💡First, I need to determine the cost of each item Laura purchased.1. **Salads**: Laura bought 2 salads, and each salad costs 3. [ 2 text{ salads} times 3 = 6 ]2. **Beef**: One kilogram of beef is two times the price of one salad. Since one salad costs 3, one kilogram of beef costs: [ 2 times 3 = 6 ] Laura bought 2 kilograms of beef: [ 2 text{ kg} times 6 = 12 ]3. **Potatoes**: One kilogram of potatoes costs one-third of the price of one salad. Therefore: [ frac{1}{3} times 3 = 1 ] Laura bought 1 kilogram of potatoes: [ 1 text{ kg} times 1 = 1 ]4. **Juice**: Laura bought 2 liters of juice, with each liter costing 1.5. [ 2 text{ liters} times 1.5 = 3 ]Finally, I add up the costs of all the items to find the total amount Laura needed to pay:[6 text{ (salads)} + 12 text{ (beef)} + 1 text{ (potatoes)} + 3 text{ (juice)} = 22]

💡Alright, so I have this geometry problem here about skew lines. Skew lines are those lines in three-dimensional space that don't intersect and aren't parallel. They just kind of go off in different directions. The question is asking which of the statements A, B, C, or D is correct when given two skew lines a and b.Let me go through each option one by one.**Option A: Through any point not on lines a and b, a plane can be made parallel to both a and b.**Hmm, okay. So, if I pick any point that's not on either of these skew lines, can I create a plane that's parallel to both lines? I know that a plane being parallel to a line means that the line doesn't intersect the plane and isn't lying on it. Since the lines are skew, they aren't parallel to each other, which might complicate things.If I try to make a plane through this point that's parallel to both a and b, I need to ensure that the direction vectors of both lines are parallel to the plane. But wait, if the lines are skew, their direction vectors aren't parallel. So, can a single plane have two different direction vectors that aren't parallel? I think a plane can have multiple direction vectors, but they have to be coplanar. Since the lines are skew, their direction vectors aren't coplanar, so it might not be possible to have a plane that's parallel to both. Maybe this isn't always possible, so A might not be correct.**Option B: Through any point not on lines a and b, a line can be made to intersect both a and b.**Okay, so from any point not on a or b, can I draw a line that intersects both a and b? I remember that in three-dimensional space, given two skew lines, there's exactly one line that intersects both and lies on the same plane as both. But does that mean from any point, I can draw such a line? I'm not sure. If the point is not on either line, can I always find a line through it that intersects both? It feels like it might not be always possible because the lines are skew, so they don't lie on the same plane, and depending on where the point is, the line might not intersect both. Maybe B is incorrect.**Option C: Through any point not on lines a and b, a line can be made parallel to both a and b.**This one says that from any point not on a or b, I can draw a line that's parallel to both lines. Wait, if a line is parallel to both a and b, then a and b must be parallel to each other, right? But the given lines are skew, which means they aren't parallel. So, if a and b aren't parallel, there can't be a line that's parallel to both. Hence, C must be incorrect.**Option D: Through line a, exactly one plane can be made parallel to line b.**Alright, so this is saying that if I take line a, there's exactly one plane that contains line a and is parallel to line b. Let me visualize this. If I have line a, and I want a plane that's parallel to line b, which is skew to a. Since they're skew, they don't intersect and aren't parallel. So, to make a plane through a that's parallel to b, I need the plane to contain a and have a direction that's parallel to b.I think this is possible because you can take a direction vector from line b and use it to construct a plane with line a. Since line a is fixed, and line b is skew, there should be exactly one such plane that contains a and is parallel to b. I can't think of more than one plane that would satisfy this condition because the direction of b is fixed relative to a. So, D seems correct.Let me just recap:- A: Doesn't seem possible because the direction vectors of a and b aren't coplanar.- B: Not necessarily possible from any point.- C: Impossible because a and b aren't parallel.- D: Seems correct because you can construct exactly one plane through a that's parallel to b.So, I think the correct answer is D.

💡First, I need to determine the initial percentage of alcohol in the 6-liter solution. Let's denote this percentage as ( P ) in decimal form.The amount of alcohol in the initial solution is ( 6 times P ) liters.When 1.8 liters of pure alcohol is added, the total volume of the solution becomes ( 6 + 1.8 = 7.8 ) liters.The final solution should be 50% alcohol, which means the amount of alcohol in the final solution should be ( 7.8 times 0.5 = 3.9 ) liters.The total alcohol in the final solution is the sum of the alcohol from the initial solution and the added pure alcohol:[6P + 1.8 = 3.9]Solving for ( P ):[6P = 3.9 - 1.8][6P = 2.1][P = frac{2.1}{6}][P = 0.35]Therefore, the initial percentage of alcohol in the solution is 35%.

💡Alright, so I've got this geometry problem about a parabola, tangents, and circumcircles. It's a bit intimidating at first, but I'll try to break it down step by step. Let me start by understanding the given information and what needs to be proven.We have a parabola Γ defined by the equation ( y^2 = 2px ) where ( p > 0 ). The focus of this parabola is denoted by F. There's a point Q outside the parabola, not on the x-axis, from which two tangents are drawn to Γ, touching it at points A and B. The lines QA and QB intersect the y-axis at points C and D, respectively. The circumcenter of triangle QAB is M. The goal is to prove that FM is a tangent to the circumcircle of triangle FCD.First, I need to recall some properties of parabolas. The standard form ( y^2 = 4ax ) has its focus at (a, 0). Comparing this to our equation ( y^2 = 2px ), it seems that 4a = 2p, so a = p/2. Therefore, the focus F is at (p/2, 0).Next, I should remember that the tangent to a parabola at a point (x₁, y₁) is given by ( yy₁ = p(x + x₁) ). Wait, let me verify that. For the parabola ( y^2 = 4ax ), the tangent at (x₁, y₁) is ( yy₁ = 2a(x + x₁) ). So, for our case, since 4a = 2p, a = p/2, the tangent equation becomes ( yy₁ = p(x + x₁) ). So, yes, that's correct.Given that, points A and B are points of tangency on the parabola, so their coordinates can be written as ( A(y₁²/(2p), y₁) ) and ( B(y₂²/(2p), y₂) ). That makes sense because plugging y₁ into the parabola equation gives x = y₁²/(2p).Now, the lines QA and QB are tangents from point Q to the parabola. The equations of these tangents are known, so I can write them down. For point A, the tangent is ( yy₁ = p(x + y₁²/(2p)) ). Simplifying, that's ( yy₁ = px + y₁²/2 ). Similarly, for point B, the tangent is ( yy₂ = px + y₂²/2 ).Since both these tangents pass through point Q, which is outside the parabola, the coordinates of Q must satisfy both tangent equations. Let me denote Q as (h, k). Then, substituting Q into both equations:1. ( k y₁ = p h + y₁²/2 )2. ( k y₂ = p h + y₂²/2 )These are two equations involving y₁ and y₂. Maybe I can solve for h and k in terms of y₁ and y₂.From equation 1: ( k y₁ - y₁²/2 = p h )From equation 2: ( k y₂ - y₂²/2 = p h )Since both equal p h, set them equal to each other:( k y₁ - y₁²/2 = k y₂ - y₂²/2 )Rearranging:( k(y₁ - y₂) = (y₁² - y₂²)/2 )Factor the right side:( k(y₁ - y₂) = (y₁ - y₂)(y₁ + y₂)/2 )Assuming y₁ ≠ y₂ (since A and B are distinct points), we can divide both sides by (y₁ - y₂):( k = (y₁ + y₂)/2 )So, the y-coordinate of Q is the average of y₁ and y₂. That's interesting.Now, substitute k back into equation 1 to find h:( (y₁ + y₂)/2 * y₁ = p h + y₁²/2 )Multiply out the left side:( (y₁² + y₁ y₂)/2 = p h + y₁²/2 )Subtract y₁²/2 from both sides:( y₁ y₂ / 2 = p h )So, h = (y₁ y₂)/(2p)Therefore, the coordinates of Q are ( (y₁ y₂)/(2p), (y₁ + y₂)/2 ). That's a good start.Next, we need to find points C and D where lines QA and QB intersect the y-axis. The y-axis is where x = 0, so let's find the y-intercepts of QA and QB.First, let's find the equations of lines QA and QB. We already have the equations of the tangents, which are QA and QB. Wait, no, the tangents are the lines QA and QB themselves because Q is the external point from which tangents are drawn.Wait, actually, in the problem statement, it says: "From a point Q outside Γ (not on the x-axis), two tangents are drawn to Γ, touching it at points A and B respectively." So, lines QA and QB are the tangents, so their equations are already given by the tangent equations we wrote earlier.But we need to find where these lines intersect the y-axis. For line QA, which is the tangent at A: ( yy₁ = px + y₁²/2 ). To find the y-intercept, set x = 0:( yy₁ = 0 + y₁²/2 ) => ( y = y₁/2 ). So, point C is (0, y₁/2).Similarly, for line QB, the tangent at B: ( yy₂ = px + y₂²/2 ). Setting x = 0:( yy₂ = y₂²/2 ) => ( y = y₂/2 ). So, point D is (0, y₂/2).Alright, so now we have points C(0, y₁/2) and D(0, y₂/2).Next, we need to find the circumcenter M of triangle QAB. The circumcenter is the intersection of the perpendicular bisectors of the sides of the triangle.Let me recall that the circumcenter is equidistant from all three vertices, so it's the intersection point of the perpendicular bisectors of any two sides.So, let's find the perpendicular bisectors of QA and QB.First, let's find the midpoints of QA and QB.Coordinates of Q: ( (y₁ y₂)/(2p), (y₁ + y₂)/2 )Coordinates of A: ( y₁²/(2p), y₁ )Midpoint of QA:x-coordinate: [ (y₁ y₂)/(2p) + y₁²/(2p) ] / 2 = [ y₁(y₂ + y₁) / (2p) ] / 2 = y₁(y₁ + y₂)/(4p)y-coordinate: [ (y₁ + y₂)/2 + y₁ ] / 2 = [ (y₁ + y₂ + 2y₁)/2 ] / 2 = (3y₁ + y₂)/4Similarly, midpoint of QB:x-coordinate: [ (y₁ y₂)/(2p) + y₂²/(2p) ] / 2 = y₂(y₁ + y₂)/(4p)y-coordinate: [ (y₁ + y₂)/2 + y₂ ] / 2 = (y₁ + 3y₂)/4Now, we need the slopes of QA and QB to find the perpendicular bisectors.Slope of QA: (y₁ - (y₁ + y₂)/2 ) / ( y₁²/(2p) - (y₁ y₂)/(2p) )Simplify numerator: (2y₁ - y₁ - y₂)/2 = (y₁ - y₂)/2Denominator: (y₁² - y₁ y₂)/(2p) = y₁(y₁ - y₂)/(2p)So, slope of QA: [ (y₁ - y₂)/2 ] / [ y₁(y₁ - y₂)/(2p) ] = (1/2) / ( y₁/(2p) ) = p / y₁Similarly, slope of QB: (y₂ - (y₁ + y₂)/2 ) / ( y₂²/(2p) - (y₁ y₂)/(2p) )Numerator: (2y₂ - y₁ - y₂)/2 = (y₂ - y₁)/2Denominator: (y₂² - y₁ y₂)/(2p) = y₂(y₂ - y₁)/(2p)Slope of QB: [ (y₂ - y₁)/2 ] / [ y₂(y₂ - y₁)/(2p) ] = (1/2) / ( y₂/(2p) ) = p / y₂Therefore, the slopes of QA and QB are p/y₁ and p/y₂ respectively.The perpendicular bisectors will have slopes that are negative reciprocals of these.So, slope of perpendicular bisector of QA: -y₁/pSimilarly, slope of perpendicular bisector of QB: -y₂/pNow, we can write the equations of the perpendicular bisectors.For the perpendicular bisector of QA:Passing through midpoint ( y₁(y₁ + y₂)/(4p), (3y₁ + y₂)/4 ) with slope -y₁/p.Equation: y - (3y₁ + y₂)/4 = (-y₁/p)(x - y₁(y₁ + y₂)/(4p))Similarly, for the perpendicular bisector of QB:Passing through midpoint ( y₂(y₁ + y₂)/(4p), (y₁ + 3y₂)/4 ) with slope -y₂/p.Equation: y - (y₁ + 3y₂)/4 = (-y₂/p)(x - y₂(y₁ + y₂)/(4p))Now, to find the circumcenter M, we need to solve these two equations simultaneously.This might get a bit messy, but let's try to proceed.First, let's write the equation for the perpendicular bisector of QA:y = (-y₁/p)x + (y₁²(y₁ + y₂))/(4p²) + (3y₁ + y₂)/4Similarly, for the perpendicular bisector of QB:y = (-y₂/p)x + (y₂²(y₁ + y₂))/(4p²) + (y₁ + 3y₂)/4Set them equal to each other:(-y₁/p)x + (y₁²(y₁ + y₂))/(4p²) + (3y₁ + y₂)/4 = (-y₂/p)x + (y₂²(y₁ + y₂))/(4p²) + (y₁ + 3y₂)/4Bring all terms to one side:[ (-y₁/p)x + (y₁²(y₁ + y₂))/(4p²) + (3y₁ + y₂)/4 ] - [ (-y₂/p)x + (y₂²(y₁ + y₂))/(4p²) + (y₁ + 3y₂)/4 ] = 0Simplify term by term:- (y₁/p - y₂/p)x + [ y₁²(y₁ + y₂) - y₂²(y₁ + y₂) ]/(4p²) + [ (3y₁ + y₂) - (y₁ + 3y₂) ]/4 = 0Factor out (y₁ + y₂) in the second term:- (y₁ - y₂)/p x + (y₁ + y₂)(y₁² - y₂²)/(4p²) + (2y₁ - 2y₂)/4 = 0Note that y₁² - y₂² = (y₁ - y₂)(y₁ + y₂), so:- (y₁ - y₂)/p x + (y₁ + y₂)^2 (y₁ - y₂)/(4p²) + (y₁ - y₂)/2 = 0Factor out (y₁ - y₂):(y₁ - y₂)[ -x/p + (y₁ + y₂)^2/(4p²) + 1/2 ] = 0Since y₁ ≠ y₂ (as A and B are distinct points), we can divide both sides by (y₁ - y₂):- x/p + (y₁ + y₂)^2/(4p²) + 1/2 = 0Solve for x:x/p = (y₁ + y₂)^2/(4p²) + 1/2Multiply both sides by p:x = (y₁ + y₂)^2/(4p) + p/2So, x-coordinate of M is (y₁ + y₂)^2/(4p) + p/2.Now, substitute this back into one of the perpendicular bisector equations to find y-coordinate.Let's use the equation from QA:y = (-y₁/p)x + (y₁²(y₁ + y₂))/(4p²) + (3y₁ + y₂)/4Substitute x:y = (-y₁/p)[ (y₁ + y₂)^2/(4p) + p/2 ] + (y₁²(y₁ + y₂))/(4p²) + (3y₁ + y₂)/4Let's compute each term:First term: (-y₁/p)[ (y₁ + y₂)^2/(4p) + p/2 ] = (-y₁ (y₁ + y₂)^2)/(4p²) - y₁ p/(2p) = (-y₁ (y₁ + y₂)^2)/(4p²) - y₁/2Second term: (y₁²(y₁ + y₂))/(4p²)Third term: (3y₁ + y₂)/4So, combining all terms:y = [ (-y₁ (y₁ + y₂)^2)/(4p²) - y₁/2 ] + (y₁²(y₁ + y₂))/(4p²) + (3y₁ + y₂)/4Let's combine the first and second terms:[ (-y₁ (y₁ + y₂)^2 + y₁²(y₁ + y₂) ) / (4p²) ] - y₁/2 + (3y₁ + y₂)/4Factor y₁(y₁ + y₂) in the numerator:y₁(y₁ + y₂)[ - (y₁ + y₂) + y₁ ] / (4p²) = y₁(y₁ + y₂)( - y₂ ) / (4p²) = - y₁ y₂ (y₁ + y₂) / (4p²)So, now we have:y = - y₁ y₂ (y₁ + y₂) / (4p²) - y₁/2 + (3y₁ + y₂)/4Combine the constants:- y₁/2 + (3y₁ + y₂)/4 = (-2y₁ + 3y₁ + y₂)/4 = (y₁ + y₂)/4So, overall:y = - y₁ y₂ (y₁ + y₂) / (4p²) + (y₁ + y₂)/4Factor out (y₁ + y₂)/4:y = (y₁ + y₂)/4 [ 1 - y₁ y₂ / p² ]Therefore, the coordinates of M are:x = (y₁ + y₂)^2/(4p) + p/2y = (y₁ + y₂)/4 [ 1 - y₁ y₂ / p² ]That's M.Now, we need to find the circumcircle of triangle FCD and show that FM is tangent to it.First, let's find the coordinates of F, C, and D.F is the focus of the parabola, which we already determined is at (p/2, 0).C is (0, y₁/2) and D is (0, y₂/2).So, triangle FCD has vertices at F(p/2, 0), C(0, y₁/2), and D(0, y₂/2).We need to find the circumcircle of triangle FCD. Let's denote its center as T.To find the circumcircle, we can find the perpendicular bisectors of two sides and find their intersection.Let's find the perpendicular bisectors of FC and FD.First, find midpoints and slopes.Midpoint of FC:x: (p/2 + 0)/2 = p/4y: (0 + y₁/2)/2 = y₁/4Slope of FC:(y₁/2 - 0)/(0 - p/2) = (y₁/2)/(-p/2) = - y₁ / pTherefore, the perpendicular bisector of FC will have slope p / y₁.Equation of perpendicular bisector of FC:Passing through (p/4, y₁/4) with slope p/y₁:y - y₁/4 = (p/y₁)(x - p/4)Similarly, midpoint of FD:x: (p/2 + 0)/2 = p/4y: (0 + y₂/2)/2 = y₂/4Slope of FD:(y₂/2 - 0)/(0 - p/2) = (y₂/2)/(-p/2) = - y₂ / pPerpendicular bisector slope: p / y₂Equation of perpendicular bisector of FD:y - y₂/4 = (p/y₂)(x - p/4)Now, we have two equations:1. y = (p/y₁)(x - p/4) + y₁/42. y = (p/y₂)(x - p/4) + y₂/4Set them equal to find the intersection point T:(p/y₁)(x - p/4) + y₁/4 = (p/y₂)(x - p/4) + y₂/4Bring all terms to one side:(p/y₁ - p/y₂)(x - p/4) + (y₁ - y₂)/4 = 0Factor p:p(1/y₁ - 1/y₂)(x - p/4) + (y₁ - y₂)/4 = 0Note that 1/y₁ - 1/y₂ = (y₂ - y₁)/(y₁ y₂), so:p( (y₂ - y₁)/(y₁ y₂) )(x - p/4) + (y₁ - y₂)/4 = 0Factor out (y₂ - y₁):(y₂ - y₁)[ -p/(y₁ y₂)(x - p/4) + 1/4 ] = 0Since y₂ ≠ y₁, we can divide both sides by (y₂ - y₁):- p/(y₁ y₂)(x - p/4) + 1/4 = 0Solve for x:- p/(y₁ y₂)(x - p/4) = -1/4Multiply both sides by -1:p/(y₁ y₂)(x - p/4) = 1/4Multiply both sides by y₁ y₂ / p:x - p/4 = y₁ y₂ / (4p)So, x = p/4 + y₁ y₂ / (4p)Now, substitute x back into one of the perpendicular bisector equations to find y.Using equation 1:y = (p/y₁)(x - p/4) + y₁/4Substitute x:y = (p/y₁)( p/4 + y₁ y₂/(4p) - p/4 ) + y₁/4Simplify inside the parentheses:p/4 - p/4 + y₁ y₂/(4p) = y₁ y₂/(4p)So,y = (p/y₁)( y₁ y₂/(4p) ) + y₁/4 = (y₂/4) + y₁/4 = (y₁ + y₂)/4Therefore, the center T of the circumcircle of triangle FCD is at:x = p/4 + y₁ y₂/(4p)y = (y₁ + y₂)/4So, T is ( p/4 + y₁ y₂/(4p), (y₁ + y₂)/4 )Now, we need to show that FM is tangent to the circumcircle of triangle FCD.First, let's find the coordinates of M again:M = ( (y₁ + y₂)^2/(4p) + p/2, (y₁ + y₂)/4 [ 1 - y₁ y₂ / p² ] )And F is at (p/2, 0).So, vector FM is from F to M:Δx = [ (y₁ + y₂)^2/(4p) + p/2 ] - p/2 = (y₁ + y₂)^2/(4p)Δy = [ (y₁ + y₂)/4 (1 - y₁ y₂ / p² ) ] - 0 = (y₁ + y₂)/4 (1 - y₁ y₂ / p² )So, vector FM is ( (y₁ + y₂)^2/(4p), (y₁ + y₂)/4 (1 - y₁ y₂ / p² ) )Now, the circumcircle of FCD has center T at ( p/4 + y₁ y₂/(4p), (y₁ + y₂)/4 ) and radius equal to the distance from T to F, C, or D.Let me compute the radius squared:Distance from T to F:Δx = p/4 + y₁ y₂/(4p) - p/2 = -p/4 + y₁ y₂/(4p)Δy = (y₁ + y₂)/4 - 0 = (y₁ + y₂)/4So, radius squared is:( -p/4 + y₁ y₂/(4p) )² + ( (y₁ + y₂)/4 )²Similarly, distance from T to C:C is (0, y₁/2)Δx = p/4 + y₁ y₂/(4p) - 0 = p/4 + y₁ y₂/(4p)Δy = (y₁ + y₂)/4 - y₁/2 = (y₁ + y₂ - 2y₁)/4 = (y₂ - y₁)/4Radius squared:( p/4 + y₁ y₂/(4p) )² + ( (y₂ - y₁)/4 )²We can verify that both expressions are equal, but perhaps it's not necessary right now.To show that FM is tangent to the circumcircle of FCD, we can use the condition that the distance from F to the circumcircle along FM is equal to the radius, or equivalently, that the power of point F with respect to the circumcircle is equal to the square of the length of the tangent from F to the circle.But since F is on the circle (wait, is F on the circle? Let me check.Wait, F is one of the points on the circumcircle of FCD, because FCD is the triangle, so F is a vertex. Therefore, the circumcircle passes through F, so the tangent at F would be the line perpendicular to the radius at F. But we are supposed to show that FM is tangent to the circumcircle of FCD, which is different from the tangent at F.Wait, no. The circumcircle of FCD passes through F, C, D. So, if FM is tangent to this circle, it must touch the circle at exactly one point, which may or may not be F.But since F is on the circle, if FM is tangent, it must be tangent at F, because a tangent from a point on the circle is unique and is the tangent at that point. So, if FM is tangent to the circumcircle of FCD, it must be tangent at F.But wait, let's think again. The tangent at F would be the line perpendicular to the radius TF at F. So, if FM is tangent at F, then FM must be perpendicular to TF.But TF is the radius, which is the vector from T to F.Wait, T is the center, so vector TF is from T to F: ( p/2 - (p/4 + y₁ y₂/(4p)), 0 - (y₁ + y₂)/4 ) = ( p/4 - y₁ y₂/(4p), - (y₁ + y₂)/4 )So, vector TF is ( p/4 - y₁ y₂/(4p), - (y₁ + y₂)/4 )Vector FM is ( (y₁ + y₂)^2/(4p), (y₁ + y₂)/4 (1 - y₁ y₂ / p² ) )If FM is tangent at F, then FM must be perpendicular to TF. So, their dot product should be zero.Compute the dot product:[ (y₁ + y₂)^2/(4p) ] * [ p/4 - y₁ y₂/(4p) ] + [ (y₁ + y₂)/4 (1 - y₁ y₂ / p² ) ] * [ - (y₁ + y₂)/4 ] = 0Let's compute each term.First term:(y₁ + y₂)^2/(4p) * ( p/4 - y₁ y₂/(4p) ) = (y₁ + y₂)^2/(4p) * [ (p² - y₁ y₂)/4p ] = (y₁ + y₂)^2 (p² - y₁ y₂) / (16 p² )Second term:(y₁ + y₂)/4 (1 - y₁ y₂ / p² ) * ( - (y₁ + y₂)/4 ) = - (y₁ + y₂)^2 / 16 (1 - y₁ y₂ / p² )So, total dot product:(y₁ + y₂)^2 (p² - y₁ y₂) / (16 p² ) - (y₁ + y₂)^2 / 16 (1 - y₁ y₂ / p² ) = 0Factor out (y₁ + y₂)^2 / 16:[ (p² - y₁ y₂)/p² - (1 - y₁ y₂ / p² ) ] = 0Simplify inside the brackets:(p² - y₁ y₂)/p² - 1 + y₁ y₂ / p² = (p² - y₁ y₂ - p² + y₁ y₂)/p² = 0So, the dot product is zero. Therefore, FM is perpendicular to TF, which means that FM is tangent to the circumcircle of FCD at point F.Wait, but earlier I thought that if F is on the circle, the tangent at F is unique and is the line perpendicular to TF. Since FM is this tangent, it must be tangent at F. Therefore, FM is tangent to the circumcircle of FCD.But the problem statement says "FM is a tangent to the circumcircle of triangle FCD". Since F is on the circle, and FM is tangent at F, that satisfies the condition.Alternatively, if F were not on the circle, we would have to show that FM touches the circle at exactly one point, but in this case, since F is on the circle, and FM is tangent at F, it's sufficient.Therefore, we've shown that FM is tangent to the circumcircle of triangle FCD.**Final Answer**boxed{FM text{ is a tangent to the circumcircle of } triangle FCD}

💡First, I need to determine the relative speed at which the two trains are approaching each other. Since they are moving in opposite directions, I will add their speeds together.The speed of the first train is 80 km/hr, and the speed of the second train is 55 km/hr.Relative speed = Speed of first train + Speed of second trainRelative speed = 80 km/hr + 55 km/hrRelative speed = 135 km/hrNext, I need to convert this relative speed from kilometers per hour to meters per second to match the units of the train lengths.1 km/hr = (1 km/hr) * (1000 m/km) / (3600 s/hr)1 km/hr = 1000/3600 m/s1 km/hr = 5/18 m/sTherefore, the relative speed in meters per second is:Relative speed = 135 km/hr * (5/18 m/s per km/hr)Relative speed = (135 * 5/18) m/sRelative speed = (15 * 5) m/sRelative speed = 75 m/sNow, I need to calculate the total distance that needs to be covered for the two trains to completely pass each other. This is the sum of their lengths.Total distance = Length of first train + Length of second trainTotal distance = 220 m + 300 mTotal distance = 520 mFinally, I can find the time it takes for the two trains to cross each other by dividing the total distance by the relative speed.Time = Total distance / Relative speedTime = 520 m / 75 m/sTime = 6.9333... secondsSo, the two trains take approximately 6.93 seconds to cross each other.

💡Okay, so I have this function here: f(x) = 2 sin²x + sin 2x. I need to find its maximum value. Hmm, trigonometric functions can sometimes be tricky, but maybe I can simplify this expression using some identities.First, I remember that sin²x can be rewritten using the double-angle identity. Let me recall: sin²x = (1 - cos 2x)/2. So if I substitute that into the function, it becomes:f(x) = 2 * (1 - cos 2x)/2 + sin 2xSimplifying that, the 2 and the denominator 2 cancel out, so I get:f(x) = (1 - cos 2x) + sin 2xWhich simplifies further to:f(x) = 1 - cos 2x + sin 2xOkay, that looks a bit simpler. Now, I have a function in terms of sin 2x and cos 2x. Maybe I can combine these into a single sine or cosine function. I remember that expressions of the form a sin θ + b cos θ can be written as R sin(θ + φ), where R is the amplitude and φ is the phase shift.Let me try that. So, in this case, my θ is 2x, and my coefficients are a = 1 for sin 2x and b = -1 for cos 2x. So, R would be sqrt(a² + b²) = sqrt(1 + 1) = sqrt(2). Then, the phase shift φ can be found using tan φ = b/a, which is tan φ = -1/1 = -1. So φ is -45 degrees or 315 degrees, which is equivalent to -π/4 radians.Therefore, I can rewrite the function as:f(x) = 1 + sqrt(2) sin(2x - π/4)Wait, let me check that. If I have a sin θ + b cos θ = R sin(θ + φ), then expanding R sin(θ + φ) gives R sin θ cos φ + R cos θ sin φ. Comparing coefficients, a = R cos φ and b = R sin φ. So in my case, a = 1 = R cos φ and b = -1 = R sin φ. Since R = sqrt(2), then cos φ = 1/sqrt(2) and sin φ = -1/sqrt(2). That corresponds to φ = -π/4. So yes, the expression becomes:f(x) = 1 + sqrt(2) sin(2x - π/4)Now, to find the maximum value of f(x), I need to find the maximum value of sin(2x - π/4). The sine function oscillates between -1 and 1, so the maximum value of sin(2x - π/4) is 1. Therefore, the maximum value of f(x) is:1 + sqrt(2) * 1 = 1 + sqrt(2)Wait, but let me double-check. Did I handle the signs correctly when combining the sine and cosine terms? Let me go back.Original expression after substitution: 1 - cos 2x + sin 2x. So that's 1 + sin 2x - cos 2x. So when I write this as R sin(2x + φ), I need to make sure that R is positive and φ is adjusted accordingly.Alternatively, maybe I can write it as R cos(2x + φ). Let me try that approach to see if I get the same result.Expressing sin 2x - cos 2x as R cos(2x + φ). Expanding R cos(2x + φ) gives R cos 2x cos φ - R sin 2x sin φ. Comparing coefficients:- For cos 2x: coefficient is -1, so R cos φ = -1- For sin 2x: coefficient is 1, so -R sin φ = 1So, R cos φ = -1 and -R sin φ = 1. Then, R² = (-1)² + (1)² = 1 + 1 = 2, so R = sqrt(2). Then, cos φ = -1/sqrt(2) and sin φ = -1/sqrt(2). That corresponds to φ = 5π/4. So, the expression becomes:sin 2x - cos 2x = sqrt(2) cos(2x + 5π/4)Therefore, f(x) = 1 + sqrt(2) cos(2x + 5π/4)The maximum value of cos is 1, so the maximum of f(x) is 1 + sqrt(2). Same result as before. So that seems consistent.Alternatively, I could have used calculus to find the maximum. Let me try that method as a check.Given f(x) = 2 sin²x + sin 2x. Let's find its derivative:f'(x) = 4 sin x cos x + 2 cos 2xWait, let me compute that again. The derivative of 2 sin²x is 4 sin x cos x (using chain rule: 2*2 sin x cos x). The derivative of sin 2x is 2 cos 2x. So f'(x) = 4 sin x cos x + 2 cos 2x.We can set this equal to zero to find critical points:4 sin x cos x + 2 cos 2x = 0Factor out 2 cos x:2 cos x (2 sin x + cos x) = 0Wait, no, that's not quite right. Let me see. Alternatively, express everything in terms of sin and cos:4 sin x cos x + 2 cos 2x = 0I can use the double-angle identity for cos 2x: cos 2x = 1 - 2 sin²x or 2 cos²x - 1. Alternatively, maybe express 4 sin x cos x as 2 sin 2x.So, f'(x) = 2 sin 2x + 2 cos 2xSet equal to zero:2 sin 2x + 2 cos 2x = 0Divide both sides by 2:sin 2x + cos 2x = 0Then, sin 2x = -cos 2xDivide both sides by cos 2x (assuming cos 2x ≠ 0):tan 2x = -1So, 2x = arctan(-1) + kπ, where k is integer.arctan(-1) is -π/4, so 2x = -π/4 + kπThus, x = -π/8 + kπ/2So the critical points are at x = -π/8 + kπ/2.Now, let's evaluate f(x) at these points.Take x = -π/8:f(-π/8) = 2 sin²(-π/8) + sin(2*(-π/8)) = 2 sin²(π/8) + sin(-π/4) = 2 sin²(π/8) - sin(π/4)I know that sin(π/8) = sqrt(2 - sqrt(2))/2, so sin²(π/8) = (2 - sqrt(2))/4Thus, 2 sin²(π/8) = 2*(2 - sqrt(2))/4 = (2 - sqrt(2))/2And sin(π/4) = sqrt(2)/2, so f(-π/8) = (2 - sqrt(2))/2 - sqrt(2)/2 = (2 - sqrt(2) - sqrt(2))/2 = (2 - 2 sqrt(2))/2 = 1 - sqrt(2)That's a minimum, since 1 - sqrt(2) is approximately -0.414.Now, take x = 3π/8 (which is -π/8 + π/2):f(3π/8) = 2 sin²(3π/8) + sin(2*(3π/8)) = 2 sin²(3π/8) + sin(3π/4)sin(3π/8) = sqrt(2 + sqrt(2))/2, so sin²(3π/8) = (2 + sqrt(2))/4Thus, 2 sin²(3π/8) = 2*(2 + sqrt(2))/4 = (2 + sqrt(2))/2sin(3π/4) = sqrt(2)/2So f(3π/8) = (2 + sqrt(2))/2 + sqrt(2)/2 = (2 + sqrt(2) + sqrt(2))/2 = (2 + 2 sqrt(2))/2 = 1 + sqrt(2)That's the maximum value.Similarly, checking x = 7π/8 (which is -π/8 + π):f(7π/8) = 2 sin²(7π/8) + sin(2*(7π/8)) = 2 sin²(7π/8) + sin(7π/4)sin(7π/8) = sin(π - π/8) = sin(π/8) = sqrt(2 - sqrt(2))/2, so sin²(7π/8) = (2 - sqrt(2))/4Thus, 2 sin²(7π/8) = (2 - sqrt(2))/2sin(7π/4) = -sqrt(2)/2So f(7π/8) = (2 - sqrt(2))/2 - sqrt(2)/2 = (2 - sqrt(2) - sqrt(2))/2 = (2 - 2 sqrt(2))/2 = 1 - sqrt(2)Again, a minimum.Similarly, x = 11π/8 (which is -π/8 + 3π/2):f(11π/8) = 2 sin²(11π/8) + sin(2*(11π/8)) = 2 sin²(11π/8) + sin(11π/4)sin(11π/8) = sin(π + 3π/8) = -sin(3π/8) = -sqrt(2 + sqrt(2))/2, so sin²(11π/8) = (2 + sqrt(2))/4Thus, 2 sin²(11π/8) = (2 + sqrt(2))/2sin(11π/4) = sin(3π/4) = sqrt(2)/2Wait, no: 11π/4 is equivalent to 11π/4 - 2π = 11π/4 - 8π/4 = 3π/4. So sin(11π/4) = sin(3π/4) = sqrt(2)/2Wait, but 11π/4 is actually 2π + 3π/4, so sin(11π/4) = sin(3π/4) = sqrt(2)/2But wait, 11π/4 is in the third quadrant, so sin is negative. Wait, no: 11π/4 is 2π + 3π/4, which is the same as 3π/4, which is in the second quadrant where sine is positive. So sin(11π/4) = sin(3π/4) = sqrt(2)/2Wait, but 11π/4 is actually 2π + 3π/4, so it's equivalent to 3π/4, which is in the second quadrant, so sine is positive. So yes, sin(11π/4) = sqrt(2)/2Thus, f(11π/8) = (2 + sqrt(2))/2 + sqrt(2)/2 = (2 + sqrt(2) + sqrt(2))/2 = (2 + 2 sqrt(2))/2 = 1 + sqrt(2)Wait, but that seems contradictory because 11π/8 is in the third quadrant, so sin(11π/8) is negative, but when squared, it becomes positive. So f(11π/8) is indeed 1 + sqrt(2). Hmm, but that seems like a maximum, but earlier at x = 3π/8, we also got 1 + sqrt(2). So both x = 3π/8 and x = 11π/8 give the maximum value.Wait, but 11π/8 is more than 2π, right? Wait, no, 11π/8 is less than 2π, since 2π is 16π/8. So 11π/8 is in the third quadrant, but when we plug into f(x), we get a positive value because of the squares and the sine term.Wait, but let me check f(11π/8):sin(11π/8) = sin(π + 3π/8) = -sin(3π/8) = -sqrt(2 + sqrt(2))/2So sin²(11π/8) = (2 + sqrt(2))/4Thus, 2 sin²(11π/8) = (2 + sqrt(2))/2sin(2*(11π/8)) = sin(11π/4) = sin(3π/4) = sqrt(2)/2So f(11π/8) = (2 + sqrt(2))/2 + sqrt(2)/2 = (2 + sqrt(2) + sqrt(2))/2 = (2 + 2 sqrt(2))/2 = 1 + sqrt(2)Yes, that's correct. So both x = 3π/8 and x = 11π/8 give the maximum value of 1 + sqrt(2).Wait, but 11π/8 is greater than π, so it's in the third quadrant, but the function f(x) still reaches the maximum there because of the combination of terms.So, from both methods—using trigonometric identities to rewrite the function and calculus to find critical points—we arrive at the conclusion that the maximum value of f(x) is 1 + sqrt(2).I think that's solid. I don't see any mistakes in my calculations, and both methods agree. So I'm confident that the maximum value is indeed 1 + sqrt(2).