answer:: Let's consider the problem in two parts: part (a) and part (b). Part (a) To prove that the line ( C_1 F ) bisects the perimeter of triangle ( ABC ). 1. Extend segment ( AC ) beyond point ( C ) to a point ( B' ) such that ( CB' = CB ). Therefore, triangle ( BCB' ) is isosceles. 2. Since ( E ) is the midpoint of the arc ( AB ) of the circumcircle of triangle ( ABC ) that contains ( C ), we have ( angle AEB = angle ACB = 2angle CBB' ). 3. Thus, ( E ) is the center of the circumcircle of triangle ( ABB' ). 4. Drop a perpendicular from ( E ) to ( AC ), and let it intersect ( AC ) at ( F ). 5. Given that ( F ) divides segment ( AB' ) into two equal parts due to the symmetry of the circle and position of ( E ). 6. Therefore, line ( C_1F ) bisects the perimeter of triangle ( ABC ). Conclusion for part (a): [ boxed{text{Line} C_1F text{bisects the perimeter of triangle} ABC.} ] Part (b) To prove that three such lines for each side of the triangle intersect at one point. 1. Consider the directness and parallelism. The line through point ( C ) parallel to ( BB' ) serves as the bisector of ( angle ACB ). 2. Given that lines ( C_{1} F ) are parallel to ( BB' ), the line ( C_{1} F ) is actually the angle bisector of the triangle formed by the midpoints of the sides of triangle ( ABC ). 3. Since these bisectors from the midpoints of the sides of the original triangle ( ABC ) intersect at one common point (the circumcenter of the medial triangle), they are concurrent. 4. Therefore, the lines constructed for each side of triangle ( ABC ) intersect at one common point. Conclusion for part (b): [ boxed{text{The three lines constructed for each side of triangle} ABC text{intersect at one point.}} ]

answer:We begin with the given quadratic expression x^2 - 4x - 2019. 1. Complete the square by adding and subtracting the square of half the coefficient of x. This gives us: x^2 - 4x - 2019 = x^2 - 4x + 4 - 4 - 2019 = (x - 2)^2 - 2023. 2. Now, notice that (x - 2)^2 is always greater than or equal to zero, as it is a perfect square. So, (x - 2)^2 - 2023 geq -2023. 3. Therefore, the minimum value of the quadratic expression x^2 - 4x - 2019 is -2023. The answer is: boxed{D. ; -2023}. The problem is solved by completing the square and utilizing the property that a perfect square is always non-negative. Understanding the complete square formula and the properties of non-negative numbers is essential for solving this problem.

answer:According to the given condition, we have overrightarrow{OA} + overrightarrow{OB} + overrightarrow{OC} = overrightarrow{0}. This implies that point O is the centroid of triangle ABC. By the properties of a centroid, the distance from O to BC is one-third of the distance from A to BC. Consequently, the area of triangle OBC is one-third of the area of triangle ABC. Using the geometric probability formula, the probability that the bean will land inside triangle OBC is frac{1}{3}. To summarize, the centroid property allows us to determine the ratio of the areas of the two triangles, which in turn helps us calculate the probability of the bean landing inside triangle OBC. This problem showcases the application of geometric representations to solve real-world problems by transforming random events into mathematical problems and deriving the position of O using vector relations. Therefore, the probability that a bean dropped within triangle ABC will land inside triangle OBC is boxed{frac{1}{3}}.

answer:Let the two consecutive integers be n and n + 1. Then, their product is n(n + 1) < 500. First, solve the inequality for n: [ n^2 + n - 500 < 0 ] We seek integer solutions to this inequality. The roots of the equation n^2 + n - 500 = 0 can be found using the quadratic formula: [ n = frac{-1 pm sqrt{1 + 2000}}{2} = frac{-1 pm sqrt{2001}}{2} ] Since sqrt{2001} is approximately 44.72, we have: [ n = frac{-1 + 44.72}{2} approx 21.86 text{ and } n = frac{-1 - 44.72}{2} approx -22.86 ] However, we need the largest integer n such that n(n+1) < 500. Testing n = 21: [ 21 times 22 = 462 < 500 ] Testing n = 22: [ 22 times 23 = 506 > 500 ] Thus, the largest integer n satisfying the inequality is 21. The sum of n and n+1 is: [ 21 + 22 = boxed{43} ]