💡To determine the set of all possible values for ( x ) such that ( y ) is a real number in the equation ( 9y^2 - 3xy + x + 8 = 0 ), we can follow these steps:1. **Identify the type of equation**: The given equation is a quadratic in terms of ( y ).2. **Coefficients of the quadratic equation**: - ( a = 9 ) - ( b = -3x ) - ( c = x + 8 )3. **Condition for real values of ( y )**: For ( y ) to be real, the discriminant ( D ) of the quadratic equation must be non-negative. [ D = b^2 - 4ac ] Substituting the coefficients: [ D = (-3x)^2 - 4 cdot 9 cdot (x + 8) = 9x^2 - 36(x + 8) ] Simplifying: [ D = 9x^2 - 36x - 288 ]4. **Factor the quadratic expression**: [ D = 9(x^2 - 4x - 32) ] Factoring the quadratic expression inside the parentheses: [ D = 9(x - 8)(x + 4) ]5. **Analyze the sign of the discriminant**: The discriminant ( D ) must be non-negative for ( y ) to be real. Therefore, we need: [ (x - 8)(x + 4) geq 0 ] This inequality holds when: - ( x leq -4 ) or ( x geq 8 )6. **Conclusion**: The values of ( x ) for which ( y ) is real are when ( x leq -4 ) or ( x geq 8 ).Thus, the final answer is A) ( x leq -4 ) or ( x geq 8 ).[boxed{x leq -4 text{ or } x geq 8}]

💡To compare the sizes of (-frac{3}{4}) and (-0.8), we first compare their absolute values to understand their positions relative to zero on the number line.1. Calculate the absolute values: - The absolute value of (-frac{3}{4}) is (|-frac{3}{4}| = frac{3}{4}). - The absolute value of (-0.8) is (|-0.8| = 0.8).2. Compare the absolute values: - We see that (frac{3}{4} < 0.8).3. Since a larger absolute value means a number is further from zero, and since we are dealing with negative numbers (where the number with a larger absolute value is actually smaller), we conclude that: - (-frac{3}{4} > -0.8).Therefore, the correct comparison symbol to fill in is (>).(boxed{>})

💡Okay, so I have this problem where there are (2^n) coins given to a couple of kids. The rule is that an interchange happens when one of the kids has at least half of all the coins. Then, from that kid's coins, each of the other kids gets as many coins as they already had. If all the coins are with one kid, no interchange can happen. The question is asking for the greatest possible number of consecutive interchanges.Hmm, let me try to understand this step by step. First, let's break down the problem. We have (2^n) coins, which is a power of two. That might be important because powers of two often relate to binary systems or something that can be split evenly multiple times.So, an interchange occurs when a kid has at least half of all the coins. Since the total number of coins is (2^n), half of that would be (2^{n-1}). So, if any kid has (2^{n-1}) or more coins, an interchange can happen.When an interchange happens, the kid with at least half the coins gives to each of the other kids as many coins as they already have. So, if there are (k) kids, and one kid has (c) coins (where (c geq 2^{n-1})), then each of the other (k-1) kids will receive (c_i) coins, where (c_i) is the number of coins they already have. Wait, so if a kid has (c_i) coins, they receive another (c_i) coins from the kid with the most. That means their amount doubles. So, after the interchange, the kid who gave away coins will have (c - sum_{i=1}^{k-1} c_i) coins left, and each of the other kids will have (2c_i) coins.But hold on, the total number of coins is (2^n), so the kid who gave away coins must have given away exactly (2^n - c) coins because the total is fixed. So, (c - (2^n - c) = 2c - 2^n) coins remain with the giver. Wait, that doesn't seem right. Let me think again. If the total number of coins is (2^n), and the giver has (c) coins, then the rest of the kids have (2^n - c) coins in total. When the giver gives each of the other kids as many coins as they already have, that means each kid (i) gets (c_i) coins, so the total given away is (sum_{i=1}^{k-1} c_i = 2^n - c). Therefore, the giver is left with (c - (2^n - c) = 2c - 2^n) coins.So, the giver's coins after the interchange are (2c - 2^n), and each of the other kids has (2c_i) coins. This seems like a transformation on the distribution of coins. Maybe we can model this as some kind of linear transformation or matrix operation?But before getting too abstract, let's try a small example to see how this works. Let's take (n=1), so (2^1 = 2) coins. Suppose we have two kids. If each kid has 1 coin, then neither has at least half (which is 1 coin). Wait, actually, each has exactly half. So, does that count? The problem says "at least half," so yes, each has at least half. So, an interchange can happen.But if each kid has 1 coin, and we perform an interchange, what happens? The kid with 1 coin gives 1 coin to the other kid. But wait, both have 1 coin, so if we pick one kid to give, they give 1 coin to the other. Then the giver has 0 coins, and the receiver has 2 coins. Then, in the next step, the kid with 2 coins has all the coins, so no more interchanges can happen. So, for (n=1), we can have 1 interchange.Wait, but the problem says "a couple of kids," which might imply more than two kids? Or is two kids acceptable? The problem doesn't specify the number of kids, just that it's a couple. So, maybe two is fine.Let's try (n=2), so (2^2 = 4) coins. Let's say we have two kids again. If each has 2 coins, then each has exactly half. So, an interchange can happen. If one kid gives 2 coins to the other, then the giver has 0 and the receiver has 4. Then, no more interchanges. So, only 1 interchange.But maybe with more kids, we can have more interchanges. Let's try three kids. Suppose we have three kids with 2, 1, and 1 coins. The kid with 2 coins has half of the total (which is 4 coins). So, that kid can give 1 coin to each of the other two kids. After the interchange, the giver has 2 - 1 - 1 = 0 coins, and the other two kids have 2 and 2 coins. Now, each of those two kids has 2 coins, which is half of the total. So, another interchange can happen.If we pick one of the kids with 2 coins, they give 2 coins to the other kid. Wait, but there are two kids with 2 coins. If we pick one, they give 2 coins to the other kid, but the other kid already has 2 coins, so they would receive 2 coins, making them have 4 coins, and the giver has 0. Then, no more interchanges. So, that's only 2 interchanges.Alternatively, maybe the first interchange leads to a different distribution. Let me check again.Starting with 2, 1, 1. The kid with 2 coins gives 1 to each of the other two, resulting in 0, 2, 2. Then, each of the two kids with 2 coins can give 2 coins to the other. But if we do that, one kid goes from 2 to 0, and the other goes from 2 to 4. So, only 2 interchanges.Is there a way to have more interchanges? Maybe with a different initial distribution.Suppose we have three kids with 3, 1, 0 coins. The kid with 3 coins has more than half (which is 2 coins). So, they can give 1 coin to the kid with 1 coin and 0 coins to the kid with 0 coins. Wait, but the rule says "to all other kids are given that much coins as the kid already had." So, the kid with 3 coins gives 1 coin to the kid with 1 coin (since that kid had 1 coin) and 0 coins to the kid with 0 coins (since that kid had 0 coins). So, after the interchange, the distribution is 3 - 1 - 0 = 2, 2, 0.Now, the two kids with 2 coins each have exactly half. So, another interchange can happen. If we pick one of them, they give 2 coins to the other kid, resulting in 0, 4, 0. Then, no more interchanges. So, again, only 2 interchanges.Hmm, seems like with three kids, we can get 2 interchanges. Maybe with more kids, we can get more?Let's try four kids. Suppose we have four kids with 2, 1, 1, 0 coins. The kid with 2 coins has half of the total (which is 4 coins). So, they can give 1 coin to each of the other three kids. Wait, but the rule says "to all other kids are given that much coins as the kid already had." So, the kid with 2 coins gives 1 coin to the kid with 1 coin, 1 coin to another kid with 1 coin, and 0 coins to the kid with 0 coins. So, after the interchange, the distribution is 2 - 1 - 1 - 0 = 0, 2, 2, 0.Now, the two kids with 2 coins each have exactly half. So, another interchange can happen. If we pick one of them, they give 2 coins to the other kid, resulting in 0, 0, 4, 0. Then, no more interchanges. So, only 2 interchanges again.Wait, maybe I need a different initial distribution. Let's try starting with one kid having 3 coins and the others having 1 each. So, 3, 1, 1, 1. The kid with 3 coins has more than half (which is 2 coins). So, they give 1 coin to each of the other three kids. After the interchange, the distribution is 3 - 1 - 1 - 1 = 0, 2, 2, 2. Now, each of the three kids has 2 coins, which is half of the total. So, another interchange can happen.If we pick one of the kids with 2 coins, they give 2 coins to each of the other two kids. Wait, but the rule says "to all other kids are given that much coins as the kid already had." So, the kid with 2 coins gives 2 coins to each of the other two kids. But each of those kids already has 2 coins, so they would receive 2 coins each, making them have 4 coins each. But the total coins would be 2 (from the giver) + 4 + 4 = 10, which is more than 4. That can't be right.Wait, no, the total number of coins is fixed at 4. So, if the giver has 2 coins and gives 2 coins to each of the other two kids, that would require giving away 4 coins, but the giver only has 2 coins. So, that's not possible. Maybe I misunderstood the rule.Let me read it again: "from the coins of one of those kids to the all other kids are given that much coins as the kid already had." So, the giver gives to each other kid as many coins as that kid already had. So, if the giver has (c) coins, and the other kids have (c_1, c_2, ..., c_{k-1}) coins, then the giver gives (c_1) coins to the first kid, (c_2) coins to the second, etc. So, the total given away is (c_1 + c_2 + ... + c_{k-1}), and the giver must have at least that much.In the case where we have 0, 2, 2, 2, the kid with 2 coins can give 2 coins to each of the other two kids. But wait, the other two kids each have 2 coins, so the giver would have to give 2 coins to each, totaling 4 coins. But the giver only has 2 coins, so that's not possible. Therefore, maybe the giver can't perform the interchange in this case.Wait, but the giver has 2 coins, and the other kids have 2 each. So, the giver needs to give 2 coins to each of the other two kids, but that would require 4 coins, which the giver doesn't have. Therefore, the interchange can't happen. So, in this case, after the first interchange, we have 0, 2, 2, 2, and no further interchanges can happen because the giver doesn't have enough coins to give each kid as many as they already have.So, in this case, only 1 interchange happened. Hmm, that's worse than before.Maybe I need a different approach. Instead of trying specific examples, let's think about the problem more abstractly.We have (2^n) coins, and an interchange happens when a kid has at least (2^{n-1}) coins. When that happens, the kid gives each of the other kids as many coins as they already have. So, if a kid has (c) coins, and the other kids have (c_1, c_2, ..., c_{k-1}), then the giver gives (c_1) to the first kid, (c_2) to the second, etc., totaling (c_1 + c_2 + ... + c_{k-1}) coins given away. Therefore, the giver must have at least (c_1 + c_2 + ... + c_{k-1}) coins.But since the total number of coins is (2^n), the giver has (c) coins, and the rest have (2^n - c) coins. So, (c geq 2^{n-1}), and (c geq 2^n - c), which implies (c geq 2^{n-1}).After the interchange, the giver has (c - (2^n - c) = 2c - 2^n) coins, and each of the other kids has (2c_i) coins.So, the new distribution is (2c - 2^n) for the giver and (2c_i) for each of the other kids.Now, let's think about what happens in terms of binary representations or something related to powers of two.If we start with one kid having (2^n - 1) coins and the rest having 1 coin each, then the kid with (2^n - 1) coins can give 1 coin to each of the other kids. After that, the giver has (2^n - 1 - (k - 1)) coins, where (k) is the number of kids. But this might not be the most efficient way to maximize the number of interchanges.Wait, maybe the key is to have the number of kids be as large as possible so that each interchange only requires giving away a small number of coins, allowing more interchanges before someone accumulates all the coins.But the problem doesn't specify the number of kids, just that it's a couple. So, maybe the number of kids can be chosen to maximize the number of interchanges.Alternatively, perhaps the number of kids is fixed, but the problem doesn't specify, so maybe we can assume the number of kids is variable, and we need to find the maximum number of interchanges possible regardless of the number of kids.Wait, the problem says "a couple of kids," which is a bit vague. It could mean two or more. Maybe the maximum number of interchanges is achieved when the number of kids is as large as possible, but since the number of coins is (2^n), the number of kids can't exceed (2^n), but that's probably too many.Alternatively, maybe the number of kids is two, as in the simplest case. Let's see.If we have two kids, starting with one kid having (2^n - 1) coins and the other having 1 coin. Then, the first kid can give 1 coin to the second kid, resulting in (2^n - 2) and 2 coins. Then, the first kid still has more than half, so another interchange can happen. The first kid gives 2 coins to the second kid, resulting in (2^n - 4) and 4 coins. This can continue until the first kid has 0 coins, which would take (2^{n-1}) interchanges. But that seems too high.Wait, let's test with (n=2). So, (2^2=4) coins. Starting with 3 and 1. First interchange: 3 gives 1 to 1, resulting in 2 and 2. Then, both have half, so another interchange: 2 gives 2 to the other, resulting in 0 and 4. So, only 2 interchanges. But (2^{n-1} = 2), which matches.For (n=3), (2^3=8) coins. Starting with 7 and 1. First interchange: 7 gives 1 to 1, resulting in 6 and 2. Second interchange: 6 gives 2 to 2, resulting in 4 and 4. Third interchange: 4 gives 4 to 4, resulting in 0 and 8. So, 3 interchanges, which is (n=3).Wait, so it seems like for two kids, the number of interchanges is (n). Let me check for (n=4). Starting with 15 and 1. First interchange: 15 gives 1 to 1, resulting in 14 and 2. Second: 14 gives 2 to 2, resulting in 12 and 4. Third: 12 gives 4 to 4, resulting in 8 and 8. Fourth: 8 gives 8 to 8, resulting in 0 and 16. So, 4 interchanges, which is (n=4).So, it seems that with two kids, the maximum number of interchanges is (n). But earlier, with three kids, I could only get 2 interchanges for (n=2), which is less than (n). So, maybe the maximum number of interchanges is achieved when there are two kids.But wait, let's think again. If we have more kids, maybe we can have more interchanges. For example, with three kids, can we have more than (n) interchanges?Let's try (n=2) with three kids. Starting with 2,1,1. First interchange: 2 gives 1 to each of the other two, resulting in 0,2,2. Then, each of the two kids with 2 coins can give 2 coins to the other, but that would require 4 coins, which they don't have. So, only 1 interchange. Hmm, less than (n=2).Alternatively, starting with 3,1,0 for (n=2). First interchange: 3 gives 1 to 1 and 0 to 0, resulting in 2,2,0. Then, each of the two kids with 2 coins can give 2 coins to the other, but again, they don't have enough. So, only 1 interchange.Wait, maybe with four kids. Starting with 2,1,1,0 for (n=2). First interchange: 2 gives 1 to each of the other two, resulting in 0,2,2,0. Then, each of the two kids with 2 coins can give 2 coins to the other, but again, they don't have enough. So, only 1 interchange.Hmm, seems like with more kids, the number of interchanges doesn't increase beyond (n). So, maybe the maximum number of interchanges is indeed (n), achieved when there are two kids.But let's think about the process. Each interchange effectively halves the number of coins the giver has, because they give away half of the total coins. Wait, no, not exactly. Let's see.When a kid has (c) coins, which is at least (2^{n-1}), they give away (2^n - c) coins, because the total is (2^n). So, the giver ends up with (2c - 2^n) coins. So, if (c = 2^{n-1}), then the giver ends up with (2*2^{n-1} - 2^n = 0). So, if a kid has exactly half, they give away all their coins.But if a kid has more than half, say (2^{n-1} + k), then they give away (2^n - (2^{n-1} + k) = 2^{n-1} - k) coins, ending up with (2*(2^{n-1} + k) - 2^n = 2^{n} + 2k - 2^n = 2k) coins.So, the number of coins the giver has after the interchange is (2k), where (k) is the amount over half.So, if we start with one kid having (2^n - 1) coins, which is just one less than the total, then (k = 2^n - 1 - 2^{n-1} = 2^{n-1} - 1). So, after the first interchange, the giver has (2*(2^{n-1} - 1) = 2^n - 2) coins. Then, the next interchange, they give away (2^n - (2^n - 2) = 2) coins, ending up with (2*(2^n - 2) - 2^n = 2^{n+1} - 4 - 2^n = 2^n - 4). Wait, that doesn't seem right.Wait, let's do it step by step for (n=3). Starting with 7 and 1.First interchange: 7 gives 1 to 1, resulting in 6 and 2.Second interchange: 6 gives 2 to 2, resulting in 4 and 4.Third interchange: 4 gives 4 to 4, resulting in 0 and 8.So, each time, the giver's coins are halved (approximately). So, 7 -> 6 -> 4 -> 0. But 6 is not exactly half of 7, but in terms of the process, it's reducing by the amount given to the other kid.Wait, maybe the number of interchanges is related to the number of times you can halve the coins before reaching zero. Since we're dealing with powers of two, it makes sense that the number of interchanges is (n).So, for (n=1), 1 interchange.For (n=2), 2 interchanges.For (n=3), 3 interchanges.And so on.Therefore, the greatest possible number of consecutive interchanges is (n).I think that's the answer. It seems consistent with the examples I tried, and the reasoning about halving the coins each time aligns with the number of interchanges being equal to (n).

💡Okay, so I've got this problem about a square cake with triangular chocolates placed on it, and they don't touch each other. The question is asking if it's always possible to cut the cake into convex polygons so that each polygon contains exactly one piece of chocolate. Hmm, interesting.First, I need to understand what a convex polygon is. A convex polygon is a shape where all its interior angles are less than 180 degrees, and every line segment between two vertices stays inside the polygon. So, shapes like triangles, squares, and rectangles are convex, but something like a star shape isn't because it has indentations.Now, the cake is a square, and the chocolates are triangular and don't touch each other. So, each chocolate is separate, and there's space between them. The goal is to cut the cake into pieces, each of which is a convex polygon, and each piece should have exactly one chocolate.I guess the first thing to consider is how the chocolates are arranged. Since they don't touch each other, there must be some space between them. Maybe I can use that space to make the cuts. But I need to make sure that each cut results in a convex shape.Let me think about a simple case first. Suppose there's only one chocolate on the cake. Then, obviously, the whole cake is a convex polygon, and it contains the one chocolate. So, that's easy.What if there are two chocolates? Let's say they're placed in opposite corners of the square. If I make a straight cut from one side of the square to the other, avoiding the chocolates, I can divide the cake into two convex polygons, each containing one chocolate. That seems doable.But what if the chocolates are placed in such a way that making a straight cut would result in one of the pieces not being convex? For example, if one chocolate is near the center and another is near the edge, maybe the cut would have to curve around the center chocolate, making the resulting piece non-convex.Wait, but the problem specifies that the chocolates don't touch each other. So, there should be enough space to make straight cuts without having to curve around them. Maybe I can use the space between the chocolates to make the necessary cuts.Another thought: perhaps I can use Voronoi diagrams. A Voronoi diagram divides a space into regions based on proximity to points. In this case, the points would be the chocolates, and each region would contain all the points closer to that chocolate than any other. Since the chocolates don't touch, the Voronoi regions would be convex polygons.But does a Voronoi diagram always result in convex polygons? I think so, because each region is defined by perpendicular bisectors between points, which are straight lines, and the intersection of half-planes defined by these lines results in convex shapes.So, if I construct a Voronoi diagram around the chocolates, each region would be a convex polygon containing exactly one chocolate. That sounds promising.But wait, the Voronoi diagram is usually applied to points, and in this case, the chocolates are triangles. Does that change anything? Maybe, because the chocolates have area, not just being points. So, the proximity regions might be different.Perhaps I need to adjust the Voronoi approach to account for the shapes of the chocolates. Instead of just considering the centers of the chocolates, I might need to consider their entire area. But I'm not sure how that would affect the convexity of the resulting regions.Alternatively, maybe I can treat each chocolate as a point and then adjust the regions to ensure that each chocolate is entirely within its region. Since the chocolates don't touch, there should be enough space to create boundaries around each chocolate that don't interfere with others.Another idea: if I can draw lines around each chocolate such that each line is equidistant from the nearest chocolates, similar to a Voronoi diagram, then each resulting region should be convex. Since the chocolates are triangles and don't touch, these equidistant lines should form convex polygons.But I'm not entirely sure if this works in all cases. What if the chocolates are arranged in a way that makes the equidistant lines intersect in a non-convex manner? Maybe if chocolates are placed very close to the edges or corners, the resulting regions could become non-convex.Wait, but the problem states that the chocolates don't touch each other. So, there must be some minimum distance between them. Maybe that ensures that the Voronoi regions remain convex.I think I need to look into the properties of Voronoi diagrams more carefully. In a plane with a set of points, the Voronoi diagram partitions the plane into regions where each region consists of all points closer to its associated point than to any other. These regions are always convex polygons or unbounded convex regions.In our case, the chocolates are not points but triangles. However, since they don't touch, we can consider their centers or some representative point within each triangle. Then, constructing a Voronoi diagram around these points should give us convex regions.But I'm still concerned about the actual shape of the chocolates affecting the convexity. Maybe if a chocolate is very large or placed in a corner, the region around it might not be convex. However, since the chocolates don't touch, their regions should have enough space to form convex shapes.Another approach: instead of using Voronoi diagrams, maybe I can use a different method to partition the cake. For example, I can draw lines from the edges of the cake towards the chocolates, ensuring that each line separates one chocolate from the others while maintaining convexity.But this might get complicated, especially if there are many chocolates. I need a more systematic way to ensure that each region is convex and contains exactly one chocolate.Going back to the Voronoi idea, I think it's a good starting point. Since the chocolates don't touch, their Voronoi regions should be well-defined and convex. Each region would contain exactly one chocolate, and the entire cake would be partitioned into these regions.However, I need to confirm if this holds true even when the chocolates are not points but have area. Maybe I need to adjust the Voronoi diagram to account for the size of the chocolates. If I treat each chocolate as a point, the regions might not fully contain the chocolates, especially if they're large.Perhaps instead of using the centers, I can use the farthest points of the chocolates from the edges. Or maybe I can offset the Voronoi diagram by the size of the chocolates to ensure that each region fully contains its chocolate.This is getting a bit complex, but I think the key idea is that as long as the chocolates don't touch, there's a way to partition the cake into convex regions around each chocolate. The exact method might involve some geometric constructions, but the principle seems sound.In conclusion, I believe that it is always possible to cut the cake into convex polygons, each containing exactly one piece of chocolate, as long as the chocolates don't touch each other. The Voronoi diagram approach, adjusted for the size and shape of the chocolates, should provide the necessary partition.

💡First, I'll define the variables:- Let ( O ) represent the number of orchestra tickets sold.- Let ( B ) represent the number of balcony tickets sold.From the problem, I have two key pieces of information:1. The total number of tickets sold is 370:[ O + B = 370 ]2. The total cost of the tickets is 3,320:[ 12O + 8B = 3,320 ]To solve this system of equations, I'll start by multiplying the first equation by 8 to align the coefficients of ( B ):[ 8O + 8B = 2,960 ]Now, I'll subtract this new equation from the second equation to eliminate ( B ):[ (12O + 8B) - (8O + 8B) = 3,320 - 2,960 ][ 4O = 360 ]Solving for ( O ):[ O = frac{360}{4} = 90 ]With ( O ) known, I'll substitute it back into the first equation to find ( B ):[ 90 + B = 370 ][ B = 370 - 90 = 280 ]Finally, to determine how many more balcony tickets were sold than orchestra tickets:[ B - O = 280 - 90 = 190 ]

💡Okay, so I'm trying to figure out if there's a non-convex polygon where the opposite sides are pairwise parallel and equal. Hmm, let's start by recalling what a convex polygon is. A convex polygon is one where all the interior angles are less than 180 degrees, and every line segment between two vertices stays inside the polygon. On the other hand, a non-convex polygon has at least one interior angle greater than 180 degrees, and at least one line segment between two vertices goes outside the polygon.Now, the question is about a polygon where opposite sides are both parallel and equal in length. I know that in a convex polygon, if opposite sides are both parallel and equal, it's called a parallelogram. But the question is specifically about a non-convex polygon. So, can a non-convex polygon have this property?Let me think about some examples. The simplest polygon is a triangle, but it doesn't have opposite sides. A quadrilateral is the next one. In a convex quadrilateral, if opposite sides are parallel and equal, it's a parallelogram. But what about a non-convex quadrilateral?I remember that a non-convex quadrilateral can be self-intersecting, like a bowtie shape. Maybe that's a candidate. If I can construct a self-intersecting quadrilateral where opposite sides are parallel and equal, that would answer the question.Let me try to visualize this. Suppose I have a quadrilateral with vertices A, B, C, D. If I arrange them in such a way that AB is parallel and equal to CD, and AD is parallel and equal to BC, but the quadrilateral is self-intersecting, then it might work.Wait, but in a self-intersecting quadrilateral, the sides cross each other. Does that affect the parallelism or equality of the sides? I think the sides can still be parallel and equal even if the polygon is self-intersecting. The key is just the relationship between the sides, not the overall shape.Maybe I can assign coordinates to the vertices to make this more concrete. Let's say point A is at (0,0). If AB is parallel and equal to CD, and AD is parallel and equal to BC, I need to choose points B, C, and D such that these conditions are satisfied.Let me try assigning coordinates:- Let A be at (0,0).- Let B be at (1,0).- Then, since AB is parallel and equal to CD, CD should also be a horizontal line segment of length 1.- Let's place C at (2,1).- Then D should be at (1,1) to make CD parallel and equal to AB.- Now, check if AD is parallel and equal to BC.- AD is from (0,0) to (1,1), which has a slope of 1 and length √2.- BC is from (1,0) to (2,1), which also has a slope of 1 and length √2.- So, AD is parallel and equal to BC.But wait, is this quadrilateral non-convex? Let me plot these points:- A(0,0), B(1,0), C(2,1), D(1,1).Connecting A to B to C to D to A. Hmm, this seems to form a convex quadrilateral, actually. Because all the interior angles are less than 180 degrees. So, this doesn't work.Maybe I need to arrange the points differently to make it non-convex. Let's try another set of coordinates.- Let A be at (0,0).- Let B be at (1,1).- Let C be at (2,0).- Let D be at (1,-1).Now, let's check the sides:- AB is from (0,0) to (1,1), which has a slope of 1 and length √2.- CD is from (2,0) to (1,-1), which also has a slope of 1 and length √2.- AD is from (0,0) to (1,-1), which has a slope of -1 and length √2.- BC is from (1,1) to (2,0), which has a slope of -1 and length √2.So, AB is parallel and equal to CD, and AD is parallel and equal to BC. Now, is this quadrilateral non-convex?Plotting these points:- A(0,0), B(1,1), C(2,0), D(1,-1).Connecting A to B to C to D to A. This shape crosses itself at the center, forming a bowtie or hourglass shape. So, it's a self-intersecting quadrilateral, which is non-convex.Therefore, this quadrilateral satisfies the conditions: it's non-convex, and its opposite sides are pairwise parallel and equal.So, yes, such a polygon exists.