answer:1. **Identify Inversion Center and Properties**: - Consider point (A) as the inversion center for performing inversion. Denote the inverse of each point by appending a star (∗) to the original point, e.g., the inverse of (O_{1}) is (O_{1}^{*}), and so on. 2. **Properties of Inversion for Circles**: - Since circles (O_{1}) and (O_{2}) intersect at point (A), their inverses under the inversion centered at (A) will be parallel lines as intersecting circles through the center transform to lines through the center. 3. **Transformation of Circle (n) and (k)**: - The circumcircle (n) of (triangle AB_{1}B_{2}) transforms into another circle (n^{*}) that intersects lines (O_{1}^{*}) and (O_{2}^{*}). Circle (k) becomes (k^{*}), and similarly, (n^{*} parallel k^{*}). 4. **Formation of Parallel Lines and Quadrilateral**: - The quadrilateral formed (B_{1}^{*}B_{2}^{*}D_{2}^{*}D_{1}^{*}) will be a parallelogram because lines (B_{1}^{*}B_{2}^{*}) and (D_{1}^{*}D_{2}^{*}) are the images of circles intersecting at (A), consequently remain parallel. (1) **Common Circle or Common Line for (C_{1}, C_{2}, D_{1}, D_{2})**: 1. The lines (B_{1}^{*}B_{2}^{*}) are parallel to (D_{1}^{*}D_{2}^{*}). 2. Since circle (O^{*}) passes through points (C_{1}^{*}), (C_{2}^{*}), (B_{1}^{*}), and (B_{2}^{*}), then points (C_{1}^{*}, C_{2}^{*}, D_{1}^{*}, D_{2}^{*}) are concyclic (lie on the same circle). 3. Therefore, the original points (C_{1}, C_{2}, D_{1}, D_{2}) will either lie on the same circle or line. (2) **Concyclicity of (B_{1}, B_{2}, D_{1}, D_{2})**: - To show that points (B_{1}, B_{2}, D_{1}, D_{2}) lie on the same circle, consider the equivalent condition after inversion: - (B_{1}, B_{2}, D_{1}, D_{2}) are concyclic if and only if (B_{1}^{*}, B_{2}^{*}, D_{1}^{*}, D_{2}^{*}) are concyclic. - Since (B_{1}^{*}B_{2}^{*}D_{2}^{*}D_{1}^{*}) forms a rectangle (equivalently their diagonals bisect each other perpendicularly), it follows: [ angle C_{1}^{*}D_{1}^{*}D_{2}^{*}=90^{circ} ] which yields, [ angle C_{1}^{*}C_{2}^{*}D_{2}^{*}=90^{circ} ] implying (l^{*} perp O_{1}^{*}). This, in turn, implies (AC_{1}) and (AC_{2}) to be perpendicular to (O_{1}) and (O_{2})'s tangents, confirming they are the diameters of their respective circles. Conclusion: Thus, the proof is complete. blacksquare

answer:To begin solving the problem, we analyze the given function: begin{aligned} f(x) &= cos(alpha_1 + x) + frac{1}{2}cos(alpha_2 + x) + frac{1}{2^2}cos(alpha_3 + x) + cdots + frac{1}{2^{n-1}}cos(alpha_n + x). end{aligned} We need to prove that from f(x_1) = f(x_2) = 0 , it implies that x_2 - x_1 = mpi where m is an integer. First, evaluate ( f left(-alpha_1right) ): begin{aligned} fleft(-alpha_1right) &= cosleft(alpha_1 - alpha_1right) + frac{1}{2} cosleft(alpha_2 - alpha_1right) + cdots + frac{1}{2^{n-1}} cosleft(alpha_n - alpha_1right) &= 1 + frac{1}{2}cos(alpha_2 - alpha_1) + cdots + frac{1}{2^{n-1}}cos(alpha_n - alpha_1). end{aligned} Notice that geqslant 1 - frac{1}{2} - frac{1}{2^2} - cdots - frac{1}{2^{n-1}} Using the sum of a geometric series, 1 - frac{1}{2} - frac{1}{2^2} cdots - frac{1}{2^{n-1}} = 1 - left(frac{1 - frac{1}{2^{n-1}}}{1 - frac{1}{2}}right) = 1 - left(frac{1 - frac{1}{2^{n-1}}}{frac{1}{2}}right) = 1 - (1 - frac{1}{2^{n-1}}) = frac{1}{2^{n-1}} Hence, geqslant frac{1}{2^{n-1}} > 0 So ( f left( x right) ) is not identically zero, because ( fleft(-alpha_1right) > 0 ). Next, consider transforming the given series of cosines to a trigonometric form involving cosine of a sum. Use the angle sum identity: cos(a + b) = cos(a) cos(b) - sin(a) sin(b) Thus, [ begin{aligned} f(x) &= left( cos alpha_1 cos x - sin alpha_1 sin x right) + left( frac{1}{2} cos alpha_2 cos x - frac{1}{2} sin alpha_2 sin x right) + cdots + &quad frac{1}{2^{n-1}} left( cos alpha_n cos x - sin alpha_n sin x right) &= left( cos alpha_1 + frac{1}{2} cos alpha_2 + cdots + frac{1}{2^{n-1}} cos alpha_n right) cos x - left( sin alpha_1 + frac{1}{2} sin alpha_2 + cdots + frac{1}{2^{n-1}} sin alpha_n right) sin x end{aligned} ] Define ( c ) and ( s ): c = cos alpha_1 + frac{1}{2} cos alpha_2 + cdots + frac{1}{2^{n-1}} cos alpha_n s = sin alpha_1 + frac{1}{2} sin alpha_2 + cdots + frac{1}{2^{n-1}} sin alpha_n Then, f(x) = c cos x - s sin x Now, recognize that this can be rewritten using the phase angle ( alpha ): r = sqrt{c^2 + s^2} neq 0 There exists an angle ( alpha ) such that: cos alpha = frac{c}{r}, quad sin alpha = frac{s}{r} Thus, [ begin{aligned} f(x) &= r cos(alpha) cos(x) - r sin(alpha) sin(x) &= r left[ cos(alpha + x) right] end{aligned} ] Hence, ( f(x) = r cos(alpha + x) ). Given that ( f left( x_1 right) = f left( x_2 right) = 0 ), it implies: r cos(alpha + x_1) = 0 quad text{and} quad r cos(alpha + x_2) = 0 which simplifies to: cos(alpha + x_1) = 0 quad text{and} quad cos(alpha + x_2) = 0 This occurs when ( alpha + x_1 ) and ( alpha + x_2 ) are odd multiples of ( frac{pi}{2} ): alpha + x_1 = kpi + frac{pi}{2}, quad alpha + x_2 = lpi + frac{pi}{2} where ( k ) and ( l ) are integers. Thus, x_2 - x_1 = (l - k)pi which implies ( x_2 - x_1 = mpi ) for some integer ( m = l - k ). Conclusion: [ boxed{x_2 - x_1 = mpi} ]

answer:Let ( W ) be the set of dogs liking watermelon and ( S ) the set of dogs liking salmon. We know: - ( |W| = 15 ) - ( |S| = 55 ) - ( |W cap S| = 10 ) Using the principle of inclusion-exclusion to find ( |W cup S| ): [ |W cup S| = |W| + |S| - |W cap S| = 15 + 55 - 10 = 60 ] This indicates 60 dogs like either or both. Now, determining the number that likes neither: [ text{Total dogs} - |W cup S| = 80 - 60 = 20 ] Therefore, (boxed{20}) dogs in the kennel do not eat either watermelon or salmon.

answer:This problem involves applying the cofunction identity and the double angle formula. First, we use the cofunction identity sin(pi - theta) = sin theta. This gives us sin theta = frac{1}{3}. Next, we apply the double angle formula for cosine, which is cos 2theta = 1 - 2sin^2 theta. Substituting sin theta = frac{1}{3} into this formula, we get: cos 2theta = 1 - 2(frac{1}{3})^2 = 1 - 2 cdot frac{1}{9} = 1 - frac{2}{9} = frac{7}{9}. So, the value of cos 2theta is boxed{frac{7}{9}}.