answer:1. Identify the key geometric elements provided in the problem: - Circle ( k ) with center ( O ) intersects line ( e ) at points ( A ) and ( B ). - Line segment ( OB ) is bisected perpendicularly at points ( C ) and ( D ). 2. Recognize that triangle ( OCB ) is equilateral: - ( OC = OB ) since ( OC ) and ( OB ) are both radii of the circle. - ( OC = CB ) because ( C ) lies on the perpendicular bisector of ( OB ). - Thus, triangle ( OCB ) conforms to the properties of an equilateral triangle, making ( angle BOC = 60^circ ). 3. Establish that triangle ( OAB ) is isosceles: - ( OA ) and ( OB ) are radii of the circle and thus are equal. - Let ( angle OAB = angle OBA = alpha ). 4. Determine the angles within triangle ( OAB ): - The sum of the interior angles of triangle ( OAB ) must be ( 180^circ ): [ angle OAB + angle OBA + angle AOB = 180^circ ] - Hence: [ alpha + alpha + angle AOB = 180^circ ] [ 2alpha + angle AOB = 180^circ ] [ angle AOB = 180^circ - 2alpha ] 5. Introduce the angle bisector ( COA ) of ( angle COA ): - Let ( angle COA = beta ). - Note that because ( angle BOC = 60^circ ), ( angle COA ) bisects ( angle AOB ), making ( beta = 60^circ/2 = 30^circ ). 6. Utilize the external angle theorem to connect the interior angles to the external angle: - By the external angle theorem ( angle OEB = 60^circ ): [ angle OEB = alpha + beta = 60^circ ] 7. Conclude that the bisector of ( angle COA ) intersects line ( e ) forming a ( 60^circ ) angle: - Given the aforementioned calculations and angle relationships, the bisector creates the specified ( 60^circ ) intersection angle with ( e ). [ boxed{60^circ} ]

answer:To find the maximum number of unique planes that can be determined by 15 points in space, we use the combination formula to calculate the number of ways to choose 3 points from these 15 points, as each group of three non-collinear points determines a plane. The formula for combinations is given by: [ binom{n}{k} = frac{n!}{k!(n-k)!} ] For our problem, [ n = 15 quad text{and} quad k = 3 ] [ binom{15}{3} = frac{15!}{3!(15-3)!} = frac{15 times 14 times 13}{3 times 2 times 1} = 455 ] Thus, the maximum number of unique planes determined by 15 points is boxed{455}. Conclusion: The maximum possible number of planes determined by 15 points, under the condition that not all points lie on the same plane, is 455. This solution accounts for every set of three non-collinear points forming a unique plane.

answer:Let's denote the side length of the square XYZW as s. Thus, the area of square XYZW is s^2. From the problem statement, square XYZW shares 25% of its area with rectangle LMNP: [ 0.25s^2 = text{Area of overlap between square XYZW and rectangle LMNP} ] Rectangle LMNP shares 40% of its area with square XYZW. Let LM = x and LP = y, the area of rectangle LMNP is xy. Therefore: [ 0.4xy = text{Area of overlap between square XYZW and rectangle LMNP} ] Set these areas of overlap equal to each other: [ 0.25s^2 = 0.4xy ] Analyzing the placement of the rectangle relative to the square, assume the simplest case where part of rectangle LMNP matches in one dimension with square XYZW. Suppose the rectangle spans the entire length of the square along one side and part-way across its width, the overlap ratio and total area hints could suggest: Assume LP = s, matching the square's side completely. Since the rectangle shares 40% of its wider area, let's break it up: [ xy = 0.25s^2 / 0.4 = frac{0.25}{0.4} s^2 = frac{5}{8}s^2 ] [ x = frac{5}{8}s^2 / s = frac{5}{8}s ] Thus, the width of the rectangle can be gleaned from how the square's area percentage is configured: [ frac{LM}{LP} = frac{frac{5}{8}s}{s} = frac{5}{8} ] So, the ratio frac{LM}{LP} in rectangle LMNP is frac{5{8}}. The final answer is B) boxed{frac{5}{8}}

answer:1. **Setup the Problem:** - We are given that every point in the plane is colored either red or blue. - For any real number ( x > 0 ), we must show there exists a color such that we can find two points of this color that are exactly ( x ) units apart. 2. **Consider a Geometric Configuration:** - Take an equilateral triangle with side length ( x ). An equilateral triangle ensures that each side has the same length, which is ( x ). 3. **Analyze the Points on the Triangle:** - Label the vertices of the equilateral triangle as ( A ), ( B ), and ( C ). - Each point ( A ), ( B ), and ( C ) must be either red or blue. 4. **Pigeonhole Principle Application:** - According to the Pigeonhole Principle, if you have three points (pigeons) and two colors (holes), at least two of these points must share the same color. - Therefore, among the three vertices, there must be at least two vertices that are of the same color. 5. **Conclusion: Argument for Color:** - Suppose these vertices are colored as such: ( A ) and ( B ) are red, and ( C ) is blue. Alternatively, ( A ) and ( C ) are red, and ( B ) is blue, or ( B ) and ( C ) are red, and ( A ) is blue. - In any case, we can always find two same-colored points out of ( A, B, C ) that are ( x ) units apart because any pair of vertices in an equilateral triangle is exactly ( x ) units apart. 6. **Illustrate the Final Argument:** - Thus, regardless of how the vertices are colored, we ensure that there is a pair of points (either both red or both blue) that are exactly ( x ) units apart. # Conclusion: Therefore, for any ( x > 0 ), there exists a color such that we can find two points of that color which are at a distance of ( x ). [ boxed{} ]