answer:1. **Given:** - A sphere with center ( O ) and radius ( r ). - Points ( A ) and ( B ) lie outside the sphere. - Two tangents ( AP ) and ( BQ ) are drawn from ( A ) and ( B ) respectively, intersecting at point ( M ). - The points of tangency are ( P ) and ( Q ). 2. **Assumptions and Notations:** - Let ( |OA| = a ), ( |OB| = b ), and ( |PM| = |QM| = x ). 3. **Key Relations:** Using the Pythagorean theorem in the right triangle ( OMP ) and ( OMQ ): [ |OM|^2 = r^2 + x^2 ] For the segments ( |AM| ) and ( |BM| ): [ |AM| = sqrt{a^2 - r^2} pm x ] [ |BM| = sqrt{b^2 - r^2} pm x ] We need to determine the conditions under which the products of these distances form a consistent geometric relationship. 4. **Case: Same Signs in ( |AM| ) and ( |BM| ):** [ |AM| = sqrt{a^2 - r^2} + x quad text{or} quad |AM| = sqrt{a^2 - r^2} - x ] [ |BM| = sqrt{b^2 - r^2} + x quad text{or} quad |BM| = sqrt{b^2 - r^2} - x ] By plugging these into the equations of the distances and using the square relations: [ left( sqrt{a^2 - r^2} pm x right)^2 = (a^2 - r^2) + 2x sqrt{a^2 - r^2} + x^2 ] [ left( sqrt{b^2 - r^2} pm x right)^2 = (b^2 - r^2) + 2x sqrt{b^2 - r^2} + x^2 ] This follows that: [ sqrt{b^2 - r^2} cdot |AM|^2 - sqrt{a^2 - r^2} cdot |BM|^2 = left( sqrt{b^2 - r^2} - sqrt{a^2 - r^2} right) |OM|^2 ] Equals a constant, say ( l_1 ): [ sqrt{b^2 - r^2} cdot |AM|^2 - sqrt{a^2 - r^2} cdot |BM|^2 + left( sqrt{a^2 - r^2} - sqrt{b^2 - r^2} right) |OM|^2 = l_1 ] 5. **Case: Different Signs in ( |AM| ) and ( |BM| ):** [ |AM| = sqrt{a^2 - r^2} + x quad text{or} quad |AM| = sqrt{a^2 - r^2} - x ] [ |BM| = sqrt{b^2 - r^2} - x quad text{or} quad |BM| = sqrt{b^2 - r^2} + x ] By plugging these into the equations, we consider different arrangements: [ left( sqrt{a^2 - r^2} pm x right)^2 = (a^2 - r^2) pm 2x sqrt{a^2 - r^2} + x^2 ] [ left( sqrt{b^2 - r^2} mp x right)^2 = (b^2 - r^2) mp 2x sqrt{b^2 - r^2} + x^2 ] This gives: [ sqrt{b^2 - r^2} cdot |AM|^2 + sqrt{a^2 - r^2} cdot |BM|^2 = left( sqrt{a^2 - r^2} + sqrt{b^2 - r^2} right) |OM|^2 ] Again, equals a constant, say ( l_2 ): [ sqrt{b^2 - r^2} cdot |AM|^2 + sqrt{a^2 - r^2} cdot |BM|^2 - left( sqrt{a^2 - r^2} + sqrt{b^2 - r^2} right) |OM|^2 = l_2 ] 6. **Conclusion:** Since the sum of the coefficients of ( |AM|^2 ), ( |BM|^2 ), and ( |OM|^2 ) in these expressions are zero, the geometric place of the points ( M ), for which one of these relationships holds, is a plane. In both cases, this plane is perpendicular to the plane formed by points ( O ), ( A ), and ( B ). Therefore, the intersection point ( M ) lies in one of the two fixed planes. [ blacksquare ]

answer:First, we complete the square on the quadratic term in the equation: [ y = frac{x^2 - 8x + 12}{16} = frac{(x^2 - 8x + 16) - 4}{16} = frac{(x-4)^2}{16} - frac{4}{16} ] [ y = frac{(x-4)^2}{16} - frac{1}{4} ] Then, to find the equation of the directrix: 1. The vertex form of the parabola is ( y = frac{1}{16}(x-4)^2 - frac{1}{4} ), with the vertex ( (4, -frac{1}{4}) ). 2. The standard form of a parabola ( y = a(x-h)^2 + k ) centered at ( (h,k) ) with a vertical opening has its focus at ( (h, k + frac{1}{4a}) ) and directrix at ( y = k - frac{1}{4a} ) where ( a = frac{1}{16} ). 3. Substituting into the formula for directrix: [ y = -frac{1}{4} - frac{1}{4 times frac{1}{16}} = -frac{1}{4} - 1 = -frac{5}{4} ] Thus, the equation of the directrix of the given parabola is (boxed{y = -frac{5}{4}}).

answer:1. **Expression for S_n**: [ S_n = frac{(n-1)n(n+1)(3n+2)}{24} ] This formula is given directly in the problem. 2. **Divisibility by 5**: To check when S_n is divisible by 5, investigate when the numerator (n-1)n(n+1)(3n+2) is divisible by 120 (since the denominator is 24). 3. **Finding the conditions**: The term (n-1)n(n+1) is the product of three consecutive integers, which ensures the divisibility by 6. For divisibility by 5, either n, n-1, or n+1 must be a multiple of 5. Further, 3n+2 divisible by 5 implies 3n equiv 3 pmod{5}, or n equiv 1 pmod{5}. 4. **Values of n and Summation**: The least values of n that satisfy both conditions (n equiv 1 pmod{5} and contributes to divisibility by 5) are 1, 6, 11, 16, 21, 26, 31, 36, summing them: [ 1 + 6 + 11 + 16 + 21 + 26 + 31 + 36 = 148 ] Conclusion: The sum of the 8 smallest values of n such that S_n is divisible by 5 is 148. The final answer is boxed{textbf{(C)} 148}.

answer:1. **Identifying Outer Center Properties**: Given that (O) is the circumcenter of the acute-angled (triangle ABC), we know: [ overrightarrow{AO} cdot overrightarrow{AC} = |overrightarrow{AO}| cdot |overrightarrow{AC}| cdot cos angle OAC. ] Also, since (O) is the circumcenter: [ cos angle OAC = frac{1}{2}. ] Therefore, [ overrightarrow{AO} cdot overrightarrow{AC} = frac{1}{2} |overrightarrow{AC}|^2. ] 2. **Substituting Known Lengths**: Given ( AC = 10 ): [ |overrightarrow{AC}|^2 = 10^2 = 100. ] So, [ overrightarrow{AO} cdot overrightarrow{AC} = frac{1}{2} times 100 = 50. ] 3. **Expressing ( overrightarrow{AO} ) in Vector Form**: Given: [ overrightarrow{AO} = x overrightarrow{AB} + y overrightarrow{AC}. ] Substitute this into the dot product expression: [ left(x overrightarrow{AB} + y overrightarrow{AC}right) cdot overrightarrow{AC} = 50. ] 4. **Utilizing the Distributive Property of Dot Products**: [ x (overrightarrow{AB} cdot overrightarrow{AC}) + y (overrightarrow{AC} cdot overrightarrow{AC}) = 50. ] [ x |overrightarrow{AB}| |overrightarrow{AC}| cos angle BAC + y |overrightarrow{AC}|^2 = 50. ] Given ( AB = 6 ) and ( AC = 10 ): [ x cdot 6 cdot 10 cdot cos angle BAC + y cdot 10^2 = 50. ] [ 60x cos angle BAC + 100y = 50. ] 5. **Using the Constraint Equation**: Given the provided equation: [ 2x + 10y = 5. ] Solve for ( x ): [ 10y = 5 - 2x. ] [ y = frac{5 - 2x}{10}. ] 6. **Substituting ( y ) in the Dot Product Equation**: [ 60x cos angle BAC + 100 left(frac{5 - 2x}{10}right) = 50. ] Simplify: [ 60x cos angle BAC + 10(5 - 2x) = 50. ] [ 60x cos angle BAC + 50 - 20x = 50. ] [ 60x cos angle BAC - 20x + 50 = 50. ] [ 60x cos angle BAC - 20x = 0. ] 7. **Isolate ( cos angle BAC )**: Factor out ( x ): [ x(60 cos angle BAC - 20) = 0. ] Since the triangle is acute-angled and ( x neq 0 ): [ 60 cos angle BAC - 20 = 0. ] [ 60 cos angle BAC = 20. ] [ cos angle BAC = frac{20}{60} = frac{1}{3}. ] # Conclusion: [ boxed{frac{1}{3}} ]