answer:1. **Statement Analysis:** - We have a set of n points P_1, P_2, cdots, P_n in a plane, with no three points being collinear. - Each pair of points is connected by a line segment, with all segments having distinct lengths. - A triangle is defined as a "middle-side triangle" if all three of its sides have lengths that appear as the middle length among triangles formed by points from the set. - A line l not passing through any point P_i splits the set D into two subsets D_1 and D_2. - We want to find the smallest value of n such that no matter how we choose l, at least one of the subsets contains a "middle-side triangle". 2. **Establishing Minimum n:** When n geq 11, regardless of the selected line l, one of the subsets D_1 or D_2 will have at least 6 points: [ text{If } n = 11 text{ and } l text{ splits } {P_1, P_2, ldots, P_{11}} text{ into } D_1 text{ and } D_2, text{ then } |D_1| geq 6 text{ or } |D_2| geq 6. ] Suppose |D_1| geq 6. 3. **Using Ramsey's Theorem:** Apply Ramsey's theorem to colored triangles formed by these points. - Consider all possible triangles within D_1. - Color the side with the middle length in each triangle red and the other sides blue. - According to Ramsey's theorem, there is at least one monochromatic triangle. Since every triangle has a middle side, the monochromatic red triangle is a "middle-side triangle". 4. **Verification for Smaller n:** - We need to prove that for n leq 10, there exists a configuration where no "middle-side triangle" is formed in D_1 or D_2. Assume n = 10 and split into D_1 and D_2 each containing 5 points: [ text{Consider these 5 points: } P_1, P_2, P_3, P_4, P_5 text{ on the circumference of a circle}. ] Assign distinct arc lengths so that any pairwise distance is unique. - Arc lengths (in radians): [ begin{aligned} overparen{P_1P_2} &= frac{pi}{10}, overparen{P_2P_3} &= frac{3pi}{5}, overparen{P_3P_4} &= frac{3pi}{10}, overparen{P_4P_5} &= frac{3pi}{4}, overparen{P_5P_1} &= frac{pi}{4}. end{aligned} ] - Distances are thus unique, preventing the formation of any "middle-side triangles". This configuration shows no middle-side triangles are formed. 5. **Concluding Minimum n:** From the above observations and tests, the minimum value of n for which the subset necessarily contains a "middle-side triangle" whenever n geq 11 is proved. [ boxed{11} ]

answer:Let's denote the three consecutive odd numbers as follows: First number: 7 (given) Second number: 7 + 2 = 9 (since they are consecutive odd numbers, we add 2) Third number: 9 + 2 = 11 (again, adding 2 to get the next odd number) According to the problem, 8 times the first number is equal to the addition of thrice the third number and a certain number added to twice the second number. We can write this as an equation: 8 * (first number) = 3 * (third number) + (certain number) + 2 * (second number) Now, let's plug in the values we have: 8 * 7 = 3 * 11 + (certain number) + 2 * 9 Simplify and solve for the certain number: 56 = 33 + (certain number) + 18 Combine like terms: 56 = 51 + (certain number) Subtract 51 from both sides to find the certain number: (certain number) = 56 - 51 (certain number) = 5 Therefore, the number added to twice the second number is boxed{5} .

answer:: Let's denote the number of white balls as (a) and the number of black balls as (b), with the condition (a neq b). To find which event has a greater probability, we'll examine the two given events. **Event A**: At some point, the number of white balls drawn equals the number of black balls drawn. **Event B**: At some point, the number of remaining white balls equals the number of remaining black balls. To compare these probabilities, we proceed step-by-step: 1. **Consider the Total Number of Balls**: The total number of draws is (a + b). We denote the sequence of draws and represent the difference between the drawn white balls and black balls after (k) draws as (y_k). 2. **Paths in a Lattice**: Each possible sequence of draws can be represented as a path on a 2-dimensional plane starting from the point ((0,0)) and ending at ((a+b, a-b)). The number of such paths is given by the binomial coefficient (C_{a+b}^{a}). 3. **Probability of Event A**: Event A occurs if at some point in the sequence, the number of white balls drawn equals the number of black balls drawn, i.e., (y_k = 0) for some (0 < k < a + b). This is equivalent to finding a path that crosses the x-axis. The number of paths that cross the x-axis at least once can be found using the reflection principle. The count of such paths is (2C_{a+b-1}^{a}). Therefore, the probability of Event A is: [ P(A) = frac{2C_{a+b-1}^{a}}{C_{a+b}^{a}} = frac{2b}{a+b} ] 4. **Probability of Event B**: Event B occurs if at some point the number of remaining white balls equals the number of remaining black balls, i.e., (a - text{(number of white balls drawn)} = b - text{(number of black balls drawn)}). Similar to Event A, we consider that the event occurs if the metric (y_k = k - 2b). This implies: [ P(B) = frac{2b}{a+b} ] 5. **Conclusion**: Both probabilities for Event A and Event B are found to be the same: [ P(A) = P(B) = frac{2 min {a,b}}{a+b} ] Thus, the probabilities of both events are equal. Therefore, there is no difference in the probabilities of these two events. [ boxed{P(A) = P(B)} ]

answer:If Jungkook stands third in line and there are 5 students between him and Yoongi, then Yoongi is standing in the 3rd position + 5 students + 1 (for Yoongi himself) = 9th position in line. Since there are 20 students in total and Yoongi is the 9th, to find out how many students are standing behind Yoongi, we subtract the 9 students in front of him from the total number of students: 20 students (total) - 9 students (in front of Yoongi) = boxed{11} students standing behind Yoongi.