answer:1. **Labeling the Triangle:** Let the vertices of the triangle be labeled as ( A, B, ) and ( C ), and the sides opposite these vertices as ( a, b, ) and ( c ), respectively. 2. **Defining the Endpoints and Incenter:** The endpoints of the segments measured from each vertex on the triangle opposite sides are denoted as ( A', B', ) and ( C' ), and the second set of measured points as ( A'', B'' ) and ( C'' ). The incenter of ( triangle ABC ), which is the center of the inscribed circle, is labeled as ( O ). 3. **Relationship between the Triangles:** Consider the triangles ( AA'C' ), ( BB'A' ), and ( CC'B' ). These triangles are isosceles because the segments measured along the sides are equal. 4. **Perpendicular Bisectors and Angle Bisectors:** The perpendicular bisectors of the sides of such isosceles triangles coincide with the angle bisectors of ( angle BAC ), ( angle ABC ), and ( angle ACB ). These perpendicular bisectors (angle bisectors) must pass through ( O ), as ( O ) is the incenter of ( triangle ABC ), thereby dividing the angles equally. 5. **Tangents from Triangle Vertices:** From point ( A ) to the inscribed circle, the tangent segments have equal length. Let this length be ( x ). Therefore, we can define: - The length from ( B ) to the circle is ( c - x ), - From ( C ) to the circle is ( b - x ). We thus have the relationship: [ BC = a = (c - x) + (b - x) ] Simplifying this, we find: [ a = b + c - 2x implies x = frac{1}{2}(b + c - a) ] 6. **Radial Distances to Tangential Points:** The tangents from other points similarly yield: [ text{The tangent length from } A' text{ to the circle} = text{The combined lengths} = frac{1}{2}(a + b + c) ] Hence, the points ( A' ), ( B' ), and ( C'' ) are at a distance: [ sqrt{left( frac{1}{2}(a + b + c) right)^2 + r^2} ] where ( r ) is the radius of the inscribed circle. 7. **Symmetry and Point Concurrency:** Since ( O ) is equidistant to ( A' ), ( B' ), ( C'' ), and the distances from ( O ) to ( A' ), ( B' ), ( C' ), ( A'' ), ( B'' ), ( C'' ) are equal, all these points lie on a circle with center ( O ). Conclusively, the endpoints of the segments measured from the vertices of the triangle ( A, B, ) and ( C ) are concyclic. Thus, we have shown that ( A', B', C', A'', B'', C'' ) all lie on a circle with center ( O ). [ boxed{} ]

answer:(1) Since the center of the ellipse is at the origin, and one focus is at F_1(0,-1), we have c = 1. Given that the eccentricity is frac{sqrt{3}}{3}, we have frac{c}{a} = frac{sqrt{3}}{3}. Solving for a, we get a = sqrt{3}. Then, using b = sqrt{a^2 - c^2}, we find b = sqrt{2}. Therefore, the standard equation of the ellipse is frac{y^2}{3} + frac{x^2}{2} = 1. (2) Let the equation of line AB be: y = kx - 1. By substituting y in the equation of the ellipse, we get: (3 + 2k^2)x^2 - 4kx - 4 = 0. Hence, x_1 + x_2 = frac{4k}{3 + 2k^2} and x_1x_2 = frac{-4}{3 + 2k^2}. The distance between x_1 and x_2 is given by |x_1 - x_2| = sqrt{(x_1 + x_2)^2 - 4x_1x_2} = 4sqrt{3} cdot frac{sqrt{1 + k^2}}{3 + 2k^2}. Let t = sqrt{1 + k^2} (t geq 1). Then, |x_1 - x_2| = 4sqrt{3} cdot frac{t}{1 + 2t^2}. Since (2t + frac{1}{t})' = 2 - frac{1}{t^2} > 0 for t geq 1, we have 2t + frac{1}{t} geq 3. This implies that the range of |x_1 - x_2| is (0, frac{4sqrt{3}}{3}]. Finally, the area of triangle ABF_2 is given by S_{triangle ABF_2} = frac{1}{2}|x_1 - x_2| cdot 2c = |x_1 - x_2|. Therefore, the range of values for S_{triangle ABF_2} is boxed{(0, frac{4sqrt{3}}{3}]}.

answer:If Sammy has 8 bottle caps and Sammy has 2 more bottle caps than Janine, then Janine has 8 - 2 = 6 bottle caps. Since Janine has 3 times as many bottle caps as Billie, we can find out how many bottle caps Billie has by dividing the number of bottle caps Janine has by 3. So, Billie has 6 / 3 = boxed{2} bottle caps.

answer:To solve for the square root of frac{4}{9}, we follow the property that the square root of a fraction is the square root of the numerator divided by the square root of the denominator. Therefore, we have: [ sqrt{frac{4}{9}} = frac{sqrt{4}}{sqrt{9}} = frac{2}{3} ] However, considering the principle of square roots which can have both positive and negative roots, we acknowledge both possibilities: [ sqrt{frac{4}{9}} = pm frac{2}{3} ] For the arithmetic square root of sqrt{16}, we first find the square root of 16, which is: [ sqrt{16} = 4 ] The arithmetic square root refers to the positive square root, thus: [ sqrt{16} = 4 ] Therefore, the answers are: boxed{pm frac{2}{3}, 4}. However, there seems to be a discrepancy in the interpretation of the second part of the solution. The correct interpretation, following the initial solution provided, should be that the arithmetic square root of sqrt{16} is indeed 4, but considering the context of arithmetic square roots being positive, the final answer should be adjusted to reflect the correct interpretation: Therefore, the corrected answers are: boxed{pm frac{2}{3}, 4}.