answer:To prove that at least one side of the rectangle R has an integer length, we proceed as follows: 1. **Setup**: Let the rectangle ( R ) be denoted as ( ABCD ) with ( A ) as the origin in a coordinate system where ( AB ) and ( AD ) are parallel to the ( x )-axis and ( y )-axis, respectively. 2. **Identifying the Problem**: We need to show that at least one of the points ( B ), ( C ), or ( D ) has integer coordinates. Since each small rectangle ( R_i ) has sides parallel to the sides of ( R ), and at least one side length of ( R_i ) is an integer, each ( R_i ) is aligned with the coordinate axes. 3. **Properties of Small Rectangles**: Each small rectangle ( R_i ) has its sides parallel to the coordinate axes and at least one integer side. This means the vertices of ( R_i ) lie on the grid points (lattice points). 4. **Counting Vertices**: Each rectangle ( R_i ) has four vertices, which are potentially grid points. Since each ( R_i ) has at least one integer side, the count of grid points among the vertices will be either 0, 2, or 4. This means the total number of grid points among all vertices of all rectangles ( R_1, R_2, ldots, R_n ) will be even. 5. **Overall Rectangle ( R )**: We now consider rectangle ( R ) formed by the union of all ( R_i ). The vertices of ( R ) will be ( A ), ( B ), ( C ), and ( D ). 6. **Counting and Parity**: Since the sum of the grid points for all vertices across all small rectangles is even and each internal grid point is counted an even number of times, the total number of integer coordinates among the vertices of ( R ) must also be even. 7. **Edge Case Including A**: The point ( A ) is at the origin, which is a grid point and thus counted once. This imposes a requirement that the sum of the grid points among ( B ), ( C ), and ( D ) is odd since their combined total with ( A ) must be an even number. 8. **Conclusion**: As ( A ) is already a grid point counted once, at least one of the points among ( B ), ( C ), or ( D ) must be a grid point to satisfy the parity requirement. Thus, we have established that the rectangle ( R ) must have at least one side length that is an integer. boxed{}

answer:To find 99^{-1} pmod{101}, note that: 1. Since 99 equiv -2 pmod{101}, we have 99^2 equiv 4 pmod{101}. 2. Continuing, 99^3 equiv -8 pmod{101}. 3. We need 99 cdot x equiv 1 pmod{101}. From our earlier finding, 99^3 equiv -8 pmod{101} suggests that multiplying both sides by the inverse of -8 modulo 101 could help. 4. The inverse of -8 modulo 101 needs to be calculated. By trial or using extended Euclidean algorithm, -8 cdot 13 = 104 equiv 3 pmod{101}; further -8 cdot 38 = -304 equiv 1 pmod{101}. So, the inverse of -8 modulo 101 is 38. 5. So 99^{-1} equiv 99^2 cdot 38 equiv 4 cdot 38 equiv 152 equiv boxed{51} pmod{101}.

answer:Since the vertex of the hyperbola is at (0, 2), it follows that a=2, and the standard equation of the hyperbola is frac {y^{2}}{4} - frac {x^{2}}{b^{2}} = 1. According to the problem, 2a+2b= sqrt {2} cdot 2c, which simplifies to a+b= sqrt {2}c. Also, a^{2}+b^{2}=c^{2}, and given a=2, solving these equations yields b^{2}=4. Therefore, the equation of the hyperbola that meets the conditions is frac {y^{2}}{4} - frac {x^{2}}{4} = 1. Hence, the correct option is boxed{text{B}}. From the given information, the standard equation of the hyperbola is frac {y^{2}}{4} - frac {x^{2}}{b^{2}} = 1, and 2a+2b= sqrt {2} cdot 2c, from which the equation of the hyperbola can be determined. This question tests the method of finding the equation of a hyperbola, which is a basic problem. When solving, it is important to carefully read the problem and appropriately apply the properties of hyperbolas.

answer:To find the unknown coordinate (x) of the vertex (C(x, 5)) of the triangle (ABC) with given vertices (A(0, 3)) and (B(4, 0)), we will use the formula for the area of a triangle using its vertices' coordinates. The area (mathcal{A}) of a triangle with vertices ((x_1, y_1)), ((x_2, y_2)), and ((x_3, y_3)) can be calculated as follows: [ mathcal{A} = frac{1}{2} left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) right| ] Given: - Vertex (A = (0, 3)) - Vertex (B = (4, 0)) - Vertex (C = (x, 5)) - The area of the triangle (mathcal{A} = 8) square units 1. Substitute the coordinates of (A), (B), and (C) into the area formula: [ mathcal{A} = frac{1}{2} left| 0(0 - 5) + 4(5 - 3) + x(3 - 0) right| ] 2. Simplify the expression inside the absolute value: [ mathcal{A} = frac{1}{2} left| 0 + 4 cdot 2 + x cdot 3 right| ] 3. Continue simplifying: [ mathcal{A} = frac{1}{2} left| 8 + 3x right| ] 4. Given that the area (mathcal{A}) is 8, we get: [ 8 = frac{1}{2} left| 8 + 3x right| ] 5. Multiply both sides by 2 to eliminate the fraction: [ 16 = left| 8 + 3x right| ] 6. Solve the absolute value equation by considering both positive and negative scenarios: [ 8 + 3x = 16 quad text{or} quad 8 + 3x = -16 ] 7. Solve each linear equation separately: For the first case: [ 8 + 3x = 16 ] Subtract 8 from both sides: [ 3x = 8 ] Divide by 3: [ x = frac{8}{3} ] For the second case: [ 8 + 3x = -16 ] Subtract 8 from both sides: [ 3x = -24 ] Divide by 3: [ x = -8 ] 8. Since we are given (0 < x < 4), we discard (x = -8), as it does not satisfy this condition. Thus, we are left with: [ x = frac{8}{3} ] # Conclusion: The unknown coordinate (x) for the vertex (C) is: [ boxed{frac{8}{3}} ]