answer:1. **Counting Face Diagonals**: - There are six faces on a rectangular prism. - Each rectangular face has four vertices. - The number of diagonals per face can be calculated as binom{4}{2} - 4 = 2 diagonals per face (the total number of ways to choose two vertices minus the edges). - Thus, six faces each with 2 diagonals gives 6 times 2 = 12 face diagonals. 2. **Counting Space Diagonals**: - There are twelve vertices in total in a rectangular prism. - Space diagonals are those which connect vertices not in the same face. - From each vertex, diagonals can go to all other vertices except those in the same face or connected by an edge. Since vertices in the same face are already considered (including edges), each vertex connects with vertices in opposite corners. - Each vertex in a face has three space diagonals (connecting to vertices in the opposite face), so each face contributes 4 times 3 = 12 space diagonals. However, this count must be divided by two because each diagonal is counted twice, once from each end. - 12 div 2 = 6 unique space diagonals from one face pair perspective, and there are two unique face pairs (considering direction), each contributing an additional 6, so 6 + 6 = 12. The total number of diagonals (both space and face diagonals) is 12 + 12 = boxed{24}.

answer:According to the problem, 2a+b+c = (a+c) + (a+b). Since a, b, c > 0, we have (a+c) > 0 and (a+b) > 0. Therefore, 2a+b+c = (a+c) + (a+b) geq 2sqrt{(a+c)(a+b)} = 2sqrt{6-2sqrt{5}} = 2(sqrt{5} - 1) = 2sqrt{5} - 2. Thus, the minimum value of 2a+b+c is 2sqrt{5} - 2. Hence, the correct option is: boxed{text{D}}. According to the problem, transforming 2a+b+c yields 2a+b+c = (a+c) + (a+b). By analyzing with the basic inequality, we obtain 2a+b+c = (a+c) + (a+b) geq 2sqrt{(a+c)(a+b)} = 2sqrt{6-2sqrt{5}}, and the calculation gives the answer. This problem examines the application of the basic inequality, focusing on analyzing the relationship between 2a+b+c and (a+c) cdot (a+b).

answer:Given the convex quadrilateral (ABCD), points (P) and (Q) on sides (AD) and (DC) respectively, such that (angle ABP = angle CBQ). Segments (AQ) and (CP) intersect at point (E). We need to prove that (angle ABE = angle CBD). 1. **Trilinear Coordinates Setup** Let ( (x: y: z) ) denote the trilinear coordinates relative to triangle (ABC). Given (angle ABP = angle CBQ), we conclude that points (P) and (Q) have trilinear coordinates of the form: [ P (p:u:q) quad text{and} quad Q (q:v:p) ] 2. **Equations of Lines AP and CQ** The line (AP) is given by the equation (y:z = u:q), and the line (CQ) is given by the equation (x:y = q:v). 3. **Intersection Points** To find the trilinear coordinates of point (D), we solve the simultaneous equations of (AP) and (CQ): [ frac{y}{z} = frac{u}{q} quad text{and} quad frac{x}{y} = frac{q}{v} ] This yields: [ x = frac{v q}{u}, quad y = frac{u}{q} y, quad z = frac{y q}{u} ] Normalizing to trilinear coordinates, we get: [ D left( frac{1}{v} : frac{1}{q} : frac{1}{u} right) ] 4. **Equations of Lines AQ and CP** Similarly, the lines (AQ) and (CP) have the equations: [ frac{y}{z} = frac{v}{p} quad text{and} quad frac{x}{y} = frac{p}{u} ] 5. **Intersection at Point E** Solving these equations for point (E), we obtain: [ x = frac{p}{u} y, quad frac{x}{frac{y p}{v}} = y = frac{y v}{p} ] Normalizing, we have: [ E left( frac{1}{u} : frac{1}{p} : frac{1}{v} right) ] 6. **Conclusion** From the trilinear coordinates of points (D) and (E): [ D left( frac{1}{v} : frac{1}{q} : frac{1}{u} right) quad text{and} quad E left( frac{1}{u} : frac{1}{p} : frac{1}{v} right) ] We see that since the coordinates are cyclic permutations, the configuration of the points implies that (angle ABD = angle ABE). Hence: [ boxed{angle ABE = angle CBD} ]

answer:The equations of the asymptotes are x pm ay = 0. Let P(m, n) be any point on the hyperbola. The equation of the line through P parallel to OB: x + ay = 0 is x + ay - m - an = 0. This line intersects the line OA: x - ay = 0 at point A(frac{m + an}{2}, frac{m + an}{2a}). The distance between point O and point A is |OA| = |frac{m + an}{2}| sqrt{1 + frac{1}{a^{2}}}. The distance between point P and line OA is d = frac{|m - an|}{sqrt{1 + a^{2}}}. Since the area of the parallelogram OBPA is 1, we have |OA| cdot d = 1. Therefore, |frac{m + an}{2}| sqrt{1 + frac{1}{a^{2}}} cdot frac{|m - an|}{sqrt{1 + a^{2}}} = 1. Given that frac{m^{2}}{a^{2}} - n^{2} = 1, we find a = 2. Thus, the eccentricity of the hyperbola is boxed{frac{sqrt{5}}{2}}. To solve this problem, we find |OA| and the distance from point P to line OA. Using the area of parallelogram OBPA, we find a and subsequently c, which allows us to find the eccentricity of the hyperbola. This problem tests the understanding of hyperbola equations and properties, as well as computational skills, making it a moderately difficult problem.