answer:**Analysis** This problem mainly examines geometric probability. Based on the probability, the length that constitutes event M is frac{1}{3} of the length of segment CD. By symmetry, when PD= frac{1}{3} CD, AB=PB. Using the Pythagorean theorem, the conclusion can be drawn. **Solution** Let's denote the event "Randomly select a point P on the side CD of rectangle ABCD, such that the longest side of triangle APB is AB" as event M, The total possible outcomes constitute the length of segment CD, and the length that constitutes event M is frac{1}{3} of the length of segment CD, Let AB=3x, AD=y, then By symmetry, when PD= frac{1}{3} CD, AB=PB, By the Pythagorean theorem, we get (3x)^{2}=y^{2}+(2x)^{2}, therefore y= sqrt{5}x , therefore frac{AD}{AB}= frac{y}{3x}= frac{ sqrt{5}}{3} . Therefore, the correct choice is boxed{C}.

answer:The problem requires us to build an electrical circuit from unit resistances such that the total resistance of the circuit is equal to the rational number frac{a}{b}. We will achieve this by using a method analogous to the division of a rectangle with dimensions a times b into squares. Let's go through the steps in a detailed manner. 1. **Starting with a Rational Number**: Given the rational number frac{a}{b}, where a and b are positive integers, we aim to realize this value as the total resistance of an electrical circuit composed of unit resistances. 2. **Conversion to Continued Fraction**: Convert the rational number frac{a}{b} into its continued fraction representation. A continued fraction for a rational number frac{a}{b} can be expressed in the form: [ frac{a}{b} = [a_0; a_1, a_2, ldots, a_n] ] where a_0, a_1, a_2, ldots, a_n are non-negative integers. 3. **Constructing the Continued Fraction Example**: Consider, for example, the rational number frac{71}{31}. Its continued fraction representation is: [ frac{71}{31} = [2; 3, 2, 4] ] This implies: [ 71/31 = 2 + frac{1}{3 + frac{1}{2 + frac{1}{4}}} ] 4. **Corresponding Graphical Representation**: In the graphical representation, each term in the continued fraction corresponds to the side length of the squares used to subdivide the rectangle a times b. In our example, the rectangle is 31 times 71. Illustrating this, we would see a series of nested squares as depicted below:  5. **Mapping to Electrical Circuit**: Each square in the graphical representation corresponds to a unit resistance in the circuit. To achieve a total resistance of frac{a}{b} in the electrical circuit: - Connect the resistances corresponding to horizontal partitions (squares) in series. - Connect the resistances corresponding to vertical partitions (squares) in parallel. This approach effectively models the continued fraction decomposition of the rational number frac{a}{b} into an equivalent electrical circuit of unit resistances. 6. **Constructing the Circuit**: Using our example of frac{71}{31} = [2; 3, 2, 4]: - The first term '2' implies two unit resistances connected in series. - The next term '3' implies three unit resistances connected in parallel, which yields a total resistance of frac{1}{3}. - Combining these using the series-parallel combinations for each continued fraction term results in the complete circuit. Here is the illustrative diagram of the resultant circuit:  7. **Conclusion**: This method showcases how to realize the given rational number frac{a}{b} as the total resistance of an electrical circuit using unit resistances, by leveraging the continuous fraction representation and corresponding the construction of the circuit based on the graphical subdivision of rectangles into squares. For the proof part of the problem, consider that for any infinite continued fraction [a_0; a_1, ldots, a_n, ldots], there exists a limit alpha which is an irrational number. By the properties of continued fractions from problem 60606, this alpha can be uniquely represented as such an infinite continued fraction, ensuring that the approximation converges to the original irrational number through successive finite continued fractions. [ boxed{text{Unit resistances can effectively model any rational }frac{a}{b}text{ as shown using continued fractions.}} ]

answer:1. **Understand the Problem:** - We need to place numbers from 1 to 4 in a 3x3 grid. - Each row, each column, and both main diagonals must contain exactly three distinct numbers from these four. 2. **Analyze the Diagonals:** - Consider one of the diagonals, for instance, the main diagonal (the one from the top left to the bottom right). Suppose we place the numbers 1, 2, and 3 on this diagonal. - The positions on the main diagonal are (1,1), (2,2), and (3,3). 3. **Implication on the Remaining Cells:** - The cells (1,3), (3,1), and (3,3) will now have only one number left, which is 4 (since 1, 2, and 3 are already on the main diagonal). 4. **Effect on the Second Diagonal:** - Consider the second diagonal (from the top right to the bottom left), specifically the positions (1,3), (2,2), and (3,1). - We previously filled (1,3) and (3,1) with the number 4. - Since the number at the intersection of both diagonals (i.e., (2,2)) is also on the other diagonal, it must be different from 4 because we need distinct numbers. 5. **Conflict and Conclusion:** - As per our assumption, we see that the placement of these numbers violates the condition since we can't maintain distinct numbers in the diagonals while simultaneously keeping rows and columns distinct. - Therefore, no possible arrangement will satisfy all conditions (rows, columns, and diagonals having different number arrangements with three unique numbers). Thus, it is impossible to arrange the numbers from 1 to 4 in the given 3x3 grid under the specified conditions. boxed{text{No, it is not possible.}}

answer:Given the exchange rate, 2 Australian dollars are worth 1 American dollar, hence 1 Australian dollar is worth frac{1}{2} American dollars. The damage in Melbourne was 45 million Australian dollars. To find the damage in American dollars: [ left( frac{1}{2} right) times 45,000,000 = 22,500,000 ] Thus, the damage in American dollars is boxed{22,500,000}.