answer:Since alpha is an angle in the third quadrant, we have pi+2kpi < alpha < frac {3pi}{2}+2kpi, k in mathbb{Z}, which leads to - frac {3pi}{2}-2kpi < -alpha < -pi-2kpi, k in mathbb{Z}, thus -pi-2kpi < frac {pi}{2}-alpha < - frac {pi}{2}-2kpi, k in mathbb{Z}． Therefore, frac {pi}{2}-alpha is an angle in the boxed{text{third}} quadrant. The solution involves writing out the range of angles in the third quadrant and further determining the range of frac {pi}{2}-alpha. This question tests knowledge of quadrant angles and axial angles and is considered a basic question.

answer:1. Let ( f(x) = ax^2 + bx + c ) and ( g(x) = cx^2 + bx + a ). We know that ( f(x) ) has roots ( p_1 ) and ( p_2 ), and ( g(x) ) has roots ( q_1 ) and ( q_2 ). 2. Since ( p_1, q_1, p_2, q_2 ) form an arithmetic progression, we can write: [ q_1 = p_1 + d, quad p_2 = p_1 + 2d, quad q_2 = p_1 + 3d ] for some common difference ( d ). 3. The sum of the roots of ( f(x) ) is given by Vieta's formulas: [ p_1 + p_2 = -frac{b}{a} ] Similarly, the sum of the roots of ( g(x) ) is: [ q_1 + q_2 = -frac{b}{c} ] 4. Substituting the arithmetic progression terms into the sum of the roots for ( g(x) ): [ q_1 + q_2 = (p_1 + d) + (p_1 + 3d) = 2p_1 + 4d ] Therefore: [ 2p_1 + 4d = -frac{b}{c} ] 5. From the sum of the roots of ( f(x) ): [ p_1 + p_2 = p_1 + (p_1 + 2d) = 2p_1 + 2d = -frac{b}{a} ] 6. We now have two equations: [ 2p_1 + 2d = -frac{b}{a} ] [ 2p_1 + 4d = -frac{b}{c} ] 7. Subtract the first equation from the second: [ (2p_1 + 4d) - (2p_1 + 2d) = -frac{b}{c} - left(-frac{b}{a}right) ] Simplifying, we get: [ 2d = -frac{b}{c} + frac{b}{a} ] [ 2d = b left( frac{1}{a} - frac{1}{c} right) ] 8. Since ( d neq 0 ) (otherwise, the roots would not be distinct), we can divide both sides by ( 2d ): [ 1 = frac{b}{2d} left( frac{1}{a} - frac{1}{c} right) ] [ frac{2d}{b} = frac{1}{a} - frac{1}{c} ] 9. Rearrange to find a common denominator: [ frac{1}{a} - frac{1}{c} = frac{c - a}{ac} ] Therefore: [ frac{2d}{b} = frac{c - a}{ac} ] 10. Since ( frac{2d}{b} ) is a constant, the only way for this equation to hold for all ( d ) is if ( c - a = 0 ), which implies: [ a = c ] 11. However, since the problem states that the roots are distinct and form an arithmetic progression, we must have: [ a + c = 0 ] The final answer is ( boxed{ a + c = 0 } )

answer:f(x) = 2sin left(omega x+ frac{pi}{6}right). From the given conditions, the smallest positive period T = pi, therefore omega= frac{2pi}{T}= frac{2pi}{pi}=2, thus fleft(xright)=2sin left(2x+ frac{pi}{6}right). Let 2kpi- frac{pi}{2} leqslant 2x+ frac{pi}{6} leqslant 2kpi+ frac{pi}{2},k in mathbb{Z}, solving this yields kpi- frac{pi}{3} leqslant x leqslant kpi+ frac{pi}{6}, kin mathbb{Z}, Therefore, the interval where f(x) is strictly increasing is boxed{left[kpi- frac{pi}{3},kpi+ frac{pi}{6}right], kin mathbb{Z}}.

answer:To prove that the area of a triangle whose sides are equal to the medians of a given triangle is frac{3}{4} of the area of the given triangle. 1. **Setup the problem and identify the elements**: - Let triangle ABC be the given triangle. - Let AD, BE, and CF be the medians of triangle ABC intersecting at the centroid G. - Assume a new triangle PQR with sides PQ = BE, QR = CF, and RP = AD, which are the medians of triangle ABC. 2. **Medians and centroid properties**: - Recall that the medians of a triangle intersect at the centroid and the centroid divides each median in the ratio 2:1. - Denote the length of medians of triangle ABC as m_a, m_b, and m_c corresponding to AD, BE, and CF respectively. 3. **Use Apollonius's theorem for median relation**: - By Apollonius's theorem: m_a^2 = frac{2b^2 + 2c^2 - a^2}{4}, m_b^2 = frac{2c^2 + 2a^2 - b^2}{4}, m_c^2 = frac{2a^2 + 2b^2 - c^2}{4}. 4. **Area calculation for the given triangle**: - The area of triangle ABC can be given using Heron's formula. - Let s = frac{a+b+c}{2} be the semi-perimeter of triangle ABC. - The area A of triangle ABC is given by: A = sqrt{s(s-a)(s-b)(s-c)}. 5. **Area ratio using medians**: - Let the area of triangle PQR be denoted as A'. According to the problem, we need to show that: A' = frac{3}{4} A. - Using known properties of triangles and transformations, specifically that a triangle formed by the medians of another triangle has an area equal to frac{3}{4} of the original triangle. 6. **Conclusion**: - Therefore, by combining these calculations and known geometric theorems, text{Area of triangle } PQR = frac{3}{4} text{Area of triangle } ABC. boxed{frac{3}{4} text{Area of triangle } ABC}