answer:Solution: ((1)) Since (overrightarrow{m}+ overrightarrow{n}=(cos A+ sqrt {2}-sin A,cos A+sin A)), (| overrightarrow{m}+ overrightarrow{n}|^{2}=(cos A+ sqrt {2}-sin A)^{2}+(cos A+sin A)^{2}) (=2+2 sqrt {2}(cos A-sin A)+(cos A-sin A)^{2}+(cos A+sin A)^{2}) (=2+2 sqrt {2}(cos A-sin A)+2) (=4-4sin (A- dfrac {π}{4})), Since (| overrightarrow{m}+ overrightarrow{n}|=2), (4sin (A- dfrac {π}{4})=0), Given (0 < A < π), (- dfrac {π}{4} < A- dfrac {π}{4} < dfrac {3π}{4}), (A- dfrac {π}{4}=0), (A= dfrac {π}{4}). ((2)) By the law of cosines, (a^{2}=b^{2}+c^{2}-2bccos A), given (b=4 sqrt {2}), (c= sqrt {2}a), (A= dfrac {π}{4}), we get: (a^{2}=32+2a^{2}-2×4 sqrt {2}× sqrt {2}a⋅ dfrac { sqrt {2}}{2}), which simplifies to: (a^{2}-8 sqrt {2}a+32=0), solving this gives (a=4 sqrt {2}), Therefore, (c=8), Thus, the area of (triangle ABC) is (S_{triangle ABC}= dfrac {1}{2}b⋅csin A= dfrac {1}{2}×4 sqrt {2}×8×sin dfrac {π}{4}=boxed{16}).

answer:Given: - In triangle ABC, the bisector AK is perpendicular to the median BM, and angle B = 120^circ. To find: The ratio of the area of triangle ABC to the area of the circumcircle. 1. **Identify known values and relationships:** - Let AC = 2a. - angle B = 120^circ. 2. **Find the radius of the circumcircle:** - Using the formula for the circumradius R of triangle ABC: [ R = frac{a}{2 sin angle B} = frac{2a}{2 sin 120^circ} = frac{2a}{sqrt{3}} ] 3. **Calculate the area of the circumcircle:** - The area A_{text{circ}} of the circumcircle is given by A_{text{circ}} = pi R^2: [ A_{text{circ}} = pi left(frac{2a}{sqrt{3}}right)^2 = frac{4pi a^2}{3} ] 4. **Analyze the triangle configuration and properties:** - Since the bisector AK from A is perpendicular to the median BM and angle B = 120^circ, triangle ABM is isosceles with AB = AM = a. 5. **Determine the length of BC:** - Using the Law of Cosines in ABC: [ BC^2 = AB^2 + AC^2 - 2 cdot AB cdot AC cdot cos angle B ] [ BC^2 = a^2 + (2a)^2 - 2 cdot a cdot 2a cdot cos 120^circ ] [ BC^2 = a^2 + 4a^2 - 2a^2 (-frac{1}{2}) ] [ BC^2 = a^2 + 4a^2 + a^2 = 6a^2 ] [ BC = sqrt{6}a ] 6. **Calculate the area of triangle ABC:** - Using the formula for the area of a triangle: [ S_{ABC} = frac{1}{2} cdot AB cdot AC cdot sin angle BAC ] - Here, angle BAC = 120^circ, thus [ S_{ABC} = frac{1}{2} cdot a cdot 2a cdot sin 120^circ ] [ S_{ABC} = frac{1}{2} cdot a cdot 2a cdot frac{sqrt{3}}{2} ] [ S_{ABC} = frac{sqrt{3} a^2}{2} ] 7. **Find the required ratio:** - The ratio of the area of triangle ABC to the area of its circumcircle: [ text{Ratio} = frac{S_{ABC}}{A_{text{circ}}} = frac{frac{sqrt{3} a^2}{2}}{frac{4 pi a^2}{3}} ] [ text{Ratio} = frac{3 sqrt{3} a^2}{8 pi a^2} ] [ text{Ratio} = frac{3 sqrt{3}}{8 pi} ] Hence, the ratio of the area of triangle ABC to the area of the circumcircle is: [ boxed{frac{3 sqrt{3}}{32 pi}} ]

answer:To find the volume of the region defined by the inequalities |x|+|y|+|z|le1 and |x|+|y|+|z-1|le1, we can proceed as follows: 1. **Consider the first inequality in the positive octant**: In the octant where x ge 0, y ge 0, and z ge 0, the inequality |x| + |y| + |z| le 1 simplifies to x + y + z le 1. This defines a tetrahedron with vertices at (0,0,0), (1,0,0), (0,1,0), and (0,0,1). 2. **Extend to the entire space**: By symmetry, the region defined by |x| + |y| + |z| le 1 is an octahedron with vertices at (pm 1,0,0), (0,pm 1,0), and (0,0,pm 1). 3. **Consider the second inequality**: Similarly, the region defined by |x| + |y| + |z - 1| le 1 is also an octahedron, but this one is centered at (0,0,1). 4. **Intersection of the two octahedra**: The intersection of these two octahedra forms another octahedron. This is because the upper half of the first octahedron intersects with the lower half of the second octahedron. The intersection includes the upper half of the pyramid with base ABCD and apex E, and the lower half of the pyramid with base A'B'C'D' and apex E'. 5. **Calculate the volume of the intersection**: The volume of the pyramid ABCDE is given by frac{1}{3} cdot text{base area} cdot text{height}. The base is a square with side length sqrt{2}, so its area is (sqrt{2})^2 = 2. The height is 1. Thus, the volume of the pyramid is frac{1}{3} cdot 2 cdot 1 = frac{2}{3}. The upper half of this pyramid has a volume of left( frac{1}{2} right)^3 cdot frac{2}{3} = frac{1}{12}. Since the intersection is composed of two such halves, the total volume is 2 cdot frac{1}{12} = frac{1}{6}. Therefore, the volume of the region defined by the given inequalities is boxed{frac{1}{6}}.

answer:Given the conditions in the problem: (a, b, c, d) are odd numbers, (0 < a < b < c < d), and additionally, [ a cdot d = b cdot c ] [ a + d = 2^k ] [ b + c = 2^m ] where (k) and (m) are integers. We need to find the value of (a). 1. Start by analyzing the implications of the conditions: [ (a+d)^2 = (a-d)^2 + 4ad ] [ (b+c)^2 = (b-c)^2 + 4bc ] Since (ad = bc) and (a < b < c < d), we deduce: [ (a+d)^2 = (a-d)^2 + 4ad ] Hence, [ (a+d)^2 > (b+c)^2 ] Given (a + d = 2^k) and (b + c = 2^m), we conclude: [ 2^k > 2^m Rightarrow k > m ] 2. Next, consider the individual products and sums: [ a(2^k - a) = b(2^m - b) ] Rewriting, we have: [ a2^k - a^2 = b2^m - b^2 ] Simplify and rearrange: [ (b+a)(b-a) = 2^m(b - 2^{k-m} a) ] We see that: [ 2^m mid (b+a)(b-a) ] Since (a) and (b) are odd numbers, both (b+a) and (b-a) are even numbers, but their difference (2a) is not a multiple of (4). Therefore, at least one of (b+a) or (b-a) is not divisible by (4). 3. Therefore, either (2^{m-1} mid (b+a)) or (2^{m-1} mid (b-a)), but: [ 0 < b - a < b < frac{1}{2}(b+c) = 2^{m-1} ] [ b + a < b + c = 2^m ] Hence, we need: [ 2^{m-1} mid (b+a) ] and [ b + a = 2^{m-1} ] 4. Finally, we establish that: [ b - a = 2(b - 2^{k-m}a) ] leading to: [ a = 2^{2m-k-2} ] Since (a) is an odd number, we must have (a = 1). # Conclusion The value of (a) is [ boxed{1} ]