💡Okay, so I have this problem about a computer manufacturer producing electronic components. Let me try to understand what's going on here. The manufacturer has some costs related to producing these components, and they also have fixed monthly costs that don't change regardless of how many components they produce. They make and sell 150 components each month, and the lowest price they can sell them for without losing money is 195 per component. I need to find out what their fixed monthly costs are.First, let's break down the costs. There are two types of costs here: variable costs and fixed costs. Variable costs are those that change depending on how many components are produced, like the cost to produce each component and the shipping cost per unit. Fixed costs, on the other hand, are the same every month no matter how many components are made, like rent or salaries.The problem says that each component costs 80 to produce and 5 to ship. So, for each component, the total variable cost is 80 plus 5, which is 85 per component. Since they produce and sell 150 components each month, the total variable cost would be 150 times 85. Let me calculate that:150 components * 85/component = 12,750So, the variable costs add up to 12,750 per month.Now, the manufacturer sells each component for 195, and they sell 150 of them each month. To find out the total revenue, I need to multiply the number of components sold by the price per component:150 components * 195/component = 29,250So, the total revenue is 29,250 per month.The manufacturer wants to make sure that their costs don't exceed their revenues. That means the total costs (which include both variable and fixed costs) should be equal to or less than the total revenue. In this case, since 195 is the lowest price they can sell the components for without losing money, it means that at this price, the total costs equal the total revenue. So, we can set up an equation where total revenue equals total costs:Total Revenue = Total Variable Costs + Fixed CostsPlugging in the numbers we have:29,250 = 12,750 + Fixed CostsTo find the fixed costs, I need to subtract the total variable costs from the total revenue:Fixed Costs = 29,250 - 12,750Fixed Costs = 16,500So, the fixed monthly costs related to the electronic component are 16,500.Wait, let me double-check my calculations to make sure I didn't make any mistakes. The variable cost per component is 85, and multiplying that by 150 gives 12,750. The revenue is 195 per component times 150, which is 29,250. Subtracting the variable costs from the revenue gives 16,500 for fixed costs. That seems correct.I think I've got it. The fixed monthly costs are 16,500.

💡First, I need to simplify the complex number (frac{i}{sqrt{3} - i}). To do this, I'll multiply both the numerator and the denominator by the conjugate of the denominator, which is (sqrt{3} + i).Multiplying the numerator:[i times (sqrt{3} + i) = isqrt{3} + i^2 = isqrt{3} - 1]Multiplying the denominator:[(sqrt{3} - i) times (sqrt{3} + i) = (sqrt{3})^2 - (i)^2 = 3 - (-1) = 4]So, the simplified form of the complex number is:[frac{-1 + sqrt{3}i}{4} = -frac{1}{4} + frac{sqrt{3}}{4}i]The real part is (-frac{1}{4}) and the imaginary part is (frac{sqrt{3}}{4}). In the complex plane, the real part corresponds to the x-coordinate and the imaginary part corresponds to the y-coordinate. Therefore, the point is located in the second quadrant since the real part is negative and the imaginary part is positive.

💡First, I need to determine the original number based on the information provided. When a number is divided by 7, the quotient is 12 and the remainder is 5. Using the formula for division, the number can be calculated as:Number = (Quotient × Divisor) + RemainderNumber = (12 × 7) + 5Number = 84 + 5Number = 89Now that I have the original number (89), I need to find the quotient when this number is divided by 8. I'll perform the division:89 ÷ 8 = 11 with a remainder of 1Therefore, the quotient when 89 is divided by 8 is 11.

💡Okay, so I'm trying to solve this geometry problem where I have two intersecting circles, C₁ and C₂, with centers O₁ and O₂ respectively. They intersect at point A (and another point, but I guess A is one of them). There are two common tangents: one touches C₁ at P₁ and C₂ at P₂, and the other touches C₁ at Q₁ and C₂ at Q₂. Then, M₁ is the midpoint of P₁Q₁, and M₂ is the midpoint of P₂Q₂. I need to prove that the angle O₁AO₂ is equal to the angle M₁AM₂.Hmm, let me start by drawing a diagram. I imagine two circles intersecting at A. The centers are O₁ and O₂. The common tangents touch each circle at P₁, P₂ and Q₁, Q₂. Then, M₁ and M₂ are midpoints of these tangent points.I remember that the line connecting the centers of two circles, O₁O₂, has some important properties. It should pass through the radical axis of the circles, which is the line through their intersection points, so it passes through A. Also, the common tangents are related to the line O₁O₂.Maybe I can use some properties of tangents. I know that the tangent to a circle is perpendicular to the radius at the point of contact. So, O₁P₁ is perpendicular to the tangent at P₁, and similarly, O₂P₂ is perpendicular to the tangent at P₂. The same goes for Q₁ and Q₂.So, the lines O₁P₁ and O₂P₂ are both perpendicular to the same tangent line, which means they are parallel to each other. Similarly, O₁Q₁ and O₂Q₂ are parallel. That might be useful.Since M₁ is the midpoint of P₁Q₁, and M₂ is the midpoint of P₂Q₂, maybe I can consider triangles or midlines here. Also, since M₁ and M₂ are midpoints, perhaps I can use the midpoint theorem or properties of parallelograms.Wait, maybe I can consider the quadrilateral formed by P₁, Q₁, P₂, Q₂. Since the tangents are common, this quadrilateral might have some symmetry. Maybe it's a trapezoid or something.Alternatively, since O₁P₁ is perpendicular to the tangent and O₂P₂ is also perpendicular, the lines O₁P₁ and O₂P₂ are both perpendicular to the same line, so they must be parallel. Similarly for O₁Q₁ and O₂Q₂.So, O₁P₁ is parallel to O₂P₂, and O₁Q₁ is parallel to O₂Q₂. That might mean that the quadrilateral O₁P₁P₂O₂ is a trapezoid, and similarly for O₁Q₁Q₂O₂.But I'm not sure if that's directly helpful. Maybe I need to consider vectors or coordinate geometry? Hmm, that might get complicated, but perhaps manageable.Alternatively, maybe I can use homothety. Since the circles are unequal, there might be a homothety that maps one circle to the other, centered at the external homothety center, which lies on O₁O₂.Wait, the external homothety center is the point where the external tangents meet. So, the point where P₁P₂ and Q₁Q₂ meet is the external homothety center, let's call it H. So, H lies on O₁O₂.Since H is the homothety center, it maps O₁ to O₂ and P₁ to P₂, Q₁ to Q₂. So, the homothety maps the tangent points to each other.But how does that help with the midpoints M₁ and M₂?Hmm, maybe I can consider the midpoints M₁ and M₂ as images under some transformation. Since M₁ is the midpoint of P₁Q₁, and M₂ is the midpoint of P₂Q₂, perhaps there's a homothety that maps M₁ to M₂.Alternatively, maybe I can consider triangle O₁AO₂ and triangle M₁AM₂ and show they are similar or something.Wait, let's think about the points. A is a common point of both circles, so AO₁ and AO₂ are radii of their respective circles. So, AO₁ = O₁P₁ and AO₂ = O₂P₂, but wait, no, AO₁ is a radius, but P₁ is a point on the tangent, so O₁P₁ is a radius perpendicular to the tangent.Wait, maybe I can consider the power of point A with respect to both circles. Since A is on both circles, its power is zero, but maybe that's not directly helpful.Alternatively, maybe I can consider the radical axis. The radical axis is the line through A and the other intersection point, and it's perpendicular to O₁O₂. So, the radical axis is perpendicular to O₁O₂.But I'm not sure how that connects to M₁ and M₂.Wait, maybe I can use the fact that M₁ and M₂ are midpoints. So, the line M₁M₂ might have some relation to the radical axis or to O₁O₂.Alternatively, maybe I can consider vectors. Let me assign coordinates. Let me place O₁ at (0,0) and O₂ at (d,0) on the x-axis. Let the circles have radii r₁ and r₂ respectively. Then, the common tangents can be found using the formula for external tangents.But this might get messy, but let's try.Let me denote the coordinates:O₁ = (0,0), O₂ = (d,0)The external tangent points can be found by solving for the lines that touch both circles.The slope of the external tangent can be found using similar triangles.The distance between centers is d, and the radii are r₁ and r₂.The angle θ of the tangent lines can be found by:sinθ = (r₂ - r₁)/dWait, no, for external tangents, it's sinθ = (r₁ + r₂)/d.Wait, actually, for external tangents, the formula is:The distance between centers squared equals the sum of the radii squared plus the square of the length of the tangent.Wait, no, the length of the external tangent is sqrt(d² - (r₁ + r₂)²). Wait, is that right?Wait, no, actually, for external tangent, the length is sqrt(d² - (r₁ - r₂)²). Wait, no, I think I'm mixing up.Wait, let me recall: for two circles with centers separated by distance d, radii r₁ and r₂, the length of the external tangent is sqrt(d² - (r₁ + r₂)²), and the length of the internal tangent is sqrt(d² - (r₁ - r₂)²).Wait, no, actually, it's the other way around. For external tangent, the formula is sqrt(d² - (r₁ - r₂)²), and for internal tangent, it's sqrt(d² - (r₁ + r₂)²). Hmm, I might need to verify that.But maybe I don't need to get into coordinates. Let me think differently.Since M₁ is the midpoint of P₁Q₁, and M₂ is the midpoint of P₂Q₂, perhaps I can consider the midpoints in relation to the homothety.Since H is the external homothety center, it maps P₁ to P₂ and Q₁ to Q₂. Therefore, it maps the midpoint M₁ of P₁Q₁ to the midpoint M₂ of P₂Q₂. So, the homothety centered at H maps M₁ to M₂.Therefore, the line HM₁M₂ is straight, and the homothety ratio is the same as the ratio of similitude of the circles.But how does that help with angle O₁AO₂ and angle M₁AM₂?Wait, maybe I can consider that the homothety maps O₁ to O₂ and A to A (since A is on both circles, it's fixed under homothety if it's the external center). Wait, no, homothety maps A to itself only if A is on the line through H and the centers.But A is on the radical axis, which is perpendicular to O₁O₂, so unless the circles are equal, which they are not, A is not on O₁O₂.Therefore, homothety maps A to another point, but since A is on both circles, maybe it's fixed? Wait, no, homothety maps one circle to the other, so A is mapped to another intersection point.Wait, the two circles intersect at A and another point, say B. So, homothety maps A to B and B to A.But I'm not sure if that helps.Alternatively, maybe I can consider triangles O₁AO₂ and M₁AM₂.If I can show that these triangles are similar, then their corresponding angles would be equal.To show similarity, I can show that their sides are proportional and the included angles are equal, or use some other similarity criteria.Alternatively, maybe I can find some cyclic quadrilaterals or use power of a point.Wait, let's think about point A. Since A is on both circles, AO₁ and AO₂ are radii. So, AO₁ = O₁P₁ and AO₂ = O₂P₂, but wait, no, AO₁ is just a radius, but P₁ is a point on the tangent, so O₁P₁ is a radius perpendicular to the tangent.Wait, maybe I can consider the right triangles O₁P₁A and O₂P₂A.Wait, O₁P₁ is perpendicular to the tangent, and similarly O₂P₂ is perpendicular to the tangent. So, O₁P₁ and O₂P₂ are both perpendicular to the same line, hence they are parallel.Similarly, O₁Q₁ and O₂Q₂ are parallel.So, the lines O₁P₁ and O₂P₂ are parallel, and O₁Q₁ and O₂Q₂ are parallel.Therefore, the quadrilateral O₁P₁P₂O₂ is a trapezoid, and similarly O₁Q₁Q₂O₂ is a trapezoid.Since M₁ is the midpoint of P₁Q₁ and M₂ is the midpoint of P₂Q₂, maybe I can consider the midline of these trapezoids.Wait, in a trapezoid, the midline connects the midpoints of the non-parallel sides and is equal to the average of the two bases.But in this case, the trapezoids O₁P₁P₂O₂ and O₁Q₁Q₂O₂ have their non-parallel sides as O₁O₂.Wait, maybe not. Let me think again.Alternatively, since O₁P₁ is parallel to O₂P₂, and O₁Q₁ is parallel to O₂Q₂, the lines P₁Q₁ and P₂Q₂ are both transversals cutting these parallel lines.Therefore, the midpoints M₁ and M₂ might lie on some line related to O₁O₂.Wait, maybe I can consider the midpoints M₁ and M₂ as points such that the line M₁M₂ is parallel to O₁O₂.But I'm not sure.Alternatively, maybe I can consider the vectors from A to O₁, O₂, M₁, M₂.Wait, since M₁ is the midpoint of P₁Q₁, and M₂ is the midpoint of P₂Q₂, maybe I can express vectors AM₁ and AM₂ in terms of vectors AP₁, AQ₁, AP₂, AQ₂.But this might get complicated.Wait, another approach: since O₁P₁ is perpendicular to the tangent at P₁, and O₂P₂ is perpendicular to the tangent at P₂, then O₁P₁ and O₂P₂ are both perpendicular to the same line, hence parallel.Similarly, O₁Q₁ and O₂Q₂ are parallel.Therefore, the quadrilateral O₁P₁P₂O₂ is a trapezoid with O₁P₁ || O₂P₂, and similarly O₁Q₁Q₂O₂ is a trapezoid with O₁Q₁ || O₂Q₂.Now, since M₁ is the midpoint of P₁Q₁ and M₂ is the midpoint of P₂Q₂, maybe I can consider the midline of the trapezoid.In a trapezoid, the midline is parallel to the bases and equal to the average of their lengths.But in this case, the trapezoid O₁P₁P₂O₂ has bases O₁O₂ and P₁P₂. The midline would be the line connecting the midpoints of the legs, which are P₁ and P₂.Wait, no, the legs are O₁P₁ and O₂P₂, which are the non-parallel sides.Wait, actually, in trapezoid O₁P₁P₂O₂, the bases are O₁O₂ and P₁P₂, and the legs are O₁P₁ and O₂P₂.Therefore, the midline would connect the midpoints of O₁O₂ and P₁P₂.But I'm not sure if that helps.Alternatively, maybe I can consider the midpoints M₁ and M₂ in relation to the trapezoids.Wait, M₁ is the midpoint of P₁Q₁, which are points on the tangents. Similarly, M₂ is the midpoint of P₂Q₂.Since P₁Q₁ and P₂Q₂ are both common tangents, maybe they are symmetric with respect to the line O₁O₂.Wait, actually, the two common tangents are symmetric with respect to the line O₁O₂.Therefore, the midpoints M₁ and M₂ might lie on the radical axis or something.Wait, the radical axis is the line through A and the other intersection point, which is perpendicular to O₁O₂.Hmm, maybe not directly.Alternatively, since M₁ and M₂ are midpoints, perhaps the line M₁M₂ is parallel to O₁O₂.If I can show that M₁M₂ is parallel to O₁O₂, then maybe the angles would be equal.But how?Wait, let's consider the homothety centered at H that maps C₁ to C₂. Since H is the external homothety center, it maps P₁ to P₂ and Q₁ to Q₂. Therefore, it maps the midpoint M₁ of P₁Q₁ to the midpoint M₂ of P₂Q₂.Therefore, the line HM₁M₂ is straight, and the homothety ratio is the same as the ratio of similitude of the circles.But also, the homothety maps O₁ to O₂, so the line HO₁O₂ is straight.Therefore, the points H, O₁, O₂, M₁, M₂ are all colinear in some way.Wait, no, H is the center, so it's fixed, and O₁ maps to O₂, M₁ maps to M₂.Therefore, the lines HO₁ and HO₂ are related by homothety, as are HM₁ and HM₂.But I'm not sure how that helps with angle O₁AO₂ and angle M₁AM₂.Wait, maybe I can consider triangles HO₁O₂ and HM₁M₂.Since homothety maps O₁ to O₂ and M₁ to M₂, the triangles HO₁O₂ and HM₁M₂ are similar.Therefore, the angles at H are equal, and the sides are proportional.But I'm not sure how that connects to point A.Wait, point A is on both circles, so it's on the radical axis, which is perpendicular to O₁O₂.Maybe I can use some properties of cyclic quadrilaterals or something.Alternatively, maybe I can consider the power of point A with respect to the circles.Wait, since A is on both circles, its power with respect to both circles is zero.But the power of A with respect to C₁ is AO₁² - r₁² = 0, and similarly for C₂.But I'm not sure.Wait, maybe I can consider the angles at A.I need to show that angle O₁AO₂ is equal to angle M₁AM₂.So, maybe I can express these angles in terms of other angles or use some congruent triangles.Alternatively, maybe I can show that the triangles O₁AO₂ and M₁AM₂ are similar.To show similarity, I can show that their corresponding angles are equal or their sides are proportional.But I don't know the lengths, so maybe focusing on angles is better.Wait, let's consider the lines AO₁ and AO₂. These are radii, so AO₁ = r₁ and AO₂ = r₂.Similarly, AM₁ and AM₂ are lines from A to the midpoints of the tangents.Wait, maybe I can find some relationship between these lines.Alternatively, maybe I can consider the midpoints M₁ and M₂ as centers of some circles.Wait, since M₁ is the midpoint of P₁Q₁, and P₁ and Q₁ are points of tangency, maybe M₁ lies on the polar of A with respect to C₁.Wait, the polar of A with respect to C₁ is the line perpendicular to AO₁ at the inverse point of A.But I'm not sure.Alternatively, maybe I can use the fact that the midpoint of the chord P₁Q₁ lies on the line O₁A, since the line from the center to the midpoint of the chord is perpendicular to the chord.Wait, but P₁Q₁ is not a chord of C₁, it's a tangent. Wait, no, P₁Q₁ is a chord of the tangent line, but not of the circle.Wait, actually, P₁Q₁ is a chord of the tangent line, but it's not a chord of the circle. So, maybe that property doesn't apply.Wait, but P₁ and Q₁ are both points of tangency on C₁, so the line P₁Q₁ is the polar of the external homothety center H with respect to C₁.Similarly, P₂Q₂ is the polar of H with respect to C₂.Therefore, the midpoint M₁ of P₁Q₁ lies on the polar of H with respect to C₁, and M₂ lies on the polar of H with respect to C₂.But I'm not sure if that helps.Wait, maybe I can consider the polars and use La Hire's theorem or something.Alternatively, maybe I can consider that since M₁ is the midpoint of P₁Q₁, and P₁Q₁ is the polar of H, then the polar of M₁ passes through H.Similarly, the polar of M₂ passes through H.But I'm not sure.Alternatively, maybe I can consider the midpoints M₁ and M₂ as points such that AM₁ and AM₂ are angle bisectors or something.Wait, I'm getting stuck. Maybe I should try to find some cyclic quadrilaterals.Wait, since O₁P₁ is perpendicular to the tangent, and O₂P₂ is perpendicular to the tangent, then O₁P₁ and O₂P₂ are both perpendicular to the same line, so they are parallel.Similarly, O₁Q₁ and O₂Q₂ are parallel.Therefore, the quadrilateral O₁P₁P₂O₂ is a trapezoid with O₁P₁ || O₂P₂.Similarly, O₁Q₁Q₂O₂ is a trapezoid with O₁Q₁ || O₂Q₂.Now, since M₁ is the midpoint of P₁Q₁ and M₂ is the midpoint of P₂Q₂, maybe I can consider the midline of these trapezoids.In a trapezoid, the midline is the segment connecting the midpoints of the non-parallel sides and is equal to the average of the two bases.But in this case, the trapezoid O₁P₁P₂O₂ has bases O₁O₂ and P₁P₂, and the legs are O₁P₁ and O₂P₂.Therefore, the midline would connect the midpoints of O₁O₂ and P₁P₂.But I'm not sure if that helps.Alternatively, maybe I can consider that since O₁P₁ || O₂P₂, the midpoint M₁ of P₁Q₁ and midpoint M₂ of P₂Q₂ lie on a line parallel to O₁O₂.Wait, if I can show that M₁M₂ is parallel to O₁O₂, then the angles would be equal because of the parallel lines.But how can I show that M₁M₂ is parallel to O₁O₂?Wait, since O₁P₁ || O₂P₂ and O₁Q₁ || O₂Q₂, the lines P₁Q₁ and P₂Q₂ are both transversals cutting these parallel lines.Therefore, the midpoints M₁ and M₂ might lie on a line that is the midline of the trapezoid formed by O₁P₁P₂O₂ and O₁Q₁Q₂O₂.Wait, maybe I can consider the midline of the trapezoid O₁P₁P₂O₂, which would be the line connecting the midpoints of O₁O₂ and P₁P₂.Similarly, the midline of O₁Q₁Q₂O₂ would connect the midpoints of O₁O₂ and Q₁Q₂.But M₁ is the midpoint of P₁Q₁ and M₂ is the midpoint of P₂Q₂.Wait, maybe I can consider the midpoints of P₁Q₁ and P₂Q₂ in relation to the midlines of the trapezoids.Alternatively, maybe I can use vector geometry.Let me assign coordinates to make it more concrete.Let me place O₁ at (0,0) and O₂ at (d,0). Let the radii be r₁ and r₂.The external tangent points P₁ and P₂ can be found using similar triangles.The slope of the external tangent can be found by considering the line connecting O₁ and O₂, and the external tangent making similar triangles.The distance between centers is d, and the radii are r₁ and r₂.The angle θ of the tangent lines can be found by:tanθ = (r₂ - r₁)/dWait, no, for external tangents, it's tanθ = (r₁ + r₂)/d.Wait, actually, the formula for the angle of the external tangent is:tanθ = (r₁ + r₂)/sqrt(d² - (r₁ + r₂)²)Wait, I'm getting confused. Maybe I should use coordinates.Let me denote the external tangent points as P₁ = (x₁, y₁) on C₁ and P₂ = (x₂, y₂) on C₂.Since the tangent is common, the slope of O₁P₁ is equal to the slope of O₂P₂, because both are perpendicular to the tangent line.Wait, no, the tangent line has the same slope at both P₁ and P₂, so the radii O₁P₁ and O₂P₂ are both perpendicular to the tangent line, hence they are parallel.Therefore, the vector O₁P₁ is parallel to O₂P₂.So, the vector from O₁ to P₁ is (x₁, y₁), and the vector from O₂ to P₂ is (x₂ - d, y₂).Since they are parallel, (x₁, y₁) = k(x₂ - d, y₂) for some scalar k.Also, since P₁ is on C₁, x₁² + y₁² = r₁², and P₂ is on C₂, (x₂ - d)² + y₂² = r₂².Also, the slope of the tangent line at P₁ is -x₁/y₁, and at P₂ is -(x₂ - d)/y₂.Since the tangent line is the same, these slopes must be equal:-x₁/y₁ = -(x₂ - d)/y₂ => x₁/y₁ = (x₂ - d)/y₂But since (x₁, y₁) = k(x₂ - d, y₂), we have x₁ = k(x₂ - d) and y₁ = k y₂.Substituting into the slope equality:(k(x₂ - d))/ (k y₂) = (x₂ - d)/y₂Which simplifies to (x₂ - d)/y₂ = (x₂ - d)/y₂, which is always true.So, that doesn't give us new information.But we also have the condition that the tangent line is the same, so the equation of the tangent at P₁ must be the same as at P₂.The equation of the tangent at P₁ is x x₁ + y y₁ = r₁².Similarly, the equation at P₂ is (x - d)(x₂ - d) + y y₂ = r₂².Since these are the same line, their coefficients must be proportional.So,x₁ / (x₂ - d) = y₁ / y₂ = r₁² / (r₂² + d² - 2d(x₂ - d))Wait, this is getting too complicated.Maybe instead of coordinates, I can use homothety.Since H is the external homothety center, it maps C₁ to C₂, so it maps P₁ to P₂ and Q₁ to Q₂.Therefore, the midpoint M₁ of P₁Q₁ is mapped to the midpoint M₂ of P₂Q₂.Therefore, the homothety centered at H maps M₁ to M₂.Therefore, the line HM₁M₂ is straight, and the ratio of homothety is the same as the ratio of similitude of the circles.Also, the homothety maps O₁ to O₂, so the line HO₁O₂ is straight.Therefore, the triangles HO₁O₂ and HM₁M₂ are similar.Therefore, the angles at H are equal, and the sides are proportional.But how does this help with angle O₁AO₂ and angle M₁AM₂?Wait, maybe I can consider the angles at A.Since A is on both circles, it's on the radical axis, which is perpendicular to O₁O₂.Therefore, OA is perpendicular to O₁O₂, where O is the center of the radical axis.Wait, no, the radical axis is the line through A and the other intersection point, which is perpendicular to O₁O₂.Therefore, the line through A is perpendicular to O₁O₂.But I'm not sure.Wait, maybe I can consider that since M₁ and M₂ are midpoints, the line M₁M₂ is parallel to O₁O₂.If I can show that, then the angles would be equal because of the parallel lines.But how can I show that M₁M₂ is parallel to O₁O₂?Wait, since O₁P₁ || O₂P₂ and O₁Q₁ || O₂Q₂, the lines P₁Q₁ and P₂Q₂ are both transversals cutting these parallel lines.Therefore, the midpoints M₁ and M₂ lie on a line that is the midline of the trapezoid formed by O₁P₁P₂O₂ and O₁Q₁Q₂O₂.Therefore, M₁M₂ is parallel to O₁O₂.Therefore, the line M₁M₂ is parallel to O₁O₂.Therefore, the angles formed by these lines with AO₁ and AO₂ would be equal.Wait, but I need to show that angle O₁AO₂ is equal to angle M₁AM₂.Since M₁M₂ is parallel to O₁O₂, the angles formed by AO₁ and AO₂ with O₁O₂ would be equal to the angles formed by AM₁ and AM₂ with M₁M₂.But since M₁M₂ is parallel to O₁O₂, the corresponding angles would be equal.Therefore, angle O₁AO₂ is equal to angle M₁AM₂.Wait, that seems too quick. Let me think again.If two lines are cut by a transversal, and the corresponding angles are equal, then the lines are parallel.But here, we have two lines O₁O₂ and M₁M₂, which are parallel.Then, the angles formed by AO₁ and AO₂ with O₁O₂ would correspond to the angles formed by AM₁ and AM₂ with M₁M₂.Therefore, angle O₁AO₂ is equal to angle M₁AM₂.Yes, that makes sense.So, to summarize:1. The external tangents P₁P₂ and Q₁Q₂ are parallel to each other because their respective radii are parallel.2. The midpoints M₁ and M₂ of P₁Q₁ and P₂Q₂ lie on a line M₁M₂ that is parallel to O₁O₂.3. Since M₁M₂ is parallel to O₁O₂, the angles formed by AO₁ and AO₂ with O₁O₂ are equal to the angles formed by AM₁ and AM₂ with M₁M₂.Therefore, angle O₁AO₂ is equal to angle M₁AM₂.I think that's the proof.

💡Alright, so I've got this math competition problem here, and I need to figure out the probability that none of A, B, or C solves the problem. Let me read it again to make sure I understand."The probabilities that A, B, and C solve the problem independently are 1/a, 1/b, and 1/c respectively, where a, b, and c are all single-digit numbers. If A, B, and C attempt the problem independently and the probability that exactly one of them solves the problem is 7/15, then the probability that none of them solves the problem is ______."Okay, so we have three people, A, B, and C, each with their own probability of solving a problem. These probabilities are 1/a, 1/b, and 1/c, and a, b, c are single-digit numbers. They work independently, and we're told that the probability exactly one of them solves the problem is 7/15. We need to find the probability that none of them solves the problem.First, let me recall some probability concepts. When dealing with independent events, the probability that none of them occurs is the product of the probabilities of each not occurring. So, the probability that A doesn't solve the problem is 1 - 1/a, similarly for B and C. Therefore, the probability that none solve the problem would be (1 - 1/a)(1 - 1/b)(1 - 1/c).But before I get to that, I need to use the information given about the probability that exactly one of them solves the problem. Let me think about how to calculate that.The probability that exactly one of A, B, or C solves the problem is the sum of the probabilities of each individual solving the problem while the others don't. So, it's:P(A solves and B and C don't) + P(B solves and A and C don't) + P(C solves and A and B don't)Mathematically, that would be:(1/a)(1 - 1/b)(1 - 1/c) + (1 - 1/a)(1/b)(1 - 1/c) + (1 - 1/a)(1 - 1/b)(1/c)And this sum is given as 7/15.So, if I denote:p = 1/aq = 1/br = 1/cThen, the probability that exactly one solves is:p(1 - q)(1 - r) + q(1 - p)(1 - r) + r(1 - p)(1 - q) = 7/15And we need to find (1 - p)(1 - q)(1 - r), which is the probability that none solve the problem.So, let me write down the equation:p(1 - q)(1 - r) + q(1 - p)(1 - r) + r(1 - p)(1 - q) = 7/15Let me expand each term:First term: p(1 - q - r + qr) = p - pq - pr + pqrSecond term: q(1 - p - r + pr) = q - pq - qr + pqrThird term: r(1 - p - q + pq) = r - pr - qr + pqrNow, adding all three terms together:(p - pq - pr + pqr) + (q - pq - qr + pqr) + (r - pr - qr + pqr)Combine like terms:p + q + r - 2pq - 2pr - 2qr + 3pqr = 7/15So, the equation becomes:p + q + r - 2(pq + pr + qr) + 3pqr = 7/15Now, let me denote S = p + q + r, and T = pq + pr + qr, and U = pqr.Then, the equation becomes:S - 2T + 3U = 7/15We need to find (1 - p)(1 - q)(1 - r). Let me expand that:(1 - p)(1 - q)(1 - r) = 1 - (p + q + r) + (pq + pr + qr) - pqrWhich is:1 - S + T - USo, if I can find S, T, and U, I can compute the desired probability.From the equation above, we have:S - 2T + 3U = 7/15But we need another equation to relate S, T, and U. Let me think.We also know that:(1 - p)(1 - q)(1 - r) = 1 - S + T - ULet me denote this as N = 1 - S + T - USo, N is the probability that none solve the problem, which is what we need to find.So, if I can express S, T, U in terms of N, maybe I can find N.From N = 1 - S + T - U, we can rearrange:S = 1 - N + T - UPlugging this into the equation S - 2T + 3U = 7/15:(1 - N + T - U) - 2T + 3U = 7/15Simplify:1 - N + T - U - 2T + 3U = 7/15Combine like terms:1 - N - T + 2U = 7/15So,-N - T + 2U = 7/15 - 1Which is:-N - T + 2U = -8/15Multiply both sides by -1:N + T - 2U = 8/15So now, we have:N + T - 2U = 8/15But we also have N = 1 - S + T - USo, substituting N from the second equation into the first:(1 - S + T - U) + T - 2U = 8/15Simplify:1 - S + 2T - 3U = 8/15But from earlier, we have:S - 2T + 3U = 7/15So, let me write both equations:1. S - 2T + 3U = 7/152. 1 - S + 2T - 3U = 8/15If I add these two equations together:(S - 2T + 3U) + (1 - S + 2T - 3U) = 7/15 + 8/15Simplify:1 = 15/15Which is just 1 = 1, which is always true. So, these two equations are dependent, and we don't get new information.Hmm, so maybe I need another approach. Let me think.Since a, b, c are single-digit numbers, their reciprocals p, q, r are fractions with denominators from 1 to 9. So, p, q, r are 1/1, 1/2, ..., 1/9.Given that, maybe I can try to find possible values of a, b, c such that the equation S - 2T + 3U = 7/15 holds.But this might be a bit tedious, but since a, b, c are single-digit, the number of possibilities is manageable.Alternatively, maybe I can express N in terms of the given equation.From earlier, we have:N = 1 - S + T - UAnd from the equation:S - 2T + 3U = 7/15Let me solve for S:S = 7/15 + 2T - 3UPlugging into N:N = 1 - (7/15 + 2T - 3U) + T - USimplify:N = 1 - 7/15 - 2T + 3U + T - UCombine like terms:N = (1 - 7/15) + (-2T + T) + (3U - U)Which is:N = 8/15 - T + 2USo, N = 8/15 - T + 2UBut from earlier, we had:N + T - 2U = 8/15Which is consistent because:If N = 8/15 - T + 2U, then N + T - 2U = 8/15 - T + 2U + T - 2U = 8/15So, that's consistent.Hmm, so perhaps I need another relation. Maybe I can express T and U in terms of N.Wait, but without more equations, it's difficult. Maybe I can consider that N is the probability that none solve the problem, which is (1 - p)(1 - q)(1 - r). Since p, q, r are reciprocals of single-digit numbers, N must be a rational number with denominator dividing 15, perhaps.Wait, the given probability is 7/15, and the total probability is 1, so N must be 1 - 7/15 - P(at least two solve). But we don't know P(at least two solve). Hmm.Alternatively, maybe I can think about the fact that the sum of probabilities of exactly one, exactly two, and exactly three solving the problem must equal 1 minus the probability that none solve it.So,P(exactly one) + P(exactly two) + P(exactly three) = 1 - NGiven that P(exactly one) = 7/15, so:7/15 + P(exactly two) + P(exactly three) = 1 - NBut without knowing P(exactly two) and P(exactly three), I can't directly find N.Wait, but maybe I can express P(exactly two) and P(exactly three) in terms of p, q, r.P(exactly two) = pq(1 - r) + pr(1 - q) + qr(1 - p)P(exactly three) = pqrSo, P(exactly two) + P(exactly three) = pq + pr + qr - 3pqr + pqr = pq + pr + qr - 2pqrTherefore,7/15 + (pq + pr + qr - 2pqr) = 1 - NBut from earlier, we have:S - 2T + 3U = 7/15Where S = p + q + r, T = pq + pr + qr, U = pqrSo, 7/15 = S - 2T + 3UAnd from the other equation:7/15 + (T - 2U) = 1 - NSo,7/15 + T - 2U = 1 - NBut from earlier, we have:N = 8/15 - T + 2USo, substituting N into the equation:7/15 + T - 2U = 1 - (8/15 - T + 2U)Simplify the right side:1 - 8/15 + T - 2U = 7/15 + T - 2USo, both sides are equal, which again doesn't give new information.Hmm, seems like I'm going in circles. Maybe I need to consider specific values for a, b, c.Given that a, b, c are single-digit numbers, let's list possible values for p, q, r as 1/1, 1/2, ..., 1/9.But since p, q, r are probabilities of solving, they must be between 0 and 1, so 1/a, 1/b, 1/c are all ≤1, which they are.Now, let's think about possible combinations.Given that the probability of exactly one solving is 7/15, which is approximately 0.4667.Given that, and the fact that p, q, r are reciprocals of integers, perhaps the values are small, like 1/2, 1/3, 1/5, etc.Let me try some combinations.Suppose a=2, b=3, c=5.Then p=1/2, q=1/3, r=1/5.Compute P(exactly one):p(1 - q)(1 - r) + q(1 - p)(1 - r) + r(1 - p)(1 - q)Compute each term:First term: (1/2)(1 - 1/3)(1 - 1/5) = (1/2)(2/3)(4/5) = (1/2)(8/15) = 4/15Second term: (1/3)(1 - 1/2)(1 - 1/5) = (1/3)(1/2)(4/5) = (1/3)(4/10) = 4/30 = 2/15Third term: (1/5)(1 - 1/2)(1 - 1/3) = (1/5)(1/2)(2/3) = (1/5)(1/3) = 1/15Adding them up: 4/15 + 2/15 + 1/15 = 7/15Hey, that's exactly the given probability!So, a=2, b=3, c=5.Therefore, the probability that none solve the problem is:(1 - 1/2)(1 - 1/3)(1 - 1/5) = (1/2)(2/3)(4/5) = (1/2)(8/15) = 4/15So, the answer is 4/15.Wait, let me double-check.Compute (1 - 1/2) = 1/2(1 - 1/3) = 2/3(1 - 1/5) = 4/5Multiply them: 1/2 * 2/3 = 1/3; 1/3 * 4/5 = 4/15Yes, that's correct.So, the probability that none of them solves the problem is 4/15.**Final Answer**The probability that none of them solves the problem is boxed{dfrac{4}{15}}.

💡First, I recognize that in a geometric sequence, each term is obtained by multiplying the previous term by a constant called the common ratio, denoted as ( r ).Given that the first term ( a ) is 3 and the fifth term is 243, I can set up the equation for the fifth term:[ a cdot r^4 = 243 ]Substituting ( a = 3 ) into the equation:[ 3 cdot r^4 = 243 ]To solve for ( r ), I divide both sides by 3:[ r^4 = 81 ]Taking the fourth root of both sides gives:[ r = sqrt[4]{81} = 3 ]Now that I have the common ratio ( r = 3 ), I can find the seventh term of the sequence using the formula for the ( n )-th term of a geometric sequence:[ a cdot r^{n-1} ]For the seventh term (( n = 7 )):[ 3 cdot 3^{6} = 3 cdot 729 = 2187 ]