💡Okay, so I have this problem about a geometric sequence. Let me try to figure it out step by step. First, the problem says that in the geometric sequence {a_n}, the common ratio q is -2, and a_3 multiplied by a_7 equals 4 times a_4. I need to find the arithmetic mean of a_8 and a_11. Hmm, okay.I remember that in a geometric sequence, each term is the previous term multiplied by the common ratio q. So, the nth term can be written as a_n = a_1 * q^{n-1}, where a_1 is the first term. Given that q is -2, so every term is -2 times the previous term. Let me write down the expressions for a_3, a_4, a_7, a_8, and a_11 using the formula.So, a_3 = a_1 * q^{2} = a_1 * (-2)^2 = a_1 * 4.Similarly, a_4 = a_1 * q^{3} = a_1 * (-2)^3 = a_1 * (-8).a_7 = a_1 * q^{6} = a_1 * (-2)^6 = a_1 * 64.a_8 = a_1 * q^{7} = a_1 * (-2)^7 = a_1 * (-128).a_11 = a_1 * q^{10} = a_1 * (-2)^10 = a_1 * 1024.Wait, but I don't know the value of a_1 yet. The problem gives me a condition: a_3 * a_7 = 4 * a_4. Let me use that to find a_1.So, substituting the expressions I have:a_3 * a_7 = (a_1 * 4) * (a_1 * 64) = a_1^2 * 256.And 4 * a_4 = 4 * (a_1 * (-8)) = 4 * (-8) * a_1 = -32 * a_1.So, according to the problem, a_3 * a_7 = 4 * a_4, which gives:a_1^2 * 256 = -32 * a_1.Hmm, let me write that equation:256 * a_1^2 = -32 * a_1.I can divide both sides by a_1, but I have to be careful because if a_1 is zero, that would be a problem. But in a geometric sequence, if a_1 is zero, all terms are zero, which would make a_3 * a_7 = 0 and 4 * a_4 = 0, so 0 = 0, which is true. But then, all terms would be zero, and the arithmetic mean would also be zero. However, let me check if a_1 can be non-zero.If a_1 ≠ 0, then I can divide both sides by a_1:256 * a_1 = -32.Then, solving for a_1:a_1 = -32 / 256 = -1/8.So, a_1 is -1/8. Okay, that makes sense. So now I can find a_8 and a_11.Let me compute a_8:a_8 = a_1 * (-2)^7 = (-1/8) * (-128) = (-1/8) * (-128).Calculating that: (-1/8) * (-128) = (128)/8 = 16. So, a_8 is 16.Now, a_11:a_11 = a_1 * (-2)^10 = (-1/8) * 1024.Calculating that: (-1/8) * 1024 = -128. So, a_11 is -128.Now, the arithmetic mean of a_8 and a_11 is (a_8 + a_11)/2.So, plugging in the values:(16 + (-128))/2 = (16 - 128)/2 = (-112)/2 = -56.So, the arithmetic mean is -56.Wait, let me double-check my calculations to make sure I didn't make a mistake.First, a_1 is -1/8. Then, a_8 is (-1/8)*(-128) = 16. That seems correct because (-1/8)*(-128) is positive, and 128 divided by 8 is 16.a_11 is (-1/8)*(1024). 1024 divided by 8 is 128, so with the negative sign, it's -128. That seems correct.Then, adding 16 and -128 gives -112, and dividing by 2 gives -56. That seems right.Alternatively, I can think about the terms in the geometric sequence without calculating a_1 first. Let me see.Given that a_3 * a_7 = 4 * a_4.In a geometric sequence, a_n = a_1 * q^{n-1}.So, a_3 = a_1 * q^2, a_4 = a_1 * q^3, a_7 = a_1 * q^6.So, a_3 * a_7 = (a_1 * q^2) * (a_1 * q^6) = a_1^2 * q^{8}.And 4 * a_4 = 4 * (a_1 * q^3).So, setting them equal:a_1^2 * q^{8} = 4 * a_1 * q^3.Divide both sides by a_1 (assuming a_1 ≠ 0):a_1 * q^{8} = 4 * q^3.Then, solving for a_1:a_1 = (4 * q^3) / q^{8} = 4 * q^{-5} = 4 / q^5.Given that q = -2, so q^5 = (-2)^5 = -32.Therefore, a_1 = 4 / (-32) = -1/8. That's the same result as before.So, that confirms that a_1 is indeed -1/8. So, the earlier calculations are correct.Therefore, a_8 is 16, a_11 is -128, and their arithmetic mean is -56.I think that's solid. I don't see any mistakes in my reasoning.**Final Answer**The arithmetic mean of (a_8) and (a_{11}) is boxed{-56}.

💡Alright, so I have this problem: "How many distinct four-digit positive integers have only odd digits and exactly one digit repeating?" Hmm, okay, let's break this down step by step. I need to find four-digit numbers, so numbers from 1000 to 9999. But not just any four-digit numbers—only those that have odd digits, and exactly one digit repeats. That means three digits are unique, and one digit is repeated twice. First, let me clarify what counts as an odd digit. The odd digits are 1, 3, 5, 7, and 9. So, each digit in the four-digit number must be one of these five. Got it. Now, exactly one digit repeats. So, in the four-digit number, one digit appears twice, and the other two digits appear once each, and all digits are odd. Let me think about how to approach this. It seems like a combinatorics problem where I need to count the number of such numbers by considering different cases or using permutations and combinations. Maybe I can break it down into steps:1. **Choose the digit that will repeat.** Since all digits must be odd, there are 5 choices for this digit (1, 3, 5, 7, 9).2. **Choose the positions where this digit will be placed in the four-digit number.** There are 4 positions (thousands, hundreds, tens, units), and I need to choose 2 positions out of these 4 for the repeating digit. The number of ways to choose 2 positions out of 4 is given by the combination formula: C(4,2) = 6.3. **Choose the remaining two digits.** Since the remaining two digits must be different and also odd, and they can't be the same as the repeating digit. So, after choosing the repeating digit, there are 4 remaining odd digits to choose from for the first unique digit and 3 remaining for the second unique digit. So, that's 4 * 3 = 12 ways.4. **Arrange the remaining two digits in the remaining two positions.** Since the two unique digits are different, the number of ways to arrange them in the two remaining positions is 2! = 2.Wait, so putting it all together: - 5 choices for the repeating digit,- 6 ways to place the repeating digit,- 12 ways to choose the other two digits,- 2 ways to arrange the other two digits.So, total number of such numbers would be 5 * 6 * 12 * 2. Let me calculate that: 5 * 6 is 30, 30 * 12 is 360, 360 * 2 is 720. Hmm, that seems high. Wait, did I overcount somewhere?Let me think again. Maybe the arrangement of the two unique digits is already accounted for when I choose their positions. Wait, no, because when I choose the positions for the repeating digit, the remaining positions are fixed, and then I assign the unique digits to those positions. So, actually, once I've chosen the two unique digits, their order matters, so I need to multiply by 2! to account for their arrangement.But wait, is that correct? Let me see. Suppose I have digits A, B, C, where A is the repeating digit, and B and C are the unique digits. The number of ways to arrange A, A, B, C is 4! / 2! = 12. But in my earlier approach, I did 5 * C(4,2) * 4 * 3 * 2. Wait, that's 5 * 6 * 12 * 2 = 720. But 4! / 2! is 12, so for each choice of A, B, C, there are 12 numbers. So, the total should be 5 * (number of ways to choose B and C) * 12.Wait, number of ways to choose B and C is P(4,2) = 12, since after choosing A, we have 4 remaining digits, and we need to choose 2 distinct ones and arrange them. So, that would be 5 * 12 * 12 = 720. Hmm, but that seems high because the total number of four-digit numbers with only odd digits is 5^4 = 625, which is less than 720. So, clearly, something is wrong here.Wait, hold on. The total number of four-digit numbers with only odd digits is 5 * 5 * 5 * 5 = 625, as you said. But 720 is more than that, which is impossible. So, my approach must be wrong.Let me re-examine my steps. Maybe I overcounted the arrangements. Let's think differently. First, choose the digit to repeat: 5 choices.Then, choose the two positions for this digit: C(4,2) = 6.Now, for the remaining two positions, we need to choose two different digits from the remaining four odd digits, and arrange them. So, the number of ways to choose and arrange two different digits from four is P(4,2) = 4 * 3 = 12.Therefore, the total number should be 5 * 6 * 12 = 360.Ah, okay, so I think I was overcounting by multiplying by an extra 2 earlier. Because when I choose the two unique digits and assign them to specific positions, that's already accounted for by the permutation P(4,2). So, I shouldn't multiply by 2 again. So, the correct total is 5 * 6 * 12 = 360.Let me verify this with another approach. The total number of four-digit numbers with only odd digits is 5^4 = 625. Now, among these, how many have exactly one digit repeating? That is, numbers with exactly two of one digit and the other two digits distinct.Another way to calculate this is:- Choose the digit to repeat: 5 choices.- Choose the two positions for this digit: C(4,2) = 6.- Choose the remaining two digits: since they must be distinct and different from the repeating digit, we have 4 choices for the first and 3 for the second, so 4 * 3 = 12.- Assign these two digits to the remaining two positions: since the positions are distinct, we can arrange the two digits in 2! = 2 ways.Wait, but that would give 5 * 6 * 12 * 2 = 720 again, which is more than 625. So, clearly, this approach is flawed.Wait, no, because when we choose the two unique digits and assign them to positions, we're already considering their order. So, perhaps the correct calculation is:- Choose the repeating digit: 5.- Choose positions for it: C(4,2) = 6.- Choose the two unique digits: C(4,2) = 6, since we're choosing 2 distinct digits from the remaining 4.- Arrange these two unique digits in the remaining two positions: 2! = 2.So, total is 5 * 6 * 6 * 2 = 360.Yes, that makes sense. So, first, choose the digit to repeat (5), then choose where it goes (6), then choose which two other digits to use (6), and then arrange them (2). So, 5 * 6 * 6 * 2 = 360.Alternatively, using permutations:The number of ways to arrange the digits where one digit repeats twice and the other two are distinct is given by:Number of choices = (number of ways to choose the repeating digit) * (number of ways to choose the two unique digits) * (number of ways to arrange all four digits).So, that would be:- Choose repeating digit: 5.- Choose two unique digits: C(4,2) = 6.- Arrange the four digits: since there are two identical and two distinct, the number of distinct arrangements is 4! / 2! = 12.So, total number is 5 * 6 * 12 = 360.Yes, that matches.So, I think the correct answer is 360.But wait, let me think again to make sure I'm not missing something. The problem says "exactly one digit repeating." Does that mean that exactly one digit is repeated, and the other two are unique? Yes, that's what it means. So, in other words, the number has the form AABC, ABAC, or ABBA, etc., where A is the repeating digit, and B and C are distinct digits different from A.So, the count should be 5 (choices for A) * C(4,2) (positions for A) * P(4,2) (choices and arrangements for B and C). Wait, P(4,2) is 12, so 5 * 6 * 12 = 360.Yes, that seems consistent.Alternatively, thinking in terms of multinomial coefficients:The number of four-digit numbers with exactly two of one digit and one each of two others is:Number of ways = (number of choices for the digit to repeat) * (number of ways to choose the other two digits) * (number of distinct arrangements).So, that's 5 * C(4,2) * (4! / 2!) = 5 * 6 * 12 = 360.Yes, that's the same as before.I think I've convinced myself that 360 is the correct answer.

💡First, I need to determine the number of regular tickets sold. I'll start by defining the variables: let S represent the number of senior citizen tickets and R represent the number of regular tickets.From the problem, I know that the total number of tickets sold is 65. This gives me the first equation:S + R = 65I also know that the total sales amount to 855, with senior tickets costing 10 each and regular tickets costing 15 each. This leads to the second equation:10S + 15R = 855To solve this system of equations, I'll use the substitution method. I'll solve the first equation for S:S = 65 - RNext, I'll substitute this expression for S into the second equation:10(65 - R) + 15R = 855Expanding the equation:650 - 10R + 15R = 855Combining like terms:5R = 205Finally, I'll solve for R:R = 41Therefore, 41 regular tickets were sold.

💡Alright, so I'm trying to solve this geometry problem about a tetrahedron and an inscribed sphere. The problem states that there's a sphere inside a tetrahedron, touching all its faces. The areas of these faces are given as ( t_1, t_2, t_3, t_4 ). Then, there are planes parallel to each face and tangent to the inscribed sphere. These planes intersect the tetrahedron and form smaller triangles with areas ( tau_1, tau_2, tau_3, tau_4 ). I need to prove that the sum ( S = frac{tau_1}{t_1} + frac{tau_2}{t_2} + frac{tau_3}{t_3} + frac{tau_4}{t_4} ) is greater than or equal to 1, and determine when equality holds.Okay, first, let me visualize this. A tetrahedron has four triangular faces. An inscribed sphere (inradius) touches each face at exactly one point. The planes that are parallel to each face and tangent to the sphere will cut the tetrahedron, creating smaller, similar triangles on each face. The areas of these smaller triangles are ( tau_1 ) to ( tau_4 ), and I need to relate them to the original areas ( t_1 ) to ( t_4 ).I remember that when you have similar figures, the ratio of their areas is the square of the ratio of their corresponding linear dimensions. So, if the smaller triangles are similar to the original faces, the ratio ( frac{tau_i}{t_i} ) should be the square of the scaling factor between the original face and the smaller triangle.But how do I find this scaling factor? Well, since the planes are tangent to the inscribed sphere and parallel to the faces, the distance between each face and the corresponding tangent plane must be related to the radius of the sphere. Let me denote the radius of the inscribed sphere as ( r ).In a tetrahedron, the inradius ( r ) can be found using the formula ( r = frac{3V}{F} ), where ( V ) is the volume of the tetrahedron and ( F ) is the total surface area. But I'm not sure if that's directly useful here.Wait, maybe I should think about the distance from the center of the sphere to each face. Since the sphere is tangent to each face, the distance from the center to each face is equal to the radius ( r ). Now, the tangent plane parallel to a face will be at a distance of ( 2r ) from the original face because the sphere has radius ( r ), and the tangent plane is on the other side of the sphere relative to the face.So, the distance between the original face and the tangent plane is ( 2r ). Since the planes are parallel, the scaling factor for the similar triangles should be related to this distance. Let me denote the height (altitude) of the tetrahedron corresponding to face ( t_i ) as ( h_i ). Then, the distance from the center of the sphere to face ( t_i ) is ( r ), so the distance from the tangent plane to face ( t_i ) is ( 2r ).Therefore, the scaling factor ( k_i ) for the smaller triangle ( tau_i ) relative to ( t_i ) is ( 1 - frac{2r}{h_i} ). Wait, is that correct? If the distance between the original face and the tangent plane is ( 2r ), and the total height is ( h_i ), then the remaining height from the tangent plane to the opposite vertex is ( h_i - 2r ). So, the scaling factor should be ( frac{h_i - 2r}{h_i} = 1 - frac{2r}{h_i} ). Therefore, the area ratio ( frac{tau_i}{t_i} ) is ( left(1 - frac{2r}{h_i}right)^2 ).So, ( S = sum_{i=1}^{4} left(1 - frac{2r}{h_i}right)^2 ). Hmm, that seems a bit complicated. Maybe I can express ( h_i ) in terms of the volume and the area ( t_i ). The volume ( V ) of the tetrahedron can be expressed as ( V = frac{1}{3} t_i h_i ) for each face ( t_i ). Therefore, ( h_i = frac{3V}{t_i} ).Substituting this into the scaling factor, we get ( 1 - frac{2r}{h_i} = 1 - frac{2r t_i}{3V} ). So, ( frac{tau_i}{t_i} = left(1 - frac{2r t_i}{3V}right)^2 ).Therefore, ( S = sum_{i=1}^{4} left(1 - frac{2r t_i}{3V}right)^2 ). Now, I need to simplify this expression and show that it's at least 1.Let me expand the square: ( left(1 - frac{2r t_i}{3V}right)^2 = 1 - frac{4r t_i}{3V} + frac{4r^2 t_i^2}{9V^2} ).So, summing over all ( i ), we have:( S = 4 - frac{4r}{3V} sum_{i=1}^{4} t_i + frac{4r^2}{9V^2} sum_{i=1}^{4} t_i^2 ).Now, the total surface area ( F = sum_{i=1}^{4} t_i ). So, the second term becomes ( frac{4r}{3V} F ).But from the formula for the inradius, ( r = frac{3V}{F} ). Substituting this into the second term:( frac{4r}{3V} F = frac{4 cdot frac{3V}{F}}{3V} F = 4 ).So, the second term is 4. Therefore, ( S = 4 - 4 + frac{4r^2}{9V^2} sum_{i=1}^{4} t_i^2 ).Simplifying, ( S = frac{4r^2}{9V^2} sum_{i=1}^{4} t_i^2 ).But ( r = frac{3V}{F} ), so ( r^2 = frac{9V^2}{F^2} ). Substituting this in:( S = frac{4 cdot frac{9V^2}{F^2}}{9V^2} sum_{i=1}^{4} t_i^2 = frac{4}{F^2} sum_{i=1}^{4} t_i^2 ).So, ( S = frac{4(t_1^2 + t_2^2 + t_3^2 + t_4^2)}{(t_1 + t_2 + t_3 + t_4)^2} ).Now, I need to show that this fraction is at least 1. Let me denote ( F = t_1 + t_2 + t_3 + t_4 ). So, ( S = frac{4(t_1^2 + t_2^2 + t_3^2 + t_4^2)}{F^2} ).I recall that for any set of positive numbers, the following inequality holds: ( frac{t_1^2 + t_2^2 + t_3^2 + t_4^2}{4} geq left( frac{t_1 + t_2 + t_3 + t_4}{4} right)^2 ). This is the Cauchy-Schwarz inequality or the QM-AM inequality.Multiplying both sides by 4, we get ( t_1^2 + t_2^2 + t_3^2 + t_4^2 geq frac{F^2}{4} ).Therefore, ( 4(t_1^2 + t_2^2 + t_3^2 + t_4^2) geq F^2 ), which implies ( frac{4(t_1^2 + t_2^2 + t_3^2 + t_4^2)}{F^2} geq 1 ).Hence, ( S geq 1 ).Equality holds when ( t_1 = t_2 = t_3 = t_4 ), meaning all the faces of the tetrahedron have equal areas. In such a case, the tetrahedron is regular, and the inscribed sphere is centered at the centroid.Wait, but is a tetrahedron with all faces of equal area necessarily regular? I think not necessarily, because the areas can be equal without the edges being equal. However, in this case, since the scaling factors are uniform due to the sphere being tangent and the planes being parallel, equality would hold when all the faces are congruent, which would imply a regular tetrahedron.So, to summarize, I transformed the problem into expressing the ratio ( S ) in terms of the areas of the faces and used inequalities to show that ( S ) is at least 1, with equality when all face areas are equal, implying a regular tetrahedron.I think this makes sense, but let me double-check my steps. Starting from the areas, I related them to the heights, used the inradius formula, substituted into the scaling factor, expanded, and then applied the Cauchy-Schwarz inequality. It seems consistent.Another way to think about it is using homothety. The tangent planes are homothetic transformations of the original faces, scaled down by a factor related to the inradius. The sum of the squares of these scaling factors relates to the original areas, leading to the inequality.Yes, I feel confident with this approach.

💡Okay, so I have this problem where there are two circles, C₁ and C₂, and they're symmetric about a line l. I need to find the equation of line l. Let me see how I can approach this.First, let me write down the equations of the circles again to make sure I have them right.Circle C₁ is given by:[ x^2 + y^2 = 4 ]That's a standard equation of a circle centered at the origin (0,0) with a radius of 2.Circle C₂ is given by:[ x^2 + y^2 - 4x + 4y + 4 = 0 ]Hmm, this one looks a bit more complicated. I think I need to rewrite this equation in the standard form of a circle to find its center and radius. To do that, I can complete the squares for both x and y terms.Let me rearrange the equation:[ x^2 - 4x + y^2 + 4y = -4 ]Now, completing the square for the x terms:Take the coefficient of x, which is -4, divide by 2 to get -2, then square it to get 4. So I add and subtract 4 for the x terms.Similarly, for the y terms:The coefficient of y is 4, divide by 2 to get 2, square it to get 4. So I add and subtract 4 for the y terms.Adding these to the equation:[ (x^2 - 4x + 4) - 4 + (y^2 + 4y + 4) - 4 = -4 ]Simplify:[ (x - 2)^2 + (y + 2)^2 - 8 = -4 ]Bring the -8 to the right side:[ (x - 2)^2 + (y + 2)^2 = 4 ]So, circle C₂ is centered at (2, -2) with a radius of 2 as well.Alright, so both circles have the same radius, which is 2. That might be helpful because if they are symmetric about a line, the line should be the perpendicular bisector of the line segment connecting their centers.Let me find the centers again:- Center of C₁: (0, 0)- Center of C₂: (2, -2)So, the line segment connecting (0,0) and (2,-2). The midpoint of this segment will lie on the line of symmetry, which is line l.Calculating the midpoint:Midpoint formula is (left( frac{x_1 + x_2}{2}, frac{y_1 + y_2}{2} right))Plugging in the values:Midpoint = (left( frac{0 + 2}{2}, frac{0 + (-2)}{2} right) = (1, -1))So, the midpoint is (1, -1). Therefore, line l must pass through this point.Next, I need to find the slope of the line segment connecting the two centers to determine the slope of the perpendicular bisector.Slope of the segment connecting (0,0) and (2,-2):Slope formula is (frac{y_2 - y_1}{x_2 - x_1})So, slope = (frac{-2 - 0}{2 - 0} = frac{-2}{2} = -1)The slope of the line segment is -1. Since line l is the perpendicular bisector, its slope will be the negative reciprocal of -1, which is 1.So, line l has a slope of 1 and passes through the midpoint (1, -1). Now, I can write the equation of line l using the point-slope form.Point-slope form is:[ y - y_1 = m(x - x_1) ]Where m is the slope and (x₁, y₁) is the point.Plugging in the values:[ y - (-1) = 1(x - 1) ]Simplify:[ y + 1 = x - 1 ]Subtract 1 from both sides:[ y = x - 2 ]So, the equation of line l is y = x - 2.Wait, let me double-check if this makes sense. If I reflect circle C₁ over the line y = x - 2, would I get circle C₂?Reflecting the center of C₁, which is (0,0), over the line y = x - 2. The reflection of a point over a line can be found using some formulas, but since I know the midpoint between (0,0) and (2,-2) is (1,-1), which lies on the line y = x - 2, that makes sense because the reflection should map (0,0) to (2,-2).Alternatively, I can subtract the equations of the two circles to find the line of symmetry. Let me try that method as a verification.Subtracting the equation of C₁ from C₂:C₂: (x^2 + y^2 - 4x + 4y + 4 = 0)C₁: (x^2 + y^2 = 4)Subtracting C₁ from C₂:[(x^2 + y^2 - 4x + 4y + 4) - (x^2 + y^2) = 0 - 4]Simplify:[-4x + 4y + 4 = -4]Divide both sides by 4:[-x + y + 1 = -1]Simplify:[-x + y = -2]Multiply both sides by -1:[x - y = 2]Which can be rewritten as:[y = x - 2]So, that confirms the equation of line l is indeed y = x - 2.I think that makes sense. Both methods give me the same result, so I'm confident that the line of symmetry is y = x - 2.**Final Answer**The equation of line ( l ) is boxed{y = x - 2}.

💡First, I need to understand the problem. We have a rectangular piece of plywood that measures 3 feet by 9 feet. The goal is to cut this plywood into 3 congruent rectangles without any waste. After cutting, I need to find the positive difference between the greatest possible perimeter of a single piece and the least possible perimeter of a single piece.To approach this, I'll consider the two main ways to make the cuts:1. **Cuts parallel to the longer side (9 feet):** - If I make cuts parallel to the longer side, I'll be dividing the 9-foot length into 3 equal parts. - Each resulting rectangle will have dimensions 3 feet by 1 foot. - The perimeter of each smaller rectangle can be calculated using the formula for the perimeter of a rectangle: ( P = 2 times (length + width) ). - So, the perimeter would be ( 2 times (3 + 1) = 8 ) feet.2. **Cuts parallel to the shorter side (3 feet):** - If I make cuts parallel to the shorter side, I'll be dividing the 3-foot length into 3 equal parts. - Each resulting rectangle will have dimensions 1 foot by 9 feet. - Using the same perimeter formula, the perimeter would be ( 2 times (9 + 1) = 20 ) feet.Finally, to find the positive difference between the greatest and least perimeters, I'll subtract the smaller perimeter from the larger one: ( 20 - 8 = 12 ) feet.