💡Okay, so I have this problem about cutting a cube with edge length 1 into parts using four planes. The goal is to make sure that in each part, the distance between any two points is less than a certain value, either 4/5 or 4/7. Hmm, interesting. I need to figure out if it's possible for both parts a) and b).Let me start with part a). The question is whether we can cut the cube into parts with four planes such that each part has all points within less than 4/5 units apart. First, I should visualize the cube. It's a unit cube, so each edge is 1 unit long. The maximum distance within the cube is the space diagonal, which is sqrt(3) ≈ 1.732. But we want each part to have a maximum distance less than 4/5, which is 0.8. So, we need to partition the cube into smaller pieces where each piece is "small enough."Four planes can divide the cube into up to 16 parts, but I don't think we need that many. Maybe fewer parts would suffice if each is small enough. How can I use four planes to cut the cube? Planes can be oriented in different ways—maybe along the axes or diagonally. If I use planes parallel to the faces, I can divide the cube into smaller rectangular prisms. Let me think about dividing each edge into equal parts. If I divide each edge into three parts, each part would be 1/3 ≈ 0.333. Then, the space diagonal of each small prism would be sqrt((1/3)^2 + (1/3)^2 + (1/3)^2) = sqrt(1/9 + 1/9 + 1/9) = sqrt(1/3) ≈ 0.577, which is less than 4/5. But wait, I only used three planes for each axis, but the problem allows four planes. Maybe I can do a better division.Alternatively, if I divide each edge into two parts, each part is 1/2. The space diagonal would be sqrt((1/2)^2 + (1/2)^2 + (1/2)^2) = sqrt(3/4) ≈ 0.866, which is more than 4/5. So that's too big.Hmm, maybe a combination. If I divide some edges into three parts and others into two. Let's say I divide one edge into three parts and the others into two. Then, the resulting prisms would have dimensions 1/3, 1/2, 1/2. The space diagonal would be sqrt((1/3)^2 + (1/2)^2 + (1/2)^2) = sqrt(1/9 + 1/4 + 1/4) = sqrt(1/9 + 2/4) = sqrt(1/9 + 1/2) = sqrt(11/18) ≈ 0.784, which is less than 4/5 (0.8). So, if I use four planes: two planes along one axis to divide it into three parts, and one plane along each of the other two axes to divide them into two parts each. That would give me 3*2*2 = 12 parts, each with dimensions 1/3, 1/2, 1/2, and each with a space diagonal of approximately 0.784, which is less than 0.8. Therefore, for part a), it seems possible. Now, moving on to part b). The question is whether we can cut the cube into parts with four planes such that each part has all points within less than 4/7 ≈ 0.571 units apart.Using the same approach as part a), if I divide the cube into 12 parts with dimensions 1/3, 1/2, 1/2, the space diagonal is approximately 0.784, which is larger than 4/7. So, that doesn't work. Is there a way to make the parts smaller? Maybe by using more divisions, but we are limited to four planes. Each plane can potentially double the number of parts, but with four planes, the maximum number of parts is 16. However, arranging four planes optimally might not necessarily give 16 parts because some planes could intersect in a way that doesn't maximize the number of regions.Alternatively, maybe using non-axis-aligned planes? For example, cutting along space diagonals or face diagonals. But I'm not sure how that would affect the maximum distance within each part.Wait, another idea. If I use four planes to create smaller cubes or smaller rectangular prisms. For instance, if I divide each edge into four parts, each part would be 1/4. The space diagonal would be sqrt(3*(1/4)^2) = sqrt(3)/4 ≈ 0.433, which is less than 4/7. But to divide each edge into four parts, I would need three planes per axis, totaling nine planes, which is more than four. So that's not feasible.Alternatively, maybe a combination of divisions. If I divide one edge into four parts, that would require three planes, leaving one plane for the other edges. But then, the resulting parts would have dimensions 1/4, 1, 1, which is a long prism, and the space diagonal would be sqrt((1/4)^2 + 1^2 + 1^2) = sqrt(1/16 + 1 + 1) = sqrt(33/16) ≈ 1.46, which is way too big.Hmm, maybe another approach. If I use four planes to create smaller regions, perhaps not all axis-aligned. For example, using planes that cut through the cube diagonally. But I'm not sure how that would work. Maybe if I use planes that are not parallel to any face, but intersecting inside the cube.Wait, another thought. If I use four planes to create a sort of grid inside the cube, but not necessarily along the axes. For example, if I use two planes along the x-axis, two along the y-axis, and none along the z-axis, but that would still be four planes. But then, the resulting parts would be 3x3x1, which is 9 parts, each with dimensions 1/3, 1/3, 1. The space diagonal would be sqrt((1/3)^2 + (1/3)^2 + 1^2) = sqrt(1/9 + 1/9 + 1) = sqrt(11/9) ≈ 1.105, which is still too big.Alternatively, if I use four planes to create smaller regions by intersecting them in the middle. For example, using four planes that each cut the cube into halves along different axes. But that would create 16 smaller cubes, each with edge length 1/2, but as we saw earlier, the space diagonal is about 0.866, which is too big.Wait, but if I use four planes, maybe not all orthogonal. For example, using four planes that are not all aligned with the cube's edges. Maybe using planes that are at 45 degrees or something. But I'm not sure how that would affect the maximum distance within each part.Alternatively, maybe using four planes to create a sort of 3D cross inside the cube, but I'm not sure how that would partition the cube.Wait, another idea. If I use four planes to divide the cube into smaller regions, each region being a tetrahedron or something. But I'm not sure if that would result in smaller maximum distances.Alternatively, maybe using four planes to create smaller rectangular prisms with smaller dimensions. For example, if I divide the cube into 8 smaller cubes with edge length 1/2, but that would require three planes along each axis, totaling nine planes, which is more than four.Hmm, this is tricky. Maybe it's not possible to get all parts to have a maximum distance less than 4/7 with only four planes. Because even if I divide the cube into 12 parts as in part a), the maximum distance is about 0.784, which is still larger than 4/7 ≈ 0.571.Wait, but maybe there's a smarter way to arrange the four planes to create smaller regions. For example, using four planes that intersect in such a way that each region is a smaller polyhedron with all points within 4/7.Alternatively, maybe using four planes to create a grid that divides the cube into smaller cells, each with dimensions that result in a space diagonal less than 4/7.Let me calculate what dimensions would be needed for the space diagonal to be less than 4/7. Suppose we have a rectangular prism with dimensions a, b, c. Then, the space diagonal is sqrt(a^2 + b^2 + c^2). We need sqrt(a^2 + b^2 + c^2) < 4/7 ≈ 0.571.So, a^2 + b^2 + c^2 < (4/7)^2 ≈ 0.595.Since the cube has edge length 1, we need to divide it into smaller prisms where the sum of the squares of their dimensions is less than 0.595.If we divide each edge into n parts, then each dimension is 1/n, and the space diagonal is sqrt(3*(1/n)^2) = sqrt(3)/n. We need sqrt(3)/n < 4/7, so n > sqrt(3)*7/4 ≈ 3.03. So, n needs to be at least 4. But dividing each edge into four parts would require three planes per axis, totaling nine planes, which is more than four.Alternatively, maybe dividing edges into different numbers of parts. For example, divide one edge into four parts, and the others into two parts. Then, the space diagonal would be sqrt((1/4)^2 + (1/2)^2 + (1/2)^2) = sqrt(1/16 + 1/4 + 1/4) = sqrt(1/16 + 2/4) = sqrt(1/16 + 1/2) = sqrt(9/16) = 3/4 = 0.75, which is still larger than 4/7.Alternatively, divide one edge into three parts, and the others into two parts. Then, the space diagonal is sqrt((1/3)^2 + (1/2)^2 + (1/2)^2) ≈ 0.784, which is still larger than 4/7.Hmm, maybe if I divide two edges into three parts and one edge into two parts. Then, the space diagonal would be sqrt((1/3)^2 + (1/3)^2 + (1/2)^2) = sqrt(1/9 + 1/9 + 1/4) = sqrt(2/9 + 1/4) = sqrt(8/36 + 9/36) = sqrt(17/36) ≈ 0.702, still larger than 4/7.Alternatively, divide all three edges into three parts. Then, each dimension is 1/3, and the space diagonal is sqrt(3*(1/3)^2) = sqrt(1/3) ≈ 0.577, which is just slightly larger than 4/7 ≈ 0.571. So, that's very close.But to divide each edge into three parts, we need two planes per axis, totaling six planes, which is more than four. So, we can't do that with only four planes.Wait, but maybe if we don't divide all edges into three parts, but just some of them. For example, divide one edge into three parts, and the others into two parts. Then, as before, the space diagonal is about 0.784, which is too big.Alternatively, if we divide two edges into three parts and one edge into two parts, as before, the space diagonal is about 0.702, still too big.Hmm, maybe another approach. Instead of dividing edges into equal parts, maybe use planes that are not axis-aligned. For example, use planes that cut through the cube diagonally, creating smaller regions with smaller maximum distances.But I'm not sure how to calculate the maximum distance in such regions. It might be complicated.Alternatively, maybe using four planes to create a sort of 3D grid that divides the cube into smaller regions, each with dimensions that result in a space diagonal less than 4/7.Wait, let me think about the maximum number of regions four planes can create. The maximum number of regions created by n planes in 3D is given by the formula R(n) = (n^3 + 5n + 6)/6. For n=4, R(4) = (64 + 20 + 6)/6 = 90/6 = 15. So, four planes can create up to 15 regions. But we need to see if we can arrange four planes such that each region has a space diagonal less than 4/7.But arranging four planes optimally to create regions with such small maximum distances might not be straightforward. It might require some specific configuration.Alternatively, maybe it's not possible. Because even if we divide the cube into 15 regions, each region would still have a significant portion of the cube, and the maximum distance within each region might still exceed 4/7.Wait, let me think about the volume. The cube has volume 1. If we divide it into 15 regions, each region would have volume 1/15 ≈ 0.0667. The maximum distance within a region would depend on its shape, but even a small volume doesn't necessarily mean a small maximum distance.For example, a very thin rectangular prism with volume 1/15 could have a long diagonal even if its volume is small.So, maybe it's not possible to ensure that all regions have a maximum distance less than 4/7 with only four planes.Alternatively, maybe using four planes to create smaller regions by intersecting them in a way that each region is a smaller cube or a smaller rectangular prism with dimensions that result in a space diagonal less than 4/7.But as we saw earlier, dividing each edge into four parts would require nine planes, which is too many. Dividing into three parts requires six planes. So, with four planes, we can't get the necessary divisions to make the space diagonal less than 4/7.Therefore, maybe it's not possible for part b).Wait, but let me double-check. Maybe there's a way to arrange four planes such that each region is a tetrahedron or some other shape with smaller maximum distances.But I'm not sure. The problem specifies that the planes are introduced simultaneously, and the cube isn't moved. So, we have to find a static arrangement of four planes that partition the cube into regions each with maximum distance less than 4/7.Alternatively, maybe using four planes to create a sort of 3D grid that divides the cube into smaller regions, each with dimensions that result in a space diagonal less than 4/7.But I'm not sure how to calculate that without knowing the exact configuration.Wait, another idea. If I use four planes to create a sort of "cross" inside the cube, dividing it into smaller regions. For example, using two planes along the x-axis and two along the y-axis, creating a grid of 3x3x1, but as before, the space diagonal is too big.Alternatively, using four planes to create smaller regions by intersecting them in the middle. For example, using four planes that each cut the cube into halves along different axes, but that would create 16 regions, each with edge length 1/2, which is too big.Hmm, I'm stuck. Maybe it's not possible to achieve part b) with only four planes.Wait, let me think about the minimal number of planes needed to achieve a certain maximum distance. For part a), we could do it with four planes by dividing one edge into three parts and the others into two, resulting in 12 regions each with space diagonal less than 4/5. For part b), we need a smaller maximum distance, so we need more divisions, but we are limited to four planes.Therefore, it's likely that part b) is not possible with only four planes.So, my conclusion is:a) Yes, it's possible.b) No, it's not possible.

💡Okay, so I have this problem where every two players play each other once, and each game results in a win for one player. For each player ( n ), they have ( a_n ) wins and ( b_n ) losses. I need to prove that the sum of the squares of the wins equals the sum of the squares of the losses, that is, ( sum a_n^2 = sum b_n^2 ).Hmm, let me think about how to approach this. First, I should probably figure out how many players there are in total. Since every two players play each other once, the total number of games played is ( frac{N(N-1)}{2} ) where ( N ) is the number of players. But I don't know if I need that right away.Each player plays ( N-1 ) games because they play against every other player once. So for each player ( n ), the number of wins ( a_n ) plus the number of losses ( b_n ) should equal ( N-1 ). That gives me the equation:[a_n + b_n = N - 1]So, that means ( b_n = (N - 1) - a_n ). Maybe I can express ( b_n^2 ) in terms of ( a_n ) and then see what happens when I sum them up.Let me write down ( b_n^2 ):[b_n^2 = left( (N - 1) - a_n right)^2]Expanding that, I get:[b_n^2 = (N - 1)^2 - 2(N - 1)a_n + a_n^2]Okay, so if I sum ( b_n^2 ) over all players, it should be:[sum b_n^2 = sum left[ (N - 1)^2 - 2(N - 1)a_n + a_n^2 right]]I can split this sum into three separate sums:[sum b_n^2 = sum (N - 1)^2 - 2(N - 1) sum a_n + sum a_n^2]Now, let's analyze each term. The first term is ( sum (N - 1)^2 ). Since ( (N - 1)^2 ) is a constant for all players, summing it over ( N ) players gives:[sum (N - 1)^2 = N(N - 1)^2]The second term is ( -2(N - 1) sum a_n ). I need to figure out what ( sum a_n ) is. Since each game has exactly one win, the total number of wins across all players is equal to the total number of games played, which is ( frac{N(N - 1)}{2} ). So:[sum a_n = frac{N(N - 1)}{2}]Substituting that back into the second term:[-2(N - 1) sum a_n = -2(N - 1) cdot frac{N(N - 1)}{2} = -N(N - 1)^2]So now, putting it all together:[sum b_n^2 = N(N - 1)^2 - N(N - 1)^2 + sum a_n^2]Wait, the first two terms cancel each other out:[N(N - 1)^2 - N(N - 1)^2 = 0]So, we're left with:[sum b_n^2 = sum a_n^2]Which is exactly what we needed to prove! That's pretty neat. By expressing ( b_n ) in terms of ( a_n ) and then expanding and summing, the extra terms canceled out, leaving the equality of the sums of squares.I think that makes sense. Let me just recap to make sure I didn't skip any steps. We started by noting that each player's total games are ( N - 1 ), so losses are just the complement of wins. Then, we expressed the square of losses in terms of wins, expanded it, and summed over all players. The key was recognizing that the total number of wins is equal to the total number of games, which allowed us to simplify the middle term. After simplifying, the only term left was the sum of the squares of wins, which proved the equality.Yeah, that seems solid. I don't see any mistakes in the reasoning. It's interesting how the algebra worked out so neatly, with the extra terms canceling each other. I guess that's the beauty of these kinds of problems—they often have elegant solutions once you approach them the right way.I wonder if there's another way to think about this problem, maybe using linear algebra or something else. But I think the method I used is straightforward and effective. It just required breaking down the problem into manageable parts and using the relationships between the variables.Another thought: maybe if I consider the variance or something related to statistics, but I don't think that's necessary here. The problem is purely combinatorial, so sticking with the algebraic approach was the right move.Alright, I feel confident that this proof works. It's always a bit nerve-wracking when you're dealing with sums and squares, but in this case, everything lined up nicely. I just need to make sure I present each step clearly so that someone else can follow along without getting lost.One thing I might have glossed over is why the total number of wins is equal to the total number of games. Let me think about that again. Each game has exactly one winner and one loser, so for every game played, there's exactly one win recorded. Therefore, the total number of wins across all players is equal to the total number of games, which is ( frac{N(N - 1)}{2} ). That makes sense because each of the ( N ) players plays ( N - 1 ) games, but each game is counted twice in that product, so we divide by 2.Yeah, that's correct. So, ( sum a_n = frac{N(N - 1)}{2} ) is accurate. Without that, the rest of the proof wouldn't hold. It's crucial to establish that relationship because it allows us to substitute and simplify the expression properly.I also want to make sure that I didn't make any sign errors when expanding the square. Let me double-check that:[left( (N - 1) - a_n right)^2 = (N - 1)^2 - 2(N - 1)a_n + a_n^2]Yes, that's correct. The middle term is negative because it's subtracting twice the product of ( (N - 1) ) and ( a_n ). So, when I expanded it, the signs were right.Another point to verify is the substitution step. I had:[-2(N - 1) sum a_n = -2(N - 1) cdot frac{N(N - 1)}{2}]Simplifying that:The 2 in the numerator and denominator cancel out, leaving:[- (N - 1) cdot N(N - 1) = -N(N - 1)^2]Which is correct. So, that term simplifies as expected.Putting it all together, the steps are logically consistent and mathematically sound. I think I've covered all the necessary points and haven't missed any critical steps. It's a good feeling when everything comes together like that.I guess the key takeaway here is that sometimes, expressing one variable in terms of another and then expanding can reveal hidden symmetries or cancellations that make the problem easier to solve. In this case, expressing ( b_n ) in terms of ( a_n ) allowed me to relate the two sums and show their equality through algebraic manipulation.It's also a good reminder of how important it is to understand the relationships between variables in a problem. Without recognizing that ( a_n + b_n = N - 1 ), I wouldn't have been able to express ( b_n ) in terms of ( a_n ), and the rest of the proof would have been much more complicated, if not impossible.Overall, I'm satisfied with this approach. It was a bit of a brain teaser at first, but breaking it down step by step made it manageable. I think this is a solid proof, and I can't find any flaws in the logic. It's always reassuring when different parts of the problem interlock so neatly to give the desired result.

💡Alright, so I have this problem here with two inequalities, p and q, and I need to find the range of values for a real number a based on some logical conditions. Let me try to unpack this step by step.First, let's understand what p and q represent.p is the inequality ( x^2 - 8x - 20 < 0 ). I need to solve this to find the range of x where this inequality holds true.Similarly, q is the inequality ( x^2 - 2x + 1 - a^2 leq 0 ) with the condition that a is greater than 0. I'll also need to solve this to find the range of x for q.Once I have the ranges for p and q, I need to analyze the logical conditions given: ¬p is a necessary but not sufficient condition for ¬q. That sounds a bit abstract, but I think it relates to the subsets of x values that satisfy these inequalities.Let me start by solving p.**Solving p: ( x^2 - 8x - 20 < 0 )**To solve this quadratic inequality, I can factor it or find its roots. Let me try factoring first.Looking for two numbers that multiply to -20 and add to -8. Hmm, -10 and +2 work because (-10) * 2 = -20 and (-10) + 2 = -8.So, the quadratic can be factored as:( (x - 10)(x + 2) < 0 )Now, to find where this product is less than zero, I can use a sign chart.The roots are x = 10 and x = -2. These divide the number line into three intervals:1. ( x < -2 )2. ( -2 < x < 10 )3. ( x > 10 )Testing each interval:- For ( x < -2 ), say x = -3: ( (-3 - 10)(-3 + 2) = (-13)(-1) = 13 > 0 )- For ( -2 < x < 10 ), say x = 0: ( (0 - 10)(0 + 2) = (-10)(2) = -20 < 0 )- For ( x > 10 ), say x = 11: ( (11 - 10)(11 + 2) = (1)(13) = 13 > 0 )So, the inequality ( (x - 10)(x + 2) < 0 ) holds true in the interval ( -2 < x < 10 ).Therefore, p is true when ( -2 < x < 10 ), and ¬p (not p) is true when ( x leq -2 ) or ( x geq 10 ).**Solving q: ( x^2 - 2x + 1 - a^2 leq 0 ) with ( a > 0 )**Let me rewrite this inequality:( x^2 - 2x + 1 - a^2 leq 0 )Notice that ( x^2 - 2x + 1 ) is a perfect square:( (x - 1)^2 - a^2 leq 0 )This can be factored as a difference of squares:( (x - 1 - a)(x - 1 + a) leq 0 )So, the inequality becomes:( (x - (1 + a))(x - (1 - a)) leq 0 )Since a is positive, ( 1 + a > 1 - a ). Therefore, the roots are x = 1 + a and x = 1 - a.Again, using a sign chart, the intervals are:1. ( x < 1 - a )2. ( 1 - a < x < 1 + a )3. ( x > 1 + a )Testing each interval:- For ( x < 1 - a ), say x = 1 - a - 1: ( (negative)(negative) = positive )- For ( 1 - a < x < 1 + a ), say x = 1: ( (0 - a)(0 + a) = (-a)(a) = -a^2 < 0 )- For ( x > 1 + a ), say x = 1 + a + 1: ( (positive)(positive) = positive )Since the inequality is less than or equal to zero, the solution includes the interval where the product is negative and the endpoints where it's zero.Therefore, q is true when ( 1 - a leq x leq 1 + a ), and ¬q (not q) is true when ( x < 1 - a ) or ( x > 1 + a ).**Understanding the Logical Condition: ¬p is a necessary but not sufficient condition for ¬q**This is a bit tricky, but let me break it down.In logic, if ¬p is a necessary condition for ¬q, it means that whenever ¬q is true, ¬p must also be true. In other words, ¬q implies ¬p. Symbolically, this is ¬q → ¬p.Additionally, it's stated that ¬p is not a sufficient condition for ¬q. This means that ¬p does not necessarily imply ¬q. In other words, there are cases where ¬p is true, but ¬q is false.Translating this into set theory terms:- The set of x where ¬q is true must be a subset of the set where ¬p is true. So, every x that satisfies ¬q must also satisfy ¬p.- However, the set where ¬p is true is larger than the set where ¬q is true. There are x values that satisfy ¬p but do not satisfy ¬q.So, in terms of intervals:- ¬q: ( x leq 1 - a ) or ( x geq 1 + a )- ¬p: ( x leq -2 ) or ( x geq 10 )For ¬q to be a subset of ¬p, the intervals of ¬q must lie entirely within the intervals of ¬p.This gives us two conditions:1. The left interval of ¬q (( x leq 1 - a )) must be to the left of or equal to the left interval of ¬p (( x leq -2 )). So, ( 1 - a leq -2 ).2. The right interval of ¬q (( x geq 1 + a )) must be to the right of or equal to the right interval of ¬p (( x geq 10 )). So, ( 1 + a geq 10 ).Let me write these inequalities:1. ( 1 - a leq -2 )2. ( 1 + a geq 10 )Solving the first inequality:( 1 - a leq -2 )Subtract 1 from both sides:( -a leq -3 )Multiply both sides by -1 (remembering to reverse the inequality sign):( a geq 3 )Solving the second inequality:( 1 + a geq 10 )Subtract 1 from both sides:( a geq 9 )So, from the first inequality, a must be at least 3, and from the second inequality, a must be at least 9. Since 9 is greater than 3, the more restrictive condition is ( a geq 9 ).Additionally, we were given that ( a > 0 ), which is already satisfied if ( a geq 9 ).Now, we need to ensure that ¬p is not a sufficient condition for ¬q. This means that there must be some x where ¬p is true, but ¬q is false. In other words, there exists x such that x is in ¬p but not in ¬q.Looking at the intervals:¬p is ( x leq -2 ) or ( x geq 10 ).¬q is ( x leq 1 - a ) or ( x geq 1 + a ).If a is exactly 9, then:¬q becomes ( x leq 1 - 9 = -8 ) or ( x geq 1 + 9 = 10 ).So, ¬q is ( x leq -8 ) or ( x geq 10 ).But ¬p is ( x leq -2 ) or ( x geq 10 ).So, in this case, the interval ( -8 < x leq -2 ) is part of ¬p but not part of ¬q. Therefore, there are x values where ¬p is true but ¬q is false, which satisfies the condition that ¬p is not sufficient for ¬q.If a were greater than 9, say a = 10, then ¬q would be ( x leq -9 ) or ( x geq 11 ). Then, the interval ( -9 < x leq -2 ) and ( 10 leq x < 11 ) would be in ¬p but not in ¬q, again satisfying the condition.Therefore, the range of a is all real numbers greater than or equal to 9.**Final Answer**The range of values for the real number ( a ) is boxed{[9, +infty)}.

💡First, I need to shift the graph of ( y = sin 2x ) to the left by ( frac{pi}{4} ) units. To do this, I replace ( x ) with ( x + frac{pi}{4} ) in the function, resulting in ( y = sinleft(2left(x + frac{pi}{4}right)right) ). Simplifying this, I get ( y = sinleft(2x + frac{pi}{2}right) ).Next, I recognize that ( sinleft(2x + frac{pi}{2}right) ) can be rewritten using the sine addition formula as ( cos 2x ). This simplifies the function to ( y = cos 2x ).Now, I need to stretch the ( x )-coordinates of all points on the graph of ( y = cos 2x ) by a factor of 2. To achieve this, I replace ( x ) with ( frac{x}{2} ) in the function, resulting in ( y = cosleft(2 cdot frac{x}{2}right) ). Simplifying this, I obtain ( y = cos x ).Therefore, after performing both transformations, the equation of the resulting function graph is ( y = cos x ).

💡Alright, so I have this problem where I need to find the equations of the tangents drawn from the point M(3, 0) to the parabola 8y = x² + 16. Then, I have to determine the angle between these tangents and find the area of the triangle ABM, where A and B are the points of tangency.First, I need to understand the given parabola equation. It's 8y = x² + 16. Maybe I should rewrite it in a more familiar form. If I divide both sides by 8, I get y = (x²)/8 + 2. So, this is a parabola that opens upwards, with its vertex at (0, 2). The standard form of a parabola is y = ax² + bx + c, so in this case, a = 1/8, b = 0, and c = 2.Now, I need to find the equations of the tangents from the point M(3, 0) to this parabola. I remember that the equation of a tangent to a parabola can be found using the point-slope form, but I need to recall the exact method.Let me think. For a general parabola y = f(x), the tangent at a point (x₀, y₀) on the parabola can be found using the derivative f’(x₀), which gives the slope of the tangent at that point. Then, using the point-slope form, the equation of the tangent is y - y₀ = f’(x₀)(x - x₀).So, let's apply this. First, I need to find the derivative of y = (x²)/8 + 2. The derivative dy/dx is (2x)/8, which simplifies to x/4. So, f’(x) = x/4.Let’s denote a general point on the parabola as (x₀, y₀). Then, the slope of the tangent at this point is m = x₀/4. The equation of the tangent line is:y - y₀ = (x₀/4)(x - x₀)But since (x₀, y₀) lies on the parabola, y₀ = (x₀²)/8 + 2. So, substituting y₀ into the equation:y - [(x₀²)/8 + 2] = (x₀/4)(x - x₀)Let me simplify this equation:y = (x₀²)/8 + 2 + (x₀/4)x - (x₀²)/4Combine like terms:y = 2 + (x₀/4)x - (x₀²)/8So, the equation of the tangent line is y = (x₀/4)x + 2 - (x₀²)/8.Now, this tangent line must pass through the point M(3, 0). So, substituting x = 3 and y = 0 into the equation:0 = (x₀/4)(3) + 2 - (x₀²)/8Let me write that out:0 = (3x₀)/4 + 2 - (x₀²)/8To make this easier, I can multiply both sides by 8 to eliminate the denominators:0 = 6x₀ + 16 - x₀²Rearranging terms:x₀² - 6x₀ - 16 = 0So, I have a quadratic equation in terms of x₀: x₀² - 6x₀ - 16 = 0.I can solve this quadratic equation using the quadratic formula. The quadratic formula is x = [-b ± √(b² - 4ac)]/(2a). Here, a = 1, b = -6, c = -16.Plugging in the values:x₀ = [6 ± √(36 + 64)]/2x₀ = [6 ± √100]/2x₀ = [6 ± 10]/2So, x₀ can be (6 + 10)/2 = 16/2 = 8, or (6 - 10)/2 = (-4)/2 = -2.Therefore, the x-coordinates of the points of tangency are x₀ = 8 and x₀ = -2.Now, let's find the corresponding y-coordinates for these x-values on the parabola.For x₀ = 8:y₀ = (8²)/8 + 2 = 64/8 + 2 = 8 + 2 = 10So, one point of tangency is (8, 10).For x₀ = -2:y₀ = (-2)²/8 + 2 = 4/8 + 2 = 0.5 + 2 = 2.5So, the other point of tangency is (-2, 2.5).Now, I can write the equations of the two tangent lines.First, for the point (8, 10):We already know the slope is x₀/4 = 8/4 = 2.Using point-slope form:y - 10 = 2(x - 8)y = 2x - 16 + 10y = 2x - 6Second, for the point (-2, 2.5):The slope is x₀/4 = (-2)/4 = -0.5.Using point-slope form:y - 2.5 = -0.5(x - (-2))y - 2.5 = -0.5(x + 2)y = -0.5x - 1 + 2.5y = -0.5x + 1.5So, the two tangent lines are y = 2x - 6 and y = -0.5x + 1.5.Next, I need to determine the angle between these two tangents. To find the angle between two lines, I can use the formula:tanθ = |(m₂ - m₁)/(1 + m₁m₂)|where m₁ and m₂ are the slopes of the two lines.Here, m₁ = 2 and m₂ = -0.5.Plugging into the formula:tanθ = |(-0.5 - 2)/(1 + (2)(-0.5))| = |(-2.5)/(1 - 1)| = |(-2.5)/0|Wait, division by zero? That means tanθ is undefined, which implies that θ is 90 degrees because the tangent of 90 degrees is undefined. So, the angle between the two tangents is 90 degrees.Now, I need to find the area of triangle ABM, where A and B are the points of tangency, which are (-2, 2.5) and (8, 10), and M is (3, 0).To find the area of triangle ABM, I can use the shoelace formula. The shoelace formula for three points (x₁, y₁), (x₂, y₂), (x₃, y₃) is:Area = |(x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂))/2|Let me assign the points:A = (-2, 2.5) = (x₁, y₁)B = (8, 10) = (x₂, y₂)M = (3, 0) = (x₃, y₃)Plugging into the formula:Area = |(-2)(10 - 0) + 8(0 - 2.5) + 3(2.5 - 10)| / 2Calculate each term:First term: (-2)(10 - 0) = (-2)(10) = -20Second term: 8(0 - 2.5) = 8(-2.5) = -20Third term: 3(2.5 - 10) = 3(-7.5) = -22.5Adding them up:-20 + (-20) + (-22.5) = -62.5Take the absolute value and divide by 2:| -62.5 | / 2 = 62.5 / 2 = 31.25But 31.25 is equal to 125/4. So, the area is 125/4.Wait, let me verify this because sometimes shoelace can be tricky. Alternatively, I can compute the area using vectors or base and height.Alternatively, I can compute the lengths of sides AM, BM, and AB and then use Heron's formula.First, compute the distance between A and M:A = (-2, 2.5), M = (3, 0)Distance AM = sqrt[(3 - (-2))² + (0 - 2.5)²] = sqrt[(5)² + (-2.5)²] = sqrt[25 + 6.25] = sqrt[31.25] = (5√5)/2Similarly, distance BM:B = (8, 10), M = (3, 0)Distance BM = sqrt[(3 - 8)² + (0 - 10)²] = sqrt[(-5)² + (-10)²] = sqrt[25 + 100] = sqrt[125] = 5√5Distance AB:A = (-2, 2.5), B = (8, 10)Distance AB = sqrt[(8 - (-2))² + (10 - 2.5)²] = sqrt[(10)² + (7.5)²] = sqrt[100 + 56.25] = sqrt[156.25] = 12.5So, sides are:AM = (5√5)/2 ≈ 5.59BM = 5√5 ≈ 11.18AB = 12.5Now, using Heron's formula:First, compute the semi-perimeter, s = (AM + BM + AB)/2s = [(5√5)/2 + 5√5 + 12.5]/2Convert all terms to have a common denominator:(5√5)/2 + (10√5)/2 + 12.5 = (15√5)/2 + 12.5So, s = [(15√5)/2 + 12.5]/2 = (15√5)/4 + 12.5/2 = (15√5)/4 + 6.25This seems complicated. Maybe Heron's formula isn't the best approach here. Alternatively, since I know the coordinates, maybe using vectors or the shoelace formula is better.Wait, earlier with shoelace, I got 125/4, which is 31.25. Let me check the shoelace formula again.Shoelace formula:Arrange the points in order: A(-2, 2.5), B(8, 10), M(3, 0), and back to A(-2, 2.5).Compute the sum of x_i y_{i+1}:(-2)(10) + 8(0) + 3(2.5) = (-20) + 0 + 7.5 = -12.5Compute the sum of y_i x_{i+1}:(2.5)(8) + 10(3) + 0(-2) = 20 + 30 + 0 = 50Subtract the two sums: | -12.5 - 50 | = | -62.5 | = 62.5Divide by 2: 62.5 / 2 = 31.25, which is 125/4. So, that's correct.Therefore, the area is 125/4.So, summarizing:Equations of tangents: y = 2x - 6 and y = -0.5x + 1.5Angle between tangents: 90 degreesArea of triangle ABM: 125/4**Final Answer**The equations of the tangents are boxed{y = -frac{1}{2}x + 1.5} and boxed{y = 2x - 6}. The angle between the tangents is boxed{90^circ}, and the area of triangle ( ABM ) is boxed{dfrac{125}{4}}.

💡Alright, so I have this problem about a rhombus on a Cartesian plane. The vertices are given as A(0,0), B(P,1), C(u,v), and D(1,P). I need to find the value of Q, which is the sum of u and v. Hmm, okay, let's break this down step by step.First, I remember that a rhombus is a type of quadrilateral where all sides are equal in length. Also, in a rhombus, the diagonals bisect each other at right angles. But wait, do they have to be at right angles? I think that's a property of a square, which is a special type of rhombus. Maybe in a general rhombus, the diagonals just bisect each other but aren't necessarily perpendicular. I should double-check that.But for now, let's focus on the fact that in a rhombus, the diagonals bisect each other. So, the midpoint of diagonal AC should be the same as the midpoint of diagonal BD. That seems like a useful property to apply here.Okay, so let's find the midpoints of both diagonals. The midpoint of a line segment with endpoints (x1, y1) and (x2, y2) is given by ((x1+x2)/2, (y1+y2)/2). First, let's find the midpoint of diagonal AC. The coordinates of A are (0,0) and C are (u,v). So, the midpoint of AC is ((0 + u)/2, (0 + v)/2) = (u/2, v/2).Next, let's find the midpoint of diagonal BD. The coordinates of B are (P,1) and D are (1,P). So, the midpoint of BD is ((P + 1)/2, (1 + P)/2).Since the midpoints must be the same in a rhombus, we can set these two midpoints equal to each other:(u/2, v/2) = ((P + 1)/2, (1 + P)/2)This gives us two equations:1. u/2 = (P + 1)/22. v/2 = (1 + P)/2Simplifying both equations by multiplying both sides by 2:1. u = P + 12. v = 1 + PWait, that's interesting. So, u equals P plus 1, and v also equals 1 plus P. That means u and v are the same, right? Because P + 1 is the same as 1 + P. So, u = v.But wait, does that make sense? If u equals v, then the point C(u, v) would lie on the line y = x. Is that necessarily true for a rhombus? Hmm, not sure. Maybe, maybe not. Let's not get bogged down here. Let's just proceed with what we have.So, from the equations, we have:u = P + 1v = P + 1Therefore, u + v = (P + 1) + (P + 1) = 2P + 2So, Q = u + v = 2P + 2Wait, that seems straightforward. But let me verify if I missed anything.I used the property that the midpoints of the diagonals are the same in a rhombus, which is correct. I calculated the midpoints correctly, set them equal, and solved for u and v. That gives me u = P + 1 and v = P + 1, leading to Q = 2P + 2.Is there another way to approach this problem? Maybe by using the slopes of the sides or the lengths of the sides?Let's try that as a verification.In a rhombus, all sides are equal in length. So, the distance from A to B should be equal to the distance from B to C, which should be equal to the distance from C to D, and so on.Let's compute the distance between A(0,0) and B(P,1):Distance AB = sqrt[(P - 0)^2 + (1 - 0)^2] = sqrt(P^2 + 1)Similarly, distance BC = sqrt[(u - P)^2 + (v - 1)^2]Since AB = BC, we have:sqrt(P^2 + 1) = sqrt[(u - P)^2 + (v - 1)^2]Squaring both sides:P^2 + 1 = (u - P)^2 + (v - 1)^2Expanding the right side:(u - P)^2 + (v - 1)^2 = (u^2 - 2Pu + P^2) + (v^2 - 2v + 1)So, P^2 + 1 = u^2 - 2Pu + P^2 + v^2 - 2v + 1Simplify:P^2 + 1 = u^2 + v^2 - 2Pu - 2v + P^2 + 1Subtract P^2 + 1 from both sides:0 = u^2 + v^2 - 2Pu - 2vSo, u^2 + v^2 - 2Pu - 2v = 0Hmm, that's one equation involving u and v.Now, let's compute the distance from B(P,1) to C(u,v):We already did that, and set it equal to AB, which gave us the equation above.Now, let's compute the distance from C(u,v) to D(1,P):Distance CD = sqrt[(1 - u)^2 + (P - v)^2]Since all sides are equal, CD should also equal AB, which is sqrt(P^2 + 1). So,sqrt[(1 - u)^2 + (P - v)^2] = sqrt(P^2 + 1)Squaring both sides:(1 - u)^2 + (P - v)^2 = P^2 + 1Expanding the left side:(1 - 2u + u^2) + (P^2 - 2Pv + v^2) = P^2 + 1Simplify:1 - 2u + u^2 + P^2 - 2Pv + v^2 = P^2 + 1Subtract P^2 + 1 from both sides:-2u + u^2 - 2Pv + v^2 = 0Which is:u^2 + v^2 - 2u - 2Pv = 0Now, we have two equations:1. u^2 + v^2 - 2Pu - 2v = 02. u^2 + v^2 - 2u - 2Pv = 0Let's subtract equation 2 from equation 1:(u^2 + v^2 - 2Pu - 2v) - (u^2 + v^2 - 2u - 2Pv) = 0 - 0Simplify:-2Pu - 2v - (-2u - 2Pv) = 0Which is:-2Pu - 2v + 2u + 2Pv = 0Factor terms:(-2Pu + 2u) + (-2v + 2Pv) = 0Factor out 2u and 2v:2u(-P + 1) + 2v(-1 + P) = 0Factor out 2:2[ u(1 - P) + v(P - 1) ] = 0Divide both sides by 2:u(1 - P) + v(P - 1) = 0Notice that (P - 1) = -(1 - P), so:u(1 - P) - v(1 - P) = 0Factor out (1 - P):(1 - P)(u - v) = 0So, either (1 - P) = 0 or (u - v) = 0Case 1: 1 - P = 0 => P = 1Case 2: u - v = 0 => u = vLet's analyze both cases.Case 1: P = 1If P = 1, then from our earlier midpoint calculations:u = P + 1 = 1 + 1 = 2v = P + 1 = 1 + 1 = 2So, u = 2, v = 2, and Q = u + v = 4But let's check if this satisfies the distance conditions.Compute AB: sqrt(1^2 + 1^2) = sqrt(2)Compute BC: sqrt[(2 - 1)^2 + (2 - 1)^2] = sqrt(1 + 1) = sqrt(2)Compute CD: sqrt[(1 - 2)^2 + (1 - 2)^2] = sqrt(1 + 1) = sqrt(2)Compute DA: sqrt[(1 - 0)^2 + (1 - 0)^2] = sqrt(1 + 1) = sqrt(2)So, all sides are equal, which is consistent with a rhombus.Case 2: u = vIf u = v, then from our midpoint calculations:u = P + 1v = P + 1So, u = v = P + 1Therefore, Q = u + v = 2(P + 1) = 2P + 2But wait, this is the same as what we got earlier. So, whether P = 1 or u = v, we end up with Q = 2P + 2. However, in Case 1, when P = 1, Q = 4, which is a specific case of Q = 2P + 2 when P = 1.Therefore, the general solution is Q = 2P + 2.But let's make sure there are no other constraints or possibilities. For example, could there be another configuration where the diagonals don't bisect each other but still form a rhombus? I don't think so, because in a rhombus, the diagonals must bisect each other by definition.Also, we can check if the sides are indeed equal when u = v = P + 1.Compute AB: sqrt(P^2 + 1)Compute BC: sqrt[(u - P)^2 + (v - 1)^2] = sqrt[(P + 1 - P)^2 + (P + 1 - 1)^2] = sqrt[1^2 + P^2] = sqrt(1 + P^2)Similarly, CD: sqrt[(1 - u)^2 + (P - v)^2] = sqrt[(1 - (P + 1))^2 + (P - (P + 1))^2] = sqrt[(-P)^2 + (-1)^2] = sqrt(P^2 + 1)And DA: sqrt[(1 - 0)^2 + (P - 0)^2] = sqrt(1 + P^2)So, all sides are equal, confirming that it's a rhombus.Therefore, the value of Q is 2P + 2.