🧠:[<think>Okay, let me try to solve this geometry problem. So, first, the problem says we have an isosceles right triangle ABC, where AB = 1, angle A is 90 degrees. Then point E is the midpoint of leg AC, and point F is on the base BC such that FE is perpendicular to BE. We need to find the area of triangle CEF.Hmm, let me start by visualizing the triangle. Since it's an isosceles right triangle with angle A being 90°, sides AB and AC should be equal. Wait, but the problem says AB = 1. If it's an isosceles right triangle with legs AB and AC, then AC should also be 1. But then point E is the midpoint of AC, so E divides AC into two equal parts, each of length 0.5. Okay, that makes sense.But wait, let me confirm. In an isosceles right triangle with legs of length 1, the hypotenuse BC would be √2, right? Because by Pythagoras, BC² = AB² + AC² = 1² + 1² = 2, so BC = √2. Got that.Now, coordinates might help here. Let me assign coordinates to each point to make this easier. Let's place point A at the origin (0,0). Since AB and AC are legs of the right triangle, let me put point B at (1,0) and point C at (0,1). Wait, but hold on, the problem says AB = 1 and angle A is 90°. So if AB is one leg and AC is the other leg, then yes, coordinates would be A(0,0), B(1,0), C(0,1). Then the hypotenuse BC would be from (1,0) to (0,1).Point E is the midpoint of AC. Since AC goes from (0,0) to (0,1), midpoint E would be at (0, 0.5). Got that.Now, point F is on BC such that FE is perpendicular to BE. So, we need to find coordinates of F on BC such that the line FE is perpendicular to BE. Then, once we have F, we can find the coordinates of C, E, F and compute the area of triangle CEF.Okay, let's write coordinates:A(0,0), B(1,0), C(0,1), E(0, 0.5).Now, BC is from (1,0) to (0,1), so parametric equations for BC can be written. Let's parameterize BC. Let me use parameter t, where t goes from 0 to 1. Then, any point F on BC can be expressed as F(t) = (1 - t, t), where t is between 0 and 1. When t=0, F is at B(1,0), and when t=1, F is at C(0,1). So that's a good parameterization.So coordinates of F are (1 - t, t) for some t between 0 and 1.Now, we need FE to be perpendicular to BE. Let's find coordinates of E and B first. E is (0, 0.5), B is (1,0). So vector BE is E - B = (0 - 1, 0.5 - 0) = (-1, 0.5). Similarly, FE is E - F. Wait, FE is from F to E. So coordinates of FE would be E - F = (0 - (1 - t), 0.5 - t) = (-1 + t, 0.5 - t).Wait, but direction matters for the vectors. If FE is the vector from F to E, then yes, it's E - F. Similarly, BE is from B to E, which is E - B. So, if FE is perpendicular to BE, their dot product should be zero.So, the vector FE is E - F = (0 - (1 - t), 0.5 - t) = (-1 + t, 0.5 - t). The vector BE is E - B = (0 - 1, 0.5 - 0) = (-1, 0.5). So the dot product of FE and BE should be zero.So, FE • BE = [(-1 + t)(-1) + (0.5 - t)(0.5)] = 0.Let me compute that:First component: (-1 + t)(-1) = 1 - tSecond component: (0.5 - t)(0.5) = 0.25 - 0.5tSo sum them up:(1 - t) + (0.25 - 0.5t) = 1 + 0.25 - t - 0.5t = 1.25 - 1.5tSet this equal to zero:1.25 - 1.5t = 0Solving for t:1.5t = 1.25t = 1.25 / 1.5 = (5/4) / (3/2) = (5/4) * (2/3) = 10/12 = 5/6 ≈ 0.8333...So t = 5/6. Therefore, point F has coordinates (1 - t, t) = (1 - 5/6, 5/6) = (1/6, 5/6). So F is at (1/6, 5/6).Now, we need to find the area of triangle CEF. Points C, E, F.Coordinates:C is (0,1)E is (0, 0.5)F is (1/6, 5/6)Let me plot these points to visualize. C is at the top of the triangle, E is halfway down AC, and F is somewhere on BC closer to C.To find the area of triangle CEF, we can use the shoelace formula.First, list the coordinates:C(0,1), E(0, 0.5), F(1/6, 5/6)Shoelace formula:Area = | (x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)) / 2 |Plugging in:x1 = 0, y1 = 1x2 = 0, y2 = 0.5x3 = 1/6, y3 = 5/6Compute each term:First term: x1(y2 - y3) = 0*(0.5 - 5/6) = 0Second term: x2(y3 - y1) = 0*(5/6 - 1) = 0Third term: x3(y1 - y2) = (1/6)*(1 - 0.5) = (1/6)*(0.5) = 1/12Sum of terms: 0 + 0 + 1/12 = 1/12Take absolute value and divide by 2: |1/12| / 2 = 1/24Wait, that can't be right. Because 1/24 seems very small. Let me check my calculation again.Wait, maybe I messed up the shoelace formula. Let me recall the formula correctly.Shoelace formula for three points (x1,y1), (x2,y2), (x3,y3):Area = | (x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)) / 2 |So, plugging in:x1(y2 - y3) = 0*(0.5 - 5/6) = 0*( -1/3 ) = 0x2(y3 - y1) = 0*(5/6 - 1) = 0*(-1/6) = 0x3(y1 - y2) = (1/6)*(1 - 0.5) = (1/6)*(0.5) = 1/12Total sum: 0 + 0 + 1/12 = 1/12Divide by 2: 1/24Hmm, so area is 1/24? But let me think, maybe the area is indeed 1/24. Let me confirm with another method.Alternatively, since points C, E, F are given, we can find vectors and compute the area via cross product.Coordinates:C(0,1), E(0, 0.5), F(1/6, 5/6)Vectors CE and CF.Vector CE = E - C = (0 - 0, 0.5 - 1) = (0, -0.5)Vector CF = F - C = (1/6 - 0, 5/6 - 1) = (1/6, -1/6)Area = 1/2 | CE x CF | (cross product)Cross product in 2D is scalar magnitude: (0)(-1/6) - (-0.5)(1/6) = 0 + 0.5/6 = 1/12Thus, area = 1/2 * 1/12 = 1/24. Okay, same result. So 1/24.But let me check if that makes sense. Let's see, the entire triangle ABC has area 0.5*1*1=0.5. Then, triangle CEF has area 1/24 ≈ 0.0417. Seems small, but maybe correct.Alternatively, maybe I made a mistake in the parameterization of point F.Wait, let's verify step by step.First, coordinates: A(0,0), B(1,0), C(0,1). Correct.Midpoint E of AC: AC is from (0,0) to (0,1), so midpoint E is (0, 0.5). Correct.Parametrization of BC: from (1,0) to (0,1). So F(t) = (1 - t, t). Correct.Vectors FE and BE:FE is E - F = (0 - (1 - t), 0.5 - t) = (-1 + t, 0.5 - t). BE is E - B = (0 - 1, 0.5 - 0) = (-1, 0.5). Correct.Dot product FE • BE = (-1 + t)(-1) + (0.5 - t)(0.5). Calculated as 1 - t + 0.25 - 0.5t = 1.25 - 1.5t. Correct.Set to zero: 1.25 -1.5t=0 => t=5/6. So F=(1 -5/6,5/6)=(1/6,5/6). Correct.Then coordinates of C(0,1), E(0,0.5), F(1/6,5/6). Correct.Shoelace formula gives 1/24, cross product gives same. So answer is 1/24? But let me think again.Wait, let's compute the base and height of triangle CEF.Looking at points C(0,1), E(0,0.5), F(1/6,5/6).Since points C and E are both on the y-axis (x=0), the line CE is vertical, from (0,1) to (0,0.5). So the length of CE is 0.5. So if we consider CE as the base, then the base length is 0.5.To find the area, we need the height corresponding to this base. The height would be the horizontal distance from point F to the line CE (which is the y-axis). Since F is at (1/6,5/6), the horizontal distance to the y-axis is 1/6. Therefore, the area is 1/2 * base * height = 1/2 * 0.5 * (1/6) = 1/2 * 0.5 * 1/6 = 1/24. Yep, same result.So that confirms the area is 1/24.But wait, just to ensure, maybe there's another approach. Let's see.Alternatively, since FE is perpendicular to BE, perhaps we can find the equations of lines BE and FE and find their intersection or something.First, find the equation of BE. Points B(1,0) and E(0,0.5). The slope of BE is (0.5 - 0)/(0 -1) = 0.5 / (-1) = -0.5. So equation is y - 0 = -0.5(x -1), which simplifies to y = -0.5x + 0.5.Then FE is perpendicular to BE, so its slope is the negative reciprocal of -0.5, which is 2. So FE has slope 2.FE passes through point F(1/6,5/6) as we found earlier. But since FE is perpendicular to BE at point F, but wait, FE is a line passing through F and E. Wait, actually, FE is a line segment from F to E. Wait, no, FE is the segment from F to E, and it's perpendicular to BE. So we already used that fact to find point F.But maybe using coordinate geometry again. Let's think.If FE is perpendicular to BE, then the product of their slopes is -1.Slope of BE is -0.5, so slope of FE should be 2.But FE connects F(1/6,5/6) to E(0,0.5). Let's compute the slope of FE:(0.5 - 5/6)/(0 - 1/6) = (-1/3)/(-1/6) = 2. So slope is 2. Correct, which is perpendicular to BE's slope of -0.5. So 2 * (-0.5) = -1. Correct.Therefore, the coordinates of F are correct.Therefore, the area is indeed 1/24.But wait, the problem says "Find the area of triangle CEF". So 1/24? Hmm, that seems small, but given the positions, maybe it's right. Let me check once more.Alternatively, perhaps using vectors. Coordinates:C(0,1), E(0,0.5), F(1/6,5/6)Vectors CE = E - C = (0, -0.5)Vectors CF = F - C = (1/6, -1/6)Area is 1/2 | CE x CF |. The cross product in 2D is calculated as (0)(-1/6) - (-0.5)(1/6) = 0 + 0.5/6 = 1/12. Then 1/2 * 1/12 = 1/24. Yes, same result.Alternatively, using base FC and height.Compute FC. Coordinates from F(1/6,5/6) to C(0,1). The distance FC is sqrt( (1/6)^2 + (1 - 5/6)^2 ) = sqrt(1/36 + 1/36 ) = sqrt(2/36) = sqrt(1/18) = 1/(3√2). Then to find the height from E to FC.But this seems more complicated. Maybe not necessary since multiple methods have given 1/24.Alternatively, use coordinate grid. Since all coordinates are known, plotting might help.But since all calculations check out, I think the answer is 1/24.Wait, but let me check again with another person's perspective. Suppose someone else did this problem and got a different answer, like 1/12. Where could I have gone wrong?Wait, in the shoelace formula, did I do it correctly?The shoelace formula for three points is indeed as I applied. Let me write the points in order: C(0,1), E(0,0.5), F(1/6,5/6), and back to C(0,1).Compute sum of x1y2, x2y3, x3y1:(0*0.5) + (0*5/6) + (1/6*1) = 0 + 0 + 1/6 = 1/6Compute sum of y1x2, y2x3, y3x1:(1*0) + (0.5*1/6) + (5/6*0) = 0 + 0.5/6 + 0 = 1/12Subtract the two sums: 1/6 - 1/12 = 1/12. Take absolute value and divide by 2: |1/12| /2 = 1/24. Yep, same answer.So, all methods lead to 1/24. Therefore, I think that's correct.But let me check the problem statement again to make sure I didn't misread anything.Problem states: in isosceles right triangle ABC, AB=1, angle A=90°, E is the midpoint of AC, F is on BC such that FE ⊥ BE. Find area of triangle CEF.Yes, that's what I solved. So the answer is 1/24.Wait a second, but the problem says ABC is an isosceles right triangle with AB=1. If angle A is 90°, then legs AB and AC are equal. So AB=AC=1, making BC the hypotenuse of length √2. So coordinates are correct.But in my mind, I initially thought ABC is an isosceles right triangle with legs AB and AC, which is correct. So all the steps are correct.Therefore, I think the area is indeed 1/24. So I should box that.Final Answerboxed{dfrac{1}{24}}</think>Given an isosceles right triangle ( triangle ABC ) with ( AB = 1 ), ( angle A = 90^circ ), point ( E ) is the midpoint of leg ( AC ), and point ( F ) is on the base ( BC ) such that ( FE perp BE ). We need to find the area of ( triangle CEF ).1. Coordinates Assignment: - Place ( A ) at the origin ((0,0)). - ( B ) at ((1,0)) and ( C ) at ((0,1)). - Midpoint ( E ) of ( AC ) is ((0, 0.5)).2. Parameterization of ( F ) on ( BC ): - Parametric coordinates of ( F ) on ( BC ) are ((1 - t, t)) for ( t ) between 0 and 1.3. Vectors and Dot Product: - Vector ( BE ) is ((-1, 0.5)). - Vector ( FE ) is ((-1 + t, 0.5 - t)). - Dot product ( FE cdot BE = 0 ) gives equation ( 1.25 - 1.5t = 0 ). - Solving for ( t ), we get ( t = frac{5}{6} ).4. Coordinates of ( F ): - Substituting ( t = frac{5}{6} ), ( F ) is (left(frac{1}{6}, frac{5}{6}right)).5. Area Calculation using Shoelace Formula: - Coordinates of ( C(0,1) ), ( E(0,0.5) ), ( Fleft(frac{1}{6}, frac{5}{6}right) ). - Shoelace formula gives area: [ text{Area} = frac{1}{2} left| 0 cdot (0.5 - frac{5}{6}) + 0 cdot (frac{5}{6} - 1) + frac{1}{6} cdot (1 - 0.5) right| = frac{1}{24} ]Thus, the area of ( triangle CEF ) is (boxed{dfrac{1}{24}}).,<think>Okay, let me try to solve this geometry problem. Hmm, it's about an isosceles right triangle ABC where AB = 1, angle A is 90 degrees. Point E is the midpoint of leg AC, and point F is on the base BC such that FE is perpendicular to BE. We need to find the area of triangle CEF. First, let me visualize the triangle. Since it's an isosceles right triangle with angle A being 90 degrees, sides AB and AC should be equal. Wait, but the problem says AB = 1. So, AB = AC = 1? Because in an isosceles right triangle, the legs are equal. Wait, but if angle A is 90 degrees, then the legs are AB and AC. Since it's isosceles, AB = AC. So both AB and AC are 1. Then the hypotenuse BC would be sqrt(2), right? Because in a right triangle with legs of length 1, the hypotenuse is sqrt(1^2 + 1^2) = sqrt(2).Now, point E is the midpoint of leg AC. Since AC is 1, then E divides AC into two segments of length 0.5 each. So AE = EC = 0.5. Okay, got that.Point F is on BC such that FE is perpendicular to BE. So we have to find the point F on BC where the line FE is perpendicular to BE. Then, once we have F, we can find the coordinates of C, E, F and compute the area of triangle CEF.Maybe coordinate geometry would be a good approach here. Let's assign coordinates to the triangle. Let me place point A at the origin (0,0). Then since AB and AC are both length 1, and angle A is 90 degrees, let's put point B at (1,0) and point C at (0,1). Then the hypotenuse BC would go from (1,0) to (0,1). Point E is the midpoint of AC. Since AC is from (0,0) to (0,1), the midpoint E would be at (0, 0.5).Now, point F is somewhere on BC. Let's parametrize BC. The line BC connects (1,0) to (0,1). So its parametric equations can be written as x = 1 - t, y = 0 + t, where t ranges from 0 to 1. So any point F on BC can be expressed as (1 - t, t) for some t between 0 and 1.We need to find t such that FE is perpendicular to BE. Let's find the coordinates of BE and FE.First, point B is at (1,0) and point E is at (0,0.5). So vector BE is E - B = (0 - 1, 0.5 - 0) = (-1, 0.5). Similarly, vector FE would be E - F or F - E? Wait, FE is the vector from F to E. Wait, FE is the line segment from F to E. So vector FE would be E - F = (0 - (1 - t), 0.5 - t) = (-1 + t, 0.5 - t). Alternatively, if FE is perpendicular to BE, then their dot product should be zero. Let me confirm:Vector BE is from B to E: E - B = (-1, 0.5).Vector FE is from F to E: E - F = (0 - (1 - t), 0.5 - t) = (-1 + t, 0.5 - t).So FE = (-1 + t, 0.5 - t). BE is (-1, 0.5). Their dot product should be zero.So (-1 + t)(-1) + (0.5 - t)(0.5) = 0.Let me compute that:First term: (-1 + t)(-1) = 1 - tSecond term: (0.5 - t)(0.5) = 0.25 - 0.5tAdding them up: (1 - t) + (0.25 - 0.5t) = 1 + 0.25 - t - 0.5t = 1.25 - 1.5tSet equal to zero: 1.25 - 1.5t = 0Solving for t: 1.5t = 1.25 => t = 1.25 / 1.5 = (5/4)/(3/2) = (5/4)*(2/3) = 10/12 = 5/6So t = 5/6. Therefore, point F is at (1 - t, t) = (1 - 5/6, 5/6) = (1/6, 5/6).Now, we need to find the area of triangle CEF. Points C, E, F are:C is at (0,1)E is at (0, 0.5)F is at (1/6, 5/6)We can use the formula for the area of a triangle given coordinates of its vertices. The formula is:Area = |(x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))/2|Let me plug in the coordinates:x1 = 0, y1 = 1 (point C)x2 = 0, y2 = 0.5 (point E)x3 = 1/6, y3 = 5/6 (point F)Plug into the formula:Area = |0*(0.5 - 5/6) + 0*(5/6 - 1) + (1/6)*(1 - 0.5)| / 2Simplify each term:First term: 0*(0.5 - 5/6) = 0Second term: 0*(5/6 - 1) = 0Third term: (1/6)*(1 - 0.5) = (1/6)*(0.5) = 1/12So Area = |0 + 0 + 1/12| / 2 = (1/12)/2 = 1/24Wait, that can't be right. Wait, 1/12 divided by 2 is 1/24? Let me check.Wait, the formula is |sum| / 2. So sum is 1/12, then divide by 2, so 1/24. But 1/24 seems very small. Let me verify.Alternatively, perhaps I made a mistake in coordinates.Wait, let's check the coordinates again.Point C is at (0,1)Point E is at (0, 0.5)Point F is at (1/6, 5/6)So triangle CEF has vertices at (0,1), (0,0.5), and (1/6, 5/6). Let me plot these mentally.From point C (0,1) to E (0,0.5) is a vertical line down to (0,0.5). Then from E to F (1/6,5/6) is a line going to the right and up a bit, then from F back to C.Alternatively, maybe we can compute the base and height.Since CE is vertical from (0,1) to (0,0.5), so the length of CE is 0.5. Then the base could be CE, which is 0.5, and the height would be the horizontal distance from F to CE. Since CE is along the y-axis (x=0), the horizontal distance from F (which is at x=1/6) to CE is 1/6. Therefore, the area would be (base * height)/2 = (0.5 * 1/6)/2 = (1/12)/2 = 1/24. Wait, same result. So perhaps that's correct?But intuitively, in a triangle of area 0.5 (since the whole triangle ABC has area (1*1)/2 = 0.5), having a sub-triangle with area 1/24 seems small, but maybe it's correct.Alternatively, let's compute using vectors or determinants.Coordinates:C: (0,1)E: (0,0.5)F: (1/6,5/6)The vectors CE and CF can be formed. Wait, but maybe using the shoelace formula is better.Shoelace formula for coordinates (x1,y1), (x2,y2), (x3,y3):Area = 1/2 |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|Plugging in:x1 = 0, y1 = 1x2 = 0, y2 = 0.5x3 = 1/6, y3 = 5/6Compute:0*(0.5 - 5/6) + 0*(5/6 - 1) + (1/6)*(1 - 0.5) = 0 + 0 + (1/6)*(0.5) = 1/12Take absolute value and multiply by 1/2: (1/12)*1/2 = 1/24. So same result.Hmm. So the area is 1/24? That seems very small, but let's check the steps again.Alternatively, maybe there's a miscalculation in the value of t. Let's check how we found t.We had FE · BE = 0.We parametrized F as (1 - t, t). Then vector FE is E - F = (0 - (1 - t), 0.5 - t) = (t - 1, 0.5 - t). Vector BE is E - B = (0 - 1, 0.5 - 0) = (-1, 0.5). So their dot product is (t - 1)(-1) + (0.5 - t)(0.5) = (-t + 1) + (0.25 - 0.5t) = 1 + 0.25 - t - 0.5t = 1.25 - 1.5t. Set equal to zero gives t = 1.25 / 1.5 = 5/6. So that's correct.Therefore, F is at (1 - 5/6, 5/6) = (1/6, 5/6). That seems correct.Then using coordinates, the area is 1/24. Hmm.Alternatively, maybe the problem is in the coordinate assignments. Let me check the initial coordinate setup.The problem says triangle ABC is an isosceles right triangle with AB = 1 and angle A = 90 degrees. Wait, if AB is one of the legs, then if angle A is 90 degrees, then legs are AB and AC. So if AB = 1, then AC should also be 1, since it's isosceles. Therefore, BC is the hypotenuse of length sqrt(2). So coordinates assigned as A(0,0), B(1,0), C(0,1) are correct.Midpoint E is at (0, 0.5). Correct.Parametrizing BC as (1 - t, t) where t from 0 to 1. Correct.So FE is from F(1 - t, t) to E(0, 0.5), which is vector (-1 + t, 0.5 - t). BE is from B(1,0) to E(0, 0.5), which is vector (-1, 0.5). Dot product is (-1 + t)(-1) + (0.5 - t)(0.5) = (1 - t) + (0.25 - 0.5t) = 1.25 - 1.5t = 0 => t = 5/6. So point F is (1 - 5/6, 5/6) = (1/6, 5/6). That's correct.Then area of CEF: coordinates are (0,1), (0, 0.5), (1/6, 5/6). Shoelace formula gives 1/24. Maybe that's correct.Wait, let me check with another approach. Let's compute the vectors.Vector CE is E - C = (0 - 0, 0.5 - 1) = (0, -0.5)Vector CF is F - C = (1/6 - 0, 5/6 - 1) = (1/6, -1/6)The area of triangle CEF is half the magnitude of the cross product of CE and CF.Cross product in 2D is scalar magnitude |CE_x * CF_y - CE_y * CF_x|So CE_x = 0, CE_y = -0.5CF_x = 1/6, CF_y = -1/6Cross product = 0 * (-1/6) - (-0.5) * (1/6) = 0 + 0.5 * 1/6 = 1/12Area is half of that: 1/24. So same result.Therefore, despite seeming small, the area is indeed 1/24.But just to be thorough, let me visualize this. The triangle CEF is a small triangle near the bottom of the original triangle. CE is half of AC, and F is closer to C on BC. The area being 1/24 makes sense as it's a small triangle. Let me also check the original triangle area is 0.5, and 1/24 is about 0.0417, which is plausible.Alternatively, maybe there's a mistake in the problem interpretation. The problem says "in isosceles right triangle ABC, AB = 1, angle A = 90°". So AB is one leg, AC is the other leg, both equal. So AB = AC = 1. Then BC is hypotenuse. E is midpoint of AC, so AE = EC = 0.5. Then FE perpendicular to BE. So from point E, we draw BE, which connects to B, and then FE is perpendicular to BE. So FE is a line from F on BC such that FE is perpendicular to BE.Alternatively, maybe using coordinate geometry was the right approach, and the answer is indeed 1/24. But let me check once more.Alternatively, use coordinate system with A at (0,0), B at (1,0), C at (0,1). Then line BE is from (1,0) to (0,0.5). The equation of BE: slope is (0.5 - 0)/(0 - 1) = -0.5. So equation is y = -0.5x + 0.5. FE is perpendicular to BE, so its slope is reciprocal and opposite sign, so 2. So FE has slope 2. Since FE passes through E(0,0.5), then equation of FE is y - 0.5 = 2x. But FE also passes through F, which is on BC. Equation of BC is x + y = 1. So F is the intersection of FE (y = 2x + 0.5) and BC (x + y = 1).Wait, let's check this. If FE is perpendicular to BE and passes through E, then FE is the line passing through E with slope 2 (since BE has slope -0.5, so perpendicular slope is 2). Therefore, equation of FE: y - 0.5 = 2(x - 0), so y = 2x + 0.5.But BC is the line from (1,0) to (0,1), equation x + y = 1. So solve 2x + 0.5 = y and x + y = 1. Substitute y = 1 - x into first equation:2x + 0.5 = 1 - x2x + x = 1 - 0.53x = 0.5x = 0.5 / 3 = 1/6Then y = 1 - x = 1 - 1/6 = 5/6. So F is at (1/6, 5/6), which matches our previous result. Therefore, coordinates are correct.Thus, the area of triangle CEF is 1/24. Therefore, the answer is boxed{dfrac{1}{24}}.Final AnswerThe area of triangle CEF is boxed{dfrac{1}{24}}.</think>Given an isosceles right triangle ( triangle ABC ) with ( AB = 1 ), ( angle A = 90^circ ), point ( E ) is the midpoint of leg ( AC ), and point ( F ) is on the base ( BC ) such that ( FE perp BE ). We need to find the area of ( triangle CEF ).1. Coordinates Assignment: - Place ( A ) at the origin ((0,0)). - ( B ) at ((1,0)) and ( C ) at ((0,1)) since ( AB = AC = 1 ). - Midpoint ( E ) of ( AC ) is ((0, 0.5)).2. Parametrization of ( BC ): - The line ( BC ) can be parametrized as ( (1 - t, t) ) where ( t ) ranges from 0 to 1. - Point ( F ) is ((1 - t, t)).3. Vectors and Dot Product: - Vector ( BE ) is ((-1, 0.5)). - Vector ( FE ) is ((t - 1, 0.5 - t)). - For ( FE perp BE ), their dot product must be zero: [ (-1 + t)(-1) + (0.5 - t)(0.5) = 0 implies 1.25 - 1.5t = 0 implies t = frac{5}{6} ] - Thus, ( F ) is at ((1 - frac{5}{6}, frac{5}{6}) = (frac{1}{6}, frac{5}{6})).4. Area Calculation: - Coordinates of ( C ), ( E ), and ( F ) are ((0, 1)), ((0, 0.5)), and ((frac{1}{6}, frac{5}{6})) respectively. - Using the shoelace formula: [ text{Area} = frac{1}{2} left| 0(0.5 - frac{5}{6}) + 0(frac{5}{6} - 1) + frac{1}{6}(1 - 0.5) right| = frac{1}{2} left| 0 + 0 + frac{1}{12} right| = frac{1}{24} ]The area of ( triangle CEF ) is (boxed{dfrac{1}{24}}).]

🔑:Three, 11. Construct(D) perp C E and the extension of E F intersect at Gamma.begin{array}{l} because angle A B E+ angle A E B=90^{circ}, angle C E D+angle A E B =90^{circ}, angle A B E=angle C E D .end{array}Therefore, Rt triangle A B E sim R mathrm{R} triangle C E D.begin{array}{l}therefore frac{S_{triangle C D E}}{S_{triangle E A B}}=left(frac{C E}{A B}right)^{2}=frac{1}{4} . frac{C E}{C D}=frac{A B}{A E}=2 .end{array}Also, angle E C F=angle D C F=45^{circ}, so C F is the angle bisector of angle D C E, and the distances from point F to C E and C D are equal.begin{aligned}therefore frac{S_{triangle C F F}}{S_{triangle C F F}} & =frac{C E}{C D}=2 . therefore S_{triangle C Y F} & =frac{2}{3} S_{triangle C F E}=frac{2}{3} times frac{1}{4} S_{triangle A B E} & =frac{2}{3} times frac{1}{4} times frac{1}{2} S_{triangle A B C}=frac{1}{24} .end{aligned}

🧠:[<think>Okay, so I need to find the ratio of the 8th term of sequence {a_n} to the sum of the 5th and 11th terms of sequence {b_n}. Both {a_n} and {b_n} are arithmetic sequences, and we know that the ratio of their sums of the first n terms, S_n over T_n, is (3n -1)/(n +3). Hmm, let me start by recalling some properties of arithmetic sequences and their sums.An arithmetic sequence has a common difference between consecutive terms. So, for {a_n}, the nth term can be written as a_n = a_1 + (n-1)d_a, where a_1 is the first term and d_a is the common difference. Similarly, for {b_n}, it's b_n = b_1 + (n-1)d_b.The sum of the first n terms of an arithmetic sequence is given by S_n = n/2 [2a_1 + (n-1)d_a] or also S_n = n(a_1 + a_n)/2. Similarly, T_n = n/2 [2b_1 + (n-1)d_b] = n(b_1 + b_n)/2.The problem gives S_n / T_n = (3n -1)/(n +3). So, let's write that out:S_n / T_n = [n/2 (2a_1 + (n -1)d_a)] / [n/2 (2b_1 + (n -1)d_b)] = [2a_1 + (n -1)d_a] / [2b_1 + (n -1)d_b] = (3n -1)/(n +3)Since the n/2 cancels out in numerator and denominator, the ratio simplifies to [2a_1 + (n -1)d_a] / [2b_1 + (n -1)d_b] = (3n -1)/(n +3). So, this equality holds for all n. Therefore, the expressions in the numerator and denominator must be proportional in a way that the ratio simplifies to that linear fraction for all n.This suggests that 2a_1 + (n -1)d_a is proportional to (3n -1) and 2b_1 + (n -1)d_b is proportional to (n +3). Alternatively, we can set up the equation:[2a_1 + (n -1)d_a] / [2b_1 + (n -1)d_b] = (3n -1)/(n +3)Cross-multiplying gives:(2a_1 + (n -1)d_a)(n +3) = (2b_1 + (n -1)d_b)(3n -1)This equation must hold true for all values of n, so the coefficients of corresponding powers of n on both sides must be equal. Let's expand both sides.First, the left-hand side (LHS):(2a_1 + (n -1)d_a)(n +3) = [2a_1 + n d_a - d_a](n +3)= (2a_1 - d_a + n d_a)(n +3)= (n d_a + (2a_1 - d_a))(n +3)= n d_a * n + n d_a *3 + (2a_1 - d_a)*n + (2a_1 - d_a)*3= d_a n² + 3d_a n + (2a_1 - d_a)n + 3(2a_1 - d_a)= d_a n² + [3d_a + 2a_1 - d_a]n + 3(2a_1 - d_a)= d_a n² + (2d_a + 2a_1)n + 6a_1 - 3d_aNow, the right-hand side (RHS):(2b_1 + (n -1)d_b)(3n -1) = [2b_1 + n d_b - d_b](3n -1)= (2b_1 - d_b + n d_b)(3n -1)= n d_b *3n + n d_b*(-1) + (2b_1 - d_b)*3n + (2b_1 - d_b)*(-1)= 3d_b n² - d_b n + 6b_1 n - 3d_b n - 2b_1 + d_b= 3d_b n² + (-d_b -3d_b +6b_1)n + (-2b_1 + d_b)= 3d_b n² + (6b_1 -4d_b)n + (-2b_1 + d_b)So, equating coefficients of n², n, and the constant term from LHS and RHS:For n² term:d_a = 3d_bFor n term:2d_a + 2a_1 = 6b_1 -4d_bFor constant term:6a_1 -3d_a = -2b_1 + d_bNow, we have three equations:1. d_a = 3d_b2. 2d_a + 2a_1 = 6b_1 -4d_b3. 6a_1 -3d_a = -2b_1 + d_bLet me substitute equation 1 into equations 2 and 3.Equation 2 becomes:2*(3d_b) + 2a_1 = 6b_1 -4d_b6d_b + 2a_1 = 6b_1 -4d_bBring all terms to one side:6d_b +4d_b +2a_1 -6b_1 =010d_b +2a_1 -6b_1=0Divide both sides by 2:5d_b +a_1 -3b_1=0So, equation 2a: a_1 = 3b_1 -5d_bEquation 3 becomes:6a_1 -3*(3d_b) = -2b_1 +d_b6a_1 -9d_b = -2b_1 +d_bBring all terms to left side:6a_1 -9d_b +2b_1 -d_b =06a_1 +2b_1 -10d_b =0Divide both sides by 2:3a_1 +b_1 -5d_b =0Now, from equation 2a, a_1 =3b_1 -5d_b. Let's substitute this into equation 3a:3*(3b_1 -5d_b) +b_1 -5d_b=09b_1 -15d_b +b_1 -5d_b=010b_1 -20d_b=0Divide both sides by 10:b_1 -2d_b=0Thus, b_1=2d_bNow, substitute back into equation 2a:a_1=3b_1 -5d_b=3*(2d_b) -5d_b=6d_b -5d_b=d_bSo now, we have:b_1=2d_ba_1=d_bd_a=3d_bSo, all terms are expressed in terms of d_b. Let's let d_b =k for simplicity. Then:d_b =kb_1=2ka_1=kd_a=3kSo, now, we can express the terms of the sequences {a_n} and {b_n} in terms of k.For {a_n}:a_n =a_1 + (n-1)d_a =k + (n-1)*3k =k +3k(n-1)=k[1 +3(n-1)]=k(3n -2)For {b_n}:b_n =b_1 + (n-1)d_b =2k + (n-1)k =k[2 +n -1]=k(n +1)Now, we need to find a_8 / (b_5 +b_11)First, compute a_8:a_8 =k(3*8 -2)=k(24 -2)=22kThen, compute b_5 and b_11:b_5 =k(5 +1)=6kb_11 =k(11 +1)=12kThus, b_5 +b_11=6k +12k=18kTherefore, the ratio a_8 / (b_5 +b_11)=22k /18k=22/18=11/9Wait, but 22 divided by 18 is 11/9, which simplifies to 1 and 2/9. But the question probably wants the fraction in simplest form, so 11/9. Let me check if that's correct.Wait, let me verify the steps again.First, when we found a_1 =d_b, and d_a=3d_b, then for a_n:a_n =a_1 + (n-1)d_a =d_b + (n-1)*3d_b =d_b[1 +3(n-1)]=d_b(3n -2). So, a_8= d_b(3*8 -2)=d_b*22=22d_b. Since d_b is a common difference, which is a constant.Similarly, for b_n =2d_b + (n-1)d_b =d_b(2 +n -1)=d_b(n +1). So, b_5 =d_b(5 +1)=6d_b, b_11 =d_b(11 +1)=12d_b. So, sum is 18d_b. Therefore, ratio is (22d_b)/(18d_b)=22/18=11/9. Yes, that's correct. So the answer is 11/9.Wait, but let me check if substituting back into the original sum ratio gives (3n -1)/(n +3). Let me verify that with the expressions for S_n and T_n.Compute S_n for {a_n}:Since a_n =k(3n -2), so first term a_1= k(3*1 -2)=k(3 -2)=k. Which matches a_1=k.Sum S_n =n/2 [2a_1 + (n -1)d_a] =n/2 [2k + (n -1)*3k] =n/2 [2k +3k(n -1)] =n/2 [2k +3kn -3k] =n/2 [3kn -k] =n/2 *k(3n -1)= (kn/2)(3n -1)Similarly, T_n for {b_n}:First term b_1=2k, common difference d_b=k.Sum T_n =n/2 [2b_1 + (n -1)d_b] =n/2 [4k + (n -1)k] =n/2 [4k +kn -k] =n/2 [kn +3k] =n/2 *k(n +3)= (kn/2)(n +3)Therefore, S_n / T_n = [ (kn/2)(3n -1) ] / [ (kn/2)(n +3) ] = (3n -1)/(n +3), which matches the given ratio. So, the expressions are correct.Therefore, the answer is 11/9.Final Answerboxed{dfrac{11}{9}}</think>Given arithmetic sequences ({a_n}) and ({b_n}) with sums of the first (n) terms (S_n) and (T_n) respectively, we know that (frac{S_n}{T_n} = frac{3n - 1}{n + 3}). We need to find (frac{a_8}{b_5 + b_{11}}).First, recall the sum formulas for arithmetic sequences:[ S_n = frac{n}{2} [2a_1 + (n - 1)d_a] ][ T_n = frac{n}{2} [2b_1 + (n - 1)d_b] ]Given the ratio:[ frac{S_n}{T_n} = frac{3n - 1}{n + 3} ]Simplifying, we equate:[ frac{2a_1 + (n - 1)d_a}{2b_1 + (n - 1)d_b} = frac{3n - 1}{n + 3} ]Cross-multiplying and expanding both sides, we equate coefficients of (n^2), (n), and the constant terms. Solving these, we find:1. (d_a = 3d_b)2. (a_1 = d_b)3. (b_1 = 2d_b)Expressing terms in terms of (d_b):- For ({a_n}): (a_n = d_b(3n - 2))- For ({b_n}): (b_n = d_b(n + 1))Calculating specific terms:- (a_8 = d_b(3 cdot 8 - 2) = 22d_b)- (b_5 = d_b(5 + 1) = 6d_b)- (b_{11} = d_b(11 + 1) = 12d_b)Summing (b_5) and (b_{11}):[ b_5 + b_{11} = 6d_b + 12d_b = 18d_b ]Thus, the ratio:[ frac{a_8}{b_5 + b_{11}} = frac{22d_b}{18d_b} = frac{22}{18} = frac{11}{9} ][boxed{dfrac{11}{9}}],<think>Okay, so I have this problem here about two arithmetic sequences, {a_n} and {b_n}. Their sums of the first n terms are S_n and T_n respectively. The ratio of these sums is given as (3n - 1)/(n + 3), and I need to find the value of a_8 divided by (b_5 + b_11). Hmm, arithmetic sequences. Let me recall some formulas.First, an arithmetic sequence has a common difference. So the nth term of an arithmetic sequence can be written as a_n = a_1 + (n - 1)d, where a_1 is the first term and d is the common difference. Similarly for {b_n}, it's b_n = b_1 + (n - 1)e, where e is the common difference for that sequence.The sum of the first n terms of an arithmetic sequence is given by S_n = n/2 * [2a_1 + (n - 1)d], right? Or equivalently, S_n = n(a_1 + a_n)/2. Same for T_n, the sum for {b_n}, it's T_n = n/2 * [2b_1 + (n - 1)e] or T_n = n(b_1 + b_n)/2.So the problem says that S_n / T_n = (3n - 1)/(n + 3). I need to find a_8 / (b_5 + b_11). Let me write down expressions for S_n and T_n in terms of their first terms and common differences.Let me start with S_n. For {a_n}, the sum S_n = n/2 [2a_1 + (n - 1)d]. Similarly, T_n = n/2 [2b_1 + (n - 1)e]. So the ratio S_n / T_n would be [n/2 (2a_1 + (n - 1)d)] / [n/2 (2b_1 + (n - 1)e)] = [2a_1 + (n - 1)d] / [2b_1 + (n - 1)e]. The n/2 cancels out.So according to the problem, [2a_1 + (n - 1)d] / [2b_1 + (n - 1)e] = (3n - 1)/(n + 3). Therefore, for all n, the ratio of these linear expressions in n is equal to (3n - 1)/(n + 3). That suggests that the two linear expressions are proportional to 3n -1 and n +3.So 2a_1 + (n - 1)d = k(3n -1) and 2b_1 + (n -1)e = k(n +3) for some constant k. Because their ratio is (3n -1)/(n +3). Alternatively, since the ratio is given as (3n -1)/(n +3), the numerator and denominator must be scalar multiples of those expressions. But since the sequences are fixed, the constants would be fixed.Alternatively, perhaps the ratio of the sums is equal to (3n -1)/(n +3). Let me think. Let me denote S_n = (3n -1) * m and T_n = (n +3) * m for some m that may depend on n? But wait, m can't depend on n because S_n and T_n are functions of n. Wait, but S_n and T_n are given as the sums for each n, so their ratio is (3n -1)/(n +3). Therefore, their ratio is a rational function in n. Since S_n and T_n are both quadratic functions of n (because sum of arithmetic sequence is quadratic in n), their ratio is a rational function. But in this case, the ratio simplifies to a linear over linear, which is a rational function of degree 1 over degree 1. However, if S_n and T_n are quadratic, their ratio would generally be a quadratic over quadratic, unless certain coefficients cancel out. So perhaps in this case, the quadratic terms in the numerator and denominator of S_n/T_n cancel out, leaving a linear/linear ratio. That must mean that the leading coefficients (the coefficients of n^2) in S_n and T_n are in the same ratio as 3/1, since (3n -1)/(n +3) tends to 3 as n approaches infinity. Let me check.Suppose S_n is a quadratic function: S_n = An^2 + Bn + C. Similarly, T_n = Dn^2 + En + F. Then S_n / T_n = (An^2 + Bn + C)/(Dn^2 + En + F). For this ratio to simplify to (3n -1)/(n +3), the quadratic terms must be proportional. So the leading coefficients A and D must satisfy A/D = 3/1, since the leading term of the numerator is 3n and the denominator is n. But when n is large, the ratio S_n/T_n ~ (A/D)n^2 / (D/n^2) = A/D. But in the given ratio, as n approaches infinity, S_n/T_n ~ 3n / n = 3. Therefore, A/D = 3. So that tells us that A = 3D.Similarly, the next terms. Let me write S_n = 3Dn^2 + Bn + C and T_n = Dn^2 + En + F. Then, (3Dn^2 + Bn + C)/(Dn^2 + En + F) = (3n -1)/(n +3). Let's perform polynomial division or equate coefficients.Cross multiplying: (3Dn^2 + Bn + C)(n +3) = (3n -1)(Dn^2 + En + F). Let me expand both sides.Left side: 3Dn^2 * n + 3Dn^2 *3 + Bn *n + Bn *3 + C *n + C *3= 3Dn^3 + 9Dn^2 + Bn^2 + 3Bn + Cn + 3C= 3Dn^3 + (9D + B)n^2 + (3B + C)n + 3CRight side: 3n * Dn^2 + 3n * En + 3n * F -1 * Dn^2 -1 * En -1 * F= 3Dn^3 + 3En^2 + 3Fn - Dn^2 - En - F= 3Dn^3 + (3E - D)n^2 + (3F - E)n - FNow, equate coefficients for each power of n.For n^3: Left side has 3D, right side has 3D. So equal.For n^2: Left side (9D + B) = Right side (3E - D)For n^1: Left side (3B + C) = Right side (3F - E)For n^0: Left side 3C = Right side (-F)So we have the following equations:1. 9D + B = 3E - D2. 3B + C = 3F - E3. 3C = -FAdditionally, we need to consider the original expressions for S_n and T_n. Since S_n and T_n are sums of arithmetic sequences, their quadratic forms are determined by the arithmetic sequences.Recall that for an arithmetic sequence, the sum S_n = n/2 [2a_1 + (n -1)d] = (n/2)(2a_1 + dn - d) = (n/2)(dn + 2a_1 - d) = (d/2)n^2 + (a_1 - d/2)n.Similarly, T_n = (e/2)n^2 + (b_1 - e/2)n.Therefore, S_n = (d/2)n^2 + (a_1 - d/2)n and T_n = (e/2)n^2 + (b_1 - e/2)n.Therefore, comparing to S_n = An^2 + Bn + C, we have A = d/2, B = (a_1 - d/2), and C = 0. Similarly, for T_n = Dn^2 + En + F, D = e/2, E = (b_1 - e/2), and F = 0.Wait, but earlier we considered C and F, but according to the formula for S_n and T_n, they don't have constant terms. So C = 0 and F = 0. Therefore, in equation 3: 3C = -F => 0 = 0. So that's okay. So equations become:1. 9D + B = 3E - D2. 3B + C = 3F - E => 3B = -E (since C = 0 and F =0)3. 3C = -F => 0=0But from the expressions of S_n and T_n, since S_n = (d/2)n^2 + (a1 - d/2)n, so A = d/2, B = a1 - d/2, and C=0. Similarly, T_n = (e/2)n^2 + (b1 - e/2)n, so D = e/2, E = b1 - e/2, F=0.So substituting into equation 1: 9D + B = 3E - D.Replace D with e/2, B with (a1 - d/2), E with (b1 - e/2):9*(e/2) + (a1 - d/2) = 3*(b1 - e/2) - e/2Simplify:Left side: (9e/2) + a1 - d/2Right side: 3b1 - 3e/2 - e/2 = 3b1 - 4e/2 = 3b1 - 2eSo equation 1 becomes:9e/2 + a1 - d/2 = 3b1 - 2eMultiply both sides by 2 to eliminate denominators:9e + 2a1 - d = 6b1 - 4eBring all terms to left side:9e + 2a1 - d -6b1 +4e =0Combine like terms:(9e +4e) +2a1 -d -6b1=013e +2a1 -d -6b1=0Equation 2: 3B = -EAgain, B = a1 -d/2, E = b1 - e/2So 3(a1 -d/2) = - (b1 - e/2)3a1 -3d/2 = -b1 + e/2Multiply both sides by 2:6a1 -3d = -2b1 + eBring all terms to left side:6a1 -3d +2b1 -e =0So now we have two equations:1. 13e +2a1 -d -6b1=02. 6a1 -3d +2b1 -e =0Hmm. So two equations with variables a1, d, b1, e. That's four variables. Hmm, but we need to find a_8/(b5 +b11). Let's express a8 and b5 + b11 in terms of a1, d, b1, e.a8 = a1 +7db5 +b11 = [b1 +4e] + [b1 +10e] = 2b1 +14eTherefore, the ratio a8/(b5 +b11) = (a1 +7d)/(2b1 +14e)So if I can express (a1 +7d) and (2b1 +14e) in terms of the variables, perhaps using the equations above, maybe relate them through the two equations we have.So let's see. Let's write equations 1 and 2 again:Equation 1: 13e +2a1 -d -6b1=0Equation 2: 6a1 -3d +2b1 -e =0Let me try to solve these equations. Let's denote variables:Let me rearrange Equation 1:2a1 - d -6b1 +13e =0 --> 2a1 -d =6b1 -13eEquation 2: 6a1 -3d +2b1 -e =0Let me express 2a1 - d from Equation 1 as 6b1 -13e. So 2a1 -d =6b1 -13e. Let's call this Equation 1a.Equation 2: 6a1 -3d +2b1 -e =0We can express 6a1 -3d as 3*(2a1 -d) = 3*(6b1 -13e) from Equation 1a. So substitute:3*(6b1 -13e) +2b1 -e =018b1 -39e +2b1 -e =0(18b1 +2b1) + (-39e -e) =020b1 -40e=0Divide both sides by 20:b1 -2e=0 --> b1=2eOkay, so from Equation 2 and substitution, we get b1=2e. Now, substitute back into Equation 1a.Equation 1a: 2a1 -d =6b1 -13e. But b1=2e, so:2a1 -d =6*(2e) -13e =12e -13e= -eThus, 2a1 -d = -e --> 2a1 =d -e --> a1=(d -e)/2So now we have expressions for a1 and b1 in terms of d and e. So let's write that:a1=(d - e)/2b1=2eTherefore, now we can express a8/(b5 +b11) in terms of d and e.From earlier:a8 =a1 +7d = (d -e)/2 +7d = (d -e +14d)/2 = (15d -e)/2b5 +b11 =2b1 +14e =2*(2e) +14e=4e +14e=18eTherefore, the ratio is (15d -e)/2 divided by 18e. So:[(15d -e)/2] / [18e] = (15d -e)/(36e)So the ratio is (15d -e)/(36e). Now, we need to find this value. To do that, we need another relation between d and e. Let's see if we can find a relation from the previous equations.We had b1=2e and a1=(d -e)/2. Let me recall that the original ratio S_n / T_n is given by (3n -1)/(n +3). Also, S_n = (d/2)n^2 + (a1 -d/2)n and T_n=(e/2)n^2 + (b1 -e/2)n.Given that S_n / T_n = (3n -1)/(n +3). Let's check for a specific value of n to find a relation between d and e.Alternatively, since we have expressions in terms of d and e, maybe substitute a1 and b1 in terms of d and e into the original ratio equation and find a relation.Let me write S_n and T_n using a1=(d -e)/2 and b1=2e.So S_n = (d/2)n^2 + [ ( (d -e)/2 - d/2 ) ]n = (d/2)n^2 + [ (d -e -d)/2 ]n = (d/2)n^2 + ( -e/2 )nSimilarly, T_n = (e/2)n^2 + [2e - e/2 ]n = (e/2)n^2 + ( (4e - e)/2 )n = (e/2)n^2 + (3e/2)nTherefore, S_n = (d/2)n^2 - (e/2)nT_n = (e/2)n^2 + (3e/2)nTherefore, S_n / T_n = [ (d/2)n^2 - (e/2)n ] / [ (e/2)n^2 + (3e/2)n ] = [ d n^2 - e n ] / [ e n^2 + 3e n ] = [n(d n - e) ] / [e n(n + 3) ] = (d n - e)/(e(n + 3))But according to the problem, S_n / T_n = (3n -1)/(n +3). Therefore:(d n - e)/(e(n +3)) = (3n -1)/(n +3)Multiply both sides by (n +3):(d n - e)/e = 3n -1Multiply both sides by e:d n - e = 3e n - eBring all terms to left side:d n - e -3e n + e =0Simplify:(d -3e)n =0This equation must hold for all n. Therefore, the coefficient of n must be zero, so d -3e =0. Thus, d=3e.Therefore, we have d=3e. Now, substitute back into the ratio (15d -e)/(36e). Since d=3e:15d -e =15*(3e) -e=45e -e=44eTherefore, the ratio becomes 44e /36e =44/36=11/9.So the answer is 11/9. Let me check my steps again to make sure I didn't make a mistake.First, we started by expressing S_n and T_n as quadratic functions. Then, equated their ratio to (3n -1)/(n +3). Cross-multiplied and expanded both sides, leading to equations relating coefficients. Then, by considering that sums of arithmetic sequences have specific forms (quadratic without constant terms), we substituted A = d/2, B = a1 -d/2, etc., leading to two equations. Then, solved those equations to express a1 and b1 in terms of d and e. Then, expressed the desired ratio in terms of d and e, and found another relation between d and e by substituting back into the original ratio equation, leading to d=3e. Then substituted d=3e into the ratio, simplifying to 11/9. Seems solid.Alternatively, maybe there's a quicker way. Let me see.Alternative approach: For arithmetic sequences, the sum S_n is of the form (n/2)(2a1 + (n-1)d). So S_n is a quadratic in n. Similarly for T_n.Given S_n / T_n = (3n -1)/(n +3). Let me denote S_n = k(3n -1) and T_n =k(n +3) for some k? But k might depend on n. Wait, but S_n and T_n are specific functions of n, so their ratio is given as (3n -1)/(n +3). Therefore, S_n = (3n -1) * c and T_n = (n +3)*c, where c is a constant (since if you take the ratio, c cancels out). Wait, but S_n and T_n are quadratic in n, while (3n -1) and (n +3) are linear. Therefore, multiplying by c would still be linear. Hence, this can't hold unless c is a linear function of n, but c can't depend on n. Therefore, this approach might not be valid.Alternatively, perhaps express S_n and T_n in terms of their average terms. S_n = n*(a1 + a_n)/2 and T_n =n*(b1 + b_n)/2. Therefore, S_n / T_n = (a1 + a_n)/(b1 + b_n) = (3n -1)/(n +3). Therefore, (a1 + a_n)/(b1 + b_n) = (3n -1)/(n +3). Therefore, (a1 + a_n) and (b1 + b_n) are linear functions proportional to (3n -1) and (n +3). Therefore, a1 + a_n = k(3n -1) and b1 + b_n =k(n +3) for some constant k. But since a1 + a_n is a linear function of n, and similarly for b1 + b_n. Let's verify.For an arithmetic sequence, a_n =a1 + (n -1)d. Therefore, a1 +a_n =a1 +a1 + (n -1)d=2a1 + (n -1)d. Similarly, b1 +b_n=2b1 + (n -1)e.So (2a1 + (n -1)d)/(2b1 + (n -1)e) = (3n -1)/(n +3). So this is the same as earlier. Therefore, setting 2a1 + (n -1)d = m(3n -1) and 2b1 + (n -1)e =m(n +3). But m must be a constant, not depending on n. Therefore, these equations must hold for all n. Therefore, the coefficients of n and the constants must match.Therefore, let's write 2a1 + (n -1)d =3m n -m. Similarly, 2b1 + (n -1)e =m n +3m.Therefore, equating coefficients:For the first equation:Coefficient of n: d =3mConstant term: 2a1 -d = -mFor the second equation:Coefficient of n: e =mConstant term: 2b1 -e =3mTherefore, from first equation: d=3m, and 2a1 -d = -m => 2a1 =d -m. Since d=3m, 2a1=3m -m=2m => a1=m.From second equation: e=m, and 2b1 -e=3m => 2b1=3m +e=3m +m=4m => b1=2m.Therefore, in terms of m, we have:a1 =md=3mb1=2me=mTherefore, the sequences are:a_n =m + (n -1)*3m= m +3m(n -1)=3m n -2mSimilarly, b_n=2m + (n -1)*m=2m +m(n -1)=m n +mNow, compute a8 and b5 +b11.a8=3m*8 -2m=24m -2m=22mb5=5m +m=6mb11=11m +m=12mTherefore, b5 +b11=6m +12m=18mThus, a8/(b5 +b11)=22m /18m=22/18=11/9.So the answer is 11/9. So this alternative approach leads to the same result, confirming the answer. Therefore, regardless of the method, the answer is 11/9. This makes me confident that the answer is correct.Final Answerboxed{dfrac{11}{9}}</think>Given two arithmetic sequences ({a_n}) and ({b_n}) with sums of the first (n) terms (S_n) and (T_n) respectively, where (frac{S_n}{T_n} = frac{3n - 1}{n + 3}), we need to find (frac{a_8}{b_5 + b_{11}}).1. Sum of Arithmetic Sequences: - The sum of the first (n) terms of an arithmetic sequence ({a_n}) is (S_n = frac{n}{2} [2a_1 + (n - 1)d]). - The sum of the first (n) terms of an arithmetic sequence ({b_n}) is (T_n = frac{n}{2} [2b_1 + (n - 1)e]).2. Ratio of Sums: - Given (frac{S_n}{T_n} = frac{3n - 1}{n + 3}), we equate the linear forms: [ frac{2a_1 + (n - 1)d}{2b_1 + (n - 1)e} = frac{3n - 1}{n + 3} ] - This implies (2a_1 + (n - 1)d = k(3n - 1)) and (2b_1 + (n - 1)e = k(n + 3)) for some constant (k).3. Solving for Coefficients: - By expanding and equating coefficients, we find: - For ({a_n}): (d = 3m) and (2a_1 - d = -m), leading to (a_1 = m). - For ({b_n}): (e = m) and (2b_1 - e = 3m), leading to (b_1 = 2m).4. Expressing Terms: - The (n)-th terms of the sequences are: - (a_n = a_1 + (n - 1)d = m + (n - 1)3m = 3mn - 2m) - (b_n = b_1 + (n - 1)e = 2m + (n - 1)m = mn + m)5. Calculating Specific Terms: - (a_8 = 3m cdot 8 - 2m = 22m) - (b_5 = 5m + m = 6m) - (b_{11} = 11m + m = 12m) - Sum (b_5 + b_{11} = 6m + 12m = 18m)6. Final Ratio: - (frac{a_8}{b_5 + b_{11}} = frac{22m}{18m} = frac{22}{18} = frac{11}{9})Thus, the final answer is (boxed{dfrac{11}{9}}).]

🔑:Given that {a_{n}} and {b_{n}} are arithmetic sequences, and S_{n} and T_{n} represent the sums of the first n terms of these sequences respectively, we are asked to find the value of frac{a_{8}}{b_{5}+b_{11}}. Given that frac{S_{n}}{T_{n}}=frac{3n-1}{n+3}, we proceed as follows:1. We start by expressing frac{a_{8}}{b_{5}+b_{11}} in terms of the sums S_{n} and T_{n}. Since a_{8} is a term in the arithmetic sequence {a_{n}} and b_{5}+b_{11} are terms in the arithmetic sequence {b_{n}}, we can utilize the properties of arithmetic sequences and their sums.2. We know that for any arithmetic sequence, the average of two terms equidistant from the beginning and end is equal to the average of the first and last term. Therefore, we can write: [ frac{a_{8}}{b_{5}+b_{11}} = frac{1}{2} cdot frac{2a_{8}}{b_{5}+b_{11}} = frac{1}{2} cdot frac{a_{1}+a_{15}}{b_{1}+b_{15}} ] This step uses the property that the average of the first and last term of an arithmetic sequence is equal to the term exactly in the middle of these two terms.3. We then relate this to the sums S_{15} and T_{15}, which are the sums of the first 15 terms of {a_{n}} and {b_{n}} respectively. Since the sum of an arithmetic sequence can be expressed as S_{n} = frac{n}{2}(first term + last term), we get: [ frac{1}{2} cdot frac{a_{1}+a_{15}}{b_{1}+b_{15}} = frac{1}{2} cdot frac{frac{15(a_{1}+a_{15})}{2}}{frac{15(b_{1}+b_{15})}{2}} = frac{1}{2} cdot frac{S_{15}}{T_{15}} ]4. Substituting the given relation frac{S_{n}}{T_{n}}=frac{3n-1}{n+3} for n=15, we find: [ frac{1}{2} cdot frac{S_{15}}{T_{15}} = frac{1}{2} times frac{3 times 15 - 1}{15 + 3} = frac{1}{2} times frac{44}{18} = frac{11}{9} ]Therefore, the final answer is boxed{frac{11}{9}}.

🧠:[<think>Okay, so I need to figure out how many different 2x2x2 cubes can be formed using four black and four white 1x1x1 cubes. The catch is that rotations of the same cube should be considered identical. Hmm, this sounds like a combinatorics problem involving group theory, maybe using Burnside's lemma? Let me recall how that works.Burnside's lemma says that the number of distinct colorings under a group of symmetries is equal to the average number of colorings fixed by each group element. So first, I need to understand the symmetry group of a cube. The cube has several rotational symmetries: rotations around axes through faces, edges, and vertices.Let me list all the rotational symmetries of a cube. There's the identity rotation (doing nothing), rotations by 90°, 180°, and 270° around the three axes through opposite faces, rotations by 180° around the six axes through opposite edges, and rotations by 120° and 240° around the four axes through opposite vertices. Let me count them:- Identity: 1- 90°, 180°, 270° around face axes: 3 axes × 3 rotations = 9- 180° around edge axes: 6 axes × 1 rotation = 6- 120°, 240° around vertex axes: 4 axes × 2 rotations = 8Total: 1 + 9 + 6 + 8 = 24 elements in the rotation group of the cube. That matches my memory; the cube has 24 rotational symmetries.Now, applying Burnside's lemma, the number of distinct colorings is (1/24) × [number of colorings fixed by each symmetry]. So I need to calculate, for each type of symmetry, how many colorings with four black and four white cubes remain unchanged when that symmetry is applied.First, the identity symmetry. Every coloring is fixed by the identity, so the number here is just the total number of colorings without considering symmetry. Since there are 8 cubes and we need to choose 4 to be black, it's C(8,4) = 70. So that's 70 fixed colorings for the identity.Next, let's consider the 90° and 270° rotations around the face axes. Let's pick one such axis, say the vertical axis (through the top and bottom faces). A 90° rotation cycles four cubes around this axis. For a coloring to be fixed by this rotation, all four cubes in each cycle must be the same color. But since we have exactly four black and four white cubes, each cycle would have to consist of one black and one white cube? Wait, no, the cycles here are the sets of cubes that are rotated into each other.Wait, in a 2x2x2 cube, a 90° rotation around a face axis actually cycles four cubes in a loop. Let me visualize the cube. Imagine looking at the top face. The four cubes on the top layer are arranged in a square. A 90° rotation would cycle the four cubes around the top face. Similarly, the four cubes on the bottom face would also be cycled. Wait, but actually, in a 2x2x2 cube, there's no distinction between layers; it's just eight cubes. Wait, maybe not. Let me think again.Wait, a 2x2x2 cube has 8 small cubes. If we rotate the entire cube 90° around a face axis, each of the four cubes on one face will move to the position of the next cube in the rotation. But actually, each 90° rotation around a face axis will form two cycles of four cubes each? Wait, no. Let's be precise.Take the cube with coordinates (x,y,z) where x, y, z ∈ {0,1}. Rotating 90° around the z-axis (the axis going through the centers of the top and bottom faces) will map (x,y,z) to (y, 1−x, z). Let's check: starting with (0,0,0), rotating 90° around z-axis would take it to (0,1,0). Then (0,1,0) would go to (1,1,0), (1,1,0) to (1,0,0), and (1,0,0) back to (0,0,0). Similarly, the top layer (z=1) has (0,0,1), (0,1,1), (1,1,1), (1,0,1) cycling in the same way.Wait, so each 90° rotation around a face axis actually cycles four cubes in a single cycle? Wait, no, each 90° rotation moves four cubes in a cycle of length 4. Wait, actually, for the entire cube, a 90° rotation around a face axis will cycle four cubes in a 4-cycle and the other four in another 4-cycle. Let me verify:Take the rotation around the z-axis. The four cubes with z=0 (bottom layer) are cycled: (0,0,0) → (0,1,0) → (1,1,0) → (1,0,0) → (0,0,0). Similarly, the four cubes with z=1 (top layer) are cycled: (0,0,1) → (0,1,1) → (1,1,1) → (1,0,1) → (0,0,1). So each 90° rotation around a face axis splits the cube into two 4-cycles.Therefore, for a coloring to be fixed by a 90° rotation, the four cubes in each cycle must be the same color. But since each cycle has four cubes, and we need exactly four black and four white cubes, each cycle would have to be entirely black or entirely white, but with both cycles being the same color? Wait, no. Wait, if we have two cycles of four cubes each, and the coloring is fixed by the rotation, then each cycle must be monochromatic. So each cycle is either all black or all white. However, since we have exactly four black and four white cubes, each cycle must be exactly half black and half white. But each cycle is four cubes. So if each cycle must be monochromatic, then one cycle would have to be all black and the other all white. Therefore, there are two possible colorings fixed by a 90° rotation: one where the bottom cycle is black and top is white, and vice versa. But wait, but in the problem statement, we need exactly four black and four white cubes. So each cycle is four cubes, so if one cycle is black and the other is white, that gives exactly four black and four white. So, for each 90° rotation, the number of colorings fixed by it is 2.But wait, let me check. Suppose we have a rotation that splits the cube into two 4-cycles. For the coloring to be fixed, each 4-cycle must be a single color. Since we have exactly four black and four white, one cycle must be black and the other white. There are two possibilities: which cycle is black and which is white. So indeed, for each 90° rotation, there are two fixed colorings.But here's a problem: in our case, the coloring must have exactly four black and four white cubes. However, in this case, if each 4-cycle is monochromatic, then yes, each 90° rotation fixes two colorings. So for each 90° rotation, the fixed colorings are two.But wait, in the 2x2x2 cube, each face rotation (90°, 180°, 270°) around each axis. Wait, so for each face axis, there are three non-identity rotations: 90°, 180°, 270°. We need to compute for each of these rotations how many colorings they fix.Wait, we considered 90°, which splits into two 4-cycles. What about 180° rotation around the face axis? Let's analyze that.A 180° rotation around the z-axis swaps (x,y,z) with (1−x,1−y,z). So, each pair of opposite cubes in the layer are swapped. Let's see:For the bottom layer (z=0), (0,0,0) swaps with (1,1,0), and (0,1,0) swaps with (1,0,0). Similarly, in the top layer (z=1), (0,0,1) swaps with (1,1,1), and (0,1,1) swaps with (1,0,1). So each 180° rotation splits the cube into four 2-cycles. Therefore, for a coloring to be fixed under this rotation, each pair must be the same color. So each 2-cycle must have both cubes the same color.Therefore, the coloring is determined by choosing a color for each pair. There are four pairs, each can be black or white. However, we need the total number of black cubes to be four. Each pair is two cubes, so if we set k pairs to black, then total black cubes are 2k. We need 2k = 4, so k=2. Therefore, the number of colorings fixed by a 180° rotation is the number of ways to choose 2 out of the 4 pairs to be black, which is C(4,2) = 6.Therefore, each 180° rotation fixes six colorings.What about 270° rotations? Wait, a 270° rotation is equivalent to a -90° rotation, so it's similar to the 90° rotation but in the opposite direction. However, in terms of cycle structure, it's the same as a 90° rotation: it splits the cube into two 4-cycles. Therefore, the number of colorings fixed by a 270° rotation is also two, just like the 90° rotation.So for each face axis, we have:- 90°: 2 fixed colorings- 180°: 6 fixed colorings- 270°: 2 fixed coloringsSince there are three face axes (x, y, z), each contributing these rotations, we need to multiply by three. So total fixed colorings for face rotations:- 3 axes × (2 + 6 + 2) = 3 × 10 = 30. Wait, no. Wait, each axis has three non-identity rotations: 90°, 180°, 270°, each of which we have calculated fixed colorings. So for each axis, it's 2 (90°) + 6 (180°) + 2 (270°) = 10. Then three axes give 30. Then adding the identity (70), so far 70 + 30 = 100.But wait, we also need to consider the edge rotations and vertex rotations. Let's continue.Next, consider the 180° rotations around edge axes. An edge axis goes through the midpoints of opposite edges. For example, take the axis going through the midpoint of the front-top edge and the midpoint of the back-bottom edge. A 180° rotation around this axis swaps certain cubes.To analyze this, let's figure out how the rotation affects the cube positions. Let me pick a specific edge axis. Let's choose the axis through the midpoints of the edge connecting (0,0,0) to (0,0,1) and the edge connecting (1,1,0) to (1,1,1). Wait, actually, edge axes go through midpoints of opposite edges. Let's consider a rotation around an axis through the midpoint of the edge between (0,0,0) and (0,1,0) and the midpoint of the edge between (1,0,1) and (1,1,1). Wait, this might be complex. Alternatively, perhaps using coordinate transformations.Alternatively, consider that a 180° rotation around an edge axis will swap pairs of cubes. For example, such a rotation might swap two pairs of cubes and leave two other pairs swapped as well. Let me check.Wait, let's take the edge axis along the line connecting the midpoint of the front top edge and the midpoint of the back bottom edge. Wait, maybe it's easier to note that a 180° rotation around an edge axis will decompose the cube into four 2-cycles. Let me try to find how.Suppose we have a cube, and we rotate 180° around an edge axis. For each cube, its position is moved to another position. Let me label the cubes as follows:Positions (x,y,z) where x, y, z ∈ {0,1}.Let's take an edge axis, say, the axis through the midpoint of the edge between (0,0,0) and (0,1,0) and the midpoint of the edge between (1,0,1) and (1,1,1). A 180° rotation around this axis. Let's see how it affects the coordinates.Alternatively, maybe use a specific rotation matrix. But perhaps it's easier to visualize.After a 180° rotation around an edge axis, each cube is either fixed or swapped with another cube. Since the rotation is 180°, the order is 2, so cycles can be of length 1 or 2. However, in a 2x2x2 cube, there's no central cube, so no fixed points. Therefore, all cubes are in 2-cycles.So the rotation decomposes the 8 cubes into four 2-cycles. Therefore, similar to the 180° face rotation, the number of colorings fixed by this rotation is the number of colorings where each pair in the 2-cycles are the same color. So each 2-cycle must be monochromatic. Therefore, each pair is either both black or both white, and we need exactly four black and four white cubes. So the number of colorings fixed is the number of ways to choose two pairs to be black (each contributing two cubes) and the other two pairs to be white. Since there are four pairs, we need to choose two pairs to be black, which is C(4,2)=6. Therefore, each 180° edge rotation fixes six colorings.Since there are six edge axes, each contributing one 180° rotation, that's 6 rotations × 6 fixed colorings = 36.Wait, hold on. Wait, no. Wait, in the cube, the edge axes are axes through midpoints of opposite edges. There are six edges on a cube, but they are grouped into three pairs of opposite edges. Wait, actually, in the cube, how many edge axes are there? Each edge axis is determined by a pair of opposite edges. Since each edge has an opposite edge, the number of edge axes is 6 edges / 2 = 3. Wait, no, wait: for a cube, each edge axis is determined by a line through the midpoints of two opposite edges. Since the cube has 12 edges, but each axis goes through two opposite edges. Wait, actually, no. Wait, each edge is part of a pair with one opposite edge, so the number of edge axes is 6. Wait, this is confusing.Wait, let me recall: the cube has 6 faces, 12 edges, 8 vertices. The axes of rotation for the cube are:- 3 axes through centers of opposite faces,- 6 axes through midpoints of opposite edges,- 4 axes through opposite vertices.Wait, no, that's incorrect. Wait, the rotational symmetry axes for a cube are:- 3 axes through the centers of opposite faces (each axis corresponds to a pair of opposite faces),- 6 axes through the midpoints of opposite edges (each axis corresponds to a pair of opposite edges),- 4 axes through opposite vertices (each axis goes through two opposite vertices).But wait, actually, in standard cube symmetry descriptions, there are three types of rotation axes:1. Face axes: 3 axes, each through the centers of opposite faces.2. Edge axes: 6 axes, each through the midpoints of opposite edges.3. Vertex axes: 4 axes, each through opposite vertices.Wait, but in our case, we considered the face axes (3), edge axes (6), and vertex axes (4). But when we counted the group elements earlier, the 180° edge rotations correspond to the edge axes. However, each edge axis allows a 180° rotation. So there are 6 edge axes, each contributing one non-identity rotation (180°). Thus, there are 6 such rotations in the rotation group. Similarly, the vertex axes allow 120° and 240° rotations. So that matches the prior count: 1 (identity) + 9 (face rotations: 3 axes × 3 rotations) + 6 (edge rotations: 6 axes × 1 rotation) + 8 (vertex rotations: 4 axes × 2 rotations) = 24 elements.Therefore, for each of the six edge axes, we have a 180° rotation. Each such rotation decomposes the cube into four 2-cycles. As we saw earlier, for each such rotation, the number of fixed colorings is C(4,2)=6. Therefore, total fixed colorings from edge rotations: 6 × 6 = 36.Moving on to vertex axes. These are axes through opposite vertices, allowing 120° and 240° rotations. Let's analyze these.A 120° rotation around a vertex axis cycles three cubes at a time. Wait, but in a 2x2x2 cube, let's see.Take a vertex axis, say through the vertices (0,0,0) and (1,1,1). A 120° rotation around this axis will cycle three coordinates. Let me think of how the rotation affects the cube positions.Alternatively, perhaps it's easier to note that a 120° rotation around a vertex axis will decompose the cube into two cycles of length 3 and fix two cubes? Wait, no. Wait, in a 2x2x2 cube, all eight cubes are involved in the rotation.Wait, actually, maybe in a 120° rotation around a vertex axis, the cubes are partitioned into cycles. Let's take the rotation axis through (0,0,0) and (1,1,1). A 120° rotation around this axis will rotate the three coordinates in a cyclic manner. Let me parametrize the rotation. For example, such a rotation might map (x,y,z) to (z,x,y) or some permutation. Let's check.Suppose we perform a 120° rotation around the axis from (0,0,0) to (1,1,1). This rotation cycles the x, y, and z axes. So a point (x,y,z) is rotated to (z,x,y). Let's verify how this affects the cube positions.Take the cube at (0,0,0). Applying the rotation, it becomes (0,0,0) since all coordinates are 0. Wait, but (0,0,0) is on the rotation axis, so it should be fixed. Similarly, (1,1,1) is also on the axis and remains fixed. Wait, but in a 2x2x2 cube, the only cubes on the axis are (0,0,0) and (1,1,1). The other cubes are off the axis. Let's check another cube: (1,0,0). Applying the rotation (x,y,z) → (z,x,y), we get (0,1,0). Then (0,1,0) becomes (0,0,1). Then (0,0,1) becomes (1,0,0). So these three cubes form a 3-cycle: (1,0,0) → (0,1,0) → (0,0,1) → (1,0,0). Similarly, the cube (0,1,0) is part of this cycle.Similarly, the cube (1,1,0) would rotate to (0,1,1), then to (1,0,1), then to (1,1,0). So that's another 3-cycle: (1,1,0) → (0,1,1) → (1,0,1) → (1,1,0).Therefore, a 120° rotation around a vertex axis fixes two cubes ((0,0,0) and (1,1,1)) and cycles the other six in two 3-cycles.Wait, but in a 2x2x2 cube, all eight cubes are at the corners. If two cubes are fixed, and the remaining six are divided into two 3-cycles. But each 3-cycle would contain three cubes. However, there are only six non-fixed cubes, so two 3-cycles of three each. But 3+3=6, which works. But each 3-cycle must consist of three cubes. Let's check with coordinates:First 3-cycle: (1,0,0) → (0,1,0) → (0,0,1) → (1,0,0)Second 3-cycle: (1,1,0) → (0,1,1) → (1,0,1) → (1,1,0)Yes, that accounts for all six non-axis cubes. The two fixed cubes are (0,0,0) and (1,1,1).Therefore, a 120° rotation around a vertex axis fixes two cubes and has two 3-cycles. For a coloring to be fixed by this rotation, the two fixed cubes must be their own color, and each 3-cycle must be monochromatic. However, since we have exactly four black and four white cubes, let's see:The two fixed cubes can be colored arbitrarily, but to maintain the coloring under rotation, each 3-cycle must be all the same color. Let's compute the number of colorings fixed by such a rotation.The two fixed cubes can be:- Both black: Then the remaining two black cubes must come from the two 3-cycles. But each 3-cycle must be either all black or all white. However, each 3-cycle has three cubes. If we color a 3-cycle black, that would contribute three black cubes, but we only have two remaining black cubes. Similarly, coloring a 3-cycle white would contribute zero. Since 3 is more than 2, it's impossible. Therefore, both fixed cubes cannot be black.- Both white: Similarly, remaining four black cubes would have to come from the two 3-cycles. But each 3-cycle can only contribute 0 or 3 black cubes. So 3a + 3b = 4, where a and b are 0 or 1. But 3(a + b) = 4 has no integer solutions. Therefore, both fixed cubes cannot be white.- One black and one white: Suppose one fixed cube is black and the other is white. Then we need three more black cubes and three more white cubes. These must come from the two 3-cycles. Each 3-cycle must be monochromatic. Let’s denote:Let’s say the fixed cubes are B and W. Then, each 3-cycle can be either B or W. Let’s denote the two 3-cycles as C1 and C2. Each can be B or W. The total number of black cubes would then be 1 (fixed) + 3*(number of 3-cycles colored black). Similarly for white.We need the total black cubes to be 4. So:1 + 3*(number of black 3-cycles) = 4Therefore, 3*(number of black 3-cycles) = 3Thus, number of black 3-cycles = 1Similarly, number of white 3-cycles = 1Therefore, the only possibility is that one 3-cycle is black and the other is white. Since there are two 3-cycles, there are 2 possibilities: C1 is black and C2 is white, or vice versa. However, the 3-cycles are distinguishable by their positions, so swapping their colors would give different colorings. But wait, under rotation, are the 3-cycles distinguishable?Wait, in the rotation, the two 3-cycles are separate, so coloring one black and the other white is a distinct coloring. However, the question is about colorings fixed by the rotation. So if we swap the colors of the two 3-cycles, would that be a different coloring? But if we fix the rotation, then for the coloring to be fixed, each 3-cycle must be monochromatic. Therefore, the coloring must assign a color to each 3-cycle such that the entire coloring is fixed. Therefore, if we have one 3-cycle black and the other white, then under rotation, since the rotation cycles the 3-cycles, wait, does the rotation swap the 3-cycles?Wait, no. Wait, the 120° rotation cycles the cubes within each 3-cycle but keeps the 3-cycles themselves intact. Wait, no. Wait, when you rotate the cube 120° around the vertex axis, each 3-cycle is mapped to itself. Because the rotation cycles the cubes within each 3-cycle. So for example, in the first 3-cycle (1,0,0), (0,1,0), (0,0,1), the rotation moves each cube to the next position in the cycle. Similarly for the second 3-cycle. Therefore, the 3-cycles are preserved as sets. Therefore, to have a coloring fixed by the rotation, each 3-cycle must be monochromatic. Additionally, the fixed cubes can be colored independently.But given that we need one fixed cube black and one white, and each 3-cycle must be monochromatic. Then, as we calculated, the only way to get four black cubes is: 1 (fixed black) + 3 (one 3-cycle black) + 0 (the other 3-cycle white). Similarly, the fixed white cube plus the other 3-cycles. Wait, let me rephrase.If one fixed cube is black and the other is white, then the remaining three black cubes must come from one 3-cycle (which has three cubes), and the remaining three white cubes must come from the other 3-cycle. So in this case, the number of colorings is 2 (choices for which fixed cube is black) multiplied by 2 (choices for which 3-cycle is black). Wait, no. The fixed cubes are specific: one is (0,0,0) and the other is (1,1,1). So if we decide to color (0,0,0) black and (1,1,1) white, then we need to color one 3-cycle black and the other white. Similarly, if we color (0,0,0) white and (1,1,1) black, then we also need to color one 3-cycle black and the other white. Therefore, the total number of colorings fixed by a 120° rotation is 2 (choices for which fixed cube is black) multiplied by 2 (choices for which 3-cycle is black), totaling 4 colorings.Wait, let me check again. The fixed cubes are (0,0,0) and (1,1,1). So:Case 1: (0,0,0) is black, (1,1,1) is white. Then we need three more black cubes and three more white cubes. So one 3-cycle must be black (adding 3) and the other 3-cycle must be white (adding 0). Wait, but the 3-cycles each have three cubes. If we set one 3-cycle to black, that gives 3 blacks, and the other to white, giving 3 whites. Combined with the fixed cubes (1 black, 1 white), total blacks: 1 + 3 = 4, whites: 1 + 3 = 4. Perfect. So for this case, there are two possibilities: which 3-cycle is black.Similarly, Case 2: (0,0,0) is white, (1,1,1) is black. Then again, we need three more whites and three more blacks. So again, one 3-cycle must be white and the other black. Again, two possibilities.Therefore, in total, for each 120° rotation, the number of fixed colorings is 2 (cases for fixed cubes) × 2 (choices of 3-cycle) = 4.Wait, but hold on. The two 3-cycles are distinguishable. For example, in the rotation axis we considered, the first 3-cycle is (1,0,0), (0,1,0), (0,0,1), and the second is (1,1,0), (0,1,1), (1,0,1). So choosing which 3-cycle to color black is a distinct choice. Therefore, for each case of fixed cubes (either (0,0,0) black and (1,1,1) white or vice versa), there are two colorings: one where the first 3-cycle is black and the second is white, and another where the first is white and the second is black. But wait, no, actually, if we fix the rotation, the two 3-cycles are fixed as well. So the choice is whether to color the first 3-cycle black or the second. However, in the problem, we are counting colorings up to rotation, but here we are counting colorings fixed by a specific rotation. So if a coloring is fixed by the rotation, then the assignment of colors to the 3-cycles must be consistent with the rotation. However, since the rotation cycles the positions within each 3-cycle, as long as each 3-cycle is monochromatic, the coloring is fixed. Therefore, for each 120° rotation, the number of colorings fixed is 2 (fixed cube colorings) × 2 (3-cycle colorings) = 4. So 4 colorings fixed per 120° rotation.Similarly, a 240° rotation is just a 120° rotation in the opposite direction. However, the cycle structure is the same: two fixed cubes and two 3-cycles. Therefore, the number of fixed colorings is also 4 per 240° rotation.Each vertex axis has two non-identity rotations: 120° and 240°, so for each of the four vertex axes, we have two rotations, each fixing four colorings. Therefore, total fixed colorings for vertex rotations: 4 axes × 2 rotations × 4 colorings = 32.Wait, but let's confirm this. Each vertex axis contributes two rotations (120° and 240°), each fixing four colorings. There are four vertex axes, so total fixed colorings: 4 × 2 × 4 = 32.Therefore, adding up all contributions:- Identity: 70- Face rotations (90°, 180°, 270°): 3 axes × (2 + 6 + 2) = 3 × 10 = 30- Edge rotations (180°): 6 axes × 6 = 36- Vertex rotations (120°, 240°): 4 axes × 2 rotations × 4 = 32Total fixed colorings: 70 + 30 + 36 + 32 = 168.Then, applying Burnside's lemma, the number of distinct colorings is 168 divided by 24 (the order of the rotation group), so 168 / 24 = 7.Wait, so the answer is 7? Let me verify this because that seems low, but considering the constraints, maybe.But let me cross-check. Let's think if there's another way to compute this. Alternatively, perhaps using the formula for cube colorings with two colors considering symmetries. However, in our case, we have exactly four black and four white cubes, so it's a specific case.But according to Burnside's lemma, if all calculations are correct, it's 7. Let me check if each step was correct.First, identity: 70. Correct.Face rotations:- 90° and 270°: 2 each. Three axes, each with two non-180° rotations (90° and 270°). So 3 axes × 2 rotations × 2 colorings = 12. Wait, but earlier I calculated 3 axes × (2 + 6 + 2) = 30. Wait, no, let me redo:Each face axis has three rotations: 90°, 180°, 270°. For each axis:- 90°: 2 fixed colorings- 180°: 6 fixed colorings- 270°: 2 fixed coloringsTotal per axis: 10. Three axes: 30. Yes.Edge rotations:6 axes, each 180°, each fixes 6 colorings. 6 × 6 = 36.Vertex rotations:4 axes, each with two rotations (120°, 240°), each fixing 4 colorings. 4 × 2 × 4 = 32.Total: 70 + 30 + 36 + 32 = 168.168 / 24 = 7. Yes.Hmm, that seems plausible. Let me see if there's any mistake in cycle structures or fixed colorings.For the 120° vertex rotations, we concluded that there are four fixed colorings per rotation. Let me confirm with an example. Suppose we fix (0,0,0) as black and (1,1,1) as white. Then one 3-cycle is black and the other is white. Since each 3-cycle has three cubes, that's 1 + 3 = 4 black cubes and 1 + 3 = 4 white cubes. Correct. Similarly, swapping the colors of the fixed cubes and the 3-cycles gives another valid coloring. So four colorings per rotation.Another check: For edge rotations (180°), each splits the cube into four 2-cycles. Choosing two pairs to be black gives C(4,2)=6. Correct.Face rotations: 90° and 270° rotations split into two 4-cycles. Each 4-cycle must be monochromatic, so two colorings (one cycle black, the other white). Correct.So, unless there's a miscalculation in the arithmetic, the total is 168, leading to 7 distinct colorings.Alternatively, let's check OEIS or other references. Wait, I can't access external resources, but I recall that the number of distinct colorings of a cube with four black and four white faces is 7. Wait, no, wait, but this is about coloring the entire cube's vertices or cubes. Wait, the question is about a 2x2x2 cube composed of eight small cubes, each colored black or white, with exactly four of each. The number of distinct colorings under rotation is 7. That seems familiar. For example, the number of distinct colorings of a cube with four black and four white faces is known to be 7. Wait, but here we're talking about the entire cube, composed of small cubes, not just the faces. Wait, but perhaps the count is similar. Wait, actually, in the case of the entire cube, it's equivalent to coloring the vertices of a cube with two colors, up to rotation, with exactly four of each color. Yes, that is a classic problem, and the answer is indeed 7. Therefore, this corroborates the result.Therefore, the answer should be 7.Final Answerboxed{7}</think>To determine the number of distinct 2 times 2 times 2 cubes that can be formed using four black and four white 1 times 1 times 1 cubes, considering rotations as identical, we use Burnside's lemma. The cube has 24 rotational symmetries, which include rotations around face axes, edge axes, and vertex axes. 1. Identity rotation: All 70 colorings (choosing 4 out of 8 cubes to be black) are fixed.2. 90° and 270° face rotations: Each splits the cube into two 4-cycles. For each rotation, there are 2 fixed colorings (one 4-cycle black, the other white). There are 3 axes, each with 2 non-180° rotations, contributing 3 times 2 times 2 = 12 fixed colorings.3. 180° face rotations: Each splits the cube into four 2-cycles. For each rotation, there are 6 fixed colorings (choosing 2 out of 4 pairs to be black). There are 3 axes, contributing 3 times 6 = 18 fixed colorings.4. 180° edge rotations: Each splits the cube into four 2-cycles. For each rotation, there are 6 fixed colorings. There are 6 axes, contributing 6 times 6 = 36 fixed colorings.5. 120° and 240° vertex rotations: Each splits the cube into two 3-cycles and fixes two cubes. For each rotation, there are 4 fixed colorings (choosing one 3-cycle to be black and the other white, with one fixed cube black and the other white). There are 4 axes, each with 2 rotations, contributing 4 times 2 times 4 = 32 fixed colorings.Summing all fixed colorings:- Identity: 70- Face rotations: 12 + 18 = 30- Edge rotations: 36- Vertex rotations: 32Total fixed colorings: 70 + 30 + 36 + 32 = 168. Applying Burnside's lemma, the number of distinct colorings is frac{168}{24} = 7.[boxed{7}],<think>Okay, so the problem is about figuring out how many different 2x2x2 cubes can be formed using four black and four white 1x1x1 cubes. And we have to consider that two cubes are the same if one can be rotated to look like the other. Hmm, this sounds like a combinatorics problem with some group theory involved, maybe using Burnside's lemma? Let me recall how that works.Burnside's lemma states that the number of distinct colorings under group actions (rotations, in this case) is equal to the average number of colorings fixed by each group action. So, first, I need to find all the possible rotations of a cube and then determine how many colorings remain unchanged by each rotation. Finally, average those numbers over all the rotations. That should give the total number of distinct cubes.Alright, step one: enumerate all the rotational symmetries of a cube. A cube has 24 rotational symmetries. Let me verify that. The cube can be rotated in various ways:- Identity rotation: doing nothing. That's 1.- Rotations about axes through centers of opposite faces. There are three pairs of opposite faces. For each pair, we can rotate 90°, 180°, or 270°. So for each axis, 3 non-identity rotations, totaling 3*3=9.- Rotations about axes through opposite edges. There are six such axes (each axis goes through the midpoints of opposite edges). For each axis, we can rotate 180°. That's 6 rotations.- Rotations about axes through opposite vertices. There are four such axes (each passing through a pair of opposite vertices). For each axis, we can rotate 120° and 240°, so 2 rotations per axis, totaling 4*2=8.Adding those up: 1 (identity) + 9 (face axes) + 6 (edge axes) + 8 (vertex axes) = 24. Yep, that checks out.So there are 24 rotational symmetries. Now, according to Burnside, the number of distinct colorings is equal to 1/24 times the sum over each symmetry of the number of colorings fixed by that symmetry.But first, let's confirm the total number of colorings without considering rotations. Since we have eight 1x1x1 cubes, four black and four white, the total number of colorings is the combination of 8 choose 4, which is 70. So there are 70 colorings before considering rotations.But we need to account for rotations. So now, we need to compute for each of the 24 rotations, how many colorings are fixed by that rotation. Then sum those numbers and divide by 24.Let me break down each type of rotation and compute the number of fixed colorings.Starting with the identity rotation. The identity rotation doesn't change anything, so all colorings are fixed. Therefore, fixed colorings here are 70.Next, rotations about face axes. Let's consider a rotation by 90° and 270°. For a 90° rotation, a cube has a cycle structure that cycles four faces. Similarly for 270°, which is just the inverse of 90°, so it has the same cycle structure. Then a 180° rotation swaps pairs of faces.Wait, hold on. Actually, when considering the permutation of the individual cubes (the 1x1x1 cubes) under a rotation, we need to see how the rotation affects each small cube's position.Wait, the 2x2x2 cube is made of eight small cubes. Each rotation will permute these eight cubes. To find the number of colorings fixed by a rotation, the coloring must be the same after the permutation. That is, all the cycles in the permutation must have the same color. Since we have two colors, each cycle can be either all black or all white. But we need the total number of black cubes to be four. So for each rotation, we need to compute the cycle structure of the permutation induced on the eight small cubes, then determine how many color assignments satisfy the cycle constraints and sum to four black cubes.Therefore, for each rotation, we have to find the cycle decomposition of the permutation of the eight positions. Then, for each cycle, all positions in a cycle must be the same color. So the number of fixed colorings is the number of ways to assign a color to each cycle such that the total number of black cubes (sum of the lengths of cycles colored black) is four.Therefore, the key is to find the cycle structure for each type of rotation.Let me categorize the rotations and find their cycle decompositions.First, identity rotation: all eight positions are fixed, so eight cycles of length 1. Therefore, the number of fixed colorings is C(8,4)=70, which matches our previous statement.Next, 90° and 270° rotations about face axes. Let's take an axis through the centers of the top and bottom faces. Rotating 90° around this axis. How does this affect the small cubes?In a 2x2x2 cube, each face has four small cubes. A 90° rotation will cycle these four cubes on each layer. But there are two layers (top and bottom). Wait, actually, in a 2x2x2 cube, rotating the top face 90° would actually cycle four cubes. Wait, but hold on, the entire cube is being rotated. So if we rotate the entire cube 90° around the vertical axis (through top and bottom faces), then each of the four side faces will move. Let me visualize this.Imagine looking at the top face. The four cubes on the top layer will each move to the next position clockwise. Similarly, the four cubes on the bottom layer will each move to the next position clockwise when viewed from the top. But since the cube is 2x2x2, each 90° rotation about the vertical axis actually cycles four cubes in a single cycle? Wait, no.Wait, in a 2x2x2 cube, a 90° rotation about a face axis (say, the vertical axis) will permute the four vertical columns. Each column is two cubes (top and bottom). So, rotating 90° will cycle the four columns. Each column is a pair of cubes. Wait, but if we rotate the entire cube 90°, each corner of the top face moves to the next corner, and similarly for the bottom face. Therefore, each corner cube is part of a cycle.Wait, perhaps it's better to label the positions of the 2x2x2 cube. Let's assign coordinates to the small cubes. Let the cube have coordinates (x,y,z) where x, y, z ∈ {0,1}. So there are eight positions.A rotation about the z-axis (vertical axis) by 90° would map (x,y,z) to (y, 1-x, z). Wait, is that right? Let me think. If we rotate the cube 90° around the z-axis, then the x and y coordinates will change. For a 90° rotation, the positive x-axis becomes the positive y-axis. So, the point (1,0,0) moves to (0,1,0). Similarly, (1,1,0) moves to (1,1,0)? Wait, no. Wait, in a 2x2x2 cube, rotating the entire cube 90° around the z-axis will cycle four positions in the top layer and similarly four positions in the bottom layer.Wait, perhaps each 90° rotation about a face axis decomposes into two cycles of four cubes each? Wait, that can't be, since 4 + 4 = 8, which is the total number of cubes. Let me check.Take the top layer: positions (0,0,1), (1,0,1), (1,1,1), (0,1,1). Rotating 90° around the z-axis would cycle these four positions. Similarly, the bottom layer: (0,0,0), (1,0,0), (1,1,0), (0,1,0) would also cycle. So each 90° rotation about the z-axis decomposes into two cycles of length 4: one cycle on the top layer and one cycle on the bottom layer.Similarly, a 180° rotation about the z-axis would decompose into four cycles of length 2: swapping opposite positions on each layer. For example, in the top layer, (0,0,1) swaps with (1,1,1), and (1,0,1) swaps with (0,1,1). Similarly for the bottom layer.Similarly, a 270° rotation is the inverse of 90°, so it's also two cycles of length 4, but cycled in the opposite direction.So, summarizing:- 90° and 270° rotations about a face axis: cycle decomposition is two 4-cycles.- 180° rotations about a face axis: cycle decomposition is four 2-cycles.Now, how about rotations about edge axes? These are axes through the midpoints of opposite edges. For example, take the axis through the midpoints of the front-top and back-bottom edges. A 180° rotation around this axis swaps certain cubes.Wait, perhaps it's better to figure out the cycle decomposition for such a rotation. Let's consider a specific edge axis. Let me pick an axis going through the midpoint of the edge between (0,0,0) and (0,0,1) and the midpoint of the edge between (1,1,0) and (1,1,1). A 180° rotation about this axis.How does this affect the cubes? Let me try to visualize. The rotation will swap some cubes. For example, the cube at (0,0,0) will move to (1,1,1), and vice versa. Similarly, (0,0,1) swaps with (1,1,0). Then the other cubes: maybe (0,1,0) swaps with (1,0,1), and (0,1,1) swaps with (1,0,0). So, this would be four 2-cycles.Wait, let me check. If we rotate 180° around this edge axis, each pair of opposite cubes are swapped. So, four pairs, hence four 2-cycles. Therefore, cycle decomposition is four 2-cycles.Similarly, each 180° edge rotation has four 2-cycles.Lastly, rotations about vertex axes. These are axes passing through opposite vertices. For example, take the axis through (0,0,0) and (1,1,1). A 120° rotation around this axis. How does this affect the cube?A 120° rotation around a body diagonal (vertex axis) cycles three axes. Let's see. The rotation would cycle three coordinates. For example, a 120° rotation around the axis from (0,0,0) to (1,1,1) would map (1,0,0) to (0,1,0), then to (0,0,1), and back. Similarly, (0,1,0) to (0,0,1) to (1,0,0), etc. So, this results in two cycles of length 3? Wait, but there are eight cubes. Wait, maybe not.Wait, in a 2x2x2 cube, a 120° rotation around a body diagonal will cycle three cubes and another three cubes, leaving two cubes fixed? Wait, let me think.Wait, the axis goes through (0,0,0) and (1,1,1). The point (0,0,0) is fixed, as is (1,1,1). Then, the other six points form two cycles of three each. Let's check:Take the point (1,0,0). Rotating 120° around the body diagonal, where does it go? The rotation matrix for 120° around the body diagonal (1,1,1) can be represented as permuting the coordinates. Specifically, (x,y,z) → (z,x,y). So (1,0,0) becomes (0,1,0). Then (0,1,0) becomes (0,0,1), and then (0,0,1) becomes (1,0,0). So that's a 3-cycle: (1,0,0) → (0,1,0) → (0,0,1) → (1,0,0). Similarly, the point (0,0,1) is in this cycle. Wait, but (0,0,1) was supposed to be part of the cycle. Wait, but (0,0,1) goes to (1,0,0), which is correct. Then the other three points: (0,1,1), (1,1,0), and (1,0,1). Let's apply the rotation. (0,1,1) becomes (1,0,1). Then (1,0,1) becomes (1,1,0). Then (1,1,0) becomes (0,1,1). So that's another 3-cycle: (0,1,1) → (1,0,1) → (1,1,0) → (0,1,1).Therefore, a 120° rotation around a body diagonal axis results in two 3-cycles and two fixed points. Similarly, a 240° rotation (which is the inverse) would have the same cycle structure, just the cycles reversed.But wait, in our case, the 2x2x2 cube only has eight small cubes. The fixed points under this rotation are (0,0,0) and (1,1,1). The other six are split into two 3-cycles. However, 3+3+2=8. So each 120° or 240° rotation about a vertex axis has two 3-cycles and two fixed points.But wait, 3-cycles? But our cube has only eight small cubes. So each 3-cycle would involve three cubes, but how does that work? Wait, but in a 2x2x2 cube, each corner is a small cube. So the coordinates (0,0,0) is a corner, (1,0,0) is another, etc. So when we do a 120° rotation around the body diagonal, we are permuting these corners.But in the 2x2x2 cube, each corner is a distinct small cube. So the permutation is as described: two fixed corners and two 3-cycles of the other six. But wait, 3+3+2=8. So each 3-cycle has three elements. Wait, 3+3+2=8? 3+3 is 6, plus 2 is 8. Yes, correct. So each 120° or 240° rotation has two 3-cycles and two fixed points.But hold on, for Burnside's lemma, when we have cycles in the permutation, each cycle must be colored uniformly. So for a rotation with cycle structure of two 3-cycles and two fixed points, the number of colorings fixed by this rotation would be calculated by assigning a color to each cycle. Each 3-cycle must be all black or all white, and each fixed point can be independently colored black or white. But we have the constraint that exactly four cubes are black and four are white.Wait, but if we have two 3-cycles and two fixed points, the cycles of length 3 must be colored either black or white. However, the fixed points can each be colored independently. So the total number of colorings fixed by such a rotation would be the number of color assignments where:- Each 3-cycle is assigned a color (black or white).- Each fixed point is assigned a color (black or white).But the total number of black cubes would be 3*a + b + c, where a is 0 or 1 (color of the first 3-cycle), and b and c are 0 or 1 (colors of the fixed points). Similarly, the total number of white cubes would be 3*(1 - a) + (1 - b) + (1 - c).But since the total number of black cubes must be 4, we have:3a + b + c = 4.Given that a is either 0 or 1, and b and c are either 0 or 1.Let's consider possible cases:Case 1: a = 0 (both 3-cycles are white). Then, b + c = 4. But since b and c can be 0 or 1, their sum can be at most 2. Impossible.Case 2: a = 1 (both 3-cycles are black). Then, 3*1 + b + c = 4 => b + c = 1. So the possible (b, c) pairs are (0,1) and (1,0). Therefore, there are 2 possibilities.Therefore, the number of colorings fixed by a 120° or 240° rotation is 2.Wait, but let's check: if both 3-cycles are black (each contributing 3 black cubes), then total black cubes from the cycles would be 6, but we need only 4. Wait, hold on! Wait, a 3-cycle has three cubes. If a 3-cycle is colored black, that's three black cubes. If both 3-cycles are colored black, that's 6 black cubes. Then, the fixed points would need to be colored white to not exceed 4. But in the equation above, if a=1 (both 3-cycles black), then b + c = 4 - 6 = -2, which is impossible. Wait, there's a miscalculation here.Wait, no. Wait, the equation 3a + b + c = 4 is incorrect. Let's re-examine.Each 3-cycle is assigned a color. If a 3-cycle is colored black, it contributes 3 black cubes. If it's colored white, it contributes 0. Similarly, each fixed point is colored black or white, contributing 1 or 0. But in the problem, we have exactly four black and four white cubes. So the total number of black cubes must equal 4.If we have two 3-cycles and two fixed points. Let me denote:- Let C1 and C2 be the two 3-cycles.- Let F1 and F2 be the two fixed points.Each cycle (C1, C2) can be colored either black (B) or white (W). Each fixed point (F1, F2) can be colored B or W.Total black cubes = 3*(number of 3-cycles colored B) + (number of fixed points colored B).We need this total to equal 4.Let x be the number of 3-cycles colored B (x = 0, 1, or 2).Let y be the number of fixed points colored B (y = 0, 1, or 2).Then, 3x + y = 4.Possible solutions:x=0: y=4 → Not possible, since y can be at most 2.x=1: 3*1 + y = 4 → y=1. Possible.x=2: 3*2 + y = 4 → 6 + y =4 → y= -2. Impossible.Thus, the only solution is x=1 and y=1. So we need to color exactly one of the two 3-cycles black, and one of the two fixed points black.How many ways is this possible?First, choose which 3-cycle to color black: 2 choices (C1 or C2).Then, choose which fixed point to color black: 2 choices (F1 or F2).Therefore, total number of colorings fixed by such a rotation is 2*2=4.Wait, but that contradicts my previous conclusion. So where did I go wrong before?Earlier, I considered a=1 (both 3-cycles colored black), but that gives 6 black cubes, which is too many. But the correct approach is to color only one 3-cycle black and one fixed point black, resulting in 3 + 1 = 4 black cubes.Therefore, the number of fixed colorings is 2 (choices for 3-cycle) * 2 (choices for fixed point) = 4.So, for each 120° or 240° rotation, the number of fixed colorings is 4.Okay, that makes sense. So, that's the correction.Now, moving on. Let's summarize the cycle structures and fixed coloring counts:1. Identity: 1 rotation, cycle type 1^8. Fixed colorings: C(8,4)=70.2. 90° and 270° rotations about face axes: 6 rotations (3 axes, each with two non-identity rotations: 90° and 270°). Each has cycle type 4^2. For each such rotation, the cycle decomposition is two 4-cycles. To have fixed colorings, each 4-cycle must be uniformly colored. So we have two cycles, each can be black or white, but total black cubes must be 4. Therefore, each 4-cycle has 4 cubes, so if both are colored black, that's 8 black cubes, which is too many. If one is colored black and the other white, that's 4 + 0 = 4. So only one possibility: color one 4-cycle black and the other white. How many ways? There are two cycles, so two choices (which cycle to color black). But since the two cycles are distinguishable only by their position, which is determined by the rotation axis. Wait, but in the case of a 90° rotation, the two 4-cycles are the top layer and the bottom layer. However, if we color the top cycle black and bottom cycle white, or vice versa. But since the rotation is part of the cube's symmetry, the two 4-cycles are in different orientations. Wait, but actually, under the rotation, the two 4-cycles are swapped? No, a 90° rotation cycles the top layer and cycles the bottom layer, but they remain separate.Wait, let me think. If we have a rotation that decomposes into two 4-cycles (top and bottom), then coloring the top 4-cycle black and the bottom 4-cycle white is different from coloring the bottom black and top white. But under the rotation, these two colorings would be equivalent? Wait, no. Because the rotation itself is a symmetry; if a coloring is fixed by the rotation, then applying the rotation doesn't change the coloring. Therefore, if we color the top 4-cycle black and the bottom white, after a 90° rotation, the top cycle (which was black) would rotate to itself, but since it's a cycle, all its cubes must be the same color. Wait, but if the coloring is fixed by the rotation, then the color of each cycle must be preserved. So if the top cycle is black, after rotation, it must still be black. Similarly for the bottom.Therefore, for a fixed coloring under a 90° rotation, the two 4-cycles (top and bottom) must each be uniformly colored, but the colors can be chosen independently, as long as the total number of black cubes is 4. Since each 4-cycle has 4 cubes, coloring one cycle black and the other white gives exactly 4 black cubes. Therefore, there are two such colorings: top black, bottom white; or top white, bottom black.But wait, but the problem is that the entire cube is rotated, so top and bottom are relative. However, in the case of a rotation about the vertical axis (say, z-axis), the top and bottom are fixed. Wait, but actually, the rotation is around the axis, so the top and bottom layers are preserved as layers. Therefore, the two 4-cycles are distinguishable (top vs. bottom). Therefore, the number of colorings fixed by a 90° rotation is 2.Wait, but each 4-cycle is a separate cycle, so you can assign color to each cycle. But since we have four black cubes, each 4-cycle can be either black or white. If one is black (4 cubes) and the other is white (4 cubes), that gives exactly four black cubes. There are two possibilities: top black, bottom white; or top white, bottom black. Therefore, fixed colorings per 90° or 270° rotation: 2.But wait, but in a 2x2x2 cube, the top and bottom layers under a 90° rotation are separate. So, if you color the top layer black and the bottom white, this is fixed by the rotation because rotating the top layer 90° doesn't change its color (all black), and same for the bottom. Similarly for the reverse. So indeed, two colorings fixed per 90° or 270° rotation.So, for each of the 90° and 270° rotations (6 in total: 3 axes, 2 rotations each), fixed colorings per rotation: 2. So total fixed colorings for these: 6*2=12.Next, 180° rotations about face axes: 3 rotations (one per face axis). Each has cycle type 2^4 (four 2-cycles). For such a rotation, each 2-cycle must be uniformly colored. Therefore, each 2-cycle can be black or white. We need the total number of black cubes to be 4. Since each 2-cycle has 2 cubes, if we color k 2-cycles black, total black cubes would be 2k. We need 2k =4, so k=2. Therefore, the number of colorings fixed by a 180° face rotation is the number of ways to choose 2 out of 4 cycles to color black. That's C(4,2)=6. Therefore, each 180° face rotation fixes 6 colorings. Since there are 3 such rotations, total fixed colorings: 3*6=18.Next, 180° edge rotations: 6 rotations (axes through midpoints of opposite edges). Each has cycle type 2^4 (four 2-cycles), similar to the 180° face rotations. Therefore, the same logic applies: to have 4 black cubes, we need to color 2 of the four 2-cycles black. Number of ways: C(4,2)=6. Therefore, each edge rotation fixes 6 colorings. Since there are 6 edge rotations, total fixed colorings: 6*6=36.Finally, 120° and 240° vertex rotations: 8 rotations (4 axes, each with two rotations: 120° and 240°). Each has cycle type 3^2 + 1^2 (two 3-cycles and two fixed points). As established earlier, the number of fixed colorings for each such rotation is 4. Therefore, total fixed colorings: 8*4=32.Now, summing up all the fixed colorings:- Identity: 70- 90° and 270° face rotations: 12- 180° face rotations: 18- 180° edge rotations: 36- 120° and 240° vertex rotations: 32Total sum: 70 + 12 + 18 + 36 + 32 = let's compute step by step.70 + 12 = 8282 + 18 = 100100 + 36 = 136136 + 32 = 168Total sum is 168. Then, divide by 24 (number of symmetries) to get the number of distinct colorings: 168 / 24 = 7.Wait, 168 divided by 24 is 7. So the answer is 7.But let me verify once again to make sure I didn't make a mistake in the calculations.Recalculating:Identity: 70Face rotations (90°, 270°): 6 rotations * 2 =12Face rotations (180°): 3 rotations *6=18Edge rotations (180°):6 rotations *6=36Vertex rotations (120°,240°):8 rotations *4=32Total sum:70+12=82; 82+18=100; 100+36=136; 136+32=168. 168/24=7.Yes, that's correct.Therefore, the number of distinct 2x2x2 cubes that can be formed with four black and four white 1x1x1 cubes, considering rotations as identical, is 7.Final Answerboxed{7}</think>To determine the number of distinct 2 times 2 times 2 cubes that can be formed using four black and four white 1 times 1 times 1 cubes, considering rotations as identical, we use Burnside's lemma. The key steps are as follows:1. Enumerate the rotational symmetries of a cube: - Identity: 1 rotation - 90° and 270° rotations about face axes: 6 rotations - 180° rotations about face axes: 3 rotations - 180° rotations about edge axes: 6 rotations - 120° and 240° rotations about vertex axes: 8 rotations2. Calculate the number of colorings fixed by each type of rotation: - Identity rotation: All 70 colorings are fixed. - 90° and 270° face rotations: Each rotation cycles two 4-cycles. Only 2 colorings (one 4-cycle black, the other white) are fixed. Total: (6 times 2 = 12). - 180° face rotations: Each rotation cycles four 2-cycles. Choosing 2 out of 4 cycles to be black gives ( binom{4}{2} = 6 ) colorings. Total: (3 times 6 = 18). - 180° edge rotations: Similar to 180° face rotations, total: (6 times 6 = 36). - 120° and 240° vertex rotations: Each rotation has two 3-cycles and two fixed points. Coloring one 3-cycle and one fixed point black gives 4 colorings. Total: (8 times 4 = 32).3. Sum the fixed colorings and divide by the number of symmetries: - Total fixed colorings: (70 + 12 + 18 + 36 + 32 = 168) - Number of distinct colorings: ( frac{168}{24} = 7 )Thus, the number of distinct 2 times 2 times 2 cubes is (boxed{7}).]

🔑:【Analysis】First, analyze the case where a color is on the same face. Then, enumerate the cases where the white on the same face becomes 3 and then 2.【Solution】Solution: According to the problem:(1) White at the bottom positions 5, 6, 7, 8 is 1 case (same face).(2) White at the bottom positions 5, 6, 7, the fourth block can be at positions 1,2,4, a total of 3 cases.(3) White at the bottom positions 5, 6, the top can be at positions 1,4 or 1,3, a total of 2 cases.(4) White at the bottom positions 5, 7, the top can be at position 1,3, a total of 1 case.1+3+2+1=7 (cases).Therefore, the answer is: 7

🧠:[<think>Okay, so I need to solve this problem where the volume and surface area of a sphere are numerically equal. Then, under that condition, find the surface area of a cube inscribed in the sphere. Hmm, let's break this down step by step.First, let me recall the formulas for the volume and surface area of a sphere. The volume ( V ) of a sphere is given by ( V = frac{4}{3}pi r^3 ), and the surface area ( S ) is ( S = 4pi r^2 ). The problem states that these two are numerically equal. So, even though they have different units in real life, here we're just considering their numerical values. That means I can set the two equations equal to each other:[frac{4}{3}pi r^3 = 4pi r^2]Alright, now I need to solve for the radius ( r ). Let's simplify this equation. Both sides have ( 4pi r^2 ), so maybe I can divide both sides by ( 4pi r^2 ) to cancel those terms out. Let me try that:Divide both sides by ( 4pi r^2 ):[frac{frac{4}{3}pi r^3}{4pi r^2} = frac{4pi r^2}{4pi r^2}]Simplifying the left side: The 4 cancels with 4, ( pi ) cancels with ( pi ), and ( r^3 ) divided by ( r^2 ) leaves ( r ). So, left with ( frac{1}{3}r ).On the right side, everything cancels out, leaving 1.So:[frac{1}{3}r = 1]Multiplying both sides by 3 gives ( r = 3 ). So the radius of the sphere is 3 units. Okay, that was straightforward. Let me just check my steps again. Yes, dividing both sides by ( 4pi r^2 ) makes sense because ( r ) can't be zero (otherwise, there's no sphere), so division is valid. So, radius is indeed 3.Now, the next part is about a cube inscribed in this sphere. I need to find the surface area of the cube. First, let's visualize this. A cube inscribed in a sphere means that all the cube's vertices lie on the sphere's surface. So, the sphere's diameter is equal to the cube's space diagonal. The space diagonal of a cube relates to its side length. Let me recall that formula. For a cube with side length ( a ), the space diagonal ( d ) is ( d = asqrt{3} ). This comes from the Pythagorean theorem extended to three dimensions. So, the space diagonal of the cube is the diameter of the sphere. Since the radius of the sphere is 3, the diameter is twice that, so 6 units. Therefore, the space diagonal of the cube is 6. So:[asqrt{3} = 6]Solving for ( a ):[a = frac{6}{sqrt{3}} = frac{6sqrt{3}}{3} = 2sqrt{3}]So, the side length of the cube is ( 2sqrt{3} ). Now, the surface area of a cube is ( 6a^2 ). Let's compute that. Plugging in ( a = 2sqrt{3} ):First, square ( a ):[(2sqrt{3})^2 = 4 times 3 = 12]Then multiply by 6:[6 times 12 = 72]So, the surface area of the cube is 72. Wait a second, let me verify all the steps again to make sure I didn't make any mistakes. Starting with the sphere's volume and surface area equal, solving for radius ( r = 3 ). Then, cube inscribed in the sphere: space diagonal equals sphere's diameter (which is 6). Space diagonal of cube is ( asqrt{3} ), so ( a = 6/sqrt{3} ), rationalizing gives ( 2sqrt{3} ). Then surface area ( 6a^2 = 6*(12) = 72 ). Seems solid. But let me just cross-check the surface area. If the cube has side length ( 2sqrt{3} ), each face is a square with area ( (2sqrt{3})^2 = 4*3=12 ). There are 6 faces, so 6*12=72. Yep, that checks out. I think that's all. So the answer should be 72.Final AnswerThe surface area of the cube is boxed{72}.</think>Given that the volume and surface area of a sphere are numerically equal, we start by setting the formulas for volume and surface area equal to each other:[frac{4}{3}pi r^3 = 4pi r^2]Dividing both sides by (4pi r^2):[frac{1}{3}r = 1]Solving for (r):[r = 3]The radius of the sphere is 3 units. The diameter of the sphere, which is twice the radius, is 6 units. This diameter is equal to the space diagonal of the inscribed cube. For a cube with side length (a), the space diagonal is (asqrt{3}). Setting this equal to the diameter of the sphere:[asqrt{3} = 6]Solving for (a):[a = frac{6}{sqrt{3}} = 2sqrt{3}]The surface area of a cube is given by (6a^2). Substituting (a = 2sqrt{3}):[(2sqrt{3})^2 = 4 times 3 = 12][6 times 12 = 72]Thus, the surface area of the cube is (boxed{72}).,<think>Okay, so I have this problem where the volume and surface area of a sphere are numerically equal. I need to find the surface area of a cube that's inscribed in that sphere. Hmm, let's break this down step by step.First, let me recall the formulas for the volume and surface area of a sphere. The volume ( V ) of a sphere is given by ( frac{4}{3}pi r^3 ), and the surface area ( S ) is ( 4pi r^2 ). The problem says these two are numerically equal, so I can set them equal to each other:[ frac{4}{3}pi r^3 = 4pi r^2 ]Okay, let's solve for ( r ). Let me simplify the equation. First, divide both sides by ( 4pi r^2 ) to reduce it. Assuming ( r neq 0 ), which makes sense because a sphere with radius 0 wouldn't have any volume or surface area.Dividing both sides by ( 4pi r^2 ):[ frac{r}{3} = 1 ]So multiplying both sides by 3 gives ( r = 3 ). That seems straightforward. So the radius of the sphere is 3 units.Now, the next part is about a cube inscribed in the sphere. When a cube is inscribed in a sphere, the sphere's diameter is equal to the cube's space diagonal. Let me visualize that. The cube is perfectly fit inside the sphere, so the farthest two vertices of the cube are at the ends of the sphere's diameter.The space diagonal of a cube relates to its side length. If the cube has a side length ( a ), then the space diagonal ( d ) is ( asqrt{3} ). I remember this from the Pythagorean theorem in three dimensions. So the space diagonal is equal to the sphere's diameter, which is ( 2r ). Since we found ( r = 3 ), the diameter is ( 6 ).So setting up the equation:[ asqrt{3} = 2r ][ asqrt{3} = 6 ][ a = frac{6}{sqrt{3}} ]Wait, simplifying that:[ a = frac{6}{sqrt{3}} ]But it's better to rationalize the denominator. Multiply numerator and denominator by ( sqrt{3} ):[ a = frac{6sqrt{3}}{3} = 2sqrt{3} ]So the side length of the cube is ( 2sqrt{3} ).Now, the surface area of a cube is ( 6a^2 ). Let me compute that.First, compute ( a^2 ):( (2sqrt{3})^2 = 4 times 3 = 12 )Then multiply by 6:( 6 times 12 = 72 )So the surface area of the cube is 72.Wait, let me double-check my steps. The sphere's radius is 3, so diameter is 6. The space diagonal of the cube is 6, so using ( asqrt{3} = 6 ), solving for ( a ), get ( 2sqrt{3} ). Then surface area is ( 6a^2 = 6*(4*3) = 6*12 = 72 ). That seems right.But let me verify the initial condition again. The volume and surface area of the sphere are numerically equal. Volume is ( frac{4}{3}pi (3)^3 = frac{4}{3}pi*27 = 36pi ). Surface area is ( 4pi (3)^2 = 36pi ). Yep, they are equal. So radius 3 is correct. Then cube inscribed in sphere with radius 3: space diagonal is 6, leading to side length ( 2sqrt{3} ), surface area 72. So the answer should be 72. Therefore, the surface area is ( boxed{72} ).Final AnswerThe surface area of the inscribed cube is boxed{72}.</think>Given that the volume and surface area of a sphere are numerically equal, we start by setting the formulas for volume and surface area equal to each other:[frac{4}{3}pi r^3 = 4pi r^2]Solving for ( r ), we divide both sides by ( 4pi r^2 ):[frac{r}{3} = 1 implies r = 3]The radius of the sphere is 3 units. The diameter of the sphere, which is twice the radius, is ( 2r = 6 ).For a cube inscribed in the sphere, the space diagonal of the cube equals the diameter of the sphere. Let ( a ) be the side length of the cube. The space diagonal of the cube is ( asqrt{3} ), so we have:[asqrt{3} = 6 implies a = frac{6}{sqrt{3}} = 2sqrt{3}]The surface area of a cube is given by ( 6a^2 ). Substituting ( a = 2sqrt{3} ):[a^2 = (2sqrt{3})^2 = 4 times 3 = 12]Thus, the surface area of the cube is:[6 times 12 = 72]The surface area of the inscribed cube is (boxed{72}).]

🔑:If r is the radius of the sphere, then according to the given conditionfrac{4}{3} pi r^{3}=4 pi r^{2} quad ; quad r=3 text{LE}(where LE denotes the unit of length). For the side length a of the cube inscribed in the sphere with radius r, we havea=frac{2}{3} sqrt{3} r=2 sqrt{3}(this follows immediately from the fact that the space diagonal of the cube is equal to the diameter of the sphere: sqrt{3} a=d=2 r ).Thus, for the surface area A of the cube, we have: A=6 a^{2}=6 cdotleft(2 sqrt{3}right)^{2}=72 text{LE}^{2}.

🧠:[<think>Okay, let's try to tackle this problem step by step. First, let me make sure I understand the problem correctly.We have an ellipse Γ with two foci, F₁ and F₂. There's a line that passes through F₁ and intersects the ellipse at points P and Q. We are given two conditions:1. The distance from P to F₂ is equal to the distance between F₁ and F₂, so |PF₂| = |F₁F₂|.2. Three times the distance from P to F₁ is equal to four times the distance from Q to F₁, so 3|PF₁| = 4|QF₁|.We need to find the ratio of the minor axis to the major axis of the ellipse Γ.Alright, let's recall some basic properties of an ellipse. The standard equation of an ellipse centered at the origin with major axis along the x-axis is (x²/a²) + (y²/b²) = 1, where a is the semi-major axis, b is the semi-minor axis, and the distance between the foci is 2c, where c² = a² - b². The foci are located at (±c, 0).In this problem, the foci are F₁ and F₂. Let's denote F₁ as (-c, 0) and F₂ as (c, 0). The major axis length is 2a, the minor axis length is 2b, so the ratio we need is (2b)/(2a) = b/a. Therefore, our target is to find b/a.Given that the line passes through F₁ and intersects the ellipse at P and Q. Let's denote the line as passing through F₁, so we can parametrize it. But first, maybe it's better to set up coordinates.Let me set up a coordinate system where the center of the ellipse is at the origin (0,0), the major axis is along the x-axis, so F₁ is at (-c, 0) and F₂ is at (c, 0). The line passes through F₁, so its equation can be written in the form y = m(x + c), where m is the slope. However, since we don't know the slope, maybe it's better to parametrize the points P and Q.Alternatively, since the line passes through F₁, we can write parametric equations for the line. Let's let the direction vector of the line be arbitrary. Let me think.But maybe using the properties of an ellipse would be better. Remember that for any point on the ellipse, the sum of distances to the two foci is 2a. So for both points P and Q, |PF₁| + |PF₂| = 2a and |QF₁| + |QF₂| = 2a.We are given that |PF₂| = |F₁F₂|. Since |F₁F₂| is the distance between the foci, which is 2c. Therefore, |PF₂| = 2c.But we also know from the ellipse definition that |PF₁| + |PF₂| = 2a. So substituting |PF₂| = 2c, we have |PF₁| + 2c = 2a => |PF₁| = 2a - 2c.Similarly, for point Q, we don't have such a specific condition, but we know that 3|PF₁| = 4|QF₁|. Since |PF₁| = 2a - 2c, then 3(2a - 2c) = 4|QF₁| => |QF₁| = (3/4)(2a - 2c) = (3/2)(a - c).But from the ellipse definition, |QF₁| + |QF₂| = 2a, so |QF₂| = 2a - |QF₁| = 2a - (3/2)(a - c) = 2a - (3a/2 - 3c/2) = (4a/2 - 3a/2) + 3c/2 = (a/2) + (3c)/2.So now, we have expressions for |PF₁|, |PF₂|, |QF₁|, |QF₂|. But how do we connect this to the ratio b/a?Perhaps we need to use coordinates. Let's consider the parametric equations of the ellipse. Alternatively, since the line passes through F₁ and intersects the ellipse at P and Q, we can write the equation of the line and solve for the intersection points with the ellipse.Let me proceed step by step.First, since the line passes through F₁ = (-c, 0) and has some slope m, its equation is y = m(x + c). We can substitute this into the ellipse equation and find the points P and Q.But maybe using parametric coordinates or vector approach would be better. Let's see.The ellipse equation is (x²/a²) + (y²/b²) = 1. The line passes through (-c, 0) with slope m: y = m(x + c). Let's substitute y into the ellipse equation.So substituting, we get:x²/a² + [m²(x + c)²]/b² = 1Multiply through by a²b²:b²x² + a²m²(x + c)^2 = a²b²Expand the (x + c)^2 term:b²x² + a²m²(x² + 2cx + c²) = a²b²Expand the left-hand side:b²x² + a²m²x² + 2a²m²c x + a²m²c² = a²b²Combine like terms:x²(b² + a²m²) + 2a²m²c x + (a²m²c² - a²b²) = 0This is a quadratic in x. Let's denote coefficients:A = b² + a²m²B = 2a²m²cC = a²m²c² - a²b²So quadratic equation: Ax² + Bx + C = 0The solutions will give us the x-coordinates of P and Q. Since the line passes through F₁ (-c, 0), which is one of the foci, but since F₁ is not on the ellipse (unless the ellipse is degenerate, which it isn't because it has two foci), so F₁ is inside the ellipse. Therefore, the line passes through F₁ and intersects the ellipse at two points P and Q. Hence, the quadratic equation should have two roots, corresponding to x-coordinates of P and Q.But since the line passes through (-c, 0), perhaps one of the roots is x = -c? Wait, but substituting x = -c into the ellipse equation:At x = -c, y = m(-c + c) = 0, so the point (-c, 0) is on the line. But (-c, 0) is F₁, which is inside the ellipse. So if we substitute x = -c into the ellipse equation:(-c)²/a² + 0²/b² = c²/a². But since for an ellipse, c < a, so c²/a² < 1, so the point F₁ is inside the ellipse, not on it. Therefore, when we solve the quadratic equation, the two solutions correspond to points P and Q on the ellipse, and the line passes through F₁ which is inside the ellipse.Therefore, the quadratic equation in x should have two distinct real roots. The sum and product of roots can be related to the coefficients:For quadratic Ax² + Bx + C = 0, sum of roots x₁ + x₂ = -B/A, product x₁x₂ = C/A.But we need to relate this to the distances |PF₁| and |QF₁|.Alternatively, maybe we can parametrize the line with a parameter t and express points P and Q in terms of t, then apply the given conditions.Alternatively, perhaps using parametric equations of the ellipse. The parametric equations of the ellipse are x = a cosθ, y = b sinθ. Then, the line passing through F₁ (-c, 0) can be expressed in parametric form as:x = -c + t cosφy = 0 + t sinφwhere φ is the angle the line makes with the x-axis, and t is a parameter. Then, substituting into the ellipse equation:[(-c + t cosφ)^2]/a² + [ (t sinφ)^2 ] /b² = 1Expanding this:(c² - 2ct cosφ + t² cos²φ)/a² + (t² sin²φ)/b² = 1Multiply through by a²b²:b²(c² - 2ct cosφ + t² cos²φ) + a² t² sin²φ = a²b²Expand:b²c² - 2b²c t cosφ + b² t² cos²φ + a² t² sin²φ = a²b²Bring all terms to left-hand side:-2b²c t cosφ + t² (b² cos²φ + a² sin²φ) + b²c² - a²b² = 0Factor the terms:t² (b² cos²φ + a² sin²φ) - 2b²c t cosφ + b²(c² - a²) = 0But c² = a² - b², so substitute:t² (b² cos²φ + a² sin²φ) - 2b²c t cosφ + b²( (a² - b²) - a² ) = 0Simplify the constant term:b²( -b² ) = -b⁴So the equation becomes:t² (b² cos²φ + a² sin²φ) - 2b²c t cosφ - b⁴ = 0This is a quadratic in t. Let me denote coefficients:A = b² cos²φ + a² sin²φB = -2b²c cosφC = -b⁴So quadratic equation: A t² + B t + C = 0Solutions:t = [2b²c cosφ ± sqrt( (2b²c cosφ)^2 - 4*A*(-b⁴) ) ] / (2A)Simplify discriminant:D = (4b⁴c² cos²φ) + 4A b⁴= 4b⁴ [c² cos²φ + A]But A = b² cos²φ + a² sin²φ, so:D = 4b⁴ [c² cos²φ + b² cos²φ + a² sin²φ]= 4b⁴ [ cos²φ(c² + b²) + a² sin²φ ]But c² = a² - b², so substituting:D = 4b⁴ [ cos²φ(a² - b² + b²) + a² sin²φ ]= 4b⁴ [ cos²φ a² + a² sin²φ ]= 4b⁴ a² [ cos²φ + sin²φ ]= 4b⁴ a² (1)= 4a²b⁴Therefore, sqrt(D) = 2ab²Thus, solutions:t = [2b²c cosφ ± 2ab²] / (2A)Simplify numerator and denominator:t = [2b²(c cosφ ± a)] / [2(b² cos²φ + a² sin²φ)]Cancel 2:t = [b²(c cosφ ± a)] / [b² cos²φ + a² sin²φ]Therefore, the two solutions for t are:t₁ = [b²(c cosφ + a)] / [b² cos²φ + a² sin²φ]t₂ = [b²(c cosφ - a)] / [b² cos²φ + a² sin²φ]These correspond to the parameters t for points P and Q. Since t represents the distance from F₁ along the line, the positive t would be in one direction and negative t in the other. However, depending on the angle φ, the points could be on either side.But perhaps we can take t₁ as the positive root and t₂ as the negative. Let's check:If φ = 0, the line is along the x-axis towards positive x. Then, t₁ = [b²(c + a)] / [b² + 0] = [b²(c + a)] / b² = c + a. The point would be at x = -c + (c + a) = a, y = 0, which is the vertex (a, 0). The other root t₂ = [b²(c - a)] / [b²] = c - a. Then x = -c + (c - a) = -a, y = 0, which is the other vertex (-a, 0). But in reality, the line along the x-axis from F₁ (-c, 0) would intersect the ellipse at (a, 0) and (-a, 0). Wait, but (-a, 0) is the other vertex, but starting from F₁ (-c, 0), moving in the negative x-direction would reach (-a, 0), which is a distance of |-a - (-c)| = c - a. But since a > c, this is negative, so maybe the parameter t is signed. Hmm, perhaps the parametrization allows t to be positive or negative, so points P and Q can be on either side of F₁.But in our problem, points P and Q are both on the line passing through F₁ and intersecting the ellipse. So, one of them is on one side of F₁, and the other is on the opposite side.But how to determine which is which? Maybe it doesn't matter; we can just assign t₁ and t₂ as the two parameters, and then compute |PF₁| and |QF₁|.But since t is the parameter in the parametric equation of the line, |PF₁| would be |t|, because the parametric equation is x = -c + t cosφ, y = 0 + t sinφ. So the distance from F₁ (-c, 0) to a point (x, y) on the line is sqrt[ (x + c)^2 + y^2 ] = sqrt[ (t cosφ)^2 + (t sinφ)^2 ] = |t| sqrt( cos²φ + sin²φ ) = |t|.Therefore, |PF₁| = |t₁| and |QF₁| = |t₂|. But in the quadratic solution, t₁ and t₂ can be positive or negative. Let's see.Looking at t₁ = [b²(c cosφ + a)] / [b² cos²φ + a² sin²φ]Given that a > c (since c = sqrt(a² - b²)), c cosφ + a is always positive because a > c, and cosφ can be at most 1. So c cosφ + a > c*(-1) + a. Since a > c, even if cosφ is -1, c cosφ + a = a - c > 0. Therefore, t₁ is positive.For t₂ = [b²(c cosφ - a)] / [b² cos²φ + a² sin²φ]Here, c cosφ - a. Since c < a, and cosφ <= 1, c cosφ - a <= c - a < 0. Therefore, t₂ is negative.Therefore, |PF₁| = t₁ (since t₁ is positive) and |QF₁| = |t₂| = -t₂ (since t₂ is negative). Therefore, given 3|PF₁| = 4|QF₁|, this becomes 3t₁ = 4(-t₂).So let's write that:3t₁ = 4(-t₂)But t₁ and t₂ are the roots from above. Let's substitute t₁ and t₂.First, compute t₁ and t₂:t₁ = [b²(c cosφ + a)] / D, where D = b² cos²φ + a² sin²φt₂ = [b²(c cosφ - a)] / DSo, 3t₁ = 4(-t₂)=> 3 [b²(c cosφ + a)/D ] = 4 [ - b²(c cosφ - a)/D ]Multiply both sides by D:3b²(c cosφ + a) = -4b²(c cosφ - a)Divide both sides by b²:3(c cosφ + a) = -4(c cosφ - a)Expand both sides:3c cosφ + 3a = -4c cosφ + 4aBring all terms to left-hand side:3c cosφ + 3a + 4c cosφ - 4a = 0Combine like terms:(3c cosφ + 4c cosφ) + (3a - 4a) = 07c cosφ - a = 0Thus,7c cosφ = a=> cosφ = a/(7c)Alright, so this gives us a relation between cosφ and a, c. Let's keep this in mind: cosφ = a/(7c)Now, the other condition is |PF₂| = |F₁F₂| = 2c.Let's compute |PF₂|. Since F₂ is at (c, 0), and P is a point on the ellipse. Let's express P in terms of the parametric equations.From the parametric line equations:x = -c + t₁ cosφy = 0 + t₁ sinφSo coordinates of P are:x_P = -c + t₁ cosφy_P = t₁ sinφThen, distance |PF₂| is sqrt[(x_P - c)^2 + y_P^2]Compute this:sqrt[ ( -c + t₁ cosφ - c )² + ( t₁ sinφ )² ]= sqrt[ ( -2c + t₁ cosφ )² + ( t₁ sinφ )² ]Square both sides:( -2c + t₁ cosφ )² + ( t₁ sinφ )² = (2c)^2 = 4c² (since |PF₂| = 2c)Expand the left-hand side:(4c² - 4c t₁ cosφ + t₁² cos²φ) + t₁² sin²φ= 4c² - 4c t₁ cosφ + t₁² (cos²φ + sin²φ)= 4c² - 4c t₁ cosφ + t₁²Therefore:4c² - 4c t₁ cosφ + t₁² = 4c²Subtract 4c² from both sides:-4c t₁ cosφ + t₁² = 0Factor:t₁(-4c cosφ + t₁) = 0Since t₁ ≠ 0 (as P is not F₁), we have:-4c cosφ + t₁ = 0=> t₁ = 4c cosφBut we already have an expression for t₁ from earlier:t₁ = [b²(c cosφ + a)] / [b² cos²φ + a² sin²φ]Therefore:4c cosφ = [b²(c cosφ + a)] / [b² cos²φ + a² sin²φ]Cross-multiplying:4c cosφ [b² cos²φ + a² sin²φ] = b²(c cosφ + a)Let me denote cosφ as k for simplicity. So let k = cosφ, then sin²φ = 1 - k².From earlier, we had cosφ = a/(7c), so k = a/(7c).Therefore, let's substitute k into the equation:4c k [b² k² + a²(1 - k²)] = b²(c k + a)First, compute the left-hand side:4c k [b² k² + a² - a² k²] = 4c k [a² + k²(b² - a²)]But since in an ellipse, b² = a² - c², so substitute b² = a² - c²:Left-hand side becomes:4c k [a² + k²(a² - c² - a²)] = 4c k [a² + k²(-c²)] = 4c k [a² - c² k²]Similarly, right-hand side:b²(c k + a) = (a² - c²)(c k + a)So equation becomes:4c k (a² - c² k²) = (a² - c²)(c k + a)Now, substitute k = a/(7c):Left-hand side:4c * (a/(7c)) [a² - c²*(a/(7c))²] = (4a/7)[a² - c²*(a²)/(49c²)] = (4a/7)[a² - (a²)/49] = (4a/7)*(48a²/49) = (4a/7)*(48a²/49) = (4*48 a³)/(7*49) = (192 a³)/(343)Right-hand side:(a² - c²)[c*(a/(7c)) + a] = (a² - c²)[a/7 + a] = (a² - c²)*(8a/7) = (8a/7)(a² - c²)So set left-hand side equal to right-hand side:(192 a³)/343 = (8a/7)(a² - c²)Multiply both sides by 343/(8a):(192 a³)/343 * 343/(8a) = (8a/7)(a² - c²) * 343/(8a)Simplify left-hand side:(192 a³)/(8a) = 24a²Right-hand side:(8a/7) * (a² - c²) * (343)/(8a) = (343/7)(a² - c²) = 49(a² - c²)Therefore:24a² = 49(a² - c²)Simplify:24a² = 49a² - 49c²Bring all terms to left-hand side:24a² - 49a² + 49c² = 0=> -25a² + 49c² = 0=> 49c² = 25a²=> c² = (25/49)a²=> c = (5/7)aSince c = sqrt(a² - b²), we have:sqrt(a² - b²) = (5/7)aSquare both sides:a² - b² = (25/49)a²=> b² = a² - (25/49)a² = (49/49 - 25/49)a² = (24/49)a²Therefore, the ratio b/a is sqrt(24/49) = (2*sqrt(6))/7But the problem asks for the ratio of the minor axis to the major axis, which is 2b/(2a) = b/a. So the answer is (2√6)/7.Wait, but let me check sqrt(24/49). sqrt(24)/7 = 2*sqrt(6)/7. Yes, that's correct. So the ratio is 2√6/7.Therefore, the ratio of the minor axis to the major axis is 2√6⁄7.Final AnswerThe ratio of the minor axis to the major axis of the ellipse Γ is boxed{dfrac{2sqrt{6}}{7}}.</think>Given an ellipse (Gamma) with foci (F_1) and (F_2), a line passing through (F_1) intersects (Gamma) at points (P) and (Q). We know (|PF_2| = |F_1F_2|) and (3|PF_1| = 4|QF_1|). We need to find the ratio of the minor axis to the major axis of the ellipse.1. Ellipse Properties: - The distance between the foci (|F_1F_2| = 2c). - The sum of distances from any point on the ellipse to the foci is (2a). - The relationship between semi-major axis (a), semi-minor axis (b), and distance (c) is (c^2 = a^2 - b^2).2. Given Conditions: - (|PF_2| = 2c) implies (|PF_1| = 2a - 2c). - (3|PF_1| = 4|QF_1|) gives (|QF_1| = frac{3}{4}(2a - 2c) = frac{3}{2}(a - c)).3. Using Parametric Equations: - The line through (F_1) intersects the ellipse at (P) and (Q). Using parametric equations and solving the quadratic equation for the line intersecting the ellipse, we find the parameters (t_1) and (t_2) corresponding to points (P) and (Q).4. Key Equations: - From the condition (3|PF_1| = 4|QF_1|), we derive (cosphi = frac{a}{7c}). - Using the distance condition (|PF_2| = 2c), we solve for (c) in terms of (a) and find (c = frac{5}{7}a).5. Calculating (b/a): - Using (c^2 = a^2 - b^2), we substitute (c = frac{5}{7}a) to get (b^2 = frac{24}{49}a^2). - Therefore, (frac{b}{a} = sqrt{frac{24}{49}} = frac{2sqrt{6}}{7}).The ratio of the minor axis to the major axis of the ellipse (Gamma) is (boxed{dfrac{2sqrt{6}}{7}}).,<think>Okay, let's see. I need to find the ratio of the minor axis to the major axis of an ellipse given some conditions about distances from the foci. Hmm. Let me recall some properties of ellipses. The standard equation of an ellipse is (x^2/a^2) + (y^2/b^2) = 1, where a is the semi-major axis, b is the semi-minor axis, and the distance between the center and each focus is c, with c^2 = a^2 - b^2. So the ratio they want is 2b/2a = b/a. So I need to find b/a.Given that the foci are F1 and F2. A line passes through F1 and intersects the ellipse at points P and Q. They tell me that |PF2| = |F1F2|, which is 2c, since the distance between the foci is 2c. Also, 3|PF1| = 4|QF1|. So I need to use these conditions to find b/a.First, maybe I should set up coordinates. Let me place the ellipse in standard position, centered at the origin. Then F1 is at (-c, 0), F2 at (c, 0). The line passes through F1, so its equation is y = m(x + c), where m is the slope. This line intersects the ellipse at points P and Q, which include F1? Wait, the line passes through F1, but F1 is not on the ellipse unless... Wait, no, the ellipse has foci at F1 and F2, but the foci are inside the ellipse, not on it. So the line passes through F1 and intersects the ellipse at P and Q. So F1 is not one of the intersection points. Wait, but if the line passes through F1, could F1 be one of the points? Let me check. If I plug F1 into the ellipse equation: ( (-c)^2 ) / a^2 + 0 = c^2 / a^2. But since c < a, this is less than 1, so F1 is inside the ellipse. Therefore, the line passes through F1 and intersects the ellipse at two points P and Q. So P and Q are two points on the ellipse, on a line passing through F1. Got it.Given |PF2| = |F1F2| = 2c. So the distance from P to F2 is equal to 2c. Also, 3|PF1| = 4|QF1|. So the distance from P to F1 is 4/3 times the distance from Q to F1. Wait, no: 3|PF1| = 4|QF1| => |PF1| = (4/3)|QF1|. So PF1 is longer than QF1. Since both P and Q are on the line through F1, maybe P is further away from F1 than Q? Or maybe the line passes through F1, goes out to P, then comes back through Q? Wait, lines intersect the ellipse at two points. Since the line passes through F1, which is inside the ellipse, the line must enter the ellipse at one point, go through F1, and exit at another point. So maybe the two intersection points are on opposite sides of F1? Hmm. So let's imagine the line passing through F1, entering the ellipse at Q, passing through F1, and exiting at P. So then QF1 is the distance from Q to F1, and PF1 is the distance from P to F1. So PF1 would be longer than QF1, which matches the given ratio: PF1 = (4/3) QF1. So maybe Q is between F1 and P.But perhaps it's better to parametrize the line and find coordinates for P and Q. Let me try that approach.Let me set up coordinates with the ellipse centered at the origin, major axis along the x-axis. So equation is (x^2/a^2) + (y^2/b^2) = 1. Foci at (-c, 0) and (c, 0), where c = sqrt(a^2 - b^2).The line passes through F1 = (-c, 0). Let me parametrize the line with a parameter t. Let me set up parametric equations. Since the line passes through (-c, 0), I can write it as:x = -c + t * cosθy = 0 + t * sinθWhere θ is the angle the line makes with the x-axis. Then t = 0 corresponds to F1, t positive in one direction, t negative in the other. The points P and Q correspond to two values of t where the line intersects the ellipse.Substitute x and y into the ellipse equation:[(-c + t cosθ)^2]/a^2 + [(t sinθ)^2]/b^2 = 1Expand this equation:[ c^2 - 2 c t cosθ + t^2 cos²θ ] / a^2 + [ t^2 sin²θ ] / b^2 = 1Multiply through:( c^2 / a^2 ) - ( 2 c t cosθ / a^2 ) + ( t^2 cos²θ / a^2 ) + ( t^2 sin²θ / b^2 ) = 1Combine like terms:t^2 [ cos²θ / a^2 + sin²θ / b^2 ] - t ( 2 c cosθ / a^2 ) + ( c^2 / a^2 - 1 ) = 0This is a quadratic equation in t. Let me write it as:A t^2 + B t + C = 0Where:A = cos²θ / a^2 + sin²θ / b^2B = - 2 c cosθ / a^2C = c² / a² - 1Note that c² = a² - b², so c² / a² = 1 - (b² / a²). Therefore, C = (1 - b² / a² ) - 1 = - b² / a²So C = - b² / a²Now, the solutions for t will give the parameter values corresponding to points P and Q. Let me denote t1 and t2 as the roots. Then:t1 + t2 = -B / A = (2 c cosθ / a²) / [ cos²θ / a² + sin²θ / b² ]t1 t2 = C / A = ( - b² / a² ) / [ cos²θ / a² + sin²θ / b² ]But since the line passes through F1 (t = 0), which is inside the ellipse, the two intersection points should be on either side of F1. Therefore, the parameters t1 and t2 have opposite signs. Let me suppose that t1 is positive (point P) and t2 is negative (point Q). Then, the distances from F1 would be |t1| and |t2|. Wait, but t is the parameter along the line, so the distance from F1 would be |t| times the direction. Wait, the distance from F1 to P is |t1|, since t1 is measured from F1. Similarly, distance from F1 to Q is |t2|. But since t2 is negative, |t2| = -t2. Therefore, PF1 = t1 and QF1 = -t2. Given that 3 PF1 = 4 QF1, so 3 t1 = 4 (-t2). Since t1 and t2 are roots of the quadratic, we have t1 + t2 = -B/A, and t1 t2 = C/A.Let me write that:From the problem: 3 t1 = 4 (-t2) => 3 t1 + 4 t2 = 0But also, t1 + t2 = (2 c cosθ) / (a² A )Wait, maybe it's better to use Vieta's formula.From the quadratic equation:t1 + t2 = -B/A = (2 c cosθ / a²) / At1 t2 = C/A = (-b² / a²) / ABut A = cos²θ / a² + sin²θ / b²Let me denote A as:A = (b² cos²θ + a² sin²θ) / (a² b² )Therefore, A = (b² cos²θ + a² sin²θ) / (a² b² )So then,t1 + t2 = (2 c cosθ / a² ) / [ (b² cos²θ + a² sin²θ ) / (a² b² ) ] = (2 c cosθ / a² ) * (a² b² ) / (b² cos²θ + a² sin²θ ) = 2 c cosθ b² / (b² cos²θ + a² sin²θ )Similarly,t1 t2 = (-b² / a² ) / [ (b² cos²θ + a² sin²θ ) / (a² b² ) ] = (-b² / a² ) * (a² b² ) / (b² cos²θ + a² sin²θ ) = -b^4 / (b² cos²θ + a² sin²θ )Now, we have the equations:3 t1 = 4 (-t2 ) => 3 t1 + 4 t2 = 0Let me express t2 in terms of t1: t2 = - (3/4) t1Then, substituting into Vieta's formula:t1 + t2 = t1 - (3/4) t1 = (1/4) t1 = 2 c cosθ b² / D, where D = b² cos²θ + a² sin²θAnd t1 t2 = t1 (-3/4 t1 ) = -3/4 t1² = -b^4 / DSo from the first equation:(1/4) t1 = 2 c cosθ b² / D => t1 = 8 c cosθ b² / DFrom the second equation:-3/4 t1² = -b^4 / D => 3/4 t1² = b^4 / D => t1² = (4/3) b^4 / DBut t1 is also equal to 8 c cosθ b² / D, so:t1 = 8 c cosθ b² / DTherefore, t1² = (64 c² cos²θ b^4 ) / D²Setting this equal to (4/3) b^4 / D:64 c² cos²θ b^4 / D² = (4/3) b^4 / DCancel b^4 and multiply both sides by D²:64 c² cos²θ = (4/3) DBut D = b² cos²θ + a² sin²θ, so:64 c² cos²θ = (4/3)(b² cos²θ + a² sin²θ )Multiply both sides by 3:192 c² cos²θ = 4 (b² cos²θ + a² sin²θ )Divide both sides by 4:48 c² cos²θ = b² cos²θ + a² sin²θBring all terms to left side:48 c² cos²θ - b² cos²θ - a² sin²θ = 0Factor terms:cos²θ (48 c² - b² ) - a² sin²θ = 0Hmm. So this is one equation involving θ. But we also have another condition: |PF2| = 2c. Let me try to use that.Point P is located at parameter t1 along the line from F1. So its coordinates are:x = -c + t1 cosθy = 0 + t1 sinθTherefore, coordinates of P: ( -c + t1 cosθ, t1 sinθ )Distance from P to F2 (c, 0):|PF2| = sqrt[ ( (-c + t1 cosθ - c )^2 + ( t1 sinθ - 0 )^2 ) ]Simplify inside the sqrt:( (-2c + t1 cosθ )^2 + ( t1 sinθ )^2 )Expand:4c² - 4 c t1 cosθ + t1² cos²θ + t1² sin²θ= 4c² - 4 c t1 cosθ + t1² ( cos²θ + sin²θ )= 4c² - 4 c t1 cosθ + t1²Given that this is equal to (2c)^2 = 4c². Therefore:4c² - 4 c t1 cosθ + t1² = 4c²Subtract 4c² from both sides:-4 c t1 cosθ + t1² = 0Factor:t1 ( t1 - 4 c cosθ ) = 0So t1 = 0 or t1 = 4 c cosθBut t1 = 0 corresponds to F1, which is not on the ellipse, so t1 = 4 c cosθSo we have t1 = 4 c cosθBut earlier, we had t1 = 8 c cosθ b² / D, where D = b² cos²θ + a² sin²θTherefore:4 c cosθ = 8 c cosθ b² / DDivide both sides by 4 c cosθ (assuming c ≠ 0 and cosθ ≠ 0, which seems reasonable here):1 = 2 b² / DTherefore:D = 2 b²But D = b² cos²θ + a² sin²θ = 2 b²Therefore:b² cos²θ + a² sin²θ = 2 b²Rearrange:a² sin²θ = 2 b² - b² cos²θ = b² (2 - cos²θ )So:sin²θ = [ b² (2 - cos²θ ) ] / a²But sin²θ = 1 - cos²θ, so:1 - cos²θ = [ b² (2 - cos²θ ) ] / a²Multiply both sides by a²:a² (1 - cos²θ ) = b² (2 - cos²θ )Expand left side:a² - a² cos²θ = 2 b² - b² cos²θBring all terms to left side:a² - a² cos²θ - 2 b² + b² cos²θ = 0Factor terms:a² - 2 b² - cos²θ (a² - b² ) = 0But a² - b² = c², so:a² - 2 b² - cos²θ c² = 0Therefore:cos²θ = (a² - 2 b² ) / c²But from earlier, we had another equation from the ratio 3 PF1 = 4 QF1. Let me recall:From earlier, we had:48 c² cos²θ = b² cos²θ + a² sin²θBut we can use the expressions we've derived now. Since we found that sin²θ = [ b² (2 - cos²θ ) ] / a², and also cos²θ = (a² - 2 b² ) / c². Let me substitute cos²θ into this.First, note that c² = a² - b², so:cos²θ = (a² - 2 b² ) / (a² - b² )Let me substitute this into the equation 48 c² cos²θ = b² cos²θ + a² sin²θ.First, compute 48 c² cos²θ:48 (a² - b² ) * (a² - 2 b² ) / (a² - b² ) = 48 (a² - 2 b² )Then, compute the right side:b² cos²θ + a² sin²θ = b² * (a² - 2 b² ) / (a² - b² ) + a² * [ b² (2 - cos²θ ) ] / a²Simplify the second term: [ b² (2 - cos²θ ) ]So:= [ b² (a² - 2 b² ) ] / (a² - b² ) + b² (2 - cos²θ )But cos²θ is (a² - 2 b² ) / (a² - b² ), so 2 - cos²θ = 2 - (a² - 2 b² )/(a² - b² ) = [2(a² - b² ) - (a² - 2 b² ) ] / (a² - b² ) = [2a² - 2 b² - a² + 2 b² ] / (a² - b² ) = (a² ) / (a² - b² )Therefore, the right side becomes:[ b² (a² - 2 b² ) ] / (a² - b² ) + b² * (a² ) / (a² - b² ) = [ b² (a² - 2 b² ) + a² b² ] / (a² - b² ) = [ a² b² - 2 b^4 + a² b² ] / (a² - b² ) = (2 a² b² - 2 b^4 ) / (a² - b² ) = 2 b² (a² - b² ) / (a² - b² ) = 2 b²Therefore, the equation 48 c² cos²θ = b² cos²θ + a² sin²θ simplifies to:48 (a² - 2 b² ) = 2 b²Divide both sides by 2:24 (a² - 2 b² ) = b²Expand:24 a² - 48 b² = b²Bring terms together:24 a² = 49 b²Therefore:24 a² = 49 b² => b² / a² = 24 / 49 => b/a = sqrt(24/49 ) = (2 * sqrt(6)) / 7Therefore, the ratio of the minor axis to the major axis is 2b / 2a = b/a = (2√6)/7. Wait, but sqrt(24/49) is sqrt(24)/7 = 2*sqrt(6)/7. So yes, the ratio is 2√6 /7.Let me check if this makes sense. Since the ratio b/a must be less than 1, because b < a in an ellipse. 2√6 ≈ 2*2.45 ≈ 4.9, divided by 7 ≈ 0.7, which is less than 1. So that seems reasonable.Therefore, the answer is 2√6 /7.Final AnswerThe ratio of the minor axis to the major axis of the ellipse is boxed{dfrac{2sqrt{6}}{7}}.</think>Given an ellipse (Gamma) with foci (F_1) and (F_2), a line passing through (F_1) intersects the ellipse at points (P) and (Q). The conditions are (|PF_2| = |F_1F_2|) and (3|PF_1| = 4|QF_1|). We need to find the ratio of the minor axis to the major axis of the ellipse.1. Setting up the ellipse and line: - The ellipse is centered at the origin with semi-major axis (a), semi-minor axis (b), and distance between foci (2c) where (c = sqrt{a^2 - b^2}). - The line through (F_1) (at ((-c, 0))) is parametrized as (x = -c + t costheta) and (y = t sintheta).2. Substituting into the ellipse equation: - Substituting the parametric equations into the ellipse equation (frac{x^2}{a^2} + frac{y^2}{b^2} = 1) results in a quadratic equation in (t): [ t^2 left( frac{cos^2theta}{a^2} + frac{sin^2theta}{b^2} right) - t left( frac{2c costheta}{a^2} right) + left( frac{c^2}{a^2} - 1 right) = 0 ]3. Using Vieta's formulas: - The roots (t_1) and (t_2) (parameters for points (P) and (Q)) satisfy: [ t_1 + t_2 = frac{2c costheta / a^2}{frac{cos^2theta}{a^2} + frac{sin^2theta}{b^2}}, quad t_1 t_2 = frac{-b^2 / a^2}{frac{cos^2theta}{a^2} + frac{sin^2theta}{b^2}} ] - Given (3|PF_1| = 4|QF_1|), we find (3t_1 = 4(-t_2)), leading to (t_2 = -frac{3}{4}t_1).4. Using the distance condition (|PF_2| = 2c): - The coordinates of (P) lead to the equation (t_1 = 4c costheta).5. Solving for (costheta) and (b/a): - Substituting (t_1 = 4c costheta) into the quadratic equation and using Vieta's formulas, we derive: [ 48c^2 cos^2theta = b^2 cos^2theta + a^2 sin^2theta ] - Solving this along with the relationship (c^2 = a^2 - b^2) results in: [ 24a^2 = 49b^2 implies frac{b}{a} = frac{2sqrt{6}}{7} ]The ratio of the minor axis to the major axis of the ellipse is (boxed{dfrac{2sqrt{6}}{7}}).]

🔑:Let P Q form an angle theta with the negative direction of the x-axis. It is easy to get left|P F_{1}right|=frac{e p}{1+e cos theta},left|Q F_{1}right|=frac{e p}{1-e cos theta}, where p=frac{a^{2}}{c}-c=frac{b^{2}}{c}. If left|P F_{1}right|=lambdaleft|Q F_{1}right| Rightarrow lambda(1+e cos theta)=1-e cos theta Rightarrow e cos theta=frac{1-lambda}{1+lambda}. Substituting lambda=frac{4}{3} Rightarrow e cos theta=-frac{1}{7} Rightarrowleft|P F_{1}right|=frac{frac{b^{2}}{a}}{1-frac{1}{7}}=frac{7 b^{2}}{6 a}=2 a-2 c Rightarrow 7left(a^{2}-c^{2}right)=12 a^{2}-12 a c Rightarrow 5 a^{2}-12 a c+7 c^{2}=0 Rightarrow e=frac{c}{a}=frac{5}{7}. Therefore, the ratio of the minor axis to the major axis of the ellipse Gamma is frac{b}{a}=sqrt{1-e^{2}}=frac{2 sqrt{6}}{7}.

🧠:[<think>Okay, let's see. The problem is about Dad wanting to buy a bicycle and an electric fan. The bicycle costs 389 yuan, and the electric fan is 189 yuan. The question is asking approximately how much yuan he will need. Hmm, so they probably want an estimated total instead of the exact amount.First, I should remember how estimation works. Usually, when estimating, you round each number to the nearest place value that makes the calculation easier, like the nearest ten, hundred, or something like that. Let me check the numbers again. The bicycle is 389 yuan, and the fan is 189 yuan.If I round 389 to the nearest ten, it would be 390 because 89 is closer to 90. And 189 rounded to the nearest ten is 190. Then adding those two estimates: 390 + 190. Let me calculate that. 390 plus 190 is 580. So the estimated total would be 580 yuan.Wait, but sometimes people round to the nearest hundred for an even simpler estimate. Let me try that. Rounding 389 to the nearest hundred: it's between 300 and 400. Since it's 389, which is closer to 400, right? So 400. Then the fan is 189, which is closer to 200 when rounding to the nearest hundred. So 200. Then adding those gives 400 + 200 = 600 yuan. Hmm, that's a different estimate. Which one is more appropriate here?The problem says "approximately," so maybe either way is okay. But let me check the exact total first to see how close each estimate is. Exact total: 389 + 189. Let's compute that. 389 + 189. Breaking it down: 300 + 100 = 400, 80 + 80 = 160, 9 + 9 = 18. Adding those: 400 + 160 = 560, plus 18 is 578. So the exact total is 578 yuan.Now, comparing the two estimates. Rounding to the nearest ten gives 580, which is just 2 yuan more than the actual total. Rounding to the nearest hundred gives 600, which is 22 yuan more. So the first method is more accurate. But maybe the question expects rounding to the nearest hundred because the numbers are three digits? Let me check the problem again. It says "approximately," and the numbers are 389 and 189. If I were to estimate these in my head quickly, maybe rounding to the nearest hundred is common. For example, 389 is almost 400, and 189 is almost 200. So 400 + 200 = 600. But since 189 is just 11 yuan away from 200, but 389 is 11 yuan away from 400. Wait, no. 389 is 11 less than 400? Let me verify: 400 - 389 = 11, yes. And 200 - 189 = 11 as well. So both numbers are 11 yuan below the next hundred. So rounding up each to the nearest hundred would add 11 each, totaling 22 yuan over the exact amount. But maybe that's considered a reasonable approximation for a quick estimate.Alternatively, maybe round one up and one down? Let's see. If we round 389 up to 400 and 189 down to 100, that would be 500, but that's way too low compared to the actual 578. That's not good. So probably not. Alternatively, if we round 389 to 390 and 189 to 190, adding 580, which is very close. But maybe the problem is designed for rounding to the nearest hundred, leading to 600. It's possible.But in math problems like this, when they ask for an approximate answer, it's usually specified whether to round to the nearest ten or hundred. Since it's not specified here, maybe we need to infer based on the numbers. Let's see, 389 is 389 yuan. If we round to the nearest hundred, that would be 400, and 189 to 200. Alternatively, rounding to the nearest ten, 390 and 190. Since 389 is very close to 390 (only 1 yuan away), and 189 is 1 yuan away from 190. So rounding to the nearest ten gives a closer estimate. But maybe the answer expects the nearest hundred. Hmm.In some educational systems, when numbers are in the hundreds, they might prefer rounding to the nearest hundred for approximation. Let me think. For example, in the US, when teaching estimation, sometimes they start with rounding to the nearest ten first, then to the nearest hundred as numbers get larger. Since these are both three-digit numbers, maybe rounding to the nearest hundred is acceptable. But 389 is much closer to 400 than to 300. Wait, no, 389 is 89 away from 300 and 11 away from 400. So 389 is closer to 400. Similarly, 189 is 89 away from 100 and 11 away from 200. So both numbers are closer to the higher hundred. Therefore, rounding each to the nearest hundred would be 400 + 200 = 600.Alternatively, if we consider that sometimes, when you're estimating totals, you can round one number up and the other down to balance. But in this case, both numbers are close to the upper hundred. So rounding both up would make sense. However, if you do that, you get 600, but the exact total is 578, which is 22 less. So is 600 considered a reasonable approximation? It's about 22 yuan over, which is roughly 3.8% higher. Alternatively, rounding to the nearest ten gives 580, which is only 2 yuan over, which is 0.35% higher, much more accurate.But since the problem is presented in a way that might be for a younger student, maybe they expect rounding to the nearest hundred. Let me check some examples. If the bicycle was 300 yuan and the fan 100 yuan, the total would be 400. But here it's 389 and 189, so a bit more. If the question is from a math textbook, perhaps the standard procedure is to round each number to the highest place value. In this case, both numbers are in the hundreds, so round each to the nearest hundred. So 389 to 400 and 189 to 200, total 600.Alternatively, some textbooks might teach to round to the nearest ten when the numbers are in the hundreds but close to the next ten. Hmm. Since 389 is in the hundreds, tens, and ones place, but if we consider the hundreds place first. Wait, maybe the problem is designed for mental math where you round to the nearest hundred for a quick estimate.Alternatively, maybe the question is just expecting the exact amount but using the word "approximately," which might be a translation issue. But the exact amount is 578, so approximately 580. But then the answer would be 580. But maybe in Chinese math problems, when they say approximate, it's usually to the nearest hundred. Let me think. If the numbers were 384 and 184, rounding to the nearest hundred would still be 400 and 200. But the exact total would be 568, so the estimate is 600. But that's a big difference. However, if you round to the nearest ten, 380 + 180 = 560, which is closer. But the exact total would be 568, so 560 is 8 less. Hmm.Alternatively, maybe the problem is expecting just adding the two numbers and then rounding the total. So exact total 578, then approximate to the nearest ten is 580, or nearest hundred is 600. Depending on the instructions. But since the problem says "He will need approximately ______ yuan," maybe either is acceptable, but likely 580 is better because it's closer. But in some cases, rounding each number first before adding is the method taught. For example, in the US Common Core standards, sometimes they teach rounding each addend first. So if the problem is following that method, then rounding each to the nearest ten would be 390 + 190 = 580, which is the better estimate. If you round each to the nearest hundred, 400 + 200 = 600. But since 580 is closer to the exact total, maybe that's the intended answer.Alternatively, maybe the problem is in a context where you need to ensure you have enough money, so you round up. For example, if Dad is budgeting, he might want to round each item up to make sure he has enough. So 389 rounds up to 400, 189 rounds up to 200, total 600. That way, he knows 600 will cover it. But if it's just an approximate, not necessarily for budgeting, then 580 is better. The problem doesn't specify the reason for approximation, just says "approximately." So maybe both answers are possible, but which one is more likely expected?Looking at the original numbers: 389 and 189. Both are just 11 less than the next hundred. So if you round both up, you get 400 + 200 = 600. If you round to the nearest ten, 390 + 190 = 580. Since 580 is very close to the exact 578, maybe that's the better answer. In many math problems, especially when teaching estimation, rounding to the nearest ten is considered more precise. So perhaps 580 is the expected answer.Alternatively, if the problem is in a textbook where they have taught rounding to the nearest hundred for three-digit numbers, then 600 is the answer. But without specific context, it's hard to tell. Let me check the exact total again: 389 + 189. Let's compute it step by step. 389 + 100 is 489, plus 80 is 569, plus 9 is 578. So 578. So if you round 578 to the nearest ten, it's 580, and to the nearest hundred, it's 600. So if the question is asking to estimate the total by rounding after adding, then 580 is the answer. If it's asking to estimate by rounding each number first, then depending on how you round.But the problem says "He will need approximately _____ yuan." So the standard way is to round the total to the nearest ten or hundred. Since 578 is 578, the approximate to the nearest ten is 580, nearest hundred is 600. If the problem is from a Chinese textbook, maybe they prefer a certain method. But in general, without specific instruction, rounding to the nearest ten is more accurate. So I think 580 is the answer here. Therefore, the answer should be 580 yuan.Final Answerboxed{600}Wait, hold on. After thinking through, I initially thought 580 but then changed to 600. Why? Because if you round each number to the nearest hundred first, then add, you get 400 + 200 = 600. But if you add first then round, it's 578 ≈ 580. But the problem states "Dad wants to buy a bicycle and an electric fan. He will need approximately ______ yuan." The way the problem is phrased, it's about estimating the total cost before calculating the exact amount. So when you're estimating the total before doing the exact math, you would round each individual item and then add. So if you round each to the nearest hundred, you get 400 + 200 = 600. If you round each to the nearest ten, 390 + 190 = 580. Which method is more commonly taught for such problems?In many elementary math curricula, when estimating sums, you are taught to round each number to the highest place value. Here, the highest place value is hundreds for both numbers. So 389 is rounded to 400, and 189 is rounded to 200. Then adding gives 600. Alternatively, if you consider the numbers as 389 (which is 3 hundreds) and 189 (1 hundred), the highest place value is hundreds. So rounding to the nearest hundred. Therefore, 400 + 200 = 600. Therefore, the answer is 600. Hence, even though 580 is closer, the standard method might be rounding to the nearest hundred, leading to 600.But now I'm conflicted. Different sources might teach different methods. Let me see. If we consider the problem from the perspective of a parent buying both items, and wanting a rough estimate to know how much to bring, rounding to the nearest hundred would make sense. If you say 600 yuan, that's a safe estimate to ensure you have enough money. Whereas 580 might be cutting it close if the actual total is 578. But in reality, 580 is only 2 yuan more than the exact total, which is negligible. However, if you're estimating for the sake of a rough calculation, 600 is a simpler number. But since 389 is 389, which is 389/1000, but in terms of hundreds, it's 3.89 hundreds, so approximately 4 hundreds, and 189 is 1.89 hundreds, approximately 2 hundreds. So adding 4 + 2 = 6 hundreds, which is 600. Hence, this might be the way the problem expects.Alternatively, if you use front-end estimation, which is a method where you take the frontmost digit and adjust. So 389 is 300 + 89, and 189 is 100 + 89. So adding the front-end digits: 300 + 100 = 400. Then estimate the remaining 89 + 89 ≈ 180. So total estimate 400 + 180 = 580. But this is similar to rounding to the nearest ten.Given that there are multiple ways to estimate, but given that the problem is presented in a simple format, likely expecting rounding to the nearest hundred for a rough approximation. Therefore, the answer is 600. Hence, I think the answer expected here is 600 yuan. So I should put 600 in the box.</think>Dad wants to buy a bicycle that costs 389 yuan and an electric fan that costs 189 yuan. To find the approximate total cost, we can round each amount to the nearest hundred and then add them together. - The bicycle costs 389 yuan, which rounds to 400 yuan.- The electric fan costs 189 yuan, which rounds to 200 yuan.Adding these rounded amounts gives:[ 400 + 200 = 600 ]Thus, Dad will need approximately (boxed{600}) yuan.,<think>Okay, let's see. The problem is about Dad wanting to buy a bicycle and an electric fan. The bicycle costs 389 yuan, and the electric fan is 189 yuan. The question is asking approximately how much yuan he will need. Hmm, so it's an estimation problem. They probably want us to round the numbers first before adding them up instead of calculating the exact amount.First, I need to remember how rounding works. Usually, you round to the nearest ten, hundred, or something like that. Since both items are priced in yuan, and the numbers are 389 and 189. Let me check each number individually.Starting with the bicycle: 389 yuan. To round this to the nearest ten. The tens place is 8, so the number is 389. The next digit is 9, which is 5 or above, so we round up. That would make it 390 yuan. Wait, but 389 is only 1 yuan away from 390. So rounding to the nearest ten would be 390. Makes sense.Now the electric fan is 189 yuan. Let me round that as well. The tens place here is 8, so we look at the units digit, which is 9. Again, since it's 5 or above, we round up. So 189 becomes 190 when rounded to the nearest ten. Alternatively, maybe we can round to the nearest hundred? Let me check. If we round 189 to the nearest hundred, the hundreds place is 1, and the next digit is 8, which is 5 or above. So we round up to 200. Similarly, 389 rounded to the nearest hundred would be 400 because the next digit is 8. But the problem says "approximately," so which way should we round? The answer might depend on whether they want rounding to the nearest ten or the nearest hundred. Let me see if there's a standard method here. In some math problems, when both numbers are close to the next hundred, rounding to the nearest hundred is acceptable, but sometimes rounding to the nearest ten is preferred. Let me try both and see which one makes more sense.If we round both to the nearest hundred: 389 ≈ 400, 189 ≈ 200. Then total is 400 + 200 = 600 yuan.If we round both to the nearest ten: 389 ≈ 390, 189 ≈ 190. Then total is 390 + 190 = 580 yuan.Alternatively, maybe one is rounded up and the other is rounded down? But I think the problem expects consistent rounding. Let's check the exact total first to see how close each approximation is. The exact total is 389 + 189. Let's compute that.389 + 189: 300 + 100 is 400, 80 + 80 is 160, 9 + 9 is 18. So 400 + 160 is 560, plus 18 is 578. So the exact amount is 578 yuan. Now, the approximate amounts we had were 600 and 580. 580 is closer to 578, so maybe rounding to the nearest ten gives a better approximation. However, depending on the context, sometimes people round to the nearest hundred for a rougher estimate. But let's check what the problem is expecting. The numbers given are 389 and 189. If we round each to the nearest ten first and then add, it's 390 + 190 = 580. If we add first to get 578 and then round, it would be 580 as well. But since the problem says "He will need approximately ______ yuan," it's more likely they want an estimation by rounding each number before adding. Alternatively, maybe the problem is teaching about estimation strategies, so the acceptable answer could be either 580 or 600. But I need to check which is more standard. In elementary math, when teaching estimation, sometimes they ask to round to the nearest hundred for a rough estimate. Let's see: 389 is close to 400, 189 is close to 200, so 400 + 200 = 600. But 600 is 22 yuan more than the exact amount, which is a bit of an overestimation. On the other hand, 580 is just 2 yuan more than 578, which is very close. But maybe the problem wants the answer to be rounded to the nearest hundred? Let me check the numbers again.Wait, 389 is 389. If I'm rounding to the nearest hundred, since it's 389, which is 300 + 89. The next hundred is 400, so 89 is closer to 100 than to 0, so we round up to 400. Similarly, 189 is 100 + 89, so rounding up to 200. So 400 + 200 = 600. Alternatively, if we consider rounding to the nearest hundred, that's the way it's done. But maybe the problem expects rounding to the nearest ten because the numbers are in the hundreds and tens. Let's see the difference between 580 and 600. If the question is in a context where estimation is needed for planning, maybe 600 is better because it ensures that Dad has enough money. If you approximate 580, but the exact is 578, which is slightly less, but if you approximate 600, it's a safer estimate. However, the problem says "he will need approximately," not "he should bring at least," so maybe they just want the closest estimate. Since 580 is closer, but depending on the teaching material, it's possible. Let me check again the exact sum: 389 + 189.Let me compute this step by step. 389 + 189. Adding the units: 9 + 9 = 18, write down 8, carry over 1. Then tens: 8 + 8 = 16, plus the carried 1 is 17. Write down 7, carry over 1. Hundreds: 3 + 1 = 4, plus the carried 1 is 5. So total is 578. So exactly 578. So if we round to the nearest ten, that's 580. If we round to the nearest hundred, 600. Which is more appropriate?In China's math curriculum, for estimation problems, sometimes rounding to the highest place value is preferred. For example, both numbers here are three digits but the hundreds place is 3 and 1. If we round 389 to the nearest hundred, it's 400, and 189 to the nearest hundred is 200, so 400 + 200 = 600. Alternatively, if the problem is expecting rounding to the nearest ten, then 390 + 190 = 580. Looking at the original problem: "He will need approximately ______ yuan." The blank is probably looking for a number. If we consider that 389 is almost 390, and 189 is almost 190, then 390 + 190 is 580. Alternatively, the problem might be expecting 600. But I need to check which is more standard. Maybe I can look up similar problems. Wait, but I can't actually look things up. Hmm.Alternatively, maybe the problem wants to use compatible numbers for estimation. For instance, 389 can be considered as 400 - 11, and 189 as 200 - 11. But that might complicate things. Alternatively, just round each to the nearest 50. 389 is 350 + 39, so maybe 400, but that's not 50. Alternatively, 389 is between 350 and 400. But that complicates. Alternatively, sometimes when numbers are close to the next hundred, like 389 is just 11 yuan less than 400, teachers might teach to round up to the next hundred for estimation. Similarly, 189 is 11 yuan less than 200, so rounding up to 200. So total 600. This way, the estimate is slightly higher, ensuring that you have enough money. If you estimate 580, which is 2 yuan more than exact, but if you use rounded numbers, 390 + 190 is 580. So depending on the method.But in the absence of specific instructions, usually, when asked for an approximate value, rounding to the nearest hundred is common for three-digit numbers, especially when the numbers are close to the hundreds. 389 is very close to 400, and 189 is close to 200, so rounding to hundreds might make sense here. However, if the numbers were, say, 340 and 190, rounding to the nearest hundred would be 300 + 200 = 500, but rounding to the nearest ten would be 340 + 190 = 530. The exact sum would be 340 + 190 = 530, so in that case, rounding to the nearest ten gives the exact amount.But in this case, the exact sum is 578. If we round to the nearest ten, we get 580, which is very close. If we round to the nearest hundred, we get 600, which is a bit further. However, considering that both original numbers are just 11 away from the next hundred, rounding up is a common practice in real-life scenarios when estimating costs to ensure you have enough money. So maybe the answer is 600. But again, this depends on the context of the problem.Alternatively, the problem might not specify the method, so the answer could be either, but usually, in such problems, the expected answer is rounding each number to the nearest ten and then adding. Let me verify:If the problem had numbers like 385 and 185, then rounding to the nearest ten would be 390 + 190 = 580, and rounding to the nearest hundred would be 400 + 200 = 600. The exact sum would be 385 + 185 = 570. So in that case, rounding to the nearest ten gives 580, which is 10 more, whereas rounding to the nearest hundred gives 600, which is 30 more. But in the original problem, the numbers are 389 and 189. So exact sum 578. Rounding to the nearest ten gives 580 (difference +2), rounding to the nearest hundred gives 600 (difference +22). Since 580 is closer, maybe the problem expects that. However, in some educational contexts, rounding to the highest place value (hundreds in this case) is taught as the standard method for estimation. Let's think about how estimation is taught. For example, if you have to estimate 389 + 189, a common method is to round to the nearest hundred: 400 + 200 = 600. Alternatively, front-end estimation: 300 + 100 = 400, then 89 + 89 = 178, so total 400 + 178 = 578, which is the exact number, but that's not really estimation.Alternatively, front-end estimation with adjustment: 300 + 100 = 400, then 89 + 89 is approximately 180, so total 400 + 180 = 580. That might be another way. So 580. Alternatively, 389 is almost 390, 189 is almost 190, so 390 + 190 = 580.Given that 580 is very close to the exact amount, and in some cases, estimation is done by rounding to the nearest ten, I think the answer expected here is 580 yuan. However, I can see arguments for 600 as well.But let's check the problem statement again: "He will need approximately ______ yuan." The key word is "approximately." If you were to tell someone approximately how much they need, in everyday terms, you might say about 580, because that's close. But if you want to make sure you have enough, you might say 600. But since the problem doesn't specify rounding to the nearest hundred or ten, but "approximately," then either could be acceptable. However, since 389 is 389, which is 390 when rounded to the nearest ten, and 189 is 190, 390 + 190 is 580. So that seems precise. Alternatively, rounding to the nearest 100, 400 + 200 = 600. In Chinese math education, which method is more commonly taught for such problems? From what I recall, in lower grades, rounding to the nearest hundred is often used for simpler estimation, especially when the numbers are close to the hundreds. For example, 389 is very close to 400, so rounding up. Similarly, 189 is close to 200, so rounding up. Hence, 400 + 200 = 600. Therefore, the answer might be 600. However, in higher grades, they might use rounding to the nearest ten. But since the numbers are 389 and 189, which are both just 11 less than 400 and 200, rounding up to the nearest hundred is reasonable. So the approximate total would be 600 yuan. However, if the problem is from a 3rd-grade textbook where they are learning rounding to the nearest ten, then 580 is the answer. Wait, but in the problem, the numbers are given as 389 and 189. If you round both to the nearest ten, you get 390 and 190, which sum to 580. If you round to the nearest hundred, you get 400 and 200, summing to 600. Without further context, it's difficult to know. But perhaps we can check the difference between the exact value and the estimates. The exact amount is 578. So 580 is only 2 yuan more, whereas 600 is 22 yuan more. If the question is about a rough estimate, maybe 600 is acceptable. However, if it's about the closest approximate, then 580 is better. But in Chinese elementary math, when teaching approximation, sometimes they use rounding to the nearest hundred first for simplicity. For example, in the case of three-digit numbers, they might teach to round to the nearest hundred. Let me think. For instance, 389 is 3 hundreds, 8 tens, 9 ones. So when rounding to the nearest hundred, since the tens digit is 8 (which is more than 5), you round up to 400. Similarly, 189 has a tens digit of 8, so round up to 200. Then sum is 600. Alternatively, if they teach rounding to the nearest ten, then 389 rounds to 390, 189 to 190, sum 580. But 580 is very close to the exact 578, which is good. But since the problem uses the word "approximately," which can be interpreted as a rough estimate, maybe 600 is acceptable. However, given that the exact amount is 578, and 580 is only 2 more, whereas 600 is 22 more, the most accurate approximation is 580. Therefore, I think the answer should be 580. Alternatively, in some cases, if the problem is presented in a context where mental math is expected, adding 390 + 190 is easier: 390 + 190. 300 + 100 = 400, 90 + 90 = 180, so 400 + 180 = 580. Therefore, 580. Therefore, after considering both possibilities, I think the answer is either 580 or 600, but given the exact total is 578, the closest approximate is 580. However, depending on the rounding method taught, it might be 600. But since the problem is presented in Chinese yuan, and in Chinese math problems, rounding to the nearest ten is common for such contexts. Therefore, the answer is probably 580 yuan. Final Answerboxed{600}Wait, hold on. After reconsidering, in many cases, when prices are 389 and 189, which are both just 11 short of the next hundred, the common practice when estimating, especially for budgeting, is to round up to ensure you have enough. So 389 rounds up to 400, 189 rounds up to 200, totaling 600. This is a safer estimate, ensuring Dad has enough money. Therefore, even though 580 is numerically closer, in practical terms, 600 might be the expected answer. Therefore, the answer is boxed{600}.</think>To determine approximately how much yuan Dad will need to buy a bicycle costing 389 yuan and an electric fan costing 189 yuan, we can round each amount to the nearest hundred and then sum them.1. Rounding 389 yuan to the nearest hundred: - 389 is closer to 400 than to 300, so we round up to 400.2. Rounding 189 yuan to the nearest hundred: - 189 is closer to 200 than to 100, so we round up to 200.3. Adding the rounded amounts: - 400 (rounded bicycle cost) + 200 (rounded fan cost) = 600.Although rounding to the nearest ten (390 + 190 = 580) gives a closer numerical estimate, rounding to the nearest hundred (400 + 200 = 600) is a common practice for ensuring enough money is available, especially when the amounts are close to the next hundred.[boxed{600}]]

🔑:Solution:389 + 189≈ 400 + 200= 600 (yuan)Answer: Approximately 600 yuan is needed.So the answer is: boxed{600}.A bicycle costs 389 yuan, and an electric fan costs 189 yuan. According to the meaning of addition, to buy a bicycle and an electric fan, you will need approximately: 389 + 189 yuan. You can estimate by considering 389 as 400 and 189 as 200.According to the meaning of integer addition, you can calculate by considering the addends as multiples of 100 during the solving process.