💡Okay, so I have this geometry problem here. Let me try to visualize it first. There are two circles centered at points A and B, each with a radius of 3 units. Point O is the midpoint of segment AB, and the length OA is given as 3√2. So, since O is the midpoint, OB should also be 3√2, right? That means the entire length of AB is twice OA, which would be 6√2. Now, there are segments OC and OD that are tangent to the circles at A and B, respectively. So, OC is tangent to the circle at A, and OD is tangent to the circle at B. Also, there's a common tangent EF. The shaded region is ECODF, which I need to find the area of.Hmm, okay. Let me try to break this down. Maybe I can draw a diagram in my mind. Points A and B are centers of the circles, each with radius 3. O is the midpoint of AB, so OA = OB = 3√2. Since OC and OD are tangent to the circles, that means OC is tangent to the circle at A, and OD is tangent to the circle at B.I remember that the tangent to a circle is perpendicular to the radius at the point of tangency. So, OC is perpendicular to the radius at the point where it touches the circle centered at A, and similarly, OD is perpendicular to the radius at the point where it touches the circle centered at B.Wait, so if OC is tangent to the circle at A, then OC is perpendicular to OA. Similarly, OD is perpendicular to OB. So, triangles OAC and OBD are right triangles with right angles at C and D, respectively.Given that OA is 3√2 and the radius AC is 3, I can use the Pythagorean theorem to find the length of OC. Let me write that down:In triangle OAC,OA² = AC² + OC²(3√2)² = 3² + OC²9*2 = 9 + OC²18 = 9 + OC²So, OC² = 9Therefore, OC = 3Similarly, OD will also be 3 units because triangle OBD is congruent to triangle OAC.Okay, so OC and OD are each 3 units long. That helps.Now, the shaded region is ECODF. Let me figure out what that region consists of. It seems like it's bounded by the two tangent segments OC and OD, the common tangent EF, and the arcs EC and FD on the circles.Wait, actually, maybe ECODF is a region that includes parts of the two circles and the area between the two tangent lines and the common tangent EF. Hmm, I need to be precise here.Perhaps I should think about the figure as being made up of several parts: the two tangent segments OC and OD, the common tangent EF, and the areas between these lines and the circles. Maybe the shaded region is the area bounded by the two tangent lines OC and OD, the common tangent EF, and the two circles.Alternatively, maybe ECODF is a polygon with vertices at E, C, O, D, and F. So, it's a five-sided figure, but with some curved sides because of the circles.Wait, no. Since EF is a common tangent, it's a straight line. So, ECODF is a region bounded by the two tangent segments EC and FD, the common tangent EF, and the two radii OC and OD. Hmm, not sure.Wait, let me think again. The shaded region is ECODF. So, starting from E, going to C, then to O, then to D, then to F, and back to E. So, it's a quadrilateral with vertices at E, C, O, D, F, but since EF is a common tangent, maybe E and F are points on the circles?Wait, perhaps E and F are the points where the common tangent EF touches the circles. So, EF is a common external tangent to both circles. So, E is a point on the circle centered at A, and F is a point on the circle centered at B.So, ECODF is a region bounded by the tangent segments EC and FD, the common tangent EF, and the lines OC and OD. Hmm, maybe it's a sort of lens-shaped area, but I need to figure out exactly.Alternatively, perhaps it's the area between the two tangent lines OC and OD, the common tangent EF, and the arcs EC and FD. So, it's a region that is bounded by two straight lines (OC and OD), a common tangent (EF), and two circular arcs (EC and FD).Alternatively, maybe it's a quadrilateral with two straight sides (OC and OD) and two curved sides (EC and FD), connected by the common tangent EF.This is getting a bit confusing. Maybe I should try to compute the area step by step.First, let me consider the coordinates. Maybe assigning coordinates to points A, B, O, C, D, E, F will help me compute the area.Let me place point O at the origin (0,0). Since O is the midpoint of AB, and OA = 3√2, then point A is at (-3√2, 0) and point B is at (3√2, 0). Each circle has a radius of 3.Now, OC is tangent to the circle at A, and OD is tangent to the circle at B. Since OC is tangent to the circle at A, and OA is the radius, which is along the x-axis from (-3√2, 0) to (0,0). The tangent at A must be perpendicular to OA, so it's vertical? Wait, OA is along the x-axis, so the tangent at A would be vertical, but since OA is horizontal, the tangent would be vertical.Wait, no. If OA is along the x-axis, then the radius OA is horizontal, so the tangent at A would be vertical. Similarly, the tangent at B would be vertical as well.But in that case, OC and OD would be vertical lines. But point C is on the tangent at A, so if OA is from (-3√2, 0) to (0,0), then the tangent at A is vertical, so point C would be somewhere along the vertical line x = -3√2.Similarly, point D would be along the vertical line x = 3√2.But then OC is the segment from O(0,0) to C(-3√2, y), and OD is from O(0,0) to D(3√2, y). But since OC and OD are both length 3, as we found earlier, then the coordinates of C and D can be determined.Wait, let's compute that. Since OC is 3 units long and is a vertical line from A(-3√2, 0) upwards or downwards. But since O is at (0,0), and OC is a tangent from O to the circle at A, which is at (-3√2, 0). So, the tangent from O to the circle at A is a line from (0,0) to a point C on the circle.Wait, but earlier, I thought OC was tangent at A, but actually, OC is tangent to the circle centered at A, but not necessarily at point A. So, point C is a point on the circle centered at A where the tangent from O touches.Similarly, point D is a point on the circle centered at B where the tangent from O touches.So, in that case, OC is a tangent from O to the circle at A, and OD is a tangent from O to the circle at B.Given that, we can compute the coordinates of points C and D.Since O is at (0,0), and circle A is centered at (-3√2, 0) with radius 3, the tangent from O to circle A will touch the circle at point C.Similarly, the tangent from O to circle B will touch at point D.The formula for the length of a tangent from a point to a circle is sqrt(d² - r²), where d is the distance from the point to the center, and r is the radius.In this case, the distance from O to A is 3√2, and the radius is 3, so the length of the tangent OC is sqrt((3√2)^2 - 3^2) = sqrt(18 - 9) = sqrt(9) = 3. So, OC = 3, which matches what we found earlier.Now, to find the coordinates of point C, we can use the fact that the tangent from O to circle A is perpendicular to the radius AC at point C.So, the line AC is perpendicular to OC.Given that, let's denote point C as (x, y). Since C lies on the circle centered at A(-3√2, 0) with radius 3, we have:(x + 3√2)^2 + (y - 0)^2 = 3^2=> (x + 3√2)^2 + y^2 = 9Also, since OC is tangent to the circle at C, the vector from A to C is perpendicular to the vector from O to C.So, the vector AC is (x + 3√2, y), and the vector OC is (x, y). Their dot product should be zero:(x + 3√2)*x + y*y = 0=> x(x + 3√2) + y^2 = 0So, we have two equations:1. (x + 3√2)^2 + y^2 = 92. x(x + 3√2) + y^2 = 0Let me expand equation 1:(x + 3√2)^2 + y^2 = x^2 + 6√2 x + 18 + y^2 = 9So, x^2 + 6√2 x + 18 + y^2 = 9=> x^2 + 6√2 x + y^2 = -9But from equation 2:x^2 + 3√2 x + y^2 = 0Subtracting equation 2 from equation 1:(x^2 + 6√2 x + y^2) - (x^2 + 3√2 x + y^2) = -9 - 0=> 3√2 x = -9=> x = -9 / (3√2) = -3 / √2 = (-3√2)/2So, x = (-3√2)/2Now, plug this back into equation 2:x^2 + 3√2 x + y^2 = 0Compute x^2:(-3√2 / 2)^2 = (9*2)/4 = 18/4 = 9/2Compute 3√2 x:3√2 * (-3√2)/2 = (3*3*2)/2 * (-1) = (18)/2 * (-1) = -9So, equation 2 becomes:9/2 - 9 + y^2 = 0=> (9/2 - 18/2) + y^2 = 0=> (-9/2) + y^2 = 0=> y^2 = 9/2=> y = ± (3√2)/2Since the tangent can be above or below the x-axis, but in the context of the problem, I think we can assume it's above, so y = (3√2)/2.Therefore, point C is at (-3√2/2, 3√2/2).Similarly, point D will be symmetric with respect to the y-axis, so D is at (3√2/2, 3√2/2).Okay, so now we have coordinates for points C and D.Now, the common tangent EF. Since EF is a common tangent to both circles, it must touch each circle at one point. Let's denote E as the point on circle A and F as the point on circle B.Given that, EF is a common external tangent. The length of the common external tangent between two circles can be calculated, but maybe we can find the coordinates of E and F.Alternatively, since we have the coordinates of C and D, and we know that EF is a common tangent, perhaps EF is parallel to CD or something.Wait, let me think. Since OC and OD are both length 3, and points C and D are at (-3√2/2, 3√2/2) and (3√2/2, 3√2/2), respectively, the line CD is horizontal at y = 3√2/2.But EF is another common tangent, so it must be parallel to CD? Or maybe not.Wait, actually, the common external tangent EF will be parallel to the line connecting the centers A and B if the circles are equal in radius. But in this case, the circles are equal, so yes, the external tangent EF will be parallel to AB, which is along the x-axis.Wait, but EF is a common tangent, so it can be either above or below the line AB. Given that OC and OD are above AB, perhaps EF is the lower common tangent.Wait, but in the problem statement, it's just mentioned as a common tangent, so it could be either. But since the shaded region is ECODF, which includes points E, C, O, D, F, it's likely that EF is the lower tangent, and ECODF is the region bounded by the two upper tangents OC and OD, the lower tangent EF, and the arcs EC and FD.Alternatively, maybe EF is the upper tangent, but given the coordinates of C and D, which are above AB, perhaps EF is the lower tangent.Wait, let me clarify. If EF is a common tangent, it can be either above or below the line AB. Since the shaded region is ECODF, which includes points C and D above AB, and points E and F, which are likely on the lower parts of the circles, connected by the tangent EF.So, EF is the lower common tangent, and ECODF is the region bounded by the upper tangents OC and OD, the lower tangent EF, and the arcs EC and FD.So, to find the area of ECODF, perhaps I can compute the area of the quadrilateral ECODF, which is bounded by the two tangent lines OC and OD, the common tangent EF, and the two arcs EC and FD.Alternatively, maybe it's the area between the two tangent lines OC and OD, the common tangent EF, and the two circles.Wait, perhaps it's easier to compute the area as the area of the rectangle formed by ABFE minus the areas of the two triangles and the two sectors.Wait, let me think. If I consider the rectangle ABFE, where E and F are points on the circles, connected by the common tangent EF, and AE and BF are the radii perpendicular to EF.But in this case, since EF is a common tangent, AE and BF would be the radii to the points of tangency E and F, and would be perpendicular to EF.Given that, the length of AE and BF is 3, which is the radius.The distance between A and B is 6√2, as we found earlier. So, the rectangle ABFE would have length AB = 6√2 and width equal to the radius, which is 3. So, the area of ABFE would be 6√2 * 3 = 18√2.But wait, is ABFE a rectangle? Because AE and BF are both perpendicular to EF, so yes, ABFE would be a rectangle.Now, within this rectangle, the shaded region ECODF is the area between the two tangent lines OC and OD, the common tangent EF, and the two arcs EC and FD.So, perhaps the area of ECODF is equal to the area of the rectangle ABFE minus the areas of triangles ACO and BDO, and the areas of the sectors ECA and FDB.Wait, let me visualize this. The rectangle ABFE has area 18√2. The shaded region ECODF is inside this rectangle, but excludes the areas of triangles ACO and BDO, and the sectors ECA and FDB.So, to compute the area of ECODF, I can subtract the areas of triangles ACO and BDO, and the areas of sectors ECA and FDB from the area of the rectangle ABFE.So, let's compute each of these components.First, the area of rectangle ABFE is 18√2.Next, the area of triangle ACO. Since ACO is a right triangle with legs OA = 3√2 and AC = 3, but wait, no. Wait, OA is 3√2, but AC is the radius, which is 3. Wait, but in triangle ACO, OA is the hypotenuse, because OC is the tangent, which we found to be 3 units.Wait, no. Wait, in triangle ACO, OA is the distance from A to O, which is 3√2, AC is the radius, which is 3, and OC is the tangent, which is 3 units. So, triangle ACO is a right triangle with legs AC = 3 and OC = 3, and hypotenuse OA = 3√2.Wait, that makes sense because 3² + 3² = 9 + 9 = 18 = (3√2)².So, the area of triangle ACO is (1/2)*AC*OC = (1/2)*3*3 = 9/2.Similarly, the area of triangle BDO is also 9/2.So, total area of triangles ACO and BDO is 9/2 + 9/2 = 9.Next, the area of sectors ECA and FDB. Since the tangent lines OC and OD make a 45-degree angle with the radii AC and BD, respectively, because in triangle ACO, the legs are equal (both 3), so it's a 45-45-90 triangle. Therefore, the angle at A is 45 degrees.Therefore, the sector ECA is a 45-degree sector of the circle centered at A, with radius 3. Similarly, sector FDB is a 45-degree sector of the circle centered at B, with radius 3.The area of a sector is (θ/360)*π*r², where θ is the central angle in degrees.So, for each sector, θ = 45 degrees, r = 3.Area of one sector = (45/360)*π*3² = (1/8)*π*9 = (9/8)π.Since there are two such sectors, the total area is 2*(9/8)π = (9/4)π.Therefore, the total area to subtract from the rectangle ABFE is 9 (from the triangles) plus (9/4)π (from the sectors).So, the area of the shaded region ECODF is:Area of ABFE - (Area of triangles ACO and BDO + Area of sectors ECA and FDB)= 18√2 - (9 + (9/4)π)= 18√2 - 9 - (9/4)πSo, the final answer is 18√2 - 9 - (9/4)π.Wait, let me double-check my steps.1. Assigned coordinates correctly: O at (0,0), A at (-3√2, 0), B at (3√2, 0).2. Found points C and D correctly using the tangent formula and solving the equations.3. Determined that EF is the lower common tangent, forming a rectangle ABFE with area 18√2.4. Calculated areas of triangles ACO and BDO as 9/2 each, totaling 9.5. Calculated areas of sectors ECA and FDB as (9/8)π each, totaling (9/4)π.6. Subtracted these areas from the rectangle to get the shaded region.Everything seems to check out. So, the area of the shaded region ECODF is 18√2 - 9 - (9/4)π.

💡Alright, so I've got this problem about moving an "angle" shape on a grid board. The board is m by n, and both m and n are at least 2. The angle consists of three cells: a central cell and two adjacent side cells. It starts covering the top-left corner and its two neighboring cells. The goal is to move this angle so that its central cell ends up covering the bottom-right corner of the board. The only operation allowed is rotating the angle 90 degrees around one of its side cells, making sure the angle stays entirely on the board after each rotation.First, I need to visualize this. The angle starts at the top-left corner, so its central cell is at position (1,1), and the two side cells are either (1,2) and (2,1) or maybe (1,2) and (2,1). Wait, actually, since it's an angle, it's probably covering two adjacent cells in one direction. So, maybe it's covering (1,1), (1,2), and (2,1). That makes sense.Now, the operation is to rotate the angle 90 degrees around one of its side cells. So, if I choose one of the side cells, say (1,2), and rotate the angle 90 degrees around that cell, where does the central cell move? Let me think. If I rotate around (1,2), the central cell (1,1) would move to (2,2), and the other side cell (2,1) would move to (2,2) as well? Wait, that can't be right because two cells can't occupy the same spot. Maybe I'm misunderstanding the rotation.Perhaps when you rotate around a side cell, the entire angle pivots around that cell. So, if I rotate around (1,2), the central cell (1,1) would move to (2,2), and the other side cell (2,1) would move to (1,2). Hmm, that seems possible. So after the rotation, the angle would now be covering (1,2), (2,2), and (2,1). Wait, no, that's not an angle anymore because (1,2) and (2,1) are both adjacent to (2,2), but (1,2) and (2,1) are diagonal to each other. So maybe that's not the right way to visualize it.Alternatively, maybe the angle can rotate in such a way that it changes its orientation. So, if it was originally covering (1,1), (1,2), and (2,1), rotating around (1,2) would make it cover (1,2), (2,2), and (1,3). Is that possible? But then (1,3) might be outside the board if n is small. Wait, n is at least 2, but if n is 2, then (1,3) would be outside. So maybe the rotation is constrained to keep the entire angle within the board.This is getting a bit confusing. Maybe I should try to think of specific examples. Let's say m=3 and n=3. So, a 3x3 grid. The angle starts at (1,1), covering (1,1), (1,2), and (2,1). If I rotate around (1,2), where does it go? Maybe it would cover (1,2), (2,2), and (1,3). But (1,3) is within the 3x3 grid. Then, if I rotate around (2,2), maybe it can move further down. Hmm, but I'm not sure.Wait, maybe I should consider the parity of the grid. If I color the grid like a chessboard, alternating black and white squares, then the central cell starts on a specific color. Each rotation might change the position of the central cell in a way that alternates its color. So, if the starting position is, say, black, then after one rotation, it might be on white, and so on. The bottom-right corner is also a specific color, depending on whether m and n are odd or even.If both m and n are odd, then the bottom-right corner is the same color as the top-left corner. If either m or n is even, then the bottom-right corner is a different color. So, if the central cell alternates colors with each rotation, it can only reach the bottom-right corner if both m and n are odd. Otherwise, if one is even, the bottom-right corner is a different color, and the central cell can't reach it.Let me test this idea with a small grid. Take m=2 and n=2. The angle starts at (1,1), covering (1,1), (1,2), and (2,1). The bottom-right corner is (2,2). If I try to rotate around (1,2), the angle would try to cover (1,2), (2,2), and (1,3), but (1,3) is outside the grid. Similarly, rotating around (2,1) would try to cover (2,1), (2,2), and (3,1), which is also outside. So, in a 2x2 grid, it's impossible to move the angle to cover (2,2) as the central cell.What about a 3x3 grid? Starting at (1,1), if I rotate around (1,2), the angle moves to (1,2), (2,2), and (1,3). Then, rotating around (2,2), it could move to (2,2), (3,2), and (2,3). Then, rotating around (2,3), it could move to (2,3), (3,3), and (2,2). But (3,3) is the bottom-right corner, but it's only covered as a side cell, not the central cell. Hmm, maybe I need another rotation.Alternatively, maybe I can move the angle down first. From (1,1), rotate around (2,1) to cover (2,1), (2,2), and (3,1). Then, rotate around (2,2) to cover (2,2), (3,2), and (2,3). Then, rotate around (3,2) to cover (3,2), (3,3), and (2,2). Again, (3,3) is a side cell, not the central cell. It seems tricky, but maybe with more rotations, I can get the central cell to (3,3).Wait, maybe I'm overcomplicating it. The key insight is the parity. If both m and n are odd, the bottom-right corner is the same color as the top-left corner, so it's possible. If either is even, it's not. So, the answer should be that it's possible if and only if both m and n are odd.I think that makes sense. So, the final answer is that it's possible when both m and n are odd.

💡Okay, so I have this inequality to prove. It says that if I have two sequences of positive real numbers, ( a_1, a_2, ldots, a_n ) and ( b_1, b_2, ldots, b_n ), and the sums of these sequences are equal, that is, ( sum_{k=1}^{n} a_k = sum_{k=1}^{n} b_k ), then the sum ( sum_{k=1}^{n} frac{a_k^2}{a_k + b_k} ) is at least half of the sum of the ( a_k )'s. Hmm, let me write that down to make it clearer:Given:[ sum_{k=1}^{n} a_k = sum_{k=1}^{n} b_k ]Prove:[ sum_{k=1}^{n} frac{a_k^2}{a_k + b_k} geqslant frac{1}{2} sum_{k=1}^{n} a_k ]Alright, so I need to show that this particular sum involving ( a_k ) and ( b_k ) is at least half of the total sum of ( a_k )'s. Let me think about what techniques I know for proving inequalities. Maybe Cauchy-Schwarz? Or perhaps the AM-GM inequality? Or maybe some other method like rearrangement or Titu's lemma? Wait, Titu's lemma is a form of the Cauchy-Schwarz inequality, right? It might be applicable here because we have fractions with squares in the numerator. Let me recall Titu's lemma: it states that for positive real numbers, ( sum frac{x_k^2}{y_k} geq frac{(sum x_k)^2}{sum y_k} ). But in our case, the denominators are ( a_k + b_k ), which are related to both ( a_k ) and ( b_k ). So maybe I can apply Titu's lemma here. Let me try that.If I set ( x_k = a_k ) and ( y_k = a_k + b_k ), then according to Titu's lemma:[ sum_{k=1}^{n} frac{a_k^2}{a_k + b_k} geq frac{left( sum_{k=1}^{n} a_k right)^2}{sum_{k=1}^{n} (a_k + b_k)} ]Okay, that seems promising. Let me compute the denominator in the right-hand side. Since ( sum a_k = sum b_k ), let's denote ( S = sum a_k = sum b_k ). Then, the denominator becomes ( sum (a_k + b_k) = sum a_k + sum b_k = S + S = 2S ).So, substituting back into the inequality from Titu's lemma:[ sum_{k=1}^{n} frac{a_k^2}{a_k + b_k} geq frac{S^2}{2S} = frac{S}{2} ]Which is exactly what we needed to prove! So, that seems straightforward. But wait, let me double-check if I applied Titu's lemma correctly.Yes, Titu's lemma is applicable here because all terms are positive. The ( a_k ) and ( b_k ) are positive real numbers, so ( a_k + b_k ) is positive, and ( a_k^2 ) is positive. Therefore, all the conditions for Titu's lemma are satisfied.Alternatively, maybe I can approach this problem using another method to verify. Let's consider the function ( f(x) = frac{x^2}{x + y} ) where ( y > 0 ). Is this function convex or concave? If it's convex, maybe I can apply Jensen's inequality.Taking the second derivative of ( f(x) ) with respect to ( x ):First derivative:[ f'(x) = frac{2x(x + y) - x^2}{(x + y)^2} = frac{2x^2 + 2xy - x^2}{(x + y)^2} = frac{x^2 + 2xy}{(x + y)^2} ]Second derivative:[ f''(x) = frac{(2x + 2y)(x + y)^2 - (x^2 + 2xy)(2(x + y))}{(x + y)^4} ]This looks complicated, but let me compute it step by step.Numerator:First term: ( (2x + 2y)(x + y)^2 )Second term: ( - (x^2 + 2xy)(2(x + y)) )Let me factor out ( 2(x + y) ) from both terms:[ 2(x + y)[(x + y)(x + y) - (x^2 + 2xy)] ][ = 2(x + y)[(x + y)^2 - x^2 - 2xy] ][ = 2(x + y)[x^2 + 2xy + y^2 - x^2 - 2xy] ][ = 2(x + y)[y^2] ][ = 2y^2(x + y) ]So the second derivative is:[ f''(x) = frac{2y^2(x + y)}{(x + y)^4} = frac{2y^2}{(x + y)^3} ]Since ( y > 0 ) and ( x + y > 0 ), ( f''(x) > 0 ). Therefore, ( f(x) ) is convex in ( x ). Given that ( f ) is convex, by Jensen's inequality, we have:[ frac{1}{n} sum_{k=1}^{n} frac{a_k^2}{a_k + b_k} geq frac{left( frac{1}{n} sum_{k=1}^{n} a_k right)^2}{frac{1}{n} sum_{k=1}^{n} (a_k + b_k)} ]Multiplying both sides by ( n ):[ sum_{k=1}^{n} frac{a_k^2}{a_k + b_k} geq frac{left( sum_{k=1}^{n} a_k right)^2}{sum_{k=1}^{n} (a_k + b_k)} ]Which is the same as before, leading to:[ sum_{k=1}^{n} frac{a_k^2}{a_k + b_k} geq frac{S^2}{2S} = frac{S}{2} ]So, whether I use Titu's lemma or Jensen's inequality, I end up with the same result. That gives me more confidence that the proof is correct.Alternatively, maybe I can approach this by considering the difference between the sum and the desired bound. Let me define:[ M = sum_{k=1}^{n} frac{a_k^2}{a_k + b_k} ]And we need to show that ( M geq frac{1}{2} S ), where ( S = sum a_k ).Another approach might be to consider pairing each term ( frac{a_k^2}{a_k + b_k} ) with ( frac{b_k^2}{a_k + b_k} ). Let me define:[ N = sum_{k=1}^{n} frac{b_k^2}{a_k + b_k} ]Then, notice that ( M + N = sum_{k=1}^{n} frac{a_k^2 + b_k^2}{a_k + b_k} ).Also, ( M - N = sum_{k=1}^{n} frac{a_k^2 - b_k^2}{a_k + b_k} = sum_{k=1}^{n} (a_k - b_k) ).But since ( sum a_k = sum b_k ), this implies ( sum (a_k - b_k) = 0 ). Therefore, ( M - N = 0 ), so ( M = N ).Therefore, ( M + N = 2M ). So, ( 2M = sum_{k=1}^{n} frac{a_k^2 + b_k^2}{a_k + b_k} ).Now, I can use the inequality ( frac{a_k^2 + b_k^2}{a_k + b_k} geq frac{a_k + b_k}{2} ). Let me verify this.Starting with ( a_k^2 + b_k^2 geq frac{(a_k + b_k)^2}{2} ) by the Cauchy-Schwarz inequality or by expanding ( (a_k - b_k)^2 geq 0 ):[ a_k^2 + b_k^2 geq frac{(a_k + b_k)^2}{2} ]Dividing both sides by ( a_k + b_k ):[ frac{a_k^2 + b_k^2}{a_k + b_k} geq frac{a_k + b_k}{2} ]Yes, that's correct. So, applying this to each term in the sum:[ sum_{k=1}^{n} frac{a_k^2 + b_k^2}{a_k + b_k} geq sum_{k=1}^{n} frac{a_k + b_k}{2} = frac{1}{2} sum_{k=1}^{n} (a_k + b_k) ]But ( sum (a_k + b_k) = 2S ), since ( sum a_k = sum b_k = S ). Therefore:[ sum_{k=1}^{n} frac{a_k^2 + b_k^2}{a_k + b_k} geq frac{1}{2} times 2S = S ]But we know that ( 2M = sum frac{a_k^2 + b_k^2}{a_k + b_k} geq S ), so:[ 2M geq S implies M geq frac{S}{2} ]Which is exactly what we needed to prove. So, this method also works and perhaps gives a different perspective on why the inequality holds.I think I've approached this problem from a couple of different angles now: using Titu's lemma, using Jensen's inequality, and considering the sum and difference of ( M ) and ( N ). All of them lead to the same conclusion, which is reassuring.Just to make sure I haven't missed anything, let me consider a simple case where ( n = 1 ). If ( n = 1 ), then ( a_1 = b_1 ) because their sums are equal. So, ( frac{a_1^2}{a_1 + b_1} = frac{a_1^2}{2a_1} = frac{a_1}{2} ), which matches the right-hand side. So, equality holds when ( n = 1 ).What about ( n = 2 )? Let me take ( a_1 = 3 ), ( a_2 = 1 ), so ( S = 4 ). Then ( b_1 + b_2 = 4 ). Let me choose ( b_1 = 1 ), ( b_2 = 3 ).Compute ( M = frac{3^2}{3 + 1} + frac{1^2}{1 + 3} = frac{9}{4} + frac{1}{4} = frac{10}{4} = 2.5 ). The right-hand side is ( frac{1}{2} times 4 = 2 ). So, ( 2.5 geq 2 ), which holds.Another example: ( a_1 = 2 ), ( a_2 = 2 ), so ( S = 4 ). Let ( b_1 = 1 ), ( b_2 = 3 ). Then, ( M = frac{4}{3} + frac{4}{5} = frac{20}{15} + frac{12}{15} = frac{32}{15} approx 2.13 ), which is still greater than 2.Wait, what if ( a_k = b_k ) for all ( k )? Then, each term becomes ( frac{a_k^2}{2a_k} = frac{a_k}{2} ), so the sum ( M = frac{1}{2} sum a_k ), which is exactly the bound. So, equality holds when all ( a_k = b_k ).Another case: suppose ( a_1 = 4 ), ( a_2 = 0 ), but wait, the problem states that all ( a_k ) and ( b_k ) are positive real numbers. So, ( a_k ) and ( b_k ) must be positive, so I can't have ( a_2 = 0 ). Let me adjust: ( a_1 = 3 ), ( a_2 = 1 ); ( b_1 = 2 ), ( b_2 = 2 ). Then, ( M = frac{9}{5} + frac{1}{3} = frac{27}{15} + frac{5}{15} = frac{32}{15} approx 2.13 ), which is still greater than 2.So, in all these examples, the inequality holds, and equality is achieved when all ( a_k = b_k ). That makes sense because if ( a_k = b_k ), then each term in the sum ( M ) is exactly half of ( a_k ), so the total sum is half of ( S ).I think I've covered multiple approaches and checked with specific examples. All of them support the inequality, so I feel confident that the proof is correct.**Final Answer**boxed{sum_{k=1}^{n} frac{a_{k}^{2}}{a_{k}+b_{k}} geqslant frac{1}{2} sum_{k=1}^{n} a_{k}}

💡Alright, so I have this problem where I need to find a 3x3 matrix P such that when I multiply it by another 3x3 matrix Q, it does two things: it swaps the second and third rows of Q, and it triples the first row of Q. The example given is that if Q is a matrix with entries a, b, c in the first row; d, e, f in the second row; and g, h, i in the third row, then PQ should result in a matrix where the first row is 3a, 3b, 3c; the second row is g, h, i; and the third row is d, e, f. Okay, so I need to figure out what matrix P would do that. I remember that when you multiply two matrices, the resulting matrix's rows are combinations of the original matrix's rows based on the rows of the first matrix. So, if I think about P as a transformation matrix, each row of P will determine how the rows of Q are combined to form the rows of PQ.Let me denote the rows of P as r1, r2, r3. So, P is:[ r1 ][ r2 ][ r3 ]And Q is:[ a b c ][ d e f ][ g h i ]When I multiply P and Q, the first row of PQ will be r1 multiplied by each column of Q. Similarly, the second row of PQ will be r2 multiplied by each column of Q, and the third row will be r3 multiplied by each column of Q.Given that the first row of PQ is 3a, 3b, 3c, that means r1 must be such that when it's multiplied by the columns of Q, it gives three times the first row of Q. So, if I think about how matrix multiplication works, the first element of the first row of PQ is r1 · [a, d, g], which should equal 3a. Similarly, the second element is r1 · [b, e, h] = 3b, and the third element is r1 · [c, f, i] = 3c.So, for r1 · [a, d, g] = 3a, that suggests that the first element of r1 is 3, and the other elements are 0 because if r1 was [3, 0, 0], then 3*a + 0*d + 0*g = 3a, which is correct. Similarly, for the second and third elements, since r1 is [3, 0, 0], multiplying by [b, e, h] gives 3b, and multiplying by [c, f, i] gives 3c. So, r1 must be [3, 0, 0].Now, moving on to the second row of PQ, which is supposed to be the third row of Q, that is [g, h, i]. So, r2 multiplied by each column of Q should give g, h, i. Let's think about how that would work. The first element of the second row of PQ is r2 · [a, d, g] = g. Similarly, the second element is r2 · [b, e, h] = h, and the third element is r2 · [c, f, i] = i.So, if I set r2 to be [0, 0, 1], then multiplying by the first column [a, d, g] gives 0*a + 0*d + 1*g = g, which is correct. Similarly, multiplying by the second column [b, e, h] gives 0*b + 0*e + 1*h = h, and multiplying by the third column [c, f, i] gives 0*c + 0*f + 1*i = i. So, r2 should be [0, 0, 1].Finally, the third row of PQ is supposed to be the second row of Q, which is [d, e, f]. So, r3 multiplied by each column of Q should give d, e, f. Let's see. The first element is r3 · [a, d, g] = d. The second element is r3 · [b, e, h] = e, and the third element is r3 · [c, f, i] = f.If I set r3 to be [0, 1, 0], then multiplying by the first column [a, d, g] gives 0*a + 1*d + 0*g = d, which is correct. Similarly, multiplying by the second column [b, e, h] gives 0*b + 1*e + 0*h = e, and multiplying by the third column [c, f, i] gives 0*c + 1*f + 0*i = f. So, r3 should be [0, 1, 0].Putting it all together, the matrix P should have rows [3, 0, 0], [0, 0, 1], and [0, 1, 0]. So, written out, P is:[ 3 0 0 ][ 0 0 1 ][ 0 1 0 ]Let me double-check to make sure this works. If I multiply P by Q, let's see:First row of P is [3, 0, 0], so multiplying by Q's columns:- First element: 3*a + 0*d + 0*g = 3a- Second element: 3*b + 0*e + 0*h = 3b- Third element: 3*c + 0*f + 0*i = 3cThat's correct for the first row.Second row of P is [0, 0, 1], so multiplying by Q's columns:- First element: 0*a + 0*d + 1*g = g- Second element: 0*b + 0*e + 1*h = h- Third element: 0*c + 0*f + 1*i = iThat's correct for the second row.Third row of P is [0, 1, 0], so multiplying by Q's columns:- First element: 0*a + 1*d + 0*g = d- Second element: 0*b + 1*e + 0*h = e- Third element: 0*c + 1*f + 0*i = fThat's correct for the third row.So, yes, this matrix P should work as intended. It triples the first row and swaps the second and third rows of Q when multiplied from the left.

💡Okay, so I have this problem here: Given the function ( fleft( frac{x+1}{2} right) = x^2 - 2x ), I need to find the range of the function ( f(x) ) on the interval ([-1, 2)). The options are A: ([-1,15]), B: ([-1,3)), C: ([-3,3)), and D: ((3,15]). Hmm, let me try to figure this out step by step.First, I think I need to find an expression for ( f(x) ) in terms of ( x ) so that I can analyze its range. The given function is ( fleft( frac{x+1}{2} right) = x^2 - 2x ). Maybe I can make a substitution to simplify this. Let me let ( t = frac{x+1}{2} ). That way, I can express ( x ) in terms of ( t ) and substitute back into the equation.So, if ( t = frac{x+1}{2} ), then solving for ( x ) gives me ( x = 2t - 1 ). Okay, that seems straightforward. Now, I can substitute this expression for ( x ) back into the original function. That is, ( f(t) = (2t - 1)^2 - 2(2t - 1) ). Let me expand that out.First, expanding ( (2t - 1)^2 ): that would be ( 4t^2 - 4t + 1 ). Then, subtracting ( 2(2t - 1) ): which is ( 4t - 2 ). So putting it all together, ( f(t) = 4t^2 - 4t + 1 - 4t + 2 ). Combining like terms, the ( -4t ) and ( -4t ) give me ( -8t ), and the constants ( 1 + 2 ) give me ( 3 ). So, ( f(t) = 4t^2 - 8t + 3 ).Wait, so that means ( f(x) = 4x^2 - 8x + 3 ). Hmm, that's a quadratic function. Quadratic functions have a parabola shape, so their range depends on whether they open upwards or downwards. Since the coefficient of ( x^2 ) is positive (4), this parabola opens upwards, meaning it has a minimum point and the range will be from that minimum to infinity. But in this case, we're only concerned with the range on the interval ([-1, 2)). So, I need to find the minimum and maximum values of ( f(x) ) on that interval.Let me write down the function again: ( f(x) = 4x^2 - 8x + 3 ). To find the minimum, I can complete the square or find the vertex. Completing the square might be a good idea here. Let me try that.Starting with ( f(x) = 4x^2 - 8x + 3 ). Factor out the 4 from the first two terms: ( f(x) = 4(x^2 - 2x) + 3 ). Now, to complete the square inside the parentheses, I take the coefficient of ( x ), which is -2, divide it by 2 to get -1, and then square it to get 1. So, I add and subtract 1 inside the parentheses: ( f(x) = 4[(x^2 - 2x + 1 - 1)] + 3 ). That simplifies to ( f(x) = 4[(x - 1)^2 - 1] + 3 ). Distribute the 4: ( f(x) = 4(x - 1)^2 - 4 + 3 ), which simplifies to ( f(x) = 4(x - 1)^2 - 1 ).Okay, so in vertex form, the function is ( f(x) = 4(x - 1)^2 - 1 ). The vertex is at ( (1, -1) ). Since the parabola opens upwards, the minimum value of ( f(x) ) is -1, occurring at ( x = 1 ). Now, I need to find the maximum value on the interval ([-1, 2)). Since the parabola opens upwards, the maximum will occur at one of the endpoints of the interval.Let me evaluate ( f(x) ) at the endpoints. First, at ( x = -1 ): ( f(-1) = 4(-1 - 1)^2 - 1 = 4(-2)^2 - 1 = 4*4 - 1 = 16 - 1 = 15 ). Next, at ( x = 2 ): ( f(2) = 4(2 - 1)^2 - 1 = 4(1)^2 - 1 = 4 - 1 = 3 ). But wait, the interval is ([-1, 2)), which means it includes -1 but does not include 2. So, the function approaches 3 as ( x ) approaches 2 from the left, but doesn't actually reach 3.Therefore, the range of ( f(x) ) on ([-1, 2)) is from the minimum value of -1 up to, but not including, 3. So, the range is ([-1, 3)). Hmm, looking back at the options, that would be option B.Wait, but hold on a second. Let me double-check my calculations because sometimes I might make a mistake. Let me recalculate ( f(-1) ) and ( f(2) ).Calculating ( f(-1) ) again: ( f(-1) = 4(-1)^2 - 8(-1) + 3 = 4*1 + 8 + 3 = 4 + 8 + 3 = 15 ). That's correct. Now, ( f(2) = 4(2)^2 - 8(2) + 3 = 16 - 16 + 3 = 3 ). So, that's also correct. But since the interval is up to 2, not including 2, the maximum value isn't actually attained; it just approaches 3. So, the range is all values from -1 up to, but not including, 3. So, ([-1, 3)).Wait, but looking back at the options, option A is ([-1,15]), which is a much larger range. Did I do something wrong here? Let me think again.Wait, no. The function ( f(x) = 4x^2 - 8x + 3 ) is defined on the interval ([-1, 2)). The vertex is at ( x = 1 ), which is within this interval, so the minimum is indeed -1. The maximum occurs at the endpoints. At ( x = -1 ), it's 15, and as ( x ) approaches 2, it approaches 3. So, the function's range should include all values from -1 up to 15, but wait, no, because the interval is only from -1 to 2, and the function is continuous on this interval.Wait, hold on. Maybe I confused the domain of ( f(x) ) with the domain of the original function. Let me clarify.The original function is given as ( fleft( frac{x+1}{2} right) = x^2 - 2x ). So, the input to ( f ) is ( frac{x+1}{2} ). The domain of this expression is all real numbers, but we're specifically looking at the range of ( f(x) ) on the interval ([-1, 2)). So, ( x ) is in ([-1, 2)), but what does that mean for the input to ( f )?Wait, no. The function ( f ) is defined such that ( fleft( frac{x+1}{2} right) = x^2 - 2x ). So, if ( x ) is in ([-1, 2)), then ( frac{x+1}{2} ) is in (left[ frac{-1 + 1}{2}, frac{2 + 1}{2} right)), which is ([0, 1.5)). So, actually, the input to ( f ) is in ([0, 1.5)), not ([-1, 2)). Wait, that might be where I went wrong.So, maybe I misinterpreted the interval. The problem says the range of ( f(x) ) on the interval ([-1, 2)). Hmm, so is the interval referring to the domain of ( f ) or the domain of the original function? Let me read the problem again."Given the function ( fleft( frac{x+1}{2} right) = x^2 - 2x ), then the range of the function ( f(x) ) on the interval ([-1,2)) is..."So, it's the range of ( f(x) ) on the interval ([-1,2)). So, ( x ) is in ([-1,2)), and we need to find the range of ( f(x) ) for ( x ) in that interval.But earlier, I found that ( f(x) = 4x^2 - 8x + 3 ), which is a quadratic function. So, on the interval ([-1,2)), the function ( f(x) ) will have a minimum at the vertex and maximums at the endpoints.Wait, but earlier, I calculated ( f(-1) = 15 ) and ( f(2) = 3 ). But since the interval is ([-1,2)), 2 is excluded, so the function approaches 3 as ( x ) approaches 2 from the left. So, the maximum value is 15 at ( x = -1 ), and the minimum is -1 at ( x = 1 ). So, the range is from -1 to 15, including both endpoints because ( x = -1 ) is included and ( x = 1 ) is included in the interval.Wait, but earlier, I thought the range was ([-1, 3)), but that was based on a misunderstanding of the interval. Let me clarify.If ( x ) is in ([-1,2)), then ( f(x) ) is defined as ( 4x^2 - 8x + 3 ). The vertex is at ( x = 1 ), which is within the interval, so the minimum value is -1. The maximum value occurs at the endpoints. At ( x = -1 ), ( f(-1) = 15 ), and as ( x ) approaches 2, ( f(x) ) approaches 3. Since 2 is not included, 3 is not attained, but 15 is attained at ( x = -1 ). So, the range should be from -1 to 15, including 15 but not including 3? Wait, that doesn't make sense because 15 is attained, but 3 is not.Wait, no. The function ( f(x) ) on the interval ([-1,2)) takes all values from its minimum at -1 up to its maximum at 15, but since the interval doesn't include 2, the function doesn't reach 3. Wait, but 3 is less than 15, so that doesn't make sense.Wait, I'm getting confused. Let me plot the function or think about it differently. The function ( f(x) = 4x^2 - 8x + 3 ) is a parabola opening upwards with vertex at (1, -1). So, on the interval ([-1,2)), the function starts at ( x = -1 ) with ( f(-1) = 15 ), goes down to the minimum at ( x = 1 ) with ( f(1) = -1 ), and then increases again towards ( x = 2 ), where it would be 3, but since 2 is excluded, it approaches 3 but doesn't reach it.Therefore, the function's range on ([-1,2)) is from -1 up to 15, but wait, no. Because as ( x ) increases from -1 to 1, ( f(x) ) decreases from 15 to -1, and then as ( x ) increases from 1 to 2, ( f(x) ) increases from -1 to 3. So, the function covers all values from -1 up to 15, but actually, when ( x ) is between -1 and 1, ( f(x) ) goes from 15 down to -1, and when ( x ) is between 1 and 2, ( f(x) ) goes from -1 up to 3. So, the overall range is from -1 to 15, because the function reaches 15 at ( x = -1 ) and goes down to -1, then back up to 3, but 15 is the highest value.Wait, but 15 is only at ( x = -1 ). So, the range is all values from -1 up to 15, because the function can take any value between -1 and 15 on that interval. So, the range is ([-1,15]), which is option A.But earlier, I thought it was ([-1,3)), but that was because I was confused about the interval. So, actually, the correct range is ([-1,15]), which is option A.Wait, but let me double-check. If ( x ) is in ([-1,2)), then ( f(x) ) is ( 4x^2 - 8x + 3 ). Let me plug in some values:At ( x = -1 ): ( f(-1) = 4*(-1)^2 - 8*(-1) + 3 = 4 + 8 + 3 = 15 ).At ( x = 0 ): ( f(0) = 0 - 0 + 3 = 3 ).At ( x = 1 ): ( f(1) = 4 - 8 + 3 = -1 ).At ( x = 1.5 ): ( f(1.5) = 4*(2.25) - 8*(1.5) + 3 = 9 - 12 + 3 = 0 ).So, as ( x ) increases from -1 to 1, ( f(x) ) decreases from 15 to -1, and then as ( x ) increases from 1 to 2, ( f(x) ) increases from -1 to 3. So, the function covers all values from -1 up to 15, because at ( x = -1 ), it's 15, and it goes down to -1, then back up to 3. So, the maximum value is 15, and the minimum is -1. Therefore, the range is ([-1,15]), which is option A.Wait, but earlier I thought the interval was for the input to ( f ), but no, the problem says the range of ( f(x) ) on the interval ([-1,2)). So, ( x ) is in ([-1,2)), and ( f(x) ) is defined as ( 4x^2 - 8x + 3 ). So, the range is indeed ([-1,15]), because the function reaches 15 at ( x = -1 ) and -1 at ( x = 1 ), and all values in between are covered.Therefore, the correct answer is A: ([-1,15]).

💡Alright, so I have this problem where I need to figure out how many ways I can arrange 4 different math books and 6 different history books on my bookshelf. The constraints are that there must be a math book on both ends of the shelf, and a history book must be in the middle position. Okay, let's break this down step by step.First, I need to visualize the bookshelf. It has a certain number of positions where the books can be placed. Since I have 4 math books and 6 history books, that's a total of 10 books. So, the bookshelf has 10 positions. Now, the first thing I need to do is handle the constraints.The first constraint is that there must be a math book on both ends. That means the first position and the last position (positions 1 and 10) must be occupied by math books. I have 4 different math books, so I need to choose which two of these will go on the ends.For the first position, I have 4 choices because there are 4 different math books. Once I've placed a math book on the first position, I can't use that same book again for the last position. So, for the last position, I have 3 remaining choices.Okay, so for the two end positions, I have 4 choices for the first and 3 choices for the last, which gives me 4 * 3 = 12 possible ways to arrange the math books on both ends.Next, there's another constraint: a history book must occupy the middle position. Since there are 10 positions in total, the middle position would be the 5th position. I have 6 different history books, so I need to choose one of these to place in the 5th position.For the middle position, I have 6 choices because there are 6 different history books. So, after placing a history book in the middle, I have 5 history books left.Now, I've already placed 2 math books on the ends and 1 history book in the middle. That accounts for 3 books. So, there are 10 - 3 = 7 positions left to fill. These remaining positions can be filled with the remaining books, which are 4 - 2 = 2 math books and 6 - 1 = 5 history books, totaling 7 books.Since all the books are different, the number of ways to arrange these 7 books in the 7 remaining positions is 7 factorial, which is 7! = 5040 ways.But wait, I think I might have made a mistake here. Let me double-check. I have 4 math books and 6 history books, making a total of 10 books. I've placed 2 math books on the ends and 1 history book in the middle, so that's 3 books placed. That leaves 7 books: 2 math and 5 history. So, the remaining 7 positions can be filled by these 7 books in 7! ways.But actually, the books are all different, so the number of ways to arrange them is indeed 7!.So, putting it all together, the total number of ways is the number of ways to choose the math books for the ends multiplied by the number of ways to choose the history book for the middle multiplied by the number of ways to arrange the remaining books.That would be 4 (choices for first end) * 3 (choices for last end) * 6 (choices for middle history book) * 7! (arrangements for the remaining books).Calculating that:4 * 3 = 1212 * 6 = 727! = 5040So, 72 * 5040 = 362,880Wait, that seems too high. Let me think again.Hold on, I think I missed something. The total number of books is 10, and I've already placed 3 of them, so the remaining 7 can be arranged in 7! ways. But I also have to consider that the remaining math books and history books are different, so their arrangements are all unique.But let me check the calculation:4 * 3 = 1212 * 6 = 727! = 504072 * 5040 = 362,880Hmm, that seems correct. But I'm a bit unsure because 362,880 feels like a large number. Maybe I should think about it differently.Alternatively, I could consider the entire arrangement process step by step:1. Choose a math book for the first position: 4 choices.2. Choose a different math book for the last position: 3 choices.3. Choose a history book for the middle position: 6 choices.4. Arrange the remaining 2 math books and 5 history books in the remaining 7 positions: 7! ways.So, the total number of arrangements is 4 * 3 * 6 * 7! = 4 * 3 * 6 * 5040 = 362,880.Yes, that seems consistent. So, the total number of ways is 362,880.But wait, I'm still a bit confused because I thought the answer might be different. Let me try another approach to verify.Another way to think about it is:- Total positions: 10- Positions 1 and 10 must be math books.- Position 5 must be a history book.So, first, choose math books for positions 1 and 10: 4 choices for position 1, 3 choices for position 10.Then, choose a history book for position 5: 6 choices.Now, we have 7 positions left (positions 2-4 and 6-9), and 7 books left (2 math and 5 history).The number of ways to arrange these 7 books in 7 positions is 7!.So, total arrangements: 4 * 3 * 6 * 7! = 4 * 3 * 6 * 5040 = 362,880.Yes, that's the same result as before. So, I think that's correct.I guess my initial concern about the number being too high was unfounded. After double-checking, it seems that 362,880 is indeed the correct number of ways to arrange the books under the given constraints.So, to summarize:- 4 choices for the first math book.- 3 choices for the last math book.- 6 choices for the middle history book.- 7! ways to arrange the remaining books.Total: 4 * 3 * 6 * 7! = 362,880.