answer:We need to find the number of ways a chess player can score 5 points after 7 games, given that a win is worth 1 point, a draw is ( frac{1}{2} ) point, and a loss is 0 points. Consider the following cases for achieving 5 points: 1. **5 Wins and 2 Losses:** For this scenario, we need to count the permutations of the sequence ( (1, 1, 1, 1, 1, 0, 0) ). The number of permutations of this sequence with repeated elements can be calculated using the multinomial coefficient: [ frac{7!}{5!2!} ] Let's compute this step-by-step: - ( 7! = 5040 ) - ( 5! = 120 ) - ( 2! = 2 ) Therefore, [ frac{7!}{5!2!} = frac{5040}{120 times 2} = frac{5040}{240} = 21 ] Alternatively, we can consider choosing 2 specific games out of 7 for the losses. This is calculated using the binomial coefficient: [ binom{7}{2} = frac{7 cdot 6}{2 cdot 1} = 21 ] 2. **4 Wins, 2 Draws, and 1 Loss:** Here, we need to arrange the sequence ( (1, 1, 1, 1, frac{1}{2}, frac{1}{2}, 0) ). The number of permutations is given by: [ frac{7!}{4!2!1!} ] Let's break down the calculations: - ( 7! = 5040 ) - ( 4! = 24 ) - ( 2! = 2 ) - ( 1! = 1 ) Therefore, [ frac{7!}{4!2!1!} = frac{5040}{24 times 2 times 1} = frac{5040}{48} = 105 ] Another way to calculate this is to choose 4 games to win out of 7, 2 games to draw out of the remaining 3: [ binom{7}{4} cdot binom{3}{2} = binom{7}{3} cdot binom{3}{1} = 35 cdot 3 = 105 ] 3. **3 Wins and 4 Draws:** For this scenario, we need to arrange the sequence ( (1, 1, 1, frac{1}{2}, frac{1}{2}, frac{1}{2}, frac{1}{2}) ). The number of permutations is given by: [ frac{7!}{3!4!} ] Let’s break down the calculations: - ( 7! = 5040 ) - ( 3! = 6 ) - ( 4! = 24 ) Therefore, [ frac{7!}{3!4!} = frac{5040}{6 times 24} = frac{5040}{144} = 35 ] Alternatively, we can choose 3 specific games out of 7 for the wins. This is simply: [ binom{7}{3} = frac{7 cdot 6 cdot 5}{3 cdot 2 cdot 1} = 35 ] By summing up all the possible scenarios: [ 21 + 105 + 35 = 161 ] # Conclusion: Thus, the result 5 points from 7 games can be achieved in (boxed{161}) different ways.

answer:Since a_1=1 and 2S_n=a_n cdot a_{n+1} (N in mathbb{N}^*), we have 2 times 1 = 1 times a_2, which gives a_2=2. For n geqslant 2, we have 2a_n=2(S_n-S_{n-1})=a_n(a_{n+1}-a_{n-1}), and since a_n > 0, this simplifies to a_{n+1}-a_{n-1}=2. Therefore, the sequence {a_n} has its odd and even terms forming arithmetic sequences separately, both with a common difference of 2. Thus, a_{2k-1}=1+2(k-1)=2k-1 and a_{2k}=2+2(k-1)=2k for k in mathbb{N}^*. Hence, a_n=n. Therefore, (1+ frac{1}{a_1}) cdot (1+ frac{1}{a_2}) cdots (1+ frac{1}{a_n})= frac{2}{1} times frac{3}{2} times cdots times frac{n+1}{n}=n+1. Thus, the inequality log_2(1+ frac{1}{a_1})+log_2(1+ frac{1}{a_2})+cdots+log_2(1+ frac{1}{a_n}) > 5 becomes log_2(n+1) > 5, which solves to n+1 > 2^5. Therefore, the minimum value of the positive integer n that satisfies the inequality log_2(1+ frac{1}{a_1})+log_2(1+ frac{1}{a_2})+cdots+log_2(1+ frac{1}{a_n}) > 5 is boxed{32}. This problem involves using the recursive relationship of the sequence and the formula for the general term of an arithmetic sequence, as well as the properties of "cumulative multiplication" and logarithms. It tests reasoning and computational skills and is considered a medium-level problem.

answer:Given a triangle ( ABC ) inscribed with a circle of radius ( r ), and given the circles inscribed in the three smaller triangles formed by the tangents to this incircle being parallel to the sides ( BC, CA, ) and ( AB ), we need to prove ( r_1 + r_2 + r_3 = r ) where ( r_1, r_2, ) and ( r_3 ) are the inradii of the smaller triangles. **Step-by-step proof:** 1. **Understanding the Task:** - We need to show that the sum of the radii of the inscribed circles in three smaller triangles is equal to the radius ( r ) of the original triangle's incircle. 2. **Initial Setup:** - Consider a triangle ( ABC ) with an incircle of radius ( r ). - The tangents to the incircle parallel to sides ( BC, CA, ) and ( AB ) meet at points which form three smaller triangles in ( ABC ). 3. **Analysis of Smaller Triangles:** - Let the tangents parallel to ( BC ) intersect at ( A_1 ) representing the triangle formed by vertex ( A ) and the tangents. Similarly, define points ( B_1 ) and ( C_1 ). - The three smaller triangles ( A A_1A_2 ), ( B B_1B_2 ), and ( C C_1C_2 ) are formed where ( A_1, A_2, ) etc. are points of tangency. 4. **Perimeter Relationships:** - The perimeter of triangle ( ABC ) can be expressed in terms of the points of tangency with the inradius. - For illustration, let the tangent segments from ( A ) to the points of tangency on ( AB ) and ( AC ) be equal (similarly for other vertices). Therefore, each smaller triangle’s perimeter plus the lengths of parts cut by the tangents will equal the original triangle’s perimeter. 5. **Calculation of the Perimeters:** [ text{Sum of perimeters of smaller triangles} = text{Perimeter of triangle ABC} ] Assuming ( P_{ABC} = text{Perimeter of triangle ABC} ), we get: [ P_{1} + P_{2} + P_{3} + 3r = P_{ABC} ] 6. **Using Area Summation Property:** - The area of triangle ( ABC ) can be considered as the sum of the areas of the three smaller triangles, each being directly proportional to the corresponding radius and the semi-perimeter of the smaller triangle. 7. **Inradius Relationship:** - Since the perpendicular distances to the sides from the tangents are equal to the radius of the incircle at the respective smaller triangles: [ r_{1}+r_{2}+r_{3}=r ] **Conclusion:** [ boxed{r_{1}+r_{2}+r_{3}=r} ]

answer:To evaluate the expression sqrt{2 -!sqrt{2 - !sqrt{2 - !sqrt{2 - cdots}}}}, let's denote it by x. Thus, we have: [x = !sqrt{2 - !sqrt{2 - !sqrt{2 - !sqrt{2 - cdots}}}}] Given this, we can observe that the expression inside the square root repeats itself, leading to the equation: [x = !sqrt{2 - x}] To solve for x, we square both sides of the equation to eliminate the square root: [x^2 = (2 - x)^2] [x^2 = 2 - x] Rearranging the equation to bring all terms to one side gives us a quadratic equation: [x^2 + x - 2 = 0] This quadratic equation can be factored into: [(x+2)(x-1) = 0] Setting each factor equal to zero gives us the possible solutions for x: [x + 2 = 0 quad text{or} quad x - 1 = 0] [x = -2 quad text{or} quad x = 1] However, since x represents a length in the context of the problem and must be positive, we discard x = -2 as a solution. Therefore, the solution to the equation, and thus the value of the original expression, is: [x = boxed{1}]