💡Okay, so I have this problem about a sequence {a_n} where the first term a_1 is 1. The sum of the first n terms, S_n, satisfies this equation: 2tS_n - (2t + 1)S_{n-1} = 2t, where t is a positive constant and n is 2, 3, 4, and so on. Part 1 asks me to prove that {a_n} is a geometric sequence. Hmm, okay. So, a geometric sequence is one where each term is a constant multiple of the previous term, right? That constant is called the common ratio. So, if I can show that a_{n+1}/a_n is constant for all n, then I can prove it's geometric.Let me start by writing down the given equation: 2tS_n - (2t + 1)S_{n-1} = 2t. Maybe I can find a recursive relationship between S_n and S_{n-1}, and then relate that to a_n.I know that S_n = a_1 + a_2 + ... + a_n, so S_n - S_{n-1} = a_n. Similarly, S_{n+1} - S_n = a_{n+1}. Maybe I can write the equation for n and n+1 and subtract them or something.Wait, let's see. If I take the given equation for n and for n+1:For n: 2tS_n - (2t + 1)S_{n-1} = 2t.For n+1: 2tS_{n+1} - (2t + 1)S_n = 2t.Now, subtract the first equation from the second:2tS_{n+1} - (2t + 1)S_n - [2tS_n - (2t + 1)S_{n-1}] = 0.Simplify that:2tS_{n+1} - (2t + 1)S_n - 2tS_n + (2t + 1)S_{n-1} = 0.Combine like terms:2tS_{n+1} - [ (2t + 1) + 2t ] S_n + (2t + 1)S_{n-1} = 0.Simplify the coefficients:(2t + 1) + 2t = 4t + 1.So, 2tS_{n+1} - (4t + 1)S_n + (2t + 1)S_{n-1} = 0.Hmm, not sure if that helps directly. Maybe I should express S_{n+1} in terms of S_n and S_{n-1}.Wait, maybe another approach. Let's consider the original equation:2tS_n - (2t + 1)S_{n-1} = 2t.Let me solve for S_n:2tS_n = (2t + 1)S_{n-1} + 2t.Divide both sides by 2t:S_n = [(2t + 1)/2t] S_{n-1} + 1.So, S_n = (1 + 1/(2t)) S_{n-1} + 1.Hmm, that seems like a linear recurrence for S_n. Maybe I can solve this recurrence relation.Let me write it as S_n - (1 + 1/(2t)) S_{n-1} = 1.This is a nonhomogeneous linear recurrence. The homogeneous part is S_n - (1 + 1/(2t)) S_{n-1} = 0, which has the solution S_n^{(h)} = C (1 + 1/(2t))^n.For the particular solution, since the nonhomogeneous term is constant, let's assume S_n^{(p)} = A, a constant.Substitute into the recurrence:A - (1 + 1/(2t)) A = 1.Simplify:A [1 - (1 + 1/(2t))] = 1.Which is A [ -1/(2t) ] = 1.So, A = -2t.Therefore, the general solution is S_n = C (1 + 1/(2t))^n - 2t.Now, apply the initial condition. For n=1, S_1 = a_1 = 1.So, S_1 = C (1 + 1/(2t))^1 - 2t = 1.Thus, C (1 + 1/(2t)) = 1 + 2t.Therefore, C = (1 + 2t)/(1 + 1/(2t)).Simplify denominator:1 + 1/(2t) = (2t + 1)/(2t).So, C = (1 + 2t) * (2t)/(2t + 1) = 2t.So, C = 2t.Therefore, the sum S_n = 2t (1 + 1/(2t))^n - 2t.Simplify:S_n = 2t [ (1 + 1/(2t))^n - 1 ].Okay, so S_n is expressed in terms of n. Now, to find a_n, since a_n = S_n - S_{n-1}.Compute a_n:a_n = S_n - S_{n-1} = 2t [ (1 + 1/(2t))^n - 1 ] - 2t [ (1 + 1/(2t))^{n-1} - 1 ].Simplify:= 2t (1 + 1/(2t))^n - 2t - 2t (1 + 1/(2t))^{n-1} + 2t.The -2t and +2t cancel:= 2t (1 + 1/(2t))^n - 2t (1 + 1/(2t))^{n-1}.Factor out 2t (1 + 1/(2t))^{n-1}:= 2t (1 + 1/(2t))^{n-1} [ (1 + 1/(2t)) - 1 ].Simplify inside the brackets:(1 + 1/(2t)) - 1 = 1/(2t).Thus,a_n = 2t (1 + 1/(2t))^{n-1} * (1/(2t)) = (1 + 1/(2t))^{n-1}.So, a_n = (1 + 1/(2t))^{n-1}.Therefore, {a_n} is a geometric sequence with common ratio r = 1 + 1/(2t).Cool, that proves part 1.Moving on to part 2. Let f(t) be the common ratio of {a_n}, so f(t) = 1 + 1/(2t). We need to define another sequence {b_n} where b_1 = 1 and for n >= 2, b_n = f(1/(b_{n-1} + 2)) - 2. We need to find the general formula for {b_n}.Alright, let's write down the recursion:b_n = f(1/(b_{n-1} + 2)) - 2.But f(t) = 1 + 1/(2t), so substitute:b_n = [1 + 1/(2 * (1/(b_{n-1} + 2)))] - 2.Simplify inside the brackets:1/(2 * (1/(b_{n-1} + 2))) = (b_{n-1} + 2)/2.So,b_n = 1 + (b_{n-1} + 2)/2 - 2.Simplify:1 + (b_{n-1} + 2)/2 - 2 = (b_{n-1} + 2)/2 - 1.So,b_n = (b_{n-1} + 2)/2 - 1 = (b_{n-1} + 2 - 2)/2 = b_{n-1}/2.Wait, that simplifies to b_n = (b_{n-1}) / 2.So, the recursion is b_n = (1/2) b_{n-1}.With b_1 = 1, this is a geometric sequence with common ratio 1/2.Therefore, the general term is b_n = (1/2)^{n-1}.So, b_n = 1/(2^{n-1}).Alright, that seems straightforward.Now, part 3: Define c_n = n b_n, and find the sum of the first n terms, T_n.So, c_n = n b_n = n / 2^{n-1}.We need to compute T_n = sum_{k=1}^n c_k = sum_{k=1}^n k / 2^{k-1}.Hmm, this is a finite sum of the form sum_{k=1}^n k x^{k-1}, where x = 1/2.I remember that the sum of k x^{k-1} from k=1 to infinity is 1/(1 - x)^2, but here we have a finite sum.Wait, let me recall the formula for the finite sum.The sum S = sum_{k=1}^n k x^{k-1}.I think the formula is S = (1 - (n + 1) x^n + n x^{n + 1}) / (1 - x)^2.Let me verify that.Yes, the sum can be derived by considering the derivative of the geometric series.Let me recall:sum_{k=0}^n x^k = (1 - x^{n+1}) / (1 - x).Differentiate both sides with respect to x:sum_{k=0}^n k x^{k - 1} = [ (0 - (n + 1) x^n ) (1 - x) - (1 - x^{n+1})(-1) ] / (1 - x)^2.Simplify numerator:- (n + 1) x^n (1 - x) + (1 - x^{n+1}).= - (n + 1) x^n + (n + 1) x^{n + 1} + 1 - x^{n + 1}.= 1 - (n + 1) x^n + n x^{n + 1}.Therefore,sum_{k=0}^n k x^{k - 1} = [1 - (n + 1) x^n + n x^{n + 1}] / (1 - x)^2.But our sum starts at k=1, so subtract the k=0 term, which is 0.Thus,sum_{k=1}^n k x^{k - 1} = [1 - (n + 1) x^n + n x^{n + 1}] / (1 - x)^2.In our case, x = 1/2.So,T_n = sum_{k=1}^n k / 2^{k - 1} = sum_{k=1}^n k (1/2)^{k - 1} = [1 - (n + 1)(1/2)^n + n (1/2)^{n + 1}] / (1 - 1/2)^2.Simplify denominator:(1 - 1/2)^2 = (1/2)^2 = 1/4.So,T_n = [1 - (n + 1)/2^n + n / 2^{n + 1}] / (1/4) = 4 [1 - (n + 1)/2^n + n / 2^{n + 1}].Simplify the terms inside:Let me write all terms with denominator 2^{n + 1}:1 = 2^{n + 1}/2^{n + 1}.(n + 1)/2^n = 2(n + 1)/2^{n + 1}.n / 2^{n + 1} remains as is.So,1 - (n + 1)/2^n + n / 2^{n + 1} = [2^{n + 1} - 2(n + 1) + n] / 2^{n + 1}.Simplify numerator:2^{n + 1} - 2n - 2 + n = 2^{n + 1} - n - 2.Thus,T_n = 4 * [ (2^{n + 1} - n - 2) / 2^{n + 1} ] = 4 [1 - (n + 2)/2^{n + 1}].Wait, let me check that:Wait, 2^{n + 1} - n - 2 is the numerator, so:[2^{n + 1} - n - 2] / 2^{n + 1} = 1 - (n + 2)/2^{n + 1}.Thus,T_n = 4 [1 - (n + 2)/2^{n + 1}].Alternatively, we can write it as:T_n = 4 - 4(n + 2)/2^{n + 1} = 4 - (n + 2)/2^{n - 1}.Wait, let's see:4(n + 2)/2^{n + 1} = (n + 2)/2^{n - 1}.Yes, because 4 / 2^{n + 1} = 1 / 2^{n - 1}.So, T_n = 4 - (n + 2)/2^{n - 1}.Alternatively, we can write it as:T_n = 4 - (n + 2)/2^{n - 1}.But let me check my steps again to make sure.Starting from:T_n = sum_{k=1}^n k / 2^{k - 1}.We used the formula for the finite sum:sum_{k=1}^n k x^{k - 1} = [1 - (n + 1)x^n + n x^{n + 1}] / (1 - x)^2.With x = 1/2, so:[1 - (n + 1)(1/2)^n + n (1/2)^{n + 1}] / (1 - 1/2)^2.Which is:[1 - (n + 1)/2^n + n / 2^{n + 1}] / (1/4) = 4 [1 - (n + 1)/2^n + n / 2^{n + 1}].Then, combining terms:1 - (n + 1)/2^n + n / 2^{n + 1}.Express 1 as 2^{n + 1}/2^{n + 1}.Then,2^{n + 1}/2^{n + 1} - 2(n + 1)/2^{n + 1} + n / 2^{n + 1}.Combine numerator:2^{n + 1} - 2(n + 1) + n = 2^{n + 1} - 2n - 2 + n = 2^{n + 1} - n - 2.Thus,[2^{n + 1} - n - 2]/2^{n + 1} = 1 - (n + 2)/2^{n + 1}.Multiply by 4:4[1 - (n + 2)/2^{n + 1}] = 4 - 4(n + 2)/2^{n + 1}.Simplify 4 / 2^{n + 1} = 1 / 2^{n - 1}.Thus,T_n = 4 - (n + 2)/2^{n - 1}.Yes, that seems correct.Alternatively, we can write it as:T_n = 4 - (n + 2)/2^{n - 1}.So, that's the sum of the first n terms of {c_n}.Let me double-check with small n to see if it makes sense.For n=1: c_1 = 1 * b_1 = 1 * 1 = 1. So, T_1 = 1.Using the formula: 4 - (1 + 2)/2^{0} = 4 - 3/1 = 1. Correct.For n=2: c_1 = 1, c_2 = 2 * b_2 = 2 * (1/2) = 1. So, T_2 = 1 + 1 = 2.Using the formula: 4 - (2 + 2)/2^{1} = 4 - 4/2 = 4 - 2 = 2. Correct.For n=3: c_3 = 3 * b_3 = 3 * (1/4) = 3/4. So, T_3 = 1 + 1 + 3/4 = 2.75.Using the formula: 4 - (3 + 2)/2^{2} = 4 - 5/4 = 11/4 = 2.75. Correct.Looks good.So, summarizing:1. {a_n} is geometric with common ratio 1 + 1/(2t).2. {b_n} is geometric with common ratio 1/2, so b_n = (1/2)^{n-1}.3. {c_n} has sum T_n = 4 - (n + 2)/2^{n - 1}.**Final Answer**1. The sequence {a_n} is geometric with common ratio 1 + frac{1}{2t}.2. The general formula for {b_n} is boxed{b_n = left(frac{1}{2}right)^{n-1}}.3. The sum of the first n terms of {c_n} is boxed{T_n = 4 - frac{n + 2}{2^{n-1}}}.

💡First, I need to determine the total distance the car has traveled. The car goes 150 kilometers to city X and then an additional 220 kilometers to city Y. Adding these distances together gives a total of 370 kilometers.Next, I'll calculate the total amount of fuel used for the entire journey. The car uses 20 liters of fuel to reach city X and an additional 15 liters to get to city Y. Summing these fuel amounts results in a total of 35 liters used.Finally, to find the average fuel consumption per kilometer, I'll divide the total fuel used by the total distance traveled. This means dividing 35 liters by 370 kilometers, which equals approximately 0.0946 liters per kilometer.

💡Okay, so I have this problem about a parabola in the Cartesian coordinate system. The equation of the parabola is given as ( y = ax^2 + (2m - 6)x + 1 ), and it passes through the point ( (1, 2m - 4) ). There are three parts to this problem: finding the value of ( a ), finding the axis of symmetry in terms of ( m ), and then determining the range of values for ( m ) given some conditions on the points lying on the parabola.Starting with part (1), I need to find the value of ( a ). Since the parabola passes through the point ( (1, 2m - 4) ), I can substitute ( x = 1 ) and ( y = 2m - 4 ) into the equation of the parabola. That should give me an equation that I can solve for ( a ).So, substituting those values in:( 2m - 4 = a(1)^2 + (2m - 6)(1) + 1 )Simplifying the right side:( 2m - 4 = a + 2m - 6 + 1 )Combine like terms:( 2m - 4 = a + 2m - 5 )Hmm, now I can subtract ( 2m ) from both sides to get:( -4 = a - 5 )Then, adding 5 to both sides:( a = 1 )Wait, that seems straightforward. So, ( a = 1 ). Let me double-check that. If I plug ( a = 1 ) back into the equation:( y = 1x^2 + (2m - 6)x + 1 )At ( x = 1 ), ( y = 1 + (2m - 6) + 1 = 2m - 4 ), which matches the given point. So, yes, ( a = 1 ) is correct.Moving on to part (2), I need to find the axis of symmetry of the parabola in terms of ( m ). The general formula for the axis of symmetry of a parabola ( y = ax^2 + bx + c ) is ( x = -frac{b}{2a} ). In this case, ( a = 1 ) and ( b = 2m - 6 ). So, plugging those into the formula:( x = -frac{2m - 6}{2(1)} = -frac{2m - 6}{2} )Simplify that:( x = -m + 3 ) or ( x = 3 - m )So, the axis of symmetry is ( x = 3 - m ). That seems right. Let me just think if there's another way to get this. Maybe by completing the square? Let's try that.Starting with ( y = x^2 + (2m - 6)x + 1 ). To complete the square, take the coefficient of ( x ), which is ( 2m - 6 ), divide by 2 to get ( m - 3 ), and square it to get ( (m - 3)^2 ). So, adding and subtracting this inside the equation:( y = (x^2 + (2m - 6)x + (m - 3)^2) - (m - 3)^2 + 1 )Simplify:( y = (x + m - 3)^2 - (m^2 - 6m + 9) + 1 )Which simplifies to:( y = (x + m - 3)^2 - m^2 + 6m - 9 + 1 )( y = (x + m - 3)^2 - m^2 + 6m - 8 )So, the vertex form is ( y = (x + m - 3)^2 - m^2 + 6m - 8 ). The vertex is at ( x = 3 - m ), which confirms the axis of symmetry is ( x = 3 - m ). Great, that checks out.Now, part (3) is a bit more involved. We have three points on the parabola: ( (-m, y_1) ), ( (m, y_2) ), and ( (m + 2, y_3) ). The conditions given are ( y_2 < y_3 leq y_1 ). I need to find the range of values for ( m ).First, let me write expressions for ( y_1 ), ( y_2 ), and ( y_3 ) using the parabola equation ( y = x^2 + (2m - 6)x + 1 ).Calculating ( y_1 ) at ( x = -m ):( y_1 = (-m)^2 + (2m - 6)(-m) + 1 )Simplify:( y_1 = m^2 - 2m^2 + 6m + 1 )Combine like terms:( y_1 = -m^2 + 6m + 1 )Calculating ( y_2 ) at ( x = m ):( y_2 = m^2 + (2m - 6)m + 1 )Simplify:( y_2 = m^2 + 2m^2 - 6m + 1 )Combine like terms:( y_2 = 3m^2 - 6m + 1 )Calculating ( y_3 ) at ( x = m + 2 ):( y_3 = (m + 2)^2 + (2m - 6)(m + 2) + 1 )First, expand ( (m + 2)^2 ):( m^2 + 4m + 4 )Then, expand ( (2m - 6)(m + 2) ):( 2m(m) + 2m(2) - 6(m) - 6(2) = 2m^2 + 4m - 6m - 12 = 2m^2 - 2m - 12 )Now, add all terms together:( y_3 = (m^2 + 4m + 4) + (2m^2 - 2m - 12) + 1 )Combine like terms:( y_3 = 3m^2 + 2m - 7 )So, now we have:( y_1 = -m^2 + 6m + 1 )( y_2 = 3m^2 - 6m + 1 )( y_3 = 3m^2 + 2m - 7 )Given the conditions ( y_2 < y_3 leq y_1 ), let's break this down into two inequalities:1. ( y_2 < y_3 )2. ( y_3 leq y_1 )Let's handle the first inequality: ( y_2 < y_3 )Substitute the expressions:( 3m^2 - 6m + 1 < 3m^2 + 2m - 7 )Subtract ( 3m^2 ) from both sides:( -6m + 1 < 2m - 7 )Bring all terms to one side:( -6m + 1 - 2m + 7 < 0 )Combine like terms:( -8m + 8 < 0 )Simplify:( -8m < -8 )Divide both sides by -8, remembering to reverse the inequality sign:( m > 1 )Okay, so the first condition gives ( m > 1 ).Now, the second inequality: ( y_3 leq y_1 )Substitute the expressions:( 3m^2 + 2m - 7 leq -m^2 + 6m + 1 )Bring all terms to one side:( 3m^2 + 2m - 7 + m^2 - 6m - 1 leq 0 )Combine like terms:( 4m^2 - 4m - 8 leq 0 )Simplify:Divide all terms by 4:( m^2 - m - 2 leq 0 )Factor the quadratic:Looking for two numbers that multiply to -2 and add to -1. Those are -2 and 1.So, ( (m - 2)(m + 1) leq 0 )To solve this inequality, we can consider the critical points ( m = 2 ) and ( m = -1 ). These divide the number line into three intervals:1. ( m < -1 )2. ( -1 leq m leq 2 )3. ( m > 2 )We can test each interval:1. For ( m < -1 ), say ( m = -2 ): ( (-2 - 2)(-2 + 1) = (-4)(-1) = 4 > 0 )2. For ( -1 leq m leq 2 ), say ( m = 0 ): ( (0 - 2)(0 + 1) = (-2)(1) = -2 < 0 )3. For ( m > 2 ), say ( m = 3 ): ( (3 - 2)(3 + 1) = (1)(4) = 4 > 0 )Since the inequality is ( leq 0 ), the solution is ( -1 leq m leq 2 ).But from the first condition, we have ( m > 1 ). So, combining both conditions, ( m ) must satisfy both ( m > 1 ) and ( -1 leq m leq 2 ). Therefore, the overlap is ( 1 < m leq 2 ).Wait, but I need to make sure that these points are on the parabola and that the axis of symmetry is considered. Let me think about the axis of symmetry, which is ( x = 3 - m ). The points are at ( x = -m ), ( x = m ), and ( x = m + 2 ). Depending on the value of ( m ), these points can be on either side of the axis of symmetry, which affects the ordering of ( y )-values.If ( m > 0 ), then ( -m ) is on the left side, ( m ) is somewhere, and ( m + 2 ) is further to the right. The axis of symmetry is ( x = 3 - m ). So, depending on where ( 3 - m ) is relative to these points, the parabola's increasing or decreasing nature will affect the ( y )-values.Let me sketch this out mentally. If ( m ) is between 1 and 2, then ( 3 - m ) is between 1 and 2. So, the axis of symmetry is somewhere between 1 and 2. The points ( -m ) would be between -2 and -1, ( m ) is between 1 and 2, and ( m + 2 ) is between 3 and 4. So, the points are spread out with ( -m ) on the far left, ( m ) near the axis, and ( m + 2 ) on the far right.Since the parabola opens upwards (as ( a = 1 > 0 )), the further a point is from the axis of symmetry, the higher its ( y )-value. So, ( y_1 ) is at ( x = -m ), which is the farthest from the axis if ( m ) is positive. ( y_3 ) is at ( x = m + 2 ), which is also far from the axis, but depending on ( m ), it might be on the same side as ( m ) or not.Wait, actually, since ( m ) is between 1 and 2, ( m + 2 ) is between 3 and 4, and the axis is between 1 and 2. So, ( m + 2 ) is on the right side of the axis, and ( -m ) is on the left side. The distance from ( -m ) to the axis is ( | -m - (3 - m) | = | -3 | = 3 ). The distance from ( m + 2 ) to the axis is ( | m + 2 - (3 - m) | = | 2m - 1 | ). Since ( m ) is between 1 and 2, ( 2m - 1 ) is between 1 and 3. So, ( y_1 ) is 3 units away, and ( y_3 ) is between 1 and 3 units away. Therefore, ( y_1 ) is farther than ( y_3 ), so ( y_1 ) should be greater than ( y_3 ), which aligns with the condition ( y_3 leq y_1 ).But wait, from our earlier calculations, ( y_1 = -m^2 + 6m + 1 ) and ( y_3 = 3m^2 + 2m - 7 ). Let me plug in ( m = 1.5 ) to see:( y_1 = -(2.25) + 9 + 1 = 7.75 )( y_3 = 3*(2.25) + 3 - 7 = 6.75 + 3 - 7 = 2.75 )So, ( y_3 = 2.75 leq y_1 = 7.75 ), which holds.Also, ( y_2 = 3*(2.25) - 9 + 1 = 6.75 - 9 + 1 = -1.25 ). So, ( y_2 = -1.25 < y_3 = 2.75 ), which also holds.What about at ( m = 2 ):( y_1 = -4 + 12 + 1 = 9 )( y_3 = 12 + 4 - 7 = 9 )So, ( y_3 = 9 leq y_1 = 9 ), which is okay.( y_2 = 12 - 12 + 1 = 1 ), so ( y_2 = 1 < y_3 = 9 ), which holds.At ( m = 1 ):( y_1 = -1 + 6 + 1 = 6 )( y_3 = 3 + 2 - 7 = -2 )But ( y_3 = -2 leq y_1 = 6 ), which is true, but from our earlier inequality, ( m > 1 ), so ( m = 1 ) is not included.Wait, but if ( m = 1 ), let's check the axis of symmetry: ( x = 3 - 1 = 2 ). The points are ( x = -1 ), ( x = 1 ), and ( x = 3 ). So, ( x = -1 ) is 3 units left, ( x = 1 ) is 1 unit left, and ( x = 3 ) is 1 unit right. So, ( y_1 ) is at ( x = -1 ), which is 3 units from the axis, ( y_2 ) is at ( x = 1 ), which is 1 unit from the axis, and ( y_3 ) is at ( x = 3 ), which is 1 unit from the axis. So, ( y_1 ) should be greater than both ( y_2 ) and ( y_3 ), but in this case, ( y_3 = -2 ) is less than ( y_1 = 6 ), but ( y_2 = 1 ) is also less than ( y_1 ). However, the condition is ( y_2 < y_3 leq y_1 ). At ( m = 1 ), ( y_2 = 1 ) and ( y_3 = -2 ), so ( y_2 > y_3 ), which violates ( y_2 < y_3 ). Hence, ( m = 1 ) is not acceptable, which is why our solution is ( m > 1 ) up to ( m = 2 ).What if ( m ) is less than 1? Let's say ( m = 0.5 ). Then, the axis of symmetry is ( x = 3 - 0.5 = 2.5 ). The points are ( x = -0.5 ), ( x = 0.5 ), and ( x = 2.5 ). So, ( x = -0.5 ) is 3 units left, ( x = 0.5 ) is 2 units left, and ( x = 2.5 ) is at the axis. So, ( y_1 ) is at ( x = -0.5 ), which is 3 units from the axis, ( y_2 ) is at ( x = 0.5 ), which is 2 units from the axis, and ( y_3 ) is at the axis, so it's the vertex. Therefore, ( y_3 ) should be the minimum, so ( y_3 leq y_2 ) and ( y_3 leq y_1 ). But our condition is ( y_2 < y_3 leq y_1 ), which would require ( y_2 < y_3 ), but in reality, ( y_3 ) is the minimum, so ( y_3 leq y_2 ). Hence, this contradicts the condition, so ( m ) cannot be less than 1.Similarly, if ( m ) is negative, say ( m = -1 ), the axis of symmetry is ( x = 3 - (-1) = 4 ). The points are ( x = 1 ), ( x = -1 ), and ( x = 1 ). Wait, ( x = m + 2 = -1 + 2 = 1 ). So, two points at ( x = 1 ) and one at ( x = -1 ). The axis is at ( x = 4 ), so ( x = -1 ) is 5 units left, ( x = 1 ) is 3 units left. So, ( y_1 ) is at ( x = 1 ), which is closer to the axis, and ( y_3 ) is also at ( x = 1 ), same as ( y_1 ). ( y_2 ) is at ( x = -1 ), which is farther. So, ( y_2 ) would be greater than ( y_1 ) and ( y_3 ). But our condition is ( y_2 < y_3 leq y_1 ), which would require ( y_2 < y_3 ), but in reality, ( y_2 > y_1 = y_3 ). So, this doesn't hold either.Therefore, considering all these cases, the only valid range for ( m ) is ( 1 < m leq 2 ).I think that's thorough enough. I considered different cases for ( m ), checked the axis of symmetry, and ensured that the conditions on ( y )-values hold true. It seems consistent that ( m ) must be greater than 1 and at most 2.

💡Okay, so I have this problem here: A regular polygon has an interior angle that measures 120 degrees. How many sides does this polygon have? Hmm, let me see. I remember that regular polygons have all sides and angles equal, so that should help. First, I think about what I know regarding polygons and their angles. I recall that the sum of the interior angles of a polygon depends on the number of sides. There's a formula for that, right? It's something like (n-2) multiplied by 180 degrees, where n is the number of sides. So, the sum of all interior angles is (n-2)*180. But wait, the problem gives me just one interior angle, not the sum. Since it's a regular polygon, all interior angles are equal. So, if I can find the measure of one interior angle, I can relate it to the total sum and find n. Let me write down the formula for the measure of each interior angle of a regular polygon. I think it's [(n-2)*180]/n. Yeah, that sounds right. So, each interior angle is equal to (n-2)*180 divided by n. Given that each interior angle is 120 degrees, I can set up the equation: 120 = [(n-2)*180]/nNow, I need to solve for n. Let me write that equation again:120 = [(n - 2) * 180] / nTo solve for n, I can multiply both sides by n to get rid of the denominator:120n = (n - 2) * 180Now, let's distribute the 180 on the right side:120n = 180n - 360Hmm, now I have 120n equals 180n minus 360. I need to get all the terms with n on one side. Let me subtract 180n from both sides:120n - 180n = -360That simplifies to:-60n = -360Now, I can divide both sides by -60 to solve for n:n = (-360)/(-60)The negatives cancel out, so:n = 360/60n = 6So, the polygon has 6 sides. Let me just double-check to make sure I didn't make any mistakes. If n is 6, then the sum of interior angles is (6-2)*180 = 4*180 = 720 degrees. Each interior angle would be 720/6 = 120 degrees, which matches the given information. Yep, that seems correct. So, the polygon is a hexagon with 6 sides.

💡Okay, so I have this determinant to evaluate, and it's a 3x3 matrix. Let me write it down again to make sure I have it right:[begin{vmatrix} 0 & cos theta & sin theta -cos theta & 0 & cos phi -sin theta & -cos phi & 0 end{vmatrix}]Alright, determinants can sometimes be tricky, especially with trigonometric functions involved. I remember that one way to calculate a 3x3 determinant is by expanding along a row or column using minors and cofactors. Since the first row has a zero, that might make things a bit simpler because multiplying by zero will eliminate one term in the expansion.Let me recall the formula for expanding a determinant along the first row. It goes like this:[text{det}(A) = a_{11} cdot M_{11} - a_{12} cdot M_{12} + a_{13} cdot M_{13}]Where ( M_{ij} ) is the minor of the element ( a_{ij} ). The minor is the determinant of the submatrix that remains after removing the (i)-th row and (j)-th column.Looking at the matrix, the first element is 0, so the first term will be zero. That leaves me with two terms to compute: the second and the third elements of the first row multiplied by their respective minors and cofactors.So, breaking it down:1. The first term is (0 cdot M_{11}), which is 0. I can ignore this.2. The second term is (-cos theta cdot M_{12}).3. The third term is (sin theta cdot M_{13}).Now, I need to find the minors ( M_{12} ) and ( M_{13} ).Starting with ( M_{12} ): This is the determinant of the submatrix obtained by removing the first row and the second column. So, removing the first row and the second column, the submatrix is:[begin{vmatrix} -cos theta & cos phi -sin theta & 0 end{vmatrix}]Calculating this determinant: ( (-cos theta)(0) - (cos phi)(-sin theta) = 0 + cos phi sin theta = cos phi sin theta ).So, ( M_{12} = cos phi sin theta ).Next, ( M_{13} ): This is the determinant of the submatrix obtained by removing the first row and the third column. So, removing the first row and the third column, the submatrix is:[begin{vmatrix} -cos theta & 0 -sin theta & -cos phi end{vmatrix}]Calculating this determinant: ( (-cos theta)(-cos phi) - (0)(-sin theta) = cos theta cos phi - 0 = cos theta cos phi ).So, ( M_{13} = cos theta cos phi ).Now, plugging these back into the expansion:[text{det}(A) = 0 - cos theta cdot (cos phi sin theta) + sin theta cdot (cos theta cos phi)]Simplifying each term:- The first term is 0.- The second term is (- cos theta cos phi sin theta).- The third term is ( sin theta cos theta cos phi ).Looking at the second and third terms, they are:[- cos theta cos phi sin theta + sin theta cos theta cos phi]Notice that these two terms are actually negatives of each other. Let me write them out:- First: ( - cos theta cos phi sin theta )- Second: ( + cos theta cos phi sin theta )Adding these together:[- cos theta cos phi sin theta + cos theta cos phi sin theta = 0]So, the entire determinant simplifies to 0.Wait, does that make sense? Let me double-check my calculations to make sure I didn't make a mistake.First, minors:- ( M_{12} ): Correct, it's ( cos phi sin theta ).- ( M_{13} ): Correct, it's ( cos theta cos phi ).Then, substituting back:- Second term: ( - cos theta cdot (cos phi sin theta) )- Third term: ( + sin theta cdot (cos theta cos phi) )Yes, these are indeed negatives of each other because multiplication is commutative, so ( cos theta cos phi sin theta = sin theta cos theta cos phi ). Therefore, subtracting and adding the same term results in zero.Alternatively, I can think about the properties of determinants. This matrix looks antisymmetric because the transpose of the matrix is equal to its negative. Let me check:The transpose of the matrix is:[begin{vmatrix} 0 & -cos theta & -sin theta cos theta & 0 & -cos phi sin theta & cos phi & 0 end{vmatrix}]Comparing this with the original matrix:Original:[begin{vmatrix} 0 & cos theta & sin theta -cos theta & 0 & cos phi -sin theta & -cos phi & 0 end{vmatrix}]Yes, the transpose is indeed the negative of the original matrix. So, it's an antisymmetric matrix. I remember that for antisymmetric matrices of odd order, the determinant is zero. Since this is a 3x3 matrix, which is odd, the determinant should be zero. That aligns with my earlier calculation.So, both methods confirm that the determinant is zero. I feel confident about this result now.**Final Answer**The determinant is boxed{0}.

💡Alright, so I have this problem about convex quadrilaterals and hexagons, and I need to prove some things about them. Let me try to understand part (a) first.Part (a) says that each of the diagonals of a convex quadrilateral ABCD divides its area in half. I need to prove that ABCD is a parallelogram. Hmm, okay. So, in a convex quadrilateral, the diagonals intersect each other, right? Let me draw a rough sketch in my mind: convex quadrilateral ABCD with diagonals AC and BD intersecting at point O.Now, if each diagonal divides the area in half, that means the area of triangle ABC is equal to the area of triangle ADC when considering diagonal AC. Similarly, the area of triangle ABD is equal to the area of triangle BCD when considering diagonal BD. So, both diagonals split the quadrilateral into two equal areas.Wait, so if the areas are equal, does that mean the triangles formed by the diagonals are congruent? Not necessarily, because congruence requires more than just equal area. But maybe the heights from the vertices to the diagonals are equal? Let me think.If diagonal AC divides the quadrilateral into two equal areas, then the heights from B and D to AC must be equal. Similarly, the heights from A and C to BD must be equal. So, does that mean that the diagonals bisect each other? Because if the heights are equal, then the points where the diagonals intersect must be the midpoints.Yes, that makes sense. If the heights from B and D to AC are equal, then point O, where the diagonals intersect, must be the midpoint of both diagonals. Because if O wasn't the midpoint, then one triangle would have a larger height than the other, leading to unequal areas. So, if both diagonals are bisected by their intersection, then ABCD must be a parallelogram. Because in a parallelogram, diagonals bisect each other.Okay, so that seems to make sense. I think I can formalize this argument. Let me try.Let O be the intersection point of diagonals AC and BD. Since diagonal AC divides the quadrilateral into two equal areas, the area of triangle ABC is equal to the area of triangle ADC. These triangles share the base AC, so their areas are proportional to their heights from B and D to AC. Since the areas are equal, the heights must be equal, meaning that B and D are equidistant from AC. Therefore, O must be the midpoint of BD.Similarly, considering diagonal BD, which also divides the quadrilateral into two equal areas, the area of triangle ABD is equal to the area of triangle BCD. These triangles share the base BD, so their areas are proportional to their heights from A and C to BD. Since the areas are equal, the heights must be equal, meaning that A and C are equidistant from BD. Therefore, O must be the midpoint of AC.Since O is the midpoint of both diagonals AC and BD, the diagonals bisect each other. This is a defining property of a parallelogram. Therefore, ABCD is a parallelogram.Alright, that seems solid. Now, moving on to part (b).Part (b) is about a convex hexagon ABCDEF. It's given that each of the diagonals AD, BE, and CF divides its area in half. I need to prove that these diagonals intersect at a single point.Hmm, okay. So, in a convex hexagon, the diagonals AD, BE, and CF are given. Each of these diagonals divides the hexagon into two regions of equal area. I need to show that these three diagonals meet at one point.Let me try to visualize this. A convex hexagon has six sides, and the diagonals AD, BE, and CF are connecting every other vertex. So, AD connects A to D, BE connects B to E, and CF connects C to F.Since each diagonal divides the hexagon into two equal areas, maybe there's some symmetry or concurrency point where all three diagonals meet. I remember that in some polygons, certain diagonals can be concurrent, especially if they have equal area properties.Let me think about the areas. If AD divides the hexagon into two equal areas, then the area of ABCD is equal to the area of DEFA. Similarly, BE divides the hexagon into two equal areas, so the area of ABCE is equal to the area of CDEF. And CF divides it into two equal areas as well, so the area of ABCF is equal to the area of CDEF.Wait, but how does this help me? Maybe I can use these area equalities to show that the diagonals must intersect at a common point.Alternatively, maybe I can use coordinate geometry. Let me assign coordinates to the hexagon and see if I can derive the concurrency.But that might be complicated. Maybe there's a more geometric approach.I recall that in a convex quadrilateral, if both diagonals bisect each other, it's a parallelogram. Maybe in a hexagon, if three diagonals each divide the area equally, they must intersect at a common point.Alternatively, perhaps I can use the concept of barycenters or centers of mass. If each diagonal divides the area equally, maybe the intersection point is the centroid of the hexagon.Wait, but I'm not sure if that's directly applicable. Let me think differently.Suppose that diagonals AD and BE intersect at point P. Then, since AD divides the hexagon into two equal areas, point P must lie somewhere such that the areas on either side of AD are equal. Similarly, since BE also divides the hexagon into two equal areas, point P must also balance the areas for BE.If I can show that CF must also pass through P, then all three diagonals intersect at P.Alternatively, maybe I can use the principle that if three lines divide the area equally, they must concur at a point.Wait, I think I've heard of a theorem related to this. Maybe it's called the Ham-Sandwich theorem, but that's more about dividing measures simultaneously. Or perhaps it's a result in geometry about concurrent area bisectors.Alternatively, maybe I can use the concept of Ceva's theorem, which deals with concurrent lines in a triangle. But this is a hexagon, so I'm not sure.Wait, maybe I can model the hexagon as two triangles or something. Let me try to decompose the hexagon.Alternatively, let me consider the areas. Since AD, BE, and CF each divide the hexagon into two equal areas, the intersection point of any two diagonals must also lie on the third diagonal.Wait, let's suppose that diagonals AD and BE intersect at point P. Then, since AD divides the hexagon into two equal areas, and BE also divides it into two equal areas, point P must be such that it balances both areas. Therefore, CF must also pass through P to maintain the area balance.Alternatively, maybe I can use the concept of ratios. Let me denote the areas.Let the total area of the hexagon be S. Then, each diagonal divides it into two regions of area S/2.Suppose diagonals AD and BE intersect at point P. Then, the area of quadrilateral ABP is equal to the area of quadrilateral EPD, because AD divides the hexagon into two equal areas. Similarly, the area of quadrilateral BCP is equal to the area of quadrilateral FPC, because BE divides the hexagon into two equal areas.Wait, maybe I can write equations for the areas.Let me denote the areas of certain triangles or quadrilaterals.Let me consider triangle ABP. Since AD divides the hexagon into two equal areas, the area of triangle ABP plus the area of triangle APD must equal S/2. Similarly, the area of triangle EPD plus the area of triangle EPC must equal S/2.But I'm getting confused. Maybe I need to think in terms of ratios.If diagonals AD and BE intersect at P, then the ratio of areas of triangles APB and EPD must be equal to the ratio of AP to PD, assuming they share the same height.Wait, maybe I can use the concept that if two lines divide the area equally, their intersection must lie on the third line that also divides the area equally.Alternatively, maybe I can use the concept of mass points or area ratios.Wait, let me try to think in terms of barycentric coordinates. If each diagonal divides the area equally, then the intersection point must be the centroid, which is the intersection of the medians. But in a hexagon, the concept of medians is not straightforward.Alternatively, maybe I can use the fact that if three lines divide the area equally, they must concur at a point.Wait, I think I've heard of a result that in a convex polygon, if three lines each divide the area into two equal parts, then they must intersect at a single point. Is that true?Yes, I think that's a theorem. It's similar to the idea that in a convex shape, any two area bisectors must intersect, and if you have three area bisectors, they must concur at a single point.So, applying that theorem here, since AD, BE, and CF each divide the hexagon into two equal areas, they must intersect at a single point.Therefore, the diagonals AD, BE, and CF intersect at a single point.Hmm, that seems too quick. Maybe I need to provide a more detailed proof.Alternatively, let me try to use the concept of Ceva's theorem in a triangle. But since this is a hexagon, maybe I can decompose it into triangles.Wait, let me consider triangle ABC. No, that's not directly helpful. Alternatively, maybe I can consider the hexagon as a combination of triangles and apply Ceva's theorem.Alternatively, let me think about the areas again. Suppose diagonals AD and BE intersect at point P. Then, since AD divides the hexagon into two equal areas, the area of quadrilateral ABP must be equal to the area of quadrilateral EPD. Similarly, since BE divides the hexagon into two equal areas, the area of quadrilateral BCP must be equal to the area of quadrilateral FPC.Wait, but how does this relate to CF?Alternatively, maybe I can use the fact that if three lines divide the area equally, then their intersection must be the same point.Wait, I think I need to use a more formal approach. Let me consider the areas created by each diagonal.Let the total area of the hexagon be S. Then, each diagonal divides it into two regions of area S/2.Suppose diagonals AD and BE intersect at point P. Then, the area of quadrilateral ABP is equal to the area of quadrilateral EPD, which is S/2. Similarly, the area of quadrilateral BCP is equal to the area of quadrilateral FPC, which is also S/2.Wait, but that can't be right because the areas would overlap. Maybe I'm miscalculating.Alternatively, perhaps I should consider the areas of the triangles formed by the diagonals.Let me denote the area of triangle ABP as x, then the area of triangle APD would be S/2 - x. Similarly, the area of triangle BCP would be y, and the area of triangle FPC would be S/2 - y.But I'm not sure how this helps me. Maybe I need to relate these areas to the third diagonal CF.Alternatively, maybe I can use the concept that if three lines divide the area equally, then their intersection must be the same point.Wait, I think I can use the following argument: Suppose diagonals AD and BE intersect at point P. Then, since AD divides the hexagon into two equal areas, P must lie on AD such that the areas on either side are equal. Similarly, since BE divides the hexagon into two equal areas, P must lie on BE such that the areas on either side are equal.Now, consider diagonal CF. Since CF also divides the hexagon into two equal areas, the point where CF intersects AD and BE must also balance the areas. Therefore, CF must pass through P, because P is already balancing the areas for AD and BE. Hence, all three diagonals intersect at point P.Yes, that makes sense. So, the intersection point of any two diagonals must also lie on the third diagonal, because it's the only point that can balance all three area divisions.Therefore, the diagonals AD, BE, and CF must intersect at a single point.Alright, I think that's a reasonable argument. Maybe I can formalize it a bit more.Let me denote the intersection of AD and BE as point P. Since AD divides the hexagon into two equal areas, the area of quadrilateral ABP is equal to the area of quadrilateral EPD. Similarly, since BE divides the hexagon into two equal areas, the area of quadrilateral BCP is equal to the area of quadrilateral FPC.Now, consider diagonal CF. Since CF divides the hexagon into two equal areas, the area of quadrilateral ABCF must be equal to the area of quadrilateral CDEF. But quadrilateral ABCF includes quadrilateral ABP and triangle BCP, while quadrilateral CDEF includes quadrilateral EPD and triangle FPC.Since the areas of ABP and EPD are equal, and the areas of BCP and FPC are equal, the areas of ABCF and CDEF are equal only if the intersection point P lies on CF as well. Therefore, CF must pass through P.Hence, all three diagonals AD, BE, and CF intersect at point P.Yes, that seems to work. So, the key idea is that the intersection point of two area-bisecting diagonals must also lie on the third area-bisecting diagonal, ensuring concurrency.Alright, I think I've got a good grasp on both parts now. Let me summarize my thoughts.For part (a), the fact that both diagonals divide the quadrilateral into equal areas implies that the diagonals bisect each other, which is a property of parallelograms. Therefore, ABCD must be a parallelogram.For part (b), the three diagonals each dividing the hexagon into equal areas must intersect at a single point because the intersection of any two such diagonals already satisfies the area condition for the third diagonal, forcing all three to concur at that point.I think that covers both parts adequately.