💡Okay, so I have this problem about a right tetrahedron (ABCD) where the edges (DA), (DB), and (DC) are mutually perpendicular. I need to find the maximum value of the ratio (frac{S}{R^2}), where (S) is the surface area and (R) is the radius of the circumscribed sphere.First, I should make sure I understand the problem correctly. A right tetrahedron means that the three edges meeting at vertex (D) are perpendicular to each other. So, if I imagine point (D) at the origin of a 3D coordinate system, then points (A), (B), and (C) would lie along the x, y, and z-axes respectively. That makes (ABCD) a rectangular prism, right? Wait, no, actually, a tetrahedron has four triangular faces, so it's not a prism. Hmm, maybe I need to clarify that.Wait, no, in a right tetrahedron, the three edges from one vertex are perpendicular, so it's like a corner of a rectangular prism. So, (ABCD) is a tetrahedron with three mutually perpendicular edges at (D). So, the faces opposite (D) are right triangles.Okay, so let me denote the lengths of (DA), (DB), and (DC) as (a), (b), and (c) respectively. So, (DA = a), (DB = b), (DC = c). Then, the other edges can be found using the Pythagorean theorem. For example, edge (AB) would have length (sqrt{a^2 + b^2}), edge (AC) would be (sqrt{a^2 + c^2}), and edge (BC) would be (sqrt{b^2 + c^2}).Now, the surface area (S) of the tetrahedron. Since it's a tetrahedron, it has four triangular faces. Each of the three faces meeting at (D) is a right triangle, and the fourth face is a triangle connecting points (A), (B), and (C). So, the surface area (S) is the sum of the areas of these four triangles.Calculating each area:1. Area of triangle (DAB): (frac{1}{2}ab)2. Area of triangle (DAC): (frac{1}{2}ac)3. Area of triangle (DBC): (frac{1}{2}bc)4. Area of triangle (ABC): This is a bit more complicated. Since (ABC) is a triangle with sides (sqrt{a^2 + b^2}), (sqrt{a^2 + c^2}), and (sqrt{b^2 + c^2}), I need to find its area. Maybe I can use Heron's formula? But that might be messy. Alternatively, since (ABCD) is a right tetrahedron, maybe there's a simpler way.Wait, actually, triangle (ABC) is the face opposite vertex (D). Since (DA), (DB), and (DC) are mutually perpendicular, the coordinates of (A), (B), and (C) can be considered as ((a, 0, 0)), ((0, b, 0)), and ((0, 0, c)) respectively. So, the vectors (AB), (AC), and (BC) can be found, and the area of triangle (ABC) can be calculated using the cross product.Let me try that. The vectors:- Vector (AB = B - A = (-a, b, 0))- Vector (AC = C - A = (-a, 0, c))The cross product (AB times AC) is:[begin{vmatrix}mathbf{i} & mathbf{j} & mathbf{k} -a & b & 0 -a & 0 & c end{vmatrix}= mathbf{i}(b cdot c - 0 cdot 0) - mathbf{j}(-a cdot c - 0 cdot (-a)) + mathbf{k}(-a cdot 0 - b cdot (-a))= bcmathbf{i} + acmathbf{j} + abmathbf{k}]The magnitude of this cross product is (sqrt{(bc)^2 + (ac)^2 + (ab)^2}). Therefore, the area of triangle (ABC) is half of that:[frac{1}{2} sqrt{a^2b^2 + a^2c^2 + b^2c^2}]So, the total surface area (S) is:[S = frac{1}{2}ab + frac{1}{2}ac + frac{1}{2}bc + frac{1}{2} sqrt{a^2b^2 + a^2c^2 + b^2c^2}]Hmm, that seems a bit complicated. Maybe I made a mistake somewhere. Let me check.Wait, actually, in a right tetrahedron, the face opposite the right-angled vertex is called the "hypotenuse face," and its area can be calculated using the formula:[text{Area} = frac{1}{2} sqrt{(ab)^2 + (ac)^2 + (bc)^2}]Yes, that's correct. So, the surface area (S) is the sum of the three right triangular faces and the hypotenuse face:[S = frac{1}{2}ab + frac{1}{2}ac + frac{1}{2}bc + frac{1}{2} sqrt{a^2b^2 + a^2c^2 + b^2c^2}]Okay, that seems right. Now, moving on to the radius (R) of the circumscribed sphere. For a tetrahedron, the circumscribed sphere (circum sphere) passes through all four vertices. The radius (R) can be found using the formula:[R = frac{sqrt{a^2 + b^2 + c^2}}{2}]Wait, is that correct? Let me think. In a right tetrahedron, the circumradius is indeed half the space diagonal of the rectangular prism formed by edges (a), (b), and (c). So yes, since the space diagonal is (sqrt{a^2 + b^2 + c^2}), the radius is half of that.So, (R = frac{sqrt{a^2 + b^2 + c^2}}{2}).Now, I need to compute the ratio (frac{S}{R^2}) and find its maximum value. Let's write down the expressions:First, (R^2 = left( frac{sqrt{a^2 + b^2 + c^2}}{2} right)^2 = frac{a^2 + b^2 + c^2}{4}).So, (frac{S}{R^2} = frac{ frac{1}{2}ab + frac{1}{2}ac + frac{1}{2}bc + frac{1}{2} sqrt{a^2b^2 + a^2c^2 + b^2c^2} }{ frac{a^2 + b^2 + c^2}{4} } )Simplify numerator and denominator:Multiply numerator and denominator by 4:[frac{S}{R^2} = frac{2ab + 2ac + 2bc + 2sqrt{a^2b^2 + a^2c^2 + b^2c^2}}{a^2 + b^2 + c^2}]So, (frac{S}{R^2} = frac{2(ab + ac + bc) + 2sqrt{a^2b^2 + a^2c^2 + b^2c^2}}{a^2 + b^2 + c^2})Hmm, this seems a bit complicated. Maybe I can factor out the 2:[frac{S}{R^2} = frac{2[ab + ac + bc + sqrt{a^2b^2 + a^2c^2 + b^2c^2}]}{a^2 + b^2 + c^2}]I need to find the maximum value of this expression. To do that, perhaps I can use some inequality or optimization technique.Let me consider using the AM-GM inequality or Cauchy-Schwarz inequality.First, let me denote (x = ab), (y = ac), and (z = bc). Then, the numerator becomes (2(x + y + z + sqrt{x^2 + y^2 + z^2})), and the denominator is (a^2 + b^2 + c^2).But I'm not sure if this substitution helps. Maybe I can express (a^2 + b^2 + c^2) in terms of (x), (y), and (z). However, that might complicate things further.Alternatively, maybe I can assume symmetry, i.e., set (a = b = c), and see what value I get. If the maximum occurs at this symmetric case, then that would be the answer. Otherwise, I might need to consider other cases.Let me try setting (a = b = c = k). Then, (ab = ac = bc = k^2), and (a^2 + b^2 + c^2 = 3k^2).Compute (S):Surface area (S = 3 times frac{1}{2}k^2 + frac{1}{2}sqrt{3k^4} = frac{3}{2}k^2 + frac{1}{2}k^2sqrt{3})So, (S = frac{3}{2}k^2 + frac{sqrt{3}}{2}k^2 = frac{3 + sqrt{3}}{2}k^2)Compute (R):(R = frac{sqrt{3k^2}}{2} = frac{ksqrt{3}}{2})So, (R^2 = frac{3k^2}{4})Therefore, (frac{S}{R^2} = frac{frac{3 + sqrt{3}}{2}k^2}{frac{3k^2}{4}} = frac{(3 + sqrt{3})/2}{3/4} = frac{(3 + sqrt{3})}{2} times frac{4}{3} = frac{2(3 + sqrt{3})}{3} = 2 + frac{2sqrt{3}}{3})Hmm, that's approximately (2 + 1.1547 = 3.1547). Is this the maximum? Maybe, but I need to check if this is indeed the maximum or if a higher value is possible when (a), (b), and (c) are not equal.Alternatively, perhaps I can use Lagrange multipliers to maximize (frac{S}{R^2}) with respect to (a), (b), and (c). But that might be a bit involved.Let me consider another approach. Let me denote (p = a^2), (q = b^2), (r = c^2). Then, (ab = sqrt{pq}), (ac = sqrt{pr}), (bc = sqrt{qr}), and (a^2 + b^2 + c^2 = p + q + r).So, the expression becomes:[frac{S}{R^2} = frac{2(sqrt{pq} + sqrt{pr} + sqrt{qr}) + 2sqrt{pq + pr + qr}}{p + q + r}]This still looks complicated. Maybe I can assume two variables are equal, say (p = q), and see if that simplifies things.Let me set (p = q = t), and (r) as another variable. Then, (ab = sqrt{t^2} = t), (ac = sqrt{tr}), (bc = sqrt{tr}), and (a^2 + b^2 + c^2 = 2t + r).So, the expression becomes:[frac{S}{R^2} = frac{2(t + 2sqrt{tr}) + 2sqrt{t^2 + 2tr}}{2t + r}]Simplify numerator:[2t + 4sqrt{tr} + 2sqrt{t^2 + 2tr}]Denominator:[2t + r]Hmm, not sure if this helps. Maybe I can set (r = kt), where (k) is a positive real number, and then express everything in terms of (t) and (k).Let me try that. Let (r = kt), so (ac = sqrt{t cdot kt} = tsqrt{k}), same with (bc = tsqrt{k}). Then, (a^2 + b^2 + c^2 = 2t + kt = t(2 + k)).The numerator becomes:[2t + 4sqrt{t cdot kt} + 2sqrt{t^2 + 2t cdot kt} = 2t + 4tsqrt{k} + 2sqrt{t^2 + 2kt^2}][= 2t + 4tsqrt{k} + 2tsqrt{1 + 2k}]Factor out (2t):[2t[1 + 2sqrt{k} + sqrt{1 + 2k}]]Denominator:[t(2 + k)]So, (frac{S}{R^2} = frac{2t[1 + 2sqrt{k} + sqrt{1 + 2k}]}{t(2 + k)} = frac{2[1 + 2sqrt{k} + sqrt{1 + 2k}]}{2 + k})Now, the expression depends only on (k). So, I can consider (f(k) = frac{2[1 + 2sqrt{k} + sqrt{1 + 2k}]}{2 + k}), and find its maximum for (k > 0).This seems more manageable. Let me compute the derivative of (f(k)) with respect to (k) and set it to zero to find critical points.First, let me write (f(k)) as:[f(k) = frac{2 + 4sqrt{k} + 2sqrt{1 + 2k}}{2 + k}]Let me denote the numerator as (N(k) = 2 + 4sqrt{k} + 2sqrt{1 + 2k}) and the denominator as (D(k) = 2 + k).Then, the derivative (f'(k)) is:[f'(k) = frac{N'(k)D(k) - N(k)D'(k)}{[D(k)]^2}]Compute (N'(k)):[N'(k) = frac{4}{2sqrt{k}} + frac{2 cdot 2}{2sqrt{1 + 2k}} = frac{2}{sqrt{k}} + frac{2}{sqrt{1 + 2k}}]Compute (D'(k)):[D'(k) = 1]So, putting it all together:[f'(k) = frac{left( frac{2}{sqrt{k}} + frac{2}{sqrt{1 + 2k}} right)(2 + k) - (2 + 4sqrt{k} + 2sqrt{1 + 2k})(1)}{(2 + k)^2}]This looks quite complicated, but maybe I can simplify the numerator.Let me compute each term:First term: (left( frac{2}{sqrt{k}} + frac{2}{sqrt{1 + 2k}} right)(2 + k))Second term: (-(2 + 4sqrt{k} + 2sqrt{1 + 2k}))Let me expand the first term:[frac{2(2 + k)}{sqrt{k}} + frac{2(2 + k)}{sqrt{1 + 2k}} = frac{4 + 2k}{sqrt{k}} + frac{4 + 2k}{sqrt{1 + 2k}}]Simplify:[frac{4}{sqrt{k}} + 2sqrt{k} + frac{4}{sqrt{1 + 2k}} + frac{2k}{sqrt{1 + 2k}}]So, the numerator becomes:[left( frac{4}{sqrt{k}} + 2sqrt{k} + frac{4}{sqrt{1 + 2k}} + frac{2k}{sqrt{1 + 2k}} right) - left( 2 + 4sqrt{k} + 2sqrt{1 + 2k} right)]Simplify term by term:1. (frac{4}{sqrt{k}} - 4sqrt{k})2. (2sqrt{k} - 2sqrt{k}) (Wait, no, let me check)Wait, actually, the terms are:- From the first part: (frac{4}{sqrt{k}} + 2sqrt{k} + frac{4}{sqrt{1 + 2k}} + frac{2k}{sqrt{1 + 2k}})- From the second part: (-2 - 4sqrt{k} - 2sqrt{1 + 2k})So, combining like terms:- (frac{4}{sqrt{k}} - 4sqrt{k})- (2sqrt{k} - 4sqrt{k} = -2sqrt{k})- (frac{4}{sqrt{1 + 2k}} - 2sqrt{1 + 2k})- (frac{2k}{sqrt{1 + 2k}} - 2)Wait, actually, let me group them properly:1. Terms with (frac{1}{sqrt{k}}): (frac{4}{sqrt{k}})2. Terms with (sqrt{k}): (2sqrt{k} - 4sqrt{k} = -2sqrt{k})3. Terms with (frac{1}{sqrt{1 + 2k}}): (frac{4}{sqrt{1 + 2k}})4. Terms with (sqrt{1 + 2k}): (-2sqrt{1 + 2k})5. Terms with (frac{k}{sqrt{1 + 2k}}): (frac{2k}{sqrt{1 + 2k}})6. Constant term: (-2)So, the numerator is:[frac{4}{sqrt{k}} - 2sqrt{k} + frac{4}{sqrt{1 + 2k}} - 2sqrt{1 + 2k} + frac{2k}{sqrt{1 + 2k}} - 2]This is getting really messy. Maybe there's a better approach.Alternatively, perhaps I can consider setting (k = 1), which corresponds to (a = b = c), and see if the derivative is zero there, indicating a maximum or minimum.Let me compute (f'(1)):First, compute each part at (k = 1):- (N'(1) = frac{2}{sqrt{1}} + frac{2}{sqrt{1 + 2 cdot 1}} = 2 + frac{2}{sqrt{3}})- (D(1) = 2 + 1 = 3)- (N(1) = 2 + 4sqrt{1} + 2sqrt{1 + 2 cdot 1} = 2 + 4 + 2sqrt{3} = 6 + 2sqrt{3})- (D'(1) = 1)So, numerator of (f'(1)):[(2 + frac{2}{sqrt{3}}) cdot 3 - (6 + 2sqrt{3}) cdot 1 = (6 + frac{6}{sqrt{3}}) - 6 - 2sqrt{3}][= 6 + 2sqrt{3} - 6 - 2sqrt{3} = 0]So, the derivative at (k = 1) is zero, indicating a critical point. Now, I need to check if this is a maximum or a minimum.Let me compute (f(k)) at (k = 1):[f(1) = frac{2[1 + 2sqrt{1} + sqrt{1 + 2 cdot 1}]}{2 + 1} = frac{2[1 + 2 + sqrt{3}]}{3} = frac{2(3 + sqrt{3})}{3} = 2 + frac{2sqrt{3}}{3}]Which is approximately (2 + 1.1547 = 3.1547).Now, let me check another value, say (k = 2):Compute (f(2)):[f(2) = frac{2[1 + 2sqrt{2} + sqrt{1 + 4}]}{2 + 2} = frac{2[1 + 2sqrt{2} + sqrt{5}]}{4} = frac{2 + 4sqrt{2} + 2sqrt{5}}{4} = frac{1 + 2sqrt{2} + sqrt{5}}{2}]Approximately, (1 + 2.828 + 2.236 = 6.064), divided by 2 is approximately (3.032), which is less than (3.1547).Now, check (k = 0.5):[f(0.5) = frac{2[1 + 2sqrt{0.5} + sqrt{1 + 1}]}{2 + 0.5} = frac{2[1 + sqrt{2} + sqrt{2}]}{2.5} = frac{2[1 + 2sqrt{2}]}{2.5} = frac{2 + 4sqrt{2}}{2.5} approx frac{2 + 5.656}{2.5} approx frac{7.656}{2.5} approx 3.0624]Still less than (3.1547).What about (k = 3):[f(3) = frac{2[1 + 2sqrt{3} + sqrt{1 + 6}]}{2 + 3} = frac{2[1 + 2sqrt{3} + sqrt{7}]}{5} approx frac{2[1 + 3.464 + 2.6458]}{5} approx frac{2[7.1098]}{5} approx frac{14.2196}{5} approx 2.8439]Even smaller.What about (k) approaching 0:As (k to 0), (f(k)) becomes:[frac{2[1 + 0 + sqrt{1 + 0}]}{2 + 0} = frac{2[1 + 1]}{2} = frac{4}{2} = 2]So, as (k) approaches 0, (f(k)) approaches 2.Similarly, as (k to infty):[f(k) approx frac{2[2sqrt{k} + sqrt{2k}]}{k} = frac{2[2sqrt{k} + sqrt{2}sqrt{k}]}{k} = frac{2sqrt{k}(2 + sqrt{2})}{k} = frac{2(2 + sqrt{2})}{sqrt{k}} to 0]So, (f(k)) approaches 0 as (k to infty).From these evaluations, it seems that (f(k)) reaches a maximum at (k = 1), which corresponds to (a = b = c). Therefore, the maximum value of (frac{S}{R^2}) is (2 + frac{2sqrt{3}}{3}), which simplifies to (frac{2}{3}(3 + sqrt{3})).To confirm, let me rationalize:[2 + frac{2sqrt{3}}{3} = frac{6}{3} + frac{2sqrt{3}}{3} = frac{6 + 2sqrt{3}}{3} = frac{2(3 + sqrt{3})}{3} = frac{2}{3}(3 + sqrt{3})]Yes, that's correct.Therefore, the maximum value of (frac{S}{R^2}) is (frac{2}{3}(3 + sqrt{3})).**Final Answer**The maximum value of (frac{S}{R^2}) is (boxed{frac{2}{3}(3+sqrt{3})}).

💡First, I need to determine the total area of the lot. The lot measures 120 feet by 60 feet, so the total area is 120 multiplied by 60, which equals 7200 square feet.Next, I'll calculate the area that will be covered with concrete. The concrete section is 40 feet by 40 feet, so the area is 40 multiplied by 40, resulting in 1600 square feet.To find the area that needs to be covered with grass seeds, I'll subtract the concrete area from the total area. That is, 7200 square feet minus 1600 square feet, which equals 5600 square feet.Each bag of grass seeds covers 56 square feet. To determine how many bags are needed, I'll divide the grassy area by the coverage of one bag. So, 5600 square feet divided by 56 square feet per bag equals 100 bags.Therefore, Amanda needs 100 bags of grass seeds to cover the new lot.

💡First, I need to rewrite the given equation of the circle ( x^2 + 2x + y^2 = 0 ) in the standard form. To do this, I'll complete the square for the ( x )-terms.Starting with the equation:[ x^2 + 2x + y^2 = 0 ]I'll complete the square for the ( x )-terms:[ x^2 + 2x = (x + 1)^2 - 1 ]Substituting this back into the original equation:[ (x + 1)^2 - 1 + y^2 = 0 ][ (x + 1)^2 + y^2 = 1 ]From this, I can see that the center of the circle is at ( (-1, 0) ) and the radius is ( 1 ).Next, I need to find the distance from the center of the circle ( (-1, 0) ) to the line ( x = 3 ). Since the line ( x = 3 ) is a vertical line, the distance from any point ( (x, y) ) to this line is simply the absolute difference between ( x ) and ( 3 ).Calculating the distance:[ text{Distance} = |3 - (-1)| = |3 + 1| = 4 ]Therefore, the distance from the center of the circle to the line ( x = 3 ) is ( 4 ).

💡Alright, so I've got this problem here with two inequalities involving a sequence of positive real numbers. Let me try to wrap my head around it step by step.First, the problem states that ( x_1, x_2, x_3, ldots ) is a sequence of positive real numbers, and ( s ) is the sum of these numbers up to ( x_n ). So, ( s = x_1 + x_2 + ldots + x_n ). Got that.Now, I need to show two things:1. ( frac{x_1}{s - x_1} + frac{x_2}{s - x_2} + ldots + frac{x_n}{s - x_n} geq frac{n}{n - 1} )2. ( frac{s - x_1}{x_1} + frac{s - x_2}{x_2} + ldots + frac{s - x_n}{x_n} geq n(n - 1) )Hmm, okay. Let's tackle them one by one.Starting with the second inequality because it seems a bit more straightforward. The expression is ( sum_{i=1}^n frac{s - x_i}{x_i} ). Let me rewrite that to see if I can simplify it.So, ( frac{s - x_i}{x_i} = frac{s}{x_i} - 1 ). That means the entire sum becomes:( sum_{i=1}^n left( frac{s}{x_i} - 1 right) = sum_{i=1}^n frac{s}{x_i} - sum_{i=1}^n 1 )Simplifying further, that's ( s sum_{i=1}^n frac{1}{x_i} - n ).Now, I remember there's an inequality called the harmonic mean-arithmetic mean (HM-AM) inequality which states that for positive real numbers, the harmonic mean is less than or equal to the arithmetic mean. Maybe I can apply that here.The harmonic mean of ( x_1, x_2, ldots, x_n ) is ( frac{n}{sum_{i=1}^n frac{1}{x_i}} ), and the arithmetic mean is ( frac{s}{n} ). So, according to HM-AM:( frac{n}{sum_{i=1}^n frac{1}{x_i}} leq frac{s}{n} )Rearranging this, I get:( sum_{i=1}^n frac{1}{x_i} geq frac{n^2}{s} )Okay, so plugging this back into my earlier expression:( s sum_{i=1}^n frac{1}{x_i} geq s cdot frac{n^2}{s} = n^2 )Therefore, the entire sum ( sum_{i=1}^n frac{s - x_i}{x_i} geq n^2 - n = n(n - 1) ). That seems to work out!Now, moving on to the first inequality: ( sum_{i=1}^n frac{x_i}{s - x_i} geq frac{n}{n - 1} ). This one looks a bit trickier.Let me see if I can manipulate this expression similarly. Let's rewrite ( frac{x_i}{s - x_i} ). Hmm, notice that ( frac{x_i}{s - x_i} = frac{s}{s - x_i} - 1 ). So, the sum becomes:( sum_{i=1}^n left( frac{s}{s - x_i} - 1 right) = sum_{i=1}^n frac{s}{s - x_i} - sum_{i=1}^n 1 )Which simplifies to ( s sum_{i=1}^n frac{1}{s - x_i} - n ).Again, maybe I can apply the HM-AM inequality here. Let's consider the terms ( s - x_i ). The harmonic mean of these terms would be ( frac{n}{sum_{i=1}^n frac{1}{s - x_i}} ), and the arithmetic mean is ( frac{s - x_1 + s - x_2 + ldots + s - x_n}{n} ).Calculating the arithmetic mean:( frac{(s - x_1) + (s - x_2) + ldots + (s - x_n)}{n} = frac{n s - (x_1 + x_2 + ldots + x_n)}{n} = frac{n s - s}{n} = frac{(n - 1)s}{n} )So, applying HM-AM:( frac{n}{sum_{i=1}^n frac{1}{s - x_i}} leq frac{(n - 1)s}{n} )Rearranging this gives:( sum_{i=1}^n frac{1}{s - x_i} geq frac{n^2}{(n - 1)s} )Plugging this back into our expression:( s sum_{i=1}^n frac{1}{s - x_i} geq s cdot frac{n^2}{(n - 1)s} = frac{n^2}{n - 1} )Therefore, the entire sum ( sum_{i=1}^n frac{x_i}{s - x_i} geq frac{n^2}{n - 1} - n = frac{n^2 - n(n - 1)}{n - 1} = frac{n}{n - 1} ). Perfect, that matches the inequality we needed to prove.So, both inequalities hold true using the HM-AM inequality. It's interesting how manipulating the expressions and applying the right inequality can lead to the desired result. I think I've got it!

💡Okay, so I have this problem in the Cartesian coordinate system. Let me try to understand it step by step. We have a point F at (0,1) and a line l: y = -1. There's a moving point P on the plane, and from P, we drop a perpendicular to line l, and the foot of that perpendicular is point Q. The condition given is that the dot product of vectors QP and QF is equal to the dot product of vectors FP and FQ. Part (I) asks for the equation of the trajectory C of the moving point P. Part (II) is about finding a point M on C such that the distance from M to the line y = x - 3 is the shortest, and also finding that shortest distance.Starting with part (I). Let me denote the coordinates of point P as (x, y). Since Q is the foot of the perpendicular from P to line l: y = -1, the coordinates of Q should be (x, -1). Because when you drop a perpendicular from P to the line y = -1, the x-coordinate remains the same, and the y-coordinate becomes -1.Now, let's write down the vectors involved. First, vector QP is from Q to P, so it's (x - x, y - (-1)) which simplifies to (0, y + 1). Similarly, vector QF is from Q to F, so it's (0 - x, 1 - (-1)) which is (-x, 2). Next, vector FP is from F to P, so it's (x - 0, y - 1) which is (x, y - 1). Vector FQ is from F to Q, so it's (x - 0, -1 - 1) which is (x, -2).The condition given is that the dot product of QP and QF equals the dot product of FP and FQ. So, let's compute both dot products.First, QP · QF: QP is (0, y + 1) and QF is (-x, 2). The dot product is (0)(-x) + (y + 1)(2) = 0 + 2(y + 1) = 2y + 2.Next, FP · FQ:FP is (x, y - 1) and FQ is (x, -2). The dot product is (x)(x) + (y - 1)(-2) = x² - 2(y - 1) = x² - 2y + 2.According to the condition, these two dot products are equal:2y + 2 = x² - 2y + 2.Let me simplify this equation:2y + 2 = x² - 2y + 2Subtract 2 from both sides:2y = x² - 2yAdd 2y to both sides:4y = x²So, the equation is x² = 4y. That's a parabola opening upwards with vertex at the origin. So, the trajectory C of point P is the parabola x² = 4y.Wait, let me double-check my steps to make sure I didn't make a mistake.1. Coordinates of Q: (x, -1) – correct.2. Vectors: - QP: (0, y + 1) – correct. - QF: (-x, 2) – correct. - FP: (x, y - 1) – correct. - FQ: (x, -2) – correct.3. Dot products: - QP · QF: 0*(-x) + (y + 1)*2 = 2(y + 1) – correct. - FP · FQ: x*x + (y - 1)*(-2) = x² - 2(y - 1) – correct.4. Setting them equal: 2(y + 1) = x² - 2(y - 1) - 2y + 2 = x² - 2y + 2 - Subtract 2: 2y = x² - 2y - Add 2y: 4y = x² - So, x² = 4y – correct.Okay, seems solid. So, part (I) is done. The equation is x² = 4y.Moving on to part (II). We need to find a point M on the trajectory C such that the distance from M to the line y = x - 3 is the shortest, and find that distance.I remember that the shortest distance from a point to a line is the perpendicular distance. So, to minimize the distance, the line connecting M and the given line y = x - 3 should be perpendicular to y = x - 3.Alternatively, another method is to parametrize the point M on the parabola and then express the distance as a function, then find its minimum using calculus.Let me try both approaches.First, let me parametrize point M on the parabola x² = 4y. So, any point M can be written as (x, x²/4).The distance from M(x, x²/4) to the line y = x - 3 can be found using the distance formula:Distance d = |Ax + By + C| / sqrt(A² + B²)First, let me write the line y = x - 3 in standard form: x - y - 3 = 0. So, A = 1, B = -1, C = -3.So, the distance is |1*x + (-1)*(x²/4) - 3| / sqrt(1 + 1) = |x - x²/4 - 3| / sqrt(2)Simplify the numerator:x - x²/4 - 3 = (-x²/4 + x - 3)So, d = | -x²/4 + x - 3 | / sqrt(2)To minimize d, we can minimize the numerator since sqrt(2) is a positive constant.Let me denote f(x) = -x²/4 + x - 3We need to find the minimum of |f(x)|. However, since the parabola f(x) is a downward opening parabola (because the coefficient of x² is negative), the expression inside the absolute value will have a maximum at its vertex.Wait, but we are dealing with the absolute value. So, the minimum of |f(x)| could be either at the vertex or where f(x) = 0.Alternatively, perhaps it's better to square the distance to make it easier, since the square will also be minimized at the same point.Let me compute d² = ( (-x²/4 + x - 3)^2 ) / 2To minimize d², we can minimize the numerator: (-x²/4 + x - 3)^2Let me set g(x) = (-x²/4 + x - 3)^2To find the minimum, take the derivative of g(x) with respect to x and set it to zero.First, compute g'(x):g'(x) = 2*(-x²/4 + x - 3)*(- (2x)/4 + 1 + 0) = 2*(-x²/4 + x - 3)*(-x/2 + 1)Set g'(x) = 0:Either (-x²/4 + x - 3) = 0 or (-x/2 + 1) = 0First, solve (-x/2 + 1) = 0:-x/2 + 1 = 0 => x = 2Now, solve (-x²/4 + x - 3) = 0:Multiply both sides by -4: x² - 4x + 12 = 0Discriminant D = 16 - 48 = -32 < 0, so no real solutions.Therefore, the only critical point is x = 2.So, the minimum occurs at x = 2.Compute y-coordinate: y = (2)^2 / 4 = 4 / 4 = 1. So, point M is (2, 1).Compute the distance d:d = | - (2)^2 /4 + 2 - 3 | / sqrt(2) = | -4/4 + 2 - 3 | / sqrt(2) = | -1 + 2 - 3 | / sqrt(2) = | -2 | / sqrt(2) = 2 / sqrt(2) = sqrt(2)So, the shortest distance is sqrt(2), achieved at point M(2,1).Alternatively, another method is to realize that the shortest distance occurs when the line connecting M and the given line is perpendicular. So, the slope of the given line y = x - 3 is 1, so the slope of the perpendicular line is -1.Therefore, the tangent to the parabola at point M should have a slope of -1.Given the parabola x² = 4y, its derivative is dy/dx = x/2.Set this equal to -1:x/2 = -1 => x = -2Wait, that would give a point at x = -2, but when I plug into the parabola, y = (-2)^2 /4 = 1, so point (-2, 1). But wait, earlier I found point (2,1). Hmm, conflicting results.Wait, perhaps I made a mistake here. Let me think again.If the tangent line at M is perpendicular to the given line y = x - 3, which has slope 1, then the tangent line should have slope -1.But for the parabola x² = 4y, the derivative is dy/dx = x/2. So, setting x/2 = -1 gives x = -2. So, point M would be (-2, 1). But earlier, using calculus, I found x = 2. So, which one is correct?Wait, perhaps my second method is flawed.Wait, no, actually, the distance is minimized when the line connecting M to the given line is perpendicular. So, the direction from M to the given line is along the direction of the normal vector of the line y = x - 3, which is (1, -1). So, the line connecting M to the given line should be in the direction (1, -1).But for the parabola, the tangent line at M has slope dy/dx = x/2. So, if the normal vector is (1, -1), then the tangent vector should be (1,1), because the tangent is perpendicular to the normal.Wait, maybe I confused the direction. Let me clarify.The given line is y = x - 3, which has a slope of 1. A line perpendicular to this would have a slope of -1.So, the line connecting M to the given line should have a slope of -1.But the tangent line at M on the parabola has slope x/2. So, if the line connecting M to the given line is perpendicular, then the tangent line at M should have slope -1.Wait, no, that's not necessarily the case. The line connecting M to the given line is the shortest distance, which is perpendicular. So, the direction of this line is (1, -1), as the normal vector.But the tangent line at M is a different concept. The tangent line is the direction of the parabola at M, which is given by the derivative.So, perhaps the two are not directly related. So, maybe my first method is correct, and the second method is not applicable here.Wait, but in my first method, I found that the minimum occurs at x = 2, giving point (2,1). Let me check the distance from (2,1) to the line y = x - 3.Compute the distance:|2 - 1 - 3| / sqrt(1 + 1) = | -2 | / sqrt(2) = 2 / sqrt(2) = sqrt(2). So, that's correct.If I take point (-2,1), distance is |-2 -1 -3| / sqrt(2) = |-6| / sqrt(2) = 6 / sqrt(2) = 3*sqrt(2), which is larger. So, indeed, (2,1) is the correct point.So, perhaps my second method was incorrect because I confused the tangent slope with the direction of the normal vector.Alternatively, another approach is to parametrize the line y = x - 3 and find the point on the parabola closest to it.But I think the first method using calculus is correct, giving point (2,1) with distance sqrt(2).Alternatively, another way is to realize that the shortest distance from a point to a line is achieved when the line segment is perpendicular. So, if we consider the line y = x - 3, any point M on the parabola x² = 4y, the shortest distance is along the line perpendicular to y = x - 3, which has slope -1.So, the line perpendicular to y = x - 3 passing through M(x, x²/4) is y - x²/4 = -1(x - x). Wait, that's not correct. Wait, the line perpendicular to y = x - 3 has slope -1, so the equation is y = -x + c, where c is a constant.This line should pass through M(x, x²/4), so x²/4 = -x + c => c = x²/4 + x.So, the equation is y = -x + x²/4 + x = x²/4.Wait, that can't be right because that would mean y = x²/4, which is the original parabola. So, perhaps this approach isn't helpful.Alternatively, the intersection point between the line y = x - 3 and the line perpendicular to it passing through M(x, x²/4) should give the closest point.So, the line perpendicular to y = x - 3 is y = -x + c, passing through M(x, x²/4). So, c = x²/4 + x.So, the intersection point is where y = x - 3 and y = -x + x²/4 + x.Set them equal:x - 3 = -x + x²/4 + xSimplify:x - 3 = x²/4Bring all terms to one side:x²/4 - x + 3 = 0Multiply by 4:x² - 4x + 12 = 0Discriminant D = 16 - 48 = -32 < 0, so no real solutions. That means the perpendicular line from M to y = x - 3 does not intersect y = x - 3, which is impossible because y = x - 3 is a straight line and the perpendicular should intersect it.Wait, that suggests an error in my approach. Maybe I made a mistake in setting up the equations.Wait, the line perpendicular to y = x - 3 passing through M(x, x²/4) is y = -x + c, and it should intersect y = x - 3 at some point (a, a - 3). So, substituting into y = -x + c:a - 3 = -a + c => c = 2a - 3But since the line passes through M(x, x²/4), we have:x²/4 = -x + c => x²/4 = -x + 2a - 3But also, the point (a, a - 3) lies on y = x - 3, so it's on both lines.But I'm getting confused here. Maybe it's better to stick with the first method where I found x = 2, y = 1, giving the shortest distance sqrt(2).Alternatively, another approach is to use Lagrange multipliers, but that might be overkill.Wait, let me try another way. The distance from M(x, x²/4) to the line y = x - 3 is |x - x²/4 - 3| / sqrt(2). To minimize this, we can minimize the numerator |x - x²/4 - 3|.Let me define f(x) = x - x²/4 - 3. We need to find the minimum of |f(x)|.Since f(x) is a quadratic function, it's a downward opening parabola (because the coefficient of x² is negative). The vertex of this parabola is at x = -b/(2a) where a = -1/4, b = 1.So, x = -1/(2*(-1/4)) = -1 / (-1/2) = 2.So, the vertex is at x = 2. Since it's a downward opening parabola, the maximum value of f(x) is at x = 2. But we are looking for the minimum of |f(x)|.So, the maximum of f(x) is at x = 2, f(2) = 2 - (4)/4 - 3 = 2 - 1 - 3 = -2. So, f(2) = -2, which is the maximum value because the parabola opens downward.Wait, but we are looking for the minimum of |f(x)|. So, the function f(x) = -x²/4 + x - 3 has its maximum at x = 2, which is -2. So, |f(x)| at x = 2 is 2.But since the parabola opens downward, as x moves away from 2, f(x) decreases towards negative infinity. So, |f(x)| increases as x moves away from 2 in either direction. Therefore, the minimum of |f(x)| occurs at the point where f(x) is closest to zero.But wait, the maximum of f(x) is -2, so the closest f(x) gets to zero is at x = 2, where f(x) = -2. So, |f(x)| is minimized at x = 2, giving |f(2)| = 2.Wait, but earlier, when I computed the distance, I got |f(2)| / sqrt(2) = 2 / sqrt(2) = sqrt(2). So, that's correct.But wait, if f(x) is maximum at x = 2, which is -2, then |f(x)| is minimized at x = 2, because f(x) is negative and getting more negative as x moves away from 2. So, the closest point is at x = 2.Therefore, the point M is (2,1), and the shortest distance is sqrt(2).So, both methods confirm that the shortest distance is sqrt(2) at point M(2,1).I think that's solid. So, part (II) is solved.

💡Okay, so I need to find the smallest three-digit number that has 3, 7, and 11 as factors. Hmm, let's see. First, I remember that if a number has multiple factors, it means that it's a multiple of each of those factors. So, if a number has 3, 7, and 11 as factors, it must be a multiple of all three.Since 3, 7, and 11 are all prime numbers, their least common multiple (LCM) would just be their product. That makes sense because prime numbers don't have any common factors other than 1. So, to find the LCM, I can just multiply them together.Let me calculate that: 3 times 7 is 21, and then 21 times 11 is 231. Okay, so 231 is the smallest number that is a multiple of 3, 7, and 11. But wait, is 231 a three-digit number? Yes, it is. So, that seems like a candidate.But I need to make sure that there isn't a smaller three-digit number that also has 3, 7, and 11 as factors. Since 231 is the product of these three primes, any smaller number would have to be a multiple of 231, but 231 is already a three-digit number. If I try to find a multiple of 231 that's smaller than 231, that would be less than 231, which would be a two-digit number or smaller, which doesn't fit the requirement.Let me double-check by dividing 231 by each of the factors to ensure it's divisible by all of them. 231 divided by 3 is 77, which is a whole number. 231 divided by 7 is 33, also a whole number. And 231 divided by 11 is 21, which is again a whole number. So, yes, 231 is divisible by 3, 7, and 11.Is there any other three-digit number smaller than 231 that could be divisible by all three? Let's think about it. If I take 3 times 7, that's 21, and then look for multiples of 21 that are also divisible by 11. The multiples of 21 are 21, 42, 63, 84, 105, 126, 147, 168, 189, 210, 231, and so on. Now, among these, which ones are divisible by 11?Let's check 21 divided by 11 is approximately 1.909, not a whole number. 42 divided by 11 is about 3.818, still not a whole number. 63 divided by 11 is around 5.727, nope. 84 divided by 11 is roughly 7.636, not whole. 105 divided by 11 is about 9.545, still not whole. 126 divided by 11 is approximately 11.454, not a whole number. 147 divided by 11 is around 13.364, nope. 168 divided by 11 is about 15.273, not whole. 189 divided by 11 is roughly 17.182, nope. 210 divided by 11 is approximately 19.09, still not a whole number. Finally, 231 divided by 11 is exactly 21, which is a whole number.So, the first multiple of 21 that is also divisible by 11 is 231. Therefore, 231 is indeed the smallest three-digit number that has 3, 7, and 11 as factors.Just to be thorough, let me consider if there's any other way to approach this problem. Maybe by finding the LCM of 3, 7, and 11 using prime factorization. Since all three numbers are prime, their LCM is simply their product, which again gives me 231. So, that confirms my earlier calculation.Another way could be to list the multiples of each number and find the smallest common multiple that's a three-digit number. Starting with multiples of 3: 3, 6, 9, 12, ..., up to 999. Multiples of 7: 7, 14, 21, 28, ..., up to 994. Multiples of 11: 11, 22, 33, 44, ..., up to 990. Looking for the smallest common multiple in all three lists that's a three-digit number. The first common multiple would be 231, as we've already determined.I think I've covered all possible angles here. I've calculated the product of the primes, checked divisibility, considered listing multiples, and even thought about the LCM method. All paths lead me to 231 as the smallest three-digit number with 3, 7, and 11 as factors. I don't see any smaller three-digit number fulfilling all these conditions.