🧠:[<think>Okay, so I need to find the diameter of the circumscribed sphere (circumsphere) of a tetrahedron formed by folding a triangle with side lengths 10, 12, and 14 along its three midlines. Hmm, let me break this down step by step. First, I should visualize what's happening here. When we fold a triangle along its midlines, we're essentially bringing the midpoints of each side together to form a tetrahedron. The midlines are the lines connecting the midpoints of the sides, right? So each midline is parallel to the opposite side and half its length.Let me recall that when you connect the midpoints of a triangle, you create four smaller triangles, each similar to the original one and each with sides half the length. But in this case, instead of just creating smaller triangles, we're folding along those midlines to form a three-dimensional shape, which is a tetrahedron. So the original triangle is divided into four smaller triangles by the midlines, and folding each of those smaller triangles up along the midlines until the midpoints meet. That should form a tetrahedron where the original triangle's vertices become the vertices of the tetrahedron, but wait, actually, no. Wait, when folding along the midlines, the midpoints become the new vertices of the tetrahedron? Or maybe the original vertices remain as part of the tetrahedron?Wait, let me think. If I have a triangle ABC with midlines connecting midpoints of the sides, say D, E, F where D is the midpoint of AB, E is midpoint of BC, and F is midpoint of AC. Then the midlines are DE, EF, FD. Folding along DE, EF, FD would bring points D, E, F together? Hmm, maybe. So when folding along the midlines, the original triangle is divided into four smaller triangles, each of which is a face of the tetrahedron? Wait, no. A tetrahedron has four triangular faces, but if we start with one triangle and fold along three midlines, how does that create a tetrahedron?Alternatively, maybe the tetrahedron is formed by connecting the midpoints and the original vertices. Wait, perhaps when you fold along the midlines, the original triangle's vertices fold in towards each other, and the midpoints become the vertices of the tetrahedron. Hmm. Let me try to visualize this. If you have triangle ABC, and you fold along the midline DE (connecting midpoints of AB and BC), then point A would fold over to meet point C? Or maybe not. Wait, folding along DE would make the part of the triangle on one side of DE fold over onto the other side. Similarly for the other midlines. If we fold along all three midlines, perhaps the three midpoints D, E, F come together at a single point, forming a tetrahedron with vertices at D, E, F, and the original centroid? Wait, maybe. Alternatively, perhaps the tetrahedron has vertices at the original triangle's midpoints and the original vertices. Hmm, this is confusing.Alternatively, maybe the tetrahedron is formed by considering the original triangle and the three smaller triangles created by the midlines. When folded, each of the three smaller triangles becomes a face of the tetrahedron, and the fourth face is the central smaller triangle. Wait, but if you fold along the midlines, the three outer smaller triangles would each fold upwards, meeting at a common vertex. So in this case, the original triangle's midpoints (D, E, F) would form the base, and the folded-up vertices would meet at a point above, forming the apex of the tetrahedron. So the tetrahedron would have vertices at D, E, F, and the apex point. Wait, but how is the apex point determined? Maybe it's the centroid of the original triangle?Alternatively, perhaps when you fold along all three midlines, each of the three corners of the original triangle gets folded towards the center, and their vertices meet at a point. So the tetrahedron would have vertices at the midpoints D, E, F and the original vertices A, B, C? But that would create more than four vertices. Wait, no, because folding along the midlines would identify certain points. For example, folding along DE would bring point A to point C, or something like that. Maybe each original vertex is folded to the midpoint of the opposite side? Hmm, this is getting a bit unclear.Perhaps it's better to approach this problem mathematically. Let me recall that when you fold a triangle along its midlines to form a tetrahedron, the resulting tetrahedron is called a medial tetrahedron. The midlines divide the original triangle into four smaller triangles, each similar to the original. Folding along the midlines brings the midpoints together, forming a tetrahedron where each edge of the tetrahedron is half the length of the corresponding edge in the original triangle. Wait, but the original triangle has sides 10, 12, 14. If we connect midpoints, the edges of the smaller triangles would be half of those, so 5, 6, 7. But the tetrahedron formed by folding would have edges that are those lengths? Maybe not directly.Wait, perhaps the tetrahedron has edges equal to the lengths of the midlines. Wait, the midlines in a triangle are equal to half the length of the opposite side. So in triangle ABC with sides a=10, b=12, c=14, the midlines opposite each side would be half of the respective side. So midline opposite side a=10 would be 5, opposite b=12 would be 6, and opposite c=14 would be 7. But when folding along these midlines, how does that translate into the edges of the tetrahedron?Alternatively, maybe the edges of the tetrahedron are the midlines themselves. Wait, if you fold the triangle along its midlines, the edges of the resulting tetrahedron would be the midlines. But the midlines have lengths 5, 6, 7. However, the original triangle's sides are 10, 12, 14, so the midlines (which connect midpoints) are each half the length of the third side. So midline connecting midpoints of AB and AC is half of BC, which is 14/2=7, midline connecting midpoints of AB and BC is half of AC=12/2=6, and midline connecting midpoints of BC and AC is half of AB=10/2=5. So the three midlines are 5, 6, 7.But when folding the triangle along these midlines, the tetrahedron is formed. So each midline becomes an edge of the tetrahedron. But how many edges does a tetrahedron have? A tetrahedron has 6 edges. So perhaps the three midlines become three edges, and the other three edges are formed by the folding process. Wait, maybe each folded edge brings together two midpoints, so each midpoint is connected to two others via the original midlines, and then the folding creates the other three edges. This is getting a bit complicated.Alternatively, maybe the tetrahedron formed by folding along the midlines has its edges as the midlines and the segments from the midpoints to the original vertices. Wait, but folding would identify those original vertices. Hmm.Wait, perhaps a better approach is to model this mathematically. Let's assign coordinates to the original triangle, then compute the coordinates of the tetrahedron after folding, and then compute the circumradius.Let me place the original triangle in a coordinate system to make calculations easier. Let's choose triangle ABC with coordinates such that:- Let’s place vertex A at (0, 0, 0)- Vertex B at (14, 0, 0), since the side opposite to A is BC with length 14. Wait, but the sides are 10, 12, 14. Wait, actually, we need to assign the sides correctly. Let me check: in a triangle with sides 10, 12, 14, the sides opposite to vertices A, B, C would correspond to lengths a, b, c. Wait, actually, let's properly define the triangle.Let me denote the triangle with sides 10, 12, 14. Let me first confirm which side is opposite which angle. Wait, let's denote the sides as follows: suppose the triangle is ABC with BC = 10, AC = 12, AB = 14. Wait, but this might complicate the calculations. Alternatively, perhaps it's easier to use the standard notation where a, b, c are the lengths opposite to angles A, B, C respectively. So if we take side a as BC, side b as AC, and side c as AB. Then the sides would be:- a = length of BC = 10- b = length of AC = 12- c = length of AB = 14But then we can place the triangle in coordinate system with vertex A at (0, 0, 0), vertex B at (c, 0, 0) = (14, 0, 0), and vertex C somewhere in the plane. Then, coordinates of C can be determined using the law of cosines. Let me compute coordinates for point C.Given triangle ABC with AB=14, AC=12, BC=10. Coordinates of A(0,0,0), B(14,0,0). Let’s find coordinates of C(x,y,0). The distance from A to C is 12, so x² + y² = 12² = 144. The distance from B to C is 10, so (x -14)² + y² = 10² = 100. Subtracting the first equation from the second:(x -14)² + y² - x² - y² = 100 - 144Expanding (x -14)²: x² -28x +196So x² -28x +196 + y² -x² - y² = -44Simplify: -28x +196 = -44Then -28x = -44 -196 = -240So x = (-240)/(-28) = 240/28 = 60/7 ≈ 8.571Then y² = 144 - x² = 144 - (60/7)² = 144 - 3600/49 = (144*49 - 3600)/49 = (7056 - 3600)/49 = 3456/49Thus y = sqrt(3456/49) = (sqrt(3456))/7. sqrt(3456): 3456 = 64*54 = 64*9*6 = 8^2*3^2*6, so sqrt(3456) = 8*3*sqrt(6) = 24√6. Therefore y = 24√6 /7 ≈ 24*2.449/7 ≈ 58.776/7 ≈ 8.397So coordinates of C are (60/7, 24√6/7, 0). So the original triangle is placed with A(0,0,0), B(14,0,0), C(60/7, 24√6/7, 0).Now, the midpoints of the sides:Midpoint D of AB: ((0 +14)/2, (0 +0)/2, 0) = (7, 0, 0)Midpoint E of BC: Coordinates of B(14,0,0), C(60/7, 24√6/7, 0). Midpoint E: ((14 +60/7)/2, (0 +24√6/7)/2, 0). Compute 14 as 98/7, so 98/7 +60/7 = 158/7, divided by 2: 79/7. The y-coordinate: (24√6/7)/2 = 12√6/7. So E is (79/7, 12√6/7, 0).Midpoint F of AC: Coordinates of A(0,0,0) and C(60/7, 24√6/7,0). Midpoint F: ((0 +60/7)/2, (0 +24√6/7)/2,0) = (30/7, 12√6/7,0).So midpoints D(7,0,0), E(79/7,12√6/7,0), F(30/7,12√6/7,0).Now, the midlines are DE, EF, FD. Let me compute these vectors.But before that, when we fold the triangle along the midlines DE, EF, FD, we have to figure out the coordinates of the new points in 3D space. Folding along a line would mean reflecting part of the triangle over that line into the third dimension. Since we are folding along all three midlines, the resulting figure is a tetrahedron. Each fold along a midline will create a dihedral angle between the original triangle and the folded part. Since all three midlines are being folded, the three flaps (smaller triangles) will meet at a common vertex above the original plane, forming the tetrahedron.Thus, the tetrahedron will have four vertices: the three midpoints D, E, F and the apex point where the three flaps meet. Wait, but when folding, the original midpoints D, E, F are part of the original triangle. If we fold the triangle along DE, EF, and FD, each fold will create a new vertex in 3D space. Alternatively, maybe the apex is the point where the three original vertices (A, B, C) are brought together? Hmm.Wait, perhaps when folding along the midlines, each original vertex (A, B, C) is folded towards the opposite midline. For example, folding along DE would bring vertex A to the position of vertex C, but given the midline DE is between B and C. Wait, maybe not. Alternatively, each flap (the three smaller triangles adjacent to the original vertices) is folded over the midline, which would bring the original vertices to a common point above the original triangle.Yes, that makes sense. So the original triangle is divided into four smaller triangles by the midlines: the central triangle DEF and three outer triangles adjacent to the original vertices A, B, C. Folding along DE, EF, FD would fold each of the outer triangles over the central triangle DEF, bringing vertices A, B, C together at a point above DEF. Thus, the tetrahedron has vertices D, E, F, and the apex point O which is the folded position of A, B, C. So all three original vertices A, B, C are identified with point O in the folded tetrahedron.Therefore, the tetrahedron is formed by connecting the midpoints D, E, F with the apex O. So the edges of the tetrahedron are DO, EO, FO, as well as DE, EF, FD. Wait, but DE, EF, FD are edges in the base triangle DEF. So the tetrahedron has base triangle DEF and apex O connected to each of D, E, F.But then, to find the coordinates of O, we need to determine where A, B, C are mapped after folding. Since folding along the midlines DE, EF, FD, each original vertex is reflected over the corresponding midline to meet at O. Therefore, O is the reflection of A over DE, the reflection of B over EF, and the reflection of C over FD. Since all three reflections must coincide at the same point O, that should give the apex.Therefore, to compute the coordinates of O, we can compute the reflection of, say, point A over the midline DE. Alternatively, since after folding, the distance from O to each of D, E, F should be equal to the distance from the original vertex to the midpoint. Wait, when you fold a point over a line, the distance from the original point to the line is equal to the distance from the reflected point to the line. So the reflection of A over DE is a point O such that AO is perpendicular to DE and twice the distance from A to DE.But perhaps an easier way is to note that in the folded tetrahedron, the edges DO, EO, FO correspond to the segments from the midpoints to the original vertices, but lifted into 3D space. Since folding brings A, B, C to O, the lengths of DO, EO, FO should be equal to the lengths from the original vertices to the midpoints. Wait, in the original triangle, the distance from A to D (midpoint of AB) is half of AB, which is 7. Wait, AB is 14, so midpoint D is at 7 units from A. Similarly, distance from B to E (midpoint of BC) is half of BC, which is 5. Wait, BC is 10? Wait, in our notation earlier, we had sides BC=10, AC=12, AB=14. Therefore, midpoint of BC is E, so BE=5. Similarly, midpoint of AC is F, so AF=6.Wait, but in the folded tetrahedron, the edges from O to D, E, F would correspond to these lengths. So OD = distance from A to D in original triangle, which is 7, OE = distance from B to E, which is 5, and OF = distance from C to F, which is 6 (since CF is midpoint of AC, which is 12, so 6). However, in the tetrahedron, these edges OD, OE, OF would be in 3D space, and we need to ensure that the folding doesn't stretch or compress the edges. Therefore, the lengths OD=7, OE=5, OF=6.But wait, in reality, when folding the triangle into 3D, the distances from O to D, E, F must equal the original distances from A, B, C to D, E, F respectively. Therefore, yes, OD=AD=7, OE=BE=5, OF=CF=6. So we can model the tetrahedron with vertices D, E, F, O where OD=7, OE=5, OF=6. Additionally, the edges DE, EF, FD are known from the original triangle's midlines.Wait, but DE, EF, FD can be computed from the coordinates we had earlier. Let's compute DE, EF, FD.First, coordinates:- D is (7, 0, 0)- E is (79/7, 12√6/7, 0)- F is (30/7, 12√6/7, 0)So DE is the distance between D(7,0,0) and E(79/7,12√6/7,0). Let's compute that:First, convert 7 to 49/7. So D is (49/7, 0, 0). E is (79/7, 12√6/7, 0).Difference in x-coordinates: 79/7 - 49/7 = 30/7Difference in y-coordinates: 12√6/7 - 0 = 12√6/7So DE distance: sqrt[(30/7)^2 + (12√6/7)^2] = (1/7)sqrt(30^2 + (12√6)^2)Compute 30^2 = 900(12√6)^2 = 144*6 = 864So total inside sqrt: 900 + 864 = 1764sqrt(1764) = 42Therefore, DE = 42/7 = 6. So DE = 6.Similarly, EF is the distance between E(79/7,12√6/7,0) and F(30/7,12√6/7,0). The y-coordinates are the same, so distance is just the difference in x-coordinates.EF = |79/7 - 30/7| = 49/7 = 7.Wait, that's 49/7 = 7? Wait, 79 -30 =49? 79-30=49? Wait, 79-30 is 49? Wait, 79-30=49? 79-30=49? No, 79-30=49? 79-30=49 is incorrect. Wait, 79-30=49? Wait, 79-30=49. 30+49=79. Yes, 30+49=79. So EF = 49/7 =7.Wait, but 79/7 -30/7 = (79 -30)/7 =49/7=7. So EF=7.Then FD is the distance between F(30/7,12√6/7,0) and D(7,0,0). Convert 7 to 49/7. So FD's x-coordinate difference: 49/7 -30/7 =19/7. Y-coordinate difference: 0 -12√6/7 = -12√6/7.Distance FD: sqrt[(19/7)^2 + (-12√6/7)^2] = (1/7)sqrt(19^2 + (12√6)^2)Compute 19^2=361(12√6)^2= 864Total inside sqrt: 361 +864=1225sqrt(1225)=35Therefore FD=35/7=5.So DE=6, EF=7, FD=5. So the base triangle DEF has sides 5,6,7. Interesting, the same as the original triangle's midlines. Now, the edges from O to D, E, F are 7,5,6 respectively.Therefore, the tetrahedron has vertices D(7,0,0), E(79/7,12√6/7,0), F(30/7,12√6/7,0), and O(x,y,z). The edges:- OD=7: distance from O to D(7,0,0) is 7- OE=5: distance from O to E(79/7,12√6/7,0) is 5- OF=6: distance from O to F(30/7,12√6/7,0) is 6We need to find coordinates of O(x,y,z) such that:sqrt[(x -7)^2 + y^2 + z^2] =7sqrt[(x -79/7)^2 + (y -12√6/7)^2 + z^2] =5sqrt[(x -30/7)^2 + (y -12√6/7)^2 + z^2] =6So we have three equations:1. (x -7)^2 + y^2 + z^2 =492. (x -79/7)^2 + (y -12√6/7)^2 + z^2 =253. (x -30/7)^2 + (y -12√6/7)^2 + z^2 =36Subtract equation 1 from equation 2:[(x -79/7)^2 - (x -7)^2] + [(y -12√6/7)^2 - y^2] =25 -49= -24Similarly, expand each term:First term: (x -79/7)^2 - (x -7)^2= [x² - (158/7)x + (79/7)^2] - [x² -14x +49]= -158x/7 + (6241/49) +14x -49= (-158x/7 +14x) + (6241/49 -49)Convert 14x to 98x/7:= (-158x/7 +98x/7) + (6241/49 -2401/49)= (-60x/7) + (3840/49)Second term: (y -12√6/7)^2 - y^2= y² - (24√6/7)y + (144*6)/49 - y²= - (24√6/7)y + 864/49So combining both terms:-60x/7 +3840/49 -24√6 y /7 +864/49 = -24Combine constants:3840 +864 =4704; 4704/49 =96So:-60x/7 -24√6 y /7 +96 = -24Multiply both sides by7:-60x -24√6 y +672 = -168Bring constants to right:-60x -24√6 y = -168 -672 = -840Divide both sides by -12:5x +2√6 y =70So equation A: 5x +2√6 y =70Similarly, subtract equation1 from equation3:[(x -30/7)^2 - (x -7)^2] +[(y -12√6/7)^2 - y^2] =36 -49= -13Expand:First term:(x -30/7)^2 - (x -7)^2= [x² - (60/7)x +900/49] -[x² -14x +49]= -60x/7 +900/49 +14x -49= (-60x/7 +98x/7) + (900/49 -2401/49)=38x/7 -1501/49Second term:(y -12√6/7)^2 - y^2 = same as before: -24√6 y /7 +864/49So total:38x/7 -1501/49 -24√6 y /7 +864/49 = -13Combine constants:-1501 +864 = -637; -637/49= -13So:38x/7 -24√6 y /7 -13 = -13Thus:38x/7 -24√6 y /7 =0Multiply both sides by7:38x -24√6 y=0So equation B:38x -24√6 y=0Now we have two equations:A:5x +2√6 y =70B:38x -24√6 y=0Let me solve these equations for x and y.From equation B:38x =24√6 y => x= (24√6 /38) y= (12√6 /19) ySubstitute into equation A:5*(12√6 /19)y +2√6 y =70Compute:(60√6 /19)y + (38√6 /19)y =70(60√6 +38√6)/19 y=70(98√6)/19 y=70Thus y=70*(19)/(98√6)= (70/98)*(19/√6)= (5/7)*(19/√6)= (95)/(7√6)Rationalizing denominator:95√6/(7*6)=95√6/42So y=95√6/42Then x= (12√6 /19)* y= (12√6 /19)*(95√6 /42)Compute:12*95=1140√6*√6=619*42=798So x= (1140*6)/798= 6840/798Simplify:Divide numerator and denominator by 6: 6840/6=1140;798/6=133So 1140/133. Let's divide 1140 by133:133*8=1064, 1140-1064=76. So 8 and 76/133. Simplify 76/133: divide numerator and denominator by 19: 76/19=4, 133/19=7. So 4/7. Thus x=8 +4/7=60/7Wait, 60/7 is approximately 8.571, which matches the x-coordinate of point C in the original triangle. Interesting. So x=60/7, y=95√6/42.So now we have x=60/7, y=95√6/42. Now we can find z from equation1:(x -7)^2 + y^2 + z^2=49Compute x-7=60/7 -49/7=11/7So (11/7)^2 + (95√6/42)^2 + z^2=49Compute each term:(121/49) + (9025*6)/(42²) + z²=49First, 9025*6=5415042²=1764Thus 54150/1764 = divide numerator and denominator by 6: 9025/294Simplify 9025/294: 294*30=8820, 9025-8820=205, so 30 +205/294. 205 and294 share a common factor? 205=5*41, 294=2*3*7². No common factors. So 205/294.Therefore, the second term is 30 +205/294 ≈30.697But let's keep it as fractions:Total left-hand side:121/49 +54150/1764 +z²=49Convert 121/49 to (121*36)/1764=4356/176454150/1764 remains.So total: (4356 +54150)/1764 +z²=49Sum 4356 +54150=58506Thus 58506/1764 +z²=49Simplify 58506/1764: divide numerator and denominator by 6: 9751/294. Let's check:58506 ÷6=9751; 1764 ÷6=294. 9751/294. Let's divide 9751 by294:294*33=9702, 9751-9702=49. So 33 +49/294=33 +7/42=33 +1/6=33.1666...So 58506/1764=33 +1/6=199/6? Wait, 33.1666 is 199/6. Because 199/6≈33.1666. 199/6=33.1666...Wait, 199 divided by6 is 33*6=198, remainder1, so 33 1/6. So 199/6. Thus 58506/1764=199/6.Therefore, 199/6 +z²=49Therefore, z²=49 -199/6= (294/6 -199/6)=95/6Thus z=±√(95/6)=±√(570)/6≈±sqrt(570)/6≈±23.87/6≈±3.978. Since the tetrahedron is formed by folding upwards, we can take z positive. So z=√(95/6)=√570/6.Therefore, coordinates of O are (60/7, 95√6/42, √570/6).Now, we have all four vertices of the tetrahedron:D(7,0,0), E(79/7,12√6/7,0), F(30/7,12√6/7,0), O(60/7,95√6/42,√570/6).Now, we need to find the circumradius of this tetrahedron. The circumradius of a tetrahedron is the radius of the sphere passing through all four vertices. The diameter would be twice that radius.To find the circumradius, we can use the formula involving the coordinates of the vertices. The circumradius can be found by solving the system of equations based on the sphere equation:(x - a)^2 + (y - b)^2 + (z - c)^2 = R^2where (a,b,c) is the center of the sphere, and R is the radius. We need to plug in the coordinates of all four vertices into this equation and solve for a, b, c, R. Then, the diameter is 2R.Alternatively, there is a formula for the circumradius of a tetrahedron in terms of its edge lengths and volume. The formula is:R = frac{abc}{4V}Wait, no, that's for a triangle. For a tetrahedron, the formula is:R = frac{|vec{AB} cdot (vec{AC} times vec{AD})|}{6V}But perhaps it's better to use coordinates. Let me set up the equations.Let the sphere have center (a,b,c) and radius R. Then:For vertex D(7,0,0):(7 -a)^2 + (0 -b)^2 + (0 -c)^2 = R^2 ...(1)For vertex E(79/7,12√6/7,0):(79/7 -a)^2 + (12√6/7 -b)^2 + (0 -c)^2 = R^2 ...(2)For vertex F(30/7,12√6/7,0):(30/7 -a)^2 + (12√6/7 -b)^2 + (0 -c)^2 = R^2 ...(3)For vertex O(60/7,95√6/42,√570/6):(60/7 -a)^2 + (95√6/42 -b)^2 + (√570/6 -c)^2 = R^2 ...(4)We have four equations with four variables a,b,c,R. Let's subtract equation (1) from equations (2), (3), (4) to eliminate R².Subtract (1) from (2):[(79/7 -a)^2 - (7 -a)^2] + [(12√6/7 -b)^2 - (0 -b)^2] + [0 -0] =0Compute each term:First term: (79/7 -a)^2 - (7 -a)^2= [ (79/7 -a) - (7 -a) ] [ (79/7 -a) + (7 -a) ]= (79/7 -a -7 +a)(79/7 -a +7 -a)= (79/7 -7)(79/7 +7 -2a)Convert 7 to 49/7:= (79/7 -49/7)(79/7 +49/7 -2a)= (30/7)(128/7 -2a)= (30/7)(128/7 -14a/7)= (30/7)( (128 -14a)/7 )= 30*(128 -14a)/49Second term: (12√6/7 -b)^2 - (-b)^2= [ (12√6/7 -b)^2 - b^2 ]= (12√6/7)^2 - 2*(12√6/7)*b + b^2 - b^2= (144*6)/49 - (24√6/7)b= 864/49 - (24√6/7)bSo total equation (2)-(1):30*(128 -14a)/49 +864/49 - (24√6/7)b =0Multiply through by49 to eliminate denominators:30*(128 -14a) +864 -24√6*7*b =0Compute:30*128=384030*(-14a)= -420aSo 3840 -420a +864 -168√6 b=0Combine constants:3840 +864=4704Thus: -420a -168√6 b +4704=0Divide equation by -42: 10a +4√6 b -112=0So equation (2)-(1):10a +4√6 b =112 ...(A)Similarly, subtract (1) from (3):[(30/7 -a)^2 - (7 -a)^2] + [(12√6/7 -b)^2 - (0 -b)^2] +0=0First term:(30/7 -a)^2 - (7 -a)^2= [30/7 -a -7 +a][30/7 -a +7 -a]= (30/7 -49/7)(30/7 +7 -2a)= (-19/7)( (30/7 +49/7) -2a )= (-19/7)(79/7 -2a )= (-19/7)(79 -14a)/7= -19*(79 -14a)/49Second term:Same as before: (12√6/7 -b)^2 -b^2=864/49 -24√6/7 bSo equation (3)-(1):-19*(79 -14a)/49 +864/49 -24√6/7 b=0Multiply by49:-19*(79 -14a) +864 -24√6*7 b=0Compute:-19*79= -1501-19*(-14a)=266aThus: -1501 +266a +864 -168√6 b=0Combine constants: -1501 +864= -637Thus:266a -168√6 b -637=0Divide by 7:38a -24√6 b -91=0 ...(B)Similarly, subtract (1) from (4):[(60/7 -a)^2 - (7 -a)^2] + [(95√6/42 -b)^2 - (0 -b)^2] + [(√570/6 -c)^2 - (0 -c)^2] =0This one is more complex. Let's compute each term.First term: (60/7 -a)^2 - (7 -a)^2= [60/7 -a -7 +a][60/7 -a +7 -a]= (60/7 -49/7)(60/7 +7 -2a)= (11/7)( (60/7 +49/7) -2a )= (11/7)(109/7 -2a )= 11*(109 -14a)/49Second term: (95√6/42 -b)^2 - (-b)^2= (95√6/42)^2 - 2*(95√6/42)*b +b^2 -b^2= (9025*6)/42² - (190√6/42)b= 54150/1764 - (190√6/42)bSimplify 54150/1764: divide numerator and denominator by 6: 9025/294 ≈30.697But let's keep it as is for now.Third term: (√570/6 -c)^2 - (-c)^2= (570/36) - (2√570/6)c +c² -c²= 570/36 - (√570/3)cSo putting it all together:[11*(109 -14a)/49] + [54150/1764 - (190√6/42)b] + [570/36 - (√570/3)c ]=0Multiply all terms by 1764 (LCM of 49, 1764, 42, 36):1764=49*36=42²=36*49. So:First term:11*(109 -14a)/49 *1764=11*(109 -14a)*36Second term:54150/1764 *1764=54150Third term:-190√6/42*b*1764= -190√6*b*(1764/42)= -190√6*b*42= -7980√6 bFourth term:570/36 *1764=570*(1764/36)=570*49=27930Fifth term:-√570/3*c*1764= -√570*c*(1764/3)= -√570*c*588= -588√570 cSo equation:11*36*(109 -14a) +54150 -7980√6 b +27930 -588√570 c=0Compute 11*36=396396*(109 -14a) +54150 -7980√6 b +27930 -588√570 c=0Compute constants:54150 +27930=82080Thus:396*(109 -14a) +82080 -7980√6 b -588√570 c=0Compute 396*109=396*(100 +9)=39600 +3564=43164396*(-14a)= -5544aThus:43164 -5544a +82080 -7980√6 b -588√570 c=0Combine constants:43164 +82080=125244So:-5544a -7980√6 b -588√570 c +125244=0Divide entire equation by -84:5544/84=66; 7980/84=95; 588/84=7; 125244/84=1491Thus:66a +95√6 b +7√570 c -1491=0 ...(C)Now, we have three equations:A:10a +4√6 b =112B:38a -24√6 b =91C:66a +95√6 b +7√570 c =1491Let me solve equations A and B first for a and b.From equation A:10a +4√6 b=112Equation B:38a -24√6 b=91Let me multiply equation A by6 to make the coefficients of b opposites:60a +24√6 b=672Equation B:38a -24√6 b=91Add both equations:60a +38a +24√6 b -24√6 b=672 +9198a=763Thus a=763/98=109/14 (since 763 divided by7=109, 98=14*7)Thus a=109/14≈7.7857Then from equation A:10*(109/14) +4√6 b=112Compute 10*(109/14)=1090/14=545/7≈77.857So 545/7 +4√6 b=1124√6 b=112 -545/7= (784/7 -545/7)=239/7Thus b= (239/7)/(4√6)=239/(28√6)=239√6/(28*6)=239√6/168≈239*2.449/168≈586.111/168≈3.489So b=239√6/168Now, substitute a=109/14 and b=239√6/168 into equation C to find c.Equation C:66a +95√6 b +7√570 c =1491Compute each term:66a=66*(109/14)= (66*109)/14. 66*100=6600, 66*9=594, so total=6600+594=7194. 7194/14=3597/7≈513.85795√6 b=95√6*(239√6/168)=95*239*(√6*√6)/168=95*239*6/168Compute numerator:95*239= Let's compute 95*200=19000, 95*39=3705, total=19000+3705=22705Then 22705*6=136230Denominator:168Thus 136230/168= divide numerator and denominator by 6:22705/28≈810.892So 95√6 b≈810.892Therefore, equation C:3597/7 +810.892 +7√570 c=1491Convert 3597/7≈513.857So total so far:513.857 +810.892≈1324.749Thus 1324.749 +7√570 c=1491So 7√570 c=1491 -1324.749≈166.251Thus c≈166.251/(7√570)Compute denominator:7√570≈7*23.874≈167.118Thus c≈166.251/167.118≈0.995But let's compute exactly.First, 3597/7 +95√6 b= 3597/7 +95√6*(239√6/168)=3597/7 + (95*239*6)/168=3597/7 + (95*239*6)/(168)Calculate 95*239=22705 as before.22705*6=136230136230/168= divide numerator and denominator by 6:22705/28≈810.892857So total:3597/7 +810.8928573597/7=513.85714285513.85714285 +810.892857≈1324.75So 7√570 c=1491 -1324.75=166.25Thus c=166.25/(7√570)= (166.25/7)/√570≈23.75/√570≈23.75/23.874≈0.994So c≈0.994. But let's keep it exact.166.25=665/4, so:c= (665/4)/(7√570)=665/(28√570)Rationalize the denominator:665/(28√570) *√570/√570=665√570/(28*570)=665√570/15960Simplify 665/15960: divide numerator and denominator by5:133/3192. 133 and3192: 3192÷133=24. So 133=7*19, 3192=133*24=7*19*24. Therefore, 133/3192=1/24. So c=√570/24Wait, wait: 665/15960=665/(665*24)=1/24. Because 665*24=15960. Yes. So 665/15960=1/24. Thus c=√570/24Therefore, c=√570/24Hence, center of the circumscribed sphere is at (109/14,239√6/168,√570/24)Now, to find the radius R, we can substitute back into equation (1):(7 -a)^2 + (0 -b)^2 + (0 -c)^2 = R^2Compute each term:7 -a=7 -109/14=98/14 -109/14= -11/14-b= -239√6/168-c= -√570/24So:(-11/14)^2 + (-239√6/168)^2 + (-√570/24)^2=R^2Calculate each term:1. (-11/14)^2=121/196≈0.6172. (-239√6/168)^2= (239² *6)/168²= (57121*6)/28224=342726/28224≈12.1433. (-√570/24)^2=570/576=95/96≈0.990Sum:121/196 +342726/28224 +95/96Convert all to denominator 28224:121/196= (121*144)/28224=17424/28224342726/28224 remains95/96= (95*294)/28224=27930/28224Total=17424 +342726 +27930=17424+342726=360150+27930=388080Thus R²=388080/28224Simplify:Divide numerator and denominator by 48:388080/48=8085;28224/48=588So 8085/588. Divide numerator and denominator by21:8085/21=385;588/21=28Thus R²=385/28=13.75Wait, 385 divided by28=13.75. Because 28*13=364, 385-364=21, 21/28=0.75. So yes, R²=13.75=55/4.Thus R=√(55/4)=√55/2Therefore, the diameter is 2R=√55.Wait, this seems off. Wait, R squared is 55/4, so R is sqrt(55)/2, so diameter is sqrt(55). But let me verify the calculations.Wait, let's check the computation of R²:First term:121/196Second term:239²*6/(168²)= (57121*6)/(28224)=342726/28224≈12.143Third term:570/576=95/96≈0.990Adding these:121/196≈0.617 +12.143≈12.76 +0.990≈13.75. So R²=13.75=55/4, so R=√(55)/2≈3.708. So diameter is 2*R=√55≈7.416.But the problem states that the answer should be a whole number or a simple fraction. Hmm. Maybe there is an error in the calculation.Wait, let me verify the steps:1. Found coordinates of original triangle correctly.2. Midpoints calculated correctly.3. Folded apex O found by solving the reflection over midlines, leading to coordinates (60/7,95√6/42,√570/6). Is this correct?4. Then set up equations for the circumsphere center (a,b,c) by equating the distances from the center to each vertex.Solved equations A and B to find a and b, then equation C to find c.Then calculated R squared as 55/4. Hmm. But 55/4 is 13.75. However, sqrt(55) is approximately 7.416, which seems reasonable.But let me verify if the process was correct.Alternatively, perhaps there's a smarter way to compute the circumradius.Given a tetrahedron with four vertices, the formula for the circumradius is |OP| where O is the centroid, but no, it's not necessarily the centroid.Alternatively, there's a formula using determinants or the scalar triple product.The formula for the circumradius R of a tetrahedron with vertices at points A, B, C, D is given by:R = frac{|vec{AB} cdot (vec{AC} times vec{AD})|}{6V}where V is the volume of the tetrahedron.Alternatively, using coordinates, we can compute the volume via the scalar triple product.But in our case, the tetrahedron has vertices D, E, F, O. Let's compute the volume.First, compute vectors DE, DF, DO.Wait, vectors from D to E, D to F, D to O.Coordinates:D(7,0,0)E(79/7,12√6/7,0)= (approx11.2857, approx4.695,0)F(30/7,12√6/7,0)= (approx4.2857, approx4.695,0)O(60/7,95√6/42,√570/6)= (approx8.571, approx5.389, approx3.978)Compute vectors:DE = E - D = (79/7 -7,12√6/7 -0,0 -0)= (79/7 -49/7,12√6/7,0)=30/7,12√6/7,0DF = F - D = (30/7 -7,12√6/7 -0,0 -0)= (30/7 -49/7,12√6/7,0)= (-19/7,12√6/7,0)DO = O - D = (60/7 -7,95√6/42 -0,√570/6 -0)= (60/7 -49/7,95√6/42,√570/6)= (11/7,95√6/42,√570/6)Now, the volume V is (1/6)| scalar triple product of DE, DF, DO |Compute the scalar triple product DE ⋅ (DF × DO)First, compute DF × DO:DF = (-19/7,12√6/7,0)DO = (11/7,95√6/42,√570/6)Compute cross product DF × DO:i component: (12√6/7 * √570/6 - 0 *95√6/42)= (12√6 * √570)/(7*6)= (12√(6*570))/42=12√3420/42Simplify √3420:3420=36*95=36*5*19. So √3420=6√(5*19)=6√95Thus i component:12*6√95 /42=72√95/42=12√95/7j component: - [ (-19/7 * √570/6 -0 *11/7 ) ]= - [ -19√570/(7*6) ]=19√570/(42)k component: (-19/7 *95√6/42 -12√6/7 *11/7)= (-19*95√6)/(7*42) - (132√6)/49Compute first term:19*95=1805, so -1805√6/(294)Second term: -132√6/49= -132√6/49Convert to common denominator 294:-132√6/49= -132*6√6/294= -792√6/294Thus k component= -1805√6/294 -792√6/294= (-1805 -792)√6/294= -2597√6/294= -2597/294*√6Simplify 2597/294: divide numerator and denominator by7:2597/7=371, 294/7=42. So -371/42*√6Therefore, DF × DO = (12√95/7,19√570/42, -371√6/42)Now, compute DE ⋅ (DF × DO )DE = (30/7,12√6/7,0)Dot product:30/7 *12√95/7 +12√6/7 *19√570/42 +0*(-371√6/42)First term: (30*12√95)/49=360√95/49Second term: (12√6 *19√570)/(7*42)= (228√(6*570))/294=228√3420/294As before, √3420=6√95, so:228*6√95/294=1368√95/294=228√95/49Thus total dot product=360√95/49 +228√95/49=588√95/49=12√95Therefore, scalar triple product=12√95Volume V=(1/6)|12√95|=2√95Thus V=2√95Then, using the formula R = frac{|vec{AB} cdot (vec{AC} times vec{AD})|}{6V} = |12√95|/(6*2√95)=12√95/(12√95)=1. Wait, this gives R=1, which contradicts the previous result. Hmm, something is wrong here.Wait, no. Wait, the formula is R = | (AB ⋅ (AC × AD)) | / (6V). But AB, AC, AD are vectors from the same vertex. However, in our case, we computed DE ⋅ (DF × DO ), which is the scalar triple product for vectors from D. So using those vectors, the scalar triple product is 6V. Wait, the volume is (1/6)| scalar triple product |, so V=(1/6)|12√95|=2√95. Then, the formula for R is | scalar triple product | / (6V )? Wait, no. Let me check.Actually, the formula is:R = frac{|vec{AB} cdot (vec{AC} times vec{AD})|}{6V}But in our case, AB, AC, AD are DE, DF, DO. The scalar triple product is DE ⋅ (DF × DO )=12√95, which is 6V. Therefore, R= (12√95)/(6*2√95)=12√95/(12√95)=1. But this contradicts the previous calculation where R=√55/2≈3.708. There must be an error.Alternatively, perhaps I mixed up the scalar triple product. Let me double-check.The scalar triple product of three vectors a, b, c is a ⋅ (b × c), and the volume is (1/6)|scalar triple product| when the vectors are from the same vertex.In our case, vectors DE, DF, DO are from vertex D. Therefore, Volume V=(1/6)| DE ⋅ (DF × DO ) | = (1/6)|12√95|=2√95. Correct.Then, the formula for the circumradius R is given by:R = frac{|vec{AB} cdot (vec{AC} times vec{AD})|}{6V} = frac{12√95}{6*2√95} = 12√95 / (12√95) =1. But this can't be, since we have a tetrahedron with edge lengths more than 1. This suggests an error in calculation.Alternatively, perhaps the formula is different. Wait, no, the formula for the circumradius of a tetrahedron is:R = frac{abc}{4V} for a triangle, but for a tetrahedron, there isn't a direct formula like that. The correct formula is:R = frac{|vec{a} cdot (vec{b} times vec{c})|}{6V}But wait, this is the same as the volume formula. Wait, no, I think there is confusion here. The scalar triple product already includes the volume. The actual formula for the circumradius involves the distances between the vertices and solving the system, which we did earlier, leading to R=√55/2.But why does the scalar triple product approach give R=1? That must be due to miscalculating the scalar triple product.Wait, let's recompute the scalar triple product DE ⋅ (DF × DO ).First, DF × DO:DF = (-19/7, 12√6/7, 0)DO = (11/7,95√6/42,√570/6)Cross product components:i: (12√6/7 * √570/6 - 0 *95√6/42) = (12√6 * √570)/(7*6) = (12√(6*570))/42Compute 6*570=3420√3420=√(4*855)=2√855=2√(9*95)=6√95Therefore, i component:12*6√95 /42=72√95/42=12√95/7j component: - [ (-19/7 * √570/6 -0 *11/7 ) ]= - [ -19√570/(7*6) ]=19√570/(42)k component: (-19/7 *95√6/42 -12√6/7 *11/7)First term: -19*95√6/(7*42)= -1805√6/294Second term: -12*11√6/(7*7)= -132√6/49= -792√6/294Thus k component: (-1805√6 -792√6)/294= -2597√6/294= -2597/294 *√6Now, DE ⋅ (DF × DO ):DE = (30/7,12√6/7,0)Dot product:30/7 *12√95/7 +12√6/7 *19√570/42 +0*(-2597√6/294)First term: (360√95)/49Second term: (228√6 *√570)/(7*42)=228√(6*570)/2946*570=3420, √3420=6√95Thus second term:228*6√95/294=1368√95/294=228√95/49Total dot product:360√95/49 +228√95/49=588√95/49=12√95Hence scalar triple product=12√95, Volume=(1/6)*12√95=2√95. Correct.Then, formula for R is:R = frac{|vec{DE} cdot (vec{DF} times vec{DO})|}{6V} = |12√95| / (6*2√95) =12√95 / 12√95=1. This suggests R=1, but this conflicts with the coordinate calculation where R=√55/2≈3.708. There is a contradiction here.Therefore, there must be a mistake in one of the calculations. Let's trace back.First, the coordinates method gave center at (109/14,239√6/168,√570/24) and R=√55/2.The scalar triple product method gives R=1. This inconsistency indicates an error in one of the methods.Let me check the scalar triple product result. If R=1, then the distance from the center to vertex D should be 1. But according to coordinates:Center (109/14,239√6/168,√570/24)Distance from center to D(7,0,0):sqrt[(7 -109/14)^2 + (0 -239√6/168)^2 + (0 -√570/24)^2]= sqrt[(98/14 -109/14)^2 + ( -239√6/168)^2 + (-√570/24)^2]= sqrt[(-11/14)^2 + (239√6/168)^2 + (√570/24)^2]= sqrt[121/196 + (239^2 *6)/(168^2) +570/(24^2)]Calculate:239^2=57121, 57121*6=342726168^2=28224342726/28224≈12.143570/(24^2)=570/576≈0.9896121/196≈0.617Total≈0.617+12.143+0.9896≈13.749, sqrt of that≈3.708≈√55/2≈3.708. Hence, R=√55/2≈3.708, not 1. Therefore, the scalar triple product method must have been applied incorrectly.Wait, the formula R = frac{|vec{AB} cdot (vec{AC} times vec{AD})|}{6V} is not the correct formula for the circumradius. That formula is actually the volume formula. My mistake. The correct formula for the circumradius requires knowing the edge lengths and using the formula:R = frac{abc}{4V} for a triangle, but for a tetrahedron, it's more complex. The general formula involves the determinant of a matrix constructed from the coordinates.Alternatively, using the coordinates found earlier, we can calculate R as sqrt(55)/2, which is approximately3.708. So the diameter would be sqrt(55). However, the problem asks for the diameter of the circumscribed sphere, which is 2R= sqrt(55). But sqrt(55) is an irrational number, approximately7.416. But the problem might expect an exact answer in boxed radical form.But let me verify if the calculation of the center coordinates was correct.Earlier, solving the system gave center (109/14,239√6/168,√570/24). Then, substituting into equation (1):(7 -109/14)^2 + (0 -239√6/168)^2 + (0 -√570/24)^2= (-11/14)^2 + ( -239√6/168)^2 + (-√570/24)^2=121/196 + (239^2 *6)/168^2 +570/24^2=121/196 +342726/28224 +570/576=121/196 +342726/28224 +95/96. Converting to a common denominator:121/196=17424/28224342726/28224=342726/2822495/96=27930/28224Sum=17424+342726+27930=388080/28224=388080÷28224=13.75=55/4. So yes, R=√(55/4)=√55/2. So diameter is 2*(√55/2)=√55.Therefore, the diameter of the circumscribed sphere is √55. However, the problem might expect a simplified radical form. √55 is already simplified, so the answer is√55.But let me check once more for any miscalculations. The key steps were:1. Placing the original triangle in coordinate system and finding midpoints.2. Finding coordinates of apex O by solving the system of equations based on distances from O to D, E, F.3. Solving for the center of the circumsphere and verifying the radius.All steps led to R=√55/2, diameter√55. Therefore, the answer should be√55.Final Answerboxed{sqrt{55}}</think>To find the diameter of the circumscribed sphere of the tetrahedron formed by folding a triangle with side lengths 10, 12, and 14 along its midlines, we proceed as follows:1. Coordinates of the Original Triangle: - Place vertex ( A ) at ( (0, 0, 0) ), vertex ( B ) at ( (14, 0, 0) ), and vertex ( C ) at ( left( frac{60}{7}, frac{24sqrt{6}}{7}, 0 right) ).2. Midpoints of the Sides: - Midpoint ( D ) of ( AB ): ( (7, 0, 0) ) - Midpoint ( E ) of ( BC ): ( left( frac{79}{7}, frac{12sqrt{6}}{7}, 0 right) ) - Midpoint ( F ) of ( AC ): ( left( frac{30}{7}, frac{12sqrt{6}}{7}, 0 right) )3. Coordinates of the Apex ( O ): - Solving the system of equations for the distances from ( O ) to ( D ), ( E ), and ( F ), we find ( O ) at ( left( frac{60}{7}, frac{95sqrt{6}}{42}, frac{sqrt{570}}{6} right) ).4. Circumradius Calculation: - Using the coordinates of the vertices ( D ), ( E ), ( F ), and ( O ), we set up the equations for the circumscribed sphere and solve for the center ( (a, b, c) ) and radius ( R ). - The radius ( R ) is found to be ( frac{sqrt{55}}{2} ).5. Diameter of the Circumscribed Sphere: - The diameter is twice the radius, resulting in ( sqrt{55} ).Thus, the diameter of the circumscribed sphere of the tetrahedron is (boxed{sqrt{55}}).,<think>Okay, so I need to find the diameter of the circumscribed sphere (circumsphere) of a tetrahedron formed by folding a triangle with sides 10, 12, and 14 along its three midlines. Hmm, let me break this down step by step. First, I should visualize what's happening here. Folding a triangle along its midlines... Midlines are the lines connecting the midpoints of the sides, right? So, if I have a triangle, and I connect the midpoints, that divides the original triangle into four smaller triangles, each similar to the original one. Then folding along those midlines would bring the corners of the original triangle together to form a tetrahedron. Let me confirm: when you fold along the midlines, each of the three midlines becomes an edge where the triangle is folded, and the midpoints become the vertices of the tetrahedron. Wait, but the original triangle has three vertices. If we fold along the midlines, the midpoints (there are three of them) and maybe the original vertices? Hmm, no, the original vertices would meet at a point when folded. So actually, the tetrahedron's vertices would be the midpoints of the original triangle's sides. Because when you fold the triangle along each midline, the original vertices come together, and the midpoints become the new vertices of the tetrahedron. So the tetrahedron has four vertices: the three midpoints of the original triangle and the centroid where the original vertices meet?Wait, no. Wait, when you fold a triangle along its midlines, you create a three-dimensional shape. Each midline is a fold line, so each of the three smaller triangles (each adjacent to a vertex) gets folded up. The three midpoints of the original triangle are each connected by the midlines, and when folded, those midpoints become the vertices of the tetrahedron, along with the point where the original vertices meet. Wait, but a tetrahedron has four vertices. So yes, the three midpoints and the centroid? Hmm, maybe. Let me think.Alternatively, perhaps the tetrahedron is formed by connecting the three midpoints and the original vertices? But when you fold along the midlines, the original vertices are brought together. So perhaps the folded tetrahedron has four vertices: the three midpoints and the folded vertex (the original three vertices are now coinciding at one point). So the tetrahedron has vertices at the three midpoints and the common folded vertex. That makes sense. So each of the three original edges is divided into two segments of length 5, 6, and 7 (since midlines split the sides into halves). So each midline is half the length of the original side? Wait, no. The midline in a triangle is a segment connecting the midpoints of two sides, and its length is half the length of the third side. Wait, yes, in a triangle, the midline is parallel to the third side and half its length. So for a triangle with sides 10, 12, 14, the midlines would be 5, 6, and 7. Wait, is that right?Wait, let's recall: in a triangle, the midline connecting the midpoints of two sides is parallel to the third side and half its length. So if the original triangle has sides of length 10, 12, and 14, then the midlines would be 5, 6, and 7. For example, the midline connecting the midpoints of the sides of length 10 and 12 would be parallel to the side of length 14 and have length 7. Similarly, the midline connecting the midpoints of sides 10 and 14 would be parallel to side 12 and have length 6, and the midline connecting midpoints of sides 12 and 14 would be parallel to side 10 and have length 5. So the three midlines are 5, 6, and 7. So when we fold the original triangle along these midlines, the resulting tetrahedron would have edges equal to these midlines? Wait, but how exactly?Alternatively, perhaps when folding along the midlines, the edges of the tetrahedron correspond to the midlines. So the tetrahedron has edges of lengths 5, 6, and 7? But a tetrahedron has six edges, so maybe each of the midlines becomes an edge, and the other edges are the segments from the midpoints to the folded vertex. Hmm. This is getting a bit confusing. Maybe I need to approach this more methodically.First, let's consider the original triangle. Let's name the triangle ABC, with side lengths BC = 10, AC = 12, and AB = 14. Then, the midpoints of the sides would be: midpoint of BC is M, midpoint of AC is N, midpoint of AB is P. The midlines would be MN, NP, and PM. Then, folding along MN, NP, and PM would bring the corners A, B, C together to a single point, let's say O. So the tetrahedron formed would have vertices at M, N, P, and O. So the edges of the tetrahedron would be OM, ON, OP, and the midlines MN, NP, PM. Wait, but OM, ON, OP would each be half the length of the original triangle's medians. Wait, maybe not. Wait, when folding, the distance from O to each midpoint would be equal to half the length of the corresponding median? Hmm, perhaps.Alternatively, perhaps the edges OM, ON, OP are equal to the distances from the original vertices to the midpoints, which are half the lengths of the sides. Wait, no. The distance from a vertex to the midpoint of a side is half the length of that side, but when folded, that distance becomes an edge in the tetrahedron. Wait, but if we fold the triangle such that vertices A, B, C meet at point O, then OA, OB, OC would be the distances from O to each original vertex. But since we are folding along the midlines, maybe O is the centroid of the original triangle? Because the centroid is the intersection of the medians, which are the lines from the vertices to the midpoints of the opposite sides. But the midlines are different; they connect midpoints of sides.Wait, perhaps not. Let me think again. If we have triangle ABC, and we fold it along the midlines MN, NP, PM, then each fold would bring vertex A to the midpoint of BC, vertex B to the midpoint of AC, and vertex C to the midpoint of AB. Wait, no. Folding along a midline would reflect a part of the triangle over the midline. So folding along MN (the midline connecting midpoints of BC and AC) would reflect the part of the triangle beyond MN over MN. But I might be overcomplicating.Alternatively, when you fold a triangle along its three midlines, you are essentially creating a three-dimensional figure where each of the four smaller triangles (the medial triangle and the three corner triangles) are folded such that the corners meet at a common point. The resulting figure is a tetrahedron with the medial triangle as one face and the other three faces being the folded corner triangles. Therefore, the vertices of the tetrahedron are the midpoints of the original triangle and the common point where the original vertices meet.So, in this case, the tetrahedron has four vertices: M, N, P (midpoints of BC, AC, AB) and O (the folded point where A, B, C meet). Therefore, the edges of the tetrahedron are OM, ON, OP, MN, NP, and PM. The lengths of MN, NP, PM are the midlines, which are half the length of the opposite sides. As established earlier, those midlines are 5, 6, and 7. So MN = 5, NP = 6, PM = 7? Wait, no, wait. Let's clarify.Original triangle ABC with sides BC = 10, AC = 12, AB = 14. The midline MN connects midpoints of BC and AC. Since midline is parallel to AB and half its length. So AB is 14, so midline MN is 7. Similarly, midline NP connects midpoints of AC and AB, which is parallel to BC and half its length, so NP = 5. Midline PM connects midpoints of AB and BC, parallel to AC and half its length, so PM = 6. Therefore, the medial triangle MNP has sides MN = 7, NP = 5, PM = 6. So edges MN, NP, PM are 7, 5, 6. Then, the edges OM, ON, OP are the distances from the folded point O to the midpoints. Since O is the result of folding the original vertices together, the distances from O to each midpoint should be equal to half the length of the original edges. Wait, because when you fold a vertex to the midpoint, the distance from the folded point to the midpoint is half the original edge length. Hmm.Wait, let's imagine folding vertex A over the midline. The original vertex A is being folded over to meet vertices B and C at point O. The distance from O to midpoint P (midpoint of AB) would be half of AB, which is 7. Similarly, distance from O to midpoint N (midpoint of AC) is half of AC, which is 6, and distance from O to midpoint M (midpoint of BC) is half of BC, which is 5. Therefore, edges OM = 5, ON = 6, OP = 7. So the tetrahedron O-M-N-P has edges:- OM = 5, ON = 6, OP = 7- MN = 7, NP = 5, PM = 6Wait, but in a tetrahedron, each pair of vertices is connected by an edge. So we need to check all the edges. Let's list all six edges:1. OM: 52. ON: 63. OP: 74. MN: 75. NP: 56. PM: 6Wait, but this seems inconsistent. Because if you have points O, M, N, P, then the edges between O and each midpoint are 5, 6, 7, and the edges between midpoints are 7, 5, 6. So that's the complete edge list. So the tetrahedron has edges of lengths 5, 5, 6, 6, 7, 7. Wait, OM = 5, ON = 6, OP = 7, and MN = 7, NP = 5, PM = 6. So edges are indeed 5, 5, 6, 6, 7, 7.So, now, to find the diameter of the circumscribed sphere (circumsphere) of this tetrahedron. The diameter would be twice the circumradius of the tetrahedron. So I need to find the circumradius of this tetrahedron and then double it. To find the circumradius of a tetrahedron, I can use the formula:R = frac{abc}{6V}Wait, no, that's for a regular tetrahedron. For a general tetrahedron, the formula is:R = frac{|vec{a} cdot (vec{b} times vec{c})|}{6V}Wait, no, perhaps I need to recall the general formula for the circumradius of a tetrahedron. The formula involves the edge lengths and the volume. Let me recall.The formula for the circumradius of a tetrahedron with edge lengths a, b, c, d, e, f can be complex, but there's a formula using the determinant of a matrix based on coordinates. Alternatively, if I can assign coordinates to the vertices of the tetrahedron, then compute the circumradius from there, that might be easier.Given that we know all the edge lengths, setting up coordinate system might help. Let me try that.First, let's assign coordinates to the original triangle. Let me place the original triangle ABC such that it's convenient for calculation. Let's put vertex A at the origin (0, 0, 0), vertex B at (14, 0, 0), and vertex C somewhere in the plane. Then, we can compute the coordinates of C using the side lengths: AB = 14, AC = 12, BC = 10.Using coordinates: Let’s denote A(0,0,0), B(14,0,0). Let’s find coordinates of C(x,y,0). Then, AC = 12, so sqrt(x² + y²) = 12, and BC = 10, so sqrt((x -14)² + y²) = 10.So, we have:x² + y² = 144(x -14)^2 + y² = 100Subtract first equation from the second:(x -14)^2 + y² - x² - y² = 100 - 144Expand (x -14)^2: x² -28x +196 -x² = -44Simplify: -28x +196 = -44 → -28x = -240 → x = 240 /28 = 60/7 ≈ 8.571Then, from x = 60/7, substitute into x² + y² =144:(60/7)^2 + y² =1443600/49 + y² =144y² =144 - 3600/49 = (144*49 -3600)/49 = (7056 -3600)/49 = 3456/49Thus, y = sqrt(3456/49) = (sqrt(3456))/7 = (24*sqrt(6))/7 ≈ 24*2.449/7 ≈ 58.776/7 ≈ 8.396So coordinates of C are (60/7, (24√6)/7, 0)Now, midpoints:Midpoint M of BC: Coordinates of B(14,0,0) and C(60/7, (24√6)/7,0). So midpoint M is:x: (14 + 60/7)/2 = (98/7 +60/7)/2 = (158/7)/2 = 79/7 ≈11.286y: (0 + (24√6)/7)/2 = (24√6)/14 = (12√6)/7 ≈4.198z: 0Midpoint N of AC: Coordinates of A(0,0,0) and C(60/7, (24√6)/7,0). Midpoint N:x: (0 +60/7)/2 =30/7 ≈4.286y: (0 + (24√6)/7)/2 =12√6/7 ≈4.198z:0Midpoint P of AB: Coordinates of A(0,0,0) and B(14,0,0). Midpoint P is (7,0,0)Now, the folded point O is where the original vertices A, B, C meet after folding. Since folding along midlines MN, NP, PM brings A, B, C to O, we need to find the coordinates of O in 3D space. Wait, but folding the triangle into a tetrahedron would result in a three-dimensional figure, so the coordinates of O would not lie in the original plane. Therefore, we need to determine the coordinates of O such that the distances from O to M, N, P are equal to half the lengths of the original edges, which are 5, 6, 7. Wait, as previously thought, OM = 5, ON =6, OP=7?Wait, let's think. When folding the triangle along midlines, the distances from O to each midpoint should correspond to the original distances from the vertices to the midpoints. For example, the distance from A to midpoint P (of AB) is 7, which is half of AB=14. Similarly, distance from B to midpoint N (of AC) is 6, which is half of AC=12, and distance from C to midpoint M (of BC) is 5, which is half of BC=10. Therefore, when folding the triangle such that A, B, C meet at O, the distances from O to the midpoints M, N, P should be 5, 6, 7 respectively. Therefore, O is a point in 3D space such that:- Distance from O to M is 5- Distance from O to N is 6- Distance from O to P is7Additionally, the coordinates of M, N, P are known in the original plane (z=0). So we can model this as a system of equations to find coordinates of O (x, y, z).Given:M is at (79/7, 12√6/7, 0)N is at (30/7, 12√6/7, 0)P is at (7, 0, 0)So, let’s denote O as (a, b, c). Then:Distance from O to M: sqrt[(a -79/7)^2 + (b -12√6/7)^2 + c²] =5Distance from O to N: sqrt[(a -30/7)^2 + (b -12√6/7)^2 + c²] =6Distance from O to P: sqrt[(a -7)^2 + b² + c²] =7So we have three equations:1. (a -79/7)^2 + (b -12√6/7)^2 + c² =252. (a -30/7)^2 + (b -12√6/7)^2 + c² =363. (a -7)^2 + b² + c² =49Let me subtract equation 1 and equation 2 to eliminate some variables. Subtract equation 1 from equation 2:[(a -30/7)^2 - (a -79/7)^2] + [(b -12√6/7)^2 - (b -12√6/7)^2] + [c² - c²] =36 -25Simplify:First term: (a -30/7)^2 - (a -79/7)^2Let’s compute this difference. Let’s denote u = a -30/7 and v = a -79/7, so u - v = (a -30/7) - (a -79/7) = 49/7 =7Then, u² -v² = (u -v)(u +v) =7*(2a -109/7)Wait, let's compute:Expand (a -30/7)^2 - (a -79/7)^2:= [a² -60a/7 + (30/7)^2] - [a² -158a/7 + (79/7)^2]= (-60a/7 + 900/49) - (-158a/7 + 6241/49)= (-60a/7 + 900/49 +158a/7 -6241/49)= (98a/7) + (900 -6241)/49=14a + (-5341)/49So this is equal to 14a -5341/49.Therefore, the first term is 14a -5341/49.Second and third terms are zero, since the b and c terms cancel out.So 14a -5341/49 =11Therefore:14a =11 +5341/49Convert 11 to 539/49:14a =539/49 +5341/49 =5880/49=120So 14a =120 → a=120/14=60/7≈8.571So a=60/7.Now, substitute a=60/7 into equation 3:(60/7 -7)^2 + b² + c²=49Compute 60/7 -7=60/7 -49/7=11/7So (11/7)^2 +b² +c²=49121/49 +b² +c²=49Thus, b² +c²=49 -121/49= (2401 -121)/49=2280/49≈46.5306Now, substitute a=60/7 into equation 1:(a -79/7)^2 + (b -12√6/7)^2 + c²=25Compute a -79/7=60/7 -79/7= -19/7So (-19/7)^2 + (b -12√6/7)^2 +c²=25361/49 + (b -12√6/7)^2 +c²=25Thus, (b -12√6/7)^2 +c²=25 -361/49= (1225 -361)/49=864/49≈17.6327But we also have from above that b² +c²=2280/49So subtract the two equations:(b -12√6/7)^2 +c² - (b² +c²) =864/49 -2280/49Expand left side:(b² -24√6/7 b + (12√6/7)^2) +c² -b² -c²= -1416/49Simplify:-24√6/7 b + (144*6)/49 = -1416/49Compute (144*6)/49=864/49Thus:-24√6/7 b +864/49= -1416/49Bring 864/49 to the right:-24√6/7 b= -1416/49 -864/49= -2280/49Multiply both sides by -7/(24√6):b= (-2280/49)*(-7)/(24√6)= (2280*7)/(49*24√6)= (2280*7)/(49*24√6)Simplify:2280 divided by 24 is 95, and 7 divided by 49 is 1/7.Wait:2280 ÷24=95 (24*95=2280)7/49=1/7Thus:b= (95 *1)/(7√6)=95/(7√6)Rationalize denominator:95√6/(7*6)=95√6/42Therefore, b=95√6/42Then, from equation b² +c²=2280/49:Compute b²= (95√6/42)^2= (9025*6)/(1764)=54150/1764= Divide numerator and denominator by 6: 9025/294Wait, 54150 ÷6=9025, 1764 ÷6=294. So 9025/294.So c²=2280/49 -9025/294Convert 2280/49 to denominator 294: 2280*6/(49*6)=13680/294Thus, c²=13680/294 -9025/294=(13680 -9025)/294=4655/294Simplify 4655/294: divide numerator and denominator by 7: 4655 ÷7=665, 294 ÷7=42. So 665/42Thus, c=±√(665/42). Since the tetrahedron is formed by folding upwards, we can take c>0, so c=√(665/42). Simplify sqrt(665/42). Let's see if 665 and 42 have common factors. 665=5*133=5*7*19, 42=6*7. So sqrt((5*7*19)/(6*7))=sqrt((5*19)/6)=sqrt(95/6)=sqrt(95)/sqrt(6)=√(95)/√6= (√570)/6≈sqrt(570)/6≈23.87/6≈3.978. So approximately 3.98. Let me check:Wait, 665/42=15.8333...sqrt(15.8333)= approx 3.98. Yes.Therefore, coordinates of O are (60/7, 95√6/42, sqrt(665/42)).So now, we have coordinates for all four vertices of the tetrahedron:O: (60/7, 95√6/42, sqrt(665/42))M: (79/7, 12√6/7, 0)N: (30/7, 12√6/7, 0)P: (7, 0, 0)Now, to find the circumradius of tetrahedron O-M-N-P. The circumradius can be found by solving the equation of the sphere passing through all four points. The sphere equation is:(x - a)^2 + (y - b)^2 + (z - c)^2 = R^2We need to find a, b, c, R such that all four points satisfy the equation. Since this is a system of four equations, solving it will give the center (a,b,c) and radius R.Alternatively, use the formula for the circumradius in terms of coordinates. The formula is:R = frac{|vec{OM} cdot (vec{ON} times vec{OP})|}{6V}But perhaps it's easier to compute the coordinates of the circumcenter by solving the system.Let me denote the center of the sphere as (h, k, l). Then, the equations are:For point O: (60/7 - h)^2 + (95√6/42 - k)^2 + (sqrt(665/42) - l)^2 = R^2For point M: (79/7 - h)^2 + (12√6/7 - k)^2 + (0 - l)^2 = R^2For point N: (30/7 - h)^2 + (12√6/7 - k)^2 + (0 - l)^2 = R^2For point P: (7 - h)^2 + (0 - k)^2 + (0 - l)^2 = R^2This gives four equations. Subtract equations pairwise to eliminate R².First, subtract equation for P from equation for M:[(79/7 - h)^2 + (12√6/7 - k)^2 + l²] - [(7 - h)^2 + k² + l²] =0Expand:(79/7 - h)^2 - (7 - h)^2 + (12√6/7 - k)^2 -k² =0Compute each term:First term: (79/7 - h)^2 - (7 - h)^2Let’s compute 79/7 ≈11.2857, 7=49/7So (79/7 - h)^2 - (49/7 -h)^2 = [(79/7 -h) - (49/7 -h)][(79/7 -h) + (49/7 -h)] = (30/7)(128/7 -2h)Wait, expand:Let’s let A =79/7 -h, B=49/7 -h. Then A² -B² = (A -B)(A +B)A -B = (79/7 -h) - (49/7 -h) =30/7A +B = (79/7 -h) + (49/7 -h) =128/7 -2hThus, first term: 30/7*(128/7 -2h)Second term: (12√6/7 -k)^2 -k² = (12√6/7)^2 -24√6/7 k +k² -k²= (144*6)/49 -24√6/7 k=864/49 -24√6/7 kPutting it all together:30/7*(128/7 -2h) +864/49 -24√6/7 k=0Multiply out:30/7*128/7 -30/7*2h +864/49 -24√6/7 k=0Compute each term:30*128=3840, so 3840/4930/7*2h=60h/7864/49 remains as isThus:3840/49 -60h/7 +864/49 -24√6/7 k=0Combine 3840/49 +864/49=4704/49=96Thus:96 -60h/7 -24√6/7 k=0Multiply all terms by7:672 -60h -24√6 k=0Divide by 12:56 -5h -2√6 k=0 → 5h +2√6 k=56 ...(A)Similarly, subtract equation for P from equation for N:[(30/7 - h)^2 + (12√6/7 - k)^2 + l²] - [(7 - h)^2 + k² + l²]=0Expand:(30/7 -h)^2 - (7 -h)^2 + (12√6/7 -k)^2 -k²=0First term:30/7 -h - (7 -h)=30/7 -h -7 +h=30/7 -49/7= -19/7Sum terms:(30/7 -h +7 -h)= (30/7 +49/7 -2h)=79/7 -2hThus, first term: (-19/7)(79/7 -2h)Second term: same as before, (12√6/7 -k)^2 -k²=864/49 -24√6/7 kThus, equation:(-19/7)(79/7 -2h) +864/49 -24√6/7 k=0Compute:-19/7*(79/7 -2h)= -19*79/49 +38h/7=-1501/49 +38h/7Adding 864/49:-1501/49 +864/49 +38h/7= (-637/49)+38h/7= -13 +38h/7Thus:-13 +38h/7 -24√6/7 k=0Multiply all terms by7:-91 +38h -24√6 k=0 →38h -24√6 k=91 ...(B)Now, we have two equations:From (A):5h +2√6 k=56From (B):38h -24√6 k=91Let me solve these equations. Let's call equation (A) and equation (B).First, let's solve equation (A) for h:5h=56 -2√6 k → h=(56 -2√6 k)/5Substitute into equation (B):38*(56 -2√6 k)/5 -24√6 k=91Multiply both sides by5:38*(56 -2√6 k) -120√6 k=455Compute 38*56=212838*(-2√6 k)= -76√6 kThus:2128 -76√6 k -120√6 k=455Combine like terms:2128 -196√6 k=455Subtract2128:-196√6 k=455 -2128= -1673Thus:196√6 k=1673 →k=1673/(196√6)=1673/(196√6)Simplify:Divide numerator and denominator by7:1673 ÷7=239, 196 ÷7=28So k=239/(28√6)=239√6/(28*6)=239√6/168≈239*2.449/168≈585.311/168≈3.484So k=239√6/168Now, substitute k into equation (A):5h +2√6*(239√6/168)=56Compute 2√6*(239√6/168)=2*239*6/168= (2868)/168=2868 ÷12=239, 168 ÷12=14 →239/14Thus:5h +239/14=56 →5h=56 -239/14= (784 -239)/14=545/14 →h=545/(14*5)=109/14≈7.7857So h=109/14Therefore, the coordinates of the circumcenter are (h, k, l)=(109/14, 239√6/168, l)Now, we need to find l. Use equation from point P:(7 - h)^2 + (0 - k)^2 + (0 - l)^2=R²But we can use another equation to solve for l. Let's use equation from point N:(30/7 - h)^2 + (12√6/7 -k)^2 + (0 - l)^2=R²But since R² is the same, set equal to equation from P:(7 - h)^2 +k² +l²=(30/7 - h)^2 + (12√6/7 -k)^2 +l²Cancel l²:(7 - h)^2 +k²=(30/7 - h)^2 + (12√6/7 -k)^2Expand both sides:Left side: (7 -h)^2 +k²=49 -14h +h² +k²Right side: (30/7 -h)^2 + (12√6/7 -k)^2= (900/49 -60h/7 +h²) + (864/49 -24√6 k/7 +k²)=900/49 +864/49 -60h/7 -24√6 k/7 +h² +k²=1764/49 -60h/7 -24√6 k/7 +h² +k²=36 -60h/7 -24√6 k/7 +h² +k²Set equal to left side:49 -14h +h² +k²=36 -60h/7 -24√6 k/7 +h² +k²Cancel h² and k²:49 -14h=36 -60h/7 -24√6 k/7Bring all terms to left:49 -14h -36 +60h/7 +24√6 k/7=0Simplify:13 -14h +60h/7 +24√6 k/7=0Convert -14h to -98h/7:13 -98h/7 +60h/7 +24√6 k/7=0Combine h terms:(-38h/7) +13 +24√6 k/7=0Multiply all terms by7:-38h +91 +24√6 k=0But we already know h=109/14 and k=239√6/168. Substitute these values:-38*(109/14) +91 +24√6*(239√6/168)=0Compute each term:First term: -38*(109)/14= -4142/14= -2071/7≈-295.857Second term:91=91Third term:24√6*(239√6)/168=24*239*6/168= (24*239*6)/168Simplify:24/168=1/7, so 239*6/7= (239*6)/7=1434/7≈204.857Thus:-2071/7 +91 +1434/7= (-2071 +1434)/7 +91= (-637)/7 +91= -91 +91=0Which checks out. So the equation is satisfied, which means our values for h and k are correct.Now, we can use the equation from point P to find l:(7 - h)^2 +k² +l²=R²First, compute (7 - h):7 -109/14=98/14 -109/14= -11/14Thus, ( -11/14 )²=121/196Compute k²=(239√6/168)^2=(239² *6)/(168²)239²=57121, so 57121*6=342,726168²=28,224Thus, k²=342726/28224=57121/4704≈12.14Compute l²=R² -121/196 -57121/4704But we need another equation to find l. Let's use equation for point O:(60/7 - h)^2 + (95√6/42 -k)^2 + (sqrt(665/42) -l)^2=R²But since we already know h, k, and l is related to R via equation from point P, maybe substitute R² from point P into this.Alternatively, compute R² from point P:R²=(7 - h)^2 +k² +l²=121/196 +57121/4704 +l²Convert to common denominator 4704:121/196=121*24/4704=2904/470457121/4704 remains as isThus, R²=(2904 +57121)/4704 +l²=60025/4704 +l²≈12.76 +l²Now, use equation from point O:(60/7 - h)^2 + (95√6/42 -k)^2 + (sqrt(665/42) -l)^2=R²=60025/4704 +l²Compute left side:First term: (60/7 -109/14)^2= (120/14 -109/14)^2=(11/14)^2=121/196= same as (7 -h)^2Wait, 60/7 -109/14= (120 -109)/14=11/14, so square is 121/196= same as above.Second term: (95√6/42 -239√6/168)^2=√6*(95/42 -239/168)=√6*( (95*4 -239)/168 )Compute 95*4=380, 380 -239=141Thus, √6*(141/168)=√6*(47/56)Therefore, the term is (47√6/56)^2= (47² *6)/56²= (2209*6)/3136=13254/3136≈4.226Third term: (sqrt(665/42) -l)^2=665/42 -2l*sqrt(665/42)+l²Thus, left side total:121/196 +13254/3136 +665/42 -2l*sqrt(665/42)+l²Convert all terms to denominator 3136:121/196=121*16/3136=1936/313613254/3136 remains665/42=665*74.6667≈ wait, 3136/42=74.6667. Wait, 665/42= (665*74.6667)/3136? Wait, maybe better to compute:665/42= (665* 3136/42)/3136=665*(3136/42)/3136=665*74.6667/3136. Hmm, this is messy. Alternatively, compute numerical values.Alternatively, recognize that:121/196≈0.61713254/3136≈4.226665/42≈15.833So left side≈0.617 +4.226 +15.833 -2l*sqrt(665/42) +l²≈20.676 -2l*3.98 +l²Right side R²≈12.76 +l²Thus, set equal:20.676 -7.96 l +l²=12.76 +l²Cancel l²:20.676 -7.96 l=12.76Thus, -7.96 l=12.76 -20.676= -7.916Thus, l= -7.916 / -7.96≈0.994But let's do exact calculation.From equation:121/196 +13254/3136 +665/42 -2l*sqrt(665/42)+l²=60025/4704 +l²Cancel l²:121/196 +13254/3136 +665/42 -2l*sqrt(665/42)=60025/4704Convert all terms to denominator 4704:121/196= (121*24)/4704=2904/470413254/3136= (13254*1.5)/4704=19881/4704665/42= (665*112)/4704= since 42*112=4704, so 665*112=74480Thus, 665/42=74480/470460025/4704 remains as is.Therefore:2904/4704 +19881/4704 +74480/4704 -2l*sqrt(665/42)=60025/4704Sum numerators:2904 +19881=22785; 22785 +74480=97265Thus:97265/4704 -2l*sqrt(665/42)=60025/4704Subtract 60025/4704:(97265 -60025)/4704 -2l*sqrt(665/42)=037240/4704 -2l*sqrt(665/42)=0Simplify 37240/4704= divide numerator and denominator by 4:9310/1176≈7.914Thus:7.914=2l*sqrt(665/42)Thus, l=7.914/(2*sqrt(665/42))≈7.914/(2*3.98)≈7.914/7.96≈0.994≈1.0But let's compute exactly:sqrt(665/42)=sqrt(665)/sqrt(42)=sqrt(665/42)=sqrt(15.8333)=≈3.98But perhaps there is an exact value.Wait, note that 665/42= (665 ÷7)/(42 ÷7)=95/6. So sqrt(95/6)=sqrt(95)/sqrt(6). Therefore, sqrt(665/42)=sqrt(95/6).So l= (37240/4704)/(2*sqrt(95/6))= (37240/4704)*(1/(2*sqrt(95/6)))Simplify:37240 ÷4704= same as 9310/1176=4655/588=665/84≈7.9167Thus,l=(665/84)/(2*sqrt(95/6))=(665)/(168*sqrt(95/6))=(665)/(168*sqrt(95)/sqrt(6))=(665*sqrt(6))/(168*sqrt(95))Simplify numerator and denominator:665=5*7*1995=5*19sqrt(95)=sqrt(5*19)=sqrt(5)*sqrt(19)Thus:(5*7*19*sqrt(6))/(168*sqrt(5)*sqrt(19)) )= (7*sqrt(6))/ (168/5*7*19 ÷ (5*19)))Wait, maybe cancel terms:Numerator:5*7*19*sqrt(6)Denominator:168*sqrt(5)*sqrt(19)Cancel 5, 7, 19:Numerator: sqrt(6)Denominator:168/(5*7*19)=168/(665)=24/95Wait, perhaps not. Let me see:Numerator:5*7*19*sqrt(6)Denominator:168*sqrt(5)*sqrt(19)=168*sqrt(5*19)=168*sqrt(95)Thus,l= (5*7*19*sqrt(6))/(168*sqrt(95))= (665*sqrt(6))/(168*sqrt(95))Rationalize sqrt(6)/sqrt(95)=sqrt(6/95)Thus,l=665/(168)*sqrt(6/95)= (665/168)*sqrt(6/95)Simplify 665/168: divide numerator and denominator by7: 95/24Thus, l=95/24*sqrt(6/95)=95/24*sqrt(6)/sqrt(95)=95/24*sqrt(6)/sqrt(95)sqrt(95)/sqrt(95)=1, so sqrt(6)/sqrt(95)=sqrt(6/95)Thus, l=95/24*sqrt(6/95)=sqrt(6/95)*(95/24)=sqrt(6*95)/24= sqrt(570)/24Therefore, l= sqrt(570)/24Check: sqrt(570)=sqrt(5*114)=sqrt(5*2*57)=sqrt(10*57)≈sqrt(570)≈23.874Thus, sqrt(570)/24≈23.874/24≈0.994≈1.0, which matches the approximate value.So exact value is sqrt(570)/24.Thus, the coordinates of the circumcenter are (109/14,239√6/168,sqrt(570)/24)Then, the radius R is computed from point P:R²=(7 -109/14)^2 +k² +l²= ( -11/14 )² + (239√6/168)^2 + (sqrt(570)/24)^2Compute each term:(-11/14)^2=121/196=0.617(239√6/168)^2= (239² *6)/168²=57121*6/28224=342726/28224≈12.14(sqrt(570)/24)^2=570/576=95/96≈0.9896Sum:≈0.617 +12.14 +0.9896≈13.746Which is R²≈13.746, so R≈sqrt(13.746)≈3.708. But we need to check exact value.Alternatively, compute R²=121/196 + (239² *6)/168² +570/576Compute each term:121/196=121/196239²=57121, so 57121*6=342726168²=28224570/576=95/96Thus,R²=121/196 +342726/28224 +95/96Convert to a common denominator. Let's use 28224.121/196=121*144/28224=17424/28224342726/28224=342726/2822495/96=95*294/28224=27930/28224Sum:17424 +342726 +27930=17424+342726=360,150+27930=388,080Thus, R²=388,080/28,224= divide numerator and denominator by 48:388,080 ÷48=8085, 28,224 ÷48=5888085/588= divide by21:8085 ÷21=385, 588 ÷21=28Thus, R²=385/28=13.75Ah! So R²=385/28 → R= sqrt(385/28)=sqrt(385)/sqrt(28)=sqrt(385)/(2*sqrt(7))=sqrt(385)/ (2*sqrt(7)) =sqrt(385/7)/2=sqrt(55)/2≈3.708Therefore, the radius R is sqrt(55)/2, so the diameter is sqrt(55).Wait, sqrt(55)≈7.416.Wait, let's confirm:If R²=385/28, then R=sqrt(385/28)=sqrt(13.75)=approx 3.708, so diameter is approx7.416.But sqrt(55)=sqrt(5*11)=approx7.416, correct.But how did R²=385/28?From the calculation:R²=388,080/28,224=388,080 ÷28,224=13.75=55/4Wait, 55/4=13.75. Wait, 385/28=13.75. Wait, 385 divided by28=13.75. 28*13=364, 385-364=21, 21/28=0.75. So yes, 385/28=13.75=55/4.Wait, but 388,080/28,224= (388,080 ÷28,224)= (388,080 ÷ 28,224)=13.75=55/4. Therefore, R²=55/4, so R= sqrt(55)/2, so diameter is sqrt(55).Wait, that is much simpler. So in fact, R²=55/4, so R= sqrt(55)/2, diameter is sqrt(55).How did I miss that? Because when simplifying 388,080/28,224:Divide numerator and denominator by 48: 388,080 ÷48=8085, 28,224 ÷48=588Then 8085/588= Divide by21:8085 ÷21=385, 588 ÷21=28. Thus 385/28=13.75=55/4.Yes, because 385=55*7, 28=4*7, so 385/28=55/4.Therefore, R²=55/4, so R=√(55)/2, so diameter is √55≈7.416. Therefore, the diameter of the circumscribed sphere is √55, so the answer is boxed{sqrt{55}}.But wait, let me verify once more. Because this seems inconsistent with my previous step where R²=385/28=55/4. Wait, 385/28=13.75=55/4=13.75. Yes, correct. So R= sqrt(55)/2. Hence, the diameter is sqrt(55). Therefore, the answer is sqrt(55).But let me confirm with another point. Let's check the distance from the center (109/14,239√6/168,sqrt(570)/24) to point O(60/7,95√6/42,sqrt(665/42)).Compute the distance squared:(60/7 -109/14)^2 + (95√6/42 -239√6/168)^2 + (sqrt(665/42)-sqrt(570)/24)^2First term: 60/7 -109/14=120/14 -109/14=11/14 → (11/14)^2=121/196Second term:95√6/42 -239√6/168= (95*4√6 -239√6)/168=(380√6 -239√6)/168=141√6/168=47√6/56 → (47√6/56)^2= (47²*6)/56²=2209*6/3136=13254/3136= approx4.226Third term: sqrt(665/42)-sqrt(570)/24But sqrt(665/42)=sqrt(665)/sqrt(42)=sqrt(665)/ (sqrt(6)*sqrt(7))=sqrt(665)/(sqrt(6)*sqrt(7))sqrt(570)/24= sqrt(570)/24= sqrt(570)/24. Not sure if they can be simplified.But compute their difference squared:= (sqrt(665/42) - sqrt(570)/24)^2=665/42 -2*sqrt(665/42)*sqrt(570)/24 +570/576=665/42 +570/576 -2*sqrt((665/42)*(570))/24Simplify:665/42≈15.8333, 570/576≈0.9896So≈15.8333 +0.9896≈16.8229 -2*sqrt(665*570/(42*24²))Compute the product inside sqrt:665*570=379,05042*24²=42*576=24,192Thus, sqrt(379050/24192)=sqrt(15.666)=≈3.96Thus, term≈16.8229 -2*3.96≈16.8229 -7.92≈8.9029Thus, total distance squared≈121/196 +4.226 +8.9029≈0.617 +4.226 +8.903≈13.746, which is≈55/4=13.75. Close enough. So this confirms R²=55/4. Therefore, diameter is sqrt(55). So the answer is boxed{sqrt{55}}.Final Answerboxed{sqrt{55}}</think>To find the diameter of the circumscribed sphere of the tetrahedron formed by folding a triangle with side lengths 10, 12, and 14 along its midlines, we follow these steps:1. Identify the Midlines and Medial Triangle: - The midlines of the original triangle are 5, 6, and 7, which form the edges of the medial triangle. - The vertices of the tetrahedron are the midpoints of the original triangle (M, N, P) and the folded point (O) where the original vertices meet.2. Determine Coordinates of Midpoints: - Place the original triangle in a coordinate system with vertices A(0, 0, 0), B(14, 0, 0), and C(60/7, (24√6)/7, 0). - Midpoints M, N, P are calculated as: - M(79/7, (12√6)/7, 0) - N(30/7, (12√6)/7, 0) - P(7, 0, 0)3. Find Coordinates of Folded Point O: - Using the distances from O to midpoints M, N, P (5, 6, 7 respectively), solve the system of equations to find O(60/7, 95√6/42, √(665/42)).4. Calculate Circumradius: - Set up the equations for the circumscribed sphere passing through points O, M, N, P. - Solve the system to find the circumcenter coordinates (109/14, 239√6/168, √570/24). - Calculate the radius using the distance from the circumcenter to any vertex (e.g., point P): - Radius ( R = frac{sqrt{55}}{2} ).5. Determine Diameter: - The diameter of the circumscribed sphere is twice the radius, resulting in ( sqrt{55} ).Thus, the diameter of the circumscribed sphere of the tetrahedron is (boxed{sqrt{55}}).]

🔑:4. sqrt{55}.The tetrahedron has equal opposite edges, and the lengths of the three pairs of opposite edges are 5, 6, and 7.As shown in Figure 4, let A B=C D=5, A C=B D=7,A D=B C=6 text {. }Take the midpoints M and N of B C and A D respectively.It is easy to see that M N is the common perpendicular segment of A D and B C, and noting that A D=B C, the midpoint O of M N is equidistant from points A, B, C, D, i.e., O is the center of the circumscribed sphere of the tetrahedron A B C D, and the radius of the sphere isR=O A=sqrt{left(frac{d}{2}right)^{2}+left(frac{A D}{2}right)^{2}}(d=M N) .Draw N E perp M C. It is easy to see that C E perp plane A E D.Since N A=N D=N E=3, we know angle A E D=90^{circ}.Then A E^{2}=A C^{2}-C E^{2}=7^{2}-d^{2},D E^{2}=D C^{2}-C E^{2}=5^{2}-d^{2},A E^{2}+D E^{2}=A D^{2}=6^{2}.Thus 7^{2}-d^{2}+5^{2}-d^{2}=6^{2} Rightarrow d^{2}=19Rightarrow 2 R=sqrt{d^{2}+A D^{2}}=sqrt{55} text {. }

🧠:[<think>Okay, let's try to solve this system of equations. So we have:1. ( y^{x} = 1.5 + y^{-x} )2. ( y^{2.5 + x} = 64 ) with ( y > 0 )Hmm, the second equation might be easier to handle first because it's a single equation with two variables, but maybe we can express one variable in terms of the other. Let me think.Starting with the second equation: ( y^{2.5 + x} = 64 ). Since ( y > 0 ), we can take logarithms if needed, but maybe we can write 64 as a power of something. 64 is ( 2^6 ), ( 4^3 ), ( 8^2 ), or even ( 64^1 ). Not sure yet. Let's see.Alternatively, if I let ( t = y^x ), then maybe I can substitute into the first equation. Let's try that.From the first equation: ( y^{x} = 1.5 + y^{-x} ). Let's let ( t = y^x ), so ( y^{-x} = 1/t ). Then the equation becomes:( t = 1.5 + frac{1}{t} )Multiply both sides by t to eliminate the denominator:( t^2 = 1.5t + 1 )Bring all terms to one side:( t^2 - 1.5t - 1 = 0 )This is a quadratic equation. Let's solve for t.Using quadratic formula: ( t = [1.5 pm sqrt{(1.5)^2 + 4}]/2 )Calculating discriminant:( (1.5)^2 + 4 = 2.25 + 4 = 6.25 )Square root of 6.25 is 2.5. So:( t = [1.5 pm 2.5]/2 )So two solutions:1. ( t = (1.5 + 2.5)/2 = 4/2 = 2 )2. ( t = (1.5 - 2.5)/2 = (-1)/2 = -0.5 )But ( t = y^x ), and since ( y > 0 ), ( y^x ) is always positive. So we discard the negative solution. Thus, ( t = 2 ). Therefore, ( y^x = 2 ).Now, let's use the second equation. From the second equation: ( y^{2.5 + x} = 64 ). Let's express this as ( y^{x} cdot y^{2.5} = 64 ).But we already know ( y^x = 2 ), so substitute that in:( 2 cdot y^{2.5} = 64 )Divide both sides by 2:( y^{2.5} = 32 )Hmm, 32 is 2^5. So, ( y^{2.5} = 2^5 ). Let's write both sides with exponent 2.5:Alternatively, let's solve for y.( y^{2.5} = 32 )Take both sides to the power of 1/2.5:( y = 32^{1/2.5} )But 2.5 is 5/2, so 1/2.5 is 2/5. Thus:( y = 32^{2/5} )Now, 32 is 2^5. So:( y = (2^5)^{2/5} = 2^{(5 * 2/5)} = 2^2 = 4 )So y is 4. Then, from earlier, ( y^x = 2 ). So:( 4^x = 2 )Express 4 as 2^2:( (2^2)^x = 2 )Which is 2^{2x} = 2^1. Therefore, 2x = 1, so x = 0.5.Let me check if these values satisfy both equations.First equation: ( y^x = 2 ), and ( y^{-x} = 1/2 ). So 1.5 + 1/2 = 2, which is correct.Second equation: ( y^{2.5 + x} = 4^{2.5 + 0.5} = 4^{3} = 64 ). That's correct too.So the solution is y = 4 and x = 0.5.Final AnswerThe solution is ( boxed{x = frac{1}{2}} ) and ( boxed{y = 4} ).</think>Given the system of equations:1. ( y^{x} = 1.5 + y^{-x} )2. ( y^{2.5 + x} = 64 ) with ( y > 0 )We start by letting ( t = y^x ). Substituting this into the first equation, we get:[ t = 1.5 + frac{1}{t} ]Multiplying both sides by ( t ) to clear the denominator:[ t^2 = 1.5t + 1 ]Rearranging into a standard quadratic form:[ t^2 - 1.5t - 1 = 0 ]Solving this quadratic equation using the quadratic formula ( t = frac{-b pm sqrt{b^2 - 4ac}}{2a} ), where ( a = 1 ), ( b = -1.5 ), and ( c = -1 ):[ t = frac{1.5 pm sqrt{(1.5)^2 + 4}}{2} ][ t = frac{1.5 pm 2.5}{2} ]This gives two solutions:[ t = frac{4}{2} = 2 quad text{and} quad t = frac{-1}{2} = -0.5 ]Since ( y > 0 ), ( t = y^x ) must be positive, so we discard ( t = -0.5 ). Thus, ( t = 2 ), meaning ( y^x = 2 ).Next, using the second equation ( y^{2.5 + x} = 64 ), we substitute ( y^x = 2 ):[ y^x cdot y^{2.5} = 64 ][ 2 cdot y^{2.5} = 64 ][ y^{2.5} = 32 ]Expressing 32 as ( 2^5 ):[ y^{2.5} = 2^5 ]Taking both sides to the power of ( frac{2}{5} ):[ y = (2^5)^{2/5} = 2^{2} = 4 ]With ( y = 4 ), we substitute back into ( y^x = 2 ):[ 4^x = 2 ][ (2^2)^x = 2 ][ 2^{2x} = 2^1 ][ 2x = 1 ][ x = 0.5 ]Both values satisfy the original equations. Therefore, the solution is:[ boxed{x = frac{1}{2}} quad text{and} quad boxed{y = 4} ],<think>Okay, so I have this system of equations to solve. Let me write them down again to make sure I got them right:1. ( y^x = 1.5 + y^{-x} )2. ( y^{2.5 + x} = 64 ) with ( y > 0 )Alright, the second equation seems a bit more straightforward because it's an exponential equation, and maybe I can solve for one variable in terms of the other. Let's start with that one.Looking at the second equation: ( y^{2.5 + x} = 64 ). Since ( y > 0 ), I can take logarithms if needed, but maybe I can express ( y ) in terms of ( x ) or vice versa. Let's try solving for ( y ).If I take both sides to the power of ( 1/(2.5 + x) ), that should give me ( y = 64^{1/(2.5 + x)} ). Alternatively, if I prefer natural logs, take the natural logarithm of both sides:( ln(y^{2.5 + x}) = ln(64) )Which simplifies to:( (2.5 + x) ln(y) = ln(64) )Therefore,( ln(y) = frac{ln(64)}{2.5 + x} )So,( y = expleft( frac{ln(64)}{2.5 + x} right) )But 64 is 2^6, so maybe we can write that as:( y = expleft( frac{6 ln(2)}{2.5 + x} right) = 2^{6/(2.5 + x)} )Hmm, that's another way to write it. Maybe this exponent manipulation is helpful. Let's note that ( y = 2^{6/(2.5 + x)} ). Let's keep that in mind.Now, let's look at the first equation: ( y^x = 1.5 + y^{-x} ). Hmmm. This looks a bit more complicated, but maybe we can manipulate it.Let me denote ( z = y^x ). Then, ( y^{-x} = 1/z ). So substituting into the first equation:( z = 1.5 + frac{1}{z} )Multiply both sides by z to eliminate the denominator:( z^2 = 1.5 z + 1 )Bring all terms to one side:( z^2 - 1.5 z - 1 = 0 )Now, we have a quadratic equation in z. Let's solve this quadratic equation.Using the quadratic formula: ( z = frac{1.5 pm sqrt{(1.5)^2 + 4}}{2} )Calculating the discriminant:( (1.5)^2 = 2.25 )So,Discriminant: ( 2.25 + 4 = 6.25 )Square root of 6.25 is 2.5. So,( z = frac{1.5 pm 2.5}{2} )So two solutions:1. ( z = frac{1.5 + 2.5}{2} = frac{4}{2} = 2 )2. ( z = frac{1.5 - 2.5}{2} = frac{-1}{2} = -0.5 )But z was defined as ( y^x ), and since ( y > 0 ), ( y^x ) is always positive. Therefore, z cannot be negative. Therefore, z = 2 is the only valid solution.So, we have ( y^x = 2 ). So, that's our first equation simplified: ( y^x = 2 ).So now, we have two equations:1. ( y^x = 2 )2. ( y^{2.5 + x} = 64 )Hmm. Maybe we can relate these two. Let's see. The second equation can be written as ( y^{2.5} cdot y^x = 64 ). But we know from the first equation that ( y^x = 2 ). Therefore, substituting that into the second equation:( y^{2.5} cdot 2 = 64 )Divide both sides by 2:( y^{2.5} = 32 )So, ( y^{2.5} = 32 ). Let's solve for y.First, note that 32 is 2^5. So,( y^{2.5} = 2^5 )Taking both sides to the power of ( 1/2.5 ):( y = (2^5)^{1/2.5} = 2^{5/2.5} = 2^{2} = 4 )Because 5 divided by 2.5 is 2. So, y = 4.Now that we have y = 4, we can substitute back into the first equation to find x.From equation 1: ( y^x = 2 ). Since y = 4,( 4^x = 2 )Express 4 as 2^2:( (2^2)^x = 2 )Simplify the left side:( 2^{2x} = 2^1 )Therefore, equate exponents:( 2x = 1 )Thus,( x = 1/2 )So, x = 0.5 and y = 4.Let me check these values in both equations to make sure they work.First equation:( y^x = 4^{0.5} = 2 )And ( y^{-x} = 4^{-0.5} = 1/2 )So, 1.5 + y^{-x} = 1.5 + 0.5 = 2, which matches y^x. Good.Second equation:( y^{2.5 + x} = 4^{2.5 + 0.5} = 4^{3} = 64 ), which is correct.So both equations are satisfied. Therefore, the solution is x = 0.5 and y = 4.Final AnswerThe solution is ( boxed{x = frac{1}{2}} ) and ( boxed{y = 4} ).</think>Given the system of equations:1. ( y^x = 1.5 + y^{-x} )2. ( y^{2.5 + x} = 64 ) with ( y > 0 )We start by solving the second equation for ( y ):[ y^{2.5 + x} = 64 ][ y = 64^{1/(2.5 + x)} ][ y = 2^{6/(2.5 + x)} ]Next, we substitute ( z = y^x ) into the first equation:[ z = 1.5 + frac{1}{z} ]Multiplying both sides by ( z ):[ z^2 = 1.5z + 1 ][ z^2 - 1.5z - 1 = 0 ]Solving this quadratic equation using the quadratic formula:[ z = frac{1.5 pm sqrt{(1.5)^2 + 4}}{2} ][ z = frac{1.5 pm 2.5}{2} ]This gives two solutions:1. ( z = 2 )2. ( z = -0.5 ) (discarded since ( z = y^x ) must be positive)Thus, ( y^x = 2 ).Substituting ( y^x = 2 ) into the second equation:[ y^{2.5} cdot 2 = 64 ][ y^{2.5} = 32 ][ y = 32^{1/2.5} ][ y = 2^{5/2.5} ][ y = 4 ]With ( y = 4 ), we solve for ( x ):[ 4^x = 2 ][ (2^2)^x = 2 ][ 2^{2x} = 2^1 ][ 2x = 1 ][ x = frac{1}{2} ]Verification in both equations confirms the solution is correct.The solution is ( boxed{x = frac{1}{2}} ) and ( boxed{y = 4} ).]

🔑:## Solution.Multiply the first equation by y^{x}, we have y^{2 x}-1.5 y^{x}-1=0. Solving this equation as a quadratic in terms of y^{x}, we get y^{x}=-frac{1}{2} (no solutions), or y^{x}=2. From the second equation of the system y^{2.5} cdot y^{x}=64 Rightarrow y^{2.5} cdot 2=64, y^{2.5}=32, y=4. Therefore, 4^{x}=2, x=frac{1}{2}.text { Answer: }left(frac{1}{2} ; 4right)

🧠:[<think>Okay, so I need to solve this problem where I have the function f(x) = sqrt(3)cos(2x) - sin(2x), and then they define y = f(x - m) where m is greater than 0. The question is asking for the minimum value of m such that y is an odd function. The options are A through D with some pi fractions. Let me think step by step.First, I remember that an odd function satisfies the property y(-x) = -y(x) for all x. So, for y = f(x - m) to be odd, we need f(-x - m) = -f(x - m) for all x. Hmm, let me check that. If y(x) = f(x - m), then y(-x) = f(-x - m). For it to be odd, this must equal -y(x) = -f(x - m). So yes, the equation is f(-x - m) = -f(x - m) for all x. Alternatively, maybe substituting variables would help. Let me let t = x - m. Then, if x = t + m. Wait, but I'm not sure if that substitution will help. Let me see.Alternatively, maybe we can express f(x) in a different form to make it easier to analyze. Since f(x) is a combination of cosine and sine with the same argument 2x, maybe we can write it as a single sinusoidal function. The expression sqrt(3)cos(2x) - sin(2x) seems like a candidate for the amplitude-phase form, like A cos(2x + φ) or A sin(2x + φ). Let's try converting it.The general formula is a cos θ + b sin θ = C cos(θ - φ), where C = sqrt(a^2 + b^2) and tan φ = b/a. Wait, or is it a sin θ + b cos θ? Wait, maybe I should recall the exact identity.Alternatively, we can write it as R sin(2x + φ) or R cos(2x + φ). Let me try both approaches.Let me try to express sqrt(3) cos(2x) - sin(2x) as R cos(2x + φ). Using the identity:R cos(2x + φ) = R cos(2x)cos φ - R sin(2x) sin φ.Comparing coefficients with sqrt(3) cos(2x) - sin(2x):So R cos φ = sqrt(3) and R sin φ = 1. Therefore, tan φ = (R sin φ)/(R cos φ) = 1/sqrt(3), so φ = arctan(1/sqrt(3)) = π/6. Then R^2 = (sqrt(3))^2 + 1^2 = 3 + 1 = 4, so R = 2. Therefore, f(x) = 2 cos(2x + π/6).Wait, hold on. If R cos φ = sqrt(3) and R sin φ = 1, then tan φ = 1/sqrt(3), which is 30 degrees or π/6. So φ is π/6, so expanding cos(2x + π/6) gives cos(2x)cos(π/6) - sin(2x)sin(π/6). Which is cos(2x)(sqrt(3)/2) - sin(2x)(1/2). Multiplying by R=2 gives sqrt(3) cos(2x) - sin(2x), which matches. So f(x) = 2 cos(2x + π/6). Alternatively, maybe it's written as a sine function. Let's check.Alternatively, we can write f(x) as R sin(2x + φ). Let's see:sqrt(3) cos(2x) - sin(2x) = R sin(2x + φ).Expanding R sin(2x + φ) = R sin(2x) cos φ + R cos(2x) sin φ.Comparing coefficients:Coefficient of sin(2x): R cos φ = -1.Coefficient of cos(2x): R sin φ = sqrt(3).Therefore, R cos φ = -1 and R sin φ = sqrt(3). Then, tan φ = (sqrt(3)/(-1)) = -sqrt(3). So φ is in the second quadrant since sin φ is positive and cos φ is negative. Therefore, φ = 2π/3. Then R^2 = (-1)^2 + (sqrt(3))^2 = 1 + 3 = 4, so R = 2. Therefore, f(x) = 2 sin(2x + 2π/3).Either way, both forms are correct, but maybe the cosine form is more useful here. Let me note both:f(x) = 2 cos(2x + π/6) = 2 sin(2x + 2π/3).But maybe the sine form is better for shifting. Hmm. Wait, if we use the sine form, then f(x - m) would be 2 sin[2(x - m) + 2π/3] = 2 sin(2x - 2m + 2π/3). For this to be an odd function, sin(2x - 2m + 2π/3) must satisfy the oddness condition.But maybe using the cosine form is better. Let's try that.f(x) = 2 cos(2x + π/6). Then f(x - m) = 2 cos[2(x - m) + π/6] = 2 cos(2x - 2m + π/6). For this to be an odd function, we need f(-x - m) = -f(x - m).Wait, hold on, actually, if y(x) = f(x - m), then y(-x) = f(-x - m). For y to be odd, we need y(-x) = -y(x), so f(-x - m) = -f(x - m) for all x. So, substituting into f, which is 2 cos(2x + π/6):Left side: f(-x - m) = 2 cos[2(-x - m) + π/6] = 2 cos(-2x - 2m + π/6).But cosine is even, so cos(-θ) = cosθ, so this becomes 2 cos(2x + 2m - π/6).Right side: -f(x - m) = -2 cos(2(x - m) + π/6) = -2 cos(2x - 2m + π/6).So setting left side equal to right side:2 cos(2x + 2m - π/6) = -2 cos(2x - 2m + π/6)Divide both sides by 2:cos(2x + 2m - π/6) = -cos(2x - 2m + π/6)Recall that cos(θ) = -cos(φ) implies that θ = φ + (2k + 1)π for some integer k. But this has to hold for all x, so the arguments inside the cosines must differ by an odd multiple of π. Let's write:For all x, 2x + 2m - π/6 = - (2x - 2m + π/6) + (2k + 1)πWait, because cos(α) = -cos(β) can also be written as cos(α) = cos(β + π), so α = ±(β + π) + 2πk. Let me check.But since this has to hold for all x, the coefficients of x and the constants must match appropriately. Let me write the equation:cos(2x + 2m - π/6) = -cos(2x - 2m + π/6)But -cos(θ) = cos(θ + π). So:cos(2x + 2m - π/6) = cos(2x - 2m + π/6 + π)Therefore, we have two possibilities for the arguments:1. 2x + 2m - π/6 = 2x - 2m + π/6 + π + 2πkor2. 2x + 2m - π/6 = - (2x - 2m + π/6 + π) + 2πkwhere k is any integer. Since this must hold for all x, the coefficients of x must be equal, and the constants must also be equal. Let's check the first case.Case 1:2x + 2m - π/6 = 2x - 2m + π/6 + π + 2πkSubtract 2x from both sides:2m - π/6 = -2m + π/6 + π + 2πkBring all terms to the left side:2m + 2m - π/6 - π/6 - π = 2πk4m - (2π/6 + π) = 2πkSimplify 2π/6 to π/3:4m - (π/3 + π) = 2πk4m - (4π/3) = 2πkThen 4m = 4π/3 + 2πkDivide by 4:m = π/3 + (π/2)kSince m > 0, we can take k = 0, 1, 2, ... etc. The minimum positive m would be when k=0: m = π/3. Let's check case 2.Case 2:2x + 2m - π/6 = -2x + 2m - π/6 - π + 2πkBring all terms to the left:2x + 2m - π/6 + 2x - 2m + π/6 + π = 2πkCombine like terms:4x + π = 2πkBut this must hold for all x, which is impossible unless the coefficient of x is zero, which is 4 here. Since 4 ≠ 0, this case is impossible. Therefore, only Case 1 gives possible solutions.Therefore, m = π/3 + (π/2)k. The minimal positive m is π/3. But wait, the options have π/3 as option C. But let me check again. Wait, maybe there's a mistake here.Wait, when I set up the equation, I considered that cos(α) = -cos(β) implies α = β + π + 2πk or α = -β - π + 2πk. Let me confirm this. Because cos(α) = -cos(β) can be rewritten as cos(α) = cos(β + π), which leads to α = β + π + 2πk or α = - (β + π) + 2πk. So the two cases are:1. α = β + π + 2πk2. α = -β - π + 2πkSo in our case, α = 2x + 2m - π/6 and β = 2x - 2m + π/6. Therefore:Case 1:2x + 2m - π/6 = (2x - 2m + π/6) + π + 2πkSimplify:2x + 2m - π/6 = 2x - 2m + π/6 + π + 2πkCancel 2x:2m - π/6 = -2m + π/6 + π + 2πkBring terms to left:2m + 2m = π/6 + π + π/6 + 2πk4m = (2π/6 + π) + 2πkSimplify 2π/6 to π/3:4m = π/3 + π + 2πk = (4π/3) + 2πkSo m = (4π/3)/(4) + (2πk)/4 = π/3 + (π/2)kCase 2:2x + 2m - π/6 = - (2x - 2m + π/6) - π + 2πkSimplify:2x + 2m - π/6 = -2x + 2m - π/6 - π + 2πkBring all terms to left:2x + 2m - π/6 + 2x - 2m + π/6 + π = 2πkCombine:4x + π = 2πkAgain, this can't hold for all x, so only Case 1 is valid.Thus, the solutions are m = π/3 + (π/2)k. Since m > 0, the minimal positive solution is π/3 when k=0. But wait, the answer options include π/3 as option C. But wait, let me check if this is correct.Wait, but let's test m = π/3. Let's see if f(x - m) is indeed odd.Given f(x) = sqrt(3) cos(2x) - sin(2x). If m = π/3, then f(x - π/3) = sqrt(3) cos[2(x - π/3)] - sin[2(x - π/3)].Compute 2(x - π/3) = 2x - 2π/3.So f(x - π/3) = sqrt(3) cos(2x - 2π/3) - sin(2x - 2π/3).Let me compute this expression. Maybe express it as a single sine or cosine function again.Alternatively, check if it's odd. Let's compute f(-x - π/3). Wait, y(x) = f(x - π/3). To check if it's odd, we need y(-x) = f(-x - π/3) = -y(x) = -f(x - π/3).Compute f(-x - π/3):sqrt(3) cos[2(-x - π/3)] - sin[2(-x - π/3)] = sqrt(3) cos(-2x - 2π/3) - sin(-2x - 2π/3).Using even-odd properties: cos is even, sin is odd.= sqrt(3) cos(2x + 2π/3) + sin(2x + 2π/3).Now, let's compute -f(x - π/3):= -[sqrt(3) cos(2x - 2π/3) - sin(2x - 2π/3)] = -sqrt(3) cos(2x - 2π/3) + sin(2x - 2π/3).So, for f(-x - π/3) to equal -f(x - π/3), we need:sqrt(3) cos(2x + 2π/3) + sin(2x + 2π/3) = -sqrt(3) cos(2x - 2π/3) + sin(2x - 2π/3).Is this true for all x? Let's check for a specific x, say x = 0.Left side: sqrt(3) cos(2π/3) + sin(2π/3) = sqrt(3)*(-1/2) + (sqrt(3)/2) = (-sqrt(3)/2 + sqrt(3)/2) = 0.Right side: -sqrt(3) cos(-2π/3) + sin(-2π/3) = -sqrt(3) cos(2π/3) - sin(2π/3) = -sqrt(3)*(-1/2) - (-sqrt(3)/2) = (sqrt(3)/2) + sqrt(3)/2 = sqrt(3). But 0 ≠ sqrt(3), so this is not valid. Therefore, m = π/3 does not satisfy the condition. Hmm, that's a problem. So my previous conclusion was wrong. There must be an error in my reasoning.Wait, maybe I messed up the equations when solving for m. Let me go back.We had the equation:cos(2x + 2m - π/6) = -cos(2x - 2m + π/6)Which led to m = π/3 + (π/2)k. But when testing m = π/3, it didn't work. So where's the mistake?Perhaps because in the equation 4m = 4π/3 + 2πk, I divided by 4 and got m = π/3 + (π/2)k. That seems correct. So why is m = π/3 not working?Alternatively, maybe the problem is that when we set the equality for all x, the phase shift requires more constraints. Wait, perhaps in addition to the phase, the function's periodicity must be considered. Let me think.Alternatively, maybe instead of converting f(x) to a cosine function, using the sine form would make it easier. Let's try that.Earlier, we found that f(x) = 2 sin(2x + 2π/3). Then f(x - m) = 2 sin[2(x - m) + 2π/3] = 2 sin(2x - 2m + 2π/3).For this to be an odd function, we need f(-x - m) = -f(x - m). Let's compute both sides.Left side: f(-x - m) = 2 sin[2(-x - m) + 2π/3] = 2 sin(-2x - 2m + 2π/3) = 2 sin(-2x - 2m + 2π/3). Since sin(-θ) = -sinθ, this becomes 2*(-sin(2x + 2m - 2π/3)) = -2 sin(2x + 2m - 2π/3).Right side: -f(x - m) = -2 sin(2x - 2m + 2π/3).Therefore, the equation is:-2 sin(2x + 2m - 2π/3) = -2 sin(2x - 2m + 2π/3)Divide both sides by -2:sin(2x + 2m - 2π/3) = sin(2x - 2m + 2π/3)For this equality to hold for all x, the arguments must satisfy either:1. 2x + 2m - 2π/3 = 2x - 2m + 2π/3 + 2πk, or2. 2x + 2m - 2π/3 = π - (2x - 2m + 2π/3) + 2πkfor some integer k. Let's analyze both cases.Case 1:2x + 2m - 2π/3 = 2x - 2m + 2π/3 + 2πkSubtract 2x from both sides:2m - 2π/3 = -2m + 2π/3 + 2πkBring variables to the left and constants to the right:2m + 2m = 2π/3 + 2π/3 + 2πk4m = 4π/3 + 2πkDivide by 2:2m = 2π/3 + πkTherefore, m = π/3 + (π/2)kCase 2:2x + 2m - 2π/3 = π - 2x + 2m - 2π/3 + 2πkSimplify:2x + 2m - 2π/3 = π - 2x + 2m - 2π/3 + 2πkBring like terms to the left:2x + 2m - 2π/3 + 2x - 2m + 2π/3 - π = 2πkCombine terms:4x - π = 2πkAgain, this must hold for all x, which is only possible if the coefficient of x is zero, but 4x term can't be canceled unless 4=0, which is impossible. Thus, Case 2 is invalid.Therefore, only Case 1 gives solutions m = π/3 + (π/2)k. Since m > 0, the smallest positive m is π/3 when k = 0. But when we tested m = π/3 earlier, it didn't satisfy the condition. So there must be a problem here.Wait, maybe when we converted f(x) into sine form, we made a mistake. Let me recheck.Original f(x) = sqrt(3) cos(2x) - sin(2x). Let me confirm the sine form. We have:sqrt(3) cos(2x) - sin(2x) = R sin(2x + φ)Expanding RHS: R sin(2x) cos φ + R cos(2x) sin φEquating coefficients:Coefficient of sin(2x): R cos φ = -1Coefficient of cos(2x): R sin φ = sqrt(3)So, R cos φ = -1 and R sin φ = sqrt(3). Then,tan φ = (R sin φ)/(R cos φ) = sqrt(3)/(-1) = -sqrt(3)Therefore, φ is in the second quadrant (since sin φ is positive and cos φ is negative). So φ = π - π/3 = 2π/3.Then R^2 = (-1)^2 + (sqrt(3))^2 = 1 + 3 = 4, so R = 2. Therefore, f(x) = 2 sin(2x + 2π/3). So that seems correct.Therefore, f(x - m) = 2 sin[2(x - m) + 2π/3] = 2 sin(2x - 2m + 2π/3). Then, f(-x - m) = 2 sin[2(-x - m) + 2π/3] = 2 sin(-2x - 2m + 2π/3) = -2 sin(2x + 2m - 2π/3). For this to equal -f(x - m) = -2 sin(2x - 2m + 2π/3), we have:-2 sin(2x + 2m - 2π/3) = -2 sin(2x - 2m + 2π/3)Divide both sides by -2:sin(2x + 2m - 2π/3) = sin(2x - 2m + 2π/3)Which leads to the previous equation. So solving gives m = π/3 + (π/2)k. So m = π/3 is the minimal solution. But when tested with x = 0, it didn't hold.Wait, let's compute y(x) = f(x - π/3) = 2 sin(2x - 2π/3 + 2π/3) = 2 sin(2x). Which is 2 sin(2x). And 2 sin(2x) is indeed an odd function because sin(-2x) = -sin(2x). So y(-x) = 2 sin(-2x) = -2 sin(2x) = -y(x). Therefore, it is odd. Wait, but earlier when I tested with x = 0, I thought there was a discrepancy. Let me check again.Wait, maybe I made an error in substitution earlier. Let me recompute f(x - m) when m = π/3.Original f(x) = sqrt(3) cos(2x) - sin(2x). Then f(x - π/3) = sqrt(3) cos[2(x - π/3)] - sin[2(x - π/3)] = sqrt(3) cos(2x - 2π/3) - sin(2x - 2π/3).If we convert this to the sine form, we saw that f(x - π/3) = 2 sin(2x). Wait, how?From the earlier calculation, using the sine form: f(x) = 2 sin(2x + 2π/3). Then f(x - π/3) = 2 sin[2(x - π/3) + 2π/3] = 2 sin(2x - 2π/3 + 2π/3) = 2 sin(2x). So indeed, f(x - π/3) = 2 sin(2x), which is an odd function because sin is odd. Therefore, y(x) = 2 sin(2x) is odd, so m = π/3 should be correct. So why did my previous test with x = 0 show a discrepancy?Wait, I think I messed up the substitution when using the cosine form. Let's check again.Using f(x) = sqrt(3) cos(2x) - sin(2x). Then f(x - π/3) = sqrt(3) cos(2(x - π/3)) - sin(2(x - π/3)) = sqrt(3) cos(2x - 2π/3) - sin(2x - 2π/3).Let me compute this for x = 0: f(-π/3) = sqrt(3) cos(-2π/3) - sin(-2π/3) = sqrt(3) cos(2π/3) + sin(2π/3) = sqrt(3)*(-1/2) + (sqrt(3)/2) = (-sqrt(3)/2 + sqrt(3)/2) = 0.Then y(0) = f(0 - π/3) = f(-π/3) = 0.If we compute -y(0) = -0 = 0. Then y(-0) = y(0) = 0. So that's consistent. Wait, but when I tested with x = 0 earlier, I think I miscalculated f(-x - m) and -f(x - m). Let's take x = π/6.Compute y(π/6) = f(π/6 - π/3) = f(-π/6) = sqrt(3) cos(-π/3) - sin(-π/3) = sqrt(3)*(1/2) + (sqrt(3)/2) = sqrt(3)/2 + sqrt(3)/2 = sqrt(3).Then y(-π/6) = f(-π/6 - π/3) = f(-π/2) = sqrt(3) cos(-π) - sin(-π) = sqrt(3)*(-1) - 0 = -sqrt(3). Which is equal to -y(π/6) = -sqrt(3). So that works.Wait, earlier I think I substituted into the wrong function. So maybe my initial test was incorrect. Therefore, m = π/3 is actually correct, and the answer is C.But wait, the options have C: π/3 and B: 5π/6. Let me check with m = 5π/6. Maybe there's a larger m that also works.Alternatively, maybe the minimal m is π/3. But according to the problem statement, the answer options include π/3 as C. However, I need to ensure that when I derived m = π/3 + (π/2)k, the minimal positive m is indeed π/3, which works. So that would make the answer C. But let me check again with another x.Take x = π/4.y(π/4) = f(π/4 - π/3) = f(-π/12) = sqrt(3) cos(-π/6) - sin(-π/6) = sqrt(3)*(sqrt(3)/2) - (-1/2) = (3/2) + 1/2 = 2.y(-π/4) = f(-π/4 - π/3) = f(-7π/12) = sqrt(3) cos(-7π/6) - sin(-7π/6). Cos(-7π/6) = cos(7π/6) = -sqrt(3)/2. Sin(-7π/6) = 1/2. So:sqrt(3)*(-sqrt(3)/2) - (1/2) = (-3/2) - 1/2 = -2, which is -y(π/4). So it works. Therefore, m = π/3 does satisfy the condition, and my initial test with x=0 was correct, but maybe I miscalculated.Therefore, the minimal m is π/3, which is option C. But wait, according to the previous calculation, when k=0, m=π/3. But in the options given, C is π/3, which should be the answer. But I need to cross-verify with the answer choices.Wait, let me check again.Original problem options:A: π/6B: 5π/6C: π/3D: 2π/3So according to my derivation, the minimal m is π/3, which is option C. But why is there an option D: 2π/3? Maybe I missed something.Wait, when I was using the cosine form, I derived m = π/3 + (π/2)k. So with k=0, m=π/3; k=1, m=π/3 + π/2 = 5π/6; k=2, m=π/3 + π = 4π/3, etc. So the possible positive m's are π/3, 5π/6, 4π/3, etc. Therefore, the minimal m is π/3, but 5π/6 is also a solution. However, since the question asks for the minimum value of m, the answer should be π/3. But let me check if m=5π/6 also works.Take m=5π/6. Then y(x) = f(x - 5π/6) = sqrt(3) cos[2(x - 5π/6)] - sin[2(x - 5π/6)] = sqrt(3) cos(2x - 5π/3) - sin(2x - 5π/3). Let's express this in sine form using the previous conversion.From the sine form: f(x) = 2 sin(2x + 2π/3). Therefore, f(x - 5π/6) = 2 sin[2(x - 5π/6) + 2π/3] = 2 sin(2x - 5π/3 + 2π/3) = 2 sin(2x - π). Since sin(2x - π) = -sin(2x), so y(x) = 2*(-sin(2x)) = -2 sin(2x), which is also an odd function. So m=5π/6 also works. But since 5π/6 is larger than π/3, the minimal m is π/3. Therefore, answer C.But let me check why the original test with m=π/3 worked. So in this case, the function becomes y(x)=2 sin(2x), which is odd. Therefore, m=π/3 is indeed correct. The confusion earlier was because of a miscalculation on my part. Therefore, the correct answer is C: π/3.But wait, according to the options given, C is π/3. But when I initially thought the answer was B, which is 5π/6, but that's incorrect. So the minimum m is π/3. Therefore, the answer is C.Final Answerboxed{C}</think>Given ( f(x) = sqrt{3} cos 2x - sin 2x ), we need to find the minimum value of ( m ) such that ( y = f(x - m) ) is an odd function. First, we convert ( f(x) ) into a sinusoidal form. Using the identity for combining sine and cosine terms, we find:[ f(x) = 2 sin(2x + frac{2pi}{3}) ]Next, we consider the function ( y(x) = f(x - m) ):[ y(x) = 2 sin(2(x - m) + frac{2pi}{3}) = 2 sin(2x - 2m + frac{2pi}{3}) ]For ( y(x) ) to be an odd function, it must satisfy ( y(-x) = -y(x) ). This implies:[ 2 sin(-2x - 2m + frac{2pi}{3}) = -2 sin(2x - 2m + frac{2pi}{3}) ]Using the property ( sin(-theta) = -sin(theta) ), we simplify the left-hand side:[ -2 sin(2x + 2m - frac{2pi}{3}) = -2 sin(2x - 2m + frac{2pi}{3}) ]This simplifies to:[ sin(2x + 2m - frac{2pi}{3}) = sin(2x - 2m + frac{2pi}{3}) ]For this equality to hold for all ( x ), the arguments of the sine functions must differ by an integer multiple of ( 2pi ) or be supplementary angles. Solving the resulting equations, we find:[ 2m - frac{2pi}{3} = -2m + frac{2pi}{3} + 2kpi ]Solving for ( m ), we get:[ 4m = frac{4pi}{3} + 2kpi ][ m = frac{pi}{3} + frac{kpi}{2} ]The minimum positive value of ( m ) occurs when ( k = 0 ):[ m = frac{pi}{3} ]Thus, the minimum value of ( m ) is (boxed{C}).,<think>Okay, so I need to find the minimum value of m > 0 such that the function y = f(x - m) is an odd function. The original function f(x) is given as sqrt(3) cos 2x - sin 2x. Let me recall what an odd function is: a function g(x) is odd if g(-x) = -g(x) for all x in its domain. So, for y = f(x - m) to be odd, it must satisfy f(-x - m) = -f(x - m) for all x. First, maybe I should express f(x) in a simpler form. The expression sqrt(3) cos 2x - sin 2x looks like a combination of sine and cosine terms with the same argument. I remember that such expressions can be written as a single sine or cosine function using the amplitude-phase form. Let me try that.The general formula is A cos θ + B sin θ = C cos(θ - φ), where C = sqrt(A² + B²) and tan φ = B/A. Alternatively, it can also be written as C sin(θ + φ'), depending on the phase shift. Let me compute that.Given f(x) = sqrt(3) cos 2x - sin 2x. Let's treat this as A cos 2x + B sin 2x, where A = sqrt(3) and B = -1. So, the amplitude C is sqrt(A² + B²) = sqrt((sqrt(3))² + (-1)^2) = sqrt(3 + 1) = sqrt(4) = 2. Then, the phase shift φ can be found using tan φ = B/A = (-1)/sqrt(3). Hmm, tan φ = -1/sqrt(3). The angle φ whose tangent is -1/sqrt(3) is -30 degrees or 330 degrees, which is -π/6 radians. So, we can write f(x) as 2 cos(2x - (-π/6)) = 2 cos(2x + π/6). Wait, let me check that. Wait, actually, the formula is A cos θ + B sin θ = C cos(θ - φ). So, in this case, since B is negative, maybe I need to adjust the phase accordingly. Let me verify the formula again. If we have A cos θ + B sin θ = C cos(θ - φ), then expanding the right-hand side gives C cos θ cos φ + C sin θ sin φ. Therefore, matching coefficients:A = C cos φB = C sin φSo, A = sqrt(3), B = -1, C = 2. Then, cos φ = sqrt(3)/2 and sin φ = -1/2. The angle φ that satisfies this is -30 degrees, which is equivalent to 11π/6 in radians (since angles are periodic with 2π). But in terms of positive angles, it's 11π/6, but in terms of the standard phase shift, it's often expressed as a positive angle between 0 and 2π. Alternatively, since cos φ is positive and sin φ is negative, φ is in the fourth quadrant, so φ = 2π - π/6 = 11π/6. However, when we write it in the form C cos(θ - φ), the phase shift would be φ. So, putting it all together:f(x) = 2 cos(2x - 11π/6). Alternatively, since cos is an even function, cos(2x - 11π/6) = cos(11π/6 - 2x). Hmm, but maybe another approach is better. Alternatively, using the formula for A cos θ + B sin θ = C sin(θ + φ'). Let's see if that's easier.Alternatively, we can write it as a sine function. Let me try that. A cos θ + B sin θ = C sin(θ + φ'). Then, expanding sin(θ + φ') gives sin θ cos φ' + cos θ sin φ', so:A cos θ + B sin θ = C sin θ cos φ' + C cos θ sin φ'Therefore, matching coefficients:A = C sin φ'B = C cos φ'Wait, that might not be standard. Let me check again. Alternatively, maybe I should use a different identity. Let me recall that A cos θ + B sin θ can also be written as C sin(θ + α), where C is the same sqrt(A² + B²). Let's compute that.If we write sqrt(3) cos 2x - sin 2x = 2 sin(2x + α). Then expanding the right-hand side:2 sin(2x + α) = 2 sin 2x cos α + 2 cos 2x sin α. Comparing coefficients:sqrt(3) cos 2x - sin 2x = 2 sin α cos 2x + 2 cos α sin 2xSo, coefficient of cos 2x: sqrt(3) = 2 sin αCoefficient of sin 2x: -1 = 2 cos αTherefore:sin α = sqrt(3)/2cos α = -1/2Which angle α satisfies this? sin α = sqrt(3)/2 at α = π/3 or 2π/3, but cos α = -1/2, so α must be in the second quadrant. So, α = 2π/3. Therefore:f(x) = 2 sin(2x + 2π/3). Hmm, that seems right. Let me check:2 sin(2x + 2π/3) = 2 [sin 2x cos 2π/3 + cos 2x sin 2π/3] = 2 [ sin 2x*(-1/2) + cos 2x*(sqrt(3)/2) ] = -sin 2x + sqrt(3) cos 2x. Yes, that matches the original f(x). So, f(x) can be written as 2 sin(2x + 2π/3). Alternatively, 2 sin[2(x + π/3)]. Because 2x + 2π/3 = 2(x + π/3). So, f(x) is 2 sin[2(x + π/3)]. This might be helpful for understanding the shift. So, the original function f(x) is a sine wave with amplitude 2, period π (since the coefficient of x inside sin is 2), shifted to the left by π/3. Now, the problem is to shift this function to the right by m units (since it's f(x - m)), and find the smallest m > 0 such that the shifted function is odd. So, y = f(x - m) = 2 sin[2((x - m) + π/3)] = 2 sin[2x - 2m + 2π/3]. So, simplifying, that's 2 sin(2x + (2π/3 - 2m)). Alternatively, perhaps another way to write it is 2 sin[2x + (2π/3 - 2m)]. For this function to be odd, it must satisfy y(-x) = -y(x) for all x. So, let's write this condition:y(-x) = 2 sin(-2x + (2π/3 - 2m)) = -2 sin(2x - (2π/3 - 2m)) Because sin(-θ) = -sinθ, so sin(-2x + (2π/3 - 2m)) = sin[-(2x) + (2π/3 - 2m)] = sin[-(2x - (2π/3 - 2m))] = -sin(2x - (2π/3 - 2m))Therefore, y(-x) = -2 sin(2x - (2π/3 - 2m)). On the other hand, -y(x) = -2 sin(2x + (2π/3 - 2m)). For y(-x) = -y(x), we need:-2 sin(2x - (2π/3 - 2m)) = -2 sin(2x + (2π/3 - 2m))Divide both sides by -2:sin(2x - (2π/3 - 2m)) = sin(2x + (2π/3 - 2m))For this to hold for all x, the arguments of the sine functions must differ by an integer multiple of 2π, or be supplementary angles differing by an integer multiple of 2π. But since this must hold for all x, the only possibility is that the difference between the arguments is a multiple of 2π. Let's compute the difference:[2x + (2π/3 - 2m)] - [2x - (2π/3 - 2m)] = 2*(2π/3 - 2m)So, the difference is 2*(2π/3 - 2m). For the sines to be equal for all x, this difference must be an integer multiple of 2π. Therefore:2*(2π/3 - 2m) = 2π*k, where k is an integer.Divide both sides by 2:2π/3 - 2m = π*kSolving for m:2m = 2π/3 - π*km = π/3 - (π*k)/2We need m > 0, so π/3 - (π*k)/2 > 0. Let's solve for k:π/3 > (π*k)/2Multiply both sides by 2/π:2/3 > kSince k is an integer, the possible values of k are k ≤ 0.Let's consider k = 0:m = π/3 - 0 = π/3. That's positive.k = -1:m = π/3 - (π*(-1))/2 = π/3 + π/2 = (2π + 3π)/6 = 5π/6. Also positive.k = -2:m = π/3 - (π*(-2))/2 = π/3 + π = 4π/3. That's larger, but since we need the minimum m > 0, the smallest positive m would be π/3 when k = 0, but wait, hold on.Wait, but let's check if these values of m actually satisfy the condition. Because sometimes when dealing with trigonometric identities, especially when dealing with equations that hold for all x, there might be more constraints.Wait, let's consider k = 0: m = π/3. Then, substitute back into the original shifted function:y(x) = 2 sin(2x + (2π/3 - 2m)) = 2 sin(2x + 2π/3 - 2*(π/3)) = 2 sin(2x + 2π/3 - 2π/3) = 2 sin(2x). Is 2 sin(2x) an odd function? Yes, because sin is an odd function, so 2 sin(2*(-x)) = 2 sin(-2x) = -2 sin(2x) = -y(x). So, yes, m = π/3 makes y(x) an odd function.Similarly, for k = -1, m = 5π/6:y(x) = 2 sin(2x + 2π/3 - 2*(5π/6)) = 2 sin(2x + 2π/3 - 5π/3) = 2 sin(2x - π). But sin(2x - π) = sin(2x)cosπ - cos(2x)sinπ = -sin(2x). So, 2 sin(2x - π) = -2 sin(2x), which is also an odd function. Wait, but hold on. If y(x) = -2 sin(2x), then y(-x) = -2 sin(-2x) = 2 sin(2x) = -y(x). So, yes, that works as well. So, m = 5π/6 also works. However, since the problem asks for the minimum value of m > 0, π/3 is smaller than 5π/6, so π/3 would be the answer. But wait, the options given include π/3 as option C and 5π/6 as option B. But let's check the options again: the choices are A: π/6, B:5π/6, C:π/3, D:2π/3.Wait, but according to our calculation, m can be π/3, 5π/6, 4π/3, etc., but the minimum positive m is π/3, which is option C. However, wait, the answer options might have different values. Let me check the problem again.Wait, the original function f(x) is sqrt(3) cos 2x - sin 2x. When we shift it to become an odd function, perhaps the phase shift required is different. Alternatively, maybe my approach has an error. Let me double-check.Alternatively, perhaps using another method. Let's recall that shifting a function to make it odd. For the function f(x - m) to be odd, it must satisfy f(-x - m) = -f(x - m) for all x. Let's substitute into the original f.So, f(-x - m) = sqrt(3) cos[2(-x - m)] - sin[2(-x - m)] = sqrt(3) cos(-2x - 2m) - sin(-2x - 2m) = sqrt(3) cos(2x + 2m) + sin(2x + 2m). Because cos is even and sin is odd.On the other hand, -f(x - m) = -[sqrt(3) cos(2(x - m)) - sin(2(x - m))] = -sqrt(3) cos(2x - 2m) + sin(2x - 2m).Therefore, setting f(-x - m) = -f(x - m):sqrt(3) cos(2x + 2m) + sin(2x + 2m) = -sqrt(3) cos(2x - 2m) + sin(2x - 2m)We need this equality to hold for all x. Let's denote θ = 2x, so that the equation becomes:sqrt(3) cos(θ + 2m) + sin(θ + 2m) = -sqrt(3) cos(θ - 2m) + sin(θ - 2m)This must hold for all θ (since x is arbitrary, θ can be any real number). Let's expand both sides using trigonometric identities.Left-hand side (LHS):sqrt(3)[cosθ cos2m - sinθ sin2m] + [sinθ cos2m + cosθ sin2m]= sqrt(3) cosθ cos2m - sqrt(3) sinθ sin2m + sinθ cos2m + cosθ sin2mRight-hand side (RHS):- sqrt(3)[cosθ cos2m + sinθ sin2m] + [sinθ cos(-2m) + cosθ sin(-2m)]But cos(-2m) = cos2m and sin(-2m) = -sin2m. So:= -sqrt(3) cosθ cos2m - sqrt(3) sinθ sin2m + sinθ cos2m - cosθ sin2mNow, let's collect terms for LHS and RHS.LHS:cosθ [sqrt(3) cos2m + sin2m] + sinθ [-sqrt(3) sin2m + cos2m]RHS:cosθ [-sqrt(3) cos2m - sin2m] + sinθ [-sqrt(3) sin2m + cos2m]Wait, let me re-express both sides step by step.Starting with LHS:sqrt(3) cosθ cos2m + (-sqrt(3) sinθ sin2m) + sinθ cos2m + cosθ sin2mGrouping cosθ terms:cosθ [sqrt(3) cos2m + sin2m]Grouping sinθ terms:sinθ [ -sqrt(3) sin2m + cos2m ]RHS:- sqrt(3) cosθ cos2m - sqrt(3) sinθ sin2m + sinθ cos2m - cosθ sin2mGrouping cosθ terms:cosθ [ -sqrt(3) cos2m - sin2m ]Grouping sinθ terms:sinθ [ -sqrt(3) sin2m + cos2m ]So, equating LHS and RHS:For cosθ terms:sqrt(3) cos2m + sin2m = -sqrt(3) cos2m - sin2mAnd for sinθ terms:(-sqrt(3) sin2m + cos2m) = (-sqrt(3) sin2m + cos2m)Wait, the sinθ terms are the same on both sides. So that equation is automatically satisfied. But the cosθ terms must satisfy:sqrt(3) cos2m + sin2m = -sqrt(3) cos2m - sin2mBring all terms to the left side:sqrt(3) cos2m + sin2m + sqrt(3) cos2m + sin2m = 0Combine like terms:2 sqrt(3) cos2m + 2 sin2m = 0Divide both sides by 2:sqrt(3) cos2m + sin2m = 0So, sqrt(3) cos2m + sin2m = 0Let me write this as:sin2m + sqrt(3) cos2m = 0This is similar to the original expression for f(x), but here with variable m. Let me solve this equation for m.Divide both sides by 2 (but actually, we can express this as a single sine function again). Let me factor out a 2:Wait, actually, the equation is sin2m + sqrt(3) cos2m = 0. Let's write this as:sin2m = -sqrt(3) cos2mDivide both sides by cos2m (assuming cos2m ≠ 0):tan2m = -sqrt(3)So, tan2m = -sqrt(3). The solutions for 2m are:2m = arctan(-sqrt(3)) + kπ, where k is an integer.The principal value of arctan(-sqrt(3)) is -π/3, since tan(-π/3) = -sqrt(3). Therefore, all solutions:2m = -π/3 + kπTherefore, solving for m:m = (-π/6) + (kπ)/2Since m > 0, we need to find the smallest positive m. Let's solve for m:m = (kπ)/2 - π/6 > 0(kπ)/2 > π/6Multiply both sides by 6/π:3k > 1k > 1/3Since k is an integer, the smallest integer k satisfying this is k = 1.Therefore, substituting k = 1:m = (1*π)/2 - π/6 = π/2 - π/6 = (3π/6 - π/6) = 2π/6 = π/3.Next integer k = 2:m = (2π)/2 - π/6 = π - π/6 = 5π/6.k = 3:m = (3π)/2 - π/6 = (9π/6 - π/6) = 8π/6 = 4π/3, etc.Therefore, the minimal positive m is π/3. So, the answer should be option C: π/3. But wait, in the options given, π/3 is option C. But wait, when we first considered writing f(x) as a sine function, we found that shifting m = π/3 makes y(x) = 2 sin(2x), which is indeed odd. So that seems to check out. However, let me verify this with the original function.Given f(x - m) = sqrt(3) cos[2(x - m)] - sin[2(x - m)] = sqrt(3) cos(2x - 2m) - sin(2x - 2m). If m = π/3, then:f(x - π/3) = sqrt(3) cos(2x - 2π/3) - sin(2x - 2π/3)Let me compute this expression. Let's compute each term:cos(2x - 2π/3) = cos2x cos2π/3 + sin2x sin2π/3 = cos2x*(-1/2) + sin2x*(sqrt(3)/2)Similarly, sin(2x - 2π/3) = sin2x cos2π/3 - cos2x sin2π/3 = sin2x*(-1/2) - cos2x*(sqrt(3)/2)Therefore,sqrt(3) cos(2x - 2π/3) = sqrt(3)[ (-1/2) cos2x + (sqrt(3)/2) sin2x ] = -sqrt(3)/2 cos2x + (3/2) sin2x- sin(2x - 2π/3) = -[ (-1/2 sin2x - sqrt(3)/2 cos2x ) ] = (1/2) sin2x + sqrt(3)/2 cos2xAdding these two results:(-sqrt(3)/2 cos2x + 3/2 sin2x) + (1/2 sin2x + sqrt(3)/2 cos2x) = [(-sqrt(3)/2 + sqrt(3)/2) cos2x] + [(3/2 + 1/2) sin2x] = 0 cos2x + 2 sin2x = 2 sin2xSo, f(x - π/3) = 2 sin2x, which is indeed an odd function, since sin(-2x) = -sin2x. Therefore, f(-x - π/3) = 2 sin(-2x) = -2 sin2x = -f(x - π/3). Therefore, it satisfies the odd function condition. So, m = π/3 works.But wait, the answer options include D: 2π/3. So why is that? Did I make a mistake here? Wait, according to the previous calculation, m = π/3 is the minimal solution. But maybe there is an error in the process. Let me check again.Alternatively, perhaps when I shifted f(x) to f(x - m), I might have m in the opposite direction. Wait, the function is shifted to the right by m, so f(x - m) corresponds to a shift of m units to the right. However, when we considered the expression as a sine function, we had f(x) = 2 sin[2(x + π/3)]. Therefore, shifting right by m would give 2 sin[2((x - m) + π/3)] = 2 sin[2x - 2m + 2π/3]. For this to be an odd function, which is 2 sin[2x + φ], we need the phase shift φ to make it an odd function. Since sin is odd, if the phase shift is 0, then it's odd. So, we need:-2m + 2π/3 = 0 + k*2π, but wait, actually, sin(2x + φ) is odd if φ is a multiple of π. Wait, no. Wait, sin(2x + φ) is odd if sin(-2x + φ) = -sin(2x + φ). Let's check:sin(-2x + φ) = -sin(2x - φ). So, setting this equal to -sin(2x + φ):-sin(2x - φ) = -sin(2x + φ)So, sin(2x - φ) = sin(2x + φ). For this to hold for all x, the arguments must satisfy 2x - φ = 2x + φ + 2πk, which implies -φ = φ + 2πk => -2φ = 2πk => φ = -πk. Alternatively, they could be supplementary angles: 2x - φ = π - (2x + φ) + 2πk. But this would lead to 4x = π + 2πk, which can't hold for all x. Therefore, the only possibility is that φ = -πk. Therefore, φ must be an integer multiple of π. Therefore, in our case, the phase after shifting is φ = 2π/3 - 2m. Therefore, to have φ = -πk, we set 2π/3 - 2m = -πk => 2m = 2π/3 + πk => m = π/3 + (πk)/2. Since m > 0, the minimal m is when k=0: m=π/3. If k=-1, m=π/3 - π/2= -π/6, which is negative. So, the minimal positive m is π/3. Hence, the answer should be C: π/3.But the options also include D: 2π/3. Wait, perhaps the question is written differently? Wait, let me check again:The problem states: y = f(x - m) is an odd function, m > 0. So, after shifting f(x) to the right by m, it becomes odd. So, according to all the calculations, m=π/3 is the minimal solution. However, in the answer options, C is π/3 and D is 2π/3. So, the answer should be C. But let me check another approach to be safe.Another method: For f(x - m) to be odd, it must satisfy f(-x - m) = -f(x - m) for all x. Let me take specific values of x to find m. For example, let x = 0:f(-0 - m) = -f(0 - m) => f(-m) = -f(-m). So, f(-m) = -f(-m) => 2f(-m) = 0 => f(-m) = 0.Therefore, a necessary condition is that f(-m) = 0. Let's compute f(-m):f(-m) = sqrt(3) cos(-2m) - sin(-2m) = sqrt(3) cos2m + sin2m. Therefore:sqrt(3) cos2m + sin2m = 0Which is the same equation we derived earlier. So, solving sqrt(3) cos2m + sin2m = 0 leads to tan2m = -sqrt(3), which gives 2m = -π/3 + kπ, so m = -π/6 + kπ/2. For m > 0, minimal m is when k=1: m = -π/6 + π/2 = ( -π/6 + 3π/6 ) = 2π/6 = π/3. So again, the minimal m is π/3. Thus, the answer is C.But perhaps there is a mistake here? Let me check with m=π/3:Compute f(-m) where m=π/3: f(-π/3) = sqrt(3) cos(-2π/3) - sin(-2π/3) = sqrt(3) cos(2π/3) + sin(2π/3). cos(2π/3) = -1/2, sin(2π/3)=sqrt(3)/2. Therefore:sqrt(3)*(-1/2) + sqrt(3)/2 = -sqrt(3)/2 + sqrt(3)/2 = 0. So, f(-π/3)=0, which is required. Then, also check another x value, say x=π/6:Compute f(-π/6 - m) and -f(π/6 - m). Let m=π/3:Left side: f(-π/6 - π/3) = f(-π/2) = sqrt(3) cos(-π) - sin(-π) = sqrt(3)(-1) - 0 = -sqrt(3)Right side: -f(π/6 - π/3) = -f(-π/6) = -[sqrt(3) cos(-π/3) - sin(-π/3)] = -[sqrt(3)(1/2) + sqrt(3)/2] = -[sqrt(3)/2 + sqrt(3)/2] = -sqrt(3)So, equality holds here. Another test: x=π/4.Left side: f(-π/4 - π/3) = f(-7π/12) = sqrt(3) cos(-7π/6) - sin(-7π/6) = sqrt(3) cos(7π/6) + sin(7π/6)cos(7π/6)= -sqrt(3)/2, sin(7π/6)= -1/2. Therefore:sqrt(3)*(-sqrt(3)/2) + (-1/2) = (-3/2) -1/2 = -2Right side: -f(π/4 - π/3) = -f(-π/12) = -[sqrt(3) cos(-π/6) - sin(-π/6)] = -[sqrt(3)(sqrt(3)/2) + 1/2] = -[3/2 + 1/2] = -[2] = -2Equality holds. Therefore, m=π/3 is correct. Therefore, the answer is C: π/3.But wait, in the options given in the problem statement:A: π/6B:5π/6C:π/3D:2π/3Therefore, the correct answer is C. However, I recall that sometimes phase shifts in trigonometric functions can have periodicity considerations. Wait, the original function has a period of π (since it's a function of 2x). So, the minimal period is π. Therefore, shifting by π would repeat the function, but we are looking for the minimal positive m. Since m=π/3 is less than π/2, which is less than the period π, it is indeed the minimal. So, why is there an option D: 2π/3? Maybe the answer is different. Wait, let me check the process again.Wait, when I solved 2m = -π/3 + kπ, and found m = -π/6 + kπ/2. For k=1, m=π/3, which is positive. For k=2, m= (2π)/2 - π/6 = π - π/6 = 5π/6. For k=3, m= (3π)/2 - π/6 = 4π/3. So, these are the possible positive m's. But the minimal is π/3, so the answer should be C. But maybe the problem is in the fact that f(x - m) being odd requires more than just f(-m) = 0. We checked for x=0, x=π/6, x=π/4, and it worked. But to be thorough, let's check another x.Take x=π/12:Left side: f(-π/12 - m) = f(-π/12 - π/3) = f(-5π/12)Compute f(-5π/12) = sqrt(3) cos(-5π/6) - sin(-5π/6) = sqrt(3) cos(5π/6) + sin(5π/6)cos(5π/6) = -sqrt(3)/2, sin(5π/6) = 1/2Thus, sqrt(3)*(-sqrt(3)/2) + 1/2 = (-3/2) + 1/2 = -1Right side: -f(π/12 - π/3) = -f(-π/4) = -[sqrt(3) cos(-π/2) - sin(-π/2)] = -[sqrt(3)*0 + 1] = -1Equality holds. Seems consistent. Thus, I think the correct answer is C: π/3. So why is there a discrepancy with the options? Wait, in the problem statement, the options are:A: π/6B:5π/6C:π/3D:2π/3Therefore, the answer should be C. However, maybe there's a mistake in the problem statement or my solution.Wait, let me verify once more using the original expression for f(x - m). If m=π/3, then:f(x - π/3) = sqrt(3) cos[2(x - π/3)] - sin[2(x - π/3)] = sqrt(3) cos(2x - 2π/3) - sin(2x - 2π/3)As computed earlier, this simplifies to 2 sin2x, which is odd. Thus, m=π/3 is correct. Alternatively, if m=2π/3, then:f(x - 2π/3) = sqrt(3) cos[2x - 4π/3] - sin[2x - 4π/3]Let me compute this:sqrt(3) cos(2x - 4π/3) - sin(2x - 4π/3)cos(2x - 4π/3) = cos2x cos4π/3 + sin2x sin4π/3 = cos2x*(-1/2) + sin2x*(-sqrt(3)/2)sin(2x - 4π/3) = sin2x cos4π/3 - cos2x sin4π/3 = sin2x*(-1/2) - cos2x*(-sqrt(3)/2) = -sin2x/2 + sqrt(3)/2 cos2xTherefore,sqrt(3)[ -1/2 cos2x - sqrt(3)/2 sin2x ] - [ -sin2x/2 + sqrt(3)/2 cos2x ]= -sqrt(3)/2 cos2x - 3/2 sin2x + sin2x/2 - sqrt(3)/2 cos2xCombine like terms:cos2x: [ -sqrt(3)/2 - sqrt(3)/2 ] = -sqrt(3) cos2xsin2x: [ -3/2 + 1/2 ] = -1 sin2xThus, f(x - 2π/3) = -sqrt(3) cos2x - sin2x. Is this an odd function?Let's check if it's odd:f(-x - 2π/3) = sqrt(3) cos[2(-x) - 4π/3] - sin[2(-x) - 4π/3] = sqrt(3) cos(-2x - 4π/3) - sin(-2x - 4π/3) = sqrt(3) cos(2x + 4π/3) + sin(2x + 4π/3)Compare to -f(x - 2π/3) = sqrt(3) cos2x + sin2xSo, we need sqrt(3) cos(2x + 4π/3) + sin(2x + 4π/3) = sqrt(3) cos2x + sin2x. Is this true?Let's compute the left-hand side:sqrt(3) cos(2x + 4π/3) + sin(2x + 4π/3)Expand using angle addition:sqrt(3)[cos2x cos4π/3 - sin2x sin4π/3] + [sin2x cos4π/3 + cos2x sin4π/3]cos4π/3 = -1/2, sin4π/3 = -sqrt(3)/2.So:sqrt(3)[cos2x*(-1/2) - sin2x*(-sqrt(3)/2)] + [sin2x*(-1/2) + cos2x*(-sqrt(3)/2)]= sqrt(3)[ -1/2 cos2x + (sqrt(3)/2) sin2x ] - 1/2 sin2x - sqrt(3)/2 cos2x= -sqrt(3)/2 cos2x + (3/2) sin2x - 1/2 sin2x - sqrt(3)/2 cos2xCombine terms:cos2x: [ -sqrt(3)/2 - sqrt(3)/2 ] = -sqrt(3) cos2xsin2x: [ 3/2 - 1/2 ] = sin2xThus, left-hand side simplifies to -sqrt(3) cos2x + sin2x. But the right-hand side is sqrt(3) cos2x + sin2x. Therefore, unless cos2x =0, these are not equal. Hence, unless for specific x, but since the equality must hold for all x, m=2π/3 does not satisfy the condition. Therefore, m=2π/3 is not a solution, confirming that the minimal m is indeed π/3.Therefore, the correct answer is C: π/3. So why does option D:2π/3 exist? Maybe a common mistake when solving the equation tan2m = -sqrt(3), getting 2m = 2π/3, hence m=π/3, but perhaps others might erroneously solve it differently. Alternatively, perhaps in the problem statement there's a typo, but based on the calculations, the answer is C.Final Answerboxed{C}</think>Given ( f(x) = sqrt{3} cos 2x - sin 2x ), we need to find the minimum value of ( m > 0 ) such that ( y = f(x - m) ) is an odd function. An odd function satisfies ( g(-x) = -g(x) ) for all ( x ). Therefore, ( f(-x - m) = -f(x - m) ) must hold for all ( x ).First, we rewrite ( f(x) ) in a simplified form using the amplitude-phase identity:[ f(x) = 2 sinleft(2x + frac{2pi}{3}right) ]To determine the shift ( m ) such that ( f(x - m) ) is odd, we consider the shifted function:[ f(x - m) = 2 sinleft(2(x - m) + frac{2pi}{3}right) = 2 sinleft(2x - 2m + frac{2pi}{3}right) ]For this to be odd, the phase shift must satisfy:[ -2m + frac{2pi}{3} = kpi ]where ( k ) is an integer. Solving for ( m ):[ -2m + frac{2pi}{3} = kpi ][ 2m = frac{2pi}{3} - kpi ][ m = frac{pi}{3} - frac{kpi}{2} ]We need ( m > 0 ). The smallest positive ( m ) occurs when ( k = 0 ):[ m = frac{pi}{3} ]Verification shows that ( f(x - frac{pi}{3}) = 2 sin(2x) ), which is indeed an odd function. Thus, the minimum value of ( m ) is ( frac{pi}{3} ).[boxed{C}]]

🔑:Since f(x)= sqrt {3}cos 2x-sin 2x=2left( dfrac { sqrt {3}}{2}cos 2x- dfrac {1}{2}sin 2xright)=2cos left(2x+ dfrac {pi}{6}right), then y=f(x-m)=2cos left[2(x-m)+ dfrac {pi}{6}right], Since y=f(x-m)(m > 0) is an odd function, then dfrac {pi}{6}-2m=kpi+ dfrac {pi}{2}(kinmathbb{Z}), thus 2m=-kpi- dfrac {pi}{3}(kinmathbb{Z}), and since m > 0, obviously, when k=-1, m reaches its minimum value, which is dfrac {pi}{3}. Therefore, the answer is: boxed{C}. By applying trigonometric identities, we can derive f(x)=2cos left(2x+ dfrac {pi}{6}right), and thus f(x-m)=2cos left[2(x-m)+ dfrac {pi}{6}right]. Using the fact that y=f(x-m)(m > 0) is an odd function, we can find the answer. This question examines the application of trigonometric identities and the properties of the cosine function, highlighting the examination of the conversion of odd and even properties of sine and cosine functions, and is considered a medium-level question.

🧠:[<think>Okay, let's see. I need to solve these two parts about the function f(x) = 4 cos x sin(x + π/6). Hmm, first, the axis of symmetry. Then, the maximum and minimum values on a specific interval. Alright, let me start with part (1). Axis of symmetry for a function usually refers to a vertical line x = a such that the function is symmetric about that line. So, for any h, f(a + h) = f(a - h). That means the graph of the function mirrors around x = a. I need to find such a line for this trigonometric function. First, maybe I should simplify f(x) using trigonometric identities. The product of sine and cosine can be converted into a sum. Let me recall the identity: sin A cos B = [sin(A + B) + sin(A - B)] / 2. Wait, but here it's 4 cos x sin(x + π/6). Let me set A = x + π/6 and B = x. Then, sin(A) cos(B) = [sin(A + B) + sin(A - B)] / 2. So substituting in A and B: sin(x + π/6) cos x = [sin((x + π/6) + x) + sin((x + π/6) - x)] / 2. Simplify that:= [sin(2x + π/6) + sin(π/6)] / 2. Therefore, f(x) = 4 * [sin(2x + π/6) + sin(π/6)] / 2 = 2 sin(2x + π/6) + 2 sin(π/6). Since sin(π/6) is 1/2, so 2 sin(π/6) = 2*(1/2) = 1. So f(x) simplifies to 2 sin(2x + π/6) + 1. That seems better. Now, this is a sine function with some transformations. The standard form is A sin(Bx + C) + D. Let's see:Amplitude is 2, period is 2π / 2 = π, phase shift is -C/B = -(π/6)/2 = -π/12, and vertical shift is 1. But how does this help me find the axis of symmetry? For a sine function, which is normally symmetric about its midline, but here it's shifted and scaled. However, sine functions themselves are not symmetric about a vertical axis unless they are shifted. Wait, maybe if there's a phase shift, the axis of symmetry would be at the point where the sine curve reaches its maximum or midline? Hmm, not sure. Alternatively, perhaps the function f(x) has an axis of symmetry if it's a transformed sine function that is symmetric about some line. For example, if we have f(a + h) = f(a - h), then x = a is the axis. Let me suppose that the axis of symmetry is at x = a, then 2 sin(2(a + h) + π/6) + 1 = 2 sin(2(a - h) + π/6) + 1. So, sin(2a + 2h + π/6) = sin(2a - 2h + π/6). For this to hold for all h, the arguments inside the sine must either be equal modulo 2π or be supplementary angles modulo 2π. But since this has to hold for all h, the only possibility is that the arguments are equal for all h, which would require that 2a + 2h + π/6 = 2a - 2h + π/6 + 2π k, where k is an integer. Simplifying:2a + 2h + π/6 = 2a - 2h + π/6 + 2π kSubtract 2a and π/6 from both sides:2h = -2h + 2π kSo 4h = 2π k => h = (π k)/2. But this needs to hold for all h, which is only possible if k=0, which gives 4h = 0 => h=0. But that's trivial. Therefore, the only solution is when the sine function is symmetric about some axis. Alternatively, maybe the axis of symmetry is at the peak or trough of the sine wave?Wait, another approach. Since the function is a sinusoidal function, its graph is a sine wave with vertical shift and phase shift. The axis of symmetry would correspond to the line around which the function is symmetric. For a standard sine wave, it's symmetric about its midline, but that's a horizontal line, not a vertical one. Wait, no—the question is about a vertical line, an axis of symmetry, which is a vertical line x = a.Wait, perhaps the function is symmetric about x = a if shifting the function left or right by a certain amount makes it an even function. For example, if we can write f(a + h) = f(a - h), which is the definition. So, to find such an a, maybe set up the equation f(a + h) = f(a - h) for all h, then solve for a. Given that f(x) = 2 sin(2x + π/6) + 1, then f(a + h) = 2 sin(2(a + h) + π/6) + 1 and f(a - h) = 2 sin(2(a - h) + π/6) + 1. Set them equal:sin(2a + 2h + π/6) = sin(2a - 2h + π/6)For this identity to hold for all h, the arguments must satisfy either:1. 2a + 2h + π/6 = 2a - 2h + π/6 + 2πk, or2. 2a + 2h + π/6 = π - (2a - 2h + π/6) + 2πkfor some integer k, for all h. Let's check the first case:1. 2a + 2h + π/6 = 2a - 2h + π/6 + 2πkSubtract 2a and π/6:2h = -2h + 2πk4h = 2πk => 2h = πkBut this needs to hold for all h, which is only possible if k=0, which would imply 2h = 0 => h=0. So only the trivial solution, which is not valid for all h. Second case:2a + 2h + π/6 = π - 2a + 2h - π/6 + 2πkWait, let me do it step by step. The supplementary angle identity:sin(θ) = sin(π - θ). So, if sin(α) = sin(β), then either α = β + 2πk or α = π - β + 2πk. So applying that here:2a + 2h + π/6 = π - (2a - 2h + π/6) + 2πkSimplify the right-hand side:= π - 2a + 2h - π/6 + 2πkCombine constants:π - π/6 = 5π/6, so:= 5π/6 - 2a + 2h + 2πkNow, set equal to left-hand side:2a + 2h + π/6 = 5π/6 - 2a + 2h + 2πkSubtract 2h from both sides:2a + π/6 = 5π/6 - 2a + 2πkBring terms with a to left and constants to right:2a + 2a = 5π/6 - π/6 + 2πk4a = 4π/6 + 2πkSimplify 4π/6 to 2π/3:4a = 2π/3 + 2πkDivide both sides by 4:a = π/6 + (πk)/2So, the possible values for a are π/6 + (πk)/2 for integer k. But since we are looking for an axis of symmetry, it's typically the principal value. Let me check for k=0: a = π/6. For k=1: a = π/6 + π/2 = 2π/3. For k=-1: a = π/6 - π/2 = -π/3. But does this hold for all h? Wait, in the equation we derived, a is expressed in terms of k, but the equation was derived from the condition that f(a + h) = f(a - h) for all h. Therefore, the axis of symmetry must be at x = π/6 + (πk)/2 for some integer k. However, the function is periodic with period π (since the original function f(x) = 2 sin(2x + π/6) + 1 has period π). Therefore, it's going to have infinitely many axes of symmetry, each separated by π/2. But since the question asks for "the" axis of symmetry, perhaps the principal one, the one closest to the origin or within a standard interval. Wait, but maybe the problem is expecting a specific answer. Let me check. If k=0, then a = π/6. Let's test if x = π/6 is an axis of symmetry. Let's pick h = π/6. Then f(π/6 + π/6) = f(π/3) and f(π/6 - π/6) = f(0). Let's compute both:First, f(0) = 4 cos 0 sin(0 + π/6) = 4*1*(1/2) = 2.f(π/3) = 4 cos(π/3) sin(π/3 + π/6) = 4*(1/2) sin(π/2) = 4*(1/2)*1 = 2. So in this case, f(π/6 + π/6) = f(π/6 - π/6) = 2. That works. Let's try another h. Let's take h = π/12. Then f(π/6 + π/12) = f(π/4), and f(π/6 - π/12) = f(π/12). Compute both:f(π/4) = 4 cos(π/4) sin(π/4 + π/6). Compute π/4 + π/6 = (3π/12 + 2π/12) = 5π/12. So sin(5π/12) is sin(75°), which is (√6 + √2)/4. Then cos(π/4) is √2/2. So 4*(√2/2)* (√6 + √2)/4 = (√2)*(√6 + √2)/2 = (√12 + √4)/2 = (2√3 + 2)/2 = √3 + 1 ≈ 1.732 + 1 = 2.732.f(π/12) = 4 cos(π/12) sin(π/12 + π/6) = 4 cos(π/12) sin(π/4). Compute cos(π/12) = cos(15°) = (√6 + √2)/4. So sin(π/4) is √2/2. So 4*( (√6 + √2)/4 )*(√2/2) = (√6 + √2)*(√2/2) = (√12 + √4)/2 = (2√3 + 2)/2 = √3 + 1 ≈ same as above. So yes, f(π/4) = f(π/12) = √3 + 1. So x = π/6 seems to be an axis of symmetry. Let me check h = π/3. f(π/6 + π/3) = f(π/2), f(π/6 - π/3) = f(-π/6). Compute both:f(π/2) = 4 cos(π/2) sin(π/2 + π/6) = 4*0*sin(2π/3) = 0.f(-π/6) = 4 cos(-π/6) sin(-π/6 + π/6) = 4*(√3/2)*sin(0) = 4*(√3/2)*0 = 0. So f(π/2) = f(-π/6) = 0. So that works as well. Therefore, x = π/6 is indeed an axis of symmetry. Similarly, if we take a = π/6 + π/2 = 2π/3, would that also be an axis of symmetry? Let's test with h=π/6:f(2π/3 + π/6) = f(5π/6), f(2π/3 - π/6) = f(π/2). Compute f(5π/6) = 4 cos(5π/6) sin(5π/6 + π/6) = 4*(-√3/2) sin(π) = 4*(-√3/2)*0 = 0.f(π/2) = 0 as before. So that's equal. Let's try h=π/12:f(2π/3 + π/12) = f(3π/4), f(2π/3 - π/12) = f(7π/12). Compute both:f(3π/4) = 4 cos(3π/4) sin(3π/4 + π/6) = 4*(-√2/2) sin(11π/12). Sin(11π/12) is sin(165°) = sin(15°) = (√6 - √2)/4. So 4*(-√2/2)*(√6 - √2)/4 = (-√2/2)*(√6 - √2) = (-√12 + √4)/2 = (-2√3 + 2)/2 = -√3 + 1.f(7π/12) = 4 cos(7π/12) sin(7π/12 + π/6) = 4 cos(7π/12) sin(3π/4). Cos(7π/12) is cos(105°) = -cos(75°) = -(√6 - √2)/4. Sin(3π/4) is √2/2. So 4*(-(√6 - √2)/4)*(√2/2) = -(√6 - √2)*(√2/2) = -(√12 - √4)/2 = -(2√3 - 2)/2 = -√3 + 1. So yes, they are equal. Therefore, x = 2π/3 is also an axis of symmetry. But since the problem is asking for an axis of symmetry, maybe it's expecting the first one, which is x = π/6. The problem might not specify multiple axes because trigonometric functions have infinitely many, but perhaps in the context of the problem, they just want the principal one. Alternatively, maybe the function has only one axis of symmetry in its fundamental period? Wait, the period is π, so between -π/2 to π/2 or something? Hmm, not sure. But since when k=0, we get x = π/6, which is a valid axis of symmetry, and the problem probably expects that. Let me check if there's another way. Alternatively, maybe we can use the original expression f(x) = 4 cos x sin(x + π/6). Let me expand sin(x + π/6):sin(x + π/6) = sin x cos π/6 + cos x sin π/6 = sin x (√3/2) + cos x (1/2). Therefore, f(x) = 4 cos x [ (√3/2) sin x + (1/2) cos x ] = 4 [ (√3/2 sin x cos x) + (1/2 cos² x) ] = 4*(√3/2 sin x cos x + 1/2 cos² x) = 2√3 sin x cos x + 2 cos² x. Hmm, now we can use double-angle identities. Let's see:sin x cos x = (sin 2x)/2, and cos² x = (1 + cos 2x)/2. So substituting:f(x) = 2√3*(sin 2x / 2) + 2*( (1 + cos 2x)/2 ) = √3 sin 2x + 1 + cos 2x. So f(x) = √3 sin 2x + cos 2x + 1. Alternatively, this can be written as a single sine or cosine function. Let me consider √3 sin 2x + cos 2x. The amplitude would be √( (√3)^2 + 1^2 ) = √(3 + 1) = 2. So √3 sin 2x + cos 2x = 2 sin(2x + φ), where φ is the phase shift. Let's compute φ:We have sin φ = adjacent / amplitude = 1/2, cos φ = √3/2. Therefore, φ = π/6. So:√3 sin 2x + cos 2x = 2 sin(2x + π/6). Therefore, f(x) = 2 sin(2x + π/6) + 1, which matches what we had earlier. So f(x) is 2 sin(2x + π/6) + 1. Now, to find the axis of symmetry, we can think of the sine function's properties. A standard sine function y = sin x is symmetric about the origin, but when shifted, the symmetry might change. However, our function is a transformed sine function. If we can write it as a sine function with a phase shift, then perhaps its axis of symmetry is where it reaches its maximum or midline. Wait, but symmetry about a vertical line. For example, the standard sine function has no vertical axis of symmetry, but a cosine function is symmetric about the y-axis. Wait, but cosine is just a shifted sine. Alternatively, if we have a function like y = sin(Bx + C), it can be symmetric about x = -C/B if we adjust the phase shift. Wait, for example, sin(2x + π/6) can be written as sin[2(x + π/12)]. So this is a sine function with horizontal compression by factor 2 and shifted left by π/12. The standard sine function is symmetric about the origin, but after shifting left by π/12, the point of symmetry (origin) moves to x = -π/12. However, sine functions are odd functions, so if you shift them, they may not have vertical symmetry axes unless specific conditions are met. Alternatively, if we can write the function as an even function after a horizontal shift. That is, if f(a + h) = f(a - h) for all h, then x = a is the axis of symmetry. To find such a, let's set f(a + h) = f(a - h). Using the expression f(x) = 2 sin(2x + π/6) + 1, we have:2 sin(2(a + h) + π/6) + 1 = 2 sin(2(a - h) + π/6) + 1Subtract 1 and divide by 2:sin(2a + 2h + π/6) = sin(2a - 2h + π/6)As before, for this equality to hold for all h, which only occurs if the function is symmetric. The general solution is either:1. 2a + 2h + π/6 = 2a - 2h + π/6 + 2πk (which leads to 4h = 2πk, which can't hold for all h unless k=0, leading to h=0, trivial), or2. 2a + 2h + π/6 = π - (2a - 2h + π/6) + 2πmWhich simplifies to:2a + 2h + π/6 = π - 2a + 2h - π/6 + 2πmCancel 2h:2a + π/6 = π - 2a - π/6 + 2πmBring terms with a to left, constants to right:2a + 2a = π - π/6 - π/6 + 2πm4a = π - (2π/6) + 2πmSimplify:4a = π - π/3 + 2πm = (2π/3) + 2πmTherefore,a = (2π/3 + 2πm)/4 = π/6 + π m/2Which is the same result as before. So a = π/6 + π m/2, where m is any integer. So the axes of symmetry are at x = π/6 + π m/2. Hence, the function has infinitely many axes of symmetry spaced by π/2, starting at π/6. Therefore, the primary axis of symmetry would be x = π/6. Therefore, part (1)'s answer is x = π/6. Let me confirm once more. When we checked specific instances earlier, like h = π/6, h = π/12, h = π/3, it worked. So yes, x = π/6 is indeed an axis of symmetry. Moving on to part (2): Find the maximum and minimum values of f(x) in the interval [-π/6, π/4]. Given that f(x) = 2 sin(2x + π/6) + 1, the maximum and minimum of this function would depend on the range of the sine function. Since the amplitude is 2, the maximum of sin is 1, so maximum of f(x) is 2*1 + 1 = 3, and minimum is 2*(-1) + 1 = -1. However, these are the global maxima and minima. But since we're restricted to the interval [-π/6, π/4], we need to check if the function attains these extrema within that interval. First, let's find the critical points by taking the derivative and setting it to zero. Then evaluate f(x) at critical points and endpoints. Given f(x) = 2 sin(2x + π/6) + 1. The derivative f'(x) = 2*cos(2x + π/6)*2 = 4 cos(2x + π/6). Set derivative to zero:4 cos(2x + π/6) = 0 => cos(2x + π/6) = 0. Solutions for 2x + π/6 = π/2 + π k, where k is integer. Therefore,2x + π/6 = π/2 + π k=> 2x = π/2 - π/6 + π k=> 2x = (3π/6 - π/6) + π k = (2π/6) + π k = π/3 + π k=> x = π/6 + π k/2So critical points are at x = π/6 + π k/2. Let's find which of these critical points lie within the interval [-π/6, π/4]. First, list possible k values:For k = 0: x = π/6 ≈ 0.523. But π/4 ≈ 0.785, so π/6 is within the interval.For k = -1: x = π/6 - π/2 = -π/3 ≈ -1.047, which is less than -π/6 ≈ -0.523, so outside the interval.For k = 1: x = π/6 + π/2 = 2π/3 ≈ 2.094, which is greater than π/4 ≈ 0.785, so outside the interval.Thus, the only critical point within the interval is x = π/6. Therefore, we need to evaluate f(x) at the endpoints x = -π/6 and x = π/4, and at the critical point x = π/6. Compute f(-π/6):f(-π/6) = 2 sin(2*(-π/6) + π/6) + 1 = 2 sin(-π/3 + π/6) + 1 = 2 sin(-π/6) + 1 = 2*(-1/2) + 1 = -1 + 1 = 0.Compute f(π/6):f(π/6) = 2 sin(2*(π/6) + π/6) + 1 = 2 sin(π/3 + π/6) + 1 = 2 sin(π/2) + 1 = 2*1 + 1 = 3.Compute f(π/4):f(π/4) = 2 sin(2*(π/4) + π/6) + 1 = 2 sin(π/2 + π/6) + 1 = 2 sin(2π/3) + 1 = 2*(√3/2) + 1 = √3 + 1 ≈ 1.732 + 1 = 2.732.Therefore, among these values:f(-π/6) = 0,f(π/6) = 3,f(π/4) ≈ 2.732.Therefore, the maximum value is 3 at x = π/6, and the minimum value is 0 at x = -π/6. However, wait a second—is there a possibility of another critical point within the interval? The critical points are x = π/6 + π k/2. For k = 0, x = π/6 is inside. For k = -1, x = -π/3 is outside. For k =1, x = 2π/3 is outside. So only x = π/6 is the critical point inside the interval. So the minimum is at x = -π/6 (0), and the maximum at x = π/6 (3). However, wait, f(x) is 2 sin(2x + π/6) +1. The range of sin is between -1 and 1, so f(x) is between -1 and 3. But maybe within the interval, the function doesn't reach -1. Let's check if there's a point where f(x) = -1. For that, 2 sin(2x + π/6) +1 = -1 => sin(2x + π/6) = -1. So 2x + π/6 = 3π/2 + 2π k => 2x = 3π/2 - π/6 + 2π k = (9π/6 - π/6) + 2π k = 8π/6 + 2π k = 4π/3 + 2π k => x = 2π/3 + π k. Within our interval [-π/6, π/4], x = 2π/3 ≈ 2.094 is outside, and x = 2π/3 - π = -π/3 ≈ -1.047 is also outside. So the minimum of f(x) is not attained in the interval; the lowest value inside the interval is 0 at x = -π/6, and the next candidate is the critical point x = π/6 which gives the maximum. But just to make sure, let's see if the function decreases from x = -π/6 to x = π/6, or increases. Since the critical point at x = π/6 is a maximum. Let's check the derivative around that point. Take x slightly less than π/6, say x = π/6 - ε. Then 2x + π/6 = 2*(π/6 - ε) + π/6 = π/3 - 2ε + π/6 = π/2 - 2ε. So cos(π/2 - 2ε) = sin(2ε) > 0, so derivative is positive. Therefore, function is increasing before x = π/6. After x = π/6, take x = π/6 + ε. Then 2x + π/6 = π/3 + 2ε + π/6 = π/2 + 2ε. cos(π/2 + 2ε) = -sin(2ε) < 0. So derivative is negative. Therefore, x = π/6 is a local maximum. Since the function increases from x = -π/6 to x = π/6 and then decreases from x = π/6 to x = π/4, the maximum is at x = π/6 (3) and the minimum is at the left endpoint x = -π/6 (0). However, wait, let's compute f(x) at the right endpoint x = π/4. We did that earlier and got √3 + 1 ≈ 2.732. So that's less than 3. Therefore, yes, the maximum is 3, the minimum is 0. But hold on, let me check the other expression of f(x) to compute f(-π/6). Original function was 4 cos x sin(x + π/6). So at x = -π/6:f(-π/6) = 4 cos(-π/6) sin(-π/6 + π/6) = 4*(√3/2)*sin(0) = 4*(√3/2)*0 = 0. Correct.At x = π/6:f(π/6) = 4 cos(π/6) sin(π/6 + π/6) = 4*(√3/2)*sin(π/3) = 4*(√3/2)*(√3/2) = 4*(3/4) = 3. Correct.At x = π/4:f(π/4) = 4 cos(π/4) sin(π/4 + π/6) = 4*(√2/2)*sin(5π/12). Sin(5π/12) is sin(75°) = (√6 + √2)/4. So 4*(√2/2)*(√6 + √2)/4 = (√2/2)*(√6 + √2) = (√12 + √4)/2 = (2√3 + 2)/2 = √3 + 1 ≈ 2.732. Correct.So yes, within the interval, the function reaches its maximum at x = π/6, which is 3, and the minimum at x = -π/6, which is 0. Therefore, the answer for part (2) is maximum 3 and minimum 0. Wait, but hold on. Let's think again. The function f(x) = 2 sin(2x + π/6) +1. The sine function normally has a range of [-1,1], so 2 sin(θ) +1 has range [-1,3]. But is there a possibility that in the interval [-π/6, π/4], the argument 2x + π/6 ranges from 2*(-π/6) + π/6 = -π/6 + π/6 = 0 to 2*(π/4) + π/6 = π/2 + π/6 = 2π/3. So the argument goes from 0 to 2π/3. So sin(θ) for θ from 0 to 2π/3. The maximum sin(θ) is at θ = π/2 (which is within 0 to 2π/3), sin(π/2) = 1. The minimum sin(θ) in this interval is at θ = 0, sin(0) = 0, but wait, θ goes from 0 to 2π/3. The minimum sin(theta) is actually at theta = 0, which is 0. So the minimum value of 2 sin(theta) +1 would be 2*0 +1 =1, but wait, but theta ranges from 0 to 2π/3. Wait, sin(theta) in [0, 2π/3] ranges from 0 to 1 (since at theta=0, sin=0; theta=pi/2, sin=1; theta=2pi/3, sin=sqrt(3)/2). So in this interval, the minimum of sin(theta) is 0, maximum is 1. Therefore, f(x) ranges from 2*0 +1 =1 to 2*1 +1=3. Wait, but earlier calculation shows that f(-pi/6) is 0. Wait, there's a discrepancy here. Wait, but wait. If theta = 2x + pi/6, and x is in [-pi/6, pi/4], then theta is in:When x = -pi/6: theta = 2*(-pi/6) + pi/6 = -pi/3 + pi/6 = -pi/6When x = pi/4: theta = 2*(pi/4) + pi/6 = pi/2 + pi/6 = 2pi/3Therefore, theta ranges from -pi/6 to 2pi/3. So sin(theta) over this interval. The minimum sin(theta) occurs at theta = -pi/6 (since sin(-pi/6) = -1/2), and the maximum at theta = pi/2 (sin(pi/2) =1). Therefore, sin(theta) ranges from -1/2 to 1 in this interval. Therefore, f(x) = 2 sin(theta) +1 would range from 2*(-1/2) +1 = -1 +1 = 0 to 2*1 +1 =3. Therefore, the minimum is 0 and maximum is 3, which matches our previous calculation. But wait, theta starts at -pi/6, goes through 0, pi/2, and up to 2pi/3. So the minimum sin(theta) is at theta = -pi/6, which is -1/2. Therefore, the function f(x) can indeed reach 0, which is at x = -pi/6. So the minimum is 0, the maximum is 3. So that's correct. So part (2)'s answer is maximum 3 at x = pi/6, minimum 0 at x = -pi/6. Therefore, summarizing:1) Axis of symmetry is x = pi/6.2) Maximum value is 3, minimum is 0 on the interval [-pi/6, pi/4].Final Answer(1) The axis of symmetry is boxed{x = frac{pi}{6}}.(2) The maximum value is boxed{3} and the minimum value is boxed{0}.</think>Given the function ( f(x) = 4 cos x sinleft(x + frac{pi}{6}right) ):1. Finding the axis of symmetry: We start by simplifying the function using trigonometric identities. Using the product-to-sum identity: [ sin A cos B = frac{sin(A + B) + sin(A - B)}{2} ] we rewrite ( f(x) ) as: [ f(x) = 4 cos x sinleft(x + frac{pi}{6}right) = 2 sinleft(2x + frac{pi}{6}right) + 1 ] This is a sine function with amplitude 2, period ( pi ), phase shift ( -frac{pi}{12} ), and vertical shift 1. To find the axis of symmetry, we solve ( f(a + h) = f(a - h) ) for all ( h ). This leads to the condition: [ sinleft(2(a + h) + frac{pi}{6}right) = sinleft(2(a - h) + frac{pi}{6}right) ] Solving this, we find the axis of symmetry at ( x = frac{pi}{6} + frac{pi k}{2} ) for integer ( k ). The principal value within the interval is ( x = frac{pi}{6} ).2. Finding the maximum and minimum values in the interval (left[-frac{pi}{6}, frac{pi}{4}right]): The function ( f(x) = 2 sinleft(2x + frac{pi}{6}right) + 1 ) has critical points where the derivative is zero: [ f'(x) = 4 cosleft(2x + frac{pi}{6}right) = 0 implies 2x + frac{pi}{6} = frac{pi}{2} + kpi ] Solving for ( x ), we find critical points at ( x = frac{pi}{6} + frac{pi k}{2} ). Within the interval (left[-frac{pi}{6}, frac{pi}{4}right]), the only critical point is ( x = frac{pi}{6} ). Evaluating ( f(x) ) at the endpoints and the critical point: - ( fleft(-frac{pi}{6}right) = 0 ) - ( fleft(frac{pi}{6}right) = 3 ) - ( fleft(frac{pi}{4}right) = sqrt{3} + 1 approx 2.732 ) The maximum value is 3 at ( x = frac{pi}{6} ), and the minimum value is 0 at ( x = -frac{pi}{6} ). Final Answer(1) The axis of symmetry is (boxed{x = frac{pi}{6}}).(2) The maximum value is (boxed{3}) and the minimum value is (boxed{0}).,<think>Okay, let's tackle this problem step by step. First, we have the function f(x) = 4 cos x sin(x + π/6). There are two parts: finding the axis of symmetry and then determining the maximum and minimum values on the interval [-π/6, π/4]. Let me start with part (1).Hmm, axis of symmetry. I remember that for functions, an axis of symmetry is a vertical line x = a such that the function is symmetric around that line. So, for any point (a + h, f(a + h)), there should be a corresponding point (a - h, f(a - h)) which is equal. So, f(a + h) = f(a - h). I need to find such a value 'a' for the given function.But how do I find that? Maybe it's easier if I first simplify the function using trigonometric identities. Let me see, f(x) = 4 cos x sin(x + π/6). There's a product of sine and cosine, perhaps I can use the sine addition formula or convert the product into a sum.The product-to-sum identity: sin A cos B = [sin(A + B) + sin(A - B)] / 2. Wait, here it's cos x sin(x + π/6). Let me adjust the identity accordingly. Let's set A = x + π/6 and B = x. Then sin(A) cos(B) = [sin(A + B) + sin(A - B)] / 2.So substituting A and B:sin(x + π/6) cos x = [sin((x + π/6) + x) + sin((x + π/6) - x)] / 2Simplify the arguments inside the sines:First term: sin(2x + π/6)Second term: sin(π/6)So, sin(x + π/6) cos x = [sin(2x + π/6) + sin(π/6)] / 2Therefore, f(x) = 4 * [sin(2x + π/6) + sin(π/6)] / 2 = 2 [sin(2x + π/6) + sin(π/6)]Simplifying further, sin(π/6) is 1/2, so:f(x) = 2 sin(2x + π/6) + 2*(1/2) = 2 sin(2x + π/6) + 1Wait, let me check that calculation again. The original expression is 4 cos x sin(x + π/6). Using the identity, we have:4 * [sin(2x + π/6) + sin(π/6)] / 2 = 2 sin(2x + π/6) + 2 sin(π/6). Since sin(π/6) = 1/2, this becomes 2 sin(2x + π/6) + 2*(1/2) = 2 sin(2x + π/6) + 1. Yes, that's correct.So, f(x) = 2 sin(2x + π/6) + 1. That simplifies the function a lot. Now, perhaps this form is easier to analyze for symmetry.Now, the function is a sine function with amplitude 2, shifted vertically up by 1, and with a phase shift. Let's see, the argument inside the sine is 2x + π/6. Let me write it as sin(2(x + π/12)). Wait, because 2x + π/6 = 2(x + π/12). So, the phase shift is -π/12. That means the graph of sin(2x) is shifted to the left by π/12.But how does the phase shift affect the axis of symmetry? For a sine function of the form sin(Bx + C), the phase shift is -C/B. In this case, B is 2, so the phase shift is -π/12. So, the function is shifted left by π/12. But sine functions themselves are symmetric about their midline, but their axes of symmetry (if any) would depend on their phase shift.Wait, but sine functions are periodic and have multiple axes of symmetry. For example, the basic sin(x) has an axis of symmetry at every peak and trough, but those are points, not vertical lines. Wait, maybe not. Wait, no, the sine function is symmetric about the origin, but that's rotational symmetry, not reflectional. Hmm. So perhaps it's not symmetric about any vertical line. Wait, but if we have a phase-shifted sine function, maybe there's a vertical line where it's symmetric.Alternatively, since we have the function in terms of 2 sin(2x + π/6) + 1, maybe we can find its maximum points and see if there's a vertical line that serves as an axis of symmetry.Alternatively, perhaps the original function before simplifying has some symmetry. Let's think. The original function was 4 cos x sin(x + π/6). Maybe expanding that expression could give another perspective. Let's try expanding sin(x + π/6):sin(x + π/6) = sin x cos(π/6) + cos x sin(π/6) = sin x (√3/2) + cos x (1/2)So, substituting back into f(x):f(x) = 4 cos x [ (√3/2) sin x + (1/2) cos x ] = 4 [ (√3/2 sin x cos x ) + (1/2 cos²x ) ]Simplify the terms:= 4*(√3/2) sin x cos x + 4*(1/2) cos²x = 2√3 sin x cos x + 2 cos²xNow, 2 sin x cos x is sin(2x), so:= 2√3*(1/2) sin(2x) + 2 cos²x = √3 sin(2x) + 2 cos²xAnd 2 cos²x can be written as 2*(1 + cos(2x))/2 = 1 + cos(2x). So:f(x) = √3 sin(2x) + 1 + cos(2x) = √3 sin(2x) + cos(2x) + 1Now, combining the terms √3 sin(2x) + cos(2x). This resembles a single sine or cosine function with some amplitude. Let me recall that A sin θ + B cos θ = C sin(θ + φ), where C = √(A² + B²) and tan φ = B/A.Wait, actually, it can be written as C sin(θ + φ) or C cos(θ - φ). Let me check:√3 sin(2x) + cos(2x). Let's factor out √( (√3)^2 + 1^2 ) = √(3 + 1) = 2. So, factor 2:= 2 [ (√3/2 sin(2x) + 1/2 cos(2x) ) ]Notice that √3/2 = cos(π/6) and 1/2 = sin(π/6). Therefore:= 2 [ cos(π/6) sin(2x) + sin(π/6) cos(2x) ] = 2 sin(2x + π/6) [Wait, is that right? Wait, sin(A + B) = sin A cos B + cos A sin B. Here, it's cos(π/6) sin(2x) + sin(π/6) cos(2x) = sin(2x + π/6). Because sin(a + b) = sin a cos b + cos a sin b. So yes, if we take a = 2x and b = π/6, then sin(2x + π/6) = sin(2x) cos(π/6) + cos(2x) sin(π/6). But in our case, it's cos(π/6) sin(2x) + sin(π/6) cos(2x), which is the same as sin(2x) cos(π/6) + cos(2x) sin(π/6) = sin(2x + π/6). Therefore, the expression becomes 2 sin(2x + π/6). Therefore, f(x) = 2 sin(2x + π/6) + 1. Wait, which is the same as before. So, yes, that checks out.So, we have f(x) = 2 sin(2x + π/6) + 1. Now, to find the axis of symmetry. Let's recall that for a sinusoidal function like A sin(Bx + C) + D, the function has a vertical shift D, amplitude |A|, period 2π / |B|, and phase shift -C / B.But symmetry axis. For a sine function, typically, it's symmetric about its midline, which is the horizontal line y = D. But that's horizontal symmetry. But the question is asking for a vertical axis of symmetry, i.e., a vertical line x = a such that f(a + h) = f(a - h) for all h.So, perhaps the function has a vertical line of symmetry where it reaches its maximum or minimum? Let's consider.The function is 2 sin(2x + π/6) + 1. Let's consider its critical points. The maximum of sin is 1, so maximum of f(x) is 2*1 + 1 = 3, and minimum is 2*(-1) + 1 = -1.The function will have maxima where 2x + π/6 = π/2 + 2π k, for integer k.Solving for x: 2x = π/2 - π/6 + 2π k = (3π/6 - π/6) + 2π k = (2π/6) + 2π k = π/3 + 2π k.Thus, x = π/6 + π k.Similarly, minima occur where 2x + π/6 = 3π/2 + 2π k => 2x = 3π/2 - π/6 + 2π k = (9π/6 - π/6) + 2π k = 8π/6 + 2π k = 4π/3 + 2π k => x = 2π/3 + π k.So the critical points are at x = π/6 + π k and x = 2π/3 + π k.If we can find a vertical line x = a such that for any h, f(a + h) = f(a - h), then that is the axis of symmetry.Alternatively, maybe the function is symmetric around one of its maxima or minima.Suppose we take x = π/6. Let's check if x = π/6 is an axis of symmetry. Let's test f(π/6 + h) vs f(π/6 - h).Compute f(π/6 + h) = 2 sin(2*(π/6 + h) + π/6) + 1 = 2 sin(π/3 + 2h + π/6) + 1 = 2 sin(π/2 + 2h) + 1.Similarly, f(π/6 - h) = 2 sin(2*(π/6 - h) + π/6) + 1 = 2 sin(π/3 - 2h + π/6) + 1 = 2 sin(π/2 - 2h) + 1.Now, sin(π/2 + 2h) = cos(2h), and sin(π/2 - 2h) = cos(2h). Therefore, f(π/6 + h) = 2 cos(2h) + 1 and f(π/6 - h) = 2 cos(2h) + 1. Therefore, they are equal. Therefore, x = π/6 is an axis of symmetry.Wait, that seems to work. Therefore, the axis of symmetry is x = π/6. Let me confirm with another point. Take h = π/6.Compute f(π/6 + π/6) = f(π/3) = 2 sin(2*(π/3) + π/6) +1 = 2 sin(2π/3 + π/6) = 2 sin(5π/6) = 2*(1/2) = 1 +1=2.f(π/6 - π/6) = f(0) = 2 sin(0 + π/6) +1 = 2*(1/2) +1=1 +1=2. So they are equal.Another test: h = π/12.f(π/6 + π/12)=f(π/4)=2 sin(2*(π/4)+π/6)+1=2 sin(π/2 + π/6)+1=2 sin(2π/3)+1=2*(√3/2)+1=√3 +1.f(π/6 - π/12)=f(π/12)=2 sin(2*(π/12)+π/6)+1=2 sin(π/6 + π/6)=2 sin(π/3)=2*(√3/2)=√3. Then add 1: √3 +1. So equal again. So that seems correct.Therefore, part (1) answer is x = π/6. Let's see if there are others. But since the function is periodic with period π (since the coefficient of x inside sine is 2), the axes of symmetry would be at x = π/6 + kπ/2, but the problem probably asks for the principal one, which is x = π/6.Hence, part (1) is solved.Now part (2): Find the maximum and minimum values of f(x) in the interval [-π/6, π/4].First, let's note that the function f(x) = 2 sin(2x + π/6) + 1. The maximum of sin is 1, the minimum is -1, so f(x) has maximum 3 and minimum -1, but these extrema might not lie within the given interval. So we need to check the critical points within the interval and the endpoints.First, let's find the critical points by taking the derivative.f(x) = 2 sin(2x + π/6) + 1f'(x) = 2 cos(2x + π/6) * 2 = 4 cos(2x + π/6)Set derivative to zero:4 cos(2x + π/6) = 0 => cos(2x + π/6) = 0Solutions are 2x + π/6 = π/2 + π k, where k is integer.Solving for x:2x = π/2 - π/6 + π k = (3π/6 - π/6) + π k = (2π/6) + π k = π/3 + π kTherefore, x = π/6 + (π/2) k.So critical points at x = π/6 + (π/2)k.Now, find all critical points within the interval [-π/6, π/4].Let me compute x for different integer k:For k = 0: x = π/6 ≈ 0.523. Is π/6 within [-π/6, π/4]? π/4 ≈ 0.785, so π/6 ≈ 0.523 is less than π/4, and -π/6 ≈ -0.523. So yes, π/6 is within the interval.For k = -1: x = π/6 - π/2 = -π/3 ≈ -1.047. That's less than -π/6 ≈ -0.523, so outside the interval.For k = 1: x = π/6 + π/2 = 2π/3 ≈ 2.094. Which is greater than π/4 ≈ 0.785, so outside.Therefore, the only critical point within the interval is x = π/6.Therefore, the extrema can occur at x = π/6 or at the endpoints x = -π/6 and x = π/4.Therefore, we need to evaluate f(x) at these three points:1. x = -π/6:f(-π/6) = 2 sin(2*(-π/6) + π/6) +1 = 2 sin(-π/3 + π/6) +1 = 2 sin(-π/6) +1 = 2*(-1/2) +1 = -1 +1=0.2. x = π/6:f(π/6) = 2 sin(2*(π/6) + π/6) +1 = 2 sin(π/3 + π/6) +1 = 2 sin(π/2) +1 = 2*1 +1=3.3. x = π/4:f(π/4) = 2 sin(2*(π/4) + π/6) +1 = 2 sin(π/2 + π/6) +1 = 2 sin(2π/3) +1 = 2*(√3/2) +1= √3 +1 ≈ 1.732 +1=2.732.Therefore, comparing the three values:f(-π/6)=0, f(π/6)=3, f(π/4)=√3 +1 ≈2.732.Hence, the maximum value is 3 at x=π/6, and the minimum value is 0 at x=-π/6.Wait, but hold on. Is there a possibility that the function could attain a lower value within the interval? Because the critical point at x=π/6 gives a maximum, but maybe the minimum is at the endpoint. Since we only have one critical point in the interval, which is a maximum, the minimum must be at one of the endpoints. Since f(-π/6)=0 and f(π/4)=√3 +1≈2.732, so 0 is the minimum. Therefore, the answers are maximum 3, minimum 0.But let me verify once more to be sure. Let's check the value at x=0, which is inside the interval. Wait, x=0 is between -π/6 and π/4.f(0)=2 sin(0 + π/6) +1=2*(1/2)+1=1 +1=2. That's between 0 and 3, so not extremum. Therefore, our previous conclusion holds.So part (2) maximum is 3, minimum is 0.Therefore, summarizing:(1) The axis of symmetry is x = π/6.(2) The maximum value is 3, and the minimum value is 0 on the interval [-π/6, π/4].Final Answer(1) The axis of symmetry is boxed{x = dfrac{pi}{6}}.(2) The maximum value is boxed{3} and the minimum value is boxed{0}.</think>Given the function ( f(x) = 4 cos x sinleft(x + frac{pi}{6}right) ):1. Finding the axis of symmetry: First, we simplify the function using trigonometric identities. Using the product-to-sum formula: [ 4 cos x sinleft(x + frac{pi}{6}right) = 2 left[ sin(2x + frac{pi}{6}) + sinleft(frac{pi}{6}right) right] ] Simplifying further, we get: [ f(x) = 2 sinleft(2x + frac{pi}{6}right) + 1 ] To find the axis of symmetry, we check if the function is symmetric around a vertical line. The critical points for the sine function occur at ( x = frac{pi}{6} + pi k ). Testing symmetry around ( x = frac{pi}{6} ): [ fleft(frac{pi}{6} + hright) = 2 sinleft(2left(frac{pi}{6} + hright) + frac{pi}{6}right) + 1 = 2 cos(2h) + 1 ] [ fleft(frac{pi}{6} - hright) = 2 sinleft(2left(frac{pi}{6} - hright) + frac{pi}{6}right) + 1 = 2 cos(2h) + 1 ] Since ( fleft(frac{pi}{6} + hright) = fleft(frac{pi}{6} - hright) ), the axis of symmetry is ( x = frac{pi}{6} ).2. Finding the maximum and minimum values in the interval (left[-frac{pi}{6}, frac{pi}{4}right]): The function ( f(x) = 2 sinleft(2x + frac{pi}{6}right) + 1 ) has critical points where the derivative is zero. The derivative is: [ f'(x) = 4 cosleft(2x + frac{pi}{6}right) ] Setting ( f'(x) = 0 ): [ cosleft(2x + frac{pi}{6}right) = 0 implies 2x + frac{pi}{6} = frac{pi}{2} + kpi implies x = frac{pi}{6} + frac{kpi}{2} ] Within the interval (left[-frac{pi}{6}, frac{pi}{4}right]), the critical point is ( x = frac{pi}{6} ). Evaluating ( f(x) ) at the critical point and endpoints: - At ( x = -frac{pi}{6} ): [ fleft(-frac{pi}{6}right) = 2 sinleft(-frac{pi}{6}right) + 1 = 0 ] - At ( x = frac{pi}{6} ): [ fleft(frac{pi}{6}right) = 2 sinleft(frac{pi}{2}right) + 1 = 3 ] - At ( x = frac{pi}{4} ): [ fleft(frac{pi}{4}right) = 2 sinleft(frac{2pi}{3}right) + 1 = sqrt{3} + 1 ] The maximum value is ( 3 ) and the minimum value is ( 0 ). Final Answer(1) The axis of symmetry is (boxed{x = dfrac{pi}{6}}).(2) The maximum value is (boxed{3}) and the minimum value is (boxed{0}).]

🔑:(1) The function f(x)=4cos x sinleft(x+ frac{pi}{6}right) can be written asf(x) = 4cos x left(frac{sqrt{3}}{2}sin x + frac{1}{2}cos xright)= sqrt{3}sin 2x + 2cos^2 x - 1 + 1= sqrt{3}sin 2x + cos 2x + 1.Using the identity sin(A + B) = sin A cos B + cos A sin B, we can rewrite f(x) asf(x) = 2sinleft(2x+ frac{pi}{6}right) + 1.To find the axis of symmetry, we set 2x+ frac{pi}{6} = frac{pi}{2}+kpi where kinmathbb{Z}. Solving for x gives us the axes of symmetryx = frac{pi}{6} + frac{kpi}{2}, quad kinmathbb{Z}.(2) When xinleft[- frac{pi}{6}, frac{pi}{4}right], the expression 2x+ frac{pi}{6}inleft[- frac{pi}{6}, frac{2pi}{3}right]. We set 2x+ frac{pi}{6}= frac{pi}{2} and solve for x, obtaining x= frac{pi}{6}.Therefore, in the interval xinleft[- frac{pi}{6}, frac{pi}{6}right], f(x) is increasing; while in the interval xinleft[ frac{pi}{6}, frac{pi}{4}right], f(x) is decreasing. Thus, when x= frac{pi}{6}, f(x) reaches its maximum value, which isboxed{3}.And when 2x+ frac{pi}{6}= -frac{pi}{6}, i.e., x= -frac{pi}{6}, f(x) reaches its minimum value, which isboxed{0}.

🧠:[<think>Okay, so I need to figure out how many subsets a set with n elements has. Hmm, let's start from the basics. A subset is any selection of elements from the original set, right? That includes everything from the empty set to the entire set itself. Let me think about small values of n first. Maybe that'll help me see a pattern. If n is 0, that's the empty set. How many subsets does the empty set have? Well, just one subset, which is the empty set itself. So when n=0, the number of subsets is 1. What about n=1? Let's say the set is {a}. The subsets would be the empty set and {a}, so that's 2 subsets. Moving on to n=2. Suppose the set is {a, b}. The subsets here are {}, {a}, {b}, and {a, b}. That's 4 subsets. For n=3, let's take {a, b, c}. The subsets would be:- The empty set {},- Single elements: {a}, {b}, {c},- Pairs: {a, b}, {a, c}, {b, c},- The entire set {a, b, c}.Counting those, that's 1 + 3 + 3 + 1 = 8 subsets. So the pattern here seems like 1, 2, 4, 8... which is 2^n. For each n, the number of subsets doubles. So when n=0, 2^0=1; n=1, 2^1=2; n=2, 2^2=4; n=3, 2^3=8. That checks out. But why does this happen? Let me think more formally. Each element in the set can be either included or excluded from a subset. So for each element, there are two choices: in or out. For example, with the set {a, b}, for element a, we can choose to include it or not, and similarly for element b. So the total number of combinations is 2 * 2 = 4, which matches the subsets we listed. Extending this, if there are n elements, each has two choices: included or excluded. Therefore, the total number of subsets should be 2 multiplied by itself n times, which is 2^n. But wait, let me make sure I'm not missing something. The empty set is counted as a subset, right? And in this reasoning, when all elements are excluded, we get the empty set, which is accounted for. Similarly, when all elements are included, we get the entire set, which is also a subset. So every combination is covered. Another way to think about it is through combinations. The number of subsets can be calculated by summing the number of subsets of each possible size. That is, the number of subsets of size 0 plus subsets of size 1, up to subsets of size n. The number of subsets of size k is given by the combination formula C(n, k) = n! / (k!(n - k)!). So the total number of subsets is the sum from k=0 to k=n of C(n, k). I remember that the sum of combinations C(n, k) for k=0 to n is equal to 2^n. This is from the binomial theorem, where (1 + 1)^n = sum_{k=0}^n C(n, k) * 1^{n-k} * 1^k} = sum_{k=0}^n C(n, k) = 2^n. So that confirms it another way. The total number of subsets is indeed 2^n. Wait, but let me verify with an example. Take n=3 again. The sum would be C(3,0) + C(3,1) + C(3,2) + C(3,3) = 1 + 3 + 3 + 1 = 8, which is 2^3. Yep, that works. Alternatively, thinking recursively. Suppose I have a set with n elements. Let's pick one element, say x. Then, all subsets either include x or they don't. The subsets that don't include x are exactly the subsets of the remaining n-1 elements. The subsets that do include x are the subsets of the remaining n-1 elements with x added in. Therefore, the total number of subsets is twice the number of subsets of a set with n-1 elements. So, if S(n) is the number of subsets for a set of size n, then S(n) = 2 * S(n-1). With the base case S(0) = 1. Solving this recurrence relation: S(n) = 2 * S(n-1) = 2 * 2 * S(n-2) = ... = 2^n * S(0) = 2^n * 1 = 2^n. That makes sense. Each time you add an element, you double the number of subsets because you can add the new element to all existing subsets. Another angle: binary representation. Each subset can be represented by a binary string of length n, where each bit corresponds to the presence (1) or absence (0) of an element. For example, for n=3, the binary numbers from 000 to 111 (0 to 7 in decimal) represent all subsets. There are 2^3 = 8 such numbers. So each binary number maps to a unique subset, hence 2^n subsets. This is similar to the earlier reasoning where each element is a bit with two choices. Is there a way this could go wrong? Suppose someone argues that some subsets are counted multiple times. But if each element is considered independently, and the presence or absence is tracked uniquely, then each combination is unique. So no duplicates. Alternatively, maybe overlapping elements? But in a set, all elements are distinct, so each combination is a unique subset. Therefore, all the different approaches lead to the same conclusion: the number of subsets is 2^n. Let me check one more example for n=4. The set {a, b, c, d}. The number of subsets should be 16. Listing them:- Size 0: 1 (empty set)- Size 1: 4 ({a}, {b}, {c}, {d})- Size 2: 6 ({a,b}, {a,c}, {a,d}, {b,c}, {b,d}, {c,d})- Size 3: 4 ({a,b,c}, {a,b,d}, {a,c,d}, {b,c,d})- Size 4: 1 ({a,b,c,d})Total: 1 + 4 + 6 + 4 + 1 = 16 = 2^4. Perfect, that's correct. So through multiple methods: enumerating small cases, using combinations, recurrence relations, binary representations—all confirm that the number of subsets of a set with cardinality n is 2^n. I think that settles it. The answer should be 2 to the power of n, written as 2^n, which in boxed form is boxed{2^n}.Final AnswerThe number of subsets of a set of cardinality ( n ) is boxed{2^n}.</think>To determine the number of subsets of a set with cardinality ( n ), we can use several methods to arrive at the conclusion:1. Enumeration of Small Cases: - For ( n = 0 ) (empty set), there is 1 subset. - For ( n = 1 ), there are 2 subsets. - For ( n = 2 ), there are 4 subsets. - For ( n = 3 ), there are 8 subsets. - The pattern suggests ( 2^n ) subsets.2. Inclusion-Exclusion Principle: - Each element in the set can be either included or excluded from a subset. - For each of the ( n ) elements, there are 2 choices, leading to ( 2^n ) combinations.3. Combination Sum: - The number of subsets is the sum of combinations ( sum_{k=0}^{n} binom{n}{k} ). - By the binomial theorem, this sum is ( (1 + 1)^n = 2^n ).4. Recursive Reasoning: - If ( S(n) ) is the number of subsets for a set of size ( n ), then ( S(n) = 2 cdot S(n-1) ). - Solving this recurrence with ( S(0) = 1 ) gives ( S(n) = 2^n ).5. Binary Representation: - Each subset corresponds to a binary string of length ( n ), where each bit represents the presence (1) or absence (0) of an element. - There are ( 2^n ) such binary numbers, hence ( 2^n ) subsets.All methods confirm that the number of subsets of a set with cardinality ( n ) is ( 2^n ).[boxed{2^n}],<think>Okay, so the question is asking how many subsets a set with cardinality n has. Hmm, I remember that this is a classic combinatorics problem. Let me try to work through it step by step. First, let me recall what a subset is. A subset of a set S is any set that contains elements from S, including the empty set and S itself. So, for example, if S is {1, 2}, the subsets are {}, {1}, {2}, and {1,2}. That's 4 subsets. Wait, 2 elements, 4 subsets. So for a set with n elements, maybe the number of subsets is 2^n? That seems familiar. But I need to verify this.Let me try with a smaller n. If n is 0, then the set is empty. The only subset is the empty set itself. So 1 subset. 2^0 is 1, which matches. If n is 1, say the set {a}, the subsets are {} and {a}, so 2 subsets. 2^1 is 2. That works. For n=2, we already saw 4 subsets. 2^2 is 4. For n=3, let's take {a, b, c}. The subsets would be: {}, {a}, {b}, {c}, {a,b}, {a,c}, {b,c}, {a,b,c}. That's 8 subsets. 2^3 is 8. So this pattern holds. So for each n, the number of subsets is 2^n. But why is that the case? Maybe there's a reasoning behind this. Let me think. When constructing a subset, for each element in the original set, we have a choice: either include it in the subset or not. So for each element, there are two choices. Since the choices are independent, the total number of subsets would be the product of the number of choices for each element. For example, with n=3: for each of the three elements, we decide yes or no. So 2*2*2 = 8. That makes sense. So generalizing, for n elements, each has two choices, leading to 2^n subsets. Alternatively, maybe we can think in terms of combinations. The number of subsets of size k is C(n,k), so the total number of subsets is the sum over k from 0 to n of C(n,k). And there's a binomial theorem identity that says the sum of C(n,k) for k=0 to n is 2^n. Yes, because (1 + 1)^n = 2^n. So that's another way to see it. But perhaps there's another approach. Let me try induction. Suppose that for a set with n elements, the number of subsets is 2^n. Let's check the base case again. If n=0, as before, 1 subset. 2^0 =1. So holds. Assume true for n=k. Now consider a set with k+1 elements. Let's pick an element, say x. The subsets can be divided into those that contain x and those that don't. The subsets that don't contain x are exactly the subsets of the original set without x, which by induction hypothesis has 2^k subsets. The subsets that do contain x are the subsets of the remaining k elements with x added in, which is also 2^k. So total subsets are 2^k + 2^k = 2^(k+1). Hence, by induction, the formula holds for all n≥0. That seems solid. So whether we approach it combinatorially, by counting choices per element, using binomial coefficients, or induction, we arrive at 2^n. Wait, let me make sure there's no catch here. Is there any possibility that I'm missing something? For example, does this count all subsets correctly, including empty set and the set itself? Let's see. When we make the choice for each element, if we choose 'no' for all elements, we get the empty set. If we choose 'yes' for all, we get the entire set. So yes, they are included. And all other subsets are combinations of some yes's and no's. Alternatively, when using the combination approach, summing C(n,0) + C(n,1) + ... + C(n,n) gives all possible sizes, which should account for all subsets. Therefore, multiple approaches confirm that the number of subsets is 2^n. I don't see any flaws in the reasoning here.Final AnswerThe number of subsets of a set with cardinality ( n ) is boxed{2^n}.</think>To determine the number of subsets of a set with cardinality ( n ), we can consider the following reasoning:1. Choice per Element: For each element in the set, there are two choices: either include it in the subset or not. Since there are ( n ) elements, the total number of subsets is ( 2 times 2 times cdots times 2 ) (n times), which is ( 2^n ).2. Binomial Coefficients: The number of subsets of size ( k ) is given by the binomial coefficient ( binom{n}{k} ). Summing over all possible subset sizes from 0 to ( n ) gives the total number of subsets: [ sum_{k=0}^{n} binom{n}{k} = 2^n ] This follows from the binomial theorem where ( (1 + 1)^n = 2^n ).3. Induction: - Base Case: For ( n = 0 ), the set is empty and has 1 subset (the empty set), which is ( 2^0 = 1 ). - Inductive Step: Assume a set with ( k ) elements has ( 2^k ) subsets. For a set with ( k + 1 ) elements, we can divide the subsets into those that include a specific element and those that do not. Each group has ( 2^k ) subsets, leading to ( 2^k + 2^k = 2^{k+1} ) subsets.All these approaches confirm that the number of subsets of a set with cardinality ( n ) is ( 2^n ).[boxed{2^n}]]

🔑:The problem can be solved in several ways.We can simply content ourselves with linking it to sequences of 0s and 1s of length n (there are 2^{n} of them) which are a form of representation of a subset, since we can, for example, say that the subset associated with a sequence is constructed such that the i^{text{th}} element is in the subset if and only if the i^{text{th}} term of the sequence is 1. For each sequence, there is therefore a set, and for each set, there is a unique sequence. There are therefore exactly 2^{n} subsets of a set of cardinality n.Another method of proof involves reasoning by induction on n. We leave it to the reader to write the proof, which should not present any difficulty.

🧠:[<think>Okay, let's see. I have two problems to solve here. The first one is a logarithmic expression, and the second one involves exponents with a logarithm in the exponent. Hmm, let me tackle them one by one.Starting with problem (1): (lg2 + lg5 - lg8) divided by (lg50 - lg40). Alright, so all these terms are logarithms with base 10, I assume, since it's written as "lg" which sometimes stands for log base 10. Let me confirm that notation. Yeah, in many contexts, especially in some countries, "lg" is base 10, "ln" is natural log, and "log" without a base might be base 10 or natural depending on the context. But here, since it's specified as "lg", I'll go with base 10.So, the numerator is lg2 + lg5 - lg8, and the denominator is lg50 - lg40. I need to simplify both numerator and denominator using logarithmic properties. Let me recall the properties: log(a) + log(b) = log(ab), log(a) - log(b) = log(a/b), and also n*log(a) = log(a^n). Maybe I can combine the terms in the numerator and denominator first.Starting with the numerator: lg2 + lg5 - lg8. Let me combine lg2 + lg5. Using the addition property, that becomes lg(2*5) = lg10. Then subtract lg8: lg10 - lg8. So that simplifies to lg(10/8) because of the subtraction property. 10 divided by 8 is 5/4. So the numerator simplifies to lg(5/4).Now the denominator: lg50 - lg40. Similarly, that's lg(50/40) = lg(5/4) because 50 divided by 40 simplifies to 5/4. Wait, so both the numerator and the denominator simplify to lg(5/4). So then the entire expression is lg(5/4) divided by lg(5/4), which is 1. That seems straightforward. Let me check my steps again to make sure.Numerator:lg2 + lg5 = lg(2*5) = lg10. Then lg10 - lg8 = lg(10/8) = lg(5/4). Correct.Denominator:lg50 - lg40 = lg(50/40) = lg(5/4). Correct.Therefore, the expression is lg(5/4) / lg(5/4) = 1. So the answer to part (1) is 1. That seems too simple, but maybe that's the case. Let me see if there's another approach or if I made a mistake in combining the logs.Alternatively, maybe using different log properties. Let me check the numerator again. lg2 + lg5 is lg(10), since 2*5=10. Then subtract lg8. So lg10 - lg8. Since lg10 is 1, because log base 10 of 10 is 1. Then 1 - lg8. Wait, but the denominator is lg50 - lg40. Let's compute the numerical values to check.If I compute numerator: lg2 ≈ 0.3010, lg5 ≈ 0.69897, lg8 ≈ 0.90309. So 0.3010 + 0.69897 - 0.90309 ≈ (0.3010 + 0.69897) = 0.99997 - 0.90309 ≈ 0.09688.Denominator: lg50 ≈ 1.69897, lg40 ≈ 1.60206. So 1.69897 - 1.60206 ≈ 0.09691.So 0.09688 / 0.09691 ≈ approximately 0.9997, which is roughly 1. So the exact value is 1, which matches my earlier conclusion. So that seems correct.Okay, so problem (1) answer is 1.Moving on to problem (2): 2^{2 + log_{sqrt(2)} (1/4)}. Hmm, this looks a bit trickier. Let me parse the exponent first. The exponent is 2 + log base sqrt(2) of 1/4. So, the entire expression is 2 raised to that power.First, let's simplify the logarithm part: log_{sqrt(2)} (1/4). Let me recall that log_b(a) = ln(a)/ln(b) or, alternatively, can be expressed in terms of exponents. Let me try converting the logarithm to exponent form.Let’s set log_{sqrt(2)} (1/4) = x. Then, by definition, (sqrt(2))^x = 1/4. Let's solve for x. sqrt(2) is 2^{1/2}, so (2^{1/2})^x = 2^{x/2}. And 1/4 is 2^{-2}. Therefore, 2^{x/2} = 2^{-2}. Since the bases are equal, the exponents must be equal. So x/2 = -2 => x = -4. Therefore, log_{sqrt(2)} (1/4) = -4.So the exponent in the original expression is 2 + (-4) = -2. Therefore, the entire expression is 2^{-2} = 1/4. So the answer is 1/4. Wait, let me verify.Alternatively, use the change of base formula. log_{sqrt(2)} (1/4) = [log_2 (1/4)] / [log_2 sqrt(2)]. Since log_b a = log_c a / log_c b. Let's compute that.log_2 (1/4) = log_2 2^{-2} = -2.log_2 sqrt(2) = log_2 2^{1/2} = 1/2.Therefore, log_{sqrt(2)} (1/4) = (-2)/(1/2) = -4. Which matches the earlier result. So yes, that's correct.Therefore, the exponent is 2 + (-4) = -2. So 2^{-2} = 1/4. So the answer is 1/4.Wait, but let me check if there's another way to approach this, maybe using properties of exponents. Let me see.Original expression: 2^{2 + log_{sqrt(2)} (1/4)}. Let's recall that a^{b + c} = a^b * a^c. So, this can be written as 2^2 * 2^{log_{sqrt(2)} (1/4)}. Then, 2^2 is 4, and then we have 4 multiplied by 2^{log_{sqrt(2)} (1/4)}.Now, let's focus on simplifying 2^{log_{sqrt(2)} (1/4)}. Let me write the exponent as log_{sqrt(2)} (1/4). Let me denote that as x. So x = log_{sqrt(2)} (1/4). Then, as before, (sqrt(2))^x = 1/4, which we solved as x = -4.So, 2^{log_{sqrt(2)} (1/4)} = 2^{-4} = 1/16. Therefore, the entire expression becomes 4 * (1/16) = 1/4. Same result. So that checks out.Alternatively, perhaps use the identity that a^{log_b c} = c^{log_b a}. Hmm, not sure if that's helpful here, but let's try.Wait, 2^{log_{sqrt(2)} (1/4)}. Let’s let a = 2, b = sqrt(2), c = 1/4. Then according to that identity, a^{log_b c} = c^{log_b a}. Let's check:Left-hand side: 2^{log_{sqrt(2)} (1/4)}.Right-hand side: (1/4)^{log_{sqrt(2)} 2}.Compute log_{sqrt(2)} 2: let’s call this y. Then (sqrt(2))^y = 2. sqrt(2) is 2^{1/2}, so 2^{y/2} = 2^1. Therefore, y/2 = 1 => y = 2. So log_{sqrt(2)} 2 = 2. Therefore, right-hand side is (1/4)^2 = 1/16. Which matches 2^{-4} = 1/16. So the identity works here. Therefore, the expression is indeed 1/16, multiplied by 4 gives 1/4. So that's another way to see it.Therefore, problem (2) answer is 1/4.Let me just recap to make sure I didn't miss anything.Problem (1): Both numerator and denominator simplified to lg(5/4), so their ratio is 1.Problem (2): Calculated the log base sqrt(2) of 1/4 as -4, added 2 to get -2, so 2^{-2} is 1/4.Yes, both seem correct. I think that's it.Final Answer(1) boxed{1}(2) boxed{dfrac{1}{4}}</think>To solve the given problems, we proceed as follows:(1) Simplify the expression (frac{lg 2 + lg 5 - lg 8}{lg 50 - lg 40}):- Numerator: [ lg 2 + lg 5 - lg 8 = lg(2 cdot 5) - lg 8 = lg 10 - lg 8 = lgleft(frac{10}{8}right) = lgleft(frac{5}{4}right) ]- Denominator: [ lg 50 - lg 40 = lgleft(frac{50}{40}right) = lgleft(frac{5}{4}right) ]- Result: [ frac{lgleft(frac{5}{4}right)}{lgleft(frac{5}{4}right)} = 1 ]Thus, the answer to part (1) is (boxed{1}).(2) Simplify the expression (2^{2 + log_{sqrt{2}} frac{1}{4}}):- First, evaluate (log_{sqrt{2}} frac{1}{4}): [ text{Let } log_{sqrt{2}} frac{1}{4} = x implies (sqrt{2})^x = frac{1}{4} implies 2^{x/2} = 2^{-2} implies frac{x}{2} = -2 implies x = -4 ]- Substitute back into the exponent: [ 2 + (-4) = -2 ]- Simplify the expression: [ 2^{-2} = frac{1}{4} ]Thus, the answer to part (2) is (boxed{dfrac{1}{4}}).,<think>Okay, so I have two problems to solve here. Let me start with the first one. It's a logarithmic expression: (lg2 + lg5 - lg8) divided by (lg50 - lg40). Hmm, logarithms can sometimes be tricky, but maybe I can simplify both the numerator and the denominator using logarithm properties.First, let me recall some logarithm rules. The logarithm of a product is the sum of the logarithms, right? So lg(a) + lg(b) = lg(ab). Similarly, the logarithm of a quotient is the difference: lg(a) - lg(b) = lg(a/b). Also, lg(a^n) = n*lg(a). Maybe these can help simplify things.Looking at the numerator: lg2 + lg5 - lg8. Let me see if I can combine those first two terms. If I add lg2 and lg5, that becomes lg(2*5) = lg10. So the numerator simplifies to lg10 - lg8. Then, using the quotient rule, that's lg(10/8). Simplifying 10/8 reduces to 5/4, so the numerator is lg(5/4).Now the denominator: lg50 - lg40. Using the same quotient rule, that's lg(50/40). Simplifying 50/40 gives 5/4, so the denominator is also lg(5/4). So the whole expression becomes lg(5/4) divided by lg(5/4), which should be 1. Wait, that seems straightforward. Let me double-check to make sure I didn't make a mistake.Numerator: lg2 + lg5 = lg(2*5) = lg10. Then subtract lg8: lg10 - lg8 = lg(10/8) = lg(5/4). Denominator: lg50 - lg40 = lg(50/40) = lg(5/4). So yeah, numerator over denominator is 1. Okay, that seems right. Maybe I was overcomplicating it at first, but I think that's correct.Moving on to the second problem: 2^{2 + log_{sqrt(2)} (1/4)}. This is an exponent expression with a logarithm in the exponent. Let me break this down step by step.First, let's handle the exponent: 2 + log_{sqrt(2)} (1/4). I need to compute log base sqrt(2) of 1/4 and then add 2 to it. Let's focus on the logarithm part.Let me recall that log_b(a) = c means that b^c = a. So here, log_{sqrt(2)} (1/4) = x means that (sqrt(2))^x = 1/4. Let me solve for x.Expressed as exponents, sqrt(2) is 2^{1/2}. So (2^{1/2})^x = 2^{(x/2)}. This is equal to 1/4, which is 2^{-2}. So we have 2^{x/2} = 2^{-2}. Since the bases are equal, the exponents must be equal: x/2 = -2 => x = -4. Therefore, log_{sqrt(2)} (1/4) = -4.So the exponent in the original expression is 2 + (-4) = -2. Therefore, the entire expression becomes 2^{-2} = 1/4. Let me check this again.Alternative approach: Maybe use logarithm properties to rewrite the exponent. Remember that a^{log_b(c)} = c^{log_b(a)} or something like that? Not sure. Alternatively, use the change of base formula. The exponent is 2 + log_{sqrt(2)}(1/4). Let's compute log_{sqrt(2)}(1/4):Using change of base formula: log_b(a) = ln(a)/ln(b). So log_{sqrt(2)}(1/4) = ln(1/4)/ln(sqrt(2)) = (-ln4)/(0.5 ln2) = (-2 ln2)/(0.5 ln2) = (-2)/(0.5) = -4. So that's the same result as before. So exponent is 2 - 4 = -2. So 2^{-2} = 1/4. Yes, that seems correct.Alternatively, maybe manipulate the original expression directly. Let's see:2^{2 + log_{sqrt(2)} (1/4)} = 2^2 * 2^{log_{sqrt(2)} (1/4)}. Because when you have a^{b + c} = a^b * a^c. So 4 * 2^{log_{sqrt(2)} (1/4)}.Now, compute 2^{log_{sqrt(2)} (1/4)}. Let me let y = log_{sqrt(2)} (1/4). Then sqrt(2)^y = 1/4. As before, sqrt(2)^y = 2^{y/2} = 1/4 = 2^{-2}, so y/2 = -2 => y = -4. So 2^y = 2^{-4} = 1/16. Wait, but that contradicts the previous step. Wait, hold on.Wait, if 2^{log_{sqrt(2)} (1/4)} = 2^{-4} = 1/16? Then the total expression would be 4 * 1/16 = 1/4. Which matches the previous result. So that's correct. But let me make sure. Let me see:Suppose I have 2^{log_{sqrt(2)} (1/4)}. Let me express 2 as (sqrt(2))^2. So (sqrt(2))^2^{log_{sqrt(2)} (1/4)} = (sqrt(2))^{2 * log_{sqrt(2)} (1/4)}. Then, using the property that a^{log_a(b)} = b. So if I have (sqrt(2))^{log_{sqrt(2)} (1/4)}^2. Wait, that's (1/4)^2 = 1/16. So yes, that works. Therefore, 2^{log_{sqrt(2)} (1/4)} = 1/16. Then 4 * 1/16 = 1/4. So that's the same as before. Okay, so that confirms the answer is 1/4.Alternatively, maybe use natural logarithm properties. Let me think. Let’s denote log_{sqrt(2)}(1/4) as x. Then (sqrt(2))^x = 1/4. Taking natural logs: x * ln(sqrt(2)) = ln(1/4). So x = ln(1/4)/ln(sqrt(2)) = (-ln4)/(0.5 ln2) = (-2 ln2)/(0.5 ln2) = -4. So x = -4. Therefore, exponent is 2 + (-4) = -2. So 2^{-2} = 1/4. Yep, same answer.Okay, so both problems seem to check out. The first one simplifies to 1, and the second one simplifies to 1/4. Let me just recap to make sure I didn't skip any steps or make any miscalculations.For problem (1):Numerator: lg2 + lg5 - lg8= lg(2*5) - lg8= lg10 - lg8= lg(10/8)= lg(5/4)Denominator: lg50 - lg40= lg(50/40)= lg(5/4)Therefore, (lg(5/4)) / (lg(5/4)) = 1. Perfect, that's straightforward.For problem (2):2^{2 + log_{sqrt(2)}(1/4)}First compute log_{sqrt(2)}(1/4):Let sqrt(2)^x = 1/4sqrt(2) = 2^{1/2}, so 2^{x/2} = 2^{-2}Therefore, x/2 = -2 => x = -4So exponent is 2 + (-4) = -2Therefore, 2^{-2} = 1/4Yep, both solutions seem solid. I don't see any mistakes here. So the answers should be 1 and 1/4.Final Answer(1) boxed{1}(2) boxed{dfrac{1}{4}}</think>(1) To simplify the expression (frac{lg 2 + lg 5 - lg 8}{lg 50 - lg 40}):- Numerator: [ lg 2 + lg 5 - lg 8 = lg(2 cdot 5) - lg 8 = lg 10 - lg 8 = lgleft(frac{10}{8}right) = lgleft(frac{5}{4}right) ]- Denominator: [ lg 50 - lg 40 = lgleft(frac{50}{40}right) = lgleft(frac{5}{4}right) ]- Therefore, the expression simplifies to: [ frac{lgleft(frac{5}{4}right)}{lgleft(frac{5}{4}right)} = 1 ](2) To evaluate the expression (2^{2 + log_{sqrt{2}} frac{1}{4}}):- First, compute (log_{sqrt{2}} frac{1}{4}): [ text{Let } log_{sqrt{2}} frac{1}{4} = x implies (sqrt{2})^x = frac{1}{4} ] [ sqrt{2} = 2^{1/2} implies (2^{1/2})^x = 2^{-2} implies 2^{x/2} = 2^{-2} implies frac{x}{2} = -2 implies x = -4 ]- Substitute back into the exponent: [ 2 + (-4) = -2 ]- Therefore, the expression simplifies to: [ 2^{-2} = frac{1}{4} ] Final Answer(1) (boxed{1})(2) (boxed{dfrac{1}{4}})]

🔑:Solution: (1) The original expression can be rewritten as frac {lg5+lg4}{1+lg5-(1+lg4)} which simplifies to boxed{1}, (2) The original expression can be rewritten as 2^2 cdot 2^{log_{ sqrt {2}} frac {1}{4}} which simplifies to 4 times 2^{log_{2} frac {1}{16}} which further simplifies to 4 times frac {1}{16}, resulting in boxed{frac {1}{4}}.