answer:1. **Problem Interpretation and Setup**: - We are given a trapezoid with sides a, b, c where a parallel c. - The angle gamma between sides b and c is twice the angle alpha opposite to it. - We need to determine under what conditions this trapezoid can be constructed and how many solutions exist. 2. **Assumptions and Initial Analysis**: - Assume the problem is already solved and give a labeling based on the provided diagram. Note the critical point: gamma = 2alpha. - Label the sides such that side a is the bottom base, side c is the top base, and sides b are the non-parallel sides of the trapezoid. - Understand the problem constraints and approach: - Since a parallel c, the difference between bases must explain the height h formed by sides b. - Let's consider the relationships given by trigonometry in the trapezoid. 3. **Analyzing the Relationship Between the Sides**: - Given gamma = 2alpha, we use properties of triangles formed within the trapezoid. - The given configuration also implies a specific trigonometric relationship: [ cos(gamma) = cos(2alpha) = 2cos^2(alpha) - 1 quad text{(Double-Angle Formula for Cosine)} ] 4. **Examining Whether the Condition Holds**: - For the trapezoid construction, we evaluate side b based on given angles. From geometry and the cosine of an angle, if gamma = 2alpha, the side b can relate to a and c - Considering the trapezoid properties, particularly its parallel sides: [ b sin(gamma) = a - c quad text{(Using relationships between sides in trapezoids and triangles)} ] 5. **Critical Condition for Side b**: - If we solve for b, simplify the problem to check: [ b = a - c ] - This suggests that for the trapezoid: - ( b ) should be exactly equal to ( a - c ). - If b neq a - c, the sides contradict each other. 6. **Solution Constraints and Uniqueness**: - If the given conditions are met, specifically b = a - c, the task of constructing such a trapezoid reveals: - There is redundancy (not enough independent information). - Infinite configurations can exist as the item is not sufficiently defined. 7. **Conclusion**: - **No Solution**: If b neq a - c. - **Infinite Solutions**: If b = a - c due to lack of enough constraints to form a unique trapezoid. Final conclusion: [ boxed{text{No unique solution or construction possible}} ]

answer:For all vectors mathbf{a} and mathbf{b}, by the Triangle Inequality, [|mathbf{a} + mathbf{b}| le |mathbf{a}| + |mathbf{b}|.] In particular, [left| mathbf{w} + begin{pmatrix} 4 -2 end{pmatrix} right| le |mathbf{w}| + left| begin{pmatrix} 4 -2 end{pmatrix} right|.] Therefore, [|mathbf{w}| ge left| mathbf{w} + begin{pmatrix} 4 -2 end{pmatrix} right| - left| begin{pmatrix} 4 -2 end{pmatrix} right| = 10 - sqrt{20} = 10 - 2sqrt{5}.] Equality occurs when we take [mathbf{w} = frac{10 - 2sqrt{5}}{sqrt{20}} begin{pmatrix} 4 -2 end{pmatrix} = frac{10}{sqrt{20}} begin{pmatrix} 4 -2 end{pmatrix} - begin{pmatrix} 4 -2 end{pmatrix},] so the smallest possible value of |mathbf{w}| is boxed{10 - 2sqrt{5}}.

answer:1. Given the tetrahedron (ABCD), all faces of which are similar right triangles with acute angles at vertices (A) and (B), and (angle CBD = alpha) and (angle CDB = beta) with (alpha + beta = 90^circ), we start by identifying some initial properties: - We know that (AB = 1). 2. In the right triangles (DBC) and (ABC), since (angle ABC = alpha), it follows that the triangles cannot be congruent because if they were, (CD) would equal (AC). 3. We thus identify that (angle ABC = beta) and (angle BAC = alpha). The segment (BC) is shared between the triangles (ABC) and (DBC). Thus, (BC < AC) in triangle (ABC) and (BC > CD) in triangle (DBC). It follows that: [ CD < BC < AC ] 4. Next, to show that (CD) is the shortest edge, we note: - From triangle (ABC), (BC = 1cdotcosalpha). - From triangle (ACD), (CD < AD = AC cdot cosalpha). 5. Given the equality of the edges due to the right angle properties and similarity criteria: [ AC = frac{1}{cosalpha} ] 6. Considering the relationship between (alpha) and (beta), we now have (cosbeta = sinalpha) and thus: [ AD = cosbeta = sinalpha ] 7. Now consider (CD): [ CD = BD cdot cosbeta = BC cdot cosalpha cdot cosbeta = cosalpha cdot sinalpha = cosalpha cdot (frac{sqrt{5} - 1}{2}) ] 8. Solving the quadratic equation (cosbeta = cos^2alpha): [ cos^2alpha = 1 - sin^2alpha ] Since ( alpha + beta = 90^circ), we substitute and solve: [ sinalpha = frac{sqrt{5} - 1}{2} ] 9. Substituting (sinalpha = frac{sqrt{5} - 1}{2}) back in: [ CD = (cosalpha cdot sinalpha)^{3/2} = left(frac{sqrt{5} - 1}{2}right)^{3/2} ] Therefore, the length of the shortest edge of the tetrahedron is: [ boxed{left(frac{sqrt{5}-1}{2}right)^{frac{3}{2}}} ]

answer:We start by determining the centers and radii of the two given circles. - The center of the circle x^2+y^2-2ax+a^2(1-b)=0 is (a, 0), and its radius is asqrt{b}. - The center of the circle x^2+y^2-2y+1-a^2b=0 is (0, 1), and its radius is also asqrt{b}, since the two circles are tangent to each other. For the two circles to be externally tangent, the distance between their centers should be equal to the sum of their radii. The distance between the centers (a, 0) and (0, 1) is given by: sqrt{(a-0)^2 + (0-1)^2} = sqrt{a^2 + 1}. Equating this distance to the sum of the radii, we get: sqrt{a^2 + 1} = 2asqrt{b}. Squaring both sides, we have: a^2 + 1 = 4a^2b. From this, we can express ab in terms of a: ab = frac{a^2+1}{4a} = frac{a}{4} + frac{1}{4a}. Next, we use the AM-GM inequality, which states that for non-negative numbers, the arithmetic mean is greater than or equal to the geometric mean. Applying this to frac{a}{4} and frac{1}{4a}: frac{a}{4} + frac{1}{4a} geq 2sqrt{left(frac{a}{4}right)left(frac{1}{4a}right)} = frac{1}{2}. The equality holds when frac{a}{4} = frac{1}{4a}, which happens when a=1. Since a and b are positive integers, the minimum value of ab is when a=1. Hence, the minimum value of ab is boxed{frac{1}{2}}.