💡Okay, so I have this problem where I need to prove that given a set of n points (where n is at least 3) that are not all lying on a straight line, there exists a convex polygon that contains all these points, and the vertices of this polygon are exactly the given points. Hmm, interesting. Let me try to break this down step by step.First, I know that a convex polygon is a polygon where all its interior angles are less than 180 degrees, and every line segment between two vertices stays inside the polygon. So, if I can form such a polygon using the given points, that would solve the problem.Since the points are not all collinear, that means there's at least one triangle that can be formed by three of these points. A triangle is always a convex polygon, so that gives me a starting point. But I need to make sure that all the points are inside or on the boundary of this polygon.Maybe I can use induction here. Let me try that approach. For the base case, when n=3, since the points are not collinear, they form a triangle, which is convex. So, the base case holds.Now, assuming that for some k >= 3, any set of k non-collinear points has a convex polygon containing all of them with vertices among the points. I need to show that this holds for k+1 points.So, take a set S of k+1 points. If I can find a convex polygon for k points, then adding the (k+1)th point should still keep the polygon convex. But how?Wait, maybe I need to consider the convex hull of the set S. The convex hull is the smallest convex polygon that contains all the points. The vertices of the convex hull are a subset of the given points. So, if I can show that the convex hull of S is a polygon with vertices among the given points, then I'm done.But how do I know that the convex hull isn't just a line or something? Well, since the points are not all collinear, the convex hull must have at least three vertices, forming a triangle or a larger polygon.Let me think about how the convex hull is constructed. It starts by finding the point with the lowest y-coordinate (or some other method) and then proceeds to build the hull by checking each point to see if it makes a non-left turn. Points that make a left turn are kept, and those that don't are discarded.So, in this process, only the points that are on the "outer edge" of the set are included as vertices of the convex hull. The points inside the hull are not vertices but are still contained within the polygon.Therefore, the convex hull of S is a convex polygon whose vertices are among the given points, and it contains all the points. That seems to fit the requirement.But wait, is the convex hull always a polygon? Yes, because with at least three non-collinear points, the convex hull will have at least three vertices, forming a triangle or more sides if there are more points on the hull.So, putting it all together, given any set of n >= 3 non-collinear points, their convex hull is a convex polygon with vertices among the points, and it contains all the points. Therefore, such a convex polygon exists.I think that makes sense. The key idea is using the concept of the convex hull, which naturally forms a convex polygon from a set of points, ensuring all points are either on the hull or inside it. Since the points are not all collinear, the hull will have enough vertices to form a proper polygon.Maybe I should also consider if there's a case where adding a point could somehow make the hull non-convex, but I don't think so because the convex hull algorithm ensures that all turns are non-left, maintaining convexity.Alright, I feel confident that this approach works. The convex hull serves as the required convex polygon, and its vertices are indeed among the given points.

💡First, I need to round the number 3.8963 to the nearest hundredth. To do this, I'll look at the digit in the thousandths place, which is 6.Since 6 is greater than or equal to 5, I'll round up the digit in the hundredths place from 9 to 10. This means the number in the tenths place will increase by 1 as well.Finally, after rounding, the number becomes 3.90.

💡First, I need to select a student to serve as the Mathematics class representative. Since student A cannot take this position, I have 4 other students to choose from.Next, I need to select 2 students from the remaining 4 students (including student A) to serve as the Chinese and English class representatives.For each of the 4 choices for the Mathematics representative, there are 4 students left for the Chinese representative and 3 students left for the English representative.Therefore, the total number of different selection methods is 4 multiplied by 4 multiplied by 3, which equals 48.

💡First, I'll denote Sophia's income as ( S ).Juan's income is 70% of Sophia's income:[J = 0.70S]Tim's income is 40% less than Juan's income, which means it's 60% of Juan's income:[T = 0.60J = 0.60 times 0.70S = 0.42S]Mary's income is 50% more than Tim's income:[M = 1.50T = 1.50 times 0.42S = 0.63S]To find what percent of Juan's income is Mary's income, I'll set up the equation:[M = text{percent of } J times J][0.63S = text{percent of } J times 0.70S]Solving for the percent:[text{percent of } J = frac{0.63S}{0.70S} = 0.9][text{percent of } J = 0.9 times 100 = 90%]

💡First, calculate the total number of sea stars and seashells Harry initially collected.Next, determine the total number of sea creatures Harry had before losing any.Then, find out how many sea creatures Harry had left after losing 25 of them.Finally, subtract the remaining sea creatures from the total items left to find the number of snails Harry collected initially.

💡Alright, so I've been given this problem about Poncelet's Theorem, and I need to prove it using the result from the previous problem. Let me try to understand what exactly Poncelet's Theorem is saying. It states that if there exists one n-gon inscribed in a given circle α and circumscribed around another circle β, then there are infinitely many such n-gons. Moreover, any point on the circle α can be taken as one of the vertices of such an n-gon.Okay, so first, let me visualize this. There's a circle α, and inside it, there's another circle β. If I can find one polygon that touches both circles—meaning all its vertices lie on α, and all its sides are tangent to β—then Poncelet's Theorem says I can actually find infinitely many such polygons. That's pretty cool, but how do I prove it?The problem mentions using the result from the previous problem. I don't have the exact statement of the previous problem, but I can assume it's related to Poncelet's Theorem or some stepping stone towards it. Maybe it was about the existence of such polygons under certain conditions or some property that can be iterated to generate more polygons.Alright, let's start by recalling what Poncelet's Theorem is about. It's a result in projective geometry, and it's often demonstrated with an ellipse and a circle, but in this case, both are circles. The theorem essentially says that if you can draw one polygon that is both inscribed in one conic (here, circle α) and circumscribed around another conic (here, circle β), then you can draw infinitely many such polygons.I think the key idea here is to use some kind of iterative process or transformation that can generate new polygons from an existing one. Maybe something like rotating the polygon or shifting its vertices in a controlled manner while maintaining the tangency condition with circle β.Let me try to break it down step by step.1. **Assumption**: There exists at least one n-gon inscribed in α and circumscribed around β. Let's denote this polygon as P1, P2, ..., Pn.2. **Goal**: Show that for any point Q on circle α, there exists an n-gon Q, Q1, Q2, ..., Qn-1 inscribed in α and circumscribed around β.3. **Idea**: If I can show that starting from any point Q on α, I can construct such an n-gon, then since there are infinitely many points on α, there are infinitely many such n-gons.But how do I construct the rest of the polygon from Q? I need to ensure that each side is tangent to β. That seems tricky. Maybe I can use some property of tangents and circles.Wait, if I have a point Q on α, I can draw two tangents from Q to β. These tangents will touch β at two points, say T1 and T2. But since the polygon is circumscribed around β, each side must be tangent to β at exactly one point. So, for each vertex, the two adjacent sides must be tangents from that vertex to β.Hmm, so starting from Q, I can draw two tangents to β, and each tangent will correspond to a side of the polygon. But how do I ensure that the entire polygon closes back to Q after n sides?This sounds like it might involve some kind of closure condition. Maybe if I can show that after n steps, the process brings me back to the starting point, then the polygon is closed.I think this is where the previous problem's result comes into play. Perhaps the previous problem established that if one such polygon exists, then the closure condition holds for any starting point, ensuring that the polygon can be completed.Let me try to formalize this.Suppose I start at a point Q on α. From Q, I draw a tangent to β, which touches β at some point T1. Then, from T1, I draw another tangent to β, which touches at T2, and so on, until after n steps, I should end up back at Q.But wait, that might not necessarily bring me back to Q unless certain conditions are met. This is where the previous problem's result is crucial. If the previous problem showed that the mapping from a point on α to the next point after drawing a tangent to β is a bijection or has some periodicity, then iterating this process n times would bring me back to the starting point.Alternatively, maybe the previous problem established that the number of such polygons is infinite by showing that there's a continuous family of them parameterized by the starting point on α.Let me think about the properties of tangents to circles. For a given circle β, the set of points from which tangents can be drawn to β forms another circle, called the director circle. But in this case, all our vertices lie on α, so α must be the director circle of β. Wait, is that necessarily true?No, not exactly. The director circle of β consists of all points from which the tangents to β are perpendicular. But in our case, the vertices lie on α, and the sides are tangents to β, but the angles between the sides aren't necessarily right angles.Hmm, maybe I'm overcomplicating it. Let's think about the dual problem. If I have a polygon circumscribed around β, its dual polygon (in some sense) is inscribed in α. But I'm not sure if that helps directly.Another approach could be to use complex numbers or coordinate geometry to parametrize the points on α and enforce the tangency condition. But that might get too involved without knowing the exact setup.Wait, maybe inversion could help here. Inversion is a powerful tool in circle geometry. If I invert the figure with respect to a circle, perhaps I can transform the problem into something more manageable.But I don't know if that's necessary. Let's stick to more elementary methods.Suppose I fix circle α and circle β inside it. Given a point Q on α, I can draw two tangents to β. Let's say these tangents touch β at points T1 and T2. Now, from T1, I can draw another tangent to β, which will touch at T3, and so on. If I can show that after n such steps, I end up back at Q, then the polygon Q, T1, T2, ..., Tn is both inscribed in α and circumscribed around β.But how do I ensure that after n steps, I return to Q? This seems to hinge on some periodicity or symmetry in the configuration.Alternatively, maybe I can use the fact that the existence of one such polygon implies that the configuration is "closed" in some sense, allowing for the generation of infinitely many polygons by rotating or otherwise transforming the initial one.Wait, if I can rotate the initial polygon around the center of α, keeping it inscribed in α and circumscribed around β, then each rotated position would give a new polygon. But does rotation preserve the tangency condition?Hmm, if β is concentric with α, then rotating the polygon would indeed preserve the tangency, because all points would just be rotated versions, maintaining equal distances from the center. But if β is not concentric, rotation might disrupt the tangency.So, concentricity might be a special case where this works, but the theorem is more general, allowing β to be any circle inside α.Therefore, rotation might not be the right approach unless β is concentric with α.Maybe instead of rotation, I can use some kind of projective transformation or a specific mapping that preserves the tangency condition.Alternatively, perhaps the key lies in the number of solutions to the tangency condition. If there's one solution, there might be infinitely many due to the nature of the equations involved.Wait, let's think algebraically. Suppose I parametrize the points on α using an angle θ. Then, the condition that the side from θ to θ' is tangent to β would impose some relationship between θ and θ'. If I can express this relationship as a function, then iterating this function n times should bring me back to the starting angle.If this function is such that iterating it n times is the identity function, then starting from any θ would give a closed polygon. But I need to ensure that this function is well-defined and that iterating it n times indeed brings me back.This seems plausible, but I need to formalize it.Alternatively, maybe I can use the concept of a Poncelet transverse, which is a polygon that is both inscribed in one conic and circumscribed around another. The theorem states that if one such polygon exists, then infinitely many do, and they can start at any point on the outer conic.In our case, the outer conic is circle α, and the inner conic is circle β.I recall that Poncelet's Theorem can be proven using the concept of projective invariance or by using complex analysis, but I'm not sure about the exact steps.Wait, maybe I can use induction. Suppose I can show that if there's one n-gon, then there's another one starting from a different point, and so on, leading to infinitely many.But induction might not directly apply here since we need to show the existence for any starting point, not just building up from one.Another idea: If I can show that the set of all such n-gons is in bijection with the points on α, then since α is a circle (and hence has infinitely many points), there would be infinitely many n-gons.But to establish this bijection, I need a way to uniquely associate each point on α with an n-gon starting at that point.This brings me back to the earlier idea of starting at a point Q on α, drawing tangents to β, and constructing the polygon step by step. If this process can be completed for any Q, then we have our bijection.But how do I ensure that the polygon closes after n steps? That seems to require some condition on the initial configuration.Wait, maybe the previous problem established that if one polygon exists, then the closure condition holds for any starting point. So, using that result, I can argue that starting from any Q, the polygon will close after n steps, giving me the desired n-gon.If that's the case, then since there are infinitely many Q on α, there are infinitely many such n-gons.Okay, so putting it all together:1. Assume there exists one n-gon inscribed in α and circumscribed around β.2. Use the result from the previous problem to argue that starting from any point Q on α, the process of drawing tangents to β and constructing the polygon will close after n steps, forming a valid n-gon.3. Since there are infinitely many points Q on α, there are infinitely many such n-gons.4. Therefore, Poncelet's Theorem is proven.I think this makes sense, but I'm not entirely sure if I've covered all the necessary steps. Maybe I need to be more precise about how the previous problem's result ensures the closure condition for any starting point.Perhaps the previous problem showed that the mapping from a point on α to the next point after drawing a tangent to β is a bijection with finite order, meaning that after n applications, it returns to the starting point. If that's the case, then starting from any point, applying this mapping n times would close the polygon, giving the desired result.Alternatively, the previous problem might have established that the set of such polygons forms a continuous family, parameterized by the starting point on α, which would also imply infinitely many polygons.In any case, combining the existence of one polygon with the result from the previous problem, which likely ensures the existence of infinitely many by some iterative or parameterization method, allows us to conclude that there are indeed infinitely many n-gons inscribed in α and circumscribed around β, with any point on α serving as a vertex.I think I've got a rough idea, but I might need to look up the exact statement of the previous problem to make sure I'm applying it correctly. However, based on what I know about Poncelet's Theorem and the reasoning above, this seems like a plausible path to the proof.