answer:Let us solve the problem step by step. 1. **Identifying key points and segments:** - The segment MN is divided into three equal parts by the diagonals of trapezoid ABCD. - Points M and N lie on AB and CD, respectively. - Let the intersections of MN with diagonals AC and BD be points K and L. Therefore, MK = KL = LN = x. 2. **Triangles similarity setup:** - Consider the triangles ( DNL ) and ( DCB ): - The triangles ( DNL ) and ( DCB ) are similar because ( DNCparallel MN parallel AD ). - Similarly, consider the triangles ( CNK ) and ( CDA ): - The triangles ( CNK ) and ( CDA ) are similar as well for the same reason. 3. **Setting up proportions (using triangle similarity):** - From the similarity of triangles ( DNL ) and ( DCB ): [ frac{x}{b} = frac{DN}{DC} ] - From the similarity of triangles ( CNK ) and ( CDA ): [ frac{2x}{a} = frac{CN}{CD} ] 4. **Solving for proportions and summing up:** - Given that DN + NC = CD, we add the proportions: [ frac{x}{b} + frac{2x}{a} = 1 ] - Finding a common denominator and solving for ( x ): [ x left( frac{1}{b} + frac{2}{a} right) = 1 ] [ x left( frac{a + 2b}{ab} right) = 1 ] [ x = frac{ab}{a + 2b} ] 5. **Calculating the length of segment (MN):** - Since MN consists of three segments of length x, we have: [ MN = 3x = 3 times frac{ab}{a + 2b} ] - Simplifying the expression for MN: [ MN = frac{3ab}{a + 2b} ] # Conclusion: The length of segment (MN) is ( boxed{frac{3ab}{a + 2b}} ).

answer:To find the length of the rectangle, we need to use the perimeter formula for a rectangle, which is: Perimeter = 2 * (length + width) We know the perimeter (the piece of ribbon) is 70 feet long, and the width is 15 feet. Let's plug these values into the formula: 70 = 2 * (length + 15) Now, let's solve for the length: 70 = 2 * length + 2 * 15 70 = 2 * length + 30 Subtract 30 from both sides to isolate the term with the length: 70 - 30 = 2 * length 40 = 2 * length Now, divide both sides by 2 to solve for the length: 40 / 2 = length 20 = length The length of the rectangle is boxed{20} feet.

answer:1. Group terms into pairs: (1-2), (3-4), dots, (99-100), leaving 101 unpaired. 2. Each pair sums to -1, and we have 50 pairs, so sum of pairs is -1 cdot 50 = -50. 3. Add the unpaired number 101 to -50: -50 + 101 = 51. Thus, the sum is boxed{51}.

answer:Given the set ( S subseteq {1, 2, ldots, 2023} ) and the condition that the sum of the squares of any two elements in ( S ) is not a multiple of 9, we aim to determine the maximum possible size of ( S ), denoted ( |S| ). 1. **Identifying quadratic residues mod 9**: - For an integer ( x ): - ( x equiv 0 pmod{9} implies x^2 equiv 0 pmod{9} ) - ( x equiv pm 1 pmod{9} implies x^2 equiv 1 pmod{9} ) - ( x equiv pm 2 pmod{9} implies x^2 equiv 4 pmod{9} ) - ( x equiv pm 3 pmod{9} implies x^2 equiv 0 pmod{9} ) - ( x equiv pm 4 pmod{9} implies x^2 equiv 7 pmod{9} ) 2. **Exclusion of elements**: - To satisfy the condition that the sum of the squares of any two elements in ( S ) is not a multiple of 9, elements whose squares are congruent to 0 mod 9 (i.e., multiples of 3) must be carefully controlled. 3. **Counting multiples of 3 in ({1, 2, ldots, 2023})**: - The multiples of 3 within this range are ( 3, 6, 9, ldots, 2022 ). - Number of terms in the arithmetic sequence ( a_n = 3n leq 2023 ): [ 3n leq 2023 implies n leq leftlfloor frac{2023}{3} rightrfloor = 674 ] - Thus, there are 674 multiples of 3 in ({1, 2, ldots, 2023}). 4. **Determining the maximal size**: - We can include at most one element for which ( x^2 equiv 0 pmod{9} ) in ( S ). - If we exclude all but one of the multiples of 3, then we exclude 673 elements (since there are 674 multiples). - Hence, the maximum number of elements we can include is: [ 2023 - 673 = 1350 ] 5. **Conclusion**: - Therefore, the maximum possible size of ( |S| ) is ( 1350 ). [ boxed{1350} ]