🧠:[<think>Okay, so I need to find the shortest distance from a point on the curve y = ln(2x - 1) to the line 2x - y + 8 = 0. The options are given as A to D, with some multiples of sqrt(5). Let me think about how to approach this.First, I remember that the shortest distance from a point to a line can be found using the formula: Distance = |Ax + By + C| / sqrt(A² + B²)Where the line is Ax + By + C = 0, and the point is (x, y). So in this case, the line is 2x - y + 8 = 0, which means A = 2, B = -1, and C = 8. But here, the point is not fixed; it's somewhere on the curve y = ln(2x - 1). So the point is (x, ln(2x - 1)) where 2x -1 > 0, so x > 1/2. So, the distance from (x, ln(2x -1)) to the line 2x - y +8 =0 would be |2x - ln(2x -1) +8| / sqrt(2² + (-1)²) = |2x - ln(2x -1) +8| / sqrt(5).Therefore, to find the minimal distance, we need to minimize the numerator |2x - ln(2x -1) +8| with respect to x, where x > 1/2. Since sqrt(5) is a positive constant, minimizing the numerator will minimize the distance. Also, since the expression inside the absolute value could be positive or negative, but the absolute value complicates things. However, since we are looking for the minimal distance, it might be that the minimal occurs where the expression inside is zero (if possible), but if not, then where the derivative is zero.Wait, but maybe the expression inside the absolute value is always positive? Let's check for x > 1/2. Let's take some sample x. Let me pick x=1. Then 2(1) - ln(2*1 -1) +8 = 2 - ln(1) +8 = 2 - 0 +8 =10, which is positive. What if x approaches 1/2 from the right? As x approaches 1/2+, ln(2x -1) approaches ln(0+) which is -infinity. So 2x approaches 1, and ln(2x -1) approaches -infty, so 2x - ln(2x -1) +8 approaches 1 - (-infty) +8 = +infty. So at x approaching 1/2, the expression inside absolute value is +infty, and as x increases, let's see what happens. Let's take x very large. Then ln(2x -1) is ln(2x) approximately, which is ln2 + lnx. So 2x - ln(2x -1) +8 ≈ 2x - ln(2x) +8. As x increases, 2x dominates, so this tends to +infty. Therefore, the expression inside the absolute value is always positive. Therefore, the absolute value can be removed, and we just need to minimize (2x - ln(2x -1) +8)/sqrt(5). Since sqrt(5) is a constant, we can just minimize the numerator, which is f(x) = 2x - ln(2x -1) +8.So we need to find the minimum of f(x) = 2x - ln(2x -1) +8 for x > 1/2. To do this, we can take the derivative of f(x) with respect to x, set it equal to zero, and solve for x.Let's compute f'(x):f'(x) = d/dx [2x - ln(2x -1) +8] = 2 - [ (2)/(2x -1) ] +0 = 2 - 2/(2x -1).Set f'(x) = 0:2 - 2/(2x -1) = 0.Let's solve for x:2 = 2/(2x -1)Multiply both sides by (2x -1):2*(2x -1) = 2Divide both sides by 2:2x -1 = 1So 2x = 2 => x =1.Therefore, the critical point is at x=1. Now we need to check if this is a minimum. Since the function f(x) tends to infinity as x approaches 1/2 and as x approaches infinity, and there's only one critical point, it must be a minimum.So the minimal distance occurs at x=1. Now compute f(1):f(1) = 2*1 - ln(2*1 -1) +8 = 2 - ln(1) +8 = 2 -0 +8 =10.Therefore, the minimal distance is 10 / sqrt(5) = 10 / sqrt(5) = 2*sqrt(5). Because 10/sqrt(5) = 2*sqrt(5) (since 10/sqrt(5) = (10*sqrt(5))/5 = 2*sqrt(5)).Therefore, the answer is B: 2√5.But wait, let me double-check. Let me verify the computation:f(1) = 2*1 - ln(2*1 -1) +8 = 2 - ln(1) +8 = 10. Then 10 divided by sqrt(5) is indeed 10/sqrt(5) = 2*sqrt(5). Because sqrt(5) is approximately 2.236, so 10/2.236 ≈4.472, and 2*sqrt(5) is also 4.472.Alternatively, rationalizing the denominator: 10/sqrt(5) = (10*sqrt(5))/(sqrt(5)*sqrt(5)) = (10*sqrt(5))/5 = 2*sqrt(5). Yes, correct.Therefore, the minimal distance is 2*sqrt(5), which is option B.Alternatively, maybe I made a mistake in assuming that the minimal occurs at x=1. Let's check another point. For example, x=2.At x=2, f(x)=2*2 - ln(3) +8=4 - ln3 +8≈12 -1.0986≈10.9014, which is larger than 10. So x=1 gives a smaller value.At x=0.6 (which is greater than 0.5), f(x)=2*0.6 - ln(1.2 -1) +8. Wait, 2x -1=1.2 -1=0.2. So ln(0.2)≈-1.6094. Then f(x)=1.2 - (-1.6094) +8≈1.2 +1.6094 +8≈10.8094, which is still higher than 10. So indeed x=1 seems to be the minimum.Therefore, the answer is B.Final Answerboxed{B}</think>To find the shortest distance from a point on the curve ( y = ln(2x - 1) ) to the line ( 2x - y + 8 = 0 ), we use the distance formula from a point ((x, y)) to the line ( Ax + By + C = 0 ), which is ( frac{|Ax + By + C|}{sqrt{A^2 + B^2}} ).For the given line ( 2x - y + 8 = 0 ), the distance from a point ((x, ln(2x - 1))) on the curve is:[frac{|2x - ln(2x - 1) + 8|}{sqrt{2^2 + (-1)^2}} = frac{|2x - ln(2x - 1) + 8|}{sqrt{5}}]We need to minimize the numerator ( |2x - ln(2x - 1) + 8| ). Since the expression inside the absolute value is always positive for ( x > frac{1}{2} ), we can ignore the absolute value and minimize ( f(x) = 2x - ln(2x - 1) + 8 ).Taking the derivative of ( f(x) ):[f'(x) = 2 - frac{2}{2x - 1}]Setting the derivative to zero to find critical points:[2 - frac{2}{2x - 1} = 0 implies 2 = frac{2}{2x - 1} implies 2x - 1 = 1 implies x = 1]Evaluating ( f(x) ) at ( x = 1 ):[f(1) = 2(1) - ln(2(1) - 1) + 8 = 2 - ln(1) + 8 = 10]Thus, the minimum distance is:[frac{10}{sqrt{5}} = frac{10sqrt{5}}{5} = 2sqrt{5}]The shortest distance is (boxed{B}).,<think>Okay, so I need to find the shortest distance from a point on the curve y = ln(2x - 1) to the line 2x - y + 8 = 0. The options are given as A to D, with some multiples of sqrt(5). Let me think about how to approach this.First, I remember that the shortest distance from a point to a line can be found using the formula: the distance from a point (x₀, y₀) to the line Ax + By + C = 0 is |Ax₀ + By₀ + C| / sqrt(A² + B²). So, maybe I can use this formula here.The line given is 2x - y + 8 = 0, so A = 2, B = -1, and C = 8. The curve is y = ln(2x - 1). So any point on this curve can be represented as (x, ln(2x - 1)). Then, the distance from this point to the line would be |2x - ln(2x - 1) + 8| / sqrt(2² + (-1)²) = |2x - ln(2x - 1) + 8| / sqrt(5).To find the minimum distance, I need to minimize this expression with respect to x. Since the denominator sqrt(5) is a constant, I can just focus on minimizing the numerator |2x - ln(2x - 1) + 8|. However, dealing with absolute value can be tricky, so maybe instead I can minimize the square of the distance, which would be (2x - ln(2x - 1) + 8)^2 / 5. But even without squaring, since the expression inside the absolute value is a continuous function, the minimum will occur either where the derivative is zero or at the endpoints of the domain.But first, let's check the domain of the curve y = ln(2x - 1). The argument of the logarithm must be positive, so 2x - 1 > 0 => x > 1/2. So x must be greater than 1/2. So the domain is (1/2, ∞).Now, let's consider the function f(x) = |2x - ln(2x - 1) + 8|. Since we're looking for the minimum distance, we can consider the expression without the absolute value first. Let's define D(x) = (2x - ln(2x - 1) + 8)/sqrt(5). The absolute value of D(x) is the distance, but since sqrt(5) is positive, the sign of D(x) depends on the numerator. Let's check if 2x - ln(2x -1 ) + 8 is always positive. For x > 1/2, 2x -1 is positive, so ln(2x -1) is defined. Let's check for x near 1/2. As x approaches 1/2 from the right, 2x -1 approaches 0, so ln(2x -1) approaches negative infinity. Therefore, 2x - ln(2x -1) +8 approaches 1 (since 2*(1/2) = 1) minus negative infinity, which is positive infinity. So near x = 1/2, the expression is positive. For larger x, 2x increases linearly, ln(2x -1) increases logarithmically, so 2x - ln(2x -1) +8 will be dominated by 2x, which is positive. Therefore, the expression inside the absolute value is always positive for x > 1/2. Therefore, the distance is simply (2x - ln(2x -1) +8)/sqrt(5), and we can drop the absolute value.So, to minimize D(x) = (2x - ln(2x -1) +8)/sqrt(5), since sqrt(5) is a constant multiplier, we can instead minimize the numerator f(x) = 2x - ln(2x -1) +8.So, let's find the derivative of f(x) with respect to x and set it to zero to find critical points.f'(x) = derivative of 2x is 2, derivative of -ln(2x -1) is - (2)/(2x -1) [using chain rule], and derivative of 8 is 0. So:f'(x) = 2 - [2/(2x -1)]Set this equal to zero:2 - (2)/(2x -1) = 0Solve for x:2 = 2/(2x -1)Multiply both sides by (2x -1):2*(2x -1) = 24x -2 = 24x = 4x = 1So critical point at x = 1. Let's check if this is a minimum.Take the second derivative:f''(x) = derivative of f'(x) = derivative of 2 - 2/(2x -1) is 0 - [ -2*2/(2x -1)^2 ] = 4/(2x -1)^2. Since x >1/2, 2x -1 >0, so f''(x) is positive. Therefore, x=1 is a local minimum. Since the function f(x) tends to infinity as x approaches 1/2 and as x approaches infinity, this critical point must be the global minimum.Therefore, the minimum distance occurs at x=1. Let's compute the distance at x=1.First, find the point on the curve:x =1, y = ln(2*1 -1) = ln(1) = 0. So the point is (1, 0).Now compute the distance from (1,0) to the line 2x - y +8 =0.Using the distance formula:|2*1 -0 +8| / sqrt(2² + (-1)^2) = |2 +8| / sqrt(4 +1) = |10| / sqrt(5) = 10 / sqrt(5) = 2*sqrt(5).Wait, that's 10 divided by sqrt(5). Let me confirm:10 / sqrt(5) = (10*sqrt(5))/5 = 2*sqrt(5). Yes, correct.But wait, the options given are A: sqrt(5), B: 2*sqrt(5), C: 3*sqrt(5), D:0. So 2*sqrt(5) is option B. But let me verify my steps again to be sure.Wait, when I calculated the distance, the formula is |Ax + By + C| / sqrt(A² + B²). But the line is 2x - y +8 =0, so A=2, B=-1, C=8. So plugging in (1,0):|2*1 + (-1)*0 +8| / sqrt(4 +1) = |2 +0 +8| / sqrt(5) = 10 / sqrt(5) = 2*sqrt(5). That's correct.But wait, let me check the calculation of f(x) at x=1. f(x) =2x - ln(2x -1) +8. At x=1, that's 2*1 - ln(1) +8 =2 -0 +8=10. So f(x)=10, divide by sqrt(5) gives the distance as 10/sqrt(5)=2*sqrt(5). So yes, the answer is B.But just to make sure there are no mistakes in the differentiation steps:f'(x) = derivative of 2x is 2, derivative of -ln(2x -1) is - [ derivative of ln(u) where u=2x-1 is (2)/(2x-1) ], so -2/(2x -1). Then derivative of 8 is 0. So f'(x)=2 -2/(2x -1). Setting to zero gives 2=2/(2x -1). Multiply both sides by (2x -1): 2*(2x -1)=2 =>4x -2=2 =>4x=4 =>x=1. Correct.Second derivative: f''(x)= derivative of 2 -2/(2x -1). The derivative of 2 is 0, derivative of -2/(2x -1) is -2*(-2)/(2x -1)^2)=4/(2x -1)^2. Which is positive for all x>1/2. So x=1 is a minimum. Correct.Therefore, the minimal distance is 2*sqrt(5), which is option B.Wait, but before finalizing, let me think again. Is there a possibility that the minimal distance could be achieved at a point where the line connecting the point on the curve to the given line is perpendicular to the given line?Yes, actually, the shortest distance from a point to a line is along the perpendicular direction. So, another way to approach this problem is to find a point on the curve where the tangent line to the curve is parallel to the given line. Because the direction of the shortest distance should be along the perpendicular, so the line connecting the point on the curve to the given line should be perpendicular to the given line. Therefore, the tangent line at that point on the curve should be parallel to the given line. Wait, is that correct?Wait, let me clarify. The shortest distance from the curve to the line will be along the direction perpendicular to the line. So, if we can find a point on the curve where the line connecting that point to the closest point on the given line is perpendicular to the given line. Alternatively, the shortest distance is achieved when the line segment connecting the point on the curve to the given line is perpendicular to the given line.But in terms of calculus, that is equivalent to minimizing the distance function, which is what I did earlier. However, another approach is that the gradient of the curve at the closest point should be parallel to the direction of the given line. Wait, maybe not. Let me think.Wait, the direction of the given line is determined by its slope. The line 2x - y +8 =0 can be rewritten as y =2x +8, which has a slope of 2. The curve y = ln(2x -1). The derivative of the curve at any point is y' = (2)/(2x -1). If the shortest distance occurs where the tangent to the curve is perpendicular to the given line, then the product of their slopes would be -1. The slope of the given line is 2, so the slope of the tangent line should be -1/2. So set y' = -1/2:2/(2x -1) = -1/2Multiply both sides by (2x -1):2 = (-1/2)(2x -1)Multiply both sides by 2:4 = - (2x -1)4 = -2x +1-2x = 4 -1 =3x= -3/2But x must be greater than 1/2. So this gives x negative, which is not in the domain. Therefore, there is no point on the curve where the tangent is perpendicular to the given line. Therefore, the minimal distance must occur at a critical point found by the derivative method, which we found at x=1.Alternatively, perhaps the direction from the point on the curve to the closest point on the line should be perpendicular to the line. The direction of the line is (1, 2) because the line is y=2x +8, so direction vector is (1, 2). The direction perpendicular to this line is (2, -1), since the normal vector to the line 2x - y +8 =0 is (2, -1).So, the vector from the point on the curve (x, y) to the closest point on the line should be in the direction of the normal vector (2, -1). So, parametrize the line from (x, y) in the direction of (2, -1) until it intersects the line 2x - y +8 =0. That intersection point is the closest point.Alternatively, perhaps using parametric equations. Let me see. Let me try this method to confirm.Suppose we have a point P(x, ln(2x -1)) on the curve. The closest point on the line 2x - y +8 =0 to P lies along the line through P in the direction of the normal vector (2, -1). So, parametric equations of the line through P in direction (2, -1):x = x + 2ty = ln(2x -1) - tWe need this to intersect the line 2x - y +8 =0. Substitute x and y into the line equation:2(x + 2t) - (ln(2x -1) - t) +8 =0Simplify:2x +4t - ln(2x -1) + t +8 =0Combine like terms:2x - ln(2x -1) +8 +5t =0But from the previous calculation, when we found the distance, we had the expression 2x - ln(2x -1) +8 = something. Wait, but here we have 2x - ln(2x -1) +8 +5t =0. Therefore:5t = - (2x - ln(2x -1) +8 )t = - (2x - ln(2x -1) +8 ) /5But the distance from P to the line is the length of this vector, which is sqrt( (2t)^2 + (-t)^2 ) = sqrt(4t² + t²) = sqrt(5t²) = |t|sqrt(5). But according to the distance formula, the distance is |2x - y +8| / sqrt(5). Since y = ln(2x -1), that's |2x - ln(2x -1) +8| / sqrt(5). Which is equal to | -5t | / sqrt(5) = | -5t | / sqrt(5) = 5|t| / sqrt(5) = |t|sqrt(5). So both methods agree.Therefore, to find the minimal distance, we need to minimize |t|sqrt(5), which is equivalent to minimizing |t|. From the expression for t:t = - (2x - ln(2x -1) +8 ) /5So to minimize |t|, we need to minimize |2x - ln(2x -1) +8|, which is exactly what we did earlier. Therefore, the critical point found at x=1 is indeed the point of minimal distance.Therefore, the minimal distance is 2*sqrt(5), which is option B.Just to be thorough, let's check if there's any other critical point or if x=1 is the only one.We found f'(x) = 2 - 2/(2x -1). Setting to zero gave x=1. Let's check the behavior of f'(x):For x approaching 1/2 from the right, 2x -1 approaches 0, so 2/(2x -1) approaches infinity, so f'(x) =2 - infinity approaches -infinity. Therefore, the derivative starts at -infty, increases as x increases. At x=1, derivative is 0. For x >1, 2/(2x -1) becomes less than 2, so f'(x) becomes positive. Therefore, the function f(x) decreases from x=1/2 to x=1, reaching a minimum at x=1, then increases afterwards. Therefore, x=1 is indeed the only critical point and it's a global minimum.Therefore, the answer must be B: 2√5.Final Answerboxed{B}</think>To find the shortest distance from a point on the curve ( y = ln(2x - 1) ) to the line ( 2x - y + 8 = 0 ), we use the distance formula from a point to a line. The distance from a point ((x_0, y_0)) to the line ( Ax + By + C = 0 ) is given by:[frac{|Ax_0 + By_0 + C|}{sqrt{A^2 + B^2}}]For the line ( 2x - y + 8 = 0 ), we have ( A = 2 ), ( B = -1 ), and ( C = 8 ). A point on the curve ( y = ln(2x - 1) ) can be represented as ((x, ln(2x - 1))). Substituting this into the distance formula, we get:[frac{|2x - ln(2x - 1) + 8|}{sqrt{5}}]To minimize this distance, we need to minimize the numerator ( |2x - ln(2x - 1) + 8| ). Since the expression inside the absolute value is always positive for ( x > frac{1}{2} ), we can consider the function ( f(x) = 2x - ln(2x - 1) + 8 ).Taking the derivative of ( f(x) ):[f'(x) = 2 - frac{2}{2x - 1}]Setting the derivative to zero to find critical points:[2 - frac{2}{2x - 1} = 0 implies 2 = frac{2}{2x - 1} implies 2(2x - 1) = 2 implies 4x - 2 = 2 implies x = 1]The second derivative ( f''(x) = frac{4}{(2x - 1)^2} ) is positive for all ( x > frac{1}{2} ), confirming a local minimum at ( x = 1 ).Substituting ( x = 1 ) into the curve equation gives the point ( (1, 0) ). The distance from this point to the line ( 2x - y + 8 = 0 ) is:[frac{|2(1) - 0 + 8|}{sqrt{5}} = frac{10}{sqrt{5}} = 2sqrt{5}]Thus, the shortest distance is ( 2sqrt{5} ), corresponding to option B.[boxed{B}]]

🔑:Let's assume a point on the curve y=ln(2x-1) is P(m,n).The tangent line passing through P must be parallel to the line 2x-y+8=0.Given y'=frac{2}{2x-1}, the slope of the tangent line frac{2}{2m-1}=2.Solving this, we get m=1, n=ln(2-1)=0.Therefore, the shortest distance from P(1,0) to the line is d=frac{|2+8|}{sqrt{2^2+(-1)^2}}=2sqrt{5}.Hence, the correct answer is boxed{B}.This problem involves setting a point on the curve and then calculating the derivative at that point. By equating this derivative to the slope of the given line, we find the coordinates of the point. Then, we use the formula for the distance from a point to a line to solve the problem. This question tests the formula for the distance from a point to a line and the mathematical transformation method. The key is to understand that the distance from the tangent point, where the line is parallel to the given line and tangent to the curve, to the line is the smallest. It is a medium-level question.

🧠:[<think>Okay, let's see. I need to figure out which of these four statements about quadrilaterals are true. The options are ①, ②, ③, ④, and the answer choices are combinations of these. Let me take each statement one by one and analyze them.Starting with statement ①: "A quadrilateral with equal diagonals that bisect each other is a rectangle." Hmm. I remember that in a parallelogram, the diagonals bisect each other. So if a quadrilateral has diagonals that bisect each other, it's at least a parallelogram. Now, if the diagonals are also equal in length, does that make it a rectangle? Let me recall. Yes, I think that's right. In a parallelogram, if the diagonals are equal, then it must be a rectangle. Because in a rectangle, the diagonals are equal and bisect each other. So if a quadrilateral has both properties—equal diagonals and bisecting each other—it should be a rectangle. So ① seems true.Next, statement ②: "A quadrilateral with perpendicular diagonals is a rhombus." Wait, rhombus is a quadrilateral with all sides equal, and its diagonals are perpendicular. But does having perpendicular diagonals automatically make it a rhombus? Hmm. Let me think. For example, a kite also has perpendicular diagonals, but a kite isn't necessarily a rhombus unless all sides are equal. So if a quadrilateral just has perpendicular diagonals, it could be a kite or a rhombus. Therefore, the statement says it is a rhombus, which might not always be true. So ② might be false. Wait, but maybe there's more context. Let me check. If a quadrilateral has perpendicular diagonals, then it's a rhombus only if the diagonals bisect each other. Because in a rhombus, the diagonals bisect each other and are perpendicular. If a quadrilateral has perpendicular diagonals but they don't bisect each other, then it's a kite. So unless we know the diagonals bisect each other, we can't say it's a rhombus. Therefore, statement ② is missing the bisecting condition, so it's false. So ② is not necessarily true.Moving on to statement ③: "A quadrilateral with diagonals that bisect each other is a parallelogram." Yes, this is one of the fundamental theorems. If in a quadrilateral the diagonals bisect each other, then it must be a parallelogram. That's a key property. So ③ is definitely true.Statement ④: "A rhombus with equal diagonals is a square." Let me unpack that. A rhombus has all sides equal, and its diagonals are perpendicular bisectors. Normally, in a rhombus, the diagonals are not equal unless it's a square. So if a rhombus has equal diagonals, then all sides are equal (because it's a rhombus), and the diagonals are equal and perpendicular. That should make it a square. Because in a square, the diagonals are equal and perpendicular. So yes, if a rhombus has equal diagonals, it must be a square. Therefore, ④ is true.So summarizing my thoughts:- ① True- ② False- ③ True- ④ TrueLooking at the answer choices:A: ①②③ – Incorrect because ② is falseB: ①③④ – Correct, since ①, ③, ④ are trueC: ②③④ – Incorrect because ② is falseD: ①②④ – Incorrect because ② is falseTherefore, the correct answer should be B: ①③④. Let me double-check to make sure I didn't make a mistake.Rechecking statement ②: The user says "A quadrilateral with perpendicular diagonals is a rhombus." If a quadrilateral just has perpendicular diagonals, it's a kite or a rhombus. But only if the diagonals also bisect each other would it be a rhombus. Since the statement doesn't mention bisecting each other, it's not necessarily a rhombus. So ② is indeed false. For ④, rhombus with equal diagonals: since all sides are equal and diagonals are equal, all angles must be 90 degrees, making it a square. That seems right.Yes, so B is the correct answer.Final Answerboxed{B}</think>To determine which statements are true, we analyze each statement individually:1. Statement ①: A quadrilateral with equal diagonals that bisect each other is a rectangle. - A quadrilateral with diagonals that bisect each other is a parallelogram. If the diagonals are also equal, it must be a rectangle. This is true.2. Statement ②: A quadrilateral with perpendicular diagonals is a rhombus. - A quadrilateral with perpendicular diagonals could be a kite or a rhombus. For it to be a rhombus, the diagonals must also bisect each other. Since this condition is not mentioned, the statement is false.3. Statement ③: A quadrilateral with diagonals that bisect each other is a parallelogram. - This is a fundamental property of parallelograms. If the diagonals bisect each other, the quadrilateral is a parallelogram. This is true.4. Statement ④: A rhombus with equal diagonals is a square. - A rhombus has all sides equal. If its diagonals are equal, all angles must be 90 degrees, making it a square. This is true.Summarizing the results:- ①: True- ②: False- ③: True- ④: TrueThe correct answer is B: ①③④.boxed{B},<think>Okay, let's see. I have to figure out which of these four statements about quadrilaterals are true. The options are ①, ②, ③, ④, and the answer choices are combinations of these. Let me go through each statement one by one.Starting with statement ①: "A quadrilateral with equal diagonals that bisect each other is a rectangle." Hmm. I remember that in a parallelogram, the diagonals bisect each other. So if a quadrilateral has diagonals that bisect each other, it's a parallelogram. Now, if those diagonals are also equal in length, does that make it a rectangle? I think in a rectangle, the diagonals are indeed equal and bisect each other. So if a parallelogram (which it is because the diagonals bisect) has equal diagonals, then it must be a rectangle. That makes sense. So ① should be true.Next, statement ②: "A quadrilateral with perpendicular diagonals is a rhombus." Wait, rhombus has perpendicular diagonals, but does the converse hold? If a quadrilateral has perpendicular diagonals, is it necessarily a rhombus? Let me recall. A kite also has perpendicular diagonals, but a kite isn't necessarily a rhombus. Unless the quadrilateral is both a kite and a parallelogram, then it's a rhombus. So if we just have perpendicular diagonals, it could be a kite or a rhombus. Therefore, statement ② might not always be true. So ② is false.Moving on to statement ③: "A quadrilateral with diagonals that bisect each other is a parallelogram." Yes, I remember this is a theorem. If in a quadrilateral the diagonals bisect each other, then it must be a parallelogram. So ③ is definitely true.Statement ④: "A rhombus with equal diagonals is a square." Let's think. A rhombus has all sides equal, and its diagonals are perpendicular bisectors. In a square, the diagonals are equal and perpendicular. So if a rhombus has equal diagonals, then it must have all angles equal to 90 degrees as well, which would make it a square. Because in a rhombus, if the diagonals are equal, then all angles are right angles. So yes, ④ should be true.So, summarizing:① True② False③ True④ TrueLooking at the answer choices:A: ①②③ – Incorrect because ② is false.B: ①③④ – Correct, all three are true.C: ②③④ – Incorrect because ② is false.D: ①②④ – Incorrect because ② is false.Wait, but let me double-check statement ④ again. If a rhombus has equal diagonals, does that make it a square?In a rhombus, the diagonals are perpendicular bisectors. The area is (d1*d2)/2. If the diagonals are equal, then d1 = d2, so area would be (d1²)/2. But in a rhombus, the diagonals also split it into four right triangles. If the diagonals are equal, then those triangles would be isosceles right triangles, which would make each angle 90 degrees. Therefore, all angles are right angles, so it's a square. Yeah, that seems right. So ④ is definitely true.Therefore, the correct answer is B: ①③④.Final Answerboxed{B}</think>To determine which statements are true, we analyze each statement step by step:1. Statement ①: A quadrilateral with equal diagonals that bisect each other is a rectangle. - A quadrilateral with diagonals that bisect each other is a parallelogram. If the diagonals are also equal, it must be a rectangle. This is true.2. Statement ②: A quadrilateral with perpendicular diagonals is a rhombus. - A quadrilateral with perpendicular diagonals could be a kite or a rhombus. Since it is not necessarily a rhombus, this statement is false.3. Statement ③: A quadrilateral with diagonals that bisect each other is a parallelogram. - This is a known theorem; if diagonals bisect each other, the quadrilateral is a parallelogram. This is true.4. Statement ④: A rhombus with equal diagonals is a square. - A rhombus with equal diagonals has all angles equal to 90 degrees, making it a square. This is true.Summarizing the results:- ① is true.- ② is false.- ③ is true.- ④ is true.The correct answer is (boxed{B}).]

🔑:To analyze each statement step-by-step:① Statement: A quadrilateral with equal diagonals that bisect each other is a rectangle. - Analysis: For a quadrilateral to be a rectangle, it must have four right angles. The property of having equal diagonals that bisect each other is indeed a characteristic of rectangles. Therefore, statement ① is correct.② Statement: A quadrilateral with perpendicular diagonals is a rhombus. - Analysis: While rhombi do have perpendicular diagonals, not all quadrilaterals with perpendicular diagonals are rhombi. The key property of a rhombus is that all sides are of equal length, which is not guaranteed by just having perpendicular diagonals. Therefore, statement ② is incorrect.③ Statement: A quadrilateral with diagonals that bisect each other is a parallelogram. - Analysis: One of the defining properties of parallelograms is indeed that their diagonals bisect each other. This statement does not require the diagonals to be equal or perpendicular, just that they bisect each other. Therefore, statement ③ is correct.④ Statement: A rhombus with equal diagonals is a square. - Analysis: A rhombus is defined as having all sides of equal length. If a rhombus also has equal diagonals, this implies that the angles are right angles, making it a square. Therefore, statement ④ is correct.Given the analysis above, the correct statements are ①, ③, and ④. This matches option B.Therefore, the answer is: boxed{B}.

🧠:[<think>Okay, so I need to solve this system of equations:1. (3y - sqrt{frac{y}{x}} - 6sqrt{xy} + 2 = 0)2. (x^{2} + 81x^{2}y^{4} = 2y^{2})And the answer is supposed to be (left(frac{1}{3}; frac{1}{3}right)) and (left(frac{sqrt[4]{31}}{12}; frac{sqrt[4]{31}}{3}right)). Let me see how to approach this.First, both equations look a bit complicated with radicals and high exponents. Maybe I can manipulate them to substitute variables or find relations between x and y. Let me start with the first equation.Looking at the first equation: (3y - sqrt{frac{y}{x}} - 6sqrt{xy} + 2 = 0). There are square roots here, so maybe I can make substitutions for the radicals. Let me denote (a = sqrt{frac{y}{x}}) and (b = sqrt{xy}). Then, maybe express the equation in terms of a and b.First, compute a and b:- (a = sqrt{frac{y}{x}}), so squaring both sides, (a^2 = frac{y}{x}) → (y = a^2 x)- (b = sqrt{xy}), squaring gives (b^2 = xy)Now, substitute y from the first into the second: (b^2 = x(a^2 x) = a^2 x^2), so (b = a x). Hmm, so (b = a x). Let me write down the first equation in terms of a and b.Original first equation:(3y - a - 6b + 2 = 0)But y is (a^2 x), so substitute:(3a^2 x - a - 6b + 2 = 0)But since b = a x, substitute that in:(3a^2 x - a - 6(a x) + 2 = 0)Simplify:(3a^2 x - a - 6a x + 2 = 0)Hmm, maybe factor out terms with a:Let me see:(3a^2 x - 6a x - a + 2 = 0)Factor a from the first three terms:a(3a x - 6x - 1) + 2 = 0Hmm, not sure if that helps. Alternatively, maybe express x in terms of a or b. Since y = a^2 x, and b = a x, then y = a^2 x = a*(a x) = a*b. So y = a*b.Alternatively, maybe express x in terms of a. From y = a^2 x, so x = y / (a^2). But also, b = sqrt(xy) = sqrt(x*(a^2 x)) = sqrt(a^2 x^2) = a x. So b = a x, so x = b / a. Then, from x = y / a², so b / a = y / a² → y = b / a * a² = a b. So y = a b. So this seems consistent.So maybe express the first equation in terms of a and b. Let's see:First equation: 3y - a - 6b + 2 = 0But y = a b, so:3a b - a - 6b + 2 = 0Hmm, maybe factor terms:a(3b - 1) - 6b + 2 = 0Not sure. Alternatively, rearrange terms:3a b - 6b - a + 2 = 0Factor b from the first two terms:b(3a - 6) - a + 2 = 0Hmm, 3a - 6 = 3(a - 2), so:3b(a - 2) - a + 2 = 0Still not obvious. Maybe let me think about possible substitutions. Alternatively, perhaps look for specific values of a and b that satisfy the equation. For example, maybe a = 1 or 2? Let me test a = 1.If a = 1, then the equation becomes:3*1*b - 1 - 6b + 2 = 0 → 3b - 1 -6b + 2 = (-3b) +1 = 0 → -3b +1=0 → b=1/3So if a=1, then b=1/3. Then, since b = a x, and a=1, so x = b / a = (1/3)/1 = 1/3. Then y = a^2 x = 1^2*(1/3) = 1/3. So (x,y) = (1/3, 1/3). Which is one of the given solutions. That works.So that gives the first solution. Maybe that's a possible solution. Let me check if that satisfies the second equation.Second equation: x² +81x² y⁴ = 2y²Plugging x=1/3, y=1/3:(1/3)² +81*(1/3)²*(1/3)^4 = 2*(1/3)^2Compute each term:First term: 1/9Second term:81*(1/9)*(1/81) =81*(1/9*1/81)=81*(1/729)=1/9Third term:2*(1/9)=2/9So total left side:1/9 +1/9=2/9, right side:2/9. So yes, that works. So (1/3,1/3) is a solution.Great. So that's one solution. Now, need to find the other one.Let me see if there are other possible a and b. Let's suppose a ≠1. Let's try to solve 3ab - a -6b +2=0. Let me rearrange this equation:3ab -6b -a +2=0Let me group terms:3ab -6b = a -2Factor out 3b from left side:3b(a - 2) = a -2Hmm, so if a -2 ≠0, then 3b = (a -2)/(a -2) =1 → 3b=1 → b=1/3. But then if a -2=0, then a=2, and the equation becomes 3*2*b -2 -6b +2=0 →6b -2 -6b +2=0→0=0. So in that case, if a=2, then equation is satisfied for any b?Wait, let me check:If a=2, then original equation is 3*2*b -2 -6b +2=0→6b -2 -6b +2=0→0=0. So it's an identity. So if a=2, then b can be any value? But in reality, a and b are related through the variables x and y.Since a= sqrt(y/x) and b= sqrt(xy). Let's see, if a=2, then sqrt(y/x)=2 → y/x=4 → y=4x. Then b= sqrt(xy)=sqrt(x*4x)=sqrt(4x²)=2x. So b=2x.But from above, when a=2, we had that the first equation becomes 0=0, so no constraints from the first equation. So in that case, we have to use the second equation to solve for x and y.So, if a=2, then y=4x, and from the second equation:x² +81x² y⁴ =2 y²Substitute y=4x:x² +81x²*(4x)^4 =2*(4x)^2Compute (4x)^4=256x⁴, so:x² +81x²*256x⁴ =2*16x²Simplify:x² +81*256x^6 =32x²Subtract x²:81*256x^6 -31x²=0Factor x²:x²(81*256x^4 -31)=0Since x is positive, x≠0, so:81*256x^4 -31=0 →81*256x^4=31 →x^4=31/(81*256)=31/(81*256)Thus x= [31/(81*256)]^{1/4}= [31]^{1/4}/(81^{1/4}*256^{1/4})= 31^{1/4}/(3*4)=31^{1/4}/12Then x= 31^{1/4}/12. Then y=4x=4*(31^{1/4}/12)=31^{1/4}/3. Which is the second solution given. So that works.Therefore, when a=2, we get the other solution. So that makes sense. So the two solutions correspond to a=1 and a=2.Wait, but when a=2, we had to use the second equation. But when a=1, we found b=1/3, leading to x=1/3 and y=1/3. When a=2, we had to use the second equation to find x and y. So that's how the two solutions arise.But let me check if there are other possibilities. Suppose a -2=0, so a=2, which we did. If a -2≠0, then 3b=1, so b=1/3. Then, with b=1/3 and a≠2, but we had a=1 leading to b=1/3. Wait, if a≠2 and 3b=1, then b=1/3, and from the first equation, when we set 3b=1, then 3ab -a -6b +2=0 becomes a*1 -a -2 +2=0 →0=0. Wait, same thing. So if 3b=1, then regardless of a, the equation becomes 0=0? Wait, let's check.Wait, original equation after rearrangement:3ab -6b -a +2=0 → (3ab -6b) -a +2=0 →3b(a -2) -a +2=0If 3b=1, then substitute 3b=1 into the equation:1*(a -2) -a +2=0 →a -2 -a +2=0→0=0. So yes, if 3b=1, then the equation is satisfied for any a. So if 3b=1, then equation is an identity. So in that case, similar to when a=2, if 3b=1, then we have to use the second equation.But 3b=1 →b=1/3. So if b=1/3, then from b = sqrt(xy) =1/3. Also, since a= sqrt(y/x), so y= a²x. So combining with sqrt(xy)=1/3:sqrt(x*(a²x))=1/3 →sqrt(a²x²)=1/3 →a x=1/3.But since a= sqrt(y/x)=sqrt(a²x /x)=sqrt(a²)=a. Wait, that's confusing. Wait, if y= a²x, then sqrt(y/x)=sqrt(a²x /x)=sqrt(a²)=a. Which is consistent with the definition.So if b= sqrt(xy)=1/3, then sqrt(xy)=1/3. Also, a= sqrt(y/x). Let me write variables in terms of a.From y= a²x and sqrt(xy)=1/3:sqrt(x*(a²x))=1/3 →sqrt(a²x²)=1/3 →a x=1/3. So a x=1/3.But a= sqrt(y/x)= sqrt(a²x /x)=sqrt(a²)=a. Which just gives a=a, no new info.So we have:From b=1/3: a x=1/3 (since b= sqrt(xy)=sqrt(a²x²)=a x=1/3)So a x=1/3. Also, from the second equation:x² +81x²y⁴=2y². Substitute y= a²x:x² +81x²*(a²x)^4=2*(a²x)^2Simplify:x² +81x²*(a^8 x^4)=2*a^4 x²→x² +81a^8 x^6=2a^4 x²Subtract x²:81a^8 x^6 +x² -2a^4 x²=0→81a^8 x^6 - (2a^4 -1)x²=0Factor x²:x²(81a^8 x^4 -2a^4 +1)=0Since x>0, x²≠0, so:81a^8 x^4 -2a^4 +1=0But from a x=1/3, x=1/(3a). Substitute x=1/(3a) into the equation:81a^8*(1/(3a))^4 -2a^4 +1=0Compute (1/(3a))^4=1/(81a^4)Thus:81a^8*(1/(81a^4)) -2a^4 +1=0 →(81/81)a^{8-4} -2a^4 +1= a^4 -2a^4 +1= -a^4 +1=0So -a^4 +1=0 →a^4=1 →a=1 (since a is positive, as it's sqrt(y/x))So a=1. Then x=1/(3a)=1/3, y= a²x=1*1/3=1/3. Which is the first solution. So the only solution when 3b=1 is a=1, giving the first solution. So that exhausts that case.Therefore, the two cases when solving the first equation are:1. When a=2, leading to solution via the second equation.2. When 3b=1, leading to a=1 and the first solution.Hence, all solutions are covered. So that's how the two solutions are found.But maybe there's another approach. Let me check the second equation.Second equation: x² +81x²y⁴=2y²Perhaps factor x²:x²(1 +81y⁴)=2y² →x²=2y²/(1 +81y⁴)So x= sqrt(2y²/(1 +81y⁴)) = y*sqrt(2/(1 +81y⁴))So x is expressed in terms of y. Then substitute this into the first equation. Let me try that.Express x in terms of y: x= y*sqrt(2/(1 +81y⁴))Then, substitute into first equation:3y - sqrt(y/x) -6sqrt(xy) +2=0First, compute sqrt(y/x):sqrt(y/x)=sqrt(y / (y*sqrt(2/(1 +81y⁴))))=sqrt(1 / sqrt(2/(1 +81y⁴)))=sqrt( sqrt( (1 +81y⁴)/2 ))= [ (1 +81y⁴)/2 ]^{1/4}Wait, that seems complicated. Let me compute step by step.sqrt(y/x) = sqrt( y / x ) = sqrt( y / (y*sqrt(2/(1 +81y⁴))) ) = sqrt(1 / sqrt(2/(1 +81y⁴))) )Which is sqrt( sqrt( (1 +81y⁴)/2 ) )Which is the same as [ (1 +81y⁴)/2 ]^{1/4}Similarly, compute sqrt(xy):sqrt(xy)=sqrt( y*sqrt(2/(1 +81y⁴)) * y )=sqrt(y² sqrt(2/(1 +81y⁴))) = y * [2/(1 +81y⁴)]^{1/4}So the first equation becomes:3y - [ (1 +81y⁴)/2 ]^{1/4} -6* y*[2/(1 +81y⁴)]^{1/4} +2=0This looks very complicated. Maybe not the best approach. Alternatively, perhaps substitute t = y² or something else.Alternatively, notice that the second equation is x² +81x²y⁴=2y². Let me divide both sides by y²:(x²/y²) +81x²y²=2Let me denote z = x/y. Then x = z y. Substitute into the equation:(z² y²)/y² +81(z y)^2 y²=2 →z² +81z² y^4=2Factor z²:z²(1 +81y^4)=2 →z²=2/(1 +81y^4)So z= sqrt(2/(1 +81y^4))But z= x/y, so x= y*sqrt(2/(1 +81y^4))Same as before. So maybe substitution isn't helpful here. Let me think again.Alternatively, let's look at the given solutions. The first one is (1/3,1/3). Let's verify how that works in the first equation:First equation: 3*(1/3) - sqrt( (1/3)/ (1/3) ) -6*sqrt( (1/3)*(1/3) ) +2Simplify:1 - sqrt(1) -6*sqrt(1/9) +2 =1 -1 -6*(1/3)+2=0 -2 +2=0. Correct.The second solution is (31^{1/4}/12, 31^{1/4}/3). Let me check if that satisfies the first equation.Let me compute each term:Let’s denote k=31^{1/4}, so x=k/12, y=k/3.Compute 3y: 3*(k/3)=kCompute sqrt(y/x): sqrt( (k/3)/(k/12) )=sqrt( (k/3)*(12/k) )=sqrt(4)=2Compute 6*sqrt(xy):6*sqrt( (k/12)*(k/3) )=6*sqrt( k²/(36) )=6*(k/6)=kSo the first equation becomes:k -2 -k +2=0. Simplify: 0=0. Correct.So that works. So the second solution works. But how would one derive that?From the earlier approach, when a=2, we derived the second solution. So if we let a=2, which gives us a relation between x and y, then substitute into the second equation. Let me try that again.Set a=2: sqrt(y/x)=2 →y=4x. Substitute into the second equation: x² +81x²*(4x)^4=2*(4x)^2Compute (4x)^4=256x^4, so:x² +81x²*256x^4=2*16x²Simplify:x² +20736x^6=32x²Subtract x²:20736x^6 -31x²=0Factor x²:x²(20736x^4 -31)=0Since x>0, x≠0, so:20736x^4=31 →x^4=31/20736 →x= (31)^{1/4}/ (20736)^{1/4}Compute 20736= 12^4? Let's check: 12^4=12*12*12*12=20736. Yes. So 20736=12^4, so x= (31)^{1/4}/12. Then y=4x= (31)^{1/4}/3. So that's the second solution.Therefore, the two solutions come from setting a=1 and a=2 in the first equation, leading to two different substitutions into the second equation.Alternatively, since when we manipulated the first equation, we found that either a=2 (leading to second solution) or 3b=1 (leading to first solution). Hence, only two solutions.Therefore, the solutions are:1. When a=1: x=1/3, y=1/32. When a=2: x=31^{1/4}/12, y=31^{1/4}/3Thus, these are the only two positive solutions.Final AnswerThe pairs of positive numbers are (boxed{left(frac{1}{3}, frac{1}{3}right)}) and (boxed{left(frac{sqrt[4]{31}}{12}, frac{sqrt[4]{31}}{3}right)}).</think>To solve the system of equations:1. (3y - sqrt{frac{y}{x}} - 6sqrt{xy} + 2 = 0)2. (x^{2} + 81x^{2}y^{4} = 2y^{2})we start by making substitutions for the radicals in the first equation. Let (a = sqrt{frac{y}{x}}) and (b = sqrt{xy}). This gives us (y = a^2 x) and (b = a x). Substituting these into the first equation, we get:[3a^2 x - a - 6ax + 2 = 0]Rearranging and factoring terms, we find:[3ab - a - 6b + 2 = 0]where (a = sqrt{frac{y}{x}}) and (b = sqrt{xy}). Testing specific values for (a) and solving for (b), we find:1. If (a = 1), then (3b - 1 - 6b + 2 = 0) leads to (b = frac{1}{3}). This gives (x = frac{1}{3}) and (y = frac{1}{3}).2. If (a = 2), the equation becomes an identity, leading us to use the second equation. Substituting (y = 4x) into the second equation, we solve for (x) and find (x = frac{sqrt[4]{31}}{12}) and (y = frac{sqrt[4]{31}}{3}).Verifying these solutions in both equations confirms they are correct. The pairs of positive numbers that satisfy the system are:[boxed{left(frac{1}{3}, frac{1}{3}right)} text{ and } boxed{left(frac{sqrt[4]{31}}{12}, frac{sqrt[4]{31}}{3}right)}],<think>Okay, so I need to solve this system of equations where both x and y are positive numbers. Let me write down the equations again to make sure I have them right:First equation:3y - √(y/x) - 6√(xy) + 2 = 0Second equation:x² + 81x²y⁴ = 2y²Hmm. Both equations look a bit complicated. Let me see if I can find a substitution or manipulate them to make things simpler.Starting with the first equation. There are terms with square roots involving both x and y. Maybe I can let t = √(x) or something like that. Alternatively, maybe set variables for the radicals. Let's see.Let me note that √(xy) is the same as √x * √y. Similarly, √(y/x) is √y / √x. Maybe if I set u = √x and v = √y, then those terms become v/u and uv. Let me try that substitution.Let u = √x and v = √y. Then x = u² and y = v². Let me substitute these into the equations.First equation:3v² - (v / u) - 6(u v) + 2 = 0Second equation:(u²)² + 81(u²)(v²)^4 = 2(v²)^2Simplify the second equation:u⁴ + 81u²v⁸ = 2v⁴Hmm, maybe that's not too helpful yet. Let's look back at the first equation.First equation after substitution:3v² - (v/u) - 6uv + 2 = 0That still has both u and v, but perhaps we can manipulate it. Let me try to multiply through by u to eliminate the denominator:3v²u - v - 6u²v + 2u = 0Hmm, not sure if that helps. Maybe factor terms?Looking at 3v²u -6u²v, that's 3uv(v - 2u). Then we have -v + 2u. So:3uv(v - 2u) - (v - 2u) = 0Wait, let's see:3uv(v - 2u) - (v - 2u) = (v - 2u)(3uv - 1) = 0Wait, is that possible? Let me check:Original expression after multiplying by u: 3v²u - v -6u²v +2uGroup terms:(3v²u -6u²v) + (-v + 2u)Factor 3uv from the first group: 3uv(v - 2u) and factor -1 from the second group: -1(v - 2u)So indeed:3uv(v - 2u) -1(v - 2u) = (v - 2u)(3uv -1) = 0So either v - 2u = 0 or 3uv -1 = 0So two cases:Case 1: v = 2uCase 2: 3uv = 1 => uv = 1/3So let's analyze both cases.Case 1: v = 2uSince u = √x and v = √y, this implies √y = 2√x => y = 4xNow substitute y = 4x into the second equation:x² + 81x²(4x)^4 = 2(4x)^2Compute (4x)^4 = 256x^4, so:x² + 81x²*256x^4 = 2*16x²Simplify:x² + 81*256x^6 = 32x²Compute 81*256: 81*256 = (80*256) + (1*256) = 20480 + 256 = 20736So:x² + 20736x^6 = 32x²Subtract x² from both sides:20736x^6 = 31x²Divide both sides by x² (since x is positive, x ≠ 0):20736x^4 = 31Thus, x^4 = 31 / 20736Take fourth root:x = (31 / 20736)^(1/4) = (31)^(1/4) / (20736)^(1/4)Compute 20736: 12^4 = 20736 because 12^2 = 144, 144^2 = 20736So 20736^(1/4) = 12Thus, x = (31)^(1/4) / 12Then y = 4x = 4*(31)^(1/4)/12 = (31)^(1/4)/3So one solution is (x, y) = (31^(1/4)/12, 31^(1/4)/3). That matches one of the given answers.Case 2: 3uv = 1 => uv = 1/3Since u = √x and v = √y, then uv = √(xy) = 1/3Therefore, √(xy) = 1/3 => xy = 1/9So xy = 1/9. Let me use this in the second equation.Original second equation: x² + 81x²y⁴ = 2y²But we know that xy = 1/9, so y = 1/(9x). Let's substitute that into the second equation.First, substitute y = 1/(9x) into the second equation:x² + 81x²*(1/(9x))^4 = 2*(1/(9x))²Compute each term:First term: x²Second term: 81x²*(1/(9x))^4 = 81x²*(1/(6561x^4)) = 81/(6561x²) = (81/6561)/x² = 1/81x²Third term: 2*(1/(9x))² = 2/(81x²)So equation becomes:x² + (1)/(81x²) = 2/(81x²)Subtract 2/(81x²) from both sides:x² - (1)/(81x²) = 0Multiply both sides by 81x² to eliminate denominators:81x^4 -1 = 0So 81x^4 =1 => x^4=1/81 => x= (1/81)^(1/4)Since x is positive, x= (1/3^4)^(1/4)=1/3Thus, x=1/3. Then y=1/(9x)=1/(9*(1/3))=1/3.So another solution is (x, y)=(1/3, 1/3). That's the other given answer.So both cases give the two solutions provided. Let me check if there are any other possibilities, but in the first equation, we factored and got only two cases, so these should be the only solutions.Therefore, the solutions are (1/3, 1/3) and (31^(1/4)/12, 31^(1/4)/3). The second one can be written as (∜31/12, ∜31/3), which matches the given answer.Final AnswerThe pairs of positive numbers are boxed{left(frac{1}{3}, frac{1}{3}right)} and boxed{left(frac{sqrt[4]{31}}{12}, frac{sqrt[4]{31}}{3}right)}.</think>Given the system of equations:[left{begin{array}{l}3y - sqrt{frac{y}{x}} - 6sqrt{xy} + 2 = 0 x^2 + 81x^2y^4 = 2y^2end{array}right.]we start by substituting ( u = sqrt{x} ) and ( v = sqrt{y} ). This transforms the equations into:1. ( 3v^2 - frac{v}{u} - 6uv + 2 = 0 )2. ( u^4 + 81u^2v^8 = 2v^4 )For the first equation, multiplying through by ( u ) gives:[3v^2u - v - 6u^2v + 2u = 0]Factoring, we get:[(v - 2u)(3uv - 1) = 0]This results in two cases:Case 1: ( v = 2u )Substituting ( v = 2u ) implies ( sqrt{y} = 2sqrt{x} ) or ( y = 4x ). Substituting ( y = 4x ) into the second equation:[x^2 + 81x^2(4x)^4 = 2(4x)^2]Simplifying, we find:[x^2 + 20736x^6 = 32x^2 implies 20736x^4 = 31 implies x = frac{sqrt[4]{31}}{12}]Thus, ( y = 4x = frac{sqrt[4]{31}}{3} ).Case 2: ( 3uv = 1 )This implies ( sqrt{xy} = frac{1}{3} ) or ( xy = frac{1}{9} ). Substituting ( y = frac{1}{9x} ) into the second equation:[x^2 + 81x^2left(frac{1}{9x}right)^4 = 2left(frac{1}{9x}right)^2]Simplifying, we find:[81x^4 = 1 implies x = frac{1}{3}]Thus, ( y = frac{1}{9x} = frac{1}{3} ).The pairs of positive numbers that satisfy the system are:[boxed{left(frac{1}{3}, frac{1}{3}right)} text{ and } boxed{left(frac{sqrt[4]{31}}{12}, frac{sqrt[4]{31}}{3}right)}]]

🔑:Solution. Let sqrt{frac{y}{x}}=u, sqrt{x y}=v quad (with u>0, quad v>0). Then quad u v=sqrt{frac{y}{x}} cdot sqrt{x y}=sqrt{y^{2}}=|y|=y,frac{v}{u}=sqrt{x y}: sqrt{frac{y}{x}}=sqrt{x^{2}}=|x|=x, since by the condition x and y are positive. The system takes the formleft{begin{array} { l } { 3 u v - u - 6 v + 2 = 0 } { frac { v ^ { 2 } } { u ^ { 2 } } + 8 1 v ^ { 6 } u ^ { 2 } = 2 u ^ { 2 } v ^ { 2 } }end{array} Leftrightarrow left{begin{array}{l}(u-2)(3 v-1)=0 1+81 v^{4} u^{4}=2 u^{4}end{array}right.right.From the first equation, it follows that u=2 or v=frac{1}{3}.If u=2, then 1+81 cdot 16 v^{4}=32, from which v=frac{sqrt[4]{31}}{6}; then x=frac{v}{u}=frac{sqrt[4]{31}}{12}, y=u v=frac{sqrt[4]{31}}{3}.If v=frac{1}{3}, then 1+u^{4}=2 u^{4}, from which u=1; then x=frac{v}{u}=frac{1}{3}, y=u v=frac{1}{3}.

🧠:[<think>Okay, let's see. The problem is about a rectangle ABCD where AB is 2 units and AD is 4 units. Points E and F are on AD and BC respectively, with AE = 1 and BF = 3. Then, the quadrilateral AEFB is folded along EF, and when this happens, the projection of point B onto the plane CDEF is point H on line DE. We need to find the dihedral angle A-DE-F.First, let me try to visualize the rectangle. ABCD is a rectangle with AB = 2 (so the length from A to B is 2), and AD = 4 (so the height from A to D is 4). Points E and F are on sides AD and BC. Since AE = 1, E is 1 unit above A on AD. Similarly, BF = 3, so since BC is also 4 units (since AD and BC are both sides of the rectangle), F is 3 units below B on BC. So, the coordinates might help here. Let me assign coordinates to the rectangle.Let me place point A at the origin (0, 0, 0). Then, since AB is 2 units, point B would be at (2, 0, 0). AD is 4 units, so point D is at (0, 4, 0). Therefore, point C, which is opposite to A, would be at (2, 4, 0). Now, point E is on AD, 1 unit from A. Since AD goes from (0,0,0) to (0,4,0), moving along the y-axis, E would be at (0,1,0). Similarly, point F is on BC, 3 units from B. BC goes from (2,0,0) to (2,4,0). Moving 3 units from B along BC would reach (2,3,0). So, F is at (2,3,0).Quadrilateral AEFB is folded along EF. So, before folding, AEFB is a quadrilateral with points A(0,0,0), E(0,1,0), F(2,3,0), and B(2,0,0). When folded along EF, points A and B (which are on either side of EF) will move in 3D space. The projection of point B onto the plane CDEF after folding is point H on line DE. We need to find the dihedral angle along the line DE between the two planes: the original plane containing A-D-E and the plane containing F-D-E. The dihedral angle is the angle between these two planes.Wait, but the dihedral angle A-DE-F is the angle between the two planes that share the common line DE. One plane is the plane containing points A, D, E, and the other is the plane containing points F, D, E. However, after folding, the plane containing F might have changed. Wait, but actually, when folding along EF, the position of F remains the same, right? Because folding is along EF, so F is on the fold line. So, point B is being folded over to a new position, but F remains fixed. Similarly, point A is being folded over. Wait, but the problem says the projection of point B on plane CDEF is H on DE. So, after folding, when we project the folded point B (let's call it B') onto plane CDEF, which is the original plane containing C, D, E, F, the projection is H on DE.So, first, I need to figure out the coordinates of point B' after folding. Then, find its projection H on plane CDEF, which lies on DE. Then, with this information, perhaps calculate the dihedral angle between the two planes.To calculate the dihedral angle between two planes, one way is to find the angle between their normal vectors. So, if we can find the normal vectors of the two planes (plane ADE and plane FDE), then the dihedral angle can be found using the dot product.But wait, after folding, the plane FDE might not be the same as before. Wait, no. Wait, DE is a common edge. Plane ADE is before folding, but after folding, the position of point A would change. Wait, maybe not. Wait, when we fold along EF, the part of the quadrilateral AEFB that's on one side of EF is folded over to the other side. So, points A and B are moved to new positions in 3D space, but points E and F remain fixed.So, after folding, point A is moved to a new position A', and point B is moved to a new position B'. The projection of B' onto plane CDEF is H on DE.Given that H is on DE, which is the line from D(0,4,0) to E(0,1,0). So, DE is the vertical line along x=0, z=0, from (0,4,0) to (0,1,0). So, any point on DE has coordinates (0, y, 0) where y ranges from 1 to 4.Since H is the projection of B' onto plane CDEF, which is the plane containing points C, D, E, F. The projection is orthogonal, so H is the closest point on plane CDEF to B', lying on DE.But since H is on DE, and DE is part of plane CDEF, then H is the foot of the perpendicular from B' to DE.Wait, but plane CDEF is a plane in 3D space. The projection of B' onto plane CDEF is the point H such that the vector from H to B' is perpendicular to the plane CDEF. However, since H is on DE, this gives a constraint.Alternatively, maybe since after folding, B' is a point in 3D space such that folding over EF maps B to B', maintaining the distance from EF. So, EF is the fold line, so B and B' are symmetric with respect to the line EF. Wait, but folding along EF would reflect B across the line EF to get B'. But since EF is in 3D space, but we need to consider the reflection across the folding line EF in the plane of the quadrilateral AEFB.Wait, maybe it's better to model this folding in 3D coordinates.Original coordinates:A(0,0,0), E(0,1,0), F(2,3,0), B(2,0,0).When folded along EF, the quadrilateral AEFB is split into two triangles: AEF and EFB. After folding, triangle EFB is lifted off the plane, and triangle AEF is also lifted? Wait, no. Folding typically would create a dihedral angle between the two planes. However, since we are folding over EF, the points A and B are on different sides of EF, so folding would bring them together? Maybe.Alternatively, perhaps folding the paper along EF causes the part containing A to fold over onto the part containing B or vice versa. However, the problem states that the projection of B onto plane CDEF is H on DE. So, after folding, B has moved to B', and the projection (orthogonal) of B' onto plane CDEF is H, which is on DE.So, let's consider that folding along EF will create a new position for B, called B', such that B' is the reflection of B over the line EF. But reflection over a line in 3D space is not straightforward. Alternatively, in the plane of the quadrilateral AEFB, folding over EF would reflect B across EF to a new point B'.Wait, the quadrilateral AEFB is in the plane of the original rectangle ABCD. However, ABCD is a rectangle, so it's a flat 2D plane. When folding along EF, which is a line in this plane, the part of the quadrilateral on one side of EF is folded over into 3D space, creating a dihedral angle. So, after folding, points A and B are moved to positions A' and B' such that EF remains fixed, and the segments AE and BF are now connected over the fold.But the problem mentions the projection of point B (probably after folding, the new position B') onto plane CDEF is H on DE. So, we need to find the coordinates of B' such that when we fold along EF, B maps to B', and the projection of B' onto plane CDEF is H on DE.To find this, maybe we can model EF as a hinge, and when folding, the angle between the two halves of the quadrilateral AEFB along EF becomes a dihedral angle. Then, the projection of B' onto plane CDEF is H.Alternatively, since after folding, B is moved to B', and H is the projection, we can use coordinates to model this.Let me set up a coordinate system. Let me assign coordinates as before:A(0,0,0), B(2,0,0), C(2,4,0), D(0,4,0), E(0,1,0), F(2,3,0).Now, EF is the line from E(0,1,0) to F(2,3,0). Let's parametrize EF. The vector from E to F is (2,2,0). So, any point on EF can be written as E + t(F - E) = (0,1,0) + t(2,2,0) = (2t, 1 + 2t, 0), where t ∈ [0,1].When folding along EF, points on one side of EF will be reflected over EF. Since quadrilateral AEFB is being folded, which side is being folded? Points A and B are on opposite sides of EF.To determine the reflection over line EF in 3D space, we need to find the reflection of point B across line EF. However, reflection over a line in 3D is a bit complicated. Alternatively, since the original figure is in the plane z=0, and folding along EF would create a dihedral angle, so points above the fold (EF) would stay, and points below would be lifted into the z-direction or vice versa. However, the exact direction depends on the folding.Alternatively, maybe it's easier to consider the dihedral angle directly. The dihedral angle between the two planes (original plane ADE and the folded plane FDE) along the line DE. Wait, but DE is part of both planes. Wait, plane ADE is the plane containing A, D, E, and plane FDE is the plane containing F, D, E. So, the dihedral angle is between these two planes along their intersection line DE. However, after folding, plane FDE might have changed? Wait, no. Wait, folding is along EF, which is not part of DE, so maybe DE remains fixed. Wait, DE is from D(0,4,0) to E(0,1,0). EF is from E(0,1,0) to F(2,3,0). So, DE and EF meet at E.Wait, perhaps after folding, the position of F remains the same, since it's on BC which is part of the original plane. Wait, but BC is part of the rectangle, but when folding, F is on the fold line EF, so F remains fixed. So, points E and F stay in place, while points A and B are moved.Alternatively, when folding along EF, the part of the paper containing A is folded over to the other side of EF. Similarly, the part containing B is folded over. However, since we are dealing with a dihedral angle between planes ADE and FDE, perhaps we need to consider the angle between these two planes in the folded structure.But I think the key here is that after folding, the projection of B onto plane CDEF is H on DE. So, perhaps we can find the coordinates of H, then use that to find the position of B', and from there compute the dihedral angle.So, let's denote the projection of B' onto plane CDEF as H. Since H is on DE, which is the line from D(0,4,0) to E(0,1,0), so H has coordinates (0, h, 0) for some h between 1 and 4.The projection of B' onto plane CDEF is H, which means that B' lies along the line perpendicular to plane CDEF at H. However, plane CDEF is the original plane of the rectangle, which is the z=0 plane. Wait, no. Wait, plane CDEF is part of the original rectangle, so it's also in the z=0 plane. So, if we consider that after folding, B' is lifted off the z=0 plane, then the projection of B' onto the z=0 plane is H. But H is on DE, so coordinates of H are (0, h, 0). Therefore, the coordinates of B' would be (0, h, k) for some k, since the projection onto z=0 is (0, h, 0).But wait, the projection is orthogonal, so B' is directly above H in the z-direction? If plane CDEF is the z=0 plane, then yes. But in that case, folding along EF would move B out of the plane, giving it a z-coordinate. But in reality, folding would reflect B over EF into a new position B', which is in 3D space. However, the projection of B' onto the original plane (z=0) is H.Alternatively, maybe the coordinate system is such that after folding, the plane CDEF remains in the z=0 plane, and B' is above or below with some z-coordinate, such that when you project orthogonally (i.e., drop a perpendicular) from B' to CDEF, you land on H.So, B' has coordinates (0, h, k), with H=(0, h, 0). Then, the distance from B' to H is |k|. But we need to find h and k such that the folding over EF results in B' being the reflection of B over EF.But how does folding affect the coordinates? Folding over a line in a plane reflects points across that line. However, in 3D, reflecting over a line can be more complex. Alternatively, since we are folding the paper along EF, the image of B after folding (B') should satisfy that the distance from B to EF is the same as the distance from B' to EF. Also, the line segment BB' is perpendicular to EF and is bisected by EF.But EF is in the z=0 plane. So, if we can find the reflection of B over EF in the plane of the rectangle, then B' would be that reflection. However, since folding takes B from one side of EF to the other in 3D space, but maintaining the same distance from EF.Alternatively, maybe the reflection in 3D space. Let's think.First, let's find the reflection of point B(2,0,0) over the line EF in the plane of the rectangle (z=0). To do this, we can use the formula for reflection over a line in 2D.In 2D, reflecting a point over a line can be done by finding the perpendicular projection of the point onto the line, then moving the point the same distance on the other side.So, let's consider the line EF in 2D (since we're in the z=0 plane). The line EF goes from E(0,1) to F(2,3). The equation of line EF can be found as follows.The slope of EF is (3 - 1)/(2 - 0) = 2/2 = 1. So, the line EF is y = x + 1.Wait, when x=0, y=1 (point E), and when x=2, y=3 (point F). So, equation is y = x + 1.Now, reflecting point B(2,0) over the line y = x + 1.The formula for reflection over a line ax + by + c = 0 is:If the line is ax + by + c = 0, then the reflection of a point (x0, y0) is:(x', y') = (x0 - 2a(ax0 + by0 + c)/(a² + b²), y0 - 2b(ax0 + by0 + c)/(a² + b²))First, let's write the line EF in standard form. y = x + 1 can be rewritten as x - y + 1 = 0. So, a = 1, b = -1, c = 1.So, for point B(2,0):Compute ax0 + by0 + c = 1*2 + (-1)*0 + 1 = 2 + 0 + 1 = 3.Then,x' = 2 - 2*1*(3)/(1 + 1) = 2 - 6/2 = 2 - 3 = -1y' = 0 - 2*(-1)*(3)/(1 + 1) = 0 + 6/2 = 3So, the reflection of B over line EF in the plane z=0 is (-1, 3, 0). However, this point is not in the original rectangle. But folding along EF would map B to this reflected point. However, this point (-1, 3, 0) is outside the original rectangle. But since we are folding into 3D space, maybe B is lifted into the z-axis.Wait, but the problem says that after folding, the projection of B (i.e., B') onto plane CDEF is H on DE. If B were reflected to (-1, 3, 0), its projection would be itself, but that's not on DE. So, this suggests that the reflection is not in the plane z=0, but in 3D space. Therefore, perhaps the reflection is not in the plane of the rectangle, but over the line EF as a line in 3D space, keeping EF fixed.Alternatively, since folding is along EF, the crease line EF remains in the plane, and the two parts of the quadrilateral AEFB are folded over each other. So, the dihedral angle between the two planes (the original plane and the folded plane) is the angle we need to find.But the problem states that the projection of B' onto CDEF is H on DE. So, perhaps B' is somewhere above DE such that when you project it down, you get H on DE. So, we need to find the coordinates of B' such that:1. The distance from B' to EF is the same as the distance from B to EF (since folding preserves distances).2. The projection of B' onto plane CDEF is H, which is on DE.First, let's compute the distance from B to EF in the original plane.In 2D, the distance from point B(2,0) to line EF (y = x + 1) is |ax0 + by0 + c| / sqrt(a² + b²) = |1*2 + (-1)*0 + 1| / sqrt(1 + 1) = |3| / sqrt(2) = 3/√2.Therefore, the distance from B to EF is 3/√2. So, after folding, B' should be at the same distance from EF but on the other side.However, since we are in 3D, the distance from B' to EF is still 3/√2, but in 3D space. The projection of B' onto plane CDEF is H on DE. So, the coordinates of H are (0, h, 0), as DE is the line x=0, z=0, from (0,4,0) to (0,1,0). Therefore, h is between 1 and 4.Now, since H is the projection of B' onto CDEF (z=0 plane), B' must be vertically above H, so B' has coordinates (0, h, k), where k is the height. Then, the distance from B' to EF should be 3/√2.First, let's find the coordinates of B' as (0, h, k). Then, the distance from B' to line EF in 3D space should be 3/√2.But line EF in 3D is from E(0,1,0) to F(2,3,0). So, parametric equations for EF are x = 2t, y = 1 + 2t, z = 0, where t ∈ [0,1].The distance from point B'(0, h, k) to line EF can be calculated using the formula for the distance from a point to a line in 3D:Distance = |(P - Q) × v| / |v|, where P is the point, Q is a point on the line, and v is the direction vector of the line.Here, Q can be E(0,1,0), and v is the direction vector from E to F, which is (2,2,0). So, vector v = (2,2,0).Vector P - Q is B' - E = (0 - 0, h - 1, k - 0) = (0, h - 1, k).Then, cross product (P - Q) × v is determinant:|i   j   k||0 h-1 k||2   2   0|= i*((h - 1)*0 - k*2) - j*(0*0 - k*2) + k*(0*2 - (h - 1)*2)= i*(-2k) - j*(-2k) + k*(-2(h - 1))= (-2k, 2k, -2(h - 1))The magnitude of this cross product is sqrt[ (-2k)^2 + (2k)^2 + (-2(h - 1))^2 ] = sqrt[4k² + 4k² + 4(h - 1)^2] = sqrt[8k² + 4(h - 1)^2] = 2*sqrt[2k² + (h - 1)^2]The magnitude of vector v is sqrt(2² + 2² + 0²) = sqrt(8) = 2√2.Therefore, the distance from B' to EF is [2*sqrt(2k² + (h - 1)^2)] / (2√2) ) = sqrt(2k² + (h - 1)^2) / √2But we know this distance should equal 3/√2.Therefore:sqrt(2k² + (h - 1)^2) / √2 = 3/√2Multiply both sides by √2:sqrt(2k² + (h - 1)^2) = 3Square both sides:2k² + (h - 1)^2 = 9 --- Equation (1)Additionally, since H is the projection of B' onto plane CDEF, which is the z=0 plane, H is (0, h, 0). But DE is the line from D(0,4,0) to E(0,1,0), so H lies on DE. Therefore, the coordinates of H must satisfy x=0, z=0, and y between 1 and 4. Therefore, h is between 1 and 4.Moreover, since folding maps B to B', the fold line EF is the perpendicular bisector of the segment BB'. However, since we are folding in 3D, the line segment BB' should be perpendicular to EF and bisected by EF.Wait, in 2D, reflecting a point over a line gives that the line is the perpendicular bisector of the segment joining the point and its reflection. In 3D, this might not hold, but given that the folding is along EF, perhaps the line EF is the axis of rotation, and the segment BB' is perpendicular to EF and lies in the plane of the fold.But perhaps in this case, the distance condition is already captured by Equation (1). Additionally, we need another condition to find h and k.Another condition is that the folding preserves the distances from the fold line EF. Since folding is an isometry, the distance from B to EF is equal to the distance from B' to EF, which we already used.But perhaps there's another condition: the original segment BF and the folded segment B'F should have the same length. Wait, BF is part of the original rectangle, BF = 3 units (given in the problem: BF=3). After folding, point B moves to B', but point F remains fixed. So, the distance from B' to F should still be the same as BF, which is 3 units. Therefore, |B'F| = 3.Coordinates of F are (2,3,0), coordinates of B' are (0, h, k). Therefore, distance B'F is sqrt[(2 - 0)^2 + (3 - h)^2 + (0 - k)^2] = sqrt[4 + (3 - h)^2 + k²] = 3.Therefore:sqrt[4 + (3 - h)^2 + k²] = 3Squaring both sides:4 + (3 - h)^2 + k² = 9So, (3 - h)^2 + k² = 5 --- Equation (2)Now, we have two equations:Equation (1): 2k² + (h - 1)^2 = 9Equation (2): (3 - h)^2 + k² = 5We can solve these two equations for h and k.First, expand Equation (2):(3 - h)^2 + k² = 5=> 9 - 6h + h² + k² = 5=> h² -6h + k² + 9 -5 = 0=> h² -6h + k² +4 =0 --- Equation (2a)Similarly, expand Equation (1):2k² + (h -1)^2 =9=> 2k² + h² - 2h +1 =9=> h² - 2h + 2k² +1 -9 =0=> h² - 2h + 2k² -8 =0 --- Equation (1a)Now, we have:Equation (1a): h² - 2h + 2k² -8 =0Equation (2a): h² -6h + k² +4 =0Let’s subtract Equation (2a) from Equation (1a):(h² - 2h + 2k² -8) - (h² -6h + k² +4) = 0 -0h² -2h +2k² -8 -h² +6h -k² -4 =0( -2h +6h ) + (2k² -k²) + (-8 -4) =04h + k² -12 =0So, 4h + k² =12 --- Equation (3)From Equation (2a): h² -6h +k² = -4But from Equation (3): k² =12 -4hSubstitute into Equation (2a):h² -6h + (12 -4h) = -4h² -6h +12 -4h = -4h² -10h +12 = -4h² -10h +16 =0Solving quadratic equation:h² -10h +16=0Discriminant D = 100 -64=36Solutions: h=(10 ±6)/2= (16)/2=8 or (4)/2=2Thus, h=8 or h=2. But since H is on DE, which is from (0,1,0) to (0,4,0), h must be between 1 and4. Therefore, h=2 is the valid solution.So, h=2. Then from Equation (3): k²=12 -4h=12 -8=4. Thus, k=±2.But since folding is upwards or downwards? The problem doesn't specify, but the dihedral angle is a positive angle between 0 and 180 degrees, so the sign might not matter. Let's assume k=2 (upwards). So, coordinates of B' are (0,2,2).Therefore, H is (0,2,0), which is on DE (since DE is from (0,4,0) to (0,1,0)), h=2 is between 1 and4.Now, with B' at (0,2,2), we can find the dihedral angle between the planes ADE and FDE. The dihedral angle is the angle between these two planes along their common line DE.To find the dihedral angle, we can find the angle between the normal vectors of the two planes.First, let's find the normal vector of plane ADE. Plane ADE contains points A(0,0,0), D(0,4,0), and E(0,1,0). However, all these points are in the z=0 plane. Wait, but A, D, E are colinear? Wait, no. Wait, A(0,0,0), D(0,4,0), E(0,1,0). These three points are along the y-axis at x=0, z=0. So, they are colinear. Wait, that can't be. Wait, if three points are colinear, they don't form a plane. Wait, this is a problem.Wait, hold on. Plane ADE: points A(0,0,0), D(0,4,0), and E(0,1,0) are all along the y-axis, so they are colinear. Therefore, the plane ADE is not defined; a plane requires three non-colinear points. This is a mistake. Wait, but in the problem statement, the dihedral angle is A-DE-F. So, the two planes are ADE and FDE. But if ADE is a line, then that doesn't make sense. Wait, perhaps there's a misunderstanding.Wait, maybe after folding, point A is moved to a new position A', and plane A'DE is the other plane. Because before folding, points A, D, E are colinear, so the plane ADE is degenerate. However, after folding, point A is moved to A', forming a non-degenerate plane A'DE. Similarly, plane FDE is the plane containing F, D, E. But F is at (2,3,0), so plane FDE is non-degenerate.Wait, the problem says "the dihedral angle A-DE-F". In dihedral angle notation, the common edge is DE, and the two planes are A-D-E and F-D-E. But initially, A-D-E is a line, but after folding, if A is moved to A', then plane A'-D-E is a valid plane. Therefore, perhaps the dihedral angle is between planes A'-DE and F-DE.Yes, this must be the case. Because before folding, plane ADE is degenerate (a line), but after folding, point A is moved to A', creating a valid plane A'-DE. Therefore, the dihedral angle is between planes A'-DE and F-DE along their common edge DE.Similarly, since we found the coordinates of B', maybe we can find the coordinates of A' as well. But the problem only mentions the projection of B', so maybe we don't need A' for this calculation. However, since the dihedral angle is between A'-DE and F-DE, we need to find the normals of these two planes.Alternatively, since folding over EF swaps points A and B in some way. But since we have found B'(0,2,2), maybe we can find the coordinates of A' as well.Wait, when folding along EF, points A and B are mirrored over EF. So, if B is mapped to B'(0,2,2), then A should be mapped to a point A', such that folding over EF reflects A to A'.But let's check. In the original quadrilateral AEFB, folding along EF would swap points A and B across the fold line EF. However, in reality, folding would create two mirrored images. But since we have the coordinates of B', maybe we can find A' by ensuring that the fold preserves distances.Alternatively, note that in the original quadrilateral AEFB, the crease line EF is the same, and after folding, the positions of A and B are swapped. But given that the paper is folded over EF, the positions of A and B in 3D space will be such that their distances to EF are preserved, and the segments AE and BF are now connected over the fold.Alternatively, since we found B', maybe we can find the equation of the plane FDE (original plane) and the plane A'DE (after folding), then compute the dihedral angle.Plane FDE contains points F(2,3,0), D(0,4,0), and E(0,1,0). Let's find the normal vector of this plane.Vectors in plane FDE:FD = D - F = (0 - 2, 4 - 3, 0 - 0) = (-2, 1, 0)FE = E - F = (0 - 2, 1 - 3, 0 - 0) = (-2, -2, 0)Cross product FD × FE:|i   j   k||-2   1   0||-2  -2   0|= i*(1*0 - 0*(-2)) - j*(-2*0 - 0*(-2)) + k*(-2*(-2) - 1*(-2))= i*0 - j*0 + k*(4 + 2) = 6kSo, the normal vector of plane FDE is (0, 0, 6), or (0, 0, 1) when normalized.Similarly, plane A'DE contains points A'(which we need to find), D(0,4,0), and E(0,1,0). Wait, but if we know that after folding, quadrilateral AEFB is folded over EF, and given that B is mapped to B'(0,2,2), perhaps A is mapped to a point A' such that the folding preserves the structure.Alternatively, since EF is the fold line, the segment AE is folded over to meet the segment FB. But this is getting a bit unclear. Maybe since we know B', we can find the normal vector of the plane A'DE.Alternatively, note that after folding, the plane A'DE contains points A', D, E. Since D and E are on the original plane CDEF (z=0), and A' is the folded position of A. However, we might not need the exact coordinates of A' if we can find the normal vector using other information.Alternatively, the dihedral angle can be found using the angle between the normals of planes FDE and A'DE. Since plane FDE has a normal vector (0,0,1) as calculated earlier, plane A'DE would have a normal vector that we can calculate once we know A'.But we need to find A'. However, given that folding along EF maps A to A', and B to B', with B'(0,2,2), perhaps we can find A' by using the fact that folding preserves distances and the structure.Alternatively, since the original quadrilateral AEFB is connected via AE and BF, after folding, the points A and B are connected over the fold EF. Given that B is mapped to B'(0,2,2), and folding preserves the distances from EF, we can find A'.Alternatively, since folding over EF swaps the positions of A and B with respect to EF, maintaining their distances. However, since we know B', perhaps A' is the reflection of A over EF in 3D space.But this might be complicated. Alternatively, note that in the folded structure, the points A' and B' are connected via the fold EF, preserving the lengths of AE and BF.Originally, AE = 1, BF = 3. After folding, these lengths should remain the same. So, A'E = AE =1, and B'F = BF =3. We already used B'F=3 to find B', so similarly, A'E=1.If we can find A', then we can determine the normal vector of plane A'DE.Alternatively, since we know that folding along EF results in B being mapped to B'(0,2,2), and since folding is along EF, we can model EF as a hinge line, and the dihedral angle is the angle between the two halves.Alternatively, the dihedral angle can be found by considering the angle between the original plane (z=0) and the folded plane containing B', E, F.Wait, plane FDE is in the z=0 plane, with normal vector (0,0,1). The plane A'DE would have a normal vector that we need to find. However, if we can find two vectors in plane A'DE, we can compute the normal.Plane A'DE contains points A', D(0,4,0), E(0,1,0). Since D and E are in the z=0 plane, and A' is somewhere in 3D space, we need to find A'.Wait, but perhaps instead of finding A', since we know that after folding, the projection of B' is H(0,2,0), which is the midpoint of DE? Because DE is from (0,1,0) to (0,4,0), so H at (0,2,0) is not the midpoint; midpoint would be at (0,2.5,0). Wait, H is at (0,2,0), so it's 1 unit above E.Alternatively, maybe there's a relation between A' and B' due to the folding. Since quadrilateral AEFB is folded along EF, the images of A and B are related.Alternatively, perhaps the line segment A'B' passes through the fold line EF, and EF is the perpendicular bisector of A'B'.But this is getting too vague. Let me think differently.We need the dihedral angle between planes A'-DE and F-DE. Since DE is the common line.To find the dihedral angle, we can take a point on DE and find two lines, one in each plane, perpendicular to DE. The angle between these two lines is the dihedral angle.Let’s choose point E on DE. In plane F-DE, a line perpendicular to DE at E would be in the direction of the projection of EF onto the plane perpendicular to DE. Similarly, in plane A'-DE, a line perpendicular to DE at E would be in the direction of the projection of EA' onto the plane perpendicular to DE.Wait, perhaps. Let me recall the method: to find the dihedral angle between two planes along a common line, you can take a point on the common line and draw a line in each plane perpendicular to the common line. The angle between these two lines is the dihedral angle.So, let's take point E(0,1,0) on DE. In plane F-D-E, we need a vector perpendicular to DE at E. DE is along the direction from D to E, which is (0, -3,0). So, DE is vertical in the y-axis direction. Therefore, a line perpendicular to DE at E would be horizontal (in the x-z plane).In plane F-D-E, which includes point F(2,3,0). So, vector EF is from E(0,1,0) to F(2,3,0): (2,2,0). To find a vector in plane F-D-E that's perpendicular to DE at E, we can take the projection of EF onto the plane perpendicular to DE.Since DE is along the y-axis, the direction of DE is (0,1,0) (from E to D), but actually from E to D is (0,3,0). Wait, DE is from D(0,4,0) to E(0,1,0), so direction vector is (0, -3,0). So, DE is along the negative y-axis. Therefore, a line perpendicular to DE at E would be in the x-z plane.In plane F-D-E, the vector EF is (2,2,0). To find a vector perpendicular to DE in plane F-D-E, we can take the component of EF perpendicular to DE. The direction of DE is (0,-3,0), so unit vector is (0,-1,0). The component of EF perpendicular to DE is EF - (EF ⋅ DE_unit) * DE_unit.EF ⋅ DE_unit = (2,2,0) ⋅ (0,-1,0) = -2. Therefore, the component of EF perpendicular to DE is (2,2,0) - (-2)*(0,-1,0) = (2,2,0) - (0,2,0) = (2,0,0). So, vector (2,0,0) is in plane F-D-E and perpendicular to DE at E.Similarly, in plane A'-D-E, we need a vector perpendicular to DE at E. Plane A'-D-E contains points A', D, E. Since we need a vector in this plane perpendicular to DE at E, we can use the direction from E to A', provided that this direction is perpendicular to DE.But DE is along the y-axis, so a direction perpendicular to DE would be in the x-z plane. Therefore, the vector from E to A' should lie in the x-z plane.But we need to find the coordinates of A'. However, perhaps since folding maps A to A', and B to B', and given the reflection across EF, we can find A' using similar steps as we did for B'.But in the case of point A(0,0,0), projecting its reflection over EF.But wait, earlier for point B, we found B'(0,2,2). Maybe for point A, we can perform the same steps.First, the distance from A to EF.Using the same line EF: y =x +1.Distance from A(0,0,0) to EF in the plane z=0 is |ax0 + by0 + c| / sqrt(a² + b²) = |1*0 + (-1)*0 +1| / sqrt(2) = 1/√2.So, after folding, the distance from A' to EF should also be 1/√2.Also, the projection of A' onto plane CDEF (z=0) would be a point on DE? Wait, the problem only mentions the projection of B is H on DE. It doesn't specify about A. So, perhaps the projection of A' could be somewhere else, but we don't have that information.Alternatively, since folding preserves the distances, and after folding, quadrilateral AEFB becomes a 3D shape with A' and B' connected over EF.However, since we already found B', maybe we can use the fact that folding preserves the distance between A and B. Originally, AB = 2 units. After folding, the distance between A' and B' should still be 2 units.Coordinates of A' are unknown, let's denote them as (x, y, z). Coordinates of B' are (0,2,2). So, distance A'B' should be 2:sqrt[(x - 0)^2 + (y - 2)^2 + (z - 2)^2] = 2=> x² + (y - 2)^2 + (z - 2)^2 = 4 --- Equation (4)Additionally, the distance from A' to EF should be 1/√2. Using similar steps as for B':Coordinates of A'(x,y,z). The distance from A' to EF is computed as follows.Line EF in 3D: parametrized as (2t, 1 + 2t, 0). The vector along EF is (2,2,0).Vector from E to A' is (x - 0, y -1, z -0) = (x, y -1, z).Cross product (A' - E) × v:|i   j   k||x   y-1  z||2   2   0|= i*((y -1)*0 - z*2) - j*(x*0 - z*2) + k*(x*2 - (y -1)*2)= (-2z, 2z, 2x - 2(y -1))Magnitude of cross product:sqrt[ (-2z)^2 + (2z)^2 + (2x - 2(y -1))^2 ] = sqrt[4z² +4z² + (2x -2y +2)^2 ] = sqrt[8z² + (2(x - y +1))^2 ] = sqrt[8z² +4(x - y +1)^2 ] = 2*sqrt[2z² + (x - y +1)^2 ]Magnitude of vector v is 2√2, so distance from A' to EF is [2*sqrt(2z² + (x - y +1)^2 )]/(2√2) = sqrt(2z² + (x - y +1)^2 ) / √2This distance should equal 1/√2:sqrt(2z² + (x - y +1)^2 ) / √2 = 1/√2Multiply both sides by √2:sqrt(2z² + (x - y +1)^2 ) =1Square both sides:2z² + (x - y +1)^2 =1 --- Equation (5)Additionally, since folding preserves the length of AE, which is 1. Original AE is from A(0,0,0) to E(0,1,0), length 1. After folding, A'E should still be length 1.Coordinates of E(0,1,0). So, distance from A'(x,y,z) to E is sqrt[(x -0)^2 + (y -1)^2 + (z -0)^2] = sqrt[x² + (y -1)^2 + z²] =1Therefore:x² + (y -1)^2 + z² =1 --- Equation (6)Now, we have three equations: Equation (4), (5), and (6):Equation (4): x² + (y - 2)^2 + (z - 2)^2 =4Equation (5): 2z² + (x - y +1)^2 =1Equation (6): x² + (y -1)^2 + z² =1This seems complex, but perhaps we can solve them.First, expand Equation (6):x² + y² -2y +1 + z² =1=> x² + y² -2y + z² =0 --- Equation (6a)Equation (4):x² + y² -4y +4 + z² -4z +4 =4Simplify:x² + y² -4y + z² -4z +8 =4=> x² + y² -4y + z² -4z +4 =0 --- Equation (4a)Subtract Equation (6a) from Equation (4a):(x² + y² -4y + z² -4z +4) - (x² + y² -2y + z²) =0 -0(-4y + (-4z) +4) - (-2y) =0-4y -4z +4 +2y =0-2y -4z +4 =0Divide by -2:y +2z -2 =0=> y = 2 -2z --- Equation (7)Now, substitute y =2 -2z into Equation (6a):x² + (2 -2z)^2 -2(2 -2z) + z² =0Expand:x² +4 -8z +4z² -4 +4z +z²=0Simplify:x² + (4z² + z²) + (-8z +4z) + (4 -4) =0x² +5z² -4z =0 --- Equation (8)Also, substitute y =2 -2z into Equation (5):2z² + (x - (2 -2z) +1)^2 =1Simplify:2z² + (x -2 +2z +1)^2 =1=>2z² + (x +2z -1)^2 =1But from Equation (8), x² = -5z² +4zPlug into Equation (5):2z² + (sqrt(-5z² +4z) +2z -1)^2 =1Wait, this might not be straightforward. Alternatively, express x² from Equation (8): x² = -5z² +4z. Then, x = ±sqrt(-5z² +4z). Then, substitute into Equation (5):2z² + ( ±sqrt(-5z² +4z) +2z -1 )^2 =1This seems complicated. Maybe we can find z by trial.Let’s suppose that z is a small positive number, given that k for B' was 2.Alternatively, note that this system might have symmetric solutions. Let’s attempt to express x in terms of z from Equation (8):x² = -5z² +4zTherefore, x = ±sqrt(-5z² +4z)Substitute into Equation (5):2z² + (x +2z -1)^2 =1Replace x with ±sqrt(-5z² +4z):2z² + [ ±sqrt(-5z² +4z) +2z -1 ]^2 =1Let’s square the term:[ ±sqrt(-5z² +4z) +2z -1 ]^2 = [2z -1 ±sqrt(-5z² +4z)]^2= (2z -1)^2 ±2(2z -1)sqrt(-5z² +4z) + (-5z² +4z)=4z² -4z +1 ±2(2z -1)sqrt(-5z² +4z) -5z² +4z= (4z² -5z²) + (-4z +4z) +1 ±2(2z -1)sqrt(-5z² +4z)= -z² +1 ±2(2z -1)sqrt(-5z² +4z)Therefore, Equation (5) becomes:2z² + (-z² +1 ±2(2z -1)sqrt(-5z² +4z)) =1Simplify:2z² -z² +1 ±2(2z -1)sqrt(-5z² +4z) =1z² +1 ±2(2z -1)sqrt(-5z² +4z) =1Subtract 1:z² ±2(2z -1)sqrt(-5z² +4z) =0This is quite complicated. Perhaps try specific values.Suppose z=0. Then, from Equation (7), y=2. From Equation (8), x²=0. So x=0. Check Equation (5):2*0 + (0 -2 +1)^2 =0 +1=1, which equals1. So (x,y,z)=(0,2,0) is a solution. But this is point H, which is the projection of B'. But A' cannot be H because H is the projection of B'.But if z=0, then A'(0,2,0) is in plane CDEF, which would mean the projection is itself, but folding should move A off the plane. Therefore, this solution might correspond to the original point, but we need the folded point.Alternatively, take z=1. Then from Equation (7), y=2 -2*1=0. From Equation (8), x²=-5*1 +4*1= -1. Not possible. Disregard.Try z=0.5:From Equation (7): y=2 -2*0.5=1From Equation (8): x²=-5*(0.5)^2 +4*0.5= -1.25 +2=0.75. So x=±√0.75=±√(3)/2≈±0.866Check Equation (5):2*(0.5)^2 + (x +2*0.5 -1)^2= 2*0.25 + (x +1 -1)^2=0.5 +x²=0.5 +0.75=1.25≠1. Not valid.Try z=0.4:From Equation (7): y=2 -2*0.4=1.2From Equation (8): x²=-5*(0.4)^2 +4*0.4= -5*0.16 +1.6= -0.8 +1.6=0.8. x=±√0.8≈±0.894Equation (5):2*(0.4)^2 + (x +2*0.4 -1)^2=2*0.16 + (x +0.8 -1)^2=0.32 + (x -0.2)^2=0.32 + (0.8 -0.2)^2 or 0.32 + (-0.894 -0.2)^2Wait, with x=±√0.8≈±0.894:For x=0.894:0.32 + (0.894 -0.2)^2≈0.32 + (0.694)^2≈0.32 +0.481=0.801≈0.8≠1For x=-0.894:0.32 + (-0.894 -0.2)^2≈0.32 + (-1.094)^2≈0.32 +1.196≈1.516≠1Not valid.Try z=0.2:y=2 -0.4=1.6x²=-5*(0.2)^2 +4*0.2= -0.2 +0.8=0.6, x=±√0.6≈±0.775Equation (5):2*(0.2)^2 + (x +0.4 -1)^2=0.08 + (x -0.6)^2For x=0.775:0.08 + (0.775 -0.6)^2≈0.08 +0.0306≈0.1106≠1For x=-0.775:0.08 + (-0.775 -0.6)^2≈0.08 + (-1.375)^2≈0.08 +1.89≈1.97≠1Not valid.This trial and error isn't working well. Maybe there's another approach.Recall that the folding along EF maps A to A' and B to B', such that EF is the fold line. Therefore, the line segment AA' should be perpendicular to EF and bisected by EF. Similarly for BB'.But in 3D, this might be more complex. However, in the plane of the original quadrilateral (z=0), the fold line EF is y =x +1. The midpoint of BB' should lie on EF.Original point B(2,0,0), folded point B'(0,2,2). The midpoint of BB' is ((2+0)/2, (0+2)/2, (0+2)/2)=(1,1,1). Does this midpoint lie on EF?Line EF in 3D is parametrized as (2t,1 +2t,0). For the midpoint (1,1,1) to lie on EF, there must exist a t such that:2t =1,1 +2t =1,0 =1.But 2t =1 implies t=0.5, then 1 +2*(0.5)=2≠1. Also, z-coordinate 0≠1. Therefore, the midpoint does not lie on EF, which contradicts the folding assumption. Therefore, our previous approach might be incorrect.This suggests that reflecting B over EF in 2D does not give the correct B', because in 3D folding, the midpoint should lie on EF and the segment BB' should be perpendicular to EF.Therefore, our initial assumption that B' is (0,2,2) might be incorrect.This means we need to revisit the problem.Alternative approach:When folding along EF, the crease line EF remains fixed, and points A and B are mirrored over EF in 3D space. The key is that the fold preserves distances from the fold line.To find the coordinates of B', we can use the fact that the fold maps B to B' such that EF is the perpendicular bisector of BB'. In 3D, this means:1. The midpoint of BB' lies on EF.2. The line BB' is perpendicular to EF.Similarly for A and A'.Let’s use these conditions to find B'.Let’s denote B'(x, y, z). The midpoint of BB' is ((2 +x)/2, (0 + y)/2, (0 + z)/2).This midpoint must lie on EF. EF is parametrized as (2t, 1 +2t,0), t ∈ [0,1].So, we have:(2 +x)/2 =2t,(y)/2 =1 +2t,z/2 =0.From the z-coordinate: z/2=0 => z=0. But this contradicts our previous result where B' had z=2. This suggests an inconsistency.Wait, if folding is along EF in 3D space, and the midpoint of BB' lies on EF with BB' perpendicular to EF, then z-coordinate of B' must be 0? But that would mean B' is in the z=0 plane, but then the projection H would be B' itself, which must lie on DE. However, the problem states that the projection of B' is H on DE. If B' is already in the plane CDEF (z=0), then H would be B', which must lie on DE.But B is originally at (2,0,0), which is not on DE. After folding, if B' is on DE, then its projection is itself. But DE is from (0,1,0) to (0,4,0). So, if B' is on DE, then its x-coordinate is 0, y between 1 and4, z=0.But in this case, the midpoint of BB' is ((2 +0)/2, (0 + y)/2, (0 +0)/2)=(1, y/2,0). This must lie on EF, which has parametrization (2t,1 +2t,0). So,1 =2t,y/2 =1 +2t,From first equation: t=0.5Second equation: y/2 =1 +2*(0.5)=1 +1=2 => y=4.Therefore, midpoint is (1,2,0), which is t=0.5 on EF: (2*0.5,1 +2*0.5,0)=(1,2,0). Correct.Therefore, B' must be (0,4,0). Wait, but (0,4,0) is point D.But B' is supposed to be the reflection of B over EF with midpoint on EF. But according to this, B' would be (0,4,0), which is point D. However, distance from B(2,0,0) to B'(0,4,0) is sqrt[(2)^2 + (4)^2] = sqrt(4 +16)=sqrt(20)=2√5. Which is not equal to twice the distance from B to EF (which is 2*(3/√2)=3√2). This doesn't match.This indicates a contradiction, which suggests that the folding in 3D cannot have both the midpoint on EF and BB' perpendicular to EF if restricted to z=0 plane. Therefore, folding must occur out of the z=0 plane, allowing B' to have a z-coordinate.Therefore, the previous assumption that the midpoint lies on EF with BB' perpendicular to EF in 3D is the correct approach, but we need to consider the z-coordinate.So, let's correct this.Let B'(x, y, z). Midpoint M of BB' is ((2 +x)/2, (0 + y)/2, (0 + z)/2). This must lie on EF, which is parametrized as (2t,1 +2t,0). Therefore:(2 +x)/2 =2t,(y)/2 =1 +2t,(z)/2 =0.From the third equation: z=0. But this again forces z=0, which brings us back to the previous contradiction. Therefore, this suggests that in strict 3D folding, maintaining midpoint on EF and BB' perpendicular to EF is impossible unless B' is in the plane CDEF, which contradicts the problem's statement that the projection of B' is H on DE (implying B' is not in CDEF).Therefore, there must be an error in my understanding of the projection.Wait, the problem states "the projection of point B on plane CDEF is point H on line DE". Wait, is it the projection of the original point B or the folded point B'?Reading the problem again: "Quadrilateral AEFB is folded along EF, such that the projection of point B on plane CDEF is point H on line DE." So, after folding, when they refer to "point B", it is the new position B', and its projection is H on DE.Therefore, B' is in 3D space, not on DE, and its orthogonal projection onto plane CDEF is H on DE.Therefore, the coordinates of B'(x,y,z) satisfy:1. The projection of B' onto CDEF (z=0) is H(0, h,0), so x=0, y=h, z=0. Wait, no. Projection orthogonally onto plane CDEF (z=0) would be (x, y,0). But the problem states that this projection is H on DE. DE is x=0, z=0, 1 ≤ y ≤4. Therefore, H has coordinates (0, h,0). Therefore, the projection of B' onto CDEF is (0, h,0), which requires that B'(x,y,z) projects orthogonally to (0, h,0). Orthogonal projection implies that B'(x,y,z) = (0, h,0) + (0,0,z). Wait, no. Orthogonal projection onto CDEF (z=0) would set the z-coordinate to 0, keeping x and y the same. However, the problem states that the projection is H on DE, which requires x=0. Therefore, the projection of B' must have x=0 and z=0, hence B' must lie along the line parallel to the z-axis passing through H(0, h,0). Therefore, B' has coordinates (0, h, k), as we initially thought.Therefore, even though folding might imply that the midpoint of BB' lies on EF and BB' is perpendicular to EF, the problem's condition that the projection of B' is H on DE gives us B'(0, h, k). Therefore, the previous approach with B'(0, h, k) is correct, and the midpoint and perpendicularity conditions might not apply in the same way due to the 3D nature.Therefore, returning to the previous equations:We had for B'(0, h, k):Equation (1): 2k² + (h -1)^2 =9Equation (2): (3 - h)^2 + k² =5Which gave h=2, k=±2.Given that the projection is H(0,2,0), and B'(0,2,2) or B'(0,2,-2). Since folding is typically considered as folding upwards, let's take k=2, so B'(0,2,2).Now, returning to the dihedral angle.The dihedral angle is between planes A'-DE-F and F-D-E. Wait, but we need to confirm the planes.Plane F-D-E is the same as before, with normal vector (0,0,1). Plane A'-D-E is the plane containing points A', D(0,4,0), and E(0,1,0). Since A' is the folded position of A, we need to find A'.But perhaps instead of finding A', since we know B', we can find the dihedral angle using another method.Alternatively, the dihedral angle can be found using the angle between the two planes: the original plane (which contains F, D, E) and the folded plane (which contains A', D, E). The dihedral angle is the angle between these two planes along their intersection DE.To find the dihedral angle, we can use the normal vectors of the two planes.Plane F-D-E has normal vector n1 = (0,0,1) as calculated before.Plane A'-D-E: To find its normal vector, we need three points: A', D, E. We need coordinates of A'.But since we have not found A' yet, maybe we can relate it to B'.Alternatively, since quadrilateral AEFB is folded along EF, which connects E(0,1,0) to F(2,3,0), the segments AE and BF are folded over EF. So, after folding, segments A'E and B'F are in 3D space.Given that AE =1 and BF=3, after folding, A'E=1 and B'F=3. We already used B'F=3 to find B'.Now, we can find A' by using the fact that A'E=1 and the distance from A' to EF is 1/√2.However, this leads us back to the previous system of equations which was complex. Alternatively, since we know B'(0,2,2), maybe we can find A' using the fact that the folding preserves the structure.Alternatively, consider that folding along EF maps A to A' and B to B', such that the quadrilateral A'EFB' is the folded version. Since EF is the fold line, the angles around EF should be preserved.Alternatively, since we know coordinates of B', and the dihedral angle is between planes A'-DE and F-DE, and since DE is common, we can compute the angle between the normals.Plane F-D-E has normal vector n1=(0,0,1).Plane A'-D-E: To find its normal vector, we need two vectors in the plane. For example, vectors DE and DA'.DE is from D(0,4,0) to E(0,1,0): (0,-3,0).DA' is from D(0,4,0) to A'(unknown). But we don't know A'.Alternatively, since we know that after folding, the dihedral angle is determined by the angle between the original plane (with normal n1) and the folded plane (with normal n2). If we can find a vector in the folded plane A'-D-E that is perpendicular to DE, then we can find n2.Alternatively, since DE is along the y-axis, and plane F-D-E is the z=0 plane, the normal vector is (0,0,1). The folded plane A'-D-E will have a normal vector that has a z-component, since A' is out of the z=0 plane.Given that B'(0,2,2) is known, perhaps we can find a relationship between A' and B'.Wait, considering that quadrilateral AEFB when folded results in A' and B' connected over EF. The lengths EA' =1 and FB' =3 are preserved. The fold line EF is the same.But maybe the dihedral angle can be found using the angle between vectors EA' and FB'.Alternatively, another approach: since we know point B'(0,2,2) and H(0,2,0), which is the projection, we can consider the dihedral angle as the angle between the original plane (z=0) and the plane containing B', H, and E.The dihedral angle is the angle between the two planes along DE. Since DE is in the z=0 plane, and the other plane contains B'(0,2,2), H(0,2,0), and E(0,1,0).The normal vector of the second plane can be found using points E, H, B'.Vectors in this plane:EH = H - E = (0,2,0) - (0,1,0) = (0,1,0)EB' = B' - E = (0,2,2) - (0,1,0) = (0,1,2)The cross product of EH and EB' will give the normal vector:EH × EB' = |i   j   k|             0   1   0             0   1   2= i*(1*2 -0*1) - j*(0*2 -0*0) + k*(0*1 -0*1)= (2, 0, 0)So, the normal vector is (2,0,0), or (1,0,0) when normalized.The normal vector of the original plane (z=0) is (0,0,1).The dihedral angle θ between the two planes is the angle between their normal vectors:cosθ = (n1 • n2) / (|n1||n2|)n1 • n2 = (0,0,1) • (1,0,0) =0Therefore, cosθ=0, which implies θ=90 degrees.But this is the angle between the plane containing E, H, B' and the z=0 plane. However, the dihedral angle in the problem is between planes A'-D-E and F-D-E. The plane F-D-E is the z=0 plane, and the plane A'-D-E is the plane containing A', D, E. However, we calculated the angle between the z=0 plane and the plane EHB', which is 90 degrees, but this might not be the same as the dihedral angle A-DE-F.Wait, perhaps there's an error here. The plane EHB' is different from plane A'-DE. Unless A' lies in plane EHB', which is not necessarily the case.Given that we might have made a miscalculation earlier, perhaps the dihedral angle is 90 degrees, but this seems too straightforward. Alternatively, let's consider another method.The dihedral angle can be found using the angle between the two planes, which can also be calculated using the angle between their normals.Plane F-D-E has normal vector n1=(0,0,1).Plane A'-D-E has normal vector n2. To find n2, we need coordinates of A'.However, we can relate A' to B' via the fold line EF. Since folding along EF swaps A and B, maintaining distances to EF, there might be a relationship between A' and B'.Alternatively, since we know B', perhaps we can consider the triangle DEB' and find the normal vector.Points D(0,4,0), E(0,1,0), B'(0,2,2).Vectors in plane DEB':ED = D - E = (0,3,0)EB' = B' - E = (0,1,2)Cross product ED × EB':|i   j   k||0   3   0||0   1   2|= i*(3*2 -0*1) - j*(0*2 -0*0) + k*(0*1 -3*0)= (6,0,0)So, normal vector is (6,0,0) or (1,0,0).Therefore, plane DEB' has normal vector (1,0,0).Plane F-D-E has normal vector (0,0,1).The angle between (1,0,0) and (0,0,1) is 90 degrees, as their dot product is 0. Therefore, the dihedral angle is 90 degrees.But wait, plane DEB' is the same as plane A'-DE if A' is in the same plane as DEB'. However, since A' is the folded image of A over EF, which is related to B', maybe plane A'-DE is the same as plane DEB', hence the dihedral angle is 90 degrees.However, the problem states it's a dihedral angle A-DE-F, which would be between planes ADE and FDE. But before folding, plane ADE is degenerate. After folding, plane A'-DE is non-degenerate. Therefore, the dihedral angle is between planes A'-DE and F-DE, which have normal vectors (1,0,0) and (0,0,1), respectively. The angle between these normals is 90 degrees, so the dihedral angle is 90 degrees.Therefore, the answer is 90 degrees, or π/2 radians.But let's verify this with another approach.Consider the line DE, which is along the y-axis. In plane F-D-E (z=0), a direction perpendicular to DE is in the x-direction. In plane A'-DE, which has normal vector (1,0,0), a direction perpendicular to DE is in the z-direction. Therefore, the angle between the x-direction and z-direction is 90 degrees.Therefore, the dihedral angle is 90 degrees.Thus, the dihedral angle A-DE-F is 90 degrees.Final Answerboxed{90^circ}</think>Given a rectangle (ABCD) with (AB = 2) and (AD = 4), points (E) and (F) are on (AD) and (BC) respectively, such that (AE = 1) and (BF = 3). When quadrilateral (AEFB) is folded along (EF), the projection of point (B) onto plane (CDEF) is point (H) on line (DE). We need to find the dihedral angle (A-DE-F).1. Coordinates Assignment: - (A(0, 0, 0)), (B(2, 0, 0)), (C(2, 4, 0)), (D(0, 4, 0)) - (E(0, 1, 0)), (F(2, 3, 0))2. Projection of (B') onto Plane (CDEF): - After,<think>Alright, let's see. So we have this rectangle ABCD where AB is 2 and AD is 4. Points E and F are on AD and BC respectively, with AE = 1 and BF = 3. Then the quadrilateral AEFB is folded along EF, and when we do that, the projection of point B onto the plane CDEF is point H on line DE. We need to find the dihedral angle A-DE-F.First, let me visualize the rectangle. ABCD with AB=2 and AD=4. So AB is the base, 2 units long, and AD is the side, 4 units. So if it's a rectangle, all angles are 90 degrees. Coordinates might help here. Let me assign coordinates to the points to make this easier.Let’s place point A at the origin (0,0). Then, since AB is 2, point B would be at (2,0). AD is 4, so point D would be at (0,4), and since it's a rectangle, point C is at (2,4).Points E and F are on AD and BC respectively. AE = 1, so since A is (0,0), E is 1 unit up AD. Since AD goes from (0,0) to (0,4), moving up 1 unit, E is at (0,1). Similarly, BF = 3. BC is from B (2,0) to C (2,4). So moving up 3 units from B, F would be at (2,3).So quadrilateral AEFB has vertices at A(0,0), E(0,1), F(2,3), and B(2,0). Then, folding along EF. When folding, point B is going to move somewhere, and its projection on plane CDEF is point H on DE. So after folding, B's projection H is on DE. The dihedral angle in question is between the two planes meeting along DE, specifically the angle between plane ADE and plane DEF (or maybe plane CDEF? Wait, the dihedral angle A-DE-F would be along edge DE, between face A-D-E and face D-E-F. But after folding, the face D-E-F would be part of the folded figure. Hmm.Wait, perhaps the dihedral angle is between the original plane ADE and the folded plane DEF. When we fold along EF, the plane AEFB is folded over EF onto the other side. So the dihedral angle along DE would be the angle between the original plane (which includes ADE) and the new position of the folded plane (which now includes DEF). So to find this dihedral angle, we might need to find the angle between the two planes along their line of intersection DE.To compute a dihedral angle between two planes along their line of intersection, we can use the normal vectors of the two planes. The dihedral angle is equal to the angle between the normal vectors. So we need to find the normal vectors of the planes ADE and DEF.First, let's find coordinates of points involved.Original points before folding:A(0,0,0) – Wait, maybe we need to consider this in 3D? Because when we fold the quadrilateral AEFB along EF, point B moves into 3D space. The projection H is on DE, which is in the plane CDEF. So perhaps we need to model this in 3D coordinates.Let me think. Let's set up a coordinate system where the original rectangle ABCD is in the xy-plane. So:A(0,0,0)B(2,0,0)C(2,4,0)D(0,4,0)E is on AD: AE =1, so E(0,1,0)F is on BC: BF=3, so F(2,3,0)Quadrilateral AEFB is folded along EF, so we need to fold the paper such that AEFB is lifted off the plane and folded over EF. After folding, point B is moved to some point B' in 3D space. The projection of B' onto the plane CDEF (which is the original plane minus the folded part) is H on DE.So projection of B' onto plane CDEF is H, which is on DE. So H lies somewhere along DE. DE is from D(0,4,0) to E(0,1,0). Wait, no: D is (0,4,0) and E is (0,1,0), so DE is the line from (0,4,0) to (0,1,0). But H is the projection of B' onto CDEF, which is a point on DE. So H is along DE.So first, let's figure out where H is. Since H is the projection of B' onto CDEF, and H is on DE, then BH' is perpendicular to CDEF. Since CDEF is the original plane (xy-plane here), then actually the projection of B' onto CDEF is just the shadow of B' onto the xy-plane. But since folding is along EF, and B is moved to B', which is folded over EF. So the projection H is the foot of the perpendicular from B' to CDEF (which is the xy-plane). But H is on DE. Therefore, the x and y coordinates of H are the same as B', and z=0, and since H is on DE, which is the line from D(0,4,0) to E(0,1,0). So DE is the vertical line x=0, z=0, from y=1 to y=4.Wait, but DE in the original figure is from D(0,4,0) to E(0,1,0). So DE is along the y-axis at x=0, from (0,4,0) to (0,1,0). So H must be a point on that line. Therefore, coordinates of H are (0, h, 0) where 1 ≤ h ≤4.But H is the projection of B' onto CDEF, so B' has coordinates (x, y, z), and its projection is (x, y, 0). But H is (0, h, 0). Therefore, x=0, y=h. So B' is (0, h, z), since projection is (0, h, 0). Therefore, B' is (0, h, z) for some z.But when we fold along EF, the point B is reflected over EF to B'. So reflection over the line EF in 3D? Wait, EF is a line in the original plane. When folding along EF, the reflection would be across the line EF, but in 3D space. So B is at (2,0,0), and after folding, B' is the reflection of B over the line EF. Then, the projection of B' onto the xy-plane (CDEF) is H on DE. So reflection over EF in 3D.But perhaps we can model this folding by considering EF as a hinge, and folding the quadrilateral AEFB over EF into the third dimension. Then, B moves to B', such that the distance from B to EF is the same as from B' to EF. Since it's a reflection over the line EF.Alternatively, since folding over EF, the segment BF is folded onto F's side. Wait, BF is length 3. After folding, B is brought to some point B' such that EF is the crease. So in the folded position, the quadrilateral AEFB becomes AEFB', with B' on the other side of EF. The projection of B' onto the plane CDEF (original plane) is H on DE.To find H, we can consider that the line BH is perpendicular to the plane CDEF (i.e., along the z-axis), so H is the projection. But H is on DE. Therefore, B' must lie directly above H in the z-direction. So coordinates of H are (0, h, 0), and B' is (0, h, k) for some k. Now, since folding preserves distances, the distance from B to EF should be equal to the distance from B' to EF.Wait, maybe we need to find coordinates of B' such that EF is the fold line, so B and B' are symmetric with respect to EF. So in 3D, EF is a line, and B and B' are mirror images over EF. The projection of B' onto the plane is H on DE.Alternatively, since after folding, the image of B is B', and the projection is H. So the line BB' is perpendicular to the plane CDEF (since projection is H), so BB' is vertical. But if the projection is H, then BB' is a vertical line (if CDEF is the xy-plane). But in reality, when folding along EF, the movement of B to B' is not necessarily vertical. Wait, perhaps the projection is orthogonal projection onto CDEF. So yes, if we consider CDEF as the xy-plane, then the projection of B' onto CDEF is the point H where you drop a perpendicular from B' to CDEF, landing at H on DE.But folding along EF would mean that B' is the reflection of B across EF. Wait, but reflection over a line in 3D is not straightforward. Maybe we need to consider the fold as a reflection over the line EF in 3D space. Alternatively, since EF is in the original plane (xy-plane), folding over EF would create a dihedral angle between the original plane and the folded plane. The dihedral angle we are to find is between the planes ADE and DEF. Wait, but DEF is part of the folded figure.Alternatively, maybe we can find the dihedral angle by considering the angle between the two planes that meet along DE. The original plane ADE is part of the original rectangle, and after folding, the plane DEF (which was originally part of the plane of CDEF) is now raised up. The dihedral angle between these two planes is determined by the folding.Alternatively, since after folding, the point B is moved to B', which is above H on DE. The dihedral angle can be found by considering the angle between the normals of the two planes. So plane ADE is the same as before, and plane DEF is now the plane containing D, E, and F', where F' might be a folded point. Wait, maybe not. Wait, DEF is in the original plane, but after folding, the plane containing DE and F (but F is still on BC). Hmm, maybe the plane DEF is unchanged? Wait, DEF is part of the original plane CDEF, so DEF is still in the xy-plane. Then, the dihedral angle between ADE and DEF would be the angle between the planes ADE and DEF, which are both in the original rectangle? But that doesn't make sense, because ADE and DEF meet along DE, and in the original flat figure, the dihedral angle would be 180 degrees. But after folding, since we folded AEFB over EF, the plane ADE is now folded, so the dihedral angle between ADE and DEF is no longer 180 degrees. Wait, maybe.Wait, perhaps I need to clarify which planes we are talking about. The dihedral angle A-DE-F is the angle between the two planes that share the edge DE: one plane is A-D-E and the other is D-E-F. In the original, unfolded figure, both planes A-D-E and D-E-F are part of the same plane (the rectangle), so the dihedral angle is 180 degrees. But after folding, when we fold quadrilateral AEFB over EF, the plane A-D-E is now angled relative to D-E-F, creating a dihedral angle less than 180 degrees.Therefore, to find this dihedral angle, we need to find the angle between the original plane ADE (which is now part of the folded figure) and the plane DEF (which remains in the original plane). Wait, but DEF is part of the original plane CDEF, which hasn't been folded. So after folding, plane ADE is now above the original plane, forming a dihedral angle with DEF along DE.Alternatively, maybe both planes ADE and DEF are part of the folded figure. Wait, when folding AEFB over EF, point A is connected to E, which is on AD, and F is on BC. So when folding, the part AEFB is lifted, making the dihedral angle along DE between the original ADE and the folded part.Hmm, perhaps another approach is needed. Let me consider the dihedral angle. The dihedral angle between two planes can be found by the angle between their normals. If we can find the normal vectors of the planes ADE and DEF after folding, then compute the angle between them.But to do that, we need coordinates of points in 3D space after folding. Since folding occurs along EF, points A and B are moved to new positions in 3D, while points E and F remain fixed (since EF is the fold line). So E and F stay in place, but A and B are moved to A' and B' such that the quadrilateral AEFB becomes A'EFB' after folding.However, in the problem statement, it's mentioned that the projection of point B on plane CDEF is H on line DE. So after folding, B is moved to B', whose projection is H on DE. So we can use this information to find the coordinates of B'.Let me try to model this step by step.Original coordinates:A(0,0,0)B(2,0,0)C(2,4,0)D(0,4,0)E(0,1,0)F(2,3,0)Fold along EF. So EF is the line from E(0,1,0) to F(2,3,0). Let's parametrize EF.Parametric equations for EF: since it goes from (0,1,0) to (2,3,0), the direction vector is (2,2,0). So any point on EF can be written as (0 + 2t, 1 + 2t, 0) for t ∈ [0,1].When folding along EF, points on one side of EF are reflected to the other side. So point B(2,0,0) is on one side of EF, and after folding, it is reflected over EF to a point B' in 3D space. The projection of B' onto the plane CDEF (the original plane) is H on DE.So first, let's find the reflection of B over the line EF in 3D space. Wait, but reflection over a line in 3D is more complex than in 2D. However, since EF is in the original plane (z=0), perhaps the fold is a reflection over the line EF in the plane, but lifted into 3D. Hmm, maybe not. Alternatively, since folding over EF would create a dihedral angle between the two halves.Alternatively, since folding over EF, the segment FB is folded over to the other side. The length of FB is 3 units. After folding, FB' is congruent to FB. The projection of B' onto the plane is H on DE. So perhaps we can find coordinates of B' such that FB' = FB = 3, and the projection H is on DE.But how?Alternatively, since H is the projection of B' onto the plane CDEF (which is the xy-plane), then H is the foot of the perpendicular from B' to CDEF, so H has coordinates (x, y, 0), and since H is on DE, which is from D(0,4,0) to E(0,1,0). Therefore, DE is the line x=0, z=0, from y=1 to y=4. Therefore, H must be (0, h, 0) where 1 ≤ h ≤4. So coordinates of H are (0, h, 0).Therefore, coordinates of B' are (0, h, k), since the projection is (0, h, 0). So B' is vertically above H by some distance k.Now, since folding over EF preserves the distance from B to EF, and the same applies to B'. So the distance from B to EF should be equal to the distance from B' to EF.Wait, but EF is in the plane z=0. So the distance from B to EF is the minimal distance from point B(2,0,0) to line EF. Let's compute that.The distance from a point (x0, y0, z0) to a line given by two points (x1, y1, z1) and (x2, y2, z2) can be computed using the formula:Distance = |(P2 - P1) × (P1 - P0)| / |P2 - P1|Where P0 is the point, P1 and P2 are points on the line.So here, line EF is from E(0,1,0) to F(2,3,0). Let's take P1 = E(0,1,0), P2 = F(2,3,0), and P0 = B(2,0,0).Vector P2 - P1 = (2,2,0)Vector P1 - P0 = (0 - 2, 1 - 0, 0 - 0) = (-2,1,0)Cross product (P2 - P1) × (P1 - P0) = determinant:|i   j   k||2   2   0||-2  1   0|Calculates to i*(2*0 - 0*1) - j*(2*0 - 0*(-2)) + k*(2*1 - (-2)*2) = i*(0) - j*(0) + k*(2 +4) = 6kThe magnitude is 6. The magnitude of P2 - P1 is sqrt(2² + 2² + 0²) = sqrt(8) = 2√2.Therefore, the distance from B to EF is |6| / 2√2 = 6 / (2√2) = 3/√2 ≈ 2.121.Similarly, the distance from B' to EF should be the same. But B' is at (0, h, k). Let's compute the distance from B'(0, h, k) to EF.Again, using the same formula. Line EF is from E(0,1,0) to F(2,3,0). Let's take P1 = E(0,1,0), P2 = F(2,3,0), P0 = B'(0, h, k).Vector P2 - P1 = (2,2,0)Vector P1 - P0 = (0 - 0, 1 - h, 0 - k) = (0, 1 - h, -k)Cross product (P2 - P1) × (P1 - P0):|i   j   k||2   2   0||0  1 - h  -k|Calculates to:i*(2*(-k) - 0*(1 - h)) - j*(2*(-k) - 0*0) + k*(2*(1 - h) - 2*0)= i*(-2k) - j*(-2k) + k*(2 - 2h)= (-2k, 2k, 2 - 2h)The magnitude of this cross product is sqrt( (-2k)^2 + (2k)^2 + (2 - 2h)^2 ) = sqrt(4k² + 4k² + (2 - 2h)^2) = sqrt(8k² + 4(1 - h)^2 )The magnitude of P2 - P1 is still 2√2. Therefore, the distance from B' to EF is sqrt(8k² + 4(1 - h)^2) / (2√2) = sqrt(8k² + 4(1 - h)^2) / (2√2).But this distance must equal the distance from B to EF, which is 3/√2. Therefore,sqrt(8k² + 4(1 - h)^2) / (2√2) = 3/√2Multiply both sides by 2√2:sqrt(8k² + 4(1 - h)^2) = 6Square both sides:8k² + 4(1 - h)^2 = 36Divide both sides by 4:2k² + (1 - h)^2 = 9So that's one equation relating h and k.Additionally, since B' is the reflection of B over EF in the folding, the line segment BB' should be perpendicular to EF and bisected by EF. Wait, in 3D folding, but since EF is a line in the plane, maybe not. Alternatively, since folding over EF maps B to B', the line EF is the crease, so the fold should place B such that the path from B to B' is perpendicular to EF and the midpoint lies on EF.But in 3D, this might not hold directly. However, considering the fold as a reflection over the line EF, in the plane containing B and EF. Wait, perhaps the reflection occurs in the plane perpendicular to the paper (xy-plane) along EF. This is getting complicated.Alternatively, since after folding, the quadrilateral AEFB becomes AEFB', with B' in 3D space. The distances from A to E and E to F and F to B should be preserved. So AE = 1, EF can be calculated, FB = 3. Let's compute EF. Original EF is from (0,1,0) to (2,3,0). The length of EF is sqrt((2-0)^2 + (3-1)^2 + 0^2) = sqrt(4 + 4) = sqrt(8) = 2√2. So EF is 2√2.After folding, E and F remain fixed, so EF is still 2√2. The distances from A to E and from F to B' should be preserved. AE is 1, so after folding, AE remains 1. FB was 3, so FB' should also be 3. So FB' = 3.But F is at (2,3,0), and B' is at (0, h, k). So distance FB' is sqrt( (0 - 2)^2 + (h - 3)^2 + (k - 0)^2 ) = sqrt(4 + (h - 3)^2 + k²). This should equal 3.Therefore,sqrt(4 + (h - 3)^2 + k²) = 3Square both sides:4 + (h - 3)^2 + k² = 9Therefore, (h - 3)^2 + k² = 5So now we have two equations:1) 2k² + (1 - h)^2 = 92) (h - 3)^2 + k² = 5Let me write them down:Equation 1: 2k² + (1 - h)^2 = 9Equation 2: (h - 3)^2 + k² = 5Let’s solve these equations. Let’s expand both.Equation 1: 2k² + (1 - 2h + h²) = 9 → 2k² + h² - 2h + 1 = 9 → 2k² + h² - 2h = 8Equation 2: (h² - 6h + 9) + k² = 5 → h² - 6h + 9 + k² = 5 → h² - 6h + k² = -4Let’s denote equation 1 as 2k² + h² - 2h = 8Equation 2 as h² - 6h + k² = -4Let’s denote equation 2 multiplied by 2: 2h² - 12h + 2k² = -8Then subtract equation 1 from this:(2h² - 12h + 2k²) - (2k² + h² - 2h) = -8 - 8Simplify:2h² -12h + 2k² -2k² -h² + 2h = -16(2h² - h²) + (-12h + 2h) + (2k² -2k²) = -16h² -10h = -16h² -10h +16 =0Solve this quadratic equation:h = [10 ± sqrt(100 -64)] /2 = [10 ± sqrt(36)] /2 = [10 ±6]/2Thus, h = (10 +6)/2=16/2=8 or h=(10-6)/2=4/2=2But H is on DE, which is from D(0,4,0) to E(0,1,0). Therefore, h must satisfy 1 ≤ h ≤4. So h=8 is invalid. Thus, h=2.Therefore, h=2. Then from equation 2: (2 -3)^2 + k² =5 → 1 +k²=5 → k²=4 →k=±2. But since folding is upwards, k should be positive. So k=2.Therefore, coordinates of B' are (0, 2, 2), since H is (0,2,0).So B' is at (0,2,2). Now, let's verify equation 1:2k² + (1 - h)^2 = 2*(4) + (1 -2)^2 =8 +1=9, which matches equation 1.Good, so B'(0,2,2). Now, we need to find the dihedral angle A-DE-F, which is the angle between planes ADE and DEF along edge DE. Since DE is from (0,1,0) to (0,4,0). Wait, DE is from D(0,4,0) to E(0,1,0). Wait, DE is the line segment connecting D(0,4,0) to E(0,1,0). So DE is vertical along the y-axis at x=0, z=0, from y=1 to y=4.Plane ADE consists of points A(0,0,0), D(0,4,0), E(0,1,0). So plane ADE is the vertical line x=0, z=0, but A is at (0,0,0). Wait, actually, points A, D, and E are colinear? Wait, A is (0,0,0), D is (0,4,0), E is (0,1,0). So they are all along the y-axis. Wait, that can't be. Wait, in the original rectangle, AD is from A(0,0,0) to D(0,4,0). E is on AD at (0,1,0). So points A, D, E are colinear. Therefore, the plane ADE is just the line AD (since three colinear points define a line, not a plane). Wait, that can't be right. There must be a miscalculation here.Wait, no. If three points are colinear, they don't define a unique plane. So perhaps the problem statement is referring to the dihedral angle along DE between two planes that share DE. But in the original figure, DE is a line segment, and the planes on either side of DE would be the plane containing DE and A, and the plane containing DE and F. Wait, but in the original figure, DE is part of the rectangle, so all points are coplanar. After folding, DE is still in the original plane CDEF, but the plane containing A (which is now folded) is no longer coplanar.Wait, perhaps after folding, point A is also moved? Wait, in the problem statement, it's quadrilateral AEFB that is folded along EF. So points A, E, F, B are part of the quadrilateral being folded. E and F are on AD and BC, so when folding along EF, points A and B are moved. However, in our previous consideration, we only moved B to B', but actually, point A should also be moved to a new position A'.But the problem states that the projection of point B (not B') is H. Wait, the problem says: "the projection of point B on plane CDEF is point H on line DE". Wait, but point B is in the original plane, which is part of CDEF (since B is on BC, which is part of CDEF). So unless after folding, point B is moved to B', and the projection of B' is H. The problem statement says: "projection of point B on plane CDEF is point H on line DE". Wait, but point B is originally on BC, which is part of plane CDEF. Therefore, its projection is itself. So perhaps the problem means after folding, when B is moved to B', the projection of B' is H. The wording is a bit confusing. Let me check again:"Quadrilateral AEFB is folded along EF, such that the projection of point B on plane CDEF is point H on line DE."Yes, so after folding, point B is now in 3D space, and its projection onto plane CDEF is H on DE. Therefore, as we computed earlier, B' is (0,2,2), H is (0,2,0). So that's correct.But then, what about point A? When folding quadrilateral AEFB along EF, point A is also moved. Let's find the coordinates of A'.Similarly to B', point A is part of the folded quadrilateral, so when folding along EF, point A is reflected over EF to A'. The projection of A' onto plane CDEF would be some point, but the problem doesn't mention it. However, since we need to find the dihedral angle A-DE-F, which involves the plane containing A, D, E and the plane containing D, E, F.Wait, but after folding, point A is moved to A', so the plane ADE is actually the plane A'-D-E. Similarly, the plane DEF remains as it is, since F is on BC, which is part of CDEF.Therefore, the dihedral angle is between planes A'-DE and DEF. To compute this angle, we can find the normal vectors of the two planes and then find the angle between them.First, let's find coordinates of A'.When folding quadrilateral AEFB over EF, both A and B are reflected over EF. We already found B'(0,2,2). Let's find A'.Similarly, the projection of A' onto plane CDEF would be a point on some line? But the problem doesn't specify, but since we might need A' for computing the plane A'-D-E.Alternatively, since folding preserves distances and angles, the quadrilateral AEFB is congruent to A'EFB'. So distances from A to E and from A to F should be preserved.Original distance AE is 1. After folding, distance A'E should still be 1. Similarly, AF is distance from A(0,0,0) to F(2,3,0): sqrt((2)^2 + (3)^2 +0) = sqrt(13). So after folding, distance A'F should also be sqrt(13).Moreover, EF is still 2√2, and FB is now FB' which we found as 3.But maybe we can find A' similarly to how we found B'.Since folding over EF, A is reflected over EF to A'. The projection of A' onto CDEF would be a point on...? The problem doesn't say, but since we need to find the dihedral angle involving A', perhaps we need coordinates of A'.Alternatively, since we have B', and since the quadrilateral AEFB is folded over EF, which connects E(0,1,0) and F(2,3,0). So the fold line is EF, and A and B are on the same side of EF, which is being folded over.Alternatively, since we know coordinates of B', perhaps we can find coordinates of A' by ensuring that the quadrilateral A'EFB' is congruent to AEFB.Original quadrilateral AEFB:A(0,0,0), E(0,1,0), F(2,3,0), B(2,0,0)After folding: A', E(0,1,0), F(2,3,0), B'(0,2,2)We need A'E = AE =1, A'F = AF = sqrt(13), and the quadrilateral should maintain the distances.Let’s denote A’ as (x, y, z). Then:Distance from A’ to E(0,1,0) is 1:sqrt( (x -0)^2 + (y -1)^2 + (z -0)^2 ) =1 → x² + (y -1)^2 + z² =1.Distance from A’ to F(2,3,0) is sqrt(13):sqrt( (x -2)^2 + (y -3)^2 + (z -0)^2 ) = sqrt(13) → (x -2)^2 + (y -3)^2 + z² =13.Also, since the folding preserves the structure, the angle at E should be preserved. But perhaps it's easier to solve the two equations.Let’s write the two equations:1) x² + (y -1)^2 + z² =12) (x -2)^2 + (y -3)^2 + z² =13Subtract equation 1 from equation 2:[(x -2)^2 + (y -3)^2 + z²] - [x² + (y -1)^2 + z²] =13 -1Expand:(x² -4x +4) + (y² -6y +9) + z² - x² - (y² -2y +1) - z² =12Simplify:-4x +4 -6y +9 +2y -1 =12Combine like terms:-4x -4y +12 =12-4x -4y =0 → x + y =0So x = -y.Now substitute x = -y into equation 1:(-y)^2 + (y -1)^2 + z² =1 → y² + y² -2y +1 + z² =1 → 2y² -2y + z² =0Similarly, we can use equation 2 with x = -y:(-y -2)^2 + (y -3)^2 + z² =13Expand:(y +2)^2 + (y -3)^2 + z² =13(y² +4y +4) + (y² -6y +9) + z² =132y² -2y +13 + z² =132y² -2y + z² =0Which is the same as equation 1. Therefore, we only have the condition x + y =0 and 2y² -2y + z² =0.Therefore, there are infinitely many solutions, but since we are folding over EF, which is in the xy-plane, and the fold should be a reflection over EF, the coordinates of A’ should be symmetric to A with respect to EF.Alternatively, perhaps we can use the fact that folding over EF maps A to A’ and B to B’, maintaining the structure. Since we know B’ is (0,2,2), maybe we can find a relation between A’ and B’.Alternatively, considering that EF is the fold line, the line segment AA’ should be perpendicular to EF and bisected by EF.Wait, in 2D reflection over a line, the segment connecting a point and its reflection is perpendicular to the line and is bisected by the line. So perhaps in 3D, the same applies? So the line EF is the perpendicular bisector of AA’ and BB’.But in 3D, reflecting over a line is more complex, but since EF is in the xy-plane, and the folding is likely a reflection over the line EF within the plane of the paper, lifted into 3D. But since we are dealing with projection H on DE, which is in the original plane, perhaps the z-coordinate is involved.Alternatively, perhaps we can use vector methods.Given that we have B’(0,2,2), we can find the equation of the fold line EF and then find A’.Wait, another approach: since quadrilateral AEFB is folded over EF to A’EFB’, with B’ known, we can compute A’ by ensuring that the quadrilateral A’EFB’ is congruent to AEFB.Original quadrilateral has sides AE=1, EF=2√2, FB=3, and AB=2. Wait, but AB is not part of the quadrilateral; AEFB is a quadrilateral with sides AE, EF, FB, and BA? Wait, AEFB connects A to E to F to B to A. So AE, EF, FB, and BA. But BA is 2, but after folding, BA becomes B’A’.But perhaps not. Maybe the connections are A to E to F to B to A. But after folding, it's A’ to E to F to B’ to A’.Therefore, the lengths should be preserved: A’E = AE =1, EF=2√2, FB’=FB=3, and B’A’ = BA=2.Wait, BA is 2 in original, but after folding, B’A’ should be 2. So distance between B’(0,2,2) and A’ should be 2.So we have another equation: distance from A’(x, y, z) to B’(0,2,2) is 2.So sqrt( (x -0)^2 + (y -2)^2 + (z -2)^2 ) =2 → x² + (y -2)^2 + (z -2)^2 =4.Previously, we had:1) x + y =02) 2y² -2y + z² =03) x² + (y -2)^2 + (z -2)^2 =4Let’s substitute x = -y into equation 3:(-y)^2 + (y -2)^2 + (z -2)^2 =4 → y² + y² -4y +4 + z² -4z +4 =4Combine terms:2y² -4y + z² -4z +8 =4 → 2y² -4y + z² -4z = -4From equation 2: 2y² -2y + z² =0 → z² = -2y² +2ySubstitute z² into equation 3:2y² -4y + (-2y² +2y) -4z = -4Simplify:2y² -4y -2y² +2y -4z = -4 → (-2y) -4z = -4 → 2y +4z =4 → y +2z =2 → y =2 -2zNow, substitute y =2 -2z into equation 2: 2y² -2y + z² =0First, compute y²:y =2 -2z → y² = (2 -2z)^2 =4 -8z +4z²So equation 2 becomes:2*(4 -8z +4z²) -2*(2 -2z) + z² =0 → 8 -16z +8z² -4 +4z +z² =0Combine like terms:(8z² + z²) + (-16z +4z) + (8 -4) =0 →9z² -12z +4 =0Solve quadratic equation 9z² -12z +4=0:Discriminant D=144 - 144=0 → z=(12)/(18)=2/3.Therefore, z=2/3.Then y=2 -2*(2/3)=2 -4/3=2/3.Since x=-y, x= -2/3.Therefore, coordinates of A’ are (-2/3, 2/3, 2/3).So A’(-2/3, 2/3, 2/3). Let's verify equations:1) x + y = -2/3 +2/3=0 ✔️2) 2y² -2y + z² =2*(4/9) -2*(2/3) + (4/9)=8/9 -4/3 +4/9= (8 +4)/9 -4/3=12/9 -12/9=0 ✔️3) Distance from A’ to B’:sqrt( (-2/3 -0)^2 + (2/3 -2)^2 + (2/3 -2)^2 )= sqrt( (4/9) + ( (-4/3)^2 ) + ( (-4/3)^2 ) )= sqrt(4/9 + 16/9 +16/9 )=sqrt(36/9)=sqrt(4)=2 ✔️Good, so A’ is (-2/3, 2/3, 2/3).Now, we need to find the dihedral angle between planes A’-DE-F and DEF.Wait, the dihedral angle A-DE-F is between the planes ADE and DEF. But after folding, plane ADE is now plane A’-D-E, and plane DEF remains as D-E-F.Therefore, the dihedral angle is between planes A’-D-E and D-E-F.To find this dihedral angle, we can compute the angle between the normal vectors of the two planes.First, let's find the normal vector of plane A’-D-E.Points in plane A’-D-E:A’(-2/3, 2/3, 2/3), D(0,4,0), E(0,1,0)Vectors in this plane:Vector DA’ = A’ - D = (-2/3 -0, 2/3 -4, 2/3 -0)= (-2/3, -10/3, 2/3)Vector DE = E - D = (0 -0,1 -4,0 -0)= (0,-3,0)Compute the normal vector by taking the cross product of DA’ and DE.DA’ = (-2/3, -10/3, 2/3)DE = (0, -3,0)Cross product:|i       j         k     ||-2/3 -10/3   2/3||0      -3       0     |= i*( (-10/3)*0 - 2/3*(-3) ) - j*( (-2/3)*0 - 2/3*0 ) + k*( (-2/3)*(-3) - (-10/3)*0 )= i*(0 + 2) - j*(0 - 0) + k*(2 - 0)= 2i +0j +2kSo normal vector N1 = (2,0,2)Similarly, find normal vector of plane DEF.Points in plane DEF: D(0,4,0), E(0,1,0), F(2,3,0)Vectors in this plane:Vector ED = D - E = (0 -0,4 -1,0 -0)=(0,3,0)Vector EF = F - E = (2 -0,3 -1,0 -0)=(2,2,0)Compute cross product of ED and EF.ED = (0,3,0)EF = (2,2,0)Cross product:|i   j   k||0   3   0||2   2   0|= i*(3*0 -0*2) - j*(0*0 -0*2) +k*(0*2 -3*2)= 0i -0j -6kSo normal vector N2 = (0,0,-6)Now, the dihedral angle between the two planes is the angle between N1 and N2.The angle θ between two vectors is given by:cosθ = (N1 • N2) / (|N1| |N2|)Compute the dot product:N1 • N2 = (2)(0) + (0)(0) + (2)(-6) = -12|N1| = sqrt(2² +0² +2²)=sqrt(8)=2√2|N2| = sqrt(0² +0² +(-6)²)=6Therefore,cosθ = -12 / (2√2 *6) = -12 / (12√2) = -1/√2Therefore, θ = arccos(-1/√2) = 135 degrees.Since dihedral angles are measured between 0 and 180 degrees, the angle is 135 degrees.Final AnswerThe dihedral angle (A-DE-F) is (boxed{135^circ}).</think>To solve the problem of finding the dihedral angle (A-DE-F) after folding the quadrilateral (AEFB) along (EF), we start by assigning coordinates to the points in the original rectangle (ABCD):- (A(0,0,0))- (B(2,0,0))- (C(2,4,0))- (D(0,4,0))- (E(0,1,0)) (since (AE = 1))- (F(2,3,0)) (since (BF = 3))When folding along (EF), point (B) is reflected to a new point (B') such that the projection of (B') onto the plane (CDEF) is (H) on line (DE). We determine the coordinates of (B') by solving the equations derived from the distances and constraints given. We find (H(0,2,0)) and (B'(0,2,2)).Next, we find the coordinates of (A') (the reflection of (A) over (EF)) by ensuring the distances (A'E), (A'F), and (A'B') are preserved. Solving the system of equations, we find (A'(-2/3, 2/3, 2/3)).To find the dihedral angle between the planes (A'-DE) and (DEF), we calculate the normal vectors of these planes. For plane (A'-D-E), the normal vector is found using the cross product of vectors (DA') and (DE), resulting in (N_1 = (2,0,2)). For plane (DEF), the normal vector is found using the cross product of vectors (ED) and (EF), resulting in (N_2 = (0,0,-6)).The angle between these normal vectors is calculated using the dot product formula:[costheta = frac{N_1 cdot N_2}{|N_1| |N_2|} = frac{-12}{2sqrt{2} cdot 6} = -frac{1}{sqrt{2}}]Thus, (theta = arccos(-1/sqrt{2}) = 135^circ).The dihedral angle (A-DE-F) is (boxed{135^circ}).]

🔑:9. As shown in Figure 3, draw E R / / D C, and draw E S perp plane E F C D. Establish a spatial rectangular coordinate system with E R, E D, and E S as the x, y, and z axes, respectively.Notice that the projection of point B on plane C D E F is point H on line D E. Let point B(0, y, z)(y>0, z>0).Since F(2,2,0), B E=sqrt{5}, and B F=3, we have:left{begin{array}{l}y^{2}+z^{2}=5, 4+(y-2)^{2}+z^{2}=9end{array} Rightarrowleft{begin{array}{l}y=1, z=2 .end{array}right.right.Thus, B(0,1,2).Therefore, overrightarrow{F B}=(-2,-1,2).Hence, overrightarrow{E A}=frac{1}{3} overrightarrow{F B}=left(-frac{2}{3},-frac{1}{3}, frac{2}{3}right).Let a normal vector of plane A D E be boldsymbol{n}=(x, y, z).Also, overrightarrow{E D}=(0,3,0). Thenleft{begin{array}{l}-frac{2}{3} x-frac{1}{3} y+frac{2}{3} z=0, 3 y=0end{array} Rightarrow n=(1,0,1) .right.Notice that a normal vector of plane C D E F is m=(0,0,1).Let the dihedral angle A-D E-F be theta.Clearly, theta is an obtuse angle.Then cos theta=-|cos |=-frac{sqrt{2}}{2}.Thus, theta=135^{circ}.

🧠:[<think>Okay, let's try to figure out this problem. We need to find all positive integers ( n ) such that the cube root of ( n ) is equal to the number you get when you remove the last three digits of ( n ). Hmm, let me break that down.First, the cube root of ( n ) is a positive integer. Let's call that integer ( k ). So, ( sqrt[3]{n} = k ), which means ( n = k^3 ). Now, removing the last three digits of ( n ) should give us ( k ). How does removing digits work? If we have a number and we remove the last three digits, it's equivalent to dividing the number by 1000 and taking the integer part. For example, if ( n ) is 123456, then removing the last three digits gives us 123, which is 123456 divided by 1000 and floored.But in this case, the problem states that the result of removing the last three digits is exactly ( k ), not just the floor. So, ( n ) divided by 1000 should be exactly ( k ), right? Wait, but if we remove three digits, it's not necessarily exactly division by 1000 unless ( n ) ends with three zeros. Hmm, maybe I need to clarify.Actually, removing the last three digits of ( n ) would mean that ( n ) is at least a four-digit number. Because if it's a three-digit number, removing three digits would leave you with nothing, which doesn't make sense here. So ( n ) must have at least four digits. But since ( n = k^3 ), ( k ) must be such that ( k^3 ) is at least a four-digit number. That means ( k geq 10 ), since ( 10^3 = 1000 ).Wait, but if ( n = k^3 ), and removing the last three digits of ( n ) gives ( k ), then ( k ) is the number formed by all digits of ( n ) except the last three. So if ( n ) is written as ( k ) followed by three digits, then ( n = k times 1000 + m ), where ( m ) is a three-digit number (possibly with leading zeros, but since ( n ) is a positive integer, ( m ) can be from 000 to 999). But since ( n ) is also equal to ( k^3 ), we have:( k^3 = 1000k + m )But ( m ) must be a three-digit number (including leading zeros) so ( 0 leq m leq 999 ).Therefore, substituting ( n = k^3 ) into the equation, we have:( k^3 = 1000k + m )But ( m = k^3 - 1000k ), and since ( 0 leq m leq 999 ), we can write:( 0 leq k^3 - 1000k leq 999 )So, this gives us inequalities:1. ( k^3 - 1000k geq 0 ) => ( k^3 geq 1000k ) => ( k^2 geq 1000 ) => ( k geq sqrt{1000} approx 31.62 ). Since ( k ) is an integer, ( k geq 32 ).2. ( k^3 - 1000k leq 999 ) => ( k^3 - 1000k - 999 leq 0 )So we need to solve for integers ( k geq 32 ) such that ( k^3 - 1000k leq 999 ).Let me write this inequality as ( k^3 - 1000k - 999 leq 0 ).We can try to find the range of ( k ) where this holds. Let's approximate.Let’s first find when ( k^3 - 1000k = 999 ).So, ( k^3 - 1000k - 999 = 0 ). Hmm, solving this equation would give the upper bound for ( k ).Alternatively, we can try plugging in values of ( k ) starting from 32 upwards until the inequality flips.Let’s compute for ( k = 32 ):( 32^3 - 1000*32 = 32768 - 32000 = 768 ). Then subtract 999: 768 - 999 = -231. So, -231 <= 0. So 32 satisfies the inequality.Wait, wait. Wait, the original inequality is ( k^3 - 1000k leq 999 ). So for ( k = 32 ):Left side: 32^3 - 1000*32 = 32768 - 32000 = 768. Then 768 <= 999? Yes, that's true. So 32 is okay.For ( k = 33 ):33^3 = 35937, 1000*33 = 33000, so 35937 - 33000 = 2937. 2937 <= 999? No, 2937 is greater than 999. So 33 does not satisfy the inequality.Wait, that's a big jump. So from k=32 to k=33, the value jumps from 768 to 2937. So the inequality ( k^3 - 1000k <= 999 ) only holds for k=32?Wait, let's check k=31. Wait, but earlier we said k >= 32 because k^2 >= 1000. Let me confirm.Wait, k^3 - 1000k >= 0 => k(k^2 - 1000) >=0. Since k is positive, this implies k^2 - 1000 >=0 => k >= sqrt(1000) ≈ 31.62, so k >=32.Therefore, k must be 32 or higher. However, when k=32, the value of k^3 - 1000k is 768, which is within 0 <= m <=999, so acceptable. For k=33, it's 2937, which is too big (since m must be less than 1000). Therefore, the only possible value is k=32.But let me check k=10, even though k=10 would give n=1000, removing last three digits gives 1, but cube root of 1000 is 10, which is not equal to 1. So k=10 is invalid. Similarly, other lower k's are invalid.But according to the inequality, k must be at least 32. So only k=32 is valid? Let me check.Wait, n = k^3 = 32^3 = 32768. Now, removing the last three digits of 32768. Let's see: 32768. The last three digits are 768. Removing them gives 32. Which is equal to k=32. So that works.Wait, but according to this, n=32768, remove last three digits gives 32, which is k. Then cube root of n is 32, which is equal to k. So that works.But is that the only solution? Let me check k=31, even though k=31 is less than 32, but let's see:k=31, n=31^3=29791. Removing last three digits: 29791 -> last three digits 791, removing gives 29. Which is not equal to 31. So that's invalid. So k=31 doesn't work. Similarly, k=32 works.But let's check k=33: n=35937. Remove last three digits: 35. Cube root of 35937 is 33, which is not equal to 35. Wait, that's not correct. Wait, 35937 divided by 1000 is 35.937, so removing last three digits would give 35, but cube root is 33, so 35 ≠33. So not equal.But according to our inequality, k=33 doesn't satisfy the equation. So only k=32 works. Thus, the only solution is n=32768.But wait, let me check if there might be other solutions. Maybe my initial inequality is wrong?Wait, the problem states that the cube root of n is equal to the positive integer obtained by removing the last three digits. So, more formally, if n is a number where if you take away the last three digits, you get k, and k is the cube root of n. So n is k followed by three digits. So n can be written as:n = k * 1000 + mwhere m is a three-digit number (from 000 to 999). But also, n = k^3. Therefore:k^3 = 1000k + mWhich implies that:m = k^3 - 1000kSince m must be between 0 and 999, inclusive, we have:0 ≤ k^3 - 1000k ≤ 999So solving for k:First inequality: k^3 - 1000k ≥ 0 ⇒ k(k^2 - 1000) ≥ 0. Since k is positive, this requires k^2 ≥ 1000 ⇒ k ≥ sqrt(1000) ≈ 31.62. So k must be at least 32.Second inequality: k^3 - 1000k ≤ 999 ⇒ k^3 ≤ 1000k + 999.Let’s solve this inequality. Let's try k=32: 32^3 = 32768. 1000*32 +999= 32000 + 999=32999. 32768 ≤32999: yes.So k=32 is valid.k=33: 33^3=35937. 1000*33 +999=33000+999=33999. 35937 ≤33999? No. So invalid.k=34: 34^3=39304. 1000*34 +999=34000+999=34999. 39304 <=34999? No.Similarly, higher k will make k^3 even larger, so the inequality won't hold.What about k=31? 31^3=29791. 1000*31 +999=31000+999=31999. 29791 <=31999? Yes, but k=31 doesn't satisfy the first inequality. Because k=31, k^3 -1000k =29791 -31000= -1209, which is negative. So m would be negative, which is invalid. So k=31 is out.Therefore, only k=32 satisfies both inequalities. Therefore, the only solution is n=32^3=32768. Let me confirm once again.n=32768. Removing the last three digits: 32768 → 32 and 768. So removing 768 gives 32. Cube root of 32768 is 32. Perfect, that works.So is there any other k? Let me check k=10. n=1000. Remove last three digits: 1. Cube root of 1000 is 10. 1≠10. So no.k=99: n=970299. Remove last three digits: 970. Cube root of 970299 is 99. 970≠99. Not equal.Wait, so the structure is that n must be k^3, and when you remove the last three digits, you get k. So unless k^3 is a number that starts with k followed by three digits, it won't work. So, in essence, k^3 must be k followed by three digits. So k^3 = k * 1000 + m, where m is three digits. So m = k^3 - 1000k, and m must be between 0 and 999.But when we solved this, only k=32 works. Therefore, n=32768 is the only solution.But just to be thorough, let me check k=32 again.32^3 = 32*32*32 = 32*1024=32768. Correct. Remove last three digits: 32768 → 32 (since 768 is the last three digits). Cube root is 32. So yes, that's a valid solution.What about k=100? n=1,000,000. Removing last three digits gives 1000. Cube root of 1,000,000 is approximately 100, but 100^3 is 1,000,000. So cube root is 100. Removing last three digits gives 1000. 100≠1000. Doesn't work.Similarly, k=5: n=125. Removing last three digits would require n to have at least four digits, but 125 is only three digits. So invalid.Therefore, the only possible n is 32768. So the answer is boxed{32768}Final Answerboxed{32768}</think>Given a positive integer ( n ), we need to find all such ( n ) where the cube root of ( n ) equals the positive integer obtained by removing the last three digits of ( n ).1. Let ( k ) be the cube root of ( n ). Then ( n = k^3 ).2. Removing the last three digits of ( n ) gives ( k ). This implies ( n ) can be written as ( k times 1000 + m ) where ( 0 leq m leq 999 ).3. Therefore, we have the equation ( k^3 = 1000k + m ).4. Rearranging, we get ( m = k^3 - 1000k ). Since ( m ) must be between 0 and 999, we have the inequalities: [ 0 leq k^3 - 1000k leq 999 ]5. Solving the first inequality ( k^3 - 1000k geq 0 ): [ k(k^2 - 1000) geq 0 implies k geq sqrt{1000} approx 31.62 ] Thus, ( k geq 32 ).6. Solving the second inequality ( k^3 - 1000k leq 999 ): - For ( k = 32 ): [ 32^3 - 1000 times 32 = 32768 - 32000 = 768 quad (text{which is } leq 999) ] - For ( k = 33 ): [ 33^3 - 1000 times 33 = 35937 - 33000 = 2937 quad (text{which is } > 999) ]7. Only ( k = 32 ) satisfies both inequalities. Checking ( n = 32^3 = 32768 ): - Removing the last three digits of 32768 gives 32, which matches the cube root of 32768.Thus, the only solution is (boxed{32768}).,<think>Alright, let's try to tackle this problem. So, we need to find all positive integers n where the cube root of n is equal to the number you get when you remove the last three digits of n. Hmm, okay. Let's break this down step by step.First, let's make sure I understand the problem correctly. If we have a number n, and we remove its last three digits, we get another positive integer. The cube root of n should be exactly that integer. So, for example, if n was a number like 123456, removing the last three digits would give us 123, and the cube root of 123456 should be 123. But of course, in reality, the cube of 123 is probably not 123456. So we need to find all such n where this condition holds.Let me formalize this. Let's let k be the positive integer obtained by removing the last three digits of n. Then according to the problem, we have:k = ∛nWhich implies that:n = k³But wait, removing the last three digits of n should give us k. How does removing the last three digits translate mathematically? If we remove the last three digits of a number, that's equivalent to dividing the number by 1000 and taking the floor of it. So, n divided by 1000, rounded down to the nearest integer, is k. But since k is an integer, this means that:k = floor(n / 1000)But we also have n = k³. So substituting, we get:k = floor(k³ / 1000)Hmm, okay. So k must satisfy the inequality:k ≤ k³ / 1000 < k + 1Multiplying all parts by 1000 to eliminate the denominator:1000k ≤ k³ < 1000(k + 1)So, we can split this into two inequalities:1. k³ ≥ 1000k2. k³ < 1000(k + 1)Let me solve these inequalities one by one.Starting with the first inequality:k³ ≥ 1000kDivide both sides by k (since k is a positive integer, we can safely do this):k² ≥ 1000So, k ≥ √1000 ≈ 31.62. Since k is an integer, k must be at least 32.Second inequality:k³ < 1000(k + 1)Let's rearrange this:k³ - 1000k - 1000 < 0Hmm, this seems a bit more complicated. Let's see if we can find the values of k where this holds.Alternatively, perhaps we can write it as:k³ < 1000k + 1000So, k³ < 1000(k + 1)We can factor out 1000 on the right side, but maybe it's better to approximate or solve for k.Alternatively, let's note that when k is large, the left side (k³) grows much faster than the right side (1000k + 1000). So, there must be some upper bound for k beyond which the inequality doesn't hold. So we need to find the maximum k for which k³ < 1000(k + 1).Let me try plugging in some values. Let's start from k = 32 upwards until the inequality flips.For k = 32:Left side: 32³ = 32 * 32 * 32 = 32 * 1024 = 32768Right side: 1000*(32 + 1) = 1000*33 = 33000So, 32768 < 33000. That's true.k = 33:Left side: 33³ = 33*33*33. Let's calculate 33*33 = 1089, then 1089*33. 1089*30 = 32670, 1089*3 = 3267, total is 32670 + 3267 = 35937.Right side: 1000*(33 + 1) = 1000*34 = 34000So, 35937 < 34000? No, 35937 is greater. So the inequality doesn't hold here. Thus, k = 33 doesn't satisfy the second inequality. Therefore, the upper bound for k is 32.Wait, so k must be between 32 and 32? Because at k = 32, the first inequality starts to hold (since k ≥32), but the second inequality holds only up to k = 32. So then the only possible k is 32.Therefore, n = k³ = 32³ = 32768.Wait, but let me check if removing the last three digits of 32768 gives 32. Let's see, 32768 divided by 1000 is 32.768, so floor(32.768) is 32. Yes, that works. So n = 32768 is a solution.But wait, the problem says "find all positive integers n". So maybe there's only one solution? Let me verify.Wait, let's check if there might be another k that I missed. For k = 10, for example, even though the first inequality isn't satisfied. Let's see, suppose k = 10. Then n = 1000. Removing the last three digits of 1000 gives 1, but the cube root of 1000 is 10, which is not 1. So that's not a solution. Similarly, if k is less than 32, n = k³ would be less than 32³ = 32768, so removing the last three digits would give a number less than 32.768, which would be floor(k³ / 1000). But if k is less than 32, then k³ is less than 32768, so floor(k³ / 1000) would be less than 32. But according to the first inequality, k needs to be at least 32. So there are no solutions below k = 32.And for k greater than 32, like 33, as we saw, k³ becomes larger than 1000*(k + 1), so the second inequality fails. Therefore, the only solution is k = 32, hence n = 32768.But wait, let me check k = 31. If k = 31, then n = 29791. Removing the last three digits would give floor(29791 / 1000) = 29, which is not 31. So that doesn't work. Similarly, k = 32, n = 32768. Removing last three digits gives 32, which matches. Then k = 33, n = 35937. Removing the last three digits gives 35, which is not 33. So that doesn't work. So indeed, only k = 32.Therefore, the answer should be n = 32768.But let me check once again to ensure I haven't made a mistake.First, cube of 32 is 32*32*32.32*32 = 1024.1024*32 = 32768. Correct.Now, removing the last three digits of 32768. The number is 32768. Last three digits are 768. So removing them gives 32. So yes, 32 is the cube root. Perfect.Is there another number that satisfies this?Suppose n is a number such that n = k³, and when we remove the last three digits, we get k. Let's think about how numbers are structured. If removing the last three digits gives k, then n must be k followed by three digits. Wait, but actually, n could be k followed by three digits, but not necessarily. For example, if k is a two-digit number, then n would be k followed by three digits, making it a five-digit number. If k is a three-digit number, n would be k followed by three digits, making it a six-digit number, etc.Wait, but in our case, k = 32, which is a two-digit number. So n is 32 followed by three digits, which is 32768. But 32768 is a five-digit number. If k was a three-digit number, say 100, then n would be 100³ = 1,000,000. Removing the last three digits gives 1000, which is not 100. So that's not a solution.Wait, but in this case, when k is a three-digit number, the number n would be k³, which is a nine-digit number if k is 100 (100³ = 1,000,000). Wait, no, 100³ is 1,000,000, which is a seven-digit number. Removing the last three digits would give 1000 (since 1,000,000 divided by 1000 is 1000). But 1000 ≠ 100, so that's not a solution.Similarly, let's check for k = 10. Then n = 1000, removing last three digits gives 1, which isn't 10. So no.Thus, the structure is that n is a number where the first part is k, and the last three digits can be anything, but in our case, n = k³, so the last three digits are determined by k³. So, for n to satisfy the condition, k³ must end with three digits such that when we strip them off, we get k. But in the problem statement, it's not that the last three digits are arbitrary, but rather that n is exactly k³, so the last three digits are fixed once k is chosen. Therefore, the only way this can happen is if k³ is a number that starts with k followed by three digits. But that's not necessarily the case unless the cube of k is of the form k*1000 + m, where m is some three-digit number (possibly with leading zeros). But in reality, since n = k³, the number n has a certain number of digits. For k = 32, n is 32768, which is 5 digits. So removing the last three digits would leave the first two digits, which is 32. For k = 100, n = 1,000,000, which is seven digits. Removing the last three digits gives 1000, which is different from 100.So perhaps the only k where k³, when divided by 1000, gives a quotient of k, is when k³ is exactly equal to k*1000 + m, where m is between 0 and 999. But since k³ = k*1000 + m, rearranged, we have k³ - k*1000 = m. Therefore, m must be an integer between 0 and 999. Therefore, k³ - 1000k must be between 0 and 999. Which is exactly the second inequality we had earlier: k³ < 1000(k + 1). Wait, but k³ - 1000k < 1000. So, that's the same as k³ < 1000k + 1000.Earlier, we found that for k = 32, k³ = 32768, and 1000k + 1000 = 32000 + 1000 = 33000. So, 32768 < 33000 holds. Then for k = 33, k³ = 35937, and 1000k + 1000 = 33000 + 1000 = 34000. But 35937 > 34000, so the inequality doesn't hold. So, only k = 32 satisfies both inequalities. Therefore, the only solution is n = 32768.But just to be thorough, let's check if there are any other possible k's that might have been missed.Suppose k is larger. Let's test k = 100. Then n = 1,000,000. Removing the last three digits gives 1000. The cube root of 1,000,000 is 100, but 1000 ≠ 100. So that's not a solution.k = 99. Then n = 970299. Removing last three digits gives 970. Cube root of 970299 is 99. So 970 ≠ 99. Not a solution.k = 50. n = 125000. Removing last three digits gives 125. Cube root is 50. Not equal.k = 31. n = 29791. Removing last three digits gives 29. Cube root is 31. Not equal.k = 32. n = 32768. Removing last three digits gives 32. Cube root is 32. Equal. So that works.k = 1. n = 1. Removing last three digits... but n is only 1 digit. So removing three digits would result in 0, which is not a positive integer. So not a solution.k = 0. Not positive, so invalid.What about k = 10? As before, n = 1000. Removing three digits gives 1, which is not 10.How about k = 5. n = 125. Removing three digits, but 125 is only three digits. Removing three digits gives 0. Not valid.So indeed, all other k's either result in n being too short (so that removing three digits leads to 0 or a number with fewer digits) or the quotient when dividing by 1000 is not equal to k. Hence, the only solution is k = 32, n = 32768.Therefore, I think the answer is n = 32768. Let me just confirm once more.Yes, 32³ = 32*32*32 = 32*1024 = 32768. The number 32768 has five digits. Removing the last three digits (768) gives us 32. The cube root of 32768 is indeed 32. So it's a perfect match. And there's no other number that satisfies this condition because for k > 32, the cube exceeds the 1000(k + 1) limit, and for k < 32, the cube isn't large enough to give k when divided by 1000. Therefore, the only solution is 32768.Final AnswerThe positive integer ( n ) that satisfies the condition is boxed{32768}.</think>Let ( n ) be a positive integer such that the cube root of ( n ) equals the positive integer obtained by removing the last three digits of ( n ). Let ( k ) be the positive integer obtained by removing the last three digits of ( n ). This implies:[ k = sqrt[3]{n} ]Thus, we have:[ n = k^3 ]Removing the last three digits of ( n ) is equivalent to dividing ( n ) by 1000 and taking the floor value, which gives:[ k = leftlfloor frac{n}{1000} rightrfloor ]Substituting ( n = k^3 ), we get:[ k = leftlfloor frac{k^3}{1000} rightrfloor ]This leads to the inequalities:[ 1000k leq k^3 < 1000(k + 1) ]Solving the first inequality:[ k^3 geq 1000k ][ k^2 geq 1000 ][ k geq sqrt{1000} approx 31.62 ]Thus, ( k geq 32 ).Solving the second inequality:[ k^3 < 1000(k + 1) ]Testing values of ( k ):- For ( k = 32 ): [ 32^3 = 32768 ] [ 1000(32 + 1) = 33000 ] [ 32768 < 33000 ]- For ( k = 33 ): [ 33^3 = 35937 ] [ 1000(33 + 1) = 34000 ] [ 35937 not< 34000 ]Thus, ( k = 32 ) is the only solution. Verifying, ( n = 32^3 = 32768 ). Removing the last three digits of 32768 gives 32, which matches the cube root of 32768.Therefore, the positive integer ( n ) that satisfies the condition is (boxed{32768}).]

🔑:26. Let n=1000 x+y, where x is a positive integer, y is an integer, and 0 leqslant y leqslant 999. According to the problem,x^{3}=1000 x+yFrom 0 leqslant y leqslant 999, we know1000 x leqslant x^{3}<1000 x+1000=1000(x+1)Thus,x^{2} geqslant 1000, x^{3}+1 leqslant 1000(x+1)Therefore,x^{2} geqslant 1000, x^{2}-x+1 leqslant 1000So,32 leqslant x<33Hence,x=32,Then,y=768Therefore,n=32768

🧠:[<think>Okay, let's try to figure out this problem. So, there's a vessel that can hold 1000 liters. Initially, it's being filled at 30 liters per second. Then, at some time t, the flow rate decreases to 15 liters per second. After 40 seconds from the start, the vessel is full. We need to find out what fraction of the vessel was filled by time t.Hmm. Let me break this down. The total time taken to fill the vessel is 40 seconds. For the first t seconds, the rate is 30 L/s, and then from t to 40 seconds, the rate drops to 15 L/s. The total volume filled in these two periods should add up to 1000 liters.So, the total water added is the sum of the water added in the first t seconds and the water added in the remaining (40 - t) seconds. Let me write that equation out.Total volume = (30 L/s * t) + (15 L/s * (40 - t)) = 1000 liters.Let me compute that:30t + 15(40 - t) = 1000.First, multiply out the terms:30t + 600 - 15t = 1000.Combine like terms:15t + 600 = 1000.Subtract 600 from both sides:15t = 400.Divide both sides by 15:t = 400 / 15.Hmm, 400 divided by 15. Let's calculate that. 15*26 = 390, so 400 - 390 = 10. So t = 26 and 10/15 seconds, which simplifies to 26 and 2/3 seconds. So t is 26.666... seconds.But wait, the question is asking for the fraction of the vessel's content that was filled at time t. So, at time t, the amount filled would be the initial rate times t, which is 30 L/s * t.So, the volume filled by time t is 30t liters. Then, the fraction would be 30t / 1000.Since we know t is 400/15 seconds, let's compute 30*(400/15) / 1000.First, 30 divided by 15 is 2. So, 2*400 = 800. Then, 800 / 1000 = 0.8. So 0.8 is 4/5. Therefore, the fraction is 4/5.Wait, that seems straightforward, but let me check again. Maybe I made a mistake in the calculations.Starting over:Total volume when full is 1000 liters.Filling starts at 30 L/s for t seconds, then 15 L/s for (40 - t) seconds.So total filled is 30t + 15(40 - t) = 1000.Simplify:30t + 600 - 15t = 100015t + 600 = 100015t = 400t = 400/15 = 80/3 ≈ 26.6667 seconds.Then, the volume filled by time t is 30*(80/3) = 30*(80)/3 = 10*80 = 800 liters.Therefore, 800 liters out of 1000 liters is 800/1000 = 4/5. So the fraction is 4/5.Wait, but the answer seems too straightforward. Let me verify if the time makes sense. If t is 80/3 seconds, which is approximately 26.67 seconds, then the remaining time is 40 - 80/3 = (120/3 - 80/3) = 40/3 ≈ 13.33 seconds.So, the first period fills 30 L/s * 80/3 s = 800 liters.Second period fills 15 L/s * 40/3 s = 15*40/3 = 200 liters.800 + 200 = 1000 liters. That checks out.So, the fraction filled at time t is 800/1000 = 4/5. Therefore, the answer is 4/5.But let me check the problem statement again. It says "the vessel (without a drain)" so no water is leaving, just the inflow. "initially filled with exactly 30 liters per second... from a later time t onwards, exactly 15 liters per second. After exactly 40s, measured from the beginning, the vessel was full." So, yes, the two rates add up over their respective times. Therefore, the calculations are correct.So, the fraction is 4/5. As a box, that would be boxed{dfrac{4}{5}}.Final Answerboxed{dfrac{4}{5}}</think>The vessel has a capacity of 1000 liters and was filled initially at 30 liters per second. From time ( t ) onwards, the flow rate was reduced to 15 liters per second. The vessel was full after 40 seconds. We need to determine the fraction of the vessel's content filled by time ( t ).First, we set up the equation for the total volume of water filled:[30t + 15(40 - t) = 1000]Simplifying this equation:[30t + 600 - 15t = 1000][15t + 600 = 1000][15t = 400][t = frac{400}{15} = frac{80}{3} text{ seconds}]Next, we calculate the volume filled by time ( t ):[30t = 30 left( frac{80}{3} right) = 800 text{ liters}]The fraction of the vessel's content filled by time ( t ) is:[frac{800}{1000} = frac{4}{5}]Thus, the fraction of the vessel's content filled at time ( t ) is (boxed{dfrac{4}{5}}).,<think>Okay, let's see. I need to figure out what fraction of the vessel's total capacity was filled by time t, where after time t, the flow rate decreases from 30 liters per second to 15 liters per second. The total time taken to fill the vessel is 40 seconds. The vessel's capacity is 1000 liters. So, let me parse the problem again. Initially, the vessel is being filled at 30 L/s, and then at some time t, they switch to filling it at 15 L/s. The total time from the start until the vessel is full is 40 seconds. So, the filling process has two phases: the first phase from 0 to t seconds at 30 L/s, and the second phase from t to 40 seconds at 15 L/s. The total water added in these two phases should equal the vessel's capacity, which is 1000 liters.So, the total water added is (30 L/s * t) + (15 L/s * (40 - t)) = 1000 liters. Hmm, that seems like an equation I can set up. Let me write that down:30t + 15(40 - t) = 1000Let me compute this. First, expand the terms:30t + 15*40 - 15t = 1000Compute 15*40: that's 600. So,30t - 15t + 600 = 1000Combine like terms:15t + 600 = 1000Subtract 600 from both sides:15t = 400Then, divide both sides by 15:t = 400 / 15Simplify that: 400 divided by 15. Let's compute it. 15*26 = 390, so 400 - 390 = 10. So, t = 26 and 10/15 seconds, which simplifies to 26 and 2/3 seconds, or 26.666... seconds.So, time t is 80/3 seconds (since 26.666... is 80/3). Wait, 80 divided by 3 is approximately 26.666..., yes. So t is 80/3 seconds.But the question is asking for the fraction of the vessel's content that was filled at time t. So, up to time t, the flow rate was 30 L/s, so the amount filled in that time is 30 L/s * t. Then, we need to compute that amount divided by the total capacity (1000 L) to get the fraction.So, let's compute the amount filled by time t:30 * t = 30 * (80/3) = 30*(80)/3 = 10*80 = 800 liters.Wait, that's 800 liters. So, 800 liters out of 1000 liters, which is 0.8, or 4/5. So the fraction is 4/5.Wait, hold on, but let me double-check. If t is 80/3 seconds (~26.666 s), then the first phase fills 30 L/s * 80/3 s = 800 L, right. Then the second phase is from 80/3 s to 40 s, which is 40 - 80/3 = (120/3 - 80/3) = 40/3 seconds. So, 15 L/s * 40/3 s = 15*(40/3) = 5*40 = 200 L. Then 800 + 200 = 1000 L, which matches the total capacity. So that checks out.Therefore, the fraction filled at time t is 800 / 1000 = 0.8 = 4/5. So the answer should be 4/5. But let me make sure that the problem is interpreted correctly. The vessel was initially filled at 30 L/s, and from time t onwards, 15 L/s. So the total time is 40 seconds. So the first phase is t seconds, second phase is (40 - t) seconds. The calculation seems correct. So, yes, the fraction is 4/5.But let me check if there's another way to approach this. Maybe using average rates or something else? Alternatively, let's suppose that t is the time when the rate changes. The total volume is 30t +15(40 - t) =1000. So, solving for t gives t= (1000 -15*40)/15. Wait, let's see:30t +15(40 - t) =100030t +600 -15t=100015t=400t=400/15=80/3≈26.666...So that's the same as before. So the first period is 80/3 seconds, which is approximately 26.666 seconds. Then, the amount filled in that time is 30*(80/3)=800 liters. So the fraction is 800/1000=4/5. So yes, 4/5.Alternatively, maybe the problem is tricking me into thinking that the time after t is 40 seconds? Wait, the problem says: "after exactly 40 s, measured from the beginning, the vessel was full." So the total time from start to finish is 40 seconds. So the time after t is (40 - t) seconds. So our calculation is correct.Therefore, the fraction is 4/5. So boxed answer is boxed{dfrac{4}{5}}.Final AnswerThe fraction of the vessel's content filled at time ( t ) is boxed{dfrac{4}{5}}.</think>To determine the fraction of the vessel's content filled at time ( t ), we start by noting that the vessel is filled at 30 liters per second initially and then at 15 liters per second from time ( t ) onwards. The total time taken to fill the vessel is 40 seconds, and the vessel's capacity is 1000 liters.We set up the equation for the total volume of water filled:[30t + 15(40 - t) = 1000]Expanding and simplifying this equation:[30t + 600 - 15t = 1000][15t + 600 = 1000][15t = 400][t = frac{400}{15} = frac{80}{3}]The volume of water filled by time ( t ) is:[30t = 30 left( frac{80}{3} right) = 800 text{ liters}]The fraction of the vessel's content filled at time ( t ) is:[frac{800}{1000} = frac{4}{5}]Thus, the fraction of the vessel's content filled at time ( t ) is boxed{dfrac{4}{5}}.]

🔑:The time elapsed from the beginning until time t, during which exactly 30 liters per second flowed into the container, is x. Then, during the time when exactly 15 liters per second flowed into the container, exactly (40-x) seconds have passed, and it holds that:30 x+(40-x) cdot 15=1000thus 15 x=400, from which we get x=frac{80}{3}.During these frac{80}{3} mathrm{~s}, exactly frac{80}{3} cdot 30 mathrm{l}, which is 800 liters of water, flowed into the container. Since frac{800}{1000}=frac{4}{5}, exactly frac{4}{5} of the container was filled at time t.