answer:1. **Define the Intersection Point**: Let ( M ) be the intersection point of the circumcircles of triangles ( AB_1C_1 ) and ( CA_1B_1 ), distinct from ( B_1 ). We need to show that point ( M ) lies on the circumcircle of the third triangle ( A_1BC_1 ). 2. **Internal Position Analysis**: Suppose point ( M ) lies inside triangle ( ABC ). - Using the Angle Sum Properties for Cyclic Quadrilaterals, we consider the angles: [ angle A_1BC_1 + angle A_1MC_1 = 180^circ ] because ( A_1BC_1 ) and ( A_1MC_1 ) are cyclic quadrilaterals. - We break it down: [ angle A_1BC_1 + angle A_1MC_1 = angle B + 360^circ - angle B_1MC_1 - angle A_1MB_1 ] Given that: [ angle B_1MC_1 = 180^circ - angle A ] and [ angle A_1MB_1 = 180^circ - angle C ] Therefore: [ angle A_1BC_1 + angle A_1MC_1 = angle B + 360^circ - (180^circ - angle A) - (180^circ - angle C) = angle B + angle A + angle C = 180^circ ] This implies that points ( A_1, B, M, C_1 ) lie on the same circle. 3. **External Position Analysis**: Now assume point ( M ) lies outside triangle ( ABC ). Let's particularly consider when ( M ) is inside angle ( angle BAC ). - We have: [ angle A_1MC_1 = angle B_1MC_1 - angle B_1MA_1 ] with: [ angle B_1MC_1 = 180^circ - angle A ] and: [ angle B_1MA_1 = 180^circ - angle C ] Therefore: [ angle A_1MC_1 = 180^circ - angle A - (180^circ - angle C) = angle C - angle A ] If the angle is considered such that: [ angle C - angle A = angle B ] This confirms our conclusion that: [ angle A_1MC_1 = angle B ] - Therefore points ( A_1, M, B, C_1 ) lie on the same circle. 4. **General Case Analysis**: Similarly, we can verify this argument for different positions of point ( M ) relative to triangle ( ABC ). Each configuration will confirm that point ( M ) lies on the circumcircle of ( A_1BC_1 ). # Conclusion: Hence, it is proven that the intersection points of the circles described around the given triangles ( AB_1C_1, A_1B_1C, A_1BC_1 ) all lie on the circumcircle of the third triangle. [ boxed{} ] [ blacksquare ]

answer:Given that in triangle ABC, frac{1}{sin A} + frac{2}{sin B} = 3(frac{1}{tan A} + frac{1}{tan B}), we can rewrite this as frac{1}{sin A} + frac{2}{sin B} = frac{3cos A}{sin A} + frac{3cos B}{sin B}. Simplifying further, we get frac{1 - 3cos A}{sin A} = frac{3cos B - 2}{sin B}, which simplifies to sin B + 2sin A = 3sin(A + B) = 3sin C. Applying the Law of Sines, we get b + 2a = 3c. Therefore, cos C = frac{a^2 + b^2 - c^2}{2ab} = frac{a^2 + b^2 - (frac{b + 2a}{3})^2}{2ab} = frac{5a^2 + 8b^2 - 4ab}{18ab} geq frac{2sqrt{40a^2b^2} - 4ab}{18ab} = frac{2sqrt{10} - 2}{9}. The equality holds if and only if sqrt{5}a = 2sqrt{2}b. Hence, the minimum value of cos C is boxed{frac{2sqrt{10} - 2}{9}}. This problem primarily tests the understanding of basic trigonometric identities, the Law of Sines and Cosines, and the application of basic inequalities. It is a moderately difficult problem.

answer:1. **Calculate ( b^{-1} ) and ( b^{-2} ):** Since ( b^{-1} ) is the reciprocal of ( b ), and ( b = 3 ): [ b^{-1} = left(3right)^{-1} = frac{1}{3}. ] And: [ b^{-2} = left(3right)^{-2} = left(frac{1}{3}right)^2 = frac{1}{9}. ] 2. **Substitute ( b^{-1} ) and ( b^{-2} ) into the expression:** [ frac{3 cdot frac{1}{3} + frac{frac{1}{9}}{3}}{3} = frac{1 + frac{1}{27}}{3} = frac{frac{28}{27}}{3}. ] 3. **Simplify the expression:** Simplify the fraction by multiplying the numerator by the reciprocal of the denominator: [ frac{frac{28}{27}}{3} = frac{28}{27} cdot frac{1}{3} = frac{28}{81}. ] 4. **Conclude with the final answer:** The value of the expression when ( b = 3 ) is frac{28{81}}. The final answer is boxed{textbf{(B)} frac{28}{81}}

answer:For condition (1), by the Law of Sines, we derive the equality frac{sin A}{a} = frac{sin B}{b} = frac{sin C}{c}. Combining this with the given condition, we have: frac{sin A}{a}cdotfrac{a}{cos A} = frac{sin B}{b}cdotfrac{b}{cos B} = frac{sin C}{c}cdotfrac{c}{cos C} which simplifies to: tan A = tan B = tan C Considering that angles A, B, C are all within (0, pi), we can conclude that: A = B = C Thus, triangle ABC must be an equilateral triangle. So, boxed{A = B = C} For condition (2), given that sin A=2sin Bcos C, we can rewrite this as: sin (B+C)=2sin Bcos C. Using the sum of angles identity for sine, we have: sin Bcos C + cos Bsin C = 2sin Bcos C. Simplifying, we find that: cos Bsin C = sin Bcos C, which implies: boxed{tan B = tan C}. Since B and C are in the interval (0, pi) we deduce that B = C, and hence triangle ABC must be an isosceles triangle with AB = AC. Therefore, from the second condition, we have: boxed{B = C}.