answer:1. **Initial Definitions and Objective**: Consider the game where the initial pair of positive integers is ((x_1, y_1)). The objective is for the player to transform this pair into ((x_{n+1}, y_{n+1})) such that (y_{n+1} = 0), via permissible moves. We need to find the conditions on the ratio (frac{x_1}{y_1}) that guarantee the first player has a winning strategy. 2. **Game Rules Recap**: Each move consists of: - Setting (x_{n+1} = min(x_n, y_n)) - Setting (y_{n+1} = max(x_n, y_n) - k cdot x_{n+1}), for some positive integer (k) 3. **Setup of Key Variables**: Let: - (m = min(x, y)) - (M = max(x, y)) We need to determine under which conditions on (m) and (M) the first player can always win. 4. **Introduction of Golden Ratio (alpha)**: Define (alpha) as the positive root of the quadratic equation (t^2 - t - 1 = 0): [ alpha = frac{1 + sqrt{5}}{2} ] This value is approximately (1.618) and is known as the golden ratio. 5. **Main Theorem and Initial Analysis**: We want to show that a player has a winning strategy if and only if: [ frac{x_1}{y_1} = 1 quad text{or} quad frac{x_1}{y_1} > alpha quad text{or} quad frac{x_1}{y_1} < frac{1}{alpha} ] In terms of (m) and (M): - If (m = M), i.e., (x_1 = y_1), the player first to move already wins by sending ((m, 0)). - If (M > alpha m): - We need to analyze the next possible moves. 6. **Analysis of M > (alpha m)**: - If (M) is a multiple of (m), i.e., (M = km) for some integer (k): - The player can always win by setting (y_{n+1} = 0) since (k cdot m - k cdot m = 0). - If (M) is not a multiple of (m): - Write (M) as (M = qm + r), where (0 < r < m). - If (q geq 2), there are two sub-cases: - (i) Passing ((m, r)) might win if ((m, r)) is a losing position. - (ii) Otherwise, if ((m, r)) is a winning position for the next player, the first player can iteratively force the next player into a losing position by leveraging the golden ratio properties. 7. **Special Case Analysis When (q = 1)**: - Here, we write (M = m + r) with (0 < r < m). - We must then verify: - (m neq r) - (m < alpha r) - Since (m + r > alpha m), we derive: [ (alpha - 1)m < r Rightarrow m < alpha r ] 8. **General Conclusion**: By combining all the steps and analysis, we conclude that the first player has a winning strategy if and only if: [ frac{x_1}{y_1} = 1 quad text{or} quad frac{x_1}{y_1} > alpha quad text{or} quad frac{x_1}{y_1} < frac{1}{alpha} ] Thus, we conclude the winning condition based on the initial ratio (frac{x_1}{y_1}). [ boxed{frac{x_1}{y_1} = 1 quad text{or} quad frac{x_1}{y_1} > alpha quad text{or} quad frac{x_1}{y_1} < frac{1}{alpha}} ]

answer:First, let's calculate the total time John spent repairing the cars. For the first 3 cars, each taking 40 minutes to repair: 3 cars * 40 minutes/car = 120 minutes For the remaining 2 cars, which take 50% longer than 40 minutes each: 50% of 40 minutes = 0.5 * 40 = 20 minutes longer So each of these cars takes 40 minutes + 20 minutes = 60 minutes to repair. For the 2 cars taking 60 minutes each: 2 cars * 60 minutes/car = 120 minutes Now, let's add the time for all 5 cars: 120 minutes (first 3 cars) + 120 minutes (remaining 2 cars) = 240 minutes To find out how much John makes per hour, we need to convert the total minutes to hours: 240 minutes / 60 minutes/hour = 4 hours Now we know that John made 80 in 4 hours. To find his hourly rate: 80 / 4 hours = 20/hour John makes boxed{20} per hour repairing cars.

answer:1. **Identify Circles and Triangles**: - C_2 is the circumcircle of triangles triangle XOZ and triangle OYZ. - O is the center of C_1 and lies on C_2. - X and Y are the intersection points of C_1 and C_2. 2. **Use Heron's formula and circumradius relation**: - For triangle XOZ, sides are r, 15, and 17. Semi-perimeter s_1 = frac{r + 15 + 17}{2} = 16 + frac{r}{2}. - For triangle OYZ, sides are r, 8, and 17. Semi-perimeter s_2 = frac{r + 8 + 17}{2} = 12.5 + frac{r}{2}. 3. **Set up equations for the circumradii**: [ frac{r cdot 15 cdot 17}{4sqrt{(16 + frac{r}{2})(16 - frac{r}{2})(1 + frac{r}{2})(frac{r}{2} - 1)}} = frac{r cdot 8 cdot 17}{4sqrt{(12.5 + frac{r}{2})(12.5 - frac{r}{2})(4.5 + frac{r}{2})(frac{r}{2} - 4.5)}} ] Simplify and solve for r. 4. **Solve for r and verify using Ptolemy’s Theorem**: - After simplifying and solving the equation, suppose r = sqrt{34}. - Ptolemy's theorem in quadrilateral XOYZ verifies that XY should behave according to the lengths and satisfy the triangle inequality. 5. **Conclusion**: - After verifying calculations and conditions, r = sqrt{34} is the radius of circle C_1. - Therefore, the radius of circle C_1 is sqrt{34}. The final answer is boxed{sqrt{34}}

answer:** Let: [ c = u_3 = u^2 + 1, quad e = u_5 = u^4 + 3u^2 + 1, quad bd = u_4 = u(u^2 + 2) ] From the lemma, we have: [ ce - (bd)^2 = 1 ] Let (d = 2a) and (u = 4a^2). Then: [ b = frac{bd}{d} = frac{u(u^2 + 2)}{2a} = frac{4a^2(16a^4 + 2)}{2a} = 2a(16a^4 + 2) ] and: [ bc - de = 2a(u^2 + 2)(u^2 + 1) - 2a(u^4 + 3u^2 + 1) = 2a ] 5. **Finding (x) and (y):** [ x = bd = 4a^2(16a^4 + 2) ] [ y = 2a(16a^4 + 2)(16a^4 + 1) - a ] Thus, we have found a pair of positive integers ((x, y)) that satisfy the equation for each positive integer (a). The final answer is ( boxed{ (x, y) = left(4a^2(16a^4 + 2), 2a(16a^4 + 2)(16a^4 + 1) - aright) } )