🧠:Okay, so I need to find the volume of the circumscribed sphere of the triangular prism A-BCD, where the side edges AB, AC, AD are mutually perpendicular. The areas of the triangles ABC, ACD, and ADB are given as √2/2, √3/2, and √6/2 respectively. Hmm, let's start by understanding the problem and visualizing the figure.First, a triangular prism usually has two triangular bases connected by three rectangles. But in this case, it's called a triangular prism A-BCD. Maybe the base is triangle BCD and the apex is A? But the edges AB, AC, AD are mutually perpendicular. That suggests that A is like a corner where three edges meet perpendicularly, similar to a 3D coordinate system. So, if I imagine point A at the origin, then points B, C, D would lie along the x, y, z axes respectively. Then the prism would be formed by translating the triangle BCD along the vector from A to some other point? Wait, I need to clarify the structure.Wait, perhaps the prism is formed by two congruent triangles connected by parallelograms. But given that AB, AC, AD are side edges and are mutually perpendicular, it might not be a standard prism. Maybe it's a type of prism where the lateral edges are perpendicular to each other. Hmm, maybe it's better to model this using coordinates.Let me try to assign coordinates to the points. Let’s place point A at the origin (0,0,0). Since AB, AC, AD are mutually perpendicular, we can assign coordinates to B, C, D along the x, y, z axes. Let’s say AB is along the x-axis, AC along the y-axis, and AD along the z-axis. So, point B is (b, 0, 0), point C is (0, c, 0), and point D is (0, 0, d). Then, the prism A-BCD would have vertices A, B, C, D and the other base triangle BCD connected by the edges. Wait, but in a prism, the two bases should be congruent triangles connected by rectangles. However, here triangle BCD is a triangle in 3D space. Let me confirm if this structure is indeed a prism.Alternatively, maybe the prism is formed by translating triangle ABC along AD, but given the problem statement, I need to be careful. Wait, the problem says "triangular prism A-BCD". Maybe the two triangular bases are ABC and another triangle, but I need to clarify.Wait, perhaps the prism is constructed such that the base is triangle ABC and the top face is a translated version along AD, but since AB, AC, AD are mutually perpendicular, this might form a prism where the lateral edges are along AD. Hmm, this is getting a bit confusing. Let me look up the definition of a triangular prism.A triangular prism is a three-dimensional shape with two congruent triangular bases and three rectangular faces connecting the corresponding sides. The sides are parallelograms, which in a right prism are rectangles. So, if A-BCD is a triangular prism, then there must be two triangular bases. Maybe the base is triangle ABC and the top face is triangle BCD? But in that case, BCD would need to be congruent to ABC and connected by rectangles. However, given that AB, AC, AD are mutually perpendicular, and the areas of the triangles ABC, ACD, ADB are given, perhaps the prism is formed in a different way.Alternatively, maybe the prism is formed by extruding triangle ABC along the AD direction. But since AD is perpendicular to both AB and AC, that extrusion would form a rectangular prism. Wait, but the problem mentions a triangular prism. Maybe the prism has a triangular base and three rectangular sides. If AB, AC, AD are mutually perpendicular, then the base could be triangle ABC, and the three rectangles would be ABXD, ACYD, and BCZD? Hmm, maybe not. This is getting a bit tangled. Let me try to approach this step by step.First, assign coordinates. Let’s set point A at (0,0,0). Since AB, AC, AD are mutually perpendicular, let's have AB along the x-axis, AC along the y-axis, and AD along the z-axis. Then, coordinates:A: (0, 0, 0)B: (x, 0, 0)C: (0, y, 0)D: (0, 0, z)Now, the areas of triangles ABC, ACD, and ADB are given. Let's compute those areas in terms of x, y, z.First, triangle ABC. Since AB is along x-axis and AC along y-axis, triangle ABC is in the xy-plane. The area of triangle ABC is (1/2)*AB*AC = (1/2)*x*y. According to the problem, this area is √2/2. So,(1/2)*x*y = √2/2 ⇒ x*y = √2.Similarly, triangle ACD. Points A, C, D. A is (0,0,0), C is (0, y, 0), D is (0,0, z). This triangle is in the y-z plane. The area is (1/2)*AC*AD = (1/2)*y*z. Given as √3/2, so:(1/2)*y*z = √3/2 ⇒ y*z = √3.Next, triangle ADB. Points A, D, B. A is (0,0,0), D is (0,0,z), B is (x,0,0). This triangle is in the x-z plane. The area is (1/2)*AB*AD = (1/2)*x*z. Given as √6/2, so:(1/2)*x*z = √6/2 ⇒ x*z = √6.So now, we have three equations:1. x*y = √22. y*z = √33. x*z = √6Our goal is to find x, y, z. Then, once we have those, we can determine the dimensions of the prism and then find the radius of the circumscribed sphere.Let me solve these equations. Let's denote:Equation 1: x*y = √2Equation 2: y*z = √3Equation 3: x*z = √6We can solve for x, y, z. Let's divide equation 3 by equation 2: (x*z)/(y*z) = √6/√3 ⇒ x/y = √2 ⇒ x = y*√2.Similarly, from equation 1: x*y = √2. Substitute x = y*√2:y*√2 * y = √2 ⇒ y²*√2 = √2 ⇒ y² = 1 ⇒ y = 1 (since lengths are positive)Then, from x = y*√2 ⇒ x = √2*1 = √2.From equation 2: y*z = √3 ⇒ 1*z = √3 ⇒ z = √3.Check equation 3: x*z = √2*√3 = √6, which matches. So, x = √2, y = 1, z = √3.So the coordinates are:A: (0,0,0)B: (√2, 0, 0)C: (0, 1, 0)D: (0, 0, √3)Now, the prism A-BCD. Wait, the problem mentions "triangular prism A-BCD". In a standard prism, there are two triangular bases. However, in this case, the points are A, B, C, D. Wait, maybe the prism is formed by triangles ABC and ABD connected by rectangles? Or is it ACD and something else? Wait, perhaps the prism has triangles ABC, ABD, and ACD as faces? But no, a prism has two triangular bases and three rectangular faces.Wait, given the points A, B, C, D, maybe the prism is A-BCD with base triangle BCD and the other base triangle connected through the edges AB, AC, AD. Hmm, this is getting confusing again. Let's think differently.In a triangular prism, there are two triangular bases and three rectangular lateral faces. If we have points A, B, C, D, then perhaps the two triangular bases are ABC and another triangle, say, DEF, but the problem says A-BCD. Maybe the two bases are ABC and BCD? Wait, but BCD is a triangle. Let me check if BCD is a triangle.Points B: (√2, 0, 0), C: (0, 1, 0), D: (0, 0, √3). So triangle BCD is formed by these three points. Let's compute the sides of triangle BCD to see if it's a triangle.Compute distance BC: between B(√2,0,0) and C(0,1,0):√[(√2 - 0)^2 + (0 - 1)^2 + (0 - 0)^2] = √[2 + 1] = √3.Distance BD: between B(√2,0,0) and D(0,0,√3):√[(√2 - 0)^2 + (0 - 0)^2 + (0 - √3)^2] = √[2 + 3] = √5.Distance CD: between C(0,1,0) and D(0,0,√3):√[(0 - 0)^2 + (1 - 0)^2 + (0 - √3)^2] = √[1 + 3] = 2.So triangle BCD has sides of length √3, √5, and 2. That's a valid triangle.So perhaps the prism has two bases: one is triangle ABC and the other is triangle BCD? But wait, how are they connected? In a prism, the corresponding vertices are connected by edges. But in this case, if ABC and BCD are the two bases, then how are the vertices connected? The edges would need to connect B to B, C to C, and D to D? That doesn't make sense. Maybe I need to reconsider.Alternatively, maybe the prism is not a right prism, but given that the edges AB, AC, AD are mutually perpendicular, perhaps it's a right prism. Wait, in a right prism, the lateral edges are perpendicular to the base. So if the base is triangle ABC, then the lateral edges would be perpendicular to the plane of ABC. But in this case, AB, AC, AD are mutually perpendicular. So if ABC is in the xy-plane, then AD is along the z-axis, which is perpendicular to the xy-plane. So then the prism is a right prism with base ABC and height AD. Then, the other base would be a translated version of ABC along the z-axis by length AD. But in the problem, the other base is BCD? Wait, perhaps not.Wait, maybe the prism is not formed by translating ABC but instead has ABC as one face and BCD as another face. Hmm. This is confusing. Let's think of all the points involved.Points:A: (0,0,0)B: (√2, 0, 0)C: (0,1,0)D: (0,0,√3)So, the edges AB, AC, AD are indeed mutually perpendicular. The triangles mentioned are ABC, ACD, and ADB.Wait, but the prism is A-BCD. Maybe the prism is formed by the points A, B, C, D and another triangle connected to them. Wait, maybe the prism has vertices A, B, C, D and another three points? But the problem mentions "triangular prism A-BCD", which might imply that the vertices are A, B, C, D, and perhaps two more? Wait, a triangular prism has six vertices: two triangular bases with three vertices each. So if the prism is A-BCD, maybe one base is triangle A-B-C and the other is triangle B-C-D? But triangle BCD is a separate triangle.Wait, but triangle BCD has three vertices B, C, D, which are connected to A. If it's a prism, the two bases should be congruent triangles connected by rectangles. But triangle ABC and triangle BCD are not congruent. ABC has sides AB=√2, AC=1, BC=√3; BCD has sides BC=√3, BD=√5, CD=2.Therefore, they are not congruent. So perhaps this is not a standard prism? Hmm, the problem states "triangular prism A-BCD". Maybe the notation is different. Maybe it's a tetrahedron? But a tetrahedron has four triangular faces, while a prism has five faces. Wait, perhaps the problem is using non-standard terminology.Alternatively, perhaps the prism is formed by extruding triangle ABD along AC, but this is just speculation. Since the problem mentions "triangular prism A-BCD" with side edges AB, AC, AD being mutually perpendicular, maybe it's a three-dimensional figure where A is connected to B, C, D, and the prism is formed by connecting these edges appropriately.Alternatively, maybe the prism is formed by three parallelograms: AB-CD, AC-BD, AD-BC. But that might form a parallelepiped. However, a prism requires two triangular bases. Given the confusion here, perhaps the key is that regardless of the exact structure, the circumscribed sphere (circum sphere) of the prism would have all its vertices lying on the sphere. Therefore, the center of the sphere is equidistant from all vertices of the prism. So, if we can find the coordinates of all the vertices, we can find the center (h, k, l) such that the distance from (h, k, l) to each vertex is equal, then the radius R would be that distance, and the volume would be (4/3)πR³.But wait, first, we need to confirm all the vertices of the prism. If it's a triangular prism, there should be six vertices. However, the problem mentions "triangular prism A-BCD", which might have four vertices? That doesn't make sense. Wait, no. A triangular prism must have two triangular bases, each with three vertices, so six vertices in total. If the prism is A-BCD, maybe the vertices are A, B, C, D and two other points. Wait, perhaps the prism is A-BCD-EFG or something? The problem statement is a bit unclear.Wait, looking back at the problem statement: "In the triangular prism A-BCD, the side edges AB, AC, AD are mutually perpendicular." So, the prism is named A-BCD, which might indicate that the vertices are A, B, C, D, but a prism needs two triangular bases, so maybe A is part of both bases? For example, one base is triangle ABC and the other is triangle ADE, but the problem doesn't mention E. Hmm.Alternatively, perhaps the prism is formed by moving triangle ABC along the vector AD to form another triangle A'B'C', but since AD is perpendicular to AB and AC, this would make a right prism with rectangular sides. However, in this case, the other base would be A’B’C’ where A’=D, B’=B + AD vector, but this is speculative.Wait, given that the areas of triangles ABC, ACD, and ADB are given, which are all faces meeting at vertex A. Maybe the prism is a three-dimensional figure with vertex A connected to B, C, D, and the other vertices connected accordingly. However, without a clear diagram, it's challenging, but perhaps if we assume that all the vertices of the prism are A, B, C, D, and another point, but I need to clarify.Alternatively, maybe the prism has two triangular bases: one is ABC and the other is BCD. So, the vertices are A, B, C, D, and maybe another point? Wait, no, a prism with bases ABC and BCD would require that the sides are quadrilaterals connecting corresponding vertices. For example, AB to BC, BC to CD, etc., but this seems not standard.Alternatively, perhaps the prism is actually a tetrahedron, but the problem states it's a prism. Hmm. Wait, perhaps the key is that even though it's called a prism, the important thing is that it has vertices A, B, C, D with AB, AC, AD mutually perpendicular, and the given areas. Then, the circumscribed sphere would have these four points on it. But a sphere in 3D space is defined by four non-coplanar points, so these four points would define the sphere. Wait, but in reality, a tetrahedron has a unique circumscribed sphere. However, the problem mentions a prism, which has more points. So, perhaps there are more vertices, but the problem only mentions A-BCD. Maybe the prism is a parallelepiped? If AB, AC, AD are edges from A, mutually perpendicular, then the figure would be a rectangular box, but with ABCD as a tetrahedron. Wait, no, in a rectangular box, all edges are perpendicular, but a prism requires two triangular bases.Wait, maybe the problem has a typo, and it's supposed to be a tetrahedron A-BCD. But the user specifies it's a triangular prism. Let me check the Chinese version or the original problem, but since I can't do that, perhaps I need to proceed with the given information.Alternatively, maybe the prism is formed by extending the edges AB, AC, AD to form a prism where ABC is one base and the other base is a translated version. However, given that AB, AC, AD are mutually perpendicular, the other base would be a translated triangle along the direction perpendicular to ABC. But since AD is already perpendicular to ABC (as AB and AC are in ABC's plane), translating ABC along AD would form a prism with height AD. Then, the other base would be A’B’C’ where A’ is D, but that might not form BCD.Alternatively, let's suppose the prism has two triangular bases: ABC and DBC. Then, the vertices are A, B, C, D, B’, C’, D’? Hmm, not sure.Alternatively, given that the problem gives areas for triangles ABC, ACD, and ADB, which are all adjacent to A, perhaps the figure is a tetrahedron with vertex A connected to B, C, D, which are mutually perpendicular. If that's the case, then it's a tetrahedron with three mutually perpendicular edges at A, and the areas of the three faces. Then, the circumscribed sphere of the tetrahedron can be found. However, the problem mentions a prism, not a tetrahedron. This is conflicting.Wait, let's check the properties again. A triangular prism has two triangular bases connected by three rectangular faces. If AB, AC, AD are mutually perpendicular side edges, then perhaps each of these edges is a side edge connecting the two bases. For example, in a standard triangular prism, there are three side edges connecting the corresponding vertices of the two triangular bases. If in this case, those three side edges are AB, AC, AD, and they are mutually perpendicular, then the two triangular bases must lie in planes perpendicular to each other. Hmm, but how?Alternatively, consider that the two triangular bases are ABC and ABD. Then, the side edges would be AA (which is the same point), BB, CC, DD. Wait, no. This is getting too confusing. Let me try to think differently.Given that the problem states "triangular prism A-BCD" with side edges AB, AC, AD mutually perpendicular, and given the areas of three triangular faces, perhaps the prism is constructed such that A is a common vertex, and B, C, D are vertices of the other base. So, the prism has two triangular bases: one is triangle A (but a triangle cannot have one vertex) — this doesn't make sense. Alternatively, the prism has base triangle BCD and apex A connected via edges AB, AC, AD. But that would make it a pyramid, not a prism.Wait, a pyramid has a base and triangular faces connecting to an apex. A prism has two bases. So if it's a prism, then there must be two triangular bases connected by three rectangles. Given that AB, AC, AD are side edges, perhaps each of these edges connects a vertex from one base to the other. For example, in a standard prism, you have two triangular bases, and each vertex of the first base is connected to the corresponding vertex of the second base via a side edge. If in this case, the side edges AB, AC, AD are connecting A to B, C, D in the other base, but since AB, AC, AD are mutually perpendicular, the other base must be positioned such that moving from A to B, A to C, A to D are along perpendicular directions.But in a standard prism, the side edges are all parallel. However, here, AB, AC, AD are mutually perpendicular, which would mean they are not parallel. Therefore, this is not a standard prism. Therefore, the term "triangular prism" here might be a mistranslation or a different structure. Alternatively, maybe it's a prismatoid where the side edges are not necessarily parallel.Given the confusion about the structure, perhaps the key is that the problem is referring to a tetrahedron with vertex A and edges AB, AC, AD mutually perpendicular, and the areas of the three faces meeting at A are given. Then, to find the circumscribed sphere of this tetrahedron. However, the problem mentions a prism, not a tetrahedron. But since the user has specified it's a triangular prism, perhaps we need to reconcile this.Alternatively, perhaps in Chinese terminology, "triangular prism" might refer to a different structure. But since I can't verify that, maybe I need to proceed with the given data.Assuming that the figure is a tetrahedron with vertex A and edges AB, AC, AD mutually perpendicular, and given the areas of the three faces at A, then we can find the lengths AB, AC, AD, and then compute the circumscribed sphere's radius. However, since the problem says it's a prism, maybe there are more vertices, and the sphere is around the entire prism. But without knowing all the vertices, it's hard to compute. However, given that we can assign coordinates to A, B, C, D, and if the prism has these four points plus others, but since the problem only gives areas for triangles connected to A, maybe the other vertices are not needed.Alternatively, if it's a triangular prism with AB, AC, AD as edges, perhaps the other vertices are B', C', D' such that ABB', ACC', ADD' are rectangles. Then, the prism would have vertices A, B, C, D, B', C', D', but the problem only mentions A-BCD. Maybe the other base is BCD, and the prism is formed by translating triangle BCD along the vector from A to some point? Wait, but the edges AB, AC, AD are mutually perpendicular, which would mean that translating BCD along a vector would not necessarily make the edges perpendicular.This is getting too tangled. Let me try to proceed with the coordinates I have and see if I can find the circumscribed sphere.If we consider that the prism has vertices A, B, C, D, and perhaps other vertices, but given that AB, AC, AD are mutually perpendicular, and the problem mentions "the triangular prism A-BCD", maybe all the vertices are A, B, C, D and the other vertices are formed by moving along the edges. Wait, this is unclear. Alternatively, perhaps the prism is a parallelepiped formed by vectors AB, AC, AD, which are mutually perpendicular. A parallelepiped has six faces, all parallelograms. But the problem mentions a triangular prism, which should have two triangular bases and three rectangular faces.Wait, perhaps the prism is half of the parallelepiped? If we take the parallelepiped formed by AB, AC, AD, and then consider a prism that is a part of it. For example, a triangular prism can be formed by splitting a parallelepiped along a diagonal. But this is speculative.Alternatively, maybe the prism is a right triangular prism, with base triangle ABC and height AD, so the other base is A'B'C' where A' is D. But then the other base would be triangle DBC, since AD is perpendicular to ABC. So, the prism would have bases ABC and DBC, connected by three rectangles: ABD, ACD, and BCD. Wait, but BCD is a triangle, so this doesn't make sense. Alternatively, the lateral faces are rectangles ABDC, ACDD', and BC something. This is too vague.Given the time I've spent trying to figure out the structure without success, maybe I need to focus on the given data and assume that the circumscribed sphere is the one that passes through all the vertices of the prism. Since we know the coordinates of A, B, C, D, and if we can figure out the coordinates of the other vertices, we can find the sphere's center and radius.Wait, but if it's a prism, there should be another triangular base. Let's assume that the prism is formed by translating triangle ABC along the vector AD. Then, the translated triangle would be A’B’C’, where A’ = A + AD vector = D. But AD vector is (0,0,√3). So, translating ABC along AD would give:A’ = D: (0,0,√3)B’ = B + AD vector: (√2,0,0) + (0,0,√3) = (√2,0,√3)C’ = C + AD vector: (0,1,0) + (0,0,√3) = (0,1,√3)Therefore, the prism has vertices A, B, C, D, B’, C’, A’ (which is D). Wait, but D is already a vertex. So, the two triangular bases are ABC and DB’C’? Let's check:Base 1: ABC: A(0,0,0), B(√2,0,0), C(0,1,0)Base 2: DB’C’: D(0,0,√3), B’(√2,0,√3), C’(0,1,√3)Then, the lateral edges are AA’=AD, BB’, CC’. In this case, the lateral edges are AD, BB’, CC’. AA’ is AD, which is along the z-axis, BB’ is from (√2,0,0) to (√2,0,√3), which is along the z-axis, and CC’ is from (0,1,0) to (0,1,√3), also along the z-axis. Therefore, this is a right triangular prism with height √3, and bases ABC and DB’C’.In this case, the vertices of the prism are:A(0,0,0), B(√2,0,0), C(0,1,0), D(0,0,√3), B’(√2,0,√3), C’(0,1,√3)So six vertices in total. Therefore, the circumscribed sphere must pass through all six points.Now, to find the center and radius of the sphere passing through these six points, we can set up equations based on the distances from the center (h,k,l) to each vertex being equal.Let’s denote the center of the sphere as (h, k, l). Then, the distance from the center to each vertex should be equal to the radius R.So, for vertex A(0,0,0):(h - 0)^2 + (k - 0)^2 + (l - 0)^2 = R² ⇒ h² + k² + l² = R²For vertex B(√2,0,0):(h - √2)^2 + k² + l² = R²For vertex C(0,1,0):h² + (k - 1)^2 + l² = R²For vertex D(0,0,√3):h² + k² + (l - √3)^2 = R²For vertex B’(√2,0,√3):(h - √2)^2 + k² + (l - √3)^2 = R²For vertex C’(0,1,√3):h² + (k - 1)^2 + (l - √3)^2 = R²So we have six equations. Let’s subtract the first equation from the others to eliminate R².Subtract A from B:(h - √2)^2 + k² + l² - (h² + k² + l²) = 0 ⇒ -2√2 h + 2 = 0 ⇒ h = 2 / (2√2) = 1/√2Similarly, subtract A from C:h² + (k - 1)^2 + l² - (h² + k² + l²) = 0 ⇒ -2k + 1 = 0 ⇒ k = 1/2Subtract A from D:h² + k² + (l - √3)^2 - (h² + k² + l²) = 0 ⇒ -2√3 l + 3 = 0 ⇒ l = 3 / (2√3) = √3/2Now we have h = 1/√2, k = 1/2, l = √3/2.Let’s check if this center works for the other vertices B’ and C’.Check vertex B’(√2,0,√3):Distance squared from center (1/√2, 1/2, √3/2):(1/√2 - √2)^2 + (1/2 - 0)^2 + (√3/2 - √3)^2Compute each term:First term: (1/√2 - √2) = (1 - 2)/√2 = (-1)/√2 ⇒ squared: 1/2Second term: (1/2)^2 = 1/4Third term: (√3/2 - √3) = (-√3/2) ⇒ squared: 3/4Total: 1/2 + 1/4 + 3/4 = (2/4 + 1/4 + 3/4) = 6/4 = 3/2Compare to R², which is h² + k² + l² = (1/2) + (1/4) + (3/4) = (2/4 + 1/4 + 3/4) = 6/4 = 3/2. So it matches.Similarly, check vertex C’(0,1,√3):Distance squared from center (1/√2, 1/2, √3/2):(1/√2 - 0)^2 + (1/2 - 1)^2 + (√3/2 - √3)^2First term: 1/2Second term: (-1/2)^2 = 1/4Third term: (-√3/2)^2 = 3/4Total: 1/2 + 1/4 + 3/4 = 3/2, which matches R².Therefore, the center of the circumscribed sphere is at (1/√2, 1/2, √3/2) and radius squared is 3/2, so radius R is √(3/2) = √6/2.Therefore, the volume of the sphere is (4/3)πR³ = (4/3)π*(√6/2)^3.Compute (√6/2)^3:(√6)^3 = (√6)*(√6)*(√6) = 6*√6Divide by 2^3 = 8:6√6 / 8 = (3√6)/4Therefore, Volume = (4/3)π*(3√6)/4 = π*(√6)So the volume is π√6.But wait, let me check the calculation again:(√6/2)^3 = (√6)^3 / 8 = (6√6)/8 = (3√6)/4Then, Volume = (4/3)π * (3√6)/4 = π√6. Correct.Therefore, the volume of the circumscribed sphere is π√6.But let me verify once more. The radius is √(3/2), so R = √(3)/√2 = √6/2. Then, R³ = (√6/2)^3 = (6√6)/8 = (3√6)/4. Then, (4/3)π*(3√6)/4 = π√6. Yes, correct.Thus, the final answer is π√6.Final AnswerThe volume of the circumscribed sphere is boxed{sqrt{6} pi}.

✅:Given the triangular prism (A-BCD) with mutually perpendicular side edges (AB, AC, AD), and the areas of triangles (triangle ABC), (triangle ACD), and (triangle ADB) being (frac{sqrt{2}}{2}), (frac{sqrt{3}}{2}), and (frac{sqrt{6}}{2}) respectively, we need to find the volume of the circumscribed sphere.1. Assigning Coordinates: - Place (A) at the origin ((0,0,0)). - Assign coordinates to (B), (C), and (D) along the x, y, and z axes respectively: (B(sqrt{2}, 0, 0)), (C(0, 1, 0)), (D(0, 0, sqrt{3})).2. Solving for Coordinates: - Using the given areas of the triangles, we derive the equations: - (x cdot y = sqrt{2}) - (y cdot z = sqrt{3}) - (x cdot z = sqrt{6}) - Solving these, we find (x = sqrt{2}), (y = 1), (z = sqrt{3}).3. Identifying Vertices of the Prism: - The vertices of the prism are (A(0,0,0)), (B(sqrt{2},0,0)), (C(0,1,0)), (D(0,0,sqrt{3})), (B'(sqrt{2},0,sqrt{3})), and (C'(0,1,sqrt{3})).4. Finding the Circumscribed Sphere: - The center ((h, k, l)) of the sphere must be equidistant from all vertices. - Solving the system of equations for the distances from the center to each vertex, we find: - (h = frac{1}{sqrt{2}}) - (k = frac{1}{2}) - (l = frac{sqrt{3}}{2}) - The radius (R) is calculated as (sqrt{frac{3}{2}} = frac{sqrt{6}}{2}).5. Calculating the Volume: - The volume of the sphere is (frac{4}{3} pi R^3): - (R^3 = left(frac{sqrt{6}}{2}right)^3 = frac{3sqrt{6}}{4}) - Volume = (frac{4}{3} pi cdot frac{3sqrt{6}}{4} = sqrt{6} pi).The volume of the circumscribed sphere is (boxed{sqrt{6} pi}).

🔑:Given the tetrahedron (A-BCD) where the edges (AB), (AC), and (AD) are mutually perpendicular and the areas of triangles ( triangle ABC ), ( triangle ACD ), and ( triangle ADB ) are (frac{sqrt{2}}{2}), (frac{sqrt{3}}{2}), and (frac{sqrt{6}}{2}) respectively.# Step 1: Express the areas in terms of the side lengthsLet ( AB = a ), ( AC = b ), and ( AD = c ).For ( triangle ABC ):[S_{triangle ABC} = frac{1}{2} times AB times AC = frac{1}{2}ab = frac{sqrt{2}}{2}]From this equation:[ab = sqrt{2} quad text{(1)}]For ( triangle ACD ):[S_{triangle ACD} = frac{1}{2} times AC times AD = frac{1}{2}bc = frac{sqrt{3}}{2}]From this equation:[bc = sqrt{3} quad text{(2)}]For ( triangle ADB ):[S_{triangle ADB} = frac{1}{2} times AD times AB = frac{1}{2}ac = frac{sqrt{6}}{2}]From this equation:[ac = sqrt{6} quad text{(3)}]# Step 2: Solve for side lengths (a), (b), and (c)We have three equations:[ab = sqrt{2} quad text{(1)}][bc = sqrt{3} quad text{(2)}][ac = sqrt{6} quad text{(3)}]Multiplying all three equations together:[(ab)(bc)(ac) = (sqrt{2})(sqrt{3})(sqrt{6})][a^2b^2c^2 = (sqrt{2} cdot sqrt{3} cdot sqrt{6})^2][a^2b^2c^2 = 36][(abc)^2 = 36][abc = sqrt{36} = 6]From equations (ab = sqrt{2}), (bc = sqrt{3}), and (ac = sqrt{6}):[abc = 6]Substitute ( a = sqrt{2} ), ( b = 1 ), and ( c = sqrt{3} ):[(sqrt{2})(1)(sqrt{3}) = 6]Thus, (a = sqrt{2}), (b = 1), and (c = sqrt{3}).# Step 3: Calculate the circumradiusThe circumradius (R) of a tetrahedron with ((ABC)) mutually perpendicular edges is:[R = frac{1}{2} sqrt{a^2 + b^2 + c^2}]Substituting (a), (b), and (c):[R = frac{1}{2} sqrt{(sqrt{2})^2 + 1^2 + (sqrt{3})^2}][R = frac{1}{2} sqrt{2 + 1 + 3}][R = frac{1}{2} sqrt{6}][R = frac{sqrt{6}}{2}]# Step 4: Calculate the volume of the circumsphereThe volume (V) of the circumsphere is given by:[V = frac{4}{3} pi R^3]Substituting (R = frac{sqrt{6}}{2}):[R^3 = left(frac{sqrt{6}}{2}right)^3 = frac{(sqrt{6})^3}{8} = frac{6sqrt{6}}{8} = frac{3sqrt{6}}{4}]The volume of the circumsphere:[V = frac{4}{3} pi left(frac{3sqrt{6}}{4}right)][V = pi cdot sqrt{6}] Conclusion[boxed{sqrt{6}pi}]

🧠:Okay, so I need to prove that if the quadratic equation y = x² + bx + c is always positive for all integer values of x, then the discriminant b² - 4c must be less than or equal to zero. Hmm, let's start by recalling some basic properties of quadratic equations.First, a quadratic equation ax² + bx + c will be always positive if its graph doesn't intersect the x-axis and opens upwards. Since the coefficient of x² here is 1, which is positive, the parabola indeed opens upwards. So, for the quadratic to be positive for all real x, the discriminant must be negative. That is, b² - 4c < 0. But the problem states that y > 0 for all integer values of x, not necessarily all real numbers. So maybe the discriminant can be zero or positive, but still, the quadratic is positive at integer points. Wait, but the problem wants us to prove that the discriminant is ≤ 0. So even if the quadratic touches the x-axis (discriminant zero) or has real roots, but since it's positive for integers, maybe the roots are not integers? Hmm, not sure. Let me think.But the problem states that y > 0 for all integer x. So even if the quadratic has real roots, as long as those roots are not integers, and the quadratic is positive at all integer points, perhaps the discriminant could still be positive. But the problem claims that even in such a case, the discriminant must be ≤ 0. So maybe my initial thought is wrong. Let me try to approach this step by step.First, suppose that the discriminant is positive, i.e., b² - 4c > 0. Then the quadratic equation x² + bx + c = 0 has two distinct real roots. Let's denote the roots as α and β, where α < β. Since the parabola opens upwards, the quadratic is positive outside the interval [α, β] and negative inside (α, β). But the problem states that y is positive for all integer x. Therefore, there must be no integer x such that x is in the interval (α, β). So, if there exists an integer between α and β, then y would be negative there, which contradicts the given condition. Hence, for y to be positive at all integers, the interval (α, β) must not contain any integers. Therefore, if the discriminant is positive, the quadratic must have its two real roots such that there are no integers between them. But the question is, can such a quadratic exist where the discriminant is positive but there's no integer between the roots? If so, then the discriminant could be positive, but the problem says we have to prove it's ≤ 0, so such quadratics must not exist. Wait, maybe the roots themselves are integers? But if roots are integers, then plugging those integers into the quadratic would give y=0, which contradicts y > 0 for all integers. So, the roots can't be integers. But even if the roots are non-integers, but between two integers, then the quadratic would be negative there. Wait, but if the entire interval (α, β) lies strictly between two consecutive integers, then there would be no integer x in that interval, so y would be positive for all integers. So in that case, the discriminant could be positive. Therefore, maybe the problem's statement is not true? But the problem says to prove that b² - 4c ≤ 0, so perhaps there's a mistake in my reasoning.Wait, maybe even if the interval (α, β) lies between two integers, but since the quadratic is positive outside, but in reality, if the quadratic is positive everywhere except between the roots, then for all integers outside [α, β], y is positive. But if the interval (α, β) is between two integers, say between n and n+1 for some integer n, then the integers n and n+1 would be outside the interval, but any integer less than n or greater than n+1 would also be outside. Wait, no. For example, suppose α = 1.2 and β = 1.8. Then the interval (1.2, 1.8) is between 1 and 2, but there's no integer in that interval. Therefore, all integer x would be either ≤1 or ≥2. At x=1, y would be positive, and at x=2, y would be positive. So even if the quadratic has a positive discriminant, but the roots are between two consecutive integers, then y is positive for all integers. Hence, in such a case, b² -4c >0, but the problem says to prove that b² -4c ≤0. Therefore, there must be some error here. Maybe the problem requires that the quadratic is positive for all integers x, but even if there are non-integer roots, the quadratic might still take positive values at integers. But the problem claims that in such a case, the discriminant must be ≤0. Therefore, my initial reasoning is conflicting with the problem statement. Therefore, perhaps I need to think differently.Alternatively, maybe even if the roots are between two integers, there is some integer x where y is not positive. Let's take an example. Suppose we have a quadratic with roots at 1.2 and 1.8. Then, let's construct such a quadratic. The quadratic can be written as (x - 1.2)(x - 1.8) = x² - 3x + 2.16. So here, b = -3, c = 2.16. But c must be an integer. So that's not allowed. Hmm. So maybe if b and c are integers, then the roots can't be non-integers with such a tight interval? Wait, maybe when b and c are integers, the roots α and β must be such that either they are both integers or irrational numbers. Because the quadratic equation x² + bx + c =0 with integer coefficients would have roots that are either integers or irrationals (by Rational Root Theorem). Because the possible rational roots are factors of c over factors of 1, so integers dividing c. So if the quadratic factors into integers, then roots are integers. Otherwise, the roots are irrational or complex. Therefore, if the discriminant is a perfect square, then roots are integers; otherwise, they are irrational. But in our problem, the discriminant is b² -4c. If the discriminant is positive and a perfect square, then roots are integers, which would make y=0 at those roots, contradicting y>0 for all integer x. If the discriminant is positive but not a perfect square, then the roots are irrational. But even if the roots are irrational, there might be an integer x lying between them, which would make y negative. So, perhaps even if the roots are irrational, but between two integers, but since the quadratic is positive outside the roots, if an integer is between α and β, then y would be negative there. Therefore, for y to be positive for all integer x, there must be no integer between α and β. However, since α and β are irrational, but how can we ensure that there is no integer between them?Wait, but irrational numbers can still be between two integers. For example, sqrt(2) ≈1.414 is between 1 and 2, but sqrt(2) is irrational. So if α and β are irrational numbers, how can we be sure that there isn't an integer between them?Alternatively, maybe even with irrational roots, if the quadratic is positive for all integer x, then the quadratic must be positive for all real x, hence discriminant ≤0. But that contradicts the example I tried earlier where discriminant is positive but roots are between two integers. But in that example, c wasn't an integer. Maybe with c being integer, such quadratics cannot exist? Let's try to construct such a quadratic with integer b and c, positive discriminant, and no integer between the roots.Let me attempt to find such a quadratic. Let's suppose the roots are between 1 and 2. So, for instance, take α = 1.5 and β = 1.5 (but that would be a double root, discriminant zero). Wait, but discriminant zero would give a repeated root, and since the quadratic is positive everywhere else, but at the root it's zero. So in that case, since the root is 1.5, which is not an integer, then y would be positive for all integers. But discriminant is zero. So that case is allowed. But the problem allows discriminant ≤0. So discriminant zero is okay, but discriminant positive might not be.Wait, suppose we take roots at 1.2 and 1.8, then the quadratic is (x -1.2)(x -1.8) =x² -3x +2.16. But c must be integer. So c=2.16 is not allowed. So maybe with integer coefficients, such quadratics with roots between two consecutive integers cannot exist? Because if the roots are between two consecutive integers, then perhaps c would not be integer. Let's try to see.Suppose we have quadratic x² + bx + c with integer b, c. Suppose it has two real roots α and β, which are between two consecutive integers, say n and n+1. Then, the quadratic can be written as (x - α)(x - β) = x² - (α + β)x + αβ. So here, b = - (α + β), and c = αβ. Since b and c are integers, then α + β must be integer, and αβ must be integer.If α and β are both irrational, but their sum and product are integers. For example, take α = n + d, β = n + (1 - d), where d is a decimal between 0 and 1. Then α + β = 2n +1, which is integer. The product αβ = (n + d)(n +1 - d) = n² + n - d² + d(1 - 2n). For this to be integer, the terms involving d must cancel out. Hmm, this seems complicated. Maybe take specific numbers. Let me try n=1, so roots between 1 and 2. Let α = 1 + t, β = 1 + (1 - t) = 2 - t, where t is between 0 and 1. Then sum α + β = 3, so b = -3. Product αβ = (1 + t)(2 - t) = 2 - t + 2t - t² = 2 + t - t². For c = αβ to be integer, 2 + t - t² must be integer. Since t is between 0 and1, t - t² is between 0 - 0 =0 and 0.5 -0.25=0.25. So 2 + t -t² is between 2 and 2.25. Therefore, c would be 2 since t - t² is less than 1. Therefore, c=2. Therefore, if we take t=0.5, then α=1.5, β=1.5, which gives c=(1.5)^2=2.25, but that's not integer. Wait, but if t is such that t - t² is integer? Since t is between 0 and1, t - t² can only be between 0 and 0.25. So the only integer possible is 0. So t - t²=0 implies t=0 or t=1, but t is between 0 and1. So t=0: α=1, β=2. But then the roots are at 1 and 2, which are integers. Then c=1*2=2. Then the quadratic is (x -1)(x -2)=x² -3x +2. Then at x=1 and x=2, y=0, which contradicts the condition that y >0 for all integer x. So that's not allowed. So if we require that c is integer, then the only possibility is when t - t² is zero, which gives roots at integers, but that causes y=0 at those integers. Therefore, such a quadratic with integer coefficients and roots between two consecutive integers cannot exist? Because otherwise, the product c would not be integer. Wait, unless t is such that t - t² is an integer, but as we saw, it's only possible when t=0 or t=1, which gives roots at integers.Therefore, perhaps for quadratics with integer coefficients, if the discriminant is positive, then the roots are either integers or irrational numbers. If they are integers, then y=0 at those integers, which is not allowed. If they are irrational, then between any two irrational numbers, there might be an integer. Wait, but how can we be sure? For example, take quadratic x² - 3x + 1. Discriminant is 9 -4 =5>0. Roots are (3 ±√5)/2. √5≈2.236, so the roots are approximately (3 -2.236)/2≈0.382 and (3 +2.236)/2≈2.618. So the roots are approximately 0.382 and 2.618. So between 0.382 and 2.618, there are integers 1 and 2. Therefore, the quadratic x² -3x +1 would be negative at x=1 and x=2. Let's check: x=1, y=1 -3 +1= -1 <0. x=2, y=4 -6 +1= -1 <0. So indeed, even though the roots are irrational, there are integers between them where y is negative. Therefore, in this case, the quadratic is not positive for all integers x.But maybe there exists a quadratic with integer coefficients, positive discriminant, and roots such that there's no integer between them? Let's try. Suppose the roots are between 0 and 1. Let's take quadratic x² + bx + c. Let’s say roots are α=0.4 and β=0.6. Then sum α + β=1.0, so b=-1. Product αβ=0.24, which is not integer. So c=0.24 is invalid. If we try to adjust roots to make product integer. For example, suppose roots are α=0.5 and β=0.5. Then sum is1.0, b=-1, product=0.25, so c=0.25, not integer. If roots are α=0.2 and β=0.8, sum=1.0, product=0.16. Still not integer. So seems impossible to get integer c if roots are between 0 and1.Alternatively, take roots between -1 and 0. For example, α=-0.5 and β=-0.5. Then sum=-1.0, so b=1, product=0.25, c=0.25, not integer. Again, not possible.Another example: roots between 2 and3. Let’s say α=2.1 and β=2.9. Sum=5.0, so b=-5. Product=2.1*2.9=6.09. So c=6.09, not integer. If we adjust to get product integer. Suppose α=2.5 and β=2.5. Sum=5.0, product=6.25. Not integer. Doesn't help.So it seems that for quadratic with integer coefficients, if discriminant is positive, then either the roots are integers (leading to y=0 at those integers) or the roots are irrationals. And in the case of irrational roots, there must be some integer between them because the distance between two irrationals can be more than 1, but maybe not necessarily. Wait, let's check another example. Take quadratic x² - x -1. Discriminant=1 +4=5. Roots=(1±√5)/2≈1.618 and ≈-0.618. So roots are approximately -0.618 and 1.618. The interval between them is (-0.618,1.618). So integers in this interval are x=0 and x=1. Let's compute y at x=0: 0 +0 -1=-1 <0. At x=1:1 -1 -1=-1 <0. So again, y is negative at these integers. Hence, the quadratic is not positive for all integer x.Another example: x² -2x -1. Discriminant=4 +4=8. Roots=(2±√8)/2=1±√2≈2.414 and -0.414. Interval (-0.414,2.414). Integers 0,1,2. y at x=0:0 -0 -1=-1 <0. x=1:1 -2 -1=-2 <0. x=2:4 -4 -1=-1 <0. Still negative.Wait, maybe find a quadratic with positive discriminant and y positive at all integers. Let's try. Let's take a quadratic that is positive except between two non-integer roots. For example, roots at 0.5 and1.5. Then quadratic is (x -0.5)(x -1.5)=x² -2x +0.75. Here, b=-2, c=0.75. But c must be integer. Not allowed. If we make c=1, then maybe x² -2x +1=(x-1)^2. Discriminant zero. Y=0 at x=1. Not allowed. If c=2, quadratic x² -2x +2. Discriminant=4 -8=-4<0. So no real roots. Then y is always positive. But discriminant is negative. So this satisfies the problem's conclusion.Alternatively, suppose we have a quadratic with roots at 1.1 and1.9. Then sum=3.0, product=2.09. So quadratic x² -3x +2.09. c is not integer. If we round c to2, then quadratic x² -3x +2, which factors to (x-1)(x-2), roots at1 and2. But then y=0 at x=1 and2, which is invalid.So it seems impossible to have a quadratic with integer coefficients, positive discriminant, and y positive for all integer x. Because either the roots are integers (invalid), or irrationals with an integer between them (making y negative at that integer). Therefore, if a quadratic with integer coefficients is positive for all integer x, then it must be positive for all real x, hence discriminant ≤0. Therefore, the initial statement is true.But wait, let's test with discriminant zero. For example, quadratic x² -2x +1=(x-1)^2. At x=1, y=0, which is invalid. So discriminant zero is allowed only if the repeated root is not an integer. For example, quadratic x² +2x +1=(x+1)^2. At x=-1, y=0. If the repeated root is a non-integer, say 1.5, then quadratic is (x-1.5)^2=x² -3x +2.25. But c=2.25 is not integer. Hence, impossible. Therefore, quadratics with integer coefficients and discriminant zero must have their repeated root at a rational number (since discriminant zero implies root at -b/(2*1)). If b is integer, then root is -b/2, which is either integer or half-integer. If it's a half-integer, like 1.5, then c=(1.5)^2=2.25, which is not integer. If the root is integer, then discriminant zero quadratic would be (x -n)^2, which at x=n gives y=0. Hence, the only quadratics with integer coefficients and discriminant zero have y=0 at some integer. Therefore, discriminant zero is not allowed in the problem's condition. Therefore, the discriminant must be strictly negative. But the problem says to prove that discriminant ≤0. So maybe including discriminant zero is a mistake, but according to the problem statement, we need to prove that if y>0 for all integer x, then discriminant ≤0. But from the above, discriminant zero would imply y=0 at some x (if the root is integer) or y>0 for all integer x (if the root is non-integer). Wait, but if the root is non-integer, then y never actually touches zero, so y would be positive everywhere, but since the root is non-integer, then at integer x, y is positive. However, if the quadratic has a discriminant zero and the root is non-integer, then c would not be integer. For example, root at 1.5: quadratic is x² -3x +2.25. c=2.25 is not integer. So with integer coefficients, discriminant zero implies the root is a rational number. Specifically, since discriminant is b² -4c=0, so c=b²/4. Since c must be integer, b² must be divisible by4, so b must be even. Let b=2k. Then c=(2k)^2/4=k². Therefore, the quadratic is x² +2k x +k²=(x +k)^2. Therefore, the root is x=-k, which is integer. Therefore, if discriminant is zero, then the quadratic is a perfect square and has an integer root, leading to y=0 at that integer. Therefore, discriminant zero is impossible under the problem's condition. Hence, the discriminant must be strictly less than zero. But the problem says to prove that discriminant ≤0. Maybe the problem allows discriminant zero, but actually, in that case, there's a contradiction. Therefore, the correct conclusion is discriminant <0, but the problem states ≤0. Perhaps the problem has a typo, but according to the instructions, I need to prove that b² -4c ≤0.Given the above reasoning, even though discriminant zero would lead to y=0 at an integer (hence violating y>0), but the problem's statement says "y >0 for all integer values of x", so discriminant zero is excluded. Therefore, why does the problem say "prove that b² -4c ≤0"? It should be strictly less than zero. However, perhaps in the problem's context, considering that discriminant zero quadratics with integer coefficients necessarily have integer roots (as we saw), hence such quadratics cannot satisfy y>0 for all integer x, hence the discriminant must indeed be strictly less than zero, so the problem's statement might have a mistake. However, according to the problem, we must prove that b² -4c ≤0. Therefore, perhaps the problem considers discriminant zero as allowed if the double root is not at an integer. But as we saw, with integer coefficients, discriminant zero implies the root is -b/2, which must be a rational number with denominator 1 or 2. If it's denominator 2, then c=(b/2)^2, which requires b even. Wait, if b is odd, then -b/2 is a half-integer, but then c=(b²)/4 would not be integer unless b is even. So if b is even, say b=2k, then c=k², and the quadratic is (x +k)^2, with root at x=-k, integer. If b is odd, say b=1, then c=(1)^2/4=0.25, not integer. Therefore, discriminant zero with integer coefficients can only occur when b is even and c is a perfect square, leading to an integer root. Therefore, any quadratic with integer coefficients and discriminant zero will have an integer root, hence y=0 at that root. Therefore, in the context of the problem, discriminant zero is impossible, so discriminant must be <0, hence ≤0.Therefore, combining all this, the correct approach is to note that if y >0 for all integer x, then in particular, the quadratic must be positive for all real x, hence discriminant must be ≤0. Wait, but earlier reasoning suggested that even if discriminant is positive, but there's no integer between the roots, then y could be positive for all integer x. However, we saw that with integer coefficients, such quadratics cannot exist because the product c would not be integer unless the roots are integers or lead to a contradiction. Therefore, the conclusion is that if y>0 for all integer x, then discriminant must be ≤0. Hence, the proof is as follows:Assume y = x² +bx +c >0 for all integer x. Suppose for contradiction that the discriminant D = b² -4c >0. Then the quadratic has two distinct real roots α and β. Since the coefficients are integers, the roots are either both integers or irrational numbers. If they are integers, then y=0 at those integers, contradicting y>0. If they are irrational, then by the Intermediate Value Theorem, there exists an integer between α and β (since the distance between α and β is sqrt(D) which is at least sqrt(1) =1, as D is positive integer). However, this may not necessarily hold. Wait, sqrt(D) can be less than1? Wait, D is b² -4c >0. Since b and c are integers, D must be at least 1. Because if b² -4c >0, then the minimal possible D is when b² -4c=1. For example, b=1, c=0: D=1-0=1. So the distance between the roots is sqrt(D)/1 = sqrt(D) ≥1. So the roots are separated by at least1 unit. Therefore, if two roots are separated by at least1 unit, then there must be at least one integer between them. For instance, if α and β are two real numbers with β -α ≥1, then there exists an integer n such that α ≤n ≤β. Therefore, y(n) ≤0, contradicting the assumption that y >0 for all integer x. Hence, our assumption that D>0 leads to a contradiction. Therefore, D must be ≤0.Yes, this makes sense. Because if the distance between the roots is at least1, then there must be an integer between them, hence y would be non-positive there. Therefore, the discriminant cannot be positive. Hence, proven.

✅:To prove that ( b^2 - 4c leq 0 ) for the quadratic ( y = x^2 + bx + c ) (where ( b ) and ( c ) are integers) given that ( y > 0 ) for all integer values of ( x ), we proceed as follows:1. Assume the contrary: Suppose ( b^2 - 4c > 0 ). Then the quadratic has two distinct real roots ( alpha ) and ( beta ) ((alpha < beta)).2. Roots and integer coefficients: Since ( b, c ) are integers, the roots must either be integers or irrational (by the Rational Root Theorem). If the roots are integers, ( y = 0 ) at those integers, contradicting ( y > 0 ). If the roots are irrational, proceed to the next step.3. Distance between roots: The roots satisfy ( beta - alpha = sqrt{b^2 - 4c} geq 1 ) (since ( b^2 - 4c ) is a positive integer). Thus, the interval ( (alpha, beta) ) spans at least one unit.4. Existence of an integer in ( (alpha, beta) ): Any interval of length ≥1 contains at least one integer. Let ( n ) be such an integer. Substituting ( x = n ) into ( y ), we get ( y(n) leq 0 ) (since ( n ) lies between the roots), contradicting ( y > 0 ).5. Conclusion: The assumption ( b^2 - 4c > 0 ) leads to a contradiction. Hence, ( b^2 - 4c leq 0 ).Final Answer: For ( y = x^2 + bx + c ) to satisfy ( y > 0 ) at all integer ( x ), it must hold that ( boxed{b^2 - 4c leq 0} ).

🔑:1. Case 1: b is an even integer. For this case, we can write ( b ) as ( b = 2k ) where ( k in mathbb{Z} ). The quadratic function can now be rewritten as ( y = x^2 + 2kx + c ). The vertex of the quadratic function ( y = x^2 + 2kx + c ) is at ( x = -frac{b}{2} = -k ). At this vertex, ( y ) takes its minimum value. Substituting ( x = -k ) into the equation: [ y = (-k)^2 + b(-k) + c ] [ y = k^2 - 2k^2 + c ] [ y = -k^2 + c ] From the problem statement, we know that ( y > 0 ) for all integer values of ( x ). Thus, at ( x = -k ), we have: [ -k^2 + c > 0 ] [ c > k^2 ] Rewriting ( k ) in terms of ( b ), we have ( b = 2k ) so ( k = frac{b}{2} ). Substituting this back in: [ c > left(frac{b}{2}right)^2 ] [ c > frac{b^2}{4} ] Rearranging inequality: [ b^2 - 4c < 0 ]2. Case 2: b is an odd integer. For this case, we can write ( b ) as ( b = 2k + 1 ) where ( k in mathbb{Z} ). The quadratic function can be rewritten as ( y = x^2 + (2k + 1)x + c ). The vertex of this quadratic function occurs at ( x = -frac{b}{2} ), which is not an integer when ( b ) is odd. Thus, we evaluate at the nearest integer points, ( x = -frac{b + 1}{2} ) and ( x = -frac{b - 1}{2} ): When ( x = frac{-b - 1}{2} ): [ y = left(frac{-b - 1}{2}right)^2 + b left(frac{-b - 1}{2}right) + c ] Simplifying: [ y = left(frac{b+1}{2}right)^2 + b left(frac{-b - 1}{2}right) + c ] [ y = frac{(b+1)^2}{4} - frac{b(b+1)}{2} + c ] [ y = frac{b^2+2b+1}{4} - frac{2b^2 + 2b}{4} + c ] [ y = frac{b^2 + 2b + 1 - 2b^2 - 2b}{4} + c ] [ y = frac{-b^2 + 1}{4} + c ] Similarly, when ( x = frac{-b + 1}{2} ): [ y = left(frac{-b + 1}{2}right)^2 + b left(frac{-b + 1}{2}right) + c ] Simplifying: [ y = left(frac{b-1}{2}right)^2 + b left(frac{-b + 1}{2}right) + c ] [ y = frac{(b-1)^2}{4} - frac{b(b-1)}{2} + c ] [ y = frac{b^2-2b+1}{4} - frac{2b^2 - 2b}{4} + c ] [ y = frac{b^2 - 2b + 1 - 2b^2 + 2b}{4} + c ] [ y = frac{-b^2 + 1}{4} + c ] Hence, both calculations yield: [ y = frac{-b^2 + 1}{4} + c ] Since ( y > 0 ) for all integer values of ( x ), when evaluated at the nearest integers to the vertex: [ frac{-b^2 + 1}{4} + c > 0 ] [ -b^2 + 1 + 4c > 0 ] [ 4c > b^2 - 1 ] [ 4c geq b^2 ] Therefore: [ b^2 - 4c leq 0 ] Conclusion:Combining both cases, we conclude:[boxed{b^2 - 4c leq 0}]

🧠:Okay, so there's this problem about three friends and a monkey. They have some fruits, and each night one friend takes a third of the fruits, but there's always one left over which they give to the monkey. Then the next friend does the same, and finally the third friend. In the morning, they divide what's left and again have one extra for the monkey. We need to find the smallest number of fruits that makes all this possible.Hmm. Let me start by trying to model each step. Let's denote the initial number of fruits as N. First friend's action at night: He tries to divide N into three parts, but there's one left over. So, N = 3a + 1. He gives 1 to the monkey and takes a, so the remaining fruits would be N - a - 1 = 3a + 1 - a - 1 = 2a. Wait, that's right. So after the first friend, there are 2a fruits left.But maybe I should represent this with equations. Let me define variables step by step.Let’s denote the number of fruits after each step:1. Start with N fruits.2. First friend: Tries to divide N into three, with 1 leftover. So N ≡ 1 mod 3. Then he takes (N - 1)/3 fruits, gives 1 to the monkey. So remaining fruits: N - (N - 1)/3 - 1 = (2N - 2)/3.Wait, let's check that:If N = 3a + 1, then he takes a fruits, gives 1 to the monkey. Remaining: 3a + 1 - a - 1 = 2a. So remaining is 2a, which is (2/3)(N - 1). Hmm, so yes, (2N - 2)/3. So the remaining fruits after first friend is (2N - 2)/3.Let me note that as N1 = (2N - 2)/3.Then the second friend comes. He does the same thing: divides N1 into three parts, finds 1 extra, gives it to the monkey, takes his third. So N1 must also be ≡ 1 mod 3. Then he takes (N1 - 1)/3, gives 1 to the monkey. Remaining fruits: N1 - (N1 - 1)/3 - 1 = (2N1 - 2)/3. Let's call this N2.Similarly, the third friend comes and does the same, so N2 must be ≡ 1 mod 3. Then he takes (N2 - 1)/3, gives 1 to the monkey, leaving N3 = (2N2 - 2)/3.In the morning, they divide N3 into three parts, again with 1 left over, so N3 ≡ 1 mod 3. Then they each take (N3 - 1)/3, give 1 to the monkey, and that's the end.So our conditions are:1. N ≡ 1 mod 32. N1 = (2N - 2)/3 ≡ 1 mod 33. N2 = (2N1 - 2)/3 ≡ 1 mod 34. N3 = (2N2 - 2)/3 ≡ 1 mod 3And we need to find the smallest N such that all these conditions are satisfied.Alternatively, working backwards might be easier. Let's start from the morning. Let's denote the number of fruits in the morning as M. Then M ≡ 1 mod 3, and they divide it into three, give 1 to the monkey, and take (M - 1)/3 each. So before the morning division, there are M fruits. But M is the same as N3, the fruits left after the third friend.Wait, no. Wait, N3 is after the third friend's division. Then in the morning, they take N3 and divide into three with one left over. So N3 must ≡ 1 mod 3, and after that division, the remaining fruits would be (2N3 - 2)/3. But actually, the problem says they divided the remaining fruits equally among themselves, found one extra, which they gave to the monkey. So maybe after dividing, they take their shares and leave nothing? Wait, but the problem says "divided the remaining fruits equally among themselves, again finding one extra fruit, which they gave to the monkey." So similar to the previous steps: total fruits before morning is M, M ≡ 1 mod 3. Each takes (M - 1)/3, gives 1 to the monkey. Then remaining fruits would be M - 3*( (M -1)/3 ) -1 = M - (M -1) -1 = 0. Wait, that can't be. So maybe in the morning, after taking their shares, there's nothing left except the one given to the monkey. But the problem says "divided the remaining fruits equally among themselves, again finding one extra fruit, which they gave to the monkey." So perhaps the process is same as before: when they divide the fruits into three, there's one left over which goes to the monkey, and each takes a third of the rest. But then the remaining fruits after that would be two-thirds of (M -1). Wait, but in the problem statement, it's not clear if in the morning they actually take the fruits home or just leave them. Wait, the problem says "divided the remaining fruits equally among themselves, again finding one extra fruit, which they gave to the monkey." So similar to the previous steps: each time, they take a third of the fruits (after giving one to the monkey), so the remaining fruits after each step is two-thirds of (current fruits minus one). So in the morning, starting from N3, they do the same: N3 ≡ 1 mod 3, then remaining fruits after morning division would be (2*(N3 -1 ))/3. But the problem doesn't mention anything after the morning division, so maybe the problem ends there. However, since they divided the fruits equally among themselves and gave one to the monkey, but in this case, since there are three friends, they each take a share. So maybe the total fruits in the morning is N3, which must be ≡1 mod3, then they give 1 to the monkey, take (N3 -1)/3 each, and then there are no fruits left? Because (N3 -1)/3 * 3 +1 = N3. So yes, they take all the fruits, giving one to the monkey. So actually, after the morning division, there are zero fruits left. So the process is that each time they reduce the fruits by (current -1)/3 +1. Wait, maybe I need to model each step carefully.Let me think step by step:First night:- N fruits.- First friend: divides into 3, remainder 1. So N = 3a + 1. Gives 1 to monkey, takes a. Remaining fruits: N - a -1 = 3a +1 -a -1 = 2a. So remaining is 2a = (2N -2)/3.Second night:- Second friend finds 2a fruits. He does the same: divides into three, remainder 1. So 2a = 3b + 1. Then he gives 1 to the monkey, takes b. Remaining: 2a - b -1 = 3b +1 - b -1 = 2b. So 2b = (2*(2a) -2)/3 = (4a -2)/3. Wait, but 2a = 3b +1, so 4a = 6b +2. So (4a -2)/3 = (6b +2 -2)/3 = 6b/3 = 2b. So yes, that works.Third night:- Third friend finds 2b fruits. Divides into three, remainder 1. So 2b = 3c +1. Gives 1 to monkey, takes c. Remaining: 2b -c -1 = 3c +1 -c -1 = 2c. So 2c = (2*(2b) -2)/3 = (4b -2)/3. Similarly, since 2b =3c +1, 4b =6c +2, so (4b -2)/3 = (6c +2 -2)/3=6c/3=2c.Morning:- They find 2c fruits. Divide into three, remainder 1. So 2c =3d +1. Give 1 to monkey, take d each. Remaining would be 2c -3d -1 =0? Because they took all the fruits. So 2c -3d -1 =0 => 3d +1 =2c.But 2c must be equal to 3d +1. So that's the last condition.So now, we need to track back all these equations.Let's denote variables step by step:Starting from the morning:Let’s denote the number of fruits in the morning as M =2c. According to the morning division, M ≡1 mod3, so 2c ≡1 mod3. Which implies 2c =3d +1. So c=(3d +1)/2. Since c must be integer, 3d +1 must be even, so d must be odd. Let d =2k +1 for some integer k ≥0. Then c=(3*(2k +1)+1)/2=(6k +3 +1)/2=(6k +4)/2=3k +2. So c=3k +2. Then M=2c=6k +4.But M is also equal to the fruits left after the third friend, which is 2c. But before that, the third friend had 2b fruits, which after his division leaves 2c. From third friend's division: 2b =3c +1. Wait, no. Wait, third friend had 2b fruits. He divides into three, finds one extra, so 2b ≡1 mod3, so 2b =3c +1. Therefore, b=(3c +1)/2. Similarly, since b must be integer, 3c +1 must be even. Since c=3k +2, substituting:b=(3*(3k +2)+1)/2=(9k +6 +1)/2=(9k +7)/2. For b to be integer, 9k +7 must be even. 9k is odd if k is odd, even if k is even. 9k +7: if k is even, 9k is even, even +7=odd. If k is odd, 9k is odd, odd +7=even. So to have 9k +7 even, k must be odd. Let k=2m +1 for some integer m ≥0. Then c=3*(2m +1)+2=6m +3 +2=6m +5. Then b=(9*(2m +1) +7)/2=(18m +9 +7)/2=(18m +16)/2=9m +8. So b=9m +8.Then M=2c=2*(6m +5)=12m +10.Now, going back to the second friend. The second friend had 2a fruits. After his division, it becomes 2b. From second friend's division: 2a=3b +1. So a=(3b +1)/2. Since a must be integer, 3b +1 must be even. Since b=9m +8, substituting:a=(3*(9m +8)+1)/2=(27m +24 +1)/2=(27m +25)/2. For a to be integer, 27m +25 must be even. 27m is odd if m is odd, even if m is even. 27m +25: if m even, 27m even, even +25=odd. If m odd, 27m odd, odd +25=even. So m must be odd. Let m=2n +1 for some integer n ≥0. Then b=9*(2n +1)+8=18n +9 +8=18n +17. Then a=(27*(2n +1)+25)/2=(54n +27 +25)/2=(54n +52)/2=27n +26. So a=27n +26.Then N1=2a=2*(27n +26)=54n +52. But N1 is the remaining fruits after the first friend. The first friend had N fruits. After dividing, he left N1=2a. From first friend's division: N=3a +1. So N=3*(27n +26)+1=81n +78 +1=81n +79.So N=81n +79. Let's check for n=0: N=79. Let's see if this works.Wait, let's test n=0. So N=79.First friend: 79. 79 divided by 3 is 26 with remainder 1. So he gives 1 to the monkey, takes 26. Remaining:79 -26 -1=52. Which is 2a=52, so a=26. Which matches our earlier calculation (a=27n +26, n=0: a=26).Second friend: 52 fruits. 52 divided by 3 is 17 with remainder 1. He gives 1 to the monkey, takes 17. Remaining:52 -17 -1=34. So 2b=34, so b=17. But according to our previous step, when n=0, m=2n +1=1 (since m must be odd when we set m=2n +1). Wait, no. Wait, when we set m=2n +1, but earlier when m was introduced, we had k=2m +1. Wait, this might be getting confusing with the substitutions. Let me retrace.We had:From the morning:d was expressed in terms of k, then c in terms of k, then b in terms of m, but substituted back through. Maybe instead of substituting multiple times, let's see with N=79.Test N=79:First friend: 79 ≡1 mod3 (79-1=78, divisible by 3). Takes 26, gives 1 to monkey. Remaining:52.Second friend: 52. 52 ≡1 mod3? 52-1=51, which is divisible by 3. 51/3=17. So takes 17, gives 1 to monkey. Remaining:52 -17 -1=34.Third friend:34. 34-1=33, divisible by 3. 33/3=11. Takes 11, gives 1 to monkey. Remaining:34 -11 -1=22.Morning:22 fruits. 22 ≡1 mod3? 22-1=21, which is divisible by 3. 21/3=7. So they give 1 to the monkey, each takes 7. Remaining:22 -7*3 -1=22 -21 -1=0. That works. Wait, but 22 divided into three parts with 1 left over:22=3*7 +1. Yes. So each takes 7, gives 1 to monkey, total taken 7*3 +1=22. So all fruits are accounted for. So N=79 seems to work?But wait, according to the previous equations, when n=0, N=81*0 +79=79. Then the remaining in the morning is 22, which is 2c=22 => c=11. Then c=3k +2 =>11=3k +2 =>k=3. Then d=(2c -1)/3=(22 -1)/3=21/3=7. Which is okay.But according to our substitutions earlier, we had c=6m +5, and m=2n +1. Wait, maybe my substitutions got tangled, but testing n=0 gives N=79, which seems to work. But according to some sources, the answer is 79? Wait, but I thought the classic problem similar to this has a higher number. Wait, let me check the steps again.Wait, let's verify each step with N=79:1. First friend: 79. 79 divided by 3: 3*26=78, remainder 1. Gives 1 to monkey, takes 26. Remaining:79 -26 -1=52.2. Second friend:52. 52 divided by 3: 3*17=51, remainder 1. Gives 1 to monkey, takes 17. Remaining:52 -17 -1=34.3. Third friend:34. 34 divided by 3:3*11=33, remainder 1. Gives 1 to monkey, takes 11. Remaining:34 -11 -1=22.4. Morning:22. 22 divided by 3:3*7=21, remainder 1. Gives 1 to monkey, each takes 7. Remaining:22 -7*3 -1=0. Perfect.So 79 works. Is this the minimal? Let's check n=-1? But n must be ≥0. So 79 is the minimal. Wait, but maybe there's a smaller number. Let me check with N=79. Let's try N=79 is the answer.But wait, let's check with n=0 gives N=79. If we take n=-1, which is invalid because n must be non-negative. Therefore, 79 is the minimal. But I have a feeling I might have missed something because sometimes these problems have larger numbers, but maybe not. Let me see.Alternatively, perhaps the answer is 79, but let me check if there is a smaller number. Let's try smaller N≡1 mod3.Let's start testing numbers congruent to 1 mod3:Start from N=1: Too small. First friend can't take a third.N=4: First friend:4≡1 mod3? 4-1=3, which is divisible by 3. So takes (4-1)/3=1, gives 1 to monkey. Remaining:4 -1 -1=2. Then second friend:2. 2≡1 mod3? No. So invalid.N=7: First friend:7-1=6, divisible by 3. Takes 2, gives 1. Remaining:7 -2 -1=4. Second friend:4. 4≡1 mod3? 4-1=3, yes. Takes 1, gives 1. Remaining:4 -1 -1=2. Third friend:2. 2≡1 mod3? No. Invalid.N=10:10-1=9. Takes 3, gives 1. Remaining:10-3-1=6. Second friend:6. 6≡1 mod3? 6-1=5, not divisible by3. No.N=13:13-1=12. Takes4, gives1. Remaining:8. Second friend:8≡1 mod3? 8-1=7, no.N=16:16-1=15. Takes5, gives1. Remaining:10. 10-1=9, which is divisible by3. Second friend takes3, gives1. Remaining:10-3-1=6. Third friend:6-1=5, not divisible by3. No.N=19:19-1=18. Takes6, gives1. Remaining:12. Second friend:12-1=11, no.N=22:22-1=21. Takes7, gives1. Remaining:14. 14-1=13, no.N=25:25-1=24. Takes8, gives1. Remaining:16. 16-1=15, yes. Second friend takes5, gives1. Remaining:16-5-1=10. Third friend:10-1=9, takes3, gives1. Remaining:10-3-1=6. Morning:6-1=5, not divisible by3. No.N=28:28-1=27. Takes9, gives1. Remaining:18. Second friend:18-1=17, no.N=31:31-1=30. Takes10, gives1. Remaining:20. 20-1=19, no.N=34:34-1=33. Takes11, gives1. Remaining:22. 22-1=21, yes. Second friend takes7, gives1. Remaining:22-7-1=14. Third friend:14-1=13, no.N=37:37-1=36. Takes12, gives1. Remaining:24. 24-1=23, no.N=40:40-1=39. Takes13, gives1. Remaining:26. 26-1=25, no.N=43:43-1=42. Takes14, gives1. Remaining:28. 28-1=27, yes. Second friend takes9, gives1. Remaining:28-9-1=18. Third friend:18-1=17, no.N=46:46-1=45. Takes15, gives1. Remaining:30. 30-1=29, no.N=49:49-1=48. Takes16, gives1. Remaining:32. 32-1=31, no.N=52:52-1=51. Takes17, gives1. Remaining:34. 34-1=33, yes. Second friend takes11, gives1. Remaining:34-11-1=22. Third friend:22-1=21, takes7, gives1. Remaining:22-7-1=14. Morning:14-1=13, no.N=55:55-1=54. Takes18, gives1. Remaining:36. 36-1=35, no.N=58:58-1=57. Takes19, gives1. Remaining:38. 38-1=37, no.N=61:61-1=60. Takes20, gives1. Remaining:40. 40-1=39, yes. Second friend takes13, gives1. Remaining:40-13-1=26. Third friend:26-1=25, no.N=64:64-1=63. Takes21, gives1. Remaining:42. 42-1=41, no.N=67:67-1=66. Takes22, gives1. Remaining:44. 44-1=43, no.N=70:70-1=69. Takes23, gives1. Remaining:46. 46-1=45, yes. Second friend takes15, gives1. Remaining:46-15-1=30. Third friend:30-1=29, no.N=73:73-1=72. Takes24, gives1. Remaining:48. 48-1=47, no.N=76:76-1=75. Takes25, gives1. Remaining:50. 50-1=49, no.N=79: As before, works.So seems like 79 is the minimal number. But why in the previous equations, when we set n=0, we get N=79, which works. Therefore, the answer should be 79.Wait, but I recall a similar problem where the answer was 81*... minus something, but maybe this one is different. Let me confirm once again.Testing N=79:First step: 79 -> gives 1 to monkey, takes 26. Remaining 52.Second step:52 -> gives1, takes17. Remaining34.Third step:34->gives1, takes11. Remaining22.Morning:22->gives1, takes7 each. Remaining0. Perfect.Yes, all steps work. So 79 is the smallest such number. Therefore, the answer is 79.

✅:The smallest number of fruits that satisfies all the distribution conditions is 79. Step-by-Step Explanation:1. First Friend's Division: - Total fruits: ( N ). - ( N equiv 1 mod 3 ). Let ( N = 3a + 1 ). - Remaining after first friend: ( 2a ).2. Second Friend's Division: - Fruits: ( 2a equiv 1 mod 3 ). Let ( 2a = 3b + 1 ). - Remaining after second friend: ( 2b ).3. Third Friend's Division: - Fruits: ( 2b equiv 1 mod 3 ). Let ( 2b = 3c + 1 ). - Remaining after third friend: ( 2c ).4. Morning Division: - Fruits: ( 2c equiv 1 mod 3 ). Let ( 2c = 3d + 1 ).Backtracking to Find Minimal ( N ):- From morning: ( 2c = 3d + 1 ). Smallest ( d = 7 ) gives ( c = 11 ).- Third friend: ( 2b = 3c + 1 = 34 ) → ( b = 17 ).- Second friend: ( 2a = 3b + 1 = 52 ) → ( a = 26 ).- First friend: ( N = 3a + 1 = 79 ).Verification:- Night 1: 79 → 26 taken, 1 to monkey. Remaining: 52.- Night 2: 52 → 17 taken, 1 to monkey. Remaining: 34.- Night 3: 34 → 11 taken, 1 to monkey. Remaining: 22.- Morning: 22 → 7 each, 1 to monkey. All fruits distributed.Answer: (boxed{79})

🔑:1. Let ( x ) be the number of mangoes received by each friend in the morning after their equal division.2. According to the problem, in the morning they divided the remaining mangoes equally, and each received ( x ) mangoes. Given that one mango was left over and given to the monkey, the total remaining mangoes should be ( 3x + 1 ).3. Let's denote the initial number of mangoes as ( N ). We will work backward to find ( N ).# Analysis of the First Friend:4. The first friend went to the storeroom and attempted to take one-third of the mangoes, but found one extra fruit which he gave to the monkey. Thus we express his share as: [ frac{N - 1}{3} ] and the remaining mangoes after he took his share and gave one to the monkey: [ frac{2(N - 1)}{3} ]# Analysis of the Second Friend:5. The second friend, unaware of the first friend's action, tried to divide the remaining mangoes by three and also found one extra which he gave to the monkey. Thus the new remainder can be expressed and processed as follows: [ frac{2(N - 1)}{3} - 1 ] Divided by three again, giving: [ frac{frac{2(N - 1)}{3} - 1 - 1}{3} = frac{2(N - 1) - 3}{9} = frac{2N - 5}{9} ]6. After the second friend took his share and gave one to the monkey, the remaining mangoes are: [ 2left(frac{2N - 5}{9}right) = frac{4N - 10}{9} ]# Analysis of the Third Friend:7. The third friend, also unbeknownst to the previous actions, divided the remaining mangoes and found one extra which he gave to the monkey. We adopt the same steps: [ frac{frac{4N - 10}{9} - 1 - 1}{3} = frac{4N - 28}{27} ]8. After the third friend took his share and gave one to the monkey, the remaining mangoes are: [ 2left(frac{4N - 28}{27}right) = frac{8N - 56}{27} ]9. Finally, in the morning, the remaining mangoes ( 3x + 1 ) should be equal to the number after all three friends took their parts and gave mangoes to the monkey: [ 3x + 1 = frac{8N - 56}{27} ]# Solve the Equation:10. Rearrange and solve for ( N ) where ( x ) has integer constraints: [ 3x + 1 = frac{8N - 56}{27} ] Multiply both sides by 27 to clear the fraction: [ 27(3x + 1) = 8N - 56 ] Simplify: [ 81x + 27 = 8N - 56 ] [ 8N = 81x + 83 ] [ N = frac{81x + 83}{8} ]11. We need ( N ) to be an integer: [ 81x + 83 equiv 0 pmod{8} ] Simplify: [ 81x + 83 equiv 5x + 3 pmod{8} ] To have a multiple of 8: [ 5x + 3 equiv 0 pmod{8} ] [ 5x equiv -3 pmod{8} ] [ 5x equiv 5 pmod{8} ] Thus, ( x equiv 1 pmod{8} ): [ x = 8k + 1 ] Start with ( k = 0 ): [ x = 1 ] Compute using ( x = 1 ): [ N = frac{81(1) + 83}{8} = 164 / 8 = 20.5 quad (text{not an integer}) ]12. Now, try higher values: [ x = 7 quad (since ( 7 equiv -1 pmod{8} )) ] Compute: [ N = frac{81(7) + 83}{8} = 650 / 8 = 81.25 quad (text{not an integer}) ]13. Try higher again: [ x = 7 ] Just checking, we should try: [ x = 19 ] Compute: [ N = frac{81(19) + 83}{8} = 20 ] and checking if their units work, tries yields 79 finally matched condition so less solution concludes: tojiori less conclusion:Therefore, the smallest number of mangoes that satisfy the problem’s conditions is:[ boxed{79} ]

🧠:Okay, so I have this problem here: If 47^100 is a 168-digit number, how many digits does 47^17 have? The options are A) 27, B) 28, C) 29, D) 30. Hmm, let me try to figure this out step by step. First, I remember that the number of digits in a number can be found using logarithms. Specifically, the formula is that the number of digits d in a number N is given by d = floor(log10(N)) + 1. So, if I can find log10(47^17), then take the floor of that and add 1, I should get the number of digits. That makes sense.But wait, the problem gives me information about 47^100 being a 168-digit number. Maybe that's a hint to use that instead of calculating log10(47) directly? Because if I can relate 47^100 to its number of digits, maybe I can find log10(47) and then use that to compute log10(47^17). Let me explore that.Given that 47^100 has 168 digits, according to the formula, that means:168 = floor(log10(47^100)) + 1Which simplifies to:floor(100 * log10(47)) + 1 = 168Subtracting 1 from both sides:floor(100 * log10(47)) = 167So, 100 * log10(47) is a number between 167 and 168 because the floor of it is 167. Therefore:167 ≤ 100 * log10(47) < 168Dividing all parts by 100:1.67 ≤ log10(47) < 1.68So, log10(47) is approximately between 1.67 and 1.68. That gives me a range for log10(47). Let me note that down: log10(47) ≈ 1.67 to 1.68.Now, I need to find log10(47^17). Using logarithm properties, log10(47^17) = 17 * log10(47). So, multiplying the range we found by 17:17 * 1.67 ≤ log10(47^17) < 17 * 1.68Calculating those:17 * 1.67 = 28.3917 * 1.68 = 28.56So, log10(47^17) is between 28.39 and 28.56. Then, taking the floor of that would give us 28, and adding 1 gives 29. So, the number of digits would be 29. Hmm, but wait, the options are 27, 28, 29, 30. So, 29 is option C. But wait, let me check my calculations again because sometimes these approximations can be tricky.Wait, let's double-check. If log10(47) is between 1.67 and 1.68, then 17 * 1.67 is indeed 28.39 and 17 * 1.68 is 28.56. Therefore, the log10(47^17) is between 28.39 and 28.56, which means the floor is 28, so floor(log10(47^17)) = 28, and then digits would be 28 + 1 = 29. So, answer C.But hold on, let me make sure there isn't a mistake here. Let's think again. If 47^100 has 168 digits, then log10(47^100) is approximately 167. Something. So log10(47^100) = 100 * log10(47) ≈ 167. Something. Therefore, log10(47) ≈ 167. Something / 100 ≈ 1.67something. But the exact value of log10(47) is actually around?Wait, maybe I should check the actual value of log10(47). Let's compute it. Since 10^1.67 is 10^(167/100) = 10^1 * 10^0.67. 10^0.67 is approximately e^(0.67 * ln10) ≈ e^(0.67 * 2.302585) ≈ e^(1.5423) ≈ 4.677. So, 10^1.67 ≈ 46.77. Which is close to 47. So, 10^1.67 ≈ 46.77, and 10^1.68 would be 10^1.68 ≈ 10^(1 + 0.68) = 10 * 10^0.68. 10^0.68 ≈ e^(0.68 * 2.302585) ≈ e^(1.5657) ≈ 4.787. So, 10^1.68 ≈ 47.87. Therefore, 47 is between 10^1.67 and 10^1.68, so log10(47) is between 1.67 and 1.68, as previously established. But more accurately, since 47 is 46.77 at 1.67 and 47.87 at 1.68, so 47 is 0.23 above 46.77. The difference between 47.87 and 46.77 is 1.1, so 0.23 / 1.1 ≈ 0.209. So, log10(47) ≈ 1.67 + 0.209*(0.01) wait, no. Wait, the interval between 1.67 and 1.68 is 0.01 in log scale. Wait, no: 1.67 to 1.68 is 0.01. Wait, 1.67 is 1 + 0.67, and 1.68 is 1 + 0.68. So, the difference is 0.01 in the logarithm. But the actual values are 46.77 to 47.87. So, the value 47 is 47 - 46.77 = 0.23 above 46.77. The total interval is 47.87 - 46.77 = 1.1. So, 0.23 / 1.1 ≈ 0.209. So, the log10(47) ≈ 1.67 + 0.209*(0.01) = 1.67 + 0.00209 ≈ 1.67209. Wait, is that right?Wait, no. Let me think again. Let me consider that the log10(47) is between 1.67 and 1.68. The antilogarithm (10^x) increases exponentially, so the difference in x (log) corresponds to a multiplicative factor in the antilog. So, to interpolate linearly between 1.67 and 1.68 to find where 47 lies, we can use linear approximation. Let me set x such that 10^(1.67 + x) = 47, where x is between 0 and 0.01. Wait, maybe better to set up the equation:Let’s denote log10(47) = 1.67 + δ, where δ is between 0 and 0.01.Then, 10^(1.67 + δ) = 47.We know that 10^1.67 ≈ 46.77, as before.So, 10^(1.67 + δ) = 10^1.67 * 10^δ ≈ 46.77 * 10^δ = 47.Therefore, 10^δ = 47 / 46.77 ≈ 1.00492.Taking log10 of both sides:δ = log10(1.00492) ≈ 0.00213 (since log10(1.00492) ≈ 0.00213; because 10^0.002 ≈ 1.0046).Therefore, δ ≈ 0.00213.Therefore, log10(47) ≈ 1.67 + 0.00213 ≈ 1.67213.Therefore, log10(47) ≈ 1.6721.So, a more accurate value is approximately 1.6721.Therefore, log10(47^17) = 17 * 1.6721 ≈ 17 * 1.6721.Let me compute that:1.6721 * 10 = 16.7211.6721 * 7 = 11.7047So, 16.721 + 11.7047 = 28.4257Therefore, log10(47^17) ≈ 28.4257So, floor(28.4257) = 28, then 28 + 1 = 29 digits. So, the answer should be 29, which is option C.But let me check another way. Since 47^100 has 168 digits, so 47^100 ≈ 10^167 (since it's a 168-digit number, the first digit is between 1 and 9, but the order of magnitude is 10^167). So, 47^100 ≈ 10^167. Therefore, log10(47^100) ≈ 167, so log10(47) ≈ 167/100 = 1.67. But as we saw earlier, more precisely, log10(47) ≈ 1.6721. Therefore, 47^17 = 10^(17 * log10(47)) ≈ 10^(17 * 1.6721) ≈ 10^28.4257 ≈ 10^0.4257 * 10^28.Now, 10^0.4257 is approximately equal to what? Let's calculate 10^0.4257.We know that log10(2.66) ≈ 0.425 because 10^0.425 ≈ 2.66. Let me verify:log10(2.66) = ln(2.66)/ln(10) ≈ 0.978 / 2.3026 ≈ 0.425. So, 10^0.425 ≈ 2.66. Therefore, 10^0.4257 ≈ approximately 2.66. So, 47^17 ≈ 2.66 * 10^28. Therefore, it's a 29-digit number because 2.66 * 10^28 is 2 followed by 28 zeros, but since it's 2.66, it's 29 digits total.Alternatively, if we use a more precise calculation for 10^0.4257:Let me compute 0.4257 * ln(10) ≈ 0.4257 * 2.302585 ≈ 0.4257 * 2.3 ≈ 0.979, plus 0.4257 * 0.002585 ≈ 0.0011, so total ≈ 0.979 + 0.0011 ≈ 0.9801. Then, e^0.9801 ≈ e^0.98 ≈ 2.664. Therefore, 10^0.4257 ≈ 2.664. Therefore, 47^17 ≈ 2.664 * 10^28, which is a 29-digit number because 2.664 is between 1 and 10, and multiplied by 10^28 gives 29 digits. So, yes, 29 digits.Therefore, the answer is C) 29.But wait, just to make sure, let me check another approach. Maybe using the given data more directly. If 47^100 has 168 digits, then 10^(167) ≤ 47^100 < 10^(168). Taking the 100th root, we get 10^(167/100) ≤ 47 < 10^(168/100). Which is 10^1.67 ≤ 47 < 10^1.68, as before. Then, raising to the 17th power: 10^(1.67*17) ≤ 47^17 < 10^(1.68*17). Calculating 1.67*17: 1.67*10=16.7, 1.67*7=11.69, total=16.7+11.69=28.39. Similarly, 1.68*17=1.68*10=16.8, 1.68*7=11.76, total=28.56. Therefore, 10^28.39 ≤ 47^17 < 10^28.56. Since 10^28.39 is 10^0.39 * 10^28 ≈ 2.45 * 10^28, and 10^28.56 is 10^0.56 * 10^28 ≈ 3.63 * 10^28. So, 47^17 is between 2.45e28 and 3.63e28. Both of these are 29-digit numbers because 2.45e28 is 2 followed by 28 zeros, which is 29 digits, and 3.63e28 is 3 followed by 28 zeros, also 29 digits. Hence, regardless of the exact value, it must have 29 digits. So, the answer is C) 29.Therefore, after considering multiple approaches and checking the calculations, I'm confident that the correct answer is 29 digits, which is option C.

✅:To determine the number of digits in (47^{17}), we use logarithms. The number of digits (d) in a number (N) is given by (d = lfloor log_{10} N rfloor + 1). Given (47^{100}) has 168 digits:[168 = lfloor 100 log_{10} 47 rfloor + 1 implies lfloor 100 log_{10} 47 rfloor = 167]Thus, (167 leq 100 log_{10} 47 < 168), which simplifies to:[1.67 leq log_{10} 47 < 1.68]For (47^{17}), we calculate:[log_{10} (47^{17}) = 17 log_{10} 47]Substituting the bounds for (log_{10} 47):[17 times 1.67 leq log_{10} (47^{17}) < 17 times 1.68][28.39 leq log_{10} (47^{17}) < 28.56]Taking the floor value:[lfloor log_{10} (47^{17}) rfloor = 28]Thus, the number of digits is:[28 + 1 = 29]Answer: (boxed{C})

🔑:1. Given 47^{100} is a 168-digit number, we can infer: [ lg(47^{100}) = 167 + a, ] where 0 leq a < 1. This relationship means: [ 47^{100} = 10^{167 + a}. ] Taking the logarithm of both sides: [ lg(47^{100}) = 100 cdot lg 47 = 167 + a, ] thus: [ lg 47 = frac{167 + a}{100} = frac{167}{100} + frac{a}{100}. ]2. Now, let's determine the number of digits in 47^{17}: [ lg(47^{17}) = 17 cdot lg 47 = 17 left(frac{167}{100} + frac{a}{100}right). ] Expanding and simplifying: [ lg(47^{17}) = frac{17 cdot 167}{100} + frac{17a}{100} = frac{2839}{100} + frac{17a}{100} = 28 + frac{39 + 17a}{100}. ]3. Given that 0 leq a < 1, then 0 leq 17a < 17. Adding 39 to this range: [ 39 leq 39 + 17a < 56. ] Dividing the inequality by 100: [ 0.39 leq frac{39 + 17a}{100} < 0.56, ] which means: [ 0 leq frac{39 + 17a}{100} < 1. ]4. From the logarithmic relationship: [ lg(47^{17}) = 28 + frac{39 + 17a}{100} ] implying that 47^{17} is a number between (10^{28}) and (10^{29}), but not (10^{29}) itself.5. Therefore, the number of digits of 47^{17} is 29.Conclusion: (boxed{C})

🧠:Okay, so I need to prove that the sum of all the vectors from the center O to each vertex of a regular polygon is zero. Let me think about how to approach this. First, let me visualize a regular polygon. Since it's regular, all sides and angles are equal. The center O is equidistant from all vertices, which means each vector OA_i has the same length. Hmm, maybe I can represent these vectors in some coordinate system to make it easier?Let me start with a simple case, like a regular triangle (n=3). If I place the triangle in a coordinate system with O at the origin, then the vertices can be represented as vectors. For a regular triangle, the angles between each vector are 120 degrees apart. Let me assign coordinates. Suppose OA1 is along the x-axis, so OA1 = (1, 0). Then OA2 would be at 120 degrees, so its coordinates would be (cos(120°), sin(120°)) which is (-1/2, √3/2). Similarly, OA3 is at 240°, which is (cos(240°), sin(240°)) = (-1/2, -√3/2). If I add these vectors:(1, 0) + (-1/2, √3/2) + (-1/2, -√3/2) = (1 - 1/2 -1/2, 0 + √3/2 -√3/2) = (0, 0). So for a triangle, it works. Let me check a square. Place the square with OA1 along the x-axis. Then OA1=(1,0), OA2=(0,1), OA3=(-1,0), OA4=(0,-1). Adding them: (1+0-1+0, 0+1+0-1) = (0,0). Okay, that also sums to zero. Maybe this pattern holds for all regular polygons? But how do I generalize this?Perhaps using complex numbers? Each vertex can be represented as a complex number on the unit circle. If the polygon is regular, the vertices are the nth roots of unity. The sum of all nth roots of unity is zero. That's a known result. But maybe I need to prove this without using complex numbers, or at least connect it back to vectors.Alternatively, using vectors and symmetry. Since the polygon is regular, rotating the polygon by 2π/n radians around the center O should map the polygon onto itself. This rotational symmetry might imply that the sum of vectors is invariant under rotation. Let's suppose S is the sum of OA1 + OA2 + ... + OAn. If we rotate the entire polygon by 2π/n, each vector OA_i becomes OA_{i+1}, with OA_{n+1} being OA1. So the sum S after rotation would be the same as S. But rotation by 2π/n is a linear transformation, which can be represented by multiplying each vector by a rotation matrix. So, the rotated sum is equal to S, but also equal to the original sum S. Therefore, S must be invariant under rotation by 2π/n. The only vector that is invariant under such a rotation is the zero vector. Hence, S = 0.Wait, that seems like a solid argument. Let me elaborate. Let R be the rotation matrix for angle 2π/n. Then applying R to each OA_i gives OA_{i+1}. So R(OA_i) = OA_{i+1}. Therefore, the sum R(S) = R(OA1) + R(OA2) + ... + R(OAn) = OA2 + OA3 + ... + OAn + OA1 = S. Therefore, R(S) = S. Since R is a rotation by 2π/n, which is not a multiple of 2π (unless n=1, which isn't a polygon), the only vector that is fixed under this rotation is the zero vector. Therefore, S must be zero.That seems to work. But let me check if there's another way to think about it, maybe using components. Suppose the polygon is centered at the origin, and one vertex is at (1,0). Then each vertex OA_k can be represented in polar coordinates as (r, θ_k), where θ_k = 2π(k-1)/n. Converting to Cartesian coordinates, each vector OA_k is (r cos θ_k, r sin θ_k). The sum of all vectors would be the sum over k=1 to n of (r cos θ_k, r sin θ_k). Since r is the same for all, factor that out: r*(sum cos θ_k, sum sin θ_k). To show the sum is zero, need to show sum cos θ_k = 0 and sum sin θ_k = 0.So, need to compute sum_{k=1}^n cos(2π(k-1)/n) and sum_{k=1}^n sin(2π(k-1)/n). These sums are known to be zero for any integer n ≥ 2. I think this can be proven using the formula for the sum of a geometric series in complex numbers. Because if we consider the sum of e^{iθ_k} where θ_k = 2π(k-1)/n, then this is the sum of the nth roots of unity, which is zero. The real part is the sum of cos θ_k and the imaginary part is the sum of sin θ_k. Hence, both sums are zero.Alternatively, using trigonometric identities. For example, sum_{k=0}^{n-1} cos(a + kd) = [sin(n d /2)/sin(d/2)] cos(a + (n-1)d/2). Similarly for sine. If we set a=0 and d=2π/n, then sum_{k=0}^{n-1} cos(0 + k*2π/n) = [sin(n*(2π/n)/2)/sin(2π/(2n))] * cos(0 + (n-1)*(2π/n)/2). Simplifying, numerator sin(π) = 0, so the sum is zero. Same for sine. Hence, sum cos θ_k = 0 and sum sin θ_k = 0.Therefore, the sum of the vectors is zero. Another approach could be using vectors and pairing opposite vectors. For even n, each vector OA_i has an opposite vector OA_{i + n/2}, which would be -OA_i. Adding them gives zero. But for odd n, there isn't an exact opposite, but the symmetry still causes the vectors to cancel out. However, this pairing argument works neatly for even n but requires more careful handling for odd n. The rotational symmetry approach works for any n ≥ 3.Wait, let's see. If n is even, say 4, then OA1 and OA3 are opposites, OA2 and OA4 are opposites. Each pair cancels. For n=6, similarly. But for n=5 (odd), each vector doesn't have a direct opposite. However, when you sum all vectors, their components in each direction cancel due to symmetry. For example, for n=5, each vector can be paired with others such that their horizontal and vertical components cancel. But maybe this is more complex. The rotational symmetry approach seems more straightforward.Another thought: the centroid of the polygon's vertices is the center O. The centroid is the average of the position vectors of the vertices. If the centroid is O, then (OA1 + OA2 + ... + OAn)/n = O. Therefore, OA1 + OA2 + ... + OAn = n*O. Wait, but O is the origin here? Wait, if O is the origin, then the centroid is (sum OA_i)/n. If the centroid is O, then (sum OA_i)/n = O, so sum OA_i = 0. Therefore, this is another way to see it: the centroid of the vertices is the center, hence the sum is zero.But why is the centroid equal to the center? Because in a regular polygon, the center is the centroid due to symmetry. Each vertex is balanced around O. So in a way, the symmetry ensures that the centroid is O. Therefore, sum OA_i = 0. But perhaps this is circular? Because we might need to prove that the centroid is O. However, given the regular polygon's symmetry, any line through O and a vertex (or midpoint of a side) is an axis of symmetry. Therefore, the centroid must lie on all these axes, which intersect only at O. Hence, centroid is O. Therefore, sum OA_i = 0. So there are multiple ways to approach this proof: using complex numbers and roots of unity, using rotational symmetry and linear transformations, using centroid properties, or using trigonometric sums. All lead to the conclusion that the sum is zero. Let me try to formalize the rotational symmetry argument. Let S = OA1 + OA2 + ... + OAn. Rotate S by 2π/n radians. Since rotation is a linear operation, rotating each vector and summing is the same as summing the rotated vectors. So, R(S) = R(OA1) + R(OA2) + ... + R(OAn). But rotating OA1 gives OA2, rotating OA2 gives OA3, ..., rotating OAn gives OA1. Therefore, R(S) = OA2 + OA3 + ... + OAn + OA1 = S. So, R(S) = S. But the only vector that is unchanged by a rotation of 2π/n (other than full rotations) is the zero vector. Therefore, S must be zero. This is because if S were non-zero, then rotating it by 2π/n would change its direction, contradicting R(S) = S. Hence, S = 0.This seems like a concise and elegant proof. It uses the symmetry of the polygon and properties of rotations. Since the polygon is regular, rotating it maps the vertices onto themselves, hence the sum must remain the same. The only vector invariant under such a rotation is the zero vector. Therefore, the sum is zero.Alternatively, using complex numbers, let each OA_k be represented by the complex number z_k = e^{2πi(k-1)/n}. The sum from k=1 to n of z_k is the sum of the nth roots of unity, which is known to be zero. Since the sum of all nth roots of unity is zero, both the real and imaginary parts sum to zero, hence the vector sum is zero.Another way to see the sum of roots of unity is zero is by considering the equation z^n = 1. The sum of all roots is the coefficient of z^{n-1} in the polynomial z^n - 1, which is zero. Hence, the sum is zero.But perhaps the problem expects a geometric proof rather than algebraic. The rotational symmetry argument is geometric and straightforward. Alternatively, breaking into x and y components. Suppose we place the polygon in the plane with OA1 along the x-axis. Each vector OA_k makes an angle θ_k = 2π(k-1)/n with the x-axis. The x-component of each vector is |OA_k| cos θ_k, and y-component is |OA_k| sin θ_k. Since all |OA_k| are equal (they are radii of the same circle), let's denote |OA_k| = r. Then the sum of x-components is r Σ cos θ_k, and sum of y-components is r Σ sin θ_k. We need to show both sums are zero.Using Euler's formula: Σ_{k=1}^n e^{iθ_k} = Σ_{k=1}^n e^{2πi(k-1)/n} = 0, as it's the sum of nth roots of unity. Therefore, both the real part (sum of cosines) and the imaginary part (sum of sines) are zero. Hence, the total sum is zero.So regardless of the approach—rotational symmetry, complex numbers, centroid, or trigonometric identities—the conclusion is that the sum of the vectors is zero. I think the rotational symmetry argument is the most elegant here because it doesn't require coordinate systems or complex numbers, just the inherent symmetry of the regular polygon. It's a concise proof that leverages the properties of rotations and invariance. Alternatively, for someone not familiar with group theory or complex numbers, the component-wise approach might be more intuitive. Let me lay that out step by step. 1. Place the regular polygon with center O at the origin.2. Assume all vectors OA_i have the same magnitude r (since it's regular).3. Express each vector OA_i in terms of its x and y components using cosine and sine of their respective angles.4. Sum all the x-components: r Σ cos(2π(k-1)/n) from k=1 to n.5. Sum all the y-components: r Σ sin(2π(k-1)/n) from k=1 to n.6. Show that both sums are zero using properties of trigonometric sums or complex exponentials.7. Conclude that the total vector sum is zero.But to do this without invoking complex numbers, one could use the identity for the sum of sines and cosines in an arithmetic progression. For angles θ_k = 2π(k-1)/n, the sum over k=1 to n of cos θ_k is zero, and similarly for sine. The formula for the sum of cosines is:Σ_{k=0}^{n-1} cos(a + kd) = [sin(n d /2) / sin(d/2)] cos(a + (n - 1)d / 2).Setting a = 0 and d = 2π/n, we get:Σ_{k=0}^{n-1} cos(0 + k*2π/n) = [sin(n*(2π/n)/2) / sin(2π/(2n))] * cos(0 + (n - 1)*(2π/n)/2).Simplifying:sin(π) / sin(π/n) * cos((n - 1)π/n) = 0 / sin(π/n) * cos(π - π/n) = 0.Similarly for the sine sum:Σ_{k=0}^{n-1} sin(a + kd) = [sin(n d /2) / sin(d/2)] sin(a + (n - 1)d / 2).Setting a = 0 and d = 2π/n:sin(π) / sin(π/n) * sin((n - 1)π/n) = 0.Therefore, both sums are zero. Hence, the total vector sum is zero.This trigonometric identity approach is more involved but also valid.In summary, the key idea is that the high symmetry of the regular polygon causes the vector contributions from each vertex to cancel out when summed. Whether approached through rotation, complex numbers, centroids, or trigonometric identities, the conclusion remains the same.

✅:To prove that the sum of the vectors ( overrightarrow{O A}_{1}, overrightarrow{O A}_{2}, ldots, overrightarrow{O A}_{n} ) from the center ( O ) of a regular polygon to its vertices is zero, we utilize the polygon's rotational symmetry:1. Rotational Symmetry Argument: - Let ( S = overrightarrow{O A}_{1} + overrightarrow{O A}_{2} + ldots + overrightarrow{O A}_{n} ). - Rotate the polygon by ( frac{2pi}{n} ). This maps each vector ( overrightarrow{O A}_i ) to ( overrightarrow{O A}_{i+1} ), with ( overrightarrow{O A}_{n+1} equiv overrightarrow{O A}_{1} ). - The rotated sum ( R(S) ) equals ( S ), since the polygon is unchanged. Thus, ( R(S) = S ). - The only vector invariant under a non-trivial rotation is the zero vector. Therefore, ( S = 0 ).2. Complex Numbers/Trigonometric Sums: - Represent vectors as complex numbers ( e^{2pi i (k-1)/n} ), which are the ( n )-th roots of unity. - The sum of all ( n )-th roots of unity is zero: ( sum_{k=1}^{n} e^{2pi i (k-1)/n} = 0 ). - Separating real and imaginary parts confirms ( sum cosleft(frac{2pi(k-1)}{n}right) = 0 ) and ( sum sinleft(frac{2pi(k-1)}{n}right) = 0 ).Final Answer: The sum of the vectors is zero due to rotational symmetry, as any non-zero sum would change under rotation, contradicting invariance. Thus, [boxed{sum_{k=1}^{n} overrightarrow{O A}_{k} = mathbf{0}}]

🔑:1. Introduce the Sum Vector: Let the sum of the vectors from the center (O) to the vertices (A_1, A_2, ldots, A_n) of the regular polygon be a vector (vec{m}). [ vec{m} = overrightarrow{OA_1} + overrightarrow{OA_2} + cdots + overrightarrow{OA_n}. ]2. Consider Rotating the Polygon: Since the polygon is regular, rotating it around its center (O) by an angle (frac{2pi}{n}) will map the polygon onto itself.3. Apply the Rotation: Under this rotation, each vertex (A_k) is mapped to (A_{k+1}) (with (A_k) shifted to (A_{k+1})), and (overrightarrow{OA_k}) becomes (overrightarrow{OA_{k+1}}).4. Sum after Rotation: The sum of the vectors (overrightarrow{OA_{k}}) remains the same after the rotation because the polygon maps onto itself. Therefore, [ overrightarrow{OA_1} + overrightarrow{OA_2} + cdots + overrightarrow{OA_n} ] must be invariant under the rotation.5. Invariance Condition: Since (vec{m}) remains the same and rotational transformation around the (n)-gon center alters all individual vectors while maintaining the symmetry, (vec{m}) itself must be a vector that is invariant to such rotation.6. Conclusion: The only vector that remains unchanged upon any non-zero angle rotation is the zero vector. Thus, (vec{m}) must be the zero vector. Hence, [ vec{m} = overrightarrow{0}. ]Therefore, the sum of the vectors (overrightarrow{OA_1}, overrightarrow{OA_2}, ldots, overrightarrow{OA_n}) is [boxed{0}.]

🧠:Alright, so I need to find all possible values of tan α where both tan α and tan 3α are integers. Hmm, okay. Let's start by recalling the trigonometric identity for tan 3α. I think it's something like tan 3α = (3 tan α - tan³ α) / (1 - 3 tan² α). Let me verify that. Yeah, the triple angle formula for tangent is tan(3α) = [3 tan α - tan³ α] / [1 - 3 tan² α]. So that's correct.So, given that both tan α and tan 3α are integers. Let me denote t = tan α. Then tan 3α becomes (3t - t³) / (1 - 3t²). Since tan 3α must be an integer, let's call that integer k. So we have (3t - t³) / (1 - 3t²) = k, where t and k are both integers.So the equation is (3t - t³) = k(1 - 3t²). Let's rearrange that. Bring everything to one side:3t - t³ - k + 3k t² = 0.Let me write that as -t³ + 3k t² + 3t - k = 0.Alternatively, multiplying both sides by -1:t³ - 3k t² - 3t + k = 0.Hmm, not sure if that helps yet. Maybe factor this equation. Let's see. Let's group terms:(t³ - 3k t²) + (-3t + k) = 0.Factor t² from the first group: t²(t - 3k) - (3t - k) = 0.Not obvious. Maybe try to factor as a cubic in t. Let's consider possible integer roots. The possible rational roots of the polynomial t³ - 3k t² - 3t + k are factors of the constant term divided by factors of the leading coefficient. The constant term is k, and leading coefficient is 1. So possible integer roots are divisors of k. But k is an integer, but since k is expressed in terms of t, maybe we can substitute possible t values?Alternatively, perhaps solve for k in terms of t. From the equation:k = (3t - t³) / (1 - 3t²). Since k must be an integer, (3t - t³) must be divisible by (1 - 3t²). Let me compute the numerator and denominator:Numerator: -t³ + 3tDenominator: -3t² + 1So, (-t³ + 3t) / (-3t² + 1) = k.Let me factor numerator and denominator:Numerator: -t(t² - 3)Denominator: -3t² + 1 = -(3t² - 1)So the expression becomes [ -t(t² - 3) ] / [ - (3t² - 1) ] = [ t(t² - 3) ] / (3t² - 1 ) = k.So, t(t² - 3) divided by (3t² - 1) must be an integer. Let's denote t as an integer. So for integer t, compute this fraction and check if it's integer.So, we need (t³ - 3t) / (3t² - 1) is integer. Let me denote this as k = (t³ - 3t) / (3t² - 1). So, for integer t, k must be integer. Let's write this as:k = (t³ - 3t)/(3t² - 1). Let's perform polynomial division to see if we can express this as a polynomial plus a remainder over the denominator. Let me divide t³ - 3t by 3t² - 1.Divide t³ by 3t², which gives (1/3)t. Multiply the divisor 3t² -1 by (1/3)t: (1/3)t * 3t² = t³; (1/3)t*(-1) = - (1/3)t. Subtract this from the dividend:(t³ - 3t) - (t³ - (1/3)t) = (0)t³ + (-3t + (1/3)t) = (-8/3)t.So the division gives (1/3)t + (-8/3)t / (3t² -1). Therefore,k = (1/3)t + [ (-8/3)t / (3t² - 1) ]Since k must be an integer, the fractional part [ (-8/3)t / (3t² -1) ] must also be an integer minus (1/3)t. Hmm, not sure if that's helpful.Alternatively, since k must be integer, the denominator 3t² -1 must divide the numerator t³ - 3t. So, 3t² -1 | t³ - 3t.Which means, there exists an integer m such that t³ - 3t = m(3t² -1). Let's write that as:t³ - 3t = 3m t² - m.Bring all terms to left side:t³ - 3m t² - 3t + m = 0.This is similar to the equation we had before. So, for integers t and m. Let's try small integer values of t and see if k is integer.Let me try t = 0: then tan α = 0, then tan 3α = 0. So k = 0. That works. So t = 0 is possible.t = 1: compute numerator: 1 - 3 = -2; denominator: 3 -1 = 2; so k = -2 / 2 = -1. So k is integer. So t =1 is possible.t = -1: numerator: -1 + 3 = 2; denominator: 3 -1 = 2; so k = 2 / 2 = 1. Integer. So t = -1 is possible.t = 2: numerator: 8 - 6 = 2; denominator: 12 -1 =11; 2 /11 is not integer. So t=2 is invalid.t = -2: numerator: -8 +6 = -2; denominator:12 -1 =11; -2/11 invalid.t=3: numerator:27 -9=18; denominator:27 -1=26; 18/26=9/13 not integer.t=-3: numerator: -27 +9= -18; denominator:27 -1=26; -18/26=-9/13 not integer.t=4: numerator:64 -12=52; denominator:48 -1=47; 52/47 ≈1.1, not integer.t=-4: numerator: -64 +12= -52; denominator:48 -1=47; -52/47≈-1.1, not integer.t=1/2? But t must be integer. The problem states that tan α is integer. So t must be integer. So only integer t's.Wait, the problem says "tg α and tg 3α are integers". So tan α is an integer, so t is integer, and tan 3α must also be integer. So yes, t must be integer, so trying integer t's is the way to go.So so far, t=0,1,-1 work. Let's check t=0. Then tan α=0, so α is a multiple of π, then tan 3α is 0, which is integer. Okay.t=1: tan α=1, which is π/4 +nπ, then tan 3α = tan(3π/4 +3nπ)= -1, which is integer. Correct.t=-1: tan α=-1, which is -π/4 +nπ, then tan 3α = tan(-3π/4 +3nπ)=1, integer. Correct.Now t=1/2 is not allowed since t must be integer. How about t=2? As above, didn't work. Let's check t=1/√3? No, t must be integer. So t must be integer.Wait, maybe there are other integer t's. Let's check larger t. For example, t=5: numerator:125 -15=110; denominator:75 -1=74; 110/74≈1.486, not integer.t= -5: numerator: -125 +15= -110; denominator:75 -1=74; -110/74≈-1.486, not integer.t=6: numerator:216 -18=198; denominator:108 -1=107; 198/107≈1.85, not integer.t= -6: numerator: -216 +18= -198; denominator:108 -1=107; -198/107≈-1.85, not integer.Hmm. Maybe t=0,1,-1 are the only solutions? Wait, let's check t=2 again.Wait t=2: numerator:8 -6=2; denominator:12 -1=11; 2/11 is not integer, so k would be 2/11, which is not integer. So invalid.t=3: 18/26=9/13, not integer. t=4:52/47, nope.How about t=1/3? But tan α must be integer, so t must be integer. So fractions are out.Wait, maybe there's a smarter way than brute force. Let's see.We have 3t² -1 divides t³ - 3t. So 3t² -1 | t³ - 3t.In number theory terms, if d divides a and d divides b, then d divides any linear combination. So 3t² -1 divides t³ - 3t. Let's compute t³ -3t - m*(3t² -1) =0 for some integer m. Wait, but that's the same equation as before.Alternatively, we can express t³ -3t = q*(3t² -1) + r, where q is the quotient and r is the remainder. Since 3t² -1 is quadratic, the remainder r would be linear or constant. Let's perform polynomial division.Divide t³ -3t by 3t² -1.First term: t³ / 3t² = (1/3)t. Multiply divisor by (1/3)t: (1/3)t*(3t² -1) = t³ - (1/3)t.Subtract this from dividend: (t³ -3t) - (t³ - (1/3)t) = -3t + (1/3)t = (-8/3)t.So remainder is (-8/3)t. Therefore, t³ -3t = (1/3)t*(3t² -1) - (8/3)t.Therefore, (t³ -3t)/(3t² -1) = (1/3)t - (8/3)t / (3t² -1).For this to be integer, the term (8/3)t / (3t² -1) must be equal to (1/3)t - k, which complicates things. Alternatively, since the entire expression must be integer, the remainder (-8/3)t must be divisible by 3t² -1. But the remainder is (-8/3)t, so (-8/3)t must be a multiple of 3t² -1. But since 3t² -1 is an integer (because t is integer), and (-8/3)t must be integer. Therefore, 3 divides 8t. Since 3 and 8 are coprime, 3 divides t. Let me check that.If (-8/3)t is an integer multiple of (3t² -1), but perhaps that's too vague. Alternatively, for the entire expression (t³ -3t)/(3t² -1) to be integer, the remainder (-8/3)t must be zero? But unless t=0, which is a solution. But when t=0, numerator is 0, denominator is -1, so k=0. Which is valid. For t≠0, (-8/3)t must be an integer multiple of (3t² -1). Hmm, maybe this approach is not straightforward.Alternatively, let's write the equation as:(t³ -3t) = k*(3t² -1)Which is:t³ -3t = 3k t² -kThen rearranged:t³ -3k t² -3t +k =0We can think of this as a cubic equation in t with integer coefficients. For each integer k, we can attempt to find integer roots t. However, since both t and k are integers, perhaps we can bound the possible values of t.Let me consider that for large |t|, the term t³ dominates, so the left side t³ -3t is roughly t³, and the right side is 3k t². So for the equation t³ ≈ 3k t², which suggests that t ≈ 3k. So if |t| is large, k is approximately t/3. But k must be integer, so t must be a multiple of 3. Let's test t=3: then k=(27 -9)/(27 -1)=18/26=9/13, not integer. t= -3: k=(-27 +9)/(27 -1)= -18/26=-9/13, not integer. So even if t is a multiple of 3, it doesn't necessarily lead to integer k. Maybe this line of thought isn't helpful.Alternatively, let's consider the equation t³ -3k t² -3t +k =0. Let's factor this equation. Maybe factor by grouping.Group terms as (t³ -3k t²) + (-3t +k) =0Factor t² from the first group: t²(t - 3k) - (3t -k) =0Not helpful. Maybe try to factor as (t - a)(t² + bt + c) =0.Suppose it factors as (t - a)(t² + bt + c) = t³ + (b -a)t² + (c -ab)t -ac =0Compare coefficients:1. Coefficient of t³:1, matches.2. Coefficient of t²: b -a = -3k3. Coefficient of t: c -ab = -34. Constant term: -ac =kSo we have:From constant term: -ac =k → ac = -kFrom coefficient of t: c -ab = -3From coefficient of t²: b -a = -3kThis system might help us find integer solutions. Let's see if we can find integers a,b,c,k such that these equations hold.Assume a is an integer root of the cubic equation. Then, for some integer a, we have:a³ -3k a² -3a +k =0.But this seems recursive. Maybe try possible small integer roots a.From the equation t³ -3k t² -3t +k =0, possible integer roots a must satisfy a | k (from the constant term: when t=a, then a³ -3k a² -3a +k =0 → k( -3a² +1 ) +a³ -3a =0 → k= (a³ -3a)/(3a² -1). So similar to before. Therefore, possible integer roots a must satisfy that (a³ -3a)/(3a² -1) is integer, which brings us back to the original problem. Hence, this approach might not yield new information.Alternatively, perhaps use the fact that both t and k are integers. Let's rearrange the equation:t³ -3k t² -3t +k =0Let me solve for k:k( -3t² +1 ) = -t³ +3tHence,k = (t³ -3t)/(3t² -1)So k is expressed in terms of t. For k to be integer, (t³ -3t) must be divisible by (3t² -1). Let me denote D = 3t² -1. Then D divides t³ -3t.Thus, D | t³ -3t. Let's compute t³ modulo D. Since D = 3t² -1, we have 3t² ≡1 mod D. Therefore, t³ = t*(t²) = t*( (1 + D)/3 ) mod D. Wait, since 3t² ≡1 mod D, then t² ≡ (1)/3 mod D. But working modulo D, fractions are tricky. Let's see:From 3t² ≡1 mod D, multiply both sides by the inverse of 3 modulo D. But D =3t² -1, so 3t² = D +1. Therefore, 3t² ≡1 mod D. Hence, t² ≡ (1)/3 mod D. Hmm, not sure if helpful.Alternatively, express t³ in terms of D:t³ = t*(t²) = t*( (D +1)/3 ). So t³ = (t(D +1))/3. Therefore, t³ ≡ (t(D +1))/3 mod D. But since D divides t³ -3t, we have t³ -3t ≡0 mod D.Therefore:(t(D +1)/3 ) -3t ≡0 mod D.Multiply through by 3 to eliminate denominator:t(D +1) -9t ≡0 mod 3D.Simplify:tD + t -9t ≡0 mod3D → tD -8t ≡0 mod3D → t(D -8) ≡0 mod3D.But D =3t² -1, so:t(3t² -1 -8) ≡0 mod3(3t² -1)Simplify inside:3t² -9 = 3(t² -3). So,t*3(t² -3) ≡0 mod3(3t² -1)Cancel 3:t(t² -3) ≡0 mod (3t² -1)Thus, (3t² -1) divides t(t² -3). But we already knew that since we had k = [t(t² -3)] / (3t² -1). So this gives the same condition. Hmm, not helpful.Alternatively, let's note that since D divides t³ -3t, and D =3t² -1, then gcd(D, t) divides the constant term of D, which is -1. Because D =3t² -1. If d divides D and t, then d divides 3t² -1 and t, so d divides 1. Therefore, gcd(D, t)=1. Therefore, since D divides t³ -3t and gcd(D, t)=1, D divides t² -3. Wait, is that correct?Yes, if D divides t(t² -3) and gcd(D, t)=1, then D divides t² -3. So, 3t² -1 divides t² -3.So, 3t² -1 divides t² -3. Let me write this as:3t² -1 | t² -3Which implies that 3t² -1 divides (t² -3) - 3*(3t² -1). Let me compute that:(t² -3) -3*(3t² -1) = t² -3 -9t² +3 = (-8t²)Therefore, 3t² -1 divides -8t². But since 3t² -1 and t² are coprime? Wait, gcd(3t² -1, t²). Let d = gcd(3t² -1, t²). Then d divides t² and 3t² -1. So d divides 3t² -1 -3*t² = -1. Therefore, gcd(3t² -1, t²)=1. Therefore, 3t² -1 divides -8.Therefore, 3t² -1 divides -8, which implies that 3t² -1 is a divisor of 8. Since 3t² -1 can be positive or negative, the possible values of 3t² -1 are the integer divisors of 8, i.e., ±1, ±2, ±4, ±8.Therefore, 3t² -1 ∈ {±1, ±2, ±4, ±8}So let's solve for each possible case:Case 1: 3t² -1 =1 → 3t² =2 → t²=2/3. Not integer t. Disregard.Case 2: 3t² -1 = -1 → 3t² =0 → t²=0 → t=0. Which we already found.Case3: 3t² -1=2 →3t²=3→t²=1→t=±1. Which we have.Case4:3t² -1=-2→3t²=-1→t²=-1/3. Disregard.Case5:3t² -1=4→3t²=5→t²=5/3. Disregard.Case6:3t² -1=-4→3t²=-3→t²=-1. Disregard.Case7:3t² -1=8→3t²=9→t²=3→t=±√3. Not integer. Disregard.Case8:3t² -1=-8→3t²=-7→t²=-7/3. Disregard.So the only valid integer solutions for t are t=0, t=1, t=-1. Therefore, the possible values of tan α are 0, 1, -1.Let me confirm these. For t=0: tan 3α =0, which is integer. For t=1: tan 3α=(3*1 -1)/(1 -3*1)= (3 -1)/(1 -3)=2/-2=-1, integer. For t=-1: tan3α=(3*(-1)-(-1)^3)/(1 -3*(-1)^2)= (-3 +1)/(1 -3)=(-2)/(-2)=1, integer. Correct.Therefore, the only possible integer values of tan α are 0, 1, -1.Final AnswerThe possible values of operatorname{tg} alpha are boxed{-1}, boxed{0}, and boxed{1}.

✅:Given that (operatorname{tg} alpha) and (operatorname{tg} 3alpha) are integers, we need to find all possible values of (operatorname{tg} alpha).We start with the trigonometric identity for (operatorname{tg} 3alpha):[operatorname{tg} 3alpha = frac{3 operatorname{tg} alpha - (operatorname{tg} alpha)^3}{1 - 3 (operatorname{tg} alpha)^2}]Let (t = operatorname{tg} alpha). Then, (operatorname{tg} 3alpha) becomes:[operatorname{tg} 3alpha = frac{3t - t^3}{1 - 3t^2}]Given that both (t) and (frac{3t - t^3}{1 - 3t^2}) must be integers, we denote this integer by (k). Thus, we have:[frac{3t - t^3}{1 - 3t^2} = k]Rearranging this equation, we get:[3t - t^3 = k(1 - 3t^2)][t^3 - 3kt^2 - 3t + k = 0]We need to find integer solutions (t) such that (frac{3t - t^3}{1 - 3t^2}) is also an integer. By analyzing the polynomial division and using number theory, we find that (3t^2 - 1) must divide (t^3 - 3t). This implies that (3t^2 - 1) must be a divisor of 8. Solving for (t) in the equation (3t^2 - 1 in {pm 1, pm 2, pm 4, pm 8}), we find the valid integer solutions for (t):- (3t^2 - 1 = -1 implies t = 0)- (3t^2 - 1 = 2 implies t = pm 1)Checking these values:- For (t = 0), (operatorname{tg} 3alpha = 0), which is an integer.- For (t = 1), (operatorname{tg} 3alpha = -1), which is an integer.- For (t = -1), (operatorname{tg} 3alpha = 1), which is an integer.Thus, the possible values of (operatorname{tg} alpha) are (boxed{-1}), (boxed{0}), and (boxed{1}).

🔑:1. Let operatorname{tg} alpha = n and operatorname{tg} 3alpha = k.2. Using the tangent triple-angle formula: [ operatorname{tg}(3alpha) = frac{3operatorname{tg} alpha - operatorname{tg}^3 alpha}{1 - 3 operatorname{tg}^2 alpha} ] Substituting operatorname{tg} alpha = n, we get: [ k = frac{3n - n^3}{1 - 3n^2} ]3. Multiplying both sides by the denominator to clear the fraction: [ k(1 - 3n^2) = 3n - n^3 ] Simplifying further: [ k - 3kn^2 = 3n - n^3 ] Rearranging the terms: [ k(1 - 3n^2) = n(3 - n^2) ] [ k (1 - 3n^2) = 3n - n^3 ]4. Assume |n| > 1.5. In this situation, the terms (1 - 3n^2) and n are coprime (mutually prime). Hence, k = pn. Substituting k into the equation: [ pn (1 - 3n^2) = 3n - n^3 ] Dividing through by n (assuming n neq 0): [ p (1 - 3n^2) = 3 - n^2 ] Rearranging: [ (3p - 1)n^2 = p - 3 ]6. Since (3p - 1) is also a positive term, we get: [ (3p - 1) leq frac{p - 3}{n^2} quad text{for} quad n^2 > 1 ] Considering n^2 > 1 and solving the inequality: [ 11p leq 1 implies p leq frac{1}{11} ] This presents a contradiction since p must be positive and greater or equal to 1 if dealing with integers.7. Therefore, the assumption |n| > 1 leads to a contradiction.8. We then consider n = -1, 0, or 1. - For n = -1: [ operatorname{tg}(-frac{pi}{4}) = -1 ] [ operatorname{tg}(3 cdot -frac{pi}{4}) = operatorname{tg}(-frac{3pi}{4}) = 1 ] Confirmed as a solution since 1 is an integer. - For n = 0: [ operatorname{tg}(0) = 0 ] [ operatorname{tg}(3 cdot 0) = operatorname{tg}(0) = 0 ] Confirmed as a solution since 0 is an integer. - For n = 1: [ operatorname{tg}(frac{pi}{4}) = 1 ] [ operatorname{tg}(3 cdot frac{pi}{4}) = operatorname{tg}(frac{3pi}{4}) = -1 ] Confirmed as a solution since -1 is an integer. Conclusion:The possible values of operatorname{tg} alpha are:[boxed{-1, 0, 1}]