answer：Okay, so I have this problem here: Given that the function ( y = f(x) ) is the inverse function of ( y = log_a x ) (where ( a > 0 ) and ( a neq 1 )), and that the graph of ( f(x) ) passes through the point ( (2, frac{1}{4}) ), I need to find the value of ( log_2 f(-1) ).Hmm, let me try to break this down step by step. First, I know that if ( f(x) ) is the inverse function of ( y = log_a x ), then ( f(x) ) must be the exponential function with base ( a ). That is, ( f(x) = a^x ). Because, generally, the inverse of a logarithmic function ( log_b x ) is the exponential function ( b^x ). So, that makes sense.Now, the graph of ( f(x) ) passes through the point ( (2, frac{1}{4}) ). That means when ( x = 2 ), ( f(2) = frac{1}{4} ). Since ( f(x) = a^x ), we can substitute these values into the equation to find ( a ).So, substituting ( x = 2 ) and ( f(2) = frac{1}{4} ), we get:[a^2 = frac{1}{4}]To solve for ( a ), we take the square root of both sides. Remembering that ( a > 0 ), we don't have to worry about the negative root. So,[a = sqrt{frac{1}{4}} = frac{1}{2}]Alright, so now we know that ( a = frac{1}{2} ). Therefore, the function ( f(x) ) is:[f(x) = left( frac{1}{2} right)^x]Alternatively, since ( left( frac{1}{2} right)^x = 2^{-x} ), we can write it as:[f(x) = 2^{-x}]Either form is correct, but maybe the second form will be more convenient later on.Now, the problem asks for ( log_2 f(-1) ). Let's first find ( f(-1) ) and then take the logarithm base 2 of that result.So, substituting ( x = -1 ) into ( f(x) ):[f(-1) = left( frac{1}{2} right)^{-1}]Wait, what is ( left( frac{1}{2} right)^{-1} )? Remember that a negative exponent means reciprocal. So,[left( frac{1}{2} right)^{-1} = frac{1}{left( frac{1}{2} right)} = 2]So, ( f(-1) = 2 ).Now, we need to compute ( log_2 f(-1) ), which is ( log_2 2 ).What's ( log_2 2 )? Well, that's asking "2 raised to what power gives 2?" And the answer is 1, because ( 2^1 = 2 ).Therefore, ( log_2 f(-1) = 1 ).Let me just recap to make sure I didn't make any mistakes. 1. ( f(x) ) is the inverse of ( log_a x ), so ( f(x) = a^x ).2. The graph passes through ( (2, frac{1}{4}) ), so ( a^2 = frac{1}{4} ) leading to ( a = frac{1}{2} ).3. Therefore, ( f(x) = left( frac{1}{2} right)^x ).4. Calculating ( f(-1) ) gives ( 2 ).5. Taking ( log_2 ) of 2 gives 1.Everything seems to check out. I don't think I made any calculation errors. It's a straightforward problem once you remember the relationship between logarithmic and exponential functions and how to find the inverse.I guess another way to think about it is that since ( f(x) ) is the inverse of ( log_a x ), their graphs are reflections over the line ( y = x ). So, if ( f(x) ) passes through ( (2, frac{1}{4}) ), then ( log_a x ) should pass through ( (frac{1}{4}, 2) ). Let me verify that. If ( log_a frac{1}{4} = 2 ), then ( a^2 = frac{1}{4} ), which again gives ( a = frac{1}{2} ). So that confirms the value of ( a ).Then, as before, ( f(-1) = left( frac{1}{2} right)^{-1} = 2 ), and ( log_2 2 = 1 ). Yeah, I think that's solid. I don't see any issues with the reasoning.**Final Answer**boxed{A}

answer：Okay, so I have this arithmetic sequence problem here. Let me try to figure it out step by step. The problem says:"In the arithmetic sequence {a_n}, a_2 = 2008, a_{2008} = a_{2004} - 16. Find the value of n for which the sum of the first n terms, S_n, is maximized. The options are A: 503, B: 504, C: 503 or 504, D: 504 or 505."Alright, so first, I know that in an arithmetic sequence, each term is obtained by adding a common difference, d, to the previous term. So, the general formula for the nth term is a_n = a_1 + (n-1)d, where a_1 is the first term.Given that a_2 = 2008, I can write that as:a_2 = a_1 + d = 2008.So, that's equation one.Next, the problem says that a_{2008} = a_{2004} - 16. Let me write that down:a_{2008} = a_{2004} - 16.But since it's an arithmetic sequence, I can express a_{2008} and a_{2004} in terms of a_1 and d.So, a_{2008} = a_1 + (2008 - 1)d = a_1 + 2007d.Similarly, a_{2004} = a_1 + (2004 - 1)d = a_1 + 2003d.So, substituting these into the equation:a_1 + 2007d = (a_1 + 2003d) - 16.Let me simplify this equation.Subtract (a_1 + 2003d) from both sides:(a_1 + 2007d) - (a_1 + 2003d) = -16.Simplify the left side:a_1 - a_1 + 2007d - 2003d = -16.So, 4d = -16.Therefore, d = -16 / 4 = -4.So, the common difference d is -4.Now, going back to the first equation:a_1 + d = 2008.We know d is -4, so:a_1 - 4 = 2008.Therefore, a_1 = 2008 + 4 = 2012.So, the first term is 2012, and the common difference is -4.Now, the problem asks for the value of n that maximizes the sum of the first n terms, S_n.I remember that the sum of the first n terms of an arithmetic sequence is given by:S_n = n/2 * (2a_1 + (n - 1)d).Alternatively, it can also be written as:S_n = n * (a_1 + a_n)/2.Since we have a_n = a_1 + (n - 1)d, we can substitute that in.But before I get into calculating S_n, maybe I should figure out when the terms of the sequence start becoming negative because the sum will be maximized just before the terms become negative, right?Since the common difference is negative (-4), the sequence is decreasing. So, the terms will start at 2012 and decrease by 4 each time.So, the first term is positive, and at some point, the terms will become zero and then negative. The sum S_n will increase as long as the terms are positive, but once the terms become negative, adding them will start decreasing the total sum.Therefore, the maximum sum occurs either at the last positive term or at the term where the term is zero.So, let's find when a_n becomes zero or negative.The nth term is:a_n = a_1 + (n - 1)d.We have a_1 = 2012 and d = -4, so:a_n = 2012 + (n - 1)(-4) = 2012 - 4(n - 1).Simplify:a_n = 2012 - 4n + 4 = 2016 - 4n.We want to find when a_n ≤ 0.So, set 2016 - 4n ≤ 0.Solving for n:2016 ≤ 4nDivide both sides by 4:504 ≤ n.So, n ≥ 504.Therefore, the 504th term is the first term that is zero or negative.Wait, let me compute a_504:a_504 = 2016 - 4*504 = 2016 - 2016 = 0.So, the 504th term is zero.Therefore, the terms from n=1 to n=503 are positive, n=504 is zero, and beyond that, they become negative.So, the sum S_n will be increasing up to n=503, and then at n=504, adding zero won't change the sum, so S_504 is equal to S_503.Wait, is that correct?Wait, let me think again.The sum S_n is the sum of the first n terms.So, S_504 = S_503 + a_504.But since a_504 is zero, S_504 = S_503 + 0 = S_503.Therefore, both S_503 and S_504 are equal, and they are the maximum sums.But wait, let me verify.Alternatively, maybe I can compute S_n and see when it's maximized.The sum S_n is given by:S_n = n/2 * (2a_1 + (n - 1)d).Plugging in the values:S_n = n/2 * (2*2012 + (n - 1)*(-4)).Simplify:S_n = n/2 * (4024 - 4(n - 1)).Simplify inside the parentheses:4024 - 4n + 4 = 4028 - 4n.So, S_n = n/2 * (4028 - 4n) = (n/2)*(4028 - 4n).Simplify further:S_n = (4028n - 4n^2)/2 = 2014n - 2n^2.So, S_n = -2n^2 + 2014n.This is a quadratic function in terms of n, and since the coefficient of n^2 is negative (-2), the parabola opens downward, meaning the maximum is at the vertex.The vertex of a parabola given by f(n) = an^2 + bn + c is at n = -b/(2a).Here, a = -2, b = 2014.So, n = -2014/(2*(-2)) = -2014/(-4) = 2014/4.Compute 2014 divided by 4:2014 ÷ 4 = 503.5.So, the vertex is at n = 503.5.But n has to be an integer, so the maximum occurs at n = 503 or n = 504.Therefore, both n = 503 and n = 504 will give the maximum sum.So, the answer should be C: 503 or 504.Wait, but let me confirm with the earlier reasoning.Since a_504 is zero, adding it to S_503 doesn't change the sum. So, S_503 = S_504.Therefore, both n=503 and n=504 give the same maximum sum.Hence, the correct answer is C.But just to make sure, let me compute S_503 and S_504.Compute S_503:Using S_n = -2n^2 + 2014n.So, S_503 = -2*(503)^2 + 2014*503.Compute 503 squared:503*503: Let's compute 500^2 = 250000, plus 2*500*3 = 3000, plus 3^2=9. So, 250000 + 3000 + 9 = 253009.So, S_503 = -2*253009 + 2014*503.Compute -2*253009 = -506018.Compute 2014*503:Let me compute 2000*503 = 1,006,000.Then, 14*503 = 7,042.So, total is 1,006,000 + 7,042 = 1,013,042.Therefore, S_503 = -506,018 + 1,013,042 = 507,024.Now, compute S_504:S_504 = -2*(504)^2 + 2014*504.Compute 504 squared:504*504: Let's compute 500^2=250,000, plus 2*500*4=4,000, plus 4^2=16. So, 250,000 + 4,000 + 16 = 254,016.So, S_504 = -2*254,016 + 2014*504.Compute -2*254,016 = -508,032.Compute 2014*504:2000*504 = 1,008,000.14*504 = 7,056.So, total is 1,008,000 + 7,056 = 1,015,056.Therefore, S_504 = -508,032 + 1,015,056 = 507,024.So, both S_503 and S_504 equal 507,024.Therefore, both n=503 and n=504 give the same maximum sum.Hence, the correct answer is C: 503 or 504.**Final Answer**boxed{C}

answer：Alright, so I have this problem here: I need to find the area in the plane contained by the graph of |x + 2y| + |2x - y| ≤ 8. Hmm, okay. I remember that absolute value equations can define regions in the plane, often polygons or diamond shapes, depending on the equation. But this one looks a bit more complicated because it's the sum of two absolute value expressions. First, I think I should try to understand what this equation represents. The expression |x + 2y| + |2x - y| ≤ 8 is like a combination of two linear expressions inside absolute values. I recall that when dealing with absolute values, it's often useful to consider different cases based on the signs of the expressions inside the absolute values. So, maybe I can break this down into different regions where the expressions inside the absolute values are positive or negative.Let me write down the expressions inside the absolute values: x + 2y and 2x - y. Depending on whether these are positive or negative, the absolute value signs can be removed by considering the sign. So, I need to find the regions where x + 2y ≥ 0 and 2x - y ≥ 0, and similarly for the other combinations.Wait, but there are four possible cases because each absolute value can be either positive or negative. So, I need to consider all four combinations:1. x + 2y ≥ 0 and 2x - y ≥ 02. x + 2y ≥ 0 and 2x - y < 03. x + 2y < 0 and 2x - y ≥ 04. x + 2y < 0 and 2x - y < 0Each of these cases will give me a different linear inequality, and the region defined by the original absolute value equation will be the union of the regions defined by these four cases. So, I can handle each case separately, find the corresponding inequalities, and then plot them to see the overall shape.Let me start with the first case: x + 2y ≥ 0 and 2x - y ≥ 0.In this case, both absolute values can be removed without changing the sign, so the equation becomes:x + 2y + 2x - y ≤ 8Simplifying that, I get:3x + y ≤ 8Okay, so that's a linear inequality. I can write it as y ≤ -3x + 8. That's a straight line with a slope of -3 and a y-intercept at 8. So, in this region where x + 2y ≥ 0 and 2x - y ≥ 0, the inequality is y ≤ -3x + 8.Now, let's move to the second case: x + 2y ≥ 0 and 2x - y < 0.Here, the first absolute value remains the same, but the second one becomes negative because 2x - y is negative. So, the equation becomes:x + 2y - (2x - y) ≤ 8Simplifying that, I get:x + 2y - 2x + y ≤ 8Which simplifies to:- x + 3y ≤ 8Or, rearranged, 3y ≤ x + 8, so y ≤ (1/3)x + 8/3.Alright, so in this region, the inequality is y ≤ (1/3)x + 8/3.Moving on to the third case: x + 2y < 0 and 2x - y ≥ 0.Here, the first absolute value becomes negative, and the second remains positive. So, the equation becomes:-(x + 2y) + 2x - y ≤ 8Simplifying that:- x - 2y + 2x - y ≤ 8Which simplifies to:x - 3y ≤ 8Or, rearranged, -3y ≤ -x + 8, so y ≥ (1/3)x - 8/3.Okay, so in this region, the inequality is y ≥ (1/3)x - 8/3.Finally, the fourth case: x + 2y < 0 and 2x - y < 0.Both absolute values become negative, so the equation becomes:-(x + 2y) - (2x - y) ≤ 8Simplifying that:- x - 2y - 2x + y ≤ 8Which simplifies to:-3x - y ≤ 8Or, rearranged, -3x - y ≤ 8, which can be written as y ≥ -3x - 8.So, in this region, the inequality is y ≥ -3x - 8.Alright, so now I have four inequalities corresponding to the four cases:1. y ≤ -3x + 8 (Case 1)2. y ≤ (1/3)x + 8/3 (Case 2)3. y ≥ (1/3)x - 8/3 (Case 3)4. y ≥ -3x - 8 (Case 4)Now, I need to find the regions where these inequalities hold, considering the conditions for each case. But this seems a bit complicated. Maybe I can plot these lines and see where they intersect to form the boundaries of the region.Alternatively, I remember that when dealing with absolute values, sometimes it's helpful to consider the equations without the absolute values and then find the intersection points. Maybe I can find the points where the expressions inside the absolute values change sign, which are the lines x + 2y = 0 and 2x - y = 0.So, x + 2y = 0 is a line with slope -1/2, and 2x - y = 0 is a line with slope 2. These lines divide the plane into four regions, each corresponding to one of the cases I considered earlier.To find the vertices of the region defined by |x + 2y| + |2x - y| ≤ 8, I can find the intersection points of the lines corresponding to the four inequalities I derived with each other and with the lines x + 2y = 0 and 2x - y = 0.Wait, but maybe a better approach is to find the intersection points of the four lines I have:1. y = -3x + 82. y = (1/3)x + 8/33. y = (1/3)x - 8/34. y = -3x - 8I can find where these lines intersect each other and also where they intersect the lines x + 2y = 0 and 2x - y = 0.Let me start by finding the intersection points of the four lines.First, find where y = -3x + 8 intersects y = (1/3)x + 8/3.Set -3x + 8 = (1/3)x + 8/3.Multiply both sides by 3 to eliminate fractions:-9x + 24 = x + 8Bring variables to one side:-9x - x = 8 - 24-10x = -16x = (-16)/(-10) = 16/10 = 8/5 = 1.6Now, plug x = 8/5 into y = -3x + 8:y = -3*(8/5) + 8 = -24/5 + 40/5 = 16/5 = 3.2So, the intersection point is (8/5, 16/5).Next, find where y = -3x + 8 intersects y = (1/3)x - 8/3.Set -3x + 8 = (1/3)x - 8/3.Multiply both sides by 3:-9x + 24 = x - 8Bring variables to one side:-9x - x = -8 - 24-10x = -32x = (-32)/(-10) = 32/10 = 16/5 = 3.2Plug x = 16/5 into y = -3x + 8:y = -3*(16/5) + 8 = -48/5 + 40/5 = -8/5 = -1.6So, the intersection point is (16/5, -8/5).Now, find where y = -3x + 8 intersects y = -3x - 8.Wait, these are parallel lines because they have the same slope. So, they never intersect.Similarly, check if y = (1/3)x + 8/3 and y = (1/3)x - 8/3 are parallel. Yes, they have the same slope, so they are parallel and don't intersect.So, the only intersections between these four lines are the two points we found: (8/5, 16/5) and (16/5, -8/5).Now, let's find where these lines intersect the lines x + 2y = 0 and 2x - y = 0.First, find where y = -3x + 8 intersects x + 2y = 0.Substitute y = -3x + 8 into x + 2y = 0:x + 2*(-3x + 8) = 0x - 6x + 16 = 0-5x + 16 = 0-5x = -16x = 16/5 = 3.2Then, y = -3*(16/5) + 8 = -48/5 + 40/5 = -8/5 = -1.6So, the intersection point is (16/5, -8/5), which we already found earlier.Next, find where y = -3x + 8 intersects 2x - y = 0.Substitute y = -3x + 8 into 2x - y = 0:2x - (-3x + 8) = 02x + 3x - 8 = 05x - 8 = 05x = 8x = 8/5 = 1.6Then, y = -3*(8/5) + 8 = -24/5 + 40/5 = 16/5 = 3.2So, the intersection point is (8/5, 16/5), which we also found earlier.Now, let's find where y = (1/3)x + 8/3 intersects x + 2y = 0.Substitute y = (1/3)x + 8/3 into x + 2y = 0:x + 2*((1/3)x + 8/3) = 0x + (2/3)x + 16/3 = 0(5/3)x + 16/3 = 0Multiply both sides by 3:5x + 16 = 05x = -16x = -16/5 = -3.2Then, y = (1/3)*(-16/5) + 8/3 = (-16/15) + 40/15 = 24/15 = 8/5 = 1.6So, the intersection point is (-16/5, 8/5).Next, find where y = (1/3)x + 8/3 intersects 2x - y = 0.Substitute y = (1/3)x + 8/3 into 2x - y = 0:2x - ((1/3)x + 8/3) = 02x - (1/3)x - 8/3 = 0(5/3)x - 8/3 = 0Multiply both sides by 3:5x - 8 = 05x = 8x = 8/5 = 1.6Then, y = (1/3)*(8/5) + 8/3 = 8/15 + 40/15 = 48/15 = 16/5 = 3.2So, the intersection point is (8/5, 16/5), which we already have.Now, let's find where y = (1/3)x - 8/3 intersects x + 2y = 0.Substitute y = (1/3)x - 8/3 into x + 2y = 0:x + 2*((1/3)x - 8/3) = 0x + (2/3)x - 16/3 = 0(5/3)x - 16/3 = 0Multiply both sides by 3:5x - 16 = 05x = 16x = 16/5 = 3.2Then, y = (1/3)*(16/5) - 8/3 = 16/15 - 40/15 = -24/15 = -8/5 = -1.6So, the intersection point is (16/5, -8/5), which we already found.Next, find where y = (1/3)x - 8/3 intersects 2x - y = 0.Substitute y = (1/3)x - 8/3 into 2x - y = 0:2x - ((1/3)x - 8/3) = 02x - (1/3)x + 8/3 = 0(5/3)x + 8/3 = 0Multiply both sides by 3:5x + 8 = 05x = -8x = -8/5 = -1.6Then, y = (1/3)*(-8/5) - 8/3 = (-8/15) - 40/15 = -48/15 = -16/5 = -3.2So, the intersection point is (-8/5, -16/5).Now, let's find where y = -3x - 8 intersects x + 2y = 0.Substitute y = -3x - 8 into x + 2y = 0:x + 2*(-3x - 8) = 0x - 6x - 16 = 0-5x - 16 = 0-5x = 16x = -16/5 = -3.2Then, y = -3*(-16/5) - 8 = 48/5 - 40/5 = 8/5 = 1.6So, the intersection point is (-16/5, 8/5), which we already have.Next, find where y = -3x - 8 intersects 2x - y = 0.Substitute y = -3x - 8 into 2x - y = 0:2x - (-3x - 8) = 02x + 3x + 8 = 05x + 8 = 05x = -8x = -8/5 = -1.6Then, y = -3*(-8/5) - 8 = 24/5 - 40/5 = -16/5 = -3.2So, the intersection point is (-8/5, -16/5), which we already found.Alright, so now I have all the intersection points:1. (8/5, 16/5)2. (16/5, -8/5)3. (-16/5, 8/5)4. (-8/5, -16/5)These four points seem to form the vertices of the region defined by the original inequality. Let me plot these points mentally:- (8/5, 16/5) is in the first quadrant.- (16/5, -8/5) is in the fourth quadrant.- (-16/5, 8/5) is in the second quadrant.- (-8/5, -16/5) is in the third quadrant.So, connecting these points should form a quadrilateral, probably a parallelogram or a diamond shape.To find the area of this quadrilateral, I can use the shoelace formula, which is a method to calculate the area of a polygon when the coordinates of the vertices are known.The shoelace formula is given by:Area = 1/2 |sum from i=1 to n of (x_i y_{i+1} - x_{i+1} y_i)|where (x_{n+1}, y_{n+1}) is (x_1, y_1).So, let's list the coordinates in order. I'll list them in a clockwise order starting from (8/5, 16/5):1. (8/5, 16/5)2. (16/5, -8/5)3. (-8/5, -16/5)4. (-16/5, 8/5)5. Back to (8/5, 16/5)Wait, actually, I think I might have made a mistake in listing the points. Let me double-check the order. The points are:- (8/5, 16/5) in the first quadrant.- (16/5, -8/5) in the fourth quadrant.- (-8/5, -16/5) in the third quadrant.- (-16/5, 8/5) in the second quadrant.So, connecting them in this order should form a quadrilateral. Alternatively, I can list them in a different order as long as they are consecutive vertices.Let me list them as:1. (8/5, 16/5)2. (16/5, -8/5)3. (-8/5, -16/5)4. (-16/5, 8/5)5. Back to (8/5, 16/5)Now, applying the shoelace formula:First, write down the coordinates:x1 = 8/5, y1 = 16/5x2 = 16/5, y2 = -8/5x3 = -8/5, y3 = -16/5x4 = -16/5, y4 = 8/5x5 = x1 = 8/5, y5 = y1 = 16/5Now, compute the sum:S = (x1 y2 - x2 y1) + (x2 y3 - x3 y2) + (x3 y4 - x4 y3) + (x4 y5 - x5 y4)Compute each term:First term: x1 y2 - x2 y1 = (8/5)*(-8/5) - (16/5)*(16/5) = (-64/25) - (256/25) = (-64 - 256)/25 = (-320)/25Second term: x2 y3 - x3 y2 = (16/5)*(-16/5) - (-8/5)*(-8/5) = (-256/25) - (64/25) = (-256 - 64)/25 = (-320)/25Third term: x3 y4 - x4 y3 = (-8/5)*(8/5) - (-16/5)*(-16/5) = (-64/25) - (256/25) = (-64 - 256)/25 = (-320)/25Fourth term: x4 y5 - x5 y4 = (-16/5)*(16/5) - (8/5)*(8/5) = (-256/25) - (64/25) = (-256 - 64)/25 = (-320)/25So, sum S = (-320/25) + (-320/25) + (-320/25) + (-320/25) = (-1280)/25Now, take the absolute value and multiply by 1/2:Area = (1/2)*|S| = (1/2)*(1280/25) = 640/25 = 25.6Wait, 640 divided by 25 is 25.6. But I think it's better to keep it as a fraction. 640/25 simplifies to 128/5, because 640 divided by 5 is 128, and 25 divided by 5 is 5. So, 128/5 is the exact value.But let me double-check my calculations because I might have made a mistake in the shoelace formula.Wait, when I computed each term, I got each term as -320/25, and there are four terms, so total sum is -1280/25. Taking absolute value gives 1280/25, and half of that is 640/25, which is indeed 128/5.But let me verify the order of the points because the shoelace formula depends on the order. I listed the points as (8/5,16/5), (16/5,-8/5), (-8/5,-16/5), (-16/5,8/5). Is this a correct cyclic order?Alternatively, maybe I should have ordered them differently. Let me try another order to see if I get the same result.Let me list them as:1. (8/5,16/5)2. (16/5,-8/5)3. (-16/5,8/5)4. (-8/5,-16/5)5. Back to (8/5,16/5)Wait, that might not be correct. Let me think about the actual positions.Alternatively, perhaps I should list them in a different sequence to ensure that the polygon is traversed correctly without crossing over itself.Let me try listing them in the order:1. (8/5,16/5)2. (16/5,-8/5)3. (-8/5,-16/5)4. (-16/5,8/5)5. Back to (8/5,16/5)This should form a convex quadrilateral without crossing.But when I calculated the terms, each term was -320/25, which seems too uniform. Maybe I made a mistake in the calculation.Let me recalculate each term step by step.First term: x1 y2 - x2 y1x1 = 8/5, y2 = -8/5x2 = 16/5, y1 = 16/5So, (8/5)*(-8/5) - (16/5)*(16/5) = (-64/25) - (256/25) = (-320)/25Second term: x2 y3 - x3 y2x2 = 16/5, y3 = -16/5x3 = -8/5, y2 = -8/5So, (16/5)*(-16/5) - (-8/5)*(-8/5) = (-256/25) - (64/25) = (-320)/25Third term: x3 y4 - x4 y3x3 = -8/5, y4 = 8/5x4 = -16/5, y3 = -16/5So, (-8/5)*(8/5) - (-16/5)*(-16/5) = (-64/25) - (256/25) = (-320)/25Fourth term: x4 y5 - x5 y4x4 = -16/5, y5 = 16/5x5 = 8/5, y4 = 8/5So, (-16/5)*(16/5) - (8/5)*(8/5) = (-256/25) - (64/25) = (-320)/25So, indeed, each term is -320/25, and sum is -1280/25. Absolute value is 1280/25, half of that is 640/25 = 128/5.So, the area is 128/5.But wait, 128/5 is 25.6, which seems reasonable, but I want to make sure that this is correct.Alternatively, maybe I can use vectors or another method to calculate the area.Another approach is to recognize that the region defined by |x + 2y| + |2x - y| ≤ 8 is a convex polygon, and the area can be found by calculating the area of the parallelogram formed by the intersection points.But since I already have the vertices, the shoelace formula should suffice.Wait, but let me check if the points are correct. I found four points:(8/5,16/5), (16/5,-8/5), (-8/5,-16/5), (-16/5,8/5)Plotting these points, they seem to form a diamond shape centered at the origin, symmetric across both axes.Alternatively, maybe I can use the fact that the region is symmetric and calculate the area in one quadrant and multiply by 4.Let me try that.In the first quadrant, the region is bounded by y ≤ -3x + 8 and y ≤ (1/3)x + 8/3.The intersection of these two lines is at (8/5,16/5), which is in the first quadrant.So, in the first quadrant, the region is a triangle with vertices at (0,0), (8/5,16/5), and (0,8).Wait, is that correct?Wait, when x=0, y ≤ 8 from the first inequality, and y ≤ 8/3 from the second inequality. Wait, no, when x=0, from y ≤ -3x +8, y ≤8, and from y ≤ (1/3)x +8/3, y ≤8/3.So, the upper bound in the first quadrant is y ≤8/3, because 8/3 is less than 8.Wait, but that contradicts the intersection point at (8/5,16/5), which is higher than 8/3.Wait, maybe I'm confusing something.Wait, in the first quadrant, both x and y are positive.From case 1: y ≤ -3x +8From case 2: y ≤ (1/3)x +8/3So, the region in the first quadrant is bounded by both inequalities.The intersection point is at (8/5,16/5), which is approximately (1.6,3.2).So, in the first quadrant, the region is a polygon with vertices at (0,0), (8/5,16/5), and (0,8/3).Wait, no, because when x=0, y ≤8/3 from case 2, and y ≤8 from case 1. So, the upper bound is y=8/3.But the intersection point is at (8/5,16/5), which is above y=8/3.Wait, that can't be. Let me calculate y=8/3, which is approximately 2.666, and 16/5 is 3.2, which is higher.So, actually, in the first quadrant, the region is bounded by y ≤ -3x +8 and y ≤ (1/3)x +8/3.But since (1/3)x +8/3 is less than -3x +8 for x >0, the upper bound is y ≤ (1/3)x +8/3.Wait, let me check at x=0: y ≤8/3 and y ≤8. So, the upper bound is y=8/3.At x=8/5, y=16/5=3.2, which is greater than 8/3≈2.666. So, that suggests that the intersection point is outside the first quadrant region defined by both inequalities.Wait, that can't be. Maybe I made a mistake in determining the regions.Wait, in case 1, we have x +2y ≥0 and 2x - y ≥0.In the first quadrant, x and y are positive, so x +2y ≥0 is always true, and 2x - y ≥0 implies y ≤2x.So, in the first quadrant, the region defined by case 1 is y ≤ -3x +8 and y ≤2x.Similarly, case 2 is x +2y ≥0 and 2x - y <0, which in the first quadrant would be y >2x, and y ≤ (1/3)x +8/3.Wait, so in the first quadrant, the region is divided into two parts by y=2x.For y ≤2x, the inequality is y ≤ -3x +8.For y >2x, the inequality is y ≤ (1/3)x +8/3.So, the intersection point of y=2x and y=-3x +8 is at:2x = -3x +85x=8x=8/5=1.6y=2*(8/5)=16/5=3.2So, the intersection point is (8/5,16/5), which is the same as before.Similarly, the intersection of y=2x and y=(1/3)x +8/3 is:2x = (1/3)x +8/3Multiply both sides by 3:6x = x +85x=8x=8/5=1.6y=2*(8/5)=16/5=3.2So, the same point.Therefore, in the first quadrant, the region is bounded by:- From (0,0) to (8/5,16/5) along y=2x.- From (8/5,16/5) to (0,8/3) along y=(1/3)x +8/3.Wait, but when x=0, y=8/3 from case 2, but from case 1, y=8.So, actually, the region in the first quadrant is a polygon with vertices at (0,8/3), (8/5,16/5), and (0,8).Wait, that doesn't seem right because (0,8) is not on y=(1/3)x +8/3.Wait, when x=0, y=8/3 from case 2, and y=8 from case 1.But since case 1 applies when y ≤2x, and at x=0, y=0, which is less than 2x=0, so case 1 applies at (0,0). But case 2 applies when y >2x, which at x=0, y>0, so case 2 applies for y>0.Wait, this is getting confusing. Maybe I should approach it differently.Alternatively, since the entire region is symmetric across both axes, I can calculate the area in the first quadrant and multiply by 4.In the first quadrant, the region is bounded by y ≤ -3x +8 and y ≤ (1/3)x +8/3.The intersection of these two lines is at (8/5,16/5).So, the region in the first quadrant is a polygon with vertices at (0,0), (8/5,16/5), and (0,8/3).Wait, but when x=0, y=8/3 from case 2, and y=8 from case 1. So, actually, the region in the first quadrant is bounded by y ≤ min(-3x +8, (1/3)x +8/3).So, the upper boundary is the lower of the two lines.At x=0, y=8/3 is less than y=8, so the upper bound is y=8/3.At x=8/5, both lines meet at y=16/5=3.2, which is greater than 8/3≈2.666.Wait, that suggests that the region in the first quadrant is bounded by y ≤ (1/3)x +8/3 from x=0 to x=8/5, and then by y ≤ -3x +8 from x=8/5 to x=8/3.Wait, no, because when x increases beyond 8/5, the line y=(1/3)x +8/3 would be above y=-3x +8.Wait, let me find where (1/3)x +8/3 = -3x +8.Set (1/3)x +8/3 = -3x +8Multiply both sides by 3:x +8 = -9x +2410x =16x=16/10=8/5=1.6Which is the same intersection point.So, in the first quadrant, the region is bounded by:- From (0,8/3) to (8/5,16/5) along y=(1/3)x +8/3.- From (8/5,16/5) to (8/3,0) along y=-3x +8.Wait, but when x=8/3≈2.666, y=-3*(8/3)+8= -8 +8=0.So, the region in the first quadrant is a polygon with vertices at (0,8/3), (8/5,16/5), and (8/3,0).So, to find the area in the first quadrant, I can calculate the area of this triangle.The coordinates are:A: (0,8/3)B: (8/5,16/5)C: (8/3,0)Using the shoelace formula for these three points:List the points in order:A: (0,8/3)B: (8/5,16/5)C: (8/3,0)Back to A: (0,8/3)Compute the sum:S = (x_A y_B - x_B y_A) + (x_B y_C - x_C y_B) + (x_C y_A - x_A y_C)Compute each term:First term: x_A y_B - x_B y_A = 0*(16/5) - (8/5)*(8/3) = 0 - 64/15 = -64/15Second term: x_B y_C - x_C y_B = (8/5)*0 - (8/3)*(16/5) = 0 - 128/15 = -128/15Third term: x_C y_A - x_A y_C = (8/3)*(8/3) - 0*0 = 64/9 - 0 = 64/9So, sum S = (-64/15) + (-128/15) + (64/9)Convert to common denominator, which is 45:-64/15 = -192/45-128/15 = -384/4564/9 = 320/45So, S = (-192 - 384 + 320)/45 = (-256)/45Take absolute value and multiply by 1/2:Area = (1/2)*|S| = (1/2)*(256/45) = 128/45So, the area in the first quadrant is 128/45.Since the region is symmetric across all four quadrants, the total area is 4*(128/45) = 512/45.Wait, but earlier I got 128/5 using the shoelace formula on all four points. There's a discrepancy here.Wait, 128/5 is 25.6, and 512/45 is approximately 11.377, which is much smaller. So, I must have made a mistake somewhere.Wait, maybe I made a mistake in identifying the vertices. Let me go back.Earlier, I found four intersection points:(8/5,16/5), (16/5,-8/5), (-8/5,-16/5), (-16/5,8/5)But when I applied the shoelace formula, I got 128/5. However, when I tried to calculate the area in the first quadrant and multiply by 4, I got 512/45, which is different.This suggests that my initial assumption about the vertices might be incorrect.Wait, perhaps the region is not just a quadrilateral but has more vertices. Maybe I missed some intersection points.Alternatively, perhaps the region is a convex polygon with eight sides, considering the symmetry across both axes and the four cases.Wait, let me think again.The original inequality is |x + 2y| + |2x - y| ≤8.This can be rewritten as |x + 2y| + |2x - y| =8, which is the boundary.To find the vertices, I need to find the points where the expressions inside the absolute values change sign, i.e., where x + 2y=0 and 2x - y=0.These lines divide the plane into four regions, and in each region, the absolute values can be removed with appropriate signs.So, in each region, the equation becomes a linear equation, and the intersection of these lines with the original equation will give the vertices.Wait, perhaps I should find all the intersection points between the lines x + 2y=0, 2x - y=0, and the lines obtained from the cases.Wait, I think I need to find all the intersection points of the lines defining the boundaries of the region.So, the boundaries are:1. y = -3x +8 (from case 1)2. y = (1/3)x +8/3 (from case 2)3. y = (1/3)x -8/3 (from case 3)4. y = -3x -8 (from case 4)Additionally, the lines where the expressions inside the absolute values change sign:5. x + 2y =06. 2x - y =0So, in total, I have six lines. The region defined by the inequality is the intersection of all the regions defined by these lines.To find the vertices, I need to find all the intersection points between these lines, but only those that satisfy the original inequality.So, let's find all possible intersection points between these six lines.First, find where y = -3x +8 intersects x + 2y=0.We already did this earlier and found (16/5, -8/5).Next, where y = -3x +8 intersects 2x - y=0.We found (8/5,16/5).Similarly, where y = (1/3)x +8/3 intersects x + 2y=0.We found (-16/5,8/5).Where y = (1/3)x +8/3 intersects 2x - y=0.We found (8/5,16/5).Where y = (1/3)x -8/3 intersects x + 2y=0.We found (16/5,-8/5).Where y = (1/3)x -8/3 intersects 2x - y=0.We found (-8/5,-16/5).Where y = -3x -8 intersects x + 2y=0.We found (-16/5,8/5).Where y = -3x -8 intersects 2x - y=0.We found (-8/5,-16/5).Additionally, we need to check where the lines y = -3x +8 and y = (1/3)x +8/3 intersect, which is at (8/5,16/5).Similarly, where y = -3x +8 and y = (1/3)x -8/3 intersect, which is at (16/5,-8/5).Where y = (1/3)x +8/3 and y = -3x -8 intersect.Set (1/3)x +8/3 = -3x -8Multiply both sides by 3:x +8 = -9x -2410x = -32x = -32/10 = -16/5 = -3.2Then, y = (1/3)*(-16/5) +8/3 = (-16/15) +40/15 =24/15=8/5=1.6So, intersection point is (-16/5,8/5), which we already have.Similarly, where y = (1/3)x -8/3 and y = -3x -8 intersect.Set (1/3)x -8/3 = -3x -8Multiply both sides by 3:x -8 = -9x -2410x = -16x = -16/10 = -8/5 = -1.6Then, y = (1/3)*(-8/5) -8/3 = (-8/15) -40/15 = -48/15 = -16/5 = -3.2So, intersection point is (-8/5,-16/5), which we already have.So, all intersection points are:(8/5,16/5), (16/5,-8/5), (-16/5,8/5), (-8/5,-16/5)These are the four vertices of the region.Therefore, the region is a quadrilateral with these four vertices.Now, applying the shoelace formula correctly.List the points in order, either clockwise or counterclockwise.Let's list them as:1. (8/5,16/5)2. (16/5,-8/5)3. (-8/5,-16/5)4. (-16/5,8/5)5. Back to (8/5,16/5)Now, compute the shoelace sum:S = (x1 y2 - x2 y1) + (x2 y3 - x3 y2) + (x3 y4 - x4 y3) + (x4 y5 - x5 y4)Compute each term:First term: x1 y2 - x2 y1 = (8/5)*(-8/5) - (16/5)*(16/5) = (-64/25) - (256/25) = (-320)/25Second term: x2 y3 - x3 y2 = (16/5)*(-16/5) - (-8/5)*(-8/5) = (-256/25) - (64/25) = (-320)/25Third term: x3 y4 - x4 y3 = (-8/5)*(8/5) - (-16/5)*(-16/5) = (-64/25) - (256/25) = (-320)/25Fourth term: x4 y5 - x5 y4 = (-16/5)*(16/5) - (8/5)*(8/5) = (-256/25) - (64/25) = (-320)/25So, sum S = (-320/25) + (-320/25) + (-320/25) + (-320/25) = (-1280)/25Take absolute value and multiply by 1/2:Area = (1/2)*|S| = (1/2)*(1280/25) = 640/25 = 128/5So, the area is 128/5.But earlier, when I tried to calculate the area in the first quadrant and multiply by 4, I got 512/45, which is different. So, I must have made a mistake in that approach.Wait, perhaps the region in the first quadrant is not a triangle but a quadrilateral, and I missed a vertex.Wait, let me think again.In the first quadrant, the region is bounded by y ≤ -3x +8 and y ≤ (1/3)x +8/3.But also, since x +2y ≥0 and 2x - y ≥0 in case 1, which in the first quadrant is always true because x and y are positive.Wait, no, in case 1, 2x - y ≥0 implies y ≤2x.So, in the first quadrant, the region is divided into two parts by y=2x.For y ≤2x, the inequality is y ≤ -3x +8.For y >2x, the inequality is y ≤ (1/3)x +8/3.So, the region in the first quadrant is bounded by:- From (0,0) to (8/5,16/5) along y=2x.- From (8/5,16/5) to (0,8/3) along y=(1/3)x +8/3.Wait, but when x=0, y=8/3 from case 2, and y=8 from case 1.But in reality, the region in the first quadrant is bounded by y ≤ min(-3x +8, (1/3)x +8/3, 2x).Wait, that's getting complicated.Alternatively, perhaps the region in the first quadrant is a polygon with vertices at (0,0), (8/5,16/5), (0,8/3).But when I calculated the area using shoelace for these three points, I got 128/45, which is approximately 2.844, and multiplying by 4 gives 512/45≈11.377, which is much less than 128/5=25.6.So, clearly, my initial approach was incorrect.Wait, perhaps the region is not just in the first quadrant but spans across all quadrants, and the four points I found are the only vertices.Therefore, the shoelace formula with these four points gives the correct area of 128/5.But to confirm, let me calculate the area using vectors.The four points are:A: (8/5,16/5)B: (16/5,-8/5)C: (-8/5,-16/5)D: (-16/5,8/5)These points form a parallelogram because the midpoints between A and C, and between B and D, are the same.Midpoint between A and C: ((8/5 + (-8/5))/2, (16/5 + (-16/5))/2) = (0,0)Midpoint between B and D: ((16/5 + (-16/5))/2, (-8/5 +8/5)/2) = (0,0)So, the figure is a parallelogram centered at the origin.The area of a parallelogram can be calculated as the magnitude of the cross product of two adjacent sides.Let's take vectors AB and AD.Vector AB = B - A = (16/5 -8/5, -8/5 -16/5) = (8/5, -24/5)Vector AD = D - A = (-16/5 -8/5,8/5 -16/5) = (-24/5, -8/5)The cross product AB × AD = (8/5)(-8/5) - (-24/5)(-24/5) = (-64/25) - (576/25) = (-640)/25The magnitude is | -640/25 | = 640/25So, the area is 640/25 = 128/5, which matches the shoelace formula result.Therefore, the area is indeed 128/5.So, after carefully going through the process, I conclude that the area is 128/5.

answer：First, identify the number of oranges each child has, which is 3 oranges per child.Next, determine the total number of children, which is 4 children.To find the total number of oranges, multiply the number of oranges each child has by the number of children: 3 oranges/child × 4 children.This calculation results in 12 oranges in total.

answer：To find a positive integer ( n ) that is divisible by 24 and whose cube root is between 9 and 9.1, we start by setting up the inequality based on the cube root condition:[9 < sqrt[3]{n} < 9.1]Cubing each part of the inequality gives:[9^3 < n < 9.1^3][729 < n < 753.571]Next, since ( n ) must be divisible by 24, we express ( n ) as ( n = 24k ) where ( k ) is an integer. Substituting this into the inequality:[729 < 24k < 753.571]Dividing the entire inequality by 24 to solve for ( k ):[frac{729}{24} < k < frac{753.571}{24}][30.375 < k < 31.399625]The only integer ( k ) within this range is 31. Therefore, substituting back to find ( n ):[n = 24 times 31 = 744]Finally, verifying the cube root of 744:[sqrt[3]{744} approx 9.02]This value is indeed between 9 and 9.1, confirming that 744 satisfies both conditions.

answer：Okay, so I have this geometry problem here about two intersecting circles, a common tangent, and some points. Let me try to visualize it first. There are two circles, Ω₁ and Ω₂, intersecting at points M and N. So, M and N are the points of intersection. There's a common tangent that touches Ω₁ at P and Ω₂ at Q. It's mentioned that P and Q are closer to N than to M, which probably means that if I draw the tangent, it's near the N side of the circles.Then, the line PN intersects Ω₂ again at R. So starting from P, drawing a line through N, and it hits Ω₂ again at R. The goal is to show that MQ is the angle bisector of angle PMR. So, MQ should split angle PMR into two equal angles.Alright, let me recall some geometry concepts that might help here. Since we're dealing with circles, tangents, and angle bisectors, properties related to circles, tangents, and angle bisectors will be useful. Maybe the power of a point, properties of cyclic quadrilaterals, or something related to similar triangles.First, since PQ is a common tangent to both circles, the lengths from P to the point of tangency on Ω₁ and from Q to the point of tangency on Ω₂ should be equal. Wait, actually, since PQ is a common external tangent, the lengths from P and Q to the points of tangency are equal. But in this case, P and Q are the points of tangency, so maybe that's not directly helpful.Let me think about the tangent-secant theorem. It states that the angle between a tangent and a chord is equal to the angle in the alternate segment. So, for point P on Ω₁, the tangent at P makes an angle with chord PM equal to the angle that PM makes with the other chord in the alternate segment. Similarly, for point Q on Ω₂, the tangent at Q makes an angle with chord QM equal to the angle in the alternate segment.Hmm, maybe I can use that to find some angle equalities. Let me try to mark some angles.Let me denote angle PMN as the angle at M between PM and MN. Since PQ is a tangent to Ω₁ at P, by the tangent-secant theorem, angle between tangent PQ and chord PM is equal to the angle that PM makes with the other chord in the alternate segment. So, angle QPM is equal to angle PNM. Wait, is that correct?Wait, maybe I should be more precise. The tangent at P makes an angle with chord PM equal to the angle that PM makes with the chord in the alternate segment. So, angle between tangent PQ and chord PM is equal to angle PNM. So, angle QPM = angle PNM.Similarly, for the tangent at Q, angle between tangent PQ and chord QM is equal to angle QNM. So, angle PQM = angle QNM.Hmm, okay, so angle QPM = angle PNM and angle PQM = angle QNM.Now, let me think about triangle PMN and triangle QMN. Maybe there's some similarity or congruence here.Wait, but I need to relate this to angle PMR. So, R is a point on Ω₂, so maybe I can use cyclic quadrilateral properties.Since R is on Ω₂, and M and N are also on Ω₂, then quadrilateral MNRQ is cyclic. So, angle MRQ is equal to angle MNQ because they subtend the same arc MQ.Wait, angle MRQ = angle MNQ. But angle MNQ is equal to angle PQM, which we found earlier is equal to angle QNM.Wait, maybe I'm getting somewhere. Let me write down the equal angles step by step.1. By tangent-secant theorem at P: angle QPM = angle PNM.2. By tangent-secant theorem at Q: angle PQM = angle QNM.3. Since R is on Ω₂, angle MRQ = angle MNQ (cyclic quadrilateral).But angle MNQ is equal to angle PQM, which is equal to angle QNM. So, angle MRQ = angle QNM.Wait, but angle QNM is equal to angle PQM, which is equal to angle MRQ. Hmm, maybe I can relate this to angle PMR.Let me think about angle PMR. It's the angle at M between PM and MR. If I can show that MQ bisects this angle, then I need to show that angle PMQ = angle QMR.So, angle PMQ is the angle between PM and MQ, and angle QMR is the angle between MQ and MR. If I can show these two angles are equal, then MQ is the bisector.From earlier, angle QPM = angle PNM, and angle PQM = angle QNM.Also, since PN intersects Ω₂ at R, maybe I can use power of a point or something related to intersecting chords.Wait, power of point P with respect to Ω₂. The power of P with respect to Ω₂ is equal to PQ², since PQ is tangent to Ω₂. Also, power of P with respect to Ω₂ is equal to PR * PN, since P lies on the secant PRN.So, PQ² = PR * PN.Similarly, power of P with respect to Ω₁ is equal to PP (which is zero since P is on Ω₁), but that might not help directly.Hmm, maybe not the power of a point, but perhaps similar triangles.Let me consider triangles PMQ and something else. Wait, maybe triangle PMQ and triangle something.Alternatively, maybe using the angle bisector theorem. If I can show that MQ divides angle PMR into two equal parts, then perhaps using the angle bisector theorem.But to use the angle bisector theorem, I need to relate the sides. Maybe I can find some ratio of sides that are equal.Wait, but I don't have much information about the sides, only about angles.Wait, going back to the angles. Let me try to express angle PMR in terms of other angles.Angle PMR is equal to angle PMN + angle NMR.From earlier, angle PMN is equal to angle QPM, which is equal to angle PNM.And angle NMR is equal to angle MRQ, which is equal to angle MNQ, which is equal to angle PQM.So, angle PMR = angle PNM + angle PQM.But angle PNM and angle PQM are both angles at N and Q respectively.Wait, but angle PNM is equal to angle QPM, and angle PQM is equal to angle QNM.Hmm, maybe I can relate these angles to angles involving MQ.Wait, angle PMQ is the angle between PM and MQ. Let me see if I can express that.Angle PMQ is equal to angle PMN + angle NMQ.From earlier, angle PMN = angle QPM, and angle NMQ is equal to angle QNM.So, angle PMQ = angle QPM + angle QNM.Similarly, angle QMR is equal to angle QMN + angle NMR.Wait, angle QMR is the angle between MQ and MR, which is equal to angle QMN + angle NMR.But angle QMN is equal to angle QNM, and angle NMR is equal to angle MRQ, which is equal to angle MNQ, which is equal to angle PQM.So, angle QMR = angle QNM + angle PQM.Wait, so angle PMQ = angle QPM + angle QNM, and angle QMR = angle QNM + angle PQM.But angle QPM is equal to angle PNM, and angle PQM is equal to angle QNM.So, angle PMQ = angle PNM + angle QNM, and angle QMR = angle QNM + angle PNM.Therefore, angle PMQ = angle QMR.Hence, MQ bisects angle PMR.Wait, that seems to be the conclusion. So, by showing that angle PMQ equals angle QMR, we've shown that MQ is the angle bisector of angle PMR.I think that's the gist of it. Maybe I should write it more formally, but the key idea is using the tangent-secant theorem to find equal angles and then using cyclic quadrilateral properties to relate those angles to the ones in the angle bisector.