answer:(1) We have f(x)=2sin xcdotdfrac {1+cos θ}{2}+cos xcdotsin θ-sin x=sin (x+θ). Since f(x) takes its minimum value at x=π, we have sin (π+θ)=-1, which implies sin θ=1. Given 0 < θ < π, we obtain θ= dfrac {π}{2}. (2) From part (1), we know f(x)=cos x. Thus, f(A)=cos A= frac { sqrt {3}}{2}, which implies A= dfrac {π}{6} since A is an angle of triangle ABC. Using the sine law, we get sin B= dfrac {bsin A}{a}= dfrac { sqrt {2}}{2}. Thus, B= dfrac {π}{4} or B= dfrac {3π}{4}. If B= dfrac {π}{4}, then C=π-A-B= dfrac {7π}{12}. If B= dfrac {3π}{4}, then C=π-A-B= dfrac {π}{12}. In summary, the possible values for angle C are boxed{C= dfrac {7π}{12}text{ or }C= dfrac {π}{12}}.

answer:1. **Initial Setup**: Consider ( triangle ABC ) with points ( D ), ( E ), and ( F ) on sides ( BC ), ( CA ), and ( AB ) respectively. Denote the intersections of segments as: - ( X ) = intersection of ( BE ) and ( CF ) - ( Y ) = intersection of ( CF ) and ( AD ) - ( Z ) = intersection of ( AD ) and ( BE ) 2. **Intersection Analysis and Circles**: Given ( triangle ABC ) and its associated points and lines, we need to prove that the circumscribed circles (( text{circumcircles} )) of triangles ( triangle XBC ), ( triangle YCA ), and ( triangle ZAB ) have a common point. 3. **Existence of a Common Point**: Suppose that the circumcircles of ( triangle YCA ) and ( triangle ZAB ) intersect at a point ( P ) different from ( A ) and ( B ). Connect ( P ) to vertices ( A ), ( B ), and ( C ). 4. **Angle Correspondences**: From our assumption that ( P ) lies on the circumcircles of ( triangle YCA ) and ( triangle ZAB ), we can infer the following angle relations: - Because ( P ) lies on the circumcircle of ( triangle YCA ): [ angle CPA = angle CYA ] - Because ( P ) lies on the circumcircle of ( triangle ZAB ): [ angle APB = angle AZB ] 5. **Derivation of ( angle BPC )**: Using the angle sum around point ( P ): [ angle BPC = 2pi - angle CPA - angle APB ] Substituting the known angle relations: [ angle BPC = 2pi - angle CYA - angle AZB ] Notice that these angles are key to our goal because: [ angle CYA + angle AZB = angle BXC ] Therefore: [ angle BPC = angle BXC ] 6. **Conclusion**: Since ( angle BPC = angle BXC ), it follows that ( P ), ( B ), ( C ), and ( X ) are concyclic. Thus, point ( P ) lies on the circumcircle of ( triangle XBC ) as well. Hence, the circumcircles of ( triangle XBC ), ( triangle YCA ), and ( triangle ZAB ) intersect at the common point ( P ). [ boxed{P} ]

answer:Let's denote the given variables clearly: - Let R be the radius of the circumscribed circle (the outer circle). - Let r be the radius of the inscribed circle (the inner circle). - Let x be the perimeter of the convex n-sided polygon. - Let A be the area of the outer circumscribed circle. - Let C be the area of the inner inscribed circle. - Let B be the area of the convex n-sided polygon. First, recall the formulas for the areas of the circles: - The area of the circumscribed circle (with radius R) is given by: [ A = pi R^2 ] - The area of the inscribed circle (with radius r) is given by: [ C = pi r^2 ] Next, we consider the perimeter of the convex n-sided polygon, x. Because the polygon fits inside the outer circle and touches it at its vertices, we know that the perimeter x is less than the circumference of the outer circle: [ x < 2 pi R ] The area B of the convex polygon can be represented using its perimeter x and the radius of the inscribed circle r: [ B = frac{1}{2} x r ] Given that x < 2 pi R, we substitute 2 pi R for x in the expression for B: [ B < frac{1}{2} (2 pi R) r ] Simplify the right side: [ B < pi R r ] Next, we multiply both sides of the inequality by 2 to compare 2B with A + C: [ 2 B < 2 pi R r ] Observe the following inequality: [ 2 pi R r < pi (R^2 + r^2) ] This inequality holds true because for any positive real numbers R and r, the arithmetic mean-geometric mean inequality (AM-GM inequality) implies: [ R^2 + r^2 geq 2 R r ] Thus it follows that: [ pi (R^2 + r^2) geq pi 2 R r ] Therefore, we can write: [ 2 pi R r < pi R^2 + pi r^2 ] By the definitions of A and C, we have A = pi R^2 and C = pi r^2, so: [ pi R^2 + pi r^2 = A + C ] Hence, combining all the inequalities together, we get: [ 2 B < pi ( R^2 + r^2 ) = A + C ] Conclusively proving the required inequality: [ 2 B < A + C ] Thus, the proof is complete: [ blacksquare ]

answer:1. **Geometry Setup**: The eight place mats create a regular octagonal formation. 2. **Circumference and Arc Length**: The circumference of the circle is (2pi times 5 = 10pi). The arc length between adjacent points (corners of place mats on the table edge) is (frac{10pi}{8} = frac{5pi}{4}). 3. **Chord Length Calculation**: The angle subtending each chord in the circle (central angle for each octagon side) is (frac{2pi}{8} = frac{pi}{4}). Using the chord length formula (2R sin(theta/2)), we have: [ text{Chord length} = 2 times 5 times sinleft(frac{pi}{8}right) = 10 times sinleft(frac{pi}{8}right). ] Evaluating (sinleft(frac{pi}{8}right) = frac{sqrt{2 - sqrt{2}}{2}), we find: [ text{Chord length} = 10 times frac{sqrt{2 - sqrt{2}}}{2} = 5sqrt{2 - sqrt{2}}. ] 4. **Finding ( x )**: Since the chord length represents the length ( x ), we have: [ x = 5sqrt{2 - sqrt{2}}. ] This is the desired length of the side of each place mat. Conclusion with boxed answer: The length ( x ) of each place mat along the edge of the table is (5sqrt{2 - sqrt{2}}). The final answer is boxed{The correct choice is (boxed{mathrm{D.} 5sqrt{2 - sqrt{2}}}).}