answer:1. **Define Points and Angles:** Let ( R = a cap b ) and ( S = c cap d ). Denote ( v = angle RAB = angle RBA ) and ( w = angle SCD = angle SDC ). 2. **Power of a Point:** To prove that ( K, L, M, N ) are concyclic, it suffices to show that: [ frac{mathrm{pow}((P),K)}{mathrm{pow}((Q),K)} = frac{mathrm{pow}((P),L)}{mathrm{pow}((Q),L)} = frac{mathrm{pow}((P),M)}{mathrm{pow}((Q),M)} = frac{mathrm{pow}((P),N)}{mathrm{pow}((Q),N)} ] By the Power of a Point theorem, this is equivalent to: [ frac{AK^2}{CK^2} = frac{AL^2}{DL^2} = frac{BM^2}{CM^2} = frac{BN^2}{DN^2} Leftrightarrow frac{AK}{CK} = frac{AL}{DL} = frac{BM}{CM} = frac{BN}{DN} ] 3. **Law of Sines:** Using the Law of Sines in (triangle AKC), we get: [ frac{AK}{CK} = frac{sin w}{sin v} ] Similarly, for the other ratios, we have: [ frac{AL}{DL} = frac{sin w}{sin v}, quad frac{BM}{CM} = frac{sin w}{sin v}, quad frac{BN}{DN} = frac{sin w}{sin v} ] Thus, we have established the desired equality, proving that ( K, L, M, N ) are concyclic. 4. **Coaxial Circles:** Since ( K, L, M, N ) are concyclic, circles ((P)), ((Q)), and ((KLMN)) are coaxial. 5. **Intersection of Tangents:** Let ( PQ ) intersect ((KLMN)) at ( H ). This gives: [ frac{P^2 - r_1^2}{Q^2 - r_2^2} = frac{mathrm{pow}((P),H)}{mathrm{pow}((Q),H)} = frac{sin^2 w}{sin^2 v} ] 6. **Law of Sines in (triangle AXB):** Let ( X ) be the intersection of ( PR ) with ((P)). By the Law of Sines in (triangle AXB), we have: [ sin(v) = frac{AB}{2r_1} ] Similarly, (sin(w) = frac{CD}{2r_2}). Therefore: [ frac{PH^2 - r_1^2}{QH^2 - r_2^2} = frac{sin^2 w}{sin^2 v} = frac{r_1^2}{r_2^2} ] This implies: [ frac{PH}{QH} = frac{r_1}{r_2} ] Using the property that if (frac{a}{b} = frac{c}{d}), then (frac{a}{b} = frac{a+c}{c+d}), we conclude that ( H ) is the center of homothety of ((P)) and ((Q)). 7. **Intersection of External Tangents:** Since ( H ) is the center of homothety, the external tangent lines of ((P)) and ((Q)) must intersect at ( H ). (blacksquare)

answer:From the equation, we have [arccos(3x) = arccos(x) + frac{pi}{6}.] Then, [cos(arccos(3x)) = cosleft(arccos(x) + frac{pi}{6}right).] Using the cosine addition formula, begin{align*} 3x &= cos(arccos x) cos frac{pi}{6} - sin(arccos x) sin frac{pi}{6} &= x frac{sqrt{3}}{2} - sqrt{1 - x^2} frac{1}{2}. end{align*} Hence, [3x = frac{sqrt{3}}{2} x - frac{1}{2} sqrt{1 - x^2}.] Rearranging terms gives us, [frac{6 - 3sqrt{3}}{2} x = - frac{1}{2} sqrt{1 - x^2}.] Squaring both sides, we find: [frac{(6 - 3sqrt{3})^2}{4} x^2 = frac{1 - x^2}{4}.] Simplifying gives us, [(36 - 36sqrt{3} + 27) x^2 = 1 - x^2,] So, [63 - 36sqrt{3} x^2 = 1 - x^2,] [x^2 = frac{1}{64 - 36sqrt{3}},] [x = pm sqrt{frac{1}{64 - 36sqrt{3}}}.] Testing the values in the original arccos equality confirms the only valid answer is x = boxed{sqrt{frac{1}{64 - 36sqrt{3}}}}.

answer:**Analysis** By using the formula for the sum of an arithmetic sequence and the given S_9=-36 and S_{13}=-104, we can find the first term and the common difference d, and then calculate a_5 and a_7. The geometric mean A satisfies A^2=a_5 cdot a_7. Substituting the values, we can find the solution. This problem mainly tests the application of the formula for the sum of an arithmetic sequence and the application of the geometric mean, which is a basic question. **Solution** Let the first term of the arithmetic sequence be a_1 and the common difference be d. From the given information, we have: begin{cases} S_9=9a_1+ frac{9 times 8}{2} times d=-36 S_{13}=13a_1+ frac{13 times 12}{2} times d=-104 end{cases} Solving these equations, we find a_1=4 and d=-2. Let the geometric mean of a_5 and a_7 be a_6. Then, we have b_6^2 =b_5 cdot b_7 =a_5 cdot a_7=(-4) times (-8)=32. Therefore, b_6=±4 sqrt{2}. Thus, the correct answer is boxed{text{D}}.

answer:1. Find all x such that g(x) = 3: ( g(x) = -0.5x^2 + x + 3 = 3 ) results in ( -0.5x^2 + x = 0 ). Factoring gives ( -0.5x(x - 2) = 0 ). Thus, x = 0 or x = 2. 2. Check if g(g(x)) = 3 for these values of x: - For g(0) = 3, we need g(3), but since g(3) = 1, it doesn't work. - For g(2) = 3, we need g(3) again, which still results in g(3) = 1. - Check g(1) since g(1) = 3.5 and g(3.5) is not defined within the domain. 3. It appears that only ( x = 0 ) and ( x = 2 ) are candidates, but neither returns a final result of 3 for g(g(x)). Therefore, g(g(x)) = 3 has boxed{0} solutions within the domain.