answer:1. **Arrangement of Points:** We start by arranging the points on a straight line from left to right. Label these points in order as (A_1, A_2, ldots, A_{30}), and assign colors as follows: - Points (A_{3k+1}) are blue. - Points (A_{3k+2}) are green. - Points (A_{3k+3}) are red for (0 leq k leq 9). 2. **Defining Distances:** Let (d_i) be the distance between (A_i) and (A_{i+1}) for (1 leq i leq 29). Our goal is to find positive values for ((d_i)_{1 leq i leq 29}) such that all pairwise distances between the points are distinct and the following constraints are satisfied: - For each blue point, the second closest point is red. - For each red point, the second closest point is green. - For each green point, the second closest point is blue. 3. **Initial Placement and Verification of Points:** Set (A_1) at position 0 on a graduated line. Place (A_i) at the position: [ A_i = d_1 + d_2 + cdots + d_i. ] This guarantees that the distance between (A_i) and (A_{i+1}) is (d_i) for (1 leq i leq 29). 4. **Ensuring Distance Conditions:** To ensure the conditions are met, assign (d_i = frac{1}{i+2}) for (1 leq i leq 27): [ d_i < d_{i-1} quad text{and} quad d_i + d_{i+1} > d_{i-1}. ] Compute (d_i + d_{i+1} - d_{i-1}): [ begin{aligned} d_i + d_{i+1} - d_{i-1} &= frac{1}{i+2} + frac{1}{i+3} - frac{1}{i+1} &= frac{(i+1)(i+3) + (i+1)(i+2) - (i+2)(i+3)}{(i+1)(i+2)(i+3)} &= frac{(i+1)^2 + 3(i+1) + (i+1)^2 + 2(i+1) - (i^2 + 5i + 6)}{(i+1)(i+2)(i+3)} &= frac{i^2 + 4i + 4}{(i+1)(i+2)(i+3)} &= frac{(i+2)^2 - 2}{(i+1)(i+2)(i+3)} > 0. end{aligned} ] 5. **Adjustments for Distinct Distances:** If some distances are not distinct, adjust (d_i) slightly. For each (i) from 1 to 27, modify (d_i) such that: [ d_i < d_{i-1} quad text{and} quad d_i + d_{i+1} > d_{i-1} ] while preserving the distinctness of previous (d_j). 6. **Placing Points A_28 and A_29:** Ensure that the distance between (A_{28}) and (A_{29}) is greater than (d_{27}), yet distinct from the other distances. Place (A_{29}) (red) at: [ d_{28} quad text{such that} quad d_{27} < d_{26} < d_{27} + d_{28}. ] 7. **Placing the Last Point A_30:** Place (A_{30}) (green) such that: [ d_{29} quad text{such that} quad d_{28} + d_{27} > d_{29} > d_{28}. ] Adjust (d_{29}) to maintain distinctness and still satisfy: [ d_{28} + d_{27} > d_{29} > d_{28}. ] 8. **Verification:** Check: - For (A_{30}), the second closest point is (A_{28}) using (d_{28} + d_{27} > d_{29}). - For (A_{29}), the second closest point is (A_{27}) using (d_{28} + d_{27} > d_{29}). # Conclusion: It is possible to arrange the points such that all conditions are satisfied. Thus, the final answer is: (boxed{text{Yes, it is possible.}})

answer:To find the values of ( A ) and ( B ) that minimize the maximum value ( M ) of the function ( F(x) ) over the interval ( 0 leq x leq frac{3}{2}pi ), let's analyze the given function: [ F(x) = left| cos^2 x + 2 sin x cos x - sin^2 x + A x + B right| ] 1. **Rewriting the Trigonometric Part:** The expression inside the absolute value can be rewritten using trigonometric identities as: [ begin{aligned} cos^2 x + 2 sin x cos x - sin^2 x &= (cos x)^2 + 2 (sin x)(cos x) - (sin x)^2 &= cos^2 x + 2 sin x cos x - sin^2 x &= cos^2 x - sin^2 x + 2 sin x cos x &= cos(2x) + sin(2x) &= sqrt{2} sin(2x + frac{pi}{4}), end{aligned} ] where we used the identity (cos(2x) = cos^2 x - sin^2 x) and (sin(2x) = 2 sin x cos x). Thus we can rewrite ( F(x) ) as: [ F(x) = left| sqrt{2} sin left(2x + frac{pi}{4}right) + A x + B right| ] 2. **Case Analysis for ( A = 0 ) and ( B = 0 ):** If ( A = 0 ) and ( B = 0 ), then: [ F(x) = left| sqrt{2} sin left(2x + frac{pi}{4}right) right| ] We know that (left| sin left(2x + frac{pi}{4} right) right| leq 1), hence: [ F(x) = sqrt{2} left|sin left(2x + frac{pi}{4}right) right| leq sqrt{2} ] The maximum value ( M ) for ( F(x) ) in the interval ( 0 leq x leq frac{3}{2}pi ) is ( sqrt{2} ). Hence, ( M = sqrt{2} ). 3. **Exploring Other Values of ( A ) and ( B ):** We will analyze how ( F(x) ) behaves for different values of ( A ) and ( B ): - **Case 1: ( A = 0 ), ( B neq 0 )** In this case, the function becomes: [ F(x) = left| sqrt{2} sin left(2 x + frac{pi}{4} right) + B right| ] Clearly, the maximum value of ( F(x) ) will be greater than ( sqrt{2} ) if ( B neq 0 ) because the absolute value would have to account for the term ( B ). - **Case 2: ( A > 0 ), ( B geq 0 )** Evaluate ( Fleft(frac{pi}{8} right) ): [ Fleft( frac{pi}{8} right) = sqrt{2} sin left( frac{pi}{4} + frac{pi}{4} right) + left( frac{pi}{8} right) A + B = sqrt{2} + frac{pi}{8} A + B ] Since ( A > 0 ) and ( B geq 0 ), it is evident that (sqrt{2} + frac{pi}{8} A + B > sqrt{2} ), thus making ( M > sqrt{2} ). - **Case 3: ( A > 0 ), ( B < 0 )** - If ( |B| < frac{9pi}{8} A ), [ left| sqrt{2} + frac{9 pi}{8} A + B right| > sqrt{2} ] - If ( |B| geq frac{9pi}{8} A ), [ left| -sqrt{2} + frac{5 pi}{8} A + B right| > sqrt{2} ] In both cases, ( M > sqrt{2} ). - **Case 4: ( A < 0 ), ( B leq 0 )** Evaluate ( Fleft(frac{5 pi}{8} right) ): [ Fleft( frac{5 pi}{8} right) = left|- sqrt{2} + frac{5 pi}{8} A + B right| > sqrt{2} ] - **Case 5: ( A < 0 ), ( B > 0 )** - If ( B > -frac{5 pi}{8} A ), [ left| -sqrt{2} + frac{5 pi}{8} A + B right| > sqrt{2} ] - If ( B geq -frac{pi}{8} A ), [ Fleft( frac{pi}{8} right) = left| sqrt{2} + frac{pi}{8} cdot A + B right| > sqrt{2} ] In conclusion, the only way to ensure the minimum ( M ) is attaining its minimum value (which is ( sqrt{2} )) is by setting ( A = 0 ) and ( B = 0 ). [ boxed{A = 0 text{, } B = 0} ]

answer:From the table, the boxes containing at least 75,000 are: 75,000, 100,000, 200,000, 300,000, 400,000, 500,000, 750,000, 1,000,000. This totals 9 boxes. To have at least a one-third chance of holding such a box, we need the condition P(text{holding at least 75,000}) geq frac{1}{3}. If there are x boxes left (including the chosen one), the probability of the chosen box containing at least 75,000 should be frac{9}{x} geq frac{1}{3}. Solving for x, we have: [ 9/x geq 1/3 implies x leq 27 ] Thus, there need to be at most 27 boxes, including the one held by the participant. Since we start with 30 boxes, the participant must eliminate at least 30 - 27 = boxed{3} boxes to have at least a one-third chance of holding a box containing at least 75,000.

answer:First distribute the negative sign and handle the expression inside the parentheses: [ (17a + 45b) + (15a + 36b) - (12a + 42b) - 3(2a + 3b) = 17a + 45b + 15a + 36b - 12a - 42b - 6a - 9b ] Now, combine like terms: [ 17a + 15a - 12a - 6a = 14a ] [ 45b + 36b - 42b - 9b = 30b ] The final simplified expression is: [ boxed{14a + 30b} ]