = 9]Finally, I'll take the square root of the sum to find the distance:[|AB| = sqrt{9} = 3]So, the length of segment AB is 3 units.

💡Okay, so I have this problem about a sequence called a "difference decreasing" sequence. Let me try to understand what that means first. The problem says that a sequence {a_n} is called "difference decreasing" if the differences between consecutive terms are decreasing. So, that means a_2 - a_1 > a_3 - a_2 > a_4 - a_3 > ... and so on. So each subsequent difference is smaller than the previous one. Got it.Now, the sequence {a_n} is given to be a "difference decreasing" sequence. Additionally, we have a relationship between the general term a_n and the sum of the first n terms, S_n. The equation given is 2S_n = 3a_n + 2λ - 1 for all natural numbers n. We need to find the range of the real number λ.Alright, let's break this down step by step.First, let's recall that S_n is the sum of the first n terms of the sequence. So, S_n = a_1 + a_2 + ... + a_n.Given the equation 2S_n = 3a_n + 2λ - 1, I can try to express this in terms of a_n and S_n.Let me write that equation again:2S_n = 3a_n + 2λ - 1.I can rearrange this to solve for S_n:S_n = (3a_n + 2λ - 1)/2.Hmm, that might be useful later.Now, let's consider the case when n = 1. For n = 1, S_1 is just a_1. So plugging n = 1 into the equation:2S_1 = 3a_1 + 2λ - 1.But S_1 = a_1, so:2a_1 = 3a_1 + 2λ - 1.Let me solve for a_1:2a_1 - 3a_1 = 2λ - 1- a_1 = 2λ - 1So, a_1 = 1 - 2λ.Alright, so a_1 is expressed in terms of λ. Good.Now, let's consider n ≥ 2. For n ≥ 2, we can write the equation for S_n and S_{n-1}:2S_n = 3a_n + 2λ - 1.Similarly, for n-1:2S_{n-1} = 3a_{n-1} + 2λ - 1.Now, subtracting these two equations:2S_n - 2S_{n-1} = 3a_n - 3a_{n-1}.But S_n - S_{n-1} is just a_n, because S_n is the sum up to n, and S_{n-1} is the sum up to n-1. So, S_n - S_{n-1} = a_n.Therefore, 2a_n = 3a_n - 3a_{n-1}.Simplify this:2a_n = 3a_n - 3a_{n-1}Subtract 2a_n from both sides:0 = a_n - 3a_{n-1}So, a_n = 3a_{n-1}.Interesting, so the sequence {a_n} is a geometric sequence with common ratio 3. That is, each term is three times the previous term.So, starting from a_1, which is 1 - 2λ, the sequence is:a_1 = 1 - 2λa_2 = 3a_1 = 3(1 - 2λ)a_3 = 3a_2 = 9(1 - 2λ)a_4 = 3a_3 = 27(1 - 2λ)And so on.So, in general, a_n = 3^{n-1}(1 - 2λ).Alright, so now we can compute the differences between consecutive terms:a_2 - a_1 = 3(1 - 2λ) - (1 - 2λ) = 2(1 - 2λ)a_3 - a_2 = 9(1 - 2λ) - 3(1 - 2λ) = 6(1 - 2λ)a_4 - a_3 = 27(1 - 2λ) - 9(1 - 2λ) = 18(1 - 2λ)And so on.So, the differences are:2(1 - 2λ), 6(1 - 2λ), 18(1 - 2λ), 54(1 - 2λ), ...Hmm, each difference is 3 times the previous difference. So, the differences themselves form a geometric sequence with common ratio 3.But wait, the problem states that the sequence is "difference decreasing," which means that each difference should be smaller than the previous one. So, the differences should be decreasing.But in our case, the differences are 2(1 - 2λ), 6(1 - 2λ), 18(1 - 2λ), etc., which are increasing by a factor of 3 each time. So, unless the common ratio is less than 1, the differences would be increasing, not decreasing.Wait, that seems contradictory. Because if the differences are increasing, then it's not a "difference decreasing" sequence. So, perhaps I made a mistake in my reasoning.Wait, let me double-check.We have a_n = 3a_{n-1}, so the sequence is geometric with ratio 3. Therefore, the differences a_{n+1} - a_n = 3a_n - a_n = 2a_n = 2*3^{n-1}(1 - 2λ).So, the differences are 2(1 - 2λ), 6(1 - 2λ), 18(1 - 2λ), etc., each time multiplied by 3. So, the differences are increasing, not decreasing.But the problem says it's a "difference decreasing" sequence, so the differences should be decreasing. Therefore, there must be something wrong here.Wait, perhaps I misapplied the condition. Let me go back.We have a_n = 3a_{n-1}, so the sequence is increasing if a_1 is positive, or decreasing if a_1 is negative.But regardless, the differences a_{n+1} - a_n = 2a_n, which is 2*3^{n-1}(1 - 2λ). So, unless 1 - 2λ is negative, the differences will be increasing in magnitude.Wait, so if 1 - 2λ is negative, then the differences would be negative, but their magnitudes would still be increasing because each difference is 3 times the previous one.But in the problem, it's stated that the differences are decreasing. So, perhaps the differences are decreasing in magnitude, but could they be negative?Wait, the problem says "difference decreasing" sequence, which is defined as a_2 - a_1 > a_3 - a_2 > a_4 - a_3 > ... So, the differences themselves are decreasing, but they don't specify whether they are positive or negative. So, if the differences are negative, they could be decreasing in the sense that each subsequent difference is less than the previous one, but in terms of magnitude, they might be increasing.Wait, let's think about this.Suppose the differences are negative and each subsequent difference is more negative, meaning each difference is less than the previous one. So, for example, if a_2 - a_1 = -2, a_3 - a_2 = -6, a_4 - a_3 = -18, etc., then each difference is less than the previous one because -6 < -2, -18 < -6, etc. So, in this case, the differences are decreasing, but their magnitudes are increasing.So, perhaps in this problem, the differences are negative and becoming more negative, hence decreasing.But in our case, the differences are 2(1 - 2λ), 6(1 - 2λ), 18(1 - 2λ), etc. So, if 1 - 2λ is positive, then the differences are positive and increasing, which would not satisfy the "difference decreasing" condition because they are increasing.If 1 - 2λ is negative, then the differences are negative and each subsequent difference is 3 times the previous one, which would make them more negative, hence decreasing.So, for the differences to be decreasing, we need 1 - 2λ < 0, which implies that λ > 1/2.Wait, let me verify that.If 1 - 2λ < 0, then 1 < 2λ, so λ > 1/2.So, in that case, the differences are negative and each subsequent difference is 3 times the previous one, making them more negative, hence decreasing.Therefore, the condition for the sequence to be "difference decreasing" is λ > 1/2.But let me make sure that this is the only condition.We also need to ensure that the differences are strictly decreasing, so each difference must be less than the previous one.Given that the differences are 2(1 - 2λ), 6(1 - 2λ), 18(1 - 2λ), etc., and since 1 - 2λ is negative, each subsequent difference is 3 times the previous one, which is more negative, hence each difference is less than the previous one.So, as long as 1 - 2λ < 0, the differences will be decreasing.Therefore, the range of λ is λ > 1/2.Wait, but let me check if there are any other constraints.We have a_n = 3^{n-1}(1 - 2λ). If λ > 1/2, then 1 - 2λ is negative, so a_n is negative for all n ≥ 1.But the problem doesn't specify any constraints on the terms of the sequence, just on the differences.So, as long as the differences are decreasing, which happens when λ > 1/2, the sequence is "difference decreasing."Therefore, the range of λ is λ > 1/2.I think that's the answer.

💡Okay, so I have this problem here about a hyperbola and a parabola. Let me try to parse it step by step. The hyperbola is given by the equation x²/a² - y²/b² = 1, where a and b are positive. The parabola is given by y² = 2px, with p also positive. The problem states two main things: first, the distance between the left vertex of the hyperbola and the focus of the parabola is 4. Second, the coordinate of the intersection point of one asymptote of the hyperbola and the axis of the parabola is (-2, -4). I need to find the equation of the hyperbola.Alright, let's start by recalling some properties of hyperbolas and parabolas. For the hyperbola x²/a² - y²/b² = 1, the vertices are at (±a, 0). So, the left vertex is at (-a, 0). The asymptotes of this hyperbola are the lines y = ±(b/a)x. For the parabola y² = 2px, the standard form is y² = 4ax, but in this case, it's y² = 2px, so comparing, 4a = 2p, which means a = p/2. The focus of this parabola is at (a, 0) which would be (p/2, 0). The axis of the parabola is the x-axis because it's a right-opening parabola. Wait, actually, the axis of the parabola y² = 2px is the x-axis. So, the axis is the line y = 0.But wait, the problem mentions the intersection point of one asymptote of the hyperbola and the axis of the parabola is (-2, -4). Hmm, the axis of the parabola is the x-axis, which is y = 0. So, how can the intersection point be (-2, -4)? That point isn't on the x-axis. Wait, maybe I'm misunderstanding something.Wait, perhaps the axis of the parabola is not the x-axis? Let me think again. The parabola y² = 2px is symmetric about the x-axis, so its axis is indeed the x-axis. Therefore, the intersection point of an asymptote and the axis should lie on the x-axis, but the given point is (-2, -4), which is not on the x-axis. That seems contradictory.Wait, hold on, maybe I misread the problem. It says the coordinate of the intersection point of one asymptote of the hyperbola and the axis of the parabola is (-2, -4). So, the asymptote intersects the axis of the parabola at (-2, -4). But the axis of the parabola is the x-axis, which is y = 0. So, if the asymptote intersects the x-axis at (-2, -4), but (-2, -4) is not on the x-axis. That doesn't make sense. There must be a mistake in my understanding.Wait, maybe the axis of the parabola is not the x-axis. Let me recall: for a parabola y² = 4ax, it's symmetric about the x-axis, opening to the right. Similarly, y² = 2px is symmetric about the x-axis, opening to the right. So, the axis is indeed the x-axis.Therefore, the intersection point of an asymptote and the axis of the parabola should lie on the x-axis. But the given point is (-2, -4), which is not on the x-axis. That seems impossible. Maybe the problem meant the directrix of the parabola? Or perhaps it's a misstatement.Wait, let me reread the problem: "the coordinate of the intersection point of one asymptote of the hyperbola and the axis of the parabola is (-2,-4)". So, it's the axis of the parabola, not the directrix. Hmm.Wait, unless the parabola is not y² = 2px but something else. Wait, no, the problem says y² = 2px. So, axis is the x-axis.Wait, maybe I made a mistake in thinking the axis is the x-axis. Let me confirm: for a parabola y² = 4ax, the axis is the x-axis. Similarly, y² = 2px is similar, just scaled, so the axis is still the x-axis.Therefore, the intersection point should be on the x-axis, but it's given as (-2, -4). That seems contradictory. Maybe the problem meant the directrix? The directrix of the parabola y² = 2px is x = -p/2. So, if the asymptote intersects the directrix at (-2, -4), that would make sense because the directrix is a vertical line.Wait, let me check: the directrix of y² = 2px is x = -p/2. So, if the asymptote intersects the directrix at (-2, -4), then x = -p/2 = -2, so p = 4. Then, the focus of the parabola is at (p/2, 0) = (2, 0). Wait, that might make sense. Maybe the problem meant the directrix instead of the axis? Or perhaps it's a translation issue. Let me think again.Alternatively, maybe the axis of the parabola is considered differently. Wait, the axis of the parabola is the line of symmetry. For y² = 2px, it's the x-axis. So, the intersection point of an asymptote and the axis is on the x-axis. But (-2, -4) is not on the x-axis. Therefore, perhaps the problem is misstated, or I'm misinterpreting it.Alternatively, maybe the axis of the parabola is not the x-axis but the line x = -2? Wait, no, the axis is the line of symmetry, which for y² = 2px is the x-axis.Wait, perhaps the problem is referring to the axis of the parabola as the line x = -2? No, that doesn't make sense. The axis is the x-axis.Wait, maybe the problem is referring to the axis of the hyperbola? No, it says the axis of the parabola.Hmm, this is confusing. Let me try to proceed with the information given, even if it seems contradictory.So, the intersection point is (-2, -4). This point lies on both an asymptote of the hyperbola and the axis of the parabola. But the axis of the parabola is the x-axis, which is y = 0. So, (-2, -4) is not on y = 0. Therefore, perhaps the problem is referring to the directrix of the parabola?If that's the case, then the directrix is x = -p/2. So, if the asymptote intersects the directrix at (-2, -4), then x = -p/2 = -2, so p = 4. Then, the focus of the parabola is at (p/2, 0) = (2, 0). Then, the distance between the left vertex of the hyperbola and the focus of the parabola is 4. The left vertex of the hyperbola is at (-a, 0). The focus of the parabola is at (2, 0). So, the distance between (-a, 0) and (2, 0) is |2 - (-a)| = |2 + a| = 4. Since a > 0, 2 + a = 4, so a = 2.So, a = 2. Therefore, the hyperbola is x²/4 - y²/b² = 1.Now, the asymptotes of the hyperbola are y = ±(b/a)x = ±(b/2)x. We know that one of these asymptotes passes through the point (-2, -4). Let's check which one it is.If we take the asymptote y = (b/2)x, plugging in (-2, -4): -4 = (b/2)(-2) => -4 = -b => b = 4.Alternatively, if we take the asymptote y = -(b/2)x, plugging in (-2, -4): -4 = -(b/2)(-2) => -4 = b => b = -4. But since b > 0, we discard this.Therefore, b = 4.So, the equation of the hyperbola is x²/4 - y²/16 = 1.Wait, but let me double-check. If the asymptote is y = (b/2)x, and it passes through (-2, -4), then:-4 = (b/2)(-2) => -4 = -b => b = 4. Correct.So, the hyperbola is x²/4 - y²/16 = 1.But wait, earlier I assumed that the intersection point was with the directrix, not the axis. Since the problem says the axis, but the point isn't on the axis, perhaps I made a wrong assumption.Alternatively, maybe the problem is correct, and I need to interpret it differently. Let me try again.The axis of the parabola y² = 2px is the x-axis. The asymptote of the hyperbola intersects the x-axis at (-2, -4). But (-2, -4) is not on the x-axis. Therefore, this seems impossible. Maybe the problem meant the directrix? Or perhaps the axis is not the x-axis?Wait, perhaps the parabola is written differently. If the parabola is y² = 2px, it's symmetric about the x-axis. But if it's written as x² = 2py, it would be symmetric about the y-axis. But the problem says y² = 2px, so it's symmetric about the x-axis.Wait, maybe the problem is referring to the axis of the parabola as the line x = -2? No, that doesn't make sense. The axis is the line of symmetry, which for y² = 2px is the x-axis.I'm stuck here. Let me try to proceed with the assumption that the intersection is with the directrix, as that makes the math work. So, if the asymptote intersects the directrix at (-2, -4), then p = 4, focus at (2, 0), distance from (-a, 0) to (2, 0) is 4, so a = 2. Then, asymptote passes through (-2, -4), so b = 4. Therefore, hyperbola is x²/4 - y²/16 = 1.Alternatively, maybe the problem is correct, and the intersection is with the axis, but the point is (-2, 0). Wait, but it's given as (-2, -4). Hmm.Wait, perhaps the axis of the parabola is not the x-axis but another line. Wait, no, for y² = 2px, the axis is the x-axis. Unless the parabola is rotated, but the equation doesn't indicate that.Wait, maybe the problem is referring to the axis of the hyperbola? No, it says the axis of the parabola.Alternatively, maybe the point (-2, -4) is the intersection of the asymptote with the axis of the parabola, but the axis is not the x-axis. Wait, that can't be, because for y² = 2px, the axis is the x-axis.I'm confused. Maybe I should proceed with the initial assumption that the intersection is with the directrix, even though the problem says axis. Because otherwise, the point (-2, -4) doesn't lie on the axis.So, assuming that, then p = 4, focus at (2, 0), distance from (-a, 0) to (2, 0) is 4, so a = 2. Then, asymptote passes through (-2, -4), so b = 4. Therefore, hyperbola is x²/4 - y²/16 = 1.But let me check if this makes sense. The asymptotes are y = ±(b/a)x = ±(4/2)x = ±2x. So, the asymptotes are y = 2x and y = -2x.Now, does the asymptote y = 2x pass through (-2, -4)? Let's see: y = 2x, plugging in x = -2, y = -4. Yes, it does. Similarly, y = -2x, plugging in x = -2, y = 4. So, the point (-2, -4) lies on the asymptote y = 2x.But wait, the problem says the intersection point is (-2, -4). So, if the asymptote is y = 2x, and the axis of the parabola is the x-axis, then the intersection point should be where y = 2x meets y = 0, which is at (0, 0). But the given point is (-2, -4), which is not on the x-axis.Therefore, my initial assumption must be wrong. Maybe the intersection is not with the axis, but with something else. Wait, the problem says the intersection point of one asymptote of the hyperbola and the axis of the parabola is (-2, -4). So, the asymptote intersects the axis at (-2, -4). But the axis is the x-axis, so (-2, -4) should be on the x-axis, which it's not. Therefore, this is impossible.Wait, unless the axis of the parabola is not the x-axis but another line. Wait, but for y² = 2px, the axis is the x-axis. So, unless the parabola is shifted, but the equation is y² = 2px, which is not shifted.Wait, maybe the problem is referring to the axis of the hyperbola? No, it says the axis of the parabola.I'm stuck. Maybe I should try a different approach. Let's denote the focus of the parabola as F. For y² = 2px, the focus is at (p/2, 0). The left vertex of the hyperbola is at (-a, 0). The distance between (-a, 0) and (p/2, 0) is |p/2 - (-a)| = |p/2 + a| = 4. So, p/2 + a = 4, since both p and a are positive.Now, the asymptote of the hyperbola intersects the axis of the parabola at (-2, -4). The axis of the parabola is the x-axis, so the intersection point should be on the x-axis. But (-2, -4) is not on the x-axis. Therefore, this seems impossible. Maybe the problem is referring to the directrix?If so, the directrix of the parabola y² = 2px is x = -p/2. So, the asymptote intersects the directrix at (-2, -4). Therefore, x = -p/2 = -2, so p = 4. Then, the focus is at (2, 0). The distance between (-a, 0) and (2, 0) is |2 - (-a)| = 2 + a = 4, so a = 2.Now, the asymptote passes through (-2, -4). The asymptotes are y = ±(b/a)x = ±(b/2)x. Let's plug in (-2, -4) into y = (b/2)x: -4 = (b/2)(-2) => -4 = -b => b = 4. Similarly, for y = -(b/2)x: -4 = -(b/2)(-2) => -4 = b => b = -4, which is invalid since b > 0.Therefore, b = 4. So, the hyperbola is x²/4 - y²/16 = 1.But wait, earlier I thought the intersection point was with the directrix, not the axis. Since the problem says the axis, but the point isn't on the axis, perhaps the problem is misstated. Alternatively, maybe I'm misinterpreting the axis.Wait, maybe the axis of the parabola is not the x-axis but another line. Wait, no, for y² = 2px, the axis is the x-axis. So, the intersection point must be on the x-axis. But it's given as (-2, -4), which is not on the x-axis. Therefore, this is impossible.Wait, unless the parabola is written differently. If it's x² = 2py, then the axis would be the y-axis, and the focus would be at (0, p/2). But the problem says y² = 2px, so it's symmetric about the x-axis.I'm stuck. Maybe I should proceed with the assumption that the intersection is with the directrix, even though the problem says axis. Because otherwise, the problem is impossible.So, assuming that, then p = 4, focus at (2, 0), a = 2, b = 4, hyperbola is x²/4 - y²/16 = 1.Alternatively, maybe the problem is correct, and the intersection is with the axis, but the point is (-2, 0). Wait, but it's given as (-2, -4). Hmm.Wait, maybe the axis of the parabola is not the x-axis but another line. Wait, no, for y² = 2px, the axis is the x-axis. Unless the parabola is rotated, but the equation doesn't indicate that.Wait, perhaps the problem is referring to the axis of the hyperbola? No, it says the axis of the parabola.I think I have to proceed with the assumption that the intersection is with the directrix, even though the problem says axis. Because otherwise, the problem is impossible.Therefore, the hyperbola is x²/4 - y²/16 = 1.But let me check again. If the asymptote intersects the directrix at (-2, -4), then p = 4, focus at (2, 0), a = 2, b = 4. Asymptotes are y = ±2x. The point (-2, -4) lies on y = 2x, which is correct. So, the hyperbola is x²/4 - y²/16 = 1.Yes, that seems consistent.

💡Alright, so I have this problem about a circle in the Cartesian coordinate system. Let me try to break it down step by step.First, the circle is given by the equation ( x^2 + y^2 = 4 ). That means it's centered at the origin (0,0) with a radius of 2. The problem mentions point A, which is where the circle intersects the negative semi-axis of x. Since the circle is centered at the origin and has a radius of 2, it should intersect the x-axis at (-2, 0) and (2, 0). But since we're talking about the negative semi-axis, point A must be (-2, 0).Now, part 1 says that lines AM and AN pass through point A and intersect the circle at points M and N respectively. The slopes of these lines are given: AM has a slope of 2, and AN has a slope of -1/2. I need to find the area of triangle AMN.Okay, so let me visualize this. Point A is at (-2, 0). Line AM has a slope of 2, so its equation should be ( y = 2(x + 2) ) because it passes through (-2, 0). Similarly, line AN has a slope of -1/2, so its equation is ( y = -frac{1}{2}(x + 2) ).Now, I need to find the points M and N where these lines intersect the circle again. So, I can solve these equations with the circle equation.Starting with line AM: ( y = 2x + 4 ). Substitute this into the circle equation:( x^2 + (2x + 4)^2 = 4 )Expanding that:( x^2 + 4x^2 + 16x + 16 = 4 )Combine like terms:( 5x^2 + 16x + 12 = 0 )Let me solve this quadratic equation. The discriminant is ( 16^2 - 4*5*12 = 256 - 240 = 16 ). So, the roots are:( x = frac{-16 pm sqrt{16}}{2*5} = frac{-16 pm 4}{10} )So, ( x = frac{-16 + 4}{10} = frac{-12}{10} = -1.2 ) and ( x = frac{-16 - 4}{10} = frac{-20}{10} = -2 ).We already know that x = -2 is point A, so the other intersection point M is at x = -1.2. Plugging back into the equation of AM:( y = 2*(-1.2) + 4 = -2.4 + 4 = 1.6 ). So, point M is (-1.2, 1.6).Similarly, let's find point N using line AN: ( y = -frac{1}{2}x - 1 ). Substitute into the circle equation:( x^2 + left(-frac{1}{2}x - 1right)^2 = 4 )Expanding that:( x^2 + frac{1}{4}x^2 + x + 1 = 4 )Combine like terms:( frac{5}{4}x^2 + x - 3 = 0 )Multiply through by 4 to eliminate the fraction:( 5x^2 + 4x - 12 = 0 )Quadratic formula: discriminant is ( 16 + 240 = 256 ). So,( x = frac{-4 pm sqrt{256}}{10} = frac{-4 pm 16}{10} )So, x = (12)/10 = 1.2 or x = (-20)/10 = -2. Again, x = -2 is point A, so point N is at x = 1.2. Plugging back into AN:( y = -frac{1}{2}(1.2) - 1 = -0.6 - 1 = -1.6 ). So, point N is (1.2, -1.6).Now, I have points A(-2, 0), M(-1.2, 1.6), and N(1.2, -1.6). To find the area of triangle AMN, I can use the shoelace formula.First, list the coordinates:A(-2, 0), M(-1.2, 1.6), N(1.2, -1.6)Shoelace formula:Area = 1/2 | (x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)) |Plugging in:1/2 | (-2)(1.6 - (-1.6)) + (-1.2)(-1.6 - 0) + 1.2(0 - 1.6) |Calculate each term:First term: (-2)(3.2) = -6.4Second term: (-1.2)(-1.6) = 1.92Third term: 1.2(-1.6) = -1.92Sum: -6.4 + 1.92 - 1.92 = -6.4Take absolute value and multiply by 1/2: (1/2)*6.4 = 3.2Wait, that's 16/5, which is 3.2. So, the area is 16/5.Hmm, seems straightforward. Let me double-check.Alternatively, since AM and AN have slopes that are negative reciprocals (2 and -1/2), they are perpendicular. Therefore, triangle AMN is a right triangle with legs AM and AN.So, if I can find the lengths of AM and AN, then the area is 1/2 * AM * AN.Let me compute the distance from A to M.Point A(-2, 0) to M(-1.2, 1.6):Distance = sqrt[(-1.2 + 2)^2 + (1.6 - 0)^2] = sqrt[(0.8)^2 + (1.6)^2] = sqrt[0.64 + 2.56] = sqrt[3.2] = sqrt(16/5) = 4/sqrt(5)Similarly, distance from A(-2, 0) to N(1.2, -1.6):sqrt[(1.2 + 2)^2 + (-1.6 - 0)^2] = sqrt[(3.2)^2 + (-1.6)^2] = sqrt[10.24 + 2.56] = sqrt[12.8] = sqrt(64/5) = 8/sqrt(5)So, lengths of AM and AN are 4/sqrt(5) and 8/sqrt(5) respectively.Area is 1/2 * (4/sqrt(5)) * (8/sqrt(5)) = 1/2 * 32/5 = 16/5. Yep, same result.So, part 1 is 16/5.Moving on to part 2. We have a point P(3√3, -5). From P, two tangents are drawn to the circle, touching the circle at E and F. We need to find the dot product of vectors PE and PF.First, I remember that the dot product of two vectors is |PE||PF|cos(theta), where theta is the angle between them.Since PE and PF are both tangents from P to the circle, their lengths are equal. The length of a tangent from a point to a circle is sqrt(OP^2 - r^2), where OP is the distance from the point to the center, and r is the radius.So, let's compute OP. Point P is (3√3, -5). The center O is (0,0). So,OP = sqrt[(3√3)^2 + (-5)^2] = sqrt[27 + 25] = sqrt[52] = 2√13.Radius r is 2, so the length of the tangent is sqrt[(2√13)^2 - 2^2] = sqrt[52 - 4] = sqrt[48] = 4√3.So, both PE and PF are 4√3 in length.Now, I need to find the angle between PE and PF. Alternatively, since vectors PE and PF have the same magnitude, the dot product is |PE|^2 * cos(theta).But how do I find theta?I recall that the angle between two tangents from a point outside the circle can be found using the formula:cos(theta) = (OP^2 - 2r^2)/OP^2Wait, actually, let me think. The angle between the tangents can be found using the cosine law in triangle OPE, where O is the center, P is the external point, and E is the point of tangency.In triangle OPE, OP is 2√13, OE is 2, and PE is 4√3. So, applying the cosine law:OE^2 = OP^2 + PE^2 - 2*OP*PE*cos(theta/2)Wait, actually, no. Let me clarify.Wait, in triangle OPE, sides are OP = 2√13, OE = 2, and PE = 4√3. So, angle at P is theta/2, because the two tangents make equal angles with the line OP.Wait, perhaps it's better to use the formula for the angle between two tangents:The angle theta between the two tangents satisfies:sin(theta/2) = r / OPSo, sin(theta/2) = 2 / (2√13) = 1/√13Therefore, theta/2 = arcsin(1/√13), so theta = 2 arcsin(1/√13)Alternatively, we can compute cos(theta) using double angle formula:cos(theta) = 1 - 2 sin^2(theta/2) = 1 - 2*(1/13) = 1 - 2/13 = 11/13So, cos(theta) = 11/13.Therefore, the dot product of vectors PE and PF is |PE||PF|cos(theta) = (4√3)(4√3)(11/13) = 16*3*(11/13) = 48*(11/13) = 528/13.So, the dot product is 528/13.Wait, let me verify. Alternatively, since the angle between PE and PF is theta, and both vectors have magnitude 4√3, their dot product is (4√3)^2 * cos(theta) = 48 * (11/13) = 528/13. Yep, same result.So, part 2 is 528/13.**Final Answer**1. The area of triangle ( AMN ) is (boxed{dfrac{16}{5}}).2. The dot product of vectors ( overrightarrow{PE} ) and ( overrightarrow{PF} ) is (boxed{dfrac{528}{13}}).

💡Alright, so I've got this problem here about a function f(x) that satisfies f(2 + x) = f(6 - x) for any real number x. Also, when x isn't equal to 4, the derivative f'(x) satisfies x f'(x) > 4 f'(x). And I need to figure out the relationship between f(2^{√a}), f(6), and f(log₃a) when 9 < a < 27. The options are A, B, C, D, each giving a different order of these function values.First, let me try to understand what the condition f(2 + x) = f(6 - x) means. It seems like a symmetry condition. If I let y = 2 + x, then the equation becomes f(y) = f(6 - (y - 2)) = f(8 - y). So, f(y) = f(8 - y). That tells me the function is symmetric about the line x = 4 because if you take any point y, its mirror image across x = 4 is 8 - y, and the function values at these points are equal. So, f is symmetric around x = 4.Next, the derivative condition: x f'(x) > 4 f'(x) for x ≠ 4. Let me rewrite this inequality:x f'(x) > 4 f'(x)Subtract 4 f'(x) from both sides:(x - 4) f'(x) > 0So, this tells me that the product of (x - 4) and f'(x) is positive. That means either both (x - 4) and f'(x) are positive, or both are negative.Case 1: If x > 4, then (x - 4) is positive. So, for the product to be positive, f'(x) must also be positive. Therefore, when x > 4, f'(x) > 0, which means f is increasing on (4, ∞).Case 2: If x < 4, then (x - 4) is negative. So, for the product to be positive, f'(x) must also be negative. Therefore, when x < 4, f'(x) < 0, which means f is decreasing on (-∞, 4).So, putting it all together, f is decreasing to the left of x = 4 and increasing to the right of x = 4. That makes sense with the symmetry about x = 4; it's like a V-shape with the vertex at x = 4.Now, I need to compare f(2^{√a}), f(6), and f(log₃a) when 9 < a < 27.First, let's figure out the ranges of 2^{√a} and log₃a when a is between 9 and 27.Starting with log₃a:Since 9 < a < 27, and 9 is 3², 27 is 3³. So, log₃9 = 2 and log₃27 = 3. Therefore, log₃a is between 2 and 3.So, log₃a ∈ (2, 3).Next, 2^{√a}:Since a is between 9 and 27, √a is between √9 = 3 and √27 ≈ 5.196. So, √a ∈ (3, 5.196).Therefore, 2^{√a} is between 2³ = 8 and 2^{5.196} ≈ 2^{5} * 2^{0.196} ≈ 32 * 1.14 ≈ 36.48.So, 2^{√a} ∈ (8, ~36.48).But wait, since a < 27, √a < √27 ≈ 5.196, so 2^{√a} < 2^{5.196} ≈ 36.48. So, 2^{√a} is between 8 and approximately 36.48.But let's see if we can get a better upper bound. Since a < 27, √a < √27 = 3√3 ≈ 5.196. So, 2^{√a} < 2^{5.196} ≈ 36.48, as before.So, 2^{√a} is between 8 and approximately 36.48.Now, let's note that 6 is in the middle of 4 and 8, but wait, 6 is actually less than 8. Wait, no, 6 is between 4 and 8.Wait, hold on. Let me think.We have f is decreasing on (-∞, 4) and increasing on (4, ∞). So, the function has a minimum at x = 4.Therefore, as we move away from x = 4 towards the right (x > 4), the function increases, and as we move away to the left (x < 4), the function decreases.So, to compare f(log₃a), f(6), and f(2^{√a}), I need to see where these arguments lie relative to 4.First, log₃a is between 2 and 3, so that's less than 4. So, log₃a ∈ (2, 3) < 4.6 is greater than 4, so 6 ∈ (4, ∞).2^{√a} is between 8 and ~36.48, so that's also greater than 4.So, log₃a is less than 4, and both 6 and 2^{√a} are greater than 4.Therefore, f(log₃a) is on the decreasing side of the function, and f(6) and f(2^{√a}) are on the increasing side.But since f is symmetric about x = 4, f(4 + t) = f(4 - t). So, for example, f(6) = f(2), because 6 is 2 units to the right of 4, and 2 is 2 units to the left of 4.Similarly, f(2^{√a}) is equal to f(8 - 2^{√a}), but since 2^{√a} is greater than 8 (since 2^{√9} = 2^3 = 8, and 2^{√a} > 8 when a > 9), 8 - 2^{√a} would be negative, which is on the decreasing side.Wait, but maybe that's complicating things. Let me think differently.Since f is decreasing on (-∞, 4) and increasing on (4, ∞), the further we go from 4 on the right side, the larger f(x) becomes, and the further we go from 4 on the left side, the smaller f(x) becomes.So, for x > 4, as x increases, f(x) increases.Similarly, for x < 4, as x decreases, f(x) decreases.So, let's compare the distances of log₃a, 6, and 2^{√a} from 4.First, log₃a is between 2 and 3, so the distance from 4 is between 1 and 2.6 is 2 units away from 4 on the right side.2^{√a} is between 8 and ~36.48, so the distance from 4 is between 4 and ~32.48.So, in terms of distance from 4:log₃a is closest to 4 on the left side, 6 is next, and 2^{√a} is the farthest on the right side.But since f is decreasing on the left and increasing on the right, the function values will behave accordingly.For the left side (x < 4), as x increases towards 4, f(x) increases because it's decreasing as x moves away from 4 to the left.Wait, no. Wait, f is decreasing on (-∞, 4). So, as x increases towards 4 from the left, f(x) increases because it's decreasing as x moves left.Similarly, on the right side (x > 4), as x increases, f(x) increases because it's increasing.So, let's see:log₃a is between 2 and 3, so as a increases from 9 to 27, log₃a increases from 2 to 3. So, log₃a is moving towards 4 from the left. Since f is decreasing on (-∞, 4), as log₃a increases, f(log₃a) decreases. Wait, no, if f is decreasing on (-∞, 4), then as x increases towards 4, f(x) increases. So, as log₃a increases from 2 to 3, f(log₃a) increases from f(2) to f(3). So, f(log₃a) is increasing as a increases.Similarly, 6 is fixed at 6, which is 2 units to the right of 4. So, f(6) is fixed.2^{√a} is increasing as a increases because √a increases, so 2^{√a} increases. Since f is increasing on (4, ∞), as 2^{√a} increases, f(2^{√a}) increases.So, as a increases from 9 to 27:- log₃a increases from 2 to 3, so f(log₃a) increases from f(2) to f(3).- 2^{√a} increases from 8 to ~36.48, so f(2^{√a}) increases from f(8) to f(36.48).But we need to compare f(log₃a), f(6), and f(2^{√a}) for a given a in (9, 27). So, for a specific a, we can say:Since log₃a < 4 < 6 < 2^{√a}, and f is decreasing on (-∞, 4) and increasing on (4, ∞), we can compare the function values based on their distances from 4.But wait, actually, for x < 4, f(x) is decreasing as x decreases, so the further left you go, the smaller f(x) becomes. Similarly, for x > 4, f(x) is increasing as x increases, so the further right you go, the larger f(x) becomes.But log₃a is between 2 and 3, which is closer to 4 than 6 is. Wait, 6 is 2 units away from 4, and log₃a is between 1 and 2 units away from 4 on the left side.But since f is symmetric about x = 4, f(6) = f(2). So, f(6) = f(2). Similarly, f(2^{√a}) = f(8 - 2^{√a}), but 8 - 2^{√a} is negative because 2^{√a} > 8 when a > 9.Wait, that might complicate things. Maybe it's better to think in terms of monotonicity.Since f is decreasing on (-∞, 4), as x increases towards 4, f(x) increases. So, the closer x is to 4 from the left, the larger f(x) is.Similarly, on the right side, as x increases beyond 4, f(x) increases.So, for log₃a, which is between 2 and 3, it's 1 to 2 units away from 4 on the left. So, f(log₃a) is less than f(4), because it's to the left of 4, and f is decreasing there.Wait, but f(4) is the minimum point because it's the vertex of the symmetry.So, f(4) is the minimum value of the function.Therefore, f(log₃a) > f(4) because as x approaches 4 from the left, f(x) increases towards f(4). Wait, no, f is decreasing on (-∞, 4), so as x increases towards 4, f(x) increases. So, f(log₃a) is greater than f(2) because log₃a > 2, and f is increasing as x approaches 4 from the left.Wait, this is getting a bit confusing. Let me try to structure it.Given that f is symmetric about x = 4, so f(4 + t) = f(4 - t) for any t.Also, f is decreasing on (-∞, 4) and increasing on (4, ∞).Therefore, for x < 4, as x increases, f(x) increases.For x > 4, as x increases, f(x) increases.So, the function has a V-shape with the minimum at x = 4.Therefore, the further x is from 4 on either side, the larger f(x) is.But wait, no. Wait, for x < 4, as x increases towards 4, f(x) increases. So, the closer x is to 4 from the left, the larger f(x) is.Similarly, for x > 4, as x increases, f(x) increases, so the further x is from 4 on the right, the larger f(x) is.So, in terms of function values:- The closer x is to 4, the smaller f(x) is.- The further x is from 4, the larger f(x) is.Wait, that makes sense because f(4) is the minimum.So, for x < 4, as x approaches 4 from the left, f(x) approaches f(4), which is the minimum.Similarly, for x > 4, as x approaches 4 from the right, f(x) approaches f(4).Therefore, for x < 4, the further left x is, the larger f(x) is (since f is decreasing on (-∞, 4)).Wait, no, if f is decreasing on (-∞, 4), then as x decreases, f(x) increases. So, the further left x is, the larger f(x) is.Similarly, for x > 4, as x increases, f(x) increases, so the further right x is, the larger f(x) is.So, in terms of function values:- The further x is from 4 in either direction, the larger f(x) is.- The closer x is to 4, the smaller f(x) is.Therefore, to compare f(log₃a), f(6), and f(2^{√a}), we need to see how far each argument is from 4.Given that:- log₃a is between 2 and 3, so the distance from 4 is between 1 and 2.- 6 is 2 units away from 4.- 2^{√a} is between 8 and ~36.48, so the distance from 4 is between 4 and ~32.48.So, in terms of distance from 4:log₃a is closest (distance 1 to 2), then 6 (distance 2), then 2^{√a} (distance 4 to ~32.48).Since f(x) increases as the distance from 4 increases, the function values will follow the same order.Therefore, f(log₃a) < f(6) < f(2^{√a}).So, the correct order is f(log₃a) < f(6) < f(2^{√a}), which corresponds to option D.Wait, let me double-check.Given that f is symmetric about x = 4, f(6) = f(2). Since 2 is less than 4, and f is decreasing on (-∞, 4), f(2) > f(3) because 2 < 3 < 4, and f is decreasing. Wait, no, if f is decreasing on (-∞, 4), then as x increases, f(x) decreases. Wait, no, if f is decreasing, then as x increases, f(x) decreases. So, f(2) > f(3) because 2 < 3 and f is decreasing.But log₃a is between 2 and 3, so f(log₃a) is between f(2) and f(3). Since f is decreasing, f(2) > f(3). Therefore, f(log₃a) is less than f(2) because log₃a > 2, and f is decreasing.Wait, this is conflicting with my earlier conclusion.Let me clarify:If f is decreasing on (-∞, 4), then as x increases, f(x) decreases.So, for x1 < x2 < 4, f(x1) > f(x2).Therefore, since log₃a is between 2 and 3, and 2 < log₃a < 3 < 4, then f(2) > f(log₃a) > f(3).But f(6) = f(2) because of the symmetry about x = 4.So, f(6) = f(2) > f(log₃a) > f(3).But f(3) is still greater than f(4), which is the minimum.On the other hand, 2^{√a} is greater than 8, which is 4 units away from 4 on the right side. Since f is increasing on (4, ∞), the further right you go, the larger f(x) is.So, f(2^{√a}) is greater than f(8), which is greater than f(6) because 8 > 6.Wait, let's see:Since f is increasing on (4, ∞), and 6 < 8 < 2^{√a} (since 2^{√a} > 8 when a > 9), then f(6) < f(8) < f(2^{√a}).But f(6) = f(2), and f(8) = f(0) because of symmetry. Wait, no, f(8) = f(0) because 8 is 4 units to the right of 4, so 4 - 4 = 0. So, f(8) = f(0).But f(0) is on the decreasing side, so f(0) > f(2) because 0 < 2 < 4 and f is decreasing.Wait, this is getting too tangled. Let me try to summarize:1. f is symmetric about x = 4, so f(4 + t) = f(4 - t).2. f is decreasing on (-∞, 4) and increasing on (4, ∞).3. Therefore, f(x) has a minimum at x = 4.4. For x < 4, as x increases, f(x) decreases.5. For x > 4, as x increases, f(x) increases.Given that:- log₃a ∈ (2, 3) < 4, so f(log₃a) is on the decreasing side.- 6 ∈ (4, ∞), so f(6) is on the increasing side.- 2^{√a} ∈ (8, ~36.48) > 4, so f(2^{√a}) is on the increasing side.Now, since f is decreasing on (-∞, 4), the larger x is (closer to 4), the smaller f(x) is.So, log₃a is between 2 and 3, which is closer to 4 than 6 is.Wait, 6 is 2 units away from 4 on the right, while log₃a is 1 to 2 units away on the left.But since f is symmetric, f(6) = f(2). And since f is decreasing on (-∞, 4), f(2) > f(3) > f(log₃a) because log₃a > 2.Wait, no, if log₃a is between 2 and 3, and f is decreasing on (-∞, 4), then as x increases from 2 to 3, f(x) decreases. So, f(2) > f(log₃a) > f(3).But f(6) = f(2), so f(6) > f(log₃a) > f(3).On the other hand, 2^{√a} is greater than 8, which is 4 units away from 4 on the right. Since f is increasing on (4, ∞), the further x is from 4, the larger f(x) is. So, f(2^{√a}) > f(8) > f(6) because 2^{√a} > 8 > 6.Wait, f(8) = f(0) because of symmetry, and f(0) is on the decreasing side, so f(0) > f(2) because 0 < 2 < 4 and f is decreasing.But f(6) = f(2), so f(8) = f(0) > f(2) = f(6).Therefore, f(2^{√a}) > f(8) > f(6) = f(2).So, putting it all together:f(log₃a) < f(6) < f(2^{√a}).Because f(log₃a) is less than f(6) because log₃a is closer to 4 on the left, and f(6) is less than f(2^{√a}) because 2^{√a} is further from 4 on the right.Therefore, the correct order is f(log₃a) < f(6) < f(2^{√a}), which is option D.

💡Okay, so I have this problem here: "If the sum of the interior angles of a polygon is 540 degrees, then the number of its sides is ( ) A: 4, B: 5, C: 6, D: 8." Hmm, I need to figure out how many sides this polygon has based on the sum of its interior angles. Alright, let's start by recalling what I know about polygons and their angles. I remember that the sum of the interior angles of a polygon depends on the number of sides it has. There's a formula for that, right? I think it's something like (n-2) multiplied by 180 degrees, where n is the number of sides. So, the formula is Sum = (n - 2) × 180°. Let me write that down to make it clearer: Sum = (n - 2) × 180. Okay, so in this problem, the sum is given as 540 degrees. That means I can set up the equation like this: 540 = (n - 2) × 180. Now, I need to solve for n to find out how many sides the polygon has.Alright, let's solve the equation step by step. First, I can divide both sides of the equation by 180 to simplify it. So, 540 divided by 180 equals (n - 2). Let me calculate that: 540 ÷ 180 = 3. So, 3 = n - 2. Now, to solve for n, I just need to add 2 to both sides of the equation. That gives me 3 + 2 = n, which simplifies to 5 = n. So, n equals 5. That means the polygon has 5 sides. Wait a second, let me double-check to make sure I didn't make any mistakes. If n is 5, then the sum of the interior angles should be (5 - 2) × 180, which is 3 × 180 = 540 degrees. Yep, that matches the given sum. So, my calculation seems correct.Just to be thorough, let me think about what a 5-sided polygon is called. I think it's a pentagon. Yeah, a pentagon has 5 sides and the sum of its interior angles is indeed 540 degrees. So, that makes sense.Let me also consider the answer choices given: A is 4, which would be a quadrilateral, and the sum of its interior angles is (4 - 2) × 180 = 360 degrees. That's not 540, so A is out. B is 5, which we've already determined gives us 540 degrees, so that's a match. C is 6, which would be a hexagon, and the sum would be (6 - 2) × 180 = 720 degrees, which is too high. D is 8, which would be an octagon, and the sum would be (8 - 2) × 180 = 1080 degrees, which is way too high. So, B is definitely the correct answer.I guess another way to think about it is by considering how each additional side adds another 180 degrees to the sum of the interior angles. Starting from a triangle, which has 3 sides and a sum of 180 degrees, each new side adds another 180 degrees. So, a quadrilateral (4 sides) has 360 degrees, a pentagon (5 sides) has 540 degrees, and so on. That helps me visualize why the formula works and why adding sides increases the sum by 180 each time.It's also interesting to note that this formula only applies to convex polygons, where all the interior angles are less than 180 degrees. If the polygon is concave, with some interior angles greater than 180 degrees, the formula still holds because it's based on the number of sides, not the specific measures of the angles. So, regardless of whether the polygon is convex or concave, as long as it's a simple polygon (not self-intersecting), the sum of the interior angles will be (n - 2) × 180 degrees.I wonder if there's a way to derive this formula from scratch, just to understand it better. Let me think. If I take a polygon and draw all the diagonals from one vertex, I can divide the polygon into (n - 2) triangles. Each triangle has an interior angle sum of 180 degrees, so multiplying that by the number of triangles gives the total sum for the polygon. That makes sense! So, that's why the formula is (n - 2) × 180. It's essentially breaking the polygon down into triangles and summing their angles.This also explains why a triangle (3 sides) has a sum of 180 degrees, as it's just one triangle. A quadrilateral can be split into two triangles, giving 2 × 180 = 360 degrees, and so on. So, each additional side adds another triangle's worth of angles, hence adding 180 degrees each time.Another thing to consider is the measure of each interior angle in a regular polygon, where all sides and angles are equal. For a regular polygon, each interior angle would be the total sum divided by the number of sides. So, for a regular pentagon, each interior angle would be 540 ÷ 5 = 108 degrees. That's something to keep in mind if the question had asked about individual angles instead of the total sum.But in this case, we were only concerned with the total sum, so the regularity of the polygon doesn't matter. Whether it's regular or irregular, as long as it's a simple polygon with 5 sides, the sum of its interior angles will be 540 degrees.I also recall that the sum of the exterior angles of any polygon is always 360 degrees, regardless of the number of sides. That's a different concept, but it's interesting to note. So, while the sum of interior angles increases with the number of sides, the sum of exterior angles remains constant at 360 degrees. That's a useful fact to remember for different types of problems involving polygons.Thinking about the exterior angles, each exterior angle is supplementary to its corresponding interior angle. So, if I know the measure of an interior angle, I can find the exterior angle by subtracting it from 180 degrees. For a regular polygon, all exterior angles are equal, so each exterior angle would be 360 divided by the number of sides. For example, in a regular pentagon, each exterior angle would be 360 ÷ 5 = 72 degrees, and each interior angle would be 180 - 72 = 108 degrees, which matches what I calculated earlier.But again, in this problem, we're only dealing with the sum of interior angles, so the exterior angles aren't directly relevant here. However, understanding both concepts can be helpful in more complex geometry problems.Let me also think about how this applies to real-world scenarios. For instance, if I were designing a building or a piece of furniture with a pentagonal shape, knowing the sum of the interior angles would help in ensuring that the structure is correctly proportioned and that the angles meet properly. It's a fundamental concept in architecture and design.In summary, to solve the problem, I used the formula for the sum of interior angles of a polygon, set it equal to 540 degrees, solved for the number of sides, and verified my answer by checking the options and considering the properties of different polygons. This step-by-step approach helped me arrive at the correct answer confidently.