answer：First, I need to find the magnitude of vector (overrightarrow{a}). Given (overrightarrow{a} = (1, sqrt{3})), the magnitude is calculated as:[|overrightarrow{a}| = sqrt{1^2 + (sqrt{3})^2} = sqrt{1 + 3} = sqrt{4} = 2]Next, I use the formula for the dot product of two vectors:[overrightarrow{a} cdot overrightarrow{b} = |overrightarrow{a}| |overrightarrow{b}| cos(theta)]where (|overrightarrow{b}| = 3) and (theta = frac{pi}{3}). Substituting the known values:[overrightarrow{a} cdot overrightarrow{b} = 2 times 3 times cosleft(frac{pi}{3}right) = 6 times frac{1}{2} = 3]

answer：Okay, so I have this problem here about a sequence {a_n} where the sum of the first n terms is S_n. They tell me that a_1 is 1, and there's this equation: 2S_n = 3a_n + m. I need to figure out which of the options A, B, C, D are correct. Let's take it step by step.First, let's understand what's given. The sum of the first n terms is S_n. So, S_n = a_1 + a_2 + ... + a_n. They also give me that a_1 = 1, which is good to know. The equation 2S_n = 3a_n + m relates the sum of the first n terms to the nth term and some constant m. I need to find out what m is, whether the sequence is arithmetic, what the nth term is, and what the sum S_n is.Let me start by trying to find m. Since a_1 is given, maybe I can plug in n=1 into the equation to find m. Let's see:When n=1, S_1 is just a_1, which is 1. So, plugging into the equation:2S_1 = 3a_1 + m2*1 = 3*1 + m2 = 3 + mSo, m = 2 - 3 = -1Okay, so m is -1. That means option A is correct. So, A: m = -1 is true.Now, let's see if the sequence is arithmetic. An arithmetic sequence has a common difference between consecutive terms. So, if I can find a pattern or a common difference, that would help. But first, maybe I can find a general formula for a_n.Given that 2S_n = 3a_n + m, and we know m is -1, so:2S_n = 3a_n - 1But S_n is the sum of the first n terms, so S_n = a_1 + a_2 + ... + a_n. Also, S_n = S_{n-1} + a_n for n >= 2.Let me write the equation for n and n-1:For n: 2S_n = 3a_n - 1For n-1: 2S_{n-1} = 3a_{n-1} - 1Now, subtract the second equation from the first:2S_n - 2S_{n-1} = 3a_n - 1 - (3a_{n-1} - 1)Simplify:2(S_n - S_{n-1}) = 3a_n - 1 - 3a_{n-1} + 1But S_n - S_{n-1} is just a_n, so:2a_n = 3a_n - 3a_{n-1}Simplify: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}Ah, so each term is 3 times the previous term. That means the sequence is geometric, not arithmetic, with a common ratio of 3.So, option B says it's an arithmetic sequence, which is false. So, B is incorrect.Now, let's find the general term a_n. Since it's a geometric sequence with a_1 = 1 and common ratio 3, the nth term is:a_n = a_1 * r^{n-1} = 1 * 3^{n-1} = 3^{n-1}So, a_n = 3^{n-1}Looking at option C: a_n = (1/3)^{n-1}. That's not correct because our common ratio is 3, not 1/3. So, C is incorrect.Now, let's find S_n. Since it's a geometric series, the sum of the first n terms is:S_n = a_1*(r^n - 1)/(r - 1) = 1*(3^n - 1)/(3 - 1) = (3^n - 1)/2So, S_n = (3^n - 1)/2Looking at option D: S_n = (3^n - 1)/2. That's exactly what we got, so D is correct.So, summarizing:A: CorrectB: IncorrectC: IncorrectD: CorrectTherefore, the correct options are A and D.**Final Answer**boxed{AD}

answer：Alright, so I have this problem where I need to construct a convex quadrilateral when all its sides and one median line are given. Hmm, okay, let's break this down step by step. First, I need to understand what exactly a median line is in a quadrilateral. I think it's a segment that connects the midpoints of two opposite sides. So, if I have a quadrilateral ABCD, the median line would connect the midpoint of AB to the midpoint of CD, for example.Now, I'm given all four sides: AB, BC, CD, and DA. Plus, one median line, say KP, where K is the midpoint of AB and P is the midpoint of CD. My goal is to construct this quadrilateral using these given lengths. Okay, so I need to figure out how to place these sides and the median line in such a way that they form a convex quadrilateral.Let me start by recalling some properties of convex quadrilaterals. A convex quadrilateral has all its interior angles less than 180 degrees, and both diagonals lie entirely inside the quadrilateral. Also, the sum of any three sides must be greater than the fourth side. I wonder if that property will help here.Since I know all the sides, maybe I can try to sketch the quadrilateral roughly and see how the median line fits into it. But without knowing the exact angles or the lengths of the diagonals, it's tricky. I need a systematic approach.I remember that in a quadrilateral, the midpoints of the sides can form another quadrilateral called the Varignon parallelogram. The median lines of the original quadrilateral are the diagonals of this Varignon parallelogram. So, if I can construct the Varignon parallelogram, maybe I can work backward to find the original quadrilateral.But wait, I only have one median line, not both. The Varignon parallelogram requires both diagonals, which correspond to the two median lines of the original quadrilateral. Since I only have one, this might not be directly applicable. Maybe I need a different approach.Let me think about the coordinates. If I assign coordinates to the vertices, I can use coordinate geometry to solve for the positions. Let's say I place point A at the origin (0,0) for simplicity. Then, I can let point B be at (AB, 0), so AB is along the x-axis. Now, I need to find coordinates for points C and D such that BC, CD, and DA have the given lengths, and the median line KP has the given length.But I also need to ensure that the midpoints K and P are connected by the median line of the given length. So, K is the midpoint of AB, which would be at ((AB)/2, 0). P is the midpoint of CD, so if I can express P in terms of C and D, that might help.Let me denote the coordinates:- A = (0, 0)- B = (c, 0), where c = AB- C = (x, y)- D = (p, q)Then, the midpoint K of AB is ((c/2), 0), and the midpoint P of CD is ((x + p)/2, (y + q)/2). The distance between K and P should be equal to the given median length, say m.So, the distance formula gives:√[((x + p)/2 - c/2)^2 + ((y + q)/2 - 0)^2] = mSimplifying:√[((x + p - c)/2)^2 + ((y + q)/2)^2] = mSquare both sides:[((x + p - c)/2)^2 + ((y + q)/2)^2] = m^2Multiply both sides by 4:(x + p - c)^2 + (y + q)^2 = 4m^2Okay, that's one equation. Now, I also know the lengths of BC, CD, and DA.Length BC:√[(x - c)^2 + (y - 0)^2] = BCSo, (x - c)^2 + y^2 = BC^2Length CD:√[(p - x)^2 + (q - y)^2] = CDSo, (p - x)^2 + (q - y)^2 = CD^2Length DA:√[(p - 0)^2 + (q - 0)^2] = DASo, p^2 + q^2 = DA^2Now, I have four equations:1. (x + p - c)^2 + (y + q)^2 = 4m^22. (x - c)^2 + y^2 = BC^23. (p - x)^2 + (q - y)^2 = CD^24. p^2 + q^2 = DA^2This looks like a system of nonlinear equations with variables x, y, p, q. It might be complicated to solve directly, but maybe I can find a way to reduce the variables.Let me see if I can express p and q in terms of x and y from equations 3 and 4.From equation 4: p^2 + q^2 = DA^2From equation 3: (p - x)^2 + (q - y)^2 = CD^2Expanding equation 3:p^2 - 2px + x^2 + q^2 - 2qy + y^2 = CD^2But from equation 4, p^2 + q^2 = DA^2, so substituting:DA^2 - 2px - 2qy + x^2 + y^2 = CD^2Rearranging:-2px - 2qy = CD^2 - DA^2 - x^2 - y^2Divide both sides by -2:px + qy = (DA^2 + x^2 + y^2 - CD^2)/2Hmm, that's one equation involving p and q. Maybe I can find another from equation 2.From equation 2: (x - c)^2 + y^2 = BC^2Expanding:x^2 - 2cx + c^2 + y^2 = BC^2So, x^2 + y^2 = BC^2 + 2cx - c^2Let me denote x^2 + y^2 = BC^2 + 2cx - c^2Going back to the previous equation:px + qy = (DA^2 + x^2 + y^2 - CD^2)/2Substitute x^2 + y^2:px + qy = (DA^2 + BC^2 + 2cx - c^2 - CD^2)/2So, px + qy = (DA^2 + BC^2 - CD^2 + 2cx - c^2)/2This is still complicated, but maybe I can find another relation from equation 1.From equation 1:(x + p - c)^2 + (y + q)^2 = 4m^2Expanding:x^2 + 2x(p - c) + (p - c)^2 + y^2 + 2yq + q^2 = 4m^2Again, from equation 4, p^2 + q^2 = DA^2, so substituting:x^2 + 2x(p - c) + (p - c)^2 + y^2 + 2yq + DA^2 = 4m^2But from equation 2, x^2 + y^2 = BC^2 + 2cx - c^2, so substituting:(BC^2 + 2cx - c^2) + 2x(p - c) + (p - c)^2 + 2yq + DA^2 = 4m^2Simplify term by term:First term: BC^2 + 2cx - c^2Second term: 2x(p - c) = 2xp - 2cxThird term: (p - c)^2 = p^2 - 2cp + c^2Fourth term: 2yqFifth term: DA^2Combine all terms:BC^2 + 2cx - c^2 + 2xp - 2cx + p^2 - 2cp + c^2 + 2yq + DA^2 = 4m^2Simplify:BC^2 + (2cx - 2cx) + (-c^2 + c^2) + 2xp + p^2 - 2cp + 2yq + DA^2 = 4m^2So, BC^2 + 2xp + p^2 - 2cp + 2yq + DA^2 = 4m^2But from equation 4, p^2 + q^2 = DA^2, so p^2 = DA^2 - q^2. Substituting:BC^2 + 2xp + (DA^2 - q^2) - 2cp + 2yq + DA^2 = 4m^2Simplify:BC^2 + 2xp + DA^2 - q^2 - 2cp + 2yq + DA^2 = 4m^2Combine like terms:BC^2 + 2DA^2 + 2xp - 2cp - q^2 + 2yq = 4m^2Factor terms:BC^2 + 2DA^2 + 2p(x - c) + q(-q + 2y) = 4m^2This is getting quite involved. Maybe I need a different strategy. Perhaps instead of using coordinates, I can use vector geometry or some geometric constructions.Let me think about the properties of the median line. The median line connects midpoints of two sides, so it's related to the midline theorem in triangles. In a triangle, the midline is parallel to the base and half its length. Maybe I can apply something similar here.If I consider triangles formed by the diagonals of the quadrilateral, perhaps I can use the midline properties. But I'm not sure. Alternatively, maybe I can construct triangles and use the given median to find the missing parts.Another idea: since I know all the sides, perhaps I can construct two triangles and then combine them. For example, construct triangle ABC with sides AB, BC, and then construct triangle ADC with sides AD, DC, and the median KP. But I'm not sure how the median fits into this.Wait, the median connects midpoints of AB and CD. So, if I can find the midpoints, I can ensure that the distance between them is as given. Maybe I can use this to adjust the position of point D relative to triangle ABC.Let me try to outline a possible construction:1. Draw side AB with length AB.2. Construct triangle ABC with side BC.3. Find the midpoint K of AB.4. Now, need to construct point D such that CD has length CD, DA has length DA, and the midpoint P of CD is at a distance m from K.Hmm, that seems plausible. So, after constructing ABC, I need to find D such that CD = given length, DA = given length, and the midpoint of CD is at distance m from K.This sounds like a problem of finding the intersection of two circles: one centered at C with radius CD, and another centered at A with radius DA. The intersection points would give possible locations for D. Then, among these, I need to choose the one such that the midpoint P of CD is at distance m from K.But how do I ensure that P is at distance m from K? Maybe I can use the midpoint condition to set up another circle.Let me formalize this:Given points A, B, C, and midpoints K of AB, I need to find D such that:- Distance from D to C is CD.- Distance from D to A is DA.- Distance from midpoint P of CD to K is m.So, D must lie at the intersection of two circles: circle centered at C with radius CD, and circle centered at A with radius DA. Let's call these intersections D1 and D2.Now, for each D1 and D2, compute the midpoint P1 and P2 of CD1 and CD2, respectively. Then, check if the distance from P1 to K or P2 to K is equal to m.If it is, then that D is the correct one. If not, maybe there's no solution, or perhaps I need to adjust something.But this seems like a trial-and-error approach. Maybe there's a more precise method.Alternatively, since I know K and need P to be at distance m from K, I can think of P lying on a circle centered at K with radius m. Also, since P is the midpoint of CD, P must lie on the perpendicular bisector of CD.Wait, no, the midpoint of CD is P, so P is determined once D is chosen. But since I don't know D yet, this might not help directly.Perhaps I can express P in terms of D and then set up the distance condition.If P is the midpoint of CD, then P = (C + D)/2. So, vectorially, P = (C + D)/2.Given that, the distance between K and P is m:|P - K| = mSubstitute P:|(C + D)/2 - K| = mMultiply both sides by 2:|C + D - 2K| = 2mSo, the vector equation is |C + D - 2K| = 2m.This gives a condition on D. So, D must satisfy:|C + D - 2K| = 2mBut D also must satisfy |D - C| = CD and |D - A| = DA.So, D lies at the intersection of three conditions:1. |D - C| = CD2. |D - A| = DA3. |C + D - 2K| = 2mThis is a system of equations in vector form. Maybe I can solve this system.Let me denote vectors:Let’s set A as the origin for simplicity, so A = (0,0). Then, vector K is the midpoint of AB, so if B is at vector b, then K = b/2.Vector C is known since we've constructed triangle ABC. Vector D is what we need to find.So, the conditions become:1. |D - C| = CD2. |D| = DA3. |C + D - 2K| = 2mExpressed in coordinates, let me assume A is at (0,0), B is at (c,0), so K is at (c/2, 0). Let C be at (x, y). Then, D is at (p, q).So, the conditions are:1. √[(p - x)^2 + (q - y)^2] = CD2. √[p^2 + q^2] = DA3. √[(x + p - c)^2 + (y + q)^2] = 2mThese are the same equations I had earlier. It seems I'm going in circles.Maybe I can use the first two equations to express p and q in terms of x and y, and then substitute into the third equation.From equation 2: p^2 + q^2 = DA^2From equation 1: (p - x)^2 + (q - y)^2 = CD^2Expanding equation 1:p^2 - 2px + x^2 + q^2 - 2qy + y^2 = CD^2But p^2 + q^2 = DA^2, so substitute:DA^2 - 2px - 2qy + x^2 + y^2 = CD^2Rearranging:-2px - 2qy = CD^2 - DA^2 - x^2 - y^2Divide by -2:px + qy = (DA^2 + x^2 + y^2 - CD^2)/2Now, from equation 3:(x + p - c)^2 + (y + q)^2 = (2m)^2Expanding:x^2 + 2x(p - c) + (p - c)^2 + y^2 + 2yq + q^2 = 4m^2Again, p^2 + q^2 = DA^2, so:x^2 + 2x(p - c) + (p - c)^2 + y^2 + 2yq + DA^2 = 4m^2But from equation 2, x^2 + y^2 = BC^2 + 2cx - c^2 (since (x - c)^2 + y^2 = BC^2)Wait, no, equation 2 is (x - c)^2 + y^2 = BC^2, which expands to x^2 - 2cx + c^2 + y^2 = BC^2, so x^2 + y^2 = BC^2 + 2cx - c^2.Substituting back:(BC^2 + 2cx - c^2) + 2x(p - c) + (p - c)^2 + 2yq + DA^2 = 4m^2Expanding:BC^2 + 2cx - c^2 + 2xp - 2cx + p^2 - 2cp + c^2 + 2yq + DA^2 = 4m^2Simplify:BC^2 + 2xp + p^2 - 2cp + 2yq + DA^2 = 4m^2But p^2 + q^2 = DA^2, so p^2 = DA^2 - q^2. Substitute:BC^2 + 2xp + DA^2 - q^2 - 2cp + 2yq + DA^2 = 4m^2Combine like terms:BC^2 + 2DA^2 + 2xp - 2cp - q^2 + 2yq = 4m^2From earlier, we have px + qy = (DA^2 + x^2 + y^2 - CD^2)/2Let me denote S = (DA^2 + x^2 + y^2 - CD^2)/2So, px + qy = SBut in the previous equation, we have 2xp - 2cp - q^2 + 2yq = 4m^2 - BC^2 - 2DA^2Let me factor out terms:2p(x - c) + 2qy - q^2 = 4m^2 - BC^2 - 2DA^2But from px + qy = S, we can express px = S - qySo, 2p(x - c) = 2px - 2pc = 2(S - qy) - 2pcSubstitute back:2(S - qy) - 2pc + 2qy - q^2 = 4m^2 - BC^2 - 2DA^2Simplify:2S - 2qy - 2pc + 2qy - q^2 = 4m^2 - BC^2 - 2DA^2The -2qy and +2qy cancel out:2S - 2pc - q^2 = 4m^2 - BC^2 - 2DA^2Now, recall that S = (DA^2 + x^2 + y^2 - CD^2)/2So, 2S = DA^2 + x^2 + y^2 - CD^2Substitute:DA^2 + x^2 + y^2 - CD^2 - 2pc - q^2 = 4m^2 - BC^2 - 2DA^2Bring all terms to one side:DA^2 + x^2 + y^2 - CD^2 - 2pc - q^2 - 4m^2 + BC^2 + 2DA^2 = 0Combine like terms:3DA^2 + x^2 + y^2 - CD^2 - 2pc - q^2 + BC^2 - 4m^2 = 0This is getting too complicated. Maybe I need to find another approach.Perhaps instead of using coordinates, I can use geometric constructions. Let me try that.1. Draw side AB with the given length.2. Construct triangle ABC with the given side BC.3. Find the midpoint K of AB.4. Now, I need to find point D such that CD = given length, DA = given length, and the midpoint P of CD is at distance m from K.So, after constructing ABC, I can attempt to construct D.To find D, I can draw two circles:- One centered at C with radius CD.- Another centered at A with radius DA.The intersection points of these circles will give possible locations for D. Let's call these intersections D1 and D2.Now, for each D1 and D2, I need to check if the midpoint P of CD is at distance m from K.So, for D1:- Find midpoint P1 of CD1.- Measure distance from P1 to K. If it's equal to m, then D1 is the correct point.- If not, check D2 similarly.If neither P1 nor P2 is at distance m from K, then perhaps the given lengths do not form a valid quadrilateral, or I might have made a mistake in construction.Assuming one of them works, say D1, then quadrilateral ABCD1 is the desired convex quadrilateral.But how do I ensure that the quadrilateral is convex? I need to make sure that all interior angles are less than 180 degrees, which might require choosing the correct intersection point D1 or D2.Alternatively, since I'm constructing it step by step, as long as I follow the given lengths and the median condition, the quadrilateral should be convex.Wait, but what if there are multiple solutions? Depending on the lengths, there might be two possible quadrilaterals, one convex and one concave. I need to ensure I choose the convex one.To do that, after constructing D1 and D2, I can check the position of D relative to triangle ABC. If D is on the same side as C relative to AB, it might form a convex quadrilateral. If it's on the opposite side, it might be concave.But I'm not entirely sure. Maybe I need to verify by checking the angles or ensuring that the points are ordered correctly.Alternatively, since I'm constructing it step by step, and assuming all given lengths are positive and satisfy the triangle inequalities, the construction should naturally lead to a convex quadrilateral.So, summarizing the steps:1. Draw side AB with the given length.2. Using point B, draw an arc with radius BC to find point C.3. Connect points B and C to complete triangle ABC.4. Find the midpoint K of AB.5. Using point C, draw an arc with radius CD.6. Using point A, draw an arc with radius DA.7. The intersection points of these arcs are potential locations for D.8. For each intersection point D, find the midpoint P of CD.9. Check if the distance from P to K is equal to the given median length m.10. If it is, connect points C and D, and D and A to complete the quadrilateral ABCD.11. Ensure the quadrilateral is convex by checking the positions of the points.This seems like a feasible construction method. I might need to use a compass and straightedge for precise drawing, especially when finding midpoints and intersections.But I'm still a bit concerned about the complexity of ensuring all conditions are met, especially the median length. Maybe there's a more straightforward geometric principle I can apply.Wait, I recall that in a quadrilateral, the line connecting the midpoints of two sides is equal to half the sum of the other two sides if the quadrilateral is a trapezoid. But this is a general quadrilateral, not necessarily a trapezoid.Alternatively, maybe I can use the formula for the length of a median in a quadrilateral. I think there's a formula that relates the lengths of the sides and the median.Let me look it up in my mind. I think the formula for the length of a median (connecting midpoints of two sides) in a quadrilateral is:m = (1/2) * sqrt(2a^2 + 2b^2 - c^2 - d^2)Where a and b are the lengths of the two sides whose midpoints are connected, and c and d are the lengths of the other two sides.Wait, is that correct? Let me verify.Actually, I think the formula for the length of a median in a quadrilateral is similar to the midline theorem in triangles but extended. It might involve the sides adjacent to the median.Alternatively, perhaps it's better to use vector methods or coordinate geometry as I tried earlier.Given that, maybe I can use the formula for the length of the median in terms of the sides.Let me denote the sides as AB = a, BC = b, CD = c, DA = d, and the median KP = m.Then, the formula for the median m connecting midpoints of AB and CD is:m^2 = (2b^2 + 2d^2 - a^2 - c^2)/4Wait, is that right? Let me derive it.Consider vectors:Let’s place the quadrilateral in a coordinate system with A at the origin, B at (a, 0), C at (x, y), and D at (p, q).Midpoint K of AB is at (a/2, 0).Midpoint P of CD is at ((x + p)/2, (y + q)/2).The vector KP is P - K = ((x + p)/2 - a/2, (y + q)/2 - 0) = ((x + p - a)/2, (y + q)/2)The length squared of KP is:[(x + p - a)/2]^2 + [(y + q)/2]^2 = m^2Multiply both sides by 4:(x + p - a)^2 + (y + q)^2 = 4m^2Now, from the sides:AB = a: distance from A(0,0) to B(a,0) is a.BC = b: distance from B(a,0) to C(x,y) is sqrt((x - a)^2 + y^2) = bCD = c: distance from C(x,y) to D(p,q) is sqrt((p - x)^2 + (q - y)^2) = cDA = d: distance from D(p,q) to A(0,0) is sqrt(p^2 + q^2) = dSo, we have:1. (x - a)^2 + y^2 = b^22. (p - x)^2 + (q - y)^2 = c^23. p^2 + q^2 = d^24. (x + p - a)^2 + (y + q)^2 = 4m^2This is the same system as before. Let me try to find a relationship between these equations.From equation 3: p^2 + q^2 = d^2From equation 4: (x + p - a)^2 + (y + q)^2 = 4m^2Expand equation 4:(x + p - a)^2 + (y + q)^2 = x^2 + 2x(p - a) + (p - a)^2 + y^2 + 2yq + q^2 = 4m^2But from equation 3, p^2 + q^2 = d^2, so:x^2 + 2x(p - a) + (p^2 - 2ap + a^2) + y^2 + 2yq + d^2 = 4m^2Simplify:x^2 + 2xp - 2ax + p^2 - 2ap + a^2 + y^2 + 2yq + d^2 = 4m^2From equation 1: (x - a)^2 + y^2 = b^2 => x^2 - 2ax + a^2 + y^2 = b^2 => x^2 + y^2 = b^2 + 2ax - a^2Substitute x^2 + y^2 into the previous equation:(b^2 + 2ax - a^2) + 2xp - 2ax + p^2 - 2ap + a^2 + 2yq + d^2 = 4m^2Simplify:b^2 + 2ax - a^2 + 2xp - 2ax + p^2 - 2ap + a^2 + 2yq + d^2 = 4m^2The -a^2 and +a^2 cancel out, as do the +2ax and -2ax:b^2 + 2xp + p^2 - 2ap + 2yq + d^2 = 4m^2From equation 3, p^2 + q^2 = d^2, so p^2 = d^2 - q^2Substitute:b^2 + 2xp + (d^2 - q^2) - 2ap + 2yq + d^2 = 4m^2Simplify:b^2 + 2xp + d^2 - q^2 - 2ap + 2yq + d^2 = 4m^2Combine like terms:b^2 + 2d^2 + 2xp - 2ap - q^2 + 2yq = 4m^2Factor terms:b^2 + 2d^2 + 2p(x - a) + q(-q + 2y) = 4m^2This is still quite complex. Maybe I can express p and q in terms of x and y from equations 1 and 2.From equation 1: (x - a)^2 + y^2 = b^2From equation 2: (p - x)^2 + (q - y)^2 = c^2From equation 3: p^2 + q^2 = d^2Let me try to solve equations 2 and 3 for p and q.From equation 3: p^2 + q^2 = d^2From equation 2: (p - x)^2 + (q - y)^2 = c^2Expanding equation 2:p^2 - 2px + x^2 + q^2 - 2qy + y^2 = c^2But p^2 + q^2 = d^2, so:d^2 - 2px - 2qy + x^2 + y^2 = c^2Rearranging:-2px - 2qy = c^2 - d^2 - x^2 - y^2Divide by -2:px + qy = (d^2 + x^2 + y^2 - c^2)/2Let me denote this as equation 5: px + qy = S, where S = (d^2 + x^2 + y^2 - c^2)/2Now, from equation 4, after substitution, we have:b^2 + 2d^2 + 2p(x - a) + q(-q + 2y) = 4m^2But from equation 5, px + qy = S, so px = S - qySubstitute px into the equation:b^2 + 2d^2 + 2(S - qy) + q(-q + 2y) = 4m^2Simplify:b^2 + 2d^2 + 2S - 2qy - q^2 + 2yq = 4m^2The -2qy and +2yq cancel out:b^2 + 2d^2 + 2S - q^2 = 4m^2But S = (d^2 + x^2 + y^2 - c^2)/2, so 2S = d^2 + x^2 + y^2 - c^2Substitute:b^2 + 2d^2 + d^2 + x^2 + y^2 - c^2 - q^2 = 4m^2Combine like terms:b^2 + 3d^2 + x^2 + y^2 - c^2 - q^2 = 4m^2From equation 1: (x - a)^2 + y^2 = b^2 => x^2 - 2ax + a^2 + y^2 = b^2 => x^2 + y^2 = b^2 + 2ax - a^2Substitute x^2 + y^2:b^2 + 3d^2 + (b^2 + 2ax - a^2) - c^2 - q^2 = 4m^2Simplify:b^2 + 3d^2 + b^2 + 2ax - a^2 - c^2 - q^2 = 4m^2Combine like terms:2b^2 + 3d^2 + 2ax - a^2 - c^2 - q^2 = 4m^2From equation 3: p^2 + q^2 = d^2 => q^2 = d^2 - p^2Substitute q^2:2b^2 + 3d^2 + 2ax - a^2 - c^2 - (d^2 - p^2) = 4m^2Simplify:2b^2 + 3d^2 + 2ax - a^2 - c^2 - d^2 + p^2 = 4m^2Combine like terms:2b^2 + 2d^2 + 2ax - a^2 - c^2 + p^2 = 4m^2From equation 5: px + qy = S = (d^2 + x^2 + y^2 - c^2)/2But I don't see a direct way to substitute p here. Maybe I need to express p in terms of x and y.From equation 5: px = S - qyBut q^2 = d^2 - p^2, so q = sqrt(d^2 - p^2). This introduces square roots, making it messy.Perhaps instead of trying to solve algebraically, I should stick to the geometric construction approach.So, to recap, the steps are:1. Draw side AB with length AB.2. Using point B, draw an arc with radius BC to find point C.3. Connect B to C to form triangle ABC.4. Find the midpoint K of AB.5. Using point C, draw an arc with radius CD.6. Using point A, draw an arc with radius DA.7. The intersection points of these arcs are potential locations for D.8. For each intersection point D, find the midpoint P of CD.9. Check if the distance from P to K is equal to the given median length m.10. If it is, connect points C to D and D to A to complete the quadrilateral ABCD.11. Ensure the quadrilateral is convex by verifying the positions of the points.This seems like a practical approach, even if it involves some trial and error. By constructing the possible points D and checking the median condition, I can determine the correct position of D to form the desired quadrilateral.I think I've gone through enough steps and considerations. Time to put it all together in a clear, step-by-step explanation.

answer：Alright, so I've got these two problems about covering the plane with rectangles and squares. Let me try to wrap my head around them one by one.Starting with part (a): We have an infinite sequence of paper rectangles where the area of the nth rectangle is n squared. The question is whether it's necessarily possible to cover the plane with them, allowing overlaps. Hmm, okay. So each rectangle has an area of n², but what about their dimensions? The problem doesn't specify, so I guess the rectangles could be of any shape as long as their area is n².Wait, but if the area is fixed, the dimensions can vary. For example, the nth rectangle could be very long and skinny or more square-like. I think the key here is that even though the areas are increasing quadratically, the shape of the rectangles might not be suitable for covering the plane efficiently.Maybe I can think of a counterexample. Suppose I make each rectangle very thin. Let's say the width of the nth rectangle is 1/n, then the height would be n² / (1/n) = n³. So each rectangle is 1/n wide and n³ tall. If I try to cover the plane with these, they get taller and taller but narrower and narrower. If I place them side by side along the x-axis, each subsequent rectangle would only cover a tiny sliver of width, but their heights go to infinity. However, the problem is whether they can cover the entire plane. But wait, even though they're tall, their width is getting smaller. So maybe there's a region in the plane that they can't cover because their widths are too narrow.Alternatively, maybe arranging them differently could help. But if the widths are getting too small, even if we stack them vertically, their combined width might not cover the entire plane. Hmm, I'm not sure. Maybe I need to think about the total area.The total area of all rectangles is the sum of n² from n=1 to infinity, which diverges. So in theory, if we can arrange them without too much overlap, maybe we can cover the plane. But the problem is about whether it's necessarily possible, regardless of the shapes of the rectangles.Wait, the problem says "the area of the nth rectangle is equal to n²." It doesn't specify the dimensions, so someone could choose the dimensions adversarially to make covering the plane impossible. For example, if all rectangles are made very long in one direction and very thin in the other, they might not be able to cover the plane.So maybe the answer is no, it's not necessarily possible to cover the plane with them because the rectangles could be shaped in a way that prevents complete coverage despite their areas increasing.Moving on to part (b): We have an infinite sequence of paper squares, and we know that for any number N, there exist squares with a total area greater than N. The question is whether it's necessarily possible to cover the plane with them, allowing overlaps.Okay, so unlike part (a), here we have squares, which are more regular in shape. The condition is that for any N, the total area of some squares exceeds N. That means the total area is unbounded, right? Because for any N, no matter how large, there's a collection of squares whose total area is bigger than N.But does that guarantee that we can cover the plane? I think it does, but I need to think carefully.If the total area is unbounded, that means we can cover arbitrarily large regions. But the plane is infinite, so we need to ensure that every point is covered by at least one square.One approach is to use a tiling argument. Since squares can be arranged in a grid pattern, and if we have squares of various sizes, we can cover the plane by strategically placing them.But wait, the problem doesn't specify that the squares have any particular size or that they are of different sizes. It just says that for any N, there exist squares with total area greater than N. So maybe all the squares are of the same size, but there are infinitely many of them. If they are all the same size, say unit squares, then yes, we can tile the plane with them without overlaps, but the problem allows overlaps.But if the squares are not of the same size, and their sizes can vary, but the total area is unbounded, then we can still cover the plane. For example, if we have squares of decreasing size, we can place them densely enough to cover every point.Wait, but how do we ensure that every point is covered? Maybe we can use a diagonalization argument or something similar. If we have squares of all sizes, we can cover the plane by arranging them appropriately.Alternatively, think about the plane being divided into regions, and each region can be covered by some square. Since the total area is unbounded, we can cover each region with enough squares.Hmm, I think the key here is that with squares, regardless of their sizes, as long as their total area is unbounded, we can arrange them to cover the entire plane. Unlike rectangles, which might be stretched in one direction, squares are balanced, so they can cover areas more uniformly.So putting it all together, for part (a), it's not necessarily possible because the rectangles could be shaped adversarially, but for part (b), it is necessarily possible because squares can be arranged to cover the plane given their total area is unbounded.**Final Answer**a) boxed{text{Not necessarily}}b) boxed{text{Necessarily so}}

answer：Okay, so I need to find the equation of a parabola that opens upwards and passes through the point (0, -2). Hmm, let's start by recalling what a parabola is. A parabola is a U-shaped curve, and since it opens upwards, the coefficient of the x² term should be positive. I remember that the general form of a quadratic equation, which represents a parabola, is y = ax² + bx + c. Here, 'a' determines how wide or narrow the parabola is and whether it opens upwards or downwards. Since it opens upwards, 'a' should be positive. Now, I need to find the specific equation that passes through the point (0, -2). Let's plug this point into the general equation to see what we get. So, if x = 0 and y = -2, substituting into y = ax² + bx + c gives:-2 = a*(0)² + b*(0) + cSimplifying that, we get:-2 = 0 + 0 + cSo, c = -2.Okay, so now I know that c is -2. But what about 'a' and 'b'? I only have one point, so I can't determine both 'a' and 'b' uniquely. Maybe I need to make some assumptions here. Since the problem doesn't specify anything else about the parabola, like its vertex or any other points it passes through, I might assume the simplest case where the parabola is symmetric about the y-axis. That would mean there's no 'bx' term, so b = 0. With b = 0, the equation simplifies to y = ax² + c. We already found that c = -2, so now it's y = ax² - 2. Now, I need to determine 'a'. But wait, I don't have any other points to plug in. Maybe I can choose a value for 'a' that makes the parabola open upwards. Since 'a' must be positive, I can choose 'a' = 1 for simplicity. So, if a = 1, then the equation becomes y = x² - 2. Let me check if this makes sense. When x = 0, y = 0 - 2 = -2, which matches the given point. Also, since 'a' is positive, the parabola opens upwards. But what if I choose a different value for 'a'? Let's say a = 2. Then the equation would be y = 2x² - 2. This also opens upwards and passes through (0, -2). So, there are infinitely many parabolas that satisfy the given condition, each with a different 'a' value. However, since the problem doesn't specify any additional constraints, like the width of the parabola or another point it passes through, the simplest solution is to set 'a' = 1. This gives us the equation y = x² - 2. I think that's the answer they're looking for. It's the standard parabola opening upwards with vertex at (0, -2). Just to be thorough, let's graph this in my mind. The vertex is at (0, -2), and since it opens upwards, as x increases or decreases from 0, y increases. So, for x = 1, y = 1 - 2 = -1; for x = 2, y = 4 - 2 = 2; and so on. This makes sense for a parabola opening upwards. I don't see any mistakes in my reasoning. I started with the general form, substituted the given point to find 'c', assumed symmetry to eliminate 'b', and chose the simplest positive 'a' value. This should be the correct equation.

answer：First, identify the universal set ( U = {1, 2, 3, 4, 5} ) and the subsets ( M = {1, 2} ) and ( N = {3, 4} ).Next, find the union of sets ( M ) and ( N ):[M cup N = {1, 2} cup {3, 4} = {1, 2, 3, 4}]Then, determine the complement of ( M cup N ) with respect to ( U ):[complement_U(M cup N) = U - (M cup N) = {1, 2, 3, 4, 5} - {1, 2, 3, 4} = {5}]Therefore, the complement of ( M cup N ) in ( U ) is ({5}).