answer:We are asked to find the flux of the vector field (mathbf{a} = mathbf{i} - mathbf{j} + x y z mathbf{k}) through the circle (S) obtained by the intersection of the sphere (x^{2} + y^{2} + z^{2} leqslant R^{2}) with the plane (y = x). The chosen side of the circle should be oriented towards the positive part of the (O x) axis. Let's break down the steps to find this flux. 1. **Intersection of plane and sphere:** The plane (y = x) intersects the sphere; we need to find the equation describing this intersection. [ left{ begin{aligned} x^{2} + y^{2} + z^{2} &= R^{2} y &= x end{aligned} right. ] Substituting (y = x) into the first equation: [ x^{2} + x^{2} + z^{2} = R^{2} implies 2x^{2} + z^{2} = R^{2} ] This is the equation of an ellipse in the (x z)-plane: [ frac{x^{2}}{R^{2}/2} + frac{z^{2}}{R^{2}} = 1 ] 2. **Normal vector to the plane:** The normal vector (mathbf{n}) to the plane (y = x) can be found from the gradient: [ mathbf{n} = -nabla (y - x) = mathbf{i} - mathbf{j} ] The unit normal vector is: [ mathbf{n}^{0} = frac{mathbf{n}}{|mathbf{n}|} = frac{mathbf{i} - mathbf{j}}{sqrt{2}} ] 3. **Angle between normal vector and (O y)-axis:** Since (mathbf{n}^0 = frac{1}{sqrt{2}}(mathbf{i} - mathbf{j})), the cosine of the angle (beta) between (mathbf{n}) and the (O y)-axis is: [ cos beta = -frac{1}{sqrt{2}} < 0 ] 4. **Element of the surface area (dS):** Given the angle (beta), the element of the surface area (dS) is: [ dS = frac{dx dz}{|cos beta|} = frac{dx dz}{|- frac{1}{sqrt{2}}|} = sqrt{2} dx dz ] 5. **Scalar product of (mathbf{a}) and (mathbf{n}^0):** Evaluating the scalar product (left(mathbf{a}, mathbf{n}^0right)): [ left(mathbf{a}, mathbf{n}^0right) = left(mathbf{i} - mathbf{j} + x y z mathbf{k}, frac{1}{sqrt{2}} (mathbf{i} - mathbf{j})right) = sqrt{2} ] 6. **Flux calculation:** To find the flux, we integrate: [ Pi = iint_{D_{xz}} sqrt{2} , dx , dz ] The region (D_{xz}) is the ellipse (frac{x^{2}}{R^{2}/2} + frac{z^{2}}{R^{2}} = 1). The area of this ellipse is: [ text{Area} = pi cdot frac{R}{sqrt{2}} cdot R = frac{pi R^{2}}{sqrt{2}} ] So the flux is: [ Pi = sqrt{2} cdot text{Area} = sqrt{2} cdot frac{pi R^{2}}{sqrt{2}} = pi R^{2} ] # Conclusion: The flux of the vector field through the circle is: [ boxed{pi R^{2}} ]

answer:1. **Establish the Midpoints and Connect Them**: Denote the tetrahedron by vertices (A, B, C,) and (S). Consider the midpoints of the opposite edges: - (M) is the midpoint of (AS) - (N) is the midpoint of (SC) - (P) is the midpoint of (SB) - (K) is the midpoint of (AB) - (L) is the midpoint of (BC) - (R) is the midpoint of (AC) 2. **Define the Midsegment Lines**: We connect the midpoints of these edges: - (ML) connects (M) and (L) - (KN) connects (K) and (N) - (PR) connects (P) and (R) 3. **Intersecting Midsegment Lines**: We aim to show these three line segments intersect at a single point. This stems from the fact that each of these segments is a midsegment of a parallelogram formed by connecting midpoints of triangles. 4. **Creating Parallelogram**: For example, let’s demonstrate it with (KN) and (ML): - (ML) and (KN) intersect because (ML) and (KN) are midsegments of triangles. - (KL) is the midsegment of triangle (ABC) parallel to (BC) and equal to half of (BC). - (MN) is the midsegment of triangle (ASC) parallel to (AC) and equal to half of (AC). 5. **Intersection Analysis**: - By observing the parallelogram (MNKL), the midsegments actually intersect at the point that divides each other equally. This behavior is mirrored for (PR) and the others line segment. 6. **Utilizing Triangle Inequality**: - From triangle (KLN), we have: [ KN < KL + LN = frac{SB}{2} + frac{AC}{2} ] - From triangle (PML), we have: [ ML < MP + PL = frac{AB}{2} + frac{SC}{2} ] - From triangle (PKR), we have: [ PR < KP + KR = frac{AS}{2} + frac{BC}{2} ] 7. **Summation of Inequalities**: - Adding these inequalities, we get: [ KN + ML + PR < frac{SB}{2} + frac{AC}{2} + frac{AB}{2} + frac{SC}{2} + frac{AS}{2} + frac{BC}{2} ] 8. **Final Step**: - Since each edge length is halved and summed over which includes each edge exactly once, we obtain: [ KN + ML + PR < text{half the sum of the edges of the tetrahedron} ] Thus, the summation of half the length of each edge is greater than the total length of the midsegments combining at one point. [ boxed{text{The sum of the three obtained segments is less than the half-sum of the edges of the tetrahedron.}} ]

answer:We use the exhaustive method to list all three-digit numbers that are perfect squares: 10^2 = 100, 11^2 = 121, …, 31^2 = 961. We can find that only 19^2 = 361 meets the requirements, where 36 div 1 = 6^2. Therefore, this three-digit number is 361. Hence, the answer is boxed{361}.

answer:Given: sum_{m, n=1}^{2000} frac{a_{m} b_{n}}{(sqrt{m}+sqrt{n})^{2}} leq 2left(sum_{m=1}^{2000} a_{m}^{2}right)^{frac{1}{2}} left(sum_{n=1}^{2000} b_{n}^{2}right)^{frac{1}{2}} 1. **Expression Division**: Consider the left-hand side and rewrite it in a factored form: begin{array}{c} sum_{m, n=1}^{2000} left(frac{a_{m}}{sqrt{m}+sqrt{n}} left(frac{m}{n}right)^{frac{1}{4}}right) cdot left(frac{b_{n}}{sqrt{m}+sqrt{n}} left(frac{n}{m}right)^{frac{1}{4}}right). end{array} 2. **Applying Cauchy-Schwarz Inequality**: By the Cauchy-Schwarz inequality, left(sum_{i=1}^{p} u_{i} v_{i}right)^2 leq left(sum_{i=1}^{p} u_{i}^2right) left(sum_{i=1}^{p} v_{i}^2right), therefore, begin{array}{c} left(sum_{m, n=1}^{2000} left(frac{a_{m}}{sqrt{m}+sqrt{n}} left(frac{m}{n}right)^{frac{1}{4}}right) left(frac{b_{n}}{sqrt{m}+sqrt{n}} left(frac{n}{m}right)^{frac{1}{4}}right)right) leq left(sum_{m, n=1}^{2000} left(frac{a_{m}}{sqrt{m}+sqrt{n}} left(frac{m}{n}right)^{frac{1}{4}}right)^2right)^{frac{1}{2}} left(sum_{m, n=1}^{2000} left(frac{b_{n}}{sqrt{m}+sqrt{n}} left(frac{n}{m}right)^{frac{1}{4}}right)^2right)^{frac{1}{2}}. end{array} 3. **Simplifying the Summation**: Simplifying the above inequality, observe that both summations are similar, so we consider only one, sum_{m=1}^{2000} a_{m}^2 sum_{n=1}^{2000} left(frac{1}{(sqrt{m}+sqrt{n})^2} left(frac{m}{n}right)^{frac{1}{2}}right), and similarly sum_{n=1}^{2000} b_{n}^2 sum_{m=1}^{2000} left(frac{1}{(sqrt{m}+sqrt{n})^2} left(frac{n}{m}right)^{frac{1}{2}}right). 4. **Bound of the Series**: We now need to show that: sum_{n=1}^{2000} left(frac{1}{(sqrt{m}+sqrt{n})^2} left(frac{m}{n}right)^{frac{1}{2}}right) leq 2, and similarly for sum_{m=1}^{2000} left(frac{1}{(sqrt{m}+sqrt{n})^2} left(frac{n}{m}right)^{frac{1}{2}}right) leq 2. These two inequalities are symmetric with respect to ( m ) and ( n ). 5. **Detailed Calculations for the Series Bound**: Consider, frac{1}{(sqrt{m} + sqrt{n})^2} left(frac{m}{n}right)^{frac{1}{2}} leq 2 left(frac{1}{sqrt{frac{n-1}{m}} + 1} - frac{1}{sqrt{frac{n}{m}} + 1}right). This can be rewritten as, frac{1}{(sqrt{m} + sqrt{n})^2} left(frac{m}{n}right)^{frac{1}{2}} leq 2left(frac{sqrt{m}}{sqrt{n-1} + sqrt{m}} - frac{sqrt{m}}{sqrt{n} + sqrt{m}}right). Therefore, frac{1}{(sqrt{m}+sqrt{n})^2} cdot frac{1}{sqrt{n}} leq 2 frac{sqrt{n} - sqrt{n-1}}{(sqrt{n-1} + sqrt{m})(sqrt{n} + sqrt{m})}. Since, (sqrt{n-1} + sqrt{m})(sqrt{n} + sqrt{m}) leq (sqrt{m} + sqrt{n})^2, and 2(sqrt{n} - sqrt{n-1}) = frac{2}{sqrt{n} + sqrt{n-1}} geq frac{1}{sqrt{n}}, we can infer that the inequalities hold, establishing a bound. 6. **Conclusion**: Using the sum calculations, sum_{n=1}^{2000}left( frac{1}{(sqrt{m}+sqrt{n})^2}left( frac{m}{n}right)^{frac{1}{2}}right) leq 2, and the Cauchy-Schwarz applications, sum_{m, n=1}^{2000} frac{a_{m} b_{n}}{(sqrt{m}+sqrt{n})^2} leq 2left(sum_{m=1}^{2000} a_{m}^2right)^{frac{1}{2}} left(sum_{n=1}^{2000} b_{n}^2right)^{frac{1}{2}}. Hence, [ boxed{sum_{m, n=1}^{2000} frac{a_{m} b_{n}}{(sqrt{m}+sqrt{n})^2} leq 2left(sum_{m=1}^{2000} a_{m}^{2}right)^{frac{1}{2}} left(sum_{n=1}^{2000} b_{n}^{2}right)^{frac{1}{2}}.} ]