🧠:Okay, so I need to figure out where the center of the circle defined by ((x + cos alpha)^2 + (y + sin alpha)^2 = 1) is located, given that the curve (x^2 cos alpha - y^2 sin alpha + 2 = 0) is an ellipse. The options are the four quadrants. Let me break this down step by step.First, the circle's equation is given, and I need to find its center. The standard form of a circle is ((x - h)^2 + (y - k)^2 = r^2), where ((h, k)) is the center. Comparing this with the given equation, ((x + cos alpha)^2 + (y + sin alpha)^2 = 1), the center should be at ((- cos alpha, - sin alpha)). So the coordinates of the center are ((- cos alpha, - sin alpha)). Therefore, the location of the center depends on the values of (cos alpha) and (sin alpha).Now, the key here is that the other equation (x^2 cos alpha - y^2 sin alpha + 2 = 0) represents an ellipse. For that to be an ellipse, the coefficients of (x^2) and (y^2) must be positive (since the standard form of an ellipse is (frac{x^2}{a^2} + frac{y^2}{b^2} = 1) with (a, b > 0)). However, looking at the given equation, it's (x^2 cos alpha - y^2 sin alpha = -2). Let me rearrange it to standard form:Divide both sides by -2 to get:[frac{x^2 cos alpha}{-2} - frac{y^2 sin alpha}{-2} = 1]Which simplifies to:[frac{x^2}{-2 / cos alpha} + frac{y^2}{2 / sin alpha} = 1]Wait, but for this to be an ellipse, both denominators under (x^2) and (y^2) must be positive. Therefore:- The denominator for (x^2) is (-2 / cos alpha), so this must be positive. Therefore, (-2 / cos alpha > 0 implies cos alpha < 0). - The denominator for (y^2) is (2 / sin alpha), so this must be positive. Therefore, (2 / sin alpha > 0 implies sin alpha > 0).So from the ellipse condition, we have that (cos alpha < 0) and (sin alpha > 0).Now, going back to the center of the circle, which is ((- cos alpha, - sin alpha)). Let's substitute the signs from the ellipse conditions:- Since (cos alpha < 0), then (- cos alpha > 0). So the x-coordinate of the center is positive.- Since (sin alpha > 0), then (- sin alpha < 0). So the y-coordinate of the center is negative.Therefore, the center has coordinates (positive, negative), which places it in the fourth quadrant. Because in the fourth quadrant, x is positive and y is negative. So the answer should be option D.But let me double-check my reasoning to make sure I didn't make a mistake.First, confirming the ellipse condition: The given equation is (x^2 cos alpha - y^2 sin alpha + 2 = 0). Rearranged, it's (x^2 cos alpha - y^2 sin alpha = -2). To write this in standard ellipse form, we need both terms to be positive when divided by -2. So:[frac{x^2 cos alpha}{-2} + frac{- y^2 sin alpha}{-2} = 1 implies frac{x^2}{-2 / cos alpha} + frac{y^2}{2 / sin alpha} = 1]Yes, that's correct. So the denominators must be positive. Therefore:- For (x^2): (-2 / cos alpha > 0 implies cos alpha < 0) - For (y^2): (2 / sin alpha > 0 implies sin alpha > 0)That's correct. So (cos alpha) is negative and (sin alpha) is positive. Then the center of the circle is at ((- cos alpha, - sin alpha)). Since (cos alpha) is negative, (- cos alpha) becomes positive. And since (sin alpha) is positive, (- sin alpha) becomes negative. So (positive, negative), which is the fourth quadrant. That makes sense.Therefore, the answer is D, the fourth quadrant.I don't see any errors in this reasoning. Let me just check if there's another way to approach this problem. For example, maybe analyzing the angle (alpha). If (cos alpha < 0) and (sin alpha > 0), then (alpha) is in which quadrant? Since cosine is negative and sine is positive, (alpha) must be in the second quadrant (where angles are between π/2 and π). In the second quadrant, (cos alpha) is negative and (sin alpha) is positive. Then, the center of the circle is ((- cos alpha, - sin alpha)). Let's think about (alpha) in the second quadrant. Let's say (alpha = 120^circ), which is 2π/3 radians.Then, (cos 120^circ = -1/2), so (- cos 120^circ = 1/2) (positive). (sin 120^circ = sqrt{3}/2), so (- sin 120^circ = -sqrt{3}/2) (negative). So the center is (1/2, -√3/2), which is in the fourth quadrant. Yep, that checks out.Another example: (alpha = 150^circ) (5π/6). (cos 150^circ = -sqrt{3}/2), so the x-coordinate is (sqrt{3}/2). (sin 150^circ = 1/2, so y-coordinate is -1/2. The center is ((sqrt{3}/2, -1/2)), which is also in the fourth quadrant. So regardless of the specific angle in the second quadrant, the center is in the fourth quadrant. Therefore, the answer is D.I think that's solid. I don't see any loopholes or mistakes here. So I can be confident that the correct answer is D.Final Answerboxed{D}

✅:Given the curve (x^2 cos alpha - y^2 sin alpha + 2 = 0) represents an ellipse, we need to determine the quadrant in which the center of the circle ((x + cos alpha)^2 + (y + sin alpha)^2 = 1) is located.1. Identify the center of the circle: - The given circle equation is ((x + cos alpha)^2 + (y + sin alpha)^2 = 1). - The center of the circle is ((- cos alpha, - sin alpha)).2. Conditions for the ellipse: - The equation (x^2 cos alpha - y^2 sin alpha + 2 = 0) must be an ellipse. - Rearranging the equation: (x^2 cos alpha - y^2 sin alpha = -2). - Dividing by -2: (frac{x^2}{-2 / cos alpha} + frac{y^2}{2 / sin alpha} = 1). - For this to be an ellipse, the denominators must be positive: - (-2 / cos alpha > 0 implies cos alpha < 0) - (2 / sin alpha > 0 implies sin alpha > 0)3. Determine the coordinates of the center: - Since (cos alpha < 0), (- cos alpha > 0) (positive x-coordinate). - Since (sin alpha > 0), (- sin alpha < 0) (negative y-coordinate).4. Conclusion: - The center coordinates are (positive, negative), which places it in the fourth quadrant.Thus, the center of the circle is located in the fourth quadrant.[boxed{D}]

🔑:1. Rewrite the given hyperbola equation: The problem states the equation: [ x^2 cos alpha - y^2 sin alpha + 2 = 0 ] Lets rewrite this equation to a more standard form: [ y^2 sin alpha - x^2 cos alpha = 2 ] Observe that the equation is a standard form of hyperbola: [ frac{y^2}{frac{2}{sin alpha}} - frac{x^2}{frac{2}{cos alpha}} = 1 ] 2. Determine the quadrant of ( alpha ): Since we are given that the equation forms a hyperbola, we analyze the trigonometric functions therein: - For ( sin alpha > 0 ), (alpha) must lie either in the first or second quadrant. - For ( cos alpha < 0 ), (alpha) must lie either in the second or third quadrant. Consequently, the only quadrant where ( sin alpha > 0 ) and ( cos alpha < 0 ) is the second quadrant.3. Rewrite and analyze the circle equation: We are given another equation of the circle: [ (x + cos alpha)^2 + (y + sin alpha)^2 = 1 ] This is the equation of a circle centered at ( (-cos alpha, -sin alpha) ) with radius 1.4. Determine the quadrant of the circle’s center: - Since (alpha) is in the second quadrant: - ( cos alpha < 0 ), thus ( -cos alpha > 0 ). - ( sin alpha > 0 ), thus ( -sin alpha < 0 ). Therefore, the center of the circle ( (-cos alpha, -sin alpha) ) lies in the fourth quadrant.Conclusion:Hence, the correct answer is (boxed{text{D}}).

🧠:Okay, so there's this cube, ABCDA'B'C'D'. Let me visualize it first. The cube has upper face ABCD and lower face A'B'C'D', with A above A', B above B', etc. So, edges are AA', BB', CC', DD'. Now, points X and Y are moving along the perimeters of ABCD and B'C'CB respectively, at the same constant speed. X starts at A moving towards B, and Y starts at B' moving towards C', both at the same time. The question is asking for the locus of the midpoint of XY as they move.First, let me make sure I understand the paths of X and Y. Point X is moving along the perimeter of the upper face ABCD. Starting at A, going to B, then to C, D, and back to A. Since it's a square, each side is of equal length. Let's assume the cube has edge length 1 for simplicity, so each side of the square is length 1. Therefore, the perimeter of ABCD is 4 units. The speed is constant, so the time taken to go around the entire perimeter is T = 4 units of time (if speed is 1 unit per second, then T would be 4 seconds, but since speed is same for both, maybe we can parameterize their positions with a parameter t, which can range from 0 to 4 as they go around the square.Similarly, point Y is moving along the perimeter of B'C'CB. Wait, the problem says "the perimeter of B'C'C'B'". Wait, is that a typo? Because B'C'C'B' would not form a closed loop. Let me check the original problem again. It says: "the perimeter of B'C'C'B'". Hmm. Wait, in the cube, the lower face is A'B'C'D', so B' is connected to C', which is connected to C (if going up?), but the face B'C'CB is a lateral face. Wait, perhaps the problem meant the perimeter of the face B'C'CB. Because B'C'CB is a square face of the cube. Let me confirm: B' is connected to C' via an edge on the lower face, but if we move from B' to C', then to C, then to B, and back to B', that forms a square face. Yes, that's one of the vertical faces of the cube. So, the perimeter of the face B'C'CB is also a square with side length 1, so perimeter 4. Therefore, Y is moving along this vertical face's perimeter.So, Y starts at B' moving towards C'. Then, from C' to C, then from C to B, and then back to B'. So, the path is B' -> C' -> C -> B -> B'. Since the cube edges are all length 1, each segment is length 1, so total perimeter is 4. Since both X and Y are moving at the same speed, they take the same time to traverse their respective perimeters. So, their motions are synchronized in terms of time.Now, we need to parameterize the positions of X and Y as functions of time, then compute the midpoint, and find the locus of this midpoint.Let's parameterize both points with a parameter t, which will represent time. Since the perimeter is 4, and the speed is 1 unit per second (assuming), then t ranges from 0 to 4, and after 4 units of time, both points return to their starting positions.Let's define the cube coordinates. Let's place the cube in 3D coordinate system with A at (0,0,1), B at (1,0,1), C at (1,1,1), D at (0,1,1), and the lower face A' at (0,0,0), B' at (1,0,0), C' at (1,1,0), D' at (0,1,0). So, the upper face ABCD is at z=1, and lower face at z=0.Therefore, point X is moving along ABCD: starting at A (0,0,1), moving to B (1,0,1), then to C (1,1,1), then to D (0,1,1), then back to A.Point Y is moving along B'C'CB: starting at B' (1,0,0), moving to C' (1,1,0), then to C (1,1,1), then to B (1,0,1), then back to B'. Wait, but from B (1,0,1) back to B' (1,0,0)? No, wait, the face B'C'CB is a square. Let me check the edges:From B' (1,0,0) to C' (1,1,0) – that's along the lower face. Then from C' (1,1,0) to C (1,1,1) – that's vertical up. Then from C (1,1,1) to B (1,0,1) – along the upper face. Then from B (1,0,1) back to B' (1,0,0) – vertical down. Wait, but that's actually a rectangle, not a square. Wait, but if all edges are length 1, then the vertical edges are length 1. So, B'C' is length 1 (from (1,0,0) to (1,1,0)), then C'C is length 1 (from (1,1,0) to (1,1,1)), then CB is length 1 (from (1,1,1) to (1,0,1)), and then BB' is length 1 (from (1,0,1) to (1,0,0)). So, yes, each edge is length 1, so the face B'C'CB is indeed a square, with perimeter 4.Therefore, Y moves along B' -> C' -> C -> B -> B'. So, starting at B', moving to C', then up to C, then to B, then down to B'.Now, let's parameterize both X(t) and Y(t).For X(t):X is moving along ABCD. The perimeter is 4, so each edge takes 1 unit of time. Let's break down the path:- From t=0 to t=1: X moves from A (0,0,1) to B (1,0,1)- From t=1 to t=2: X moves from B (1,0,1) to C (1,1,1)- From t=2 to t=3: X moves from C (1,1,1) to D (0,1,1)- From t=3 to t=4: X moves from D (0,1,1) back to A (0,0,1)Similarly, for Y(t):Y is moving along B'C'CB. Let's break down Y's path:- From t=0 to t=1: Y moves from B' (1,0,0) to C' (1,1,0)- From t=1 to t=2: Y moves from C' (1,1,0) to C (1,1,1)- From t=2 to t=3: Y moves from C (1,1,1) to B (1,0,1)- From t=3 to t=4: Y moves from B (1,0,1) back to B' (1,0,0)So, both X and Y spend 1 unit of time on each edge. Therefore, their positions can be parameterized linearly in each interval.Let me write parametric equations for X(t) and Y(t) in each interval.For X(t):1. t ∈ [0,1): - Moving from A (0,0,1) to B (1,0,1) - Parametric equations: x = t y = 0 z = 12. t ∈ [1,2): - Moving from B (1,0,1) to C (1,1,1) - Parametric equations: x = 1 y = t - 1 z = 13. t ∈ [2,3): - Moving from C (1,1,1) to D (0,1,1) - Parametric equations: x = 1 - (t - 2) y = 1 z = 14. t ∈ [3,4): - Moving from D (0,1,1) to A (0,0,1) - Parametric equations: x = 0 y = 1 - (t - 3) z = 1Similarly, for Y(t):1. t ∈ [0,1): - Moving from B' (1,0,0) to C' (1,1,0) - Parametric equations: x = 1 y = t z = 02. t ∈ [1,2): - Moving from C' (1,1,0) to C (1,1,1) - Parametric equations: x = 1 y = 1 z = t - 13. t ∈ [2,3): - Moving from C (1,1,1) to B (1,0,1) - Parametric equations: x = 1 y = 1 - (t - 2) z = 14. t ∈ [3,4): - Moving from B (1,0,1) to B' (1,0,0) - Parametric equations: x = 1 y = 0 z = 1 - (t - 3)Now, the midpoint M(t) of XY is given by the average of their coordinates:M(t) = [(X_x(t) + Y_x(t))/2, (X_y(t) + Y_y(t))/2, (X_z(t) + Y_z(t))/2]So, we can compute M(t) for each interval by plugging in the parametric equations of X(t) and Y(t).Let's proceed interval by interval.Interval 1: t ∈ [0,1)X(t) = (t, 0, 1)Y(t) = (1, t, 0)Midpoint M(t):x = (t + 1)/2y = (0 + t)/2 = t/2z = (1 + 0)/2 = 0.5So, M(t) = [(t + 1)/2, t/2, 0.5]This is a parametric line segment in 3D space. Let's see if we can express this in terms of x, y, z.From t=0 to t=1:At t=0: M = (0.5, 0, 0.5)At t=1: M = (1, 0.5, 0.5)So, in this interval, as t increases from 0 to 1, x goes from 0.5 to 1, y goes from 0 to 0.5, z is fixed at 0.5. This is a straight line in the plane z=0.5, from (0.5, 0, 0.5) to (1, 0.5, 0.5).Interval 2: t ∈ [1,2)X(t) = (1, t - 1, 1)Y(t) = (1, 1, t - 1)Midpoint M(t):x = (1 + 1)/2 = 1y = [(t - 1) + 1]/2 = t/2z = (1 + (t - 1))/2 = t/2So, M(t) = [1, t/2, t/2]As t goes from 1 to 2:At t=1: M = (1, 0.5, 0.5)At t=2: M = (1, 1, 1)So, in this interval, x=1, y and z both go from 0.5 to 1. This is a straight line along the diagonal in the plane x=1, from (1, 0.5, 0.5) to (1, 1, 1).Interval 3: t ∈ [2,3)X(t) = (1 - (t - 2), 1, 1) = (3 - t, 1, 1)Y(t) = (1, 1 - (t - 2), 1) = (1, 3 - t, 1)Midpoint M(t):x = [(3 - t) + 1]/2 = (4 - t)/2y = [1 + (3 - t)]/2 = (4 - t)/2z = (1 + 1)/2 = 1So, M(t) = [(4 - t)/2, (4 - t)/2, 1]As t goes from 2 to 3:At t=2: M = (2/2, 2/2, 1) = (1, 1, 1)At t=3: M = (1/2, 1/2, 1)So, this is a straight line in the plane z=1, from (1, 1, 1) to (0.5, 0.5, 1). The x and y coordinates both decrease from 1 to 0.5.Interval 4: t ∈ [3,4)X(t) = (0, 1 - (t - 3), 1) = (0, 4 - t, 1)Y(t) = (1, 0, 1 - (t - 3)) = (1, 0, 4 - t)Midpoint M(t):x = (0 + 1)/2 = 0.5y = [(4 - t) + 0]/2 = (4 - t)/2z = [1 + (4 - t)]/2 = (5 - t)/2So, M(t) = [0.5, (4 - t)/2, (5 - t)/2]As t goes from 3 to 4:At t=3: M = (0.5, (4 - 3)/2, (5 - 3)/2) = (0.5, 0.5, 1)At t=4: M = (0.5, (4 - 4)/2, (5 - 4)/2) = (0.5, 0, 0.5)So, this is a straight line from (0.5, 0.5, 1) to (0.5, 0, 0.5). Here, x=0.5, y decreases from 0.5 to 0, z decreases from 1 to 0.5.Now, let's analyze the entire locus by connecting these intervals.From interval 1: Line segment from (0.5, 0, 0.5) to (1, 0.5, 0.5)Interval 2: Line segment from (1, 0.5, 0.5) to (1, 1, 1)Interval 3: Line segment from (1, 1, 1) to (0.5, 0.5, 1)Interval 4: Line segment from (0.5, 0.5, 1) back to (0.5, 0, 0.5)So, connecting these four segments forms a closed loop. Let's check if these four lines form a regular shape.Looking at the coordinates:1. (0.5, 0, 0.5) to (1, 0.5, 0.5): This is in the plane z=0.5, moving diagonally in x-y plane.2. (1, 0.5, 0.5) to (1, 1, 1): This moves up along y and z.3. (1, 1, 1) to (0.5, 0.5, 1): This moves diagonally down in x and y, staying in z=1.4. (0.5, 0.5, 1) to (0.5, 0, 0.5): This moves down in y and z.So, connecting these points, the shape seems to be a tetrahedron? Or maybe a square in 3D space?Wait, let's calculate the distances between consecutive midpoints.From (0.5, 0, 0.5) to (1, 0.5, 0.5): distance is sqrt( (0.5)^2 + (0.5)^2 + 0 ) = sqrt(0.25 + 0.25) = sqrt(0.5)From (1, 0.5, 0.5) to (1, 1, 1): sqrt(0 + (0.5)^2 + (0.5)^2) = sqrt(0.25 + 0.25) = sqrt(0.5)From (1, 1, 1) to (0.5, 0.5, 1): sqrt( (0.5)^2 + (0.5)^2 + 0 ) = sqrt(0.5)From (0.5, 0.5, 1) to (0.5, 0, 0.5): sqrt(0 + (0.5)^2 + (0.5)^2 ) = sqrt(0.5)So, all edges are equal length sqrt(0.5). Now, the angles between the edges? Let's check if adjacent edges are perpendicular.Take the first two edges: from (0.5,0,0.5) to (1,0.5,0.5) and then to (1,1,1). The vectors are (0.5,0.5,0) and (0,0.5,0.5). The dot product is (0.5)(0) + (0.5)(0.5) + (0)(0.5) = 0.25. Since the vectors are of length sqrt(0.5), the angle between them is arccos(0.25 / (sqrt(0.5)*sqrt(0.5))) = arccos(0.25 / 0.5) = arccos(0.5) = 60 degrees. So, not right angles. Therefore, the figure is a regular tetrahedron? But a regular tetrahedron has all edges equal and all angles equal, which is 60 degrees here. Wait, but a regular tetrahedron has angles arccos(1/3) ≈ 54.7 degrees, so not the same. Hmm, maybe it's a non-regular tetrahedron.Alternatively, perhaps the figure is a square in 3D space, but skewed. However, since all edges are equal and each consecutive edge is connected by 60-degree angles, it's more like a regular tetrahedron, but in 3D.Wait, but in 3D, four points with all edges equal would form a regular tetrahedron. But here, the four midpoints have edges of sqrt(0.5) between consecutive points, but what about the diagonals?For example, distance from (0.5,0,0.5) to (1,1,1): sqrt( (0.5)^2 + (1)^2 + (0.5)^2 ) = sqrt(0.25 + 1 + 0.25) = sqrt(1.5) ≈ 1.2247Distance from (1,0.5,0.5) to (0.5,0.5,1): sqrt( (0.5)^2 + 0 + (0.5)^2 ) = sqrt(0.25 + 0.25) = sqrt(0.5) ≈ 0.7071So, the diagonals are not equal, so it's not a regular tetrahedron. Therefore, the shape is a non-regular tetrahedron. However, maybe there's another way to see this.Alternatively, perhaps the locus is a rectangle or another quadrilateral in 3D space. Wait, but since all the edges are equal and the angles between consecutive edges are 60 degrees, as calculated, it's a rhombus? But in 3D, a rhombus can be planar or not. Let's check if the four midpoints lie on a plane.To check planarity, we can use the determinant method. If the volume of the tetrahedron formed by any three points and the fourth is zero, then they are coplanar.Take the four points:P1: (0.5, 0, 0.5)P2: (1, 0.5, 0.5)P3: (1, 1, 1)P4: (0.5, 0.5, 1)Compute the volume of the tetrahedron P1P2P3P4.The volume can be calculated using the scalar triple product:V = |( (P2 - P1) ⋅ [ (P3 - P1) × (P4 - P1) ] ) / 6|First, find vectors:P2 - P1 = (0.5, 0.5, 0)P3 - P1 = (0.5, 1, 0.5)P4 - P1 = (0, 0.5, 0.5)Compute cross product (P3 - P1) × (P4 - P1):Let me compute:i component: (1)(0.5) - (0.5)(0.5) = 0.5 - 0.25 = 0.25j component: - [(0.5)(0.5) - (0.5)(0)] = - [0.25 - 0] = -0.25k component: (0.5)(0.5) - (1)(0) = 0.25 - 0 = 0.25So, cross product is (0.25, -0.25, 0.25)Then, dot product with (P2 - P1) = (0.5, 0.5, 0):0.5*0.25 + 0.5*(-0.25) + 0*0.25 = 0.125 - 0.125 + 0 = 0Therefore, the volume is 0, meaning the four points are coplanar.So, the four midpoints lie on a plane. Therefore, the locus is a quadrilateral lying on a plane. Since all sides are equal and the diagonals are equal?Wait, let's check the diagonals. The diagonals would be P1-P3 and P2-P4.Distance P1-P3: sqrt( (0.5)^2 + (1)^2 + (0.5)^2 ) = sqrt(0.25 + 1 + 0.25) = sqrt(1.5) ≈ 1.2247Distance P2-P4: sqrt( (0.5)^2 + (0)^2 + (0.5)^2 ) = sqrt(0.25 + 0 + 0.25) = sqrt(0.5) ≈ 0.7071So, the diagonals are not equal. However, in a rhombus, the diagonals are not equal but they bisect each other at right angles. Let's check if the diagonals bisect each other.Midpoint of P1-P3: [(0.5 + 1)/2, (0 + 1)/2, (0.5 + 1)/2] = (0.75, 0.5, 0.75)Midpoint of P2-P4: [(1 + 0.5)/2, (0.5 + 0.5)/2, (0.5 + 1)/2] = (0.75, 0.5, 0.75)So, both diagonals have the same midpoint. Therefore, the diagonals bisect each other. Now, check if they are perpendicular.The vectors for the diagonals:P1-P3: (0.5, 1, 0.5)P2-P4: (-0.5, 0, 0.5)Dot product: 0.5*(-0.5) + 1*0 + 0.5*0.5 = -0.25 + 0 + 0.25 = 0So, the diagonals are perpendicular. Therefore, since the quadrilateral is a parallelogram with equal sides and perpendicular diagonals, it is a rhombus, and in fact, since it's a rhombus with perpendicular diagonals, it's a square.Wait, but in a rhombus, if the diagonals are perpendicular, it's a square only if all sides are equal and angles are 90 degrees. Wait, no, a rhombus with perpendicular diagonals is a kite as well, but actually, in a rhombus, the diagonals are always perpendicular. Wait, actually, in a rhombus, the diagonals are perpendicular bisectors, but the sides are not necessarily at right angles. Wait, no: A rhombus is a quadrilateral with all sides equal, and diagonals are perpendicular. But if the angles are 90 degrees, then it's a square. So, if a rhombus has perpendicular diagonals, it's a rhombus, and if it also has right angles, it's a square. But in this case, the sides are equal, the diagonals are perpendicular and bisect each other, but are the angles 90 degrees?Wait, let's check the angles between the sides. Take the first two edges:From P1 to P2: vector (0.5, 0.5, 0)From P2 to P3: vector (0, 0.5, 0.5)The angle between these two vectors can be found by the dot product:(0.5, 0.5, 0) ⋅ (0, 0.5, 0.5) = 0*0.5 + 0.5*0.5 + 0*0.5 = 0.25The magnitude of each vector is sqrt(0.5^2 + 0.5^2 + 0^2) = sqrt(0.25 + 0.25) = sqrt(0.5) ≈ 0.7071So, cos(theta) = 0.25 / (sqrt(0.5)*sqrt(0.5)) = 0.25 / 0.5 = 0.5Thus, theta = 60 degrees. So, the angle between consecutive sides is 60 degrees, which is not 90 degrees. Therefore, it's a rhombus with angles 60 and 120 degrees, so not a square.Therefore, the locus is a rhombus lying on a plane, with all sides equal to sqrt(0.5), diagonals of lengths sqrt(1.5) and sqrt(0.5), which are perpendicular to each other.But the problem asks for the locus of the midpoint. From the parametrization, we see that the midpoint moves along four connected straight-line segments, forming a rhombus in a plane.But to confirm, perhaps there's a simpler description. Since all the midpoints lie on a plane, and the shape is a rhombus, maybe it's a square rotated in 3D space?Alternatively, perhaps the plane can be described in terms of the cube's geometry.Looking at the coordinates of the midpoints:The four points are (0.5,0,0.5), (1,0.5,0.5), (1,1,1), (0.5,0.5,1). Let's see if these lie on a specific plane.To find the equation of the plane, we can use three points and find the equation.Take points P1 (0.5,0,0.5), P2 (1,0.5,0.5), P3 (1,1,1).First, find two vectors in the plane:Vector v1 = P2 - P1 = (0.5, 0.5, 0)Vector v2 = P3 - P1 = (0.5, 1, 0.5)Compute the normal vector n = v1 × v2Calculate cross product:i component: (0.5)(0.5) - (0)(1) = 0.25 - 0 = 0.25j component: - [ (0.5)(0.5) - (0)(0.5) ] = - [0.25 - 0] = -0.25k component: (0.5)(1) - (0.5)(0.5) = 0.5 - 0.25 = 0.25So, normal vector n = (0.25, -0.25, 0.25). We can multiply by 4 to eliminate decimals: n = (1, -1, 1)Therefore, the equation of the plane is 1(x) -1(y) +1(z) = dPlug in point P1 (0.5,0,0.5):1*(0.5) -1*0 +1*0.5 = 0.5 + 0 + 0.5 = 1 = dThus, the plane equation is x - y + z = 1Check with other points:P2 (1,0.5,0.5): 1 -0.5 +0.5 = 1, correct.P3 (1,1,1): 1 -1 +1 =1, correct.P4 (0.5,0.5,1): 0.5 -0.5 +1 =1, correct.So, all four midpoints lie on the plane x - y + z =1.Therefore, the locus is a rhombus lying on the plane x - y + z =1.But can we describe this rhombus more specifically?Given that it's a rhombus with vertices at (0.5,0,0.5), (1,0.5,0.5), (1,1,1), and (0.5,0.5,1), we can perhaps note that these points are midpoints of certain edges of the cube.Alternatively, maybe this rhombus is actually a square in the plane x - y + z =1, but rotated.Wait, to check if it's a square, we need to see if all angles are 90 degrees, but earlier we saw that consecutive edges have 60-degree angles. Wait, but that was in 3D. Maybe when projected onto the plane, the angles are 90 degrees?Alternatively, since the sides are of equal length and the diagonals are perpendicular, in the plane, this would be a square. Wait, in plane geometry, a rhombus with perpendicular diagonals is a square. Wait, no: in plane geometry, a rhombus with perpendicular diagonals is a kite and a rhombus, but to be a square, the diagonals must also be equal in length. In our case, the diagonals are sqrt(1.5) and sqrt(0.5), which are not equal. Therefore, in the plane x - y + z =1, the figure is a rhombus with unequal diagonals, hence not a square.Therefore, the locus is a rhombus lying on the plane x - y + z =1, with vertices at the four midpoints calculated.But perhaps there's another way to parametrize this. Let's think about the parametrization of M(t) in each interval.For example, in interval 1, M(t) = [(t + 1)/2, t/2, 0.5]. Let's see if we can express this in terms of the plane equation x - y + z =1.Plugging in:x - y + z = ( (t +1)/2 ) - (t/2 ) + 0.5 = (t +1 - t)/2 + 0.5 = 0.5 + 0.5 =1. Correct.Similarly, in interval 2, M(t)= [1, t/2, t/2]x - y + z =1 - t/2 + t/2 =1. Correct.Interval3, M(t)= [(4 - t)/2, (4 - t)/2,1]x - y + z= (4 - t)/2 - (4 - t)/2 +1=0 +1=1. Correct.Interval4, M(t)= [0.5, (4 - t)/2, (5 - t)/2]x - y + z=0.5 - (4 - t)/2 + (5 - t)/2 =0.5 + [ - (4 - t) + (5 - t) ] /2 =0.5 + (1)/2=0.5 +0.5=1. Correct.So, all points lie on the plane x - y + z =1.Moreover, the shape is a rhombus, as established.But perhaps there's a different perspective. For example, considering symmetry.Given that both X and Y are moving at the same speed along their respective perimeters, which are both squares, and starting at the same time, their midpoints might have some symmetric properties.Alternatively, since the cube has symmetry, perhaps the locus is a square rotated in 3D space. However, our analysis shows it's a rhombus, not a square. Alternatively, it's a regular hexagon? But we have four segments, so it's a quadrilateral.Wait, but maybe when considering the entire movement, the midpoint traces out the rhombus on the plane x - y + z =1. So, the answer is that the locus is a rhombus lying on the plane x - y + z =1, with vertices at (0.5, 0, 0.5), (1, 0.5, 0.5), (1,1,1), and (0.5,0.5,1).But the problem might expect a more specific description, perhaps identifying it as a square or another specific quadrilateral. Alternatively, maybe parameterizing it differently.Alternatively, notice that the rhombus can be seen as the intersection of the plane x - y + z =1 with the cube.Let me check the intersection of the plane x - y + z =1 with the cube.The cube has vertices from (0,0,0) to (1,1,1). The plane x - y + z =1 intersects the cube. Let's find the intersection polygon.To find the intersection, we need to find edges of the cube that intersect the plane.An edge intersects the plane if the two vertices of the edge are on opposite sides of the plane.Compute for each edge:For each edge, plug the coordinates of the two endpoints into x - y + z. If one is less than 1 and the other is greater than 1, the edge intersects the plane.Let's go through all edges of the cube:1. A(0,0,1) to B(1,0,1): At A, x - y + z = 0 -0 +1=1. At B, 1 -0 +1=2. So, both endpoints on or above the plane. The point A is on the plane, B is above. So, intersection at A.2. B(1,0,1) to C(1,1,1): At B, 1 -0 +1=2; at C, 1 -1 +1=1. So, C is on the plane, B is above. Intersection at C.3. C(1,1,1) to D(0,1,1): At C, 1 -1 +1=1; at D, 0 -1 +1=0. So, D is below, intersection somewhere along CD.Wait, but the plane equation at D: 0 -1 +1=0. So, the edge from C(1,1,1) to D(0,1,1) has one point on the plane (C) and one below. So, intersection at C.4. D(0,1,1) to A(0,0,1): At D, 0 -1 +1=0; at A, 0 -0 +1=1. So, A is on the plane, D is below. Intersection at A.5. A'(0,0,0) to B'(1,0,0): At A', 0 -0 +0=0 <1; at B',1 -0 +0=1. So, intersects at B'.6. B'(1,0,0) to C'(1,1,0): At B',1 -0 +0=1; at C',1 -1 +0=0. So, intersection at B'.7. C'(1,1,0) to D'(0,1,0): At C',1 -1 +0=0; at D',0 -1 +0=-1. Both below, no intersection.8. D'(0,1,0) to A'(0,0,0): At D',0 -1 +0=-1; at A',0 -0 +0=0. Both below or on, but A' is 0, which is less than 1, so no intersection.Vertical edges:9. A(0,0,1) to A'(0,0,0): At A,1; at A',0. Intersection at A.10. B(1,0,1) to B'(1,0,0): At B,2; at B',1. Intersection at B'.11. C(1,1,1) to C'(1,1,0): At C,1; at C',0. Intersection at C.12. D(0,1,1) to D'(0,1,0): At D,0; at D',-1. No intersection.Other vertical edges:Wait, the cube has 12 edges: 4 on top, 4 on bottom, 4 vertical.We have checked all.So, the intersection points are the vertices A, B', C, and the midpoints of edges where the plane cuts through.Wait, no. From the above, the intersection points are at A, B', C, and along edges where endpoints straddle the plane.Wait, let's check edges where one endpoint is above and the other is below.Looking at edges:Edge D(0,1,1) to A(0,0,1): D is below (0 -1 +1=0 <1), A is on the plane. Intersection at A.Edge A'(0,0,0) to B'(1,0,0): A' is below, B' is on the plane. Intersection at B'.Edge C(1,1,1) to D(0,1,1): C is on the plane, D is below. Intersection at C.Edge B'(1,0,0) to C'(1,1,0): B' is on the plane, C' is below. Intersection at B'.So, the plane intersects the cube at points A, B', C, and also possibly along other edges?Wait, let's recheck. For edges that cross the plane x - y + z =1:Edge E connects (x1,y1,z1) to (x2,y2,z2). If (x1 - y1 + z1 -1) and (x2 - y2 + z2 -1) have opposite signs, then the edge intersects the plane.Let me check edges not connected to A, B', or C.Edge from B(1,0,1) to C(1,1,1): B is above (2), C is on (1). So, no crossing.Edge from C'(1,1,0) to D'(0,1,0): Both below.Edge from D'(0,1,0) to A'(0,0,0): Both below.Edge from A'(0,0,0) to D(0,1,1): A'(0,0,0) gives 0 -0 +0=0 <1; D(0,1,1) gives 0 -1 +1=0 <1. Both below, no intersection.Edge from B'(1,0,0) to B(1,0,1): B' is on the plane (1 -0 +0=1), B is above (1 -0 +1=2). So, intersection at B'.Edge from C(1,1,1) to C'(1,1,0): C is on the plane (1 -1 +1=1), C' is below. Intersection at C.Edge from D(0,1,1) to D'(0,1,0): Both below.Edge from A'(0,0,0) to A(0,0,1): A' is below, A is on the plane. Intersection at A.So, the plane x - y + z =1 intersects the cube at points A, B', C, and along no other edges because all other edges either have both endpoints on one side of the plane or intersect at these three points. Wait, but three points can't form a polygon. There must be another intersection.Wait, perhaps I missed an edge. Let's check edge from B'(1,0,0) to C'(1,1,0): B' is on the plane, C' is below. So, intersection at B'.Edge from C'(1,1,0) to C(1,1,1): C' is below, C is on the plane. Intersection at C.Edge from C(1,1,1) to B(1,0,1): C is on the plane, B is above. Intersection at C.Edge from B(1,0,1) to B'(1,0,0): B is above, B' is on the plane. Intersection at B'.Edge from A(0,0,1) to D(0,1,1): A is on the plane, D is below. Intersection at A.So, the plane x - y + z =1 intersects the cube at points A, B', C, and forms a triangle? But in our earlier analysis, the locus of midpoints is a quadrilateral. Hmm, this seems contradictory.Wait, perhaps the plane intersects the cube in a polygon with more vertices. Let me double-check.Wait, when I considered the intersection of the plane with the cube, perhaps I missed some edges where the plane cuts through.Wait, let's check edge from D'(0,1,0) to C'(1,1,0): D'(0,1,0) gives 0 -1 +0 = -1 <1, C'(1,1,0) gives 1 -1 +0 =0 <1. Both below.Edge from A'(0,0,0) to D'(0,1,0): Both below.Edge from A'(0,0,0) to B'(1,0,0): A' is below, B' is on the plane. Intersection at B'.Edge from C'(1,1,0) to D'(0,1,0): Both below.Edge from D(0,1,1) to C(1,1,1): D is below, C is on the plane. Intersection at C.Edge from D(0,1,1) to A(0,0,1): D is below, A is on the plane. Intersection at A.So, total intersection points are A, B', C, and no others. That forms a triangle. But our midpoint locus is a quadrilateral, which is different.Wait, this suggests that the midpoint locus is different from the intersection of the plane with the cube. Therefore, the locus is a quadrilateral (rhombus) lying on the plane x - y + z =1, entirely inside the cube.Therefore, the answer should be that the locus is a rhombus lying on the plane x - y + z = 1, connecting the midpoints (0.5,0,0.5), (1,0.5,0.5), (1,1,1), and (0.5,0.5,1).Alternatively, since the problem may expect a more geometric description rather than coordinates, perhaps stating it's a square or a rectangle. But based on our analysis, it's a rhombus.However, another approach: let's consider the coordinates of the midpoint in each interval and see if there's a parametric equation that can describe the entire locus.But since the movement is periodic and the locus is a closed shape, combining the four line segments, the answer is the rhombus as described.Alternatively, think about the cube's geometry: since X is on the top face and Y is on the front face (assuming B'C'CB is the front face), their midpoints would trace a path that's a combination of their movements. Given the symmetry in their paths, the locus being a rhombus makes sense.Therefore, after thorough analysis, the locus of the midpoint of XY is a rhombus lying on the plane x - y + z =1, with vertices at (0.5,0,0.5), (1,0.5,0.5), (1,1,1), and (0.5,0.5,1).

✅:The locus of the midpoint of XY is a rhombus lying on the plane ( x - y + z = 1 ), with vertices at the midpoints ((0.5, 0, 0.5)), ((1, 0.5, 0.5)), ((1, 1, 1)), and ((0.5, 0.5, 1)). This rhombus is formed by connecting four linear segments as both points move along their respective perimeters at equal speeds.Step-by-Step Explanation:1. Cube Coordinates: Assign coordinates to the cube with edge length 1. The upper face ABCD is at ( z=1 ) and the lower face A'B'C'D' at ( z=0 ). Points are: - ( A(0,0,1) ), ( B(1,0,1) ), ( C(1,1,1) ), ( D(0,1,1) ) - ( A'(0,0,0) ), ( B'(1,0,0) ), ( C'(1,1,0) ), ( D'(0,1,0) )2. Paths of X and Y: - X travels along the perimeter of ABCD: ( A rightarrow B rightarrow C rightarrow D rightarrow A ). - Y travels along the perimeter of B'C'CB: ( B' rightarrow C' rightarrow C rightarrow B rightarrow B' ).3. Parametric Equations: Both points move at the same speed, completing their paths in 4 time units. Their coordinates for each interval ( t in [0,4) ) are derived as linear functions of ( t ).4. Midpoint Calculation: The midpoint ( M(t) ) is computed by averaging the coordinates of ( X(t) ) and ( Y(t) ) in each interval: - Interval [0,1): ( M(t) = left( frac{t+1}{2}, frac{t}{2}, 0.5 right) ) - Interval [1,2): ( M(t) = left( 1, frac{t}{2}, frac{t}{2} right) ) - Interval [2,3): ( M(t) = left( frac{4-t}{2}, frac{4-t}{2}, 1 right) ) - Interval [3,4): ( M(t) = left( 0.5, frac{4-t}{2}, frac{5-t}{2} right) )5. Geometric Analysis: - Each interval corresponds to a straight-line segment in 3D space. - All midpoints lie on the plane ( x - y + z = 1 ). - Connecting these segments forms a rhombus with vertices at the calculated midpoints, confirmed by equal side lengths and perpendicular diagonals within the plane.Final Answer: The locus of the midpoint of XY is a rhombus lying on the plane ( x - y + z = 1 ), connecting the points ( left( frac{1}{2}, 0, frac{1}{2} right) ), ( (1, frac{1}{2}, frac{1}{2}) ), ( (1, 1, 1) ), and ( left( frac{1}{2}, frac{1}{2}, 1 right) ).

🔑:To solve this problem, we need to track the midpoint M of two moving points X and Y around the cube, given their starting positions and movement speeds. By breaking the problem into parts corresponding to different segments of movement for X and Y, we can analyze their coordinates and thus determine the path of the midpoint M.Let's summarize the coordinates of each relevant point in the cube and their transformations:- The cube is considered in 3D space.- The vertices A, B, C, D are on the upper face.- Vertices A', B', C', D' are on the lower face.- We set the coordinates as follows for simplicity: - A as (0, 0, 0) - B as (1, 0, 0) - C as (1, 1, 0) - D as (0, 1, 0) - A' as (0, 0, 1) - B' as (1, 0, 1) - C' as (1, 1, 1) - D' as (0, 1, 1)Let us proceed with each segment of the movements of X and Y and track the midpoint M:1. While X is on AB and Y is on B'C': - X moves from A = (0,0,0) to B = (1,0,0) so X at any time t is (t, 0, 0) for 0 leq t leq 1. - Y moves from B' = (1,0,1) to C' = (1,1,1) so Y at any time t is (1, t, 1) for 0 leq t leq 1. - Midpoint M of X and Y is: [ M = left(frac{t + 1}{2}, frac{t}{2}, frac{0 + 1}{2} right) = left(frac{t + 1}{2}, frac{t}{2}, frac{1}{2}right) ]2. While X is on BC and Y is on C'C: - X moves from B=(1,0,0) to C=(1,1,0) so X at any time t is (1, t, 0) for 0 leq t leq 1. - Y moves from C'=(1,1,1) to C = (1,1,0) so Y at any time t is (1,1, 1-t) for 0 leq t leq 1. - Midpoint M of X and Y is: [ M = left(1, frac{t + 1}{2}, frac{1 - t}{2}right) ]3. While X is on CD and Y is on CB: - X moves from C=(1,1,0) to D=(0,1,0) so X at any time t is (1 - t, 1, 0) for 0 leq t leq 1. - Y moves from C=(1,1,0) to B=(1,0,0) so Y at any time t is (1, 1 - t, 0) for 0 leq t leq 1. - Midpoint M of X and Y is: [ M = left(frac{2 - t}{2}, frac{2 - t}{2}, 0right) = left(1 - frac{t}{2}, 1 - frac{t}{2}, 0 right) ]4. While X is on DA and Y is on BB': - X moves from D=(0,1,0) to A=(0,0,0) so X at any time t is (0, 1 - t, 0) for 0 leq t leq 1. - Y moves from B=(1,0,0) to B'=(1, 0, 1), so Y at any time t is (1, 0, t) for 0 leq t leq 1. - Midpoint M of X and Y is: [ M = left(frac{1}{2}, frac{1 - t}{2}, frac{t}{2}right) ]The lines that are traced out by the midpoint M in each of these steps identify specific line segments in the cube, and together they form a rhombus. Specifically:- The midpoint M traces out the lines VW, WC, CU, and UV.Thus, the locus of the midpoint of XY is traced out to form the rhombus CUVW, where these points are:- C is (1, 1, 0).- U is the center of ABCD, which is (0.5, 0.5, 0).- V is the center of ABB'A, which is (1, 0.5, 0.5).- W is the center of B'C'CB, which is (1, 0.5, 0.5).Thus, the locus of the midpoint of XY is a rhombus in the cube:[boxed{The locus of the midpoint of XY is the rhombus CUVW}]

🧠:Okay, let's see. I need to find the parameter ( p ), the coordinates of the focus, and the equation of the directrix for the parabola given by ( y^2 = 2 p x ). The hint given is that the tangent to this parabola at some point has the intercept form equation ( frac{x}{-75} + frac{y}{15} = 1 ). Hmm, okay, so first, maybe I should recall some properties of parabolas and their tangent equations.First, the standard form of the parabola is ( y^2 = 4 a x ), where the focus is at ( (a, 0) ) and the directrix is ( x = -a ). Wait, but in the given equation here, it's ( y^2 = 2 p x ). So comparing with the standard form, ( 4 a = 2 p ), which would mean ( a = p/2 ). So then, the focus should be at ( (a, 0) = (p/2, 0) ), and the directrix is ( x = -a = -p/2 ). So that's the focus and directrix in terms of ( p ). So once I find ( p ), I can just plug into those expressions.Now, the main problem is to find ( p ), using the information about the tangent line. The tangent line given is ( frac{x}{-75} + frac{y}{15} = 1 ). Let me rewrite that equation in a more familiar form. Multiplying both sides by 75 to eliminate denominators:( -x + 5y = 75 )Wait, let me check that. If I have ( frac{x}{-75} + frac{y}{15} = 1 ), multiply each term by 75:( x/(-75) * 75 + y/15 * 75 = 1 * 75 )Which simplifies to:( -x + 5y = 75 )So the equation of the tangent is ( -x + 5y = 75 ), or rearranged as ( 5y - x = 75 ), or ( x = 5y - 75 ). But maybe I should keep it in standard linear form. Anyway, the key is that this line is tangent to the parabola ( y^2 = 2 p x ). So I need to find the value of ( p ) such that this line is tangent to the parabola. How do I find the condition for a line to be tangent to a parabola? Well, one way is to substitute the equation of the line into the equation of the parabola and then set the discriminant of the resulting quadratic equation to zero, which ensures that there's exactly one point of intersection (the tangent point). So let's try that. Let me express ( x ) from the tangent line equation. From ( -x + 5y = 75 ), solving for ( x ):( x = 5y - 75 )Now, substitute this into the parabola equation ( y^2 = 2 p x ):( y^2 = 2 p (5y - 75) )Expanding the right side:( y^2 = 10 p y - 150 p )Bring all terms to the left side:( y^2 - 10 p y + 150 p = 0 )This is a quadratic equation in ( y ). For the line to be tangent to the parabola, this quadratic must have exactly one solution, which means its discriminant should be zero. The discriminant ( D ) of ( a y^2 + b y + c = 0 ) is ( D = b^2 - 4 a c ). Here, ( a = 1 ), ( b = -10 p ), and ( c = 150 p ).Calculating the discriminant:( D = (-10 p)^2 - 4 * 1 * 150 p )( D = 100 p^2 - 600 p )Setting discriminant equal to zero for tangency:( 100 p^2 - 600 p = 0 )Factor out 100 p:( 100 p (p - 6) = 0 )So, solutions are ( p = 0 ) or ( p = 6 ). However, ( p = 0 ) would make the original parabola equation ( y^2 = 0 ), which is just the x-axis, which can't have a tangent line like the given one. Therefore, ( p = 6 ).So, ( p = 6 ). Then, the focus is at ( (p/2, 0) = (6/2, 0) = (3, 0) ), and the directrix is ( x = -p/2 = -3 ).Wait, but let me verify this. If ( p = 6 ), the parabola is ( y^2 = 12 x ). Let's check if the line ( -x + 5y = 75 ) is indeed a tangent to this parabola.Substituting ( x = 5y - 75 ) into ( y^2 = 12 x ):( y^2 = 12 (5y - 75) )( y^2 = 60 y - 900 )( y^2 - 60 y + 900 = 0 )Compute discriminant:( D = (-60)^2 - 4 * 1 * 900 = 3600 - 3600 = 0 )Yes, discriminant is zero, so it's a tangent. Thus, ( p = 6 ) is correct.Alternatively, another method to find the tangent line to a parabola. For the parabola ( y^2 = 4 a x ), the equation of the tangent at point ( (x_1, y_1) ) is ( y y_1 = 2 a (x + x_1) ). Wait, but in our case, the equation is ( y^2 = 2 p x ), which is equivalent to ( y^2 = 4 a x ) with ( 4 a = 2 p ), so ( a = p/2 ). Then the tangent equation at point ( (x_1, y_1) ) would be ( y y_1 = 2 a (x + x_1) ), substituting ( a ):( y y_1 = p (x + x_1) )So that's another way to write the tangent line. Since we have the tangent equation given, maybe we can compare coefficients.Given that the tangent line is ( frac{x}{-75} + frac{y}{15} = 1 ), which can also be written as ( y = frac{1}{15} x + frac{75}{15} ), wait, no, let me check:Starting with ( frac{x}{-75} + frac{y}{15} = 1 ), solving for y:Multiply both sides by 15: ( -x/5 + y = 15 )So, ( y = (x/5) + 15 ). Wait, but that contradicts my previous calculation where I had ( x = 5y - 75 ). Wait, let me check again.Original equation: ( frac{x}{-75} + frac{y}{15} = 1 )Multiply both sides by 15: ( -x/5 + y = 15 )So, ( y = x/5 + 15 ). Therefore, the tangent line is ( y = frac{1}{5}x + 15 ). So slope is 1/5, y-intercept 15. Alternatively, written as ( x/5 - y + 15 = 0 ).But earlier, when I solved for x, I had ( x = 5y - 75 ). Let me check that. If ( frac{x}{-75} + frac{y}{15} = 1 ), then:Multiply all terms by 75: ( -x + 5y = 75 ), so ( x = 5y - 75 ). So both expressions are equivalent.So, in either case, the line is ( y = (1/5)x + 15 ) or ( x = 5y - 75 ).Now, using the tangent equation from the parabola's point form. Suppose the tangent line touches the parabola at point ( (x_1, y_1) ). Then, the equation of the tangent is ( y y_1 = p (x + x_1) ).So, given that our tangent line is ( y = (1/5)x + 15 ), let's express this in the form ( y = m x + c ), where m is 1/5 and c is 15.Comparing with the tangent equation ( y y_1 = p (x + x_1) ). Let's rearrange the standard tangent equation:( y y_1 = p x + p x_1 )Bring all terms to left side:( p x - y y_1 + p x_1 = 0 )Alternatively, solving for y:( y = (p / y_1) x + (p x_1)/ y_1 )So, comparing with our tangent line ( y = (1/5) x + 15 ), we can equate the coefficients:Slope: ( p / y_1 = 1/5 )Intercept: ( (p x_1) / y_1 = 15 )So, from the slope equation:( p / y_1 = 1/5 implies y_1 = 5 p )From the intercept equation:( (p x_1) / y_1 = 15 implies (p x_1) / (5 p) = 15 implies x_1 / 5 = 15 implies x_1 = 75 )So, the point of tangency is ( (x_1, y_1) = (75, 5 p) ). But since this point lies on the parabola ( y^2 = 2 p x ), substituting these coordinates into the parabola equation:( (5 p)^2 = 2 p * 75 )Calculating:( 25 p^2 = 150 p )Divide both sides by p (assuming p ≠ 0, which we already established):( 25 p = 150 implies p = 6 )Same result as before. So, this confirms that ( p = 6 ). Therefore, the parameter ( p ) is 6. Then, as before, the focus is at ( (p/2, 0) = (3, 0) ), and the directrix is ( x = -3 ).Alternatively, maybe another approach to cross-validate. Let's find the point of tangency.We know the tangent line is ( y = (1/5)x + 15 ), and the parabola is ( y^2 = 12 x ). Let's find their intersection point. Substitute y from the line into the parabola equation:( [(1/5)x + 15]^2 = 12 x )Expand left side:( (1/25)x^2 + (2 * 1/5 * 15) x + 225 = 12 x )( (1/25)x^2 + (6) x + 225 = 12 x )Multiply all terms by 25 to eliminate denominator:( x^2 + 150 x + 5625 = 300 x )Bring all terms to left:( x^2 + 150 x + 5625 - 300 x = 0 )( x^2 - 150 x + 5625 = 0 )Solve for x:Discriminant D = 150² - 4 * 1 * 5625 = 22500 - 22500 = 0Thus, x = [150 ± 0]/2 = 75. So x = 75, then y = (1/5)*75 + 15 = 15 + 15 = 30. So the point of tangency is (75, 30). Checking if this point is on the parabola ( y^2 = 12 x ):Left side: 30² = 900Right side: 12 * 75 = 900Yes, correct. So the tangent point is (75,30). So this also confirms that the value of p is 6, since the parabola is ( y^2 = 12 x ), which is ( 2 p x ) with p = 6. Therefore, all previous steps are correct.Thus, the parameter ( p = 6 ), the focus is at (3, 0), and the directrix is ( x = -3 ).Final AnswerThe parameter ( p ) is ( boxed{6} ), the coordinates of the focus are ( boxed{(3, 0)} ), and the equation of the directrix is ( boxed{x = -3} ).

✅:To find the parameter ( p ), the coordinates of the focus, and the equation of the directrix of the parabola given by ( y^2 = 2px ) with the tangent line ( frac{x}{-75} + frac{y}{15} = 1 ), we follow these steps:1. Rewrite the tangent line equation: The given tangent line equation can be rewritten as ( -x + 5y = 75 ) or ( x = 5y - 75 ).2. Substitute the tangent line into the parabola equation: Substitute ( x = 5y - 75 ) into ( y^2 = 2px ): [ y^2 = 2p(5y - 75) ] Simplifying this, we get: [ y^2 - 10py + 150p = 0 ]3. Set the discriminant of the quadratic equation to zero: The discriminant ( D ) of ( y^2 - 10py + 150p = 0 ) must be zero for tangency: [ D = (-10p)^2 - 4 cdot 1 cdot 150p = 100p^2 - 600p ] Setting ( D = 0 ): [ 100p(p - 6) = 0 ] Solving this, we get ( p = 6 ) (since ( p = 0 ) is not valid).4. Determine the coordinates of the focus and the directrix: - For the parabola ( y^2 = 2px ), the focus is at ( left(frac{p}{2}, 0right) ). Substituting ( p = 6 ), the focus is ( (3, 0) ). - The directrix is the line ( x = -frac{p}{2} ). Substituting ( p = 6 ), the directrix is ( x = -3 ).5. Verification: - The tangent line ( -x + 5y = 75 ) intersects the parabola ( y^2 = 12x ) at exactly one point, confirming the discriminant is zero. - The point of tangency (75, 30) lies on the parabola and satisfies both the parabola and the tangent line equations. Final AnswerThe parameter ( p ) is ( boxed{6} ), the coordinates of the focus are ( boxed{(3, 0)} ), and the equation of the directrix is ( boxed{x = -3} ).

🔑:Given the equation of the parabola ( y^2 = 2px ) and a tangent line at a point with intercepts ( frac{x}{-75} + frac{y}{15} = 1 ).1. Rewrite the tangent line equation in intercept form: We know that the equation of the tangent to the parabola at the point ( (x_0, y_0) ) can be written as: [ frac{x}{-x_0} + frac{y}{frac{px_0}{y_0}} = 1 ] Substituting ( x_0 = frac{y_0^2}{2p} ) into the tangent line equation, we get: [ frac{x}{-frac{y_0^2}{2p}} + frac{y}{frac{py_0}{2p}} = 1 ]2. Simplify substitutions: Simplify the terms in the denominator: [ frac{x}{-frac{y_0^2}{2p}} = frac{-2px}{y_0^2} ] and [ frac{y}{frac{y_0}{2}} = frac{2y}{y_0} ] So the equation becomes: [ frac{-2px}{y_0^2} + frac{2y}{y_0} = 1 ]3. Use given tangent line intercepts: Comparing this with given intercepts: [ frac{x}{-75} + frac{y}{15} = 1 ] We equate corresponding terms, hence: [ frac{-2px}{y_0^2} = frac{x}{-75} quad text{and} quad frac{2y}{y_0} = frac{y}{15} ]4. Solve for ( y_0 ): For the second part: [ frac{2y}{y_0} = frac{y}{15} ] implies: [ 2 = frac{y_0}{15} implies y_0 = 30 ]5. Determine parameter ( p ): For the first part: [ frac{-2px}{900} = frac{x}{-75} ] simplifies to: [ -2px = frac{-x cdot 900}{75} implies -2px = -12x implies 2p = 12 implies p = 6 ]6. Equation of the direktrix and focus: The directrix for the parabola ( y^2 = 2px ) is given by ( x = -frac{p}{2} ): [ x = - frac{6}{2} = -3 ] The coordinates of the focus ( F left( frac{p}{2}, 0 right) ): [ F left( frac{6}{2}, 0 right) = F(3, 0) ] Conclusion:- The parameter ( p ) is ( 6 ).- The equation of the directrix is ( x = -3 ).- The coordinates of the focus are ( (3, 0) ).[boxed{p = 6, , x = -3, , F(3, 0)}]

🧠:Okay, let me try to figure out this problem step by step. So, we have two cryptocurrencies, Chukhoyn and Antonium. Both start at 1 on March 1st. Then, each day after that, Chukhoyn increases by 1. So, on day 1 (March 1), it's 1, day 2 (March 2) it's 2, day 3 it's 3, and so on. That part seems straightforward.Now, Antonium is a bit trickier. Its exchange rate each day is equal to the sum of the previous day's rates of Chukhoyn and Antonium divided by their product. Let me write that down as a formula to make it clearer.Let me denote:- C(n) as the exchange rate of Chukhoyn on day n.- A(n) as the exchange rate of Antonium on day n.Given the problem, we have:C(n) = 1 + (n - 1). Because on day 1 it's 1, day 2 it's 2, etc. So, C(n) = n.Wait, wait. Let me check that. If March 1 is day 1, then on day n, Chukhoyn is n dollars. So, yes, C(n) = n.For Antonium, the rate each day is the sum of the previous day's rates of Chukhoyn and Antonium divided by their product. So, the formula would be:A(n) = [C(n-1) + A(n-1)] / [C(n-1) * A(n-1)].Wait, hold on. The problem says "each day thereafter, it was equal to the sum of the previous day's rates of Chukhoyn and Antonium divided by their product." So, starting from day 2, Antonium's rate is (C(n-1) + A(n-1)) divided by (C(n-1) * A(n-1)).So, for each day n ≥ 2, A(n) = [C(n-1) + A(n-1)] / [C(n-1) * A(n-1)].Let me write that as:A(n) = [C(n-1) + A(n-1)] / [C(n-1) * A(n-1)].Simplify that expression. Let's see:First, note that C(n-1) = n - 1, since Chukhoyn increases by 1 each day starting from 1 on day 1. So, substituting C(n-1):A(n) = [(n - 1) + A(n-1)] / [(n - 1) * A(n-1)].Let me split the numerator:A(n) = (n - 1)/[(n - 1) * A(n-1)] + A(n-1)/[(n - 1) * A(n-1)].Simplifying each term:First term: (n - 1)/[(n - 1) * A(n-1)] = 1/A(n-1)Second term: A(n-1)/[(n - 1) * A(n-1)] = 1/(n - 1)Therefore, A(n) = 1/A(n-1) + 1/(n - 1)Hmm, that's an interesting recursive formula. So, each day, Antonium's exchange rate is the reciprocal of the previous day's Antonium rate plus the reciprocal of (n - 1).Given that A(1) = 1, since it starts at 1 on March 1.So, we need to compute A(92), which is May 31, the 92nd day.But calculating this recursively for 91 steps (from day 1 to day 92) might be tedious. Maybe there's a pattern or a closed-form formula?Let me try to compute the first few terms to see if a pattern emerges.Starting with:n = 1: A(1) = 1n = 2: A(2) = 1/A(1) + 1/(2 - 1) = 1/1 + 1/1 = 1 + 1 = 2n = 3: A(3) = 1/A(2) + 1/(3 - 1) = 1/2 + 1/2 = (1 + 1)/2 = 1n = 4: A(4) = 1/A(3) + 1/(4 - 1) = 1/1 + 1/3 ≈ 1 + 0.333... ≈ 1.333...n = 5: A(5) = 1/A(4) + 1/4 ≈ 1/(1.333...) + 0.25 ≈ 0.75 + 0.25 = 1Wait, interesting! So, A(1)=1, A(2)=2, A(3)=1, A(4)=1.333..., A(5)=1. Then let's do a few more.n = 6: A(6) = 1/A(5) + 1/5 = 1/1 + 0.2 = 1 + 0.2 = 1.2n = 7: A(7) = 1/A(6) + 1/6 ≈ 1/1.2 + 0.1666... ≈ 0.8333... + 0.1666... = 1n = 8: A(8) = 1/A(7) + 1/7 ≈ 1/1 + 1/7 ≈ 1 + 0.1428 ≈ 1.1428...n = 9: A(9) = 1/A(8) + 1/8 ≈ 1/1.1428 + 0.125 ≈ 0.875 + 0.125 = 1Wait a second, this seems to alternate between 1 and some other values. Let me check again.Wait, n=1:1, n=2:2, n=3:1, n=4:1.333..., n=5:1, n=6:1.2, n=7:1, n=8:1.1428..., n=9:1.So, every odd n after 1 seems to be 1? Wait, n=3,5,7,9: yes, on the odd days, starting from day 3, it's 1? Wait, but day 1 is 1, day 3 is 1, day 5 is 1, day7 is1, day9 is1. So, every even n, A(n) is some value, and every odd n, A(n)=1?Wait, let's check:At n=2: A(2)=2n=3: A(3)=1/A(2)+1/2=1/2 +1/2=1n=4: A(4)=1/A(3)+1/3=1/1 +1/3=1.333...n=5:1/A(4)+1/4= 1/(4/3) +1/4= 3/4 +1/4=1n=6:1/A(5)+1/5=1/1 +1/5=1.2n=7:1/A(6)+1/6=1/(6/5)+1/6=5/6 +1/6=1n=8:1/A(7)+1/7=1/1 +1/7≈1.1428n=9:1/A(8)+1/8=1/(8/7)+1/8=7/8 +1/8=1Yes, so it seems that for odd n≥1, A(n)=1, and for even n, A(n) = (n/2)/ (n/2 -1 +1) ? Wait, not sure.Wait, n=2:2, which is 2/1n=4:1.333...=4/3n=6:1.2=6/5n=8:1.1428≈8/7Ah! So, for even n, A(n) = n/(n - 2). Wait, let's check:n=2: 2/(2 - 2)= undefined. Wait, that can't be.Wait, n=2: 2/ (2 - 0) ? Not sure.Wait, n=2: A(2)=2=2/1n=4:4/3n=6:6/5n=8:8/7So, the pattern is for even n, A(n) = n/(n -1 +1) ? Hmm, maybe not. Wait, n=2:2/1, n=4:4/3, n=6:6/5, n=8:8/7.So, for even n, A(n) = n/(n -1). Wait, let's check:n=2:2/(2-1)=2/1=2 ✔️n=4:4/(4-1)=4/3 ✔️n=6:6/5 ✔️n=8:8/7 ✔️Yes, so for even n, A(n)=n/(n-1). So, if that's the case, then for even days, Antonium is n/(n-1), and for odd days, it's 1.But let's verify this hypothesis with the recursive formula.Assume that for even n, A(n)=n/(n-1), and for odd n ≥1, A(n)=1.Let's use mathematical induction.Base case:n=1 (odd): A(1)=1, which matches.n=2 (even): A(2)=2/1=2, which matches.Inductive step: Suppose for some even k=2m, A(k)=2m/(2m -1). Then, let's compute A(k+1), which is odd.A(k+1) =1/A(k) +1/kSince k is even, k=2m:A(k+1)=1/(2m/(2m -1)) +1/(2m)= (2m -1)/2m + 1/(2m)= [ (2m -1) +1 ] /2m= 2m /2m=1Which matches A(k+1)=1 for odd k+1.Now, for the next term, A(k+2) where k+2 is even:A(k+2)=1/A(k+1) +1/(k+1)But A(k+1)=1, so:A(k+2)=1/1 +1/(k+1)=1 +1/(k+1)But k=2m, so k+1=2m+1, and k+2=2m+2.Therefore, A(k+2)=1 +1/(2m+1)= (2m+2)/(2m+1)Which is (k+2)/(k+1). Since k+2=2m+2 and k+1=2m+1.Therefore, A(k+2)= (k+2)/(k+1). Which matches the pattern for even n.Thus, by induction, the formula holds:For odd n: A(n)=1For even n: A(n)=n/(n−1)Therefore, since May 31 is the 92nd day, which is even, A(92)=92/(92−1)=92/91≈1.010989...But the question asks for how much Antonium was worth on May 31, so we need to present the exact value, which is 92/91 dollars.So, the final answer is 92/91, which is approximately 1.010989, but as a fraction, it's 92 over 91.Wait, but let me confirm this with another approach. Let me try to compute A(92) using the formula.Since 92 is even, according to the pattern, A(92)=92/91. Let me check the recursive steps again for a larger n to ensure the pattern holds.For example, n=10:A(10)=1/A(9) +1/9=1/1 +1/9≈1.111..., which is 10/9≈1.111... So yes, 10/9.Similarly, n=12:12/11, etc.Therefore, the formula seems solid. So, since 92 is even, A(92)=92/91.Therefore, the answer is 92/91, which can be written as boxed{dfrac{92}{91}}.

✅:The exchange rate of Antonium on May 31 (the 92nd day) is determined by the recursive relation derived from the problem statement. By analyzing the pattern through induction, we find that for even days ( n ), the rate is ( frac{n}{n-1} ), and for odd days, it is 1. Since May 31 is the 92nd day (even), the value of Antonium is:[boxed{dfrac{92}{91}}]

🔑:Let ( C(n) ) and ( A(n) ) denote the value of the cryptocurrencies Chukhoyn and Antonium on day ( n ).We are given the following initial conditions and rules:1. On March 1 (day ( 1 )), both cryptocurrencies are worth 1, that is: [ C(1) = A(1) = 1 ]2. Each day, the value of Chukhoyn increases by 1. So, [ C(n) = n quad text{for any day } n. ]3. Each day, the value of Antonium is given by the sum of the previous day's values of Chukhoyn and Antonium divided by their product on the previous day: [ A(n) = frac{C(n-1) + A(n-1)}{C(n-1) cdot A(n-1)} ]Steps to find the value of Antonium on May 31 (92nd day):1. Base Case: [ C(1) = 1, quad A(1) = 1 ]2. Induction Step: Suppose ( A(n) ) follows a specific pattern for odd and even numbered days. Let's verify this by induction: - For odd days (e.g., day 1, day 3, ...): Assume ( A(n) = 1 ) for all odd ( n ). We already have ( A(1) = 1 ). Now, for an odd day, ( n = 2k + 1 ): - On the previous day ( (n-1 = 2k) ), let's suppose: [ A(2k) = frac{2k}{2k-1} ] and [ C(2k) = 2k ] Given the rule: [ A(2k + 1) = frac{C(2k) + A(2k)}{C(2k) cdot A(2k)} = frac{2k + frac{2k}{2k-1}}{2k cdot frac{2k}{2k-1}} = frac{2k + frac{2k}{2k-1}}{frac{4k^2}{2k-1}} = frac{2k(2k-1) + 2k}{4k^2} = frac{2k(2k-1) + 2k}{4k^2} = 1 ] This verifies that Antonium on odd days is ( 1 ). - For even days ( n = 2k ): [ A(2k) = frac{2k}{2k-1} ] To verify, assuming the previous ( A(2k - 1) = 1 ) and ( C(2k - 1) = 2k - 1 ): [ A(2k) = frac{C(2k-1) + A(2k-1)}{C(2k-1)A(2k-1)} = frac{2k-1 + 1}{(2k-1) cdot 1} = frac{2k}{2k-1} ]Therefore, Antonium follows this pattern:- On odd days, ( A(n) = 1 ),- On even days with number ( 2n ), ( A(2n) = frac{2n}{2n-1} ).Calculation for the 92nd day:The 92nd day is even:[A(92) = frac{2 cdot 46}{2 cdot 46 - 1} = frac{92}{91}]Conclusion:The value of Antonium on the 92nd day (May 31) is:[boxed{frac{92}{91} text{ dollars}}]

🧠:Okay, so I need to find the smallest constant C such that the inequality [ sum_{i < j} x_i x_j (x_i^2 + x_j^2) leq C left( sum_{i=1}^n x_i right)^4 ]holds for all non-negative real numbers ( x_1, x_2, ldots, x_n ) where ( n geq 2 ). Also, I have to determine when equality occurs. First, let me try to understand what the problem is asking. The left-hand side (LHS) is a sum over all pairs of terms ( x_i x_j (x_i^2 + x_j^2) ). The right-hand side (RHS) is a constant multiple of the fourth power of the sum of all ( x_i ). The goal is to find the minimal C such that this inequality is always true, regardless of the non-negative values of the ( x_i ).Since all variables are non-negative, I can maybe use some inequalities like Cauchy-Schwarz, AM-GM, or other symmetric inequalities. Also, considering symmetry, perhaps the maximum occurs when all variables are equal or when some variables are zero and others are equal. That's a common scenario in symmetric inequalities.Let me first try some small cases to get an idea. Let's take n = 2. Then, there is only one term in the LHS: ( x_1 x_2 (x_1^2 + x_2^2) ). The RHS would be ( C(x_1 + x_2)^4 ).For n = 2, let's compute C. Let me set variables to make it easier. Let’s let t = x_1/x_2 (assuming x_2 ≠ 0, since variables are non-negative). Let’s set x_2 = 1 and x_1 = t. Then the inequality becomes:( t * 1 * (t^2 + 1) ≤ C(t + 1)^4 ).Simplify LHS: ( t(t^2 + 1) ).RHS: ( C(t + 1)^4 ).So we need to find the minimal C such that for all t ≥ 0,( t(t^2 + 1) ≤ C(t + 1)^4 ).To find the minimal C, we can compute the maximum of ( frac{t(t^2 + 1)}{(t + 1)^4} ) over t ≥ 0.Let’s compute this function. Let f(t) = ( frac{t(t^2 + 1)}{(t + 1)^4} ).To find its maximum, take derivative and set to zero.First, compute f(t):f(t) = ( frac{t^3 + t}{(t + 1)^4} ).Compute derivative f’(t):Using quotient rule:f’(t) = [ (3t^2 + 1)(t + 1)^4 - (t^3 + t)(4)(t + 1)^3 ] / (t + 1)^8Factor out (t + 1)^3 in numerator:= [ (3t^2 + 1)(t + 1) - 4(t^3 + t) ] / (t + 1)^5Expand numerator:(3t^2 + 1)(t + 1) = 3t^3 + 3t^2 + t + 14(t^3 + t) = 4t^3 + 4tSubtract: 3t^3 + 3t^2 + t + 1 - 4t^3 - 4t = -t^3 + 3t^2 - 3t + 1Set numerator equal to zero:-t^3 + 3t^2 - 3t + 1 = 0Multiply both sides by -1:t^3 - 3t^2 + 3t - 1 = 0Notice that this factors as (t - 1)^3 = 0. So t = 1 is a triple root.Therefore, critical point at t = 1.Now check f(1):f(1) = (1*(1 + 1))/(2^4) = 2 / 16 = 1/8.Check behavior as t approaches 0:f(t) ≈ t / 1 → 0.As t approaches infinity:f(t) ≈ t^3 / t^4 = 1/t → 0.Thus, maximum at t = 1 is 1/8. Hence, for n=2, C = 1/8.So for n=2, the minimal C is 1/8, achieved when x1 = x2.Now, let's check n=3. Maybe the minimal C is the same? Or different?Wait, but the problem states n ≥ 2, so maybe the answer is C=1/8 for all n ≥2? But I need to verify.Wait, let's test n=3. Suppose for n=3, we set all variables equal. Let x1 = x2 = x3 = t. Then sum of variables is 3t. So RHS is C*(3t)^4 = 81 C t^4.LHS is sum over i<j of x_i x_j (x_i^2 + x_j^2). For three variables, there are three terms: each pair.Compute each term:For x1, x2: t * t (t^2 + t^2) = t^2 * 2t^2 = 2 t^4Similarly, x1, x3: 2 t^4x2, x3: 2 t^4Total LHS: 6 t^4Set LHS ≤ C*RHS:6 t^4 ≤ 81 C t^4 → 6 ≤ 81 C → C ≥ 6 /81 = 2/27 ≈ 0.07407.But for n=2, the required C was 1/8 ≈0.125, which is larger. So if we set n=3, and all variables equal, then we need a smaller C than for n=2. But the problem says n ≥2 and find the minimal C that works for all n ≥2. Wait, no. Wait, the problem is: for a fixed integer n ≥2, find the minimal C such that for all non-negative reals x1,...,xn, the inequality holds. So perhaps for each n, the minimal C is different. Wait, but the problem says "Find the smallest constant C such that for all non-negative reals x1,...,xn...". So the answer depends on n. Wait, but the problem states "Find the smallest constant C" for a fixed n. So the answer is dependent on n, but the problem wants a general answer for any n ≥2? Wait, the problem says "Let n ≥2 be a fixed integer. Find the smallest constant C...". So for each n, C is different. Wait, but maybe the answer is 1/8, regardless of n, but when variables are two, the maximum ratio occurs when only two variables are non-zero. Let's check.Suppose n=3. Let me take two variables as 1 and one as 0. Let x1 = x2 =1, x3=0. Then sum x_i =2. The LHS is sum over i<j. So pairs (1,1), (1,0), (1,0). The terms:x1x2(x1^2 + x2^2) =1*1*(1+1)=2x1x3(x1^2 +x3^2)=1*0*(1+0)=0x2x3(x2^2 +x3^2)=1*0*(1+0)=0Total LHS=2. RHS=C*(2)^4=16C. So 2 ≤16C→ C≥1/8. But in the case where all three variables are equal, we found C≥2/27≈0.074, which is smaller than 1/8≈0.125. Thus, the maximum ratio occurs when two variables are equal and the rest are zero. Therefore, for n=3, the minimal C is still 1/8.Wait, that seems conflicting. Let me check again.If in n=3, if we take two variables as 1 and one as 0, then LHS is 2 and RHS is 16C. So for inequality to hold, C must be at least 2/16=1/8. On the other hand, if we take all three variables equal, then the required C is 6/(81)=2/27≈0.074, which is less than 1/8. Therefore, the critical case is when two variables are non-zero and others are zero. Therefore, the maximum ratio LHS/RHS is 1/8, same as n=2. Therefore, for n=3, C=1/8. Similarly, perhaps for any n≥2, the maximum occurs when two variables are equal and others are zero, hence C=1/8 regardless of n≥2.But let me test n=4. Take two variables as 1 and the other two as 0. Then, LHS is sum over all pairs. How many pairs contribute? The pairs between the two 1s: (1,1) which is 2. The pairs between 1 and 0: (1,0) which contribute 0. The pairs between 0s: 0. So total LHS=2. The RHS is C*(1+1+0+0)^4= C*16. So again, 2 ≤16C→C≥1/8. If we take all variables equal, then sum x_i =4t. LHS is sum_{i<j} x_i x_j (x_i^2 +x_j^2). If each x_i =t, then each term is t*t*(t^2 + t^2)=2 t^4. The number of terms is C(4,2)=6. So total LHS=12 t^4. RHS= C*(4t)^4=256 C t^4. Thus, 12 t^4 ≤256 C t^4→ C≥12/256=3/64≈0.0469, which is smaller than 1/8. So again, the maximum ratio occurs when two variables are non-zero. Therefore, even for n=4, C=1/8.Therefore, perhaps for any n≥2, the minimal C is 1/8, achieved when two variables are equal and the rest are zero. Let me check with n=5. Suppose two variables are 1, others zero. LHS is 2, RHS is C*(2)^4=16C. So C≥1/8. If all variables equal, sum is 5t. LHS is C(5,2)*2 t^4=10*2 t^4=20 t^4. RHS= C*(5t)^4=625 C t^4. So 20 ≤625 C → C≥20/625=4/125=0.032, which is less than 1/8. So again, two variables case gives higher C.Thus, the pattern seems to hold. For any n≥2, the maximum ratio of LHS/RHS occurs when exactly two variables are non-zero and equal, and the rest are zero. Therefore, the minimal C is 1/8, regardless of n≥2.Wait, but let me think again. Suppose for some n, there's a configuration with more variables that gives a higher ratio. For example, if we take three variables non-zero. Let's check with n=3. Let me take x1 =x2 =1, x3 = t. Let's see if for some t>0, the ratio is higher than 1/8.Compute LHS:Sum over all pairs:x1x2(x1^2 +x2^2) + x1x3(x1^2 +x3^2) + x2x3(x2^2 +x3^2)=1*1*(1 +1) +1*t*(1 + t^2) +1*t*(1 + t^2)= 2 + 2t(1 + t^2)Sum x_i =1 +1 +t=2 + tRHS= C*(2 + t)^4So the ratio is [2 + 2t + 2t^3]/ (2 + t)^4. Let’s see if this can exceed 1/8.Take t approaching 0: ratio approaches [2 +0 +0]/16=1/8.Take t approaching infinity: numerator ~ 2t^3, denominator ~ t^4, so ratio ~2/t→0.At t=1: ratio= [2 +2 +2]/81=6/81=2/27≈0.074 <1/8.Perhaps the maximum is still at t=0? Wait, when t approaches 0, the ratio approaches 1/8. So maybe even when adding a small third variable, the ratio remains at most 1/8. Therefore, the maximum ratio is achieved when the third variable is zero. Thus, the case of two variables gives the maximum.Alternatively, take three variables with two variables equal to 1 and the third equal to s, and see if for some s, the ratio exceeds 1/8.But in the limit as s approaches 0, the ratio approaches 1/8. For any s>0, the ratio is less. Therefore, the maximum is achieved when s=0. Hence, for n=3, the maximum ratio is 1/8. Similarly, for higher n, adding more variables can only distribute the sum more, but the maximum ratio is still when two variables are non-zero.Therefore, conjecture: For any n ≥2, the minimal C is 1/8, achieved when two variables are equal and the rest are zero.But to confirm, let's try another configuration. Suppose for n=3, take x1 =a, x2 =a, x3 =b. Let's see if we can make the ratio LHS/RHS larger than 1/8.Compute LHS:x1x2(x1^2 +x2^2) + x1x3(x1^2 +x3^2) + x2x3(x2^2 +x3^2)= a*a*(2a^2) + a*b*(a^2 + b^2) + a*b*(a^2 + b^2)= 2a^4 + 2ab(a^2 + b^2)Sum x_i =2a + bRHS=C*(2a + b)^4We need to see if there exist a,b >0 such that [2a^4 + 2ab(a^2 + b^2)] / (2a + b)^4 >1/8.Let’s set a=1 and vary b. Let’s let b =k, so variables are x1=1, x2=1, x3=k.Compute LHS=2*1^4 + 2*1*k*(1 + k^2)=2 + 2k(1 +k^2)=2 + 2k + 2k^3.Sum x_i=2 +k.RHS=C*(2 +k)^4.So ratio is [2 +2k +2k^3]/(2 +k)^4. Let’s check if this ratio can exceed 1/8.Let’s set k=1: ratio= (2 +2 +2)/81=6/81=2/27≈0.074 <1/8≈0.125.k=0.5: ratio= [2 +1 +2*(0.125)]/(2.5)^4= [3 +0.25]/39.0625≈3.25/39.06≈0.083 <0.125.k=2: ratio= [2 +4 +16]/(4)^4=22/256≈0.0859 <0.125.k approaching 0: ratio approaches [2 +0 +0]/16=1/8≈0.125.So maximum at k approaching 0, which is the case where the third variable is negligible. Therefore, even in this configuration, the maximum ratio is 1/8.Therefore, the minimal C is indeed 1/8 for n≥2, and equality occurs when two variables are equal and the rest are zero.But to be thorough, let's consider another case where three variables are non-zero and equal. Let x1=x2=x3=t, and others zero. Then for n=3, sum x_i=3t. LHS= sum_{i<j} x_i x_j (x_i^2 +x_j^2). Each pair:t*t*(t^2 + t^2)=2 t^4. There are 3 pairs, so LHS=6 t^4.RHS=C*(3t)^4=81 C t^4.Thus, ratio=6/81=2/27≈0.074 <1/8. Hence, even with three variables, the ratio is lower.Alternatively, take one variable large and others small. Let’s take x1=M, x2=1, and others zero. Let’s see how the ratio behaves as M increases.For n=3, LHS is x1x2(x1^2 +x2^2) +x1x3(...) +x2x3(...). But x3=0, so LHS= M*1*(M^2 +1). Sum x_i= M +1. RHS= C*(M +1)^4. So ratio= [M(M^2 +1)] / (M +1)^4. Let’s analyze this as M approaches infinity.As M→infty, numerator ~M^3, denominator ~M^4, so ratio ~1/M→0. At M=1, ratio= (1*2)/16=2/16=1/8. For M>1, ratio decreases. For M<1, let’s set M approaching 0. Then ratio approaches 0. So maximum at M=1, ratio=1/8. Therefore, again, the maximum occurs when two variables are equal (M=1 and x2=1), which gives ratio 1/8.Therefore, regardless of how we set the variables, the maximum ratio of LHS/RHS is 1/8, achieved when two variables are equal and others zero. Thus, the minimal C is 1/8 for any n≥2.To formalize this, we can use the following approach:We need to show that for any n≥2 and any non-negative reals x1,...,xn,sum_{i<j} x_i x_j (x_i^2 +x_j^2) ≤ (1/8) (sum x_i)^4,with equality when two variables are equal and the rest are zero.First, note that the LHS can be rewritten as:sum_{i<j} x_i x_j (x_i^2 +x_j^2) = sum_{i<j} (x_i^3 x_j + x_i x_j^3).This is equal to sum_{i≠j} x_i^3 x_j, since each term x_i^3 x_j appears once for each pair (i,j) where i≠j. However, since in the original sum i<j, each unordered pair is counted once, so the sum over i<j is half of the sum over i≠j. Wait:Wait, sum_{i<j} (x_i^3 x_j +x_i x_j^3) = sum_{i<j} x_i^3 x_j + sum_{i<j} x_i x_j^3.But sum_{i<j} x_i^3 x_j + sum_{i<j} x_i x_j^3 = sum_{i≠j} x_i^3 x_j. Because for i≠j, if i < j, then x_i^3 x_j is in the first sum, and if j < i, then x_j^3 x_i is in the second sum. So indeed, sum_{i<j} x_i x_j (x_i^2 +x_j^2) = sum_{i≠j} x_i^3 x_j.Alternatively, note that for each pair i and j, the term is x_i x_j (x_i^2 +x_j^2) =x_i^3 x_j +x_i x_j^3. Thus, summing over all i < j gives exactly the sum over all unordered pairs of x_i^3 x_j +x_j^3 x_i. So it's equivalent to sum_{i≠j} x_i^3 x_j / 2, since each unordered pair is counted twice in the sum over i≠j.Wait, no. If we sum over i≠j, then each ordered pair (i,j) where i≠j is considered. So sum_{i≠j} x_i^3 x_j would include both (i,j) and (j,i). However, in our original sum, for i < j, we have x_i^3 x_j +x_j^3 x_i. So the sum over i < j of x_i x_j (x_i^2 +x_j^2) is equal to sum_{i < j} (x_i^3 x_j +x_j^3 x_i). Which is the same as summing over all i ≠ j, but dividing by 2 because each pair is counted twice. Hence,sum_{i < j} x_i x_j (x_i^2 +x_j^2) = (1/2) sum_{i ≠ j} x_i^3 x_j.Alternatively, not sure if that helps. Maybe another approach.Consider the inequality we need to prove:sum_{i < j} x_i x_j (x_i^2 +x_j^2) ≤ (1/8) (sum x_i)^4.We can use the Cauchy-Schwarz inequality or other symmetric inequalities. Let’s try to express both sides in terms of symmetric sums.First, note that (sum x_i)^4 expands to:(sum x_i)^4 = sum x_i^4 + 4 sum_{i < j} x_i^3 x_j + 6 sum_{i < j} x_i^2 x_j^2 + 12 sum_{i < j < k} x_i^2 x_j x_k + 24 sum_{i < j < k < l} x_i x_j x_k x_l.But this might be complicated. Alternatively, note that the LHS is sum_{i < j} x_i x_j (x_i^2 +x_j^2) = sum_{i < j} x_i^3 x_j + sum_{i < j} x_i x_j^3 = sum_{i ≠ j} x_i^3 x_j / 2.Because for each i ≠ j, we have x_i^3 x_j and x_j^3 x_i, and when we sum over i < j, each pair is counted once. So sum_{i < j} (x_i^3 x_j +x_j^3 x_i) = sum_{i ≠ j} x_i^3 x_j / 2.Wait, no. Suppose i < j: then in the sum over i < j, we have x_i^3 x_j +x_j^3 x_i. So for each unordered pair {i,j}, we have these two terms. So sum_{i < j} (x_i^3 x_j +x_j^3 x_i) = sum_{i < j} x_i^3 x_j + sum_{i < j} x_j^3 x_i = sum_{i < j} x_i^3 x_j + sum_{i < j} x_j^3 x_i.But sum_{i < j} x_j^3 x_i is the same as sum_{i < j} x_i^3 x_j with i and j swapped, which is the same as sum_{j < i} x_i^3 x_j. So overall, the total sum is sum_{i ≠ j} x_i^3 x_j.Yes, because summing over i < j and j < i covers all i ≠ j. Hence,sum_{i < j} (x_i^3 x_j +x_j^3 x_i) = sum_{i ≠ j} x_i^3 x_j.Therefore, LHS = sum_{i ≠ j} x_i^3 x_j.So the inequality is:sum_{i ≠ j} x_i^3 x_j ≤ (1/8) (sum x_i)^4.Alternatively, we can write sum_{i ≠ j} x_i^3 x_j = sum_i x_i^3 sum_{j ≠ i} x_j.But sum_{j ≠ i} x_j = S - x_i where S = sum x_i.Thus,sum_{i ≠ j} x_i^3 x_j = sum_i x_i^3 (S - x_i) = S sum_i x_i^3 - sum_i x_i^4.Therefore, the inequality becomes:S sum x_i^3 - sum x_i^4 ≤ (1/8) S^4.But I'm not sure if this helps directly. Maybe we can use Hölder's inequality. Hölder's inequality states that for sequences a_i, b_i, c_i, etc., the product sums can be bounded by the product of norms. However, I need to think of a way to apply it here.Alternatively, consider the case where two variables are non-zero. Suppose x1 = x2 = t and others zero. Then sum x_i =2t. LHS= x1x2(x1^2 +x2^2)=t*t*(2t^2)=2t^4. RHS=(1/8)(2t)^4= (1/8)16t^4=2t^4. So equality holds.If we have more variables, then the LHS can be written as the sum over all pairs. If we add more variables (even with some non-zero), the question is whether this sum can exceed 1/8 times the fourth power of the sum. From previous examples, when variables are spread out more, the LHS becomes smaller relative to the RHS.To prove the inequality in general, perhaps we can use the Cauchy-Schwarz inequality or a homogenization approach.Since the inequality is homogeneous, we can assume that sum x_i =1. Then the inequality becomes:sum_{i < j} x_i x_j (x_i^2 +x_j^2) ≤ C.We need to find the maximum of sum_{i < j} x_i x_j (x_i^2 +x_j^2) under the constraint sum x_i =1 and x_i ≥0.Using Lagrange multipliers, perhaps. Let’s set f(x1,...,xn) = sum_{i < j} x_i x_j (x_i^2 +x_j^2) and constraint g(x1,...,xn)=sum x_i -1=0.But this might be complicated due to the number of variables. Alternatively, consider that the maximum occurs at the boundary of the domain, which in this case is when as many variables as possible are zero. Because of the symmetry, the maximum should occur when only two variables are non-zero.Assume that only two variables, say x and y, are non-zero, with x + y =1. Then f(x,y)=xy(x^2 + y^2). We need to maximize this over x,y ≥0 with x + y =1.Let’s substitute y=1 -x. Then f(x)=x(1 -x)(x^2 + (1 -x)^2).Compute f(x):x(1 -x)(x^2 +1 -2x +x^2)=x(1 -x)(2x^2 -2x +1)Let’s compute this function for x in [0,1].Take derivative f’(x):Let’s denote f(x)=x(1 -x)(2x^2 -2x +1).First, expand:= x(1 -x)(2x^2 -2x +1)Let’s compute:Multiply (1 -x)(2x^2 -2x +1):=2x^2(1 -x) -2x(1 -x) +1(1 -x)=2x^2 -2x^3 -2x +2x^2 +1 -xCombine like terms:2x^2 +2x^2=4x^2-2x^3-2x -x= -3x+1So the expression becomes:x(4x^2 -2x^3 -3x +1)=4x^3 -2x^4 -3x^2 +xNow take derivative:f’(x)=12x^2 -8x^3 -6x +1Set to zero:12x^2 -8x^3 -6x +1=0This seems a bit messy. Let’s try plugging x=1/2:f’(1/2)=12*(1/4) -8*(1/8) -6*(1/2) +1=3 -1 -3 +1=0. So x=1/2 is a critical point.Evaluate f(1/2)= (1/2)(1 -1/2)(2*(1/2)^2 -2*(1/2) +1)= (1/2)(1/2)(2*(1/4) -1 +1)= (1/4)(0.5 -1 +1)= (1/4)(0.5)=1/8.Now check other critical points. Suppose there are others. Let’s attempt to solve 12x^2 -8x^3 -6x +1=0.Let me factor this polynomial. Let’s write as -8x^3 +12x^2 -6x +1=0. Multiply both sides by -1:8x^3 -12x^2 +6x -1=0.Try rational roots. Possible roots are x=1/1,1/2,1/4,1/8.Test x=1:8 -12 +6 -1=1≠0.x=1/2:8*(1/8) -12*(1/4) +6*(1/2) -1=1 -3 +3 -1=0. So x=1/2 is a root.Thus, factor the polynomial as (x -1/2)(quadratic). Use polynomial division.Divide 8x^3 -12x^2 +6x -1 by (x -1/2). Let’s use synthetic division.Root at x=1/2:Coefficients:8 | -12 |6 |-1Bring down 8Multiply by 1/2:4Add to next coefficient: -12 +4= -8Multiply by1/2: -4Add to next coefficient:6 + (-4)=2Multiply by1/2:1Add to last coefficient: -1 +1=0.Thus, the polynomial factors as (x -1/2)(8x^2 -8x +2).Set 8x^2 -8x +2=0.Discriminant=64 -64=0. So double root x=(8)/(2*8)=0.5. Hence, the polynomial factors as (x -1/2)^2(8x -4). Wait, no. Wait, 8x^2 -8x +2=0.Using quadratic formula:x=(8±sqrt(64-64))/16=8/16=1/2. So the quadratic is 8(x -1/2)^2. Hence, the original polynomial is (x -1/2)^3*8. Wait, but we factored out (x -1/2) once, so it's (x -1/2)^2*(8x -4). Wait, no. Let me check:We have:8x^3 -12x^2 +6x -1 = (x -1/2)(8x^2 -8x +2)But 8x^2 -8x +2=2(4x^2 -4x +1)=2(2x -1)^2. Hence, the factorization is (x -1/2)*2*(2x -1)^2.Thus, roots at x=1/2 (double root) and x=1/2. So the only real root is x=1/2 with multiplicity 3.Therefore, the derivative 12x^2 -8x^3 -6x +1=0 only has x=1/2 as a real root (triple root). Therefore, the function f(x) has critical point at x=1/2 and the endpoints x=0 and x=1.At x=0 and x=1, f(x)=0. At x=1/2, f(x)=1/8. Hence, the maximum is indeed 1/8 when x=y=1/2.Thus, when sum x_i=1, the maximum of LHS is 1/8, achieved when two variables are 1/2 and the rest zero. Hence, the minimal C is 1/8.Therefore, for any n≥2, the smallest constant C is 1/8, and equality occurs when two variables are equal and the others are zero.To confirm this, consider any number of variables. If we have more than two variables, the maximum ratio LHS/RHS is still achieved by setting all but two variables to zero. This is because adding more variables distributes the sum more, leading to a lower concentration in the terms of the LHS, which involve products of variables. The terms in the LHS are pairwise products with quadratic terms, so having more variables would spread out the products, reducing their sum relative to the fourth power of the total sum, which grows faster when variables are added.Therefore, the conclusion is that the minimal C is 1/8 for any n≥2, with equality when two variables are equal and the rest are zero.

✅:The smallest constant ( C ) is ( boxed{dfrac{1}{8}} ). Equality occurs when exactly two of the variables are equal and the rest are zero.Step-by-Step Explanation:1. Problem Understanding: We need to find the smallest constant ( C ) such that for all non-negative reals ( x_1, x_2, ldots, x_n ) (( n geq 2 )): [ sum_{i < j} x_i x_j (x_i^2 + x_j^2) leq C left( sum_{i=1}^n x_i right)^4. ] Determine when equality holds.2. Case Analysis for Small ( n ): - For ( n = 2 ): The inequality reduces to ( x_1 x_2 (x_1^2 + x_2^2) leq C (x_1 + x_2)^4 ). By setting ( x_1 = x_2 = t ), we find ( C = frac{1}{8} ). - For ( n geq 3 ): Testing configurations where variables are equal or some are zero shows the maximum ratio ( text{LHS}/text{RHS} ) is still ( frac{1}{8} ).3. General Case: - Symmetry and Homogeneity: The inequality is homogeneous, allowing normalization ( sum x_i = 1 ). - Maximizing the LHS: Using Lagrange multipliers or analyzing critical points, the maximum occurs when two variables are non-zero and equal (( x = y = frac{1}{2} )) and others are zero, yielding ( text{LHS} = frac{1}{8} ).4. Conclusion: - The maximum value of the left-hand side relative to the right-hand side is ( frac{1}{8} ), regardless of ( n geq 2 ). - Equality Condition: Achieved when exactly two variables are equal and all others are zero.Thus, the minimal constant is ( C = frac{1}{8} ), with equality when two variables are equal and the rest are zero.

🔑:# Initial Setup:Given the inequality:[sum_{i<j} x_i x_j (x_i^2 + x_j^2) leq C left( sum_{i=1}^{n} x_i right)^4]we aim to find the smallest constant (C) and determine when equality occurs.# Proof by a member of the Chinese team at IMO:1. Expression for the Fourth Power: The fourth power of the sum of (x_i) can be expanded as follows: [ left( sum_{i=1}^n x_i right)^4 = left( sum_{i=1}^n x_i^2 + 2 sum_{i < j} x_i x_j right)^2 ]2. Applying the Inequality: Applying the Cauchy-Schwarz inequality to the square: [ left( sum_{i=1}^n x_i^2 + 2 sum_{i < j} x_i x_j right)^2 geq 4 left( sum_{i=1}^n x_i^2 right) left( 2 sum_{i<j} x_i x_j right) ]3. Simplifying: Simplifying the expression above, we get: [ 4 left( sum_{i=1}^n x_i^2 right) left( 2 sum_{i<j} x_i x_j right) = 8 sum_{i<j} left( x_i x_j sum_{k=1}^n x_k^2 right) ] Notice that ( sum_{k=1}^n x_k^2 geq x_i^2 + x_j^2 ) for any ( i, j ).4. Relating Both Sides: Therefore, [ 8 sum_{i<j} left( x_i x_j sum_{k=1}^n x_k^2 right) geq 8 sum_{i<j} x_i x_j (x_i^2 + x_j^2) ] which shows: [ left( sum_{i=1}^n x_i right)^4 geq 8 sum_{i<j} x_i x_j (x_i^2 + x_j^2) ]5. Conclusion on (C): Thus, (C leq frac{1}{8}).# Checking for Equality:6. Testing Specific Configuration: Let (x_3 = x_4 = cdots = x_n = 0), and testing with ( n = 2 ): [ left( x_1 + x_2 right)^4 = left( x_1^2 + x_2^2 + 2 x_1 x_2 right)^2 geq 8 x_1 x_2 (x_1^2 + x_2^2) ] For equality, need (x_1^2 + x_2^2 = 2 x_1 x_2) which implies: [ (x_1 - x_2)^2 = 0 implies x_1 = x_2 ]So, equality occurs if exactly two ( x_i ) are equal and the rest are zero.# Alternative solution by Gerhard Woeginger:1. Base Case for Induction: For ( n = 2 ): [ sum_{i<j} x_i x_j (x_i^2 + x_j^2) leq frac{1}{8} left( sum_{i=1}^n x_i right)^4 ] Simplifies to: [ (x_1 - x_2)^4 geq 0 ] which is true and equality holds if (x_1 = x_2).2. Induction Step: For ( n > 2 ): Assume (x_1 + cdots + x_n = 1) and (x_1 leq x_2 leq cdots leq x_n). Replace (x_1) and (x_2) by (0) and (x_1 + x_2), respectively. The sum on the right hand side remains the same but the sum on the left increases by: [ (x_1 + x_2)^3 S - (x_1^3 + x_2^3) S - x_1 x_2 (x_1^2 + x_2^2) ] where (S = x_3 + x_4 + cdots + x_n) and (S geq frac{1}{3}). This simplifies to: [ x_1 x_2 (x_1 + x_2 - x_1^2 - x_2^2) > 0 ] unless (x_1 = 0), proving the statement by induction.# Conclusion:The smallest constant (C) is (frac{1}{8}) with equality iff exactly two ( x_i ) are equal and the rest are zero.[boxed{frac{1}{8}}]

🧠:Okay, let's try to figure out this problem step by step. So, we have two glasses, each with the same volume initially: one has water, and the other has alcohol. The question is whether we can transfer liquid back and forth between them such that eventually, the first glass ends up with a higher percentage of alcohol than the second glass. Hmm, interesting.First, let me make sure I understand the problem correctly. Both glasses start with the same amount of their respective liquids. Let's say each has 100 ml for simplicity. So, Glass A has 100 ml water, Glass B has 100 ml alcohol. We can transfer any amount from one to the other, and after each transfer, the mixture is even, right? So, if I pour some amount from Glass A to B, then Glass B will have a mixture of alcohol and water, and vice versa. Then, maybe we can do this multiple times to try to get the first glass to have a higher concentration of alcohol than the second. The question is whether this is possible.Let me start by considering a simple case. Suppose I transfer some amount from Glass B (alcohol) to Glass A (water). Let's say I transfer 10 ml of alcohol to Glass A. Now, Glass A has 110 ml total, which is 100 ml water and 10 ml alcohol. So, the concentration of alcohol in Glass A is 10/110 ≈ 9.09%. Meanwhile, Glass B now has 90 ml alcohol left. If I transfer back 10 ml from Glass A to Glass B, what happens?Wait, the mixture in Glass A is 10 ml alcohol and 100 ml water. So, if I take 10 ml from Glass A, which is a mixture, the amount of alcohol transferred back would be (10/110)*10 ≈ 0.909 ml alcohol and 9.091 ml water. So, after transferring back 10 ml to Glass B, Glass B would have 90 ml alcohol + 0.909 ml alcohol = 90.909 ml alcohol and 9.091 ml water. Total volume in B is 90 + 0.909 + 9.091 = 100 ml again. So, concentration in Glass B would be 90.909/100 ≈ 90.909%, and Glass A now has 100 ml water - 9.091 ml water = 90.909 ml water and 10 ml alcohol - 0.909 ml alcohol ≈ 9.091 ml alcohol. So, Glass A would have ≈9.091/100 ≈9.091% alcohol. So, after transferring back, Glass B has higher concentration than Glass A. So, in this case, transferring back and forth doesn't help in making Glass A have higher concentration than Glass B.Wait, but maybe if we do more transfers? Let's think. Suppose after transferring 10 ml alcohol to A, making A 10% alcohol, then transferring 10 ml back, which makes B ≈90.909% alcohol. Then, if we transfer another 10 ml from B to A. But now, B has 90.909% alcohol. So transferring 10 ml from B to A would add 9.0909 ml alcohol and 0.9091 ml water to Glass A. So Glass A would now have 9.091 + 9.0909 ≈18.182 ml alcohol and 90.909 + 0.9091 ≈91.818 ml water. Total volume in A is 110 ml again. So concentration is 18.182/110 ≈16.53%. Then, transferring back 10 ml from A to B would take away (16.53% of 10 ml) ≈1.653 ml alcohol and 8.347 ml water. So, B would then have 90.909 - 9.0909 + 1.653 ≈ let's see, original after first transfer back was 90.909 ml alcohol and 9.091 ml water. Then transferring 10 ml from B to A added 9.0909 alcohol and 0.9091 water to A, so B had 90.909 - 9.0909 = 81.8181 alcohol and 9.091 - 0.9091 = 8.1819 water. Then transferring back 10 ml from A (which had 18.182 alcohol and 91.818 water) would take 1.653 alcohol and 8.347 water. So B would now have 81.8181 + 1.653 ≈83.471 alcohol and 8.1819 + 8.347 ≈16.529 water. Total in B is 100 ml. So concentration in B is 83.471/100 ≈83.47%, while A now has 18.182 -1.653 ≈16.529 alcohol and 91.818 -8.347 ≈83.471 water. So concentration in A is 16.529/100 ≈16.53%. So, after two transfers, A is at ~16.5%, B at ~83.5%. So still, B is higher. Hmm.It seems like each time we do a transfer from B to A and back, the concentration in A increases, but the concentration in B decreases, but B is still higher. Maybe if we keep doing this multiple times? Let's see if there's a pattern here. Let's try to model this.Let me denote the amount transferred each time as x. Let's assume we transfer x amount each time. Let's start with Glass A: V volume of water, Glass B: V volume of alcohol. Let's transfer x from B to A. Now, Glass A has V water + x alcohol. Concentration in A: x/(V + x). Glass B now has V - x alcohol. Then, transfer x back from A to B. The amount of alcohol transferred back is (x/(V + x)) * x = x²/(V + x). So, Glass A now has x - x²/(V + x) = xV/(V + x) alcohol and V water. Glass B has (V - x) + x²/(V + x) alcohol and x - x²/(V + x) water. Let's compute the concentration in each.Concentration in A: [xV/(V + x)] / V = x/(V + x).Concentration in B: [ (V - x) + x²/(V + x) ] / V.Let me compute the numerator for B's alcohol:(V - x) + x²/(V + x) = [ (V - x)(V + x) + x² ] / (V + x ) = [V² - x² + x²]/(V + x) = V²/(V + x).So concentration in B is [V²/(V + x)] / V = V/(V + x).Therefore, after transferring x from B to A and then x back from A to B, the concentration in A is x/(V + x), and in B is V/(V + x). So, since V > x (assuming we transferred less than the total volume), then V/(V + x) > x/(V + x). So B still has higher concentration. So this suggests that doing this once doesn't help.But what if we repeat the process? Let's see.After the first transfer, concentrations are A: x/(V + x), B: V/(V + x).Now, transfer another x from B to A. But wait, after the first cycle, B has V/(V + x) concentration. So transferring x from B to A would transfer x*(V/(V + x)) alcohol and x*(x/(V + x)) water. Wait, but B's total volume is still V. Wait, after the first transfer back, the total volume in each glass is V again, right? Because we transferred x out and then x back. Wait, let's check:Initially, each has V. Transfer x from B to A: A has V + x, B has V - x. Then transfer x back from A to B: A has V, B has V. So yes, volumes are restored. But concentrations have changed.So, after first cycle, Glass A has xV/(V + x) alcohol and V water. Wait, no, total volume in A is V, so if it has xV/(V + x) alcohol, then water is V - xV/(V + x) = V²/(V + x). Similarly, Glass B has V²/(V + x) alcohol and xV/(V + x) water.So, in the second cycle, if we transfer x from B to A again. Now, Glass B has concentration V²/(V + x) alcohol. So transferring x from B to A would transfer x*(V²/(V + x))/V = xV/(V + x) alcohol and x*(xV/(V + x))/V = x²/(V + x) water. Wait, but B has total V volume. So concentration of alcohol in B is V²/(V + x) / V = V/(V + x). So transferring x amount from B would transfer x*(V/(V + x)) alcohol and x*(x/(V + x)) water. So, adding that to Glass A.After transferring x from B to A, Glass A has:Alcohol: xV/(V + x) + xV/(V + x) = 2xV/(V + x)Water: V²/(V + x) + x²/(V + x) = (V² + x²)/(V + x)Total volume in A: V + x.Then, transfer x back from A to B. The concentration in A is [2xV/(V + x)] / (V + x) = 2xV/(V + x)^2.So, transferring x from A to B would transfer:Alcohol: 2xV/(V + x)^2 * x = 2x²V/(V + x)^2Water: (V² + x²)/(V + x)/(V + x) * x = x(V² + x²)/(V + x)^2Therefore, after transferring back, Glass A has:Alcohol: 2xV/(V + x) - 2x²V/(V + x)^2 = 2xV(V + x - x)/(V + x)^2 = 2xV²/(V + x)^2Water: (V² + x²)/(V + x) - x(V² + x²)/(V + x)^2 = (V² + x²)(V + x - x)/(V + x)^2 = (V² + x²)V/(V + x)^2Total volume in A: V.Glass B now has:Alcohol: V²/(V + x) + 2x²V/(V + x)^2Water: xV/(V + x) + x(V² + x²)/(V + x)^2This is getting complicated. Let's see if we can find a pattern or a general formula. Alternatively, maybe there's a smarter way to approach this problem.Wait, maybe instead of discrete transfers, we can think in terms of proportions. Let's consider the total amount of alcohol and water in each glass.Initially, total alcohol is 100 ml in Glass B, total water is 100 ml in Glass A. After any number of transfers, the total alcohol in the system is still 100 ml, and total water is still 100 ml. So, globally, the concentration of alcohol in the entire system is 50%. But each glass can have varying concentrations.The key is whether we can manipulate the concentrations such that Glass A has more than 50% alcohol and Glass B has less than 50% alcohol. Because if both glasses have 50%, then they are equal. But the question is whether Glass A can have higher than Glass B.Wait, but if the total alcohol is 100 ml and total water is 100 ml, then the average concentration is 50%. So, if one glass has more than 50%, the other must have less than 50%. Therefore, it's possible only if we can have one glass above 50% and the other below. But the problem is asking if Glass A can have a higher percentage than Glass B. So, if we can get Glass A to have, say, 60% and Glass B to have 40%, then yes. But is that possible?But wait, according to the previous calculations, when we transferred alcohol to water and back, the concentrations in Glass A were increasing but still less than Glass B. So maybe there's a different approach.Alternatively, perhaps we can transfer all the liquid from one glass to the other and then transfer back. Let's try that.Suppose we transfer all 100 ml alcohol from Glass B to Glass A. Now, Glass A has 100 ml water + 100 ml alcohol = 200 ml of 50% alcohol. Glass B is empty. Then, transfer 100 ml back from Glass A to Glass B. Since the mixture is 50%, transferring 100 ml back would transfer 50 ml alcohol and 50 ml water. So, Glass A now has 50 ml alcohol and 50 ml water (50% concentration), and Glass B has 50 ml alcohol and 50 ml water (50% concentration). So, both are equal. That doesn't help.Alternatively, what if we transfer a different amount? Let's say we transfer half of Glass B to Glass A. So, transfer 50 ml alcohol to Glass A. Now, Glass A has 100 ml water + 50 ml alcohol = 150 ml, concentration 1/3 alcohol. Glass B has 50 ml alcohol. Then, transfer 50 ml back from Glass A to Glass B. The mixture in Glass A is 1/3 alcohol, so transferring 50 ml would transfer 50*(1/3) ≈16.666 ml alcohol and 33.333 ml water. So, Glass B now has 50 + 16.666 ≈66.666 ml alcohol and 33.333 ml water. Concentration in B is 66.666/100 ≈66.666%. Glass A has 150 -50 =100 ml, with 50 -16.666 ≈33.333 ml alcohol and 100 -33.333 ≈66.666 ml water. Concentration ≈33.333%. So again, B is higher.Hmm. It seems like no matter how we transfer, Glass B remains with higher concentration. Is there a way to reverse this?Wait, maybe if we do multiple transfers. Let's try another approach. Start with transferring x from A to B. But initially, A is water, so transferring water to B (alcohol). Let's see.Initially, A: 100 ml water, B:100 ml alcohol. Transfer x ml water from A to B. Now, B has 100 ml alcohol + x ml water. Concentration in B: 100/(100 + x). Then, transfer x ml back from B to A. The mixture in B is 100/(100 + x) alcohol, so transferring x ml back would transfer x*(100/(100 + x)) alcohol and x*(x/(100 + x)) water to A. So, Glass A now has x*(100/(100 + x)) alcohol and (100 - x) + x*(x/(100 + x)) water. Let's compute that.Alcohol in A: 100x/(100 + x)Water in A: (100 - x) + x²/(100 + x) = [ (100 - x)(100 + x) + x² ] / (100 + x ) = [10000 - x² + x²]/ (100 + x ) = 10000/(100 + x )So concentration in A: (100x/(100 + x )) / (100x/(100 + x ) + 10000/(100 + x )) = 100x / (100x + 10000) = x/(x + 100)Concentration in B after transfer: Original alcohol in B was 100 - x*(100/(100 + x )) = 100 - 100x/(100 + x ) = 100*(100 + x - x)/(100 + x ) = 100²/(100 + x )Volume in B is back to 100 ml, so concentration is 100²/(100 + x ) /100 = 100/(100 + x )So again, concentration in A is x/(x + 100), in B is 100/(x + 100). So, B still has higher concentration.This seems like a pattern. Whether we transfer from B to A or A to B, after a transfer and back, the concentration in B remains higher. Is there a way to make A higher?Wait, maybe if we don't transfer back the same amount. For example, transfer a large amount, then transfer a different amount back. Let's say we transfer 100 ml alcohol from B to A, making A 200 ml (100 water + 100 alcohol), then transfer 50 ml back. The 50 ml transferred back would be 25 ml alcohol and 25 ml water. So, Glass B would have 25 ml alcohol and 25 ml water, concentration 50%. Glass A would have 75 ml alcohol and 75 ml water, concentration 50%. Still equal.Alternatively, transfer 100 ml alcohol to A, then transfer 150 ml back. But Glass A only has 200 ml, so transferring 150 ml back would leave 50 ml in A. The 150 ml transferred back would be 75 ml alcohol and 75 ml water. So Glass B has 75 ml alcohol +75 ml water =150 ml. But original Glass B was 100 ml, so this is not allowed since we can only transfer back up to the amount transferred. Wait, no, after transferring 100 ml to A, B has 0 ml. Then transferring back 150 ml is impossible. So that approach doesn't work.Alternatively, transfer 50 ml alcohol to A, making A 150 ml (100 water +50 alcohol). Then transfer 25 ml from A to B. The concentration in A is 50/150 =1/3. So transferring 25 ml would transfer 25*(1/3)≈8.333 ml alcohol and 16.666 ml water. So Glass B (which had 50 ml alcohol left) now has 50 +8.333≈58.333 ml alcohol and 16.666 ml water. Concentration ≈58.333/75≈77.78%. Glass A has 50 -8.333≈41.666 ml alcohol and 100 -16.666≈83.333 ml water. Concentration ≈41.666/125≈33.33%. Still B higher.This seems frustrating. Every time we do a transfer, the concentration in B remains higher. Is there a mathematical way to prove that it's impossible?Let me think in terms of conservation. The total amount of alcohol is 100 ml. Suppose after several transfers, Glass A has a higher concentration than Glass B. Let’s denote the amount of alcohol in Glass A as a, and in Glass B as 100 - a. Similarly, the amount of water in Glass A is w, and in Glass B is 100 - w (since total water is 100 ml). The concentration in Glass A is a/(a + w), and in Glass B is (100 - a)/( (100 - a) + (100 - w) ). Wait, but total volume in each glass must be the same as initially? No, actually, the volumes can change as we transfer liquid back and forth. Wait, initially each glass has 100 ml. But when we transfer x ml from one to the other, the volumes change. However, after any transfer, you can transfer back any amount, so the volumes can vary. But the problem allows transferring any amount, so volumes can be different. Wait, but the problem statement says "it is allowed to transfer some amount of liquid from one glass to the other (while the solution is mixed evenly)". It doesn't specify that we have to end up with the same volume in each glass. So, maybe we can have different volumes in each glass, which might help.Ah! Maybe this is the key. Previously, I assumed transferring back the same amount, but maybe if we don't keep the volumes the same, we can create a situation where Glass A has a higher concentration. For example, transfer a lot from B to A, making A have a large volume with a certain concentration, then transfer a small amount back, adjusting the concentrations.Let me try this. Start with Glass A: 100 ml water, Glass B:100 ml alcohol.Transfer 100 ml alcohol from B to A. Now, A has 200 ml (100 water +100 alcohol), 50% concentration. B is empty.Now, transfer 50 ml from A to B. The mixture in A is 50%, so transferring 50 ml gives 25 ml alcohol and 25 ml water to B. Now, B has 25 ml alcohol and 25 ml water (50% concentration). A has 75 ml alcohol and 75 ml water (still 50%). So both are equal.Not helpful. What if we transfer a different amount? Suppose after transferring 100 ml alcohol to A, making A 200 ml 50%, then transfer 100 ml back. But that would just reverse the process.Alternatively, transfer 100 ml from B to A, then transfer 150 ml back. Wait, but A only has 200 ml, so transferring 150 ml back leaves 50 ml in A. The 150 ml transferred back would be 75 alcohol and 75 water. So B would have 75 alcohol and 75 water, 50% concentration. A has 25 alcohol and 25 water, also 50%. Still equal.Alternatively, transfer x ml from B to A, then transfer y ml back, where y ≠x. Let's see.Suppose we transfer 100 ml (all) alcohol from B to A, making A 200 ml 50%. Then transfer y ml back to B. The concentration in A is 50%, so y ml transferred back has 0.5y alcohol and 0.5y water. So B now has 0.5y alcohol and 0.5y water. The concentration in B is 50%. A has (100 -0.5y) alcohol and (100 -0.5y) water. Still 50% each. So no difference.Hmm. What if we don't transfer all the alcohol first? Let's try a different approach. Transfer 50 ml alcohol from B to A. Now, A has 100 water +50 alcohol =150 ml, concentration 1/3 alcohol. B has 50 alcohol. Now, transfer 50 ml from A to B. The mixture in A is 1/3 alcohol, so transferring 50 ml would transfer 50*(1/3)≈16.666 alcohol and 33.333 water. So B now has 50 +16.666=66.666 alcohol and 33.333 water. Concentration ≈66.666/100≈66.666%. A has 50 -16.666≈33.333 alcohol and 100 -33.333≈66.666 water. Concentration≈33.333/100≈33.333%. Still B higher.Wait, but what if after transferring 50 ml alcohol to A, making A 150 ml (1/3), we transfer a different amount back, say 25 ml. Then, transferring 25 ml from A (1/3 alcohol) would transfer 25*(1/3)≈8.333 alcohol and 16.666 water. So B would have 50 +8.333≈58.333 alcohol and 16.666 water. Concentration≈58.333/75≈77.78%. A has 50 -8.333≈41.666 alcohol and 100 -16.666≈83.333 water. Concentration≈41.666/125≈33.33%. Still B higher.Alternatively, transfer 10 ml from B to A, then 20 ml back. Let's compute:Transfer 10 ml alcohol to A: A has 110 ml (100 water +10 alcohol), concentration≈9.09%. B has 90 alcohol.Transfer 20 ml from A to B: concentration in A is≈9.09%, so alcohol transferred:20*0.0909≈1.818 ml, water transferred:18.182 ml. So B now has 90 +1.818≈91.818 alcohol and 18.182 water. Concentration≈91.818/110≈83.47%. A has 10 -1.818≈8.182 alcohol and 100 -18.182≈81.818 water. Concentration≈8.182/90≈9.09%. Still B higher.This seems to consistently result in B having a higher concentration. Is there a mathematical invariant here that ensures that B's concentration is always higher than A's?Let me think. Suppose after any number of transfers, let’s denote the amount of alcohol in Glass A as a, and water as w. Similarly, in Glass B, alcohol is 100 -a, and water is 100 -w. The concentration in A is a/(a +w), and in B is (100 -a)/( (100 -a)+(100 -w) )= (100 -a)/(200 -a -w).We need to check if it's possible for a/(a +w) > (100 -a)/(200 -a -w).But since the total liquid in each glass can vary, a +w and 200 -a -w are the volumes in Glass A and B respectively.But we need to see if there exists a and w such that:a/(a +w) > (100 -a)/(200 -a -w)Let’s cross multiply:a*(200 -a -w) > (100 -a)*(a +w)Expand both sides:200a -a² -aw > 100a +100w -a² -awSimplify:200a -a² -aw -100a -100w +a² +aw >0Which simplifies to:100a -100w >0 ⇒ 100(a -w) >0 ⇒ a >wSo the inequality holds iff a >w.But in Glass A, the amount of alcohol a and water w must satisfy that a ≤w + something? Wait, no. The total alcohol in the system is 100 ml, so a ≤100, and total water is 100 ml, so w ≤100. But a and w are independent except that a ≤100 and w ≤100. Wait, no. Actually, the total alcohol in both glasses is 100, so a + (100 -a) =100, and total water is w + (100 -w)=100. So actually, a and w can vary between 0 and 100, but independently? No, because when you transfer liquid, you transfer both alcohol and water. So a and w are related based on the transfers.Wait, this seems like a misunderstanding. Let's clarify.At any point, the total alcohol is 100 ml: a_A + a_B =100, where a_A is alcohol in A, a_B in B.Similarly, total water is 100 ml: w_A +w_B =100.The concentration in Glass A is a_A/(a_A +w_A), and in Glass B is a_B/(a_B +w_B).We need to see if a_A/(a_A +w_A) > a_B/(a_B +w_B).But since a_B =100 -a_A and w_B=100 -w_A, substitute:Concentration in B: (100 -a_A)/( (100 -a_A) + (100 -w_A) )= (100 -a_A)/(200 -a_A -w_A)So the inequality is:a_A/(a_A +w_A) > (100 -a_A)/(200 -a_A -w_A)Which simplifies to a_A >w_A, as previously found.So, the question reduces to: can we have a_A >w_A in Glass A?But since a_A +w_A is the total volume in Glass A, which can vary. So, if we can arrange transfers such that in Glass A, the amount of alcohol exceeds the amount of water, then its concentration would be higher than in Glass B.But given that the total alcohol is 100 and total water is 100, can we have a_A >w_A?Yes, but only if in Glass A, a_A >w_A, which would imply that in Glass B, a_B =100 -a_A <100 -w_A =w_B, meaning Glass B would have more water than alcohol. But since total alcohol equals total water (100 each), if Glass A has more alcohol than water, then Glass B must have more water than alcohol. Therefore, it's possible.Wait, but how?Wait, the total alcohol is 100, total water 100. If we can have a_A >w_A, then since w_A =100 -w_B, we have a_A >100 -w_B. But a_A +a_B=100, so a_B=100 -a_A. Similarly, w_B=100 -w_A.So, substituting, a_A >100 - (100 -w_B) ⇒a_A >w_B. But a_B=100 -a_A, and w_B=100 -w_A. This is getting tangled. Let's think differently.If in Glass A, alcohol > water, then since total alcohol = total water =100, then in Glass B, water > alcohol. Therefore, the concentration in Glass A would be higher than 50%, and in Glass B lower than 50%. So, if this is achievable, then yes, Glass A can have higher concentration.But how to achieve a_A >w_A?Let me think of an extreme case. Suppose we transfer all water from A to B. But initially, A has only water. If we transfer all 100 ml water to B, then B has 100 alcohol +100 water. Then, transfer back some amount. For example, transfer 100 ml back from B to A. The mixture in B is 50% alcohol, so transferring 100 ml back would give A 50 alcohol and 50 water. So now, A has 50 alcohol and 50 water (50%), B has 50 alcohol and 50 water (50%). Not helpful.But what if we transfer a different amount? Suppose after transferring all water to B, making B 200 ml 50% concentration, we transfer 150 ml back to A. Then, A gets 75 alcohol and 75 water, concentration 50%. B has 25 alcohol and25 water. Still 50%. Not helpful.Alternatively, transfer 100 ml from B to A, making A 100 ml (50% from before), but this doesn't change anything.Wait, maybe another approach. Let's consider that each transfer operation is a linear operation on the concentrations. There might be a conservation law here that prevents the concentration in A from exceeding that in B.Alternatively, think in terms of the difference in concentrations.Let’s define C_A = a_A / (a_A +w_A), C_B =a_B / (a_B +w_B).We need to check if C_A >C_B.But note that a_A +a_B=100, w_A +w_B=100.Let’s express C_A and C_B in terms of a_A and w_A.C_A = a_A / (a_A +w_A)C_B = (100 -a_A)/(200 -a_A -w_A)We need to check if a_A/(a_A +w_A) > (100 -a_A)/(200 -a_A -w_A)Cross multiply:a_A*(200 -a_A -w_A) > (100 -a_A)(a_A +w_A)Expand:200a_A -a_A² -a_Aw_A >100a_A +100w_A -a_A² -a_Aw_ACancel terms:200a_A >100a_A +100w_A100a_A >100w_Aa_A >w_ASo, the inequality C_A >C_B holds iff a_A >w_A.Therefore, to have a higher concentration in A than B, we need the amount of alcohol in A to exceed the amount of water in A.But given that total alcohol is 100 and total water is 100, is this possible?Yes, for example, if A has 60 alcohol and 40 water, then a_A=60 >w_A=40, and B has 40 alcohol and 60 water. Then C_A=60/100=60%, C_B=40/100=40%, so C_A >C_B.But the question is whether such a distribution is achievable through the allowed operations.The allowed operations are transferring any amount from one glass to the other, with mixing in between.So, to get a_A >w_A, we need to have more alcohol than water in Glass A. How?Let’s suppose we do the following:1. Transfer x ml from B (alcohol) to A (water). Now, A has x ml alcohol and 100 ml water. Total volume in A:100 +x. Concentration in A: x/(100 +x). B has 100 -x ml alcohol.2. Now, transfer y ml from A to B. The concentration in A is x/(100 +x), so transferring y ml transfers y*x/(100 +x) alcohol and y*100/(100 +x) water. Therefore, after transfer:- Glass A has x - y*x/(100 +x) =x*(100 +x - y)/(100 +x) alcohol and 100 - y*100/(100 +x) water.- Glass B has (100 -x) + y*x/(100 +x) alcohol and y*100/(100 +x) water.We need to choose x and y such that in Glass A, alcohol > water.So:x*(100 +x - y)/(100 +x) >100 - y*100/(100 +x)Multiply both sides by (100 +x):x*(100 +x - y) >100*(100 +x) -100yExpand left side:100x +x² -xyRight side:10000 +100x -100ySubtract right side from left side:100x +x² -xy -10000 -100x +100y =x² -xy -10000 +100ySet inequality:x² -xy -10000 +100y >0Let’s rearrange:x² -10000 +y*(-x +100) >0This is a bit complicated, but let's try plugging in some numbers. Suppose we take x=100 ml (transfer all alcohol to A). Then:x=100:(100)^2 -100* y -10000 +100y =10000 -100y -10000 +100y=0So inequality becomes 0>0, which is false. Not helpful.Try x=50:x=50:50² -50y -10000 +100y=2500 -50y -10000 +100y= -7500 +50y >0 →50y>7500→y>150. But y cannot exceed the volume in A, which after transferring x=50 is 150 ml. So y>150 is impossible. Hence, no solution.Try x=80:x=80:80² -80y -10000 +100y=6400 -80y -10000 +100y= -3600 +20y >0 →20y>3600→y>180. But after transferring x=80, A has 180 ml. So y cannot exceed 180. So y>180 impossible.Similarly, x=90:x=90:90² -90y -10000 +100y=8100 -90y -10000 +100y= -1900 +10y >0 →10y>1900→y>190. But A has 190 ml, so y>190 impossible.Wait, this suggests that for any x <100, the required y to make the inequality positive is larger than the volume available in A after transferring x. Hence, impossible.What if x=0? No, then no transfer.What if x=10:x=10:10² -10y -10000 +100y=100 -10y -10000 +100y= -9900 +90y >0 →90y>9900→y>110. But A has 110 ml after x=10, so y>110 impossible.So in all cases, the required y to make a_A >w_A exceeds the volume available in Glass A, which is 100 +x. Therefore, it's impossible.Therefore, it's impossible to have a_A >w_A through these operations, hence impossible to have C_A >C_B.Thus, the answer is no, it's not possible.

✅:Step-by-Step Explanation and Answer:Problem Analysis:We start with two glasses: Glass A (100 ml water) and Glass B (100 ml alcohol). The goal is to determine if we can transfer liquids between them such that Glass A ends up with a higher alcohol concentration than Glass B.Key Observations:1. Conservation of Totals: The total alcohol and water in the system remain constant at 100 ml each.2. Concentration Relationship: If Glass A has a higher alcohol concentration, Glass B must have a lower concentration (since the average system concentration is 50%).Mathematical Formulation:Let ( a_A ) and ( w_A ) represent the alcohol and water in Glass A, respectively. For Glass A's concentration to exceed Glass B's:[frac{a_A}{a_A + w_A} > frac{100 - a_A}{(100 - a_A) + (100 - w_A)}]Simplifying this inequality leads to ( a_A > w_A ).Transfer Operations Analysis:- Any transfer from B to A introduces alcohol into A but also increases its total volume.- Transferring back from A to B removes a mixture, but the concentration in A remains diluted by water.Critical Insight:For ( a_A > w_A ), Glass A must contain more alcohol than water. However, every transfer operation inherently links the amounts of alcohol and water moved. Detailed calculations show that the required amount of alcohol to surpass water in Glass A would necessitate transferring more liquid than available, making it impossible.Conclusion:Due to the conservation constraints and the dependency of transferred mixtures, it's impossible for Glass A to achieve a higher alcohol concentration than Glass B through allowable operations.Final Answer:No, it is not possible for the first glass to have a higher percentage of alcohol than the second glass through such operations.

🔑:1. Initial Setup: - Assume we have two glasses: the first glass contains 100g of water, and the second glass containts 100g of alcohol.2. Objective: - We need to prove that the percentage of alcohol in the first glass can never exceed 50%, and the percentage of alcohol in the second glass can never be less than 50%.3. Initial Condition: - Initially, the first glass has 100g water (0% alcohol), and the second glass has 100g alcohol (100% alcohol). The given statement holds true at the beginning since 0% ≤ 50% and 100% ≥ 50%.4. Assume True at Some Step: - Suppose, at some moment, the first glass has water content no less than the alcohol content, and the second glass has alcohol content no less than the water content.5. Transfer Operation: - Let's transfer a certain amount ( x ) from the first glass to the second glass. After the transfer, we stir the solutions thoroughly so that the liquids mix evenly.6. Content Calculation: - Let the content of alcohol in the first glass initially be ( A_1(t) ), and the content of water in the first glass initially be ( W_1(t) ). - Similarly, in the second glass, let the content of alcohol be ( A_2(t) ), and the content of water be ( W_2(t) ).7. Mixing and Transferring: - When ( x ) grams of liquid is transferred from the first glass to the second glass: - From the first glass: [ text{Alcohol transferred} = frac{A_1(t) cdot x}{A_1(t) + W_1(t)} ] [ text{Water transferred} = frac{W_1(t) cdot x}{A_1(t) + W_1(t)} ] - The content of the first glass after transferring: [ A_1(t+1) = A_1(t) - frac{A_1(t) cdot x}{A_1(t) + W_1(t)} ] [ W_1(t+1) = W_1(t) - frac{W_1(t) cdot x}{A_1(t) + W_1(t)} ] - The content of the second glass after receiving the transfer: [ A_2(t+1) = A_2(t) + frac{A_1(t) cdot x}{A_1(t) + W_1(t)} ] [ W_2(t+1) = W_2(t) + frac{W_1(t) cdot x}{A_1(t) + W_1(t)} ]8. Transfer Back to the First Glass: - Similar calculations will be applied when transferring liquid back from the second glass.9. Inductive Hypothesis: - By induction, assuming the hypothesis holds for the current step ( t ), it is true that the ratios of water to alcohol in both glasses will either remain the same or move towards the mean, but never invert. - This ensures that in both glasses, after each transfer operation, the first glass always has at least as much water as alcohol, and the second glass always has at least as much alcohol as water.10. Conclusion: - Therefore, it is impossible to have a higher percentage of alcohol in the first glass than in the second by these operations.[ boxed{text{Impossible}} ]