answer:Given a parallelogram ABCD with side lengths of 1, and angle ABC = 120^circ. Point E is on the extension of line BC, and AE intersects CD at point F. We are asked to find the angle between vectors overrightarrow{BF} and overrightarrow{ED}. 1. **Setting up the coordinate system:** - We establish a coordinate plane with origin at point B. - The coordinates of the points are as follows: [ Aleft(-frac{1}{2}, frac{sqrt{3}}{2}right), quad B(0,0), quad C(1,0), quad Dleft(frac{1}{2}, frac{sqrt{3}}{2}right) ] - Assume point E(a, 0) where a > 1. 2. **Equation of lines:** - For line CD, calculate the slope: [ text{slope of } CD = frac{frac{sqrt{3}}{2} - 0}{frac{1}{2} - 1} = -sqrt{3} ] Hence, the equation of CD is: [ y = -sqrt{3}(x - 1) ] - For line AE, calculate the slope: [ text{slope of } AE = frac{frac{sqrt{3}}{2} - 0}{-frac{1}{2} - a} = -frac{sqrt{3}}{2a + 1} ] Hence, the equation of AE is: [ y = -frac{sqrt{3}}{2a+1}(x - a) ] 3. **Finding the intersection point ( F ):** - Solve the system of equations: [ begin{cases} y = -sqrt{3}(x - 1) y = -frac{sqrt{3}}{2a + 1}(x - a) end{cases} ] - Set the equations equal to each other to find x: [ -sqrt{3}(x - 1) = -frac{sqrt{3}}{2a+1}(x - a) ] Solving for x, we get: [ sqrt{3}(x - 1) = frac{sqrt{3}}{2a+1}(x - a) ] [ x - 1 = frac{x - a}{2a + 1} ] [ (2a + 1)(x - 1) = x - a ] [ 2ax + x - 2a - 1 = x - a ] [ 2ax - 2a - 1 = -a ] [ 2ax = a + 2a + 1 ] [ 2ax = 3a + 1 ] [ x = frac{a + 1}{2a} ] - Find y: [ y = -sqrt{3} left( frac{a + 1}{2a} - 1 right) = frac{sqrt{3}(a - 1)}{2a} ] - Therefore, F has coordinates: [ F left( frac{a + 1}{2a}, frac{sqrt{3}(a - 1)}{2a} right) ] 4. **Vectors (overrightarrow{BF}) and (overrightarrow{ED}):** - Calculate overrightarrow{BF}: [ overrightarrow{BF} = F - B = left( frac{a + 1}{2a}, frac{sqrt{3}(a - 1)}{2a} right) ] - Calculate overrightarrow{ED}: [ overrightarrow{ED} = D - E = left( frac{1}{2} - a, frac{sqrt{3}}{2} - 0 right) = left( frac{1 - 2a}{2}, frac{sqrt{3}}{2} right) ] 5. **Dot product of (overrightarrow{BF}) and (overrightarrow{ED}):** [ overrightarrow{BF} cdot overrightarrow{ED} = left( frac{a + 1}{2a} right) left( frac{1 - 2a}{2} right) + left( frac{sqrt{3}(a - 1)}{2a} right) left( frac{sqrt{3}}{2} right) ] [ = frac{(a + 1)(1 - 2a)}{4a} + frac{3(a - 1)}{4a} ] [ = frac{a + 1 - 2a^2 - 2a}{4a} + frac{3(a - 1)}{4a} ] [ = frac{1 - a - 2a^2}{4a} + frac{3a - 3}{4a} ] [ = frac{1 - a - 2a^2 + 3a - 3}{4a} ] [ = frac{-2a^2 + 2a - 2}{4a} ] [ = -frac{2(a^2 - a + 1)}{4a} ] [ = -frac{a^2 - a + 1}{2a} ] 6. **Magnitudes of (overrightarrow{BF}) and (overrightarrow{ED}):** - For overrightarrow{BF}: [ left| overrightarrow{BF} right| = sqrt{ left( frac{a + 1}{2a} right)^2 + left( frac{sqrt{3}(a - 1)}{2a} right)^2 } ] [ = sqrt{ frac{(a + 1)^2}{4a^2} + frac{3(a - 1)^2}{4a^2} } ] [ = sqrt{ frac{(a + 1)^2 + 3(a - 1)^2}{4a^2} } ] [ = sqrt{ frac{a^2 + 2a + 1 + 3(a^2 - 2a + 1)}{4a^2} } ] [ = sqrt{ frac{a^2 + 2a + 1 + 3a^2 - 6a + 3}{4a^2} } ] [ = sqrt{ frac{4a^2 - 4a + 4}{4a^2} } ] [ = sqrt{ frac{4(a^2 - a + 1)}{4a^2} } ] [ = sqrt{ frac{a^2 - a + 1}{a^2} } ] [ = frac{sqrt{a^2 - a + 1}}{a} ] - For overrightarrow{ED}: [ left| overrightarrow{ED} right| = sqrt{ left( frac{1 - 2a}{2} right)^2 + left( frac{sqrt{3}}{2} right)^2 } ] [ = sqrt{ frac{(1 - 2a)^2}{4} + frac{3}{4} } ] [ = sqrt{ frac{1 - 4a + 4a^2 + 3}{4} } ] [ = sqrt{ frac{4a^2 - 4a + 4}{4} } ] [ = sqrt{a^2 - a + 1} ] 7. **Calculate the cosine of the angle (theta):** [ cos theta = frac{ overrightarrow{BF} cdot overrightarrow{ED} }{ | overrightarrow{BF} | cdot | overrightarrow{ED} | } ] [ = frac{ -frac{a^2 - a + 1}{2a} }{ frac{sqrt{a^2 - a + 1}}{a} cdot sqrt{a^2 - a + 1} } ] [ = frac{-frac{a^2 - a + 1}{2a}}{frac{a^2 - a + 1}{a}} ] [ = -frac{1}{2} ] 8. **Conclude the angle (theta):** [ cos theta = -frac{1}{2} ] Therefore, [ theta = 120^circ ] [ boxed{120^circ} ]

answer:Using the Remainder Theorem, we find the remainder by evaluating the polynomial at the root of ( 4x - 8 = 0 ), which simplifies to ( x = 2 ). Substituting ( x = 2 ) into ( 8x^3 - 20x^2 + 28x - 30 ): [ 8 cdot 2^3 - 20 cdot 2^2 + 28 cdot 2 - 30 = 8 cdot 8 - 20 cdot 4 + 28 cdot 2 - 30. ] [ = 64 - 80 + 56 - 30. ] [ = 64 + 56 - 80 - 30. ] [ = 120 - 110. ] [ = 10. ] Thus, the remainder when ( 8x^3 - 20x^2 + 28x - 30 ) is divided by ( 4x - 8 ) is (boxed{10}).

answer:Using the logarithmic identity log_2 frac{x}{y} = log_2 x - log_2 y, we can simplify each term in the sum: [ log_2 frac{2}{1} + log_2 frac{3}{2} + cdots + log_2 frac{10010}{10009} = (log_2 2 - log_2 1) + (log_2 3 - log_2 2) + cdots + (log_2 10010 - log_2 10009) ] This sum is telescopic, meaning most intermediate terms cancel: [ log_2 10010 - log_2 1 = log_2 10010 ] To find the largest integer less than this value, note that 2^{13} = 8192 and 2^{14} = 16384. Therefore, 13 < log_2 10010 < 14. The largest integer less than log_2 10010 is thus: [ boxed{13} ]

answer:Given two circles of radius 17 intersect at points A and B, with point C on the first circle and point D on the second circle, point B lies on the line segment CD, and angle CAD = 90^circ. A point F is chosen on the perpendicular to CD through point B such that BF = BD with points A and F on opposite sides of line CD, we need to find the length of segment CF and the area of triangle ACF. Part (a): 1. Let R = 17 be the radius of the circles. 2. Denote the angles as angle BAD = alpha and angle BCF = beta. 3. We know angle BAC = 90^circ - alpha since angle CAD = 90^circ. Application of the Law of Sines in triangle BAD and triangle BAC: [ BD = 2R sin alpha ] [ BC = 2R cos alpha ] Given: [ CF^2 = BC^2 + BD^2 ] Substituting the values: [ CF^2 = (2R cos alpha)^2 + (2R sin alpha)^2 = 4R^2 (cos^2 alpha + sin^2 alpha) ] Using Pythagoras identity: [ cos^2 alpha + sin^2 alpha = 1 ] Therefore: [ CF^2 = 4R^2 cdot 1 = 4R^2 ] So: [ CF = 2R = 2 times 17 = 34 ] Conclusion: [ boxed{34} ] Part (b): 1. Given BC = 16, we need to find the area of triangle ACF. 2. From the given condition B F = B D, and since tan beta = frac{BF}{BC} = frac{BD}{BC} = frac{sin alpha}{cos alpha} = tan alpha, we have beta = alpha. 3. The angles angle ADC and angle ACD span the same arc in the identical circles they are inscribed in. Thus: [ angle ADC = angle ACD = frac{pi}{4} ] Since BC = 16 and: [ cos alpha = frac{BC}{2R} = frac{8}{17} ] We calculate AC using: [ AC = 2R sin left( frac{pi}{4} + alpha right) ] Therefore, the area S_{ACF} is computed as: [ S_{ACF} = frac{1}{2} cdot AC cdot CF cdot sin angle ACF = frac{1}{2} cdot 2R sin left( frac{pi}{4} + alpha right) cdot 2R sin left( frac{pi}{4} + alpha right) ] [ S_{ACF} = 2R^2 sin^2 left( frac{pi}{4} + alpha right) ] Using the identity sin (x+y) = sin x cos y + cos x sin y: [ 2R^2 sin^2 left( frac{pi}{4} + alpha right) = R^2 (1 + cos 2alpha) ] For cos alpha = frac{8}{17}: [ sin alpha = sqrt{1 - cos^2 alpha} = sqrt{1 - left(frac{8}{17}right)^2} = sqrt{frac{225}{289}} = frac{15}{17} ] Thus: [ 2R^2 left( frac{1}{2} + sin 2alpha right) = 2 cdot 289 left( frac{1}{2} + 2 cos alpha sin alpha right) ] [ = 289 left( 1 + 2 cdot frac{8}{17} cdot frac{15}{17} right) = 289 left(1 + frac{240}{289}right) = 289 cdot 2 = 578 ] Thus the area of triangle ACF is: [ boxed{529} ]