answer:Solution: (1) Since the polar equation of the circle is rho=2cos theta, thus the Cartesian coordinate equation of circle C is x^{2}+y^{2}-2x=0, substituting begin{cases} x=-1+tcos alpha y=tsin alpha end{cases} into x^{2}+y^{2}-2x=0, we get t^{2}-4tcos alpha+3=0, and since line l intersects circle C at points A and B, therefore Delta =16cos ^{2}alpha-12 > 0, solving this gives: cos alpha > dfrac { sqrt {3}}{2} or cos alpha < - dfrac { sqrt {3}}{2}. Given alphain[0,pi), hence the range of values for alpha is alphain[0, dfrac {pi}{6})cup( dfrac {5pi}{6},pi). (2) Let the two real roots of the equation t^{2}-4tcos alpha+3=0 be t_{1} and t_{2}, then by the geometric meaning of parameter t: dfrac {1}{|PA|}+ dfrac {1}{|PB|}= dfrac {|t_{1}+t_{2}|}{t_{1}t_{2}}= dfrac {|4cos alpha|}{3}, and since dfrac { sqrt {3}}{2} < |cos alpha|leqslant 1, therefore dfrac {2 sqrt {3}}{3} < dfrac {|4cos alpha|}{3}leqslant dfrac {4}{3}, therefore the range of values for dfrac {1}{|PA|}+ dfrac {1}{|PB|} is boxed{left( dfrac {2 sqrt {3}}{3}, dfrac {4}{3}right]}.

answer:1. **Given Conditions and Setup:** - Let ABC be an acute triangle. - Construct isosceles triangles DAC, EAB, and FBC exterior to ABC such that DA = DC, EA = EB, and FB = FC. - The angles are given as: [ angle ADC = 2angle BAC, quad angle BEA = 2angle ABC, quad angle CFB = 2angle ACB. ] - Define D', E', and F' as the intersections of lines DB and EF, EC and DF, and FA and DE respectively. 2. **Objective:** - Find the value of the sum: [ frac{DB}{DD'} + frac{EC}{EE'} + frac{FA}{FF'}. ] 3. **Claim:** - We need to show that: [ frac{DB}{DD'} + frac{EC}{EE'} + frac{FA}{FF'} = 4. ] 4. **Proof:** - Let O be the circumcenter of triangle ABC and assume the circumradius R = 1 for simplicity. - We will use the properties of the isosceles triangles and the given angle conditions to relate the areas of the triangles involved. 5. **Area Relations:** - Note that triangle AEB sim triangle AOC and triangle AOB sim triangle ADC due to the given angle conditions. - This similarity gives us: [ AE = R cdot frac{c}{b} = frac{c}{b}, quad AD = frac{b}{a}. ] - The angle angle EAD = 180^circ - angle B. 6. **Law of Cosines:** - Using the Law of Cosines in triangle ADE: [ DE^2 = left(frac{b}{a}right)^2 + left(frac{c}{b}right)^2 + 2 cdot frac{b}{a} cdot frac{c}{b} cdot cos(angle EAD). ] - Since cos(180^circ - angle B) = -cos(angle B), we have: [ DE^2 = frac{b^2}{a^2} + frac{c^2}{b^2} + 2 cdot frac{b}{a} cdot frac{c}{b} cdot (-cos(angle B)). ] - Simplifying: [ DE^2 = frac{b^2}{a^2} + frac{c^2}{b^2} - 2 cdot frac{c}{a} cdot cos(angle B). ] 7. **Area Calculation:** - The area of triangle DEF can be related to triangle ABC using the ratio of their sides. - The area of triangle DEF is: [ [DEF] = frac{a^2b^2 + b^2c^2 + c^2a^2}{4abc}. ] - The area of triangle ADE using the sine area formula: [ [ADE] = frac{1}{2} cdot AE cdot AD cdot sin(angle EAD) = frac{1}{2} cdot frac{c}{b} cdot frac{b}{a} cdot sin(angle B) = frac{bc}{4a}. ] 8. **Summing Areas:** - We need to show: [ frac{bc}{4a} + frac{ca}{4b} + frac{ab}{4c} = frac{a^2b^2 + b^2c^2 + c^2a^2}{4abc}. ] - This is true by inspection and symmetry of the terms. 9. **Final Calculation:** - Using the area relations, we have: [ frac{DB}{DD'} + frac{EC}{EE'} + frac{FA}{FF'} = frac{[BFDE]}{[DEF]} + frac{[CDEF]}{[DEF]} + frac{[AEFD]}{[DEF]} = 3 + frac{[ADE] + [BEF] + [CFD]}{[DEF]} = 4. ] The final answer is boxed{4}.

answer:To solve this problem, we consider the action of moving a point on the number line by 2 units to reach the point representing -2. This movement can occur in two ways: either moving to the right (adding 2) or moving to the left (subtracting 2). 1. If moving to the right (adding 2) reaches -2, then we have the equation: A + 2 = -2. Solving for A gives us: [A = -2 - 2] [A = -4] 2. If moving to the left (subtracting 2) reaches -2, then we have the equation: A - 2 = -2. Solving for A gives us: [A = -2 + 2] [A = 0] Therefore, the number represented by point A can be either 0 or -4. So, the final answer is boxed{0 text{ or } -4}.

answer:Let's denote the total number of students in the class as ( N ) and the average of one fourth of the class as ( A ). Since one fourth of the class has an average of ( A ), then three fourths of the class has an average of 80. The total sum of the grades for the entire class can be calculated by adding the sum of the grades of one fourth of the class and the sum of the grades of the remaining three fourths of the class. Let's calculate the sum of the grades for each part of the class: - One fourth of the class: ( frac{N}{4} times A ) - Three fourths of the class: ( frac{3N}{4} times 80 ) The total sum of the grades for the entire class is the sum of the above two parts: [ text{Total sum of grades} = frac{N}{4} times A + frac{3N}{4} times 80 ] We know that the average grade for the entire class is 84, so the total sum of the grades for the entire class can also be expressed as: [ text{Total sum of grades} = N times 84 ] Now we can set the two expressions for the total sum of grades equal to each other: [ frac{N}{4} times A + frac{3N}{4} times 80 = N times 84 ] We can simplify this equation by dividing both sides by ( N ): [ frac{1}{4} times A + frac{3}{4} times 80 = 84 ] Now, let's solve for ( A ): [ frac{1}{4} times A = 84 - frac{3}{4} times 80 ] [ frac{1}{4} times A = 84 - 60 ] [ frac{1}{4} times A = 24 ] Now, multiply both sides by 4 to solve for ( A ): [ A = 24 times 4 ] [ A = 96 ] Therefore, the average of one fourth of the class is boxed{96} .