answer:Given ((overrightarrow{OB} - overrightarrow{OC}) cdot (overrightarrow{OB} + overrightarrow{OC} - 2overrightarrow{OA}) = 0), it can be rewritten as ((overrightarrow{OB} - overrightarrow{OC}) cdot (overrightarrow{OB} - overrightarrow{OA} + overrightarrow{OC} - overrightarrow{OA}) = 0), thus (overrightarrow{BC}(overrightarrow{AB} + overrightarrow{AC}) = 0), which leads to ((overrightarrow{AB} - overrightarrow{AC}) cdot (overrightarrow{AB} + overrightarrow{AC}) = 0), meaning (|overrightarrow{AB}|^2 - |overrightarrow{AC}|^2 = 0), so (|overrightarrow{AB}| = |overrightarrow{AC}|), therefore, triangle (ABC) is an isosceles triangle. Hence, the correct answer is: boxed{C}. This problem involves converting the given vector equation involving vectors (overrightarrow{OA}, overrightarrow{OB}, overrightarrow{OC}) into a relationship between the sides of the triangle, thereby determining the shape of the triangle. It tests the knowledge of determining the shape of a triangle, involving the parallelogram law of vector addition and subtraction, scalar product operation of plane vectors, modulus operation of plane vectors, and the criteria for determining an isosceles triangle. Mastery of the scalar product operation rules of plane vectors is key to solving this problem.

answer:First, let's simplify the denominator. We know that i^2 = -1, so i^3 = i cdot i^2 = i cdot -1 = -i. Now, the expression becomes: frac{1+i}{1-i} To eliminate the imaginary number in the denominator, we multiply both the numerator and denominator by the conjugate of the denominator: frac{(1+i)}{(1-i)} cdot frac{(1+i)}{(1+i)} = frac{(1+i)^2}{(1-i)(1+i)} Now, let's compute the numerator and denominator separately: - Numerator: (1+i)^2 = 1^2 + 2cdot 1 cdot i + i^2 = 1 + 2i - 1 = 2i - Denominator: (1-i)(1+i) = 1^2 - i^2 = 1 - (-1) = 2 Thus, the expression simplifies to: frac{2i}{2} Finally, we can cancel out the common factor of 2 to get the answer: boxed{i}

answer:1. **Identify I, incenter I and given conditions:** - Let ( I ) be the incenter of ( triangle ABC ). - Point ( D ) is outside the triangle such that ( DA parallel BC ) and ( DB = AC ), but ( ABCD ) is not a parallelogram. 2. **Draw perpendicular from I to BC:** - Let ( IX ) meet ( BC ) at ( Z ). 3. **Identify relations using tangents:** - By properties of equal tangents from a point to a circle, we have: [ text{BI} = text{BZ} quad text{and} quad text{CI} = text{CZ}. ] - As ( I ) is the incenter: [ text{BI} + text{CI} = overline{BC}. ] - Combining these equalities, we get: [ text{BZ} + text{CZ} = overline{BC}. ] 4. **Summing distances using AB, BC, AC:** - Express ( CZ ): [ text{AC} + text{BC} - text{AB} = 2 cdot text{CZ}. ] - Simplifying, we have: [ text{CZ} = frac{text{AC} + text{BC} - text{AB}}{2}. ] 5. **Locating the excircle contact point with BC:** - Let ( Z' ) be the point where the excircle (opposite ( D )) of ( triangle DBC ) touches ( BC ). 6. **Equal tangents from B and C:** - Similar approach as above for the tangents: [ text{DB} + (overline{BC} - text{CZ}') = text{DC} + text{CZ}'. ] - From ( DB = AC ) and ( DC = AB ), we have: [ text{CZ}' = frac{DB + overline{BC} - DC}{2} = frac{text{AC} + overline{BC} - text{AB}}{2}. ] Next up, we conclude ( Z' ) coincides with ( Z ): - So we have established: [ Z' = Z. ] 7. **Identifying X as the center of the excircle:** - ( X ) lies on the perpendicular to ( BC ) at ( Z ) and on the bisector of ( angle BDC ). - This makes ( X ) the center of the excircle. 8. **X lies on the circumcircle of triangle CDX:** - Since ( triangle CDX ) is cyclic, ( angle YDX ) is subtended by the same arc as ( angle YCX ). - Thus, ( angle YDX = angle YCX = angle XCB ). 9. **Analyzing angles using cyclic properties:** - We have: [ angle XCB = 90^circ - frac{angle BDC}{2}. ] - Using cyclic quadrilateral property ( angle BDC = angle YCD ): [ angle YXD = angle YCD = angle BDC. ] 10. **Compare angles to show isosceles condition:** - Hence, [ angle YDX = 90^circ - frac{angle YXD}{2}. ] - From which it follows: [ YX = DX. ] Therefore, ( DXY ) is isosceles as required: [ boxed{YX = DX}. ]

answer:# Step-by-Step Solution Part (1): Finding the Smallest Positive Period 1. **Given Function Transformation** The given function is f(x)=2sin x cdot cos x + 2sqrt{3}cos^2 x - sqrt{3}. We can use trigonometric identities to transform it as follows: begin{align*} f(x) & = 2sin x cdot cos x + 2sqrt{3}cos^2 x - sqrt{3} & = sin 2x + 2sqrt{3} left(frac{1+cos 2x}{2}right) - sqrt{3} & = sin 2x + sqrt{3}cos 2x & = 2 left(frac{1}{2} sin 2x + frac{sqrt{3}}{2} cos 2xright) & = 2sinleft(2x+frac{pi}{3}right). end{align*} 2. **Determining the Smallest Positive Period** Since the function has been simplified to 2sinleft(2x+frac{pi}{3}right), we recognize that the argument of the sine function is 2x + frac{pi}{3}. The coefficient of x inside the sine function indicates the rate of oscillation. The period of sin(2x) is pi, hence the smallest positive period of f(x) is also boxed{pi}. Part (2): Finding the Solution Set of the Inequality f(x) geq 1 1. **Inequality Transformation** From part (1), we have f(x) = 2sinleft(2x+frac{pi}{3}right). For f(x) geq 1, we need sinleft(2x+frac{pi}{3}right) geq frac{1}{2}. This translates to the inequality: begin{align*} 2kpi + frac{pi}{6} & leq 2x+frac{pi}{3} leq 2kpi + frac{5pi}{6}, & kin mathbb{Z}, end{align*} which simplifies to: begin{align*} -frac{pi}{12} + kpi & leq x leq frac{pi}{4} + kpi, & kin mathbb{Z}. end{align*} 2. **Determining the Solution Set within [0,pi]** - For k=0, we have the interval [-frac{pi}{12}, frac{pi}{4}]. Intersecting this with [0,pi], we get [0, frac{pi}{4}]. - For k=1, we have the interval [frac{11pi}{12}, frac{5pi}{4}]. Intersecting this with [0,pi], we get [frac{11pi}{12}, pi]. Therefore, the solution set of the inequality f(x) geq 1 on the interval [0,pi] is boxed{[0, frac{pi}{4}] cup [frac{11pi}{12}, pi]}. Analysis - **Part (1)** involved using trigonometric identities and the periodicity of the sine function to simplify the function and find its period. - **Part (2)** utilized the properties of the sine function to determine the solution set of the inequality within the given interval. This problem demonstrates the application of trigonometric identity transformations and the properties of the sine function, categorizing it as an intermediate level problem.