answer:: Given: Let alpha be a real number, not an odd multiple of pi. We need to prove that tan(alpha/2) is rational if and only if cos(alpha) and sin(alpha) are rational. 1. First, we define ( t = tan(alpha/2) ). 2. Using the half-angle formulas for sine and cosine, we get: [ cos(alpha) = frac{1-t^2}{1+t^2} ] and [ sin(alpha) = frac{2t}{1+t^2} ] 3. **(⇒ Direction)**: - Assume ( t = tan(alpha/2) ) is rational. - Substituting ( t ) into the half-angle formulas: [ cos(alpha) = frac{1-t^2}{1+t^2} ] and [ sin(alpha) = frac{2t}{1+t^2} ] - Since ( t ) is rational, it follows that ( t^2 ) is also rational. Thus, the expressions ( 1-t^2 ) and ( 1+t^2 ) are rational. - Therefore, since both the numerator and denominator are rational, ( cos(alpha) ) and ( sin(alpha) ) are both rational. 4. **(⇐ Direction)**: - Assume ( cos(alpha) ) and ( sin(alpha) ) are rational. - Denote ( cos(alpha) = k ) and ( sin(alpha) = m ), where ( k ) and ( m ) are rational numbers. - Recall the identity ( k^2 + m^2 = 1 ). Since ( k ) and ( m ) are rational, ( k^2 + m^2 ) must also be rational. Hence, it is equal to 1. - From the identity ( sin(2theta) = 2sin(theta)cos(theta) ) and ( cos(2theta) = cos^2(theta) - sin^2(theta) ), we write: [ cos(alpha) = cos^2(alpha/2) - sin^2(alpha/2) ] [ sin(alpha) = 2 sin(alpha/2) cos(alpha/2) ] - Using the substitution ( t = tan(alpha/2) ): [ t = frac{sin(alpha/2)}{cos(alpha/2)} ] - The expression ( cos(alpha) = frac{1-t^2}{1+t^2} ) implies replacing ( k ) with ( frac{1-t^2}{1+t^2} ). Rearranging for ( t^2 ): [ k = cos(alpha) = frac{1-t^2}{1+t^2} ] [ k(1+t^2) = 1-t^2 ] [ k + kt^2 = 1 - t^2 ] [ kt^2 + t^2 = 1 - k ] [ t^2(1 + k) = 1 - k ] [ t^2 = frac{1 - k}{1 + k} ] - Since ( k = cos(alpha) ) is rational and ( k neq -1 ) (as (alpha) is not an odd multiple of pi), ( frac{1 - k}{1 + k} ) is a rational number implying ( t^2 ) is rational. 5. Conclusively, since ( t^2 ) is rational, ( t = tan(alpha/2) ) must also be rational because the square root of a rational number (assuming it remains rational) is rational. The conclusion is that (boxed{tan(alpha/2) text{ is rational iff } cos(alpha) text{ and } sin(alpha) text{ are rational.}})

answer:1. **Setup**: Let (d = AB) be the distance between points (A) and (B). Consider the condition (PA + PB = frac{1}{2} cdot AB = frac{1}{2}d). 2. **Triangle Inequality Analysis**: By the triangle inequality, for any triangle formed by (P, A,) and (B), (PA + PB) must be greater than or equal to (AB). Since (frac{1}{2}d < d), there's no direct triangle configuration involving (A), (B), and (P) that satisfies this condition. 3. **Geometric Insight**: If (PA + PB = frac{1}{2}d), then (P) cannot lie on any part, inside or outside, of the segment (AB) because the sum of distances in these cases would be at least (d). 4. **Immediate Conclusion**: - Since any point (P) in the plane would generally have (PA + PB geq d), it's impossible for any point (P) to have (PA + PB = frac{1}{2}d) given (d = AB). 5. **Final Determination**: There are no points (P) that satisfy the condition (PA + PB = frac{1}{2}d) for (d = AB), implying that such a set of points is non-existent. Thus, the conclusion is that (text{No Points Satisfy This Condition}). The final answer is boxed{C) No points satisfy this condition}

answer:To solve this problem, we first understand that the weights of each section of the golden rod form an arithmetic progression (AP) because the rod changes uniformly from thick to thin. Given that when one foot is cut from the end, it weighs 2 catties, we can consider this as the first term of our AP, which we denote as a_1 = 2 catties. Similarly, when one foot is cut from the base, it weighs 4 catties. Since the rod is 5 feet long, this measurement corresponds to the fifth term in our AP, which we denote as a_5 = 4 catties. The formula for the sum of the first n terms of an arithmetic progression is given by S_n = frac{n}{2} times (a_1 + a_n), where a_1 is the first term, a_n is the nth term, and n is the number of terms. In this case, we want to find the total weight of the golden rod, which is the sum of the weights of all 5 sections. Therefore, we have n = 5, a_1 = 2, and a_5 = 4. Substituting these values into the formula for the sum of an AP, we get: [ S_5 = frac{5}{2} times (2 + 4) = frac{5}{2} times 6 = 15 ] Therefore, the total weight of the golden rod is boxed{15} catties.

answer:# Problem: Show how any quadrilateral can be divided into three trapezoids (a parallelogram can also be considered a trapezoid). 1. Identify the largest internal angle of the given quadrilateral (ABCD). Let's denote this angle as ( angle B ). 2. From vertex (B), draw a line segment (BM) parallel to the opposite side (AD), such that point (M) is inside the quadrilateral. This line segment (BM) will cut the quadrilateral into two regions. 3. Now, through point (M), draw another line segment (MN) parallel to side (BC), and another line segment (MK) parallel to side (CD), ensuring that the points (N) and (K) respectively fall inside the quadrilateral. 4. As a result, the quadrilateral (ABCD) will be divided into three regions: - The first trapezoid is formed between line segments (BM) and (AD), which are parallel. - The second trapezoid is formed between line segments (MN) and (BC), which are parallel. - The third trapezoid is formed between line segments (MK) and (CD), which are parallel. # Conclusion: By proper construction of line segments parallel to the sides of the quadrilateral, any quadrilateral can be systematically divided into three trapezoids. [ boxed{} ]