💡First, I notice that the given product alternates between fractions of the form (frac{1}{3^n}) and (frac{3^m}{1}), where (n) and (m) are integers.I observe that each pair of consecutive terms multiplies to 3:[frac{1}{3} cdot frac{9}{1} = 3, quad frac{1}{27} cdot frac{81}{1} = 3, quad text{and so on}.]Next, I determine the number of such pairs in the sequence. The sequence starts at (frac{1}{3}) and ends at (frac{19683}{1}), with each term being a power of 3:[3^1, 3^4, 3^7, dots, 3^{16}.]There are six pairs in total.Finally, I calculate the total product by raising 3 to the power of the number of pairs:[3^6 = 729.]

💡Okay, so I have this problem about an ellipse, and I need to solve two parts. Let me try to understand what's given and what I need to find.First, the ellipse is given by the equation ( C: frac{x^2}{a^2} + frac{y^2}{b^2} = 1 ) where ( a > b > 0 ). The upper vertex is at point ( A(0, b) ), which makes sense because the upper vertex of an ellipse is at ( (0, b) ). There's also a point ( Pleft(frac{4}{3}, frac{b}{3}right) ) that lies on the ellipse. So, this point must satisfy the ellipse equation.Additionally, there's a circle with diameter ( AP ) that passes through the right focus point ( F ) of the ellipse. The right focus is at ( (c, 0) ) where ( c = sqrt{a^2 - b^2} ). So, the circle with diameter ( AP ) must pass through ( F ).For part (1), I need to find the equation of the ellipse ( C ). That means I need to find the values of ( a ) and ( b ).Let me start by using the fact that point ( P ) lies on the ellipse. Plugging ( Pleft(frac{4}{3}, frac{b}{3}right) ) into the ellipse equation:[frac{left(frac{4}{3}right)^2}{a^2} + frac{left(frac{b}{3}right)^2}{b^2} = 1]Simplifying this:[frac{frac{16}{9}}{a^2} + frac{frac{b^2}{9}}{b^2} = 1][frac{16}{9a^2} + frac{1}{9} = 1][frac{16}{9a^2} = 1 - frac{1}{9} = frac{8}{9}][frac{16}{9a^2} = frac{8}{9}][frac{16}{a^2} = 8][a^2 = frac{16}{8} = 2]So, ( a^2 = 2 ), which means ( a = sqrt{2} ).Now, I know that for an ellipse, ( c^2 = a^2 - b^2 ). So, ( c = sqrt{a^2 - b^2} = sqrt{2 - b^2} ).Next, the circle with diameter ( AP ) passes through the focus ( F(c, 0) ). Let me recall that if a circle has diameter ( AP ), then any point on the circle satisfies the condition that the angle at that point is a right angle. So, triangle ( AFP ) should be a right triangle with the right angle at ( F ).Therefore, vectors ( overrightarrow{FA} ) and ( overrightarrow{FP} ) should be perpendicular, meaning their dot product is zero.Let me compute these vectors:( overrightarrow{FA} = A - F = (0 - c, b - 0) = (-c, b) )( overrightarrow{FP} = P - F = left(frac{4}{3} - c, frac{b}{3} - 0right) = left(frac{4}{3} - c, frac{b}{3}right) )Their dot product is:[(-c) left(frac{4}{3} - cright) + (b) left(frac{b}{3}right) = 0][- frac{4c}{3} + c^2 + frac{b^2}{3} = 0]Multiply both sides by 3 to eliminate denominators:[-4c + 3c^2 + b^2 = 0]But I know that ( c^2 = 2 - b^2 ), so substitute that in:[-4c + 3(2 - b^2) + b^2 = 0][-4c + 6 - 3b^2 + b^2 = 0][-4c + 6 - 2b^2 = 0]But ( c^2 = 2 - b^2 ), so ( b^2 = 2 - c^2 ). Substitute that into the equation:[-4c + 6 - 2(2 - c^2) = 0][-4c + 6 - 4 + 2c^2 = 0][2c^2 - 4c + 2 = 0]Divide the entire equation by 2:[c^2 - 2c + 1 = 0]This factors as:[(c - 1)^2 = 0]So, ( c = 1 ).Now, since ( c = 1 ), and ( c^2 = 2 - b^2 ):[1 = 2 - b^2][b^2 = 1]So, ( b = 1 ) (since ( b > 0 )).Therefore, the equation of the ellipse is:[frac{x^2}{2} + y^2 = 1]That completes part (1). Now, moving on to part (2).Part (2) says: Suppose a moving line ( l ) intersects ellipse ( C ) at only one point, and there exist two fixed points on the x-axis such that the product of their distances to line ( l ) is 1. Determine the coordinates of these two fixed points.Okay, so line ( l ) is a tangent to the ellipse because it intersects at only one point. So, ( l ) is a tangent line to ellipse ( C ).We need to find two fixed points on the x-axis such that for any tangent line ( l ), the product of the distances from these two points to ( l ) is 1.Let me denote the two fixed points as ( M_1(lambda, 0) ) and ( M_2(mu, 0) ).Given that for any tangent line ( l ), the product ( d(M_1, l) times d(M_2, l) = 1 ).I need to find ( lambda ) and ( mu ).First, let me recall the general equation of a tangent to an ellipse ( frac{x^2}{a^2} + frac{y^2}{b^2} = 1 ). The equation can be written as ( y = mx + c ), where ( c = pm sqrt{a^2 m^2 + b^2} ). But since the line is tangent, the condition is ( c^2 = a^2 m^2 + b^2 ).But in our case, the ellipse is ( frac{x^2}{2} + y^2 = 1 ), so ( a^2 = 2 ) and ( b^2 = 1 ). Thus, the condition for tangency is ( c^2 = 2 m^2 + 1 ).Alternatively, another form of the tangent line is ( frac{xx_1}{a^2} + frac{yy_1}{b^2} = 1 ), where ( (x_1, y_1) ) is the point of contact. But since the line is moving, it's better to use the slope form.So, let me consider the tangent line as ( y = mx + c ), with ( c^2 = 2 m^2 + 1 ).Now, the distance from a point ( (x_0, y_0) ) to the line ( Ax + By + C = 0 ) is ( frac{|Ax_0 + By_0 + C|}{sqrt{A^2 + B^2}} ).First, let me write the equation of the tangent line in standard form. ( y = mx + c ) can be rewritten as ( mx - y + c = 0 ). So, ( A = m ), ( B = -1 ), ( C = c ).So, the distance from ( M_1(lambda, 0) ) to the line is:[d_1 = frac{|m lambda - 1 times 0 + c|}{sqrt{m^2 + 1}}} = frac{|m lambda + c|}{sqrt{m^2 + 1}}]Similarly, the distance from ( M_2(mu, 0) ) to the line is:[d_2 = frac{|m mu + c|}{sqrt{m^2 + 1}}]The product ( d_1 times d_2 ) is:[frac{|m lambda + c|}{sqrt{m^2 + 1}} times frac{|m mu + c|}{sqrt{m^2 + 1}} = frac{|(m lambda + c)(m mu + c)|}{m^2 + 1}]We are told that this product is equal to 1 for all tangent lines ( l ). So,[frac{|(m lambda + c)(m mu + c)|}{m^2 + 1} = 1]Since ( c^2 = 2 m^2 + 1 ), we can express ( c ) in terms of ( m ). Let me substitute ( c^2 = 2 m^2 + 1 ) into the equation.But before that, let me square both sides to eliminate the absolute value:[frac{(m lambda + c)^2 (m mu + c)^2}{(m^2 + 1)^2} = 1]Wait, no, actually, the equation is:[frac{|(m lambda + c)(m mu + c)|}{m^2 + 1} = 1]So, squaring both sides:[frac{(m lambda + c)^2 (m mu + c)^2}{(m^2 + 1)^2} = 1]But this seems complicated. Maybe instead, I can express ( c ) in terms of ( m ) and substitute.Since ( c^2 = 2 m^2 + 1 ), ( c = sqrt{2 m^2 + 1} ) or ( c = -sqrt{2 m^2 + 1} ). But since the tangent line can be above or below, maybe we can consider both cases, but perhaps it's symmetric.Alternatively, maybe instead of using ( y = mx + c ), I can parametrize the tangent lines differently.But perhaps another approach is to consider the condition ( d_1 times d_2 = 1 ) and express it in terms of ( lambda ) and ( mu ).Let me denote ( d_1 times d_2 = 1 ):[frac{|(m lambda + c)(m mu + c)|}{m^2 + 1} = 1]Let me expand the numerator:[|m lambda + c| times |m mu + c| = |(m lambda + c)(m mu + c)| = |m^2 lambda mu + m c (lambda + mu) + c^2|]So,[frac{|m^2 lambda mu + m c (lambda + mu) + c^2|}{m^2 + 1} = 1]Since this must hold for all ( m ), the expression inside the absolute value must be equal to ( m^2 + 1 ) or ( - (m^2 + 1) ). But since it's inside an absolute value, we can write:[m^2 lambda mu + m c (lambda + mu) + c^2 = pm (m^2 + 1)]But this must hold for all ( m ), so the coefficients of corresponding powers of ( m ) must be equal on both sides.But ( c ) is dependent on ( m ) as ( c^2 = 2 m^2 + 1 ). So, ( c = sqrt{2 m^2 + 1} ). Hmm, this complicates things because ( c ) is not a constant but depends on ( m ).Wait, maybe I can express ( c ) in terms of ( m ) and substitute into the equation.Let me denote ( c = sqrt{2 m^2 + 1} ). Then, ( c^2 = 2 m^2 + 1 ).Substituting into the equation:[m^2 lambda mu + m sqrt{2 m^2 + 1} (lambda + mu) + (2 m^2 + 1) = pm (m^2 + 1)]This seems messy because of the square root. Maybe instead, I can square both sides to eliminate the square root, but that might complicate things further.Alternatively, perhaps I can consider specific cases for ( m ) to find ( lambda ) and ( mu ).Let me try choosing specific values of ( m ) to create equations for ( lambda ) and ( mu ).First, let me take ( m = 0 ). Then, the tangent line is horizontal.For ( m = 0 ), the tangent line is ( y = c ), with ( c^2 = 1 ), so ( c = pm 1 ). So, the tangent lines are ( y = 1 ) and ( y = -1 ).Compute ( d_1 times d_2 ) for ( m = 0 ):For ( y = 1 ):Distance from ( M_1(lambda, 0) ) is ( |0 + 1| / sqrt{0 + 1} = 1 ).Similarly, distance from ( M_2(mu, 0) ) is ( |0 + 1| / 1 = 1 ).So, product is ( 1 times 1 = 1 ), which satisfies the condition.Similarly, for ( y = -1 ):Distances are ( |0 - 1| / 1 = 1 ), so product is 1.So, for ( m = 0 ), the condition is satisfied regardless of ( lambda ) and ( mu ). So, this doesn't give us new information.Let me try another value, say ( m = 1 ).For ( m = 1 ), ( c^2 = 2(1)^2 + 1 = 3 ), so ( c = sqrt{3} ) or ( c = -sqrt{3} ).So, the tangent lines are ( y = x + sqrt{3} ) and ( y = x - sqrt{3} ).Let me compute ( d_1 times d_2 ) for ( y = x + sqrt{3} ):First, rewrite the line as ( x - y + sqrt{3} = 0 ).Distance from ( M_1(lambda, 0) ):[d_1 = frac{|lambda - 0 + sqrt{3}|}{sqrt{1 + 1}} = frac{|lambda + sqrt{3}|}{sqrt{2}}]Similarly, distance from ( M_2(mu, 0) ):[d_2 = frac{|mu + sqrt{3}|}{sqrt{2}}]Product:[d_1 times d_2 = frac{|lambda + sqrt{3}| |mu + sqrt{3}|}{2} = 1]So,[|lambda + sqrt{3}| |mu + sqrt{3}| = 2]Similarly, for the other tangent line ( y = x - sqrt{3} ):Rewrite as ( x - y - sqrt{3} = 0 ).Distance from ( M_1(lambda, 0) ):[d_1 = frac{|lambda - 0 - sqrt{3}|}{sqrt{2}} = frac{|lambda - sqrt{3}|}{sqrt{2}}]Distance from ( M_2(mu, 0) ):[d_2 = frac{|mu - sqrt{3}|}{sqrt{2}}]Product:[d_1 times d_2 = frac{|lambda - sqrt{3}| |mu - sqrt{3}|}{2} = 1]So,[|lambda - sqrt{3}| |mu - sqrt{3}| = 2]So, now I have two equations:1. ( |lambda + sqrt{3}| |mu + sqrt{3}| = 2 )2. ( |lambda - sqrt{3}| |mu - sqrt{3}| = 2 )This seems symmetric. Maybe ( lambda = -mu ) or something like that.Let me assume ( lambda = -mu ). Let me denote ( mu = -lambda ).Then, equation 1 becomes:[|lambda + sqrt{3}| | -lambda + sqrt{3}| = 2][|(lambda + sqrt{3})(- lambda + sqrt{3})| = 2][| - lambda^2 + 3 | = 2]So,Either ( - lambda^2 + 3 = 2 ) or ( - lambda^2 + 3 = -2 )Case 1: ( - lambda^2 + 3 = 2 )[- lambda^2 = -1 implies lambda^2 = 1 implies lambda = pm 1]Case 2: ( - lambda^2 + 3 = -2 )[- lambda^2 = -5 implies lambda^2 = 5 implies lambda = pm sqrt{5}]But let's check equation 2 with ( mu = -lambda ):Equation 2:[|lambda - sqrt{3}| | -lambda - sqrt{3}| = 2][|(lambda - sqrt{3})(- lambda - sqrt{3})| = 2][| - lambda^2 - 3 | = 2]So,Either ( - lambda^2 - 3 = 2 ) or ( - lambda^2 - 3 = -2 )Case 1: ( - lambda^2 - 3 = 2 )[- lambda^2 = 5 implies lambda^2 = -5 ), which is impossible.Case 2: ( - lambda^2 - 3 = -2 )[- lambda^2 = 1 implies lambda^2 = -1 ), which is also impossible.So, only Case 1 in equation 1 gives possible solutions, which are ( lambda = pm 1 ).Thus, ( lambda = 1 ) or ( lambda = -1 ), and ( mu = -1 ) or ( mu = 1 ).Therefore, the two fixed points are ( (1, 0) ) and ( (-1, 0) ).Wait, but let me verify this with another value of ( m ) to ensure consistency.Let me take ( m = 1 ) again, but with ( lambda = 1 ) and ( mu = -1 ).Compute ( d_1 times d_2 ) for the tangent line ( y = x + sqrt{3} ):[d_1 = frac{|1 + sqrt{3}|}{sqrt{2}}, quad d_2 = frac{|-1 + sqrt{3}|}{sqrt{2}}][d_1 times d_2 = frac{(1 + sqrt{3})(-1 + sqrt{3})}{2} = frac{(-1 + 3)}{2} = frac{2}{2} = 1]Similarly, for the tangent line ( y = x - sqrt{3} ):[d_1 = frac{|1 - sqrt{3}|}{sqrt{2}}, quad d_2 = frac{|-1 - sqrt{3}|}{sqrt{2}}][d_1 times d_2 = frac{(1 - sqrt{3})(-1 - sqrt{3})}{2} = frac{(-1 - 3)}{2} = frac{-4}{2} = -2]Wait, the absolute value was not considered here. Actually, the distances are absolute values, so:[d_1 = frac{|1 - sqrt{3}|}{sqrt{2}}, quad d_2 = frac{|-1 - sqrt{3}|}{sqrt{2}}][d_1 times d_2 = frac{|1 - sqrt{3}| times |-1 - sqrt{3}|}{2} = frac{( sqrt{3} - 1 )( sqrt{3} + 1 )}{2} = frac{(3 - 1)}{2} = frac{2}{2} = 1]Yes, that works.Let me test another slope, say ( m = sqrt{1} ), but that's similar to above. Maybe ( m = sqrt{2} ).For ( m = sqrt{2} ), ( c^2 = 2*(2) + 1 = 5 ), so ( c = sqrt{5} ) or ( c = -sqrt{5} ).Take the tangent line ( y = sqrt{2} x + sqrt{5} ).Compute distances from ( (1, 0) ) and ( (-1, 0) ):First, rewrite the line as ( sqrt{2} x - y + sqrt{5} = 0 ).Distance from ( (1, 0) ):[d_1 = frac{|sqrt{2} * 1 - 0 + sqrt{5}|}{sqrt{ (sqrt{2})^2 + (-1)^2 }} = frac{|sqrt{2} + sqrt{5}|}{sqrt{2 + 1}} = frac{sqrt{2} + sqrt{5}}{sqrt{3}}]Distance from ( (-1, 0) ):[d_2 = frac{|sqrt{2}*(-1) - 0 + sqrt{5}|}{sqrt{3}} = frac{| -sqrt{2} + sqrt{5}|}{sqrt{3}}]Product:[d_1 times d_2 = frac{(sqrt{2} + sqrt{5})( -sqrt{2} + sqrt{5})}{3} = frac{(5 - 2)}{3} = frac{3}{3} = 1]Perfect, it works.Similarly, for the other tangent line ( y = sqrt{2} x - sqrt{5} ):Rewrite as ( sqrt{2} x - y - sqrt{5} = 0 ).Distance from ( (1, 0) ):[d_1 = frac{|sqrt{2} - 0 - sqrt{5}|}{sqrt{3}} = frac{|sqrt{2} - sqrt{5}|}{sqrt{3}}]Distance from ( (-1, 0) ):[d_2 = frac{|sqrt{2}*(-1) - 0 - sqrt{5}|}{sqrt{3}} = frac{| -sqrt{2} - sqrt{5}|}{sqrt{3}} = frac{sqrt{2} + sqrt{5}}{sqrt{3}}]Product:[d_1 times d_2 = frac{|sqrt{2} - sqrt{5}| (sqrt{2} + sqrt{5})}{3} = frac{(5 - 2)}{3} = 1]Again, it works.Therefore, it seems consistent that the two fixed points are ( (1, 0) ) and ( (-1, 0) ).But wait, in the ellipse equation, the foci are at ( (pm c, 0) ), and earlier we found ( c = 1 ). So, the foci are at ( (1, 0) ) and ( (-1, 0) ). So, these fixed points are actually the foci of the ellipse.That makes sense because the product of distances from the foci to any tangent line is related to the properties of the ellipse. In fact, for an ellipse, the product of the distances from the foci to any tangent line is equal to ( b^2 ), but in this case, it's given as 1. Since ( b^2 = 1 ), it fits.So, the two fixed points are the foci of the ellipse, which are ( (1, 0) ) and ( (-1, 0) ).**Final Answer**(1) The equation of ellipse ( C ) is (boxed{dfrac{x^2}{2} + y^2 = 1}).(2) The coordinates of the two fixed points are (boxed{(1, 0)}) and (boxed{(-1, 0)}).

💡First, I need to determine the speed of the train using the information provided about it passing a 400-meter-long platform in 160 seconds.When the train passes the platform, it covers the combined length of the train and the platform. Therefore, the total distance covered is 1200 meters (length of the train) plus 400 meters (length of the platform), which equals 1600 meters.Using the formula for speed, which is distance divided by time, I can calculate the speed of the train:Speed = 1600 meters / 160 seconds = 10 meters per second.Now that I know the speed of the train, I can determine the time it takes to cross a tree. Since a tree can be considered a point relative to the length of the train, the train only needs to cover its own length to completely pass the tree.Using the formula for time, which is distance divided by speed, I can calculate the time it takes to cross the tree:Time = 1200 meters / 10 meters per second = 120 seconds.Therefore, it takes the train 120 seconds to cross a tree.

💡Okay, so I have this problem here: I need to find how many two-digit numbers exist such that when you add the number to its reverse, the result is a perfect square. Hmm, interesting. Let me try to break this down step by step.First, let's clarify what a two-digit number is. A two-digit number can be represented as 10a + b, where 'a' is the tens digit and 'b' is the units digit. For example, the number 34 would be 10*3 + 4 = 34. The reverse of this number would be 43, which is 10*4 + 3 = 43.So, if I have a number N = 10a + b, its reverse would be R = 10b + a. The problem states that N + R should be a perfect square. Let's write that out:N + R = (10a + b) + (10b + a) = 11a + 11b = 11(a + b)So, the sum of the number and its reverse is 11 times the sum of its digits. Now, this sum needs to be a perfect square. That means 11(a + b) should be a perfect square.Hmm, okay. Since 11 is a prime number, for 11(a + b) to be a perfect square, (a + b) must be a multiple of 11. Because 11 is prime, the only way for 11(a + b) to be a perfect square is if (a + b) itself is 11 times some perfect square. But wait, let's think about the possible values of (a + b).Since 'a' and 'b' are single digits (from 0 to 9), the maximum value of (a + b) is 9 + 9 = 18. So, (a + b) can range from 0 to 18. But since 'a' is the tens digit of a two-digit number, 'a' can't be 0. So, 'a' ranges from 1 to 9, and 'b' ranges from 0 to 9.Given that, the possible values for (a + b) are from 1 + 0 = 1 up to 9 + 9 = 18. Now, we need 11(a + b) to be a perfect square. Let's list the perfect squares that are multiples of 11 within the range of 11*1 = 11 up to 11*18 = 198.The perfect squares between 11 and 198 are:16 (4^2), 25 (5^2), 36 (6^2), 49 (7^2), 64 (8^2), 81 (9^2), 100 (10^2), 121 (11^2), 144 (12^2), 169 (13^2), 196 (14^2).Now, let's see which of these are multiples of 11:- 121 is 11^2, which is 11*11.- 196 is 14^2, which is 14*14, but 196 divided by 11 is approximately 17.818, which is not an integer, so 196 is not a multiple of 11.- Similarly, checking the others: 16, 25, 36, 49, 64, 81, 100, 121, 144, 169. Only 121 is a multiple of 11.So, 11(a + b) must equal 121. Therefore, a + b = 121 / 11 = 11.So, the sum of the digits 'a' and 'b' must be 11. Now, we need to find all two-digit numbers where the sum of the digits is 11.Let's list all possible pairs (a, b) where a + b = 11, with 'a' ranging from 1 to 9 and 'b' ranging from 0 to 9.Starting with a = 2, b = 9 (since a = 1 would make b = 10, which is not a single digit):- a = 2, b = 9 → number 29- a = 3, b = 8 → number 38- a = 4, b = 7 → number 47- a = 5, b = 6 → number 56- a = 6, b = 5 → number 65- a = 7, b = 4 → number 74- a = 8, b = 3 → number 83- a = 9, b = 2 → number 92Wait, what about a = 1? If a = 1, then b would need to be 10, but b has to be a single digit, so that's not possible. Similarly, a = 0 isn't valid because it's not a two-digit number.So, we have the numbers: 29, 38, 47, 56, 65, 74, 83, 92. That's 8 numbers in total.Let me double-check to make sure each of these numbers, when added to their reverses, gives 121, which is a perfect square.Take 29: reverse is 92. 29 + 92 = 121, which is 11^2. Good.38 + 83 = 121. Good.47 + 74 = 121. Good.56 + 65 = 121. Good.65 + 56 = 121. Good.74 + 47 = 121. Good.83 + 38 = 121. Good.92 + 29 = 121. Good.All of them check out. So, there are 8 such numbers.I think that's all. I don't think I missed any pairs because I systematically went through all possible 'a' values from 2 to 9 and found the corresponding 'b' that makes the sum 11. And since 121 is the only perfect square multiple of 11 within the possible range, these are the only numbers that satisfy the condition.

💡First, I need to determine the work rates of the two individuals involved in completing the job. Let's denote the first person as A and the second person as B.Person A can complete the job in 4 hours, so their work rate is 1/4 of the job per hour.When A and B work together, they complete the job in 1 hour. This means their combined work rate is 1 job per hour.To find B's work rate, I subtract A's work rate from the combined work rate:Work rate of B = Combined work rate - Work rate of AWork rate of B = 1 - 1/4Work rate of B = 3/4 of the job per hourNext, I need to consider the lowest fraction of the job that can be done, which is given as 0.3666666666666667. This fraction represents the smallest indivisible part of the job that B can do alone.Since B's work rate is 3/4 of the job per hour, I can calculate the time it takes for B to complete the job individually by taking the reciprocal of the work rate:Time for B to complete the job = 1 / (3/4)Time for B to complete the job = 4/3 hoursTherefore, it takes the third person 4/3 hours, or approximately 1.3333 hours, to complete the job individually.

💡Okay, so I have this problem about triangle ABC. It says that sin(2A + B) equals 2 sin B, and I need to find the minimum value of tan A + tan C + 2/tan B. Hmm, triangles and trigonometric functions. Let me try to break this down step by step.First, I remember that in any triangle, the sum of the angles is π radians, or 180 degrees. So, A + B + C = π. That might come in handy later.The given equation is sin(2A + B) = 2 sin B. I need to work with this equation to find some relationship between the angles A, B, and C. Maybe I can express everything in terms of one variable or find a ratio between the tangents.Let me try to expand sin(2A + B). Using the sine addition formula, sin(2A + B) = sin(2A)cos B + cos(2A)sin B. So, substituting that into the equation:sin(2A)cos B + cos(2A)sin B = 2 sin B.Hmm, let me write that out:sin(2A)cos B + cos(2A)sin B = 2 sin B.I can factor sin B on the right side, but maybe I can bring all terms to one side:sin(2A)cos B + cos(2A)sin B - 2 sin B = 0.Factor sin B:sin(2A)cos B + sin B (cos(2A) - 2) = 0.Hmm, not sure if that helps. Maybe I can express sin(2A) and cos(2A) in terms of sin A and cos A. Let me recall that sin(2A) = 2 sin A cos A and cos(2A) = cos² A - sin² A.Substituting these into the equation:2 sin A cos A cos B + (cos² A - sin² A) sin B = 2 sin B.Hmm, that seems a bit complicated. Maybe I can use the identity for sin(2A + B) in another way. Alternatively, perhaps I can use the fact that in a triangle, A + B + C = π, so 2A + B = A + (A + B) = A + (π - C). So, 2A + B = π + A - C.Wait, that might not be helpful. Alternatively, maybe I can write 2A + B as A + (A + B). Since A + B = π - C, so 2A + B = A + (π - C) = π + A - C.So, sin(2A + B) = sin(π + A - C) = -sin(A - C), because sin(π + x) = -sin x.So, sin(2A + B) = -sin(A - C). Therefore, the given equation becomes:-sin(A - C) = 2 sin B.But sin B is positive because B is an angle in a triangle, so 0 < B < π, so sin B > 0. Therefore, -sin(A - C) = 2 sin B implies that sin(A - C) is negative. So, A - C is negative, meaning C > A.So, sin(A - C) = -2 sin B.But sin(A - C) = -sin(C - A), so sin(C - A) = 2 sin B.Hmm, interesting. So, sin(C - A) = 2 sin B.But wait, the sine function has a maximum value of 1, so 2 sin B must be less than or equal to 1. Therefore, sin B ≤ 1/2. So, B ≤ π/6 or B ≥ 5π/6. But since B is an angle in a triangle, it must be less than π, so B ≤ π/6 or 5π/6 < B < π. But 5π/6 is 150 degrees, which is quite large. Let me see if that's possible.If B is 5π/6, then the other angles A and C would have to add up to π - 5π/6 = π/6. But since C > A, as we saw earlier, C would be greater than A, but both would have to be less than π/6. Hmm, that might complicate things. Alternatively, maybe B is less than or equal to π/6.But let's not get ahead of ourselves. Let's see if we can find a relationship between the angles.We have sin(C - A) = 2 sin B.Also, since A + B + C = π, we can write C = π - A - B.So, substituting into sin(C - A):sin(π - A - B - A) = sin(π - 2A - B) = sin(π - (2A + B)) = sin(2A + B).Wait, that's interesting. So, sin(C - A) = sin(2A + B). But from the given equation, sin(2A + B) = 2 sin B. Therefore, sin(C - A) = 2 sin B.Wait, that's the same as before. So, we have sin(C - A) = 2 sin B.But as I thought earlier, since sin(C - A) = 2 sin B, and sin B ≤ 1/2, so B ≤ π/6.So, B is at most 30 degrees.Alternatively, maybe I can express sin(C - A) in terms of A and B.Since C = π - A - B, then C - A = π - 2A - B.So, sin(C - A) = sin(π - 2A - B) = sin(2A + B), because sin(π - x) = sin x.Wait, that's the same as before. So, sin(2A + B) = 2 sin B.Which is the original equation. Hmm, so perhaps I need a different approach.Let me consider expressing everything in terms of one variable.Since A + B + C = π, I can express C as π - A - B.So, the expression we need to minimize is tan A + tan C + 2/tan B.Substituting C:tan A + tan(π - A - B) + 2/tan B.But tan(π - x) = -tan x, so tan(π - A - B) = -tan(A + B).So, the expression becomes:tan A - tan(A + B) + 2/tan B.Hmm, that might not be helpful. Alternatively, maybe I can express tan C in terms of A and B.Alternatively, perhaps I can use the sine rule or cosine rule.Wait, let's think about the given equation: sin(2A + B) = 2 sin B.Let me write 2A + B as (A + B) + A.So, sin((A + B) + A) = sin(π - C + A) because A + B = π - C.So, sin(π - C + A) = sin(π + (A - C)) = -sin(A - C).Therefore, sin(2A + B) = -sin(A - C) = 2 sin B.So, -sin(A - C) = 2 sin B.Which implies sin(A - C) = -2 sin B.But sin(A - C) is equal to sin A cos C - cos A sin C.So, sin A cos C - cos A sin C = -2 sin B.Hmm, maybe I can relate this to the sine rule.In triangle ABC, by the sine rule, we have:a / sin A = b / sin B = c / sin C = 2R,where a, b, c are the sides opposite angles A, B, C respectively, and R is the circumradius.So, perhaps I can express sin A, sin B, sin C in terms of the sides.But I'm not sure if that will help directly. Maybe I can express sin A and sin C in terms of sin B.Alternatively, let's consider expressing tan A, tan C, and tan B in terms of each other.Let me recall that tan A + tan C + 2/tan B is the expression to minimize.Let me denote x = tan A, y = tan C, z = tan B.So, we need to minimize x + y + 2/z.But we have some relationships between x, y, z.From the given equation, sin(2A + B) = 2 sin B.Let me try to express this in terms of x, y, z.First, let's recall that in a triangle, A + B + C = π, so B = π - A - C.So, 2A + B = 2A + π - A - C = A + π - C.So, sin(2A + B) = sin(A + π - C) = sin(π + A - C) = -sin(A - C).So, sin(2A + B) = -sin(A - C) = 2 sin B.Therefore, -sin(A - C) = 2 sin B.So, sin(A - C) = -2 sin B.But sin(A - C) = sin A cos C - cos A sin C.So, sin A cos C - cos A sin C = -2 sin B.Hmm, maybe I can divide both sides by cos A cos C to express this in terms of tan A and tan C.Let me try that.Divide both sides by cos A cos C:(sin A cos C)/(cos A cos C) - (cos A sin C)/(cos A cos C) = (-2 sin B)/(cos A cos C).Simplify:tan A - tan C = (-2 sin B)/(cos A cos C).Hmm, not sure if that helps. Alternatively, maybe I can express sin B in terms of A and C.Since B = π - A - C, sin B = sin(π - A - C) = sin(A + C).So, sin B = sin(A + C) = sin A cos C + cos A sin C.Therefore, sin(A - C) = -2 sin(A + C).So, sin A cos C - cos A sin C = -2(sin A cos C + cos A sin C).Let me write that:sin A cos C - cos A sin C = -2 sin A cos C - 2 cos A sin C.Bring all terms to one side:sin A cos C - cos A sin C + 2 sin A cos C + 2 cos A sin C = 0.Combine like terms:(1 + 2) sin A cos C + (-1 + 2) cos A sin C = 0.So, 3 sin A cos C + 1 cos A sin C = 0.So, 3 sin A cos C + cos A sin C = 0.Hmm, maybe I can factor this.Let me factor sin A cos C:sin A cos C (3) + cos A sin C (1) = 0.Alternatively, perhaps I can write this as:3 sin A cos C = - cos A sin C.Divide both sides by cos A cos C:3 tan A = - tan C.So, tan C = -3 tan A.Interesting. So, tan C is negative three times tan A.But in a triangle, all angles are between 0 and π, so their tangents are positive because angles are between 0 and π/2 or π/2 and π, but in a triangle, all angles are less than π, so if an angle is greater than π/2, its tangent is negative.Wait, but in a triangle, only one angle can be greater than π/2. So, if C is greater than π/2, then tan C is negative, which would make sense because tan C = -3 tan A, so if tan C is negative, tan A must be positive, which it is because A is less than π/2.So, C is greater than π/2, and A is less than π/2.So, tan C = -3 tan A.So, we have tan C = -3 tan A.So, now, let's express tan B in terms of tan A and tan C.Since A + B + C = π, so B = π - A - C.So, tan B = tan(π - A - C) = -tan(A + C).Using the tangent addition formula:tan(A + C) = (tan A + tan C)/(1 - tan A tan C).So, tan B = - (tan A + tan C)/(1 - tan A tan C).But we know tan C = -3 tan A, so let's substitute that:tan B = - (tan A - 3 tan A)/(1 - tan A (-3 tan A)).Simplify numerator and denominator:Numerator: tan A - 3 tan A = -2 tan A.Denominator: 1 + 3 tan² A.So, tan B = - (-2 tan A)/(1 + 3 tan² A) = (2 tan A)/(1 + 3 tan² A).So, tan B = (2 tan A)/(1 + 3 tan² A).So, now, we have expressions for tan C and tan B in terms of tan A.So, our expression to minimize is:tan A + tan C + 2/tan B.Substituting tan C = -3 tan A and tan B = (2 tan A)/(1 + 3 tan² A):tan A + (-3 tan A) + 2 / [(2 tan A)/(1 + 3 tan² A)].Simplify term by term:First term: tan A.Second term: -3 tan A.Third term: 2 divided by (2 tan A)/(1 + 3 tan² A) is equal to 2 * (1 + 3 tan² A)/(2 tan A) = (1 + 3 tan² A)/tan A.So, putting it all together:tan A - 3 tan A + (1 + 3 tan² A)/tan A.Simplify:(tan A - 3 tan A) = -2 tan A.So, the expression becomes:-2 tan A + (1 + 3 tan² A)/tan A.Let me write this as:-2 tan A + (1/tan A) + 3 tan A.Simplify:(-2 tan A + 3 tan A) + 1/tan A = tan A + 1/tan A.So, the expression simplifies to tan A + 1/tan A.So, we need to minimize tan A + 1/tan A.Let me denote x = tan A. Since A is an acute angle (because C is obtuse), x > 0.So, we need to minimize x + 1/x.I remember that for x > 0, the function x + 1/x has a minimum at x = 1, where the value is 2. This is by AM-GM inequality: x + 1/x ≥ 2√(x * 1/x) = 2.So, the minimum value is 2, achieved when x = 1, i.e., when tan A = 1, so A = π/4.So, the minimum value of tan A + tan C + 2/tan B is 2.Wait, let me double-check.We started with sin(2A + B) = 2 sin B.Expressed tan C = -3 tan A.Expressed tan B in terms of tan A.Substituted into the expression to get tan A + 1/tan A.Which is minimized at 2.Yes, that seems correct.So, the minimum value is 2.