💡Okay, so I have this problem where Petya drew a polygon on a square grid, and the area is 100 cells. He checked that it can be divided into two equal polygons and also into twenty-five equal polygons. The question is asking if it's necessarily possible to divide it into fifty equal polygons. Hmm, interesting.First, let me try to visualize this. The polygon is on a square grid, so all the sides are along the grid lines, and the area is 100 cells. That means it's a pretty sizable shape. Now, it can be divided into two equal polygons. So, each of those would have an area of 50 cells. Similarly, dividing it into twenty-five equal polygons would mean each has an area of 4 cells. Wait, each of the twenty-five polygons has an area of 4 cells. That makes me think of tetrominoes, which are shapes made up of four squares connected edge-to-edge. There are five types of tetrominoes: I, O, T, S, and L. Each of these covers four squares, so if the original polygon can be divided into twenty-five equal polygons, each must be one of these tetrominoes.Now, the question is about dividing it into fifty equal polygons. If each of the twenty-five polygons is a tetromino, then each tetromino could potentially be split into two dominoes, which are shapes made up of two squares. So, if each tetromino can be split into two dominoes, then the original polygon can be divided into fifty dominoes. But wait, not all tetrominoes can be split into two dominoes. For example, the T-tetromino has a central square with three squares attached to one side. If I try to split that into two dominoes, it might not work because the dominoes have to be connected edge-to-edge. Let me think about that. If I have a T-tetromino, can I split it into two dominoes? Let me visualize it. The T-shape has three squares in a row and one square attached to the center of that row, forming a T. If I try to split it into two dominoes, each domino would need to cover two squares. But because of the T-shape, it's not obvious how to split it into two dominoes without overlapping or leaving gaps. Hmm, maybe the T-tetromino can't be split into two dominoes. That would mean that if the original polygon is made up of T-tetrominoes, then it can't be split into fifty dominoes. But the problem says that the polygon can be split into two equal polygons and into twenty-five equal polygons. So, does that mean that each of the twenty-five polygons is a T-tetromino? Or could they be other types of tetrominoes?Wait, the problem doesn't specify what kind of tetrominoes are used, just that it can be divided into twenty-five equal polygons. So, it could be a mix of different tetrominoes. But if even one of them is a T-tetromino, then splitting it into two dominoes might not be possible. So, does that mean that the original polygon can't necessarily be split into fifty equal polygons?But the problem is asking if it's necessarily possible. So, if it's possible in some cases, but not necessarily in all cases, then the answer would be no. But I need to think more carefully.Let me consider the coloring argument. If I color the grid like a chessboard, alternating black and white squares, then each domino will always cover one black and one white square. Similarly, each tetromino will cover either two black and two white squares or some other combination. If the original polygon has an equal number of black and white squares, then it's possible to tile it with dominoes. But if it doesn't, then it's not possible. So, let me check the coloring of the original polygon. Since it's a polygon on a square grid with area 100, which is even, it's possible that the number of black and white squares is equal, but it's not necessarily the case.Wait, actually, for any polygon on a square grid, the difference between the number of black and white squares is related to the area. If the area is even, like 100, then the difference between black and white squares must be even as well. So, if the polygon can be tiled with dominoes, then the number of black and white squares must be equal. But if the difference is non-zero, then it can't be tiled with dominoes.But the problem doesn't specify anything about the coloring, so I can't assume it's balanced. However, the fact that it can be divided into twenty-five equal polygons, each of area 4, which are tetrominoes, might impose some constraints on the coloring.Each tetromino covers four squares. Let's see how they affect the coloring. The I-tetromino is four squares in a straight line, so it covers two black and two white squares. The O-tetromino is a 2x2 square, which also covers two black and two white squares. The T-tetromino, on the other hand, covers three squares in a row and one attached to the center. Depending on where it's placed, it might cover three of one color and one of the other, or two and two. Wait, actually, no. Let me think.If I place a T-tetromino on the grid, it will cover three squares in a row and one square attached to the center. If the row is horizontal, then the three squares in the row will alternate colors, so two of one color and one of the other. The attached square will be the same color as the middle square of the row. So, in total, the T-tetromino will cover three squares of one color and one of the other. So, it has a difference of two in coloring.Similarly, the S and Z tetrominoes, which are the two mirrored L-shapes, each cover two squares of one color and two of the other. Wait, no. Let me think again. The S-tetromino is two squares, then two squares offset by one. So, it covers two of one color and two of the other. Similarly for the Z-tetromino.Wait, actually, no. Let me visualize it. The S-tetromino is like a zig-zag. If I place it on the grid, it will cover two squares of one color and two of the other. Similarly, the Z-tetromino does the same. The L-tetromino, which is three squares in a row and one square attached to the end, will cover three of one color and one of the other, similar to the T-tetromino.So, in summary, the I, O, S, and Z tetrominoes each cover two black and two white squares, while the T and L tetrominoes cover three of one color and one of the other.Now, if the original polygon is divided into twenty-five tetrominoes, each of which is either I, O, S, Z, T, or L, then the total difference in coloring will be the sum of the differences from each tetromino.If all the tetrominoes are I, O, S, or Z, then the total difference would be zero, meaning the original polygon has an equal number of black and white squares. If some of them are T or L tetrominoes, each contributing a difference of two (either +2 or -2), then the total difference would be 2 times the number of T or L tetrominoes.But the original polygon has an area of 100, which is even, so the difference between black and white squares must be even. Therefore, the number of T or L tetrominoes must be even, because each contributes a difference of two, and an even number of them would result in an even total difference.Now, if we want to divide the polygon into fifty dominoes, each domino covers one black and one white square. Therefore, the total number of black and white squares must be equal. So, the difference must be zero. But if the original polygon has a non-zero difference, then it can't be tiled with dominoes. However, from the above, the difference is 2 times the number of T or L tetrominoes. Since the number of T or L tetrominoes must be even, the total difference is a multiple of four. Wait, no. If each T or L contributes two, and the number of them is even, say 2k, then the total difference is 4k, which is a multiple of four.But the original polygon has an area of 100, so the number of black squares plus white squares is 100. If the difference is 4k, then the number of black squares is (100 + 4k)/2 and white squares is (100 - 4k)/2. Both must be integers, which they are since 100 is even and 4k is even.But for the polygon to be tileable by dominoes, the difference must be zero. So, unless k is zero, meaning there are no T or L tetrominoes, the difference won't be zero. Therefore, if the original polygon is divided into twenty-five tetrominoes, and if any of them are T or L, then the total difference is non-zero, and thus it can't be tiled with dominoes.But the problem states that the polygon can be divided into two equal polygons. Each of those would have an area of 50. If the original polygon has a non-zero difference, then each of the two equal polygons would also have a non-zero difference. But each of those would need to have an equal number of black and white squares to be tileable by dominoes, which they aren't if the original polygon has a non-zero difference.Wait, but the problem doesn't say that the two equal polygons are tileable by dominoes, just that they are equal polygons. So, maybe they don't have to be tileable by dominoes, just that they are congruent or something. Hmm, the problem says "divided along the cell boundaries into both two equal polygons and twenty-five equal polygons." So, equal in area, but not necessarily in shape.But if the original polygon can be divided into two equal polygons, each of area 50, and also into twenty-five equal polygons, each of area 4, then perhaps the structure is such that it's compatible with both divisions.Wait, but if the original polygon has a non-zero difference in coloring, then dividing it into two equal polygons would mean each of those has half the difference. So, if the original difference is 4k, then each of the two polygons would have a difference of 2k. For them to be tileable by dominoes, their difference must be zero, so 2k must be zero, meaning k is zero. Therefore, the original polygon must have a difference of zero, meaning it can be tiled with dominoes.Wait, that seems important. If the original polygon can be divided into two equal polygons, each of which must have half the difference in coloring. If the original difference is non-zero, then each of the two polygons would have a non-zero difference, which would prevent them from being tileable by dominoes. But the problem doesn't specify that the two equal polygons are tileable by dominoes, just that they are equal.Hmm, maybe I'm overcomplicating it. Let me try a different approach.If the polygon can be divided into twenty-five equal polygons, each of area 4, and also into two equal polygons, each of area 50, then perhaps the structure is such that it's a multiple of both 2 and 25, which is 50. So, maybe it can be divided into fifty equal polygons, each of area 2.But wait, area 2 would be dominoes. So, if the original polygon can be divided into dominoes, then it must have an equal number of black and white squares. But earlier, I thought that if the original polygon has a non-zero difference, it can't be tiled with dominoes. But if it can be divided into two equal polygons, each of which can be further divided into twenty-five dominoes, then maybe it's possible.Wait, no. If the original polygon can be divided into two equal polygons, each of area 50, and each of those can be divided into twenty-five dominoes, then the original polygon can be divided into fifty dominoes. But that would require that each of the two equal polygons has an equal number of black and white squares, which would imply that the original polygon also has an equal number of black and white squares.But earlier, I thought that if the original polygon is divided into twenty-five tetrominoes, some of which are T or L, then the difference is non-zero. But if the original polygon can be divided into two equal polygons, each of which can be divided into twenty-five dominoes, then the original polygon must have an equal number of black and white squares.Therefore, the original polygon must have a difference of zero, meaning it can be tiled with dominoes. Therefore, it can be divided into fifty equal polygons, each of area 2.Wait, but earlier I thought that if the original polygon is divided into twenty-five tetrominoes, some of which are T or L, then the difference is non-zero. But if the original polygon can be divided into two equal polygons, each of which can be divided into twenty-five dominoes, then the original polygon must have a difference of zero.Therefore, the original polygon must have a difference of zero, meaning it can be tiled with dominoes. Therefore, it can be divided into fifty equal polygons, each of area 2.But wait, does that mean that the original polygon cannot have any T or L tetrominoes? Because if it did, the difference would be non-zero, contradicting the fact that it can be divided into two equal polygons, each of which can be divided into dominoes.Therefore, the original polygon must be composed of tetrominoes that don't contribute to the difference, i.e., I, O, S, and Z tetrominoes. Therefore, it can be tiled with dominoes, and hence divided into fifty equal polygons.So, putting it all together, the fact that the original polygon can be divided into two equal polygons implies that it has an equal number of black and white squares, meaning it can be tiled with dominoes. Therefore, it can be divided into fifty equal polygons, each of area 2.Therefore, the answer is yes, it is necessarily possible to divide it into fifty equal polygons.

💡Alright, so I've got this problem here about a restaurant counting their sales for the day. They sold different numbers of meals at different prices, and I need to figure out how much money they made in total. Okay, let's break this down step by step.First, I need to understand what exactly is being asked. The restaurant sold three types of meals: 10 meals at 8 each, 5 meals at 10 each, and 20 meals at 4 each. I need to find the total amount of money they made from all these sales. So, essentially, I need to calculate the revenue from each type of meal and then add them all together to get the total revenue for the day.Alright, so let's start with the first type of meal: 10 meals sold at 8 each. To find out how much money they made from these meals, I need to multiply the number of meals by the price per meal. So, that would be 10 multiplied by 8. Let me write that down: 10 * 8 = 80. Okay, so they made 80 from the first type of meal.Next, the second type of meal: 5 meals sold at 10 each. Again, I need to multiply the number of meals by the price per meal. So, that's 5 * 10. Let me calculate that: 5 * 10 = 50. Got it, they made 50 from the second type of meal.Now, onto the third type of meal: 20 meals sold at 4 each. Using the same method, I'll multiply the number of meals by the price per meal. So, that's 20 * 4. Calculating that gives me 20 * 4 = 80. Alright, they made another 80 from the third type of meal.Now that I have the revenue from each type of meal, I need to add them all together to find the total revenue for the day. So, I'll add 80 (from the first type) + 50 (from the second type) + 80 (from the third type). Let's do that step by step to make sure I don't make any mistakes.First, 80 + 50 equals 130. Then, adding the last 80 to that, 130 + 80 equals 210. So, altogether, the restaurant made 210 from all their sales that day.Wait a second, let me double-check my calculations to make sure I didn't make any errors. Starting with the first type: 10 meals at 8 each. 10 times 8 is definitely 80. Okay, that's correct. Then, 5 meals at 10 each: 5 times 10 is 50. That seems right too. Finally, 20 meals at 4 each: 20 times 4 is 80. Yep, that checks out.Now, adding them up again: 80 plus 50 is 130, and 130 plus 80 is 210. Yep, that's consistent. So, I think I've got it right. The restaurant made a total of 210 throughout the day.Just to be thorough, let me think about another way to approach this problem to confirm my answer. Maybe I can add up the number of meals first and then see if that gives me any insights. So, they sold 10 + 5 + 20 meals in total. That adds up to 35 meals. But since each meal is priced differently, I can't just multiply the total number of meals by a single price to get the total revenue. That's why I had to calculate each type separately and then add them up.Another thing I could consider is whether there are any discounts or special offers that might affect the total revenue, but the problem doesn't mention anything like that. It seems like it's a straightforward calculation based on the given numbers.I also wonder if there's a quicker way to do this calculation, maybe by grouping similar numbers or using some kind of mathematical shortcut. For example, I notice that the first and third types of meals both resulted in 80 in revenue. Maybe I could have combined those first and then added the 50 from the second type. Let's see: 80 + 80 is 160, and then adding 50 gives me 210. Yep, that's the same result.Alternatively, I could have added the number of meals sold at similar prices first. For instance, adding the 10 meals at 8 and the 20 meals at 4, which are both lower-priced meals, and then adding the 5 meals at 10 separately. Let's try that approach:First, 10 meals at 8: 10 * 8 = 80Then, 20 meals at 4: 20 * 4 = 80Adding those together: 80 + 80 = 160Now, adding the 5 meals at 10: 5 * 10 = 50Total revenue: 160 + 50 = 210Again, I arrive at the same total of 210. It seems like no matter how I approach the problem, the answer remains consistent. That gives me more confidence that 210 is indeed the correct total revenue for the day.Just to make sure I'm not missing anything, let me read the problem statement again: "They sold 10 meals at 8 each, 5 meals at 10 each, and 20 meals at 4 each. In dollars, how much money did the restaurant make throughout the day?" Yep, that's exactly what I calculated. There's no mention of taxes, tips, or any other expenses, so I don't need to factor those in. It's purely about calculating the total sales revenue.I also think about whether the restaurant might have different costs associated with each meal, but since the problem only asks for the total money made, which I assume refers to gross revenue, I don't need to consider costs or net profit. It's just about adding up the sales from each meal type.In summary, by calculating the revenue from each type of meal individually and then summing them up, I determined that the restaurant made a total of 210 throughout the day. I double-checked my calculations using different methods to ensure accuracy, and all approaches led me to the same conclusion. Therefore, I'm confident that the total revenue is indeed 210.

💡Okay, so I have this problem about an arithmetic progression. It says that for a non-constant arithmetic progression ( (a_n) ), there exists a natural number ( n ) such that ( a_n + a_{n+1} = a_1 + a_2 + dots + a_{3n-1} ). I need to prove that there are no zero terms in this progression. Hmm, let's try to unpack this step by step.First, let me recall what an arithmetic progression is. It's a sequence where each term after the first is obtained by adding a constant difference. So, if the first term is ( a ) and the common difference is ( d ), then the ( n )-th term is ( a_n = a + (n-1)d ). Since it's non-constant, ( d ) can't be zero. That makes sense.Now, the problem states that there exists a natural number ( n ) such that the sum of the ( n )-th and ( (n+1) )-th terms equals the sum of the first ( 3n - 1 ) terms. Let me write that down:( a_n + a_{n+1} = a_1 + a_2 + dots + a_{3n-1} )Let me express both sides in terms of ( a ) and ( d ).Starting with the left-hand side (LHS):( a_n = a + (n-1)d )( a_{n+1} = a + nd )So, adding them together:( a_n + a_{n+1} = [a + (n-1)d] + [a + nd] = 2a + (2n - 1)d )Okay, so the LHS simplifies to ( 2a + (2n - 1)d ).Now, the right-hand side (RHS) is the sum of the first ( 3n - 1 ) terms. The formula for the sum of the first ( m ) terms of an arithmetic progression is:( S_m = frac{m}{2} [2a + (m - 1)d] )So, substituting ( m = 3n - 1 ):( S_{3n-1} = frac{3n - 1}{2} [2a + (3n - 2)d] )Let me write that out:( S_{3n-1} = frac{3n - 1}{2} (2a + (3n - 2)d) )So, the RHS is ( frac{3n - 1}{2} (2a + (3n - 2)d) ).Now, according to the problem, LHS = RHS, so:( 2a + (2n - 1)d = frac{3n - 1}{2} (2a + (3n - 2)d) )Hmm, okay. Let me try to simplify this equation.First, multiply both sides by 2 to eliminate the denominator:( 2 times [2a + (2n - 1)d] = (3n - 1)(2a + (3n - 2)d) )Calculating the left side:( 4a + 2(2n - 1)d = 4a + (4n - 2)d )Right side:( (3n - 1)(2a + (3n - 2)d) )Let me expand this:First, distribute ( 3n - 1 ) over the terms inside the parentheses:( (3n - 1) times 2a + (3n - 1) times (3n - 2)d )Calculating each part:( 2a(3n - 1) = 6na - 2a )( (3n - 1)(3n - 2)d = [9n^2 - 6n - 3n + 2]d = (9n^2 - 9n + 2)d )So, putting it all together, the right side is:( 6na - 2a + (9n^2 - 9n + 2)d )Now, let's write the equation again:Left side: ( 4a + (4n - 2)d )Right side: ( 6na - 2a + (9n^2 - 9n + 2)d )So, setting them equal:( 4a + (4n - 2)d = 6na - 2a + (9n^2 - 9n + 2)d )Let me bring all terms to one side to simplify:( 4a + (4n - 2)d - 6na + 2a - (9n^2 - 9n + 2)d = 0 )Combine like terms:For ( a ):( 4a + 2a - 6na = 6a - 6na = 6a(1 - n) )For ( d ):( (4n - 2)d - (9n^2 - 9n + 2)d = [4n - 2 - 9n^2 + 9n - 2]d = (-9n^2 + 13n - 4)d )So, the equation becomes:( 6a(1 - n) + (-9n^2 + 13n - 4)d = 0 )Let me write it as:( 6a(1 - n) = (9n^2 - 13n + 4)d )Hmm, so we have:( 6a(1 - n) = (9n^2 - 13n + 4)d )Since the progression is non-constant, ( d neq 0 ). So, we can solve for ( a ) in terms of ( d ):( a = frac{(9n^2 - 13n + 4)d}{6(1 - n)} )Simplify the denominator:( 6(1 - n) = -6(n - 1) )So,( a = frac{(9n^2 - 13n + 4)d}{-6(n - 1)} )Let me factor the numerator if possible. Let's see:( 9n^2 - 13n + 4 )Looking for factors of ( 9 times 4 = 36 ) that add up to -13. Hmm, -9 and -4.Wait, ( 9n^2 - 9n - 4n + 4 = 9n(n - 1) - 4(n - 1) = (9n - 4)(n - 1) )Yes, that works.So,( 9n^2 - 13n + 4 = (9n - 4)(n - 1) )So, substituting back:( a = frac{(9n - 4)(n - 1)d}{-6(n - 1)} )We can cancel out ( (n - 1) ) from numerator and denominator, assuming ( n neq 1 ). But wait, if ( n = 1 ), then ( (n - 1) = 0 ), but in the original equation, if ( n = 1 ), the RHS would be ( S_{2} = a_1 + a_2 ), and the LHS would be ( a_1 + a_2 ). So, that would mean ( a_1 + a_2 = a_1 + a_2 ), which is always true, but the problem states that the progression is non-constant, so ( d neq 0 ). However, if ( n = 1 ), we might not get any specific information about ( a ) and ( d ). So, perhaps ( n ) is greater than 1.Assuming ( n neq 1 ), we can cancel ( (n - 1) ):( a = frac{(9n - 4)d}{-6} = -frac{(9n - 4)d}{6} )So, ( a = -frac{(9n - 4)d}{6} )Okay, so now we have ( a ) expressed in terms of ( d ) and ( n ). Let me write that:( a = -frac{(9n - 4)d}{6} )Now, the problem asks us to prove that there are no zero terms in this progression. So, we need to show that for all natural numbers ( m ), ( a_m neq 0 ).Recall that ( a_m = a + (m - 1)d ). So, if ( a_m = 0 ), then:( a + (m - 1)d = 0 )Substituting the expression for ( a ):( -frac{(9n - 4)d}{6} + (m - 1)d = 0 )Let me factor out ( d ):( d left( -frac{9n - 4}{6} + (m - 1) right) = 0 )Since ( d neq 0 ) (the progression is non-constant), the term in the parentheses must be zero:( -frac{9n - 4}{6} + (m - 1) = 0 )Let me solve for ( m ):( (m - 1) = frac{9n - 4}{6} )( m = 1 + frac{9n - 4}{6} = frac{6 + 9n - 4}{6} = frac{9n + 2}{6} )So, ( m = frac{9n + 2}{6} )But ( m ) must be a natural number, meaning ( frac{9n + 2}{6} ) must be an integer.Let me see if this is possible.Simplify ( frac{9n + 2}{6} ):( frac{9n + 2}{6} = frac{3(3n) + 2}{6} = frac{3n}{2} + frac{1}{3} )Wait, that doesn't seem helpful. Alternatively, let's write it as:( frac{9n + 2}{6} = frac{9n}{6} + frac{2}{6} = frac{3n}{2} + frac{1}{3} )Hmm, so ( frac{3n}{2} + frac{1}{3} ) must be an integer.But ( frac{3n}{2} ) is either an integer or a half-integer, depending on whether ( n ) is even or odd.If ( n ) is even, say ( n = 2k ), then ( frac{3n}{2} = 3k ), which is an integer. Then, ( frac{3n}{2} + frac{1}{3} = 3k + frac{1}{3} ), which is not an integer.If ( n ) is odd, say ( n = 2k + 1 ), then ( frac{3n}{2} = frac{3(2k + 1)}{2} = 3k + frac{3}{2} ), which is a half-integer. Adding ( frac{1}{3} ) gives ( 3k + frac{3}{2} + frac{1}{3} = 3k + frac{11}{6} ), which is not an integer.Therefore, in both cases, ( frac{9n + 2}{6} ) is not an integer. Hence, ( m ) cannot be a natural number.This implies that there is no natural number ( m ) such that ( a_m = 0 ). Therefore, there are no zero terms in the arithmetic progression.Wait, let me double-check my steps to make sure I didn't make a mistake.1. I expressed ( a_n ) and ( a_{n+1} ) correctly in terms of ( a ) and ( d ).2. Calculated the sum ( S_{3n-1} ) correctly using the arithmetic series formula.3. Set LHS equal to RHS and expanded both sides.4. Brought all terms to one side and combined like terms correctly.5. Factored the quadratic in the numerator correctly.6. Canceled ( (n - 1) ) assuming ( n neq 1 ), which seems reasonable because if ( n = 1 ), the equation holds trivially but doesn't give us information about ( a ) and ( d ).7. Expressed ( a ) in terms of ( d ) and ( n ).8. Set ( a_m = 0 ) and substituted ( a ) to find ( m ).9. Showed that ( m ) cannot be an integer, hence no zero terms.I think all steps are correct. Therefore, the conclusion is that there are no zero terms in the progression.**Final Answer**boxed{text{There are no zero terms in the progression.}}

💡First, I need to determine the total amount of work required to build the wall. With 60 workers taking 3 days, the total work can be calculated by multiplying the number of workers by the number of days.Next, I'll calculate the work rate of one worker by dividing the total work by the number of workers.Now, if only 40 workers are available, I'll find the new work rate by multiplying the individual worker rate by 40.Finally, to find out how many days it will take for 40 workers to complete the same amount of work, I'll divide the total work by the new work rate.

💡Alright, so I've got this problem about an ellipse, and I need to find its equation first and then figure out the range of a ratio t based on some conditions. Let me try to break this down step by step.**Part 1: Finding the Equation of the Ellipse**The ellipse is given by the equation:[frac{x^2}{a^2} + frac{y^2}{b^2} = 1 quad (a > b > 0)]We know two things:1. The eccentricity ( e = frac{sqrt{3}}{2} ).2. The area of the rhombus formed by connecting its four vertices is 4.First, let's recall that the eccentricity ( e ) of an ellipse is given by:[e = frac{c}{a}]where ( c ) is the distance from the center to each focus, and ( a ) is the semi-major axis. So, given ( e = frac{sqrt{3}}{2} ), we can write:[frac{c}{a} = frac{sqrt{3}}{2} quad Rightarrow quad c = frac{sqrt{3}}{2}a]Also, for an ellipse, we know that:[c^2 = a^2 - b^2]Substituting ( c = frac{sqrt{3}}{2}a ) into this equation:[left(frac{sqrt{3}}{2}aright)^2 = a^2 - b^2][frac{3}{4}a^2 = a^2 - b^2][b^2 = a^2 - frac{3}{4}a^2 = frac{1}{4}a^2][b = frac{a}{2}]So, ( b ) is half of ( a ).Next, the area of the rhombus formed by the four vertices of the ellipse. The vertices of the ellipse are at ( (pm a, 0) ) and ( (0, pm b) ). Connecting these four points forms a rhombus.The area of a rhombus can be calculated as:[text{Area} = frac{1}{2} times d_1 times d_2]where ( d_1 ) and ( d_2 ) are the lengths of the diagonals.In this case, the diagonals are the major and minor axes of the ellipse. So, the lengths are:- ( d_1 = 2a ) (major axis)- ( d_2 = 2b ) (minor axis)Thus, the area is:[frac{1}{2} times 2a times 2b = 2ab]We are told this area is 4:[2ab = 4 quad Rightarrow quad ab = 2]But earlier, we found that ( b = frac{a}{2} ). Substituting this into the equation:[a times frac{a}{2} = 2][frac{a^2}{2} = 2][a^2 = 4 quad Rightarrow quad a = 2]Since ( a > 0 ), we take the positive root. Then, ( b = frac{a}{2} = frac{2}{2} = 1 ).So, the equation of the ellipse is:[frac{x^2}{4} + y^2 = 1]**Part 2: Finding the Range of t**Now, we have a line ( l ) passing through point ( D(1, 0) ) with a slope ( k ) (( k neq 0 )) intersecting the ellipse at points ( M(x_1, y_1) ) and ( N(x_2, y_2) ). We need to find the ratio ( t ) of the areas of triangles ( OMD ) and ( OND ), where ( O ) is the origin. The condition given is ( k^2 < frac{5}{12} ), and we need to find the range of ( t ).First, let's write the equation of line ( l ). Since it passes through ( D(1, 0) ) with slope ( k ), its equation is:[y = k(x - 1)]This line intersects the ellipse ( frac{x^2}{4} + y^2 = 1 ). Let's substitute ( y = k(x - 1) ) into the ellipse equation:[frac{x^2}{4} + [k(x - 1)]^2 = 1][frac{x^2}{4} + k^2(x^2 - 2x + 1) = 1][frac{x^2}{4} + k^2x^2 - 2k^2x + k^2 = 1]Multiply through by 4 to eliminate the fraction:[x^2 + 4k^2x^2 - 8k^2x + 4k^2 = 4][(1 + 4k^2)x^2 - 8k^2x + (4k^2 - 4) = 0]This is a quadratic in ( x ):[(1 + 4k^2)x^2 - 8k^2x + (4k^2 - 4) = 0]Let me denote this as:[A x^2 + B x + C = 0]where:- ( A = 1 + 4k^2 )- ( B = -8k^2 )- ( C = 4k^2 - 4 )The solutions to this quadratic will give us the x-coordinates of points ( M ) and ( N ). Let's denote them as ( x_1 ) and ( x_2 ).Using Vieta's formulas, we know:[x_1 + x_2 = frac{-B}{A} = frac{8k^2}{1 + 4k^2}][x_1 x_2 = frac{C}{A} = frac{4k^2 - 4}{1 + 4k^2}]Similarly, since ( y = k(x - 1) ), we can find ( y_1 ) and ( y_2 ):[y_1 = k(x_1 - 1)][y_2 = k(x_2 - 1)]So, let's compute ( y_1 + y_2 ) and ( y_1 y_2 ):First, ( y_1 + y_2 ):[y_1 + y_2 = k(x_1 - 1) + k(x_2 - 1) = k(x_1 + x_2 - 2)][= kleft( frac{8k^2}{1 + 4k^2} - 2 right) = kleft( frac{8k^2 - 2(1 + 4k^2)}{1 + 4k^2} right)][= kleft( frac{8k^2 - 2 - 8k^2}{1 + 4k^2} right) = kleft( frac{-2}{1 + 4k^2} right) = -frac{2k}{1 + 4k^2}]Next, ( y_1 y_2 ):[y_1 y_2 = [k(x_1 - 1)][k(x_2 - 1)] = k^2(x_1 x_2 - x_1 - x_2 + 1)][= k^2left( frac{4k^2 - 4}{1 + 4k^2} - frac{8k^2}{1 + 4k^2} + 1 right)][= k^2left( frac{4k^2 - 4 - 8k^2 + (1 + 4k^2)}{1 + 4k^2} right)][= k^2left( frac{4k^2 - 4 - 8k^2 + 1 + 4k^2}{1 + 4k^2} right)][= k^2left( frac{(4k^2 - 8k^2 + 4k^2) + (-4 + 1)}{1 + 4k^2} right)][= k^2left( frac{0k^2 - 3}{1 + 4k^2} right) = -frac{3k^2}{1 + 4k^2}]So, we have:- ( y_1 + y_2 = -frac{2k}{1 + 4k^2} )- ( y_1 y_2 = -frac{3k^2}{1 + 4k^2} )Now, the ratio ( t ) is defined as the ratio of the areas of triangles ( OMD ) and ( OND ). Let's recall that the area of a triangle given by three points can be found using the determinant formula. However, since both triangles share the same base ( OD ), the ratio of their areas will be equal to the ratio of their heights from ( M ) and ( N ) to the base ( OD ).But in this case, since ( OD ) is along the x-axis from ( (0,0) ) to ( (1,0) ), the height of each triangle is simply the absolute y-coordinate of points ( M ) and ( N ). Therefore, the ratio ( t ) is:[t = frac{S_{triangle OMD}}{S_{triangle OND}} = frac{|y_1|}{|y_2|}]But since ( y_1 ) and ( y_2 ) can be positive or negative, and considering the line intersects the ellipse at two points, we need to consider the signs. However, since the ratio is of areas, it's the absolute value. But let's see if we can express this ratio in terms of ( y_1 ) and ( y_2 ).Alternatively, since both triangles share the same base ( OD ), the areas are proportional to the heights, which are ( |y_1| ) and ( |y_2| ). So, ( t = frac{|y_1|}{|y_2|} ).But let's think about the line intersecting the ellipse. Depending on the slope ( k ), the points ( M ) and ( N ) can be on different sides of the x-axis. However, since ( D ) is at ( (1,0) ), and the line passes through ( D ), it's possible that one point is above the x-axis and the other is below, or both on the same side. But given that ( k neq 0 ), it's likely that they are on opposite sides, making ( y_1 ) and ( y_2 ) have opposite signs.Therefore, ( y_1 y_2 < 0 ), which means ( t = frac{|y_1|}{|y_2|} = left| frac{y_1}{y_2} right| ). But since ( y_1 ) and ( y_2 ) have opposite signs, ( frac{y_1}{y_2} ) is negative, so ( t = -frac{y_1}{y_2} ).So, ( t = -frac{y_1}{y_2} ).Our goal is to find the range of ( t ) given ( k^2 < frac{5}{12} ).Let me denote ( t = -frac{y_1}{y_2} ). Then, ( frac{y_1}{y_2} = -t ).We can use the expressions for ( y_1 + y_2 ) and ( y_1 y_2 ) to find a relationship involving ( t ).Let me compute ( frac{(y_1 + y_2)^2}{y_1 y_2} ):[frac{(y_1 + y_2)^2}{y_1 y_2} = frac{left(-frac{2k}{1 + 4k^2}right)^2}{-frac{3k^2}{1 + 4k^2}} = frac{frac{4k^2}{(1 + 4k^2)^2}}{-frac{3k^2}{1 + 4k^2}} = frac{4k^2}{(1 + 4k^2)^2} times frac{-(1 + 4k^2)}{3k^2} = -frac{4}{3(1 + 4k^2)}]So,[frac{(y_1 + y_2)^2}{y_1 y_2} = -frac{4}{3(1 + 4k^2)}]But also, we can express ( frac{(y_1 + y_2)^2}{y_1 y_2} ) in terms of ( t ):[frac{(y_1 + y_2)^2}{y_1 y_2} = frac{y_1^2 + 2y_1 y_2 + y_2^2}{y_1 y_2} = frac{y_1^2}{y_1 y_2} + frac{2y_1 y_2}{y_1 y_2} + frac{y_2^2}{y_1 y_2} = frac{y_1}{y_2} + 2 + frac{y_2}{y_1}]Let me denote ( frac{y_1}{y_2} = -t ), so ( frac{y_2}{y_1} = -frac{1}{t} ). Therefore,[frac{(y_1 + y_2)^2}{y_1 y_2} = (-t) + 2 + left(-frac{1}{t}right) = -t - frac{1}{t} + 2]So, we have:[-t - frac{1}{t} + 2 = -frac{4}{3(1 + 4k^2)}]Let me rearrange this equation:[-t - frac{1}{t} + 2 = -frac{4}{3(1 + 4k^2)}][-t - frac{1}{t} = -frac{4}{3(1 + 4k^2)} - 2][-t - frac{1}{t} = -frac{4}{3(1 + 4k^2)} - frac{6(1 + 4k^2)}{3(1 + 4k^2)}][-t - frac{1}{t} = -frac{4 + 6(1 + 4k^2)}{3(1 + 4k^2)}][-t - frac{1}{t} = -frac{4 + 6 + 24k^2}{3(1 + 4k^2)}][-t - frac{1}{t} = -frac{10 + 24k^2}{3(1 + 4k^2)}][t + frac{1}{t} = frac{10 + 24k^2}{3(1 + 4k^2)}]Let me denote ( s = t + frac{1}{t} ). Then,[s = frac{10 + 24k^2}{3(1 + 4k^2)}]We can simplify the right-hand side:[s = frac{10 + 24k^2}{3(1 + 4k^2)} = frac{10}{3(1 + 4k^2)} + frac{24k^2}{3(1 + 4k^2)} = frac{10}{3(1 + 4k^2)} + 8 times frac{k^2}{1 + 4k^2}]But perhaps it's better to express it as:[s = frac{10 + 24k^2}{3(1 + 4k^2)} = frac{10 + 24k^2}{3 + 12k^2}]Let me factor numerator and denominator:Numerator: ( 10 + 24k^2 = 2(5 + 12k^2) )Denominator: ( 3 + 12k^2 = 3(1 + 4k^2) )So,[s = frac{2(5 + 12k^2)}{3(1 + 4k^2)} = frac{2}{3} times frac{5 + 12k^2}{1 + 4k^2}]Let me write ( frac{5 + 12k^2}{1 + 4k^2} ) as:[frac{5 + 12k^2}{1 + 4k^2} = frac{5 + 12k^2}{1 + 4k^2} = frac{5 + 12k^2}{1 + 4k^2}]Let me perform polynomial division or see if I can express this as a constant plus something:Let me write:[5 + 12k^2 = A(1 + 4k^2) + B]Expanding:[5 + 12k^2 = A + 4A k^2 + B]Comparing coefficients:- Constant term: ( 5 = A + B )- ( k^2 ) term: ( 12 = 4A ) ⇒ ( A = 3 )Then, from ( 5 = A + B ), ( 5 = 3 + B ) ⇒ ( B = 2 )So,[frac{5 + 12k^2}{1 + 4k^2} = frac{3(1 + 4k^2) + 2}{1 + 4k^2} = 3 + frac{2}{1 + 4k^2}]Therefore,[s = frac{2}{3} left( 3 + frac{2}{1 + 4k^2} right ) = frac{2}{3} times 3 + frac{2}{3} times frac{2}{1 + 4k^2} = 2 + frac{4}{3(1 + 4k^2)}]So,[s = t + frac{1}{t} = 2 + frac{4}{3(1 + 4k^2)}]But from earlier, we have:[s = t + frac{1}{t} = 2 + frac{4}{3(1 + 4k^2)}]But we also have:[frac{(y_1 + y_2)^2}{y_1 y_2} = -frac{4}{3(1 + 4k^2)}]Which we used to get to this point.Now, our goal is to find the range of ( t ) given ( k^2 < frac{5}{12} ).First, let's analyze ( frac{4}{3(1 + 4k^2)} ).Since ( k^2 < frac{5}{12} ), let's find the range of ( 1 + 4k^2 ):- Minimum value of ( 1 + 4k^2 ): When ( k^2 ) is minimum, which is approaching 0, so ( 1 + 4k^2 ) approaches 1.- Maximum value of ( 1 + 4k^2 ): When ( k^2 = frac{5}{12} ), so ( 1 + 4 times frac{5}{12} = 1 + frac{5}{3} = frac{8}{3} ).Therefore, ( 1 leq 1 + 4k^2 < frac{8}{3} ).Thus, ( frac{4}{3(1 + 4k^2)} ) will be:- When ( 1 + 4k^2 ) is minimum (1), ( frac{4}{3 times 1} = frac{4}{3} ).- When ( 1 + 4k^2 ) is maximum (( frac{8}{3} )), ( frac{4}{3 times frac{8}{3}} = frac{4}{8} = frac{1}{2} ).So, ( frac{1}{2} < frac{4}{3(1 + 4k^2)} leq frac{4}{3} ).But wait, since ( k^2 < frac{5}{12} ), ( 1 + 4k^2 < frac{8}{3} ), so ( frac{4}{3(1 + 4k^2)} > frac{4}{3 times frac{8}{3}} = frac{1}{2} ).But actually, as ( k^2 ) approaches 0, ( frac{4}{3(1 + 4k^2)} ) approaches ( frac{4}{3} ), and as ( k^2 ) approaches ( frac{5}{12} ), it approaches ( frac{1}{2} ).Therefore,[frac{1}{2} < frac{4}{3(1 + 4k^2)} < frac{4}{3}]But in our expression for ( s ):[s = t + frac{1}{t} = 2 + frac{4}{3(1 + 4k^2)}]So, substituting the range of ( frac{4}{3(1 + 4k^2)} ):[2 + frac{1}{2} < s < 2 + frac{4}{3}][frac{5}{2} < s < frac{10}{3}]So, ( frac{5}{2} < t + frac{1}{t} < frac{10}{3} ).Now, we need to solve the inequality:[frac{5}{2} < t + frac{1}{t} < frac{10}{3}]Let me denote ( u = t + frac{1}{t} ). We need to find the range of ( t ) such that ( frac{5}{2} < u < frac{10}{3} ).First, let's analyze the function ( u(t) = t + frac{1}{t} ).This function has a minimum at ( t = 1 ), where ( u(1) = 2 ). As ( t ) increases beyond 1, ( u(t) ) increases, and as ( t ) approaches 0 from the right, ( u(t) ) approaches infinity. Similarly, as ( t ) approaches infinity, ( u(t) ) approaches infinity.But in our case, ( t = -frac{y_1}{y_2} ), and since ( y_1 y_2 < 0 ), ( t ) is positive. Also, ( t ) is a ratio of areas, so it's positive.But let's think about the possible values of ( t ). Since ( t = frac{|y_1|}{|y_2|} ), and the line intersects the ellipse at two points, ( t ) can be greater than 1 or less than 1, depending on which point is farther from the x-axis.But let's solve the inequality ( frac{5}{2} < t + frac{1}{t} < frac{10}{3} ).Let me consider the equation ( t + frac{1}{t} = c ), where ( c ) is a constant. The solutions are:[t = frac{c pm sqrt{c^2 - 4}}{2}]So, for ( c = frac{5}{2} ):[t = frac{frac{5}{2} pm sqrt{left(frac{5}{2}right)^2 - 4}}{2} = frac{frac{5}{2} pm sqrt{frac{25}{4} - 4}}{2} = frac{frac{5}{2} pm sqrt{frac{9}{4}}}{2} = frac{frac{5}{2} pm frac{3}{2}}{2}]So,- ( t = frac{frac{5}{2} + frac{3}{2}}{2} = frac{8}{4} = 2 )- ( t = frac{frac{5}{2} - frac{3}{2}}{2} = frac{2}{4} = frac{1}{2} )Similarly, for ( c = frac{10}{3} ):[t = frac{frac{10}{3} pm sqrt{left(frac{10}{3}right)^2 - 4}}{2} = frac{frac{10}{3} pm sqrt{frac{100}{9} - frac{36}{9}}}{2} = frac{frac{10}{3} pm sqrt{frac{64}{9}}}{2} = frac{frac{10}{3} pm frac{8}{3}}{2}]So,- ( t = frac{frac{10}{3} + frac{8}{3}}{2} = frac{18}{6} = 3 )- ( t = frac{frac{10}{3} - frac{8}{3}}{2} = frac{2}{6} = frac{1}{3} )So, the equation ( t + frac{1}{t} = c ) has solutions ( t = 2 ) and ( t = frac{1}{2} ) when ( c = frac{5}{2} ), and ( t = 3 ) and ( t = frac{1}{3} ) when ( c = frac{10}{3} ).Now, let's analyze the inequality ( frac{5}{2} < t + frac{1}{t} < frac{10}{3} ).We know that ( t + frac{1}{t} ) is a decreasing function for ( 0 < t < 1 ) and increasing for ( t > 1 ).So, for ( t > 1 ):- When ( t = 2 ), ( u = frac{5}{2} )- When ( t = 3 ), ( u = frac{10}{3} )Since ( u ) increases as ( t ) increases beyond 1, the inequality ( frac{5}{2} < u < frac{10}{3} ) corresponds to ( 2 < t < 3 ).For ( 0 < t < 1 ):- When ( t = frac{1}{2} ), ( u = frac{5}{2} )- When ( t = frac{1}{3} ), ( u = frac{10}{3} )Since ( u ) decreases as ( t ) increases towards 1 from below, the inequality ( frac{5}{2} < u < frac{10}{3} ) corresponds to ( frac{1}{3} < t < frac{1}{2} ).Therefore, the range of ( t ) is:[frac{1}{3} < t < frac{1}{2} quad text{or} quad 2 < t < 3]So, putting it all together, the range of ( t ) is ( frac{1}{3} < t < frac{1}{2} ) or ( 2 < t < 3 ).**Final Answer**(1) The equation of the ellipse is (boxed{dfrac{x^2}{4} + y^2 = 1}).(2) The range of ( t ) is (boxed{dfrac{1}{3} < t < dfrac{1}{2} text{ or } 2 < t < 3}).

💡Alright, so I came across this interesting problem about placing ten consecutive natural numbers on the vertices and edges of a regular triangular pyramid, also known as a tetrahedron. The goal is to arrange these numbers such that for each edge, the sum of the three numbers on the edge and at its endpoints remains constant. At first glance, this seems like a challenging puzzle, but I think I can break it down step by step.First, let me visualize the structure. A regular tetrahedron has four vertices and six edges. So, we're talking about placing numbers on these four vertices and six edges, making a total of ten positions. That matches up with the ten consecutive natural numbers we need to use. Cool, so each position gets a unique number from the sequence.Now, the key condition here is that for each edge, the sum of the three numbers on the edge and at its endpoints remains constant. Let's denote this constant sum as ( S ). Since there are six edges, if I were to sum up all these six sums, I would get ( 6S ).But wait, if I just add up all the edge sums, I'm going to be counting some numbers multiple times. Specifically, the numbers on the edges are each counted once, but the numbers on the vertices are each counted three times because each vertex is connected to three edges. So, the total sum ( 6S ) is equal to the sum of all ten numbers plus twice the sum of the numbers at the vertices. Let me write that down:[6S = text{Sum of all ten numbers} + 2 times text{Sum of numbers at vertices}]Now, the sum of ten consecutive natural numbers can be calculated. Let's say the numbers start from ( n ). Then, the sum is:[text{Sum} = n + (n+1) + (n+2) + dots + (n+9) = 10n + 45]This is because the sum of the first ten consecutive numbers starting from 0 is ( 0 + 1 + 2 + dots + 9 = 45 ). So, the total sum is ( 10n + 45 ).Plugging this back into our earlier equation:[6S = 10n + 45 + 2 times text{Sum of numbers at vertices}]Now, here's where it gets tricky. The sum ( 10n + 45 ) is always an odd number because ( 10n ) is even (since 10 is even) and 45 is odd. So, adding an even and an odd number gives an odd result. Therefore, ( 10n + 45 ) is odd.On the other hand, ( 6S ) is clearly an even number because it's a multiple of 6. So, we have an equation where an even number equals an odd number plus twice the sum of the numbers at the vertices. But twice any number is even, so adding an even number to an odd number results in an odd number. Therefore, the right side of the equation is odd, while the left side is even. This is a contradiction because an even number cannot equal an odd number.This contradiction suggests that our initial assumption—that such an arrangement is possible—is incorrect. Therefore, it's impossible to place ten consecutive natural numbers on the vertices and edges of a regular tetrahedron such that the sum of the three numbers on each edge and at its endpoints remains constant.But wait, let me double-check my reasoning to make sure I didn't miss anything. I considered the total sum of all edges and accounted for the multiple counts of the vertex numbers. The key point is that the total sum ( 6S ) must be even, but the sum of the ten consecutive numbers is odd, leading to an inconsistency. This seems solid.I also thought about whether there's a way to adjust the numbers or their arrangement to avoid this contradiction, but since the sum of ten consecutive numbers is inherently odd, and ( 6S ) must be even, there's no way around it. The parity doesn't match, making such an arrangement impossible.So, after carefully analyzing the problem and checking my logic, I'm confident that the conclusion is correct.