🧠:[<think>Okay, so I need to figure out which of these sets is empty. Let's look at each option one by one.Starting with option A: {0}. Hmm, that's a set containing the element 0. Since there's clearly an element in there, this set isn't empty. So A is not the answer.Next up is option B: {x | x² + 1 = 0}. Alright, this set consists of all real numbers x such that when you square them and add 1, you get 0. Let me solve the equation x² + 1 = 0. Subtracting 1 from both sides gives x² = -1. Now, in real numbers, squaring any real number can't give a negative result. So there are no real solutions here. Therefore, the set B has no elements, which means it's empty. But wait, let me check the other options just to be sure.Option C: {x | x² - 1 = 0}. Solving x² - 1 = 0, we get x² = 1, so x = 1 or x = -1. That means the set C is {1, -1}, which definitely has elements. So C isn't empty.Option D: {x | x < 0}. This set includes all real numbers less than 0. There are plenty of negative numbers, like -1, -0.5, -100, etc. So this set has infinitely many elements and is not empty.So going back, A has one element, B has none, C has two elements, and D has infinitely many. Therefore, the empty set is option B. But wait, the problem might be considering complex numbers? The question didn't specify, but usually, unless stated otherwise, we assume we're dealing with real numbers. If we consider complex numbers, then x² + 1 = 0 would have solutions x = i and x = -i. But since complex numbers aren't typically assumed unless mentioned, I think the answer is B. Let me double-check.In most standard math problems like this, especially in multiple-choice questions without context, they're referring to real numbers. So yeah, B is empty. So the answer should be B.Final Answerboxed{B}</think>To determine which set is empty, we analyze each option:- Option A: {0} contains the element 0, so it is not empty.- Option B: {x | x² + 1 = 0}. Solving the equation x² + 1 = 0 gives x² = -1. In real numbers, there are no solutions since the square of a real number cannot be negative. Therefore, this set is empty.- Option C: {x | x² - 1 = 0}. Solving x² - 1 = 0 gives x = 1 or x = -1, so the set {1, -1} is not empty.- Option D: {x | x < 0} includes all negative real numbers, which are infinite, so it is not empty.Assuming the context of real numbers (as complex numbers are not mentioned), the empty set is B.boxed{B},<think>Okay, let's see. The question is asking which of the given sets is empty. The options are A, B, C, D. I need to check each one to see if there are any elements in them. If a set has no elements, that's the empty set. Alright, let's go through them one by one.Starting with option A: {0}. Hmm, this set contains the element 0. So there's definitely something in there. Therefore, set A is not empty. Got it. So we can eliminate option A.Moving on to option B: {x | x² + 1 = 0}. Let's analyze this. The set consists of all x such that when you square them and add 1, you get 0. So the equation is x² + 1 = 0. Solving for x, we subtract 1 from both sides to get x² = -1. To find x, we take the square root of both sides. But wait, the square root of a negative number isn't a real number. It would be an imaginary number, specifically ±i. However, unless specified otherwise, I think these sets are probably dealing with real numbers. If the problem is considering complex numbers, then there would be solutions. But generally, in these multiple-choice questions, unless stated, they expect real numbers. So if x has to be a real number, then there are no solutions. Therefore, the set B would be empty. Hmm, that seems like a candidate for the answer. But let me check the other options just to be sure.Option C: {x | x² - 1 = 0}. Let's solve this equation. x² - 1 = 0. Adding 1 to both sides gives x² = 1. Taking the square root of both sides, x = ±1. So the solutions are 1 and -1. Therefore, set C contains two elements: 1 and -1. So this set is definitely not empty. So option C is out.Option D: {x | x < 0}. This set consists of all real numbers less than 0. So that's all the negative numbers. There are infinitely many negative numbers, like -1, -0.5, -100, etc. So this set is definitely not empty. Therefore, option D is out.So recapping: A has 0, so not empty. B has no real solutions, so it's empty. C has two elements, D has infinitely many. Therefore, the answer should be option B.Wait, just to make sure, did I interpret the problem correctly? The question didn't specify the domain of x. If x is allowed to be a complex number, then B would have solutions, which are i and -i. But in most standard math problems like this, especially when options include things like {0} and {x | x <0}, which are real number sets, I think the assumption is that x is a real number. So unless stated otherwise, we consider real numbers here. Therefore, B is indeed empty. So yeah, B is the empty set. Therefore, the answer is option B.Final Answerboxed{B}</think>To determine which set is empty, we analyze each option:- Option A: {0} contains the element 0, so it is not empty.- Option B: {x | x² + 1 = 0}. Solving x² + 1 = 0 gives x² = -1. For real numbers, there are no solutions because the square of a real number cannot be negative. Thus, this set is empty.- Option C: {x | x² - 1 = 0}. Solving x² - 1 = 0 gives x = ±1, so the set contains elements 1 and -1, making it non-empty.- Option D: {x | x < 0} includes all negative real numbers, which are infinitely many, so it is not empty.Assuming the domain is real numbers (as complex numbers are not specified), the empty set is option B.boxed{B}]

🔑:Set A contains the element 0, therefore A is not empty. Since the equation x^2+1=0 has no solution, set B is empty. Since x^2-1=0 implies x=±1, set C is not empty. Set D consists of numbers less than 0, so D is not empty. Therefore, the correct answer is boxed{text{B}}.

🧠:[<think>Alright, let's try to tackle this problem step by step. First, let me make sure I understand the setup correctly. We have two squares, ABCD and EFGH. The area of square ABCD is 9 cm², so each side should be √9 = 3 cm. Similarly, the area of square EFGH is 64 cm², so each side is √64 = 8 cm. The problem states that side BC of square ABCD lies on EH of square EFGH. I need to visualize this. So, square ABCD is smaller, with side BC (which is 3 cm) lying along side EH of the larger square EFGH (which is 8 cm). So, BC is a part of EH? Or is EH the entire side, and BC is aligned with it? Hmm. Maybe the two squares are positioned such that BC is on top of EH? Let me try to sketch this mentally.Also, the area of triangle ACG is given as 6.75 cm², and we need to find the area of triangle ABE. Let me note the key points: A, B, C, D are the vertices of the smaller square; E, F, G, H are the vertices of the larger square. Since BC is on EH, that might mean that points B and C are somewhere along the side EH of the larger square. Wait, but EH is a side of the larger square, which is 8 cm. But BC is only 3 cm. So, BC is a segment along the 8 cm side EH of the larger square. So, points B and C are somewhere on EH, 3 cm apart. Then, the rest of square ABCD is built off of that BC segment?Wait, but square ABCD is a separate square. If BC is lying on EH, then perhaps square ABCD is adjacent to square EFGH along BC and EH. Maybe overlapping? Or is square ABCD placed on top of square EFGH such that BC is coinciding with part of EH? This is a bit confusing without a figure, but let's try to work through it.First, let me note down all the given data:- Area of ABCD = 9 cm² ⇒ side length 3 cm.- Area of EFGH = 64 cm² ⇒ side length 8 cm.- BC lies on EH.- Area of triangle ACG = 6.75 cm².- Find the area of triangle ABE.Since BC is on EH, and BC is 3 cm, EH is 8 cm, so BC is a portion of EH. Let me assume that point B coincides with point E, but then BC would extend 3 cm along EH from E. Alternatively, maybe BC is somewhere in the middle of EH? The problem doesn't specify the exact position, so perhaps we need to figure it out based on the given area of triangle ACG.Let me consider coordinate geometry. Maybe assigning coordinates to the points will help. Let's place square EFGH in a coordinate system. Let me set point E at the origin (0,0). Since EH is a side of length 8 cm, and assuming the square is oriented with sides horizontal and vertical, then point H would be at (8,0), F at (0,8), and G at (8,8). Wait, no. If EFGH is a square, then the points would be E, F, G, H in order. So, if E is at (0,0), then F might be at (0,8), G at (8,8), and H at (8,0). Wait, that would make EFGH a square with sides EF, FG, GH, HE. Wait, but in that case, EH would be from (8,0) to (0,0), but that's not correct. Wait, maybe I need to correct the order.Wait, the square EFGH: if E is connected to F, F to G, G to H, and H back to E. So, if E is at (0,0), then F would be at (8,0), G at (8,8), and H at (0,8). Wait, no, that would make EFGH a square rotated 90 degrees. Wait, maybe the coordinates need to be adjusted.Alternatively, let me fix the coordinate system so that square EFGH has side EH on the x-axis. So, point E is at (0,0), point H is at (8,0). Then, since it's a square, the other points F and G must be at (0,8) and (8,8), but that would make EFGH a rectangle unless the sides are vertical and horizontal. Wait, no. If EH is the base from (0,0) to (8,0), then to make it a square, the other sides must be vertical. So, point F would be at (8,0) + upward direction, but then H is at (8,0). Wait, maybe I need to think differently.Wait, maybe the square EFGH is placed such that EH is horizontal. Let's fix E at (0,0), then H is at (8,0). Then, since it's a square, F and G must be located such that all sides are 8 cm and right angles. So, from E(0,0), moving up to F(0,8), then right to G(8,8), then down to H(8,0), and back to E(0,0). Wait, that's a square with sides EF = 8, FG = 8, etc. So in this case, EH is from (8,0) to (0,0). Wait, no, if E is (0,0) and H is (8,0), then moving from E to H is along the x-axis. Then, to make the square, we need to go up from E(0,0) to F(0,8), then right to G(8,8), then down to H(8,0), and back to E. Wait, but that would make EH a diagonal? Wait, no, EH would be a side. Wait, perhaps my coordinate system is conflicting with the labeling.Alternatively, maybe the square EFGH is placed with E at (0,0), F at (8,0), G at (8,8), and H at (0,8). That way, each side is 8 cm. Then EH would be from H(0,8) to E(0,0), which is vertical. But the problem states that BC lies on EH. If EH is vertical, then BC is also vertical. But the problem says "side BC lies on EH". So BC is a side of the smaller square ABCD, which is 3 cm. So, BC is 3 cm in length, lying on EH which is 8 cm. So, if EH is vertical from (0,8) to (0,0), then BC would be a vertical segment of 3 cm along this. So, perhaps B is at (0,5) and C is at (0,2), making BC 3 cm? Then, square ABCD is constructed off of BC. Wait, but a square with vertical side BC would have AB and CD horizontal. So, if BC is vertical from (0,5) to (0,2), then AB would go from B(0,5) to A(-3,5), since the side length is 3 cm. Then, D would be at (-3,2). But then, the square ABCD is to the left of EH. Alternatively, if BC is along EH from (0,0) to (0,8), but only 3 cm, perhaps from (0,0) to (0,3), then the square ABCD would extend to the right or left. Hmm.But since the problem mentions triangle ACG and ABE, we need to figure out the coordinates of points A, C, G, as well as A, B, E.This is getting a bit complicated. Maybe assigning coordinates is the way to go. Let me try to set up a coordinate system where point E is at (0,0), and square EFGH is placed such that EH is along the x-axis from E(0,0) to H(8,0). Then, the square EFGH would have points F(0,8), G(8,8), and H(8,0). Wait, no, that would make EH a horizontal side of length 8, and EF a vertical side from E(0,0) to F(0,8). Then, FG would be from F(0,8) to G(8,8), and GH from G(8,8) to H(8,0), and HE from H(8,0) back to E(0,0). Wait, but that's actually a rectangle, not a square, because EH is 8 units horizontally, and EF is 8 units vertically. So, that is a square, because all sides are 8 units. Wait, no, in that case, EF is vertical from (0,0) to (0,8), FG is horizontal from (0,8) to (8,8), GH is vertical down from (8,8) to (8,0), and HE is horizontal back to (0,0). Wait, no, HE would be from (8,0) to (0,0), which is 8 units. So actually, each side is 8 units, so it is a square. So, in this coordinate system:- E(0,0)- F(0,8)- G(8,8)- H(8,0)But then, EH is from E(0,0) to H(8,0), which is the base. Now, the problem states that side BC of square ABCD lies on EH. Since EH is from (0,0) to (8,0), which is the x-axis. So, BC is a side of square ABCD, which is 3 cm, lying on EH (the x-axis). So, square ABCD is sitting on the x-axis with side BC as part of EH. So, points B and C are on the x-axis, 3 cm apart, and the square is either above or below the x-axis. But since it's a square, likely above or below, but given that the other square EFGH is above the x-axis (since F and G are at (0,8) and (8,8)), maybe ABCD is also above the x-axis? Or maybe not. Wait, but the problem doesn't specify the orientation, so we might need to figure that out.Assuming square ABCD is above the x-axis, then points A, B, C, D would be arranged such that BC is on the x-axis, and the square extends upward. Since BC is 3 cm, let's say point B is at (x, 0) and point C is at (x + 3, 0). Then, point A would be at (x, 3) and point D at (x + 3, 3). However, depending on where BC is placed along EH (which is from (0,0) to (8,0)), x can vary between 0 and 8 - 3 = 5. So, BC can be anywhere from (0,0) to (3,0), up to (5,0) to (8,0). The exact position might affect the area of triangle ACG.But the problem gives the area of triangle ACG as 6.75 cm², which is 27/4. So, maybe through coordinate geometry, we can find the coordinates of points A, C, G, compute the area, set it equal to 6.75, and solve for the position of square ABCD along EH.Once we have the position of ABCD, we can then find the coordinates of points A, B, E and compute the area of triangle ABE.Okay, let's formalize this.Let me assign coordinates:- Square EFGH: - E(0,0) - F(0,8) - G(8,8) - H(8,0)- Square ABCD: - Since BC is on EH (the x-axis from (0,0) to (8,0)), let's let B be at (a, 0) and C at (a + 3, 0). - Since it's a square, the other points A and D can be determined. Assuming the square is above the x-axis, then: - A is at (a, 3) - D is at (a + 3, 3) - Alternatively, if the square is below the x-axis, then: - A would be at (a, -3) - D at (a + 3, -3) - But since the problem mentions triangle ACG, and point G is at (8,8), if square ABCD is below the x-axis, then point A would be at (a, -3), and the distance between A and G would be larger. But the area of triangle ACG is given, so maybe the position (above or below) can be determined based on that. Let's assume it's above the x-axis first.So, points:- A(a, 3)- B(a, 0)- C(a + 3, 0)- D(a + 3, 3)Now, point G is at (8,8). We need to find the area of triangle ACG.First, find coordinates of points A, C, G:- A(a, 3)- C(a + 3, 0)- G(8,8)The area of triangle ACG can be calculated using the shoelace formula or determinant formula.The formula for the area of a triangle given three points (x1,y1), (x2,y2), (x3,y3) is:Area = |(x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)) / 2|Plugging in the coordinates:Area = |(a*(0 - 8) + (a + 3)*(8 - 3) + 8*(3 - 0)) / 2|Simplify each term:First term: a*(0 - 8) = -8aSecond term: (a + 3)*(8 - 3) = (a + 3)*5 = 5a + 15Third term: 8*(3 - 0) = 24So, combining:Area = |(-8a + 5a + 15 + 24) / 2| = |(-3a + 39) / 2|Given that the area is 6.75 cm², which is 27/4. So,|(-3a + 39)/2| = 27/4Multiply both sides by 2:| -3a + 39 | = 27/2 = 13.5So, two cases:1) -3a + 39 = 13.52) -3a + 39 = -13.5Case 1:-3a + 39 = 13.5-3a = 13.5 - 39 = -25.5a = (-25.5)/(-3) = 8.5But since square ABCD is along EH from (0,0) to (8,0), and BC is 3 cm, the maximum value of a is 8 - 3 = 5. So, a = 8.5 would place point C at 8.5 + 3 = 11.5, which is beyond H(8,0). Therefore, this solution is invalid.Case 2:-3a + 39 = -13.5-3a = -13.5 - 39 = -52.5a = (-52.5)/(-3) = 17.5Wait, 52.5 divided by 3 is 17.5, which is even further to the right, which is impossible. So, this can't be correct either.Wait, something's wrong here. Both solutions give a beyond 8, which is outside the square EFGH. That suggests that my assumption about the orientation of square ABCD might be incorrect. Maybe square ABCD is not above the x-axis but below?Let me try that. If square ABCD is below the x-axis, then:- A(a, -3)- B(a, 0)- C(a + 3, 0)- D(a + 3, -3)Now, points A, C, G:- A(a, -3)- C(a + 3, 0)- G(8,8)Calculating the area:Area = |(a*(0 - 8) + (a + 3)*(8 - (-3)) + 8*(-3 - 0))/2|Simplify each term:First term: a*(0 - 8) = -8aSecond term: (a + 3)*(8 + 3) = (a + 3)*11 = 11a + 33Third term: 8*(-3 - 0) = 8*(-3) = -24Combine:Area = |(-8a + 11a + 33 - 24)/2| = |(3a + 9)/2|Set equal to 6.75 = 27/4:|(3a + 9)/2| = 27/4Multiply both sides by 2:|3a + 9| = 27/2 = 13.5Again, two cases:1) 3a + 9 = 13.53a = 13.5 - 9 = 4.5a = 4.5 / 3 = 1.52) 3a + 9 = -13.53a = -13.5 - 9 = -22.5a = -22.5 / 3 = -7.5Again, considering the position along EH (from 0 to 8), a must satisfy that C(a + 3, 0) is ≤ 8, so a + 3 ≤ 8 ⇒ a ≤ 5. So, a = 1.5 is valid (since 1.5 + 3 = 4.5 ≤ 8), and a = -7.5 would place B at (-7.5, 0), which is outside EH (since EH is from 0 to 8). So, only a = 1.5 is valid.Therefore, square ABCD is below the x-axis with B at (1.5, 0), C at (4.5, 0), A at (1.5, -3), D at (4.5, -3).Now, we need to find the area of triangle ABE. Points A, B, E.Coordinates:- A(1.5, -3)- B(1.5, 0)- E(0, 0)So, triangle ABE has vertices at (1.5, -3), (1.5, 0), and (0, 0).Let me calculate its area using the shoelace formula.Coordinates in order:(1.5, -3), (1.5, 0), (0, 0), back to (1.5, -3)Area = |(1.5*(0 - 0) + 1.5*(0 - (-3)) + 0*(-3 - 0))/2|Simplify:= |(0 + 1.5*3 + 0)/2|= |(0 + 4.5 + 0)/2|= |4.5 / 2|= 2.25Which is 9/4 cm². But wait, 2.25 is 9/4. Let me check the calculation again.Alternatively, since the triangle has a vertical side from (1.5, -3) to (1.5, 0), which is length 3, and then a base from (1.5, 0) to (0, 0), which is 1.5 units. Wait, but actually, triangle ABE is a right triangle with base along the x-axis from E(0,0) to B(1.5,0), length 1.5 cm, and height from B(1.5,0) down to A(1.5,-3), which is 3 cm. So, area is (base * height)/2 = (1.5 * 3)/2 = 4.5 / 2 = 2.25 cm², which is 9/4. So, 9/4 is 2.25. But the problem asks for the answer in square centimeters, and 9/4 is 2.25. So, the area of triangle ABE is 2.25 cm², which is 9/4. However, let me confirm once again.Alternatively, using coordinates:E(0,0), B(1.5,0), A(1.5,-3). The triangle is a right triangle with legs of 1.5 cm (from E to B) and 3 cm (from B to A). So, area = (1.5 * 3)/2 = 4.5 / 2 = 2.25 cm². Yes, that's correct.Therefore, the area is 2.25 cm², which is 9/4. So, the answer should be boxed{dfrac{9}{4}} or boxed{2.25}. But the problem might expect the answer as a fraction. Let me check the calculations once more to be sure.Wait, but when I calculated the area using the shoelace formula:Coordinates: A(1.5, -3), B(1.5, 0), E(0,0)Formula:Area = |(x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))/2|Plugging in:x1 = 1.5, y1 = -3x2 = 1.5, y2 = 0x3 = 0, y3 = 0So,Area = |1.5*(0 - 0) + 1.5*(0 - (-3)) + 0*(-3 - 0)| / 2= |0 + 1.5*3 + 0| / 2= |4.5| / 2= 2.25Yes, that's correct. So, 2.25 cm² is indeed the area. Converting 2.25 to a fraction, that's 9/4. Therefore, the answer is 9/4 cm².But let me just make sure that my assumption about the square being below the x-axis is correct. If the square were above the x-axis, we saw that the value of 'a' would be 8.5 or 17.5, which are invalid. So, the square must be below the x-axis. Therefore, the coordinates are as I found, leading to the area of triangle ABE as 9/4.Wait, but before I conclude, let me check if there's another possible configuration. For example, maybe the square ABCD is placed such that BC is on EH but rotated? Wait, but the problem states that ABCD is a square, so all sides are equal and angles are 90 degrees. If BC is lying on EH, then the square must be aligned such that BC is a horizontal or vertical side. Since EH is a horizontal side (from E(0,0) to H(8,0)), then BC must also be horizontal. Wait, in my coordinate system, EH is horizontal. But in my calculation, I considered BC as horizontal (along the x-axis). So, if BC is horizontal, then the square is either above or below the x-axis, as I considered.But in my first assumption, I thought the square is above the x-axis, but that led to invalid positions. Then, considering it's below the x-axis gave a valid position. Therefore, that must be the correct configuration.Thus, the area of triangle ABE is 9/4 cm², which is 2.25 cm². So, the answer should be boxed{dfrac{9}{4}}.Final AnswerThe area of triangle ( A B E ) is boxed{dfrac{9}{4}} square centimeters.</think>Given the problem, we need to find the area of triangle ( ABE ) given the areas of squares ( ABCD ) and ( EFGH ), and the area of triangle ( ACG ).1. Determine the side lengths of the squares: - The area of square ( ABCD ) is 9 cm², so each side is ( sqrt{9} = 3 ) cm. - The area of square ( EFGH ) is 64 cm², so each side is ( sqrt{64} = 8 ) cm.2. Assign coordinates to the vertices of the squares: - Place square ( EFGH ) in a coordinate system with ( E ) at ( (0, 0) ), ( F ) at ( (0, 8) ), ( G ) at ( (8, 8) ), and ( H ) at ( (8, 0) ). - Since side ( BC ) of square ( ABCD ) lies on ( EH ), we consider ( BC ) as a segment on the x-axis. Let ( B ) be at ( (a, 0) ) and ( C ) at ( (a + 3, 0) ). Given the area of triangle ( ACG ) is 6.75 cm², we need to determine the correct position of ( a ).3. Calculate the coordinates of points ( A ), ( C ), and ( G ): - If square ( ABCD ) is below the x-axis, then ( A ) is at ( (a, -3) ), ( B ) at ( (a, 0) ), ( C ) at ( (a + 3, 0) ), and ( D ) at ( (a + 3, -3) ). - Using the coordinates ( A(a, -3) ), ( C(a + 3, 0) ), and ( G(8, 8) ), we apply the shoelace formula to find the area of triangle ( ACG ).4. Solve for ( a ) using the given area of triangle ( ACG ): - The area formula gives ( |(3a + 9)/2| = 27/4 ). - Solving this, we find ( a = 1.5 ) as the valid solution.5. Determine the coordinates of points ( A ), ( B ), and ( E ): - ( A ) is at ( (1.5, -3) ), ( B ) at ( (1.5, 0) ), and ( E ) at ( (0, 0) ).6. Calculate the area of triangle ( ABE ): - Using the coordinates ( A(1.5, -3) ), ( B(1.5, 0) ), and ( E(0, 0) ), the area is calculated as a right triangle with base 1.5 cm and height 3 cm. - The area is ( frac{1.5 times 3}{2} = 2.25 ) cm², which is ( frac{9}{4} ) cm².Thus, the area of triangle ( ABE ) is boxed{dfrac{9}{4}} square centimeters.,<think>Okay, let's see. So, there's this problem about two squares, ABCD and EFGH. The area of ABCD is 9 cm², so each side must be 3 cm because the area of a square is side squared, right? So if the area is 9, the side length is √9 = 3 cm. Similarly, the area of EFGH is 64 cm², so each side of that square must be √64 = 8 cm. Got that down.The problem mentions that side BC of square ABCD lies on EH of square EFGH. Hmm, so BC is one side of the smaller square, which is 3 cm, and EH is a side of the larger square, which is 8 cm. So, BC is part of EH. Maybe they're overlapping or something? The figure isn't shown, but I need to visualize it. Let me try to draw a mental picture.Square EFGH is bigger, with side 8 cm. Square ABCD is smaller, with side 3 cm. The side BC of the smaller square lies on the side EH of the larger square. So, BC is a segment on EH. Since BC is 3 cm and EH is 8 cm, the rest of EH would be 8 - 3 = 5 cm. So, maybe the smaller square is placed such that BC is on EH, and the rest of EH (the remaining 5 cm) is part of the larger square. But where exactly is ABCD placed on EFGH? Is it at one end, overlapping part of EH, or somewhere in the middle?The problem also states that the area of triangle ACG is 6.75 cm². We need to find the area of triangle ABE. Let me start by labeling the points to understand their positions.First, square ABCD: points A, B, C, D. Since it's a square, all sides are equal, and all angles are 90 degrees. Let's assume the standard labeling where the square is labeled clockwise. So, AB is adjacent to BC, which is adjacent to CD, and so on.Square EFGH: points E, F, G, H. Similarly, a larger square with sides 8 cm. So, EF, FG, GH, HE are all 8 cm. The problem says that BC lies on EH. So, BC is a part of EH. That suggests that point B and point C are on EH. Wait, but BC is a side of square ABCD. So, if BC is lying on EH, then points B and C are somewhere along the line EH of the larger square. Hmm.Let me try to sketch this mentally. Let's place square EFGH with E at the origin (0,0), then F at (8,0), G at (8,8), and H at (0,8). Wait, but if EH is from E(0,0) to H(0,8), that's a vertical side. Wait, but squares can be rotated. Wait, maybe the squares are placed such that their sides are horizontal and vertical? Because otherwise, the problem would mention angles or rotations. So, assuming standard orientation, with sides horizontal and vertical.Wait, but the problem mentions that BC lies on EH. So, if square EFGH has side EH, and square ABCD has side BC lying on EH. Let me confirm: in square EFGH, the sides are EF, FG, GH, HE. So, EH is one of the sides. So, EH is a side of length 8 cm. Similarly, BC is a side of ABCD, length 3 cm. So, BC is lying on EH. So, BC is a segment along EH. Therefore, points B and C must lie somewhere on EH. Since EH is 8 cm and BC is 3 cm, B and C must be 3 cm apart along EH. So, depending on where BC is placed on EH, the position of square ABCD can vary. But without the figure, it's a bit ambiguous. However, we can perhaps figure out based on the area of triangle ACG.So, triangle ACG: points A, C, G. Let me think. Point A is a vertex of square ABCD. Since ABCD is a square with side 3 cm, and BC is on EH (which is a side of EFGH, the larger square). Let's assume coordinates to model this. Coordinates might help.Let me assign coordinates to the squares. Let's place square EFGH with E at (0,0), F at (8,0), G at (8,8), and H at (0,8). So, side EH is from (0,8) to (0,0). Wait, but in the problem statement, it's mentioned that side BC lies on EH. So, BC is part of EH. Wait, in my current coordinate system, EH is vertical from (0,0) to (0,8). But BC is a horizontal side of square ABCD. Hmm, so that might not make sense unless the square ABCD is rotated. Wait, this is a conflict.Wait, maybe the squares are placed such that their sides are horizontal and vertical. Then, if BC is lying on EH, which is a horizontal side. Wait, perhaps in the problem, EH is a horizontal side. Let me check again.If square EFGH has area 64, so side 8 cm. If we place it with horizontal and vertical sides, then E could be at (0,0), F at (8,0), G at (8,8), H at (0,8). Then EH is vertical from (0,8) to (0,0). If BC is lying on EH, which is vertical, then BC must also be vertical. But BC is a side of square ABCD, which has area 9, so side 3 cm. If BC is vertical, then the square ABCD is placed with BC vertical. So, points B and C are on EH (the vertical side from (0,0) to (0,8)). So, BC is a vertical segment of 3 cm on EH. So, for example, point B could be at (0,5) and point C at (0,8), so BC is 3 cm. Then, square ABCD would extend to the right from BC. So, point D would be at (3,8), and point A at (3,5). Wait, but then ABCD would be a square with sides of 3 cm, right? So from B(0,5) to C(0,8) is 3 cm up, then from C(0,8) to D(3,8) is 3 cm to the right, then D(3,8) to A(3,5) is 3 cm down, and A(3,5) to B(0,5) is 3 cm to the left. Wait, but that last part, from A(3,5) to B(0,5) is 3 cm to the left, which is 3 cm. So, that works. Then square ABCD is attached to the side EH of square EFGH at BC, extending to the right.Alternatively, BC could be placed lower on EH. For example, point B at (0,2) and C at (0,5), then square ABCD would extend to the right from there, with D at (3,5) and A at (3,2). But regardless of where BC is placed vertically on EH, the horizontal position (the x-coordinate) would be 0 since EH is the vertical side from (0,0) to (0,8). Wait, but in my previous example, square ABCD is placed to the right of EH, so its x-coordinate goes from 0 to 3. But square EFGH is from x=0 to x=8, and y=0 to y=8. So, overlapping? Or is square ABCD outside of EFGH? Hmm, the problem doesn't specify whether the squares overlap or are adjacent. But given that the area of triangle ACG is given, which probably involves points from both squares, so maybe they are placed in such a way that parts overlap or are adjacent.Wait, but in my previous coordinate system, if square EFGH is from (0,0) to (8,8), and square ABCD is attached to the side EH (which is the left side of EFGH) and extends to the left, but then BC would be on EH, which is vertical. Wait, this is getting confusing. Maybe I need to fix the coordinate system.Let me try another approach. Let's assume that both squares are placed on a coordinate plane. Let me assign coordinates such that point E is at (0,0), and since EFGH is a square of side 8, then F is at (8,0), G at (8,8), and H at (0,8). So, EH is the left side from (0,0) to (0,8).Now, square ABCD has area 9, so side 3 cm. The side BC is lying on EH. So, BC is a vertical segment of length 3 cm on EH. So, points B and C are somewhere on EH, which is the line x=0 from (0,0) to (0,8). Let's say point B is at (0, b) and point C is at (0, b+3), since the length BC is 3 cm. Then, since ABCD is a square, from point C, we move 3 cm to the right to get point D, which would be at (3, b+3). From D, we move down 3 cm to point A, which would be at (3, b). Then from A, moving left 3 cm brings us back to B at (0, b). So, square ABCD is a rectangle? Wait, no, since all sides are 3 cm and angles are 90 degrees, so it's a square. So, coordinates are:- B: (0, b)- C: (0, b+3)- D: (3, b+3)- A: (3, b)So, square ABCD is placed to the right of EH, attached at BC on EH. The value of b can vary between 0 and 5, since EH is 8 cm, and BC is 3 cm, so the highest point C can be is at (0,8), which would mean b+3=8 => b=5. If b is less than 5, then the square ABCD is placed lower on EH.Now, we need to find the area of triangle ABE. Let's see. Points A, B, and E. Point E is at (0,0). Point B is at (0, b). Point A is at (3, b). So, triangle ABE has vertices at (0,0), (0, b), and (3, b). Let me visualize this triangle. It's a right triangle with base along the y-axis from (0,0) to (0, b), and then a horizontal line from (0, b) to (3, b), and then a slant line from (3, b) back to (0,0). Wait, actually, the triangle would have vertices at E(0,0), B(0, b), and A(3, b). So, plotting these points, the base is from E to B, vertical, and then from B to A, horizontal, and from A back to E. But actually, the triangle is a right triangle on the y-axis and the line y = b. The base is EB of length b, and the height is BA of length 3. Wait, but in reality, the area would be (base * height)/2. The base is the vertical distance from E to B, which is b, and the height is the horizontal distance from B to A, which is 3. So, area would be (b * 3)/2 = 1.5b. So, if we can find the value of b, then we can compute the area of triangle ABE.But how do we find b? The problem gives the area of triangle ACG as 6.75 cm². So, we need to find coordinates of points A, C, G and compute the area, set it equal to 6.75, solve for b, then compute the area of ABE.Let's find the coordinates of points A, C, G. Point A is at (3, b). Point C is at (0, b+3). Point G is at (8,8). So, triangle ACG is formed by these three points.To find the area of triangle ACG, we can use the shoelace formula. Let's recall the formula: for points (x1,y1), (x2,y2), (x3,y3), the area is |(x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))/2|.Plugging in the coordinates:Point A: (3, b)Point C: (0, b+3)Point G: (8, 8)Compute the area:Area = |(3*( (b+3) - 8 ) + 0*(8 - b) + 8*(b - (b+3)) ) / 2|= |(3*(b - 5) + 0 + 8*(-3)) / 2|= |(3b - 15 - 24)/2|= |(3b - 39)/2|The area is given as 6.75 cm², which is 27/4. Wait, 6.75 is 27/4? Wait, 6.75 = 27/4? Let me check: 27 divided by 4 is 6.75, yes. So:|(3b - 39)/2| = 27/4Multiply both sides by 2:|3b - 39| = 27/2So, two cases:1) 3b - 39 = 27/22) -(3b - 39) = 27/2 => -3b + 39 = 27/2Solve case 1:3b = 39 + 27/2 = (78 + 27)/2 = 105/2b = (105/2)/3 = 105/6 = 35/2 = 17.5But wait, square EFGH has side length 8 cm, and point C is at (0, b+3). Since EH is from (0,0) to (0,8), b+3 must be ≤ 8. If b = 17.5, then b+3 = 20.5, which is way beyond 8. So, this is impossible. So, case 1 is invalid.Case 2:-3b + 39 = 27/2-3b = 27/2 - 39Convert 39 to 78/2:-3b = 27/2 - 78/2 = (-51)/2Divide both sides by -3:b = (-51/2)/(-3) = (51/2)/3 = 51/6 = 17/2 = 8.5Again, b = 8.5, which would mean point B is at (0, 8.5), but EH is only up to (0,8). So, this is also impossible. Wait, so both cases give b values outside the possible range (b has to be between 0 and 5, since C is at (0, b+3) and EH is up to (0,8)). So, b+3 ≤ 8 => b ≤ 5. But both solutions give b =17.5 and 8.5, which are greater than 5. That's a problem. Which suggests that perhaps my coordinate system is incorrect.Hmm, maybe I assigned the coordinates wrong. Let me double-check.Wait, maybe I messed up the orientation of the squares. If EH is a horizontal side instead of vertical. Let me try that.Alternative coordinate system: Let me place square EFGH with E at (0,0), F at (8,0), G at (8, -8), H at (0, -8). Wait, but that would make EH a horizontal side from (0,0) to (8,0). Wait, but in the problem statement, it just says "side BC lies on EH". So, EH could be horizontal. Let me try this.Let me place square EFGH with E at (0,0), F at (8,0), G at (8,8), and H at (0,8). Wait, no, that's a vertical EH. Wait, no, EFGH can be placed with EH as a horizontal side. Wait, squares can be placed in any orientation, but unless specified, we usually assume sides are horizontal/vertical.Wait, perhaps the problem considers EH as a horizontal side. Let's try that.Let me define square EFGH with E at (0,0), H at (8,0), so EH is the base, horizontal, length 8. Then, F would be at (8,8), and G at (0,8). Wait, but that would make EFGH a square rotated 90 degrees. Wait, no. Wait, if EH is horizontal from (0,0) to (8,0), then to make a square, the next point F would be at (8,0), then G at (8,8), and H at (0,8). Wait, no, that's not a square. Wait, actually, no.Wait, a square with horizontal side EH of length 8 would have E at (0,0), H at (8,0). Then, to make a square, the other points F and G would be at (8,0) moved up 8 units vertically and left 8 units horizontally. Wait, no. Wait, confusion here.Alternatively, if E is at (0,0), H is at (8,0), then the square EFGH would have sides EH, HF, FG, GE. Wait, no, that's not a square unless HF is perpendicular to EH. So, to make a square, from E(0,0) to H(8,0), then from H(8,0), move up 8 units to F(8,8), then from F(8,8) move left 8 units to G(0,8), then from G(0,8) down 8 units to E(0,0). That makes a square. So, in this case, EH is the bottom side from (0,0) to (8,0), HF is the right side from (8,0) to (8,8), FG is the top side from (8,8) to (0,8), and GE is the left side from (0,8) to (0,0). Wait, but then GE is vertical, same as HF. Wait, but in this case, the square EFGH is actually a diamond shape when placed with EH horizontal. Wait, no, if we have E at (0,0), H at (8,0), F at (8,8), G at (0,8), then it's a square rotated 45 degrees, but with sides of length 8√2, which contradicts the area. Wait, area would be (8√2)² / 2 = 64, which matches. Wait, no. Wait, the area of a square with side length s is s². If the distance between E and H is 8 units, but if EH is a diagonal of the square, then the area would be (8²)/2 = 32. But the problem states that the area is 64, so EH must be a side, not a diagonal.Therefore, the initial coordinate system must have EH as a vertical or horizontal side of length 8. So, perhaps the initial assumption of EH as vertical was correct, but then the conflict with BC being vertical arises. Alternatively, maybe the squares are placed such that EH is horizontal.Wait, let's try a different coordinate system where EH is horizontal. Let me set E at (0,0), H at (8,0). Then, since it's a square, the other two vertices F and G must be at (8,0) and some point. Wait, no. Wait, starting from E(0,0), moving along EH to H(8,0). Then, from H(8,0), moving up to form the square. So, the next vertex would be G(8,8), and then F would be (0,8). Wait, no. Wait, EFGH is a square. So, from E(0,0) to H(8,0) is one side. Then, from H(8,0), moving perpendicularly to EH to get to G. Since EH is horizontal, the perpendicular direction is vertical. So, moving up 8 units from H(8,0) to G(8,8). Then from G(8,8) moving left 8 units to F(0,8), and then down to E(0,0). Wait, but that would make F at (0,8), so connecting F to E would be a vertical line. But in that case, the square EFGH would have sides EH (8,0), HG (8,8), GF (0,8), and FE (0,0). Wait, but GF is from (8,8) to (0,8), which is horizontal, and FE is from (0,8) to (0,0), which is vertical. So, this is actually a rectangle, not a square, unless all sides are equal. Wait, EH is 8 units, HG is 8 units (from (8,0) to (8,8)), GF is 8 units (from (8,8) to (0,8)), and FE is 8 units (from (0,8) to (0,0)). Wait, no, FE is 8 units vertically, but GF is 8 units horizontally. So, actually, EFGH is a square with side length 8, placed with sides horizontal and vertical. So, E(0,0), H(8,0), G(8,8), F(0,8). Wait, but then connecting H to G to F to E would form a square. Yes, that works. So EH is the bottom side, HG is the right side, GF is the top side, and FE is the left side.But in this case, EH is horizontal from (0,0) to (8,0). Then, the problem states that BC lies on EH. So, BC is a side of square ABCD (area 9, side 3) lying on EH (the bottom side of EFGH from (0,0) to (8,0)). So, BC is a horizontal segment of length 3 on EH. So, points B and C are on EH, which is the line from (0,0) to (8,0). Let's assign coordinates accordingly.Let me denote the coordinates again. Square EFGH:- E: (0,0)- H: (8,0)- G: (8,8)- F: (0,8)Square ABCD has area 9, so side 3. BC is a side lying on EH. So, points B and C are on EH (the bottom side from (0,0) to (8,0)). Let’s say point B is at (x, 0) and point C is at (x + 3, 0), since BC is length 3. Then, since ABCD is a square, from point C, we move up 3 units to point D( x + 3, 3 ), then left 3 units to point A( x, 3 ), and back to B( x, 0 ). So, the square ABCD is sitting on the side EH of EFGH, starting at point B( x, 0 ) and extending up and to the left.Wait, but in this case, the square ABCD is above the side EH, extending upwards. But since EFGH is a square from (0,0) to (8,8), then if ABCD is placed on EH (the bottom side), it would be inside EFGH only if x + 3 ≤ 8, i.e., x ≤ 5. Otherwise, part of ABCD would be outside. But the problem doesn't specify whether ABCD is inside or outside EFGH. However, since the area of triangle ACG is given, which is likely formed by points from both squares, we need to figure out their positions.Point A is at (x, 3), point C is at (x + 3, 0), and point G is at (8,8). So, triangle ACG is formed by these three points. Let's compute its area.Using the shoelace formula again:Coordinates:A: (x, 3)C: (x + 3, 0)G: (8,8)Area = |(x*(0 - 8) + (x + 3)*(8 - 3) + 8*(3 - 0))/2|= |(x*(-8) + (x + 3)*5 + 8*3)/2|= |(-8x + 5x + 15 + 24)/2|= |(-3x + 39)/2|Given that the area is 6.75 cm² = 27/4 cm²:|(-3x + 39)/2| = 27/4Multiply both sides by 2:| -3x + 39 | = 27/2Again, two cases:1) -3x + 39 = 27/22) -(-3x + 39) = 27/2 => 3x - 39 = 27/2Case 1:-3x = 27/2 - 39Convert 39 to 78/2:-3x = 27/2 - 78/2 = -51/2x = (-51/2)/(-3) = 51/6 = 17/2 = 8.5But x is the coordinate on EH from (0,0) to (8,0). If x = 8.5, that's outside the range [0,8], so invalid.Case 2:3x - 39 = 27/23x = 27/2 + 39Convert 39 to 78/2:3x = 27/2 + 78/2 = 105/2x = (105/2)/3 = 105/6 = 35/2 = 17.5Again, x =17.5, which is way beyond 8, invalid.Wait, this is the same problem as before. Both solutions for x are outside the possible range. What's going on? Did I mess up the coordinate assignments again?Wait, maybe the square ABCD is placed such that it extends downward from EH, but EH is already the bottom side of EFGH. That doesn't make sense. Alternatively, maybe the square is placed to the side.Wait, perhaps my coordinate system is flipped. Let me try another approach. Let me suppose that EH is the top side of square EFGH. So, E is at (0,8), H is at (8,8), G is at (8,0), and F is at (0,0). So, EH is the top horizontal side from (0,8) to (8,8). Then, placing square ABCD on EH: BC lies on EH, so BC is a horizontal segment on the top side of EFGH.So, square ABCD has side BC on EH (the top side). Then, points B and C are on EH(0,8) to (8,8). Let's say point B is at (x,8) and point C is at (x + 3,8). Then, square ABCD extends downward from BC. So, point D would be at (x + 3,8 - 3) = (x + 3,5), and point A would be at (x,5). Then, connecting A to B is vertical from (x,5) to (x,8).Now, in this coordinate system, square EFGH has vertices E(0,8), H(8,8), G(8,0), F(0,0). Square ABCD is sitting on top of EH, extending downward. Points:- B: (x,8)- C: (x + 3,8)- D: (x + 3,5)- A: (x,5)Now, triangle ACG is formed by points A(x,5), C(x + 3,8), and G(8,0). Let's compute its area.Using the shoelace formula:Area = |x*(8 - 0) + (x + 3)*(0 - 5) + 8*(5 - 8)| / 2= |8x + (x + 3)*(-5) + 8*(-3)| / 2= |8x -5x -15 -24| / 2= |3x -39| / 2Given that the area is 6.75 = 27/4:|3x - 39| / 2 = 27/4Multiply both sides by 2:|3x - 39| = 27/2Again, two cases:1) 3x - 39 = 27/23x = 27/2 + 39 = 27/2 + 78/2 = 105/2x = 105/6 = 35/2 = 17.5Again, x =17.5, which is outside the EH side from (0,8) to (8,8). Not possible.Case 2:-(3x - 39) = 27/2-3x + 39 = 27/2-3x = 27/2 - 39 = 27/2 - 78/2 = -51/2x = (-51/2)/(-3) = 51/6 = 17/2 = 8.5Still x=8.5, which is outside the range. So, same problem.Hmm. So regardless of how I assign the coordinate system, the solutions for x or b are outside the permissible range. This suggests that perhaps my initial assumption about the placement of the squares is incorrect.Wait, maybe the squares are not axis-aligned? Maybe square EFGH is rotated such that EH is a diagonal or something. But the problem states that the area of EFGH is 64, so side length 8. If EH is a diagonal, then the side length would be EH / sqrt(2) = 8 / sqrt(2) = 4*sqrt(2), but then the area would be (4*sqrt(2))² = 32, which contradicts the given area of 64. So, EH must be a side, not a diagonal.Alternatively, maybe the squares are placed such that ABCD is inside EFGH, but given that ABCD has area 9 and EFGH has area 64, that's possible. But then BC lying on EH (a side of EFGH). So, BC is part of the side EH of EFGH. Let me think.If ABCD is inside EFGH, with BC on EH. Let me try coordinates again.Let me place square EFGH with E at (0,0), H at (8,0), G at (8,8), F at (0,8). So, EH is the bottom side. Square ABCD has BC lying on EH. So, points B and C are on EH from (0,0) to (8,0). Let's say B is at (x,0), C at (x+3,0). Then, since ABCD is a square, moving upwards from C to D, which would be at (x+3,3), and then left to A at (x,3), and down to B. So, square ABCD is sitting on the bottom side EH of EFGH, extending upward into EFGH. Then, point G is at (8,8).Triangle ACG: points A(x,3), C(x+3,0), G(8,8). Let's compute the area.Using shoelace formula:Area = |x*(0 - 8) + (x + 3)*(8 - 3) + 8*(3 - 0)| / 2= |x*(-8) + (x + 3)*5 + 8*3| / 2= |-8x + 5x + 15 + 24| / 2= |-3x + 39| / 2Set equal to 6.75 = 27/4:|-3x + 39| / 2 = 27/4Multiply both sides by 2:|-3x + 39| = 27/2Which is the same equation as before, leading to x =17.5 or x=8.5, both outside the EH side (0 ≤ x ≤ 8). So, impossible. Therefore, my coordinate system must be wrong.Wait, maybe the square ABCD is placed outside of EFGH? If EH is the bottom side of EFGH, and ABCD is attached to EH but outside, so extending downward. Then, points B and C are on EH (the bottom side of EFGH), and square ABCD extends below EH.In this case, coordinates:EFGH:- E(0,0), H(8,0), G(8,8), F(0,8)ABCD:- B(x,0), C(x+3,0), D(x+3,-3), A(x,-3)Then, triangle ACG: points A(x,-3), C(x+3,0), G(8,8)Compute area using shoelace formula:Area = |x*(0 - 8) + (x + 3)*(8 - (-3)) + 8*(-3 - 0)| / 2= |x*(-8) + (x + 3)*11 + 8*(-3)| / 2= |-8x + 11x + 33 - 24| / 2= |3x + 9| / 2Set equal to 6.75 = 27/4:|3x + 9| / 2 = 27/4Multiply both sides by 2:|3x + 9| = 27/2So two cases:1) 3x + 9 = 27/23x = 27/2 - 9 = 27/2 - 18/2 = 9/2x = 9/6 = 3/2 = 1.52) -(3x + 9) = 27/2-3x -9 = 27/2-3x = 27/2 + 9 = 27/2 + 18/2 = 45/2x = -45/6 = -15/2 = -7.5Case 2 is invalid because x must be between 0 and 5 (since C is at x+3 ≤ 8 => x ≤ 5). So, x=1.5 is valid. Therefore, x=1.5. So, points:- B(1.5, 0)- C(4.5, 0)- D(4.5, -3)- A(1.5, -3)So, square ABCD is placed below EH, extending downward from x=1.5 to x=4.5 on EH.Now, we need to find the area of triangle ABE. Points A(1.5, -3), B(1.5,0), E(0,0).Using coordinates:A(1.5, -3), B(1.5,0), E(0,0)This is a triangle with vertices at (0,0), (1.5,0), and (1.5,-3). The base can be from E(0,0) to B(1.5,0), which is 1.5 cm, and the height is from B down to A(1.5,-3), which is 3 cm. So, area is (base * height)/2 = (1.5 * 3)/2 = 4.5 / 2 = 2.25 cm². Which is 9/4 cm². Alternatively, using the shoelace formula:Area = |0*(0 - (-3)) + 1.5*(-3 - 0) + 1.5*(0 - 0)| / 2= |0 + 1.5*(-3) + 1.5*0| / 2= | -4.5 | / 2= 4.5 / 2= 2.25 cm²So, 2.25 cm², which is 9/4. Therefore, the area of triangle ABE is 2.25 cm², which is 9/4. So, the answer should be boxed{dfrac{9}{4}} or boxed{2.25}.But let me confirm once more. If x=1.5, then ACG area is:Points A(1.5, -3), C(4.5,0), G(8,8)Area calculation:Using shoelace formula:1.5*(0 - 8) + 4.5*(8 - (-3)) + 8*(-3 - 0)= 1.5*(-8) + 4.5*11 + 8*(-3)= -12 + 49.5 -24= 13.5Absolute value divided by 2: |13.5| /2 = 6.75, which matches. So, correct.Therefore, the area of triangle ABE is 2.25 cm², which is 9/4. So, boxed answer: boxed{dfrac{9}{4}}.Final AnswerThe area of triangle ( A B E ) is boxed{dfrac{9}{4}} square centimeters.</think>The area of square (ABCD) is 9 square centimeters, so each side is 3 cm. The area of square (EFGH) is 64 square centimeters, so each side is 8 cm. Side (BC) of square (ABCD) lies on side (EH) of square (EFGH). We need to find the area of triangle (ABE) given that the area of triangle (ACG) is 6.75 square centimeters.1. Coordinates Assignment: - Place square (EFGH) with (E(0,0)), (H(8,0)), (G(8,8)), and (F(0,8)). - Square (ABCD) is placed such that (BC) lies on (EH). Let (B(x,0)) and (C(x+3,0)) since (BC = 3) cm. - Points (A) and (D) are determined by moving downward from (B) and (C) respectively, resulting in (A(x,-3)) and (D(x+3,-3)).2. Area Calculation for Triangle (ACG): - Points (A(x,-3)), (C(x+3,0)), and (G(8,8)). - Using the shoelace formula: [ text{Area} = frac{1}{2} left| x(0 - 8) + (x + 3)(8 - (-3)) + 8(-3 - 0) right| ] Simplifying: [ left| -8x + 11(x + 3) - 24 right| / 2 = left| 3x + 9 right| / 2 ] - Given area is 6.75 cm²: [ left| 3x + 9 right| / 2 = 27/4 implies left| 3x + 9 right| = 27/2 ] - Solving for (x): [ 3x + 9 = 27/2 implies x = 1.5 ]3. Area Calculation for Triangle (ABE): - Points (A(1.5, -3)), (B(1.5, 0)), and (E(0, 0)). - Using the shoelace formula: [ text{Area} = frac{1}{2} left| 0(0 - (-3)) + 1.5(-3 - 0) + 1.5(0 - 0) right| ] Simplifying: [ left| 0 - 4.5 + 0 right| / 2 = 4.5 / 2 = 2.25 ]The area of triangle (ABE) is (boxed{dfrac{9}{4}}) square centimeters.]

🔑:2.25

🧠:[<think>Okay, so I need to find the number of good permutations of the numbers 1 through 10. A permutation is bad if there exist three indices i, j, k with i < j < k such that σ(j) < σ(k) < σ(i). Otherwise, it's good. Hmm. So, a good permutation is one where you can't find such a triple. First, let's understand what this condition means. If a permutation is bad, there's a subsequence of three elements where the middle one is the smallest, and the last one is the next smallest. For example, if we have positions i, j, k with values like 5, 2, 3. Here, σ(i) = 5, σ(j) = 2, σ(k) = 3. Then, since 2 < 3 < 5, and i < j < k, this would be a bad permutation. So, the key is avoiding such 3-element decreasing-then-increasing patterns. Wait, actually, it's not exactly a valley or a peak, it's more specific. It's a situation where the middle element is the smallest, and the third element is in between the first and the middle. So, it's like a 3-element pattern where the middle is the minimum, and the third is the second smallest. So, this is sometimes called a 132 pattern? Let me confirm: in permutation patterns, a 132 pattern is when you have a subsequence where the first element is the largest, the second is the smallest, and the third is the middle. So yes, exactly this scenario. So, avoiding the 132 pattern. Therefore, the problem reduces to counting the number of 132-avoiding permutations of length 10.Now, permutation pattern avoidance is a classic topic in combinatorics. For certain patterns, like 123 (increasing) or 132, the number of permutations avoiding them is given by the Catalan numbers. Wait, is that the case? Let me recall. The Catalan numbers count the number of permutations avoiding any 3-letter pattern, but actually, that's not quite right. Wait, the number of 132-avoiding permutations of length n is the nth Catalan number. Similarly, Catalan numbers count the number of permutations avoiding any single 3-letter pattern. So, yes, the answer should be the 10th Catalan number.But let me verify this. Let's check for small n. For n=1, there's 1 permutation, which is trivially good. Catalan number C1 = 1. For n=2, there are 2 permutations, both good since you can't have three elements. C2 = 2. For n=3, the Catalan number is 5. Let's count the 132-avoiding permutations. There are 6 permutations of 3 elements. The only one that contains a 132 pattern is the permutation 132 itself. So, 5 permutations avoid it, which matches C3=5. For n=4, C4=14. Let's see, there are 24 permutations. The number of 132-avoiding permutations should be 14. So, if that's consistent, then the numbers do match. Therefore, the number of good permutations is the 10th Catalan number.But let me just make sure that the problem is equivalent to 132-avoiding permutations. The problem states that a permutation is bad if there exists i < j < k with σ(j) < σ(k) < σ(i). So, in other words, there is a subsequence where the middle term is the smallest, and the last term is in the middle in terms of value. So, that is exactly the 132 pattern. Because if you have positions i < j < k with σ(j) < σ(k) < σ(i), then the values correspond to σ(i) being the largest, σ(j) the smallest, and σ(k) the middle, which is the 132 pattern. So yes, avoiding such patterns is equivalent to 132-avoiding permutations, which are counted by Catalan numbers.Therefore, the number of good permutations is the 10th Catalan number. The nth Catalan number is given by the formula (1/(n+1)) * binomial(2n, n). So, for n=10, that would be (1/11) * binomial(20, 10). Let me compute that.First, binomial(20, 10) is 184756. Then, 184756 divided by 11. Let's compute 184756 / 11. 11*16796 = 184,756. Because 11*16000 = 176,000. 184,756 - 176,000 = 8,756. 11*800 = 8,800, which is too much. 11*796 = 8756. Therefore, 16000 + 796 = 16796. So, 184756 /11 =16796. So, the 10th Catalan number is 16796. Therefore, the number of good permutations is 16796.Wait, but let me just cross-check with another formula for Catalan numbers. The recursive formula is C(n) = sum_{i=0}^{n-1} C(i) C(n-1 -i). For n=10, this would be a bit tedious, but maybe we can check a known value. The 10th Catalan number is indeed 16796. Let me confirm with a list of Catalan numbers:C0 = 1C1 = 1C2 = 2C3 = 5C4 = 14C5 = 42C6 = 132C7 = 429C8 = 1430C9 = 4862C10 = 16796Yes, that's correct. So, C10 is 16796. Therefore, the answer should be 16796. So, the number of good permutations is the 10th Catalan number, which is 16796.But hold on, let me make sure there isn't a misunderstanding in the problem statement. The problem says "there exist i < j < k" such that "σ(j) < σ(k) < σ(i)". So, if such a triplet exists, the permutation is bad. Therefore, good permutations are exactly those that avoid this pattern, which is the 132 pattern. So, 132-avoiding permutations are counted by Catalan numbers, so the answer is the 10th Catalan number. Therefore, 16796.Alternatively, sometimes people might confuse 132-avoiding with other patterns, but since the problem explicitly states the forbidden pattern as a 132, then yes, Catalan number applies. Therefore, the answer is 16796.Final AnswerThe number of good permutations is boxed{16796}.</think>To determine the number of good permutations of the numbers (1, 2, ldots, 10), we need to avoid permutations that contain a specific pattern. A permutation (sigma) is bad if there exist indices (i, j, k) such that (1 leq i < j < k leq 10) and (sigma(j) < sigma(k) < sigma(i)). This pattern is equivalent to the 132 pattern in permutation terminology.Permutations avoiding the 132 pattern are counted by the Catalan numbers. The (n)-th Catalan number is given by the formula:[C_n = frac{1}{n+1} binom{2n}{n}]For (n = 10), the 10th Catalan number is calculated as follows:[C_{10} = frac{1}{11} binom{20}{10}]Calculating the binomial coefficient:[binom{20}{10} = 184756]Dividing by 11:[frac{184756}{11} = 16796]Thus, the number of good permutations is the 10th Catalan number, which is (boxed{16796}).,<think>Okay, so I need to find the number of "good" permutations of the numbers 1 through 10. A permutation is bad if there exist three indices i < j < k such that σ(j) < σ(k) < σ(i). If no such triplet exists, the permutation is good. Hmm, okay. So my task is to count all permutations that do not have this specific pattern. Let me try to understand the problem better. The forbidden pattern is when you have a subsequence (not necessarily consecutive) where the middle element is smaller than the last element, and the last element is smaller than the first. In other words, in the permutation, if you can find three elements such that the one in the middle position is the smallest, followed by the last one, then the first one, that's a bad permutation. So for example, if in the permutation, there is a subsequence like 3, 1, 2, then that's bad because 1 (the middle element) < 2 (the last element) < 3 (the first element). Wait, actually, the indices have to be i < j < k, so the positions have to be in order. So in the permutation, the values at positions i, j, k should have σ(j) < σ(k) < σ(i). So the middle position j has the smallest value, then position k has a larger value, and position i has the largest value. So the pattern is middle, then last, then first in terms of values. So like 2, 1, 3 would not be bad because here i=1, j=2, k=3, but σ(j)=1, σ(k)=3, σ(i)=2. So σ(j) < σ(i) < σ(k). That's different. So the bad pattern is specifically when σ(j) < σ(k) < σ(i) for i < j < k. So in terms of the permutation, the values have the middle one smallest, then the last, then the first. So an example would be if we have positions 1, 2, 3 with values 3, 1, 2. Then σ(2)=1 < σ(3)=2 < σ(1)=3, so that's bad. So permutations that avoid this 3-element pattern are called "good". The question is asking how many such good permutations there are for n=10.I remember that permutations avoiding specific patterns are studied in combinatorics, often related to Catalan numbers or other classical results. But the exact count depends on the specific pattern. The problem here is similar to 321-avoiding permutations or 132-avoiding permutations. Wait, let's recall: a 132-avoiding permutation is one that does not have a subsequence where the first element is the smallest, the second is the largest, and the third is the middle. So 1, 3, 2 would be the forbidden pattern. Wait, but in our case, the forbidden pattern is different. Let's compare.The forbidden pattern here is σ(j) < σ(k) < σ(i) with i < j < k. Let's assign indices. Let’s say positions are i, j, k. So in the permutation, at these positions, the values are σ(i), σ(j), σ(k). The condition is σ(j) < σ(k) < σ(i). So in terms of the actual values, it's middle position has the smallest value, followed by the last position, then the first position. So in the actual permutation, the relative order is high, low, medium. Wait, no. Let's take an example. Let's say i=1, j=2, k=3. Then σ(2) < σ(3) < σ(1). So the values at positions 1,2,3 would be, say, 3,1,2. So 3 (position 1), 1 (position 2), 2 (position 3). So the relative order is high, low, medium. So the forbidden pattern is 3-1-2. Which is similar to the 312 pattern. Wait, 312-avoiding permutations. Let me check: a 312-avoiding permutation is one that does not have a subsequence where the first element is the third smallest, the second is the first smallest, and the third is the second smallest. So exactly 3-1-2. So yes, the forbidden pattern here is 312. So the problem is equivalent to counting the number of 312-avoiding permutations of length 10.Therefore, if that is the case, the number of such permutations is given by the 10th Catalan number. Because it's known that the number of 312-avoiding permutations of length n is the nth Catalan number. Let me confirm this. Catalan numbers count the number of Dyck paths, the number of non-crossing partitions, the number of triangulations, and also the number of 312-avoiding permutations (and other permutation classes). For example, the number of 312-avoiding permutations of length n is indeed the nth Catalan number. Similarly, 321-avoiding permutations are also counted by Catalan numbers. Wait, but I need to make sure whether 312-avoiding permutations are Catalan. Let me verify with small n.For n=1, Catalan number is 1. There's only one permutation, which is trivially 312-avoiding.For n=2, Catalan number is 2. The permutations are 12 and 21. Both avoid 312 because there's no three-element subsequence. So that works.For n=3, Catalan number is 5. The total permutations are 6. The 312-avoiding permutations would be all except the one that has 312. Let's list them:123, 132, 213, 231, 321, 312.The permutation 312 is the only one containing the 312 pattern, so the number of 312-avoiding permutations is 5, which is the 3rd Catalan number. So yes, that seems to hold.Therefore, the number of 312-avoiding permutations of length n is the nth Catalan number, which is given by C_n = (1/(n+1)) * (2n choose n). Therefore, for n=10, the number should be C_10 = (1/11) * (20 choose 10).Calculating that: (20 choose 10) is 184756, then divided by 11. Let me compute 184756 ÷ 11. 11*16796 = 184756 (since 11*16000=176000, 11*700=7700, 11*96=1056; 176000+7700=183700; 183700+1056=184756). Therefore, C_10 = 16796.Therefore, the number of good permutations is the 10th Catalan number, which is 16,796.But let me just make sure that the forbidden pattern in the problem is indeed equivalent to 312-avoiding permutations. The problem states that a permutation is bad if there exist i < j < k with σ(j) < σ(k) < σ(i). Let's map this to the permutation pattern. The indices i, j, k correspond to positions 1, 2, 3 in the pattern. The values σ(i), σ(j), σ(k) must satisfy σ(j) < σ(k) < σ(i). So if we think of the pattern as the permutation 312, because in the pattern positions 1,2,3 have values 3,1,2. Wait, but in the problem statement, the pattern is in terms of the values: the middle position j has the smallest value, followed by k, then i. So in terms of the relative order, it's middle, then last, then first. Which corresponds to the permutation 312 when considering the positions. So if you have a subsequence where the first element (i) is the third highest, the second element (j) is the lowest, and the third element (k) is the second highest. So the relative order of the values is 3,1,2. So yes, that's the 312 pattern. Therefore, avoiding this pattern is equivalent to 312-avoiding permutations, counted by Catalan numbers.Therefore, the answer should be the 10th Catalan number, which is 16,796.But just to be thorough, let me check another source or example. Let's take n=3. The problem states that a permutation is bad if there exists i < j < k with σ(j) < σ(k) < σ(i). For n=3, how many bad permutations are there? There is only one: the permutation 312. Because in 312, we have σ(2)=1 < σ(3)=2 < σ(1)=3. So that's the only bad permutation. Therefore, the number of good permutations is 5, which is C_3=5. So that checks out.Similarly, for n=4, the Catalan number is 14. Let's confirm that with the logic. The total permutations are 24. The number of 312-avoiding permutations is 14. So if we exclude 24 - 14 = 10 permutations that contain the 312 pattern. Wait, but actually, the number might not be exactly total minus Catalan, but according to theory, the 312-avoiding permutations are counted by Catalan numbers. Therefore, for n=4, it's 14. So yes, if I can trust that the 312-avoiding permutations are Catalan, then this should hold.Therefore, generalizing, the number of good permutations for n=10 is the 10th Catalan number. Therefore, the answer is 16796.But just to be 100% sure, let me confirm with another approach. Suppose we try to model the permutations avoiding the 312 pattern. It's known that such permutations are counted by the Catalan numbers, which also count Dyck paths. The correspondence is often through bijections. For example, the number of 312-avoiding permutations can be shown to be Catalan via recursive arguments, where you consider the position of the element 1 and recursively build the permutation. But since the problem is about 10 elements, and the answer is a Catalan number, unless there is a misunderstanding in the problem statement, this should be correct.Wait, let me re-examine the problem statement. It says a permutation is bad if there exist i < j < k with σ(j) < σ(k) < σ(i). So, the middle index j has the smallest value, then k, then i. So the pattern is σ(j) < σ(k) < σ(i). Let's check if this is equivalent to the 312 pattern. In the 312 pattern, the first element is the third smallest, the second is the first smallest, and the third is the second smallest. So in terms of values, they are 3,1,2. So in the permutation, if you have three elements in positions i < j < k, their values are 3,1,2. But in the problem, the condition is σ(j) < σ(k) < σ(i). So in terms of the positions, the middle position j has the smallest value, the last position k has the middle value, and the first position i has the largest value. So in terms of the values, ordered by positions: σ(i), σ(j), σ(k) correspond to high, low, medium. So in the actual permutation, the subsequence is high, low, medium. Which is exactly the 312 pattern. Because 3 is high, 1 is low, 2 is medium. So yes, the forbidden pattern is 312. Therefore, avoiding this pattern gives the Catalan number.Thus, the number of good permutations is the 10th Catalan number, which is 16796. Therefore, the answer is boxed{16796}.Final Answerboxed{16796}</think>To determine the number of good permutations of the numbers 1 through 10, we need to count permutations that avoid the specific pattern where there exist indices (i < j < k) such that (sigma(j) < sigma(k) < sigma(i)). This pattern is equivalent to avoiding the 312 permutation pattern.A permutation is bad if it contains a subsequence where the middle element is the smallest, followed by the last element, and then the first element. This is known as the 312 pattern. Permutations avoiding the 312 pattern are counted by the Catalan numbers.The (n)-th Catalan number is given by:[C_n = frac{1}{n+1} binom{2n}{n}]For (n = 10), the 10th Catalan number is:[C_{10} = frac{1}{11} binom{20}{10}]Calculating this, we find:[binom{20}{10} = 184756][C_{10} = frac{184756}{11} = 16796]Thus, the number of good permutations of the numbers 1 through 10 is (boxed{16796}).]

🔑:1. Understanding the Problem: We need to find the number of permutations of the numbers (1, 2, ldots, 10) that do not contain a subsequence of the form ( sigma(j) < sigma(k) < sigma(i) ) for (1 leq i < j < k leq 10). Such permutations are known as 312-avoiding permutations.2. Characterizing 312-Avoiding Permutations: A permutation is 312-avoiding if it does not contain any subsequence of three elements where the first element is the largest, the second element is the smallest, and the third element is in between. These permutations are also known as stack-sortable permutations.3. Catalan Numbers: The number of 312-avoiding permutations of (n) elements is given by the (n)-th Catalan number. The (n)-th Catalan number (C_n) is given by: [ C_n = frac{1}{n+1} binom{2n}{n} ] This formula arises from the recursive nature of 312-avoiding permutations.4. Recursive Definition: Let (g(n)) denote the number of 312-avoiding permutations of (1, 2, ldots, n). The recursive formula for (g(n)) is: [ g(n) = sum_{j=0}^{n-1} g(j) g(n-1-j) ] This recurrence relation is characteristic of Catalan numbers.5. Calculating (g(10)): Using the formula for Catalan numbers, we calculate (C_{10}): [ C_{10} = frac{1}{10+1} binom{20}{10} = frac{1}{11} binom{20}{10} ] First, compute (binom{20}{10}): [ binom{20}{10} = frac{20!}{10! cdot 10!} ] Using factorial values: [ 20! = 2,432,902,008,176,640,000 ] [ 10! = 3,628,800 ] Therefore: [ binom{20}{10} = frac{2,432,902,008,176,640,000}{(3,628,800)^2} = 184,756 ] Now, compute (C_{10}): [ C_{10} = frac{1}{11} times 184,756 = 16,796 ]The final answer is (boxed{16,796}).

🧠:[<think>Okay, let's see. The problem is about finding the sum of a series of fractions where each denominator is the sum of consecutive integers starting from 1 up to some number. The series starts with 1/(1+2), then 1/(1+2+3), and continues all the way up to 1/(1+2+…+2022). Then, after finding this sum as a reduced fraction m/n, we need to compute m + n. Alright, so first, I need to understand the general term of this series and see if there's a pattern or formula that can help sum it up.First, let me recall that the sum of the first k positive integers is given by the formula S_k = k(k + 1)/2. So, for example, 1 + 2 = 3, which is 2(2 + 1)/2 = 3. Similarly, 1 + 2 + 3 = 6, which is 3(3 + 1)/2 = 6. So, each denominator in the series is actually S_{k} where k starts from 2 (since the first term is 1/(1+2) which is S_2) and goes up to 2022. Wait, but the problem says "1/(1+2) + 1/(1+2+3) + … + 1/(1+2+…+2022)". So the denominators are S_2, S_3, ..., S_{2022}. Therefore, the general term is 1/S_k where k ranges from 2 to 2022. Therefore, the sum is Σ_{k=2}^{2022} 1/S_k. Since S_k = k(k + 1)/2, each term is 2/(k(k + 1)). Therefore, the sum becomes Σ_{k=2}^{2022} 2/(k(k + 1)). Hmm, so maybe this is a telescoping series? Because 2/(k(k + 1)) can be expressed as a difference of two fractions. Let me recall that 1/(k(k + 1)) = 1/k - 1/(k + 1). So, multiplying by 2, we have 2/(k(k + 1)) = 2(1/k - 1/(k + 1)). Therefore, the sum Σ_{k=2}^{2022} 2/(k(k + 1)) is equal to 2 Σ_{k=2}^{2022} (1/k - 1/(k + 1)). This is indeed a telescoping series. Let's write out the terms to see how the cancellation works. Let's consider the sum from k = 2 to k = 2022 of (1/k - 1/(k + 1)):= (1/2 - 1/3) + (1/3 - 1/4) + (1/4 - 1/5) + ... + (1/2022 - 1/2023)Each negative term cancels with the positive term of the next fraction. So, all the middle terms cancel, and we're left with the first term of the first expression, which is 1/2, and the last term of the last expression, which is -1/2023. Therefore, the sum is 1/2 - 1/2023. Then, multiplying by 2 (since each term was multiplied by 2), the total sum is 2*(1/2 - 1/2023) = (1 - 2/2023). Simplifying this, 1 is 2023/2023, so 2023/2023 - 2/2023 = 2021/2023. Therefore, the sum is 2021/2023. Now, the problem states that this is equal to m/n where m and n are coprime positive integers. So we need to check if 2021 and 2023 share any common factors. First, note that 2023 is 2023 divided by 7: 7*289 = 2023? Wait, 7*288 = 2016, so 2023 - 2016 = 7, so 2023 = 7*289 + 0? Wait, 7*289 = 2023. Let me check that. 289 is 17^2, right? 17*17 = 289. So 7*17^2 = 2023. Therefore, 2023 = 7*17^2. Now, 2021: let's factor 2021. Let's check divisibility by small primes. 2021 ÷ 43 is 47, because 43*47 = 2021. Let me verify: 40*47 = 1880, 3*47 = 141, so 1880 + 141 = 2021. Yes, correct. So 2021 = 43*47. So, 2021 is 43*47, and 2023 is 7*17*17. Are there any common factors between 2021 and 2023? Let's see. 43, 47, 7, 17. All primes here. 43, 47, 7, 17 are all primes, and none are shared between the two factorizations. Therefore, 2021 and 2023 are coprime. Therefore, the fraction 2021/2023 is already in its simplest form, so m = 2021 and n = 2023. Therefore, m + n = 2021 + 2023 = 4044. Wait, let me double-check the arithmetic. 2021 + 2023: 2000 + 2000 = 4000, 21 + 23 = 44, so total is 4044. Correct. But let me just confirm each step again to make sure I didn't skip anything. First, the sum is from k=2 to k=2022 of 1/S_k, where S_k = k(k+1)/2. Therefore, each term is 2/(k(k+1)), which telescopes when summed. Then, expanding the telescoping series gives 2*(1/2 - 1/2023) = 1 - 2/2023 = 2021/2023. Then, checking if 2021 and 2023 are coprime by factoring both. 2021 factors into 43*47, and 2023 factors into 7*17*17. Since there's no overlap in prime factors, the fraction is reduced. Therefore, m + n = 4044. Yes, that seems correct. So the answer is 4044.Final Answerboxed{4044}</think>The problem requires finding the sum of the series (frac{1}{1+2} + frac{1}{1+2+3} + cdots + frac{1}{1+2+cdots+2022}) and expressing it as a reduced fraction (frac{m}{n}), then computing (m + n).1. Identify the general term: Each term in the series is (frac{1}{1+2+cdots+k}) where (k) ranges from 2 to 2022. The sum of the first (k) positive integers is given by (S_k = frac{k(k+1)}{2}). Therefore, each term can be written as (frac{2}{k(k+1)}).2. Rewrite the series: The series becomes (sum_{k=2}^{2022} frac{2}{k(k+1)}).3. Telescoping series: Recognize that (frac{2}{k(k+1)}) can be decomposed as (2 left( frac{1}{k} - frac{1}{k+1} right)). The series then transforms into a telescoping series: [ 2 sum_{k=2}^{2022} left( frac{1}{k} - frac{1}{k+1} right) ]4. Evaluate the telescoping series: The series simplifies as: [ 2 left( left( frac{1}{2} - frac{1}{3} right) + left( frac{1}{3} - frac{1}{4} right) + cdots + left( frac{1}{2022} - frac{1}{2023} right) right) ] Most terms cancel out, leaving: [ 2 left( frac{1}{2} - frac{1}{2023} right) = 2 left( frac{2023 - 2}{2 cdot 2023} right) = 1 - frac{2}{2023} = frac{2021}{2023} ]5. Check if the fraction is reduced: Factorizing 2021 and 2023, we find (2021 = 43 times 47) and (2023 = 7 times 17^2). They share no common factors, so (frac{2021}{2023}) is in simplest form.6. Compute (m + n): Here, (m = 2021) and (n = 2023), so (m + n = 2021 + 2023 = 4044).[boxed{4044}],<think>Alright, let's tackle this problem step by step. The problem is to find the sum of a series of fractions where each denominator is the sum of consecutive integers starting from 1 up to some number. The series starts with 1/(1+2) + 1/(1+2+3) + ... all the way up to 1/(1+2+...+2022). Then, we need to express this sum as a reduced fraction m/n and find m + n. First, I need to understand the structure of each term in the series. Each denominator seems to be the sum of the first k integers, starting from 1. For example, the first term is 1/(1+2), which is the sum from 1 to 2. The next term is 1/(1+2+3), which is the sum from 1 to 3, and so on until the last term, which is the sum from 1 to 2022. Wait, hold on. The first term is 1/(1+2). But the sum from 1 to 2 is (2*3)/2 = 3, right? So the first term is 1/3. The next term is 1/(1+2+3) which is 6, so 1/6. Then the next term would be 1/10, since 1+2+3+4=10, and so on. But the problem statement says the series starts with 1/(1+2), so the first term corresponds to k=2, the next to k=3, up to k=2022. Therefore, the general term in the series is 1 divided by the sum from 1 to k, where k ranges from 2 to 2022. So, the series is the sum from k=2 to k=2022 of 1/(1+2+...+k).But wait, actually, the first term is 1/(1+2) which is k=2, the next term is 1/(1+2+3) which is k=3, and so the last term is 1/(1+2+...+2022) which is k=2022. So the summation is from k=2 to k=2022 of 1/(sum from i=1 to k of i).Now, I need to recall that the sum of the first k integers is given by the formula S = k(k + 1)/2. Therefore, each term can be rewritten as 1/(k(k + 1)/2) = 2/(k(k + 1)). So, the entire sum becomes the sum from k=2 to k=2022 of 2/(k(k + 1)).Therefore, the problem reduces to calculating the sum from k=2 to k=2022 of 2/(k(k + 1)).Now, 2/(k(k + 1)) is a telescoping term. I remember that fractions of the form 1/(k(k + 1)) can be written as 1/k - 1/(k + 1). Let me verify that:1/k - 1/(k + 1) = (k + 1 - k)/(k(k + 1)) = 1/(k(k + 1)). So, yes, indeed. Therefore, 2/(k(k + 1)) = 2*(1/k - 1/(k + 1)).So, substituting back into the sum, we have:Sum from k=2 to 2022 of 2*(1/k - 1/(k + 1)) = 2*Sum from k=2 to 2022 (1/k - 1/(k + 1)).Now, telescoping series are nice because many terms cancel out. Let's write out the first few terms and the last few terms to see the pattern.For k=2: 2*(1/2 - 1/3)For k=3: 2*(1/3 - 1/4)For k=4: 2*(1/4 - 1/5)...For k=2022: 2*(1/2022 - 1/2023)When we add these up, the -1/3 from the first term cancels with the +1/3 from the second term, the -1/4 from the second term cancels with the +1/4 from the third term, and so on, all the way up to the -1/2022 from the second-to-last term cancels with the +1/2022 from the last term. Therefore, the only terms that don't cancel are the first positive term and the last negative term.So, the entire sum becomes 2*(1/2 - 1/2023).Let me check that again. Let's factor out the 2:Total sum = 2*[ (1/2 - 1/3) + (1/3 - 1/4) + ... + (1/2022 - 1/2023) ]Each negative term cancels with the positive term in the next fraction. So, all the intermediate terms cancel, leaving:2*(1/2 - 1/2023) = 2*( (1/2) - (1/2023) )Calculating that:First, compute 1/2 - 1/2023. To subtract these fractions, find a common denominator. The common denominator is 2*2023 = 4046.So:1/2 = 2023/40461/2023 = 2/4046Therefore, 1/2 - 1/2023 = (2023 - 2)/4046 = 2021/4046Then, multiplying by 2:2*(2021/4046) = (2*2021)/4046But 4046 is 2*2023, so:(2*2021)/(2*2023) = 2021/2023Therefore, the total sum simplifies to 2021/2023.Wait, but wait, let me check the arithmetic again. Let's do it step by step:Compute 1/2 - 1/2023:Convert to decimal if necessary to check:1/2 = 0.51/2023 ≈ 0.000494, so 0.5 - 0.000494 ≈ 0.499506Then 2021/4046 ≈ 2021 divided by 4046. Let's compute 2021*2=4042, so 2021/4046 = 2021/(2*2023) = (2021/2023)/2. Wait, maybe better to divide numerator and denominator by GCD(2021,2023). Since 2023 - 2021 = 2, so GCD(2021,2023)= GCD(2021,2)=1, since 2021 is odd. Similarly, 2023 is 2023 = 7*17*17, wait, 2023 divided by 7 is 289, and 289 is 17 squared. So 2023=7*17^2. 2021 is 43*47 (since 43*47=2021). So 2021 and 2023 are coprime? Because 2023 is 7*17^2 and 2021 is 43*47. There's no common prime factors. So 2021/2023 is already reduced.Wait, but in the calculation above, the sum simplifies to 2021/2023. Then the problem states that the sum is m/n where m and n are coprime, so m=2021 and n=2023, and m + n = 2021 + 2023 = 4044.But let's make sure I didn't make a miscalculation in the telescoping sum.Starting again:Sum from k=2 to 2022 of 2/(k(k+1)) = 2*Sum from k=2 to 2022 [1/k - 1/(k+1)]Expanding the sum:Sum [1/2 - 1/3 + 1/3 - 1/4 + ... + 1/2022 - 1/2023]All the middle terms cancel, leaving 1/2 - 1/2023.Then multiply by 2: 2*(1/2 - 1/2023) = 1 - 2/2023 = (2023/2023 - 2/2023) = 2021/2023.Yes, that's correct. Therefore, the sum is indeed 2021/2023. Since 2021 and 2023 are coprime (as established earlier, they share no common factors), m=2021, n=2023, so m + n = 2021 + 2023 = 4044.Wait, but let me confirm once more that 2021 and 2023 are coprime. 2021 factors: let's see, 2021 divided by 43 is 47 (since 43*47=2021). 2023 divided by 7 is 289, which is 17^2. So 2021=43*47, 2023=7*17^2. They have no common prime factors, so yes, GCD(2021,2023)=1. Therefore, the fraction is reduced, and m + n is 4044.Therefore, the answer should be 4044.Final Answerboxed{4044}</think>The problem requires finding the sum of the series (frac{1}{1+2} + frac{1}{1+2+3} + cdots + frac{1}{1+2+cdots+2022}) and expressing it as a reduced fraction (frac{m}{n}), then finding (m + n).1. Identify the general term: Each term in the series is the reciprocal of the sum of the first (k) integers, starting from (k=2) to (k=2022). The sum of the first (k) integers is given by (frac{k(k+1)}{2}).2. Rewrite each term: The general term (frac{1}{1+2+cdots+k}) can be rewritten as (frac{2}{k(k+1)}).3. Express as a telescoping series: The term (frac{2}{k(k+1)}) can be decomposed into (2left(frac{1}{k} - frac{1}{k+1}right)).4. Sum the series: [ sum_{k=2}^{2022} frac{2}{k(k+1)} = 2 sum_{k=2}^{2022} left(frac{1}{k} - frac{1}{k+1}right) ]5. Telescoping effect: Most terms cancel out, leaving: [ 2 left( left(frac{1}{2} - frac{1}{2023}right) right) ]6. Simplify the remaining terms: [ 2 left( frac{1}{2} - frac{1}{2023} right) = 2 left( frac{2023 - 2}{2 cdot 2023} right) = 2 left( frac{2021}{4046} right) = frac{2021}{2023} ]7. Check coprimality: The numerator 2021 and the denominator 2023 are coprime since they share no common prime factors.Thus, (m = 2021) and (n = 2023), leading to (m + n = 2021 + 2023 = 4044).[boxed{4044}]]

🔑:4044

🧠:[<think>Alright, let me try to figure out this probability problem. So, the question is: we have a point P with coordinates (m, n), where m and n are the results of rolling a die twice consecutively. We need to find the probability that P lies outside the circle defined by the equation x² + y² = 16. Hmm, okay.First, let me make sure I understand the problem correctly. We roll a die twice, the first roll gives us m, the second gives us n. So, m and n can each be any integer from 1 to 6, inclusive. Then, the point (m, n) is plotted on a coordinate plane, and we want to know the probability that this point is outside the circle with radius 4 (since the equation x² + y² = 16 is a circle centered at the origin with radius 4).Right. So, the first step is probably to figure out all the possible points (m, n) and then determine how many of those lie outside the circle. Then, the probability would be the number of such points divided by the total number of possible points.Since each die has 6 faces, rolling it twice gives 6 * 6 = 36 possible outcomes. So, the total number of points is 36. Therefore, the denominator of our probability will be 36.Now, the key is to find the number of points (m, n) such that m² + n² > 16. Because if a point is outside the circle x² + y² = 16, then the distance from the origin squared (which is m² + n²) must be greater than 16. So, we need to count how many pairs (m, n) satisfy m² + n² > 16.Alternatively, we could also compute the probability by subtracting the probability that the point is inside or on the circle from 1. So, P(outside) = 1 - P(inside or on). Maybe that's easier? Let's see.But either way, we need to count the number of (m, n) pairs where m and n are integers from 1 to 6, inclusive, such that m² + n² > 16.So, let me approach this systematically. Since m and n can each be 1, 2, 3, 4, 5, or 6, let's list all possible pairs and check whether they satisfy the condition. But that might take a while. Maybe there's a smarter way.Alternatively, for each possible value of m, we can determine the values of n for which m² + n² > 16.Let's consider m from 1 to 6:Case 1: m = 1Then, m² = 1. So, 1 + n² > 16 => n² > 15. Since n is an integer from 1 to 6, n² must be greater than 15. Let's see:n² > 15. The squares of n are 1, 4, 9, 16, 25, 36. So, n² must be greater than 15, which means n² >= 16. Therefore, n must be 4, 5, or 6 (since 4²=16, 5²=25, 6²=36). So, when m=1, n can be 4,5,6. That's 3 values.Case 2: m = 2m² = 4. Then, 4 + n² > 16 => n² > 12. So, n² must be greater than 12. The possible n² are as above. So, n² >12 implies n² >=16. Therefore, same as above, n=4,5,6. So, again 3 values.Case 3: m = 3m² = 9. Then, 9 + n² >16 => n² >7. So, n² >7. Looking at the squares, n² can be 9,16,25,36. Therefore, n must be 3,4,5,6. Wait, but n² >7. Let's verify:n=1: 1, n=2:4, n=3:9, etc. So, n² >7 when n² >=9, which is n >=3. So, n=3,4,5,6. That's 4 values.Case 4: m=4m²=16. Then, 16 +n² >16 => n² >0. Since n is at least 1, n² is at least 1. So, all n from 1 to 6 satisfy n² >0. Therefore, all 6 values of n are acceptable. Wait, but hold on. 16 + n² >16 implies n²>0, which is always true because n is at least 1, so n² is at least 1. So, for m=4, all n=1,2,3,4,5,6 are valid. So 6 values.Case 5: m=5Similarly, m²=25. Then, 25 +n² >16. But 25 is already greater than 16, so regardless of n, this will always be true. Therefore, all n=1 to 6 are acceptable. So, 6 values.Case 6: m=6Same as m=5. m²=36. 36 +n² >16 is always true. So, n can be 1 to 6. 6 values.Wait, let me check if that's correct. So, for m=4,5,6, regardless of n, the sum m² +n² will be greater than 16. For m=4: 16 +1=17 >16, yes. For m=5, even n=1: 25 +1=26>16. For m=6: 36 +1=37>16. So, indeed, all n are acceptable for m=4,5,6.So, summarizing:For m=1: 3 values of nFor m=2:3For m=3:4For m=4:6For m=5:6For m=6:6Total number of favorable outcomes: 3+3+4+6+6+6 = let's compute that step by step.Starting with m=1:3m=2:3 (total 6)m=3:4 (total 10)m=4:6 (total 16)m=5:6 (total 22)m=6:6 (total 28)So, 28 favorable outcomes where the point (m,n) lies outside the circle.Therefore, the probability is 28/36, which simplifies to 7/9.Wait, but hold on. Let me double-check this calculation. Because when m=1, n=4,5,6: three points. For m=2, same. For m=3, n=3,4,5,6: four points. Then for m=4,5,6, all n, so 6 each. So 3+3+4+6+6+6=3+3=6, +4=10, +6=16, +6=22, +6=28. So yes, 28. Then total is 36, so 28/36=7/9.But wait, let me verify with another approach. Maybe count the points inside or on the circle and subtract from total.So, total points:36. If the number of points inside or on the circle is 36-28=8. Let's check if that's correct. How many points (m,n) satisfy m² +n² <=16.So, m and n from 1 to6. Let's list them.For m=1:n² <=16 -1=15. So n²<=15. So n can be 1,2,3 (since 4²=16 which is >15). So n=1,2,3:3 points.For m=2:n² <=16 -4=12. So n²<=12. So n=1,2,3 (3²=9<=12, 4²=16>12). So 3 points.For m=3:n² <=16 -9=7. So n²<=7. So n=1,2 (since 2²=4<=7, 3²=9>7). So n=1,2:2 points.For m=4:n² <=16 -16=0. So n²<=0. But n is at least 1, so no solutions. Therefore, 0 points.Similarly, m=5 and m=6: m²=25 and 36, which are already greater than 16, so even with n=1, m² +1>16, so no points here.So, total points inside or on the circle: m=1:3, m=2:3, m=3:2. Total 3+3+2=8. So, 8 points inside or on the circle, 36-8=28 outside. So, same result. Therefore, probability 28/36=7/9.Therefore, the answer should be 7/9. Wait, but let me check if I missed any points.Wait, when m=3 and n=3, m² +n²=9+9=18>16, so (3,3) is outside. But in the inside count, m=3: n=1,2. So, n=3 is excluded. So, yes, that's correct.Similarly, m=3, n=3 is outside. So, for m=3, n=3,4,5,6 are outside, which is 4 points. So, that seems correct.Similarly, m=1, n=4,5,6: 3 points. m=2, n=4,5,6:3 points. The rest, m=4,5,6 with any n: 6 each.Yes, that seems to check out. So, 28/36=7/9. Therefore, the probability is 7/9.But wait, just to make sure, let's list all points inside or on the circle:For m=1:(1,1), (1,2), (1,3)For m=2:(2,1), (2,2), (2,3)For m=3:(3,1), (3,2)So, total 3+3+2=8 points. So, 8 points inside or on the circle. The remaining 36-8=28 are outside, so probability 28/36=7/9.Therefore, the answer should be 7/9.But wait, just to be thorough, let's verify each of these points:For m=1:- (1,1): 1+1=2 <=16: inside- (1,2):1+4=5 <=16: inside- (1,3):1+9=10 <=16: inside- (1,4):1+16=17 >16: outsideSimilarly for m=2:- (2,1):4+1=5 <=16: inside- (2,2):4+4=8 <=16: inside- (2,3):4+9=13 <=16: inside- (2,4):4+16=20 >16: outsideFor m=3:- (3,1):9+1=10 <=16: inside- (3,2):9+4=13 <=16: inside- (3,3):9+9=18 >16: outsideSo, yes, the points inside are exactly the ones listed. So, 8 points. So, the calculation is correct.Therefore, the probability is 28/36, which simplifies by dividing numerator and denominator by 4: 7/9.Yes, so 7/9 is the probability that the point lies outside the circle.Final AnswerThe probability is boxed{dfrac{7}{9}}.</think>To determine the probability that the point ( P(m, n) ) lies outside the circle ( x^2 + y^2 = 16 ), where ( m ) and ( n ) are the results of rolling a die twice, we need to count the number of pairs ((m, n)) such that ( m^2 + n^2 > 16 ).1. Total possible outcomes: Since each die has 6 faces, the total number of outcomes is ( 6 times 6 = 36 ).2. Counting favorable outcomes: - For ( m = 1 ): ( 1^2 + n^2 > 16 ) implies ( n geq 4 ). Possible ( n ): 4, 5, 6 (3 values). - For ( m = 2 ): ( 2^2 + n^2 > 16 ) implies ( n geq 4 ). Possible ( n ): 4, 5, 6 (3 values). - For ( m = 3 ): ( 3^2 + n^2 > 16 ) implies ( n geq 3 ). Possible ( n ): 3, 4, 5, 6 (4 values). - For ( m = 4 ): ( 4^2 + n^2 > 16 ) is always true for ( n geq 1 ). Possible ( n ): 1, 2, 3, 4, 5, 6 (6 values). - For ( m = 5 ): ( 5^2 + n^2 > 16 ) is always true for ( n geq 1 ). Possible ( n ): 1, 2, 3, 4, 5, 6 (6 values). - For ( m = 6 ): ( 6^2 + n^2 > 16 ) is always true for ( n geq 1 ). Possible ( n ): 1, 2, 3, 4, 5, 6 (6 values).3. Summing the favorable outcomes: - ( 3 + 3 + 4 + 6 + 6 + 6 = 28 )4. Calculating the probability: - The probability is ( frac{28}{36} = frac{7}{9} ).Thus, the probability that point ( P ) lies outside the circle ( x^2 + y^2 = 16 ) is (boxed{dfrac{7}{9}}).,<think>Okay, let me try to solve this probability problem. So, the question is: We have a point P with coordinates (m, n), where m and n are the numbers obtained by rolling a die twice consecutively. We need to find the probability that P lies outside the circle defined by the equation x² + y² = 16. Hmm, okay.First, let me make sure I understand the problem. We roll a die twice, so each of m and n can be any integer from 1 to 6, inclusive. So, the possible coordinates (m, n) are all pairs where m and n are each between 1 and 6. That gives us a total of 6 x 6 = 36 possible points, right? So, the sample space has 36 equally likely outcomes.The next part is about the circle x² + y² = 16. We need to find the probability that the point (m, n) lies outside this circle. So, the probability would be the number of points (m, n) where m² + n² > 16 divided by the total number of points, which is 36. So, basically, we need to count how many of these 36 points satisfy m² + n² > 16.Alright, so my task is to figure out how many pairs (m, n) with m and n each from 1 to 6 satisfy m² + n² > 16. Then, divide that count by 36 to get the probability.Let me start by recalling that the circle x² + y² = 16 has a radius of 4, because 16 is (4)². So, the radius is 4. So, any point (x, y) that is outside this circle must be at a distance greater than 4 from the origin. Since our coordinates m and n are integers from 1 to 6, the points are all in the first quadrant, right? Because m and n are positive. So, we're dealing with points in the first quadrant where both x and y are integers between 1 and 6.So, to find the points outside the circle, we need m² + n² > 16. So, for each possible value of m (from 1 to 6), I can check which values of n (from 1 to 6) make m² + n² > 16. Then, sum all those up.Alternatively, maybe it's easier to compute the number of points inside or on the circle (i.e., m² + n² ≤ 16) and subtract that from 36 to get the number of points outside. Then, divide by 36. Maybe that's a better approach, since sometimes counting the complement is easier.Let me try both ways and see which is less error-prone. Let's start by finding the number of points inside or on the circle.So, for each m from 1 to 6, let's find the number of n such that n² ≤ 16 - m². Since n must be an integer between 1 and 6.Let me list m from 1 to 6:Case 1: m = 1Then, 16 - m² = 16 - 1 = 15. So, we need n² ≤ 15. Since n is from 1 to 6, let's compute n² for each n:n=1: 1 ≤ 15 → yesn=2: 4 ≤ 15 → yesn=3: 9 ≤ 15 → yesn=4: 16 ≤ 15 → non=5: 25 ≤ 15 → non=6: 36 ≤ 15 → noWait, wait, no. Wait, 16 - m² is 15 here. So, n² must be less than or equal to 15. So, n can be 1, 2, 3 because 4² is 16, which is greater than 15. So, n=1,2,3. So, for m=1, there are 3 values of n such that (1, n) is inside or on the circle.Case 2: m=216 - m² = 16 -4 =12. So, n² ≤12. So, n can be 1, 2, 3 because:n=1:1≤12n=2:4≤12n=3:9≤12n=4:16≤12? No. Similarly, n=5 and 6 would be even bigger. So, again, 3 values.Wait, n=3 gives 9, which is less than 12, so yes. So, n=1,2,3. So, 3 values again.Case 3: m=316 - m² =16 -9 =7. So, n² ≤7. So, n can be 1 and 2 because:n=1:1≤7n=2:4≤7n=3:9≤7? No. So, n=1 and 2. So, 2 values.Case 4: m=416 - m² =16 -16=0. So, n² ≤0. But n is at least 1, so n² is at least 1. So, no solutions here. So, for m=4, there are 0 values of n such that (4, n) is inside or on the circle.Wait, hold on. Wait, m=4, 16 - m² = 0, so n² ≤0. But n is at least 1, so n² is 1 or higher. So, no points here. So, zero.Similarly, m=5:16 -25 = -9. So, n² ≤-9. But n² is always non-negative, so no solutions here either. So, zero.Similarly, m=6:16 -36 = -20. So, n² ≤-20. Again, impossible. So, zero.So, summarizing the number of points inside or on the circle:For m=1:3m=2:3m=3:2m=4:0m=5:0m=6:0Total inside or on the circle: 3+3+2+0+0+0=8Therefore, the number of points outside the circle is total points (36) minus inside points (8), so 36 -8=28.Therefore, the probability is 28/36, which simplifies to 7/9.Wait, hold on. Let me check again. Maybe I missed something.Wait, let's verify. For m=1, n=1,2,3: (1,1), (1,2), (1,3). Then m=2: (2,1), (2,2), (2,3). For m=3: (3,1), (3,2). So that's 3+3+2=8 points inside or on the circle. Then, the rest are outside.So, 36-8=28. So, 28/36=7/9. So, the probability is 7/9.But wait, let me confirm by calculating the outside points directly. Maybe I made a mistake in the complement.Alternatively, let's compute for each m from 1 to 6, how many n satisfy m² + n² >16.For m=1:m²=1. So, n² >16 -1=15. So, n² >15. Since n is from 1 to6. So, n² must be greater than 15. The squares of n:1,4,9,16,25,36. So, n²>15 means n²=16,25,36, which correspond to n=4,5,6. So, three values. So, for m=1, 3 points outside.Wait, but when m=1, earlier calculation said that inside points are n=1,2,3 (3 points), so outside points would be 6-3=3. Correct.Similarly, m=2:m²=4. So, n²>12. n² must be greater than 12. The n values where n²>12 are n=4,5,6 (since 4²=16>12, 5²=25>12, 6²=36>12). So, 3 points. So, 3 outside points. So, same as before: 6-3=3. Correct.For m=3:m²=9. So, n²>16 -9=7. So, n²>7. So, n can be 3,4,5,6 because:n=3:9>7n=4:16>7n=5:25>7n=6:36>7So, n=3,4,5,6. That's 4 values. So, 4 outside points here. Wait, but earlier, when we calculated the inside points for m=3, we had n=1 and 2 (2 points), so outside points would be 6-2=4. Correct. So, 4 points. So, that's different from before. Wait, but in the complement method, for m=3, inside points were 2, so outside is 4. But in the initial calculation, when I thought maybe 2 inside, 4 outside, but in the first approach, when I calculated the inside points as 8, which includes m=3, n=1 and 2, so 2 points, so outside points would be 6-2=4. So, correct.Similarly, m=4:m²=16. So, n²>16 -16=0. So, n²>0. But n is at least 1, so n² is at least 1, which is greater than 0. Therefore, all n=1 to6 satisfy n²>0. So, for m=4, all 6 points (4,1) to (4,6) lie outside the circle. So, 6 outside points.Similarly, m=5:m²=25. So, n²>16 -25= -9. Since n² is always non-negative, all n satisfy n² > -9. So, all 6 points (5,1) to (5,6) lie outside. So, 6 outside points.Same with m=6: m²=36. 16 -36=-20. So, n²> -20. All n=1 to6 satisfy this, so 6 points.Therefore, the total outside points are:m=1:3m=2:3m=3:4m=4:6m=5:6m=6:6So, adding these up: 3+3=6; 6+4=10; 10+6=16; 16+6=22; 22+6=28. So, total 28 points outside.Therefore, probability is 28/36=7/9.So, that's the same as before. So, that confirms that the answer is 7/9.Wait, but let me check for m=3 again. For m=3, m² + n² >16.So, 9 + n² >16 → n² >7. So, n must be 3,4,5,6. Because 3²=9>7, so yes. So, n=3,4,5,6. So, 4 values, which is 4 points. That's correct. So, 4 points outside for m=3.So, adding up all the outside points:m=1: 3 (n=4,5,6)m=2:3 (n=4,5,6)m=3:4 (n=3,4,5,6)m=4:6 (all n=1-6)m=5:6 (all n=1-6)m=6:6 (all n=1-6)Total: 3+3+4+6+6+6=28.Yes, 28. So, 28/36 simplifies to 7/9.Therefore, the probability is 7/9. So, I think that's the correct answer.But wait, just to be thorough, let me list all the points inside the circle to double-check.For m=1:(1,1): 1+1=2 ≤16 → inside(1,2):1+4=5 ≤16 → inside(1,3):1+9=10 ≤16 → inside(1,4):1+16=17 >16 → outside(1,5):1+25=26 >16 → outside(1,6):1+36=37 >16 → outsideSo, inside:3, outside:3.For m=2:(2,1):4+1=5 ≤16 → inside(2,2):4+4=8 ≤16 → inside(2,3):4+9=13 ≤16 → inside(2,4):4+16=20 >16 → outside(2,5):4+25=29 >16 → outside(2,6):4+36=40 >16 → outsideInside:3, outside:3.For m=3:(3,1):9+1=10 ≤16 → inside(3,2):9+4=13 ≤16 → inside(3,3):9+9=18 >16 → outside(3,4):9+16=25 >16 → outside(3,5):9+25=34 >16 → outside(3,6):9+36=45 >16 → outsideSo, inside:2, outside:4.For m=4:(4,1):16+1=17 >16 → outsideAll other n for m=4: 16+4=20, 16+9=25, etc. All outside. So, 6 outside.Similarly, m=5 and m=6:All points (5,n) and (6,n) will have m² ≥25 and 36, which are already greater than 16, so adding any n² (which is at least 1) will make the sum greater than 16. So, all 6 points for each m=5 and m=6 are outside.Therefore, total inside points:3+3+2=8, which matches our earlier calculation.Therefore, the number of points outside is 36 -8=28. Hence, the probability is 28/36=7/9.Therefore, the answer should be 7/9.I think this is correct. Let me check if I considered all possible m and n values correctly. Especially for m=3, n=3: 9+9=18>16, so outside. So, that's correct. Similarly, m=3,n=2:9+4=13≤16, inside. So, yes, two inside points for m=3. Alright, everything checks out. I don't see any errors here.Final AnswerThe probability is boxed{dfrac{7}{9}}.</think>To find the probability that a point ( P(m, n) ) lies outside the circle ( x^2 + y^2 = 16 ) where ( m ) and ( n ) are the results of rolling a die twice, we need to count the number of points ((m, n)) such that ( m^2 + n^2 > 16 ).1. Total possible outcomes: Since each die roll has 6 outcomes, there are ( 6 times 6 = 36 ) possible points.2. Points inside or on the circle: We calculate the number of points ((m, n)) such that ( m^2 + n^2 leq 16 ): - For ( m = 1 ): ( n^2 leq 15 ). Possible ( n ): 1, 2, 3 (3 points). - For ( m = 2 ): ( n^2 leq 12 ). Possible ( n ): 1, 2, 3 (3 points). - For ( m = 3 ): ( n^2 leq 7 ). Possible ( n ): 1, 2 (2 points). - For ( m = 4, 5, 6 ): No solutions as ( m^2 ) alone exceeds 16. Total inside points: ( 3 + 3 + 2 = 8 ).3. Points outside the circle: Subtract the inside points from the total points: - Total outside points: ( 36 - 8 = 28 ).4. Probability: The probability is the number of outside points divided by the total points: - Probability: ( frac{28}{36} = frac{7}{9} ).Thus, the probability that point ( P ) lies outside the circle is (boxed{dfrac{7}{9}}).]

🔑:This problem involves the classical probability model and its calculation formula. The key to solving the problem is to calculate the total number of basic events and the total number of basic events contained in the studied event. In this problem, the number of basic events contained in the event is calculated using the listing method, which is often used to statistically count the basic events contained in some events where the rules are not obvious.There are a total of 36 points with coordinates (m, n) obtained by rolling a dice twice consecutively.The basic events included in "point P is outside the circle x^{2}+y^{2}=16" are:(1,4), (1,5), (1,6),(2,4), (2,5), (2,6),(3,3), (3,4), (3,5), (3,6),(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),(6,1), (6,2), (6,3), (6,4), (6,5), (6,6).There are a total of 28 such points.Therefore, the probability that point P is outside the circle x^{2}+y^{2}=16 is frac {28}{36}= frac {7}{9};Hence, the answer is boxed{ frac {7}{9}}.

🧠:[<think>Okay, so I have this problem about triangle ABC. Let's start with part (I). The problem states that in triangle ABC, the sides opposite angles A, B, C are a, b, c respectively, and we have the equation ( b^2 + c^2 - a^2 = bc ). They want me to find the measure of angle A. Hmm, this looks familiar. I remember in trigonometry, the Law of Cosines relates the sides of a triangle to the cosine of one of its angles. The formula is ( a^2 = b^2 + c^2 - 2bc cos A ). Let me write that down.So, according to the Law of Cosines:( a^2 = b^2 + c^2 - 2bc cos A ).But the problem gives me ( b^2 + c^2 - a^2 = bc ). Let me rearrange the Law of Cosines equation to match this. If I move ( a^2 ) to the left side and the other terms to the right side, I get:( b^2 + c^2 - a^2 = 2bc cos A ).But the problem says that ( b^2 + c^2 - a^2 = bc ). So substituting that into the equation from the Law of Cosines, I have:( bc = 2bc cos A ).Hmm, let me check that again. Wait, so if ( b^2 + c^2 - a^2 = bc ), then according to the Law of Cosines, that left side is equal to ( 2bc cos A ). So substituting:( 2bc cos A = bc ).Then, if I divide both sides by bc (assuming bc ≠ 0, which makes sense in a triangle), we get:( 2 cos A = 1 ).So, solving for cos A:( cos A = frac{1}{2} ).Therefore, angle A must be the angle whose cosine is 1/2. From trigonometry, I know that ( cos 60° = frac{1}{2} ), so angle A is 60 degrees. In radians, that's π/3. But since the problem asks for the measure, and it's common to use degrees in such problems unless specified otherwise, I think 60° is the answer for part (I). Let me just confirm once more. The Law of Cosines step seems right. Starting with their given equation, substituting into the Law of Cosines, solving for cos A. Yeah, that seems solid. So part (I) answer is 60 degrees.Moving on to part (II). The problem states: Given a=2, let ( f(x)= sqrt{3} sin frac{x}{2} cos frac{x}{2} + cos^2 frac{x}{2} ). When x=B, f(x) reaches its maximum value. Find the area of triangle ABC.Okay, so first, a=2. We need to find the area. Let's see. The area of a triangle can be found using various formulas. One common formula is ( frac{1}{2}ab sin C ), but to use that, we need two sides and the included angle. Alternatively, if we know all three sides, we can use Heron's formula. Or maybe since we have angles involved, we can use the formula ( frac{1}{2}bc sin A ). Wait, but do we know angle A? Yes, from part (I), angle A is 60°, so that might be useful.But first, let's parse the given information. They define a function f(x) which is ( sqrt{3} sin frac{x}{2} cos frac{x}{2} + cos^2 frac{x}{2} ). And when x is angle B, this function reaches its maximum value. So we need to analyze this function, find its maximum, and use that to find angle B, then use that information to find the area of the triangle.Let me first simplify f(x). Let's see:( f(x) = sqrt{3} sin frac{x}{2} cos frac{x}{2} + cos^2 frac{x}{2} ).Hmm, I can recognize that ( sin frac{x}{2} cos frac{x}{2} ) is equal to ( frac{1}{2} sin x ). Let me verify that. Using the double-angle identity: ( sin 2θ = 2 sin θ cos θ ), so ( sin θ cos θ = frac{1}{2} sin 2θ ). Therefore, ( sin frac{x}{2} cos frac{x}{2} = frac{1}{2} sin x ). So substituting that in:( f(x) = sqrt{3} cdot frac{1}{2} sin x + cos^2 frac{x}{2} ).So, simplifying:( f(x) = frac{sqrt{3}}{2} sin x + cos^2 frac{x}{2} ).Now, the second term is ( cos^2 frac{x}{2} ). There's a double-angle identity for cosine squared: ( cos^2 θ = frac{1 + cos 2θ}{2} ). Let me apply that here, where θ is ( frac{x}{2} ). So:( cos^2 frac{x}{2} = frac{1 + cos x}{2} ).Therefore, substituting back into f(x):( f(x) = frac{sqrt{3}}{2} sin x + frac{1 + cos x}{2} ).Let me combine these terms:( f(x) = frac{sqrt{3}}{2} sin x + frac{1}{2} + frac{cos x}{2} ).So, combining the sine and cosine terms:( f(x) = frac{1}{2} + frac{sqrt{3}}{2} sin x + frac{1}{2} cos x ).Hmm, this looks like a linear combination of sine and cosine, which can be written as a single sine or cosine function. Let me recall that ( a sin x + b cos x = R sin(x + phi) ) or ( R cos(x + phi) ), where ( R = sqrt{a^2 + b^2} ) and φ is the phase shift.In this case, the coefficients are ( frac{sqrt{3}}{2} ) for sine and ( frac{1}{2} ) for cosine. Let me factor out ( frac{1}{2} ) first:Wait, actually, the expression is:( frac{sqrt{3}}{2} sin x + frac{1}{2} cos x ).Which is the same as:( sin x cdot frac{sqrt{3}}{2} + cos x cdot frac{1}{2} ).Notice that ( frac{sqrt{3}}{2} = cos 30° ) and ( frac{1}{2} = sin 30° ). So this resembles the sine addition formula:( sin(x + 30°) = sin x cos 30° + cos x sin 30° ).Therefore, ( frac{sqrt{3}}{2} sin x + frac{1}{2} cos x = sin(x + 30°) ).Therefore, substituting back into f(x):( f(x) = frac{1}{2} + sin(x + 30°) ).Wait, let me check that:( sin(x + 30°) = sin x cos 30° + cos x sin 30° = sin x cdot frac{sqrt{3}}{2} + cos x cdot frac{1}{2} ). Yes, that's exactly the combination we have. So indeed, the expression simplifies to:( f(x) = frac{1}{2} + sin(x + 30°) ).So, the function f(x) is ( frac{1}{2} + sin(x + 30°) ). Now, the problem states that when x = B, f(x) reaches its maximum value. Let's analyze this.The sine function has a maximum value of 1. Therefore, the maximum value of ( sin(x + 30°) ) is 1, so the maximum value of f(x) is ( frac{1}{2} + 1 = frac{3}{2} ). Therefore, the maximum value of f(x) is ( frac{3}{2} ), which occurs when ( sin(x + 30°) = 1 ), i.e., when ( x + 30° = 90° + 360°k ), where k is an integer. Since x is an angle in a triangle, it must be between 0° and 180°, so the only possible solution is ( x + 30° = 90° ), so ( x = 60° ). Therefore, angle B is 60°.Wait, so if x is angle B, then angle B is 60°. Let me verify that. If f(x) reaches its maximum when x=B, and the maximum occurs when x + 30° = 90°, so x=60°, then angle B=60°. So angle B is 60 degrees.But wait, in a triangle, the sum of angles is 180°. From part (I), angle A is 60°, and now angle B is 60°, so angle C would be 180° - 60° - 60° = 60°. So triangle ABC is equilateral? Because all angles are 60°, which would mean all sides are equal. But they told us that a=2. If the triangle is equilateral, then all sides are 2, so the area would be ( frac{sqrt{3}}{4} times 2^2 = sqrt{3} ). But let's check whether this is correct.Wait, but is this necessarily an equilateral triangle? Because we have angle A=60°, angle B=60°, angle C=60°, so yes, that would make it equilateral. But let's confirm using the given information. Wait, in part (II), they tell us that a=2, which is opposite angle A. If all angles are 60°, then all sides are equal, so a=2, so all sides are 2, so the triangle is equilateral. So area is indeed ( sqrt{3} ). But let's make sure that angle B is indeed 60°, because perhaps there's an error in my reasoning.Wait, let's go back to the function f(x). The problem says when x=B, f(x) reaches its maximum value. We transformed f(x) into ( frac{1}{2} + sin(x + 30°) ), which has maximum value 3/2 when ( x + 30° = 90° ), so x=60°, so angle B=60°. So that's correct. Therefore, angle B is 60°, angle A is 60°, angle C is 60°, triangle is equilateral with side a=2, so area is ( sqrt{3} ). But wait, let me check again. If angle A is 60°, and sides a, b, c are opposite angles A, B, C. If all angles are 60°, then by the Law of Sines, all sides must be equal. So since a=2, then b=2 and c=2. Therefore, the area would be ( frac{sqrt{3}}{4} times (2)^2 = sqrt{3} ). So that seems correct.But wait, maybe there's another possibility. Wait, when I considered the maximum of f(x), I assumed that the maximum occurs at x=60°, but is there any other angle in the valid range (0° to 180°) where ( sin(x + 30°) = 1 )? The sine function is 1 at 90°, so x+30°=90°, which gives x=60°, as I found. But for example, if x was 60° + 360°, but since x is an angle in a triangle, it's between 0° and 180°, so 60° is the only solution. So angle B=60°, angles sum to 180°, so angle C=60°, triangle is equilateral. So with a=2, all sides are 2. Therefore, the area is √3. Therefore, part (II) answer is √3.Wait, but let me check if there's another way to approach this without assuming it's equilateral. Maybe using the Law of Sines.Law of Sines states that ( frac{a}{sin A} = frac{b}{sin B} = frac{c}{sin C} = 2R ), where R is the radius of the circumscribed circle. Since angle A is 60°, and a=2, then:( frac{2}{sin 60°} = 2R ).So ( 2R = frac{2}{frac{sqrt{3}}{2}} = frac{4}{sqrt{3}} ), so R= ( frac{2}{sqrt{3}} ).Then, angle B=60°, so side b is:( b = 2R sin B = frac{4}{sqrt{3}} cdot sin 60° = frac{4}{sqrt{3}} cdot frac{sqrt{3}}{2} = 2 ).Similarly, side c=2. So sides are all 2, hence equilateral. Therefore, area is √3.Alternatively, if I use the formula for area: ( frac{1}{2}ab sin C ). But here, sides a and b would both be 2, angle C is 60°, so area is ( frac{1}{2} times 2 times 2 times sin 60° = 2 times frac{sqrt{3}}{2} = sqrt{3} ). Yep, same result.Wait, but the problem didn't state that the triangle is equilateral. It gave a specific condition in part (I) and then in part (II) with a=2 and angle B such that f(x) is maximized. So perhaps it's a different triangle? But according to the calculations, angle B=60°, so all angles are 60°, so triangle is equilateral. So maybe that's correct.Alternatively, maybe there's a miscalculation in transforming the function f(x). Let me check again.Original f(x): ( sqrt{3} sin frac{x}{2} cos frac{x}{2} + cos^2 frac{x}{2} ).First step: ( sqrt{3} sin frac{x}{2} cos frac{x}{2} = frac{sqrt{3}}{2} sin x ). Correct.Second term: ( cos^2 frac{x}{2} = frac{1 + cos x}{2} ). Correct.Therefore, f(x) = ( frac{sqrt{3}}{2} sin x + frac{1 + cos x}{2} ).Then, rewritten as ( frac{1}{2} + frac{sqrt{3}}{2} sin x + frac{1}{2} cos x ). Correct.Then, recognizing ( frac{sqrt{3}}{2} sin x + frac{1}{2} cos x = sin(x + 30°) ). Wait, let's verify that.Using the identity ( sin(a + b) = sin a cos b + cos a sin b ).If we have ( sin(x + 30°) ), that's ( sin x cos 30° + cos x sin 30° ).Which is ( sin x times frac{sqrt{3}}{2} + cos x times frac{1}{2} ). Which is exactly the same as the coefficients in our expression. Therefore, yes, ( frac{sqrt{3}}{2} sin x + frac{1}{2} cos x = sin(x + 30°) ). Therefore, f(x) = ( frac{1}{2} + sin(x + 30°) ).Therefore, maximum value of f(x) is ( frac{1}{2} + 1 = frac{3}{2} ), achieved when ( sin(x + 30°) = 1 ), so x + 30° = 90°, so x = 60°. Therefore, angle B = 60°. So that's correct.Therefore, angles A, B, C are all 60°, triangle is equilateral, with side a=2. Area is √3. So answer for part (II) is √3.But let me just make sure that there's no other interpretation here. For example, maybe angle B is not 60°, but the maximum occurs at another point. Wait, but the maximum of sin(x + 30°) is 1, so the maximum occurs when x + 30° = 90°, so x=60°, regardless of the triangle. So if x=B, then angle B=60°, no alternative. Therefore, the conclusion is correct.Therefore, both answers are 60° for part (I) and √3 for part (II).Final Answer(I) The measure of angle ( A ) is boxed{60^circ}.(II) The area of ( triangle ABC ) is boxed{sqrt{3}}.</think>(I) Given the equation ( b^2 + c^2 - a^2 = bc ) in (triangle ABC), we use the Law of Cosines, which states ( a^2 = b^2 + c^2 - 2bc cos A ). Rearranging this, we get ( b^2 + c^2 - a^2 = 2bc cos A ). Substituting the given equation into this, we find ( bc = 2bc cos A ), leading to ( cos A = frac{1}{2} ). Therefore, angle ( A ) is ( 60^circ ).(II) Given ( a = 2 ) and the function ( f(x) = sqrt{3} sin frac{x}{2} cos frac{x}{2} + cos^2 frac{x}{2} ), we simplify ( f(x) ). Using trigonometric identities, we rewrite ( f(x) ) as ( frac{1}{2} + sin(x + 30^circ) ). The maximum value of ( sin(x + 30^circ) ) is 1, occurring when ( x + 30^circ = 90^circ ), so ( x = 60^circ ). Thus, angle ( B = 60^circ ). With angles ( A ), ( B ), and ( C ) each ( 60^circ ), (triangle ABC) is equilateral with side ( a = 2 ). The area is calculated as ( frac{sqrt{3}}{4} times 2^2 = sqrt{3} ). Final Answer(I) The measure of angle ( A ) is boxed{60^circ}.(II) The area of ( triangle ABC ) is boxed{sqrt{3}}.,<think>Okay, so I have this triangle ABC, and the sides opposite angles A, B, and C are a, b, c respectively. The first part of the problem says that ( b^2 + c^2 - a^2 = bc ), and I need to find the measure of angle A. Hmm, this looks like something related to the Law of Cosines. Let me recall that the Law of Cosines states that ( a^2 = b^2 + c^2 - 2bc cos A ). So if I rearrange that formula, it becomes ( cos A = frac{b^2 + c^2 - a^2}{2bc} ).Wait, the problem gives ( b^2 + c^2 - a^2 = bc ). So substituting that into the formula for cosine of A, we get ( cos A = frac{bc}{2bc} = frac{1}{2} ). Therefore, angle A must be 60 degrees because cosine of 60° is 0.5. Let me double-check that. If ( cos A = 0.5 ), then yes, A is 60° or π/3 radians. That seems straightforward. So part I should be 60 degrees.Moving on to part II. They give a = 2, and a function ( f(x) = sqrt{3} sin frac{x}{2} cos frac{x}{2} + cos^2 frac{x}{2} ). They mention that when x = B, f(x) reaches its maximum value. Then we need to find the area of triangle ABC. First, let me try to simplify the function f(x). Let's look at the terms:1. ( sqrt{3} sin frac{x}{2} cos frac{x}{2} ). That looks like a double-angle identity. Since ( sin theta cos theta = frac{1}{2} sin 2theta ), so maybe this term can be written as ( sqrt{3} cdot frac{1}{2} sin x ), which is ( frac{sqrt{3}}{2} sin x ).2. The second term is ( cos^2 frac{x}{2} ). There's an identity for ( cos^2 theta = frac{1 + cos 2theta}{2} ), so substituting θ as x/2, this becomes ( frac{1 + cos x}{2} ).Putting these together, f(x) becomes:( frac{sqrt{3}}{2} sin x + frac{1}{2} (1 + cos x) )Simplify that:( frac{sqrt{3}}{2} sin x + frac{1}{2} + frac{1}{2} cos x )So, combining the trigonometric terms:( frac{1}{2} + frac{sqrt{3}}{2} sin x + frac{1}{2} cos x )Hmm, the expression ( frac{sqrt{3}}{2} sin x + frac{1}{2} cos x ) resembles the form ( sin(x + phi) ) or ( cos(x - phi) ). Let me recall that ( a sin x + b cos x = R sin(x + phi) ), where ( R = sqrt{a^2 + b^2} ). Here, a is ( frac{sqrt{3}}{2} ), and b is ( frac{1}{2} ). So R would be:( R = sqrt{ left( frac{sqrt{3}}{2} right)^2 + left( frac{1}{2} right)^2 } = sqrt{ frac{3}{4} + frac{1}{4} } = sqrt{1} = 1 )So then, ( frac{sqrt{3}}{2} sin x + frac{1}{2} cos x = sin(x + phi) ), but actually, more precisely, since the coefficients correspond to sin and cos, we can write this as:( sin x cos phi + cos x sin phi = sin(x + phi) ). Comparing coefficients:( cos phi = frac{sqrt{3}}{2} )( sin phi = frac{1}{2} )Which angle φ satisfies these? Well, cos φ = √3/2 and sin φ = 1/2. That's φ = 30°, or π/6 radians. Therefore, the expression simplifies to ( sin(x + 30°) ).Therefore, f(x) can be written as:( frac{1}{2} + sin(x + 30°) )So the maximum value of f(x) occurs when ( sin(x + 30°) ) is maximized, which is 1. Therefore, the maximum value of f(x) is ( frac{1}{2} + 1 = frac{3}{2} ). But the problem states that when x = B, f(x) reaches its maximum. Therefore, when x = B, ( sin(B + 30°) = 1 ), which implies that ( B + 30° = 90° ), so ( B = 60° ).Wait, hold on. If ( sin(B + 30°) = 1 ), then ( B + 30° = 90° + 360°k ), where k is an integer. Since B is an angle in a triangle, it must be between 0° and 180°, so the only possibility is k=0, which gives ( B = 60° ).So angle B is 60°. Now, we need to find the area of triangle ABC. We know angle A is 60°, angle B is 60°, so angle C must be 60°, since the sum of angles in a triangle is 180°. Wait, all angles are 60°, so the triangle is equilateral? But side a is given as 2. If all angles are 60°, then all sides are equal, so sides b and c would also be 2. Then the area would be ( frac{sqrt{3}}{4} times 2^2 = sqrt{3} ).Wait, but let me check again. Wait, part I gave that angle A is 60°, part II says when x=B, f(x) reaches maximum. We found angle B is 60°, so angle C would be 60°, making it equilateral. But let me confirm whether angle C is indeed 60°.Sum of angles A + B + C = 180°, so 60° + 60° + C = 180°, so C = 60°. Therefore, the triangle is equilateral. But in that case, all sides are equal. Given that side a = 2, then sides b and c are also 2. Therefore, area is ( frac{sqrt{3}}{4} times 2^2 = sqrt{3} ). So is that the answer? But let me make sure that there are no other possibilities.Wait, maybe I made a wrong assumption here. Let me go step by step again. From part I, angle A is 60°. Then in part II, a = 2. The function f(x) is given, and when x = B, f(x) reaches its maximum. We converted f(x) to ( frac{1}{2} + sin(x + 30°) ), so the maximum occurs when ( sin(x + 30°) = 1 ), so x + 30° = 90°, so x = 60°, hence angle B is 60°. Therefore, angles A and B are 60°, so angle C is 60°, making it an equilateral triangle. Thus, sides a, b, c are all equal. Since a = 2, then sides b and c are 2 as well. Therefore, the area is indeed ( frac{sqrt{3}}{4} times 2^2 = sqrt{3} ).But let me think again. Suppose that in part I, angle A is 60°, but in part II, the triangle is not necessarily equilateral. Wait, if angle A is 60°, angle B is 60°, then angle C must be 60°, so it's forced. Therefore, the triangle must be equilateral. Hence, sides are all 2, area is sqrt(3). So maybe that's correct.Wait, but let me check with Law of Sines. If angle A is 60°, a = 2. Then by Law of Sines, ( frac{a}{sin A} = frac{b}{sin B} = frac{c}{sin C} ). If angle A is 60°, then ( frac{2}{sin 60°} = frac{2}{sqrt{3}/2} = frac{4}{sqrt{3}} ). Then if angle B is 60°, then ( frac{b}{sin 60°} = frac{4}{sqrt{3}} ), so ( b = frac{4}{sqrt{3}} times sin 60° = frac{4}{sqrt{3}} times frac{sqrt{3}}{2} = 2 ). Similarly, c would be 2. Therefore, sides are all 2. So triangle is equilateral. Therefore, the area is sqrt(3). So that's consistent.Therefore, the answer for part II is sqrt(3). But let me check again if my function f(x) was simplified correctly. Original f(x) was ( sqrt{3} sin frac{x}{2} cos frac{x}{2} + cos^2 frac{x}{2} ).Yes, first term: ( sqrt{3} sin frac{x}{2} cos frac{x}{2} ). As I did before, using the identity sin θ cos θ = 1/2 sin 2θ, so this becomes ( sqrt{3} times frac{1}{2} sin x ) = ( frac{sqrt{3}}{2} sin x ).Second term: ( cos^2 frac{x}{2} ). Using the identity ( cos^2 θ = frac{1 + cos 2θ}{2} ), so substituting θ = x/2, we get ( frac{1 + cos x}{2} ). So yes, the second term is ( frac{1}{2} + frac{1}{2} cos x ). Therefore, combining both terms:( frac{sqrt{3}}{2} sin x + frac{1}{2} + frac{1}{2} cos x ). Then combining the sine and cosine terms:Let me see, ( frac{sqrt{3}}{2} sin x + frac{1}{2} cos x ). To express this as a single sine function, we can write this as R sin(x + φ), where R is the amplitude and φ is the phase shift.Calculating R: ( R = sqrt{ (frac{sqrt{3}}{2})^2 + (frac{1}{2})^2 } = sqrt{ frac{3}{4} + frac{1}{4} } = sqrt{1} = 1 ).Then, φ is such that:( cos φ = frac{sqrt{3}}{2} ) and ( sin φ = frac{1}{2} ). Which angle φ satisfies this? It's 30°, since cos 30° = √3/2 and sin 30° = 1/2. Therefore, the expression becomes:( 1 cdot sin(x + 30°) ). Therefore, f(x) = ( frac{1}{2} + sin(x + 30°) ). Thus, the maximum value of f(x) is when sin(x + 30°) = 1, so maximum value is ( frac{1}{2} + 1 = frac{3}{2} ), which occurs when x + 30° = 90°, so x = 60°. Therefore, angle B = 60°, as before.Therefore, the triangle has angles 60°, 60°, 60°, sides all 2, area sqrt(3). Therefore, the area is sqrt(3). So part II answer is sqrt(3).Wait, but the problem states "when x = B, f(x) reaches its maximum value". So x is in the domain where the function is defined. Since B is an angle of a triangle, it's between 0 and 180°, so x here is in that interval. So our calculation is correct.Alternatively, could there be another angle B where f(x) is maximum? Let's check if angle B is 60°, but suppose the triangle isn't equilateral. Wait, if angle A is 60°, angle B is 60°, then angle C is 60°, so all angles are 60°, so the triangle is equilateral. So sides must be equal. Therefore, if a = 2, then all sides are 2, so area is sqrt(3). So that's correct.Alternatively, maybe there's a different approach. For example, using area formula ( frac{1}{2}ab sin C ). But if all sides are 2, then it's equilateral, so area is sqrt(3). So yes, that's correct.Therefore, I think both answers are 60° and sqrt(3). Wait, but let me check once more for part I. The given equation is ( b^2 + c^2 - a^2 = bc ). From Law of Cosines, ( a^2 = b^2 + c^2 - 2bc cos A ). Therefore, rearranged, ( b^2 + c^2 - a^2 = 2bc cos A ). The problem states ( b^2 + c^2 - a^2 = bc ). Therefore, ( 2bc cos A = bc ), so ( 2 cos A = 1 ), so ( cos A = frac{1}{2} ), so A = 60°, correct.Therefore, part I is 60°, part II is sqrt(3). So the answers are:(I) 60°, or π/3 radians.(II) Area is sqrt(3).So the final answers should be:(I) boxed{60^circ}(II) boxed{sqrt{3}}Final Answer(I) boxed{60^circ}(II) boxed{sqrt{3}}</think>In triangle ABC, the sides opposite to angles A, B, C are a, b, c respectively, and we are given b^2 + c^2 - a^2 = bc.(I) Using the Law of Cosines, we know:[a^2 = b^2 + c^2 - 2bc cos A]Rearranging this, we get:[cos A = frac{b^2 + c^2 - a^2}{2bc}]Given b^2 + c^2 - a^2 = bc, substituting this into the equation for cos A gives:[cos A = frac{bc}{2bc} = frac{1}{2}]Thus, angle A is:[A = cos^{-1}left(frac{1}{2}right) = 60^circ](II) Given a = 2 and the function f(x) = sqrt{3} sin frac{x}{2} cos frac{x}{2} + cos^2 frac{x}{2}, we simplify f(x):[sqrt{3} sin frac{x}{2} cos frac{x}{2} = frac{sqrt{3}}{2} sin x][cos^2 frac{x}{2} = frac{1 + cos x}{2}]Combining these, we get:[f(x) = frac{sqrt{3}}{2} sin x + frac{1}{2} + frac{1}{2} cos x]This can be rewritten using a trigonometric identity:[f(x) = frac{1}{2} + sin(x + 30^circ)]The maximum value of f(x) occurs when sin(x + 30^circ) = 1, which implies x + 30^circ = 90^circ. Therefore, x = 60^circ, so angle B = 60^circ.Since angles A and B are both 60^circ, angle C must also be 60^circ, making triangle ABC equilateral with all sides equal to 2. The area of an equilateral triangle with side length a is:[text{Area} = frac{sqrt{3}}{4} a^2 = frac{sqrt{3}}{4} times 2^2 = sqrt{3}]Final answers:(I) boxed{60^circ}(II) boxed{sqrt{3}}]

🔑:(I) In triangle ABC, since b^{2}+c^{2}-a^{2}=bc,By the cosine rule a^{2}=b^{2}+c^{2}-2bccos A, we have cos A= frac {1}{2}.Since 0 < A < pi,We have A= frac {pi}{3}.(II) f(x)= sqrt {3}sin frac {x}{2}cos frac {x}{2}+cos ^{2} frac {x}{2}= frac{ sqrt {3}}{2}sin x+ frac{1}{2}cos x+ frac{1}{2}=sin (x+ frac {pi}{6})+ frac{1}{2},When x=B, f(B)=sin (B+ frac {pi}{6})+ frac{1}{2},Since A= frac {pi}{3}, Hence Bin(0, frac {2pi}{3}),Thus frac {pi}{6} < B+ frac {pi}{6} < frac {5pi}{6},So when B+ frac {pi}{6}= frac {pi}{2}, i.e., B= frac {pi}{3}, f(B) attains its maximum value, which is frac {3}{2}.Moreover, since A= frac {pi}{3}, we have C= frac {pi}{3}.Therefore, triangle ABC is an equilateral triangle.Hence, the area S = frac{1}{2}a^2 sin{frac{pi}{3}} = boxed{sqrt{3}}.