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question:For each point X in the plane, a real number r_X > 0 is assigned such that 2|r_X r_Y | le |XY | , for any two points X, Y . (Here |XY | denotes the distance between X and Y ) A frog can jump from X to Y if r_X = |XY | . Show that for any two points X and Y , the frog can jump from X to Y in a finite number of steps.
answer:1. **Define the function and initial conditions:** Let ( r(X): mathbb{R}^2 to mathbb{R}^+ ) be a continuous function that assigns a positive real number to each point ( X ) in the plane. Fix a point ( X_0 in mathbb{R}^2 ). The frog can jump from ( X ) to ( Y ) if ( r_X = |XY| ). 2. **Initial jump capability:** The frog can jump to any point on the circle ( C(X_0, r_0) ), where ( r_0 = r(X_0) ). For every ( X in C(X_0, r_0) ), we have ( r(X) in [r_0/2, 3r_0/2] ). This follows from the given condition ( 2|r_X - r_Y| le |XY| ). 3. **Sweeping the area:** As ( X ) moves along ( C(X_0, r_0) ), the circles ( C(X, r(X)) ) will cover the area between ( C(X_0, r_0/2) ) and ( C(X_0, r_0) ). This is due to the continuity of ( r(X) ). Therefore, the frog can jump to any point within this ring in two jumps. 4. **Further jumps:** For any ( X in C(X_0, r_0/2) ), it holds that ( r(X) in [3r_0/4, 5r_0/4] ). Applying the same argument, the frog can jump from ( X ) to any point on ( C(X, 5r_0/8) ). As ( X ) moves along ( C(X_0, r_0/2) ), the circles ( C(X, 5r_0/8) ) will cover an area around ( X_0 ). Thus, the frog can jump from ( X_0 ) to any point between ( C(X_0, r_0/8) ) and ( C(X_0, r_0) ) within four jumps. 5. **Reaching smaller circles:** By repeating this argument, the frog can reach any point inside ( C(X_0, r_0/4) ) within no more than eight jumps. 6. **General case:** Consider two points ( X_0 ) and ( Y_0 ) in ( mathbb{R}^2 ). Let ( D ) be the closed disk ( {X : |X - X_0| leq |X_0Y_0|} ). The function ( r(X) ) reaches its minimal value ( r_0 > 0 ) on ( D ). This means the frog can reach any point within a distance of ( r_0/4 ) from any ( X in D ). Starting from ( X_0 ), the frog can reach ( Y_0 ) in a finite number of steps. (blacksquare)
question:) Let left{a_{1}, a_{2}, ldots, a_{n}right} be a sequence of integers whose sum equals 1. Prove that exactly one of its cyclic shifts left{a_{1}, a_{2}, ldots, a_{n}right}, left{a_{2}, ldots, a_{n}, a_{1}right}, ldots, left{a_{n}, a_{1}, ldots, a_{n1}right} has all its partial sums (from the beginning to an arbitrary element) positive. b) Derive the following equalities: C_{n}=frac{C_{2n+1}^{n}}{2n+1}=frac{C_{2n}^{n}}{n+1}=frac{(4n2)!!!!}{(n+1)!}, where (4n2)!!!!=2 cdot 6 cdot 10 cdot ldots(4n2) is a product involving every fourth number. Definition of Catalan numbers C_{n} can be found in the reference book.
answer:Part (a) 1. **Define the weight of the sequence**: Let {a_1, a_2, ldots, a_n} be a sequence of integers such that the sum of the elements is 1. Define the weight of the sequence to be the number k such that the partial sums s_1 = a_1, ldots, s_k = a_1 + ldots + a_k are positive, and s_{k+1} leq 0 (if k < n). 2. **Construct the maximum weight sequence**: Consider the cyclic shift that results in the sequence {b_1, ldots, b_n} with the maximum weight. Suppose this maximum weight is less than n. 3. **Find the longest negative partial sum**: Let's examine the longest non-positive partial sum b_1 + ldots + b_k. This means that after k elements, the sum is non-positive. 4. **Rearrange to increase weight**: All sums b_{k+1}, b_{k+1} + b_{k+2}, ldots, b_{k+1} + ldots + b_n must be positive. By shifting b_{k+1}, ldots, b_n to the beginning of the sequence, the weight of the sequence would increase, which contradicts the assumption of maximum weight. Thus, the weight must be n. 5. **Prove uniqueness of satisfying cyclic shift**: Suppose there are two different cyclic shifts that satisfy the condition. Let {a_1, a_2, ldots, a_n} and {a_m, ldots, a_n, a_1, ldots, a_{m-1}} both have all partial sums positive. Then a_1 + ldots + a_{m-1} and a_m + ldots + a_n must be natural numbers whose sum is 1. This is a contradiction since no combination of positive integers will sum up to 1. Hence, only one such cyclic shift exists. Conclusion: [ blacksquare ] Part (b) 1. **Count sequences with specific sums**: The number of sequences with (n+1) ones and n minus ones is C_{2n+1}^n. From part (a), the number of such sequences where all partial sums are positive is frac{C_{2n+1}^n}{2n+1}. 2. **Relate to Catalan numbers**: By discarding the first element (which is 1) of each valid sequence from part (a), we obtain the sequences in problem (underline{60447}), which count as C_n. Thus, C_n = frac{C_{2n+1}^n}{2n+1}. 3. **Alternative representation using combinations**: We know that C_n = frac{1}{n+1}C_{2n}^n. Using this equivalence, we can rewrite it as: [ C_n = frac{(2n)!} {(n+1)!n!} ] 4. **Derive the product notation**: Recognizing that (2n)! can be factored as 1 cdot 3 cdot ldots cdot (2n-1) cdot 2 cdot 4 cdot ldots cdot 2n = (2 cdot 6 cdot 10 ldots (4n-2)) cdot (1 cdot 2 ldots n): [ (2n)! = (4n-2)!!!! cdot n! ] 5. **Combine the results**: Thus, substituting into the formula for C_n yields: [ C_n = frac{(4n-2)!!!!} {(n+1)!} ] Conclusion: [ boxed{C_n=frac{C_{2n+1}^n}{2n+1}=frac{C_{2n}^n}{n+1}=frac{(4n-2)!!!!}{(n+1)!}} ]
question:A can finish a piece of work in some days. B can do it in 15 days. They work together for two days and then A goes away. B finishes the work in 7 days. In how many days can A finish the work alone?
answer:Let's denote the number of days A takes to finish the work alone as ( x ) days. A's work rate is ( frac{1}{x} ) of the work per day. B's work rate is ( frac{1}{15} ) of the work per day. When A and B work together for two days, they complete ( 2 left( frac{1}{x} + frac{1}{15} right) ) of the work. After A leaves, B works alone for 7 more days, completing ( 7 left( frac{1}{15} right) ) of the work. The total work done when they work together and when B works alone should add up to 1 (the whole work), so we can write the equation: [ 2 left( frac{1}{x} + frac{1}{15} right) + 7 left( frac{1}{15} right) = 1 ] Now, let's solve for ( x ): [ 2 left( frac{1}{x} + frac{1}{15} right) = 1 - 7 left( frac{1}{15} right) ] [ 2 left( frac{1}{x} + frac{1}{15} right) = frac{15}{15} - frac{7}{15} ] [ 2 left( frac{1}{x} + frac{1}{15} right) = frac{8}{15} ] [ frac{2}{x} + frac{2}{15} = frac{8}{15} ] [ frac{2}{x} = frac{8}{15} - frac{2}{15} ] [ frac{2}{x} = frac{6}{15} ] [ frac{1}{x} = frac{3}{15} ] [ frac{1}{x} = frac{1}{5} ] [ x = 5 ] So, A can finish the work alone in boxed{5} days.
question:Consider three lines in a plane given by the equations 2y 3x = 4, 3x + y = 5, and 6x 4y = 8. Determine the points accounting for the intersection of at least two of these three lines.
answer:Analyzing the slopes of each line: 1. First line: 2y - 3x = 4 rearranges to y = frac{3}{2}x + 2. The slope (m) is frac{3}{2}. 2. Second line: 3x + y = 5 rearranges to y = -3x + 5. The slope (m) is -3. 3. Third line: 6x - 4y = 8 rearranges to y = frac{3}{2}x - 2. The slope (m) is frac{3}{2}. Now, checking intersections: - Lines 1 and 3 are both parallel since they have the same slope frac{3}{2}. They are not identical lines because their y-intercepts are different (2 and -2). Therefore, they don't intersect. - Lines 1 and 2: To find their intersection, solve the system: - y = frac{3}{2}x + 2 - y = -3x + 5 Setting the equations equal: frac{3}{2}x + 2 = -3x + 5 implies frac{3}{2}x + 3x = 5 - 2 implies frac{9}{2}x = 3 implies x = frac{2}{3} times 3 = 2. Plugging x = 2 into the first equation: y = frac{3}{2}(2) + 2 = 3 + 2 = 5. Therefore, the intersection point of lines 1 and 2 is (2,5). - Lines 2 and 3: Set -3x + 5 = frac{3}{2}x - 2. Solving gives: - -3x - frac{3}{2}x = -2 - 5 implies -frac{9}{2}x = -7 implies x = frac{14}{9}. Substituting x = frac{14}{9} into y = -3x + 5: y = -3(frac{14}{9}) + 5 = -frac{42}{9} + 5 = -frac{42}{9} + frac{45}{9} = frac{3}{9} = frac{1}{3}. So, the intersection point of lines 2 and 3 is (frac{14}{9}, frac{1}{3}). Thus, there are boxed{2} intersection points.