answer:1. **Define the given elements and point of intersection** Let triangle ABC be a right triangle with angle BCA = 90^circ. The sides opposite to vertices A, B, and C are denoted as a, b, and c respectively, where a and b are the perpendicular sides and c is the hypotenuse. Suppose odot K is the circumcircle of triangle ABC, and define circles odot K_1 and odot K_2 that intersect with the hypotenuse c, altitude CD (from C to AB), and the arcs overparen{BC} and overparen{AC} of odot K, respectively. Let the radii of odot K_1 and odot K_2 be r_1 and r_2. Let circles odot K_1 and odot K_2 intersect line AB at points T and S, respectively. 2. **Apply the intersecting chords theorem** By the Intersecting Chords Theorem, we know: [ r_1 + r_2 = frac{AT cdot TB}{AB} + frac{AS cdot SB}{AB} ] 3. **Relate the intersection points to triangle sides** Utilizing properties from the problem and given right triangle: [ b = AC = AT quad text{and} quad a = BC = BS ] 4. **Substitute and simplify expressions** Next, substitute these relations into the theorem: [ r_1 + r_2 = frac{AC cdot (AB - AC)}{AB} + frac{(AB - BC) cdot BC}{AB} ] Simplify the expression: [ r_1 + r_2 = frac{AC cdot (c - AC)}{c} + frac{(c - BC) cdot BC}{c} ] Since AC = b and BC = a, we substitute to get: [ r_1 + r_2 = frac{b cdot (c - b)}{c} + frac{(c - a) cdot a}{c} ] 5. **Further simplify to reach the final result** Combine the fractions: [ r_1 + r_2 = frac{b(c - b) + a(c - a)}{c} ] Expand and combine like terms: [ r_1 + r_2 = frac{bc - b^2 + ac - a^2}{c} ] Finally, rearrange the terms: [ r_1 + r_2 = frac{bc + ac - b^2 - a^2}{c} ] Recognize that a^2 + b^2 = c^2 (Pythagorean theorem for right triangles): [ r_1 + r_2 = frac{bc + ac - c^2}{c} ] Simplify the numerator: [ r_1 + r_2 = frac{(a + b)c - c^2}{c} ] Factor out c: [ r_1 + r_2 = a + b - c ] # Conclusion: [ boxed{r_1 + r_2 = a + b - c} ]

answer:1. **Define the distances and angles:** Let ( d_1 le d_2 le ldots le d_{n-1} ) be the distances from ( O ) to the center of each disc ( D_i ), arranged in increasing order. According to condition (b), we have: [ d_i le i + 1 quad text{for} quad i = 1, 2, ldots, n-1. ] 2. **Calculate the angle subtended by each disc:** The angle at which the disc ( D_i ) is seen from ( O ) is at least ( frac{2}{d_i} ). This is because the diameter of each disc is 2 (since they are unit discs), and the angle subtended by a chord of length ( 2 ) at a distance ( d_i ) is ( frac{2}{d_i} ). 3. **Define the set ( A_i ):** Assign to each disc ( D_i ) the set: [ A_i = { varphi in [0, 2pi) mid text{the line through } O text{ with argument } varphi text{ stabs } D_i }. ] Thus, ( A_i subset [0, 2pi) ) and the measure of ( A_i ), denoted ( m(A_i) ), is at least ( frac{4}{d_i} ). 4. **Sum of measures ( m(A_i) ):** We need to find the total measure of all ( A_i ): [ sum_{i=1}^{n-1} m(A_i) ge sum_{i=1}^{n-1} frac{4}{d_i}. ] Using the inequality ( d_i le i + 1 ), we get: [ sum_{i=1}^{n-1} frac{4}{d_i} ge sum_{i=1}^{n-1} frac{4}{i + 1}. ] 5. **Estimate the harmonic sum:** The harmonic sum can be approximated using the natural logarithm: [ sum_{i=1}^{n-1} frac{1}{i + 1} approx ln(n) - ln(1) = ln(n). ] Therefore: [ sum_{i=1}^{n-1} frac{4}{i + 1} ge 4 left( ln(n) - ln(1) right) = 4 ln(n). ] 6. **Common point in the sets ( A_i ):** There are at least ( frac{sum m(A_i)}{2pi} ) sets among ( A_i ) with a common point. Thus: [ sum_{i=1}^{n-1} m(A_i) ge 4 ln(n). ] Hence, there are at least: [ frac{4 ln(n)}{2pi} = frac{2 ln(n)}{pi} ] sets among ( A_i ) with a common point ( varphi ). 7. **Conclusion:** The line ( ell ) through ( O ) with ( angle(Oell, Ox) = varphi ) stabs all corresponding discs. Therefore, there exists a line through ( O ) that stabs at least ( frac{2 ln(n)}{pi} ) discs in ( mathcal{D} ). The final answer is ( boxed{ frac{2 ln(n)}{pi} } ).

answer:Since A={xinmathbb{Z}|1<x<7}, B={x|xgeq10 text{ or } xleq2}, therefore complement_{mathbb{R}}B={xinmathbb{R}|2<x<10}, A={2,3,4,5,6} therefore Acap(complement_{mathbb{R}}B)={2,3,4,5,6}cap{xinmathbb{R}|2<x<10}={3,4,5,6}, Hence, the answer is: boxed{{3,4,5,6}}.

answer:**(1) Finding the Analytical Expression of the Profit W** Given the annual fixed cost is 400,000 and an additional 16 is required for each product produced, the cost function can be represented as C(x) = 16x + 400 (in thousand dollars since x is in thousands and the fixed cost is in thousands as well). The revenue function, R(x), is given by: [R(x)=left{begin{array}{ll} 400-6x, & 0<xleq40 frac{7400}{x}-frac{40000}{{x^2}}, & x>40 end{array}right.] The profit function, W, is the difference between revenue and cost, W = xR(x) - C(x). - For 0 < x leq 40: [W = x(400-6x) - (16x + 400)] [= 400x - 6x^2 - 16x - 400] [= -6x^2 + 384x - 400] - For x > 40: [W = xleft(frac{7400}{x}-frac{40000}{x^2}right) - (16x + 400)] [= 7400 - frac{40000}{x} - 16x - 400] [= -frac{40000}{x} - 16x + 7000] Therefore, the analytical expression of the profit W (in million dollars) as a function of the annual output x (in thousand pieces) is: [W=left{begin{array}{ll} -6x^2 + 384x - 400, & 0<xleq40 -frac{40000}{x} - 16x + 7000, & x>40 end{array}right.] **(2) Finding the Maximum Profit** - For 0 < x leq 40: [W = -6x^2 + 384x - 400] [= -6(x^2 - 64x + 400/6)] [= -6(x^2 - 64x + 1024/6 - 1024/6 + 400/6)] [= -6[(x - 32)^2 - 1024/6 + 400/6]] [= -6(x - 32)^2 + 6104] Thus, when x = 32, W_{max} = 6104. - For x > 40: [W = -frac{40000}{x} - 16x + 7000] Using AM-GM inequality, we get: [W leq -2sqrt{frac{40000}{x} cdot 16x} + 7000] [= -2sqrt{640000} + 7000] [= -1600 + 7000] [= 5400] This inequality becomes an equality when frac{40000}{x} = 16x, i.e., x = 50, giving W_{max} = 5400. Comparing the two maximum values, 6104 > 5400, the maximum profit is obtained when x = 32 thousand pieces. Therefore, when the annual output is 32 thousand pieces, the company obtains the maximum profit in the production of this product, and the maximum profit is boxed{6104} million dollars.