answer:Solution: (1) Let (x_{1},y_{1}) be a point on the ellipse, which transforms into point (x,y) on curve C under the given transformation, According to the problem, we have begin{cases} x=x_{1} y= frac {1}{2}y_{1}end{cases}. From x_{1}^{2}+ frac {y_{1}^{2}}{4}=1, we get x^{2}+ frac {(2y)^{2}}{4}=1, which means the equation of curve C is x^{2}+y^{2}=1 Therefore, the parametric equation of C is begin{cases} x=cos alpha y=sin alphaend{cases}(alpha text{ is a parameter}); (2) Since the tangent line at D is perpendicular to l, OD is parallel to l, Therefore, set the parametric equation begin{cases} x=cos alpha y=sin alphaend{cases}(alpha text{ is a parameter}) with alpha= frac {pi}{4}, frac {5pi}{4}, We get D( frac { sqrt {2}}{2}, frac { sqrt {2}}{2}) or D(- frac { sqrt {2}}{2},- frac { sqrt {2}}{2}). Thus, the coordinates of D are boxed{D( frac { sqrt {2}}{2}, frac { sqrt {2}}{2}) text{ or } D(- frac { sqrt {2}}{2},- frac { sqrt {2}}{2})}.

answer:1. Let the sides of the right triangle be a and b, where without loss of generality, a = b + r. 2. Using the Pythagorean theorem, a^2 + b^2 = h^2. Substitute a = b + r into the equation: [ (b + r)^2 + b^2 = h^2 quad Rightarrow quad b^2 + 2br + r^2 + b^2 = h^2 quad Rightarrow quad 2b^2 + 2br + r^2 = h^2 ] [ 2b(b + r) + r^2 = h^2 ] 3. The formula rs = A (area of the triangle), with s being the semiperimeter (a+b+c)/2, where c is the hypotenuse h. Since r = frac{A}{s}, we substitute c = h and a = b + r: [ s = frac{(b + r) + b + h}{2} = frac{2b + r + h}{2} ] 4. Then s = b + frac{r + h}{2}. Substitute this back into the area expression: [ A = frac{pi r^2}{b + frac{r + h}{2}} = frac{2pi r^2}{2b+r+h} ] 5. The ratio of the area of the circle to the area of triangle using rs = A: [ frac{pi r^2}{rs} = frac{pi r}{s} = frac{pi r}{b +frac{r + h}{2}} = frac{2pi r}{2b + r + h} ] Conclude with the simplifications and substitutions to match the original variables used in determining sides and semiperimeter. Adjust constants and variables to create a coherent, consistent relationship. Final Answer: [ boxed{frac{2pi r}{2b + r + h}} ]

answer:**Step 1: Calculate the side length and number of smaller cubes** First, we divide the unit cube into smaller cubes. Since we are given 1985 points, let's figure out into how many smaller cubes we should divide the unit cube such that one of these smaller cubes contains at least 32 points. Notice that: [ 64 times 31 = 1984. ] Thus, if we divide the unit cube into (64) smaller cubes, each with a side length of (frac{1}{4}), one of these smaller cubes must contain at least: [ leftlceil frac{1985}{64} rightrceil = 32 text{ points}. ] **Step 2: Maximum distance between any two points in a smaller cube** Now consider any two points lying inside one of these smaller cubes. The maximum distance between any two points in a cube with side length (frac{1}{4}) is the space diagonal. The space diagonal (d) of a cube with side length (s) can be calculated using the Pythagorean theorem in 3 dimensions: [ d = sqrt{s^2 + s^2 + s^2} = ssqrt{3}. ] Setting (s = frac{1}{4}), we get: [ d = frac{1}{4} sqrt{3} = frac{sqrt{3}}{4}. ] **Step 3: Maximum perimeter of the 32-gon** Consider the worst-case scenario where the perimeter formed by these 32 points is maximum. For the perimeter to be maximal, each edge of the polygon connecting two points has the length of the space diagonal. However, forming a polygon that connects each point optimally can be complex. Instead, we provide an upper bound by considering a straightforward approach where each edge is at most (frac{sqrt{3}}{4}). The total perimeter (P) of the 32-gon, assuming the distance between each pair of consecutive points (if we constructed a cycle) is at most (frac{sqrt{3}}{4}), can be bounded by: [ P leq 32 times frac{sqrt{3}}{4} = 8 sqrt{3}. ] This proves that the perimeter of any polygon, using these 32 points as vertices, can always be constrained to ( 8 sqrt{3} ). # Conclusion: [ boxed{8sqrt{3}} ]

answer:From the parametric equations of line l, we can eliminate the parameter t to find the Cartesian equation of the line: begin{cases} x = 2t, y = 1 - 4t, end{cases} which gives us the linear equation 2x + y - 1 = 0. For curve C, represented by the parametric equations: begin{cases} x = sqrt{5}costheta, y = m + sqrt{5}sintheta, end{cases} we can square and add the two equations to find its Cartesian form. Recognizing that cos^2theta+sin^2theta=1, we get: x^2 + (y - m)^2 = 5, which is the equation of a circle with center (0, m) and radius sqrt{5}. For the line l to be tangent to the circle C, the perpendicular distance from the center of the circle to the line must equal the radius of the circle. Thus, using the distance formula for the point (0,m) to the line 2x + y - 1 = 0, we have: frac{|2(0) + m - 1|}{sqrt{2^2 + 1^2}} = sqrt{5}, simplifying this expression gives us: frac{|m - 1|}{sqrt{5}} = sqrt{5}. This leads to |m - 1| = 5, which solves to m - 1 = 5 or m - 1 = -5. Therefore, m = 6 or m = -4. Thus the correct choice is: boxed{text{A. } -4 text{ or } 6}.