answer:1. **Define the setup and the points of tangency**: Let omega be the given circle tangent to the lines (AD), (CD), the ray [BA) beyond A, and the ray [BC) beyond C. Denote the points of tangency by K, L, M, and N with the lines (BC), (AD), (CD), and (AB), respectively. 2. **Identify the intersection point of tangents**: Let T be the intersection point of the common external tangents to omega_1 and omega_2. Let r_1 and r_2 be the radii of omega_1 and omega_2, and J_1 and J_2 their centers, respectively. 3. **Apply homothety**: Considering a homothety centered at T that maps omega_1 to omega_2, we have the following relation: [ frac{T - J_1}{r_1} = frac{T - J_2}{r_2} ] This implies: [ r_1 (T - J_2) = r_2 (T - J_1) ] 4. **Solve for T**: Rearrange the equation and solve for T: [ r_1 T - r_1 J_2 = r_2 T - r_2 J_1 ] [ r_1 T - r_2 T = r_1 J_2 - r_2 J_1 ] [ T(r_1 - r_2) = r_1 J_2 - r_2 J_1 ] [ T = frac{r_1 J_2 - r_2 J_1}{r_1 - r_2} ] 5. **Define vector tau**: Let vec{tau} be the vector with affix i frac{c-a}{|c-a|}: [ tau = i frac{c-a}{|c-a|} ] Therefore: [ tau^2 = -frac{c-a}{overline{c-a}} ] 6. **Expression involving tau**: Recognize that: [ tau^2 = -frac{overline{l} + overline{n} - overline{k} - overline{m}}{l + n - k - m} ] 7. **Calculate lambda**: Let lambda = frac{1}{2} (overline{a} tau + a overline{tau}): [ lambda = overline{tau} frac{k n - l m}{k + n - l - m} ] 8. **Use collinearity**: Since O, J_1, and B are collinear: [ J_1 = k n overline{J_1} ] 9. **Distances and collinearity**: Write the condition for the distances: [ 0 = d(J_1, (AB)) + d(J_1, (AC)) ] Which gives: [ 0 = frac{1}{2} (j_1 overline{n} + n overline{j_1} - 2) + j_1 overline{tau} + tau overline{j_1} - 2 lambda ] 10. **Compute j_1 and r_1**: From the above conditions, compute j_1: - J_1, r_1 using d(J_1, (AB)) - Similarly compute J_2, r_2 11. **Final verification**: Verify that tau overline{tau} = 1, completing the proof that T, the intersection point of the common tangents, lies on omega. # Conclusion: [ blacksquare ]

answer:(1) Since a_6>0 and a_7<0, and knowing that a_1=23, we can deduce that the common difference d must be negative. The arithmetic sequence formula is a_n=a_1+(n-1)d. For n=6, we have a_6=23+5d>0, and for n=7, a_7=23+6d<0. Solving these inequalities, we find that d must be -4. Therefore, the value of the common difference d is boxed{-4}. (2) Using the arithmetic sequence formula a_n=a_1+(n-1)d and substituting a_1=23 and d=-4, we get a_n=23+(n-1)(-4)=23-4n+4=27-4n. Therefore, the general term a_n is boxed{a_n=-4n+27}.

answer:For option A, it does not satisfy the conditions of the problem; For option B, since l_1 and l_2 are intersecting lines, and m parallel l_1 implies that l_1 parallel alpha, analogously n parallel l_2 implies that l_2 parallel alpha, thus alpha parallel beta holds and sufficiency is established. However, alpha parallel beta does not necessarily lead to m parallel l_1; they could also be skew lines, which means that necessity does not hold. Therefore, option B is the correct choice; For option C, since m and n do not necessarily intersect, it is a necessary but not sufficient condition; For option D, n parallel l_2 can be transformed into n parallel beta, similar to option C, thus not fitting the problem. In conclusion, the correct answer is boxed{B}.

answer:1. Start by rewriting the given function: [ f(x) = 2 sin left( frac{pi}{4} sin left( sqrt{x-2} + x + 2 right) - frac{5pi}{2} right) ] 2. Use the trigonometric identity to transform the function for easier analysis. The identity sin(a - b) = sin(a) cos(b) - cos(a) sin(b) can be utilized, however, in this form it is simpler to use the property of sine functions shifting by multiples of pi: [ 2 sin left( theta - frac{5pi}{2} right) = -2 cos (theta) ] Given that theta = frac{pi}{4} sin left( sqrt{x-2} + x + 2 right), this transforms our function to: [ f(x) = -2 cos left( frac{pi}{4} sin left( sqrt{x-2} + x + 2 right) right) ] 3. Next, analyze the argument of the outer cosine function. Determine the bounds for sqrt{x-2} + x + 2 to understand the range of the inner sine function: [ sqrt{x-2} + x + 2: quad x geq 2 text{ (domain of the square root function)} ] When ( x = 2 ): [ sqrt{2-2} + 2 + 2 = 4 ] As (x) increases without bound: [ sqrt{x-2} approx sqrt{x} quad text{and} quad x + 2 approx x implies sqrt{x-2} + x + 2 to infty ] Therefore: [ sqrt{x-2} + x + 2 in [4, infty) ] 4. Now determine the range of the inner sine function: [ sin left( sqrt{x-2} + x + 2 right) ] Since sin(t) for t in [0, infty) always lies within [-1, 1], we have: [ sin left( sqrt{x-2} + x + 2 right) in [-1, 1] ] 5. Given the sine value, scale by the factor of frac{pi}{4}: [ frac{pi}{4} sin left( sqrt{x-2} + x + 2 right) in left[ -frac{pi}{4}, frac{pi}{4} right] ] 6. Apply cosine to these scaled values: [ cos left( frac{pi}{4} sin left( sqrt{x-2} + x + 2 right) right) in left[ cos left( -frac{pi}{4} right), cos left( frac{pi}{4} right) right] ] Since cos(-x) = cos(x): [ cos left( frac{pi}{4} right) = frac{sqrt{2}}{2} ] Therefore: [ cos left( frac{pi}{4} sin left( sqrt{x-2} + x + 2 right) right) in left[ frac{sqrt{2}}{2}, 1 right] ] 7. Finally, scale this result by -2: [ -2 cos left( frac{pi}{4} sin left( sqrt{x-2} + x + 2 right) right) in left[ -2, -sqrt{2} right] ] # Conclusion: [ boxed{[-2, -sqrt{2}]} ]